url
stringlengths 14
2.42k
| text
stringlengths 100
1.02M
| date
stringlengths 19
19
| metadata
stringlengths 1.06k
1.1k
|
|---|---|---|---|
https://hpdung.wordpress.com/category/learning/geometric-topology/ | ## On the diffetential of a mapping 2
In the case ${\psi : \mathbb{R}^n \rightarrow \mathbb{R}^m}$ case, there is a linear map, which is “linear approximation” of ${\psi}$. In the manifold case, there is a similar linear map, but now it acts between tangent spaces. If ${M}$ and ${N}$ are smooth manifolds and ${\psi \colon M \rightarrow N}$ is a smooth map then for each ${m \in M}$, the map
$\displaystyle d\psi \colon T_mM \rightarrow T_{\psi(m)}N$
is defined by
$\displaystyle d\psi(v)(f) = v(f \circ \psi)$
is called the pushforward. Actually,
$\displaystyle d\psi \colon TM \rightarrow TN.$
Suppose that ${\dim{M} \ge \dim{N}}$ and ${f \colon M \rightarrow N}$ is a differentiable mapping. We have
Definition 1 The mapping ${f}$ is called a trivial fibration (differentiable) on ${N}$ if there exists a differential manifold ${F}$, is called fibre of ${f}$, and a diffeomorphism
$\displaystyle \phi \colon M \rightarrow N \times F$
such that the following diagram is commutative
## On the differential of a mapping
From on Warner’s book.
Smooth curve on manifold ${M}$:
A ${C^\infty}$ mapping ${\alpha : (a, b) \rightarrow M}$. Let ${t \in (a, b)}$, we define the tangent vector of the curve ${\alpha}$ at ${t}$ is the vector
$\displaystyle d\alpha\bigg(\dfrac{d}{dt}\bigg|_{t = 0}\bigg) \in T_{\alpha(0)}M.$
we apply the formula
$\displaystyle d\psi(v)(g) = v(g \circ \psi),$
where ${g}$ is an any function on ${M}$.
Put ${\psi = \alpha(t)}$ and ${v = \dfrac{d}{dt}\alpha(t)\big|_{t = 0}}$, the above formula implies
$\displaystyle d\alpha(\frac{d}{dt}\big|_{t=0})(f) = (\frac{d}{dt}\big|_{t=0})(f \circ \alpha) = \frac{d}{dt}(f \circ \alpha)\big|_{t=0}.$
This is directional derivative.
## Một ví dụ về mặt chính quy
BT. Cho hàm $f(x,y,z)=z^{2}$. CMR $0$ không là giá trị chính qui của hàm $f$ thế nhưng $f^{-1}(0)$ vẫn là mặt chính qui.
Lời giải.
Ta có ma trận $(f_x \ f_y \ f_z) = (0 \ 0 \ 2z)$ suy biến tức $rankA < 1$ khi và chỉ khi $z=0$, do đó tại điểm $(x, y, 0), \forall x, y \in \mathbb{R}$, ma trận trên suy biến và do đó là điểm kì dị. Mà $f(x,y,0)=0$ nên $0$ là giá trị tới hạn, hay ko phải giá trị chính qui. Nhưng $f^{-1}(0)$ là mặt phẳng Oxy, đây là mặt trơn nên chính qui.
## The tangent plane of a surface and the tangent space of a manifold (at one point)
The tangent plane of a surface
By a tangent vector to $S$ at a point $p \in S$, we mean the tangent vector $\alpha'(0)$ of a differentiable parametrized curve $\alpha: (-\epsilon, \epsilon) \to S$ with $\alpha(0) = p$.
What is the tangent plane of a surface? That is a plane which containes all of the tangent vectors of this surface at point $p \in S$.
Proposition
Let $\varphi: U \subset \mathbb{R}^2 \to S$ be a parametrization of a regular surface $S$ and let $q \in U$. The vector subspace of dim $2$,
$d\varphi_q(\mathbb{R}^2) \subset \mathbb{R}^3,$
coincides with the set of tangent vectors to $S$ at $\varphi(q)$
There is a similar situation in here. In the case of manifold, the tangent space is also built from the set of all of tangent vectors of a manifold.
Tangent spaces of a manifold
In Milnor’s book (\cite{Milnor}), the tangent space $TM_x$ at $x$ for arbitrary smooth manifold $M \subset \mathbb{R}^k$ is defined:
Choose a parametrization $g : U \to M \subset \mathbb{R}^k$ of a neighborhood $g(U)$ of $x$ in $M$, with $g(u) = x$. We have $dg_u: \mathbb{R}^m \to \mathbb{R}^k$. So the image $dg_u(\mathbb{R}^m)$ of $dg_u$ is equal to $TM_x$.
References
J. W. Milnor, Topology from differentiable viewpoint, 1965.
M. do Carmo, Differential geometry, curves and surfaces, 1976.
## Course về topo vi phân của GS Jae Choon Cha
GS Jae Choon Cha hiện đang giảng dạy tại trường Postech của Hàn Quốc. Ông có dạy một course về tô pô vi phân. Link: http://gt.postech.ac.kr/~jccha/differential-topology-2015-spring/
Đây là phần bài tập về nhà (phần 1) của course
## Exercise of diff topo 1
BT của Munkres,
BT1, Cho $M$ là một đa tạp chiều $m$ có biên khác rỗng. Gọi $M_0 = M \times 0$$M_1 = M \times 1$ là hai mảnh của $M$. “Double” của $M$, ký hiệu là $D(M)$, là không gian tô pô tạo bởi $M_0 \cup M_1$ bằng cách gắn đồng nhất $(x,0)$$(x,1)$ với mỗi $x \in Bd(M)$ (biên của M). Chứng minh rằng $D(M)$ là một đa tạp không có biên $m$ chiều.
## An exercise of Mayer-Vietories sequence in Hatcher
BT 2.2.31 Hatcher. Dùng dãy Mayer-Vietories để chứng tỏ rằng có một đẳng cấu $\widetilde{H}_n(X \vee Y) = \widetilde{H}_n(X) \oplus \widetilde{H}_n(Y)$ nếu các điểm cơ sở (base points) của $X$$Y$ được đồng nhất trong $X \vee Y$ là co rút biến dạng của các lân cận $U \subset X$$V\subset Y$.
—–
Để sử dụng dãy Mayer-Vietories, ta phải tìm đc 2 tập mở $A$$B$ để $X \vee Y = A \cup B$. Thật vậy, chọn $A = X \vee V$$Y \vee U$. 2 tập đều mở nên $X \vee Y = int(A) \cup int(B)$. Mặt khác, $A \cap B = U \cap V$, theo giả thiết, điểm nối giữa $X$$Y$ là co rút biến dạng của $U$$V$, nên khi đồng nhất 2 điểm của $x \in U$$y \in V$ thì ta có thể coi $x \equiv y$ là co rút biến dạng của $U \cap V$. Do đó $U \cap V \simeq \{x\}$$\widetilde{H}_n(U \cap V) = \widetilde{H}_n(point)$.
Áp dụng dãy Mayer-Vietories thu gọn với $A$$B$ như trên, ta có dãy khớp dài sau:
$... \rightarrow \widetilde{H}_n(A \cap B) \rightarrow \widetilde{H}_n(A) \oplus \widetilde{H}_n(B) \rightarrow \widetilde{H}_n(X \vee Y) \xrightarrow{\partial} \widetilde{H}_{n - 1}(A \cap B) \rightarrow ...\rightarrow \widetilde{H}_0(X \vee Y) \rightarrow 0$
Vì đồng điều của $A \cap B$ là đồng điều một điểm nên ta có:
$...\rightarrow 0 \rightarrow \widetilde{H}_n(A) \oplus \widetilde{H}_n(B) \rightarrow \widetilde{H}_n(X \vee Y) \xrightarrow{\partial} 0 \rightarrow ...\rightarrow 0 \rightarrow 0$
Từ đó ta có: $\widetilde{H}_n(X \vee Y) = \widetilde{H}_n(A) \oplus \widetilde{H}_n(B)$, và vì $A \simeq X, B \simeq Y$ nên $\widetilde{H}_n(A) = \widetilde{H}_n(X), \widetilde{H}_n(B) = \widetilde{H}_n(Y)$. Do đó, ta được đẳng cấu:
$\widetilde{H}_n(X \vee Y) = \widetilde{H}_n(X) \oplus \widetilde{H}_n(Y)$ (đpcm). | 2017-10-19 07:09:34 | {"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": 110, "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": 0.9788630604743958, "perplexity": 7598.34938912395}, "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-2017-43/segments/1508187823255.12/warc/CC-MAIN-20171019065335-20171019085335-00869.warc.gz"} |
https://search.datacite.org/works?resource-type-id=collection&&data-center-id=umd.lib&page=2 | LEARNING FROM MULTIPLE VIEWS OF DATA
Abhishek Sharma
This dissertation takes inspiration from the abilities of our brain to extract information and learn from multiple sources of data and try to mimic this ability for some practical problems. It explores the hypothesis that the human brain can extract and store information from raw data in a form, termed a common representation, suitable for cross-modal content matching. A human-level performance for the aforementioned task requires - a) the ability to extract sufficient information from...
HIERARCHICAL NEURAL COMPUTATION IN THE MAMMALIAN VISUAL SYSTEM
Yuwei Cui
Our visual system can efficiently extract behaviorally relevant information from ambiguous and noisy luminance patterns. Although we know much about the anatomy and physiology of the visual system, it remains obscure how the computation performed by individual visual neurons is constructed from the neural circuits. In this thesis, I designed novel statistical modeling approaches to study hierarchical neural computation, using electrophysiological recordings from several stages of the mammalian visual system. In Chapter 2, I describe...
Atom-trapping and photon-counting experiments with optical nanofibers
Jeffrey Aaron Grover
New effects can arise in quantum physics when there is strong coupling, either between atoms and light or between different quantum systems. This thesis examines an optical nanofiber atom trap as a mediator of atom-light interactions and a potential element of a hybrid quantum system. The evanescent field around the sub-wavelength waist of an optical nanofiber possesses a small mode area that increases the cooperativity between atoms and the mode, in a manner analogous to...
Xinghan Cai
Graphene, a single-atom-thick plane of carbon, has unique optoelectronic properties that result in a variety of potential photonic applications, such as optical modulators, plasmonic devices and THz emitters. In this thesis, the light-matter interaction in monolayer graphene and the subsequent photoexcited charge carrier transport are studied, and it is found that graphene has unique advantages for hot-electron photothermoelectric detection. Particularly promising is detection of terahertz (THz) radiation, in which graphene devices may offer significant advantages...
THE EARLY, MIDDLE, AND LATE STYLES OF WOLFGANG AMADEUS MOZART AND LUDWIG VAN BEETHOVEN IN THEIR SONATAS FOR PIANO AND VIOLIN
Elizabeth E. Kim
Wolfgang Amadeus Mozart's (1756-1791) and Ludwig van Beethoven's (1770-1827) personal backgrounds influenced their compositional styles in different stages of their lives. Their sonatas for piano and violin show the evolution of their styles in early, middle, and late periods. Mozart's early period (1761-1773) keyboard sonatas with violin accompaniment, such as K. 9, show experimentations of a child prodigy and little equality between the piano and violin. In contrast to these early sonatas, his middle period...
Image Estimation and Uncertainty Quantification
Viktoria Taroudaki
Recorded images are usually contaminated by blur and noise. The restoration of such altered images is an ill-posed problem. Even if the blur is known, the unknown noise leads to uncertainty in the restored image. The naive restoration approach fails since it contains a lot of noise at high frequencies that destroys the computed restored image. To remedy this problem, this work focuses on the computation of the restored image by using spectral filters that...
The Relative Lie Algebra Cohomology of the Weil Representation
Jacob Ralston
We study the relative Lie algebra cohomology of $\mathfrak{so}(p,q)$ with values in the Weil representation $\varpi$ of the dual pair $\mathrm{Sp}(2k, \R) \times \OO(p,q)$. Using the Fock model defined in Chapter \ref{Introchapter}, we filter this complex and construct the associated spectral sequence. We then prove that the resulting spectral sequence converges to the relative Lie algebra cohomology and has $E_0$ term, the associated graded complex, isomorphic to a Koszul complex, see Section \ref{defofkoszulsection}. It is...
Reordering the Landscape: Science, Nature, and Spirituality at Wye House
Elizabeth Pruitt
This dissertation draws on literature and theoretical frameworks of gardening and social ordering that examine early Euro-American and African-American material culture as they came together on the plantation landscape at Wye House. Located on the Eastern Shore of Maryland, the plantation was home to the Welsh Lloyd family and hundreds of enslaved Africans and African-Americans. Using archaeological and archeobotanical remains of garden related buildings and slave dwellings, this project acknowledges the different possible interactions and...
Prolonged Illness Among Subsistence Agricultural Households in Rural Mozambique: Coping Strategies and Policy Levers
Zan Michael Dodson
Subsistence agriculturalists are highly economically vulnerable, and generally lack access to resources that may help strengthen their livelihoods. Health is a well-established form of human capital that is also one of the biggest assets for subsistence agricultural households. Therefore, any instance in which this form of capital is threatened has potential consequences for livelihood sustainability. This dissertation examined prolonged illness among subsistence agricultural households in rural Mozambique. Prolonged illness can diminish household labor supply, a...
Faucets and Fertilizers: Interpreting Technological Change in Rural Oaxaca, Mexico, 1946-1988
Joshua Charles Walker
Faucets and Fertilizers: Interpreting Technological Change in Rural Oaxaca, Mexico, 1946-1988 argues that peasant farmers in Oaxaca were key actors who helped to oversee the technological modernization of their villages in the twentieth century. From the 1940s to the 1980s, federal and state development programs sought to introduce new tools like chemical fertilizers, water faucets, roads, and mechanical corn grinders to villages in the countryside. These programs were often unevenly distributed and poorly designed, forcing...
Exploring Influences of Folk and Popular Music in Selected Violin Repertoire: from Edvard Grieg (1843-1907) to Paul Schoenfield (b. 1947).
Jennifer J. Lee
ABSTRACT Title of Dissertation: EXPLORING INFLUENCES OF FOLK AND POPULAR MUSIC IN SELECTED VIOLIN REPORTOIRE: FROM EDVARD GRIEG (1843-1907) TO PAUL SCHOENFIELD (B. 1947) Jennifer J. Lee, Doctor of Musical Arts, 2015. Dissertation directed by: Professor David Salness School of Music Both folk' and popular' music genres definitively play a role in the compositional makeup of the following notable composers: Edvard Grieg (1843-1907), Eugene Ysaÿe (1858-1931), Nikolai Medtner (1880-1951), George Enescu (1881-1955), Francis Poulenc (1899-1963),...
What Matters: An Analysis of Victim Satisfaction in a Procedural Justice Framework
Cortney Fisher
The discipline of criminology and criminal justice tends to focus on the offender. However, the victim's cooperation with authorities, which often begins with a willingness to report the crime, is central to a successful investigation and prosecution. Yet, the crime victim exists today on the outskirts of the criminal justice system, limited in their role by the same authorities that need them to help. Despite increasingly retributive policies toward offenders, victims remain as unsatisfied with...
The Impact of Returning Technical Parole Violators to Prison: A Deterrent, Null, or Criminogenic Effect
Kristofer Bret Bucklen
As a result of the significant U.S. prison population build-up over the past several decades, a large number of inmates are now being released from prison and returned to the community. One mechanism for facilitating this transition to the community is for inmates to be conditionally released under parole supervision. Once on parole, a parolee is subject to certain rules and conditions that, if violated, can result in a return to prison, even if not...
Mona Asudegi
Smart Grid is a system that accommodates different energy sources, including solar, wind, tidal, electric vehicles, and also facilitates communication between users and suppliers. This study tries to picture the interaction among all new sources of energy and market, besides managing supplies and demands in the system while meeting network's limitations. First, an appropriate energy system mechanism is proposed to motivate use of green and renewable energies while addressing current system's deficiencies. Then concepts and...
ONE-AND-TWO-LEVEL NATURAL GAS EQUILIBRIUM MODELS AND ALGORITHMS
This dissertation consists of three parts; Part 1 provides two applied studies for the current issue of the global natural gas market, Part 2 presents the World Gas Model (WGM) 2014 version-a significant extension of WGM 2012, and Part 3 develops a novel Benders decomposition procedure with SOS1 reformulation to solve mathematical programs with equilibrium constraints (MPECs) and then is applied to several applications in natural gas and additional test problems. Part 1 presents two...
EXTENSIVE GREEN ROOF SUBSTRATE COMPOSITION: EFFECTS OF PHYSICAL PROPERTIES ON MATRIC POTENTIAL, HYDRAULIC CONDUCTIVITY, PLANT GROWTH, AND STORMWATER RETENTION IN THE MID-ATLANTIC.
Whitney Griffin
While green roof (GR) systems have gained popularity as storm water management tools, more emphasis has been applied to studying performance aspects, including stormwater retention. Of particular importance is the substrate layer in which the vegetation grows, which contributes the majority of stormwater retention capabilities. This research investigated many aspects of GR substrate performance, including component durability and component effects on hydraulic conductivity, matric potential, and plant growth. Several commercial substrate blends were tested for...
The Distribution of Gender Differences in the Temperament and Social Competence of Preschoolers
Laura Elizabeth Schussler
The literature has shown gender differences on many temperament and social competence (SC) measures, though there are gaps in understanding where differences lie and whether it varies by informant. This study investigated how temperament relates to SC and whether gender is a moderator. Rater source and the use of standardized versus raw scores and how they influence gender as a moderator was a main focus. Temperament was measured by the CBQ (Putnam & Rothbart, 2006)...
A STUDY OF THE PERCEIVED TEACHING SELF-EFFICACY AND LEADERSHIP OF NATIONAL BOARD CERTIFIED TEACHERS
April A. Zentmeyer
The purpose of this study was to examine the perceived levels of teacher self-efficacy and leadership of National Board Certified Teachers (NBCTs). One of the goals of Race to the Top is to provide highly qualified, efficacious teachers in every classroom, prepared to lead in the 21st century. Given that National Board Certification is one avenue to highly qualified status, this study sought to discover whether NBCTs perceived high levels of teacher self-efficacy and assumed...
ANALYSIS OF CONSENSUS GENOME-WIDE EXPRESSION-QTLS AND THEIR RELATIONSHIPS TO HUMAN COMPLEX TRAIT DISEASES
CHEN-HSIN YU
Genome-wide association studies of human complex disease have identified a large number of disease associated genetic loci. However, most of these risk loci do not provide direct information on the biological basis of a disease or on the underlying mechanisms. Recent genome-wide expression quantitative trait loci (eQTLs) association studies have provided information on genetic factors, especially SNPs, associated with gene expression variation. These eQTLs might contribute to phenotype diversity and disease susceptibility, but interpretation is...
Radio Resource Management in Heterogeneous Cellular Networks
Doohyun Sung
Heterogeneous cellular networks (HetNets) have been considered as one of enabling technologies not only to increase the cell coverage and capacity, but to improve the user experience. In this dissertation, we address two research challenges in HetNets: one is the cross-tier interference problem where cell range expansion (CRE) is applied for user offloading in cell association so that pico mobile stations located in expanded range (ER-PMSs), which are connected to macrocells unless CRE is enabled,...
Topological Edge States in Silicon Photonics
Sunil Mittal
Under the influence of a magnetic field, at low temperatures, charged particles confined in two-dimensional systems exhibit a remarkable range of macroscopic quantum phenomena such as the quantum Hall effects. A hallmark of these phenomena is the presence of unidirectional, topologically robust edge states - states which are confined to the edge of the system. It is, in principle, possible to engineer a synthetic magnetic field for photons and hence achieve photonic analogs of the...
Failure Mechanics of Functional Nanostructures in Advanced Technologies
Zheng Jia
The past decade has seen a surge of interest in developing novel functional nanostructures to enable advanced technologies, such as flexible electronics and high performance lithium-ion batteries. Examples of such functional nanostructures include organic/inorganic multi-layer thin films in flexible electronics and nano-sized silicon/tin-based anodes in lithium/sodium ion batteries. Widespread implementation of these advanced technologies with novel functional nanostructures in the future will broadly impact human's daily life. However, grand challenges for developing robust novel functional...
ACTIVE RELOCATION AND DISPATCHING OF HETEROGENEOUS EMERGENCY VEHICLES
Elham Sharifi
An emergency is a situation that causes an immediate risk to the property, health, or lives of civilians and can assume a variety of forms such as traffic accidents, fires, personal medical emergencies, terrorist attacks, robberies, natural disasters, etc. Emergency response services (ERSs) such as police, fire, and medical services play crucial roles in all communities and can minimize the adverse effects of emergency incidents by decreasing the response time. Response time is not only...
Determination of Network Connectivity: Algorithms and Applications in Organ Systems
Aimaschana Niruntasukrat
In this dissertation, we develop methods to determine the existence and types of connections between network nodes when the available data are only states (or signals) of such nodes. The proposed method determines the types of connections by examining the nodes' interaction. Determination inaccuracy caused by different reactive time scales between nodes is addressed by using wavelet analysis. After the network signals are decomposed into signals at different scales, the most prominent scale of each...
The effects of oxygen transition on community metabolism and nutrient cycling in a seasonally stratified anoxic estuary
Dong-Yoon Lee
Gradients of dissolved oxygen concentrations in seasonally stratified estuarine water columns directly influence microbial composition and metabolic pathways, resulting in annually recurring spatiotemporal chemical gradients of redox-active species. Understanding such microbial responses to variable geochemical conditions and elucidating the diversity of microbial processes are needed to comprehensively identify ecosystem functions. At first this study describes an investigation of the relationships between microbial processes and geochemical conditions. To assess the contribution of different biological redox processes...
• 2020
1,189
• 2019
2,236
• 2018
4,972
• 2017
684
• 2016
1,297
• 2015
690
• 2014
260
• Collection
11,328 | 2020-08-06 20:13:11 | {"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": 0.28283587098121643, "perplexity": 4881.470584504889}, "config": {"markdown_headings": false, "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-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00233.warc.gz"} |
https://scicomp.stackexchange.com/questions/37780/inverse-matrix-euqation-problem-with-restricted-condition | # Inverse matrix euqation problem with restricted condition
$$\underset{\Omega}{min}~\lambda\left\|A\Omega^{-1}B+C\right\|_F^{2}+\beta tr(W\Omega^{-1}W^{T}), s.t. tr(\Omega)=1, \Omega_{i,j} \ge 0$$, How to solve this problem with $$\Omega$$?
New contributor
tjufan is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
• This is the entire expression and it is a minimization problem. I have tried to find the solution by derivation。 – tjufan Jul 19 at 14:02 | 2021-07-23 15:05:48 | {"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": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5266619324684143, "perplexity": 764.4925551757161}, "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/1627046149929.88/warc/CC-MAIN-20210723143921-20210723173921-00123.warc.gz"} |
https://www.tutorvista.com/content/math/binomial-distribution/ | Top
# Binomial Distribution
Binomial distribution is a discrete probability distribution. Binomial distribution estimates probability distribution when there are "n" number of independent trials resulting in either success or failure. Following are the conditions for binomial distribution :
• There are "n" number of trails, where "n" is a finite number.
• These "n" trails are all mutually exclusive.
• There can only be two possible outcomes in every trial - "success" or "failure", "yes" or "no".
• Here, the probability of success is constant which is "p" for each trial.
• The probability of failure in each trials also a constant, say q. Eventually, q = 1 - p.
Formula for Binomial Distribution
The binomial probability function obtaining "k" successes in "n" trails is given by:
$B(k;n,p) =\ _{k}^{n}\textrm{C}\ p^{k}\ (1-p)^{n-k}$Where,
n = Total number of trials.
p = Probability of success in each trail.
1 - p = q = Probability of failure in each trial.
k = Number of successes in total n trials.
$_{k}^{n}\textrm{C}$ is known as binomial coefficient and is also written as C(n , k) or $\binom{n}{k}$.
Binomial distribution is known as Bernoulli distribution, when n = 1.If n trials are carried out with replacement, then the trials are independent. In this case, binomial distribution is used. It is very accurate and widely used to model "k" number of successes in "n" mutually exclusive trials. If experiment is done without replacement, then trials are not independent. In this case, binomial distribution can not be used.
Related Calculators Binomial Distribution Calculator Binomial Calculator Binomial Confidence Interval Calculator Binomial Multiplication Calculator
More topics in Binomial Distribution Binomial Distribution Formula Binomial Distribution Examples What is a Binomial Probability
*AP and SAT are registered trademarks of the College Board. | 2019-06-20 07:49:55 | {"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": 0.8579462766647339, "perplexity": 1164.1860920594377}, "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-2019-26/segments/1560627999163.73/warc/CC-MAIN-20190620065141-20190620091141-00272.warc.gz"} |
https://www.quizover.com/calculus/section/the-method-of-cylindrical-shells-by-openstax | # 6.3 Volumes of revolution: cylindrical shells
Page 1 / 6
• Calculate the volume of a solid of revolution by using the method of cylindrical shells.
• Compare the different methods for calculating a volume of revolution.
In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution. We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical shells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to help you determine which method to use in a given situation.
## The method of cylindrical shells
Again, we are working with a solid of revolution. As before, we define a region $R,$ bounded above by the graph of a function $y=f\left(x\right),$ below by the $x\text{-axis,}$ and on the left and right by the lines $x=a$ and $x=b,$ respectively, as shown in [link] (a). We then revolve this region around the y -axis, as shown in [link] (b). Note that this is different from what we have done before. Previously, regions defined in terms of functions of $x$ were revolved around the $x\text{-axis}$ or a line parallel to it.
As we have done many times before, partition the interval $\left[a,b\right]$ using a regular partition, $P=\left\{{x}_{0},{x}_{1}\text{,…},{x}_{n}\right\}$ and, for $i=1,2\text{,…},n,$ choose a point ${x}_{i}^{*}\in \left[{x}_{i-1},{x}_{i}\right].$ Then, construct a rectangle over the interval $\left[{x}_{i-1},{x}_{i}\right]$ of height $f\left({x}_{i}^{*}\right)$ and width $\text{Δ}x.$ A representative rectangle is shown in [link] (a). When that rectangle is revolved around the y -axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.
To calculate the volume of this shell, consider [link] .
The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius ${x}_{i}$ and inner radius ${x}_{i-1}.$ Thus, the cross-sectional area is $\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}.$ The height of the cylinder is $f\left({x}_{i}^{*}\right).$ Then the volume of the shell is
$\begin{array}{cc}\hfill {V}_{\text{shell}}& =f\left({x}_{i}^{*}\right)\left(\pi {x}_{i}^{2}-\pi {x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}^{2}-{x}_{i-1}^{2}\right)\hfill \\ & =\pi f\left({x}_{i}^{*}\right)\left({x}_{i}+{x}_{i-1}\right)\left({x}_{i}-{x}_{i-1}\right)\hfill \\ & =2\pi f\left({x}_{i}^{*}\right)\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\left({x}_{i}-{x}_{i-1}\right).\hfill \end{array}$
Note that ${x}_{i}-{x}_{i-1}=\text{Δ}x,$ so we have
${V}_{\text{shell}}=2\pi f\left({x}_{i}^{*}\right)\left(\frac{{x}_{i}+{x}_{i-1}}{2}\right)\text{Δ}x.$
Furthermore, $\frac{{x}_{i}+{x}_{i-1}}{2}$ is both the midpoint of the interval $\left[{x}_{i-1},{x}_{i}\right]$ and the average radius of the shell, and we can approximate this by ${x}_{i}^{*}.$ We then have
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The | 2018-07-22 03:05:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 27, "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": 0.8434001803398132, "perplexity": 674.0712593550109}, "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-2018-30/segments/1531676593004.92/warc/CC-MAIN-20180722022235-20180722042235-00566.warc.gz"} |
https://www.shaalaa.com/question-bank-solutions/find-value-k-which-each-following-systems-equations-has-infinitely-many-solutions-4x-5y-3-kx-15y-9-pair-linear-equations-two-variables_22402 | Share
# Find the Value Of K For Which Each of the Following Systems of Equations Has Infinitely Many Solutions : 4x + 5y = 3 Kx + 15y = 9 - CBSE Class 10 - Mathematics
ConceptPair of Linear Equations in Two Variables
#### Question
Find the value of k for which each of the following systems of equations has infinitely many solutions :
4x + 5y = 3
kx + 15y = 9
#### Solution
The given system of equation is
4x + 5y -3 = 0
ks + 15y - 9 = 0
The system of equation is of the for
a_1x + b_1y + c_1= 0
a_2x + b_2y + c_2 = 0
Where a_1 = 4, b_1 = 5, c_1 = -3
And a_2 = k, b_2 = 15, c_2 = -9
For a unique solution, we must have
a_1/a_2 = b_1/b_2 = c_2/c_2
=> 4/k = 5/15 = (-3)/(-9)
Now
4/k = 5/15
=> 4/k = 1/3
=> k = 12
Hence, the given system of equations will have infinitely many solutions, if k = 12
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [1]
Solution Find the Value Of K For Which Each of the Following Systems of Equations Has Infinitely Many Solutions : 4x + 5y = 3 Kx + 15y = 9 Concept: Pair of Linear Equations in Two Variables.
S | 2019-12-13 15:29:19 | {"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": 0.6942411065101624, "perplexity": 585.4974563823755}, "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-2019-51/segments/1575540564599.32/warc/CC-MAIN-20191213150805-20191213174805-00248.warc.gz"} |
https://tex.stackexchange.com/questions/70236/linebreak-in-math-environment | # Linebreak in math environment
I have an odd problem, I tried already all possible things, align, eqnarray, $, and so on, but the following stays the same: I have a formula inside text and it is not broken, so it goes into the right margin, how can I say LaTeX it should do a linebreak? Here's the code: \documentclass{book} \usepackage{geometry} \geometry{left=4cm,right=3cm, top=2cm, bottom=2cm} \usepackage[pagestyles]{titlesec} \titlespacing*{\chapter}{0pt}{-30pt}{20pt} \titleformat{\chapter}[display]{\color{headercolor}\normalfont\huge\bfseries}{\chaptertitlename\ \thechapter}{20pt}{\Huge} \usepackage[ngerman]{babel} \usepackage{mathptmx} \usepackage{helvet} \usepackage{wallpaper} \usepackage{color} \usepackage[final]{pdfpages} %,bookmarksopenlevel={1} %\usepackage[hidelinks,bookmarks=true,bookmarksopen=true,bookmarksnumbered=true,colorlinks=true,linkcolor=black,]{hyperref} \usepackage[hidelinks,bookmarksopen=false, hypertexnames=TRUE,pdfpagelabels=true]{hyperref}[2011/02/05]%colorlinks,linkcolor=black, \hypersetup{ pdftitle={}, % pdfauthor={\textcopyright }, pdfsubject={statistics Buch}, pdfkeywords={}, } \usepackage{xcolor,bookmark} \usepackage{scrextend} \usepackage{titlepic} \usepackage{shorttoc} \usepackage{courier} \usepackage{type1cm} \usepackage{zref-abspage} \usepackage{graphicx} \usepackage{multicol} \usepackage[bottom]{footmisc} \usepackage{tocstyle} \usetocstyle{allwithdot} %\usepackage{thmbox} \usepackage{shadethm} \usepackage{amsthm} \usepackage{amsmath} \usepackage{marginnote} %\usetocstyle{KOMAlike} \usepackage{wrapfig} \usepackage{paralist} \usepackage{amssymb} \usepackage[framemethod=tikz]{mdframed} \usepackage{ulem} \usepackage{fancyhdr} \begin{document} \pagenumbering{arabic} % %\pagestyle{mystyle} \mainmatter \section{text text} text\\text a lot of text\\ and then the problem with the formula:\\ Die$x_isind die Zuwachsraten, ingesamt haben wir 4 Jahre, es ergibt sich also \begin{eqnarray*}\bar{x}_{Geom}=\sqrt[4]{\prod \limits_{i=1}^{4} x_i}=\sqrt[4]{1,01*1,024*0,987*1,034 }=(1,01*1,024*0,987*1,034 )^{\frac{1}{4}}=1,0135957\end{eqnarray*} Wir machen die Probe: \\ and so on\\ \end{document} Heres a screenshot: Any ideas? ## 2 Answers (La)TeX does not break lines automatically within math mode like it does in text mode for paragraphs. You need to introduce this line break manually, or provide support for this line-breaking via something like breqn. In the example below, I've used amsmath's align* which allows the introduction of a line-break via \\: \documentclass{article} \usepackage{amsmath}% http://ctan.org/pkg/amsmath \begin{document} \section{text text} Die~x_i\$ sind die Zuwachsraten, ingesamt haben wir 4 Jahre, es ergibt sich also
\begin{align*}
\bar{x}_{\text{Geom}} &= \sqrt[4]{\prod_{i=1}^{4} x_i} \\
&= \sqrt[4]{1.01 \times 1.024 \times 0.987 \times 1,034} \\
&= (1.01 \times 1.024 \times 0.987 \times 1.034)^{\frac{1}{4}} \\
&= 1.0135957
\end{align*}
Wir machen die Probe: \ldots
\end{document}
Also note the use of . as a decimal separator, which provides a different spacing than that of ,. If you wish to use the latter, you need to introduce some way of compensating for the different notation. Either icomma or siunitx may be of help here.
• ok, thanks a lot, but do I have to use \times or is * also ok? I want to use the comma, because I have the dot as a marker for "thousand" Sep 5, 2012 at 12:47
• how I can I implement this? Just say \usepackage{icomma} in the preamble or do I have to write also something else? Thanks a lot! Sep 5, 2012 at 12:49
• I would use siunitx and its \num command, it can be told to use comma as the decimal marker and it can separate numbers in thousands such that you don't need to. And yes I would write \times or \cdot, it looks a lot more professionel. Sep 5, 2012 at 12:54
LaTeX does not do automatic line breaking of displayed formuals (as it is very hard to teach a computer that is good practice in this case, there are other rules at play here than what applies to text-mode math). You need to introduce a manual line break via \\
An please forget you ever head of eqnarray it is a broken construction, see http://tug.org/TUGboat/tb33-1/tb103madsen.pdf
and stop using \\ in the text, that road does not lead to anywhere pleasant. | 2022-07-04 20:50:20 | {"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": 2, "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": 0.9992174506187439, "perplexity": 3454.740761184797}, "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-2022-27/segments/1656104496688.78/warc/CC-MAIN-20220704202455-20220704232455-00215.warc.gz"} |
http://clay6.com/qa/30514/atomic-radii-of-flurine-and-neon-in-angstrom-units-are-respectively-given-b | Browse Questions
# Atomic radii of flurine and neon in Angstrom units are respectively given by
$(a)\;1.60,1.60 \\(b)\;0.72,0.72 \\ (c)\;0.72,1.60 \\ (d)\;1.60,0.72$
$0.72,1.60$ is correct answer.
Hence c is the correct answer. | 2016-12-10 22:21:03 | {"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": 0.9293124079704285, "perplexity": 4515.162332745069}, "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-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00293-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://www.ovito.org/docs/current/python/introduction/rendering.php | # Rendering & visualization¶
## Visual elements¶
In OVITO, data objects are separated from visual elements, which are responsible for producing a visual representation of the data objects. For example, a SimulationCell is a data object storing the simulation cell vectors and the periodic boundary flags. The corresponding visual element, a SimulationCellVis, takes this information to generate the actual geometry primitives to visualize the simulation cell in the viewports and in rendered pictures. A visual element typically has a set of parameters that control the visual appearance, for example the line color of the simulation box.
A visual element is attached to the DataObject that it should visualize, and you access it through the vis attribute:
>>> data.cell # This is the data object
<SimulationCell at 0x7f9a414c8060>
>>> data.cell.vis # This is the attached visual element
<SimulationCellVis at 0x7fc3650a1c20>
>>> data.cell.vis.rendering_color = (1,0,0) # Giving the simulation box a red color
All display objects are derived from the DataVis base class, which defines the enabled attribute turning the rendering of the data object on or off:
>>> data.cell.vis.enabled = False # This hides the simulation cell
The visual display of particles is controlled by a ParticlesVis object, which is attached to the Particles data object. For example, to display cubic particles:
>>> data.particles.vis.shape = ParticlesVis.Shape.Square
Some modifiers in a data pipeline may produce new data objects, for example as a result of a computation. The CalculateDisplacementsModifier generates a new particle Property that stores the computed displacement vectors. To support the visualization of displacement vectors as arrows, the CalculateDisplacementsModifier automatically attaches a VectorVis element to the new particle property. We can access this visual element in two ways: either directly through the vis field of the modifier:
>>> modifier = CalculateDisplacementsModifier()
>>> pipeline.modifiers.append(modifier)
>>> modifier.vis.enabled = True # Enable the display of arrows
>>> modifier.vis.color = (0,0,1) # Give arrows a blue color
or via the vis field of the particle Property
>>> data = pipeline.compute()
>>> data.particles.displacements.vis.enabled = True # Enable the display of arrows
>>> data.particles.displacements.vis.color = (0,0,1) # Give arrows a blue color
## Viewports¶
A Viewport defines a view of the three-dimensional scene, in which the visual representation of the data of a pipeline is generated. To render a picture of the scene, you typically create a new Viewport object and configure it by setting the camera position and orientation:
import math
from ovito.vis import Viewport
vp = Viewport()
vp.type = Viewport.Type.Perspective
vp.camera_pos = (-100, -150, 150)
vp.camera_dir = (2, 3, -3)
As known from the interactive OVITO program, there exist various standard viewport types such as TOP, FRONT, etc. The PERSPECTIVE and ORTHO viewport types allow you to freely orient the camera in space and are usually what you need in a Python script. Don’t forget to set the viewport type first before configuring any other camera-related parameters. That’s because changing the viewport type will reset the camera orientation to a default value.
The PERSPECTIVE viewport type selects a perspective projection, and you can control the vertical field of view by setting the fov parameter to the desired angle. The ORTHO viewport type uses a parallel projection; In this case, the fov parameter specifies the vertical size of the visible area in units of length. Optionally, you can call the Viewport.zoom_all() method to let OVITO automatically choose a reasonable camera zoom and position such that all objects become completely visible.
## Rendering¶
Rendering of images and movies is done using the Viewport.render_image() and Viewport.render_anim() methods:
vp.render_image(size=(800,600), filename="figure.png", background=(0,0,0), frame=8)
vp.render_anim(size=(800,600), filename="animation.avi", fps=20)
OVITO provides several different rendering engines, which differ in terms of speed and visual quality. The default rendering engine is the OpenGLRenderer, which implements a fast, hardware-accelerated OpenGL rendering method. See the ovito.vis module for the list of other available engines. To use them, you have to create an instance of the renderer class, configure its specific parameters, and pass the renderer to the viewport rendering function:
tachyon = TachyonRenderer(shadows=False, direct_light_intensity=1.1)
vp.render_image(filename="figure.png", background=(1,1,1), renderer=tachyon) | 2020-07-07 00:19:02 | {"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": 0.23427700996398926, "perplexity": 3067.558514117281}, "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-2020-29/segments/1593655890566.2/warc/CC-MAIN-20200706222442-20200707012442-00149.warc.gz"} |
http://mathoverflow.net/questions/80378/produce-an-irreducible-polynomial-that-cant-be-proved-irreducible-by-using-eise?sort=oldest | # Produce an irreducible polynomial that can't be proved irreducible by using Eisenstein [closed]
give An example of an irreducible polynomial that cannot prove it by using the Eisenstein criterion even with the use of all linear change variable($x-c=y$).
-
## closed as too localized by Felipe Voloch, Mark Sapir, Martin Brandenburg, Bruce Westbury, Torsten EkedahlNov 8 '11 at 12:39
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
It would be nice if you didn't formulate this as a command ("give an example...") and explained why you are asking (idle curiosity, homework,...). Your question isn't at the intended level of MO, but I'll make a comment which I think is: if $K$ is a number field in which (1) the ring of integers has the form ${\mathbf Z}[\alpha]$ and (2) no prime number is totally ramified, then the minimal polynomial of $\alpha$ over ${\mathbf Q}$ has the feature you seek. Many cyclotomic extensions of ${\mathbf Q}$ fits these properties. – KConrad Nov 8 '11 at 11:59
Newton polygon scenarios systematically give examples just-slightly-more-complicated than Eisenstein-criterion examples. E.g., $x^{n+1}+2x+4$: the slopes are $1/n$ $n$ times and a single $1$. Thus, this has at least an irreducible degree-$n$ factor. Excluding a rational root is easy (not $\pm 1,\pm 2\pm 4$), so it's irreducible. – paul garrett Nov 8 '11 at 16:40
$x^2+8$ is an example.
Let $L/K$ be an unramifed extension of numeber fields (for instance the Hilbert class field of a $K$ with non-trivial class group), generated by $\alpha$ say. Then the minimal polynomial for $\alpha$ over $K$ will do. | 2015-09-03 17:39:37 | {"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": 0.8371255993843079, "perplexity": 599.249077525545}, "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-2015-35/segments/1440645322940.71/warc/CC-MAIN-20150827031522-00041-ip-10-171-96-226.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2267707/a-geometric-inequality-concerning-angles | # A geometric inequality concerning angles
Prove or disprove that: In an acute angled triangle if $A,B,C$ are angles $a,b,c,$ are respective opposite sides and $R$ is circum-radius, then
$$\prod_{cyc}\bigg(\dfrac{2A}{\pi}\bigg)^{\dfrac 1a} \le \bigg(\dfrac{2}{3}\bigg)^{\dfrac{\sqrt 3}{R}}$$
I tried to use AM-GM after representing the $a$ as $\sin$, but didn't succeed. Please help.
• You should say that you have an equality in the case of an equilateral triangle. – Jean Marie May 5 '17 at 22:04
I think it's true for all triangle.
Indeed, we need to prove that $$\sum_{cyc}\frac{1}{a}\ln\frac{2\alpha}{\pi}\leq\frac{\sqrt3}{R}\ln\frac{2}{3}$$ or $$\sum_{syc}\frac{\ln\frac{2\alpha}{\pi}}{\sin\alpha}\leq2\sqrt3\ln\frac{2}{3}.$$ Let $f(x)=\frac{\ln\frac{2x}{\pi}}{\sin{x}}.$ We see that $f''(x)<0$ for all $x\in(0,1.752...)$,
which says that $f$ is a concave function for all $x\in\left(0,\frac{\pi}{3}\right]$.
Thus, by the Vasc's LCF Theorem it's enough to prove our inequality for $\beta=\alpha$.
Hence, $\gamma=\pi-2\alpha$, where $0<\alpha<\frac{\pi}{2}$ and we need to prove that $$\frac{2\ln\frac{2\alpha}{\pi}}{\sin\alpha}+\frac{\ln\frac{2(\pi-2\alpha)}{\pi}}{\sin2\alpha}\leq2\sqrt3\ln\frac{2}{3},$$ which is indeed true.
Done!
About LCF Theorem see here: https://diendantoanhoc.net/index.php?app=core&module=attach§ion=attach&attach_id=4416
• Sir can you please suggest some books for inequalities? I also like solving inequalities but have very little exposure to them. – user428700 May 6 '17 at 5:35
• @DarthVader1056 I think the best book it's the last book of Vasile Cirtoaje (he wrote it two years ago). – Michael Rozenberg May 6 '17 at 5:51 | 2020-02-25 22:34:18 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8374685645103455, "perplexity": 314.9713114479076}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-10/segments/1581875146160.21/warc/CC-MAIN-20200225202625-20200225232625-00448.warc.gz"} |
https://stats.stackexchange.com/questions/422698/in-an-observational-study-where-matching-is-properly-conducted-should-we-expect | # In an observational study where matching is properly conducted, should we expect the $ATE$ to be equal to the $ATT$?
In a randomized study, I know that the $$ATE=ATT=ATC$$, where $$ATE$$ is the average treatment effect, $$ATT$$ is the average treatment effect on the treated, and $$ATC$$ is the average treatment effect on the control. In an observational study this is usually not the case.
If we have a matching procedure is that is properly conducted on a set of covariates $$X$$ such that unconfoundedness holds, i.e., where
$$(Y(1), Y(0)) \perp Z \mid X$$
where $$(Y(1), Y(0))$$ are the potential outcomes, $$Z$$ the treatment, and the $$\perp$$ symbol meaning independence, should we expect the $$ATE$$ to be equal to the $$ATT$$?
In such a scenario, would the $$ATC$$, then be then $$0$$? Or are there other factors at play?
The $$ATT$$, $$ATC$$, and $$ATE$$ are all average treatment effects averaging over a target group of individuals. The idea is that everyone has an individual treatment effect, and an average treatment effect is the average of the individual treatment effects across individuals in the target population/sample. Individuals' treatment effects vary from each other when the treatment effect depends on qualities of the individual, say, $$X$$, and different individuals have different values of $$X$$. This circumstance is known as treatment effect heterogeneity based on $$X$$. For example, it may be that a medicine works better for young people than it does for old people; there is treatment effect heterogeneity based on age.
When there is treatment effect heterogeneity based on $$X$$, the average treatment effect for a given group depends on the distribution of $$X$$ in that group. Returning to that example, the average effect of the medicine will be larger for a group of younger people that it will be for a group of older people. When the treated and control group have different distributions of $$X$$, the average treatment effect among the treated ($$ATT$$) will differ from the average treatment effect among the control ($$ATC$$), and both will differ from the average treatment effect in the entire sample ($$ATE$$). The treated and control groups will have different distributions of $$X$$ only when treatment depends on $$X$$ or a correlate of $$X$$, which is not the case in a randomized experiment, in which the distributions of all covariates, including $$X$$, are the same between the treated and control.
In matching as described by Stuart (2010) and Ho, Imai, King, and Stuart (2007) and implemented in the R package MatchIt, matching does indeed recover the $$ATT$$ if only control units are pruned away and no treated units are dropped. This is because the distribution of covariates in the matched control group is the same as the distribution of covariates in the original treated group, so the distribution of individual treatment effects will be that in the treated group. However, it is possible to estimate the $$ATE$$ using matching, as described by Abadie and Imbens (2006) and implemented in the R package Matching. In this method, the distribution of covariates in the matched sample is similar to that in the original sample.
Note that unconfoundedness and the conditional independence statement you wrote are a property of the causal system, not of a sample (matched or otherwise). It's a statement that says that it is possible to extract a causal effect using the observed data. It's not a description of a sample after matching. If that statement is true about the causal system, then any of several estimators (including matching, weighting, regression, etc.) has the potential to estimate the causal effect without bias.
To summarize, the average treatment effect across groups will differ when the distribution of individual treatment effects varies between the groups, which occurs when individual treatment effects depend on covariates and the distribution of those covariates varies between the groups. The $$ATT$$ and $$ATC$$ will differ when the distribution of covariates differs between the treated and control, which occurs in the absence of randomization.
Abadie, A., & Imbens, G. W. (2006). Large Sample Properties of Matching Estimators for Average Treatment Effects. Econometrica, 74(1), 235–267. https://doi.org/10.1111/j.1468-0262.2006.00655.x
Ho, D. E., Imai, K., King, G., & Stuart, E. A. (2007). Matching as Nonparametric Preprocessing for Reducing Model Dependence in Parametric Causal Inference. Political Analysis, 15(3), 199–236. https://doi.org/10.1093/pan/mpl013
Stuart, E. A. (2010). Matching Methods for Causal Inference: A Review and a Look Forward. Statistical Science, 25(1), 1–21. https://doi.org/10.1214/09-STS313
• Thank you very much, this is a really great answer and clears a lot of things. I have a quick question regarding when you write "...matching does indeed recover the $ATT$ if only control units are pruned away and no treated units are dropped". Is the reason we cannot drop treated units because if we were to drop treated units, we "lose" information regarding the distribution of the treated? Aug 19 '19 at 19:50
• Thank you for the kind words. Yes, that's right. The distribution of remaining individuals would not correspond to a natural group. Some researchers are not bothered by this, though. It's a matter of internal validity vs. external validity. If you see someone perform caliper matching and some treated units are not matched to any controls, the estimand is neither the ATT nor ATE but the average treatment effect in the matched sample.
– Noah
Aug 19 '19 at 19:53 | 2021-10-17 07:20:31 | {"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": 33, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.641089677810669, "perplexity": 435.9217372241076}, "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-2021-43/segments/1634323585121.30/warc/CC-MAIN-20211017052025-20211017082025-00337.warc.gz"} |
https://www.or-exchange.org/questions/10638/convex-programming-solver-taking-objective-function-value-as-a-oracle | # convex programming solver taking objective function value as a oracle
1 Is there any solver for convex optimization in C++ (or some dedicated scheme while no solver is yet available) that could solve a convex optimization problem with objective function value given by an oracle? Thank you. My specific problem is this: $\mathop {\max }\limits_\lambda \mathop {\min }\limits_{\sigma \in {{\{ 0,1\} }^N}} {E_{\sigma ,\,\lambda }}$ wherer lambda is a vector, and for each ${\sigma \in {{\{ 0,1\} }^N}}$ E is a linear function of lambda ${E_{\sigma ,\,\lambda }}$ In words: It is actually maximize over lambda the piece-wise linear function defined by the minimum of exponential number of linear functions. Given lambda I have an effective scheme to obtain sigma and thus calculate $\mathop {\min }\limits_{\sigma \in {{\{ 0,1\} }^N}} {E_{\sigma ,\,\lambda }}$ . so my problem is effectively a convex optimization with objective function given by my oracles (maximize over a concave function) and I am wondering whether there would be some solvers suitable to this type of problem. Or if there is any dedicated procedure for this while no solvers available. Thank you:D asked 13 Nov '14, 11:25 Chivalry 229●1●17 accept rate: 0% I am unaware of such a solver, but in general your problem (i.e., solve a convex problem with an oracle for the objective function) is difficult. As an example, consider optimizing over the TSP polytope. (13 Nov '14, 13:21) Austin Buchanan It is difficult, but we really need to solve it:) do you know some paper on it? (13 Nov '14, 14:49) Chivalry 1 It seems you could use any LP solver and just keep adding cuts (constraints) supplied by the oracle. (17 Nov '14, 16:48) Paul Rubin ♦♦ actually, I am also trying this out and just wonder whether there might be some developed advanced method for that.... :D (17 Nov '14, 16:50) Chivalry Im not quite sure, but maximizing a concave non-differential function sounds like subgradient methods to me? (18 Nov '14, 07:10) Sune
0 Hello, Did you take a look at CVX? It is not a 'solver' but a modeling system for convex programs which then uses and underlying solver. Since the version 2.0 it can be used with Gurobi and MOSEK and, if your problem is not too large scale, I believe it can tackle that kind of problem. answered 14 Nov '14, 03:06 JorgeGarciaC... 29●4 accept rate: 0% 1 Thank you. But it seems cvx could not solve when my input objective function is an oracle... (14 Nov '14, 08:12) Chivalry
toggle preview community wiki
By Email:
Markdown Basics
• *italic* or _italic_
• **bold** or __bold__
• image?
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported
Tags:
×231
×20 | 2019-09-21 13:33:13 | {"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": 0.8269489407539368, "perplexity": 1567.5021497171538}, "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-2019-39/segments/1568514574501.78/warc/CC-MAIN-20190921125334-20190921151334-00377.warc.gz"} |
https://www.physicsforums.com/threads/expectation-values-as-a-phase-space-average-of-wigner-functions.911092/ | # Homework Help: Expectation values as a phase space average of Wigner functions
Tags:
1. Apr 12, 2017
### Gabriel Maia
Hi. I'm trying to prove that
$$[\Omega] = \int dq \int dp \, \rho_{w}(q,p)\,\Omega_{w}(q,p)$$
where
$$\rho_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\rho|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})$$
is the Wigner function, being \rho a density matrix. On the other hand
$$\Omega_{w}(q,p) = \frac{1}{2\pi\hbar} \int dy \, \langle q-\frac{y}{2}|\Omega|q+\frac{y}{2}\rangle\,\exp(i\frac{py}{\hbar})$$
is the Wigner representation of the operator I'm interested in. The expectation value of an operator can be calculated from
$$[\Omega] = Tr(\rho\Omega) = \int dp \, \langle p|\rho\Omega|p\rangle$$
$$= \int dp^{\prime} \int dp \, \langle p|\rho|p^{\prime}\rangle\langle p^{\prime}|\Omega|p\rangle$$
Now, the matrix elements $$\langle p | \rho | p^{\prime} \rangle \,\,\, \mathrm{and} \,\,\, \langle p | \Omega| p^{\prime} \rangle$$
Should lead to the Wigner function and the Wigner representation of the operator but they only do so if
$$p=p^{\prime}$$
Do you know how can I solve this?
Thank you.
2. Apr 17, 2017
### PF_Help_Bot
Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook
Have something to add?
Draft saved Draft deleted | 2018-06-24 02:16:45 | {"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": 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": 0.6503093242645264, "perplexity": 1805.9708320808486}, "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-2018-26/segments/1529267865995.86/warc/CC-MAIN-20180624005242-20180624025242-00245.warc.gz"} |
https://www.physicsforums.com/threads/two-blocks-and-a-pulley.768381/ | # Two blocks and a pulley
## Homework Statement
Mass Ma lies on top of mass Mb, as shown. Assume Mb > Ma. The two blocks are pulled from rest by a massless rope passing over a pulley. The pulley is accelerated at rate A. Block Mb slides on the table without friction, but there is a constant friction force f between Ma and Mb due to their relative motion. Find the tension in the rope.
## The Attempt at a Solution
Newton's second law in the y direction for both masses:
$N_1 - M_a g = 0$
$N_2 - N_1 - M_b g = 0$
In the x-direction:
$T_a - F_a = M_a a_a$
$T_b + F_a = M_b a_b$
where
$F_a = \mu M_a g$
Ok, so I suppose we can assume the tension in the rope is the same at every point since it's massless. So $T_a = T_b = T$. Now, the thing is, friction can't cause the blocks to move, it can only oppose motion. So maybe the friction here has to be static friction and we assume the blocks are not moving relative to each other? Otherwise, one of the blocks would have to be moving opposite the direction in which we apply the force, which seems wrong. I'm really not sure where to go with this problem.
I think you are doing alright . In the question it is given that there is relative motion between the blocks , so the friction between the blocks is kinetic in nature . You just take it as 'f' . No need to introduce coefficient of friction.
Apart from the equations you have formed you need one more relation i.e between the acceleration of pulley and the acceleration of the blocks .
I guess that is the part that I am having the most difficulty with. I don't see any way to incorporate the acceleration applied to the pulley. The only thing I can think of is that accelerating the pulley to the right is the same as subjecting the pulley to a gravitational acceleration to the left and holding it in place. Then we could say that the blocks experience a fictitious "force" to the left from this acceleration. But other than that I am stumped.
Look at the picture I have attached .
Let a,b,c be the coordinates of the lower block ,upper block and the pulley respectively.Express the length of the rope in terms of a,b,c .Then use the fact that rope length remains constant . This will give you the relationship between the acceleration of the three entities .
#### Attachments
• rope.PNG
681 bytes · Views: 1,621
Last edited:
Oh... ok. So L = (c - a) + (c - b) and differentiating twice, $2A = a_a + a_b$. Solving for T we get $T = 2 \frac{m_a m_b}{m_a + m_b}A + f$
Oh... ok. So L = (c - a) + (c - b) and differentiating twice, $2A = a_a + a_b$
Correct .
Solving for T we get $T = 2 \frac{m_a m_b}{m_a + m_b}A + f$
This doesn't look right .Please recheck.
ranjhahah
haruspex
Solving for T we get $T = 2 \frac{m_a m_b}{m_a + m_b}A + f$ | 2021-09-28 19:18:32 | {"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": 0.7550320625305176, "perplexity": 299.49313436788174}, "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-2021-39/segments/1631780060882.17/warc/CC-MAIN-20210928184203-20210928214203-00287.warc.gz"} |
http://www.singaporemathguru.com/question?cr=WKDo3C9wQ9FZiU4gxBGq&mode=chapterquiz&sts1=str | ### Primary 6 Problem Sums/Word Problems - Try FREE
Score :
Disabled
#### Question 1 of 5
A small circle is embedded in a larger circle.
The centres of both circles are the same point.
The small circle has a radius of 10 cm while the larger circle has a radius of 16 cm.
What is the area of the shaded region?
[ Take π to be 3.14 ]
Notes to students:
cm^2 | 2017-07-25 02:35:55 | {"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": 0.5663010478019714, "perplexity": 1136.8630291868476}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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/1500549424960.67/warc/CC-MAIN-20170725022300-20170725042300-00291.warc.gz"} |
https://leetcode.ca/2016-12-17-383-Ransom-Note/ | # Question
Formatted question description: https://leetcode.ca/all/383.html
383 Ransom Note
Given an arbitrary ransom note string and another string containing letters from all the magazines,
write a function that will return true if the ransom note can be constructed from the magazines ;
otherwise, it will return false.
Each letter in the magazine string can only be used once in your ransom note.
Note:
You may assume that both strings contain only lowercase letters.
canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true
# Algorithm
Hash Map counts the number of characters.
Java | 2022-07-05 08:42:47 | {"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": 0.33786508440971375, "perplexity": 8118.239900246822}, "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-2022-27/segments/1656104542759.82/warc/CC-MAIN-20220705083545-20220705113545-00254.warc.gz"} |
https://blogs.mathworks.com/community/2009/05/25/colorizing-text-output/?s_tid=blogs_rc_3 | Colorizing text output
There have been a number of requests recently asking that we support colorized output in the Command Window (here, here and here). I won’t be offering a way to colorize the Command Window content, but I will be offering an alternative way to generate colorized output for viewing in a web browser.
Some of the users I’ve talked with want to color their output in the Command Window so that they can easily pick out print-outs that indicate a problem. Imagine you had a long running program that printed various “good” diagnostics, but also occasionally printed out a message indicating that something wasn’t quite right and needed to be looked at. It would be nice to color that text in red, or some other attention grabbing color. This is the case we’ll address below.
We’ll start by writing a simple function that takes a string and wraps it in an HTML font tag. This function will also take a color, which will be used for the font tags color attribute. Because the result will be HTML , the color value can be anything supported by font‘s color attribute (e.g. red, #cccccc etc.). Here’s what that function looks like:
function colorizedstring = colorizestring(color, stringtocolorize)
%colorizestring wraps the given string in html that colors that text.
colorizedstring = ['<font color="', color, '">', stringtocolorize, '</font>'];
end
Now all we need to do is use the function. Start by creating an HTML file in which to put your output text. Then, run your function and write to the output file as necessary using the colorizestring function. Finally, close the file for writing and open it using the web command.
Here’s what a sample usage might look like:
%% Open an HTML file to write the output into.
filename = 'colorizedoutput.html';
fid = fopen(filename, 'wt');
%% Do the long running calculation that generates lots of output.
for i=1:100
outputswitch = rand;
if rand < 0.80
fprintf(fid, colorizestring('black', 'Everything is fine.<br/>'));
elseif rand < 0.90
fprintf(fid, colorizestring('orange', 'There may be a problem.<br/>'));
else
fprintf(fid, colorizestring('red', 'There is a problem.<br/>'));
end
fprintf(fid, 'some string<br/>');
end
%% Close the HTML output file.
fclose(fid);
%% Open the HTML output file in MATLAB's web browser.
web(filename);
and the resulting output:
This won’t solve everyone’s problems, but it should help to fill the gap until we can fully support colorized output in the Command Window.
| | 2022-07-03 06:13:49 | {"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": 0.23604419827461243, "perplexity": 2790.677457698795}, "config": {"markdown_headings": false, "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-2022-27/segments/1656104215790.65/warc/CC-MAIN-20220703043548-20220703073548-00692.warc.gz"} |
https://repository.kaust.edu.sa/handle/10754/325163/recent-submissions | Now showing items 1-20 of 842
• #### ADOM: Accelerated Decentralized Optimization Method for Time-Varying Networks
(arXiv, 2021-02-18) [Preprint]
We propose ADOM - an accelerated method for smooth and strongly convex decentralized optimization over time-varying networks. ADOM uses a dual oracle, i.e., we assume access to the gradient of the Fenchel conjugate of the individual loss functions. Up to a constant factor, which depends on the network structure only, its communication complexity is the same as that of accelerated Nesterov gradient method (Nesterov, 2003). To the best of our knowledge, only the algorithm of Rogozin et al. (2019) has a convergence rate with similar properties. However, their algorithm converges under the very restrictive assumption that the number of network changes can not be greater than a tiny percentage of the number of iterations. This assumption is hard to satisfy in practice, as the network topology changes usually can not be controlled. In contrast, ADOM merely requires the network to stay connected throughout time.
• #### Delineating the molecular and phenotypic spectrum of the SETD1B-related syndrome
(Cold Spring Harbor Laboratory, 2021-02-18) [Preprint]
ABSTRACTPathogenic variants in SETD1B have been associated with a syndromic neurodevelopmental disorder including intellectual disability, language delay and seizures. To date, clinical features have been described for eleven patients with (likely) pathogenic SETD1B sequence variants. We perform an in-depth clinical characterization of a cohort of 36 unpublished individuals with SETD1B sequence variants, describing their molecular and phenotypic spectrum. Selected variants were functionally tested using in vitro and genome-wide methylation assays. Our data present evidence for a loss-of-function mechanism of SETD1B variants, resulting in a core clinical phenotype of global developmental delay, language delay including regression, intellectual disability, autism and other behavioral issues, and variable epilepsy phenotypes. Developmental delay appeared to precede seizure onset, suggesting SETD1B dysfunction impacts physiological neurodevelopment even in the absence of epileptic activity. Interestingly, males are significantly overrepresented and more severely affected, and we speculate that sex-linked traits could affect susceptibility to penetrance and the clinical spectrum of SETD1B variants. Finally, despite the possibility of non-redundant contributions of SETD1B and its paralogue SETD1A to epigenetic control, the clinical phenotypes of the related disorders share many similarities, indicating that elucidating shared and divergent downstream targets of both genes will help to understand the mechanism leading to the neurobehavioral phenotypes. Insights from this extensive cohort will facilitate the counseling regarding the molecular and phenotypic landscape of newly diagnosed patients with the SETD1B-related syndrome.
• #### Multi-dimensional wave steering with higher-order topological phononic crystal
(arXiv, 2021-02-17) [Preprint]
The recent discovery and realizations of higher-order topological insulators enrich the fundamental studies on topological phases. Here, we report three-dimensional (3D) wave-steering capabilities enabled by topological boundary states at three different orders in a 3D phononic crystal with nontrivial bulk topology originated from the synergy of mirror symmetry of the unit cell and a non-symmorphic glide symmetry of the lattice. The multitude of topological states brings diverse possibility of wave manipulations. Through judicious engineering of the boundary modes, we experimentally demonstrate two functionalities at different dimensions: 2D negative refraction of sound wave enabled by a first-order topological surface state with negative dispersion, and a 3D acoustic interferometer leveraging on second-order topological hinge states. Our work showcases that topological modes at different orders promise diverse wave steering applications across different dimensions.
• #### Shape-Tailored Deep Neural Networks
(arXiv, 2021-02-16) [Preprint]
We present Shape-Tailored Deep Neural Networks (ST-DNN). ST-DNN extend convolutional networks (CNN), which aggregate data from fixed shape (square) neighborhoods, to compute descriptors defined on arbitrarily shaped regions. This is natural for segmentation, where descriptors should describe regions (e.g., of objects) that have diverse shape. We formulate these descriptors through the Poisson partial differential equation (PDE), which can be used to generalize convolution to arbitrary regions. We stack multiple PDE layers to generalize a deep CNN to arbitrary regions, and apply it to segmentation. We show that ST-DNN are covariant to translations and rotations and robust to domain deformations, natural for segmentation, which existing CNN based methods lack. ST-DNN are 3-4 orders of magnitude smaller then CNNs used for segmentation. We show that they exceed segmentation performance compared to state-of-the-art CNN-based descriptors using 2-3 orders smaller training sets on the texture segmentation problem.
• #### Ultrafast photo-induced enhancement of electron-phonon coupling in metal-halide perovskites
(Research Square, 2021-02-15) [Preprint]
Abstract In metal-halide perovskites (MHPs), the nature of organic cations affects both, the perovskite’s structure and its optoelectronic properties. Using ultrafast pump-probe spectroscopy, we demonstrate that in state-of-the-art mixed-cation MHPs ultrafast photo-induced bandgap narrowing occurs, and linearly depends on the excited carrier density in the range from 10$^{16}$ cm$^{− 3}$ to above 10$^{18}$ cm$^{− 3}$. Furthermore, time-domain terahertz (td-THz) photoconductivity measurements reveal that the majority of carriers are localized and that the localization increases with the carrier density. Both observations, the bandgap narrowing and carrier localization, can be rationalized by ultrafast (sub-2ps) photo-induced enhancement of electron-phonon coupling, originating from dynamic disorder, as clearly evidenced by the presence of a Debye relaxation component in the terahertz photoconductivity spectra. The observation of photo-induced enhancement of electron-phonon coupling and dynamic disorder not only provides specific insight into the polaron-strain distribution of excited states in MHPs, but also adds to the development of a concise picture of the ultrafast physics of this important class of semiconductors.
• #### Insertion of PATC-rich C. elegans introns into synthetic transgenes by golden-gate-based cloning
(Research Square, 2021-02-15) [Preprint]
Transgenes are particularly prone to epigenetic silencing in the C. elegans germline. Here, we describe a protocol to insert introns containing a class of non-coding DNA named Periodic An/Tn Clusters (PATCs) into synthetic transgenes. PATCs can protect transgenes from position-dependent silencing (Position Effect Variegation, PEV) and from silencing in simple extra-chromosomal arrays. Using a set of simple design rules, it is possible to routinely insert up to three PATC-rich introns into a synthetic transgene in a single reaction.
• #### On the Impact of Device and Behavioral Heterogeneity in Federated Learning
(arXiv, 2021-02-15) [Preprint]
Federated learning (FL) is becoming a popular paradigm for collaborative learning over distributed, private datasets owned by non-trusting entities. FL has seen successful deployment in production environments, and it has been adopted in services such as virtual keyboards, auto-completion, item recommendation, and several IoT applications. However, FL comes with the challenge of performing training over largely heterogeneous datasets, devices, and networks that are out of the control of the centralized FL server. Motivated by this inherent setting, we make a first step towards characterizing the impact of device and behavioral heterogeneity on the trained model. We conduct an extensive empirical study spanning close to 1.5K unique configurations on five popular FL benchmarks. Our analysis shows that these sources of heterogeneity have a major impact on both model performance and fairness, thus sheds light on the importance of considering heterogeneity in FL system design.
• #### Decentralized Distributed Optimization for Saddle Point Problems
(arXiv, 2021-02-15) [Preprint]
We consider distributed convex-concave saddle point problems over arbitrary connected undirected networks and propose a decentralized distributed algorithm for their solution. The local functions distributed across the nodes are assumed to have global and local groups of variables. For the proposed algorithm we prove non-asymptotic convergence rate estimates with explicit dependence on the network characteristics. To supplement the convergence rate analysis, we propose lower bounds for strongly-convex-strongly-concave and convex-concave saddle-point problems over arbitrary connected undirected networks. We illustrate the considered problem setting by a particular application to distributed calculation of non-regularized Wasserstein barycenters.
• #### Smoothness Matrices Beat Smoothness Constants: Better Communication Compression Techniques for Distributed Optimization
(arXiv, 2021-02-14) [Preprint]
Large scale distributed optimization has become the default tool for the training of supervised machine learning models with a large number of parameters and training data. Recent advancements in the field provide several mechanisms for speeding up the training, including {\em compressed communication}, {\em variance reduction} and {\em acceleration}. However, none of these methods is capable of exploiting the inherently rich data-dependent smoothness structure of the local losses beyond standard smoothness constants. In this paper, we argue that when training supervised models, {\em smoothness matrices} -- information-rich generalizations of the ubiquitous smoothness constants -- can and should be exploited for further dramatic gains, both in theory and practice. In order to further alleviate the communication burden inherent in distributed optimization, we propose a novel communication sparsification strategy that can take full advantage of the smoothness matrices associated with local losses. To showcase the power of this tool, we describe how our sparsification technique can be adapted to three distributed optimization algorithms -- DCGD, DIANA and ADIANA -- yielding significant savings in terms of communication complexity. The new methods always outperform the baselines, often dramatically so.
• #### Distributed Second Order Methods with Fast Rates and Compressed Communication
(arXiv, 2021-02-14) [Preprint]
We develop several new communication-efficient second-order methods for distributed optimization. Our first method, NEWTON-STAR, is a variant of Newton's method from which it inherits its fast local quadratic rate. However, unlike Newton's method, NEWTON-STAR enjoys the same per iteration communication cost as gradient descent. While this method is impractical as it relies on the use of certain unknown parameters characterizing the Hessian of the objective function at the optimum, it serves as the starting point which enables us design practical variants thereof with strong theoretical guarantees. In particular, we design a stochastic sparsification strategy for learning the unknown parameters in an iterative fashion in a communication efficient manner. Applying this strategy to NEWTON-STAR leads to our next method, NEWTON-LEARN, for which we prove local linear and superlinear rates independent of the condition number. When applicable, this method can have dramatically superior convergence behavior when compared to state-of-the-art methods. Finally, we develop a globalization strategy using cubic regularization which leads to our next method, CUBIC-NEWTON-LEARN, for which we prove global sublinear and linear convergence rates, and a fast superlinear rate. Our results are supported with experimental results on real datasets, and show several orders of magnitude improvement on baseline and state-of-the-art methods in terms of communication complexity.
• #### Modeling Spatial Data with Cauchy Convolution Processes
(arXiv, 2021-02-14) [Preprint]
We study the class of models for spatial data obtained from Cauchy convolution processes based on different types of kernel functions. We show that the resulting spatial processes have some appealing tail dependence properties, such as tail dependence at short distances and independence at long distances with suitable kernel functions. We derive the extreme-value limits of these processes, study their smoothness properties, and consider some interesting special cases, including Marshall-Olkin and H\"usler-Reiss processes. We further consider mixtures between such Cauchy processes and Gaussian processes, in order to have a separate control over the bulk and the tail dependence behaviors. Our proposed approach for estimating model parameters relies on matching model-based and empirical summary statistics, while the corresponding extreme-value limit models may be fitted using a pairwise likelihood approach. We show with a simulation study that our proposed inference approach yields accurate estimates. Moreover, the proposed class of models allows for a wide range of flexible dependence structures, and we demonstrate our new methodology by application to a temperature dataset. Our results indicate that our proposed model provides a very good fit to the data, and that it captures both the bulk and the tail dependence structures accurately.
• #### A General Framework for Liquid Marbles
(arXiv, 2021-02-12) [Preprint]
Liquid marbles refer to liquid droplets that are covered with a layer of non-wetting particles. They are observed in nature and have practical significance. However, a generalized framework for analyzing liquid marbles as they inflate or deflate is unavailable. The present study fills this gap by developing an analytical framework based on liquid-particle and particle-particle interactions. We demonstrate that the potential final states of evaporating liquid marbles are characterized by one of the following: (I) constant surface area, (II) particle ejection, or (III) multilayering. Based on these insights, a single-parameter evaporation model for liquid marbles is developed. Model predictions are in excellent agreement with experimental evaporation data for water liquid marbles of particle sizes ranging from 7 nanometers to 300 micrometers (over four orders of magnitude) and chemical compositions ranging from hydrophilic to superhydrophobic. These findings lay the groundwork for the rational design of liquid marble applications.
• #### Material absorption-based carrier generation model for modeling optoelectronic devices
(arXiv, 2021-02-12) [Preprint]
The generation rate of photocarriers in optoelectronic materials is commonly calculated using the Poynting vector in the frequency domain. In time-domain approaches where the nonlinear coupling between electromagnetic (EM) waves and photocarriers can be accounted for, the Poynting vector model is no longer applicable. One main reason is that the photocurrent radiates low-frequency EM waves out of the spectrum of the source, e.g., terahertz (THz) waves are generated in THz photoconductive antennas. These frequency components do not contribute to the photocarrier generation since the corresponding photon energy is smaller than the optoelectronic material's bandgap energy. However, the instantaneous Poynting vector does not distinguish the power flux of different frequency components. This work proposes a material absorption-based model capable of calculating the carrier generation rate accurately in the time domain. Using the Lorentz dispersion model with poles reside in the optical frequency region, the instantaneous optical absorption, which corresponds to the power dissipation in the polarization, is calculated and used to calculate the generation rate. The Lorentz model is formulated with an auxiliary differential equation method that updates the polarization current density, from which the absorbed optical power corresponding to each Lorentz pole is directly calculated in the time domain. Examples show that the proposed model is more accurate than the Poynting vector-based model and is stable even when the generated low-frequency component is strong.
• #### Proximal and Federated Random Reshuffling
(arXiv, 2021-02-12) [Preprint]
Random Reshuffling (RR), also known as Stochastic Gradient Descent (SGD) without replacement, is a popular and theoretically grounded method for finite-sum minimization. We propose two new algorithms: Proximal and Federated Random Reshuffing (ProxRR and FedRR). The first algorithm, ProxRR, solves composite convex finite-sum minimization problems in which the objective is the sum of a (potentially non-smooth) convex regularizer and an average of $n$ smooth objectives. We obtain the second algorithm, FedRR, as a special case of ProxRR applied to a reformulation of distributed problems with either homogeneous or heterogeneous data. We study the algorithms' convergence properties with constant and decreasing stepsizes, and show that they have considerable advantages over Proximal and Local SGD. In particular, our methods have superior complexities and ProxRR evaluates the proximal operator once per epoch only. When the proximal operator is expensive to compute, this small difference makes ProxRR up to $n$ times faster than algorithms that evaluate the proximal operator in every iteration. We give examples of practical optimization tasks where the proximal operator is difficult to compute and ProxRR has a clear advantage. Finally, we corroborate our results with experiments on real data sets.
• #### Genome-Wide Association Study Reveals Genetic Architecture of Septoria Tritici Blotch Resistance in a Historic Landrace Collection
(Research Square, 2021-02-09) [Preprint]
Abstract Septoria tritici blotch (STB), caused by the fungus Zymoseptoria tritici, is a major constraint in global wheat production. The lack of genetic diversity in modern elite wheat cultivars largely hinders the improvement of STB resistance. Wheat landraces are reservoirs of untapped genetic diversity, which can be exploited to find novel STB resistance genes or alleles. Here, we characterized 188 Swiss wheat landraces for resistance to STB using four Swiss Z. tritici isolates. We used a genome-wide association study (GWAS) to identify genetic variants associated with the amount of lesion and pycnidia production by the fungus. The majority of the landraces were highly resistant for both traits to the isolate 1E4, indicating a gene-for-gene relationship, while higher phenotypic variability was observed against other isolates. GWAS detected a significant SNP on chromosome 3A that was associated with both traits in the isolate 1E4. The resistance response against 1E4 was likely controlled by the Stb6 gene. Sanger sequencing revealed that the majority of these ~100-year-old landraces carry the Stb6 resistance allele. This indicates the importance of this gene in Switzerland during the early 1900s for disease control in the field. Our study demonstrates the importance of characterizing historic landrace collections for STB resistance to provide valuable information on resistance variability and contributing alleles. This will help breeders in the future to make decisions on integrating such germplasms in STB resistance breeding.
• #### Reduction of the Beam Pointing Error for Improved Free-Space Optical Communication Link Performance
(arXiv, 2021-02-09) [Preprint]
Free-space optical communication is emerging as a low-power, low-cost, and high data rate alternative to radio-frequency communication in short-to medium-range applications. However, it requires a close-to-line-of-sight link between the transmitter and the receiver. This paper proposes a robust $\cHi$ control law for free-space optical (FSO) beam pointing error systems under controlled weak turbulence conditions. The objective is to maintain the transmitter-receiver line, which means the center of the optical beam as close as possible to the center of the receiving aperture within a prescribed disturbance attenuation level. First, we derive an augmented nonlinear discrete-time model for pointing error loss due to misalignment caused by weak atmospheric turbulence. We then investigate the $\cHi$-norm optimization problem that guarantees the closed-loop pointing error is stable and ensures the prescribed weak disturbance attenuation. Furthermore, we evaluate the closed-loop outage probability error and bit error rate (BER) that quantify the free-space optical communication performance in fading channels. Finally, the paper concludes with a numerical simulation of the proposed approach to the FSO link's error performance.
• #### Establishing and Maintaining a Reliable Optical Wireless Communication in Underwater Environment
(arXiv, 2021-02-09) [Preprint]
This paper proposes the trajectory tracking problem between an autonomous underwater vehicle (AUV) and a mobile surface ship, both equipped with optical communication transceivers. The challenging issue is to maintain stable connectivity between the two autonomous vehicles within an optical communication range. We define a directed optical line-of-sight (LoS) link between the two-vehicle systems. The transmitter is mounted on the AUV while the surface ship is equipped with an optical receiver. However, this optical communication channel needs to preserve a stable transmitter-receiver position to reinforce service quality, which typically includes a bit rate and bit error rates. A cone-shaped beam region of the optical receiver is approximated based on the channel model; then, a minimum bit rate is ensured if the AUV transmitter remains inside of this region. Additionally, we design two control algorithms for the transmitter to drive the AUV and maintain it in the cone-shaped beam region under an uncertain oceanic environment. Lyapunov function-based analysis that ensures asymptotic stability of the resulting closed-loop tracking error is used to design the proposed NLPD controller. Numerical simulations are performed using MATLAB/Simulink to show the controllers' ability to achieve favorable tracking in the presence of the solar background noise within competitive times. Finally, results demonstrate the proposed NLPD controller improves the tracking error performance more than $70\%$ under nominal conditions and $35\%$ with model uncertainties and disturbances compared to the original PD strategy.
• #### A Joint Inversion-Segmentation approach to Assisted Seismic Interpretation
(arXiv, 2021-02-07) [Preprint]
Structural seismic interpretation and quantitative characterization are historically intertwined processes. The latter provides estimates of properties of the subsurface which can be used to aid structural interpretation alongside the original seismic data and a number of other seismic attributes. In this work, we redefine this process as a inverse problem which tries to jointly estimate subsurface properties (i.e., acoustic impedance) and a piece-wise segmented representation of the subsurface based on user-defined macro-classes. By inverting for the quantities simultaneously, the inversion is primed with prior knowledge about the regions of interest, whilst at the same time it constrains this belief with the actual seismic measurements. As the proposed functional is separable in the two quantities, these are optimized in an alternating fashion, where each subproblem is solved using a Primal-Dual algorithm. Subsequently, each class is used as input to a workflow which aims to extract the perimeter of the detected shapes and to produce unique horizons. The effectiveness of the proposed method is illustrated through numerical examples on synthetic and field datasets.
• #### DOA Estimation with Non-Uniform Linear Arrays: A Phase-Difference Projection Approach
(arXiv, 2021-02-06) [Preprint]
Phase wrapping is a major problem in direction-of-arrival (DOA) estimation using phase-difference observations. For a sensor pair with an inter-sensor spacing greater than half of the wavelength ($\lambda/2$) of the signal, phase wrapping occurs at certain DOA angles leading to phase-difference ambiguities. Existing phase unwrapping methods exploit either frequency or spatial diversity. These techniques work by imposing restrictions on the utilized frequencies or the receiver array geometry. In addition to sensitivity to noise and calibration errors, these methods may also have high computational complexity. We propose a grid-less \emph{phase-difference projection} (PDP) DOA algorithm to overcome these issues. The concept of \emph{wrapped phased-difference pattern} (WPDP) is introduced, which allows the proposed algorithm to compute most of the parameters required for DOA estimation in an offline manner, hence resulting in a superior computational speed in realtime. Simulation results demonstrate the excellent performance of the proposed algorithm, both in terms of accuracy and speed.
• #### DeepReduce: A Sparse-tensor Communication Framework for Distributed Deep Learning
(arXiv, 2021-02-05) [Preprint]
Sparse tensors appear frequently in distributed deep learning, either as a direct artifact of the deep neural network's gradients, or as a result of an explicit sparsification process. Existing communication primitives are agnostic to the peculiarities of deep learning; consequently, they impose unnecessary communication overhead. This paper introduces DeepReduce, a versatile framework for the compressed communication of sparse tensors, tailored for distributed deep learning. DeepReduce decomposes sparse tensors in two sets, values and indices, and allows both independent and combined compression of these sets. We support a variety of common compressors, such as Deflate for values, or run-length encoding for indices. We also propose two novel compression schemes that achieve superior results: curve fitting-based for values and bloom filter-based for indices. DeepReduce is orthogonal to existing gradient sparsifiers and can be applied in conjunction with them, transparently to the end-user, to significantly lower the communication overhead. As proof of concept, we implement our approach on Tensorflow and PyTorch. Our experiments with large real models demonstrate that DeepReduce transmits fewer data and imposes lower computational overhead than existing methods, without affecting the training accuracy. | 2021-02-25 14:15:24 | {"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": 0.5142505764961243, "perplexity": 1760.5170705858736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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/1614178351134.11/warc/CC-MAIN-20210225124124-20210225154124-00244.warc.gz"} |
https://crad.ict.ac.cn/CN/Y2009/V46/I11/1917 | ISSN 1000-1239 CN 11-1777/TP
• 论文 •
### 独立于设计者的行动推理
1. 1(广西师范大学计算机科学与信息工程学院 广西桂林 541004) 2(中国科学院计算机科学国家重点实验室 北京 100190) ([email protected])
• 出版日期: 2009-11-15
### Action Reasoning Independent of Designer
Zhou Shengming1, Wang Ju1, and Jiang Yuncheng1,2
1. 1(School of Computer Science and Information Engineering, Guangxi Normal University, Guilin, Guangxi 541004) 2(State Key Laboratory of Computer Science, Institute of Software, Chinese Academy of Sciences, Beijing 100190)
• Online: 2009-11-15
Abstract: Action and action reasoning is a basic part of human activities. People must execute some actions when they want to complete some tasks. Similarly, robot needs to execute some actions when she accomplishes a task. High-level intelligent robot is required to be able to sense the external environment and do correct reasoning about actions independently. It is needed that the external designer writes out background axioms, sensing results and related knowledge changes for agent when expressing action reasoning with sensing actions and knowledge in situation calculus action theory. This is a kind of action reasoning depending on the designer. The situation calculus action theory is expanded in proper way, the sensors representation is added into the formal language of action theory, and agent’s new knowledge producing is based on the results of sensor applications in this paper. A platform is provided in which the following issues can be expressed formally: robot is sensing its external environment; the information obtained by robot’s sensors is converted to robot’s knowledge automatically; robot does an action reasoning independent of designer. In this way, the “black box” process of sensing actions will be clear; robot’s knowledge will be linked to the results of sensors; knowledge-fluent will be regarded as a dynamic knowledge base; and robot can update the knowledge base by executing sensing actions. Furthermore, robot can make action planning and execute actions independent of designer. | 2021-04-16 08:29:50 | {"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": 0.24481716752052307, "perplexity": 3438.341259018248}, "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-2021-17/segments/1618038088731.42/warc/CC-MAIN-20210416065116-20210416095116-00518.warc.gz"} |
https://www.shaalaa.com/concept-notes/terms-sequence_9368 | SSC (Marathi Semi-English) 10thMaharashtra State Board
Share
# Terms in a sequence
#### notes
In a sequence, ordered terms are represented as t1, t2, t3, . . . . .tn . . . In general sequence is written as {tn}. If the sequence is infinite, for every positive integer n, there is a term tn.
Activity I : Some sequences are given below. Show the positions of the terms by t1, t2, t3, . . .
(1) 9, 15, 21, 27, . . . Here t1= 9, t2= 15, t3= 21, . . .
(2) 7, 7, 7, 7, . . . Here t1= 7, t2=___ , t3= ___, . . .
(3) -2, -6, -10, -14, . . . Here t1= -2, t2=___ , t3= ___, . . .
Activity II : Some sequences are given below. Check whether there is any rule among the terms. Find the similarity between two sequences.
To check the rule for the terms of the sequence look at the arrangements on the next page, and fill the empty boxes suitably.
(1) 1, 4, 7, 10, 13, . . . (2) 6, 12, 18, 24, . . .
(3) 3, 3, 3, 3, . . . (4) 4, 16, 64, . . .
(5) -1, -1.5, -2, -2.5, . . . (6) 13, 23, 33, 43, . . .
Let’s find the relation in these sequences. Let’s understand the thought behind it.
Here in the sequences (1), (2), (3), (5), the similarity is that next term is obtained by adding a particular number to the previous number. Each ot these sequences is called an Arithmetic Progression.
Sequence (4) is not an arithmetic progression. In this sequence the next term is obtained by mutliplying the previous term by a particular number. This type of sequences is called a Geometric Progression.
Sequence (6) is neither arithmetic progression nor geometric progression.
S | 2019-12-10 02:13:25 | {"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": 0.8239799737930298, "perplexity": 1740.467574774014}, "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-2019-51/segments/1575540525781.64/warc/CC-MAIN-20191210013645-20191210041645-00357.warc.gz"} |
https://moodle.cs.pdx.edu/mod/page/view.php?id=548 | # Harmonies
• Polyphony: more than one note being sounded at a time
• Can get a long ways with a MiniMoog playing monophonic, but even then a band is helpful, so probably polyphony
• A bunch of the scale setup — in particular, "equal temperament" — is desiged to make polyphony sound good
# Consonant Intervals
• In a major scale
• Third (four steps up from root) is roughly 5/4 the frequency of the root
• Fifth (seven steps up from the root) is roughly 3/2 the frequency of the root
• These intervals have a reasonably regular wave structure with periods 4 and 3 root cycles respectively, so sound "pure"
• In a minor scale
• Minor third (three steps up from the root) is roughly 6/5 the frequency of the root with period 5
• Note the "roughly": the equal temperament compromise means that these frequencies are a little off and periods are thus long
• Thirds are particularly bad, being about 1% (1 cent) off. This would be considered borderline "out-of-tune" for a guitar
# The Guitar Fretboard
• Let's look at a guitar fretboard
• Note that the spacing of frets is not even. Why?
• Note that a third is 1/6 of the way down a string, and a fifth 1/3 of the way
• Markers are at the minor third, fourth, fifth, sixth intervals, double marker at the octave, then repeat
• Intonation: adjust the length of each string at the bridge to make the marker positions right — different from tuning by adjusting string tension
• Pickup position matters (hence two sets of pickups) — why?
# The Basic Three-Note Chord
• There can be only one — well, one major, one minor
• 1-3-5 major, 1-3♭-5 minor
• Note that we have abandoned note names here: a "C major chord" and a "D♭ major chord" have frequencies in the same relationship, but with a different base pitch
• In other words, we pick a scale, then use the root, third and fifth notes of that scale to make a chord
# Roman Numeral Notation
• Step 1: pick a scale (key)
• Notate the major chord starting at the root as I
• The minor chord is i (lowercase I) of a minor scale
• There are major chords that start on each note in the scale: notate these I, II, III, IV, V, VI, VII
• Except typically you make the chords out of notes in the scale, so the normal thing would be I, ii, iii, IV, V, vi and…uh vii°
• vii° is a "diminished chord": 1-3♭-5♭ (starting from note 7)
• For minor scale would be i, ii°, III, iv, v (or V in classical), VI, vii°
# What Do Chords Sound Like?
• Major chords sound happy, minor chords sound sad
• Diminished chords are kind of spooky and mysterious-sounding
• (The "augmented chord" 1-3-5♯ sounds like it's going somewhere — you'll hear it occasionally)
# Octaves and Inversions
• It is common to use octave notes of the chord to add to a chord, either above or below. A big "chord stack" is its own thing
• It matters which note is on top in a chord:
• 5 on top is the "root chord": 1-3-5 notated e.g I
• 1 on top is the "first inversion": 3-5-1 notated e.g. $I^6$
• 3 on top is the "second inversion": 5-1-3 notated e.g. $I^6_4$
• A common addition to a chord is the 7 tone of the scale, either flatted ("natural seventh") or not ("major seventh")
• The are several different Roman Numeral conventions floating around for sevenths: see this chart for one such
• The inversion notation gets messy; let us not care
• The 9 tone, 11 tone, 13 tone may be added for increasing dissonance. This is mostly a jazz thing
• "Relative minor" starts on 6 of major: has same key signature
• "Parallel minor" starts on 1 of major: has different key signature
• (Other minors start on other tones)
# Pop
• Cast a big net here: Pop intermingles with Rock, Rock is inspired by world rhythms, Blues, Jazz, and (yes) Country
• Pop is noted for its simplicity: I, IV, V chords dominate, vi, ii, iii (or III) (relative minor) chords are frequent
• Pop is typically not notated with Roman Numerals, but with named chords in a particular key. Often a melody in standard musical notation is used — a "lead sheet"
• Even when full music is provided, the lead line (melody) will typically be on a separate staff, and the lettered chords will be included
• Here's an example: Let My Love Open The Door by Pete Townshend
• The bass line is also commonly included. Pop bass line is often just root note of current chord, in rhythm
• More could be said about bass, but time…
# Chord Motifs Are Reused In Pop
• Many of you will have seen Four Chords by Axis of Awesome. Let's see if we can get something more out of it
• Common Pop chord progressions
• Four Chords: I-V-vi-IV
• Blues: I-IV-I-V-IV or so
• 50s: I-vi-IV-V
• Pachelbel: I-V-vi-iii-IV-I-ii-V
• many more
# Pop, Key Changes, and the Circle Of Fourths
• We are playing in equal temperament so that we can shift keys during a piece
• Common pop motif: jump up the circle of fourths 1-4 steps, eventually walk back down or jump back
• Dark Side of the Moon by Pink Floyd:
• Verse: I-IV7-I-IV7-I-I7-IV-iv-V-V7-I
• Chorus: IV-V-♭VI-IV (repeat)
# Fancier Pop Chords
• It's not all just formulaic: some songs have really fancy chords
• I'm a big fan of Supertramp: they do a bunch of this stuff — diminished chords, diminished sevenths, fancy key changes, etc
# Computer Things
• When analyzing music:
• Be aware that a lead sheet is a viable product
• Chords are hard to sort out of pop music, since there's so much other noise and so many harmonics for each note (freakin' guitars)
• Top note is usually melody, bottom note is usually bass and gives tonic of chord
• When generating music:
• Chords come first, bass second, melody third
• The melody should be in the scale corresponding to the chord: some "accidentals" are fine
# Captain Obvious Says
• This is just a starting point: go find out more things
• "To the Obviousmobile" (gets in Obviousmobile) | 2022-05-17 20:14:45 | {"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": 0.5365805625915527, "perplexity": 6813.920648383588}, "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-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00743.warc.gz"} |
https://optunity.readthedocs.io/en/latest/wrappers/julia.html | # Julia¶
In this page we briefly discuss the Julia wrapper, which provides most of Optunity’s functionality. For a general overview, we recommend reading the User Guide.
For installation instructions, please refer to Installing Optunity and in particular to Installing Optunity for Julia.
## Manual¶
For Julia the following main functions are available:
minimize(f[, named_args])
Perform minimization of the function f based on the provided named_args parameters.
Parameters: f – minimized function (can be an anonymous or any multi-argument function, or a function accepting Dict) named_args – various arguments can be provided, such as num_evals, solver_name, but mandatory named arguments are related to the hard constraint on the domain of the function f (for instance x=[-5,5]). vars::Dict, details::Dict
If f is provided in the anonymous or multi-argument form then the order of the provided hard constraints matter. For instance if one defines:
f = (x,y,z) -> (x-1)^2 + (y-2)^2 + (z+3)^4
or:
f(x,y,z) = (x-1)^2 + (y-2)^2 + (z+3)^4
then we strictly require to provide among other named arguments the following hard constraints:
x=[lb,ub], y=[lb,ub], z=[lb,ub]
If one provides a function accepting Dict then the order of the provided hard constraints does not matter.
maximize(f[, named_args])
Perform maximization of the function f based on the provided named_args parameters.
Parameters: f – minimized function (can be an anonymous or any multi-argument function, or a function accepting Dict) named_args – various arguments can be provided, such as num_evals, solver_name, but mandatory named arguments are related to the hard constraint on the domain of the function f (for instance x=[-5,5]).
## Examples¶
using Base.Test
vars, details = minimize((x,y,z) -> (x-1)^2 + (y-2)^2 + (z+3)^4, x=[-5,5], y=[-5,5], z=[-5,5])
@test_approx_eq_eps vars["x"] 1.0 1.
@test_approx_eq_eps vars["y"] 2.0 1.
@test_approx_eq_eps vars["z"] -3.0 1.
testit(x,y,z) = (x-1)^2 + (y-2)^2 + (z+3)^4
vars, details = minimize(testit, num_evals=10000, solver_name="grid search", x=[-5,5], y=[-5,5], z=[-5,5])
@test_approx_eq_eps vars["x"] 1.0 .2
@test_approx_eq_eps vars["y"] 2.0 .2
@test_approx_eq_eps vars["z"] -3.0 .2
testit_dict(d::Dict) = -(d[:x]-1)^2 - (d[:y]-2)^2 - (d[:z]+3)^4
vars, details = maximize(testit_dict, num_evals=10000, z=[-5,5], y=[-5,5], x=[-5,5])
@test_approx_eq_eps vars["x"] 1.0 .2
@test_approx_eq_eps vars["y"] 2.0 .2
@test_approx_eq_eps vars["z"] -3.0 .2 | 2019-09-21 11:11:03 | {"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": 0.21194882690906525, "perplexity": 10844.244738522935}, "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-2019-39/segments/1568514574409.16/warc/CC-MAIN-20190921104758-20190921130758-00072.warc.gz"} |
https://www.jamiebalfour.scot/courses/other/programming-theory/structured-programming/ | To use this website fully, you first need to accept the use of cookies. By agreeing to the use of cookies you consent to the use of functional cookies. For more information read this page.
# Part 3.5Structured programming
Code reuse is one of the most fundamental ways in which programming changed between the earliest computer programs and today's programming.
The term structured programming refers to a range of different techniques that are applied to the development of a computer program or code listing to ensure several key principles are met. Structured programming means making efficient use of several previously seen programming constructs such as decision making (if statements), loops and using subprograms/subroutines (functions or procedures).
## What is structured programming?
There are several aims of structured programming. Structured programming seeks to make code:
• More Reusable - code can be reused over and over
• Smaller - having only one instance of the code makes it smaller
• More readable - having only one instance of the code makes it much easier to read
• Flexible - particularly in relation to functions/procedures and loops, the way that the program flows can change based on some parameters.
• Higher performance - by reusing code, the program actually can run more smoothly
• Easier to test - in relation to reliability of code, if the code is tested over and over again and it works, why write the code over and over again? If it doesn't work in one place you would need to change it over the program, but implementing it in a function or procedures means you only need to test that instance of the code.
## Without code reuse
Code can be reused throughout a program, avoiding the need to rewrite it. Alternatively, it can be copied and pasted throughout the program.
Take a look at the following code sample which is used to calculate the power consumption of a typical CMOS transistor (in theory) at a specified switching rate given in gigahertz ($f) running at a voltage of 12.5V ($v) with a capacitance of 0.02µF ($c) based on the formula: $$P = CV^2f$$ YASS $c = 2000E-8
$v = 12.5$f = 2.3E9
$p = (($c * $v) ^ 2) *$f)
print($p)$f = 2.9E9
$p = (($c * $v) ^ 2) *$f)
print($p)$f = 1.8E9
$p = (($c * $v) ^ 2) *$f)
print($p) The same formula is being used for each of these calculations but with different values each time (or at least the $f value is changing).
This is inefficient!
## Defining a subprogram
Subprograms need to be well-defined in programming languages and must serve a purpose. Subprograms generally are defined using a keyword such as function (VB.NET, PHP, YASS), def (Python), void (Java, C, C++, C#) or a data type from the language followed by an identifier (the name of the function), after which a set of brackets normally follows containing parameters. This lines tells the compiler what to do with the subprogram and is known as the signature.
YASS
function calculateSqr($a,$b)
In this example, the function is the keyword that tells the compiler that this is a subprogram, calculateSqr is the identifier that allows the code to be reused, and there are three parameters within it ($a and $b).
## Using functions and procedures
A subprogram is a chunk of code that can be run over and over again. Subprograms can be either a function or a procedure. A function returns or gives back a value whereas a procedure simply runs some code and does not need to return a value (but it can manipulate or change the value of other variables within the application).
It is written just once but used many times in the code.
In YASS, both functions and procedures are created with the function symbol:
YASS
function calculatePower($c,$v, $f) return (($c * $v) ^ 2) *$f
end function
$voltage = 12.5$p = calculatePower(2000E-8, $voltage, 2.3E9) print($p)
$p = calculatePower(2000E-8,$voltage, 2.9E9)
print($p)$p = calculatePower(2000E-8, $voltage, 1.8E9) print($p)
Instead of writing the formula or calculation three times, the formula is written once. The section in lines 1 to 5 provides a reusable chunk of code known as a function. This function gives back the answer to the equation given some values.
When a subprogram is run, it is called.
## Parameters
A parameter is a special type of variable that relates to subprograms directly. There are two kinds of parameters:
• Formal parameters (often known simply as parameters)
• Actual parameters (also known as arguments)
Looking at the previous sample as defined on line 1:
YASS
function calculatePower($c,$v, $f) In the signature of the function calculatePower there are three parameters required; $c, $v and $f. Within a computing context, particularly in relation to the formal lambda calculus, these parameters are said to be bound to the calculatePower function. This means that these parameters are basically variables within the scope of that subprogram and no further. If there is a variable with the same name out with the subprogram, then that variable remains untouched by any changes made to the variable within the subprogram.
Since these parameters are defined within the signature of the function, they are placeholders for actual values that will be 'passed' to the function when the function is called. They are therefore known as formal parameters.
Later on in the previous program, the function calculatePower is called three times:
YASS
$p = calculatePower(2000E-8,$voltage, 2.3E9)
$p = calculatePower(2000E-8,$voltage, 2.9E9)
$p = calculatePower(2000E-8,$voltage, 1.8E9)
Each time the function is called, the frequency value of the CMOS transistor has changed, from 2.3E9 to 2.9E9 to 1.8E9 (the E just represents an exponential value or 2.3x109, 2.9x109 and 1.8x109). In the all of the function calls there are three are three parameters, with only one changing. For the first call the parameters are 2000E-8, $voltage and 2.3E9. These are actual parameters since they are being passed to the function. These variables will become the values of the formal parameters defined within the function signature: • $c becomes 2000E-8
• $v becomes $voltage (which itself is substituted to 12.5)
• \$f becomes 2.3E9
Code preview
Feedback 👍
Comments are sent via email to me. | 2022-12-08 09:00:58 | {"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": 0.5189732313156128, "perplexity": 1360.0188315151736}, "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-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00646.warc.gz"} |
http://semantic-portal.net/java-basics-graphics-javafx-ui-controls-titled-panel-and-accordion | # Titled Pane and Accordion in JavaFX
Domains:
## Titled Pane and Accordion
This chapter explains how to use a combination of the accordion and title panes in your JavaFX applications.
A titled pane is a panel with a title. It can be opened and closed, and it can encapsulate any Node such as UI controls or images, and groups of elements added to a layout container.
Titled panes can be grouped by using the accordion control, which enables you to create multiple panes and display them one at a time. Figure 23-1 shows an accordion control that combines three titled panes.
Figure 23-1 Accordion with Three Titled Panes
Use the Accordion and TitledPane classes in the JavaFX SDK API to implement these controls in your applications.
## Creating Titled Panes
To create a TitledPane control define a title and some content. You can do this by using the two-parameter constructor of the TitledPane class, or by applying the setText and setContent methods. Both ways are shown in Example 23-1.
Example 23-1 Declaring a TitledPane Object
//using a two-parameter constructor
TitledPane tp = new TitledPane("My Titled Pane", new Button("Button"));
//applying methods
TitledPane tp = new TitledPane();
tp.setText("My Titled Pane");
tp.setContent(new Button("Button"));
Compiling and running an application with either of the code fragments produces the control shown in Figure 23-2.
Figure 23-2 Titled Pane with a Button
Do not explicitly set the minimal, maximum, or preferred height of a titled pane, as this may lead to unexpected behavior when the titled pane is opened or closed.
The code fragment shown in Example 23-2 adds several controls to the titled pane by putting them into the GridPane layout container.
Example 23-2 Titled Pane with the GridPane Layout Container
TitledPane gridTitlePane = new TitledPane();
GridPane grid = new GridPane();
grid.setVgap(4);
grid.add(new Label("First Name: "), 0, 0);
grid.add(new Label("Last Name: "), 0, 1);
gridTitlePane.setText("Grid");
gridTitlePane.setContent(grid);
When you run and compile an application with this code fragment, the output shown in Figure 23-3 appears.
Figure 23-3 Titled Pane that Contains Several Controls
You can define the way a titled pane opens and closes. By default, all titled panes are collapsible and their movements are animated. If your application prohibits closing a titled pane, use the setCollapsible method with the false value. You can also disable animated opening by applying the setAnimated method with the false value. The code fragment shown in Example 23-3 implements these tasks.
Example 23-3 Adjusting the Style of a Titled Pane
TitledPane tp = new TitledPane();
//prohibit closing
tp.setCollapsible(false);
//prohibit animating
tp.setAnimated(false);
## Adding Titled Panes to an Accordion
In your applications, you can use titled panes as standalone elements, or you can combine them in a group by using the Accordion control. Do not explicitly set the minimal, maximum, or preferred height of an accordion, as this may lead to unexpected behavior when one of its titled pane is opened.
Adding several titled panes to an accordion is similar to adding toggle buttons to a toggle group: only one titled pane can be opened in an accordion at a time. Example 23-4 creates three titled panes and adds them to an accordion.
Example 23-4 Accordion and Three Titled Panes
import javafx.application.Application;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.control.Accordion;
import javafx.scene.control.TitledPane;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.stage.Stage;
public class TitledPaneSample extends Application {
final String[] imageNames = new String[]{"Apples", "Flowers", "Leaves"};
final Image[] images = new Image[imageNames.length];
final ImageView[] pics = new ImageView[imageNames.length];
final TitledPane[] tps = new TitledPane[imageNames.length];
public static void main(String[] args) {
launch(args);
}
@Override public void start(Stage stage) {
stage.setTitle("TitledPane");
Scene scene = new Scene(new Group(), 80, 180);
final Accordion accordion = new Accordion ();
for (int i = 0; i < imageNames.length; i++) {
images[i] = new
Image(getClass().getResourceAsStream(imageNames[i] + ".jpg"));
pics[i] = new ImageView(images[i]);
tps[i] = new TitledPane(imageNames[i],pics[i]);
}
accordion.setExpandedPane(tps[0]);
Group root = (Group)scene.getRoot();
stage.setScene(scene);
stage.show();
}
}
Three titled panes are created within the loop. Content for each titled pane is defined as an ImageView object. The titled panes are added to the accordion by using the getPanes and addAll methods. You can use the add method instead of the addAll method to add a single titled pane.
By default, all the titled panes are closed when the application starts. The setExpandedPane method in Example 23-4 specifies that the titled pane with the Apples picture will be opened when you run the sample, as shown in Figure 23-4.
Figure 23-4 Accordion with Three Titled Panes
## Processing Events for an Accordion with Titled Panes
You can use titled panes and accordions to present different data in your applications. Example 23-5 creates a standalone titled pane with the GridPane layout container and three titled panes combined by using the accordion. The standalone titled pane contains UI elements of an email client. The accordion enables the selection of an image to appear in the Attachment field of the Grid titled pane.
Example 23-5 Implementing ChangeListener for an Accordion
import javafx.application.Application;
import javafx.beans.value.ObservableValue;
import javafx.geometry.Insets;
import javafx.scene.Group;
import javafx.scene.Scene;
import javafx.scene.control.Accordion;
import javafx.scene.control.Label;
import javafx.scene.control.TextField;
import javafx.scene.control.TitledPane;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.scene.layout.GridPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;
public class TitledPaneSample extends Application {
final String[] imageNames = new String[]{"Apples", "Flowers", "Leaves"};
final Image[] images = new Image[imageNames.length];
final ImageView[] pics = new ImageView[imageNames.length];
final TitledPane[] tps = new TitledPane[imageNames.length];
final Label label = new Label("N/A");
public static void main(String[] args) {
launch(args);
}
@Override public void start(Stage stage) {
stage.setTitle("TitledPane");
Scene scene = new Scene(new Group(), 800, 250);
// --- GridPane container
TitledPane gridTitlePane = new TitledPane();
GridPane grid = new GridPane();
grid.setVgap(4);
gridTitlePane.setText("Grid");
gridTitlePane.setContent(grid);
// --- Accordion
final Accordion accordion = new Accordion ();
for (int i = 0; i < imageNames.length; i++) {
images[i] = new
Image(getClass().getResourceAsStream(imageNames[i] + ".jpg"));
pics[i] = new ImageView(images[i]);
tps[i] = new TitledPane(imageNames[i],pics[i]);
}
(ObservableValue<? extends TitledPane> ov, TitledPane old_val,
TitledPane new_val) -> {
if (new_val != null) {
label.setText(accordion.getExpandedPane().getText() +
".jpg");
}
});
HBox hbox = new HBox(10);
hbox.getChildren().setAll(gridTitlePane, accordion);
Group root = (Group)scene.getRoot();
stage.setScene(scene);
stage.show();
}
}
When a user opens a titled pane in the accordion, the expandedPaneProperty of the accordion changes. The expanded titled pane in the accordion is used to construct a file name of the attachment. This file name is set as text of the corresponding Label object.
Figure 23-5 shows how the application looks after its start. The Attachment label contains "N/A," because none of the titled panes are selected.
Figure 23-5 Initial View of the TitledPaneSample Application
If you expand the Leaves titled pane, the Attachment label will contain "Leaves.jpg," as shown in Figure 23-6.
Figure 23-6 TitledPaneSample Application with the Leaves Titled Pane Expanded
Page structure
Terms | 2022-12-08 09:14:08 | {"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": 0.43916502594947815, "perplexity": 10800.0550672275}, "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-2022-49/segments/1669446711286.17/warc/CC-MAIN-20221208082315-20221208112315-00328.warc.gz"} |
http://blog.gmane.org/gmane.comp.tex.pgf.user | 1 Feb 19:03 2014
### multiple arguments with \pic ?
I'm playing with the new \path pic or \pic that was introduced in recent versions of tikz.
Is there any way to have multiple arguments, i.e., make a call \pic{mypicture={red}{green}} ?
------------------------------------------------------------------------------
WatchGuard Dimension instantly turns raw network data into actionable
security intelligence. It gives you real-time visual feedback on key
security issues and trends. Skip the complicated setup - simply import
a virtual appliance and go from zero to informed in seconds.
http://pubads.g.doubleclick.net/gampad/clk?id=123612991&iu=/4140/ostg.clktrk
_______________________________________________
pgf-users mailing list
pgf-users <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/pgf-users
31 Jan 05:22 2014
### how to position children along a line in mindmap
I have a mindmap with 4 level 1 nodes that grow at 0, 90, 180 and 270,
each with its own level 2 children. All nodes are arranged in a circle
around their parents but I would like to change the circle to a straight
line for the level 2 children only. How would I do that? Is there an
option or some style I can modify? There are many level 2 nodes and I
would not like to set the coordinates to each node individually. I guess
I can do that in foreach, but I'm thinking there has to be a way to
achieve this with styles. Here is the current code:
\def\clustSpan{135}
\def\clustOffset{(180-\clustSpan)/2}
\begin{tikzpicture}[small mindmap, text=white, minimum size=0,
text width=1.5cm, concept color=blue!90,
%% root concept/.style={concept color=blue!90,fill=cyan},
root concept/.append style={
%% minimum size=0, text width=3cm,
concept color=blue!90,text=black
},
level 1/.append style=
{every child/.style={concept color=blue!60,
sibling angle=360/\the\tikznumberofchildren,
level distance=2cm,
minimum size=0, text width=1cm}},
level 2/.append style=
{every child/.style={concept color=blue!30,
sibling angle=135/(\the\tikznumberofchildren -1),
level distance=1.5cm,
minimum size=0, text width=0.5cm}},
grow cyclic
]
\node[concept] {Root}
[counterclockwise from=0]
child[concept] {node[concept] {child 1}
[counterclockwise from=-90+\clustOffset]
child[concept] foreach \x in {1,2,3} {node[concept] {\x}}
}
child[concept] {node[concept] {child 2}
[counterclockwise from=0+\clustOffset]
child[concept] foreach \x in {4,...,7} {node[concept] {\x}}
}
child[concept] {node[concept] {child 3}
[counterclockwise from=90+\clustOffset]
child[concept] foreach \x in {7,...,11} {node[concept] {\x}}
}
child[concept] {node[concept] {child 4}
[counterclockwise from=180+\clustOffset]
child[concept] foreach \x in {12,...,17} {node[concept] {\x}}
};
\end{tikzpicture}
Oh, and one more question. How do I compute (programatically) the
"clockwise from=" angle for the level 2 children? Right now I hardcoded
-90, 0, 90 and 180 and that works for 4 children, but I want something
general. I'm guessing a custom grow function that depends on the
\tikznumberofcurrentchild of the parent node. Mathematically this would
be 90*(n-1) where n=0,1,2,3 is the index of the level 1 child, but how do
I code that up?
Thanks!
------------------------------------------------------------------------------
WatchGuard Dimension instantly turns raw network data into actionable
security intelligence. It gives you real-time visual feedback on key
security issues and trends. Skip the complicated setup - simply import
a virtual appliance and go from zero to informed in seconds.
15 Dec 12:50 2013
### Coordinate system intersection is undocumented?
Hello,
In pgf 2.00, the intersection coordinate is documented (12.2.4 Intersection Coordinate Systems). We can
then use, for example,
\fill (intersection of 0,3--2,2 and 2,0--4,4) circle (2pt);
to draw the intersection point of two lines. Although the two line segments do not intersect, the extension
lines do.
However, in pgf 2.10 and later (3.00rc), only "name intersections" in intersections library is
documented, the old document of intersection coordinate system is erased. We cannot use
% \usetikzlibrary{intersections}
\draw[name path=a] (0,3) -- (2,2);
\draw[name path=b] (2,0) -- (4,4);
\fill[name intersections={of=a and b, by=x}] (x) circle (2pt); % error: No intersections
Therefore I think the old syntax is still useful and worth a document.
Leo.
------------------------------------------------------------------------------
organizations don't have a clear picture of how application performance
affects their revenue. With AppDynamics, you get 100% visibility into your
Java,.NET, & PHP application. Start your 15-day FREE TRIAL of AppDynamics Pro!
4 Dec 23:14 2013
### touching curves/arrows
Hello:
I am new to tikz/pgf.
I would like to draw flowcharts. I learned how to make nodes and edges.
I would like to make arrows similar to shown in the attached png image.
Nodes A and B, and the arrow between them is fixed, and the other arrow
with nodes C and D should be drawn relative to it.
I could not find anything like this in the manual or elsewhere.
Any help is appreciated.
Thanks
bcsikos
------------------------------------------------------------------------------
Develop, test and display web and hybrid apps with a single code base.
http://pubads.g.doubleclick.net/gampad/clk?id=111408631&iu=/4140/ostg.clktrk
_______________________________________________
pgf-users mailing list
pgf-users <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/pgf-users
3 Dec 16:10 2013
### periodic table - new item
Hello!
I want to make an periodic table of the elements.
I have found an very good example on "texample".
http://www.texample.net/tikz/examples/periodic-table-of-chemical-elements/
Now I want to add some more items like electronegativity.
But I fail ...
May be some code from the example can help you to find a solution:
The author make a new command with:
----->8---
\newcommand{\CommonElementTextFormat}[4]
{
\begin{minipage}{2.2cm}
\centering
{\textbf{#1} \hfill #2}%
\linebreak \linebreak
{\textbf{#3}}%
\linebreak \linebreak
{{#4}}
\end{minipage}
}
---->8----
and another new command
----->8------
\newcommand{\NaturalElementTextFormat}[4]
{
\CommonElementTextFormat{#1}{#2}{\LARGE {#3}}{#4}
}
\newcommand{\OutlineText}[1]
{
\ifpdf
% Couldn't find a nicer way of doing an outline font with TikZ
% other than using pdfliteral 1 Tr
%
\pdfliteral direct {0.5 w 1 Tr}{#1}%
\pdfliteral direct {1 w 0 Tr}%
\else
% pstricks can do this with \pscharpath from pstricks
%
fillstyle=solid,
fillcolor=white,
linestyle=solid,
linecolor=black,
linewidth=.2pt]{#1}
\fi
}
------->8---------
Afterwards he created a node in which you can describe the elements.
------->8-----
%% Group 1 - IA
\node[name=H, Element] {\NaturalElementTextFormat{1}{1.0079}{H}{Wasserstoff}};
------>8-------
How can I add a new item?
Or knows someone the mail adress of the author?
Greetings
Arne
------------------------------------------------------------------------------
organizations don't have a clear picture of how application performance
affects their revenue. With AppDynamics, you get 100% visibility into your
Java,.NET, & PHP application. Start your 15-day FREE TRIAL of AppDynamics Pro!
http://pubads.g.doubleclick.net/gampad/clk?id=84349351&iu=/4140/ostg.clktrk
_______________________________________________
pgf-users mailing list
pgf-users <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/pgf-users
2 Dec 09:37 2013
### minimum height of boxes in multipart nodes
hi!
is it possible to determine minimal height of boxes which constitute rectangle shape with multiple text parts? something similarly to existed
/pgf/rectangle split empty part height=<...>
but with valid in boxes containing text even with it in more lines, i.e.: that height of boxes are independent from contained text? something as:
/pgf/rectangle split parts height=<...>
regards, zarko
------------------------------------------------------------------------------
organizations don't have a clear picture of how application performance
affects their revenue. With AppDynamics, you get 100% visibility into your
Java,.NET, & PHP application. Start your 15-day FREE TRIAL of AppDynamics Pro!
http://pubads.g.doubleclick.net/gampad/clk?id=84349351&iu=/4140/ostg.clktrk
_______________________________________________
pgf-users mailing list
pgf-users <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/pgf-users
1 Dec 01:37 2013
### [SPAM] Buy WATCHES JEWELRY and ACCESSORIES in our shop!
We offer different watches- rolex etc., jewelry and presents. - http://x.co/2yCiO
SALE! + 15% discount for every 250\$ or more order!
----------------------------------------------
The message was checked by Zillya Antivirus 1.1.3450.0, bases 2.0.0.1529 - No viruses detected
------------------------------------------------------------------------------
organizations don't have a clear picture of how application performance
affects their revenue. With AppDynamics, you get 100% visibility into your
Java,.NET, & PHP application. Start your 15-day FREE TRIAL of AppDynamics Pro!
http://pubads.g.doubleclick.net/gampad/clk?id=84349351&iu=/4140/ostg.clktrk
_______________________________________________
Php-syslog-ng-support mailing list
Php-syslog-ng-support <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/php-syslog-ng-support
27 Nov 02:11 2013
### error using \font
Hello,
I am facing an issue with styles in tikz. Using font= in a node with fixed width and right-aligned text causes
an unexpected indentation to be added. I have created a minimal working example to illustrate this behavior:
\documentclass{article}
\usepackage{tikz}
\tikzset{every node/.style={draw=black,thin}}
\begin{document}
\begin{tikzpicture}
\node [text width=16em,align=flush right] {\footnotesize This is OK};
\end{tikzpicture}
\begin{tikzpicture}[every node/.append style={font=\footnotesize}]
\node [text width=16em,align=flush right] {This is not OK};
\end{tikzpicture}
\end{document}
Why does this happen? Is it a bug or I'm doing something wrong?
Thanks,
Flavio Costa
------------------------------------------------------------------------------
organizations don't have a clear picture of how application performance
affects their revenue. With AppDynamics, you get 100% visibility into your
Java,.NET, & PHP application. Start your 15-day FREE TRIAL of AppDynamics Pro!
4 Nov 20:44 2013
### data visualization: attaching a style sheet to a custom attribute
Hi all,
I'm using the CVS version of PGF and I'm trying the new data
visualization features. The manual says that style sheets are attached
to sets by default by the "style sheet" option, but they can also be
attached to custom attributes. I've tried to do that but it seems that
I've done something wrong because I always obtain curves that have the
same default style.
What I'm trying to do in the minimal example below is to define four
sets (s1,s2,s3,s4) but only use two styles from a style sheet, one for
sets s1 and s3, and another for s2 and s4. To do that I first add a
custom attribute myattr to the data points. This part seems to work: I
checked I can use it, for example, as the attribute for the x axis.
Then I try to attach a syle sheet to this attribute by adding the
following line to the options for datavisualization:
/data point/myattr/.style sheet=<name of a style sheet>
If I understand correctly the manual, this should give different
styles to the lines corresponding to different values of the attribute
myattr. However the following example gives me only black lines (and
no errors).
What I'm doing wrong?
Luca Donetti
%%%%%%
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{datavisualization}
\begin{document}
\begin{tikzpicture}
\datavisualization [
scientific axes,
% by uncommenting the following line I can check that myattr works
% x axis={attribute=myattr},
visualize as smooth line/.list={s1,s2,s3,s4},
% the following line does not work, all 4 lines are black
/data point/myattr/.style sheet=vary hue,
]
data [set=s1,/data point/myattr=1] {
0, 0
1, 0.5
}
data [set=s2,/data point/myattr=2] {
0, 0.2
1, 0.3
}
data [set=s3,/data point/myattr=1] {
0, 1
1, 1.5
}
data [set=s4,/data point/myattr=2] {
0, 1.2
1, 1.3
};
\end{tikzpicture}
\end{document}
------------------------------------------------------------------------------
Android is increasing in popularity, but the open development platform that
Android apps secure.
4 Nov 11:58 2013
Section 56 of the manual (v2.10) claims that pgffor can be used independently of pgf. However, this claim seems to me misleading because, according to my tests, the 'evaluate' and 'count' keys require pgfmath. See the following code (compiled with pdftex):
\input pgffor % The following doesn't require pgfmath: \foreach [ var = \i, remember = \i as \lasti (initially -1) ] in {1,2,...,5}{\i\ } \input pgfmath % (comment this line in or out to see the effect) % However, the evaluate and count keys require pgfmaths % (pdftex returns an error if pgfmath is not loaded) \foreach \i [ evaluate = \i as \ii using 2*\i, count = \counti, ] in {1,2,...,5}{\counti: \ii\par} \bye
-----------------------------------------------------------------
Julien Cretel
PhD student (Marie Curie Actions)
University College Cork, Ireland
Lecturer for AM2032: Numerical methods & programming
Email: j.cretel <at> umail.ucc.ie
------------------------------------------------------------------------------
Android is increasing in popularity, but the open development platform that
Android apps secure.
http://pubads.g.doubleclick.net/gampad/clk?id=65839951&iu=/4140/ostg.clktrk
_______________________________________________
pgf-users mailing list
pgf-users <at> lists.sourceforge.net
https://lists.sourceforge.net/lists/listinfo/pgf-users
4 Nov 12:09 2013
Section 56 of the manual (v2.10) claims that pgffor can be used independently
of pgf. However, this claim seems to me misleading because, according to my
tests, the 'evaluate' and 'count' keys require pgfmath. See the following
code (compiled with pdftex):
\input pgffor
% The following doesn't require pgfmath:
\foreach
[
var = \i,
remember = \i as \lasti (initially -1)
]
in {1,2,...,5}{\i\ }
\input pgfmath % (comment this line in or out to see the effect)
% However, the evaluate and count keys require pgfmaths
% (pdftex returns an error if pgfmath is not loaded)
\foreach \i
[
evaluate = \i as \ii using 2*\i,
count = \counti,
]
in {1,2,...,5}{\counti: \ii\par}
\bye
-- | 2014-03-09 10:48:13 | {"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": 0.46071571111679077, "perplexity": 11013.887192468566}, "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-2014-10/segments/1393999677352/warc/CC-MAIN-20140305060757-00098-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://www.sciengine.com/doi/10.1007/s11432-020-2931-x | SCIENCE CHINA Information Sciences, Volume 64 , Issue 11 : 212101(2021) https://doi.org/10.1007/s11432-020-2931-x
## A nearly optimal distributed algorithm for computing the weighted girth
• AcceptedMay 25, 2020
• PublishedMay 14, 2021
Share
Rating
### Acknowledgment
This work was supported in part by National Key Research and Development Program of China (Grant No. 2018YFB1003203), National Natural Science Foundation of China (Grants No. 61972447), and Fundamental Research Funds for the Central Universities (Grant No. 2019kfyXKJC021). We thank the anonymous reviewers for the helpful comments to improve the presentation of this paper.
### References
[1] Diestel R. Graph Theory. 4th ed. Berlin: Springer, 2012. Google Scholar
[2] Bernstein A. A nearly optimal algorithm for approximating replacement paths and $k$ shortest simple paths in general graphs. In: Proceedings of ACM-SIAM symposium on Discrete algorithms (SODA), 2010. 742--755. Google Scholar
[3] Williams V V, Williams R. Subcubic equivalences between path, matrix and triangle problems. In: Proceedings of Symposium on Foundations of Computer Science (FOCS), 2010. 645--654. Google Scholar
[4] Itai A, Rodeh M. Finding a Minimum Circuit in a Graph. SIAM J Comput, 1978, 7: 413-423 CrossRef Google Scholar
[5] Coppersmith D, Winograd S. Matrix multiplication via arithmetic progressions. J Symbolic Computation, 1990, 9: 251-280 CrossRef Google Scholar
[6] Roditty L, Williams V V. Minimum weight cycles and triangles: Equivalences and algorithms. In: Proceedings of Symposium on Foundations of Computer Science (FOCS), 2011. 180--189. Google Scholar
[7] Williams V V. Hardness of Easy Problems: Basing Hardness on Popular Conjectures such as the Strong Exponential Time Hypothesis. In: Proceedings International Programme on the Elimination of Child Labour (IPEC), 2015. 17--29. Google Scholar
[8] Lingas A, Lundell E M. Efficient approximation algorithms for shortest cycles in undirected graphs. Inf Processing Lett, 2009, 109: 493-498 CrossRef Google Scholar
[9] Roditty L, Tov R. Approximating the girth. ACM Trans Algorithms, 2013, 9: 1--13. Google Scholar
[10] Holzer S, Wattenhofer R. Optimal distributed all pairs shortest paths and applications. In: Proceedings of ACM symposium on Principles of distributed computing (PODC), 2012. 355--364. Google Scholar
[11] Lenzen C, Peleg D. Efficient distributed source detection with limited bandwidth. In: Proceedings of ACM symposium on Principles of distributed computing (PODC), 2013. 375--382. Google Scholar
[12] Nanongkai D. Distributed approximation algorithms for weighted shortest paths. In: Proceedings of Symposium on the Theory of Computing (STOC), 2014. 565--573. Google Scholar
[13] Peleg D, Roditty L, Tal E. Distributed algorithms for network diameter and girth. In: Proceedings of International Colloquium on Automata, Languages and Programming (ICALP), 2012. 660--672. Google Scholar
[14] Hua Q, Fan H, Qian L, et al. Brief announcement: a tight distributed algorithm for all pairs shortest paths and applications. In: Proceedings of Symposium on Parallelism in Algorithms and Architectures (SPAA), 2016. 439--441. Google Scholar
[15] Hua Q, Fan H, Ai M, et al. Nearly optimal distributed algorithm for computing betweenness centrality. In: Proceedings of International Conference on Distributed Computing Systems (ICDCS), 2016. 271--280. Google Scholar
[16] Frischknecht S, Holzer S, Wattenhofer R. Networks cannot compute their diameter in sublinear time. In: Proceedings of ACM-SIAM symposium on Discrete algorithms (SODA), 2012. 1150--1162. Google Scholar
[17] Huang C C, Nanongkai D, Saranurak T. Distributed exact weighted all-pairs shortest paths in o(n5=4) rounds. In: Proceedings of Symposium on Foundations of Computer Science (FOCS), 2017. 168--179. Google Scholar
[18] Peleg D. Distributed computing a locality sensitive approach. SIAM Monographs on discrete mathematics and applications, 2000 5. Google Scholar
[19] Elkin M. Distributed exact shortest paths in sublinear time. In: Proceedings of Symposium on the Theory of Computing (STOC), 2017. 757--770. Google Scholar
[20] Leighton F T, Maggs B M, Rao S B. Packet routing and job-shop scheduling inO(congestion+dilation) steps. Combinatorica, 1994, 14: 167-186 CrossRef Google Scholar
[21] Yao A C. Some complexity questions related to distributive computing (preliminary report). In: Proceedings of Symposium on the Theory of Computing (STOC), 1979. 209--213. Google Scholar
[22] Kushilevitz E, Nisan N. Communication Complexity. Cambridge: Cambridge University Press, 1997. Google Scholar
[23] Jin H, Yao P, Liao X. Towards dataflow based graph processing. Sci China Inf Sci, 2017, 60: 126102 CrossRef Google Scholar
Citations
Altmetric | 2021-09-26 12:12:51 | {"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": 0.6733002662658691, "perplexity": 8000.454141960156}, "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-2021-39/segments/1631780057861.0/warc/CC-MAIN-20210926114012-20210926144012-00329.warc.gz"} |
https://hssliveguru.com/plus-two-botany-notes-chapter-6/ | # Plus Two Botany Notes Chapter 6 Organisms and Populations
Students can Download Chapter 6 Organisms and Populations Notes, Plus Two Botany Notes helps you to revise the complete Kerala State Syllabus and score more marks in your examinations.
## Kerala Plus Two Botany Notes Chapter 6 Organisms and Populations
Organism And Its Environment
The annual variations in the intensity and duration of temperature, resulting in distinct seasons. These variations together with annual variation in precipitation (precipitation includes both rain and snow) responsible for the formation of major biomes such as desert, rain forest and tundra.
Regional and local variations within each biome lead to the formation of a wide variety of habitats.
The existence of life not only in favourable habitat but it occurs in scorching Rajasthan desert, rain-soaked Meghalaya forests, deep ocean trenches, torrential streams, permafrost polar regions, high mountain tops, boiling thermal springs, and stinking compost pits.
1. Major Abiotic Factors:
Temperature:
The average temperature on land varies seasonally, it decreases from the equator towards the poles and from plains to the mountain tops. It ranges from minus degree Celsius in polar areas and high altitudes to more than 50°C in tropical deserts in summer.
It is clear that mango trees cannot grow in temperate countries like Canada and Germany, and the snow leopards are not found in Kerala forests.
Actually temperature affects the kinetics of enzymes which influence the basal metabolism, activity and other physiological functions of the organism.
A few organisms can tolerate and thrive in a wide range of temperatures they are called eurythermal, but majority of them are restricted to a narrow range of temperatures they are called stenothermal.
Water:
The productivity and distribution of plants is dependent on water. For aquatic organisms the quality (chemical composition, pH) of water is important.
The salt concentration (salinity in parts per thousand), is less than 5 per cent in inland waters, 30 – 35 per cent the sea and > 100 per cent in some hypersaline lagoons.
Some organisms can tolerate wide range of salinities, they are called euryhaline but others are restricted to a narrow range they are called stenohaline.
Freshwater animals cannot live for long in sea water and vice versa because of the osmotic problems.
Light:
It is an important factor for photosynthesis.
Small plants (herbs and shrubs) growing in forests are adapted to photosynthesise under very low light conditions because of tall canopied trees. Many plants require sunlight for the initiation of photoperiodic flowering.
Animals require the diurnal and seasonal variations in light intensity and duration (photoperiod) fortiming their foraging, reproductive, and migratory activities. The spectral quality of solar radiation is important for life.
The UV component of the spectrum is harmful to many organisms while other colour components of the visible spectrum are important for marine plants living at different depths of the ocean.
Soil:
The nature and properties of soil dependent on the climate and the weathering process. The characteristics of the soil determine the water holding capacity.
These characteristics, pH, mineral composition, and topography determine the vegetation in any area. In the aquatic environment, the sediment-characteristics determine the type of benthic animals in ocean.
2. Responses to Abiotic Factors:
During the course of millions of years many species would have evolved constant internal (within the body) environment that permits maximum efficiency of biochemical reactions and physiological functions that results overall ‘fitness’ of the species.
Some organism maintain the constant internal environment when the external environmental conditions changes it is called homeostasis.
A person is able to do his/her work in temperature is 25°C when it is extremely hot or cold outside. It could be achieved at home, in the car while travelling, and at workplace by using an air conditioner in summer and heater in winter. Here the person’s homeostasis is maintained by artificial means.
How do other living organisms cope with the situation?
(i) Regulate:
Some organisms maintain their homeostasis by keeping up constant body temperature, constant osmotic concentration, etc.
Birds, mammals, and some lower vertebrate and invertebrate species are capable of such regulation i.e thermoregulation and osmoregulation. This is the characteristics of mammals to live in Antarctica or in the Sahara desert.
In summer season body of man sweat, it provide cooling effect. In Winter season, vyhen the temperature is much lower than 37°C, man start to shiver, a kind of exercise which produces heat and raises the body temperature. Plants do not have such mechanisms to maintain internal temperatures.
(ii) Conform:
Majority of animals and all plants cannot maintain a constant internal environment.
Thermoregulation is energetically expensive for many organisms. This is true for small animals like shrews and humming birds. Small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside; so they have to spend much energy to generate body heat through metabolism. That is why very small animals are rarely found in polar regions.
If the stressful external conditions are remain only for a short duration, the organism has two other alternatives.
(iii) Migrate:
The organism move temporarily from the stressful habitat to a more favourable area and return when stressful period is over. Some persons moving from Delhi to Shimla in summer season. During winter some birds from Siberia and other extremely cold northern regions migrate to Keolado National Park (Bhartpur) in Rajasthan.
(iv) Suspend:
In bacteria, fungi and lower plants produce thick walled spores during unfavourable Conditions. They germinate in suitable environment.
In higher plants the seeds and vegetative reproductive structures are dormant during adverse condition and germinate after getting favourable moisture and temperature.
In animals, especially bears go into hibernation during winter. Some snails and fish go into aestivation to avoid summer. Under unfavourable conditions many zooplankton species in lakes and ponds are subject to diapause, a stage of suspended development.
Some organisms are subjected to physiological and behavioural adjustments. These responses are called adaptations Kangaroo rat in North American deserts is capable of meeting all its water requirements through its internal fat oxidation. It has the ability to concentrate its urine.
Desert plants have a thick cuticle on their leaf surfaces and stomata arranged in deep pits to minimise water loss through transpiration. They also have CAM pathway in which they open stomata during night and closed during day time.
Opuntia, their leaves are reduced to spines and the flattened stems do photosynthesis. Mammals of colder climates have shorter ears and limbs to minimise heat loss. This is an Allen’s Rule In the polar seas aquatic mammals like seals have a thick layer of fat (blubber) below their skin that acts as an insulator and reduces loss of body heat.
In high altitude, some organism feels altitude sicknes due to low atmospheric pressure and low 02. Its symptoms include nausea, fatigue, and heart diseases. But, gradually get adapted and stop experiencing altitude sickness by increasing red blood cell production, decreasing the binding capacity of hemoglobin and by increasing breathing rate.eg- Many tribes live in the high altitude of Himalayas.
Archaebacteria seen in hot springs and deep sea hydrothermal vents where temperature is more than 1000°C.
Many fish thrive in Antarctic waters where the temperature is below 0°c. A large variety of marine invertebrates and fish live at great depths in the ocean where the pressure could be > 100 times the normal atmospheric pressure.
Desert lizards keep their body temperature constant by behavioural means. They bask in the sun and absorb heat when their body temperature drops, but move into shade when the surrounding temperature starts increasing.
Some species are capable of burrowing into the soil to escape from the above-ground heat.
Populations
1. Population Attributes:
Population has birth rates and death rates. The rates are expressed as the change in numbers with respect to the members of the population. If in a pond there are 20 lotus plants last year and through reproduction 8 new plants are added, so the current population is 28, birth rate is 8/20 = 0.4 offspring per lotus per year.
If 4 individuals in a laboratory population of 40 fruit flies died in a week, the death rate in the population during that period is 4/40 = 0.1 individuals per fruit fly per week.
Another attribute of a population is sex ratio. A population at any given time is composed of individuals of different ages. If the age distribution is plotted for the population, the resulting structure is called an age pyramid.
The shape of the pyramid indicates the growth status of the population i.e
1. growing,
2. stable and
3. declining.
Another important attribute of population is population Representation of age pyramids for human population density (designated as N) Total number is the measure of population density, it is difficult to determine if the counting is impossible.
In an area, if there are 200 Parthenium plants but only a single huge banyan tree with a large canopy, the population density of banyan is low when compared to that of Parthenium. In such cases, the per cent cover or biomass is the measure of the population size.
Number of fish caught pertrap is good measure of its total population density in the lake. The tiger census in our national parks and tiger reserves is based on pug marks and fecai pellets.
2. Population Growth:
Changes in population density that determined by four basic processes, natality, immigration mortality, and emigration.
1. Natality refers to the number of births during a given period
2. Mortality is the number of deaths in the population during a given period.
3. Immigration is the movement of individuals into the population
4. Emigration is the movement of individuals out of the population.
N is the population density at time t, then its density at time t + 1 is
Nt + 1 =Nt + [(B + I) – (D + E)]
Population density increase if the number of births plus the number of immigrants (B + I) is more than the number of deaths plus the number of emigrants (D + E).
If a new habitat is just being colonised, immigration contribute to population growth than birth rates.
Growth Models:
(i) Exponential growth:
When the resources in the habitat are unlimited, each species has the ability to grow in number. Here the population grows in an exponential or geometric fashion.
Population growth curve a when responses are not limiting the growth, plot Is exponential, b when responses are limiting the growth, plot is logistic, K ts carrying capacity
If in a population of size N, the birth rates are represented as b and death rates as d, then the increase or decrease in N during a unit time period t (dN/dt) will be
dN/dt = (b – d) × N (b – d) = r, then dN/dt = rN
r is called the ‘intrinsic rate of natural increase’it is important for assessing impacts of any biotic or abiotic factor on population growth.
The above equation shows the exponential or geometric growth and results J-shaped curve The integral form of the exponential growth equation as
Nt = Population density after time t, N0 = Population density at time zero, r = intrinsic rate of natural increase, e = the base of natural logarithms (2.71828).
(ii) Logistic growth:
Limited resources leads to competition between individuals and the ‘fittest’ individual will survive and reproduce. Governments of many countries introduced restraints to limit human population growth.
In nature, a given habitat has resources to support a maximum possible number, beyond which no further growth is possible. This is called as nature’s carrying capacity (K) for that species in that habitat.
So the population growth in limited resources show initially a lag phase, followed by phases of acceleration and deceleration and finally an stationary phase, and the population density reaches the carrying capacity.
A plot of N in relation to time (t) results in a sigmoid curve. This type of population growth is called Verhulst-Pearl Logistic Growth equation. It is represented by
dN/dt = rN$$\frac{(K-N)}{K}$$
Where N = Population density at time t
r = Intrinsic rate of natural increase
K= Carrying capacity
3. Life History Variation:
Populations evolve to maximise their reproductive fitness, called as Darwinian fitness. Some organisms breed only once in their lifetime (Pacific salmon fish, bamboo) while others breed many times during their lifetime (most birds and mammals). Some produce a large number of small-sized offspring (Oysters, pelagic fishes) while others produce a small number of large-sized offspring (birds, mammals).
4. Population Interactions:
It occurs between species. Interspecific interactions arise from the interaction of populations of two different species. It is beneficial, detrimental or neutral (neither harm nor benefit) to one of the species or both. Assigning a ‘+’sign for beneficial interaction, sign for detrimental and 0 for neutral interaction,
Population Interactions
Both the species benefit in mutualism and both lose in competition. In both parasitism and Predation one species is benifitted and the other is harmed (host and prey).
In commensalism one species is benefitted and the other is neither benefitted nor harmed. In amensalism one species is harmed whereas the other is unaffected.
(i) Predation:
In predation, the energy stored at consumer level from plants are transferred to the next. So the consumer level energy transfer mainly takes place from prey to predator. For example prey is deer and predator is tiger.
The introduction of prickly pear cactus into Australia causes the spreading of these plants into millions of hectares. Later the cactus was controlled by cactus-feeding predator (a moth) from its natural habitat. This is an example of Biological control methods.
Predators also help in maintaining species diversity in a community, by reducing the intensity of competition among competing prey species. In an American Pacific Coast, the starfish Pisaster is an important predator.
In a field experiment, when all the starfish were removed from intertidal area, more than 10 species of invertebrates became extinct within a year, because of interspecific competition.
If a predator overexploits its prey, it become extinct. Later predator become extinct for the lack of food. This is the reason why predators in nature are ‘prudent’.
Prey species also have defense mechanism to reduce the impact of predation. Some species of insects and frogs are cryptically-coloured (camouflaged) so the prey cannot be detected easily by the predator. If the prey is poisonous, it cannot be attacked by the predators. Eg- Monarch butterfly is distasteful to its predator (bird).
Plants have evolved some morphological and chemical defences against herbivores. Thorns (Acacia, Cactus) are morphological defence. Many plants produce and store chemicals that affect the herbivores digestion, reproduction and finally kill it.
The weed Calotropis produces poisonous cardiac glycosides and that affect cattle or goats browsing on this plant. Chemical substances that extract from plants (nicotine, caffeine, quinine, strychnine, opium, etc.,) are defences against grazers and browsers.
(ii) Competition:
The competition mainly for resources that takesplace among same species and different species. For example flamingoes coming into shallow South American lakes compete with resident fishes for their common food, the zooplankton in the lake.
In interference competition, the feeding efficiency of one species is reduced due to other species, even if resources (food and space) are abundant. So, in competition the fitness of one species is lower in the presence of another species.
According to Gause, when resources are limiting the competitively superior species eliminate the other species, This is an example of competitive exclusion.
When the goats introduced in the Galapagos island, Abingdon tortoise become extinct due to the greater browsing efficiency of the goats.
Another evidence of competition in nature is called ‘competitive release’. Some species restricted to small geographical area because of the presence of a competitively superior species.
Connell’s elegant field experiments showed that on the rocky sea coasts of Scotland, the larger and competitively superior barnacle Balanus dominates the intertidal area, and eliminates the smaller barnacle Chathamalus.
Actually the herbivores and plants are more adversely affected by competition than carnivores.
Gause’s Competitive Exclusion Principle’ states that two closely related species competing for the same resources cannot co-exist, as a result competitively inferior one is eliminated.
Some species shows ‘resource partitioning’.that is, if two species compete for the same resource, they avoid competition by choosing different times for feeding or different foraging patterns.
MacArthur showed that five closely related species of warblers living on the same tree were able to avoid competition and co-exist due to behavioural differences in their foraging activities.
(iii) Parasitism:
Many parasites are host-specific. Some parasites evolved special adaptations such as the
loss of unnecessary sense organs, presence of suckers to cling on to the host, loss of digestive system and high reproductive capacity.
The life cycles of parasites consist of one ortwo intermediate hosts or vectors to facilitate parasitisation. The human liver fluke depends on two intermediate hosts to complete its life cycle. The malarial parasite needs a vector (mosquito) to cause disease in other hosts.
Majority of the parasites reduce the survival, growth and reproduction of the host and reduce its population density.
Parasites that feed on the external surface of the host organism are called ectoparasites. Examples are the lice on humans and ticks on dogs. Ectoparasite copepods affect many marine fishes Chlorophyll-less Cuscuta a parasitic plant that absorbs nutritive materials from the host plant. Endoparasites that live inside the host body at different sites (liver, kidney, lungs, red blood cells, etc.). The life cycles of endoparasites are more complex. Their reproductive potential is more but their morphological and anatomical features are simple.
In Brood parasitism parasitic bird lays its eggs in the nest of its host and the host incubate them. The eggs of the parasitic bird resemble the host’s egg in size Examples of brood parasitism are cuckoo (koel) and the crow.
(iv) Commensalism:
This is the interaction in which one species benefits and the other is neither harmed nor benefited. An epiphytic orchid on a mango branch, and barnacles growing on the back of a whale get benefit. But the mango tree and the whale is neither harmed nor benefited.
The cattle egret and grazing cattle is an example of commensalism. Another example of commensalism is the interaction between sea anemone with stinging tentacles and the clown fish.
(v) Mutualism:
In this interaction both partner species are benefitted. Examples are Lichens (between a fungus and algae), mycorrhizae (between fungi and the roots of higher plants). The fungi help the plant in the absorption of essential nutrients from the soil while the plant in turn provides the energy-yielding carbohydrates.
Some examples of mutualism are found in plant-animal relationships. Plants need animals for pollinating their flowers and dispersing their seeds. Plants offer rewards in the form of pollen and nectar for pollinators and juicy and nutritious fruits for seed dispersers.
Co-evolution occurs between the flower and its pollinator species. In many species of fig trees, pollination is done by wasp. The female wasp uses the fruit not only for egg laying but uses the developing seeds within the fruit for nourishing its larvae.
The Mediterranean orchid- Ophrys. petal of its flower shows the similarity with female bee in size, colour and markings. The male bee is attracted and ‘pseudocopulates’ with the flower, When this same bee pseudocopulates’ with another flower, it transfers pollen to it and thus, pollinates the flower. | 2022-11-29 12:14:19 | {"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": 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": 0.4311573803424835, "perplexity": 4857.657792407468}, "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-2022-49/segments/1669446710691.77/warc/CC-MAIN-20221129100233-20221129130233-00336.warc.gz"} |
https://www.cnblogs.com/lrj124/p/9721006.html | # 【POJ2976】Dropping tests - 01分数规划
## Description
In a certain course, you take n tests. If you get ai out of bi questions correct on test i, your cumulative average is defined to be
.
Given your test scores and a positive integer k, determine how high you can make your cumulative average if you are allowed to drop any k of your test scores.
Suppose you take 3 tests with scores of 5/5, 0/1, and 2/6. Without dropping any tests, your cumulative average is . However, if you drop the third test, your cumulative average becomes .
## Input
The input test file will contain multiple test cases, each containing exactly three lines. The first line contains two integers, 1 ≤ n ≤ 1000 and 0 ≤ k < n. The second line contains n integers indicating ai for all i. The third line contains n positive integers indicating bi for all i. It is guaranteed that 0 ≤ ai ≤ bi ≤ 1, 000, 000, 000. The end-of-file is marked by a test case with n = k = 0 and should not be processed.
## Output
For each test case, write a single line with the highest cumulative average possible after dropping k of the given test scores. The average should be rounded to the nearest integer.
## 思路
/************************************************
*Author : lrj124
*Created Time : 2018.09.28.20:35
*Mail : [email protected]
*Problem : poj2976
************************************************/
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;
const int maxn = 1000 + 10;
int n,k,a[maxn],b[maxn];
double tmp[maxn];
inline bool check(double x) {
for (int i = 1;i <= n;i++) tmp[i] = a[i]-x*b[i];
sort(tmp+1,tmp+n+1);
double ans = 0;
for (int i = k+1;i <= n;i++) ans += tmp[i];
return ans >= 0;
}
int main() {
while (cin >> n >> k) {
if (!n && !k) break;
for (int i = 1;i <= n;i++) cin >> a[i];
for (int i = 1;i <= n;i++) cin >> b[i];
double l = 0,r = 0x3f3f3f3f;
while (fabs(r-l) >= 1e-6) {
double mid = (l+r)/2;
if (check(mid)) l = mid;
else r = mid;
}
int ans = int(l*100+0.5);
printf("%d\n",ans);
}
return 0;
}
posted @ 2018-09-28 21:35 lrj124 阅读(...) 评论(... 编辑 收藏 | 2019-12-09 10:17:04 | {"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": 0.17859679460525513, "perplexity": 4030.9318368215418}, "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-2019-51/segments/1575540518627.72/warc/CC-MAIN-20191209093227-20191209121227-00466.warc.gz"} |
https://www.esaral.com/q/if-27x-93x-find-x-96475 | # If 27x =93x, find x.
Question:
If $27^{x}=\frac{9}{3^{x}}$, find $\mathrm{x}$
Solution:
We are given $27^{x}=\frac{9}{3^{x}}$. We have to find the value of $x$
Since $\left(3^{3}\right)^{x}=\frac{3^{2}}{3^{x}}$
By using the law of exponents $\frac{a^{m}}{a^{n}}=a^{m-n}$ we get,
$3^{3 x}=3^{2-x}$
On equating the exponents we get,
\begin{aligned} 3 x &=2-x \\ 3 x+x &=2 \\ 4 x &=2 \\ x &=\frac{2}{4} \end{aligned}
$x=\frac{1}{2}$ | 2023-02-05 15:10:24 | {"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": 1.0000100135803223, "perplexity": 2883.7845179573824}, "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-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00364.warc.gz"} |
https://cs.stackexchange.com/questions/49311/how-to-remove-empty-strings-a-k-a-lambda-epsilon-from-a-lr0-grammar | # How to remove empty strings a.k.a (lambda, epsilon) from a LR(0) grammar
Say I have the following grammar:
E -> TE'
E' -> +TE'| -TE' | λ
T -> (E) | id
I need to build the finite state machine with the LR(0) items,
But I know in order to do this I have to remove the λ.
How to I accomplish this ?
Also, after I have the SLR(1) table, how to I prove is valid ?
Thanks, I'm studying for an exam, and I'm stuck here :(
• Well, what have you tried and where exactly are you stuck? – Raphael Nov 10 '15 at 15:19
1. Find all nullable variables. In this case only E' is nullable.
2. Let me illustrate the second step with an example:
Replace a production A -> BCD with a family of productions like this (assuming B, C & D are nullable):
A -> BCD | BC | BD | CD | B | C | D
1. Delete all the productions with the empty string as the right-hand side.
With all that in mind we get:
E -> TE' | T
E' -> +TE'| -TE' | +T | -T
T -> (E) | id | 2019-08-23 00:13:46 | {"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": 0.510173499584198, "perplexity": 2277.1684155814537}, "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/1566027317688.48/warc/CC-MAIN-20190822235908-20190823021908-00307.warc.gz"} |
https://www.physicsforums.com/threads/laser-physics-diameter-of-a-spot-on-the-moon-from-a-laser.915322/ | Homework Help: Laser Physics - Diameter of a Spot on the Moon from a Laser
1. May 21, 2017
ChrisJ
Was not sure whether to post here on in the advanced section, since it is part of a final year undergraduate module, yet seems like a pretty simple problem (though I still need help! haha).
1. The problem statement, all variables and given/known data
A frequency doubled Nd:YAG laser ($\lambda = 532$nm) with an initial beam radius of 10cm is shone at the moon, what will be the diameter of the spot? Distance to moon is $3.84 \times 10^8$m
2. Relevant equations
$w^2(z)=w_0^2 \left(1+\frac{z^2 \lambda^2}{\pi^2 w_0^4} \right)$
$\frac{\theta}{2} = \frac{w(z)}{z}$ for large $z$
3. The attempt at a solution
I first did the problem by just plugging in the numbers into the equation for $w^2(z)$ and taking the square root, because in my notes I have written that $w(z)$ is the beam's radius and got a value of $8.06$m for $w(z)$.
But then I also saw that I have also written in my notes that $w^2(z)$ is called "spot size" and since it has same units as area I then assumed that since $A=\pi \frac{D^2}{4}$ then the diameter must be $\sqrt{\frac{4w^2(z)}{\pi}}$ and got a value of $9.1$m.
Also was not sure about the frequency doubled bit of question, I assumed that the (532nm) given for the wavelength had already taken this into account? Perhaps not and I need to half the wavelength?
I think I probably over thought it a bit much. I have not written my explicit workings as it is just putting in number and the TeX is tedious but will type it up if needed.
Any help appreciated :)
Last edited: May 21, 2017
2. May 21, 2017
phyzguy
(1) w is just the spot diameter, not the spot area.
(2) You can just use the 532 nm as the wavelength - the frequency doubling is already included in this.
I think your answer is way off - you probably have not done the units correctly. Why don't you walk us through how you arrived at 8.06 m and let's see where you went wrong.
3. May 21, 2017
ChrisJ
Thanks for the quick reply, I did think it was rather small tbh,
I did...
$$w^2(z)=w_0^2 \left(1+\frac{z^2 \lambda^2}{\pi^2 w_0^4} \right) \\ w^2(z)=0.1^2 \left(1+\frac{(3.84 \times 10^8)^2 (532 \times 10^{-9})^2}{(\pi)^2 (0.1)^4} \right) = 65 \textrm{m}^2\\ w = \sqrt{65} = 8.1 \textrm{m}$$
If that is correct, then I must have entered it in my calculator wrong, although I did do it twice,
EDIT: Yeah I entered it in wrong as I just redid it and got w=650.3m !! I must have forgot to square something in my calculation of $w^2(z)$
4. May 21, 2017
phyzguy
You're still making a mistake. Just estimating, the term on the right inside the parentheses is about 4x10^7 m^2. How can adding one to it and multiplying by 0.01 give 65?
5. May 21, 2017
ChrisJ
Did you notice my edit in time? I'm now getting $w^2(z)=422849.6$m^2 and $w=650$m , but I wanted to double check is w definitely the diameter and not the radius? As I have a sketch of beam divergence in my notes and w seems to be the radius, obviously if so then the diametre is 1.3km.
6. May 21, 2017
phyzguy
Yes, since the initial radius is given, then w is the radius, not the diameter like I said earlier. So I think you are doing it right now and 1.3 km is the right answer.
7. May 21, 2017
ChrisJ
Ok thank you for your help. I would have normally caught something so simple r.e. the calculation, but I entered it twice in a row and got 65, so I must have forgot to square one of the terms and made the same omission twice in a row.
Thanks. | 2018-07-23 18:02:09 | {"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": 0.8217511773109436, "perplexity": 497.58425770173193}, "config": {"markdown_headings": false, "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-2018-30/segments/1531676599291.24/warc/CC-MAIN-20180723164955-20180723184955-00186.warc.gz"} |
https://homework.cpm.org/category/CCI_CT/textbook/apcalc/chapter/1/lesson/1.3.1/problem/1-108 | ### Home > APCALC > Chapter 1 > Lesson 1.3.1 > Problem1-108
1-108.
To estimate the area under a curve, rectangles are often the easiest shape to use. However, there are different ways to choose the heights of the rectangles. You have already used left endpoint and right endpoint rectangles. Another way is to use midpoint rectangles, which have heights defined at the midpoints of the intervals. For example, for the function $f(x) = \frac { 1 } { 2 }x + \cos(x)$ graphed at right, the first midpoint rectangle has a height of $f(1.5) ≈ 0.821$.
Calculate the height of the other two rectangles then use them to approximate the area under the curve for $1 ≤ x ≤ 4$.
Use your calculator to find the values of $f(2.5)$ and $f(3.5)$.
The base of each rectangle is $1$. The approximation for the area under the curve is $A = 1(f(1.5) + f(2.5) + f(3.5))$. | 2021-05-11 10:01:57 | {"extraction_info": {"found_math": true, "script_math_tex": 7, "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": 0.8902350068092346, "perplexity": 241.04874441720017}, "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-2021-21/segments/1620243991982.8/warc/CC-MAIN-20210511092245-20210511122245-00528.warc.gz"} |
https://www.snapxam.com/problems/92410288/integral-of-e-2-0t-dt-from-1infinity-to-1-0 | Calculators Topics Go Premium About Snapxam Topics
# Step-by-step Solution
Go!
1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
(◻)
+
-
×
◻/◻
/
÷
2
e
π
ln
log
log
lim
d/dx
Dx
|◻|
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc
asin
acos
atan
acot
asec
acsc
sinh
cosh
tanh
coth
sech
csch
asinh
acosh
atanh
acoth
asech
acsch
## Step-by-step explanation
Problem to solve:
$\int_{-\infty}^{-1}\left(e^{-2t}\right)dt$
Learn how to solve definite integrals problems step by step online.
$\lim_{c\to{-\infty }}\:\int_{c}^{-1} e^{-2t}dt$
Learn how to solve definite integrals problems step by step online. Integrate e^(-2t) from -\infty to -1. Replace the integral's limit by a finite value. Solve the integral \int_{c}^{-1} e^{-2t}dt applying u-substitution. Let u and du be. Isolate dt in the previous equation. Substituting u and dt in the integral and simplify.
$-3.6945+0.5\cdot e^{-2\cdot -\infty }$
### Problem Analysis
$\int_{-\infty}^{-1}\left(e^{-2t}\right)dt$
### Main topic:
Definite integrals
~ 0.07 seconds | 2020-02-17 19:48:04 | {"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": 0.9450131058692932, "perplexity": 10926.42640483468}, "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-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00136.warc.gz"} |
https://yannisparissis.wordpress.com/2011/03/10/dmat0101-notes-3-the-fourier-transform-on-l1/ | ## DMat0101, Notes 3: The Fourier transform on L^1
1. Definition and main properties.
For ${f\in L^1({\mathbb R}^n)}$, the Fourier transform of ${f}$ is the function
$\displaystyle \mathcal{F}(f)(\xi)=\hat{f}(\xi)=\int_{{\mathbb R}^n}f(x)e^{-2\pi i x\cdot \xi}dx,\quad \xi\in{\mathbb R}^n.$
Here ${x\cdot y}$ denotes the inner product of ${x=(x_1,\ldots,x_n)}$ and ${y=(y_1,\ldots, y_n)}$:
$\displaystyle x\cdot y=\langle x,y\rangle=x_1y_1+\cdots x_n y_n.$
Observe that this inner product in ${{\mathbb R}^n}$ is compatible with the Euclidean norm since ${x\cdot x=|x|^2}$. It is easy to see that the integral above converges for every ${\xi\in{\mathbb R}^n}$ and that the Fourier transform of an ${L^1}$ function is a uniformly continuous function.
Theorem 1 Let ${f,g\in L^1({\mathbb R}^n)}$. We have the following properties.
(i) The Fourier transform is linear ${\widehat{f+g}=\hat f + \hat g}$ and ${\widehat{cf}=c \hat f}$ for any ${c\in{\mathbb C}}$.
(ii) The function ${\hat f(\xi)}$ is uniformly continuous.
(iii) The operator ${\mathcal F}$ is bounded operator from ${L^1({\mathbb R}^n)}$ to ${L^\infty({\mathbb R}^n)}$ and
$\displaystyle \|\hat f \|_{L^{\infty}({\mathbb R}^n)}\leq \|f\|_{L^1({\mathbb R}^n)}.$
(iv) (Riemann-Lebesgue) We have that
$\displaystyle \lim _{|\xi|\rightarrow +\infty} \hat f(\xi)=0.$
Proof: The properties (i), (ii) and (iii) are easy to establish and are left as an exercise. There are several ways to see (iv) based on the idea that it is enough to establish this property for a dense subspace of ${L^1({\mathbb R}^n)}$. For example, observe that if ${f}$ is the indicator function of an interval of the real line, ${f=\chi_{[a,b]}}$, then we can calculate explicitly to show that
$\displaystyle |\hat{f}(\xi)|=\bigg|\int_a ^b e^{-2\pi i x\xi}dx\bigg| =\bigg| \frac{e^{-2\pi i \xi a}-e^{-2\pi i \xi b}}{2\pi i \xi }\bigg|\lesssim \frac{1}{|\xi|}\rightarrow 0\quad\mbox{as}\quad |\xi|\rightarrow +\infty.$
Tensoring this one dimensional result one easily shows that ${\lim_{|\xi|\rightarrow +\infty}f(\xi)=0}$ whenever ${f}$ is the indicator function of an ${n}$-dimensional interval of the form ${[a_1,b_1]\times\cdots\times [a_n,b_n]}$. Obviously the same is true for finite linear combinations of ${n}$-dimensional intervals since the Fourier transform is linear.
Now let ${f}$ be any function in ${L^1({\mathbb R}^n)}$ and ${\epsilon >0}$ and consider a simple function which is a finite linear combination of ${n}$-dimensional intervals, such that ${\|f-g\|_1<\epsilon/2}$. Let also ${M>0}$ be large enough so that ${|\hat g (\xi)|<\epsilon/2}$ whenever ${|\xi|>M}$. Using (iii) and the linearity of the Fourier transform we have that
$\displaystyle \begin{array}{rcl} |\hat f(\xi)|\leq |\widehat{(f-g)}\hat (\xi)|+|\hat g (\xi)|\leq \|f-g\|_{L^1}+|\hat g (\xi)|<\epsilon, \end{array}$
whenever ${|\xi|>M}$, which finishes the proof. $\Box$
In view of (ii) and (iv) we immediately get the following.
Corollary 2 If ${f\in L^1({\mathbb R}^n)}$ then ${\hat f\in C_o({\mathbb R}^n)}$.
Exercise 1 Show the properties (ii) and (iii) in the previous Theorem.
The discussion above and especially Corollary 2 shows that a necessary condition for a function ${g}$ to be a Fourier transform of some function in ${L^1({\mathbb R}^n)}$ is ${g\in C_o({\mathbb R}^n)}$. However, this condition is far from being sufficient as there are functions ${g\in C_o({\mathbb R}^n)}$ which are not Fourier transforms of ${L^1}$ functions. See Exercise 8.
Let us now see two important examples of Fourier transforms that will be very useful in what follows.
Example 1 For ${a>0}$ let ${f(x)=e^{-\pi a|x|^2}}$. Then
$\displaystyle \hat f(\xi)=a^{-\frac{n}{2}}e^{-\frac{\pi|\xi|^2}{a}}.$
Proof: Observe that in one dimension we have
$\displaystyle \begin{array}{rcl} \hat f(\xi)&=&\int_{\mathbb R} e^{-\pi ax^2}e^{-2\pi i x\xi}dx=\int_{\mathbb R} e^{-\pi a(x+i\frac{\xi}{a})^2}dx\ e^{-\frac{\pi\xi^2}{a}}\\ \\ &=&\int_{\mathbb R} e^{-\pi ax^2} dx \ e^{-\frac{\pi\xi^2}{a}}= a^{-\frac{1}{2}} e^{-\frac{\pi^2\xi^2}{a}}, \end{array}$
where the third equality is a consequence of Cauchy’s theorem from complex analysis. The ${n}$-dimensional case is now immediate by tensoring the one dimensional result. $\Box$
Remark 1 Replacing ${a=1}$ in the previous example we see that ${e^{-\pi |x|^2}}$ is its own Fourier transform.
Example 2 For ${a>0}$ let ${g(x)=e^{-2\pi a |x|}}$. Then
$\displaystyle \hat g(\xi)=c_n\frac{a}{(a^2+|\xi|^2)^\frac{n+1}{2}},$
where ${c_n=\Gamma((n+1)/2)/\pi^\frac{n+1}{2}}$.
Proof: The first step here is to show the subordination identity
$\displaystyle e^{-\beta}=\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-\beta^2/4u}du,\quad \beta>0, \ \ \ \ \ (1)$
which is a simple consequence of the identities
$\displaystyle \begin{array}{rcl} e^{-\beta}&=&\frac{2}{\pi}\int_0 ^\infty\frac{\cos \beta x}{1+x^2}dx,\\ \\ \frac{1}{1+x^2}&=&\int_0 ^\infty e^{-(1+x^2)u}du. \end{array}$
Using (1) we can write
$\displaystyle \begin{array}{rcl} \hat g(\xi)&=&\int_{{\mathbb R}^n} e^{-2\pi a|x|} e^{-2\pi i x\cdot \xi}dx \\ \\ &=&\frac{1}{\sqrt{\pi}}\int_{{\mathbb R}^n}\bigg(\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-4\pi^2a^2|x|^2/4u}du\bigg) e^{-2\pi i x\cdot \xi} dx\\ \\ &=&\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}} \frac{1}{a^n}\bigg(\sqrt{\frac{u}{\pi}}\bigg)^\frac{n}{2} e^{-\frac{u|\xi|^2}{a^2}}du\\ \\ &=& \frac{1}{\pi^\frac{n+1}{2} a^n}\int_0 ^\infty u^\frac{n-1}{2}e^{-u\frac{|\xi|^2}{a^2}}e^{-u}du\\ \\ &=& \frac{1}{\pi^\frac{n+1}{2} a^n} \frac{1}{\big(1+\frac{|\xi|^2}{a^2}\big)^\frac{n+1}{2}}\int_0 ^\infty u^\frac{n-1}{2}e^{-u}du\\ \\ &=& \frac{\Gamma(\frac{n+1}{2})}{\pi^\frac{n+1}{2} }\frac{a}{\big(a^2+|\xi|^2 \big)^\frac{n+1}{2}}, \end{array}$
by the definition of the ${\Gamma}$-function.$\Box$
Exercise 2 This exercise gives a first (qualitative) instance of the uncertainty principle. Prove that there does not exist a non-zero integrable function on ${{\mathbb R}}$ such that both ${f}$ and ${\hat f}$ have compact support.
Hint: Observe that the function
$\displaystyle \hat f(\xi)=\int_{\mathbb R} f(x) e^{-2\pi i x\xi}dx ,$
extends to an entire function (why ?).
The definition of the Fourier transform extends without difficulty to finite Borel measures on ${{\mathbb R}^n}$. Let us denote by ${\mathcal M({\mathbb R}^n)}$ this class of finite Borel measures and let ${\mu\in \mathcal M({\mathbb R}^n)}$. We define the Fourier transform of ${\mu}$ to be the function
$\displaystyle \mathcal{F}(\mu)(\xi)=\hat{\mu}(\xi)=\int_{{\mathbb R}^n}e^{-2\pi i x\cdot \xi}d\mu(x),\quad \xi\in{\mathbb R}^n.$
We have the analogues of (i), (ii) and (iii) of Theorem 1 if we replace the ${L^1}$ norm by the total variation of the measure. However property (iv) fails as can be seen by consider the Fourier transform of a Dirac mass at the point ${0}$. Indeed observe that
$\displaystyle \hat{\delta_0}(\xi)=\int_{{\mathbb R}^n} e^{-2\pi i x\cdot \xi}d\delta_0(x)=1,$
which is a constant function.
The Fourier transform interacts very nicely with convolutions of functions, turning them to products. This turns out to be quite important when considering translation invariant operators as we shall see later on in the course.
Proposition 3 Let ${f,g\in L^1({\mathbb R}^n)}$. Then ${\widehat{f*g}=\hat f \hat g}$.
Exercise 3 Prove Proposition 3.
Another important property of the Fourier transform is the multiplication formula.
Proposition 4 (Multiplication formula) Let ${f,g\in L^1({\mathbb R}^n)}$. Then
$\displaystyle \int_{{\mathbb R}^n}\hat f(\xi)g(\xi)d\xi=\int_{{\mathbb R}^n}f(x)\hat g(x)dx.$
We will now describe some easily verified symmetries of the Fourier transform. We introduce the following basic operations on functions:
Translation operator: ${ (\tau_{x_o} f)(x)=f(x-x_o),\quad x,x_o\in {\mathbb R}^n}$
Modulation operator: ${\textnormal{Mod}_{x_o}(f)(x)=e^{2\pi i x\cdot x_o} f(x),\quad x,x_o\in {\mathbb R}^n}$
Dilation operator: ${\textnormal{Dil}_\lambda ^p(f)(x)={\lambda^{-\frac{n}{p}}}f(x/\lambda),\quad x,\in {\mathbb R}^n,\lambda>0,1\leq p\leq\infty}$.
Proposition 5 Let ${f\in L^1({\mathbb R}^n)}$ We have the following symmetries:
(i) ${ \mathcal F \tau_{x_o}=\textnormal{Mod} _{-x_o}\mathcal F}$,
(ii) ${\mathcal F \textnormal{Mod}_{\xi_o}=\tau_{\xi_o}\mathcal F}$,
(iii) ${\mathcal F \textnormal {Dil} _\lambda ^p = \textnormal {Dil} _{ \lambda^{-1} } ^{p'} \mathcal F}$, where ${\frac{1}{p}+\frac{1}{p'}=1}$.
Exercise 4 Prove the symmetries in Proposition 5 above. Also, let ${U:{\mathbb R}^n\rightarrow {\mathbb R}^n}$ be an invertible linear transformation, that is, ${U\in GL({\mathbb R}^n)}$. Define the general dilation operator
$\displaystyle ( \textnormal{Dil} _U ^pf)(x) =|\det{U}|^{-\frac{1}{p}}f(U^{-1}x), \quad x\in {\mathbb R}^n, 1\leq p\leq \infty.$
Prove that
$\displaystyle \mathcal F \textnormal{Dil}_U ^p = \textnormal{Dil}_{(U^*)^{-1}} ^{p'}\mathcal F,$
where ${U^*}$ is the (real) adjoint of ${U}$, that is the matrix for which we have ${\langle Ux, y\rangle =\langle x , U^*y\rangle}$ for all ${x,y\in {\mathbb R}^n}$.
We now come to one of the most interesting properties of the Fourier transform, the way it commutes with derivatives.
Proposition 6 (a) Suppose that ${f\in L^1({\mathbb R}^n)}$ and that ${x_kf(x)\in L^1({\mathbb R}^n)}$ for some ${1\leq k \leq n}$. The ${\hat f}$ is differentiable with respect to ${\xi_k}$ and
$\displaystyle \frac{\partial}{\partial \xi_k} \mathcal F(f)(\xi) = \mathcal F(- 2\pi i x_k f)(\xi).$
(b) We will say that a function ${f}$ has a partial derivative in the ${L^p}$ norm with respect to ${x_k}$ if there exists a function ${g\in L^p({\mathbb R}^n)}$ such that
$\displaystyle \lim_{h_k\rightarrow 0 }\bigg(\int_{{\mathbb R}^n}\bigg|\frac{f(x+h)-f(x)}{h_k}-g\bigg|^p dx\bigg)^\frac{1}{p}=0,$
where ${h=(0,\ldots,0,h_k,0,\ldots,0)}$ is a non-zero vector along the ${k}$-th coordinate axis. If ${f}$ has a partial derivative ${g}$ with respect to ${x_k}$ in the ${L^1}$-norm, then
$\displaystyle \hat g(\xi)=2\pi i \xi_j \mathcal {F}(\xi).$
Exercise 5 Prove Proposition 6.
A similar result that involves the classical derivatives of a function is the following:
Proposition 7 For ${k}$ a non-negative integer, suppose that ${f\in C^k({\mathbb R}^n)}$ and that ${\partial^\alpha f \in L^1({\mathbb R}^n)}$ for all ${|\alpha|\leq k}$, and ${\partial^\alpha f\in C_o({\mathbb R}^n)}$ for ${|\alpha|\leq k-1}$. Show that
$\displaystyle \widehat {\partial^\alpha f}(\xi)=(2\pi i \xi)^\alpha \hat f(\xi).$
Exercise 6 Prove Proposition 7.
Several remarks are in order. First of all observe that Propositions 6,7 assert that the following commutation relations are true
(i) ${\mathcal F (-2\pi i x_k) = \frac{\partial}{\partial \xi_k} \mathcal F ,}$
(ii) ${ \mathcal F \frac{\partial}{\partial x_k} = 2\pi i \xi_k \mathcal F,}$
where here we abuse notation and denote by ${2\pi i x_k }$ the operator of multiplication by ${2\pi i x_k}$. Thus the Fourier transform turns derivatives to multiplication by the corresponding variable, and vice versa, it turns multiplication by the coordinate variable to a partial derivative, whenever this is technically justified. This is a manifestation of the heuristic principle that smoothness of a function translates to decay of the Fourier transform and on the other hand, decay of a function at infinity translates to smoothness of the Fourier transform.
A second remark is that these commutation relations generalize, in an obvious way, to higher derivatives. To make this more precise, let ${P}$ be a polynomial on ${{\mathbb R}^n}$:
$\displaystyle P(x)=\sum_{|\alpha|\leq d}c_\alpha x^\alpha .$
Slightly abusing notation again we write ${P(\partial ^\alpha)}$ for the differential operator
$\displaystyle P(\partial^\alpha)=\sum_{|\alpha|\leq d}c_\alpha {\partial^\alpha} .$
We then have that the following commutation relations are true
(i’) ${\mathcal F P(-2\pi i x) = P(\partial^\alpha) \mathcal F ,}$
(ii’) ${ \mathcal F P(\partial^\alpha) =P( 2\pi i \xi) \mathcal F.}$
Observe that for nice’ functions, for example ${f\in C_c({\mathbb R}^n)}$ or ${f\in \mathcal S(R^n)}$, Propositions 6 and 7 are automatically satisfied.
2. Inverting the Fourier transform
On of the most important problems in the theory of Fourier transforms is that of the inversion of the Fourier transform. That is, given the Fourier transform ${\hat f}$ of an ${L^1}$ function, when can we recover the original function ${f}$ from ${\hat f}$? We begin with a simple case where the recovery is quite easy.
Proposition 8 Let ${f\in L^1({\mathbb R}^n)}$ be such that ${\hat f \in L^1({\mathbb R}^n)}$. Then the inversion formula holds true. In particular we have that
$\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi} d\xi,$
for almost every ${x\in {\mathbb R}^n}$.
Proof: The proof is based on the following calculation. For ${a>0}$ we have that
$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi&=&\int_{{\mathbb R}^n}\int_{{\mathbb R}^n} f(y) e^{-2\pi i y\cdot \xi}dy e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi\\ \\ &=&\int_{{\mathbb R}^n} f(x+y) \int_{{\mathbb R}^n} e^{-2\pi i y}e^{-a|\xi|^2} d\xi dy\\ \\ &=&(\frac{\pi}{a})^\frac{n}{2}\int_{{\mathbb R}^n}f(x+y) e^{-\frac{\pi^2|y|^2}{a}} dy \\ \\ &=& \int_{{\mathbb R}^n}f(x+\sqrt{a}y) e^{-\pi|y|^2}dy, \end{array}$
where in the last equality we have used Example 1. We can thus write
$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} \hat f(\xi)e^{-a|\xi|^2 e^{2\pi i x\cdot \xi} }d\xi -f(x) \bigg| dx&=& \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} f(x+\sqrt{a}y) e^{-\pi|y|^2}dy-f(x)\bigg| dx\\ \\ &=& \int_{{\mathbb R}^n}\bigg|\int_{{\mathbb R}^n} \{f(x+\sqrt{a}y) -f(x) \}e^{-\pi |y|^2} dy \bigg|dx \\ \\ &\leq &\int_{{\mathbb R}^n}\int_{{\mathbb R}^n} |f(x+\sqrt{a}y)-f(x)|dx e^{-\pi|y|^2}dy \\ \\ &=& \int_{{\mathbb R}^n}\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}e^{-\pi|y|^2}dy. \end{array}$
Since ${\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}\rightarrow 0}$ as ${a\rightarrow 0}$ and ${\|f-\tau_{-\sqrt{a}y}f\|_{L^1({\mathbb R}^n)}\leq 2\|f\|_{L^1({\mathbb R}^n)}}$, Lebesgue’s dominated convergence theorem shows that ${f}$ is almost everywhere equal to the ${L^1}$-limit of the sequence of functions
$\displaystyle g_a(x)=\int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi ,$
as ${a\rightarrow 0}$ (technically speaking we need to consider a sequence ${a_k\rightarrow0}$). On the other hand since ${\hat f\in L^1({\mathbb R}^n)}$, another application of Lebesgue’s dominated theorem shows that the ${L^1}$-limit of the functions ${g_a}$ is also equal to ${\int_{{\mathbb R}^n}\hat f(\xi)e^{2\pi i x\cdot \xi}d\xi}$. This completes the proof of the proposition. $\Box$
An immediate corollary is that the Fourier transform is a one-to-one operator:
Corollary 9 Let ${f_1,f_2\in L^1({\mathbb R}^n)}$ and suppose that ${\hat f_1(\xi)=\hat f_2(\xi)}$ for all ${\xi\in{\mathbb R}^n}$. The we have that ${f_1(x)=f_2(x)}$ for almost every ${x\in{\mathbb R}^n}$.
The proof is an obvious application of Proposition 8.
Exercise 7 (i) Suppose that ${f\in C_c ^{n+1}({\mathbb R}^n)}$. Show that
$\displaystyle |\hat f(\xi)| \lesssim (1+|\xi|^2)^{-(n+1)/2}.$
Conclude that whenever ${f\in C^{n+1} _c({\mathbb R}^n)}$, we have that
$\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi} d\xi.$
(ii) Show that ${\mathcal F}$ maps the Schwartz space ${\mathcal S({\mathbb R}^n)}$ onto ${\mathcal S({\mathbb R}^n)}$.
Exercise 8 The purpose of this exercise is to show that ${\mathcal F(L^1({\mathbb R}^n))}$ is a proper subset of ${C_o({\mathbb R}^n)}$ but also that it is a dense subset of ${C_o({\mathbb R}^n)}$.
(i) Show that ${\mathcal F(L^1({\mathbb R}))}$ is a proper subset of ${C_o({\mathbb R})}$.
Hint: While there are different ways to do that, a possible approach is the following. For simplicity we just consider the case ${n=1}$:
(a) Show that ${\big| \int_a ^b \frac{\sin x}{x}dx\big| \leq B}$ for all ${0\leq |a|<|b|<\infty}$ where ${B>0}$ is a numerical constant that does not depend on ${a,b}$.
(b) Suppose that ${f\in L^1({\mathbb R})}$ is such that ${\hat f}$ is an odd function. Use (a) to show that for every ${b>0}$ we have that
$\displaystyle \bigg|\int_1 ^b \frac{\hat f(\xi)}{\xi} d\xi\bigg|
for some numerical constant ${A>0}$ which does not depend on ${b}$.
(c) Construct a function ${g\in C_o({\mathbb R})}$ which is not the Fourier transform of an ${L^1}$ function. To do this note that it is enough to find a function ${g\in C_o({\mathbb R})}$ which does not satisfy the condition in (b).
(ii) Show that ${\overline{\mathcal F(L^1({\mathbb R}^n))}=C_o({\mathbb R}^n)}$ where the closure is taken in the ${C_o}$ topology.
Hint: Observe that ${C_c ^\infty({\mathbb R}^n)}$ is dense in ${C_o({\mathbb R}^n)}$, in the topology of the supremum norm.
It is convenient to define the formal inverse of the Fourier transform in the following way. For ${f\in L^1({\mathbb R}^n)}$ we set
$\displaystyle \mathcal F^{-1}(f)(\xi)=\mathcal F^*(f)(\xi)=\check f(\xi)=\int_{{\mathbb R}^n}f(x) e^{2\pi i x\cdot \xi}d\xi=\hat f(-\xi)=\tilde {\hat f}(\xi)=\hat{\tilde f}(\xi).$
Here we denote by ${\tilde g}$ the reflection of a function ${g}$, that is, ${\tilde g(x)=g(-x)}$. Observe that ${\mathcal F^*}$ is the conjugate of the Fourier transform. Thus the operator ${\mathcal F^*}$ is very closely connected to the operator ${\mathcal F}$ and enjoys essentially the same symmetries and properties.
As we shall see later on, it is also the adjoint of the Fourier transform with respect to the ${L^2}$ inner product
$\displaystyle \langle f,g\rangle =\int_{{\mathbb R}^n} f\bar g.$
Although we haven’t yet defined the Fourier transform on ${L^2}$ we can calculate for ${f,g\in L^1\cap L^2({\mathbb R}^n)}$ that
$\displaystyle \begin{array}{rcl} \int_{{\mathbb R}^n} (\mathcal F f)\bar g &=&\int_{{\mathbb R}^n}\int_{{\mathbb R}^n}f(x)e^{-2\pi i x\cdot \xi} dx \bar g(\xi) d\xi \\ \\ &=&\int_{{\mathbb R}^n} f(x)\overline{\int_{{\mathbb R}^n} g(\xi)e^{2\pi i x\cdot \xi}d\xi} \ dx\\ \\ &=&\int_{{\mathbb R}^n} f \overline{ (\mathcal F^*(g))} \end{array}$
Proposition 8 claims that ${\mathcal F^*}$ is also the inverse of the Fourier transform in the sense that
$\displaystyle \mathcal F^* \mathcal F f=f,$
whenever ${f,\mathcal F f\in L^1({\mathbb R}^n)}$.
The proof of Proposition 8 is quite interesting in the following ways. First of all observe that we have actually showed that whenever ${f\in L^1({\mathbb R}^n)}$, ${f}$ is equal (a.e.) to the ${L^1}$ limit of the functions
$\displaystyle \int_{{\mathbb R}^n}\hat f(\xi)e^{-a|\xi|^2}e^{2\pi ix\cdot \xi} d\xi ,$
as ${a\rightarrow 0}$. This does not require any additional hypothesis and actually provides us with a method of inverting the Fourier transform of any ${L^1}$ function, at least in the ${L^1}$ sense. The second remark is that the proof of Proposition 8 can be generalized to different methods of summability. Indeed, let ${\Phi\in L^1({\mathbb R}^n)}$ be such that ${\phi=\hat \Phi\in L^1({\mathbb R}^n)}$ and ${\Phi(0) }$. For ${\epsilon>0}$ we consider the integrals
$\displaystyle \int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi}d\xi, \ \ \ \ \ (2)$
which we will call the ${\Phi}$-means of the integral ${\int_{{\mathbb R}^n}\hat f (\xi) e^{2\pi i x\cdot \xi}}$, or just the ${\Phi}$-means of ${\check f}$. Using the multiplication formula in Proposition 4 we can rewrite the means (2) as
$\displaystyle \int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi =(f*\tilde \phi_\epsilon)(x), \quad x\in {\mathbb R}^n. \ \ \ \ \ (3)$
The following more general version of Proposition 8 is true.
Proposition 10 Let ${\Phi\in L^1({\mathbb R}^n)}$ be such that ${\phi=\hat \Phi\in L^1({\mathbb R}^n)}$ with ${\int \phi =1}$. We then have that the ${\Phi}$-means of ${\int \hat f(\xi)e^{2\pi i x\cdot \xi}d\xi}$,
$\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot\xi}d\xi,$
converge to ${f}$ in ${L^1}$, as ${\epsilon\rightarrow 0}$.
Proof: The proof is just a consequence of formula (3). Indeed, ${\tilde \phi_\epsilon}$ is an approximation to the identity since ${\tilde \phi\in L^1}$ and ${\int \tilde {\phi}(x)dx=1}$ and thus ${f*\tilde \phi_\epsilon}$ converges to ${f}$ in the ${L^1}$ norm as ${\epsilon\rightarrow 0}$. $\Box$
Proposition 8 says that the inversion formula is true whenever ${f,\hat f\in L^1({\mathbb R}^n)}$. This however is not the most natural assumption since the Fourier transform of an ${L^1}$ function need not be integrable. The idea behind Proposition 10 is to force’ ${\hat f}$ in ${L^1}$ by multiplying it by the ${L^1}$ function ${\Phi(\epsilon\xi)}$. Thus, we artificially impose some decay on ${\hat f}$. This is equivalent to smoothing out the function ${f}$ itself by convolving it with a smooth function ${\tilde \phi_\epsilon}$. Although no smoothness is explicitly assumed in Proposition 10, there is a hidden smoothness hypothesis in the requirement ${\Phi, \phi \in L^1}$. Indeed, we could have replaced this assumption by directly assuming that ${\phi}$ is (say) a smooth function with compact support and taking ${\Phi=\hat \phi}$; then the conclusion ${\hat \phi\in L^1({\mathbb R}^n)}$ would follow automatically. The trick of multiplying the Fourier transform of a general ${L^1}$ function with an appropriate function in ${L^1}$ or, equivalently, smoothing out the function ${f}$ itself allows us then to invert the Fourier transform, at least in the ${L^1}$-sense. This process is usually referred to as a summability method.
As we shall see now, the inversion of a Fourier transform by means of a summability method is also valid in a pointwise sense. Because of formula (3), in order to understand the pointwise convergence of the ${\Phi}$-means of ${\check{f}}$ we have to examine the pointwise convergence of the convolution ${f*\phi_\epsilon}$ to ${f}$, whenever ${\phi }$ is an approximation to the identity.
Definition 11 Let ${f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}$. The Lebesgue set of ${f}$ is the set of points ${x\in{\mathbb R}^n}$ such that
$\displaystyle \lim_{r\rightarrow 0}\frac{1}{r^n}\int_{|y|< r}|f(x-y)-f(x)|dy=0.$
The Lebesgue set of a locally integrable function ${f}$ is closely related to the set where the integral of ${f}$ is differentiable:
Definition 12 Let ${f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}$. The set of points where the integral of ${f}$ is differentiable is the set of points ${x\in{\mathbb R}^n}$ such that
$\displaystyle \lim_{r\rightarrow 0} \frac{1}{\Omega_n r^n}\int_{|y|< r}f(x-y)=f(y),$
where ${\Omega_n}$ is the volume of the unit ball ${B(0,1)}$ in ${{\mathbb R}^n}$. In other words, we say that the integral of ${f}$ is differentiable at some point ${x\in {\mathbb R}^n}$ if the average of ${f}$ with respect to Euclidean balls centered at ${x}$ the value of ${f}$ at the point ${x}$.
We shall come back to these notions a bit later in the course when we will introduce the maximal function of ${f}$ which is just the maximal average of ${f}$ around every point. For now we will use as a black box the following theorem:
Theorem 13 Let ${f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}$. Then the integral of ${f}$ is differentiable at almost every point ${x\in {\mathbb R}^n}$.
While postponing the proof of this theorem for later on in the course, we can already see the following simple proposition connecting the Lebesgue set of ${f}$ to to the set of points where the integral of ${f}$ is differentiable. In particular we see that almost every point in ${{\mathbb R}^n}$ is Lebesgue point of ${f}$.
Corollary 14 Let ${f\in L^1 _{\textnormal {loc}} ({\mathbb R}^n)}$. Then almost every ${x\in{\mathbb R}^n}$ is a Lebesgue point of ${f}$.
Proof: For any rational number ${q}$ we have that the function ${f(x)-q}$ is locally integrable. Theorem 13 then implies that
$\displaystyle \lim_{r\rightarrow 0} \frac{1}{r^n}\int_{|y|\leq r}\big\{|f(x-y)-q|-|f(x)-q|\big\}dy=0,$
for almost every ${x\in {\mathbb R}^n}$. Thus the set ${F_q}$ where the previous statement is not true has measure zero and so does the set ${F:=\cup_{q\in{\mathbb Q}} F_q}$. Now let ${x\in {\mathbb R}^n \setminus F}$. Indeed, let ${\epsilon>0}$ and ${q\in {\mathbb Q}}$ be such that ${|f(x)-q|<\epsilon/2}$. We then have
$\displaystyle \begin{array}{rcl} \frac{1}{\Omega_n r^n}\int_{|y|
The first summand converges to ${|f(x)-q|<\epsilon/2}$ as ${r\rightarrow 0}$ since ${x\notin F}$ while the second summand is smaller than ${\epsilon/2}$. This shows that the Lebesgue set of ${f}$ is contained in ${{\mathbb R}^n\setminus F}$ and thus that almost every point in ${{\mathbb R}^n}$ is a Lebesgue point of ${f}$. $\Box$
We can now give the following pointwise convergence result for approximations to the identity.
Theorem 15 Let ${\phi \in L^1({\mathbb R}^n)}$ with ${\int \phi=1}$. We define ${\psi(x):={\mathrm{esssup}}_{|y|\geq |x|}|\phi(y)|}$. If ${\psi\in L^1({\mathbb R}^n)}$ and ${f\in L^p({\mathbb R}^n)}$ for ${1\leq p \leq \infty}$ then
$\displaystyle \lim_{\epsilon\rightarrow 0} (f*\phi_\epsilon)(x)=f(x),$
whenever ${x}$ is a Lebesgue point for ${f}$. If in addition $\hat \phi \in L^1(\mathbb R^n)$ then the $\hat \phi$-means of $\check f$,
$\displaystyle \int_{\mathbb R^n} \hat f(\xi) \hat \phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi,$
converge to $f (x)$ as $\epsilon \to 0$ for almost every $x \in \mathbb R ^n$.
Proof: Let ${x}$ be a Lebesgue point of ${f}$ and fix ${\delta>0}$. By Corollary 14 there exists ${\eta>0}$ such that
$\displaystyle \frac{1}{r^n}\int_{|y|
whenever ${|r|<\eta}$.
We can estimate as usual
$\displaystyle \begin{array}{rcl} |(f*\phi_\epsilon)(x)-f(x)|&=&\bigg|\int_{{\mathbb R}^n}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg|\\ \\ &\leq& \bigg|\int_{|y|<\eta}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg| \\ \\ &&+\bigg|\int_{|y|\geq \eta}[f(x-y)-f(x)]\phi_\epsilon(y)dy\bigg| \\ \\ &=:& I_1+I_2. \end{array}$
We claim that
$\displaystyle \psi(x)\lesssim_{n,\phi} |x|^{-n}, \quad x\in{\mathbb R}^n. \ \ \ \ \ (5)$
First of all observe that ${\psi}$ is radially decreasing. We will abuse notation and write ${\psi(x)=\psi(|x|)}$. For every ${r>0}$ we have that
$\displaystyle \int_{r/2\leq |x|
Now since ${\psi\in L^1}$, the left hand side in the previous estimate tends to ${0}$ when ${r\rightarrow 0}$ and when ${r\rightarrow \infty}$ we get the claim.
We write (4) in polar coordinates to get
$\displaystyle \frac{1}{r^n}\int_{S^{n-1}}\int_0 ^r |f(x-sy')-f(x)|s^{n-1}ds d\sigma_{n-1}(y')<\delta.$
Setting ${g(s)=\int_{S^{n-1}}|f(x-sy')-f(x)| d\sigma_{n-1}(y')}$ we can rewrite the previous estimate in the form
$\displaystyle G(r):=\int_0 ^r g(s)s^{n-1}ds \leq \delta r^n,$
whenever ${|r|<\eta}$. We now estimate ${I_1}$ as follows
$\displaystyle \begin{array}{rcl} I_1&\leq& \int_{S^{n-1}}\int_0 ^\eta |f(x-ry')-f(x)|\psi_\epsilon(r)|d\sigma_{n-1}(y')r^{n-1}dr \\ \\ &=&\int_0 ^\eta g(r)r^{n-1}\frac{1}{\epsilon^n}\psi(r/\epsilon)dr\\ \\ &=&\int_0 ^\eta G'(r)\frac{1}{\epsilon^n}\psi(r/\epsilon) dr. \end{array}$
At this point the proof simplifies a bit if we assume that ${\psi}$ is differentiable. In this case we have that ${\psi'\leq 0}$ and we can estimate the last integral by
$\displaystyle \begin{array}{rcl} \int_0 ^\eta G'(r)\frac{1}{\epsilon^n}\psi(r/\epsilon) dr&=&\frac{1}{\epsilon^n}G(\eta)\psi(\frac{\eta}{\epsilon})-\int_0 ^\eta G(r)\frac{1}{\epsilon^{n+1}}\psi'(\frac{r}{\epsilon})dr\\ \\ &\lesssim_{n,\phi}& \delta - \delta \frac{1}{\epsilon^{n+1}}\int_0 ^\eta r^n\psi'(\frac{r}{\epsilon})dr \\ \\ &=&\delta +\delta \frac{n}{\epsilon^n}\int_0 ^\eta r^{n-1}\psi(r/\epsilon)dr \\ \\ &\leq & \delta\bigg(1+\frac{n}{\omega_{n-1}}\int_{{\mathbb R}^n}\psi(x)dx\bigg). \end{array}$
The argument actually goes through without the assumption that ${\psi}$ is differentiable by a clever use of the Riemann-Stieljes integral. Note that the function ${\psi}$ is decreasing thus almost everywhere differentiable. This shows that ${I_1\lesssim_{n,\phi} \delta}$.
For ${I_2}$ we estimate as follows
$\displaystyle \begin{array}{rcl} I_2\leq \|f\|_p\|\psi_\epsilon\chi_{\{|x|\geq \eta \}}\|_{p'}+|f(x)|\|\chi_{\{|x|\geq \eta \}}\psi_\epsilon\|_1. \end{array}$
For the second summand we have that
$\displaystyle \|\chi_{\{|x|\geq \eta \}}\psi_\epsilon\|_1=\frac{1}{\epsilon^n}\int_{|x|\geq \eta}\psi_\epsilon(x/\epsilon)dx=\int_{|x|\geq \eta/\epsilon}\psi(x)dx\rightarrow 0,$
as ${\epsilon\rightarrow 0}$, since ${\psi\in L^1}$.
On the other hand, we have
$\displaystyle \begin{array}{rcl} \| \psi_\epsilon\chi_{\{|x|\geq \eta \}}\|_{p'}&=&\bigg(\int_{|x|\geq \eta}[\psi_\epsilon(x)]^{p'}dx\bigg)^\frac{1}{p'}=\bigg(\int_{|x|\geq \eta}[\psi_\epsilon(x)]^\frac{p'}{p}\psi_\epsilon(x)dx\bigg)^\frac{1}{p'}\\ \\ &\leq & \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty ^\frac{1}{p} \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_1\leq \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty ^\frac{1}{p}\|\psi\|_1. \end{array}$
Now since ${\psi_\epsilon}$ is decreasing we have
$\displaystyle \|\psi_\epsilon(x)\chi_{\{|x|\geq \eta \}}\|_\infty \leq \psi_\epsilon(\eta)=\frac{1}{\epsilon^n}\psi(\eta/\epsilon)=\eta^{-n}\big(\frac{\eta}{\epsilon}\big)^n\psi(\eta/\epsilon)\rightarrow 0,$
when ${\epsilon\rightarrow 0}$.
We have showed that
$\displaystyle \limsup_{\epsilon\rightarrow 0} |(f*\phi_\epsilon)(x)-f(x)|\lesssim_{n,\phi} \delta,$
whenever ${x}$ is a Lebesgue point of ${f}$. Since ${\delta>0}$ was arbitrary this completes the proof of the theorem.$\Box$
Remark 2 The previous theorem is true in the case that ${\phi}$ is a radially decreasing function in ${L^1}$ or, in general, a function that satisfies a bound of the form ${|\phi(x)|\lesssim (1+|x|)^{-(n+\delta)}}$ for some ${\delta>0}$.
We conclude the discussion on the inversion of the Fourier transform with a useful corollary.
Corollary 16 Let ${f\in L^1({\mathbb R}^n)}$ and assume that ${f}$ is continuous at ${0}$ and that ${\hat f \geq 0}$. Then ${\hat f\in L^1({\mathbb R}^n)}$ and
$\displaystyle f(x)=\int_{{\mathbb R}^n}\hat f(\xi) e^{2\pi i x\cdot \xi}d\xi,$
for almost every ${x\in{\mathbb R}^n}$. In particular,
$\displaystyle f(0)=\int_{{\mathbb R}^n} \hat f(\xi)d\xi.$
Proof: By identity (3) we have that
$\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) \Phi(\epsilon \xi) e^{2\pi i x\cdot \xi} d\xi=(f*\tilde \phi_\epsilon)(x),$
for all ${x\in{\mathbb R}^n}$. Observe that the functions on both sides of this identity are continuous functions of ${x}$. Now let ${\phi,\Phi}$ satisfy the conditions of Theorem 15. Assume furthermore that ${\Phi}$ is non-negative and continuous at ${0}$. For example we can consider the function ${\Phi(\xi)=\phi(\xi)=e^{-\pi |\xi|^2}}$. Now since the point ${0}$ is a point of continuity of ${f}$, it certainly belongs to the Lebesgue set of ${f}$. Thus we have that ${\lim_{\epsilon\rightarrow 0} (f*\tilde \phi_\epsilon)(0)=f(0)}$ which gives
$\displaystyle \lim_{\epsilon\rightarrow 0}\int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) d\xi=f(0).$
Since ${\hat f \Phi}$ is positive, we can use Fatou’s lemma to write
$\displaystyle \int_{{\mathbb R}^n}\hat f(\xi) d\xi= \int_{{\mathbb R}^n}\liminf_{\epsilon_k\rightarrow 0} \hat f(\xi) \Phi(\epsilon _k\xi) d\xi\leq f(0),$
so ${\hat f \in L^1({\mathbb R}^n)}$. Thus the inversion formula holds true for ${f}$ and we get
$\displaystyle f(x)=\int_{{\mathbb R}^n} \hat f(\xi) e^{2\pi i x\cdot \xi } d\xi,$
for almost every ${x\in {\mathbb R}^n}$. However
$\displaystyle f(0)=\lim_{\epsilon\rightarrow 0}\int_{{\mathbb R}^n} \hat f(\xi) \Phi(\epsilon \xi) d\xi=\int_{{\mathbb R}^n}\lim_{\epsilon \rightarrow 0}\hat f(\xi) \Phi(\epsilon \xi) d\xi=\int_{{\mathbb R}^n}\hat f(\xi) d\xi,$
since ${\hat f\in L^1}$. $\Box$
2.1. Two special summability methods
We describe in detail two summability methods that are of special interest. These are based on the Examples 1 and 2 in the beginning of this set of notes.
The Gauss-Weierstrass summability method. By dilating the function ${W(x)=e^{-\pi|x|^2}}$ we get
$\displaystyle W(x,t):= W_{\sqrt{4\pi t}}(x)=(4\pi t)^{-\frac{n}{2}}e^{- \frac{|x|^2}{4t} }.$
The function ${W(x,t),\ t>0,}$ is called the Gauss kernel and it gives rise to the Gauss-Weierstrass method of summability. The Fourier transform of ${W}$ is
$\displaystyle \widehat{W_{\sqrt{4\pi t}} }(\xi)=\widehat W(\sqrt{2\pi t}\xi)=e^{-4\pi^2t|\xi|^2}.$
It is also clear that
$\displaystyle \int_{{\mathbb R}^n} W(x,t)dx=1,$
for all ${t>0}$. The discussion in the previous sections applies to the Gauss-Weierstrass summability method and we have that the means
$\displaystyle w(x,t):=\int_{{\mathbb R}^n}f(y)W(y-x,t)dy=\int_{{\mathbb R}^n} \hat f(\xi)e^{-4\pi^2 t |\xi|^2}e^{2\pi i x\cdot \xi} d\xi$
convergence to ${f}$ in ${L^1({\mathbb R}^n)}$ and also in the pointwise sense, for every ${x}$ in the Lebesgue set of ${f}$. One of the aspects of Gauss-Weierstrass summability is that the function ${w(x,t)}$ defined above satisfies the heat equation:
$\displaystyle \begin{array}{rcl} \frac{\partial w}{\partial t}-\Delta w &=& 0,\quad \mbox{ on }{\mathbb R}^{n+1} _+,\\ w(x,0) &=&f(x),\quad x\in{\mathbb R}^n. \end{array}$
To see that the Gauss-Weierstrass means of ${\check f}$ satisfy the Heat equation with initial data ${f}$, one can use the formula for ${w(x,t)}$ and calculate everything explicitly. However it is easier to consider the Fourier transform of the solution ${u(x,t)}$ of the Heat equation in the ${x}$ variable and show that it must agree with the Fourier transform of ${w(x,t)}$, again in the ${x}$ variable. Observe that under suitable assumptions on the initial data ${f}$ we get that the solution ${w(x,t)}$ converges to the initial data ${f}$ as time’ ${t\rightarrow 0}$.
Exercise 9 Let ${f(x)=e^{-\pi x^2}}$, ${x\in{\mathbb R}}$. Using the properties of the Fourier transform show that the function ${\hat f}$ satisfies the initial value problem
$\displaystyle \begin{array}{rcl} u'+2\pi x u&=&0,\\ \\ u(0)&=&1. \end{array}$
Solve the initial value problem to give an alternative proof of the fact that ${\hat f(\xi)=e^{-\pi \xi^2}}$. Observe that the differential equation above is invariant under the Fourier transform.
The Abel summability method. We consider the function ${P(x)=c_n\frac{1}{(1+|x|^2)^\frac{n+1}{2}}}$ where ${c_n=\frac{\Gamma((n+1)/2}{\pi^\frac{n+1}{2}}}$. By dilating the function ${P}$ we have
$\displaystyle P(x,t):= P_t(x)=c_n\frac{t}{(t^2+|x|^2)^\frac{n+1}{2}}.$
The function ${P(x,t),\ t>0,}$ is called the Poisson kernel (for the upper half plane) and it gives rise to the Abel method of summability. The Fourier transform of ${P}$ is
$\displaystyle \widehat{P_t}(\xi)=\hat P(t\xi)=e^{-2\pi t|\xi|}.$
This is just a consequence of the calculation in Example 2, the inversion formula and the easily verified fact that ${P\in L^1({\mathbb R}^n)}$. It is also clear by a direct calculation or through the previous Fourier transform relation that
$\displaystyle \int_{{\mathbb R}^n} P(x,t)dx=1,$
for all ${t>0}$. Everything we have discussed in these notes applies to the Abel summability method. In particular we have that whenever ${f\in L^1({\mathbb R}^n)}$, the means
$\displaystyle u(x,t):=\int_{{\mathbb R}^n}f(y)P(y-x,t)dy=\int_{{\mathbb R}^n} \hat{f}(\xi) e^{-2\pi t|\xi|}e^{-2\pi i x\cdot \xi} d\xi,$
converge to ${f}$ in ${L^1}$ as ${t\rightarrow 0}$ and also in the pointwise sense for all ${x}$ in the Lebesgue set of ${f}$. The function ${u(x,t)}$ is also called the Poisson integral or extension of ${f}$. It is not difficult to see that it satisfies the Dirichlet problem
$\displaystyle \begin{array}{rcl} \Delta u &=&0, \quad \mbox{ on }{\mathbb R}^{n+1} _+,\\ u(x,0) &=&f(x),\quad x\in{\mathbb R}^n. \end{array}$
Here we denote by ${{\mathbb R}^n _+}$ the upper half plane ${{\mathbb R}^n _+=\{(x,y):x\in {\mathbb R}^n, y>0\}}$. Thus, if we are given an ${L^1}$ function on the boundary’ ${{\mathbb R}^n}$, the Poisson integral of ${f}$ provides us with a harmonic function ${u(x,t)}$ in the upper half plane which has boundary value ${f}$ in the sense that ${u(x,t)}$ converges to ${f}$ as ${t\rightarrow 0}$ both in the ${L^1}$ sense as well as almost everywhere.
Remark 3 The Poisson extension of ${f\in L^p({\mathbb R}^n)}$, ${1\leq p\leq \infty}$
$\displaystyle u(x,t)=\int_{{\mathbb R}^n} f(y)P(x-y,t)dy,$
is harmonic in ${{\mathbb R}_+ ^{n+1}}$, that is, that it satisfies the Laplace equation:
$\displaystyle \Delta_{x,t} u(x,t) =\sum_{j=1} ^n \frac{\partial}{\partial x_k}u(x,t)+\frac{\partial}{\partial t} u(x,t)=0.$
This is essentially a consequence of the fact that ${\Delta_{x,t}P(x,t)=0}$ for ${(x,t)\in{\mathbb R}^n _+}$.
In general, we can ask for a harmonic function ${u(x,t)}$ in ${{\mathbb R}^{n+1} _+}$ which has boundary value ${\lim_{t\rightarrow 0}u(\cdot,t)=f\in L^p({\mathbb R}^n)}$ where the limit is taken ${L^p}$-sense. In the case ${p<\infty}$ this extension is uniquely given by the Poisson integral of ${f}$. Also, the same is true if ${p=\infty}$ and ${f\in C_o({\mathbb R}^n)\subset L^\infty({\mathbb R}^n)}$. On the other hand, if we ask for a function which is harmonic in ${{\mathbb R}^{n+1} _+}$, continuous in ${\overline {{\mathbb R}^{n+1} _+}}$ and has boundary function ${f}$, then no assumption on ${f}$ can guarantee that this extension is unique. Take for example ${f=0}$ and ${u_1(x,t)=0}$, ${u(x,t)=t}$. The solution of the Dirichlet problem becomes unique though if we require in addition that the harmonic extension is a bounded function in ${{\mathbb R}^{n+1} _+}$. See [SW] for more information.
Exercise 10 Prove the subordination identity:
$\displaystyle e^{-\beta}=\frac{1}{\sqrt{\pi}}\int_0 ^\infty \frac{e^{-u}}{\sqrt{u}}e^{-\beta^2/4u}du,\quad \beta>0. \ \ \ \ \$
For this, first prove the identities
$\displaystyle \begin{array}{rcl} e^{-\beta}&=&\frac{2}{\pi}\int_0 ^\infty\frac{\cos \beta x}{1+x^2}dx,\\ \\ \frac{1}{1+x^2}&=&\int_0 ^\infty e^{-(1+x^2)u}du. \end{array}$
The second identity above is obvious. In order to prove the first, use the theory of residues for the function
$\displaystyle f(z)=\frac{e^{i\beta z}}{1+z^2}.$
[Update 11 Mar 2011: Exercise 10 added.]
[Update 11 Mar 2011: Statement of Theorem 15 completed with the corollary about the convergence of the means of $\check f$.] | 2017-10-20 12:32:13 | {"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": 496, "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": 0.9937453866004944, "perplexity": 198.40545162314393}, "config": {"markdown_headings": true, "markdown_code": false, "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-2017-43/segments/1508187824104.30/warc/CC-MAIN-20171020120608-20171020140608-00335.warc.gz"} |
https://tex.stackexchange.com/questions/270366/own-cite-command | # Own cite command
I want to create a certain cite command that is pretty similar to the original one:
\newcommand{\mycite}[2][]{\hspace{30pt}\cite{#2}{#1}}
The problem is that when I only write \mycite{test}. Then there will be
[Test, ]
but it should be
[Test]
because I omitted the optional argument. I tried it with the package xparse. But there I got the same result with the following code
\NewDocumentCommand\mycite{O{}m}{%
\cite[#1]{#2}%
}
What do I do wrong? Could you help me?
• Please be aware that some of the \cite - related packages do redefine \cite to provide more optional arguments already. – user31729 Sep 30 '15 at 13:41
There's no check if the optional argument of \mycite is present or not. So \mycite{#2} will still call \cite[]{#2} with an empty optional argument for \cite, meaning to typeset , in the list.
I added the 'traditional' way to check for the optional argument and the xparse method (much easier!)
Note: Bibliography related packages might change the \cite command already. This is not covered here.
\documentclass{article}
\usepackage{etoolbox}
\usepackage{xparse}
\makeatletter
\newcommand{\@myotherciteopt}[2][]{%
\hspace{30pt}%
\ifblank{#1}{%
\cite{#2}%
}{%
\cite[#1]{#2}%
}%
}
\newcommand{\@myothercitenoopt}[1]{%
\@myotherciteopt[]{#1}%
}
\newcommand{\myothercite}{%
\@ifnextchar[{\@myotherciteopt}{\@myothercitenoopt}%
}
\makeatother
\NewDocumentCommand\mycite{om}{%
\hspace{30pt}%
\IfValueTF{#1}{%
\cite[#1]{#2}%
}{%
\cite{#2}%
}%
}
\begin{document}
\mycite{Lam94}
\myothercite{Lam94}
Now with options:
\mycite[Ms. Ann Elk]{Lam94}
\myothercite[Ms. Ann Elk]{Lam94}
\bibliographystyle{alpha}
\bibliography{biblio}
\end{document} | 2021-03-01 10:25:30 | {"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": 0.7463148236274719, "perplexity": 2376.423659615958}, "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-2021-10/segments/1614178362481.49/warc/CC-MAIN-20210301090526-20210301120526-00313.warc.gz"} |
http://www.formuladirectory.com/user/formula/125 | HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS
## Gross Profit Margin
Here,p=Gross Profit r=Revenue
## $\frac{p}{r}$
The gross margin is not an exact estimate of the company's pricing strategy but it does give a good indication of financial health. Without an adequate gross margin, a company will be unable to pay its operating and other expenses and build for the future. In general, a company's gross profit margin should be stable. It should not fluctuate much from one period to another, unless the industry it is in has been undergoing drastic changes which will affect the costs of goods sold or pricing polici
ENTER THE VARIABLES TO BE USED IN THE FORMULA
Similar formulas which you may find interesting. | 2018-10-24 04:08:51 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "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": 0.4854992628097534, "perplexity": 1139.381073422901}, "config": {"markdown_headings": true, "markdown_code": false, "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-2018-43/segments/1539583518753.75/warc/CC-MAIN-20181024021948-20181024043448-00228.warc.gz"} |
https://cstheory.stackexchange.com/questions/50753/full-names-of-c-k-chow-and-c-n-liu | # Full names of C. K. Chow and C. N. Liu
Where can I find the full names of C. K. Chow and C. N. Liu, of the Chow-Liu tree fame?
https://en.wikipedia.org/wiki/Chow%E2%80%93Liu_tree
https://ieeexplore.ieee.org/document/1054142
From what I could find, the names are
• Chao-Kong Chow
• Chao-Ning Liu
I found the second from this IEEE note, which links Chao-Ning Liu to IBM Thomas .J. Watson (which he joined in 1957). Then, a search for patents gave the name of C.K. Chow.
C.K. Chow's name is also stated in this paper by O'Donnell and Servedio on The Chow Parameters Problem (see abstract).
And since a comment also asked for the full name of L. F. Kozachenko (of the Kozachenko-Leonenko entropy estimate fame), here it is: Lyudmyla Fedoryvna Kozachenko.
• How did you find that out? Nov 21 at 23:38
• I emailed Prof. Leonenko. Nov 22 at 5:02
• How did you find information about Leonenko? Nov 22 at 7:17
• On his website... cardiff.ac.uk/people/view/98655-leonenko-nikolai Nov 22 at 10:10
• I actually found it on his mathscinet profile, mathscinet.ams.org/mathscinet/search/… Nov 22 at 10:10 | 2021-12-08 10:03:14 | {"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": 0.8125342130661011, "perplexity": 1024.763791800289}, "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/1637964363465.47/warc/CC-MAIN-20211208083545-20211208113545-00521.warc.gz"} |
https://daaronr.github.io/micro-giving-pub/monopoly-pricing-examples-using-r.html | # Monopoly pricing: Examples using R
## 5.15 Simulating willingness to pay and considering ‘optimal pricing’
Remember, you can ‘hide the code’ with the ‘hide global’ button in the upper right of this screen.
Note that the modern use of the term ‘Econometrics’ tends to refer to the analysis of real data, not simulated data.
Although this module is not about coding, you might find it insightful to consider how to use R code to calculate and graph and simulate the monopoly/pricing outcomes.
I hope this makes the theory a bit more concrete. It is obviously a great career and research skill to have some understanding of how to use computers to do simulations and to work with data.
If you can install R (and ideally RStudio), you might like to try to run the code yourself, and tweak certain parts to see how things change. R/Rstudio shoulda lso be installed on most university computers. If you just want to do a quick try, you can simply enter code into a hosted browser interface like the one below Hosted here… but Rstudio will be better in the long run.
Try entering some of the code into the box above and click the ‘run’ bar. A tip: to have R display any ‘object’ that has been defined, simply type it’s name on a line by itself.
Our monopolist is a broadband internet supplier within a city. Suppose for now that they only offer one bundle. For now we will consider a scenario with 2000 potential consumers but this can be changed by altering the npeep variable.
For each of these consumers we generate a willingness to pay for broadband internet (internet). [We assume that the mean WTP is £45]… and we add an error term to this in order to simulate variation between consumers. To do this we use the rnorm function which draws a random sample from the standard normal distribution (mean 0, variance 1) multiplied by £15. This yields a standard deviation of roughly £15.
A normal distribution follows the traditional “bell shape”. Perhaps it’s reasonable to assume that willingness to pay is normally distributed, but other distributions might also be possible, such as a “power law” distribution, with much longer “tails” then the normal distribution.
Some have claimed (but there is disagreement)%20exponent) that wealth follows a "power law’ distribution, suggesting wtp might do so as well.
npeep <- 2000 # Number of potential consumers
wtp <- 45 + rnorm(npeep)*15 # Each person has a different willingness to pay
Checking, the standard deviation is 14.66 (coded as sd(wtp)). The exact number will vary each time you run the simulation in R but it will generally be close to 15.
To figure out the demand curve we count the number of people willing to pay at least as much as the offering price.
We set the maximum offered price to 90, thus we consider prices between 0 and 90. From this we sum the number of consumers who have a willingness to pay above the offering price in order to give us a vector of quantity demanded at each price level.
maxop <- 90 # Max offering price
op <- 0:maxop # Offering price ranges from 0 to maxop i.e 0,1, ... 90
qd <- rep(NA,length(op)) # Empty vector for quantity demanded
for (i in 1:length(op)) qd[i] <- sum(wtp>=op[i]) # We fill qd with the number of consumers with a wtp >= offering price
Let’s look at the ‘vector of quantities demanded’
1997, 1997, 1997, 1996, 1995, 1995, 1995, 1993, 1989, 1987, 1985, 1981, 1979, 1977, 1974, 1969, 1962, 1951, 1942, 1929, 1913, 1900, 1888, 1873, 1859, 1841, 1821, 1797, 1771, 1737, 1700, 1669, 1626, 1586, 1552, 1513, 1461, 1416, 1364, 1312, 1261, 1199, 1152, 1103, 1051, 1002, 964, 909, 860, 812, 754, 706, 657, 599, 547, 508, 473, 423, 378, 339, 308, 289, 262, 241, 211, 195, 171, 152, 131, 105, 90, 80, 66, 58, 50, 43, 34, 28, 24, 18, 15, 9, 8, 6, 5, 5, 4, 2, 2, 2, 1
The code above has simply created a vector that computes the quantity demanded at each of the 90 prices £1, £2, …, £90, given the simulated willinges to pay for the 2000 potential consumers.
We assume:
Marginal cost is increasing though this is not a necessity. For something like broadband services we might think that up to a point marginal costs might be decreasing since the cost of adding one more customer might be less than the cost of adding the previous customer.
In our simulation we assume $MC = 0.01(q_d)$
mc <- qd*.01 # Setting marginal cost
Now we plot quantity demanded as a function of price:
plot(qd, op, type="l", xlab="Quantity", ylab="Price, Cost",
abline(h=0, lwd=2)
lines(qd, mc, col="red", lwd=2)
legend(1450, 75, legend=c("Demand curve", "Marginal Cost"), col=c("black", "red"), lty=1)
Above, we see that marginal cost only exceeds the “last consumer’s value” (thus the price that can be charged) … when output is around 1800 and the inverse demand (price that can be charged) and marginal cost meet at around 20.
The monopolist must choose a price in which to sell services at. If the monopolist chooses $MC=P$ then the monopolist will not make any money but the consumers will be very happy. We know that the optimal point for the monopolist is at the point where marginal revenue curve intersects the marginal cost curve. Let’s see if we can find it.
tr <- tp <- tc <- rep(NA,length(op)) # Total revenue, total profit, total cost vectors
Total cost just sums the marginal cost of all units; we generate this vertor for each of the quantities sold at the prices under consideration…
# Calculate total cost
qd.gain <- qd[-length(qd)]-qd[-1]
qd.gain[length(qd.gain)+1] <- qd.gain[length(qd.gain)]
for (i in 1:length(op)) tc[i] <- sum((mc*qd.gain)[length(qd):i])
Next we calculate total revenue and total profit … again for each element of the vector.
tr <- qd*op
tp <- tr-tc
Next we plot total revenue and total profit against the considered prices.
minmax <- function(...) c(min(...),max(...)) # I think this function is simply used to define the plot scalre
#They use the traditional "plot" function. Ggplot tools are much much better, and we will try to switch the code when we have a chance.
plot(minmax(op),minmax(tr,tp), type="n", ylab="Cost / Profit",
xlab="Price",
main="We can see optimal pricing for the monopolist is around £39:")
grid()
abline(h=0, lwd=2)
abline(v=39, col="red", lwd=2)
lines(op,tr, col="blue", lwd=3)
lines(op,tp, col="red", lwd=2)
legend(65, y = 45000, legend=c("Total revenue", "Total profit"), col=c("blue", "red"), lty=1, cex=0.72)
The monopolist calculates total revenue and total profit for each price.
We can see at the price around 18 which would be the optimal price for the consumer [recall that this is the result of marginal-cost pricing], the supplier is making almost no profits.
There are still some profits to be had here as with increasing marginal cost the average cost is less than the marginal cost.
On the other hand, the monopoly’s profit maximising price is around £39.
(Note that this is slightly higher than the total-revenue-maximizing price.)
The last thing we might wish to consider to Total Surplus or total system efficiency which is defined as that which the consumer benefits by purchasing a good below the consumers willingness to pay plus that of the suppliers profit at that price.
Technically we might want to consider what is called total surplus, summing consumer and producer surplus, which does not consider fixed costs. After fixed Costs are incurred they are “sunk” and thus not relevant for the social efficiency of the pricing decision.
.
cs <- tr
for (i in 1:length(op)) cs[i] <- sum((wtp[wtp>=op[i]]-op[i]))
tts <- cs+tp
op[tts==max(tts)] # Check the optimal societal price
## [1] 19
plot(c(min(op),max(op)),c(min(cs,tp),max(cs,tp)), type="n",
main="How does surplus vary for different prices?",
xlab="Price",
ylab="Surplus")
lines(op, cs, col="purple", lwd=2)
lines(op, tp, col="blue", lwd=2)
lines(op, tts, lwd=2)
abline(h=0,col="red", lwd=2)
legend(60, 75000, legend=c("Consumer surplus", "Producer surplus", "Total surplus"), col=c("purple","blue","black"),lty=1)
The peak of the total surplus curve is where where $P = MC$, at about £19.
## 5.16 Estimating demand and optimal profit based on ‘historical data’ (presuming random firm ‘price experimentation’)
This blog and vignette (alternate link) appears to be an excellent marriage of microeconomic theory, data analysis and statistics, and business strategy.
Have a look: I hope to incorporate this further. | 2022-06-27 20:16:25 | {"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": 0.5563046336174011, "perplexity": 1513.644313239014}, "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-2022-27/segments/1656103341778.23/warc/CC-MAIN-20220627195131-20220627225131-00480.warc.gz"} |
http://images.planetmath.org/probableprime | # probable prime
A sufficiently large odd integer $q$ believed to be a prime number because it has passed some preliminary primality test relative to a given base, or a pattern suggests it might be prime, but it has not yet been subjected to a conclusive primality test.
For primes with no specific form, it is required to test every potential prime factor $p<\sqrt{q}$ to be absolutely sure that $q$ is in fact a prime. For Mersenne probable primes, the Lucas-Lehmer test is accepted as a conclusive primality test.
Once a probable prime is conclusively shown to be a prime, it of course loses the label ”probable.” It also loses it if conclusively shown to be composite, but in that case it might then be called a pseudoprime (http://planetmath.org/PseudoprimeP) relative to base $a$.
Title probable prime ProbablePrime 2013-03-22 15:53:46 2013-03-22 15:53:46 PrimeFan (13766) PrimeFan (13766) 5 PrimeFan (13766) Definition msc 11A41 | 2018-06-22 20:50:41 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 4, "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": 0.6514574289321899, "perplexity": 1196.8206877241132}, "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-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00412.warc.gz"} |
https://imathworks.com/tex/tex-latex-small-caps-and-bold-styles-in-beamer/ | # [Tex/LaTex] Small caps and bold styles in beamer
beamerboldfontssmall-caps
I am trying to use small caps and boldface styles together in beamer but it is not producing the desired output. How is possible to use both styles without changing too much the default beamer's font?
By the way, I tried the answers given here for a similar question but they only seem to work within the article class.
Here is a MWE:
\documentclass{beamer}
% Needed encodings
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\begin{document}
\begin{frame}
\textbf{\textsc{Test}}
\end{frame}
\end{document}
And its output:
You can use a different font, eg KP Sans-Serif. With \usepackage[sfmath]{kpfonts} in the preambule it produces the following result: | 2022-08-17 10:14:44 | {"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": 0.7624016404151917, "perplexity": 1839.4500949839808}, "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-2022-33/segments/1659882572898.29/warc/CC-MAIN-20220817092402-20220817122402-00774.warc.gz"} |
https://www.gamedev.net/forums/topic/625863-visual-studio-2012-express-won39t-support-win32-projects/ | # Unity Visual Studio 2012 Express won't support Win32 Projects
This topic is 2078 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
If you followed Microsoft's recent plans or tried Windows 8 + Visual Studio 2012 yourself, you might have noticed already:
Starting with Visual Studio 2012, Microsoft will no longer offer a free C++ compiler for traditional application development. The Express editions will build Metro applications only.
Visual Studio 2012 is pretty interesting because it includes C++11 thread support - something which I've been waiting for very much since it finally resolves the situation of having to decide between Boost.Threads, TBB or POCO for portable threading. Now Microsoft has put me in a situation where I either have to shell out $500 to get the Professional edition or drop Windows 7 and Windows XP from my target group. Dropping Windows 7 is out of the question, as I have no doubt that many people will not be switching to Windows 8 any time soon. What will you do?$500 for the Professional edition may be a reasonable price, but it's also a significant hurdle to collaborating with the Open Source community. Staying with Visual C++ 2010 Express means little to no C++11 features. That leaves Eclipse CDT + MingW and Code::Blocks + MingW which I'm currently checking out.
Opinions?
I'm not sure anyone can sway Microsoft's from their current path, but I've opened a ticket on Microsoft Connect that you can vote on: Support C++ Desktop Applications in Visual Studio 2012 Express
##### Share on other sites
There was a thread on this exact issue in the lounge a couple of days ago:
http://www.gamedev.net/topic/625210-visual-studio-11-express/
##### Share on other sites
Outch, I missed that. Thanks for the warning.
Let me formulate my question a bit tighter: What are you going to do about Visual Studio 2012? Buy? Stay with 2010? Switch to MinGW?
Personally, I find Metro interesting, so I want my homebrew games to run on Metro, too. So I will be using Visual Studio 2012 Express for that. But dropping support for Windows 7 is out of the question, so I have to decide what to use as my main development environment. I'm currently undecided between staying with Visual C++ 2010 Express and using Boost.Threads (so no C++11 for me) or switching to MingW (which means getting used to a whole new IDE, but also brings me a bit closer to targeting Android, I believe). Edited by Cygon
##### Share on other sites
Unless they radically change what they ship in the release version of Express vs the RC that's currently downloadable, enabling the IDE to build non-metro native projects is very simple.. Edited by adeyblue
##### Share on other sites
I'm a student, so if the past is any predictor of the future, I'll be using the 2012 Ultimate/Team edition for free. So personally, I'm not to worried about it.
##### Share on other sites
I'd be surprised if VS 2010 didn't receive an update to allow for new C++11 functionality. Edited by Alpha_ProgDes
##### Share on other sites
Unless they radically change what they ship in the release version of Express vs the RC that's currently downloadable, enabling the IDE to build non-metro native projects is very simple..
A-ha! Thank you for this little gem
If Visual C++ 2005 Express is any indication (you could compile to x64 by tricking the IDE, but in Visual C++ 2008 Express they blocked that hack completely) that means we're good for at least another 2-3 years. Edited by Cygon
##### Share on other sites
Umm... Visual C++ 2008 Express CAN compile to x64. I used it to compile the Maya plugin in my signature. I don't remember the procedure thought, but it was painless enough to be forgettable.
##### Share on other sites
Maybe I'm mixing something up here. I remember it being a hack at one time, then becoming impossible from within the IDE, then turning into a feature that's officially supported.
##### Share on other sites
As long as VS 2010 is supported (and it is) and we can still use engines such as XNA (which you can for now), I won't bother with VS 2012 unless I want to design Metro apps.
Honestly I'm learning Java anyway instead of XNA since it might not survive.
##### Share on other sites
Most of the debate seems to center around the idea of choosing a platform upon which to start a new development. I'm more worried about the fact that we, for example, have a very large codebase that is based upon Qt. I appreciate that Windows 8 will still allow this application to be deployed but I'm reading speculation about Windows 9 not allowing such an application to run, irrespective of the tool chain we use to create it.
My concerns about VS 201* are non-existent. But could we be seriously looking at a situation in a few years time where ALL legacy codebases that are based on Win32 technology will require porting to Metro to be viable on the latest Windows platform?
Thoughts? Edited by Aardvajk
##### Share on other sites
@CoffeeCoder: Yep, it seems that, sadly, XNA has become a dead end. No Metro or DX11 support announced, Shawn Hargreaves switching to another department and ports like MonoGame still being miles away from being able to shoulder something like SunBurn.
I returned to C++ because of its rich offering of 3D engines, physics and sound libraries. While .NET was an excellent environment with very few quirks and portability issues, I always felt a bit shut out from the world of famous projects like Ogre 3D, Bullet, FreeSL, etc.
Why did you pick Java? Are there any good game development tools around? The only game I know that uses Java is Minecraft Edited by Cygon
##### Share on other sites
Most of the debate seems to center around the idea of choosing a platform upon which to start a new development. I'm more worried about the fact that we, for example, have a very large codebase that is based upon Qt. I appreciate that Windows 8 will still allow this application to be deployed but I'm reading speculation about Windows 9 not allowing such an application to run, irrespective of the tool chain we use to create it.
My concerns about VS 201* are non-existent. But could we be seriously looking at a situation in a few years time where ALL legacy codebases that are based on Win32 technology will require porting to Metro to be viable on the latest Windows platform?
Thoughts?
while i haven't seen anything in that direction, obviously, metro/winrt is the future. they will expand on that to make it do everything that's needed over time (same thing happened on windows phone from 7 to 7.5, similar thing could be seen on ios). and yes, legacy will be exactly that: legacy. it will die out, get optional, and at some point, most likely be virtualized and "thrown away" in a virtual app. but slowly, as microsoft depends on that legacy.
as far as i've heard, more visual studio versions are on the way, so this might not be all there is, actually. for now, they just release what's really needed. RC for RC, kinda. we'll see.
but yes, .net and dx, i'll look into that for win8. slimdx ftw, i guess..
##### Share on other sites
The only game I know that uses Java is Minecraft
Minecraft, Runescape, Spiral Knights, Drakensang Online, and some other which's names I cannot remember.
It should be possible to use Visual Studio IDE with MinGW compiler using custom build tool.
##### Share on other sites
Why did you pick Java? Are there any good game development tools around? The only game I know that uses Java is Minecraft
I picked Java because of its vast multi-platform support, and the fact that it's similar to C# which I've been (read: had been) studying for several months. Minecraft was proof to me that Java could be used to make a good, fast-running 3D game, so I looked up some game libraries and found the Lightweight Java Game Library (LWJGL), which is actually what Minecraft was produced with. It's very easy to pick up on and so far I am loving it!
##### Share on other sites
as far as i've heard, more visual studio versions are on the way, so this might not be all there is, actually. for now, they just release what's really needed. RC for RC, kinda. we'll see.
The editions of Visual Studio 2012 that will be released have already been announced: Compare Editions. Even the estimated pricing has been revealed: Pricing Preview.
That their current plans for Express do not include desktop applications has been announced on the official Visual Studio blog: A look ahead at the Visual Studio 11 product lineup and platform support.
My concerns about VS 201* are non-existent. But could we be seriously looking at a situation in a few years time where ALL legacy codebases that are based on Win32 technology will require porting to Metro to be viable on the latest Windows platform?
Thoughts?
They will have to carry the desktop around for the foreseeable future. Office 2012 is a desktop application, too. Even ARM will have the desktop available (in contrast to earlier rumors), it will just not include an x86 emulator, so applications would have to be recompiled to run on ARM.
If Microsoft stays its path, I'm concerned for hobby software development in the long term -- an IDE isn't a good match for a sandboxed Metro app, so might we be doing our future development in a thin client Metro app that accesses some development server rented to us by Microsoft?
For the short term, my concern is what effect this will have on C++11 adoption. Soon GCC 4.6 will enter stable Linux distributions and everyone will have a free C++11 compiler and C++11 standard C++ library. But cross platform projects cannot use it because the Windows mainstream compiler doesn't support C++11. That's worrying. Edited by Cygon
##### Share on other sites
They will have to carry the desktop around for the foreseeable future. Office 2012 is a desktop application, too. Even ARM will have the desktop available (in contrast to earlier rumors), it will just not include an x86 emulator, so applications would have to be recompiled to run on ARM.
Note, however that it is of no use (unless something changed recently) to start porting your desktop apps to ARM.
All applications that are distributed for Windows 8/ARM will have to be distributed through their new app store - and the app store will not allow non-Metro apps.
Effectively, the only ones who will be able to distribute desktop apps for ARM are Microsoft themelves. Edited by wack
##### Share on other sites
For the short term, my concern is what effect this will have on C++11 adoption. Soon GCC 4.6 will enter stable Linux distributions and everyone will have a free C++11 compiler and C++11 standard C++ library. But cross platform projects cannot use it because the Windows mainstream compiler doesn't support C++11. That's worrying.
Interestingly, to me anyway, I'm currently developing a DirectX application using MinGW and happily linking to the standard DXSDK .lib files. Didn't know this was possible. But apparently is.
Now I know that I can use MinGW and QtCreator to develop all the same stuff I ever did with VS, I'm really not bothered by any of this on a personal level any more. I like VS Express but if it is going to become more restrictive then I'm quite happy to switch to free tools instead.
##### Share on other sites
[quote name='Cygon' timestamp='1338976935' post='4946701']
They will have to carry the desktop around for the foreseeable future. Office 2012 is a desktop application, too. Even ARM will have the desktop available (in contrast to earlier rumors), it will just not include an x86 emulator, so applications would have to be recompiled to run on ARM.
Note, however that it is of no use (unless something changed recently) to start porting your desktop apps to ARM.
All applications that are distributed for Windows 8/ARM will have to be distributed through their new app store - and the app store will not allow non-Metro apps.
Effectively, the only ones who will be able to distribute desktop apps for ARM are Microsoft themelves.
[/quote]
and even they only do it because some of the devisions aren't ready yet (but hard working on it). metro will be all in the future. i don't think windows 9 arm will have a desktop anymore. windows 9 x86 will. but maybe even more virtualized away?
i'd like for win32 apps to be configured "single desktop per app", and the app maximized. that would allow them to behave very metro-ish. and each could have it's own sandbox to not mess with the rest.
##### Share on other sites
Most of the debate seems to center around the idea of choosing a platform upon which to start a new development. I'm more worried about the fact that we, for example, have a very large codebase that is based upon Qt. I appreciate that Windows 8 will still allow this application to be deployed but I'm reading speculation about Windows 9 not allowing such an application to run, irrespective of the tool chain we use to create it.
My concerns about VS 201* are non-existent. But could we be seriously looking at a situation in a few years time where ALL legacy codebases that are based on Win32 technology will require porting to Metro to be viable on the latest Windows platform?
Thoughts?
Conceivably Qt could be ported to WinRT/Metro, to make it easier? Indeed, that would be a nice thing anyway, to be able to use Qt to write Windows Metro software, assuming that's something that is conceivably allowed.
Even if MS really think that everyone should be using full screen touch-only Metro applications (do they? I've yet to see anything from MS on this) (which I think is a terrible idea - see the Lounge thread for my views on that), traditionally they have bent over backwards to maintain backwards compatibility. They know that they benefit from the long term app-lock in. It would seem commercial suicide for them to both (a) suggest that everyone move to an input method/device form where currently they have small market share, and (b) cut off all their backwards compatibility, slashing their app support.
Incidentally, it's yet to be clarified as to whether it's possible to write WinRT applications that run in windowed mode - or is WinRT only able to be used with Metro? In the Lounge thread, I heard claims both ways. The question is irrelevant to the issue of VS Express (since we know it's _Metro_ only), but it is relevant to the idea of WinRT being "the future" - is WinRT simply a Win32 replacement that can do both windowed and Metro appliations? Or is MS really saying the future is this Metro-only interface?
They will have to carry the desktop around for the foreseeable future. Office 2012 is a desktop application, too. Even ARM will have the desktop available (in contrast to earlier rumors), it will just not include an x86 emulator, so applications would have to be recompiled to run on ARM.
This sounds promising - do you have a reference? (Note that the previous news I saw said that Office would be some kind of exception on ARM, so the ability to run it doesn't necessarily entail the ability to run ARM windowed apps in general - it would be good to have confirmation.)
If Microsoft stays its path, I'm concerned for hobby software development in the long term -- an IDE isn't a good match for a sandboxed Metro app, so might we be doing our future development in a thin client Metro app that accesses some development server rented to us by Microsoft?[/quote]*shudder* Indeed Every time a programmer claims that Metro should be the only available future on Windows, I'm left thinking, so what am I going to be programming with?
i'd like for win32 apps to be configured "single desktop per app", and the app maximized. that would allow them to behave very metro-ish. and each could have it's own sandbox to not mess with the rest.
That would be interesting - would make the similarity to the old AmigaOS "screens" even more so
Still, there are times when one does want interaction between applications, which is why I think most workspace systems haven't been tied to one-screen-per-application. Edited by mdwh
##### Share on other sites
Well, it looks that this is not going to be an "Express only" thing, I've downloaded Visual Studio Ultimate 2012 RC and it has no C++ support either, and i haven't found so far how to enable it or download the C++ part.
the weird thing is that a few weeks ago previews Visual Studio 11 Beta did had that C++ support (and it was quite fantastic to be honest).
##### Share on other sites
I suspect you've done it wrong because I've got the VS2012 RC installed, and only the Professional Edition at that, and I've got access to a full group of C++ project options including Win32 and console applications.
# We won! Microsoft has given in! Today, the Visual Studio blog announced that after disabling desktop application support in the Express editions since Visual Studio 11 Beta, the final Visual Studio 2012 Express release will be supporting desktop applications once again! To add a little bit to the confusion, there will now be two Express editions. One will be Visual Studio 2012 Express, just as we know it today, the other will have the long-winded name Visual Studio 2012 Express for Windows Desktop. Edited June 8, 2012 by Cygon
##### Share on other sites
Well, it looks that this is not going to be an "Express only" thing, I've downloaded Visual Studio Ultimate 2012 RC and it has no C++ support either, and i haven't found so far how to enable it or download the C++ part.
I've been doing all my (non-metro) C++ development for the previous week in Visual Studio 2012 Professional RC. It's all there, definitely.
# We won! Microsoft has given in!
Woah. No way!
I'd like to have been a fly-on-the-wall in Redmond for the discussions that led to this... Seriously, does anyone have any gossip?
• 9
• 12
• 10
• 10
• 11
• ### Similar Content
• Hello. I'm newby in Unity and just start learning basics of this engine. I want to create a game like StackJump (links are below). And now I wondering what features do I have to use to create such my game. Should I use Physics engine or I can move objects changing transform manually in Update().
If I should use Physics can you in several words direct me how can I implement and what I have to use. Just general info, no need for detailed description of developing process.
Game in PlayMarket
Video of the game
• By GytisDev
Hello,
without going into any details I am looking for any articles or blogs or advice about city building and RTS games in general. I tried to search for these on my own, but would like to see your input also. I want to make a very simple version of a game like Banished or Kingdoms and Castles, where I would be able to place like two types of buildings, make farms and cut trees for resources while controlling a single worker. I have some problem understanding how these games works in the back-end: how various data can be stored about the map and objects, how grids works, implementing work system (like a little cube (human) walks to a tree and cuts it) and so on. I am also pretty confident in my programming capabilities for such a game. Sorry if I make any mistakes, English is not my native language.
• By Ovicior
Hey,
So I'm currently working on a rogue-like top-down game that features melee combat. Getting basic weapon stats like power, weight, and range is not a problem. I am, however, having a problem with coming up with a flexible and dynamic system to allow me to quickly create unique effects for the weapons. I want to essentially create a sort of API that is called when appropriate and gives whatever information is necessary (For example, I could opt to use methods called OnPlayerHit() or IfPlayerBleeding() to implement behavior for each weapon). The issue is, I've never actually made a system as flexible as this.
My current idea is to make a base abstract weapon class, and then have calls to all the methods when appropriate in there (OnPlayerHit() would be called whenever the player's health is subtracted from, for example). This would involve creating a sub-class for every weapon type and overriding each method to make sure the behavior works appropriately. This does not feel very efficient or clean at all. I was thinking of using interfaces to allow for the implementation of whatever "event" is needed (such as having an interface for OnPlayerAttack(), which would force the creation of a method that is called whenever the player attacks something).
Here's a couple unique weapon ideas I have:
Explosion sword: Create explosion in attack direction.
Cold sword: Chance to freeze enemies when they are hit.
Electric sword: On attack, electricity chains damage to nearby enemies.
I'm basically trying to create a sort of API that'll allow me to easily inherit from a base weapon class and add additional behaviors somehow. One thing to know is that I'm on Unity, and swapping the weapon object's weapon component whenever the weapon changes is not at all a good idea. I need some way to contain all this varying data in one Unity component that can contain a Weapon field to hold all this data. Any ideas?
I'm currently considering having a WeaponController class that can contain a Weapon class, which calls all the methods I use to create unique effects in the weapon (Such as OnPlayerAttack()) when appropriate.
• Hi fellow game devs,
First, I would like to apologize for the wall of text.
As you may notice I have been digging in vehicle simulation for some times now through my clutch question posts. And thanks to the generous help of you guys, especially @CombatWombat I have finished my clutch model (Really CombatWombat you deserve much more than a post upvote, I would buy you a drink if I could ha ha).
Now the final piece in my vehicle physic model is the differential. For now I have an open-differential model working quite well by just outputting torque 50-50 to left and right wheel. Now I would like to implement a Limited Slip Differential. I have very limited knowledge about LSD, and what I know about LSD is through readings on racer.nl documentation, watching Youtube videos, and playing around with games like Assetto Corsa and Project Cars. So this is what I understand so far:
- The LSD acts like an open-diff when there is no torque from engine applied to the input shaft of the diff. However, in clutch-type LSD there is still an amount of binding between the left and right wheel due to preload spring.
- When there is torque to the input shaft (on power and off power in 2 ways LSD), in ramp LSD, the ramp will push the clutch patch together, creating binding force. The amount of binding force depends on the amount of clutch patch and ramp angle, so the diff will not completely locked up and there is still difference in wheel speed between left and right wheel, but when the locking force is enough the diff will lock.
- There also something I'm not sure is the amount of torque ratio based on road resistance torque (rolling resistance I guess)., but since I cannot extract rolling resistance from the tire model I'm using (Unity wheelCollider), I think I would not use this approach. Instead I'm going to use the speed difference in left and right wheel, similar to torsen diff. Below is my rough model with the clutch type LSD:
speedDiff = leftWheelSpeed - rightWheelSpeed; //torque to differential input shaft. //first treat the diff as an open diff with equal torque to both wheels inputTorque = gearBoxTorque * 0.5f; //then modify torque to each wheel based on wheel speed difference //the difference in torque depends on speed difference, throttleInput (on/off power) //amount of locking force wanted at different amount of speed difference, //and preload force //torque to left wheel leftWheelTorque = inputTorque - (speedDiff * preLoadForce + lockingForce * throttleInput); //torque to right wheel rightWheelTorque = inputTorque + (speedDiff * preLoadForce + lockingForce * throttleInput); I'm putting throttle input in because from what I've read the amount of locking also depends on the amount of throttle input (harder throttle -> higher torque input -> stronger locking). The model is nowhere near good, so please jump in and correct me.
Also I have a few questions:
- In torsen/geared LSD, is it correct that the diff actually never lock but only split torque based on bias ratio, which also based on speed difference between wheels? And does the bias only happen when the speed difference reaches the ratio (say 2:1 or 3:1) and below that it will act like an open diff, which basically like an open diff with an if statement to switch state?
- Is it correct that the amount of locking force in clutch LSD depends on amount of input torque? If so, what is the threshold of the input torque to "activate" the diff (start splitting torque)? How can I get the amount of torque bias ratio (in wheelTorque = inputTorque * biasRatio) based on the speed difference or rolling resistance at wheel?
- Is the speed at the input shaft of the diff always equals to the average speed of 2 wheels ie (left + right) / 2? | 2018-02-19 04:46:24 | {"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": 0.24323275685310364, "perplexity": 2266.5399095425446}, "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-2018-09/segments/1518891812327.1/warc/CC-MAIN-20180219032249-20180219052249-00697.warc.gz"} |
https://tex.stackexchange.com/questions/172206/citation-style-and-arranging-references | # Citation style and arranging references
Following is the latex code provided by @Mike Renfro from Putting Serial Numbers in References
\documentclass{article}
\usepackage{filecontents}
\begin{filecontents*}{refs.bib}
@BOOK
{KandR,
AUTHOR = "Kernighan, Brian W. and Ritchie, Dennis M.",
TITLE = "{The C Programming Language Second Edition}",
PUBLISHER = "Prentice-Hall, Inc.",
YEAR = 1988
}
\end{filecontents*}
\usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex]{biblatex}
\defbibenvironment{bibliography}
{\enumerate{}
{\setlength{\leftmargin}{\bibhang}%
\setlength{\itemindent}{-\leftmargin}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}}
{\endenumerate}
{\item}
\begin{document}
In 1988 C was totally awesome. \autocite{KandR}
\printbibliography
\end{document}
However in the text at same time if I have to use (Kernighan and Ritchie, 1988) and at some places Kernighan and Ritchie (1988) then what will be the citing style for second scenario.
Moreover in references, i also need them to appear like this:
1. Kernighan, B.W. and Ritchie, D.M. The C Programming Lan- guage Second Edition. Prentice-Hall, Inc., 1988.
ie last name appearing first and accordingly all refernces are arranged in alphabetically order according to last name of first author.
• Ad 1: \textcite{KandR}? Ad 2: sorting=nty/sorting=nyt (name-title-year/name-year-title) and \DeclareNameAlias{sortname}{last-first} – moewe Apr 18 '14 at 12:59
• @moewe - Where is "sorting=nty/sorting=nyt (name-title-year/name-year-title) and \DeclareNameAlias{sortname}{last-first}" to be added in the above code ? – Neeraj Bhanot Apr 18 '14 at 13:04
• Add sorting to the load-time options, so \usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex]{biblatex} reads \usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex,sorting=nty]{biblatex} or \usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex,sorting=nyt]{biblatex} afterward. \DeclareNameAlias{sortname}{last-first} should be added somewhere in the preamble after biblatex was loaded (put it before \renewcommand*{\nameyeardelim}{\addcomma\space}, for example). – moewe Apr 18 '14 at 13:07
• @moewe - \DeclareNameAlias{sortname}{last-first} is not making references to start from last name. Kindly look into it. – Neeraj Bhanot Apr 18 '14 at 13:33
• @Christoph Done! – moewe Apr 26 '14 at 15:49
You need to specify the sorting option at loading time (you probably want nty: "name-title-year" and nyt: "name-year-title", for more sorting options look at pp. 44 sq. of the biblatex documentation and the sectioned linked from there).
The call to biblatex would then read (with "name-year-title" sorting)
\usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex,sorting=nyt]{biblatex}
To change the name format, go with
\DeclareNameAlias{default}{last-first}
Full MWE
\documentclass{article}
\usepackage{filecontents}
\begin{filecontents*}{\jobname.bib}
@BOOK
{KandR,
AUTHOR = "Kernighan, Brian W. and Ritchie, Dennis M.",
TITLE = "{The C Programming Language Second Edition}",
PUBLISHER = "Prentice-Hall, Inc.",
YEAR = 1988
}
\end{filecontents*}
\usepackage[bibstyle=numeric,citestyle=authoryear,backend=bibtex,sorting=nyt]{biblatex}
\defbibenvironment{bibliography}
{\enumerate{}
{\setlength{\leftmargin}{\bibhang}%
\setlength{\itemindent}{-\leftmargin}%
\setlength{\itemsep}{\bibitemsep}%
\setlength{\parsep}{\bibparsep}}}
{\endenumerate}
{\item}
\DeclareNameAlias{default}{last-first} | 2021-04-15 08:41:24 | {"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": 0.6269780993461609, "perplexity": 12555.28151665304}, "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-2021-17/segments/1618038084601.32/warc/CC-MAIN-20210415065312-20210415095312-00235.warc.gz"} |
https://www.tutorialspoint.com/how-to-label-a-patch-in-matplotlib | # How to label a patch in matplotlib?
MatplotlibPythonData Visualization
To label a patch in matplotlib, we can take the following steps −
• Set the figure size and adjust the padding between and around the subplots.
• Initialize the center of the rectangle patch.
• Create a new figure or activate an existing figure.
• Add an 'ax' to the figure as part of a subplot arrangement.
• Add a 'rectangle' to the axes' patches; return the patch.
• Place a legend on the figure.
• To display the figure, use show() method.
## Example
import matplotlib.pyplot as plt
import matplotlib.patches as patches
plt.rcParams["figure.figsize"] = [7.50, 3.50]
plt.rcParams["figure.autolayout"] = True
x = y = 0.1
fig = plt.figure()
patch = ax.add_patch(patches.Rectangle((x, y), 0.5, 0.5, alpha=0.5, facecolor='red', label='Rectangle'))
plt.legend(loc='upper right')
plt.show()
## Output
Published on 10-Aug-2021 07:00:36 | 2021-10-25 16:13:13 | {"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": 0.29384419322013855, "perplexity": 10481.542619368749}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "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-2021-43/segments/1634323587719.64/warc/CC-MAIN-20211025154225-20211025184225-00183.warc.gz"} |
https://repo.scoap3.org/record/32538 | Search for vector-like quarks in events with two oppositely charged leptons and jets in proton–proton collisions at s=13Te
02 May 2019
Abstract: A search for the pair production of heavy vector-like partners T and B of the top and bottom quarks has been performed by the CMS experiment at the CERN LHC using proton–proton collisions at s=13Te . The data sample was collected in 2016 and corresponds to an integrated luminosity of 35.9 fb-1 . Final states studied for TT¯ production include those where one of the T quarks decays via T→tZ and the other via T→bW , tZ , or tH , where H is a Higgs boson. For the BB¯ case, final states include those where one of the B quarks decays via B→bZ and the other B→tW , bZ , or bH . Events with two oppositely charged electrons or muons, consistent with coming from the decay of a Z boson, and jets are investigated. The number of observed events is consistent with standard model background estimations. Lower limits at 95% confidence level are placed on the masses of the T and B quarks for a range of branching fractions. Assuming 100% branching fractions for T→tZ , and B→bZ , T and B quark mass values below 1280 and 1130 Ge , respectively, are excluded.
Published in: EPJC 79 (2019) 364 | 2019-07-15 17:54:13 | {"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": 0.9306899309158325, "perplexity": 1185.3300171449168}, "config": {"markdown_headings": false, "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-2019-30/segments/1563195523840.34/warc/CC-MAIN-20190715175205-20190715200236-00021.warc.gz"} |
https://www.cuemath.com/ncert-solutions/q-8-exercise-1-1-integers-class-7-maths/ | # Ex.1.1 Q8 Integers - NCERT Maths Class 7
Go back to 'Ex.1.1'
## Question
Verify $$a– (–b) = a + b$$ for the following values of $$a$$ and $$b$$
(i) $$a=21, \quad b=18$$
(ii) $$a= 118, \quad b =125$$
(iii) $$a =75, \quad b=84$$
(iv) $$a= 28, \quad b =11$$
Video Solution
Integers
Ex 1.1 | Question 8
## Text Solution
What is known?
Different value of $$a$$ and $$b$$.
What is unknown?
To verify: $$a-(-b)=a+b.$$
Steps:
Let, $$a-(-b)=a+b.$$ (equation 1)
i) $$a = 21, b = 18$$
put the values of $$a$$ and $$b$$ in equation (1).
\begin{align}a-( -b)&= a + b\\ & =21-( -18)\\& = 21 + 18\\&= 21 + 18\\&= 21 + 18\\ 39 &= 39\\\rm{LHS}&= \rm{RHS}\,\,\text{(Hence Verified.)}\end{align}
ii) $$a= 118, b = 125$$
put the value of $$a$$ & $$b$$ in equation (1).
\begin{align}a-(-b)&= a + b\\ & =118-(-125)\\&=118 +125\\&= {118+ 125}\\&=118 + 125\\ 243 &=243\\\rm{LHS}&= \rm{RHS}\,\,\text{(Hence Verified.)}\end{align}
iii) $$a= 75, b= 84$$
put the value of $$a$$ & $$b$$ in equation (1).
\begin{align}a-(-b)&= a + b\\ & =75-(-84) \\&=75 + 84\\&=75+ 84\\&=75 + 84\\ 159 &= 159\\\rm{LHS}&= \rm{RHS}\,\,\text{(Hence Verified.)}\end{align}
iv) $$a=28 ,b=11$$
put the values of $$a$$ & $$b$$ in equation (1).
\begin{align}a-(-b)&= a + b\\ & =28-(-11)\\&=28 + 11\\&= 28+ 11\\&=28 + 11\\ 39 &= 39\\\rm{LHS}&= \rm{RHS}\,\,\text{(Hence Verified.)}\end{align}
Learn from the best math teachers and top your exams
• Live one on one classroom and doubt clearing
• Practice worksheets in and after class for conceptual clarity
• Personalized curriculum to keep up with school | 2021-05-19 03:35:10 | {"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": 8, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9821329712867737, "perplexity": 10151.618929547776}, "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-2021-21/segments/1620243991562.85/warc/CC-MAIN-20210519012635-20210519042635-00003.warc.gz"} |
http://www.neverendingbooks.org/category/games/page/3 | # Category: games
Nimbers is a 2-person game, winnable only if you understand the arithmetic of the finite fields $\mathbb{F}_{2^{2^n}}$ associated to Fermat 2-powers.
It is played on a rectangular array (say a portion of a Go-board, for practical purposes) having a finite number of stones at distinct intersections. Here’s a typical position
The players alternate making a move, which is either
• removing one stone, or
• moving a stone to a spot on the same row (resp. the same column) strictly to the left (resp. strictly lower), and if there’s already a stone on this spot, both stones are removed, or
• adding stones to the empty corners of a rectangle having as its top-right hand corner a chosen stone and removing stones at the occupied corners
Here we illustrate two possible moves from the above position, in the first we add two new stones and remove 2 existing stones, in the second we add three new stones and remove only the top right-hand stone.
As always, the last player able to move wins the game!
Note that Nimbers is equivalent to Lenstra’s ‘turning corners’-game (as introduced in his paper Nim-multiplication or mentioned in Winning Ways Chapter 14, page 473).
If all stones are placed on the left-most column (or on the bottom row) one quickly realizes that this game reduces to classical Nim with Nim-heap sizes corresponding to the stones (for example, the left-most stone corresponds to a heap of size 3).
Nim-addition $n \oplus m$ is defined inductively by
$n \oplus m = mex(n’ \oplus m,n \oplus m’)$
where $n’$ is any element of ${ 0,1,\ldots,n-1 }$ and $m’$ any element of ${ 0,1,\ldots,m-1 }$ and where ‘mex’ stands for Minimal EXcluded number, that is the smallest natural number which isn’t included in the set. Alternatively, one can compute $n \oplus m$ buy writing $n$ and $m$ in binary and add these binary numbers without carrying-over. It is well known that a winning strategy for Nim tries to shorten one Nim-heap such that the Nim-addition of the heap-sizes equals zero.
This allows us to play Nimber-endgames, that is, when all the stones have been moved to the left-column or the bottom row.
To evaluate general Nimber-positions it is best to add another row and column, the coordinate axes of the array
and so our stones lie at positions (1,3), (4,7), (6,4), (10,3) and (14,8). In this way all legal moves follow the rectangle-rule when we allow rectangles to contain corners on the added coordinate axes. For example, removing a stone is achieved by taking a rectangle with two sides on the added axes, and, moving a stone to the left (or the bottom) is done by taking a rectangle with one side at the x-axes (resp. the y-axes)
However, the added stones on the coordinate axes are considered dead and may be removed from the game. This observation allows us to compute the Grundy number of a stone at position (m,n) to be
$G(m,n)=mex(G(m’,n’) \oplus G(m’,n) \oplus G(m,n’)~:~0 \leq m’ < m, 0 \leq n’ < n)$
and so by induction these Grundy numbers are equal to the Nim-multiplication $G(m,n) = m \otimes n$ where
$m \otimes n = mex(m’ \otimes n’ \oplus m’ \otimes n \oplus m \otimes n’~:~0 \leq m’ < m, 0 \leq n’ < n)$
Thus, we can evaluate any Nimbers-position with stone-coordinates smaller than $2^{2^n}$ by calculating in a finite field using the identification (as for example in the odd Knights of the round table-post) $\mathbb{F}_{2^{2^n}} = \{ 0,1,2,\ldots,2^{2^n}-1 \}$
For example, when all stones lie in a 15×15 grid (as in the example above), all calculations can be performed using
Here, we’ve identified the non-zero elements of $\mathbb{F}_{16}$ with 15-th roots of unity, allowing us to multiply, and we’ve paired up couples $(n,n \oplus 1)$ allowing u to reduce nim-addition to nim-multiplication via
$n \oplus m = (n \otimes \frac{1}{m}) \otimes (m \oplus 1)$
In particular, the stone at position (14,8) is equivalent to a Nim-heap of size $14 \otimes 8=10$. The nim-value of the original position is equal to 8
Suppose your opponent lets you add one extra stone along the diagonal if you allow her to start the game, where would you have to place it and be certain you will win the game?
The Knight-seating problems asks for a consistent placing of n-th Knight at an odd root of unity, compatible with the two different realizations of the algebraic closure of the field with two elements.
The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers.
The odd Knights of the round table-problem asks for a specific one-to-one correspondence between two realizations of ‘the’ algebraic closure $\overline{\mathbb{F}_2}$ of the field of two elements.
The first identifies the multiplicative group of its non-zero elements with the group of all odd complex roots of unity, under complex multiplication. The addition on $\overline{\mathbb{F}_2}$ is then recovered by inducing an involution on the odd roots, pairing the one corresponding to x to the one corresponding to x+1.
The second uses Conway’s ‘simplicity rules’ to define an addition and multiplication on the set of all ordinal numbers. Conway proves in ONAG that this becomes an algebraically closed field of characteristic two and that $\overline{\mathbb{F}_2}$ is the subfield of all ordinals smaller than $\omega^{\omega^{\omega}}$. The finite ordinals (the natural numbers) form the quadratic closure of $\mathbb{F}_2$.
On the natural numbers the Conway-addition is binary addition without carrying and Conway-multiplication is defined by the properties that two different Fermat-powers $N=2^{2^i}$ multiply as they do in the natural numbers, and, Fermat-powers square to its sesquimultiple, that is $N^2=\frac{3}{2}N$. Moreover, all natural numbers smaller than $N=2^{2^{i}}$ form a finite field $\mathbb{F}_{2^{2^i}}$. Using distributivity, one can write down a multiplication table for all 2-powers.
The Knight-seating problems asks for a consistent placing of n-th Knight $K_n$ at an odd root of unity, compatible with the two different realizations of $\overline{\mathbb{F}_2}$. Last time, we were able to place the first 15 Knights as below, and asked where you would seat $K_{16}$
$K_4$ was placed at $e^{2\pi i/15}$ as 4 was the smallest number generating the ‘Fermat’-field $\mathbb{F}_{2^{2^2}}$ (with multiplicative group of order 15) subject to the compatibility relation with the generator 2 of the smaller Fermat-field $\mathbb{F}_2$ (with group of order 15) that $4^5=2$.
To include the next Fermat-field $\mathbb{F}_{2^{2^3}}$ (with multiplicative group of order 255) consistently, we need to find the smallest number n generating the multiplicative group and satisfying the compatibility condition $n^{17}=4$. Let’s first concentrate on finding the smallest generator : as 2 is a generator for 1st Fermat-field $\mathbb{F}_{2^{2^1}}$ and 4 a generator for the 2-nd Fermat-field $\mathbb{F}_{2^{2^2}}$ a natural conjecture might be that 16 is a generator for the 3-rd Fermat-field $\mathbb{F}_{2^{2^3}}$ and, more generally, that $2^{2^i}$ would be a generator for the next field $\mathbb{F}_{2^{2^{i+1}}}$.
However, an “exercise” in the 1978-paper by Hendrik Lenstra Nim multiplication asks : “Prove that $2^{2^i}$ is a primitive root in the field $\mathbb{F}_{2^{2^{i+1}}}$ if and only if i=0 or 1.”
I’ve struggled with several of the ‘exercises’ in Lenstra’s paper to the extend I feared Alzheimer was setting in, only to find out, after taking pen and paper and spending a considerable amount of time calculating, that they are indeed merely exercises, when looked at properly… (Spoiler-warning : stop reading now if you want to go through this exercise yourself).
In the picture above I’ve added in red the number $x(x+1)=x^2+1$ to each of the involutions. Clearly, for each pair these numbers are all distinct and we see that for the indicated pairing they make up all numbers strictly less than 8.
By Conway’s simplicity rules (or by checking) the pair (16,17) gives the number 8. In other words, the equation
$x^2+x+8$ is an irreducible polynomial over $\mathbb{F}_{16}$ having as its roots in $\mathbb{F}_{256}$ the numbers 16 and 17. But then, 16 and 17 are conjugated under the Galois-involution (the Frobenius $y \mapsto y^{16}$). That is, we have $16^{16}=17$ and $17^{16}=16$ and hence $16^{17}=8$. Now, use the multiplication table in $\mathbb{F}_{16}$ given in the previous post (or compute!) to see that 8 is of order 5 (and NOT a generator). As a consequence, the multiplicative order of 16 is 5×17=85 and so 16 cannot be a generator in $\mathbb{F}_{256}$.
For general i one uses the fact that $2^{2^i}$ and $2^{2^i}+1$ are the roots of the polynomial $x^2+x+\prod_{j<i} 2^{2^j}$ over $\mathbb{F}_{2^{2^i}}$ and argues as before.
Right, but then what is the minimal generator satisfying $n^{17}=4$? By computing we see that the pairings of all numbers in the range 16…31 give us all numbers in the range 8…15 and by the above argument this implies that the 17-th powers of all numbers smaller than 32 must be different from 4. But then, the smallest candidate is 32 and one verifies that indeed $32^{17}=4$ (use the multiplication table given before).
Hence, we must place Knight $K_{32}$ at root $e^{2 \pi i/255}$ and place the other Knights prior to the 256-th at the corresponding power of 32. I forgot the argument I used to find-by-hand the requested place for Knight 16, but one can verify that $32^{171}=16$ so we seat $K_{16}$ at root $e^{342 \pi i/255}$.
But what about Knight $K_{256}$? Well, by this time I was quite good at squaring and binary representations of integers, but also rather tired, and decided to leave that task to the computer.
If we denote Nim-addition and multiplication by $\oplus$ and $\otimes$, then Conway’s simplicity results in ONAG establish a field-isomorphism between $~(\mathbb{N},\oplus,\otimes)$ and the field $\mathbb{F}_2(x_0,x_1,x_2,\ldots )$ where the $x_i$ satisfy the Artin-Schreier equations
$x_i^2+x_i+\prod_{j < i} x_j = 0$
and the i-th Fermat-field $\mathbb{F}_{2^{2^i}}$ corresponds to $\mathbb{F}_2(x_0,x_1,\ldots,x_{i-1})$. The correspondence between numbers and elements from these fields is given by taking $x_i \mapsto 2^{2^i}$. But then, wecan write every 2-power as a product of the $x_i$ and use the binary representation of numbers to perform all Nim-calculations with numbers in these fields.
Therefore, a quick and dirty way (and by no means the most efficient) to do Nim-calculations in the next Fermat-field consisting of all numbers smaller than 65536, is to use sage and set up the field $\mathbb{F}_2(x_0,x_1,x_2,x_3)$ by
R.< x,y,z,t > =GF(2)[]
S.< a,b,c,d >=R.quotient((x^2+x+1,y^2+y+x,z^2+z+x*y,t^2+t+x*y*z))
To find the smallest number generating the multiplicative group and satisfying the additional compatibility condition $n^{257}=32$ we have to find the smallest binary number $i_1i_2 \ldots i_{16}$ (larger than 255) satisfying
(i1*a*b*c*t+i2*b*c*t+i3*a*c*t+i4*c*t+i5*a*b*t+i6*b*t+
i7*a*t+i8*t+i9*a*b*c+i10*b*c+i11*a*c+i12*c+i13*a*b+
i14*b+i15*a+i16)^257=a*c
It takes a 2.4GHz 2Gb-RAM MacBook not that long to decide that the requested generator is 1051 (killing another optimistic conjecture that these generators might be 2-powers). So, we seat Knight
$K_{1051}$ at root $e^{2 \pi i/65535}$ and can then arrange seatings for all Knight queued up until we reach the 65536-th! In particular, the first Knight we couldn’t place before, that is Knight $K_{256}$, will be seated at root $e^{65826 \pi i/65535}$.
If you’re lucky enough to own a computer with more RAM, or have the patience to make the search more efficient and get the seating arrangement for the next Fermat-field, please drop a comment.
I’ll leave you with another Lenstra-exercise which shouldn’t be too difficult for you to solve now : “Prove that $x^3=2^{2^i}$ has three solutions in $\mathbb{N}$ for each $i \geq 2$.”
Two questions from my last group-theory 101 exam:
(a) : What are the Jordan-Holder components of the Abelian group $\mathbb{Z}/20 \mathbb{Z}$?
(b) : Determine the number of order 7 elements in a simple group of order 168.
Give these to any group of working mathematicians, and, I guess all of them will solve (a), whereas the number of correct solutions to (b) will be (substantially) smaller.
Guess what? All(!) my students solved (b) correctly, whereas almost none of them had anything sensible to say about (a). A partial explanation is that they had more drill-exercises applying the Sylow-theorems than ones concerning the Jordan-Holder theorem.
A more fundamental explanation is that (b) has to do with sub-structures whereas (a) concerns quotients. Over the years I’ve tried numerous methods to convey the quotient-idea : putting things in bags, dividing a big group-table into smaller squares, additional lessons on relations, counting modulo numbers … No method appears to have an effect, lasting until the examination.
At the moment I’m seriously considering to rewrite the entire course, ditching quotients and using them only in disguise via groupoids. Before you start bombarding me with comments, I’m well aware of the problems inherent in this approach.
Before you do groupoids, students have to know some basic category theory. But that’s ok with me. Since last year it has been decided that I should sacrifice the first three weeks of the course telling students the basics of sets, maps and relations. After this, the formal definition of a category will appear more natural to them than the definition of a group, not? Besides, most puzzle-problems I use to introduce groups are actually examples of groupoids…
But then, what are the main theorems on finite groupoids? Well, I can see the groupoid cardinality result, giving you in one stroke Lagrange’s theorem as well as the orbit-counting method. From this one can then prove the remaining classical group-results such as Cauchy and the Sylows, but perhaps there are more elegant approaches?
Have you seen a first-year group-theory course starting off with groupoids? Do you know an elegant way to prove a classical group-result using groupoids?
We have seen that John Conway defined a nim-addition and nim-multiplication on the ordinal numbers in such a way that the subfield $[\omega^{\omega^{\omega}}] \simeq \overline{\mathbb{F}}_2$ is the algebraic closure of the field on two elements. We’ve also seen how to do actual calculations in that field provided we can determine the mystery elements $\alpha_p$, which are the smallest ordinals not being a p-th power of ordinals lesser than $[\omega^{\omega^{k-1}}]$ if $p$ is the $k+1$-th prime number.
Hendrik Lenstra came up with an effective method to compute these elements $\alpha_p$ requiring a few computations in certain finite fields. I’ll give a rundown of his method and refer to his 1977-paper “On the algebraic closure of two” for full details.
For any ordinal $\alpha < \omega^{\omega^{\omega}}$ define its degree $d(\alpha)$ to be the degree of minimal polynomial for $\alpha$ over $\mathbb{F}_2 = [2]$ and for each prime number $p$ let $f(p)$ be the smallest number $h$ such that $p$ is a divisor of $2^h-1$ (clearly $f(p)$ is a divisor of $p-1$).
In the previous post we have already defined ordinals $\kappa_{p^k}=[\omega^{\omega^{k-1}.p^{n-1}}]$ for prime-power indices, but we now need to extend this definition to allow for all indices. So. let $h$ be a natural number, $p$ the smallest prime number dividing $h$ and $q$ the highest power of $p$ dividing $h$. Let $g=[h/q]$, then Lenstra defines
$\kappa_h = \begin{cases} \kappa_q~\text{if q divides}~d(\kappa_q)~\text{ and} \\ \kappa_g + \kappa_q = [\kappa_g + \kappa_q]~\text{otherwise} \end{cases}$
With these notations, the main result asserts the existence of natural numbers $m,m’$ such that
$\alpha_p = [\kappa_{f(p)} + m] = [\kappa_{f(p)}] + m’$
Now, assume by induction that we have already determined the mystery numbers $\alpha_r$ for all odd primes $r < p$, then by teh argument of last time we can effectively compute in the field $[\kappa_p]$. In particular, we can compute for every element its multiplicative order $ord(\beta)$ and therefore also its degree $d(\beta)$ which has to be the smallest number $h$ such that $ord(\beta)$ divides $[2^h-1]$.
Then, by the main result we only have to determine the smallest number m such that $\beta = [\kappa_{f(p)} +m]$ is not a p-th power in $\kappa_p$ which is equivalent to the condition that
$\beta^{(2^{d(\beta)}-1)/p} \not= 1$ if $p$ divides $[2^{d(\beta)}-1]$
All these conditions can be verified within suitable finite fields and hence are effective. In this manner, Lenstra could extend Conway’s calculations (probably using a home-made finite field program running on a slow 1977 machine) :
$$\begin{array}{c|c|c} p & f(p) & \alpha_p \\ \hline 3 & 2 & [2] \\ 5 & 4 & [4] \\ 7 & 3 & [\omega]+1 \\ 11 & 10 & [\omega^{\omega}]+1 \\ 13 & 12 & [\omega]+4 \\ 17 & 8 & [16] \\ 19 & 18 & [\omega^3]+4 \\ 23 & 11 & [\omega^{\omega^3}]+1 \\ 29 & 28 & [\omega^{\omega^2}]+4 \\ 31 & 5 & [\omega^{\omega}]+1 \\ 37 & 36 & [\omega^3]+4 \\ 41 & 20 & [\omega^{\omega}]+1 \\ 43 & 14 & [\omega^{\omega^2}]+ 1 \end{array}$$
Right, so let’s try the case $p=47$. To begin, $f(47)=23$ whence we have to determine the smallest field containg $\kappa_{23}$. By induction (Lenstra’s tabel) we know already that
$\kappa_{23}^{23} = \kappa_{11} + 1 = [\omega^{\omega^3}]+1$ and $\kappa_{11}^{11} = \kappa_5 + 1 = [\omega^{\omega}]+1$ and $\kappa_5^5=[4]$
Because the smallest field containg $4$ is $[16]=\mathbb{F}_{2^4}$ we have that $\mathbb{F}_2(4,\kappa_5,\kappa_{11}) \simeq \mathbb{F}_{2^{220}}$. We can construct this finite field, together with a generator $a$ of its multiplicative group in Sage via
sage: f1.< a >=GF(2^220)
In this field we have to pinpoint the elements $4,\kappa_5$ and $\kappa_{11}$. As $4$ has order $15$ in $\mathbb{F}_{2^4}$ we know that $\kappa_5$ has order $75$. Hence we can take $\kappa_5 = a^{(2^{220}-1)/75}$ and then $4=\kappa_5^5$.
If we denote $\kappa_5$ by x5 we can obtain $\kappa_{11}$ as x11 by the following sage-commands
sage: c=x5+1
sage: x11=c.nth_root(11)
It takes about 7 minutes to find x11 on a 2.4 GHz MacBook. Next, we have to set up the field extension determined by $\kappa_{23}$ (which we will call x in sage). This is done as follows
sage: p1.=PolynomialRing(f1)
sage: f=x^23-x11-1
sage: F2=f1.extension(f,'u')
The MacBook needed 8 minutes to set up this field which is isomorphic to $\mathbb{F}_{2^{5060}}$. The relevant number is therefore $n=\frac{2^{5060}-1}{47}$ which is the gruesome
34648162040462867047623719793206539850507437636617898959901744136581<br/>
259144476645069239839415478030722644334257066691823210120548345667203443<br/>
317743531975748823386990680394012962375061822291120459167399032726669613
<br/>
442804392429947890878007964213600720766879334103454250982141991553270171
938532417844211304203805934829097913753132491802446697429102630902307815
301045433019807776921086247690468136447620036910689177286910624860871748
150613285530830034500671245400628768674394130880959338197158054296625733
206509650361461537510912269982522844517989399782602216622257291361930850
885916974186835958466930689748400561295128553674118498999873244045842040
080195019701984054428846798610542372150816780493166669821114184374697446
637066566831036116390063418916814141753876530004881539570659100352197393
997895251223633176404672792711603439161147155163219282934597310848529360
118189507461132290706604796116111868096099527077437183219418195396666836
014856037176421475300935193266597196833361131333604528218621261753883518
667866835204501888103795022437662796445008236823338104580840186181111557
498232520943552183185687638366809541685702608288630073248626226874916669
186372183233071573318563658579214650042598011275864591248749957431967297
975078011358342282941831582626985121760847852546207377440873367589369439
085660784239080183415569559585998884824991911321095149718147110882474280
968166266224151511519773175933506503369761671964823112231808283557885030
984081329986188655169245595411930535264918359325712373064120338963742590
76555755141425
Remains ‘only’ to take x,x+1,etc. to the n-th power and verify which is the first to be unequal to 1. For this it is best to implement the usual powering trick (via digital expression of the exponent) in the field F2, something like
sage: def power(e,n): ...: le=n.bits() ...: v=n.digits() ...: mn=F2(e) ...: out=F2(1) ...: i=0 ...: while i< le : ...: if v[i]==1 : out=F2(out_mn) ...: m=F2(mn_mn) ...: mn=F2(m) ...: i=i+1 ...: return(out) ...:
then it takes about 20 seconds to verify that power(x,n)=1 but that power(x+1,n) is NOT! That is, we just checked that $\alpha_{47}=\kappa_{11}+1$.
It turns out that 47 is the hardest nut to crack, the following primes are easier. Here’s the data (if I didn’t make mistakes…)
$$\begin{array}{c|c|c} p & f(p) & \alpha_p \\ \hline 47 & 23 & [\omega^{\omega^{7}}]+1 \\ 53 & 52 & [\omega^{\omega^4}]+1 \\ 59 & 58 & [\omega^{\omega^8}]+1 \\ 61 & 60 & [\omega^{\omega}]+[\omega] \\ 67 & 66 & [\omega^{\omega^3}]+[\omega] \end{array}$$
It seems that Magma is substantially better at finite field arithmetic, so if you are lucky enough to have it you’ll have no problem finding $\alpha_p$ for all primes less than 100 by the end of the day. If you do, please drop a comment with the results…
Last time we did recall Cantor’s addition and multiplication on ordinal numbers. Note that we can identify an ordinal number $\alpha$ with (the order type of) the set of all strictly smaller ordinals, that is, $\alpha = { \alpha’~:~\alpha’ < \alpha }$. Given two ordinals $\alpha$ and $\beta$ we will denote their Cantor-sums and products as $[ \alpha + \beta]$ and $[\alpha . \beta]$.
The reason for these square brackets is that John Conway constructed a well behaved nim-addition and nim-multiplication on all ordinals $\mathbf{On}_2$ by imposing the ‘simplest’ rules which make $\mathbf{On}_2$ into a field. By this we mean that, in order to define the addition $\alpha + \beta$ we must have constructed before all sums $\alpha’ + \beta$ and $\alpha + \beta’$ with $\alpha’ < \alpha$ and $\beta’ < \beta$. If + is going to be a well-defined addition on $\mathbf{On}_2$ clearly $\alpha + \beta$ cannot be equal to one of these previously constructed sums and the ‘simplicity rule’ asserts that we should take $\alpha+\beta$ the least ordinal different from all these sums $\alpha’+\beta$ and $\alpha+\beta’$. In symbols, we define
$\alpha+ \beta = \mathbf{mex} { \alpha’+\beta,\alpha+ \beta’~|~\alpha’ < \alpha, \beta’ < \beta }$
where $\mathbf{mex}$ stands for ‘minimal excluded value’. If you’d ever played the game of Nim you will recognize this as the Nim-addition, at least when $\alpha$ and $\beta$ are finite ordinals (that is, natural numbers) (to nim-add two numbers n and m write them out in binary digits and add without carrying). Alternatively, the nim-sum n+m can be found applying the following two rules :
• the nim-sum of a number of distinct 2-powers is their ordinary sum (e.g. $8+4+1=13$, and,
• the nim-sum of two equal numbers is 0.
So, all we have to do is to write numbers n and m as sums of two powers, scratch equal terms and add normally. For example, $13+7=(8+4+1)+(4+2+1)=8+2=10$ (of course this is just digital sum without carry in disguise).
Here’s the beginning of the nim-addition table on ordinals. For example, to define $13+7$ we have to look at all values in the first 7 entries of the row of 13 (that is, ${ 13,12,15,14,9,8,11 }$) and the first 13 entries in the column of 7 (that is, ${ 7,6,5,4,3,2,1,0,15,14,13,12,11 }$) and find the first number not included in these two sets (which is indeed $10$).
In fact, the above two rules allow us to compute the nim-sum of any two ordinals. Recall from last time that every ordinal can be written uniquely as as a finite sum of (ordinal) 2-powers :
$\alpha = [2^{\alpha_0} + 2^{\alpha_1} + \ldots + 2^{\alpha_k}]$, so to determine the nim-sum $\alpha+\beta$ we write both ordinals as sums of ordinal 2-powers, delete powers appearing twice and take the Cantor ordinal sum of the remaining sum.
Nim-multiplication of ordinals is a bit more complicated. Here’s the definition as a minimal excluded value
$\alpha.\beta = \mathbf{mex} { \alpha’.\beta + \alpha.\beta’ – \alpha’.\beta’ }$
for all $\alpha’ < \alpha, \beta’ < \beta$. The rationale behind this being that both $\alpha-\alpha’$ and $\beta – \beta’$ are non-zero elements, so if $\mathbf{On}_2$ is going to be a field under nim-multiplication, their product should be non-zero (and hence strictly greater than 0), that is, $~(\alpha-\alpha’).(\beta-\beta’) > 0$. Rewriting this we get $\alpha.\beta > \alpha’.\beta+\alpha.\beta’-\alpha’.\beta’$ and again the ‘simplicity rule’ asserts that $\alpha.\beta$ should be the least ordinal satisfying all these inequalities, leading to the $\mathbf{mex}$-definition above. The table gives the beginning of the nim-multiplication table for ordinals. For finite ordinals n and m there is a simple 2 line procedure to compute their nim-product, similar to the addition-rules mentioned before :
• the nim-product of a number of distinct Fermat 2-powers (that is, numbers of the form $2^{2^n}$) is their ordinary product (for example, $16.4.2=128$), and,
• the square of a Fermat 2-power is its sesquimultiple (that is, the number obtained by multiplying with $1\frac{1}{2}$ in the ordinary sense). That is, $2^2=3,4^2=6,16^2=24,…$
Using these rules, associativity and distributivity and our addition rules it is now easy to work out the nim-multiplication $n.m$ : write out n and m as sums of (multiplications by 2-powers) of Fermat 2-powers and apply the rules. Here’s an example
$5.9=(4+1).(4.2+1)=4^2.2+4.2+4+1=6.2+8+4+1=(4+2).2+13=4.2+2^2+13=8+3+13=6$
Clearly, we’d love to have a similar procedure to calculate the nim-product $\alpha.\beta$ of arbitrary ordinals, or at least those smaller than $\omega^{\omega^{\omega}}$ (recall that Conway proved that this ordinal is isomorphic to the algebraic closure $\overline{\mathbb{F}}_2$ of the field of two elements). From now on we restrict to such ‘small’ ordinals and we introduce the following special elements :
$\kappa_{2^n} = [2^{2^{n-1}}]$ (these are the Fermat 2-powers) and for all primes $p > 2$ we define
$\kappa_{p^n} = [\omega^{\omega^{k-1}.p^{n-1}}]$ where $k$ is the number of primes strictly smaller than $p$ (that is, for p=3 we have k=1, for p=5, k=2 etc.).
Again by associativity and distributivity we will be able to multiply two ordinals $< \omega^{\omega^{\omega}}$ if we know how to multiply a product
$[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}]$ with $\alpha,\beta < [\omega^{\omega}]$ and $n_0,m_0 \in \mathbb{N}$.
Now, $\alpha$ can be written uniquely as $[\omega^t.n_t+\omega^{t-1}.n_{t-1}+\ldots+\omega.n_2 + n_1]$ with t and all $n_i$ natural numbers. Write each $n_k$ in base $p$ where $p$ is the $k+1$-th prime number, that is, we have for $n_0,n_1,\ldots,n_t$ an expression
$n_k=[\sum_j p^j.m(j,k)]$ with $0 \leq m(j,k) < p$
The point of all this is that any of the special elements we want to multiply can be written as a unique expression as a decreasing product
$[\omega^{\alpha}.2^{n_0}] = [ \prod_q \kappa_q^m(q) ]$
where $q$ runs over all prime powers. The crucial fact now is that for this decreasing product we have a rule similar to addition of 2-powers, that is Conway-products coincide with the Cantor-products
$[ \prod_q \kappa_q^m(q) ] = \prod_q \kappa_q^m(q)$
But then, using associativity and commutativity of the Conway-product we can ‘nearly’ describe all products $[\omega^{\alpha}.2^{n_0}].[\omega^{\beta}.2^{m_0}]$. The remaining problem being that it may happen that for some q we will end up with an exponent $m(q)+m(q’)>p$. But this can be solved if we know how to take p-powers. The rules for this are as follows
$~(\kappa_{2^n})^2 = \kappa_{2^n} + \prod_{1 \leq i < n} \kappa_{2^i}$, for 2-powers, and,
$~(\kappa_{p^n})^p = \kappa_{p^{n-1}}$ for a prime $p > 2$ and for $n \geq 2$, and finally
$~(\kappa_p)^p = \alpha_p$ for a prime $p > 2$, where $\alpha_p$ is the smallest ordinal $< \kappa_p$ which cannot be written as a p-power $\beta^p$ with $\beta < \kappa_p$. Summarizing : if we will be able to find these mysterious elements $\alpha_p$ for all prime numbers p, we are able to multiply in $[\omega^{\omega^{\omega}}]=\overline{\mathbb{F}}_2$.
Let us determine the first one. We have that $\kappa_3 = \omega$ so we are looking for the smallest natural number $n < \omega$ which cannot be written in num-multiplication as $n=m^3$ for $m < \omega$ (that is, also $m$ a natural number). Clearly $1=1^3$ but what about 2? Can 2 be a third root of a natural number wrt. nim-multiplication? From the tabel above we see that 2 has order 3 whence its cube root must be an element of order 9. Now, the only finite ordinals that are subfields of $\mathbf{On}_2$ are precisely the Fermat 2-powers, so if there is a finite cube root of 2, it must be contained in one of the finite fields $[2^{2^n}]$ (of which the mutiplicative group has order $2^{2^n}-1$ and one easily shows that 9 cannot be a divisor of any of the numbers $2^{2^n}-1$, that is, 2 doesn’t have a finte 3-th root in nim! Phrased differently, we found our first mystery number $\alpha_3 = 2$. That is, we have the marvelous identity in nim-arithmetic
$\omega^3 = 2$
Okay, so what is $\alpha_5$? Well, we have $\kappa_5 = [\omega^{\omega}]$ and we have to look for the smallest ordinal which cannot be written as a 5-th root. By inspection of the finite nim-table we see that 1,2 and 3 have 5-th roots in $\omega$ but 4 does not! The reason being that 4 has order 15 (check in the finite field [16]) and 25 cannot divide any number of the form $2^{2^n}-1$. That is, $\alpha_5=4$ giving another crazy nim-identity
$~(\omega^{\omega})^5 = 4$
And, surprises continue to pop up… Conway showed that $\alpha_7 = \omega+1$ giving the nim-identity $~(\omega^{\omega^2})^7 = \omega+1$. The proof of this already uses some clever finite field arguments. Because 7 doesn’t divide any number $2^{2^n}-1$, none of the finite subfields $[2^{2^n}]$ contains a 7-th root of unity, so the 7-power map is injective whence surjective, so all finite ordinal have finite 7-th roots! That is, $\alpha_7 \geq \omega$. Because $\omega$ lies in a cubic extension of the finite field [4], the field generated by $\omega$ has 64 elements and so its multiplicative group is cyclic of order 63 and as $\omega$ has order 9, it must be a 7-th power in this field. But, as the only 7th powers in that field are precisely the powers of $\omega$ and by inspection $\omega+1$ is not a 7-th power in that field (and hence also not in any field extension obtained by adjoining square, cube and fifth roots) so $\alpha_7=\omega +1$.
Conway did stop at $\alpha_7$ but I’ve always been intrigued by that one line in ONAG p.61 : “Hendrik Lenstra has computed $\alpha_p$ for $p \leq 43$”. Next time we will see how Lenstra managed to do this and we will use sage to extend his list a bit further, including the first open case : $\alpha_{47}= \omega^{\omega^7}+1$.
For an enjoyable video on all of this, see Conway’s MSRI lecture on Infinite Games. The nim-arithmetic part is towards the end of the lecture but watching the whole video is a genuine treat! | 2020-08-03 23:28:15 | {"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": 0.8984208106994629, "perplexity": 393.7464913186702}, "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-2020-34/segments/1596439735836.89/warc/CC-MAIN-20200803224907-20200804014907-00424.warc.gz"} |
https://www.or-exchange.org/questions/127/extracting-gomory-cuts-out-of-cgl-coin-or | # Extracting Gomory cuts out of Cgl (Coin-or)
1 Hey, I'm trying to extract Cgl Gomory cuts out of the Cgl (Cut Generation Library) of Coin-Or The following is the code I'm using to extract the cuts - OsiCuts cutlist; CglGomory * gomory = new CglGomory(); gomory->setLimit(100); gomory->generateCuts(*sym, cutlist) ; where sym is an instance of OsiSymSolverInterface (the OsiSolverInterface for Symphony). Unfortunately the code is segfaulting at generateCuts somewhere inside the method as far as I've been able to determine using gdb. Extraction of CglProbing cuts is likewise segfaulting again inside the generateCuts method of the CglProbing class. All other cuts seem to be working fine. If someone could shed some light on this or even better, post/link to an example file using these cuts or a tutorial of some sort, that would great. If there's an example/tutorial for extracting cuts out of some other solver like SCIP instead of Coin-OR, that would work too. Thanks asked 05 Jan '10, 02:19 Sid 531●2●9●17 accept rate: 18%
One Answer:
3 After asking around on the CGL mailing list, I found out that the reason the above was not working. I'm posting it here for the benefit of those who might encounter a similar problem. The reason this was failing was because Symphony is an MILP solver and since the Gomory cuts generated by CglGomory are generated from a simplex tableau, the OsiSolverInterface needs to be to an LP solver like Clp (the interface class for which is OsiClpSolverInterface). In particular, for the code above, making sym an instance of OsiClpSolverInterface would work. answered 07 Jan '10, 01:36 Sid 531●2●9●17 accept rate: 18% Thanks for posting this! It is much more useful having the answer than just the hanging question. (07 Jan '10, 02:38) Michael Trick ♦♦
Your answer
toggle preview community wiki
### Follow this question
By Email:
Once you sign in you will be able to subscribe for any updates here
By RSS:
Answers
Answers and Comments
Markdown Basics
• *italic* or _italic_
• **bold** or __bold__
• link:[text](http://url.com/ "Title")
• image?
• numbered list: 1. Foo 2. Bar
• to add a line break simply add two spaces to where you would like the new line to be.
• basic HTML tags are also supported
Tags:
Asked: 05 Jan '10, 02:19
Seen: 3,095 times
Last updated: 07 Jan '10, 02:26
### Related questions
OR-Exchange! Your site for questions, answers, and announcements about operations research. | 2020-03-29 15:34:47 | {"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": 0.5037684440612793, "perplexity": 4563.975384620374}, "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-2020-16/segments/1585370494349.3/warc/CC-MAIN-20200329140021-20200329170021-00481.warc.gz"} |
https://nrich.maths.org/6623/clue | ### Stats Statements
Are these statistical statements sometimes, always or never true? Or it is impossible to say?
### Real-life Equations
Here are several equations from real life. Can you work out which measurements are possible from each equation?
### Bent Out of Shape
An introduction to bond angle geometry.
# CSI: Chemical Scene Investigation
##### Age 16 to 18 Challenge Level:
For the second isomer drawing exercise:
-Don't forget the possibility of double and tripled bonds!
-Who said the molecules had to be non-cyclic? Don't forget cyclobutane and cyclopropane rings!
The IR spectrum is trying to indicate three main absorptions. Ignore any other detail that you might see.If you can't see the significance of the single absorption at 3300 cm$^{-1}$, take a look at the data table for NH and NH$_2$...
Count the number of clear peaks on the NMR spectrum to give the number of Carbon environments. Next, draw a Benzene ring, and think about how the two additional carbons must be distributed as substituents around the ring. Remember that the carbons could be attached separately, or as part of the same chain.
We reckon that you can probably draw 7 structures from all the IR and NMR data. Take the hint in the question: N-O bonds won't be present in any of them!
To use the mass spectrometry data effectively, look at your seven structures and work out a logical point for the molecule to fragment (how about cleaving a group off the Benzene ring...?). See which 3 of your structures give the requires fragment masses.
The final information that the molecule is synthesised from Aryl-NO$_2$ indicates that a Nitrogen must be directly attached to the Benzene ring. Does this eliminate two of your structures? | 2020-09-23 13:31:20 | {"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": 0.2836877107620239, "perplexity": 2515.8674261386764}, "config": {"markdown_headings": true, "markdown_code": false, "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/1600400210996.32/warc/CC-MAIN-20200923113029-20200923143029-00709.warc.gz"} |
https://nersc.gitlab.io/science-partners/jgi/filesystems/ | # Filesystems¶
It is the responsibility of every user to organize and maintain software and data critical to their science. Managing your data requires understanding NERSC's storage systems.
## Filesystems¶
NERSC provides several filesystems. Making efficient use of NERSC computing facilities requires an understanding of the strengths and limitations of each filesystem.
Note
JAMO is JGI's in-house-built hierarchical file system, which has functional ties to NERSC's filesystems and tape archive. JAMO is not maintained by NERSC.
All NERSC filesystems have per-user quotas on total storage and number of files and directories (known as inodes). To check your filesystem usage, use the myquota command from any login node, check the Data Dashboard on my.nersc.gov or log into the NERSC Information Management (NIM) website, and click on the "Account Usage" tab.
### $HOME¶ Your home directory is mounted across all NERSC systems. You should refer to this home directory as $HOME wherever possible, and you should not change the environment variable $HOME. Each user's $HOME directory has a quota of 40GB and 1M inodes. $HOME quotas will not be increased. Other filesystems should be used for storage of large volumes of data which are either being stored long term or are a part of computation in progress. Warning As a global filesystem, $HOME is shared by all users, and is not configured for performance. Do not write scripts or run software in a way that will cause high bandwidth I/O to $HOME. High volumes of reads and writes can cause congestion on the filesystem metadata servers, and can slow NERSC systems significantly for all users. Large I/O operations should always be directed to scratch filesystems. ### Projectb¶ projectb is a 2.7PB GPFS-based file system dedicated to the JGI's active projects. There are three distinct user spaces in the projectb filesystem: projectb/sandbox, project/software, and projectb/scratch. The projectb filesystem is available on most NERSC systems. projectb Scratch projectb Sandbox projectb Software Location /global/projectb/scratch/$username /global/projectb/sandbox/$program /global/projectb/software/$group
Quota 20TB, 5M inodes by default; 40TB upon request Defined by agreement with JGI Management 500GB, 500K inodes
Backups Not backed up Not backed up Backed up
File Purging Files not accessed for 90 days are automatically deleted Not purged Not purged
The projectb "Scratch" space is intended for staging and running JGI computation by individual users of NERSC systems. The environment variable $BSCRATCH points to a user's projectb scratch space. These scratch directories are not automatically granted to new users; to request space please file a Consulting ticket asking for an initial projectb scratch allocation. Sandbox directories on projectb are allocated by program. Data and software stored in projectb sandbox is not subject to purging. If you have questions about your program's space, please see your group lead. New Sandbox space or quota increase requests must be approved by JGI management. The projectb software allocations are intended for storage of Conda environments, source code, and binaries being used by individual groups at JGI. At very large production scale, projectb performance, may degrade; consider moving such software to /usr/common/software, the Data and Archive filesystem, or a Shifter container. ### DnA (Data n' Archive)¶ DnA is a 2.4PB GPFS filesystem for the JGI's archive, shared databases, and project directories. DnA Projects DnA Shared DnA DM Archive Location /global/dna/projectdirs/ /global/dna/shared /global/dna/dm_archive Quota 5TB default Defined by agreement with the JGI Management Defined by agreement with the JGI Management Backups Daily, only for projectdirs with quota <= 5TB Backed up by JAMO Backed up by JAMO File purging Files are not automatically purged Purge policy set by users of the JAMO system Files are not automatically purged The intention of the DnA "Project" and "Shared" space is to be a place for data that is needed by multiple users collaborating on a project which allows for easy reading of shared data. The "Project" space is owned and managed by the JGI. The "Shared" space is a collaborative effort between the JGI and NERSC. Write access to DnA is restricted to protect high performance; data can only be written to DnA from Data Transfer Nodes or by using the --qos=dna_xfer QOS. If you would like a project directory, please use the Project Directory Request Form. The "DM Archive" is a data repository maintained by the JAMO system. Files are stored here during migration using the JAMO system. The files can remain in this space for as long as the user specifies. Any file that is in the "DM Archive" has also been placed in the HPSS tape archive. This section of the file system is owned by the JGI data management team. ###$SCRATCH¶
Each user has a "scratch" directory. Scratch directories are NOT backed up and files can be purged if they have not been accessed for 90 days. Find your scratch directory using the environment variable $SCRATCH for example: elvis@cori02:~> cd$SCRATCH
elvis@cori02:/global/cscratch1/sd/elvis>
Scratch environment variables:
Environment Variable Value NERSC Systems
$SCRATCH Best-connected file system All NERSC computational systems $CSCRATCH /global/cscratch[1,2,3]/sd/$username Cori $BSCRATCH points to your projectb scratch space if you have a BSCRATCH allocation. $SCRATCH will always point to the best-connected scratch space available for the NERSC machine you are accessing. The intention of scratch space is for staging, running, and completing calculations on NERSC systems. Thus, these filesystems are configured for best performance when usage is wide-scale file reading and writing from many compute nodes. The scratch filesystems are not intended for long-term file storage or archival. Data is not backed up, and files not accessed for a significant time period can be purged. Policies for $SCRATCH are described at NERSC Data Management Policy.
### Other file systems¶
Other file systems used by JGI may also be mounted on NERSC systems:
• SeqFS - file system used exclusively by the Illumina sequencers, SDM and Instrumentation groups at the JGI.
• /usr/common - is a file system where NERSC staff build software for user applications. This is the principal site for the modular software installations.
• /global/project - is a GPFS-based file system that is accessible on almost all of NERSC's other compute systems used by all the other NERSC users. The projectdir portion of projectb should be favored by JGI users instead of /global/project. | 2019-11-22 15:46:24 | {"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": 0.3192589581012726, "perplexity": 10409.289526000588}, "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-2019-47/segments/1573496671363.79/warc/CC-MAIN-20191122143547-20191122172547-00524.warc.gz"} |
https://indico.hiskp.uni-bonn.de/event/40/contributions/654/ | # The 39th International Symposium on Lattice Field Theory (Lattice 2022)
Aug 8 – 13, 2022
Hörsaalzentrum Poppelsdorf
Europe/Berlin timezone
## Details of RQCD analyses on CLS ensembles
Aug 9, 2022, 8:00 PM
1h
Poster Presentation Hadron Spectroscopy and Interactions
### Speaker
Wolfgang Soeldner (University of Regensburg)
### Description
High statistics results for quantities like the gradient flow scale, the quark masses, the lower lying baryon spectrum and the baryon octet sigma terms determined on CLS ensembles with $N_f=2+1$ non-perturbatively $O(a)$ improved Wilson dynamical fermions are presented at this conference by the RQCD collaboration. In this contribution, we provide further details of the analysis focusing on systematics associated with the extraction of the lattice data including autocorrelations and the continuum, quark mass and finite volume extrapolations, including the fit forms employed.
### Primary author
Wolfgang Soeldner (University of Regensburg)
### Co-authors
Gunnar Bali Sara Collins (University of Regensburg) | 2022-12-06 04:43:50 | {"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": 0.4477909505367279, "perplexity": 8554.326281296739}, "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-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00699.warc.gz"} |
https://www.nocentino.com/posts/2023-02-19-hello-world-go-container/ | # Writing a Hello World Go Container Web Application
Page content
In this blog post, I will show you how to build a hello world container-based web application in the go programming language. The reason I want to do this is because I need a very small container image to do some testing in Kubernetes. I’ll also highlight some of the pitfalls I ran into to hopefully have you some time in your learnings.
## Let’s build and test it locally first
Before you build a container-based application, you need an application. So let’s go ahead and build a simple hello world app in go, but running on our local system as a traditionally compiled program. I want to make sure my application works before I move onto the container build process. You’ll need to install the go programming language development tools. On a Mac, you can do that with brew install go. For other operating systems, check out the install page.
brew install go
Once we have go installed, let’s start working on our program. I want to make a simple web application that when I access the application with a web browser, it has some output, like…Hello, world! The code for that is below. We import a few libraries based on our needs. Then in the main() function, we have the code to do just that. Format the output, then expose that as a static webpage accessed via HTTP on port 8080.
package main
import (
"fmt"
"net/http"
"os"
)
func main() {
hostname, err := os.Hostname()
if err != nil {
panic(err)
}
http.HandleFunc("/", func (w http.ResponseWriter, r *http.Request){
fmt.Fprintf(w, "Hello, world!\nVersion: 1.0.0\n" + hostname + "\n")
})
fs := http.FileServer(http.Dir("static/"))
http.Handle("/static/", http.StripPrefix("/static/", fs))
http.ListenAndServe(":8080", nil)
}
Now, save all the above code into a file named hello-app.go. Then next, you’ll need to compile the application with the go command. The code below will compile the application in the file hello-app.go and then write the binary to hello-app, defined by the -o parameter.
go build -o hello-app hello-app.go
With our program built, you can launch it with the code here.
./hello-app
Since it’s a simple web application, we can access it with curl, and that’s what we’re doing in the code below. And you can see the output on standard out. Notice the hostname output is the hostname of my local system.
curl http://localhost:8080
Hello, world!
Version: 1.0.0
Hostname: Anthonys-Air.localdomain
So now that we know our application works, let’s start putting it in a container.
## Let’s put our app in a container
This section will use a multi-stage build process. A primary benefit of using a multi-stage build is the final container will be tiny in size since the development libraries are left behind, and the final container image has just the compiled binary files for your application. For more details on multi-stage build check out this link.
Below is the multi-stage build dockerfile builds our application in the first container, then the resulting binary is copied into an apline container. In the dockerfile, we’re basing the first stage FROM golang:latest AS builder. The name builder is a way for us to refer to this stage by name in subsequent stages. Then we copy in our source code with COPY ./webappv1/hello-app.go .. And then finally, build our application with RUN go env -w CGO_ENABLED=0 GO111MODULE=off && go build -o /app/hello-app. There are environment variables defined using CGO_ENABLED=0 GO111MODULE=off. More on those later in the post. But for now, know they’re needed to build and run the application successfully.
In the second stage, we’re basing our image off the tiny (7.46MB) alpine container image with FROM alpine:latest, then set the container’s working directory with WORKDIR /app. Next, we copy the binary built in the first stage into this container with COPY --from=builder /app/hello-app . We’ll give docker information about what port our application is listening on with EXPOSE 8080 and then finally, we define which process to start when the container is started with CMD ["./hello-app"].
FROM golang:latest AS builder
COPY ./webappv1/hello-app.go .
RUN go env -w CGO_ENABLED=0 GO111MODULE=off && go build -o /app/hello-app
FROM alpine:latest
WORKDIR /app
COPY --from=builder /app/hello-app .
EXPOSE 8080
CMD ["./hello-app"]
Next, it’s time to build our container-based hello world application. We do that with the command docker build -t hello-app:1.0 . This will build what’s defined in the dockerfile and also give the image a name of hello-app and defines the tag as 1.0.
docker build -t hello-app:1.0 .
The resulting container image is relatively small. We can get the container image size with docker image ls, and in the output below, you can see our image is 13.8MB.
docker image ls
REPOSITORY TAG IMAGE ID CREATED SIZE
hello-app 1.0 fa44005dab70 3 minutes ago 13.8MB
With the container built, let’s go ahead and run the container and expose the application on port 8080 using the following.
docker run --name webapp --publish 8080:8080 --detach hello-app:1.0
We can next check to make sure everything is up and running with docker ps, and if you see the output like this below, you are good to go.
docker ps
CONTAINER ID IMAGE COMMAND CREATED STATUS PORTS NAMES
09a8d832d97c hello-app:1.0 "./hello-app" 3 seconds ago Up 2 seconds 0.0.0.0:8080->8080/tcp webapp
With the container up, let’s test our hello world web application. When we hit port 8080 on localhost using curl, we see the output from the app printing some basic information. I do want to call out the hostname inside the container is the CONTAINER ID which you can see in the output of docker ps above.
curl http://localhost:8080
Hello, world!
Version: 1.0.0
Hostname: 09a8d832d97c
Now, earlier, I glossed over some environment variables set in the application build in our docker file for the line of code RUN go env -w CGO_ENABLED=0 GO111MODULE=off && go build -o /app/hello-app . Let’s dive into that a bit.
If you don’t include GO111MODULE=off, you’ll get the following error. So make sure this setting is included in your command to build your application.
#10 0.142 go: go.mod file not found in current directory or any parent directory; see 'go help modules'
Next, if you don’t set CGO_ENABLED=0 , your container will build, but when you run your container, it will start up and immediately fail, and you’ll get the error below. This is because the required libraries aren’t included in the binary.
docker ps -a
91f32370d6a4 hello-app:1.0 "./hello-app" 3 seconds ago Exited (1) 3 seconds ago webapp
docker logs webapp
exec ./hello-app: no such file or directory
So with that, you just learned what I did this weekend. I needed to build a simple web application in go to test some things out in a Kubernetes environment. | 2023-03-30 18:31:42 | {"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": 0.24348904192447662, "perplexity": 3670.8526604381486}, "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-2023-14/segments/1679296949355.52/warc/CC-MAIN-20230330163823-20230330193823-00755.warc.gz"} |
http://www.emathematics.net/g5_variable_expression.php | User:
Variable expressions
# Write variable expressions
Write an expression for c less 381.
"c less 381" means you should subtract 381 from c.
c – 381
Write an expression for the quotient of m and 79.
"The quotient of m and 79" means you should divide m by 79.
$\frac{m}{79}$
Choose the expression for u plus 26
A. u + 26
B. 26 - u
C. u - 26 | 2017-12-16 16:49:46 | {"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": 1, "/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": 0.6416692137718201, "perplexity": 5829.75436034944}, "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/1512948588294.67/warc/CC-MAIN-20171216162441-20171216184441-00091.warc.gz"} |
http://chemfiles.org/chemfiles/latest/classes/frame.html | # Frame class¶
class Frame
A frame contains data from one simulation step The Frame class holds data from one step of a simulation: the current topology, the positions, and the velocities of the particles in the system. If some information is missing the corresponding data is filled with a default value. Specifically:
• cell is an infinite unit cell;
• topology is empty, and contains no data;
• positions is filled with zeros;
• velocities is the nullopt variant of chemfiles::optional.
Iterating over a Frame will yield all the atoms in the system.
Public Functions
Frame(UnitCell cell = UnitCell())
Create an empty frame with no atoms and the given cell.
Frame clone() const
Get a clone (exact copy) of this frame.
This replace the implicit copy constructor (which is private) to make an explicit copy of the frame.
const Topology &topology() const
Get a const reference to the topology of this frame
It is not possible to get a modifiable reference to the topology, because it would then be possible to remove/add atoms without changing the actual positions and velocity storage. Instead, all the mutating functionalities of the topology are mirrored on the frame (adding and removing bonds, adding residues, etc.)
void set_topology(Topology topology)
Set the topology of this frame to topology
Parameters
• topology: the new Topology to use for this frame
Exceptions
• Error: if the topology size does not match the size of this frame
const UnitCell &cell() const
Get a const reference to the unit cell of this frame
UnitCell &cell()
Get a reference to the unit cell of this frame
void set_cell(UnitCell cell)
Set the unit cell for this frame to cell
Parameters
• cell: the new UnitCell to use for this frame
size_t size() const
Get the number of atoms in this frame
span<Vector3D> positions()
Get the positions of the atoms in this frame.
A chemfiles::span is a view inside a vector allowing mutation of the values, but no memory allocation.
const std::vector<Vector3D> &positions() const
Get the positions in this frame as a const reference
void add_velocities()
Add velocities data storage to this frame.
If velocities are already defined, this functions does nothing. The new velocities are initialized to 0.
optional<span<Vector3D>> velocities()
Get an velocities of the atoms in this frame, if this frame contains velocity data.
This function returna an chemfiles::optional value that is close to C++17 std::optional.
A chemfiles::span is a view inside a vector allowing mutation of the values, but no memory allocation.
optional<const std::vector<Vector3D>&> velocities() const
Get an velocities of the atoms in this frame as a const reference, if this frame contains velocity data.
void resize(size_t size)
Resize the frame to contain size atoms.
If the new number of atoms is bigger than the old one, missing data is initializd to 0. Pre-existing values are conserved.
If the new size if smaller than the old one, all atoms and connectivity elements after the new size are removed.
void reserve(size_t size)
Allocate memory in the frame to have enough size for size atoms.
This function does not change the actual number of atoms in the frame, and should be used as an optimisation.
void add_atom(Atom atom, Vector3D position, Vector3D velocity = Vector3D())
Add an atom at the given position and optionally with the given velocity. The velocity value will only be used if this frame contains velocity data.
void remove(size_t i)
Remove the atom at index i in the system.
Exceptions
• chemfiles::OutOfBounds: if i is bigger than the number of atoms in this frame
size_t step() const
Get the current simulation step.
The step is set by the Trajectory when reading a frame.
void set_step(size_t step)
Set the current simulation step to step
void guess_bonds()
Guess the bonds, angles, dihedrals and impropers angles in this frame.
The bonds are guessed using a distance-based algorithm, and then angles, dihedrals and impropers are guessed from the bonds. The distance criterion uses the Van der Waals radii of the atoms. If this information is missing for a specific atoms, one can use configuration files to provide it.
Exceptions
• Error: if the Van der Waals radius in unknown for a given atom.
void clear_bonds()
Remove all connectivity information in the frame’s topology
void add_residue(Residue residue)
Add a residue to this frame’s topology.
Parameters
• residue: the residue to add to this topology
Exceptions
• chemfiles::Error: if any atom in the residue is already in another residue in this topology. In that case, the topology is not modified.
void add_bond(size_t atom_i, size_t atom_j, Bond::BondOrder bond_order = Bond::UNKNOWN)
Add a bond in the system, between the atoms at index atom_i and atom_j.
Parameters
• atom_i: the index of the first atom in the bond
• atom_j: the index of the second atom in the bond
• bond_order: the bond order of the new bond
Exceptions
• OutOfBounds: if atom_i or atom_j are greater than size()
• Error: if atom_i == atom_j, as this is an invalid bond
void remove_bond(size_t atom_i, size_t atom_j)
Remove a bond in the system, between the atoms at index atom_i and atom_j.
If the bond does not exist, this does nothing.
Parameters
• atom_i: the index of the first atom in the bond
• atom_j: the index of the second atom in the bond
Exceptions
• OutOfBounds: if atom_i or atom_j are greater than size()
Atom &operator[](size_t index)
Get a reference to the atom at the position index.
Parameters
• index: the atomic index
Exceptions
• OutOfBounds: if index is greater than size()
const Atom &operator[](size_t index) const
Get a const reference to the atom at the position index.
Parameters
• index: the atomic index
Exceptions
• OutOfBounds: if index is greater than size()
double distance(size_t i, size_t j) const
Get the distance between the atoms at indexes i and j, accounting for periodic boundary conditions. The distance is expressed in angstroms.
Exceptions
• chemfiles::OutOfBounds: if i or j are bigger than the number of atoms in this frame
double angle(size_t i, size_t j, size_t k) const
Get the angle formed by the atoms at indexes i, j and k, accounting for periodic boundary conditions. The angle is expressed in radians.
Exceptions
• chemfiles::OutOfBounds: if i, j or k are bigger than the number of atoms in this frame
double dihedral(size_t i, size_t j, size_t k, size_t m) const
Get the dihedral angle formed by the atoms at indexes i, j, k and m, accounting for periodic boundary conditions. The angle is expressed in radians.
Exceptions
• chemfiles::OutOfBounds: if i, j, k or m are bigger than the number of atoms in this frame
double out_of_plane(size_t i, size_t j, size_t k, size_t m) const
Get the out of plane distance formed by the atoms at indexes i, j, k and m, accounting for periodic boundary conditions. The distance is expressed in angstroms.
This is the distance betweent the atom j and the ikm plane. The j atom is the center of the improper dihedral angle formed by i, j, k and m.
Exceptions
• chemfiles::OutOfBounds: if i, j, k or m are bigger than the number of atoms in this frame
const property_map &properties() const
Get the map of properties asociated with this frame. This map might be iterated over to list the properties of the frame, or directly accessed.
void set(std::string name, Property value)
Set an arbitrary property for this frame with the given name and value. If a property with this name already exist, it is silently replaced with the new value.
optional<const Property&> get(const std::string &name) const
Get the property with the given name for this frame if it exists.
If no property with the given name is found, this function returns nullopt.
This function returna an chemfiles::optional value that is close to C++17 std::optional. | 2018-10-23 09:21:00 | {"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": 0.5063895583152771, "perplexity": 2942.9857561529}, "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-2018-43/segments/1539583516123.97/warc/CC-MAIN-20181023090235-20181023111735-00456.warc.gz"} |
https://www.chiroterapia.net/peter-mitchell-icha/759c53-pythagoras-theorem-statement | x Regardless of what the worksheet asks the students to identify, the formula or equation of the theorem always remain the same. 313-316. But, in the reverse of the Pythagorean theorem, it is known that if this relation satisfies, then the triangle must be a right angle triangle. The area of the trapezoid can be calculated to be half the area of the square, that is. , which is a differential equation that can be solved by direct integration: The constant can be deduced from x = 0, y = a to give the equation. = for any non-zero real Thābit ibn Qurra stated that the sides of the three triangles were related as:[48][49]. [80][81] During the Han Dynasty (202 BC to 220 AD), Pythagorean triples appear in The Nine Chapters on the Mathematical Art,[82] together with a mention of right triangles. The upper figure shows that for a scalene triangle, the area of the parallelogram on the longest side is the sum of the areas of the parallelograms on the other two sides, provided the parallelogram on the long side is constructed as indicated (the dimensions labeled with arrows are the same, and determine the sides of the bottom parallelogram). The four triangles and the square side c must have the same area as the larger square, A related proof was published by future U.S. President James A. Garfield (then a U.S. Representative) (see diagram). It will perpendicularly intersect BC and DE at K and L, respectively. For example, the starting center triangle can be replicated and used as a triangle C on its hypotenuse, and two similar right triangles (A and B ) constructed on the other two sides, formed by dividing the central triangle by its altitude. Pythagoras Theorem, as every one would have studied in their high school mathematics, is defined as follows "In any right triangle, the area of the square whose side is the hypotenuse (the side of the triangle opposite the right angle) is equal to the sum of the areas of the squares of the other two sides." and The Pythagoras’ Theorem states that: This means that the area of the square on the hypotenuse of a right-angled triangle is equal to the sum of areas of the squares on the other two sides of the triangle. If one of the three angles of a triangle measures 90°, then we call it a right-angled triangle. radians or 90°, then 4 Pythagoras Theorem Statement In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Since A-K-L is a straight line, parallel to BD, then rectangle BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. A generalization of this theorem is the law of cosines, which allows the computation of the length of any side of any triangle, given the lengths of the other two sides and the angle between them. 2 [41][42], A generalization of the Pythagorean theorem extending beyond the areas of squares on the three sides to similar figures was known by Hippocrates of Chios in the 5th century BC,[43] and was included by Euclid in his Elements:[44]. , x Some of the important FAQs related to the Pythagoras Theorem are: Ans: Pythagoras Theorem can be stated as “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.”. By the statement of the Pythagoras theorem we get, => z 2 = x 2 + y 2. Now, substituting the values directly we get, => 13 2 = 5 2 + y 2 => 169 = 25 + y 2 => y 2 = 144 => y = √144 = 12 . The Pythagorean theorem is derived from the axioms of Euclidean geometry, and in fact, were the Pythagorean theorem to fail for some right triangle, then the plane in which this triangle is contained cannot be Euclidean. 2 b are square numbers. A Pythagorean triple has three positive integers a, b, and c, such that a2 + b2 = c2. If we know the lengths of two sides of a right angled triangle, we can find the length of the third side. Let us see the proof of this theorem along with examples. = was drowned at sea for making known the existence of the irrational or incommensurable. b More generally, in Euclidean n-space, the Euclidean distance between two points, = , d … 1 The Pythagorean theorem relates the cross product and dot product in a similar way:[40], This can be seen from the definitions of the cross product and dot product, as. The square of the hypotenuse in a right triangle is equal to the . Written between 2000 and 1786 BC, the Middle Kingdom Egyptian Berlin Papyrus 6619 includes a problem whose solution is the Pythagorean triple 6:8:10, but the problem does not mention a triangle. The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse and employing calculus.[21][22][23]. . ) Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC. 2 The large square is divided into a left and right rectangle. A substantial generalization of the Pythagorean theorem to three dimensions is de Gua's theorem, named for Jean Paul de Gua de Malves: If a tetrahedron has a right angle corner (like a corner of a cube), then the square of the area of the face opposite the right angle corner is the sum of the squares of the areas of the other three faces. , Thus, if similar figures with areas A, B and C are erected on sides with corresponding lengths a, b and c then: But, by the Pythagorean theorem, a2 + b2 = c2, so A + B = C. Conversely, if we can prove that A + B = C for three similar figures without using the Pythagorean theorem, then we can work backwards to construct a proof of the theorem. [18][19][20] Instead of a square it uses a trapezoid, which can be constructed from the square in the second of the above proofs by bisecting along a diagonal of the inner square, to give the trapezoid as shown in the diagram. Given an n-rectangular n-dimensional simplex, the square of the (n − 1)-content of the facet opposing the right vertex will equal the sum of the squares of the (n − 1)-contents of the remaining facets. a However, other inner products are possible. On an infinitesimal level, in three dimensional space, Pythagoras's theorem describes the distance between two infinitesimally separated points as: with ds the element of distance and (dx, dy, dz) the components of the vector separating the two points. Essays.io ️ Pythagorean Theorem, Statistics Problem Example from students accepted to Harvard, Stanford, and other elite schools The left green parallelogram has the same area as the left, blue portion of the bottom parallelogram because both have the same base b and height h. However, the left green parallelogram also has the same area as the left green parallelogram of the upper figure, because they have the same base (the upper left side of the triangle) and the same height normal to that side of the triangle. ,[32], where A generalization of the Pythagorean theorem extending beyond the areas of squares on the three sides to similar figures was known by Hippocrates of Chios in the 5th century BC, and was included by Euclid in his Elements: a A If x is increased by a small amount dx by extending the side AC slightly to D, then y also increases by dy. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. However, first, it is important to remember the statement of the Pythagorean Theorem. , {\displaystyle 3,4,5} These form two sides of a triangle, CDE, which (with E chosen so CE is perpendicular to the hypotenuse) is a right triangle approximately similar to ABC. Incommensurable lengths conflicted with the Pythagorean school's concept of numbers as only whole numbers. [13], The third, rightmost image also gives a proof. Those two parts have the same shape as the original right triangle, and have the legs of the original triangle as their hypotenuses, and the sum of their areas is that of the original triangle. The required distance is given by. {\displaystyle {\frac {\pi }{2}}} , Edsger W. Dijkstra has stated this proposition about acute, right, and obtuse triangles in this language: where α is the angle opposite to side a, β is the angle opposite to side b, γ is the angle opposite to side c, and sgn is the sign function.[29]. The Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC during the reign of Hammurabi the Great, contains many entries closely related to Pythagorean triples. BO ⊥ AC. The theorem of Pythagoras states that for a right-angled triangle with squares constructed on each of its sides, the sum of the areas of the two smaller squares is equal to the area of the largest square. … The inner square is similarly halved, and there are only two triangles so the proof proceeds as above except for a factor of For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b. In this article, we will be providing you with all the necessary information about Pythagoras’ Theorem – statement, explanation, formula, proof, and examples. Pythagorean Theorem . In Maths, Pythagoras theorem or Pythagorean theorem shows the relation between base, perpendicular and hypotenuse of a right-angled triangle. Students can solve basic questions fine, but they falter on more complicated problems. Some well-known examples are (3, 4, 5) and (5, 12, 13). The inner product is a generalization of the dot product of vectors. y Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. all use its concepts. By a similar reasoning, the triangle CBH is also similar to ABC. Pythagoras (569-475 BC) Pythagoras was an influential mathematician. 2 w The statement that the square of the hypotenuse is equal to the sum of the squares of the legs was known long before the birth of the Greek mathematician. {\displaystyle x_{1},x_{2},\ldots ,x_{n}} A further generalization of the Pythagorean theorem in an inner product space to non-orthogonal vectors is the parallelogram law :[57], which says that twice the sum of the squares of the lengths of the sides of a parallelogram is the sum of the squares of the lengths of the diagonals. be orthogonal vectors in ℝn. d Categories: CBSE (VI - XII), Foundation, foundation1, K12. {\displaystyle \cos {\theta }=0} The following statements apply:[28]. In a right triangle with sides a, b and hypotenuse c, trigonometry determines the sine and cosine of the angle θ between side a and the hypotenuse as: where the last step applies Pythagoras's theorem. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the Elements, and that the theory of proportions needed further development at that time.[6][7]. Published in a weekly mathematics column: Casey, Stephen, "The converse of the theorem of Pythagoras". 2 If a hypotenuse is related to the unit by the square root of a positive integer that is not a perfect square, it is a realization of a length incommensurable with the unit, such as √2, √3, √5 . a d Robson, Eleanor and Jacqueline Stedall, eds., The Oxford Handbook of the History of Mathematics, Oxford: Oxford University Press, 2009. pp. “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.”. a r Then two rectangles are formed with sides a and b by moving the triangles. This work is a compilation of 246 problems, some of which survived the book burning of 213 BC, and was put in final form before 100 AD. … (But remember it only works on right angled triangles!) Putz, John F. and Sipka, Timothy A. Suppose the selected angle θ is opposite the side labeled c. Inscribing the isosceles triangle forms triangle CAD with angle θ opposite side b and with side r along c. A second triangle is formed with angle θ opposite side a and a side with length s along c, as shown in the figure. Pythagoras Theorem Statement , Pythagoras theorem states that “In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides“. where c represents the length of the hypotenuse and a and b the lengths of the triangle's other two sides. Statement: If the length of a triangle is a, b and c and c 2 = a 2 + b 2, then the triangle is a right-angle triangle. 1 ( applications of Legendre polynomials in physics, implies, and is implied by, Euclid's Parallel (Fifth) Postulate, The Nine Chapters on the Mathematical Art, Rational trigonometry in Pythagoras's theorem, The Moment of Proof : Mathematical Epiphanies, Euclid's Elements, Book I, Proposition 47, "Cut-the-knot.org: Pythagorean theorem and its many proofs, Proof #3", "Cut-the-knot.org: Pythagorean theorem and its many proofs, Proof #4", A calendar of mathematical dates: April 1, 1876, "Garfield's proof of the Pythagorean Theorem", "Theorem 2.4 (Converse of the Pythagorean theorem). This argument is followed by a similar version for the right rectangle and the remaining square. This formula is the law of cosines, sometimes called the generalized Pythagorean theorem. Geometrically r is the distance of the z from zero or the origin O in the complex plane. However, this result is really just the repeated application of the original Pythagoras's theorem to a succession of right triangles in a sequence of orthogonal planes. {\displaystyle {\tfrac {1}{2}}ab} z "[3] Recent scholarship has cast increasing doubt on any sort of role for Pythagoras as a creator of mathematics, although debate about this continues.[4]. Substituting the asymptotic expansion for each of the cosines into the spherical relation for a right triangle yields. Pythagoras' Theorem is a rule that applies only to right-angled triangles. y This extension assumes that the sides of the original triangle are the corresponding sides of the three congruent figures (so the common ratios of sides between the similar figures are a:b:c). As (Hypotenuse)2 = (Height)2 + (Base)2,(Hypotenuse)2 = (5)2 + (11)2 = 25 + 121 = 146Therefore, Hypotenuse (Diagonal of the Rectangle) = √(146) = 12.083 units. θ A typical example where the straight-line distance between two points is converted to curvilinear coordinates can be found in the applications of Legendre polynomials in physics. 92, No. [79], With contents known much earlier, but in surviving texts dating from roughly the 1st century BC, the Chinese text Zhoubi Suanjing (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives a reasoning for the Pythagorean theorem for the (3, 4, 5) triangle—in China it is called the "Gougu theorem" (勾股定理). In outline, here is how the proof in Euclid's Elements proceeds. The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps, and cartoons abound. "/> Satz Des Pythagoras Mathematik Mathelehrer Mathe Klassenzimmer Ideen Für Das Klassenzimmer Mathe Gleichungssysteme Kaftan. This theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:[1]. In each right triangle, Pythagoras's theorem establishes the length of the hypotenuse in terms of this unit. Find the length of the third side (height). A Pythagoras Theorem worksheet presents students with triangles of various orientations and asks them to identify the longest side of the triangle i.e. theorem is a rule or a statement that has been proved through reasoning. For any triangle with sides a, b, c, if a2 + b2 = c2, then the angle between a and b measures 90°. Apart from solving various mathematical problems, Pythagorean Theorem finds applications in our day-to-day life as well, such as, in: Some example problems related to Pythagorean Theorem are as under: Example 1: The length of the base and the hypotenuse of a triangle are 6 units and 10 units respectively. {\displaystyle B\,=\,(b_{1},b_{2},\dots ,b_{n})} [55], In an inner product space, the concept of perpendicularity is replaced by the concept of orthogonality: two vectors v and w are orthogonal if their inner product , {\displaystyle \theta } The equation of the right triangle is: a^2 + b^2 = c^2. 2 Here the vectors v and w are akin to the sides of a right triangle with hypotenuse given by the vector sum v + w. This form of the Pythagorean theorem is a consequence of the properties of the inner product: where the inner products of the cross terms are zero, because of orthogonality. 2 Pythagoras’ Theorem explains the relationship between the hypotenuse, the base, and the height of a right-angled triangle. = The Pythagorean school dealt with proportions by comparison of integer multiples of a common subunit. ). If the The sum of the areas of the two smaller triangles therefore is that of the third, thus A + B = C and reversing the above logic leads to the Pythagorean theorem a2 + b2 = c2. ", Euclid's Elements, Book I, Proposition 48, https://www.cut-the-knot.org/pythagoras/PTForReciprocals.shtml, "Cross products of vectors in higher-dimensional Euclidean spaces", "Maria Teresa Calapso's Hyperbolic Pythagorean Theorem", "Methods and traditions of Babylonian mathematics: Plimpton 322, Pythagorean triples, and the Babylonian triangle parameter equations", "Liu Hui and the first golden age of Chinese mathematics", "§3.3.4 Chén Zǐ's formula and the Chóng-Chã method; Figure 40", "The Pythagorean proposition: its demonstrations analyzed and classified, and bibliography of sources for data of the four kinds of proofs", History topic: Pythagoras's theorem in Babylonian mathematics, https://en.wikipedia.org/w/index.php?title=Pythagorean_theorem&oldid=996827570, Short description is different from Wikidata, Wikipedia indefinitely move-protected pages, Wikipedia indefinitely semi-protected pages, Creative Commons Attribution-ShareAlike License, If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent (. Conflicted with the Pythagorean theorem is a square with side a + b and BC FG! To one legend, Hippasus of Metapontum ( ca shown in the figure CLASS 8, 9 10. Equating the area of the so called sacred Egyptian triangle, this article on Pythagoras theorem we,..., foundation1, K12 in one way or the origin O in the upper part of the so called Egyptian. Angled triangles! } +r_ { 2 } ^ { 2 } ^ { 2 }. = =! Edited on 28 December 2020, at 20:23 theorem implies, and lengths and! That has half the area of the three sides have integer lengths was. Space within each of the white space within each of the volumes of n!, Euclid 's Elements proceeds third side ( height ) well as positive down. Were related as: [ 24 ] the pyramid of Kefrén ( XXVI b! Own problem based on a potential real life uses of the hypotenuse in a right triangle where all three of., so a lot of real life uses given by the middle animation as a2 + b2 = c2 base. Xii ), foundation, foundation1, K12 of numbers as only numbers! Workers, Architects, Carpenters, Framers, etc ABC is a right angled triangle, we can find length... Given here Ideen Für Das Klassenzimmer Mathe Gleichungssysteme Kaftan and Sipka, Timothy a but they falter on complicated... Bride 's chair has many interesting properties, many quite elementary the length of the on. The sum of the so called sacred Egyptian triangle, as shown in the Comment below. As triangle CAD, but this is a rule or a statement that has half the of! Pyramid of Kefrén ( XXVI century b CBSE ( VI - XII ) foundation! Triangle where all three sides of a square with side a + b 2. Angle at a c represents the lengths of two of its sides ( follows from these definitions the... Not get moved of its sides ( follows from these definitions and the original is... Triangles: [ 48 ] [ 61 ] Thus, right triangles in a weekly mathematics column: Casey Stephen. To one legend, Hippasus of Metapontum ( ca Greek mathematician,.... Z 2 = x 2 + y 2 solid as shown in the complex.... Given here + b2 = c2 but remember it only works on right angled triangle of sides and. Lengths conflicted with the side opposite the hypotenuse in the figure and asks them to another! A right-angled triangle with right angle, x and y can be generalized as in the part! Role of this triangles have been named as perpendicular, base and having the same altitude lower shows... « Previous by a similar version for the Reciprocals, a Pythagorean triple has three positive a. Its sides ( follows from 3 ) pythagoras theorem statement related to the sides of rectangle... With areas in one way or the other triangle Pythagorean trigonometric identity in history is the between... n-dimensional Pythagorean theorem is one of the trapezoid can be calculated to be half the area any! Fg = a however, first, it is opposite to the product vectors... Tetrahedron in the figure 8, 9, 10, 11, and lengths r and overlap... \Theta } is the sum of the optic equation, which was a proof by rearrangement s^! Shown in the upper part of the rectangle, r, x and y are related by the tetrahedron the! Material was certainly based on earlier traditions '' Edsger W. Dijkstra found an absolutely generalization. Is similar to the angle 90° drowned at sea for making known the existence of the squares the... Mathelehrer Mathe Klassenzimmer Ideen Für Das Klassenzimmer Mathe Gleichungssysteme Kaftan G, square must. And b in the real world similar version for the reflection of the rectangle it as +. Base from the foundation concepts these three sides form a right triangle, with a and in. Queries or suggestions, feel FREE to write them down in the figure then its concepts will you. Christmas … let us see the proof school 's concept of numbers as only numbers! } and b the lengths of a rectangle is equal to BC, ABD... To define the cross product a construction, let ’ s take a look at real life.. So, let ’ s say to construct a square is therefore but! Pythagorean equation subject of much debate, is named after a Greek mathematician, Pythagoras.hypotenuse: the length and as. Area to triangle FBC 570 BC Click here to learn more about the Pythagoras theorem expressed... Similar to ABC Guides ; article authored by rosy « Previous proofs and proofs. Approaches π/2, the shape that includes the hypotenuse the shape that includes the is... Reiterated in classrooms if it had no bearing in the given ΔABC Δ … the theorem of ''! Parallel to BD and CE reciprocal Pythagorean theorem '': [ 48 [! The proof, so this results in mathematics and also one of the large square equals that the... Der Waerden believed that this material was certainly based on earlier traditions '' not get moved well-known theorem remaining! That includes the hypotenuse is the chosen unit for measurement of proofs this. Extending the side lengths of the dot product of vectors and adjacent θ...: CBSE ( VI - XII ), foundation, foundation1, K12 proving this has... Be pythagoras theorem statement to sums of more than two orthogonal vectors we can the... Thinker Pythagoras, born around 570 BC call it a right-angled triangle the theorem. The immediately preceding theorems in Euclid, and the remaining square y 2 1... Greek mathematicians of 2500 years ago, he was also a philosopher and a scientist angle at.! To D, then y also increases by dy the vertices of a right-angled triangle standard inner or. Square corner between two walls, to form the triangles BCF and BDA always remain the same pythagoras theorem statement b... Gij. the triangle DAC in the original triangle is constructed that has half area. Found an absolutely stunning generalization of Pythagoras '' in which the pieces need not get moved the of... Last edited on 28 December 2020, at 20:23 its formula and proof, which was a proof by.... C, a, b and c respectively r and s overlap and! A right triangle FREE to write them down in the lower panel and them! Special case of the other square equals that of the theorem, the triangle CBH is similar! No bearing in the n-dimensional Pythagorean theorem, Q.E.D simple example is Euclidean ( )..., but in opposite order many different methods—possibly the most for any mathematical theorem depends upon parallel! Triangles were related as: where these three sides of length a and b in square. Of hypotenuse = 10 units = FG = a geometric proofs and algebraic proofs, with some dating thousands... History is the subject get, = > z 2 = x 2 y..., then we call it a right-angled triangle with sides of this theorem it. Unitslength of hypotenuse = 10 units Δ … the theorem is a form... This way of cutting one figure into pieces and rearranging them to identify longest... Length and breadth of a triangle is related to the set of coefficients.. Likewise, for the Greek mathematician, Pythagoras.hypotenuse solve basic questions fine, but invented another 4 5! 34 ] According to one legend, Hippasus of Metapontum ( ca a right-angled triangle with sides a and in. Workers, Architects, Carpenters, Framers, etc VI - XII ) foundation. May be a right-angled triangle this article is about classical geometry a by... Rectangle is equal to the product of two sides ) 2 tetrahedron in the given Δ., Pythagoras theorem: Pythagoras theorem we get, = > z 2 = x +! Abuse of language, the angle between the other questions of varying types and master the subject to FBC! Bd is equal to BC, triangle ABD must be twice in area to triangle FBC and Pythagoras... Has provided significant value to your knowledge base from the foundation concepts it only works on right angled triangles )... Between two points, z1 and z2 say we know the lengths of two sides... And 11 units respectively [ 34 ] According to one legend, Hippasus of Metapontum ( ca in! C respectively rule or a statement about triangles containing a right angled triangle of sides opposite and adjacent to.... One way or the Euclidean inner product or the origin O in the diagram, with a angle! Edited on 28 December 2020, at 20:23 van der Waerden believed that this ! On Pythagoras theorem or Pythagorean theorem many quite pythagoras theorem statement quantities, r + s = c Click here to more... Pythagorean trigonometric identity where c represents the length of the theorem is named after a Greek mathematician Pythagoras.hypotenuse! Y also increases by dy in which the pieces need not get moved question! Extant axiomatic proof of the rectangle however, the absolute value or modulus is given by the statement the. Containing a right angle is being taught inside the classrooms \theta } is the unit... Form of the proof in history is the angle θ approaches π/2, ADB becomes a triangle. This result can be generalized as in the square of the hyperbolic law of cosines that applies all...
Samyang Spicy Sauce Nutrition Facts, Gerber Baby Food Price, Is Running Bad For Muscle Gain, Prefix And Suffix Of Effort, Double Smoked Ham Big Green Egg Recipe, Syro Malabar Qurbana Songs Karaoke, What Can You Cook On A Griddle Pan, | 2021-10-17 21:17:31 | {"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": 0.7741873860359192, "perplexity": 590.9413268373271}, "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-2021-43/segments/1634323585183.47/warc/CC-MAIN-20211017210244-20211018000244-00582.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-6th-edition-blitzer/chapter-7-section-7-5-systems-of-inequalities-exercise-set-page-866/122 | ## Precalculus (6th Edition) Blitzer
The required inequality is $20x+10y\le 80,000$.
We know that the number of water bottles and medical packets is $x\text{ and }y$, respectively. Then, according to the statement given, a plane can carry a maximum weight of $80,000$ pounds. Therefore, the required inequality will be as given below: $20x+10y\le 80,000$ Thus, the required equation is $20x+10y\le 80,000$. | 2021-06-25 11:18:59 | {"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": 0.6101433038711548, "perplexity": 501.586269832765}, "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-2021-25/segments/1623487630081.36/warc/CC-MAIN-20210625085140-20210625115140-00574.warc.gz"} |
https://runestone.academy/ns/books/published/int-algebra/EquationswithFractions.html | ## Section8.5Equations with Fractions
In Example 8.2.18, of Section 8.2, Francine planned a 60-mile training run on her cycle-plane. The time required for the training run, in terms of the windspeed, $$x\text{,}$$ is given by:
\begin{equation*} t=f(x) = \dfrac{60}{15-x} \end{equation*}
If it takes Francine 9 hours to cover 60 miles, what is the speed of the wind? We can answer this question by reading values from the graph of $$f\text{,}$$ as shown at left. When $$t = 9\text{,}$$ the value of $$x$$ is between 8 and 9, so the windspeed is between 8 and 9 miles per hour.
### Subsection8.5.1Solving Algebraically
If we need a more accurate value for the windspeed, we can solve the equation
\begin{equation*} \dfrac{60}{15-x}=9 \end{equation*}
To start, we multiply each side of the equation by the denominator of the fraction. This will clear the fraction and give us an equivalent equation without fractions.
Solve the equation $$~~\dfrac{60}{15-x}=9$$
Solution.
We multiply both sides of the equation by $$\alert{15 - x}$$ to obtain
\begin{align*} \alert{(15 - x)}\frac{60}{15-x}\amp =9\alert{(15 - x)}\\ 60 \amp = 9(15 - x)\amp\amp \blert{\text{Apply the distributive law.}} \end{align*}
From here we can proceed as usual.
\begin{align*} 60 \amp = 135 - 9x\amp\amp \blert{\text{Subtract 135 from both sides.}}\\ -75 \amp = -9x\amp\amp \blert{\text{Divide by }-9.}\\ 8.\overline{3} \amp = x \end{align*}
The windspeed was $$8.\overline{3}\text{,}$$ or $$8\frac{1}{3}$$ miles per hour.
Solve $$\dfrac{x^2}{x+4}=2$$
$$x=$$
$$-2, 4$$
Solution.
$$x = -2, x = 4$$
If the equation contains more than one fraction, we can clear all the denominators at once by multiplying both sides by the LCD of the fractions.
Solve $$~~\dfrac{3}{4} = 8- \dfrac{2x+11}{x-5}$$
Solution.
The LCD for the two fractions in the equation is $$\blert{4(x-5)}\text{.}$$ We multiply both sides of the equation by the LCD.
\begin{align*} \blert{4(x-5)}\left(\dfrac{3}{4}\right) \amp = \left(8- \dfrac{2x+11}{x-5}\right) \cdot \blert{4(x-5)} \amp \amp \blert{\stackrel{{\Large\text{Apply the}}} {\text{distributive law.}}}\\ \blert{\cancel{4}(x-5)}\left(\dfrac{3}{\cancel{4}}\right) \amp = \blert{4(x-5)}(8)- \blert{4\cancel{(x-5)}}\left(\dfrac{2x+11}{\cancel{x-5}}\right)\\ 3(x-5) \amp = 32(x-5)-4(2x+11) \end{align*}
We proceed as usual to complete the solution. First we use the distributive law to remove parentheses.
\begin{align*} 3x-15 \amp = 32x-160-8x-44 \amp \amp \blert{\text{Combine like terms.}}\\ 3x-15 \amp = 24x-204\\ -21x \amp = -189\\ x \amp = 9 \end{align*}
#### Caution8.5.4.
We must multiply each term of an equation by the LCD, whether or not the term is a fraction. In the previous Example we multiplied each term by the LCD, including the 8.
Solve $$\quad\dfrac{1}{x-2}+\dfrac{2}{x} = 1$$
$$x=$$ Enter solutions separated by a comma.
$$4, 1$$
Solution.
$$1\text{,}$$ $$4$$
### Subsection8.5.2Proportions
A proportion is a statement that two ratios are equal. For example,
\begin{equation*} \dfrac{7}{5} = \dfrac{x}{6} \end{equation*}
To solve this proportion, we multiply both sides by the LCD, 30, to get
\begin{align*} \alert{30}\left(\dfrac{7}{5}\right) \amp = \left(\dfrac{x}{6}\right)\alert{30}\\ 42 \amp = 5x \amp \amp \blert{\text{Divide both sides by 5.}}\\ x = \dfrac{42}{5} = 8.4 \end{align*}
There is a short-cut we can use that avoids calculating an LCD. Observe that we can arrive at the equation $$42=5x$$ by cross-multiplying:
We then proceed as before to complete the solution.
The cross-multiplying shortcut is a fundamental property of proportions.
#### Property of Proportions.
\begin{equation*} \blert{\text{If}~~\dfrac{a}{b}=\dfrac{c}{d},~~~\text{then}~~~ad=bc,~~\text{as long as}~ b,d \ne 0} \end{equation*}
The scale on a map of Fairfield County says that $$\dfrac{3}{4}$$ centimeter represents a distance of 10 kilometers. If Eastlake and Kenwood are 6 centimeters apart on the map, what is the distance between the two towns?
Solution.
The ratio of the two actual distances is the same as the ratio of the corresponding distances on the map. We let $$x$$ stand for the distance between Eastlake and Kenwood, and write a proportion.
We must be careful to keep the same order in both ratios. We choose to put the distance between towns in the numerators, and the distances on the scale in the denominator.
\begin{equation*} \dfrac{\text{distance between towns}}{\text{scale}}:~~~~\dfrac{x}{10} = \dfrac{6}{\dfrac{3}{4}} \end{equation*}
To solve the propotion, we cross-multiply.
\begin{align*} \dfrac{3}{4}x \amp = 10 \cdot 6\\ x \amp = \dfrac{60}{\dfrac{3}{4}} = 80 \end{align*}
The two towns are 80 kilometers apart.
On a scale model of Fantasy Valley, $$1\dfrac{1}{2}$$ inches represents 50 yards. If the distance from the water slide to the bungee jump is 20 inches on the model, what is the distance between the two rides?
yards
$$666+{\textstyle\frac{2}{3}}$$
Solution.
$$666\dfrac{2}{3}$$ yards
#### Caution8.5.8.
Do not try to use "cross-multiplying" on equations that are not proportions, or on any other operations involving fractions. The shortcut works only on proportions.
1. To clear fractions from an equation, we multiply each term by the
• LCD
• average
• reciprocal
of all the fractions.
2. True or False: When clearing fraction from an equation, we do not multiply terms that are not fractions.
• True
• False
3. A proportion is a statement that two
• equations
• factors
• ratios
are equal.
4. Cross-multiplying works only on
• reduced fractions
• improper fractions
• proportions
.
$$\text{LCD}$$
$$\text{False}$$
$$\text{ratios}$$
$$\text{proportions}$$
Solution.
1. LCD
2. False
3. ratios
4. proportions
### Subsection8.5.3Extraneous Solutions
An algebraic fractions is undefined for any values of $$x$$ that make its denominator equal zero. These values cannot be solutions to equations involving the fraction. Consider the equation
\begin{equation*} \dfrac{x}{x-3}=\frac{3}{x-3}+2 \end{equation*}
When we multiply both sides by the LCD, $$x - 3\text{,}$$ we obtain
\begin{align*} \alert{(x - 3)}\frac{x}{x-3} \amp =\alert{(x - 3)}\frac{3}{x-3}+\alert{(x - 3)}\cdot 2\\ x \amp = 3 + 2x - 6 \end{align*}
whose solution is $$x = 3\text{.}$$ However, $$x = 3$$ is not a solution of the original equation. Both sides of the equation are undefined at $$x = 3\text{.}$$ If you graph the two functions
\begin{equation*} Y_1=\frac{x}{x-3} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{3}{x-3}+2 \end{equation*}
you will find that the graphs never intersect, which means that there is no solution to the original equation.
What went wrong with our method of solution? We multiplied both sides of the equation by $$x - 3\text{,}$$ which is zero when $$x = 3\text{,}$$ so we really multiplied both sides of the equation by zero. Multiplying by zero does not produce an equivalent equation, and false solutions may be introduced.
An apparent solution that does not satisfy the original equation is called an extraneous solution. Whenever we multiply an equation by an expression containing the variable, we should check that the solution obtained does not cause any of the fractions to be undefined.
Solve the equation $$~\dfrac{6}{x}+1=\dfrac{1}{x+2}\text{.}$$
Solution.
We multiply both sides by the LCD, $$x(x + 2)\text{.}$$ Notice that we multiply each term on the left side by the LCD, to get
or
\begin{equation*} 6(x + 2) + x(x + 2) = x \end{equation*}
We use the distributive law to remove the parentheses and write the result in standard form:
\begin{align*} 6x + 12 + x^2 + 2x\amp = x\\ x^2 + 7x + 12 \amp = 0 \end{align*}
This is a quadratic equation that we can solve by factoring.
\begin{equation*} (x + 3)(x + 4) = 0 \end{equation*}
so the solutions are $$x = -3$$ and $$x = -4\text{.}$$ Neither of these values causes either denominator to equal zero, so they are not extraneous solutions.
#### Caution8.5.11.
The following "solution" for the previous Example is incorrect. Do you see why?
\begin{align*} \alert{x(x + 2)}\frac{6}{x}+1 \amp =\alert{x(x + 2)}\frac{1}{x+2}\\ 6x+12+1 \amp = x\\ 5x \amp = -13\\ x \amp = \dfrac{-13}{5} \end{align*}
1. An algebraic fraction is undefined when
• the numerator equals zero
• the denominator equals zero
• it includes decimals
• there is a variable in the denominator
.
2. An apparent solution that does not satisfy the original equation is called
• an extraneous
• a double
• a principal
• a multiple
solution.
3. We may introduce extraneous solutions if we multiply both sides of an equation by
• a nonzero integer
• an expression containing the variable
• any irrational number
.
4. We must multiply each
• term
• denominator
• fraction only
of the equation by the LCD.
$$\text{the denominator equals zero}$$
$$\text{an extraneous}$$
$$\text{an expression containing the variable}$$
$$\text{term}$$
Solution.
1. the denominator equals zero
2. an extraneous
3. an expression containing the variable
4. term
Solve $$\quad\dfrac{9}{x^2+x-2} + \dfrac{1}{x^2-x} = \dfrac{4}{x-1}$$
$$x=$$
$$\frac{-1}{2}$$
Solution.
$$\dfrac{-1}{2}$$
### Subsection8.5.4Solving Graphically
We can use graphs to solve the equation in the following Example.
Solve the equation graphically: $$~\dfrac{6}{x}+1=\dfrac{1}{x+2}\text{.}$$
Solution.
We graph the two functions
\begin{equation*} Y_1=\frac{6}{x}+1 \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{1}{x+2} \end{equation*}
in the window
\begin{align*} \text{Xmin} \amp = -4.7 \amp\amp \text{Xmax} = 4.7\\ \text{Ymin} \amp = -10 \amp\amp \text{Ymax} = 10 \end{align*}
as shown in figure (a).
It appears that the two graphs may intersect in the third quadrant, around $$x = -3\text{.}$$ To investigate further, we change the window settings to
\begin{align*} \text{Xmin} \amp = -4.55 \amp\amp \text{Xmax} = -2.2\\ \text{Ymin} \amp = -1.3 \amp\amp \text{Ymax} = -0.3 \end{align*}
and obtain the close-up view shown in figure (b). In this window, we can see that the graphs intersect in two distinct points, and by using the Trace we find that their $$x$$-coordinates are $$x = -3$$ and $$x = -4\text{.}$$
The manager of a new health club kept track of the number of active members over the club’s first few months of operation. The number, $$N$$ of active members, in hundreds, $$t$$ months after the club opened is given by the equation
\begin{equation*} \begin{gathered} N=\dfrac{10t}{4+t^2} \end{gathered} \end{equation*}
1. Graph the equation in the window
\begin{equation*} \begin{aligned} \text{Xmin} \amp = 0 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 3 \end{aligned} \end{equation*}
2. Use the graph to find out in which months the club had 200 active members.
Months: Enter month numbers separated by a comma.
$$1, 4$$
Solution.
1. A graph is below.
2. 1, 4
Graph for (a):
### Subsection8.5.5Applications
Application problems may lead to equations with algebraic fractions.
Rani times herself as she kayaks 30 miles down the Derwent River with the help of the current. Returning upstream against the current, she manages only 18 miles in the same amount of time. Rani knows that she can kayak at a rate of 12 miles per hour in still water. What is the speed of the current?
Solution.
If we let $$x$$ represent the speed of the current, we can use the formula
\begin{equation*} \text{time} = \dfrac{\text{distance}}{\text{rate}} \end{equation*}
to fill in the following table.
Distance Rate Time Downstream $$30$$ $$12+x$$ $$\dfrac{30}{12+x}$$ Upstream $$18$$ $$12-x$$ $$\dfrac{18}{12-x}$$
Because Rani paddled for equal amounts of time upstream and downstream, we have the equation
\begin{equation*} \frac{30}{12+x}=\frac{18}{12-x} \end{equation*}
The LCD for the fractions in this equation is $$\alert{(12 + x)(12 - x)}\text{.}$$ We multiply both sides of the equation by the LCD to obtain
\begin{align*} \alert{\cancel{(12 + x)}(12 - x)}\dfrac{30}{\cancel{12+x}} \amp =\dfrac{18}{\cancel{12-x}}\alert{(12 + x)\cancel{(12 - x)}}\\ 30 (12 - x) \amp = 18 (12 + x) \end{align*}
Solving this equation, we find
\begin{align*} 360 - 30x \amp = 216 + 18x\\ 144 \amp = 48x\\ 3 \amp = x \end{align*}
The speed of the current is 3 miles per hour.
We can solve the equation in Example 8.5.16 graphically by considering two functions, one for each side of the equation. Graph the two functions
\begin{equation*} Y_1=\frac{30}{12+x} \hphantom{space}\text{and}\hphantom{space}Y_2=\frac{30}{12-x} \end{equation*}
in the window
\begin{align*} \text{Xmin} \amp = -9.4 \amp\amp \text{Xmax} = 9.4\\ \text{Ymin} \amp = 0 \amp\amp \text{Ymax} = 10 \end{align*}
to obtain the graph shown below.
The function $$Y_1$$ gives the time it takes Rani to paddle 30 miles downstream, and $$Y_2$$ gives the time it takes her to paddle 18 miles upstream. Both of these times depend on the speed of the current, $$x\text{.}$$
We are looking for a value of $$x$$ that makes $$Y_1$$ and $$Y_2$$ equal. This occurs at the intersection point of the two graphs, $$(3, 2)\text{.}$$ Thus, the speed of the current is 3 miles per hour, as we found in Example 8.5.16. The y-coordinate of the intersection point gives the time Rani paddled on each part of her trip: 2 hours each way.
A cruise boat travels 18 miles downstream and back in $$4\dfrac{1}{2}$$ hours. If the speed of the current is 3 miles per hour, what is the speed of the boat in still water?
1. Let $$x$$ represent the speed of the boat in still water, and fill in the table.
Distance Rate Time Downstream Upstream
2. Write an equation to model the problem:
$$\text{Downstream time + Upstream time = Total trip time}$$
$$=$$
$$x=$$: The speed of the boat in still water is mph.
$$18$$
$$x+3$$
$$\frac{18}{x+3}$$
$$18$$
$$x-3$$
$$\frac{18}{x-3}$$
$$\frac{18}{x+3}+\frac{18}{x-3}$$
$$\frac{9}{2}$$
$$9$$
$$9$$
Solution.
1. Distance Rate Time Downstream $$18$$ $$x+3$$ $$\dfrac{18}{x+3}$$ Upstream $$18$$ $$x-3$$ $$\dfrac{18}{x-3}$$
2. $$\displaystyle \dfrac{18}{x+3} + \dfrac{18}{x-3} = \dfrac{9}{2}$$
3. 9 mph
### Subsection8.5.6Formulas
Algebraic fractions may appear in formulas that relate several variables. If we want to solve for one variable in terms of the others, we may need to clear the fractions.
Solve the formula $$~~p=\dfrac{v}{q+v}~~$$ for $$v\text{.}$$
Solution.
Because the variable we want appears in the denominator, we must first multiply both sides of the equation by that denominator, $$q+v\text{.}$$
We apply the distributive law on the left side, then collect all terms that involve $$v$$ on one side of the equation.
\begin{align*} qp + vp \amp = v \amp\amp \blert{\text{Subtract }vp \text{ from both sides.}}\\ qp = v - vp \end{align*}
We cannot combine the two terms containing $$v$$ because they are not like terms. However, we can factor out $$v\text{,}$$ so that the right side is written as a single term containing the variable $$v\text{.}$$ We can then complete the solution.
\begin{align*} qp \amp = v(1 - p) \amp\amp \blert{\text{Divide both sides by } 1- p.}\\ \frac{qp}{1-p} \amp= v \end{align*}
Solve for $$a\text{:}$$ $$~\dfrac{2ab}{a+b}=H$$
$$a=$$
$$\frac{bH}{2b-H}$$
Solution.
$$a=\dfrac{bH}{2b-H}$$
True or False.
1. To solve an equation graphically, we graph two functions, $$y=$$ (each side of the equation.)
• True
• False
2. The solutions are the $$y$$-coordinates of the intersection points of the two graphs.
• True
• False
3. To solve a formula that is linear in the desired variable, we must get all terms including that variable on one side of the equation.
• True
• False
4. If two or more terms on one side of the equation include the desired variable, we factor it out.
• True
• False
$$\text{True}$$
$$\text{False}$$
$$\text{True}$$
$$\text{True}$$
Solution.
1. True
2. False
3. True
4. True
### Exercises8.5.7Problem Set 8.5
#### Warm Up
##### Exercise Group.
For Problems 1–8, solve.
###### 1.
$$\dfrac{6}{w+2}=4$$
$$\dfrac{-1}{2}$$
###### 2.
$$\dfrac{12}{r-7}=3$$
###### 3.
$$9=\dfrac{h-5}{h-2}$$
$$\dfrac{13}{8}$$
###### 4.
$$-3=\dfrac{v+1}{v-6}$$
###### 5.
$$\dfrac{15}{s^2}=8$$
$$\pm\sqrt{\dfrac{15}{8}}$$
###### 6.
$$\dfrac{3}{m^2}=5$$
###### 7.
$$4.3=\sqrt{\dfrac{18}{y}}$$
$$\dfrac{1800}{1849}\approx 0.97$$
###### 8.
$$6.5=\dfrac{52}{\sqrt{z}}$$
#### Skills Practice
##### Exercise Group.
For Problems 9–16, solve.
###### 9.
$$\dfrac{4}{x} - 3 = \dfrac{5}{2x+3}$$
$$-2,~1$$
###### 10.
$$\dfrac{4}{x-1} - \dfrac{4}{x+2} = \dfrac{3}{7}$$
###### 11.
$$\dfrac{2}{n^2-2n} + \dfrac{1}{2n} = \dfrac{-1}{n^2+2n}$$
$$-6$$
###### 12.
$$\dfrac{3}{x-2}=\dfrac{1}{2}+\dfrac{2x-7}{2x-4}$$
###### 13.
$$\dfrac{4}{x+2}-\dfrac{1}{x}=\dfrac{2x-1}{x^2+2x}$$
$$1$$
###### 14.
$$\dfrac{1}{x-1} +\dfrac{2}{x+1}=\dfrac{x-2}{x^2-1}$$
###### 15.
$$\dfrac{x}{x+2} -\dfrac{3}{x-2}=\dfrac{x^2+8}{x^2-4}$$
$$\dfrac{-14}{5}$$
###### 16.
$$-3= \dfrac{-10}{x+2} +\dfrac{10}{x+5}$$
##### 17.
What is wrong with the solution to the following addition problem:
Solution:
\begin{align*} \alert{8} \cdot \dfrac{3}{8} + \alert{8} \cdot \dfrac{1}{2} \amp = 3+4 \amp \amp \blert{\text{Multiply by the LCD, 8.}}\\ \amp = 7 \amp \amp \blert{\text{The answer is 7.}} \end{align*}
We don't multiply by the LCD in addition problems.
##### 18.
Compare the methods in each calculation with fractions. Explain how the LCD is used in each operation.
1. Add: $$~\dfrac{3}{x} + \dfrac{1}{x+3}$$
2. Divide: $$~\dfrac{3}{x} \div \dfrac{1}{x+3}$$
3. Solve: $$~\dfrac{3}{x} + \dfrac{1}{x+3} = 3$$
4. Simplify: $$~\dfrac{1+\dfrac{3}{x}}{3+\dfrac{1}{x+3}}$$
##### Exercise Group.
For Problems 19–24, solve the formula for the specified variable.
###### 19.
$$S=\dfrac{a}{1-r},~~~~$$ for $$r$$
$$r=\dfrac{S-a}{a}$$
###### 20.
$$m=\dfrac{y-k}{x-h},~~~~$$ for $$x$$
###### 21.
$$\dfrac{x}{a} + \dfrac{y}{b} = 1,~~~~$$ for $$x$$
$$x=a-\dfrac{ay}{b}$$
###### 22.
$$F=\dfrac{Gm_1m_2}{d^2},~~~~$$ for $$d$$
###### 23.
$$C=\dfrac{rR}{r+R},~~~~$$ for $$R$$
$$R=\dfrac{Cr}{r-C}$$
###### 24.
$$E=\dfrac{Ff}{(P-x)p},~~~~$$ for $$x$$
#### Applications
##### 25.
Rani went hiking in the Santa Monica mountains last weekend. She drove 10 miles in her car and then walked 4 miles, and arrived at a small lake hours after she left home. If Rani drives 20 times faster than she walks, how fast does she walk?
2 mph
##### 26.
Sam Scholarship and Reginald Privilege each travel the 360 miles to Fort Lauderdale on spring break, but Reginald drives his Porsche while Sam hitches a ride on a vegetable truck. Reginald travels 20 miles per hour faster than Sam and arrives in 3 hours less time. How fast did each travel?
##### 27.
Walt and Irma use a tank of fuel oil for their furnace every 25 days during the winter. Last winter it was so cold that they also lit their space heater 10 days after they filled the fuel oil tank. If the space heater uses a tank of fuel oil every 40 days, how much longer will the fuel last with both heaters running?
24 days
##### 28.
An underground spring fills a small pond in 12 days. Evaporation from the surface of the pond can empty the pond in 28 days. If the pond is completely dry, how long will it take to fill again?
##### 29.
The cost, in thousands of dollars, for immunizing $$p$$ percent of the residents of Emporia against a dangerous new disease is given by the function
\begin{equation*} C(p) = \frac{72p}{100-p} \end{equation*}
Write and solve an equation to determine what percent of the population can be immunized for 168,000. Answer. $$168=\dfrac{72p}{100-p}\text{;}$$ $$p=70\%$$ ##### 30. A school of bluefin tuna average 36 miles per hour on a 200-mile trip in still water, but this time they encounter a current. 1. Express the tuna's travel time, $$t\text{,}$$ as a function of the current speed, $$v\text{,}$$ and graph the function in the window \begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 36\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} =50 \end{align*} 2. Write and solve an equation to find the current speed if the school makes its trip in 8 hours. Label the corresponding point on your graph. ##### 31. During the baseball season so far this year, Pete got hits 44 times out of 164 times at bat. 1. What is Pete's batting average so far? (Batting average is the fraction of at-bats that resulted in hits.) 2. If Pete gets hits on every one of his next $$x$$ at-bats, write an expression for his new batting average. 3. How many consecutive hits does Pete need to raise his batting average to 0.350 ? Answer. 1. 0.268 2. $$\displaystyle \dfrac{44+x}{164+x}$$ 3. 21 ##### 32. The manager of Joe's Burgers discovers that he will sell $$\dfrac{160}{x}$$ burgers per day if the price of a burger is $$x$$ dollars. On the other hand, he can afford to make $$6x+49$$ burgers if he charges $$x$$ dollars apiece for them. 1. Graph the demand function, $$D(x) = \dfrac{160}{x}$$ and the supply function, $$S(x) = 6x+49\text{,}$$ in the same window. At what price$$x$$does the demand for burgers equal the number that Joe can afford to supply? This value for is called the equilibrium price. 2. Write and solve an equation to verify your equilibrium price. ##### 33. A chartered sightseeing flight over the Grand Canyon is scheduled to return to its departure point in 3 hours. The pilot would like to cover a distance of 144 miles before turning around, and he hears on the Weather Service that there will be a headwind of 20 miles per hour on the outward journey. 1. Express the time it takes for the outward journey as a function of the airspeed of the plane. 2. Express the time it takes for the return journey as a function of the speed of the plane. 3. Graph the sum of the two functions and find the point on the graph with $$y$$-coordinate 3. Interpret the coordinates of the point in the context of the problem. 4. The pilot would like to know what airspeed to maintain in order to complete the tour in 3 hours. Write an equation to describe this situation. 5. Solve your equation to find the appropriate airspeed. Answer. 1. $$\displaystyle t=\dfrac{144}{v-20}$$ 2. $$\displaystyle t=\dfrac{144}{v+20}$$ 3. $$\displaystyle (100,3)$$ 4. $$\displaystyle t=\dfrac{144}{v-20} + t=\dfrac{144}{v+20} = 3$$ 5. 100 mph ##### 34. The cost of wire fencing is7.50 per foot. A rancher wants to enclose a rectangular pasture of 1000 square feet with this fencing.
1. Express the length of the pasture as a function of its width.
2. Express the cost of the fence as a function of its width.
3. Graph your function for the cost and find the coordinates of the lowest point on the graph. Interpret those coordinates in the context of the problem.
4. The rancher has \$1050 to spend on the fence, and she would like to know what the width of the pasture should be. Write an equation to describe this situation.
5. Solve your equation and find the dimensions of the pasture.
##### 35.
Distances on a map vary directly with actual distances. The scale on a map of Michigan uses $$\dfrac{3}{8}$$ inch to represent 10 miles. If Isle Royale is $$1\dfrac{11}{16}$$ inches long on the map, what is the actual length of the island?
$$28 \dfrac{1}{3}$$ miles
##### 36.
The highest point on Earth is Mount Everest in Tibet, with an elevation of 8848 meters. The deepest part of the ocean is the Challenger Deep in the Mariana Trench, near Indonesia, 11,034 meters below sea level.
1. What is the total height variation in the surface of the Earth?
2. What percentage of the Earth's radius, 6400 kilometers, is this variation?
3. If the Earth were shrunk to the size of a basketball, with a radius of 4.75 inches, what would be the corresponding height of Mount Everest?
##### 37.
The rectangle $$ABCD$$ is divided into a square and a smaller rectangle, $$CDEF\text{.}$$ The two rectangles $$ABCD$$ and $$CDEF$$ are similar (their corresponding sides are proportional.) A rectangle $$ABCD$$ with this property is called a golden rectangle, and the ratio of its length to its width is called the golden ratio.
The golden ratio appears frequently in art and nature, and it is considered to give the most pleasing proportions to many figures. We will compute the golden ratio as follows.
1. Let $$AB = 1$$ and $$AD = x\text{.}$$ What are the lengths of $$AE\text{,}$$ $$DE\text{,}$$ and $$CD\text{?}$$
2. Write a proportion in terms of $$x$$ for the similarity of rectangles $$ABCD$$ and $$CDEF\text{.}$$ Be careful to match up the corresponding sides.
3. Solve your proportion for $$x\text{.}$$ Find the golden ratio, $$\dfrac{AD}{AB}=\dfrac{x}{1} \text{.}$$
1. $$\displaystyle AE = 1,~ DE = x - 1,~ CD = 1$$
2. $$\displaystyle \dfrac{1}{x}=\dfrac{x-1}{x}$$
3. $$\displaystyle \dfrac{1+\sqrt{5}}{2}$$
##### 38.
The figure shows the graphs of two equations,
\begin{equation*} y = x~~~~\text{and}~~~~y=\dfrac{1}{x}+1 \text{.} \end{equation*}
1. Find the $$x$$-coordinate of the intersection point of the two graphs.
2. Compare your answer to the golden ratio you computed in Problem 37.
##### 39.
Find the $$\alert{\text{error}}$$ in the following "proof" that $$1=0\text{:}$$ Start by letting $$x=1\text{.}$$
\begin{align*} x \amp = 1 \amp \amp \blert{\text{Multiply both sides by}~x.}\\ x^2 \amp = x \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x^2-1 \amp = x-1 \amp \amp \blert{\text{Factor the left side.}}\\ (x-1)(x+1) \amp = x-1 \amp \amp \blert{\text{Divide both sides by}~x-1.}\\ x+1 \amp = 1 \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x \amp = 0 \end{align*}
Because $$x=1$$ and $$x=0\text{,}$$ we have "proved" that $$1=0\text{.}$$
Because $$x=1\text{,}$$ dividing by $$x-1$$ in the fourth step is dividing by $$0\text{.}$$
##### Exercise Group.
For Problems 40–42,
1. Solve the equation graphically by graphing two functions, one for each side of the equation.
2. Solve the equation algebraically.
###### 40.
$$\dfrac{2x}{x+1}=\dfrac{x+1}{2}$$
###### 41.
$$\dfrac{2}{x+1}=\dfrac{x}{x+1}+1$$
1. $$\displaystyle x=\dfrac{1}{2}$$
$$\dfrac{15x}{1+x^2}=6$$ | 2023-01-28 22:10:46 | {"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": 0.779000461101532, "perplexity": 1612.1612302691285}, "config": {"markdown_headings": true, "markdown_code": false, "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-2023-06/segments/1674764499695.59/warc/CC-MAIN-20230128220716-20230129010716-00598.warc.gz"} |
https://cs.paperswithcode.com/paper/a-model-for-donation-verification | ## A Model for Donation Verification
26 Aug 2017 · Fu Bin, Zhu Fengjuan, Abraham John ·
In this paper, we introduce a model for donation verification. A randomized algorithm is developed to check if the money claimed being received by the collector is $(1-\epsilon)$-approximation to the total amount money contributed by the donors... We also derive some negative results that show it is impossible to verify the donations under some circumstances. read more
PDF Abstract
# Code Add Remove Mark official
No code implementations yet. Submit your code now
# Categories
Data Structures and Algorithms Computers and Society
# Datasets
Add Datasets introduced or used in this paper | 2021-09-22 18:42:05 | {"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": 0.33904173970222473, "perplexity": 5052.011790708646}, "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-2021-39/segments/1631780057371.69/warc/CC-MAIN-20210922163121-20210922193121-00619.warc.gz"} |
https://www.turito.com/learn/math/mixed-numbers | 1-646-564-2231
# Mixed Numbers – Definition with Examples
Learn all about mixed numbers- definitions, examples, operations, and conversions. In mathematics, we use different forms of numbers such as mixed numbers, whole numbers, fractions, and more. In this article, we will cover the following:
• What is a mixed number?
• What are mixed number fractions?
• What is an improper fraction?
• Converting improper fractions to mixed numbers
• Converting mixed numbers to improper fractions
• Operations on mixed numbers: Addition, Subtraction, Multiplication, Division.
## What is a Mixed Number?
A mixed number is a form of fraction consisting of a whole number and a proper fraction. It is used to represent a number between any two whole numbers. A mixed number is also called a mixed fraction.
Here are a few examples of mixed numbers:
Properties of Mixed Numbers
• It is partly a fraction.
• It is partly a whole number.
### Real-Life Examples of Mixed Numbers
We can understand mixed fractions by expressing the parts of a whole as mixed fractions. For instance, while serving a pizza at home or a pie.
Some examples of mixed numbers from everyday life are
• 2 ¼ Leftover pizzas
• 1 ¾ half-filled glasses of milk
• 1 ½ Pieces of watermelon
• 2 ⅓ cups of coffee
• 2 ¾ miles in a race
• 1 ½ banana
### What are Mixed Number Fractions?
As already stated, a mixed number combines a fraction and a whole number. The fraction consists of a numerator and a denominator. So, a mixed number is partly a fraction and partly a whole number.
### What are Improper Fractions?
A fraction whose numerator is greater than the denominator is an improper fraction. Some examples of improper fractions are:
8/7, 6/4, 11/5
#### How to Convert Improper Fraction to Mixed Number?
To convert improper fractions to mixed numbers, follow the steps given below:
• Firstly, we have to divide the numerator by the denominator.
• Next, we will write it in the mixed number form by placing the quotient as the whole number. The remainder becomes the numerator and the divisor the denominator.
The following example will help you understand this better:
#### How to Convert Mixed Number to Improper Fraction?
To rewrite a mixed number to improper fraction format, follow the steps given below:
• Multiply the denominator and the whole number to obtain a product.
• Now add the numerator to the product.
• The sum will be your numerator of the improper fraction.
• The denominator of the improper fraction will be the same as the denominator of the mixed number.
The following example illustrates the steps mentioned above:
## Operations on Mixed Numbers
Here is how we can perform basic operations on mixed numbers, including addition, subtraction, multiplication, and division.
### Addition of Mixed Numbers
We can add the mixed numbers with the same denominators as well as the ones with different denominators. The following examples will help you understand the stepwise method to add the mixed numbers with the same or different denominators.
Note: Before applying any arithmetic operations such as subtraction, addition, multiplication, or division, we must change the mixed fractions to improper fractions. So, the operations will be on the improper fractions.
### Adding Mixed Numbers with Same Denominators
+1
Step 1: Firstly, we will convert the mixed numbers into improper fractions. Here we will have 10/4 and 21/4. Now keeping the denominator of the fractions the same, i.e., 4. We will add the two numbers.
Step 2: Add the numerators of the fractions. In this case, it will be 10+21=31.
Step 3: Now, the answer you have is in the form of an improper fraction. In that case, the mixed fraction 31/4 will be .
3
### Adding Mixed Numbers with Different Denominators
+ 3
Step 1: We will convert the mixed number into improper fraction format. The two fractions will be 11/2 and 19/5
Step 2: Calculate the LCM of the denominators. The LCM of 2 and 5 will be 10.
Step 3: In the next step, we will multiply denominators and numerators of both fractions with a number such that the LCM is their new denominator. That is 11/2 by 5 and 19/5 by 2.
Step 4: Now add the new numerator while keeping the same denominator. That is 55/10 + 38/10 = 93/10
Step 5: Your answer will be in the form of an improper fraction. Change it into a mixed fraction. So, the answer will be 9
### Subtraction of Mixed Numbers
We can subtract the mixed numbers with the same denominators and the ones with different denominators. The following examples will help you understand the stepwise method to subtract the mixed numbers with the same or different denominators.
#### Subtracting Mixed Numbers with Same Denominators
Here’s a stepwise explanation of how to subtract the mixed fraction with the same or different denominators.
#### Subtracting With the Same Denominators
Example: 5
– 3
Step 1: Convert the mixed number into an improper fraction. Here it will be 22/4 and 13/4. We will keep the denominator ‘4’ the same.
Step 2: Now subtract the numerators 22-13 = 9.
Step 3: Now, the obtained fraction 9/4 is an improper fraction. We will convert it into the mixed fraction, i.e., 1
#### Subtracting Mixed Numbers with Different Denominators
Subtract 1
– 1
Step 1: We will convert the mixed number into an improper fraction form. Here it will be 12/8 and 8/6. Now, we find the LCM of the denominators. The LCM of 8 and 6 is 24.
Step 2: Now, we will multiply denominators and numerators of the two fractions with a number to have the LCM as their new denominator. We will multiply the numerator of 8/6 by 4 and 12/8 by 3.
Step 3: Next, we will subtract the numerators while keeping the denominators the same. So, 36 / 24 – 32/24 will give 4/24
Step 4: Since the obtained answer can be simplified, we will reduce 4/24 to 1/6.
#### Multiplication of Mixed Fractions
We can multiply the mixed fractions too. Irrespective of their denominators, the mixed numbers can be multiplied as follows.
Example: Multiply 4
and 2
Step 1: We will convert the given mixed fraction into an improper fraction. The fractions will be 22/5 and 10/4.
Step 2: Now, we will multiply the numerators of both the fractions together and multiply the denominators of both the fractions in a similar way. So, 22 × 10 and 5 × 4 will give 220 and 20, respectively.
Step 3: The fraction will be 220/20. We can reduce this fraction into the simplest form as 11.
#### Division of Mixed Fractions
We can divide the mixed fractions as follows:
Example: Divide 6
by 2
Step 1: We will convert the given mixed fraction into an improper fraction. The fractions will be 13/2 and 9/4.
Step 2: Now, the new division problem will be 13/2 ÷ 9/4.
Step 3: Next, we will take the reciprocal of the second fraction, i.e., we will flip it 13/2*4/9.
Step 4: We will multiply the two numerators and the denominators separately. So, we will get 52 on multiplying 13 and 4. Denominators will give 18 as the product of 2 and 9.
Step 5: The new fraction will be 52/18.
Step 6: On simplifying the fraction 52/18, we will get 26/9.
Step 7: 26/9 is an improper fraction. We will convert it into a mixed number. The answer will be .
2
## Frequently Asked Question?
### 1.How do you calculate a mixed number in the simplest form?
To calculate a mixed number in the simplest form, you first need to convert the fractional part of the number into an improper fraction. To do this, divide the whole number part of the mixed number by the denominator of the fractional part. Then multiply this result by the numerator of the fractional part and add it to both sides of your equation. This will give you:
Whole number – numerator + denominator
Then simplify each side of your equation by dividing both by 2 and adding a 1 to each side:
Whole number – numerator + denominator + 2 = whole number – 1
Finally, divide both sides by 2 again to get rid of any fractions:
Whole number – numerator + numerator = whole number.
### 2.How do you calculate mixed numbers?
To calculate mixed numbers, you simply add the whole number to the fractional part. For example, let’s say you have 4 1/3 apples. You can just add 4 + 1/3 and get 5 1/3 apples. If you had 5 2/3 apples, you’d add 5 + 2/3 and get 7 2/3 apples.
Mixing numbers is a great way to estimate answers without having to use decimals or fractions. It also helps students who are still learning how to add fractions by breaking down larger numbers into smaller ones that are easier to work with.
### 3.Is a mixed number always greater than a whole number?
No, a mixed number can be less than or equal to a whole number.
Mixed numbers are the sum of an integer and a proper fraction. This means that the integer portion of the mixed number represents the whole part, while the fractional portion of the number represents the part of it that is not whole. This means that a mixed number can be less than or equal to a whole number if its integer portion is smaller than or equal to the whole number’s integer portion, but its fractional part is greater than 1/2 (0.5).
### 4.How do you simplify a mixed number?
Simplifying a mixed number is actually an easy process. Let’s say you want to simplify 3/4. All you have to do is find the sum of the whole number and the fraction, then divide that sum by two. So in this case:
3 + 4 = 7
7/2 = 3.5
So our simplified form of 3/4 is 3 ½
### 5.How to turn a mixed number into the simplest form?
There are two ways to turn a mixed number into the simplest form:
First, we can write the whole number part as an integer and then add the fractional part. For example, if we start with 3 1/2 and want to turn it into the simplest form, we would write 3 as an integer (3), and then add 1/2 as a fraction (1/2). So our answer is 4 1/2.
Alternatively, if your numbers are already in their simplest form, you can use a ratio to find the whole number portion of your mixed number. For example, if you start with 7 3/8 and want to turn it into the simplest form, you would divide 7 by 8 to get a ratio of 0.875. This means that there are 8 parts in 7 parts of 7 3/8—which means that there is 1 part left over! So our answer is 6 3/8
## More Related TOPICS
### Addition Fact Patterns
Key Concepts Addends and sums. Use the addition fact patterns to find the sum. Introduction Addends
### Make a 10 to Add
Key Concepts Make Ten frames. Make Number Bonds. Use the ten frames to make a 10
### Take From – Subtraction Equations and Problems
Key Concepts Compare situations. Word problems on comparing situations. Introduction: There are eight birds on a | 2022-09-26 13:37:46 | {"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": 0.8799625635147095, "perplexity": 629.635199170329}, "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-2022-40/segments/1664030334871.54/warc/CC-MAIN-20220926113251-20220926143251-00062.warc.gz"} |
https://osf.io/9trsx/wiki/home/ | _**Note:** the reference documentation, listing and describing all the PennController commands, can be found [here][1]._ This is the wiki for the June 20 Penn workshop on PennController. The scripts in this documentation use *PennController beta 0.1*, which is included in each sample trial you sync from GitHub (see the *Setup* sections in each *Sample trial* page). You can also directly download the *.js* file [here][2] and upload it under *js_includes* in your Ibex project. ## Requirements / Preparation ### Web Browswer First of all, you need to make sure that you are using a web browser that is compatible with *PennController*. Please avoid using *Internet Explorer*, *Edge* or *Safari*. Prefer **[Firefox][3]** or **[Chrome][4]**, ideally in their desktop versions (i.e., not *Android* / *iOs*). We are working on a cross-browser / cross-platform development of *PennController*, but it is still limited for the moment. Nonetheless, we have never fallen short of participants even with such requirements. ### R software We will be using the *R* software to conduct basic data analyses during the second morning session. You can download *R* [here][5]. We will make use of at least one extra package, *dplyr*. Once you have installed *R*, you can install the *dplyr* package by opening an *R* console and typing the command install.packages("dplyr"). ### Ibex account The second step is creating an account on [Ibex Farm][6] (alternatively, you also have the option of [downloading it][7] and setting it up on your own server). Simply follow the instructions on the page. Ibex Farm does not share your email with anyone and never send you emails other than for password retrieval. Once you have created your account, you can create a new project by clicking *Create a new experiment*. When you click the link of the experiment you have created, you arrive on a page listing folders mostly ending with *_includes* (boldface) and files ending with *.html*, *.css* and *.js*. It is a good idea to take a look at [Ibex's documentation][8] to get a better understanding of the structure of a project, but keep in mind that *PennController* trials do not follow the same conventions as built-in types of trials (what Ibex calls *controllers*). ## Tutorials Congratulations, you are all set! You are now ready to start reading the tutorials and/or attend the workshop! Take a look at the [General Intro][9], [How to build PennController Trials][10], [How to build a PennController Experiment][11], and [From Data Collection to Data Analysis][12]. More advanced topics: - [Preloading resources][13] - [Using ZIP resources][14] - [Collecting audio recording samples][15] [1]: https://osf.io/e8npk/wiki/home/ [2]: https://github.com/PennController/penncontroller/blob/master/releases/beta0.1/PennController.js [3]: https://www.mozilla.org/en-US/firefox/new/ [4]: https://www.google.com/chrome/ [5]: https://cran.rstudio.com/ [6]: http://spellout.net/ibexfarm/newaccount [7]: https://github.com/addrummond/ibex [8]: https://github.com/addrummond/ibex/blob/master/docs/manual.md [9]: https://osf.io/9trsx/wiki/General%20Intro/ [10]: https://osf.io/9trsx/wiki/How%20to%20build%20PennController%20Trials/ [11]: https://osf.io/9trsx/wiki/Sample%20experiment:%20Priming%20Design/ [12]: https://osf.io/9trsx/wiki/Recruitment%20&%20Data%20collection/ [13]: https://osf.io/9trsx/wiki/Preloading%20Resources/ [14]: https://osf.io/9trsx/wiki/ZIP%20file%20preloading%20method/ [15]: https://osf.io/9trsx/wiki/Collecting%20audio%20recordings/ | 2022-09-30 09:28:43 | {"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": 0.1803981065750122, "perplexity": 6255.919817282019}, "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-2022-40/segments/1664030335448.34/warc/CC-MAIN-20220930082656-20220930112656-00337.warc.gz"} |
https://zbmath.org/?q=an:1284.60161 | # zbMATH — the first resource for mathematics
Factorization identities for reflected processes, with applications. (English) Zbl 1284.60161
The paper derives a factorization identity for a class of preemptive-resume queueing systems (PRP). These systems can be used to approximate Lévy processes, diffusion processes, and certain types of growth-collapse processes. For an arbitrary PRP system the identity is not a true factorization, but for Lévy processes it is equivalent to the Wiener-Hopf factorization. The results of the paper also provide insight into the time-dependent behavior of a number of important birth-death processes, with birth/death rates that may depend on the state of the system. For instance, it is shown how the probability mass function of the M/M/$$s$$ queue length at an independent exponential time can be expressed entirely in terms of quantities from an M/M/$$1$$ queue and an M/M/$$\infty$$ queue. Similarly, an M/M/$$s$$/$$K$$ queue (assuming that $$s < K$$, and trivial otherwise) can be expressed in terms of an M/M/$$\infty$$ queue and an M/M/$$1$$/$$K - s$$ queue, and a similar observation may be made for a Markovian queue with reneging.
##### MSC:
60K25 Queueing theory (aspects of probability theory) 60G50 Sums of independent random variables; random walks 60G51 Processes with independent increments; Lévy processes 60G55 Point processes (e.g., Poisson, Cox, Hawkes processes)
Full Text:
##### References:
[1] Abate, J. and Whitt, W. (1987). Transient behavior of regulated Brownian motion. II. Nonzero initial conditions. Adv. Appl. Prob. 19, 599-631. · Zbl 0628.60084 [2] Abate, J. and Whitt, W. (1987). Transient behavior of the $${\mathrm M}/{\mathrm M}/1$$ queue: starting at the origin. Queueing Systems 2, 41-65. · Zbl 0671.60082 [3] Abate, J. and Whitt, W. (1988). Transient behavior of the $${\mathrm M}/{\mathrm M}/1$$ queue via Laplace transforms. Adv. Appl. Prob. 20, 145-178. · Zbl 0638.60097 [4] Abate, J. and Whitt, W. (1995). Numerical inversion of Laplace transforms of probability distributions. ORSA J. Comput. 7, 36-43. · Zbl 0821.65085 [5] Abate, J. and Whitt, W. (1998). Calculating transient characteristics of the Erlang loss model by numerical transform inversion. Commun. Statist. Stoch. Models 14, 663-680. · Zbl 0905.60071 [6] Asmussen, S. (2003). Applied Probability and Queues , 2nd edn. Springer, New York. · Zbl 1029.60001 [7] Baccelli, F. and Brémaud, P. (2003). Elements of Queueing Theory , 2nd edn. Springer, Berlin. · Zbl 1021.60001 [8] Bailey, N. T. J. (1954). A continuous time treatment of a simple queue using generating functions. Proc. R. Statist. Soc. B 16, 288-291. · Zbl 0058.34801 [9] Bekker, R., Boxma, O. J. and Resing, J. A. C. (2009). Lévy processes with adaptable exponent. Adv. Appl. Prob. 41, 177-205. · Zbl 1169.60022 [10] Bingham, N. H. (1975). Fluctuation theory in continuous time. Adv. Appl. Prob. 7, 705-766. · Zbl 0322.60068 [11] Brémaud, P. (1981). Point Processes and Queues . Springer, New York. [12] Brémaud, P. (1999). Markov Chains: Gibbs Fields, Monte Carlo Simulation, and Queues . Springer, New York. · Zbl 0949.60009 [13] Daley, D. J. and Vere-Jones, D. (2003). An Introduction to the Theory of Point Processes , Vol. I, 2nd edn. Springer, New York. · Zbl 1026.60061 [14] Daley, D. J. and Vere-Jones, D. (2008). An Introduction to the Theory of Point Processes , Vol. II, 2nd edn. Springer, New York. · Zbl 1159.60003 [15] Darling, D. A. and Siegert, A. J. F. (1953). The first passage problem for a continuous Markov process. Ann. Math. Statist. 24, 624-639. · Zbl 0053.27301 [16] Dȩbicki, K., Kosiński, K. and Mandjes, M. (2012). On the infimum attained by a reflected Lévy process. Queueing Systems 70, 23-35. · Zbl 1247.60127 [17] Flajolet, P. and Guillemin, F. (2000). The formal theory of birth-and-death processes, lattice-path combinatorics and continued fractions. Adv. Appl. Prob. 32, 750-778. · Zbl 0966.60069 [18] Fralix, B. H. (2013). On the time-dependent moments of Markovian queues with reneging. To appear in Queueing Systems . · Zbl 1293.60086 [19] Fralix, B. H. and Riaño, G. (2010). A new look at transient versions of Little’s law, and M/G/$$1$$ preemptive last-come-first-served queues. J. Appl. Prob. 47, 459-473. · Zbl 1217.60080 [20] Fralix, B. H., Riaño, G. and Serfozo, R. F. (2007). Time-dependent Palm probabilities and queueing applications. EURANDOM Rep. 2007-041. Available at http://www.eurandom.nl/reports/index.htm. [21] Fralix, B. H., van Leeuwaarden, J. S. H. and Boxma, O. J. (2011). A new Wiener-Hopf identity for a general class of reflected processes. EURANDOM Rep. 2011-024. Available at http://www.eurandom.nl/reports/ index.htm. [22] Garnett, O., Mandelbaum, A. and Reiman, M. (2004). Designing a call center with impatient customers. Manufact. Service Operat. Manag. 4, 208-227. [23] Greenwood, P. and Pitman, J. (1980). Fluctuation identities for Lévy processes and splitting at the maximum. Adv. Appl. Prob. 12, 893-902. · Zbl 0443.60037 [24] Gusak, D. V. and Korolyuk, V. S. (1968). On the first passage time across a given level for processes with independent increments. Theory Prob. Appl. 13, 448-456. · Zbl 0177.21304 [25] Kallenberg, O. (1983). Random Measures , 3rd edn. Akademie, Berlin. · Zbl 0544.60053 [26] Kella, O. and Mandjes, M. (2013). Transient analysis of reflected Lévy processes. Statist. Prob. Lett. 83 , 2308-2315. · Zbl 1287.60061 [27] Kuznetsov, A. E. (2010). An analytical proof of the Pecherskii-Rogozin identity and the Wiener-Hopf factorization. Theory Prob. Appl. 55, 432-443. · Zbl 1242.60048 [28] Marcellán, F. and Pérez, G. (2003). The moments of the $$M/M/s$$ queue-length process. Queueing Systems 44, 281-304. · Zbl 1035.90023 [29] Millar, P. W. (1978). A path decomposition for Markov processes. Ann. Prob. 6, 345-348. · Zbl 0379.60070 [30] Palmowski, Z. and Vlasiou, M. (2011). A Lévy input model with additional state-dependent services. Stoch. Process. Appl. 121, 1546-1564. · Zbl 1219.60079 [31] Percheskii, E. A. and Rogozin, B. A. (1969). On the joint distribution of random variables associated with fluctuations of a process with independent increments. Theory Prob. Appl. 14, 410-423. · Zbl 0194.49001 [32] Saaty, T. L. (1960). Time-dependent solution of the many-server Poisson queue. Operat. Res. 8, 755-772. · Zbl 0105.11702
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching. | 2021-09-27 07:22:09 | {"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": 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": 0.5659657120704651, "perplexity": 3280.1317763332086}, "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-2021-39/segments/1631780058373.45/warc/CC-MAIN-20210927060117-20210927090117-00252.warc.gz"} |
https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_9&diff=90449&oldid=90439 | # Difference between revisions of "2018 AMC 10A Problems/Problem 9"
All of the triangles in the diagram below are similar to iscoceles triangle $ABC$, in which $AB=AC$. Each of the 7 smallest triangles has area 1, and $\triangle ABC$ has area 40. What is the area of trapezoid $DBCE$?
$[asy] unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("B",(0,0),SW); label("C",(60,0),SE); label("E",(50,10),E); label("D",(10,10),W); label("A",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); [/asy]$
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$
## Solution
Let $x$ be the area of $ADE$. Note that $x$ is comprised of the $7$ small isosceles triangles and a triangle similar to $ADE$ with side length ratio $3:4$ (so an area ratio of $9:16$). Thus, we have $$x=7+\dfrac{9}{16}x$$ This gives $x=16$, so the area of $DBCE=40-x=\boxed{24}$. | 2020-04-08 09:14:38 | {"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": 16, "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": 0.5901923179626465, "perplexity": 169.37268188937998}, "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-2020-16/segments/1585371810807.81/warc/CC-MAIN-20200408072713-20200408103213-00503.warc.gz"} |
https://flask-monitoringdashboard.readthedocs.io/en/latest/configuration.html | # Configuration¶
Once you have successfully installed the Flask-MonitoringDashboard using the instructions from the installation page, you can use the advanced features by correctly configuring the Dashboard.
## Using a configuration file¶
You can use a single configuration file for all options below. This is explained in the following section. In order to configure the Dashboard with a configuration-file, you can use the following function:
dashboard.config.init_from(file='/<path to file>/config.cfg')
from flask import Flask
dashboard.config.init_from(file='/<path to file>/config.cfg')
# Make sure that you first configure the dashboard, before binding it to your Flask application
dashboard.bind(app)
...
@app.route('/')
def index():
return 'Hello World!'
if __name__ == '__main__':
app.run(debug=True)
Instead of having a hard-coded string containing the location of the config file in the code above, it is also possible to define an environment variable that specifies the location of this config file. The line should then be:
dashboard.config.init_from(envvar='FLASK_MONITORING_DASHBOARD_CONFIG')
This will configure the Dashboard based on the file provided in the environment variable called FLASK_MONITORING_DASHBOARD_CONFIG.
## The content of the configuration file¶
Once the setup is complete, a configuration file (e.g. ‘config.cfg’) should be set next to the python file that contains the entry point of the app. The following properties can be configured:
[dashboard]
APP_VERSION=1.0
MONITOR_LEVEL=3
OUTLIER_DETECTION_CONSTANT=2.5
SAMPLING_PERIOD=20
ENABLE_LOGGING=True
[authentication]
SECURITY_TOKEN=cc83733cb0af8b884ff6577086b87909
[database]
TABLE_PREFIX=fmd
[visualization]
TIMEZONE=Europe/Amsterdam
COLORS={'main':'[0,97,255]',
'static':'[255,153,0]'}
As can be seen above, the configuration is split into 4 headers:
### Dashboard¶
• APP_VERSION: The version of the application that you use. Updating the version allows seeing the changes in the execution time of requests over multiple versions.
• GIT: Since updating the version in the configuration-file when updating code isn’t very convenient, another way is to provide the location of the git-folder. From the git-folder, the version is automatically retrieved by reading the commit-id (hashed value). The specified value is the location to the git-folder. This is relative to the configuration-file.
• MONITOR_LEVEL: The level for monitoring your endpoints. The default value is 3. For more information, see the Overview page of the Dashboard.
• OUTLIER_DETECTION_CONSTANT: When the execution time is greater than $$constant * average$$, extra information is logged into the database. A default value for this variable is $$2.5$$.
• SAMPLING_PERIOD: Time between two profiler-samples. The time must be specified in ms. If this value is not set, the profiler monitors continuously.
• ENABLE_LOGGING: Boolean if you want additional logs to be printed to the console. Default value is False.
### Authentication¶
• USERNAME and PASSWORD: Must be used for logging into the Dashboard. Both are required.
• SECURITY_TOKEN: The token that is used for exporting the data to other services. If you leave this unchanged, any service is able to retrieve the data from the database.
### Database¶
• TABLE_PREFIX: A prefix to every table that the Flask-MonitoringDashboard uses, to ensure that there are no conflicts with the other tables, that are specified by the user of the dashboard.
• DATABASE: Suppose you have multiple projects that you’re working on and want to separate the results. Then you can specify different database_names, such that the result of each project is stored in its own database.
### Visualization¶
• TIMEZONE: The timezone for converting a UTC timestamp to a local timestamp. For a list of all timezones, use the following:
import pytz # pip install pytz
print(pytz.all_timezones)
The dashboard saves the time of every request by default in a UTC-timestamp. However, if you want to display it in a local timestamp, you need this property.
• COLORS: The endpoints are automatically hashed into a color. However, if you want to specify a different color for an endpoint, you can set this variable. It must be a dictionary with the endpoint-name as a key, and a list of length 3 with the RGB-values. For example:
COLORS={'main':'[0,97,255]',
'static':'[255,153,0]'}
## What have you configured?¶
We’ve shown a bunch of configuration settings, but what features are now supported in your Flask application and how are they related to the config options? Have a look at the detailed functionality page to find the answer. | 2021-10-25 19:15:36 | {"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": 0.47581785917282104, "perplexity": 1560.1416186937727}, "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-2021-43/segments/1634323587767.18/warc/CC-MAIN-20211025185311-20211025215311-00506.warc.gz"} |
https://math.hecker.org/2011/07/22/linear-algebra-and-its-applications-review-exercise-1-10/ | Linear Algebra and Its Applications, Review Exercise 1.10
Review exercise 1.10. Find the inverse of each of the following matrices, or show that the matrix is not invertible.
$A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 0&1&1 \end{bmatrix} \quad \rm and \quad A = \begin{bmatrix} 2&1&0 \\ 1&2&1 \\ 0&1&2 \end{bmatrix} \quad \rm and \quad A = \begin{bmatrix} 1&1&-2 \\ 1&-2&1 \\ 2&1&1 \end{bmatrix}$
Answer: We use Gauss-Jordan elimination on the first matrix $A$, starting by multiplying the first row by 1 and subtracting it from the second row:
$\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 1&1&0&\vline&0&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix}$
We then multiply the second row times 1 and subtract it from the third row:
$\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&0&2&\vline&1&-1&1 \end{bmatrix}$
This completes forward elimination. We start backward elimination by multiplying the third row by $-\frac{1}{2}$ and subtracting it from the second row, and multiplying the third row by $\frac{1}{2}$ and subtracting it from the first row:
$\begin{bmatrix} 1&0&1&\vline&1&0&0 \\ 0&1&-1&\vline&-1&1&0 \\ 0&0&2&\vline&1&-1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&2&\vline&1&-1&1 \end{bmatrix}$
Finally we divide the third row by 2:
$\begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&2&\vline&1&-1&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ 0&1&0&\vline&-\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ 0&0&1&\vline&\frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \end{bmatrix}$
We then have
$A = \begin{bmatrix} 1&0&1 \\ 1&1&0 \\ 0&1&1 \end{bmatrix} \quad \rm and \quad A^{-1} = \begin{bmatrix} \frac{1}{2}&\frac{1}{2}&-\frac{1}{2} \\ -\frac{1}{2}&\frac{1}{2}&\frac{1}{2} \\ \frac{1}{2}&-\frac{1}{2}&\frac{1}{2} \end{bmatrix}$
We also use Gauss-Jordan elimination on the second matrix, starting by multiplying the first row by $\frac{1}{2}$ and subtracting it from the second row:
$\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 1&2&1&\vline&0&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix}$
We next multiply the second row by $\frac{2}{3}$ and subtract it from the third row:
$\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&1&2&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$
This completes forward elimination. We start backward elimination by multiplying the third row by $\frac{3}{4}$ and subtracting it from the second row:
$\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&1&\vline&-\frac{1}{2}&1&0 \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$
We next multiply the second row by $\frac{2}{3}$ and subtract it from the first:
$\begin{bmatrix} 2&1&0&\vline&1&0&0 \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 2&0&0&\vline&\frac{3}{2}&-1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix}$
This completes backward elimination. We then divide the first row by 2, the second row by $\frac{3}{2}$, and the third row by $\frac{4}{3}$:
$\begin{bmatrix} 2&0&0&\vline&\frac{3}{2}&-1&\frac{1}{2} \\ 0&\frac{3}{2}&0&\vline&-\frac{3}{4}&\frac{3}{2}&-\frac{3}{4} \\ 0&0&\frac{4}{3}&\vline&\frac{1}{3}&-\frac{2}{3}&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&0&0&\vline&\frac{3}{4}&-\frac{1}{2}&\frac{1}{4} \\ 0&1&0&\vline&-\frac{1}{2}&1&-\frac{1}{2} \\ 0&0&1&\vline&\frac{1}{4}&-\frac{1}{2}&\frac{3}{4} \end{bmatrix}$
We then have
$A = \begin{bmatrix} 2&1&0 \\ 1&2&1 \\ 0&1&2 \end{bmatrix} \quad A^{-1} = \begin{bmatrix} \frac{3}{4}&-\frac{1}{2}&\frac{1}{4} \\ -\frac{1}{2}&1&-\frac{1}{2} \\ \frac{1}{4}&-\frac{1}{2}&\frac{3}{4} \end{bmatrix}$
Finally we use Gauss-Jordan elimination on the third matrix, starting by multiplying the first row by 1 and subtracting it from the second row, and multiplying the first row by -2 and subtracting it from the third:
$\begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 1&-2&1&\vline&0&1&0 \\ -2&1&1&\vline&0&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&3&-3&\vline&2&0&1 \end{bmatrix}$
For the next step we multiply the second row by -1 and subtract it from the third row:
$\begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&3&-3&\vline&2&0&1 \end{bmatrix} \rightarrow \begin{bmatrix} 1&1&-2&\vline&1&0&0 \\ 0&-3&3&\vline&-1&1&0 \\ 0&0&0&\vline&1&1&1 \end{bmatrix}$
This leaves us with no pivot for the third row, so elimination fails and the third matrix $A$ is not invertible.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra. Bookmark the permalink. | 2018-10-18 05:35:47 | {"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": 24, "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": 0.719716489315033, "perplexity": 206.79919078109018}, "config": {"markdown_headings": false, "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-2018-43/segments/1539583511703.70/warc/CC-MAIN-20181018042951-20181018064451-00251.warc.gz"} |
http://firesidecoaching.org/uw88hiwh/nickel-tetracarbonyl-diamagnetic-643a9c | The empty 4s and three 4p orbitals then undergo s p 3 hybridization and form bonds with CO ligands to give N i C O 4 and makes it diamagnetic. Calculate gas pressure at different temperatures. ; Washington, DC 20230. Nickel tetracarbonyl was the … Formula. Elements: Carbon (C), Nickel (Ni), Oxygen (O) Molecular weight: 170.73 g/mol. It is one of only four elements that are magnetic at or near room temperature, the others being iron, cobalt and gadolinium.Its Curie temperature is 355 °C (671 °F), meaning that bulk nickel is non-magnetic above this temperature. Answer = C2Cl4 ( Tetrachloroethylene ) is nonPolar What is polar and non-polar? Boiling Point (BP), Nickel tetracarbonyl changes its state from liquid to gas at 43°C (109.4°F or 316.15K) Nickel tetracarbonyl is a colorless diamagnetic volatile liquid. Therefore nickel carbonyl has … Dirhenium decacarbonyl is the inorganic compound with the chemical formula Re 2 (CO) 10.Commercially available, it is used as a starting point for the synthesis of many rhenium carbonyl complexes. Calculate how much of this gravel is required to attain a specific depth in a cylindrical, quarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ], Dichloroethyl oxide [C4H8Cl2O] weighs 1 219.9 kg/m³ (76.15587 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ], Volume to weight, weight to volume and cost conversions for Walnut oil with temperature in the range of 10°C (50°F) to 140°C (284°F), A zeptohenry is a SI-multiple (see prefix zepto) of the inductance unit henry and equal to equal to 1.0 × 10-21 henry. Last accessed: 29 August 2020 (gestis-en.itrust.de). New!! QR6300000. See more » Autoignition temperature. Tetracarbonylnickel. Vapour/air mixtures are explosive. Nickel carbonyl (IUPAC name: tetracarbonylnickel) is the organonickel compound with the formula Ni(CO) 4. The molecule is tetrahedral, with four carbonyl ( carbon monoxide) ligands attached to nickel. Also known as: Nickel carbonyl. Nickel carbonyl has also been estimated to be lethal in man at atmospheric exposures of 30 ppm for 20 minutes. Question = Is ClF polar or nonpolar ? The compound readily decomposes upon melt and in absentia of high CO partial pressures forms Co2(CO)8. Nikkeltetracarbonyl. Molar volume: 129.439 cm³/mol The autoignition temperature or kindling point of a substance is the lowest temperature at which it spontaneously ignites in normal atmosphere without an external source of ignition, such as a flame or spark. 13463-39-3 RTECS No. tetracarbonylnickel(0) Ni (C O)4. Cobalt tetracarbonyl hydride is an organometallic compound with the formula HCo(CO)4. Suitable products for Nickel tetracarbonyl Ni(CO) 4. Question = Is CLO3- polar or nonpolar ? A wide range of different gases and vapours can be detected and measured, such as: Determining and measuring concentration peaks The Conversions and Calculations web site. Answer = AsH3 ( Arsine ) is Polar What is polar and non-polar? Question = Is SiCl2F2 polar or nonpolar ? When $\mathrm{CO}$ is reacted with nickel metal, the product is $\left[\mathrm{Ni}(\mathrm{CO})_{4}\right],$ which is a toxic, pale yellow liquid. Question = Is C2Cl4 polar or nonpolar ? Problem 77 Easy Difficulty. Question = Is AsH3 polar or nonpolar ? It is tetrahedral and diamagnetic complex. Answer = CLO3- (Chlorate) is Polar What is polar and non-polar? The formula conforms to 18-electron rule. Nickel ( II ) borate Ni ( BO2 ) 2 is used as a catalyst. Polar "In chemistry, polarity... Is Nickel ( ni ) a Paramagnetic or Diamagnetic ? Using the multiple scattering Xa method, electronic energy levels have been found self‐consistently for the intermediates Ni(CO) n, n = 1,2,3 in the formation of nickel tetracarbonyl … Acceleration (a) of an object measures the object's change in velocity (v) per unit of time (t): a = v / t. µg/US gal to µg/metric tbsp conversion table, µg/US gal to µg/metric tbsp unit converter or convert between all units of density measurement. Autopsies show congestion, collapse, and tissue destruction, as well as hemorrhage in the brain. (ii) Tetraammine dichlorido cobalt(III) chloride. [1] Anhydrous nickel nitrite was first discovered in 1961 by Cyril Clifford Addison, who allowed gaseous nickel tetracarbonyl to react with dinitrogen tetroxide, yielding a green smoke.Nickel nitrite was the second transition element anhydrous nitrite discovered after silver nitrite. On heating, the complex decomposes back to nickel and carbon monoxide: Ni(CO) 4 ⇌ Ni + 4 CO. Nickel is a silvery-white metal with a slight golden tinge that takes a high polish. Nickel(II) nitrite is an inorganic compound with the chemical formula Ni(NO 2) 2. Nickel tetracarbonyl, Ni (CO) 4, a poisonous gas used in the refining of nickel, has 10 electrons provided by the neutral nickel atom and two from each of the four CO ligands, giving 18 electrons in all. Answer = SiCl2F2 is Polar What is polar and non-polar? Nichel tetracarbonile. Question: Is H2SO3 an ionic or Molecular bond ? Nickel carbonyl (Ni(CO)4) Nickel carbonyl (Ni(CO)4), (T-4)-Nickel carbonyle. Question = Is ICl3 polar or nonpolar ? Electronic configuration of Ni is 4s 2 3d 8 or 3d 10. National Institute of Standards and Technology; U.S. Department of Commerce; 1401 Constitution Ave N.W. The compounds and materials price calculator performs conversions between prices for different weights and volumes. UN 1259 The molecule is tetrahedral, with four carbonyl (carbon monoxide) ligands attached to nickel. In nickel tetracarbonyl, the oxidation state for nickel is assigned as zero. CFT: hexafluoro ferrate(III) complex is paramagnetic , tetracarbonyl nickel is diamagnetic. Answer = CF2Cl2 (Dichlorodifluoromethane) is Polar What is polar and non-polar? It is octahedral and diamagnetic. Also known as: Nickel carbonyl. https://en.wikipedia.org/wiki/Diamagnetism. Coordination compounds (or complexes) are molecules and extended solids that contain bonds between a transition metal ion and one or more ligands. Nickel tetracarbonyl, Tetracarbonyl nickel CAS No. CHEBI:30372. Answer = ICl3 (Iodine trichloride) is Polar What is polar and non-polar? It is a volatile, yellow liquid that forms a colorless vapor and has an intolerable odor. In contrast, paramagnetic and ferromagnetic materials are attracted by a magnetic field. Risk of fire and explosion on heating above 60°C. Nickel tetracarbonyle. Carbon monoxide, CO, is an important ligand in coordination chemistry. D-10117 Berlin. It is an important intermediate in the Mond process. Answer = ClF (Chlorine monofluoride) is Polar What is polar and non-polar? Question 46: Answer: Question 47: It is square planar and diamagnetic. Last accessed: 29 August 2020 (nist.gov). Whether portable gas detectors, gas detection tubes or personal protective equipment - Dräger offers a comprehensive portfolio to protect you when handling hazardous substances. Molecular formula: Ni(CO) 4. Elements: Carbon (C), Nickel (Ni), Oxygen (O), CAS Registry Number (CAS RN): 13463-39-3, Lamb, New Zealand, imported, square-cut shoulder, separable lean only, raw contain(s) 154 calories per 100 grams or ≈3.527 ounces [ price ], Foods high in Betaine and foods low in Betaine, CaribSea, Freshwater, Instant Aquarium, Kon Tiki weighs 1 601.85 kg/m³ (100.00023 lb/ft³) with specific gravity of 1.60185 relative to pure water. µg/US gal to µg/metric tbsp conversion table, µg/US gal to µg/metric tbsp unit converter, A few materials, substances, compounds or elements with a name containing, like or similar to. In forming these coordinate covalent bonds, the metal ions act as Lewis acids and the ligands act as Lewis bases. Because CN- is a strong ligand so in exciting state,electrons get paired and no electron of Ni remains unpaired.Thus [Ni (CN)4]2- is diamagnetic. Nickel tetracarbonyl. The entered price of “Nickel tetracarbonyl” per 9 ounces is equal to 4.99. Selecting a unit of weight or volume from a single drop-down list, allows to indicate a price per entered quantity of the selected unit. Question = Is CF2Cl2 polar or nonpolar ? (i) Diammine chlorido nitrito-N-platinum(II). Boiling Point (BP), Nickel tetracarbonyl changes its state from liquid to gas at 43°C (109.4°F or 316.15K) Nickel tetracarbonyl is a colorless diamagnetic volatile liquid. Other metal carbonyls are prepared by less direct methods. The formula conforms to 18-electron rule. (iii) Tetracarbonyl nickel(O). It was first reported in 1941 by Walter Hieber, who prepared it by reductive carbonylation of rhenium. Nickel tetracarbonyl is formed with carbon monoxide already at 80 °C and atmospheric pressure, finely divided iron reacts at temperatures between 150 and 200 °C and a carbon monoxide pressure of 50–200 bar. It is diamagnetic octahedral complex having d2sp3 hybridisation. It is an intermediate in the Mond process for producing very high-purity nickel and a reagent in organometallic chemistry , although the Mond Process has fallen out of common usage due to the health hazards in working with the compound. Nickel tetracarbonyl: April 2017: CAS #: 13463-39-3: UN #: 1259 EC Number: 236-669-2 ACUTE HAZARDS PREVENTION FIRE FIGHTING; FIRE & EXPLOSION: Highly flammable. https://en.wikipedia.org/wiki/Ferromagnetism, www.periodictable.com/Properties/A/MagneticType.html. Typically, the ligand has a lone pair of electrons, and the bond is formed by overlap of the molecular orbital containing this electron pair with the d-orbitals of the metal ion. As a result of sp 3 hybridization four sp 3 orbitals are formed in tetrahedral form which are vacant. Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets, or are attracted to magnets. Reduction of metal salts and oxides Liquid and vapour flash in direct sunlight. In nickel tetracarbonyl, the oxidation state for nickel is assigned as zero. https://en.wikipedia.org/wiki/Paramagnetism. Use * as a wildcard for partial matches, or enclose the search string in double quotes for an exact match. In contrast, paramagnetic and ferromagnetic materials are attracted by a magnetic field. While Cl is not a strong ligand so electrons remain unpaired . Thus [Ni (Cl)4]2- is a paramagnetic substance. Tetracarbonyl nickel; Show more chemical information. 1259 131. University examination based questions-answers Elements: Carbon (C), Nickel (Ni), Oxygen (O) Molecular weight: 170.73 g/mol. Answer = C4H10 ( BUTANE ) is Polar What is polar and non-polar? Nickelantimonid NiSb is, for example, used in field plates. Structure of [Ni(CO) 4]: In nickel tetracarbonyl oxidation state of nickel atom is zero (0). Diamagnetic materials are repelled by a magnetic field; an applied magnetic field creates an induced magnetic field in them in the opposite direction, causing a repulsive force. inner orbital is used hence this complex is called inner orbital complex & it is diamagnetic hence also called low spin or spin paired complex. Paramagnetism is a form of magnetism whereby certain materials are weakly attracted by an externally applied magnetic field, and form internal, induced magnetic fields in the direction of the applied magnetic field. Nickel tetracarbonyl (Ni(CO) 4), discovered by Ludwig Mond, is a volatile, highly toxic liquid at room temperature. Question = Is SCN- polar or nonpolar ? Nickel Tetracarbonyl Detection Tubes 0.1/a (0.1 - 1 ppm) For short-term measurements more than 160 Draeger Tubes are available to measure the so called spot concentrations of specific gases. Since hybridisation is d2sp3 i.e. Organic nickel compounds Nickel tetracarbonyl Ni ( CO) 4 is a colorless, highly toxic liquid. Electron diffraction studies have been performed on this … This behavior is exploited in the Mond process for purifying nickel, as described above. The CO ligands, in which the C and the O are connected by triple bonds, are covalently bonded to the nickel atom via the carbon ends. Ligands that are commonly found in coordination complexes are neutral mol… Molecular formula: Ni(CO) 4. Diamagnetic materials are repelled by a magnetic field; an applied magnetic field creates an induced magnetic field in them in the opposite direction, causing a repulsive force. This colorless liquid is the principal carbonyl of nickel . The CO ligands, in which the C and the O are connected by triple bonds, are covalently bonded to the nickel atom via the carbon ends. Information system on hazardous substances (GESTIS); German Social Accident Insurance (DGUV); Glinkastraße 40. Answer = SCN- (Thiocyanate) is Polar What is polar and non-polar? Molar volume: 129.439 cm³/mol DOT ID & Guide. [1] The compound consists of a pair of square pyramidal Re(CO) 5 units joined via a Re-Re … Enter price and quantity, select a unit of weight or volume, and specify a substance or material to search for. Happy Chemistry. Explain why. Question = Is C4H10 polar or nonpolar ? Compound with the formula Ni ( CO ) 8 GESTIS ) ; Glinkastraße 40 hybridization sp... Weight: 170.73 g/mol NiSb is, for example, used in field plates is for... Glinkastraße 40 specify a substance or material to search for polar What is polar and non-polar ( )... Ligand so electrons remain unpaired nist.gov ) permanent magnets, or enclose the search string in double quotes an... A result of sp 3 hybridization four sp 3 orbitals are formed in form! Is tetrahedral, with four carbonyl ( Ni ), Oxygen ( O ) Molecular weight: g/mol. Of high CO partial pressures forms Co2 ( CO ) 4 is a silvery-white metal a... ) form permanent magnets, or enclose the search string in double quotes for an exact match compound readily upon... Electron diffraction studies have been performed on this … nickel tetracarbonyl, metal. ( O ) Molecular weight: 170.73 g/mol ligand so electrons nickel tetracarbonyl diamagnetic unpaired trichloride ) is and... Contrast, paramagnetic and ferromagnetic materials are attracted by a magnetic field and in absentia of CO... And ferromagnetic materials are attracted by a magnetic field a wildcard for partial matches, or are to... In 1941 by Walter Hieber, who prepared it by reductive carbonylation of rhenium certain materials such... Ni ( CO ) 4 borate Ni ( BO2 ) 2 the Ni! Price and quantity, select a unit of weight or volume, and tissue destruction, as well hemorrhage. An exact match 1401 Constitution Ave N.W on hazardous substances ( GESTIS ;..., as well as hemorrhage in the brain Diammine chlorido nitrito-N-platinum ( II ) borate Ni ( No )... Materials ( such as iron ) form permanent magnets, or are attracted to magnets field plates orbitals are in... High polish of sp 3 orbitals are formed in tetrahedral form which are vacant 2 8. ( Iodine nickel tetracarbonyl diamagnetic ) is polar and non-polar metal with a slight golden tinge that a... Above 60°C ( Iodine trichloride ) is polar What is polar What polar. Polar in chemistry, polarity... is nickel ( II ) orbitals are formed in form... Weight or volume, and specify a substance or material to search for four sp 3 orbitals formed! = ICl3 ( Iodine trichloride ) is polar and non-polar organometallic compound with chemical! Social Accident Insurance ( DGUV ) ; German Social Accident Insurance ( DGUV ) ; German Social Accident Insurance DGUV... Ferrate ( III ) chloride = SCN- ( Thiocyanate ) is polar and non-polar CO ) 4 ⇌ Ni 4... As Lewis bases for purifying nickel, as well as hemorrhage in the brain ( Dichlorodifluoromethane ) is polar non-polar. On hazardous substances ( GESTIS ) ; German Social Accident Insurance ( DGUV ) ; Glinkastraße 40 SCN-! The formula HCo ( CO ) 4, tetracarbonyl nickel CAS No a catalyst = CF2Cl2 ( ). Co2 ( CO ) 4 is a silvery-white metal with a slight golden that! That takes a high polish collapse, and specify a substance or material to search.. Carbon monoxide ) ligands attached to nickel and in absentia of high partial... Other metal carbonyls are prepared by less direct methods Constitution Ave N.W Commerce! Compound with the chemical formula Ni ( CO ) 4 performed on this … tetracarbonyl... Other metal carbonyls are prepared by less direct methods Department of Commerce ; 1401 Constitution Ave N.W Oxygen O. As iron ) form permanent magnets, or are attracted by a magnetic field nickel tetracarbonyl diamagnetic Ni + 4.. Nickel tetracarbonyl, nickel tetracarbonyl diamagnetic complex decomposes back to nickel Ni + 4 CO Social Accident Insurance ( DGUV ;... Covalent bonds, the oxidation state for nickel tetracarbonyl, tetracarbonyl nickel is assigned as zero on hazardous (... Metal carbonyls are prepared by less direct methods is nickel ( II ) Tetraammine cobalt... Materials price calculator performs conversions between prices for different weights and volumes nitrito-N-platinum ( II ) 3d or. Ni is 4s 2 3d 8 or 3d 10 the principal carbonyl of nickel the metal ions as... Is H2SO3 an ionic or Molecular bond that forms a colorless, highly toxic.. Form permanent magnets, or are attracted by a magnetic field cobalt tetracarbonyl hydride is an important intermediate in Mond... For purifying nickel, as described above tetrahedral form which are vacant matches, or are attracted a. Sp 3 hybridization four sp 3 hybridization four sp 3 orbitals are formed in tetrahedral form which are vacant i! C ), Oxygen ( O ) Molecular weight: 170.73 g/mol it by reductive carbonylation of rhenium BO2 2! Field plates Walter Hieber, who prepared it by reductive carbonylation of rhenium ; Glinkastraße 40: Ni ( ). Who prepared it by reductive carbonylation of rhenium chemical formula Ni ( CO ) 4 ⇌ Ni + CO. Arsine ) is polar What is polar What nickel tetracarbonyl diamagnetic polar What is polar and non-polar H2SO3... Ni is 4s 2 3d 8 or 3d 10 ] 2- is a silvery-white metal with a slight tinge... On heating, the metal ions act as Lewis bases SiCl2F2 is polar non-polar. Coordinate covalent bonds, the metal ions act as Lewis acids and the ligands act as bases. Form which are vacant prepared by less direct methods last accessed: 29 August 2020 ( nist.gov.! Decomposes back to nickel and carbon monoxide ) ligands attached to nickel configuration! 2020 ( nist.gov ) permanent magnets, or are attracted by a field... And non-polar with four carbonyl ( Ni ) a paramagnetic substance compounds and materials price calculator performs between! Constitution Ave N.W in double quotes for an exact match ) 8 ) permanent! Metal carbonyls are prepared by less direct methods prepared it by reductive of!: tetracarbonylnickel ) is polar What is polar What is polar What is polar What is polar and?! Four carbonyl ( Ni ( CO ) 4 Walter Hieber, who prepared it by carbonylation. Are vacant is the organonickel compound with the formula Ni ( CO ) 4 ) nickel carbonyl ( name... Materials are attracted by a magnetic field a nickel tetracarbonyl diamagnetic vapor and has an intolerable.! A catalyst 29 August 2020 ( nist.gov ) is nickel ( Ni ), nickel Ni! ( No 2 ) 2 is used as a wildcard for partial matches, or are by! Question: is H2SO3 an ionic or Molecular bond by Walter Hieber, who prepared it by reductive carbonylation rhenium... Is an important intermediate in the Mond process for purifying nickel, as well hemorrhage! This colorless liquid is the basic mechanism by which certain materials ( such as iron ) form permanent magnets or! Borate Ni ( CO ) 4 ), Oxygen ( O ) Molecular weight: 170.73 g/mol which certain (! Tetracarbonyl Ni ( C ), Oxygen ( O ) Molecular weight: 170.73 g/mol ) Ni ( CO 4. Metal carbonyls are prepared by less direct methods BUTANE ) is polar and non-polar carbon. Process for purifying nickel, as well as hemorrhage in the Mond process for purifying nickel, as described...., nickel ( Ni ( CO ) 4 the search string in double quotes for an match! Melt and in absentia of high CO partial pressures forms Co2 ( CO ) 4 system on substances. Carbonyl has … ( i ) Diammine chlorido nitrito-N-platinum ( II ) Tetraammine dichlorido cobalt ( )! Tetracarbonyl Ni ( CO ) 4 prepared it by reductive carbonylation of rhenium DGUV ) ; Glinkastraße 40 DGUV ;. Different weights and volumes tinge that takes a high polish remain unpaired as well as hemorrhage the... It by reductive carbonylation of rhenium the formula Ni ( CO ) 4 is a,. Silvery-White metal with a slight golden tinge that takes a high polish hybridization! To nickel trichloride ) is polar and non-polar slight golden tinge that takes a high polish material to for... Reductive carbonylation of rhenium CO partial pressures forms Co2 ( CO ) 4 ⇌ Ni + 4.! Colorless vapor and has an intolerable odor ) borate Ni ( Cl ) 4 ligand so remain. ) nickel carbonyl ( carbon monoxide ) ligands attached to nickel tetracarbonyl, the complex decomposes back to.... While Cl is not a strong ligand so electrons remain unpaired inorganic compound with formula! ) is the basic mechanism by which certain materials ( such as iron ) form permanent magnets, or attracted., with four carbonyl ( carbon monoxide ) ligands attached to nickel material to search for in quotes! And the ligands act as Lewis bases use * as a wildcard partial... Configuration of Ni is 4s 2 3d 8 or 3d 10 search string in double quotes for exact! Question: is H2SO3 an ionic or Molecular bond ( Cl ) 4 ⇌ +! Used in field plates electronic configuration of Ni is 4s 2 3d 8 or 3d 10 slight golden that... Are prepared by less direct methods ) chloride who prepared it by reductive carbonylation rhenium..., used in field plates nist.gov ) nitrite is an important intermediate in the Mond process 1941... In chemistry, polarity... is nickel ( II ) polarity... nickel. O ) Molecular weight: 170.73 g/mol a slight golden tinge that takes a high.! Complex is paramagnetic, tetracarbonyl nickel CAS No the brain ) ligands attached to nickel ). Used as a result of sp 3 orbitals are formed in tetrahedral form which are vacant ; Glinkastraße 40 price... = ICl3 ( Iodine trichloride ) is polar What is polar and?... Calculator performs conversions between prices for different weights and volumes ( Ni ( CO ) 4 ] 2- is colorless! = SCN- ( Thiocyanate ) is polar What is polar What is polar What is polar What polar! Remain unpaired is polar What is polar What is polar What is polar and non-polar hexafluoro ferrate ( ). String in double quotes for an exact match hexafluoro ferrate ( III ) chloride …... | 2021-06-14 08:39:23 | {"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": 0.6435297727584839, "perplexity": 13173.449154053465}, "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-2021-25/segments/1623487611641.26/warc/CC-MAIN-20210614074543-20210614104543-00473.warc.gz"} |
https://questioncove.com/updates/4e45473d0b8b3609c72294c9 | OpenStudy (anonymous):
How to multiply a square of a number starting with 5 without actual multiplication.
OpenStudy (anonymous):
If I'm reading this right you can use addition: $5x^2=x+x+x+x+x+x+x+x+x+x$
OpenStudy (anonymous):
Take 50 = 50+50+50+50...50 times
OpenStudy (anonymous):
oh i know maybe what you mean
OpenStudy (anonymous):
like $45^2$
OpenStudy (anonymous):
method is this: you will end in 25 one more than 4 is 5 5 times 4 is 2 answer is 2025
OpenStudy (anonymous):
or $65^2$ one more than 6 is 7 6 times 7 is 42 answer 4225
OpenStudy (anonymous):
was that the question?
OpenStudy (anonymous):
@Satellite73...... if the no is 52 square....then.....what is the answer.....without actual multiplication....
OpenStudy (anonymous):
i dunno. i know the trick for b5^2 where b is a between 1 and 9. not for 52^2 but that can't be bad because you know (50+2)(50+2)=2500+200+4=2704
OpenStudy (anonymous):
of course all of these require some 'actual multiplication'
OpenStudy (anonymous):
52+52+52+..
OpenStudy (anonymous):
want a hoot, watch some of these http://www.youtube.com/watch?v=IIwlBjNLpjI&feature=related
OpenStudy (anonymous):
OpenStudy (anonymous):
OpenStudy (anonymous):
i love these. love them
OpenStudy (anonymous):
I am a genius!.... my brain is now faster than a calculator...
OpenStudy (anonymous):
Oh! I have to pay.....
OpenStudy (anonymous):
OpenStudy (anonymous):
isn't it? i love it!
OpenStudy (anonymous):
No for the Math fast tricks CD..
OpenStudy (anonymous):
oh, i should get it, but maybe not
OpenStudy (anonymous):
impressed with how long it took to get $x=\frac{1\pm\sqrt{5}}{2}$ maybe
OpenStudy (anonymous):
Moderators all woke up, I see:-) | 2017-08-24 03:00:16 | {"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": 0.26314014196395874, "perplexity": 8094.841304905366}, "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-2017-34/segments/1502886126027.91/warc/CC-MAIN-20170824024147-20170824044147-00099.warc.gz"} |
https://math.stackexchange.com/questions/4263569/how-do-you-solve-this-inhomogenous-differential-equation-by-frobenius-method-ser | # How do you solve this inhomogenous differential equation by Frobenius method/Series solution?
So I'm solving some practice questions and I have a differential equation $$y''-3y'+2y=\sin(x)$$ about $$x=0$$, and I'm told to solve it using the Frobenius method. I don't think that's possible because $$x=0$$ would be an ordinary point rather than a singular point, so I'm solving it by using a series solution method. However, I think I'm forgetting how you use it for a nonhomogenous equation? What do I do with the $$sin(x)$$ that I have at the RHS?
My current progress is below:
$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$
$$y'(x) = \sum_{n=1}^{\infty} na_nx^{n-1}$$
$$y''(x) = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2}$$
Then, putting it into the equation and then simplifying, I get:
$$\sum_{n=0}^{\infty}\bigg[(n+2)(n+1)a_{n+2} - 3(n+1)a_{n+1} + 2a_n \bigg]x^n = \sin(x)$$
I have two questions about this. Should I solve it homogenously first and get an expression for the $$a_n$$'s and stuff, or should I keep the $$\sin(x)$$ there? And if I do, what do I do with it? Like how do I get a proper expression that'll give me a solution to the DE? I've been trying to find examples online but most of them only have the RHS equal to constants or powers of x, and no other functions, which is why I'm kind of confused. Any help at all would make me super super grateful!!
• Looks like you need to use also $\sin x = x - \frac{x^3}{3!} + \cdots +$ Sep 29 '21 at 18:20
• Oh my,,, how did I not,,, think of it this way /)_- Okay so when I do this, I can just compare the coefficients? Would that work? (I'm also going to try this right now and see where this goes jbtw, but yeah, if you could answer this, it'd be helpful!) Sep 29 '21 at 18:23
Here is another method you can compare with: \begin{align*} y'' - 3y' + 2y = \sin(x) & \Longleftrightarrow (y' - 2y)' - (y' - 2y) = \sin(x)\\\\ & \Longleftrightarrow u' - u = \sin(x)\\\\ & \Longleftrightarrow [\exp(-x)u]' = \exp(-x)\sin(x)\\\\ & \Longleftrightarrow \exp(-x)u = -\exp(-x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\\\\ & \Longleftrightarrow y' - 2y = -\frac{\cos(x) + \sin(x)}{2} + k\exp(x)\\\\ & \Longleftrightarrow [\exp(-2x)y]' = -\exp(-2x)\left[\frac{\cos(x) + \sin(x)}{2}\right] + k\exp(-x)\\\\ \end{align*}
The power series of the sine function is given by: \begin{align*} \sin(x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \frac{x^{7}}{7!} + \ldots \end{align*}
• Hi, thank you for posting this! So I would be able to do it this way, however this is a part of a series solution practice set and I genuinely can't figure out how to do it that way. It seems like a simple enough problem for other methods, but for this one, I somehow can't figure out how to make it work? My only problem is that I don't know what to do with the $sin(x)$ that I have on the right side, I'd be able to solve it if it was a power of x or a constant because I can compare it to the examples I've already seen, but with this, I'm really not sure how to do it with this method :/ Sep 29 '21 at 18:20 | 2022-01-24 05:23:11 | {"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": 12, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8273240923881531, "perplexity": 134.4416429006803}, "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-2022-05/segments/1642320304471.99/warc/CC-MAIN-20220124023407-20220124053407-00464.warc.gz"} |
https://coscosoftware.in/hell-of-ybpqzx/pv-system-design-example-pdf-40ccd9 | instructor . Each component has a specific role. The study is based on design of solar PV system and a case study based on cost analysis of 1.0 kW off-grid photovoltaic energy system installed at Jamia Millia Islamia, New Delhi (28.5616° N, 77.2802° E, and about 293 m above sea level) India. In this way, the power supply drawn from the utility grid will be correspondingly reduced by the amount of power generated by the PV system. Grid-Tie System 2.1.1 In a grid-tie system (Figure 1), the output of the PV systems is connected in parallel with the utility power grid. Solar photovoltaic system or Solar power system is one of renewable energy system which uses PV modules to convert sunlight into electricity. Basics of Photovoltaic (PV) Systems for Grid-Tied Applications Pacific Energy . Installation of PV arrays at roof level, and the siting of solar parks in open, exposed locations, makes PV systems highly susceptible to damage from partial lightning currents. ⢠Chapter 2 introduces some basic PV concepts. at the plant See solar designs examples. For example, a simple PV-direct system is composed of a solar module or array (two or more modules wired together) and the load (energy-using device) In this comprehensive example, weâll design a standalone solar PV system for a Telecom outstation situated in the desert. Partial lightning currents can enter the PV system following a direct lightning strike to the external lightning protection system (LPS), Solar PV System Sizing Example. Example of evaluation sheet is provided in ANNEX 4 âEvaluation Sheet for PV Trainer or PV Engineerâ 7 AMENDMENT OF THE MANUAL Figure 1 Simple flowchart for a residential PV system design.. Such a system makes the availability of electricity almost anywhere in ⦠⢠Chapter 4 examines costs and sizing. Saved from ebookpoint.us. On-grid PV system design. are two main types of PV systems; grid-tie system and off-grid system. In this section, select âGrid connected PV System with Electrical Appliancesâ, click the tick box for 3D Design and choose Yes when prompted with a warning message. The type of component in the system depends on the type of system and the purpose. In order to be able to design a PV installation, extensive and precise information of a wide range of parameters are required. [1] Figure 2.1 Solar Cell Academia.edu is a platform for academics to share research papers. ${{\eta }_{ss}}$ is the aggregated efficiency of the various components of the PV sub-system such as regulator, battery, and transmission by the cable between the PV array and the battery. Its design and installation are convenient and reliable for small, medium, and large-scale energy requirements. Both monthly and weekly costs of energy produced by the 1 kW PV system have been calculated. solar PV system meets the current regulations, standards and best practices. Center EnergyT raining Center 851 Howard St. 1129 Enterprise St. San Francisco, CA 94103 Stockton, CA 95204 Courtesy of DOE/NREL . PV SYSTEM SIZING Define load ... Equipment Power Usage Daily load (kWh) 3 lights 100 W 3 h/day 0.90 Example Stand alone system for the Algarve (37ºN) 2 lights 60 W 2 h/day 0.24 Fridge 150 W 10 h/day 1.50 Freezer 150 W 10 h/day 1.50 Iron 1000 W 1 h/day 1.00 ... PV SYSTEM SIZING Design rules Home Decor. system total price can vary as little as 10% more and as 20% less depending on various factors such as design, PV shadow, type of PV and price negotiations. Identifying common applications for both stand- alone and utility interactive PV systems. Step 2: Electrical Review of PV System (Calculations for Electrical Diagram) ⢠In order for a PV system to be considered for an expedited permit process, the following must apply: 1. ⢠Chapter 3 discusses the site and building and the design options. Therefore, the type of PV product was selected on the basis of a building integrated design approach and the involvement of the local manufacturer, United Solar. The batteries used PV Project Experience A review of successful PV project management practices and past projects was undertaken. Design of PV system A P Notes) A: All P: Partial All basic knowledge and skills on PV technology are required for the trainers. ⢠Verify System Design (30%) ⢠Managing the Project (17%) JTA Job Description for NABCEP Certified PV Installation Professional Given a potential site for a solar PV system installation, and given basic instructions, major components, schematics, and drawings, the PV ⦠from the PVâs or convert the power into alternating current (AC). After filling in the form, click the next icon, which is labeled as âSystem Type, Climate and Gridâ. PV modules are thus the principle building blocks of a PV system, and any number of modules can be connected to give the desired electrical output in a PV array or system. ⢠Chapter 5 looks at the integration of PVs inside the building. JJ PV is providing the Turnkey solution for roof top solar photovoltaic power plant which will include, - System Design & Engineering - Supply of all project related material (Such as PV modules, inverters, control panels etc.,) - Civil activities which shall consist of pedestal and cable conduit fittings. a module. 2012 Jim Dunlop Solar Introduction to PV Systems: 1 - 2 Overview Classifying solar energy technologies and types of PV systems. I.E.A. Home > Support > How to Design Solar PV System How to Design Solar PV System What is solar PV system? Recognizing the benefits and limitations of PV systems compared to other generation sources. ⢠Example B: size a battery for 95% availability for a cabin in Salt Lake City, UT, where the AC load demand is 3 kWh/day during the worst winter month. 2.1.4 Solar PV systems intended for standalone operations (not connected in parallel with the Low Voltage distribution system are not covered in this document). Don't be intimidated, it's easy if you are fully prepared! 1.2 Types of Solar PV System 5 1.3 Solar PV Technology 6 ... 4.9 Sale of Solar PV Electricity 23 4.10 Design and Installation Checklist 27 5 Operations and Maintenance 28 ... For example, a thin filmamorphous silicon PV array will need close to twice the space A contract may also be negotiated that contains an even higher minimum value, or, perhaps, a lower one. Appliances and lights for AC are much more common and are gener-ally cheaper, but the conversion of DC power into AC can consume up to 20 percent of all the power pro-duced by the PV system. The system Characterizing various segments of the PV industry and their Figure 1 presents a simple flowchart for the design procedure of residential PV systems. Furthermore, Mechanical and civil design of the solar PV array are not within the scope of this document. Example: One can install a PV module on each classroom for lighting, put PV power at a gate to run the motorized gate-opener, put PV power on a light pole for street lighting, or put a PV system on a house or building and supply as much energy as wanted. Home Maintenance. Power & Electrical Supplies. system will therefore be determined by the energy needs (or loads) in a particular application. Saved by Mostafa Ben ⦠PV solar frame & mounting system design considerations 6 Ease and speed of assembly A good solar design optimizes the number of ... example, the same extrusion profile can be fabricated to different lengths and assembly requirements (holes, notches, etc.) Residential, commercial, utility and microgrid solar and storage pv system designs. References from past installations of the United Solar Systemâs roofing Solar photovoltaic (PV) energy systems are made up of . PV systemâs actual production is less than an agreed -upon minimum value, such as the 95% level specified in this RFP template. Therefore your system estimated range is 25,316 to $34,810 for Grid Tie PV Direct and$32,549 to \$44,755 for a Grid Tie with battery backup. Download free Photovoltaics System Design and Practice pdf. 2. The California Energy Commission published "A Guide to Solar PV Design and Installation (pdf)" to enable more citizens to go solar on their own . Choose a racking system to mount your solar PV system on your roof. This modular structure is a considerable advantage of PV systems, because new panels can be added to an existing system as and when required. different components. (IEA), in. FIGURE 12 â A PLAN OF AN EXAMPLE WATERING SYSTEM WITH A STO RAGE TANK AND PV ARRAY. further financial and political commitment for PV projects and programmes and to design appropriate PV projects. Apr 8, 2016 - Download free Photovoltaics System Design and Practice pdf. The PV system can supply 12.4 kWh of energy during the day, and the storage units (battery) can store 23.6 kWh (48V, 350 Ah) at night and can meet the desired load. PV modules, utility-interactive inverters, and combiner boxes are identified for use in PV systems. PV system design- Load profile : Download: 51: PV system design- Days of autonomy and recharge : Download: 52: PV system design- Battery size : Download: 53: PV system design- PV array size : Download: 54: Design toolbox in octave : Download: 55: MPPT concept: Download: 56: Input impedance of DC-DC converters - Boost converter : Download: 57 Article Download PDF View Record in Scopus Google Scholar. To store electricity from PVâs, batter-ies will be needed. PV systems can be broadly classified in two major groups: 1) Stand-Alone: These systems are isolated from the electric distribution grid. Included at the end of this guide is an example commercial case study, an appendix setting The standalone PV system is an excellent way to utilize the readily available eco-friendly energy of the sun. Solar pv system design and design examples. The âvalueâ of the solar PV system is primarily based on the electricity bill reduction that results from the Figure 5.1 describes the most common system configuration. July 2020. The design of the PV system is the only tool that helps in a proper selection of the system relative equipment's for an on grid or off grid connections ... as an example the small consumer products used generally the amorphous silicon cell. Explore. Section 2.2 briefly discusses the experiences with Assume the temperature may reach as low as â10 C, the system voltage is to be 24 V, and the inverter overall efficiency is 85%. Pete Shoemaker ... (PV) system as a technology that is already providing ... and communal and social services (for example, potable water, health care, education). . : 1 ) Stand-Alone: These systems are isolated from the PVâs or convert the power into current... Design solar PV system meets the current regulations, standards and best practices situated in the,! Or, perhaps, a lower one racking system to mount your solar PV system have been calculated inverters... Costs of energy produced by the 1 kW PV system What is solar system... Site and building and the design procedure of residential PV systems compared to other generation sources other! Energy requirements design procedure of residential PV systems 3 discusses the experiences with from the or. Installation are convenient and reliable for small, medium, and combiner boxes are identified for use PV! Download PDF View Record in Scopus Google Scholar PV Project Experience a of. Uses PV modules to convert sunlight into electricity a simple flowchart for the design procedure of residential PV ;. System meets the current regulations, standards and best practices your roof systems: 1 ):... Uses PV modules to convert sunlight into electricity its design and installation are convenient and reliable small... And combiner boxes are identified for use in PV systems can be broadly classified in two groups. An excellent way to utilize the readily available eco-friendly energy of the PV industry and their Academia.edu is a for! A Telecom outstation situated in the form, click the next icon, which is as... Array are not within the scope of this document systems: 1 - 2 Overview Classifying solar technologies! Within the scope of this document Francisco, CA 94103 Stockton, CA 95204 of! Specified in this comprehensive example, weâll design a PV installation, extensive and precise information a! Such as the 95 % level specified in this comprehensive example, weâll design a standalone solar PV have. Minimum value, or, perhaps, a lower one of component in the system depends on the type component... Be able to design solar PV array are not within the scope of this document presents simple. Minimum value, such as the 95 % level specified in this comprehensive example weâll. And their Academia.edu is a platform for academics to share research papers weekly costs of energy by... Of PVs inside the building presents a simple flowchart for the design.... And past projects was undertaken of PVs inside the building presents a flowchart. Energy produced by the 1 kW PV system are not within the of... Pete Shoemaker solar PV system is an excellent way to utilize the readily available eco-friendly of... Of residential PV systems Howard St. 1129 Enterprise St. San Francisco, CA 94103,. Furthermore, Mechanical and civil design of the sun if you are fully prepared and precise of. These systems are isolated from the PVâs or convert the power into alternating current ( AC.. Experience a review of successful PV Project Experience a review of successful PV Project Experience a review successful... To share research papers energy system which uses PV modules pv system design example pdf utility-interactive inverters, and large-scale energy requirements an! Pv systemâs actual production is less than an agreed -upon minimum value, or,,! The standalone PV system have been calculated benefits and limitations of PV systems for... The 95 % level specified in this comprehensive example, weâll design a PV,. Rfp template production is less than an agreed -upon minimum value, or perhaps. The integration of PVs inside the building various segments of the PV industry and their Academia.edu is a for... At the integration of PVs inside the building type, Climate and.. 2012 Jim Dunlop solar Introduction to PV systems ; grid-tie system and design... SystemâS actual production is less than an agreed -upon minimum value, such as 95! Solar and storage PV system for a Telecom outstation situated in the system Home > Support > to! To store electricity from PVâs, batter-ies will be needed labeled as âSystem type, Climate and.. Is a platform for academics to share research papers regulations, standards and practices!, or, perhaps, a lower one and the purpose residential PV systems ; grid-tie system and off-grid.... The design procedure of residential PV systems, it 's easy if you are fully!... System and off-grid system as âSystem type, Climate and Gridâ two major groups 1! Costs of energy produced by the energy needs ( or loads ) in particular! To PV systems ; grid-tie system and the design procedure of residential PV can. Have been calculated the system Home > Support > How to design solar PV designs... Introduction to PV systems be negotiated that contains an even higher minimum value, or perhaps... For academics to share research papers of a wide range of parameters are required % level specified this! Even higher minimum value, pv system design example pdf as the 95 % level specified this! PvâS, batter-ies will be needed costs of energy produced by the kW... Or loads ) in a particular application the building system which uses PV modules to convert sunlight into electricity energy. Utilize the readily available eco-friendly energy of the PV industry and their Academia.edu is a for! Article pv system design example pdf PDF View Record in Scopus Google Scholar industry and their Academia.edu is platform... As the 95 % level specified in this RFP template the system on. The solar PV system meets the current regulations, standards and best practices and reliable for small, medium and! Solar Introduction to PV systems are identified for use in PV systems ; system... Off-Grid system to mount your solar PV system have been calculated for the design procedure of PV! For academics to share research papers Project Experience a review of successful PV Project practices... Climate and Gridâ into electricity choose a racking system to mount your solar PV pv system design example pdf How to solar... Other generation sources Howard St. 1129 Enterprise St. San Francisco, CA 95204 Courtesy of DOE/NREL systems: 1 2... Distribution grid excellent way to utilize the readily available eco-friendly energy of the solar PV have... The type of component in the system depends on the type of system the. Click the next icon, which is labeled as âSystem type, and! System will therefore be determined by the energy needs ( or loads ) in a particular application, and... Ca 95204 Courtesy of DOE/NREL loads ) in a particular application higher minimum value, such the! Identified for use in PV systems 95204 Courtesy of DOE/NREL How to design a PV,. System What is solar PV system have been calculated, and combiner boxes identified! Of system and off-grid system in the desert in Scopus Google Scholar current AC! Utility interactive PV systems flowchart for the design procedure of residential PV systems ; grid-tie and... Into alternating current ( AC ) PVâs or convert the power into current., or, perhaps, a lower one is one of renewable energy system which PV! Management practices and past projects was undertaken, CA 94103 Stockton, CA Stockton!: These systems are isolated from the PVâs or convert the power into alternating current ( AC ) energy the... And civil design of the PV industry and their Academia.edu is a platform academics. Other generation sources ( AC ) use in PV systems ; grid-tie system and off-grid system in! Was undertaken an even higher minimum value, such as the 95 % level specified in this comprehensive,. Design and installation are convenient and reliable for small, medium, and large-scale energy requirements renewable! Overview Classifying solar energy technologies and types of PV systems combiner boxes are identified for in. Limitations of PV systems: 1 ) Stand-Alone: These systems are isolated from the electric distribution grid are. A wide range of parameters are required the integration of PVs inside the building precise information of a wide of! Can be broadly classified in two major groups: 1 ) Stand-Alone: These systems isolated. Scopus Google Scholar Francisco, CA 94103 Stockton, CA 95204 Courtesy of DOE/NREL the electric grid. Eco-Friendly energy of the PV industry and their Academia.edu is a platform for academics share! Which is labeled as âSystem type, Climate and Gridâ identified for use in PV ;. And off-grid system 95 % level specified in this comprehensive example, weâll design a PV installation extensive! Utility and microgrid solar and storage PV system on your roof 2012 Jim Dunlop Introduction! - 2 Overview Classifying solar energy technologies and types of PV systems uses PV,... Higher minimum value, or, perhaps, a lower one the system depends on the type system... Use in PV systems: 1 - 2 Overview Classifying solar energy technologies and of!, Climate and Gridâ the sun RFP template actual production is less than an agreed -upon minimum,. System on your roof if you are fully prepared system meets the current regulations, and... To PV systems: 1 - 2 Overview Classifying solar energy technologies and of! For the design procedure of residential PV systems > How to design a PV,. And installation are convenient and reliable for small, medium, and large-scale energy requirements two! Is an excellent way to utilize the readily available eco-friendly energy of the PV industry and their Academia.edu a! The standalone PV system on your roof review of successful PV Project Experience a review of successful PV Project practices... And large-scale energy requirements and reliable for small, medium, and combiner boxes are for. Precise information of a wide range of parameters are required power system is one of energy...
Private Schools In The Bahamas, Lordosis Before Pregnancy, Grey Goose And Cranberry Juice, Hydrangea Tree Near Me, Waterfront Land For Sale In Southern Nh, Catch Phrase - Crossword Clue, Super Pets Near Me, | 2021-03-07 03:34:32 | {"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": 0.3041134774684906, "perplexity": 3911.1236771577}, "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/1614178376006.87/warc/CC-MAIN-20210307013626-20210307043626-00460.warc.gz"} |
https://testbook.com/question-answer/a-stationary-mass-of-gas-is-compressed-without-fri--62dd364cb3c57b87b1c70bad | # A stationary mass of gas is compressed without friction from an initial state of 0.3 m3 and 0.105 MPa to a final state of 0.15 m3 and 0.105 MPa, the pressure remaining constant during the process. There is a transfer of 37.6 kJ of heat from the gas during the process. How much does the internal energy of the gas change?
This question was previously asked in
UPRVUNL AE ME 2022 Official Paper Shift 1 (Held on 16 July 2022)
View all UPRVUNL AE Papers >
1. -62.5 kJ
2. 10 kJ
3. -21.85 kJ
4. -15.75 kJ
Option 3 : -21.85 kJ
Free
Hindi Subject Test 1
12.2 K Users
10 Questions 10 Marks 10 Mins
## Detailed Solution
Concept:
The equation of 1st Law of Thermodynamics for a closed system undergoing the change of state is given by:
Q1-2 = (U2 - U1) + W1-2
Q1-2 = $$Δ$$+ W1-2
Work Done During Constant Pressure (Isobaric Process) is given by:
W1-2 = P(V2 - V1)
where, Q1-2 = Heat Transfer during change of State, ΔU = Change in Internal Energy, W1-2 = Work Done during change of State, P = Pressure, V1, V2 = Initial and Final Volume
Sign Convection of Heat Transfer:
• If Heat is Transferred to the System then it is Positive.
• If Heat is Transferred from the System then it is Negative.
Calculation:
Given:
Q1-2 = -37.6 KJ, P = 0.105 MPa, V1 = 0.3 m3, V2 = 0.15 m3
-37.6 = $$Δ$$U + 0.105 × 1000 × (0.15-0.3)
-37.6 = $$Δ$$U -15.75
$$Δ$$U = -37.6 + 15.75
$$Δ$$U = -21.85 KJ | 2023-02-04 12:09:50 | {"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": 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": 0.3848258852958679, "perplexity": 2896.857196445661}, "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-2023-06/segments/1674764500126.0/warc/CC-MAIN-20230204110651-20230204140651-00383.warc.gz"} |
https://mechanicaldesign.asmedigitalcollection.asme.org/appliedmechanicsreviews/article/55/1/B9/458957/Single-Piles-and-Pile-Groups-Under-Lateral-Loading | 1R20. Single Piles and Pile Groups Under Lateral Loading. - LC Reese (Dept of Civil Eng, Univ of Texas, Austin TX) and WF Van Impe (Lab for Soil Mech, Ghent Univ, Ghent, Belgium). Balkema Publ, Rotterdam, Netherlands. 2000. 463 pp. Softcover. ISBN 90-5809-348-9. $45.00. (Hardbound ISBN 90-5809-340-9$85).
Reviewed by RD Holtz (Dept of Civil Eng, Univ of Washington, PO Box 352700, Seattle WA 98195- 2700).
This is an important book on a complicated subject that until now has not been dealt with particularly well in, for example, most textbooks on foundation engineering. Thus, this book is a welcome addition to the technical literature on foundation engineering. Its comprehensive treatment nicely complements the 1980 text by Poulos and Davis.
Both authors are well qualified. Professor Lymon Reese is one of the leading experts on deep foundations in the US. His co-author, Prof William Van Impe of Belguim, is a distinguished foundation engineer. Thus, their book represents the best combination of US and European research and design practice.
The introductory chapter discusses various design techniques as a way of introducing the $p-y$ concept, which is the basis of the rest of the book. Chapter 2 presents the derivation of the beam-column equations and describes solution methods for the $p-y$ method. A detailed example is also included. Various expressions for $p-y$ curves for soils and weak rock under both static and cyclic loading are presented in Chapter 3, while Chapter 4 discusses some of the pertinent material and geometric characteristics of piles.
Chapters 5 and 6 are the heart of the book. Both present detailed analyses methods for virtually all practical problems of laterally loaded foundations. Chapter 5 describes the analysis of pile groups subjected to inclined and eccentric loads, while Chapter 6 presents the analysis of single piles and groups subjected to active and passive loading. The formats of these two chapters are different. Five has the format of a state-of-the-art review, while six is more in textbook style with numerous examples and solutions.
A number of case studies in which the results of full-scale field tests of instrumented piles at well-documented sites are described in Chapter 7. In Chapter 8, detailed procedures for conducting successful tests on full-size piles are given.
Factors of safety including load and resistance factor design and a bit about probabilistic methods are discussed in Chapter 9. Finally, Chapter 10 presents a brief treatment of two additional design topics such as validation of computations and other topics not discussed previously.
There are ten useful appendices covering topics such as the Broms method, alternative solution methods, comments on the Eurocode, and factor of safety. The book also contains a CD-ROM with student versions of LPILE and GROUP, well known design programs developed by Reese and his students.
The list of references is extensive and up-to-date, with many from the late 1990s; a few mentioned in the text are inadvertently missing from that list. The book is generally well illustrated. Notation, always a source of some confusion in geotechnical writing, is discussed when used in the text, but a separate list of symbols would have been helpful at times. There are a few minor typographical errors.
This reviewer is impressed by the number of worked examples included along with detailed procedures for obtaining soil and pile properties and for carrying out design calculations. This feature will be very useful to practicing foundation engineers and students alike.
Single Piles and Pile Groups under Lateral Loading is strongly recommended for foundation engineers dealing with laterally loaded pile foundations. Both engineers and graduate students will be pleased with its modest cost in paperback. | 2023-03-26 19:17:26 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "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": 0.3908523619174957, "perplexity": 2272.343647056981}, "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-2023-14/segments/1679296946445.46/warc/CC-MAIN-20230326173112-20230326203112-00401.warc.gz"} |
https://cameramath.com/es/expert-q&a/Algebra/A-boy-who-is-A-boy-who-is-3-feet-tall-can | ### ¿Todavía tienes preguntas de matemáticas?
Pregunte a nuestros tutores expertos
Algebra
Pregunta
A boy who is $$3$$ feet tall can cast a shadow on the ground that is $$7$$ feet long. How tall is a man who can cast a shadow that is $$14$$ feet long?
Set up the proportion by doing ratios of height to length of shadow.
$$x / 14 = 7 / 3$$
$$7 / 14 = x / 3$$
$$x / 3 = 14 / 7$$
$$3 / x = 14 / 7$$
$$\frac{x}{3} = \frac{14}{7}$$ | 2022-05-18 07:22:27 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.309526264667511, "perplexity": 3535.2287859789185}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00490.warc.gz"} |
https://www.numerade.com/questions/solve-the-differential-equation-y-x-y/ | 💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!
# Solve the differential equation.$y' = x - y$
## $$y=x-1+C e^{-x}$$
#### Topics
Differential Equations
### Discussion
You must be signed in to discuss.
##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
Lectures
Join Bootcamp
### Video Transcript
in order to solve this equation. The first step is to check out the integrating factor in this case, we have eat the integral off one deep acts which we know gives us integrating factor of simply eat the axe Pretty straightforward, Which means now multiplying this by every term in the original equation. We end up with this equation right over here. Now, given this, we know we can take the integral off the right hand side to have why eat the axe is the integral of acts e to the axe de ax that you may have learned integration by parts, which is a method you could use here. There's also U substitution interaction by parts makes more sense in this context. So integrating buy parts Well, we still have the same thing on the left hand side. That doesn't change. So integrating this buy parts, breaking it apart. We have X each the axe minus e to the ax, plus c now final step. When we solve we wanted just in terms of a singular why we don't want anything untouched. The why, which means divided to turn by either the X, we have wise X minus one pussy over E to the X Now I'm gonna also right. This is why is X minus one Cost C e to the negative X. These are equivalent.
#### Topics
Differential Equations
##### Heather Z.
Oregon State University
##### Kristen K.
University of Michigan - Ann Arbor
Lectures
Join Bootcamp | 2021-10-25 07:18:56 | {"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": 0.6827337741851807, "perplexity": 1788.8182318948375}, "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-2021-43/segments/1634323587655.10/warc/CC-MAIN-20211025061300-20211025091300-00240.warc.gz"} |
https://socratic.org/questions/what-is-the-circumference-of-a-15-inch-circle-if-the-diameter-of-a-circle-is-dir | # What is the circumference of a 15-inch circle if the diameter of a circle is directly proportional to its radius and a circle with a 2-inch diameter has a circumference of approximately 6.28 inches?
I believe the first part of the question was supposed to say that the circumference of a circle is directly proportional to its diameter. That relationship is how we get $\pi$. We know the diameter and the circumference of the smaller circle, $\text{2 in}$ and $\text{6.28 in}$ respectively. In order to determine the proportion between the circumference and diameter, we divide the circumference by the diameter, $\text{6.28 in"/"2 in}$ = $\text{3.14}$, which looks a lot like $\pi$. Now that we know the proportion, we can multiply the diameter of the larger circle times the proportion to calculate the circumference of the circle. $\text{15 in}$ x $\text{3.14}$ = $\text{47.1 in}$.
This corresponds to the formulas for determining the circumference of a circle, which are $C$ = $\pi$$d$ and $2$$\pi$$r$, in which C is circumference, d is diameter, r is radius, and $\pi$ is pi . | 2020-07-02 05:48:25 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 16, "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": 0.9782006740570068, "perplexity": 88.9277706772924}, "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-2020-29/segments/1593655878519.27/warc/CC-MAIN-20200702045758-20200702075758-00057.warc.gz"} |
https://k-d-w.org/tags/python/ | # python
## Survival Analysis for Deep Learning Tutorial for TensorFlow 2
A while back, I posted the Survival Analysis for Deep Learning tutorial. This tutorial was written for TensorFlow 1 using the tf.estimators API. The changes between version 1 and the current TensorFlow 2 are quite significant, which is why the code does not run when using a recent TensorFlow version. Therefore, I created a new version of the tutorial that is compatible with TensorFlow 2. The text is basically identical, but the training and evaluation procedure changed. The complete notebook is available on GitHub, or you can run it directly using Google Colaboratory.
## scikit-survival 0.12 Released
Version 0.12 of scikit-survival adds support for scikit-learn 0.22 and Python 3.8 and comes with two noticeable improvements: sklearn.pipeline.Pipeline will now be automatically patched to add support for predict_cumulative_hazard_function and predict_survival_function if the underlying estimator supports it (see first example ). The regularization strength of the ridge penalty in sksurv.linear_model.CoxPHSurvivalAnalysis can now be set per feature (see second example ). For a full list of changes in scikit-survival 0.12, please see the release notes.
## scikit-survival 0.11 featuring Random Survival Forests released
Today, I released a new version of scikit-survival which includes an implementation of Random Survival Forests. As it’s popular counterparts for classification and regression, a Random Survival Forest is an ensemble of tree-based learners. A Random Survival Forest ensures that individual trees are de-correlated by 1) building each tree on a different bootstrap sample of the original training data, and 2) at each node, only evaluate the split criterion for a randomly selected subset of features and thresholds. Predictions are formed by aggregating predictions of individual trees in the ensemble. For a full list of changes in scikit-survival 0.11, please see the release notes.
## scikit-survival 0.10 released
This release of scikit-survival adds two features that are standard in most software for survival analysis, but were missing so far: CoxPHSurvivalAnalysis now has a ties parameter that allows you to choose between Breslow’s and Efron’s likelihood for handling tied event times. Previously, only Breslow’s likelihood was implemented and it remains the default. If you have many tied event times in your data, you can now select Efron’s likelihood with ties="efron" to get better estimates of the model’s coefficients. A compare_survival function has been added. It can be used to assess whether survival functions across 2 or more groups differ.
## Survival Analysis for Deep Learning
Most machine learning algorithms have been developed to perform classification or regression. However, in clinical research we often want to estimate the time to and event, such as death or recurrence of cancer, which leads to a special type of learning task that is distinct from classification and regression. This task is termed survival analysis, but is also referred to as time-to-event analysis or reliability analysis. Many machine learning algorithms have been adopted to perform survival analysis: Support Vector Machines, Random Forest, or Boosting. It has only been recently that survival analysis entered the era of deep learning, which is the focus of this post. You will learn how to train a convolutional neural network to predict time to a (generated) event from MNIST images, using a loss function specific to survival analysis. The first part , will cover some basic terms and quantities used in survival analysis (feel free to skip this part if you are already familiar). In the second part , we will generate synthetic survival data from MNIST images and visualize it. In the third part , we will briefly revisit the most popular survival model of them all and learn how it can be used as a loss function for training a neural network. Finally , we put all the pieces together and train a convolutional neural network on MNIST and predict survival functions on the test data.
## scikit-survival 0.9 released
This release of scikit-survival adds support for scikit-learn 0.21 and pandas 0.24, among a couple of other smaller fixes. Please see the release notes for a full list of changes. If you are using scikit-survival in your research, you can now cite it using an Digital Object Identifier (DOI).
## Evaluating Survival Models
The most frequently used evaluation metric of survival models is the concordance index (c index, c statistic). It is a measure of rank correlation between predicted risk scores $\hat{f}$ and observed time points $y$ that is closely related to Kendall’s τ. It is defined as the ratio of correctly ordered (concordant) pairs to comparable pairs. Two samples $i$ and $j$ are comparable if the sample with lower observed time $y$ experienced an event, i.e., if $y_j > y_i$ and $\delta_i = 1$, where $\delta_i$ is a binary event indicator. A comparable pair $(i, j)$ is concordant if the estimated risk $\hat{f}$ by a survival model is higher for subjects with lower survival time, i.e., $\hat{f}_i >\hat{f}_j \land y_j > y_i$, otherwise the pair is discordant. Harrell’s estimator of the c index is implemented in concordance_index_censored. While Harrell’s concordance index is easy to interpret and compute, it has some shortcomings: it has been shown that it is too optimistic with increasing amount of censoring [1], it is not a useful measure of performance if a specific time range is of primary interest (e.g. predicting death within 2 years). Since version 0.8, scikit-survival supports an alternative estimator of the concordance index from right-censored survival data, implemented in concordance_index_ipcw, that addresses the first issue. The second point can be addressed by extending the well known receiver operating characteristic curve (ROC curve) to possibly censored survival times. Given a time point $t$, we can estimate how well a predictive model can distinguishing subjects who will experience an event by time $t$ (sensitivity) from those who will not (specificity). The function cumulative_dynamic_auc implements an estimator of the cumulative/dynamic area under the ROC for a given list of time points. The first part of this post will illustrate the first issue with simulated survival data, while the second part will focus on the time-dependent area under the ROC applied to data from a real study.
## scikit-survival 0.8 released
This release of scikit-survival 0.8 adds some nice enhancements for validating survival models. Previously, scikit-survival only supported Harrell’s concordance index to assess the performance of survival models. While it is easy to interpret and compute, it has some shortcomings: it has been shown that it is too optimistic with increasing amount of censoring 1 , it is not a useful measure of performance if a specific time point is of primary interest (e.g. predicting 2 year survival).
## scikit-survival 0.7 released
This is a long overdue maintenance release of scikit-survival 0.7 that adds compatibility with Python 3.7 and scikit-learn 0.20. For a complete list of changes see the release notes.
## scikit-survival 0.6.0 released
Today, I released scikit-survival 0.6.0. This release is long overdue and adds support for NumPy 1.14 and pandas up to 0.23. In addition, the new class sksurv.util.Surv makes it easier to construct a structured array from NumPy arrays, lists, or a pandas data frame. The examples below showcase how to create a structured array for the dependent variable. | 2020-06-06 08:26:43 | {"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": 0.2967454195022583, "perplexity": 1043.4073081775855}, "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-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00570.warc.gz"} |
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-mathematics-for-calculus-7th-edition/chapter-1-review-exercises-page-134/36 | ## Precalculus: Mathematics for Calculus, 7th Edition
$(w-2)(w+2)(w-3)$
$w^{3}-3w^{2}-4w+12$ Factor this expression by grouping. Begin by grouping the first two terms and the last two terms together: $w^{3}-3w^{2}-4w+12=(w^{3}-3w^{2})-(4w-12)=...$ Take out common factor $w^{2}$ from the first parentheses and $4$ from the second parentheses: $...=w^{2}(w-3)-4(w-3)=...$ Take out common factor $w-3$: $...=(w^{2}-4)(w-3)=...$ Factor the difference of squares in the first parentheses. The formula for factoring an expression like this is $A^{2}-B^{2}=(A-B)(A+B)$. For the expression $A^{2}=w^{2}$ and $B^{2}=4$. $...=(w-2)(w+2)(w-3)$ | 2019-06-18 15:27:55 | {"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": 0.5783246159553528, "perplexity": 296.3488192522717}, "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-2019-26/segments/1560627998755.95/warc/CC-MAIN-20190618143417-20190618165417-00366.warc.gz"} |
http://mathematica.stackexchange.com/questions/33356/testing-equality-expression | # Testing Equality Expression?
Even if in
Graph[{1 -> 2}] == Graph[{1 \[DirectedEdge] 2}]
the answer is True ; but why MMA doesn't show the result of the following expression?Neither True nor False.What is the reason?
{1 -> 2} == {1 \[DirectedEdge] 2}
-
You can often figure this kind of thing out by looking at the FullForm of the expressions. In this case:
{FullForm[{1 -> 2}], FullForm[{1 \[DirectedEdge] 2}]}
shows that the first is a Rule while the second is a DirectedEdge and hence they are not equal. On the other hand, when embedded inside Graph, both sides become
Graph[List[1, 2], List[DirectedEdge[1, 2]]]
and hence are equal.
-
Thanks but then why it doesn't show the False? – Alex Oct 3 '13 at 0:23
It shows True when both sides are the same. It shows False when the two sides are different. To see the False use the triple equals: {1 -> 2} === {1 \[DirectedEdge] 2}does indeed return False. It's the difference between "equals" and "identical" or Equals and SameQ. – bill s Oct 3 '13 at 0:32
@Alex As bill s says, == and === are different and you probably were looking for ===. Also see: Evaluating an If condition to yield True/False – rm -rf Oct 3 '13 at 1:03 | 2014-09-02 21:40:39 | {"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": 0.7353501915931702, "perplexity": 3419.9765824141014}, "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-2014-35/segments/1409535922871.14/warc/CC-MAIN-20140901014522-00059-ip-10-180-136-8.ec2.internal.warc.gz"} |
http://zbmath.org/?q=an:1134.65032 | # zbMATH — the first resource for mathematics
##### Examples
Geometry Search for the term Geometry in any field. Queries are case-independent. Funct* Wildcard queries are specified by * (e.g. functions, functorial, etc.). Otherwise the search is exact. "Topological group" Phrases (multi-words) should be set in "straight quotation marks". au: Bourbaki & ti: Algebra Search for author and title. The and-operator & is default and can be omitted. Chebyshev | Tschebyscheff The or-operator | allows to search for Chebyshev or Tschebyscheff. "Quasi* map*" py: 1989 The resulting documents have publication year 1989. so: Eur* J* Mat* Soc* cc: 14 Search for publications in a particular source with a Mathematics Subject Classification code (cc) in 14. "Partial diff* eq*" ! elliptic The not-operator ! eliminates all results containing the word elliptic. dt: b & au: Hilbert The document type is set to books; alternatively: j for journal articles, a for book articles. py: 2000-2015 cc: (94A | 11T) Number ranges are accepted. Terms can be grouped within (parentheses). la: chinese Find documents in a given language. ISO 639-1 language codes can also be used.
##### Operators
a & b logic and a | b logic or !ab logic not abc* right wildcard "ab c" phrase (ab c) parentheses
##### Fields
any anywhere an internal document identifier au author, editor ai internal author identifier ti title la language so source ab review, abstract py publication year rv reviewer cc MSC code ut uncontrolled term dt document type (j: journal article; b: book; a: book article)
An iterative method for symmetric solutions and optimal approximation solution of the system of matrix equations ${A}_{1}X{B}_{1}={C}_{1},{A}_{2}X{B}_{2}={C}_{2}$. (English) Zbl 1134.65032
The symmetric solutions of the systems of matrix equations ${A}_{1}X{B}_{1}={C}_{1}$, ${A}_{2}X{B}_{2}={C}_{2}$ can not be easily obtained by applying matrix decompositions. The authors are proposing an iterative method to solve systems of matrix equations, where when the system of matrix equations is consistent, and its solution can be obtained within finite iterative steps, and its least-norm solution can be obtained by choosing a special kind of initial iterative matrix. Additionally, its optimal approximation solution to a given matrix can be derived by finding the least-norm symmetric solution of a new system of matrix equations ${A}_{1}\stackrel{^}{X}{B}_{1}={\stackrel{^}{C}}_{1}$, ${A}_{2}\stackrel{^}{X}{B}_{2}={\stackrel{^}{C}}_{2}$.
Finally, the author demonstrates the applicability of the proposed method on systems of matrix equations.
##### MSC:
65F30 Other matrix algorithms 15A24 Matrix equations and identities 65F10 Iterative methods for linear systems | 2014-04-24 16:17:06 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "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": 0.7576761841773987, "perplexity": 4790.441096551614}, "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-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00405-ip-10-147-4-33.ec2.internal.warc.gz"} |
https://gmatclub.com/forum/a-certain-rectangular-window-is-twice-as-long-as-it-is-wide-144270.html?sort_by_oldest=true | It is currently 13 Dec 2017, 03:22
# Decision(s) Day!:
CHAT Rooms | Ross R1 | Kellogg R1 | Darden R1 | Tepper R1
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# A certain rectangular window is twice as long as it is wide.
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
### Hide Tags
Manager
Joined: 02 Dec 2012
Posts: 178
Kudos [?]: 3648 [1], given: 0
A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
17 Dec 2012, 06:21
1
KUDOS
5
This post was
BOOKMARKED
00:00
Difficulty:
5% (low)
Question Stats:
85% (00:46) correct 15% (00:48) wrong based on 837 sessions
### HideShow timer Statistics
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
[Reveal] Spoiler: OA
Kudos [?]: 3648 [1], given: 0
Math Expert
Joined: 02 Sep 2009
Posts: 42577
Kudos [?]: 135458 [1], given: 12695
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
17 Dec 2012, 06:26
1
KUDOS
Expert's post
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
Given that P = 2l + 2w = 10, where l and w are the length and width in feet, respectively. Since, also given that l = 2w, then we have that 2l + l = 10 --> l = 10/3 --> w = l/2 = 10/6 = 5/3.
_________________
Kudos [?]: 135458 [1], given: 12695
Director
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 613
Kudos [?]: 810 [1], given: 59
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
17 Dec 2012, 06:31
1
KUDOS
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
L = 2W
Perimeter = 2(L+W) = 2(2W + W) = 6W = 10(given)
W = 10/6 = 5/3
L = 2W = 10/3
Answer will be B
Hope it helps!
_________________
Ankit
Check my Tutoring Site -> Brush My Quant
GMAT Quant Tutor
How to start GMAT preparations?
How to Improve Quant Score?
Gmatclub Topic Tags
Check out my GMAT debrief
How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities
Kudos [?]: 810 [1], given: 59
Manager
Joined: 04 Oct 2011
Posts: 215
Kudos [?]: 64 [0], given: 44
Location: India
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
Re: A certain rectangular window is twice as long as it is wide [#permalink]
### Show Tags
25 Dec 2012, 13:01
megafan wrote:
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) $$\frac{3}{2}$$ by $$\frac{7}{2}$$
(B) $$\frac{5}{3}$$ by $$\frac{10}{3}$$
(C) $$2$$ by $$4$$
(D) $$3$$ by $$6$$
(E) $$\frac{10}{3}$$ by $$\frac{20}{3}$$
Relatively easy question, except for the confusing term - "twice as long as it is wide", which I happen to screw up over. Did someone happen to compile a list of these confusing statements in word problems? If so, please do share, it would be immensely helpful for me, and I am sure for other members as well.
Ya its quite easy....
twice as long as it is wide means : l=2w
Given p=10
2(l+w)=10
l+w=5
2w+w=5
w=5/3
and l = 10/3
If u feel difficult in understanding "twice as long as it is wide", since length should be always greater than width and here it is twice l=2w
_________________
GMAT - Practice, Patience, Persistence
Kudos if u like
Kudos [?]: 64 [0], given: 44
Intern
Joined: 18 Nov 2011
Posts: 36
Kudos [?]: 15 [0], given: 0
Concentration: Strategy, Marketing
GMAT Date: 06-18-2013
GPA: 3.98
A certain rectangular window is twice as long as it is wide. If [#permalink]
### Show Tags
04 Apr 2013, 15:09
bmwhype2 wrote:
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
my answer was B. Answer Key says E.
Width $$=$$ $$x$$
Length$$=$$ $$2x$$
Perimeter $$= x+x+2x+2x = 6X$$
$$6x = 10$$
$$x= \frac{5}{3}$$
Therefore, dimensions are:
Width $$= \frac{5}{3}$$
Length $$= 2*\frac{5}{3} = \frac{10}{3}$$
B all the way
Kudos [?]: 15 [0], given: 0
Non-Human User
Joined: 09 Sep 2013
Posts: 14877
Kudos [?]: 287 [0], given: 0
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
30 May 2014, 11:04
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Kudos [?]: 287 [0], given: 0
Director
Joined: 23 Jan 2013
Posts: 602
Kudos [?]: 31 [0], given: 41
Schools: Cambridge'16
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
31 May 2014, 21:59
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
We can eliminate C,D,E as not giving 10 feeet as perimeter. In choice A, 3/2=1.5 and 7/2=3.5 so it is not double difference. The answer is B
Kudos [?]: 31 [0], given: 41
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 437
Kudos [?]: 143 [0], given: 169
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
A certain rectangular window is twice as long as it is wide. If [#permalink]
### Show Tags
19 Jan 2015, 02:02
The dimesnsions of the box are:
w = w
l = 2w
Perimeter:
2w + 2(2w) = 10
2w + 4w = 10
6w = 10
w = 10 / 6
w = 5 / 3
l = 2w
l = 2 (5/3)
l = 10 / 3
So. the dimensions are 5/3 and 10/3, ANS B
Kudos [?]: 143 [0], given: 169
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10379
Kudos [?]: 3683 [1], given: 173
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
19 Jan 2015, 12:48
1
KUDOS
Expert's post
HI All,
Many Test Takers would approach this with a combination of algebra and geometry rules, which is fine - the math is easy enough to do. On Test Day though, you have 2 goals when dealing with a question:
1) Get it correct, if possible.
2) Answer it in the fastest way possible.
Since the answer choices ARE numbers, and the prompt describes some rather specific facts, you can use the answers to your advantage and TEST THE ANSWERS.
We're told 2 things about a rectangle:
1) It's length is TWICE it's width.
2) It's perimeter is 10
We're asked for the dimensions of the rectangle.
Looking at the answer choices, there are some really easy answers to knock out....
Answer C: 2 by 4......this perimeter would = 12, not 10. ELIMINATE C.
Answer D: 3 by 6......this perimeter would = 18, not 10. ELIMINATE D.
Answer E: 10/3 by 20/3......this perimeter would be > 18, not 10. ELIMINATE E.
With the remaining 2 answers, only 1 of them has a length that is TWICE the width....
Answer A: 3/2 and 7/2.....7/2 is NOT TWICE 3/2. Eliminate A.
[Reveal] Spoiler:
B
GMAT assassins aren't born, they're made,
Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: [email protected]
# Rich Cohen
Co-Founder & GMAT Assassin
Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************
Kudos [?]: 3683 [1], given: 173
Non-Human User
Joined: 09 Sep 2013
Posts: 14877
Kudos [?]: 287 [0], given: 0
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
27 Feb 2016, 02:36
Hello from the GMAT Club BumpBot!
Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Kudos [?]: 287 [0], given: 0
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1931
Kudos [?]: 1015 [0], given: 3
Location: United States (CA)
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
21 Jun 2016, 09:23
A certain rectangular window is twice as long as it is wide. If its perimeter is 10 feet, then its dimensions in feet are
(A) 3/2 by 7/2
(B) 5/3 by 10/3
(C) 2 by 4
(D) 3 by 6
(E) 10/3 by 20/3
We set up the dimensions of the rectangular window as w = width and l = length.
We are given that the rectangular window is twice as long as it is wide. Thus we can say:
l = 2w
We also know that the perimeter is 10 feet, so we can say:
2l + 2w = 10
We can now substitute in 2w for l and we have:
2 x 2w + 2w = 10
4w + 2w = 10
6w = 10
w = 10/6 = 5/3
Because we know that the length is twice the width, we know that the length is:
2 x 5/3 = 10/3
_________________
Scott Woodbury-Stewart
Founder and CEO
GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions
Kudos [?]: 1015 [0], given: 3
Senior Manager
Joined: 22 Nov 2016
Posts: 251
Kudos [?]: 64 [0], given: 42
Location: United States
GPA: 3.4
Re: A certain rectangular window is twice as long as it is wide. [#permalink]
### Show Tags
05 Jul 2017, 20:16
Take width as x, length is 2x. Perimeter = x+2x+x+2x = 6x = 10; x=5/3
Hence length is 10/3 and width is 5/3
_________________
Kudosity killed the cat but your kudos can save it.
Kudos [?]: 64 [0], given: 42
Re: A certain rectangular window is twice as long as it is wide. [#permalink] 05 Jul 2017, 20:16
Display posts from previous: Sort by
# A certain rectangular window is twice as long as it is wide.
new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2017-12-13 11:22:07 | {"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": 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": 0.42939844727516174, "perplexity": 6799.122808071446}, "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-2017-51/segments/1512948522999.27/warc/CC-MAIN-20171213104259-20171213124259-00309.warc.gz"} |
http://caml.inria.fr/pub/ml-archives/caml-list/1994/10/2ec0abaa687630eb7f08ea2fe9b1960b.en.html | • Dominic Duggan
[ Home ] [ Index: by date | by threads ]
[ Search: ]
[ Message by date: previous | next ] [ Message in thread: previous | next ] [ Thread: previous | next ]
Date: -- (:) From: Dominic Duggan Subject: technical report available: "kinded parametric overloading"
Readers of this mailing list may be interested in the following
technical report, which has been submitted to a journal.
Although not the main point of the paper, we thought some people
might be amused by the following application, which we understand
was the subject of some discussion on this mailing list some time back.
-----------------------------------------------------------------------
Consider = defined for integers, strings, chars and sets.
Under our type system = is overloaded with the type:
K{=} >= {int, string, char, K{=} set}
= : 'a * 'a -> bool
Suppose specialized instances of = are defined for sets of integers
and sets of chars. This is possible in our type system (without
ambiguity), giving the type of = as:
K{=} >= {int, string, char, int set, char set,
(K{=} set) \ {int set, char set}}
= : 'a * 'a -> bool
After some normalization, this becomes:
K{=} >= {int, string, char, int set, char set,
(K{=} \ {int,char}) set}
= : 'a * 'a -> bool
The paper discusses a mechanism for closing up'' such a type.
Using this mechanism, = can be given the closed type'':
= : 'a * 'a -> bool
where 'a : k
k = {int, string, char, int set, char set, k' set}
k' = {string, int set, char set, k' set}
-----------------------------------------------------------------------
Details of the paper follow:
author = "Dominic Duggan and John Ophel",
institution = "University of Waterloo, Department of Computer Science",
year = 1994,
number = {CS-94-35},
month = "September",
note = {Supersedes CS-94-15 and CS-94-16, March 1994, and CS-93-32,
August 1993}
}
Abstract:
attention in the functional programming community. The main approach has been
that of Haskell type classes. An approach to the type-checking and semantics
constrain type variables. {\em Open kinds} constrain type variables by sets of
operations and are useful for the incremental development of reusable
procedures (in a similar manner to Haskell classes), while {\em closed kinds}
constrain type variables by sets of types (essentially providing a type-safe
form of dynamic typing). The type system includes a rule for closing up'' an
open kind to a closed kind. Applications of these faciities include local
module system, and an optimization which replaces call-site closure
construction with dynamic dispatching based on explicit type tags. A set
difference operation for kinds allows the unambiguous typing of overlapping
overload instances. The system of kinds is provided in some detail. A type
system and type inference algorithm are sketched. An operational semantics is
provided and used to verify semantic soundness. This operational semantics is
also used to verify the correctness of the removal of call-site closure
construction.
-------------------------------------------------------------------------------
The paper may be retrieved at either of the following addresses: | 2014-03-12 05:06:16 | {"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": 0.5315800905227661, "perplexity": 9042.398413185949}, "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-2014-10/segments/1394021365169/warc/CC-MAIN-20140305120925-00096-ip-10-183-142-35.ec2.internal.warc.gz"} |
https://openeuphoria.org/wiki/view/forum-msg-id-130069-edit.wc | ### forum-msg-id-130069-edit
Original date:2016-07-17 01:10:31 Edited by: ryanj Subject: Re: RedyCode 0.9.3
I fixed some problems in the 0.9.4 branch. Path detection should now work correctly when RedyCode is moved to a different folder. Paths relative to the Redy folder tree are re-detected, yet manually-set external paths are preserved.
In other words, default paths act as relative paths, so you can move RedyCode to a different location (or put it on a flash drive). But if you set paths to outside locations, such as C:\euphoria\include, it will preserve it even if you move redy to a different location. If any path becomes invalid, an error message will be displayed and the Preferences dialog will show automatically, so you can fix the invalid path.
I also fixed a bug in find/replace, where the first match was skipped when searching from the top.
I'll test for a few more days before making it official. Next on the todo list: get undo working! | 2022-01-21 10:34:13 | {"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": 0.8328047394752502, "perplexity": 3927.5037168466197}, "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-2022-05/segments/1642320303356.40/warc/CC-MAIN-20220121101528-20220121131528-00475.warc.gz"} |
http://sioc-journal.cn/Jwk_hxxb/CN/Y1995/V53/I7/689 | ### 锂的新萃取体系研究
1. 中国科学院上海有机化学研究所
• 发布日期:1995-07-15
### Study on new extraction system for lithium
SHENG HUAIYU;LI BEILI;CHEN YAOHUAN;YUAN QUN;XU QINGREN;YAN JINYING;YAO JIEXING;YE WEIZHEN;LONG HAIYAN;DAI PUXING
• Published:1995-07-15
o-Arylazo-β-arylols (HA) were first time chosen as the extractants for lithium (Li^+) with the formation of chelating anion (LiA~2^-), which could be extracted into inactive solvents in the presence of a quaternary ammonium base (Q^+X^-). The mechanism of this chelation-ion association synergic extraction was discussed in detail. The dynamics data and the thermodynamics parameters and the structure of the extractable complexes were determined. It was shown that the new system exhibits excellent properities, especially the larger separation coefficient β~Li~/~Na than other systems hitherto issued. | 2020-11-26 04:07:03 | {"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": 0.6202453374862671, "perplexity": 12915.286327697442}, "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-2020-50/segments/1606141186414.7/warc/CC-MAIN-20201126030729-20201126060729-00078.warc.gz"} |
http://golem.ph.utexas.edu/category/2008/02/virtually_real_or_really_virtu.html | February 2, 2008
Virtually Real or Really Virtual
Posted by David Corfield
If NASA can have a presence in Second Life with their CoLab project, maybe we at the Café should be thinking about the next step. Posted at February 2, 2008 12:34 PM UTC
TrackBack URL for this Entry: http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1591
Re: Virtually Real or Really Virtual
we at the Café should be thinking about the next step.
2-presence?
I suspect it’ll be easier to start with strict 2-presence, though obviously we’ll need to understand weak 2-presence (or at least, braided 2-presence) to think about 3-presence.
Posted by: Allen Knutson on February 2, 2008 3:51 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Didn’t have a chance to watch the whole vid, but I want to know why are all these virtual worlds only 3-d. Shouldn’t we be flying through 4-d space? The question is more serious than it sounds: the n-catsters are trying to develop higher dim’l algebra. We have a handle on dim 1, 2 and to some degree 3 because we have geometric intuition in the apparently 3-d world. We don’t have intuition in 4d. Computers can do this for us. So like Joe Cube in Rucker’s Spaceland (sorry no time to look up the URLs), we could be flying about in 4d video games and VR spaces. [Wait till you see my new 4-d Tuxedo!] After we get over the motion sickness (cubes necker reversing can be really disorienting), we might begin to train our right brain, which then can tell our left brain how to do theses things.
Posted by: Scott Carter on February 2, 2008 4:27 PM | Permalink | Reply to this
4-D games and crystals; Re: Virtually Real or Really Virtual
Since I do a lot of math and Physics in 4-D and 5-D, I think the world is ready for a true 4-D computer game. I have discussed this with several professional computer ganme designers. Kids playing it will learn Relativity while very very young, and build stardrives and time machines and Mimzies. Or something. I presume this audience’s familiarity with 4-D graphics (such as the nice interactive polytopes on MathWorld), and am frustrated by the reluctance to my pitches for such a game to companies in a hit-driven industry. I do have professional TV, fil, and CD-Rom writing experience, and book/story/play publishing, but don’t want to go into a side-bar on the skewed long-tail distributions implied by “hit-based industry”).
On a related matter, Torbjörn Larsson ( April 5, 2007) wrote:
“When the basic course in solid state physics covers phonon spectra, it typically starts out with an 1D
lattice model, as in Kittel’s “Introduction to Solid State Physics”. The lowest longitudinal mode (the
uniform mode) would be equivalent to moving the whole crystal, and is typically neglected in a Fourier
analysis with fixed or infinite boundaries. At the course only a few of us noted that, after all it is a simple model and it was outside its purpose.
Of course some excitation modes (hitting) may indeed move the crystal. And others (temperature) will
benefit from a more realistic analysis from averaging et cetera in real, finite crystals. But the interesting reaction was the teacher’s. He simply said
something equivalent to “Don’t bother” and dropped the matter, which wasn’t very helpful at the time. A common test for models is, or should be, to push it
‘til it breaks. It is done to judge if it is inherently wrong or where its domain of applicability is or what would improve it. Apparently that educator wasn’t used to think in that way.”
“The 1D lattice model, as in Kittel’s “Introduction to
Solid State Physics” is not a bad place to start. But there is no such thing (so far as I know) as 1-D magnetism. Magnetism starts in 2-D.”
“Ferromagnetism was first explained at a deep mathematical level by Ising’s model, which also explains weird phenomena such as fractional quantum Hall state accompanied by hysteresis in accord with 2D Ising ferromagnetism. and domain formation.”
“The stuff that PHysics students and EE students learn about ‘left-hand rule’ and Maxwell’s Equations (actually most students now learn Heaviside’s equations) are uniquely dependent on 3-D space plus
1-D time. Only in 3-D and 7-D is there a good cross product.”
“Which is part of why I’m playing with 4-D space + 1-D time crystals.”
“Note that actual substances that form ‘quasicrystals’ in 3-D are best explained as projections of 6-D crystals! Penrose tiles are a 2-D side-effect of 6-D geometry and crystallography.”
“I’ve been reading… a paper by John Horton Conway (yes, that Conway: Game of Life, Surreal Numbers) and Neil J. A. Sloane about multidimensional lattices. Very very weird things happen at 11-D and 24-D. Physics intuition seems to get in the way of seeing things in these spaces.”
“Very simple questions such as ‘what is a crystal?’ have shown me the depths of my ignorance in blindingly clear ways.”
Since then, I’ve twice discussed with Math/Physics professor Barry Simon (Caltech) the unfinished work by Richard Feynman on 4-dimensional ionic hypercube-symmetry crystals, and their perturbative stability, and have emailed back and forth with his son Carl Feynman on certain open questions in 4-D lattice polytope symmetry. And I’ve discussed the matter with professor Tim Poston (whose PhD was from Penrose).
See also the recent issue of Science, on 4-D phase transition in incommeasurate 3-D solid.
Posted by: Jonathan Vos Post on February 2, 2008 8:30 PM | Permalink | Reply to this
Re: 4-D games and crystals; Re: Virtually Real or Really Virtual
There is a plenty of 2n-D games. All the games there the player control party of n characters on the 2D board.
Posted by: serg271 on February 3, 2008 7:08 AM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
One of the things we found with higher dimensional algebra in the 1970s, and which always amazed me, was 2-dimensional rewriting’. This is the power behind the 2-dimensional van Kampen theorem.
Group theory workers on identities among relations use deformations of pictures’ of homotopy elements. A problem is that to go further one needs to record deformations of pictures, which seems to need a 3-dimensional language. Work with Al-Agl and Steiner uses a 3-dimensional rewriting procedure to prove a key braid relation.
It would be good if computers could go further. But for n-dimensional rewriting (general n) one seems to need symbols on a page to model what is going on! Trying to find the context to do this is a key part of the problem. This is in the background of my previous comments on the contrasting properties of crossed complexes, and of globular and cubical omega-groupoids. This variety of representations is important.
Posted by: Ronnie Brown on February 3, 2008 10:44 AM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
for deformations of pictures, is the work of e.g. Scott Carter relevant?
Posted by: jim stasheff on February 3, 2008 1:32 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Jim, thanks for the plug! I think that Alissa Crans, Mohamed Elhamdadi, and Masahico Saito have quite a lot to do with this work. The e.g. makes this implicit, but not explicit.
What we have seen on the higher dimensional algebra side is not as explicit as Ronnie Brown is mentioning but we have a lot of glimpses into this: Take a piece of classical algebra, represent it as diagrams, interpret the diagrams as a categorical internalization, for example, now watch the diagram move to express the axiom. The motion of the diagram can be projected (decategorified), and some extra-structure can be reimposed on the projection of the motion. Then the identities among relations (or similar axioms) can be visualized by watching the singular points of the projection undergo their singularities. The movie move theorem and its smarter brother (C., Rieger, and Saito) are example exploitations of this idea.
The recent stuff enhances this with some cocycles attached at the singular points. I am pretty sure (and I think Masahico agrees) that the whole picture goes through nicely in the Frobenius algebra case because all the identities among relations, and their higher order analogues are controlled by the Stasheff polytopes.
Of course, as a hobby, I try to draw things in as an explicit fashion as possible. And this relates back to Jonathan Van Post’s remark. It takes time and hard work to visualize higher dimensional stuff. I had a conversation with J. H. Conway the other day, and I was surprised at some of the questions that even he could not answer. He mentioned a geometric (4d) proof of Heron’s formula, for example. If higher dimensional problems are hard for him, do the rest of us have a chance? So 4d gaming, 4d fly throughs, and all of these things can help build intuition for higher-dimensional algebra.
Here is a homework problem. Take any suitably short book on Hopf algebras (eg Majid’s big pink) and write all the proofs in diagrammatic form. After finding the diagrammatic proof, then take the kinematics of the proof and elaborate them as “spin foam” pictures. Now are there key axioms, key proofs, etc that manifest themselves in a 3-d branched solid?
In regards to JB’s comment about sending primates into space, I think the original space program did a good thing: It took a fundamentally military competition and turned it into a humanitarian endeavor. Some spin-off technology includes the relays from my keyboard to this forum and back. Maybe not directly moon related, but some organization had to put those tin cans in space. The current proposal to send people to Mars does seem far fetched in light of the overwhelming social problems caused by humanity’s taxing this environment so. Some thought (Asimov?) that our purpose was to get the hell out of Dodge: Space exploration was what we were meant to do. As a species, we tend to foul our homes to the point that they are uninhabitable, and then move on.
NASA has no mandate to address the problems of this planet even if they have the resources to make inroads. We, as scientists, can lobby for that mandate to change.
Posted by: Scott Carter on February 3, 2008 9:18 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Scott Carter wrote:
In regards to JB’s comment about sending primates into space, I think the original space program did a good thing: It took a fundamentally military competition and turned it into a humanitarian endeavor.
Just so there’s no confusion: I completely agree!
I also think it was very important for us to see this:
The Apollo mission made good use of technologies available at the time, before computers really took off. My objections are to a manned Mars mission, which at this point would be a huge step backwards. We don’t need to shoot people into space anymore to see these wonderful things, and study them in marvelous detail:
NASA has no mandate to address the problems of this planet even if they have the resources to make inroads. We, as scientists, can lobby for that mandate to change.
For starters, we should lobby to undo some of the damage
already done by the Bush administration’s wrongheaded focus on the Mars mission, combined with their ostrich-like posture when it comes to human-caused global heating:
NASA is canceling or delaying a number of satellites designed to give scientists critical information on the earth’s changing climate and environment.
The space agency has shelved a $200 million satellite mission headed by a Massachusetts Institute of Technology professor that was designed to measure soil moisture – a key factor in helping scientists understand the impact of global warming and predict droughts and floods. The Deep Space Climate Observatory, intended to observe climate factors such as solar radiation, ozone, clouds, and water vapor more comprehensively than existing satellites, also has been canceled. And in its 2007 budget, NASA proposes significant delays in a global precipitation measuring mission to help with weather predictions, as well as the launch of a satellite designed to increase the timeliness and accuracy of severe weather forecasts and improve climate models. The changes come as NASA prioritizes its budget to pay for completion of the International Space Station and the return of astronauts to the moon by 2020 – a goal set by President Bush that promises a more distant and arguably less practical scientific payoff. Ultimately, scientists say, the delays and cancellations could make hurricane predictions less accurate, create gaps in long-term monitoring of weather, and result in less clarity about the earth’s hydrological systems, which play an integral part in climate change. Posted by: John Baez on February 5, 2008 3:06 AM | Permalink | Reply to this Re: Virtually Real or Really Virtual Which of the current candidates has a clue about this? Posted by: jim stasheff on February 5, 2008 12:57 PM | Permalink | Reply to this Re: Virtually Real or Really Virtual Here is a remark on space travel, for what it’s worth. NASA can have a presence At some point in the video the off-voice says something about “not sending us into space but getting space here to us”. I have always thought that interpreting “space exploration” as “figuring out how to make primates survive a wee bit beyond their biosphere” is mislead, and that one should be content about, and in fact grateful for, having various robots sent out there sending their data back here. Hope I don’t have to turn that statement into an $n$-categorical analogy now, somehow. Posted by: Urs Schreiber on February 2, 2008 5:42 PM | Permalink | Reply to this Re: Virtually Real or Really Virtual Urs wrote: I have always thought that interpreting “space exploration” as “figuring out how to make primates survive a wee bit beyond their biosphere” is misleading, and that one should be content about, and in fact grateful for, having various robots sent out there sending their data back here. Indeed, one of the many backwards steps taken by the current US administration is their crazy plan to send canned primates to Mars. It’ll be incredibly expensive if it ever happens: NASA guesses it’ll cost$104 billion just to get back to the Moon by 2020 — and by 2024, the Government Accounting Office guesses the total price of the manned Mars mission will hit \$230 billion.
With luck, we’ll never carry out this stupid stunt. But, even the planning to do it is already crushing many projects that are far more interesting.
For a more detailed rant on this subject, see my comments here.
Posted by: John Baez on February 3, 2008 2:17 AM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Bloomberg TV just said that NASA has hired a prominent ad agency but the story hasn’t appeared yet on bloomberg.com
Does anyone know if NASA engineers could potentially work on serious alternative energy projects or would doing so be outside of their abilities or interests?
Posted by: Charlie Stromeyer Jr on February 3, 2008 2:12 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
If you were living in Europe around 1500 would you guys have written the following?
“Indeed, one of the many backwards steps taken by the current crowned heads of Europe is their crazy plan to send boxed primates to America. It’ll be incredibly expensive if it ever happens. Spain guesses it’ll cost thousands of pounds of gold just to maintain a presence in the Carribean.
With luck, we’ll never carry out this stupid stunt. But, even the planning to do it is already crushing many projects that are far more interesting.”
Posted by: Jeffery Winkler on February 4, 2008 6:54 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Your analogy is flawed. The monarchs of Europe sent explorers around the globe to take over territory and make money. The practicality of the endeavor was obvious to them, though it certainly involved risks.
And the reason is: they knew Europeans could easily survive on other parts of the Earth. They also soon learned they could subdue the locals with superior weaponry.
Here is a quote from Columbus on his journey: “Many of the men I have seen have scars on their bodies, and when I made signs to them to find out how this happened, they indicated that people from other nearby islands come to San Salvador to capture them; they defend themselves the best they can. I believe that people from the mainland come here to take them as slaves. They ought to make good and skilled servants, for they repeat very quickly whatever we say to them… I could conquer the whole of them with 50 men, and govern them as I pleased.”
Will we see any folks on the Mars mission making similar remarks? No: they’ll be struggling to survive on a chilly airless ball of rock, for no particular good reason.
Until something changes — like genetically engineering ‘men’ that’ll enjoy life on other planets — manned space missions will remain a tired old 1950’s-science-fiction-style vision of the future. There are much cooler things to do now.
Posted by: John Baez on February 5, 2008 12:40 AM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
Perhaps one should regard space programs not as part of science, but as belonging to psychology, because people project their phantasies into it like into a big Rorschach-Test. E.g. here a book on russian SF, in which the author describes the supposed role of gnostic and eschatological thinking on the russian science community and its interest in space. I wonder if escape phantasies, e.g. for the case that consequences of global warming become desastrous, play a similar role in the way americans think about spacetravel?
Posted by: Thomas Riepe on February 6, 2008 2:02 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
I’ve often thought this, and that the distance between escape fantasies and eschatology isn’t necessarily all that large…
Posted by: Tim Silverman on February 6, 2008 11:56 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
I guess the only difference between attitudes towards human space travel is that one community thinks of it in terms of salvation, the other in terms of comfortable escape opportunities. But „salvation“ needs only to be achieved by few, representing all others, whereas „escape“ would be requested by all. So, a few cosmonauts suffice for the one community, but americans phantasy about space colonies . I wonder if such phantasies of a possible comfortable escape made the leading people in economy and politics accept irresponsible politics the past? Why should leading people in economy not imagine that they could, if necessary, transfer their way to live into space, e.g. like they can transfer their favorite sites for vacation , if the real thing is in a too unpleaseant surrounding?
Posted by: Thomas Riepe on February 7, 2008 10:08 AM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
“the next step”? Get scanned and virtual by the Harvard brain peeler.
Posted by: Thomas Riepe on February 3, 2008 6:00 PM | Permalink | Reply to this
Re: Virtually Real or Really Virtual
BTW, does anyone know if there exists a japanese-for-mathematicians book? I forgot nearly all I ever learned of it.
Posted by: Thomas Riepe on February 7, 2008 3:11 PM | Permalink | Reply to this
Post a New Comment | 2014-08-02 04:30:02 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 1, "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": 0.3901311159133911, "perplexity": 2503.1201414421316}, "config": {"markdown_headings": false, "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-2014-23/segments/1406510276353.59/warc/CC-MAIN-20140728011756-00436-ip-10-146-231-18.ec2.internal.warc.gz"} |
http://www.relativitybook.com/w/index.php?title=Einstein:Book_chapter_18_-_Special_and_General_Principle_of_Relativity&direction=prev&oldid=554 | # Einstein:Book chapter 18 - Special and General Principle of Relativity
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Albert Einstein: Relativity: The Special and the General Theory
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 - - 18 19 20 21 22 23 24 25 26 27 28 29 - - 30 31 32
## 18: Special and General Principle of Relativity
THE basal principle, which was the pivot of all our previous considerations, was the special principle of relativity, i.e. the principle of the physical relativity of all uniform motion. Let us once more analyse its meaning carefully.
It was at all times clear that, from the point of view of the idea it conveys to us, every motion must be considered only as a relative motion. Returning to the illustration we have frequently used of the embankment and the railway carriage, we can express the fact of the motion here taking place in the following two forms, both of which are equally justifiable:
(a) The carriage is in motion relative to the embankment.
(b) The embankment is in motion relative to the carriage.
In (a) the embankment, in (b) the carriage, serves as the body of reference in our statement of the motion taking place. If it is simply a question of detecting or of describing the motion involved, it is in principle immaterial to what reference-body we refer the motion. As already mentioned, this is self-evident, but it must not be confused with the much more comprehensive statement called "the principle of relativity" which we have taken as the basis of our investigations.
The principle we have made use of not only maintains that we may equally well choose the carriage or the embankment as our reference-body for the description of any event (for this, too, is self-evident). Our principle rather asserts what follows: If we formulate the general laws of nature as they are obtained from experience, by making use of
(a) the embankment as reference-body,
(b) the railway carriage as reference-body,
then these general laws of nature (e.g. the laws of mechanics or the law of the propagation of light in vacuo) have exactly the same form in both cases. This can also be expressed as follows: For the physical description of natural processes, neither of the reference-bodies $K$, $K'$ is unique (lit. "specially marked out") as compared with the other. Unlike the first, this latter statement need not of necessity hold a priori, it is not contained in the conceptions of "motion" and "reference-body" and derivable from them; only experience can decide as to its correctness or incorrectness.
Up to the present, however, we have by no means maintained the equivalence of all bodies of reference $K$ in connection with the formulation of natural laws.
Our course was more on the following lines. In the first place, we started out from the assumption that there exists a reference-body $K$, whose condition of motion is such that the Galileian law holds with respect to it: A particle left to itself and sufficiently far removed from all other particles moves uniformly in a straight line. With reference to $K$ (Galileian reference-body) the laws of nature were to be as simple as possible. But in addition to $K$, all bodies of reference $K'$ should be given preference in this sense, and they should be exactly equivalent to $K$ for the formulation of natural laws, provided that they are in a state of uniform rectilinear and non-rotary motion with respect to $K$; all these bodies of reference are to be regarded as Galileian reference-bodies. The validity of the principle of relativity was assumed only for these reference-bodies, but not for others (e.g. those possessing motion of a different kind). In this sense we speak of the special principle of relativity, or special theory of relativity.
In contrast to this we wish to understand by the "general principle of relativity" the following statement: All bodies of reference $K$, $K'$, etc., are equivalent for the description of natural phenomena (formulation of the general laws of nature), whatever may be their state of motion. But before proceeding farther, it ought to be pointed out that this formulation must be replaced later by a more abstract one, for reasons which will become evident at a later stage.
Since the introduction of the special principle of relativity has been justified, every intellect which strives after generalisation must feel the temptation to venture the step towards the general principle of relativity. But a simple and apparently quite reliable consideration seems to suggest that, for the present at any rate, there is little hope of success in such an attempt. Let us imagine ourselves transferred to our old friend the railway carriage, which is travelling at a uniform rate. As long as it is moving uniformly, the occupant of the carriage is not sensible of its motion, and it is for this reason that he can without reluctance interpret the facts of the case as indicating that the carriage is at rest, but the embankment in motion.
Moreover, according to the special principle of relativity, this interpretation is quite justified also from a physical point of view.
If the motion of the carriage is now changed into a non-uniform motion, as for instance by a powerful application of the brakes, then the occupant of the carriage experiences a correspondingly powerful jerk forwards. The retarded motion is manifested in the mechanical behaviour of bodies relative to the person in the railway carriage. The mechanical behaviour is different from that of the case previously considered, and for this reason it would appear to be impossible that the same mechanical laws hold relatively to the non-uniformly moving carriage, as hold with reference to the carriage when at rest or in uniform motion. At all events it is clear that the Galileian law does not hold with respect to the non-uniformly moving carriage.
Because of this, we feel compelled at the present juncture to grant a kind of absolute physical reality to non-uniform motion, in opposition to the general principle of relativity. But in what follows we shall soon see that this conclusion cannot be maintained. | 2022-10-05 02:36:14 | {"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": 0.6652611494064331, "perplexity": 399.7201543852884}, "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-2022-40/segments/1664030337531.3/warc/CC-MAIN-20221005011205-20221005041205-00092.warc.gz"} |
http://aspdistribution.kalanda.info/what-does-qsd/21d60d-sodium-acetate-acid-or-base | A good example of such a salt is ammonium bicarbonate, NH4HCO3; like all ammonium salts, it is highly soluble, and its dissociation reaction in water is as follows: $\text{NH}_4\text{CO}_3(\text{s})\rightarrow \text{NH}_4^+(\text{aq})+\text{HCO}_3^-(\text{aq})$. We determine whether the hydrolyzable ion is acidic or basic by comparing the Ka and Kb values for the ion; if Ka > Kb, the ion will be acidic, whereas if Kb > Ka, the ion will be basic. [citation needed] Sodium acetate (anhydrous) is widely used as a shelf-life extending agent, pH control agent It is safe to eat at low concentration. Pour autoriser Verizon Media et nos partenaires à traiter vos données personnelles, sélectionnez 'J'accepte' ou 'Gérer les paramètres' pour obtenir plus d’informations et pour gérer vos choix. What makes a basic salt basic? Ans: Sodium acetate (CH3COONa) is a solid-state salt that can not be used in anhydrous or liquid form as an acid or base. Sodium Bicarbonate: Because the bicarbonate ion is the conjugate base of carbonic acid, a weak acid, sodium bicarbonate will yield a basic solution in water. We can determine the answer by comparing Ka and Kb values for each ion. 1 Product Result The following is a more complicated scenario in which a salt contains a cation and an anion, both of which are capable of participating in hydrolysis. Acetic acid, sodium salt, trihydrate. Sodium acetate (CH3COONa) is a salt in solid state and can't be regarded as an acid or base in anhydrous or molten form. Sodium Acetate, 3M, pH 5.2, Molecular Biology Grade - CAS 127-09-3 - Calbiochem. … The first step toward answering this question is recognizing that there are two sources of the OAc-ion in this solution. Base excess is defined as the amount of strong acid that must be added to each liter of fully oxygenated blood to return the pH to 7.40 at a temperature of 37°C and a pCO 2 of 40 mmHg (5.3 kPa). When dissolved in water, acidic salts form solutions with pH less than 7.0. Let us use an acetic acid–sodium acetate buffer to demonstrate how buffers work. Now, with NaOH being a strong base and CH3COOH being a weak acid, the resulting solution is fundamental in nature. Because the strong base NaOH has been converted to the weak base sodium acetate, the pH of the solution won't rise nearly as much as if the acetic acid weren't present in the first place. The reactions are as follows: $\text{NH}_4^+(\text{aq})+\text{H}_2\text{O}(\text{l})\rightleftharpoons \text{H}_3\text{O}^+(\text{aq})+\text{NH}_3(\text{aq})\quad\quad \text{K}_\text{a}=5.6\times10^{-10}$, $\text{HCO}_3^-(\text{aq})+\text{H}_2\text{O}(\text{l})\rightleftharpoons \text{H}_2\text{CO}_3(\text{aq})+\text{OH}^-(\text{aq})\quad\quad \text{K}_\text{b}=2.4\times 10^{-8}$. In summary, when a salt contains two ions that hydrolyze, compare their Ka and Kb values: Hydrolysis of salts: This video examines the hydrolysis of an acid salt, a basic salt, and a salt in which both ions hydrolyze. In acid-base chemistry, a salt is defined as the ionic compound that results from a neutralization reaction between an acid and a base. It forms a basic solution in water. In this case, the value of Kb for bicarbonate is greater than the value of Ka for ammonium. When dissolved in water, acidic salts will yield solutions with pH less than 7.0. Nos partenaires et nous-mêmes stockerons et/ou utiliserons des informations concernant votre appareil, par l’intermédiaire de cookies et de technologies similaires, afin d’afficher des annonces et des contenus personnalisés, de mesurer les audiences et les contenus, d’obtenir des informations sur les audiences et à des fins de développement de produit. However, as we have already discussed, the ammonium ion acts as a weak acid in solution, while the bicarbonate ion acts as a weak base. Vous pouvez modifier vos choix à tout moment dans vos paramètres de vie privée. For example, sodium acetate, NaCH 3 CO 2, is a salt formed by the reaction of the weak acid acetic acid with the strong base sodium … On addition of the base, the hydroxide released by the base … This conjugate base is usually a weak base. The component ions in a salt can be inorganic; examples include chloride (Cl−), the organic acetate (CH3COO−), and monatomic fluoride (F−), as well as polyatomic ions such as sulfate (SO42−). Na+ forms NaOH in water while acetate forms … "Acetate" also describes the conjugate base or ion … Acid salts result from the neutralization of a strong acid with a weak base. Yahoo fait partie de Verizon Media. Sodium acetate salt, or simply sodium acetate, has many practical uses. Cas 127-09-3 - Calbiochem nonmetallic or radical base ) a 0.001M solution of a base... Its salt by comparing Ka and Kb values for Each ion ( Cl– ) which. Acetate buffer to demonstrate how buffers work if it contains both the weak acid meaning... Relative à la vie privée by comparing Ka and Kb values for ion... Protons in the salt is the conjugate base of a strong base NaOH and CH3COOH a! Formed in the anion in the neutralization reaction between an acid salt added acid ) ⇌ CH 3 –... Buffer because it contains both the weak acid CH3COONa is placed in water hydrolysis, compare.! Anions, both of which participate in hydrolysis acetate ion CH3COO- proton that will dissociate... Solution of sodium acetate, 3M, pH 5.2, Molecular Biology Grade - CAS 127-09-3 Calbiochem... The converse of basic salts ; they are formed in the cation are most commonly ammonium salts or. A proton that will weakly dissociate in water no bacteriostat, antimicrobial sodium acetate acid or base or added buffer base. Ch3Coona is placed in water, a salt derived from a strong acid with a weak acid and weak. Acid salts are the converse of basic salts ; they are formed in the neutralization a! Acid of the weak acid, meaning that it only partially ionizes when dissolved water. A buffer because it contains both the weak acid and their effects on a solution of sodium acetate is... Proton in the anion relative aux cookies we can determine the answer by comparing Ka and Kb values Each. A protonated amine group both of which participate in hydrolysis ion is capable deprotonating. Acid salts result from the neutralization reaction between a strong acid with a weak.! Of sodium acetate is a buffer because it contains both the weak acid and a weak CH3COOH. Antimicrobial agent or added buffer that the anion in the salt acidic both the weak acid, meaning it. With NaOH being a strong base and CH3COOH being a weak acid and base in water, a salt..., the value of Kb for bicarbonate is greater than 7.0 Molecular Biology Grade - CAS 127-09-3 -.. From buffer solution ) 2 is defined as the ionic compound that results from a strong base NaOH and.. Raising the solution is fundamental in nature ; they are formed in the cation are most commonly salts! Section we will consider basic salts ; they are formed in the anion an acid salt aux cookies fundamental nature... As the ionic compound that results from a strong base NaOH and a base base of.... Will only be basic if it contains both the weak acid CAS 127-09-3 - Calbiochem acetic... An example of an acid salt dissociate in water we have a basic salt … on Addition of and! In the cation are most commonly ammonium salts, such as ammonium bicarbonate ( )... In mind that a salt is the conjugate base of a strong base NaOH a... And Kb values for Each ion potassium bisulfate—will yield weakly acidic solutions in water, acidic will! On Addition of acid salts result from the neutralization reaction between a strong base NaOH and a base chloride!, contain cations and anions, both of which participate in hydrolysis its salt acidic in. Of Kb for bicarbonate is greater than the value of Ka for ammonium notre. Bicarbonate, NaHCO3 the resulting solution is fundamental in nature a basic solution, sodium acetate pure! Protons of acid will be removed by the acetate ions and sodium ions dissociate. Different salt solutions, and in this section we will consider basic salts ; they formed. That contain a protonated amine group also contain an acidic proton include: of. Base or an acid and its salt to produce sodium ion Na+ and acetate ion.. Contains the conjugate base of HCl greater than the value of Ka for.! That will weakly dissociate in water we have a basic salt added acid ) ⇌ 3! Greater than the value of Ka for ammonium - CAS 127-09-3 - Calbiochem with a weak acid.! Ka for ammonium acetate ions to form an acetic acid molecule choix à moment... Bicarbonate be acidic or basic a solution with pH less than 7.0 relative aux cookies CH3COO-..., both of which participate in hydrolysis in hydrolysis et notre Politique relative à la vie privée et notre relative! Solution contains no bacteriostat, antimicrobial agent or added buffer Molecular Biology Grade - CAS 127-09-3 - Calbiochem dans paramètres. Acidic solutions in water weak acid and its salt this is because it dissociates into acetate to. Sodium acetate, has many practical uses are formed in the neutralization reaction between a strong and! 5.2, Molecular Biology Grade - CAS 127-09-3 - Calbiochem it contains the conjugate base a. The resulting solution is administered after dilution by the intravenous route as an electrolyte.! Ionizes when dissolved in water, acidic salts form solutions with pH than. Politique relative aux cookies for Each ion pH less than 7.0 water and yielding a basic solution, sodium.... The value of Kb for bicarbonate is a basic solution of acid will be by. Water producing NaOH and a weak base with a weak acid and its salt the bicarbonate is., which makes it an acid because it is due to the fact that the anion meaning it... Bicarbonate, NaHCO3 with this solution of ammonium bicarbonate ( NH4HCO3 ), contain cations anions! And the acetate ion CH3COO- less than 7.0 values for Each ion - goal! The intravenous route as an electrolyte replenisher … sodium acetate in pure water mind that a derived... Undergo hydrolysis removed by the acetate ion pH greater than 7.0 and we 'll start this. A salt derived from a neutralization reaction between a strong base and CH3COOH in solution sodium. In which both cation and anion are capable of deprotonating water, acidic salts will yield solutions pH! Is a buffer because it contains both the weak acid and its.... Ions can hydrolyze, will a solution of a basic salt yields a solution sodium! Be acidic or basic between a strong acid and base contains the conjugate base of HCl commonly! Ionic compounds that contain a protonated amine group solution ’ s pH formation of acid and a weak.! The released protons of acid and its salt between a strong base and CH3COOH being a strong NaOH... Itself is not a base salt ; the acetate ions and sodium.! No bacteriostat, antimicrobial agent or added buffer + + CH 3 COO – ( from acid. + CH 3 COOH ( from added acid ) ⇌ CH 3 COOH from! Buffer because it dissociates into sodium acetate acid or base and acetate ion is the conjugate base carbonic..., acidic salts form solutions with pH less than 7.0: anilinium chloride: anilinium chloride: anilinium chloride an! Form an acetic acid molecule form an acetic acid molecule are formed in the neutralization reaction between strong... S pH will only be basic if it contains both the weak base chloride, for instance, contains (. Contains no bacteriostat, antimicrobial agent or added buffer CH3COOH being a weak acid and a base ions to an... Participate in hydrolysis, for instance, contains chloride ( Cl– ), which makes it acid! Our goal is to find the pH of a weak base yield weakly acidic solutions in water a strong NaOH. Of the weak acid a strong acid and a base or an acid the neutralization of solution. Bicarbonate, NaHCO3 dissolved in water, acidic salts will yield solutions with pH less 7.0., contain cations and anions that can both undergo hydrolysis ⇌ CH 3 COOH ( from added acid ⇌. Are capable of deprotonating water, thereby raising the solution is fundamental in.. Reaction between a strong acid with a weak acid acetate ions to form an acetic acid–sodium acetate to. Salts and their effects on a solution of ammonium bicarbonate be acidic or basic the ammonium ion contains a that! In the salt is defined as the ionic compound that results from a strong acid with a base., which is the conjugate base of a basic solution, sodium bicarbonate, NaHCO3 sodium acetate acid or base from..., nonmetallic or radical base ) a protonated amine group, acidic salts form solutions pH! Solutions with pH less than 7.0 acidic salts form solutions with pH less 7.0... Fact that the anion … on Addition of acid, a basic salt the! Solution ’ s pH conjugate base of a solution of ammonium bicarbonate NH4HCO3! That result from the neutralization reaction between a strong acid with a weak acid, meaning that it partially... That the anion in the neutralization reaction between an acid and its salt a buffer because it into! Salts in which both cation and anion are capable of hydrolysis, compare.... Ammonium bicarbonate ( NH4HCO3 ), contain cations and anions, both of which participate hydrolysis! The ionic compound that results from a strong base NaOH and CH3COOH ) which. Nh4Hco3 ), which is the conjugate acid of the weak acid CH3COOH salts ; they formed! To produce sodium ion Na+ and acetate ion is the conjugate base HCl... Kb for bicarbonate is a salt containing cations and anions, both of which participate in hydrolysis by comparing and. Converse of basic salts acetate … sodium acetate ( a basic solution, bicarbonate. Sodium ion Na+ and acetate ion is the conjugate base of a strong NaOH... Consider a 0.001M solution of a salt will only be basic if it contains the conjugate base carbonic... The salt acidic, 3M, pH 5.2, Molecular Biology Grade - CAS 127-09-3 - Calbiochem in.! | 2022-05-25 19:38:46 | {"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": 0.4592650234699249, "perplexity": 7488.497680768889}, "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-2022-21/segments/1652662593428.63/warc/CC-MAIN-20220525182604-20220525212604-00337.warc.gz"} |
https://math.stackexchange.com/questions/925919/x-equiv-k2-mod-3-iff-x-equiv-1-mod-3 | # $(x \equiv k^2 \mod 3) \iff x \equiv 1 \mod 3$
Is it true that if 3 does not divide $x$,
$$x\equiv k^2\mod 3 \iff x\equiv 1 \mod 3$$
If the above statement is correct , There are two parts to prove
$$x\equiv k^2\mod 3 \implies x\not\equiv 0 \mod 3$$ $$x\equiv k^2\mod 3 \implies x\not\equiv 2 \mod 3$$
How to prove them ?
• Your question is not clear, because $3\not|x \implies x\not\equiv0\pmod3$, i.e., if $3$ does not divide $x$, then obviously $x\not\equiv0\pmod3$. – barak manos Sep 10 '14 at 5:55
• @barakmanos yeah , thanku . – hanugm Sep 10 '14 at 6:13
You can see from this analysis that the only possible values for $k^2$ are $0,1\pmod{3}$, which is what you wanted to show. | 2019-04-23 22:21:06 | {"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": 0.8081744909286499, "perplexity": 206.5475304405032}, "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/1555578613888.70/warc/CC-MAIN-20190423214818-20190424000818-00526.warc.gz"} |
http://mathhelpforum.com/calculus/55457-limit-derivation-help-please.html | 1. ## Limit, derivation, help please
Hi! I need some help in the following problems.
1. What is the limit if n tends to infinity:
[(n+2) choose 2]/(n choose 2)]^n
2. Derivation
y=(4^x-x^4)/log(7)x; that is the seven-based log of x
3. Derivation
y=ln[e^x+e^(-x)]
4. Limit if x tends to zero
1/[e^(2x)-1]-1/2x
Thanks a lot.
2. Originally Posted by csodacsiga
[(n+2) choose 2]/(n choose 2)]^n
Hint: $\displaystyle {a\choose 2} = \frac{a(a-1)}{2}$
2. Derivation
y=(4^x-x^4)/log(7)x; that is the seven-based log of x
This is the same as,
$\displaystyle y= \frac{e^{\ln 4 x} - x^4}{\left( \frac{\ln x}{\ln 7} \right)}$
Now apply the quotient rule.
y=ln[e^x+e^(-x)]
With chain rule we get,
$\displaystyle (e^x + e^{-x})' \cdot \frac{1}{e^x + e^{-x}} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$
4. Limit if x tends to zero
1/[e^(2x)-1]-1/2x
Thanks a lot.
This one is not clear.
3. For 1., $\displaystyle {n+2\choose2}\bigg/{n\choose2} = \frac{\frac12(n+2)(n+1)}{\frac12n(n-1)} = 1+\frac{4n+2}{n(n-1)}$. So $\displaystyle \left({n+2\choose2}\bigg/{n\choose2}\right)^{\!\!n} = \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n}$. This looks as
though it should be close to $\displaystyle \bigl(1+\tfrac4n\bigr)^n$, which has the limit $\displaystyle e^4$ as $\displaystyle n\to\infty$.
To justify that guess, you could take the logarithm of $\displaystyle \left(1+\frac{4n+2}{n(n-1)}\right)^{\!\!n}$ and show that it tends to 4 as $\displaystyle n\to\infty$.
4. Thank you both for the great helps.
1/[e^(2x)-1]-1/2x is
(one over (e to the power of two x) minus one) minus one over two x
5. Originally Posted by csodacsiga
Thank you both for the great helps.
1/[e^(2x)-1]-1/2x is
(one over (e to the power of two x) minus one) minus one over two x
$\displaystyle \frac1{e^{2x}-1} - \frac1{2x} = \frac{1+2x-e^{2x}}{2x(e^{2x}-1)}$. Now apply l'Hôpital two or three times.
6. I assume this is it:
$\displaystyle \lim_{x\to 0}\frac{1}{e^{2x}-1}-\frac{1}{2x}$?.
We can be lazy and use ol' L'Hopital.
Rewrite:
$\displaystyle \frac{-1}{2}\displaystyle{\lim_{x\to 0}\frac{-2x+e^{2x}-1}{x(e^{2x}-1)}}$
L'Hopital:
$\displaystyle \displaystyle{-\lim_{x\to 0}\frac{e^{2x}-1}{2xe^{2x}+e^{2x}-1}}$
L'Hopital again:
$\displaystyle \displaystyle{-\lim_{x\to 0}\frac{1}{2x+2}}$
Now, we can see it?.
Now, try it without L'Hopital. That is 'funner'. | 2018-06-19 03:17:08 | {"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": 0.9610667824745178, "perplexity": 3051.1323366097267}, "config": {"markdown_headings": true, "markdown_code": false, "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-2018-26/segments/1529267861752.19/warc/CC-MAIN-20180619021643-20180619041643-00000.warc.gz"} |
https://physics.stackexchange.com/questions/444600/relation-of-dispersion-for-a-plasma | # Relation of dispersion for a plasma
Assuming the electric field :
$$\vec{E} = E_{0}\,e^{i(kz-\omega t)}\vec{e_{z}}$$
and the complex relation by doing $$\vec{rot}\,(\vec{rot}\times\vec{E})$$
with $$\vec{rot}\times \vec{E}= i \vec{k}\times \vec{E}$$ :
I get with Maxwell equation : $$k^2 \vec{E} = \mu_{0}\omega \vec{J}-i\dfrac{\omega^2}{c^2}\vec{E}$$
How can I find the expression of conductivity $$\sigma(\omega)$$ such $$\vec{J}=\sigma\vec{E}$$ and deduce the dispersion relation for these waves ? | 2020-04-04 19:52:29 | {"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": 6, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9163309931755066, "perplexity": 577.7252523200332}, "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-2020-16/segments/1585370524604.46/warc/CC-MAIN-20200404165658-20200404195658-00451.warc.gz"} |
http://mathoverflow.net/questions/1537/is-there-a-murnaghan-nakayama-rule-for-gln-q/2611 | # Is there a Murnaghan-Nakayama Rule for GL(n,q)?
The Murnaghan-Nakayama rule for S_n is a combinatorial rule to compute the irreducible characters of the symmetric group. Is there a q-analogue of this rule for GL(n,q) to compute the irreducible characters? For example, exhibiting that the value of the unipotent characters of GL(n,q) on a unipotent class is given by the cocharge Kostka-Foulkes polynomials, and showing other special cases.
- | 2016-04-29 20:20:51 | {"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": 0.8871144652366638, "perplexity": 697.8646181937213}, "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-2016-18/segments/1461860111396.55/warc/CC-MAIN-20160428161511-00151-ip-10-239-7-51.ec2.internal.warc.gz"} |
https://www.rtg1995.rwth-aachen.de/cms/RTG1995/Forschung/Projekte/~gjpd/Topological-solitons-in-chiral-magnets/lidx/1/ | # P4: Topological solitons in chiral magnets
## Project Description
With the detection of relevant Dzyaloshinskii-Moriya (DM) interactions in ultra-thin films deposited on metal surfaces(1), a new research chapter was started in the field of low-dimensional magnetism. The DM interaction can lead to very complex homo-chiral magnetic structures such as multidimensional solitons (skyrmions) as observed in the nano-skyrmion lattice of iron on iridium(2). Such multidimensional localised states of the size of a few nanometers are not only of interest in modern physics but also for spintronics as was outlined recently(3). Such complex magnetic structures are understood as a result of a competition between different spin interactions: the Heisenberg-, the DM, and the four-spin biquadratic interaction to name the most relevant ones. Currently, there are three research trusts approaching the investigation of topological solitons in chiral magnets in reduced dimensions and thin films: (i) atomic scale nanostructures of adatoms on surfaces exhibiting a DM interaction through a large Rashba effect, (ii) ultrathin magnetic films on substrates with large spin-orbit interactions, and (iii) films of cubic helimagnets. In reduced dimensions thermal fluctuations are of particular relevance. Our goal is therefore to investigate such chiral magnetic systems at finite temperature. Of interest are the exploration of the magnetic phase diagram $(B, T)$ of these systems, the investigation of more exotic thermodynamic phases where the vector and scalar spin chirality are order parameters, or the stability of the topological nature against thermal fluctuations. For this purpose, at first the strengths of the different spin interactions are determined by mapping the energy landscape of magnetic states calculated using the material specific density functional theory methods(4)-(7) onto the spin-models. So far this mapping was made mainly for Heisenberg interactions. In this project the process should now be extended to isotropic and anisotropic interactions of higher order. We will employ the electronic structure programs(8), particularly the FLEUR program and the KKR method developed by us. The thermodynamic properties of the chiral nanomagnets are then to be examined on the basis of the effective spin-models with realistic spin-coupling constants by means of the stochastic atomistic spin dynamic juSpinX and Monte Carlo methods(9). Depending on the expected complexity of the magnetic energy landscape and phase space structures also extended ensemble methods, such as the Wang-Landau method(10), (11) and parallel tempering(12),(13) and hybrid methods will be applied. Also effective methods for the treatment of extensive spin interactions(14) will be studied here.
This project can benefit from progress in P1 , P2, and P18 in particular in order to account for quantum effects in the effective spin interaction. The spin-quadrupol transport of P14 is also related to the four-spin biquadratic interaction studied here.
(1) M. Bode, M. Heide, K. von Bergmann, S. Heinze, G. Bihlmayer, A. Kubetzka, O. Pietzsch, S. Blügel
and R. Wiesendanger, Nature 447, 190 (2007)
(2)S. Heinze, K. von Bergmann, M. Menzel, J. Brede, A. Kubetzka, R. Wiesendanger, G. Bihlmayer and S. Blügel,
Nature Phys. 7, 713 (2011)
(3) A. Fert, V. Cross and J. Sampaio, Nature Nanotech. 8, 152 (2013)
(4) M. Ležaić, P. Mavropoulos, G. Bihlmayer and S. Blügel, Phys. Rev. B 88, 134403 (2013)
(5) A. I. Liechtenstein, M. I. Katsnelson, V. P. Antropov and V. A. Gubanov, J. Magn. Magn. Mater. 67, 65 (1987)
(6) L. Udvardi, L. Szunyogh, K. Palotas and P. Weinberger, Phys. Rev. B 68, 104436 (2002)
(7) S. V. Halilov, H. Eschrig, A. Y. Perlov and P. M. Oppeneer, Phys. Rev. B 58, 293 (1998)
(8) DFT from Forschungszentrum Jülich
(9) M. Holtschneider, S. Wessel and W. Selke, Phys. Rev. B 75, 224417 (2007)
(10) F. Wang and D. P. Landau, Phys. Rev. Lett. 86, 2050 (2001)
(11) P. Dayal, S. Trebst, S. Wessel, D. Wärtz, M. Troyer, S. Sabhapandit and S. N. Coppersmith,
Phys. Rev. Lett. 92, 097201 (2004)
(12) R. H. Swendsen and J.-S. Wang, Phys. ReV. Lett. 57, 2607 (1986)
(13) K. Hukushima and K.J. Nemoto, Phys. Soc. Jpn. 65, 1604 (1996)
(14) E. Luijten and H. W. J. Blöte, Int. J. Mod. Phys. 6, 359 (1995) | 2021-03-09 08:09:07 | {"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": 0.5747002363204956, "perplexity": 3826.4100556422786}, "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/1614178389472.95/warc/CC-MAIN-20210309061538-20210309091538-00535.warc.gz"} |
http://ams.org/bookstore-getitem/item=GSM-102 | New Titles | FAQ | Keep Informed | Review Cart | Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education
Introduction to Fourier Analysis and Wavelets
Mark A. Pinsky, Northwestern University, Evanston, IL
SEARCH THIS BOOK:
2002; 376 pp; hardcover
Volume: 102
ISBN-10: 0-8218-4797-X
ISBN-13: 978-0-8218-4797-8
List Price: US$69 Member Price: US$55.20
Order Code: GSM/102
This book provides a concrete introduction to a number of topics in harmonic analysis, accessible at the early graduate level or, in some cases, at an upper undergraduate level. Necessary prerequisites to using the text are rudiments of the Lebesgue measure and integration on the real line. It begins with a thorough treatment of Fourier series on the circle and their applications to approximation theory, probability, and plane geometry (the isoperimetric theorem). Frequently, more than one proof is offered for a given theorem to illustrate the multiplicity of approaches.
The second chapter treats the Fourier transform on Euclidean spaces, especially the author's results in the three-dimensional piecewise smooth case, which is distinct from the classical Gibbs-Wilbraham phenomenon of one-dimensional Fourier analysis. The Poisson summation formula treated in Chapter 3 provides an elegant connection between Fourier series on the circle and Fourier transforms on the real line, culminating in Landau's asymptotic formulas for lattice points on a large sphere.
Much of modern harmonic analysis is concerned with the behavior of various linear operators on the Lebesgue spaces $$L^p(\mathbb{R}^n)$$. Chapter 4 gives a gentle introduction to these results, using the Riesz-Thorin theorem and the Marcinkiewicz interpolation formula. One of the long-time users of Fourier analysis is probability theory. In Chapter 5 the central limit theorem, iterated log theorem, and Berry-Esseen theorems are developed using the suitable Fourier-analytic tools.
The final chapter furnishes a gentle introduction to wavelet theory, depending only on the $$L_2$$ theory of the Fourier transform (the Plancherel theorem). The basic notions of scale and location parameters demonstrate the flexibility of the wavelet approach to harmonic analysis.
The text contains numerous examples and more than 200 exercises, each located in close proximity to the related theoretical material. | 2013-05-24 11:36:08 | {"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": 0.21394436061382294, "perplexity": 761.9168923747965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "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-2013-20/segments/1368704645477/warc/CC-MAIN-20130516114405-00099-ip-10-60-113-184.ec2.internal.warc.gz"} |
http://starbeamrainbowlabs.com/blog/?tags=linux | Archive
## Tag Cloud
3d account algorithms announcement architecture archives arduino artificial intelligence artix assembly async audio bash batch blog bookmarklet booting c sharp c++ challenge chrome os code codepen coding conundrums coding conundrums evolved command line compilers compiling css dailyprogrammer debugging demystification distributed computing downtime electronics email embedded systems encryption es6 features event experiment external first impressions future game github github gist graphics hardware hardware meetup holiday html html5 html5 canvas infrastructure interfaces internet io.js jabber javascript js bin labs learning library linux low level lua maintenance manjaro network networking node.js operating systems performance photos php pixelbot portable privacy programming problems project projects prolog protocol protocols pseudo 3d python reddit reference release releases resource review rust secrets security series list server software sorting source code control statistics svg technical terminal textures three thing game three.js tool tutorial tutorials twitter ubuntu university update updates upgrade version control virtual reality virtualisation visual web website windows windows 10 xmpp xslt
## Getting an updated version of MonoDevelop on Ubuntu
Thanks to the AUR (The Arch User Repository), an updated version of MonoDevelop is never very far away on Manjaro, Artix, Arch Linux and friends. Unfortunately, such delights can't be found so easily in the Ubuntu and Debian ecosystem. On a check inspection of the official repositories, the latest version available is 5.10, while the AUR is offering 7.1.2!
In addition, the newer versions offered via this 'flatpak' thing are sandboxed, so you can't combined multiple technologies like Node.JS and the amazing npm in your C♯ applications with it.
Thankfully, all is not lost. A clever GitHub user by the name of cra0zy has put together a script that will install the latest version of MonoDevelop for us - without all the hassle of sandboxes or manually compiling from source. Here's a link to this installer:
cra0zy/monodevelop-run-installer
This post is mainly for my own reference, but I'd love to know if it helped you out too. If it did, leave a comment below!
## The Great Migration of Manjaro
It was just before lunch in the library, and I was checking my university emails on my travelling laptop that runs Manjaro OpenRC. While that was going on, I was inducing a few updates that it notified me about with yaourt -Syua. First mistake.
During the installation, it decided to upgrade openrc to the Briston in the AUR (Arch User Repository), but I didn't think anything of it particularly - I knew that Manjaro OpenRC was dying deprecated. Second mistake.
Once the updates were complete, I shut it down and sent on my way - or at least I tried to - it wouldn't shut down, instead proceeding to log out and leave it at that. I resolved to investigate the problem when I got home. Third mistake.
By the time I came to use it again, I was greeted with an ominous message:
[Firmware Bug]: TSC_DEADLINE disabled due to Errata; please update microcode to version: 0x52 (or later)
Failed to execute /init (error -2)
Kernel panic - not syncing: No working init found. Try passing init= option to kernel. See Linux Documentation/admin-guide/init.rst
CPU: 0 PID: 1 Comm: swapper/0 Not tainted 4.13.2-1-MANJARO #1
Hardware name: Entroware Apollo/Apollo, BIOS 1.05.05 04/27/2017
Hrm. That's odd. Maybe something went wrong in the update? Linux has what's called kernel parameters that tell it how to boot. They specify things like "here's the root partition of the system", and "please let me edit files on the system after booting". To udnerstand how this fits into the next part of the story, it's first necessary to take a quick retour and look at how, precisely the linux kernel goes about booting a system. This is best explained with a diagram:
(Rendered with Ascidia. Textual diagram source available here)
1. BIOS / UEFI POST - The starting point of the boot process. The BIOS / UEFI turns on all the devices, runs some basic hardware checks, and (usually) gives the user a choice of what they want to boot from.
2. rEFInd - grub may be used instead of rEFInd, but the basic principle is the same: it asks the user how they want to boot from the hard drive. Kernel parameters are decided on here.
3. Initialisation: The Linux kernel is executed by the bootloader, and it proceeds to initialise itself and the connected devices.
4. Mount initial RAM disk: The Linux encounters a chicken-and-egg problem rather early on: How can it start talking to the connected devices if it doesn't know how to talk to them? The initial RAM disk solves that problem: It contains a bunch of drives and other such components that the kernel needs to initialise all the connected devices. It's like a cut-down root file system, in a sense.
5. Load drivers: The Linux kernel loads the drivers from the initial RAM disk (aka initrd) and starts initialising all the connected devices.
6. Mount root (read-only): The main root file system is mounted next, but only in read-only mode while the boot process finishes.
7. Execute init: It is at this point that the very first process is executed. It usually presides at /sbin/init, but this can be changed through the init kernel parameter.
8. Mount root (read-write): The init process (under SysVinit at least) then remounts the root filesystem such that it is writeable.
9. Mount other partitions: The next job is the mounting of the other partitions in /etc/fstab. This is also done by SysVinit if I recall correctly.
10. Reach runlevels: The main runlevels managed by the service manager (e.g. OpenRC) are now executed in order by the service manager.
Phew, that took more explaining than I thought! And to think it all happens in the span of about 10 seconds....! With that out of the way, let's continue with the story.
Let's try specifying the init kernel parameter - maybe the update cleared it for some random reason....? I had no idea what I was getting myself into :P
Unexpectedly, specifying init=/sbin/init didn't work. Neither did specifying init=/bin/sh. At this point, I suspected that there was something seriously wrong. I (correctly) guessed that it was the update I performed that morning that was to blame. After a bunch of backing and forthing, I managed to get hold of a previous copy of the openrc package that was replaced by the 0.27 version from the AUR. After doing a full backup, I tried installing it and removing the new openrc-sysvinit package that was also installed.
Before we continue further, I should probably explain how I managed to install the previous package version. Didn't I just explain that my system wasn't bootable? Well, yes. But I also had the original manjaro-architect installation media that I used to build the system in the first place. With that in hand, I could use rEFInd to boot from that (my UEFI firmware makes it a bit of a pain otherwise!), and then mount the root partition of the broken system and chroot into it. This process allows me to pretend that the system is actually booted, while piggybacking off the live installation media of the boot process. It works a bit like this:
lsblk # Find the root partition
mkdir /mnt/os;
mount /dev/sdZY /mnt/os # Mount the root partition
mount /dev/sdAB /mnt/os/boot/efi # Mount the EFI partition
manjaro-chroot /mnt/os bash # Enter the chroot and execute bash
Back to the story. Sadly, valiant though my effort was to replace the openrc and openrc-sysvinit packages was, it did not solve the problem. Eventually, I ended up having to perform a blind migration to Artix Linux, the spiritual successor to both Manjaro OpenRC and Arch OpenRC (apparently the developers of both came together to create Artix Linux).
Eventually, I ended up with a successful migration that I performed inside the chroot, and the system was bootable again! Next time, I'll always run pacman -Syu before yaourt -Syua. I'll also set up a temporary backup solution for my system files (I've already got one in place for my personal files) while I figure out a more permanent one that backs up across the network.
## Manjaro OpenRC Cheat Sheet
Amidst preparations for my third year at university, I've put together a sort of reference sheet to help me remember all the common commands needed when using Manjaro with OpenRC. It's not complete, but I'll continue to update it with various useful commands I stumble upon. You can find it below.
If you have any that you find useful, post a comment below! I'd love to see what you come up with - I might even add it to this list (crediting you of course)!
### Cast List
• sudo pacman: Main package manager
• yaourt: pacman wrapper, also has AUR support. Swap out pacman for yaourt to include the AUR.
• sudo rankmirrors: Finds and selects the fastest repository mirrors.
• sudo rc-update - Enable and disable services
• sudo service - Start, stop, and query the status of services
### Commands
#### Package Management
Command Description
rankmirrors -i -m rank -d Interactively re-find the fastest mirrors
pacman -Sy Synchronise local repository metadata
pacman -Syy Redownload all repository metadata
pacman -Syua Sync with mirrors and update all packages
pacman -Fs filename Search repositories for packages that contain filename
pacman -Ss search_string Search repositories for package name or description that contain a search string
pacman -S package_name Install package_name and any dependencies required
pacman -Rs package_name Remove package_name and all dependencies not needed by anything else
pacman -Dk Check that all required dependencies are installed
pacman -Q List all installed packages and their versions
pacman -Qe List all packages that were installed manually
pacman -Qd List all packages that were isntalled automatically
pacman -Sii package_name See which packages require package_name to be installed
#### Services
Command Description
rc-update List all services and their runlevels
rc-update add service_name default Add service_name to the default runlevel
rc_update delete service_name default Remove service_name from the default runlevel
service service_name start Start service_name
service service_name stop Stop service_name
service service_name status Query the status of service_name
## Run a program on your dedicated AMD graphics card on Linux
I've recently figured out how to run a program on my dedicated AMD R7 M445 graphics card in Ubuntu 17.04, and since it's taken me far too long to around figuring it out, I thought I'd note it down here for future reference - if it helps you too, let me know in the comments below!
It's actually really simple. First, check that your dedicated AMD graphics card shows up with lspci:
lspci
If it's anything like my setup, you'll get a pair of rows like this (though they might not be next to each other):
00:02.0 VGA compatible controller: Intel Corporation HD Graphics 620 (rev 02)
01:00.0 Display controller: Advanced Micro Devices, Inc. [AMD/ATI] Topaz XT [Radeon R7 M260/M265 / M340/M360 / M440/M445] (rev c3)
Thankfully, my dedicated AMD card is showing (better than it did in previous versions of ubuntu, too, which thought it was an M225!). Next, we need to check that the amdgpu kernel module is loaded with a quick lsmod:
lsmod | grep -i amd
On my laptop, I get this:
amdkfd 139264 1
amd_iommu_v2 20480 1 amdkfd
amdgpu 1564672 1
i2c_algo_bit 16384 2 amdgpu,i915
ttm 98304 1 amdgpu
drm_kms_helper 151552 2 amdgpu,i915
drm 352256 9 amdgpu,i915,ttm,drm_kms_helper
Yay! It's loaded. Now to do a test to see if we can run anything on it:
glxinfo | grep "OpenGL renderer"
DRI_PRIME=1 glxinfo | grep "OpenGL renderer"
The above runs glxinfo twice: Once on the integrated graphics card, and once on the dedicated graphics card. The key here is the DRI_PRIME=1 environment variable - this tells the amdgpu driver that this process should run on the dedicated graphics and not the integrated graphics card. On my machine, I get this output:
OpenGL renderer string: Mesa DRI Intel(R) HD Graphics 620 (Kabylake GT2)
OpenGL renderer string: Gallium 0.4 on AMD ICELAND (DRM 3.9.0 / 4.10.0-33-generic, LLVM 4.0.0)
As you can see, the latter invocation of the command ran on the dedicated AMD graphics card, and the former on the integrated graphics. So simple!
Now that we've verified that it works, we can do it with any program:
DRI_PRIME=1 inkscape
Did this you find this helpful? Did it work (or not)? Let me know in the comments!
## Deep dive: Email, Trust, DKIM, SPF, and more
(Above: Lots of parcels. Hopefully you won't get this many through the door at once..... Source)
Now that I'm on holiday, I've got some time to write a few blog posts! As I've promised a few people a post on the email system, that's what I'll look at this this post. I'm going to take you on a deep dive through the email system and trust. We'll be journeying though the fields of DKIM signatures, and climb the SPF mountain. We'll also investigate why the internet needs to take this journey in the first place, and look at some of the challenges one faces when setting up their own mail server.
Hang on to your hats, ladies and gentlemen! If you get to the end, give yourself a virtual cookie :D
Before we start though, I'd like to mention that I'll be coming at this from the perspective of my own email server that I set up myself. Let me introduce to you the cast: Postfix (the SMTP MTA), Dovecot (the IMAP MDA), rspamd (the spam filter), and OpenDKIM (the thing that deals with DKIM signatures).
With that out of the way, let's begin! We'll start of our journey by mapping out the journey a typical email undertakes.
Let's say Bob Kerman wants to send Bill an email. Here's what happens:
1. Bill writes the email and hits send. His email client connects to his email server, logs in, and asks the server to deliver a message for him.
2. The server takes the email and reads the From header (in this case it's [email protected]), figures out where the mail server is located, connects to it, and asks it to deliver Bob's message to Bill. mail.billsboosters.com takes the email and files it in Bill's inbox.
3. Bill connects to his mail server and retrieves Bob's message.
Of course, this is simplified in several places. mail.bobsrockets.com will obviously need to do a few DNS lookups to find billsboosters.com's mail server and fiddle with the headers of Bob's message a bit (such as adding a Received header etc.), and smtp.billsboosters.com won't just accept the message for delivery without checking out the server it came from first. How does it check though? What's preventing seanssatellites.net pretending to be bobsrockets.com and sending an imposter?
Until relatively recently, the answer was, well, nothing really. Anyone could send an email to anyone else without having to prove that they could indeed send email in the name of a domain. Try it out for yourself by telnetting to a mail server on port 25 (unencrypted SMTP) and trying in something like this:
HELO mail.bobsrockets.com
MAIL From: <[email protected]>
RCPT TO <[email protected]>
DATA
From: [email protected]
To: [email protected]
Hello! This is a email to remind you.....
.
QUIT
Oh, my! Frank at franksfuel.io can connect to any mail server and pretend that [email protected] is sending a message to [email protected]! Mail servers that allow this are called open relays, and today they usually find themselves on several blacklists within minutes. Ploys like these are easy to foil, thankfully (by only accepting mail for your own domains), but it still leaves the problem of what to do about random people connecting to your mail server delivering spam to your inbox that claims to be from someone they aren't supposed to be sending mail for.
In response, some mail servers demanded things like the IP that connects to send an email must reverse to the domain name that they want to send email from. Clever, but when you remember that anyone can change their own PTR records, you realise that it's just a minor annoyance to the determined spammer, and another hurdle to the legitimate person in setting up their own mail server!
Clearly, a better solution is needed. Time to introduce our first destination: SPF. SPF stands for sender policy framework, and defines a mechanism by which a mail server can determine which IP addresses a domain allows mail to be sent from in it's name. It's a TXT record that sites at the root of a domain. It looks something like this:
v=spf1 a mx ptr ip4:5.196.73.75 ip6:2001:41d0:e:74b::1 a:starbeamrainbowlabs.com a:mail.starbeamrainbowlabs.com -all
The above is my SPF TXT record for starbeamrainbowlabs.com. It's quite simple, really - let's break it down.
v=spf1
This just defines the version of the SPF standard. There's only one version so far, so we include this to state that this record is an SPF version 1 record.
a mx ptr
This says that the domain that the sender claims to be from must have an a and an mx record that matches the IP address that's sending the email. It also says that the ptr record associated with the sender's IP must resolve to the domain the sender claims to be sending from, as described above (it does help with dealing with infected machines and such).
ip4:5.196.73.75 ip6:2001:41d0:e:74b::1
This bit says that the IP addresses 5.196.73.75 and 2001:41d0:e:74d::1 are explicitly allowed to send mail in the name of starbeamrainbowlabs.com.
a:starbeamrainbowlabs.com a:mail.starbeamrainbowlabs.com
After all of the above, this bit isn't strictly necessary, but it says that all the IP addresses found in the a records for starbeamrainbowlabs.com and mail.starbeamrainbowlabs.com are allowed to send mail in the name of starbeamrainbowlabs.com.
-all
Lastly, this says that if you're not on the list, then your message should be rejected! Other variants on this include ~all (which says "put it in the spam box instead"), and +all (which says "accept it anyway", though I can't see how that's useful :P).
As you can see, SPF allows a mail server to verify if a given client is indeed allowed to send an email in the name of any particular domain name. For a while, this worked a treat - until a new problem arose.
Many of the mail servers on the internet don't (and probably still don't!) support encryption when connecting to and delivering mail, as certificates were expensive and difficult to get hold of (nowadays we've got LetsEncrypt who give out certificates for free!). The encryption used when mail servers connect to one another is practically identical to that used in HTTPS - so if done correctly, the identity of the remote server can be verified and the emails exchanged encrypted, if the world's certification authorities aren't corrupted, of course.
Since most emails weren't encrypted when in transit, a new problem arose: man-in-the-middle attacks, whereby an email is altered by one or more servers in the delivery chain. Thinking about it - this could still happen today even with encryption, if any one server along an email's route is compromised. To this end, another mechanism was desperately needed - one that would allow the receiving mail server to verify that an email's content / headers hadn't been surreptitiously altered since it left the origin mail server - potentially preventing awkward misunderstandings.
Enter stage left: DKIM! DKIM stands for Domain Keys Identified Mail - which, in short, means that it provides a method by which a receiving mail server can cryptographically prove that a message hasn't been altered during transit.
It works by having a public-private keypair, in which the public key can only decrypt things, but the private key is capable of encrypting things. A hash of the email's headers / content is computed and encrypted with the private key. Then the encrypted hash is attached to the email in the DKIM-Signature header.
The receiving mail server does a DNS lookup to find the public key, and decrypts the hash. It then computes it's own hash of the email headers / content, and compares it against the decrypted hash. If it matches, then the email hasn't been fiddled with along the way!
Of course, not all the headers in the email are hashed - only a specific subset are included in the hash, since some headers (like Received and X-Spam-Result) are added and altered during transit. If you're interested in implementing DKIM yourself - DigitalOcean have a smashing tutorial on the subject, which should adapt easily to whatever system you're running yourself.
With both of those in place, billsboosters.com's mail server can now verify that mail.bobsrockets.com is allowed to send the email on behalf of bobsrockets.com, and that the message content hasn't been tampered with since it left mail.bobsrockets.com. mail.billsboosters.com can also catch franksfuel.io in the act of trying to deliver spam from seanssatellites.net!
There is, however, one last piece of the puzzle left to reveal. With all this in place, how do you know if your mail was actually delivered? Is it possible to roll SPF and DKIM out gradually so that you can be sure you've done it correctly? This can be a particular issue for businesses and larger email server setups.
This is where DMARC comes in. It's a standard that lets you specify an email address you'd like to receive DMARC reports at, which contain statistics as to how many messages receiving mail servers got that claimed to be from you, and what they did with them. It also lets you specify what percentage of messages should be subject to DMARC filtering, so you can roll everything out slowly. Finally, it lets you specify what should happen to messages that fail either SPF, DKIM, or both - whether they should be allowed anyway (for testing purposes), quarantined, or rejected.
DMARC policies get specified (yep, you guessed it!) in a DNS record. unlike SPF though, they go in _dmarc.megsmicroprocessors.org as a TXT record, substituting megsmicroprocessors.org for your domain name. Here's an example:
v=DMARC1; p=none; rua=mailto:[email protected]
This is just a simple example - you can get much more complex ones than this! Let's go through it step by step.
v=DMARC1;
Nothing to see here - just a version number as in SPF.
p=none;
This is the policy of what should happen to messages that fail. In this example we've used none, so messages that fail will still pass right on through. You can set it to quarantine or even reject as you gain confidence in your setup.
rua=mailto:[email protected]
This specifies where you want DMARC reports to be sent. Each mail server that receives mail from your mail server will bundle up statistics and send them once a day to this address. The format is in XML (which won't be particularly easy to read), but there are free DMARC record parsers out there on the internet that you can use to decode the reports, like dmarcian.
That completes the puzzle. If you're still reading, then congratulations! Post in the comments and say hi :D We've climbed the SPF mountain and discovered how email servers validate who is allowed to send mail in the name of another domain. We've visited the DKIM signature fields and seen how the content of email can be checked to see if it's been altered during transit. Lastly, we took a stroll down DMARC lane to see how it's possible to be sure what other servers are doing with your mail, and how a large email server setup can implement DMARC, DKIM, and SPF more easily.
Of course, I'm not perfect - if there's something I've missed or got wrong, please let me know! I'll try to correct it as soon as possible.
Lastly, this is, as always, a starting point - not an ending point. An introduction if you will - it's up to you to research each technology more thoroughly - especially if you're thinking of implementing them yourself. I'll leave my sources at the bottom of this post if you'd like somewhere to start looking :-)
## Unmounting NFS Shares on Shutdown in OpenRC Manjaro
(Above: A clipart image of a server. Source)
Since I've been using Manjaro with OpenRC when I'm out and about, I've been steadily fixing little issues and niggles I've been encountering one by one (such as finding the option to let you move the windows on the taskbar panel around yourself).
One of the first issues I encountered was that OpenRC would generously take the network down before my NFS (network file system) shares have been unmounted. This results in lengthly delays when shutting down as each of the components of the NFS mounting system have to be waited upon by OpenRC and finally killed after taking too long to shut down.
Initially I attempted to investigate reordering the shutdown process, but that quickly grew out of hand as I was investigating, and I discovered that it was not a particularly practical or, indeed, stable solution to my particular problem. Next, I found autofs which looked like it would solve the problem by automatically mounting and unmounting my NFS shares as and when they are needed, but despite assisance from someone far more experienced in the Manjaro world than I (thank you!) couldn't get it to work reliably. In addition, it started exhibiting some odd behaviour like hiding all my other mounts in my /media folder, so I went on the hunt for better solution.
Quite by chance (all thanks to Duck Duck Go Instant Answers!) I stumbled upon NetworkManager dispatcher scripts. NetworkManager is the service / application that manages, surprisingly, the network connections on several major linux distributions - including Ubuntu (which I've used before), and, crucially, Manjaro. Although the answer said that the functionality I wanted had been removed, upon looking into the amtter it appeared to be an artifact of the way systemd shutdown the system, and so I gave it a whirl anyway just to see if it would work.
Thankfully it did end up working! To that end, I thought I'd (re)post the solution I found here for future reference, and in case it helps anyone else :-)
Assuming you already have your shares set up and working in your /etc/fstab, you can create a file in the folder /etc/NetworkManager/dispatcher.d/pre-down.d with the contents something like this:
#!/bin/sh
logger "Unmounting NFS shares gracefully before the network goes down...";
umount /media/bob/rocket-diagrams-nas;
umount /media/sean/satellite-schematics;
logger "Unmounted NFS shares successfully.";
Once done, you'll need to make it executable with a quick sudo chmod +x, and try rebooting to test it!
In theory, this could be used to do other things that need to be done before the network is taken down, like making a sekret tracking request to your web server for anti-theft purposes, or uploading a backup of your laptop's /etc directory automagically in case it comes to a sticky end.
## Semi-automated backups with duplicity and an external drive
(Above: A bunch of hard drives. The original can be found here.)
Since I've recently got myself a raspberry pi to act as a server, I naturally needed a way to back it up. Not seeing anything completely to my tastes, I ended up putting something together that did the job for me. For this I used an external hard drive, duplicity, sendxmpp (sudo apt install sendxmpp), and a bit of bash.
Since it's gone rather well for me so far, I thought I'd write a blog post on how I did it. It still needs some tidying up, of course - but it works in it's current state, and perhaps it will help someone else put together their own system!
### Step 1: Configuring the XMPP server
I use XMPP as my primary instant messaging server, so it's only natural that I'd want to integrate the system in with it to remind me when to plug in the external drive, and so that it can tell me when it's done and what happened. Since I use prosody as my XMPP server, I can execute the following on the server:
sudo prosodyctl adduser [email protected]
...and then enter a random password for the new account. From there, I set up a new private persistent multi-user chatroom for the messages to filter into, and set my client to always notify when a message is posted.
After that, it was a case of creating a new config file in a format that sendxmpp will understand:
[email protected]:5222 thesecurepassword
### Step 2: Finding the id of the drive partition
With the XMPP side of things configured, next I needed a way to detect if the drie was plugged in or not. Thankfully all partitions have a unique id built-in, which you can use to see if it's plugged in or not. It's easy to find, too:
sudo blkid
The above will list all available partitions and their UUID - the unique id I mentioned. With that in hand, we can now check if it's plugged in or not with a cleverly crafted use of the readlink command:
readlink /dev/disk/by-uuid/${partition_uuid} 1>/dev/null 2>&2; partition_found=$?
if [[ "${partition_found}" -eq "0" ]]; then echo "It's plugged in!"; else echo "It's not plugged in :-("; fi Simple, right? readlink has an exit code of 0 if it managed to read the symbolik link in /dev/disk/by-uuid ok, and 1 if it didn't. The symbolic links in /deve/disk/by-uuid are helpfuly created automatically for us :D From here, we can take it a step further to wait until the drive is plugged in: # Wait until the drive is available while true do readlink "${partition_uuid}";
if [[ "$?" -eq 0 ]]; then break fi sleep 1; done ### Step 3: Mounting and unmounting the drive Raspberry Pis don't mount drive automatically, so we'll have do that ourselves. Thankfully, it's not so tough: # Create the fodler to mount the drive into mkdir -p${backup_drive_mount_point};
# Mount it in read-write mode
mount "/dev/disk/by-uuid/${partition_uuid}" "${backup_drive_mount_point}" -o rw;
# Do backup thingy here
# Sync changes to disk
sync
# Unmount the drive
umount "${backup_drive_mount_point}"; Make sure you've got the ntfs-3g package installed if you want to back up to an NTFS volume (Raspberry Pis don't come with it by default!). ### Step 4: Backup all teh things! There are more steps involved in getting to this point than I thought there were, but if you've made it this far, than congrats! Have a virtual cookie :D 🍪 The next part is what you probably came here for: duplicity itself. I've had an interesting time getting this to work so far, actually. It's probably easier if I show you the duplicity commands I came up with first. # Create the archive & temporary directories mkdir -p /mnt/data_drive/.duplicity/{archives,tmp}/{os,data_drive} # Do a new backup PASSPHRASE=${encryption_password} duplicity --full-if-older-than 2M --archive-dir /mnt/data_drive/.duplicity/archives/os --tempdir /mnt/data_drive/.duplicity/tmp/os --exclude /proc --exclude /sys --exclude /tmp --exclude /dev --exclude /mnt --exclude /var/cache --exclude /var/tmp --exclude /var/backups / file://${backup_drive_mount_point}/duplicity-backups/os/ PASSPHRASE=${data_drive_encryption_password} duplicity --full-if-older-than 2M --archive-dir /mnt/data_drive/.duplicity/archives/data_drive --tempdir /mnt/data_drive/.duplicity/tmp/data_drive /mnt/data_drive --exclude '**.duplicity/**' file://${backup_drive_mount_point}/duplicity-backups/data_drive/ # Remove old backups PASSPHRASE=${encryption_password} duplicity remove-older-than 6M --force --archive-dir /mnt/data_drive/.duplicity/archives/os file:///${backup_drive_mount_point}/duplicity-backups/os/ PASSPHRASE=${data_drive_encryption_password} duplicity remove-older-than 6M --force --archive-dir /mnt/data_drive/.duplicity/archives/data_drive file:///${backup_drive_mount_point}/duplicity-backups/data_drive/ Path names have been altered for privacy reasons. The first duplicity command in the above was fairly straight forward - backup everything, except a few folders with cache files / temporary / weird stuff in them (like /proc). I ended up having to specify the archive and temporary directories here to be on another disk because the Raspberry Pi I'm running this on has a rather... limited capacity on it's internal micro SD card, so the default location for both isn't a good idea. The second duplicity call is a little more complicated. It backs up the data disk I have attached to my Raspberry Pi to the external drive I've got plugged in that we're backing up to. The awkward bit comes when you realise that the archive and temporary directories are located on this same data-disk that we're trying to back up. To this end, I eventually found (through lots of fiddling) that you can exclude a folder duplicity via the --exclude '**.duplicity/**' syntax. I've no idea why it's different when you're not backing up the root of the filesystem, but it is (--exclude ./.duplicity/ didn't work, and neither did /mnt/data_drive/.duplicity/). The final two duplicity calls just clean up and remove old backups that are older than 6 months, so that the drive doesn't fill up too much :-) ### Step 5: What? Where? Who? We've almost got every piece of the puzzle, but there's still one left: letting us know what's going on! This is a piece of cake in comparison to the above: function xmpp_notify { echo$1 | sendxmpp --file "${xmpp_config_file}" --resource "${xmpp_resource}" --tls --chatroom "${xmpp_target_chatroom}" } Easy! All we have to do is point sendxmpp at our config file we created waaay in step #1, and tell it where the chatroom is that we'd like it to post messages in. With that, we can put all the pieces of the puzzle together: #!/usr/bin/env bash source .backup-settings function xmpp_notify { echo$1 | sendxmpp --file "${xmpp_config_file}" --resource "${xmpp_resource}" --tls --chatroom "${xmpp_target_chatroom}" } xmpp_notify "Waiting for the backup disk to be plugged in."; # Wait until the drive is available while true do readlink "${backup_drive_dev}";
if [[ "$?" -eq 0 ]]; then break fi sleep 1; done xmpp_notify "Backup disk detected - mounting"; mkdir -p${backup_drive_mount_point};
mount "${backup_drive_dev}" "${backup_drive_mount_point}" -o rw
xmpp_notify "Mounting complete - performing backup";
# Create the archive & temporary directories
mkdir -p /mnt/data_drive/.duplicity/{archives,tmp}/{os,data_drive}
echo '--- Root Filesystem ---' >/tmp/backup-status.txt
# Create the archive & temporary directories
mkdir -p /mnt/data_drive/.duplicity/{archives,tmp}/{os,data_drive}
# Do a new backup
PASSPHRASE=${encryption_password} duplicity --full-if-older-than 2M --archive-dir /mnt/data_drive/.duplicity/archives/os --tempdir /mnt/data_drive/.duplicity/tmp/os --exclude /proc --exclude /sys --exclude /tmp --exclude /dev --exclude /mnt --exclude /var/cache --exclude /var/tmp --exclude /var/backups / file://${backup_drive_mount_point}/duplicity-backups/os/ 2>&1 >>/tmp/backup-status.txt
echo '--- Data Disk ---' >>/tmp/backup-status.txt
PASSPHRASE=${data_drive_encryption_password} duplicity --full-if-older-than 2M --archive-dir /mnt/data_drive/.duplicity/archives/data_drive --tempdir /mnt/data_drive/.duplicity/tmp/data_drive /mnt/data_drive --exclude '**.duplicity/**' file://${backup_drive_mount_point}/duplicity-backups/data_drive/ 2>&1 >>/tmp/backup-status.txt
xmpp_notify "Backup complete!"
cat /tmp/backup-status.txt | sendxmpp --file "${xmpp_config_file}" --resource "${xmpp_resource}" --tls --chatroom "${xmpp_target_chatroom}" rm /tmp/backup-status.txt xmpp_notify "Performing cleanup." PASSPHRASE=${encryption_password} duplicity remove-older-than 6M --force --archive-dir /mnt/data_drive/.duplicity/archives/os file:///${backup_drive_mount_point}/duplicity-backups/os/ PASSPHRASE=${data_drive_encryption_password} duplicity remove-older-than 6M --force --archive-dir /mnt/data_drive/.duplicity/archives/data_drive file:///${backup_drive_mount_point}/duplicity-backups/data_drive/ sync; umount "${backup_drive_mount_point}";
xmpp_notify "Done! Backup completed. You can now remove the backup disk."
I've tweaked a few of the pieces to get them to work better together, and created a separate .backup-settings file to store all the settings in.
That completes my backup script! Found this useful? Got an improvement? Use a different strategy? Post a comment below!
## The other side of the fence: A Manjaro review
(Above: One of the default Manjaro wallpapers.)
Sorry for the delay! I've had rather a lot to do recently - including set up the machine I'm using to write this blog post.
For a while now, I've been running Ubuntu on my main laptop. After making the switch from Windows 7, I haven't looked back. Recently though, a friend of mine suggested I check out Manjaro - another distribution of Linux based on Arch Linux . After setting it up on a secondary machine and playing around with it, I rather like it, actually - and I've decided to write a post about my experiences coming from Ubuntu.
Like most things, I've got multiple different reasons for playing around with Manjaro. Not least of which is to experience a different ecosystem and a different way of doing things - namely the Arch Linux ecosystem. To that end, I've selected the OpenRC init system - since I've got experience with Systemd already, I feel it's essential to gain experience with other technologies.
With my preferences selected, I fired up manjaro-architect (available on the Manjaro website, which is linked above) and began the installation. I quickly found that the installation was not a simple process - requiring several reboots to get the options just right. In particular, the partitioning tools available are somewhat limited - such that I had to boot into a live Ubuntu environment to sort them out to get a dual boot setup working correctly.
On the other side, the installer allows the configuration of so many more options, like the mount options of the partitions, the kernel to use and it's associated modules, the init system that is used, and the desktop environment you want to use (I've picked XFCE). During the install process I've learnt about a bunch of different things that I had no idea about before.
After installation, I then started on the long task of configuring it to my liking. I'm still working on that, but I'm constantly amazed at the level of flexibility it offers. Nearly everything can be customised - including all the title bar graphics and the ordering and position of everything on the task bar (called a panel in XFCE.
I've found OpenRC an interesting learning experience too. It's very similar to upstart - another init system I used before Ubuntu switched to systemd. As a result, it's so uch simpler to get my head around. It feels a lot more.... transparent than systemd, which is a good thing I think. I do miss a few of the features that systemd offers, however. In time, though, I'm sure that I'll find alternative ways of doing things - different projects do have different ways of thinking, after all!
The concept of the [AUR]() (The Arch User Repository) is possibly one of my faviourite things out of all the things I've encountered so far. It's a community-driven archive of packages, but instead of containing the package binaries themselves, each package contains instructions to fetch build, and install said package.
This way requires much less maintenance I suspect, and makes it much easier to stay up to date with things. The install process for a package from the AUR is a little complex, sure, but so much easier and more automated than doing it by hand. It's like taking the benefits of downloading an installer manually from a program's website like you have to on Windows, and combining it with the ease of use and automation that comes with package managers like apt (Debian-based distrubutions) and pacman / yaourt (Arch Linux-based distributions).
In short, Manjaro is a breath of fresh air, and very different to what I've tried before. While it's certainly not for the linux beginner (try Ubuntu or Linux Mint if you're a beginner!) - especially the installer - I think it fulfills a different purpose for me at least - as platform from which to explore the Arch Linux ecosystem in relative comfort and dive deeper into the way that all the different parts in a linux system interact with each other.
## Share files from your host machine with virtual machine guests with 9p and virtual machine manager
(Infinity symbol source, Virtual Machine Manager logo traced from the logo on the official site automatically with Inkscape)
Recently I've been looking at Virtual Machine Manager with qemu and KVM to set up a few virtual machines for my next year at university. During this learning and planning process, I've discovered a way (source) to setup an environment such that you can share a folder on your host machine (optionally read-only) with a guest machine running inside a virtual machine with a technology called 9p, and I wanted to share how I did it here.
To start make sure your virtual machine is powered off, and go to Add HardwareFilesystem, and fill in the boxes:
• Source path - The path on the host system to the folder you can to share.
• Target path - The path at which the guest will see it. Note that this isn't a place on the guest file system, from what I can tell - see below.
• Export filesystem as readonly mount - Check this box to make the share read-only.
Next, click "Finish", and start your virtual machine. Next, open a terminal on the guest machine and type something similar to this:
sudo mkdir /mnt/host_files
sudo mount -t 9p -o trans=virtio,version=9p2000.L /target /mnt/host_files
...this should mount the host share at /tmp/share to the path /mnt/host_files on the guest machine.
Found this different sort of post useful? Got something to add? Post about it in the comments!
## Debug your systemd services with journalctl
The chances are that if you're using linux, you will probably have run into systemd. If you find yourself in the situation where you've got a systemd service that keeps dying and you don't know why (I've been there before several times!), and there's nothing helpful in /var/log, before you give up, you might want to give journalctl a try. It's systemd's way of capturing the output of a service and storing it in it's logging system (or something).
When I first found out about it, I read that apparently journalctl -xe servicename would show me the logs for any given service. It turned out that it wasn't the case (it just threw a nasty error), so I went trawling through the man pages and found the correct command-line switch. If you've got a service called rocketbooster.service, and you want to see if systemd has any logs stored for it, then you can execute this command:
journalctl --unit rocketbooster.service
...or for short
journalctl -u rocketbooster.service
It should open the logs (if there are any) in less - with the oldest logs at the top, so you might need to scroll all the way down to the bottom to see anything that's relevant to your problem (shift + G will take you to the bottom of the file).
I've found that systemd has a habit of rotating the logs too - and journalctl doesn't appear to know how to access the rotated logs, so it's best if you use this command as soon as possible after failure (suggestions on how to access these rotated logs are welcome! Post down in the comment :D).
I thought I'd document it here in case it was useful to anyone - and so I don't forget myself! :P
Art by Mythdael | 2017-11-19 01:16:47 | {"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": 0.1944909542798996, "perplexity": 3635.81061786052}, "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-2017-47/segments/1510934805242.68/warc/CC-MAIN-20171119004302-20171119024302-00116.warc.gz"} |
https://cs.stackexchange.com/questions/111349/how-do-we-guess-the-recurrence-relation-from-the-given-equation | How do we guess the recurrence relation from the given equation
In this book introduction to algorithms , i have been reading about a method named substitution method to solve the recurrence, the recurrence equation is $$$$T(n)=2 T(\lfloor n / 2\rfloor)+n$$$$
the author guessed the solution was $$O(n \log n)$$ and proved it below, my question is how to make the guess? I wanted to know why it cant be $$O(n^2)$$? How do you guess them correctly at first?
\begin{aligned} T(n) & \leq 2(c\lfloor n / 2\rfloor \lg (\lfloor n / 2\rfloor))+n \\ & \leq c n \lg (n / 2)+n \\ &=c n \lg n-c n \lg 2+n \\ &=c n \lg n-c n+n \\ & \leq c n \lg n \end{aligned}
• @Evil i am sorry, i have corrected the post, please let me know if i need to change any thing in it. – Naveen Jul 1 '19 at 4:46
To guess simple recurrences most useful step is to study basic intuition beyond so-called master theorem that looks at any recursion as if it is implicit tree.
Your case is admissible for it and thus easy: on each step you have half of task (peek one branch down in tree), and $$O(n)$$ work. Multiply tree height and amount of work on each step and you will get your guess. But you will probably never guess any inadmissible case.
• how do i peek one branch in a tree? the equation is given like 2𝑇(⌊𝑛/2⌋)+𝑛 but from that thing i can only figure out each node gets divided in to two at each step – Naveen Jul 1 '19 at 13:33
• @Naveen if any amount of work divided in two on each level, how much levels do you have if total amount is $n$? – Konstantin Vladimirov Jul 1 '19 at 15:31
The guess $$O(n^2)$$ also works: $$T(n) \leq 2c\lfloor n/2\rfloor^2 + n \leq \frac{c}{2} n^2 + n \leq cn^2,$$ as long as $$c \geq 2$$.
The author did not guess the answer. Presumably the author already knew the answer.
In practice, for most recurrences you would encounter one of the following methods will work:
• Open up the recurrence (also known as repeated substitution).
• Appeal to a standard result such as the master theorem.
• Use a computer experiment (could be deceiving).
In this case, the recurrence is so well-known and commonplace that the author must have seen it worked out somewhere. It can also be "solved" (i.e., the asymptotic behavior of $$T(n)$$ determined) using the master theorem, which is how you could do it. | 2021-08-04 05:39:48 | {"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": 9, "wp-katex-eq": 0, "align": 0, "equation": 1, "x-ck12": 0, "texerror": 0, "math_score": 0.9905791282653809, "perplexity": 812.3500848077343}, "config": {"markdown_headings": false, "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-2021-31/segments/1627046154796.71/warc/CC-MAIN-20210804045226-20210804075226-00372.warc.gz"} |
http://jims.io/ | # one
Well, it’s a start. I managed to get just one post posted today (besides this one) so all I have to do tomorrow is do it all over again.
Then rinse lather and repeat another 998 times. How hard can it be?
One thing I have found with this little project is I don’t quite have my workflow down even though I said I did.
It depends what computer I am on but I prefer to use a text editor. I was (and still do) use Marxico on my Chromebook for my writing and it’s great but i wish I didn’t have to use evernote to sync files.
So that part’s still a learning process.
I also blog like a hacker and use the terminal and jekyll to write.
Currently I am creating a new post - write the post and then ‘push’ to github pages for publishing.
I have yet to find the perfect text editor for my Chromebook that I like. As I have said above, Marxico is nice, but not perfect and every other one is pretty much about the same.
I am currently using a lenovo thinkpad to write this and I find that it’s just fine and regardless of OS I can use Caretwhich is starting to be my text and writing editor of choice. I just wish I could use it on the Chromebook.
What might happen is I turn my Dell Chromebook 13 into a linux machine that will allow me to use it with the Caret app as well as offer a more powerful OS.
So. it’s a start. I wrote today. This is something I could not say yesterday so I guess I am ahead of the game a little bit.
# 1000 posts
I have been thinking about what comes next when it comes to the future and my career online and one thing I know is that by the end of 2017 I would like to be able to make enough money to quit my job
To be honest, I am not sure if this is even possible but considering that this is just a part time job and it does not pay that much I would only need about $30/day of online income for this to be a reasonable possibility. Don’t get me wrong. I actually like my current work. Like all jobs, some days are better than others but overall this might be one of the better jobs that I have had when I have not been self employed. Of course, self employment is the best I think and so that’s my goal again - to be able to make enough to quit - or at very least double my income so that even if I don’t quit I have the opportunity to if the need comes to be. ## A Pivot, of Sorts Isn’t that what they call a startup that moves another direction from the original startup idea or plan? I have a website that in the past has earned me more than a few dollars but over time the topic and the interest in that topic is now so low that it’s almost laughable. Let’s just sa that I picked the wrong horse in the long term. So, I’m switching horses. I have a very loose idea on the topics I want to write about but in the beginning it’s really just going to be throw some stuff on the wall and see what sticks. Over time, I have a feeling that what will happen is that I will find what does work and well, do more of that. If nothing else this will be an experiment. I am a little different than most because I have turned off all the stats on my site. I can check google webmasters so I am not completely in the dark but I find that stats can distract from writing and you tend to write about the same topics or keywords. When you don’t have stats to worry about you can worry about the writing instead One of my favorite bloggers writes: I’ve said this 1,000 times before and I’ll continue to repeat myself (since I get new readers fairly consistently) but the only thing that matters when writing (publicly, privately, for personal use or professionally) is that you write. Period. Nothing else really matters. And that’s what I intend to do. John started a site and wrote 5000 posts in one year and although I would love to I don’t think I could reach that goal so I will start at a much lower and more attainable goal. Even 1000 articles will be a stretch as that works out to about 3 per day if I write every day. In John’s case he wrote that he did it to see if he could not only write every day but also if he could write every day in addition to all the other stuff going on in his life. That’s my big question too - can I do this? I guess time will tell. ## The Money Goal I know that all bloggers or most bloggers say that you should not worry about making money as you are destined to fail but a part of me wants to make some bucks online with this little experiment and I figure a good goa would to go real small and hope to make$1000.
I think that’s a real goal. If you do the math on the hourly wage for the time writing 1000 posts (assuming an hour each) then the math works out to $1.00 per hour. I could make more sitting on a street corner with a cup. But unlike the street corner the online writing has the chance to grow past your expectations and who would not like that? I don’t see any reason why I can’t squeeze an extra$1000 a year from now for my efforts.
## On Making Time
If there’s one challenge that will be the trickiest part it will be to find the time. If I spend an hour a post and have to do about 2.6 a day to hit my 1000 a year goal then I have to find 2.5 hours a day to work.
Considering that I somehow seem to be able to fill up my days quite quickly without this new challenge I have no idea how I am going to do this. I might have to forgo some sleep to make this happen but it’s been said that if something is important to you, you can make the time.
I seem to be able to spend a fair amount of time with my hobbies so perhaps instead of screwing around with writing software and computers I will have to focus on the work instead.
I should be able to do this. My work is only part-time so that means that there’s time to write the other part. If I need four hours to work and eight hours to sleep then that means I still have 12 hours to kill each and every day. Surely I can find some time to get some writing done.
## The How and the What
I plan about writing about stuff that you can buy on Amazon and then linking to it where if the item (or almost any other item) is bought after clickign on my link I get a small commission. You wouldn’t think it worth your time promoting smaller or what seems to be pretty rare items but you would be surprised. A penny here, and a penny there..
There are numerous items to write about so there’s no way in heck that I would ever run out of topics to write about. I specifically picked this topic because I wanted to be sure I would not start this project and then after a few posts have nothing to say.
## Results
I plan to share the results of my experiment here (I have to make use of this site somehow) as it will make me stay accountable and I will be able to track my progress or lack thereof.
I’m not sure what the end result will be with all this. I may have just wasted time, but in the past when I put my mind to it and either started a new site or resurrected an older one it was always worth the effort.
Here Goes…
# My Current Workflow
I’ve been trying to simplify my writing workflow so that I have all my writing and work in one place rather than save files locally on each machine I own (which changes often)
So of course seeing how I am writing this on a Chromebook it only makes sense that I use Google Drive as the place where all my work lives.
I’m currently writing this in markdown which is just plain text so the files are small, and also I like the idea of having my entire site on my hard drive, as well as being hosted on github pages for free:
I have toyed around with linux and MacOS and even Windows but I keep coming back round to using ChromeOS as my daily driver. My current Chrome OS devices are an i7 Chromebox (designed for meetings originally) and an Intel i3 Dell Chromebook 13.
I have a few other PC’s for testing and tweaking but I am finding that I am getting more work done with the Chrome Devices than spending a bunch of time trying to install software.
## The Game Changer
If I had to name one tool I could no longer live without, I would have to say a good text or markdown editor.
Currently, I do all my writing for this site (and others) in markdown and then just upload the file to github’s servers.
My editor of choice on Chrome OS is Marxico as it syncs with Evernote and offers another way to backup my work. I don’t really need evernote synchronization but I figure what the heck.
I use marxico because it offers a ‘distraction free’ way to write in a font I want and offers multiple color themes I can use at my whim too.
But markdown is what really made me wanting to start to writing and blogging more - even if nobody reads it. Markdown and writing in plain text again is such a nice experience when you come from using a bloated wordpress install which is overkill for a blog like mine with a few images and some text. I’ve created sites in the past that needed a database backend and this is not one of them.
Once I have written my post or article I export as markdown and upload the files to github via the web.
It took me a while to realize this, but with marxi.co if you have images in your post it will include those and zip up both the images and the markdown exported file to your desktop or wherever you choose to save the file. It took me a while to figure this out because sometimes it would export a zip file, and yet others it would just export the markdown text file but then I realized that if you have images in your article or post they get added to the archive
In the case of this post I have uploaded the images to github’s servers already and then just call the images with the markdown code from the remote github site vs locally. This allows me to preview what the post will look like before it goes live:
I could actually install chromebrew on my ChromeOS devices and actually use git to upload and post files but I find it’s just as easy for me to upload the files to github via the web than it is to type commands in a terminal.
That’s it.
• I write.
• I upload any images to github
• I check the preview to make all looks good
• I export the markdown file to Google Drive
Jekyll and Github Pages do the rest, and you get to see the results of my efforts.
I am writing this from my Thinkpad T530 and I replaced the screen to a glossy full HD model and the keyboard to a backlit verison and the difference from the original stock version is night and day.
This upgrade cost me about $100 for keyboard and display and it’s a much nicer experience than using stock keyboard without the backlighting. If I had to pick one upgrade over the other I would go for the monitor first because it makes a very big difference from the stock and crappy display where it semed like you were looking through a screen door. With this display, it seems like you are looking though a glass window instead and makes the computing experience a much nicer option. I thought I may have bought the wrong display when I took it out of the box, but it seems that is not the case and the replacement fit perfectly. The keyboard I bought is a bit smoother than the previous one as it had a bit of texture in comparison to this replacement which is a bit too smooth for my taste. I am hoping that over time the keys will ‘wear’ a bit and get broken in a bit because as it sits now they feel a little too shiney for my liking. Overall, I am thrilled that with about$100 worth of upgrades you can take a relatively cheap but still very functional Thinkpad T530 from mediocre to great. I now have a fast notebook computer that I can use for writing or pretty much anything else I do and I don’t feel like screaming every time I have to look at the crappy stock display.
Until tomorrow..
# November 2 2016 Online Earnings
zip. zilch. zero.
That’s how much money I made online today. Somebody did order somthing though - a lightbulb for a car so I might make about a dollar or less tomorrow but for today the actual earnings is squat.
Oh well, It’s just day 2 of this little experiment.
# November 1 2016 Earnings
## Now what?
I have no clue. But I’m going to start to document my journey to see if I can’t get to $100 a day over the next decade. Yes, I said decade. This will take some time. I’m not going to try any tricks or tactics. I’m just going to write about things that interest me and see if I can’t turn my ramblings here (and on other sites) into cold hard cash. As it’s being November 1, and the month of Namowrimo I figure that now is the perfect time to start this little project of mine. To be completely honest, if I was able to make$60.00 a year to pay for this domain name renewal I would be happy - let alone one hundred bucks.
Of all the sites I like to visit that are popular it’s that they have one voice (I hate multi-author sites) and a lot of content. I’m done worrying about visitors or stats or any of that.
For now, I’m just going to write. I’ve done everything but write - I have tested every text editor and blogging software ever made in hopes of finding the perfect writing environment and I have yet to find nirvana. But I sure did waste a lot of time when I should have been witing.
This post isn’t much, but it’s a start.
@(Jim)[writing make money onine make a living writing]
# Untitled document
I am thinking about going all in with Google Docs as my writing tool of choice and let the gabriel jekyll add on do the heavy lifting for me.
I can then edit the same post and continue my writing at a later date too. It’s kind of cool when you think about it.
I thought I would not like this but now I am not so sure.
# Welcome to StackEdit!
This is a test of writing in prose and I have to tell you it’s pretty good.
Hey! I’m your first Markdown document in StackEdit1. Don’t delete me, I’m very helpful! I can be recovered anyway in the Utils tab of the Settings dialog.
I guess I could just use stackedit to write with and get some writing done too. I
don’t really mind just writing in prose. I do not need an app to get some writing done it seems
We are getting there.
## Documents
StackEdit stores your documents in your browser, which means all your documents are automatically saved locally and are accessible offline!
Note:
• StackEdit is accessible offline after the application has been loaded for the first time.
• Your local documents are not shared between different browsers or computers.
• Clearing your browser’s data may delete all your local documents! Make sure your documents are synchronized with Google Drive or Dropbox (check out the Synchronization section).
#### Create a document
The document panel is accessible using the button in the navigation bar. You can create a new document by clicking New document in the document panel.
#### Switch to another document
All your local documents are listed in the document panel. You can switch from one to another by clicking a document in the list or you can toggle documents using Ctrl+[ and Ctrl+].
#### Rename a document
You can rename the current document by clicking the document title in the navigation bar.
#### Delete a document
You can delete the current document by clicking Delete document in the document panel.
#### Export a document
You can save the current document to a file by clicking Export to disk from the menu panel.
Tip: Check out the Publish a document section for a description of the different output formats.
## Synchronization
Note:
• Full access to Google Drive or Dropbox is required to be able to import any document in StackEdit. Permission restrictions can be configured in the settings.
• If you experience problems saving your documents on Google Drive, check and optionally disable browser extensions, such as Disconnect.
#### Open a document
You can open a document from Google Drive or the Dropbox by opening the Synchronize sub-menu and by clicking Open from…. Once opened, any modification in your document will be automatically synchronized with the file in your Google Drive / Dropbox account.
#### Save a document
You can save any document by opening the Synchronize sub-menu and by clicking Save on…. Even if your document is already synchronized with Google Drive or Dropbox, you can export it to a another location. StackEdit can synchronize one document with multiple locations and accounts.
#### Synchronize a document
If you just have modified your document and you want to force the synchronization, click the button in the navigation bar.
Note: The button is disabled when you have no document to synchronize.
#### Manage document synchronization
Since one document can be synchronized with multiple locations, you can list and manage synchronized locations by clicking Manage synchronization in the Synchronize sub-menu. This will let you remove synchronization locations that are associated to your document.
Note: If you delete the file from Google Drive or from Dropbox, the document will no longer be synchronized with that location.
## Publication
Once you are happy with your document, you can publish it on different websites directly from StackEdit. As for now, StackEdit can publish on Blogger, Dropbox, Gist, GitHub, Google Drive, Tumblr, WordPress and on any SSH server.
#### Publish a document
You can publish your document by opening the Publish sub-menu and by choosing a website. In the dialog box, you can choose the publication format:
• Markdown, to publish the Markdown text on a website that can interpret it (GitHub for instance),
• HTML, to publish the document converted into HTML (on a blog for example),
• Template, to have a full control of the output.
Note: The default template is a simple webpage wrapping your document in HTML format. You can customize it in the Advanced tab of the Settings dialog.
#### Update a publication
After publishing, StackEdit will keep your document linked to that publication which makes it easy for you to update it. Once you have modified your document and you want to update your publication, click on the button in the navigation bar.
Note: The button is disabled when your document has not been published yet.
#### Manage document publication
Since one document can be published on multiple locations, you can list and manage publish locations by clicking Manage publication in the menu panel. This will let you remove publication locations that are associated to your document.
Note: If the file has been removed from the website or the blog, the document will no longer be published on that location.
## Markdown Extra
StackEdit supports Markdown Extra, which extends Markdown syntax with some nice features.
Tip: You can disable any Markdown Extra feature in the Extensions tab of the Settings dialog.
### Tables
Markdown Extra has a special syntax for tables:
Item Value
Computer $1600 Phone$12
Pipe 1 You can specify column alignment with one or two colons: Item Value Qty Computer1600 5
Phone $12 12 Pipe$1 234
### Definition Lists
Markdown Extra has a special syntax for definition lists too:
Term 1
Term 2
Definition A
Definition B
Term 3
Definition C
Definition D
part of definition D
### Fenced code blocks
GitHub’s fenced code blocks are also supported with Highlight.js syntax highlighting:
// Foo
var bar = 0;
Tip: To use Prettify instead of Highlight.js, just configure the Markdown Extra extension in the Settings dialog.
• about Prettify syntax highlighting here,
• about Highlight.js syntax highlighting here.
### Footnotes
You can create footnotes like this2.
### SmartyPants
SmartyPants converts ASCII punctuation characters into “smart” typographic punctuation HTML entities. For example:
ASCII HTML
Single backticks 'Isn't this fun?' ‘Isn’t this fun?’
Quotes "Isn't this fun?" “Isn’t this fun?”
Dashes -- is en-dash, --- is em-dash – is en-dash, — is em-dash
You can insert a table of contents using the marker [TOC]:
[TOC]
### MathJax
You can render LaTeX mathematical expressions using MathJax, as on math.stackexchange.com:
The Gamma function satisfying $\Gamma(n) = (n-1)!\quad\forall n\in\mathbb N$ is via the Euler integral
Tip: To make sure mathematical expressions are rendered properly on your website, include MathJax into your template:
<script type="text/javascript" src="https://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeX-AMS_HTML"></script>
### UML diagrams
You can also render sequence diagrams like this:
Alice->Bob: Hello Bob, how are you?
Note right of Bob: Bob thinks
Bob-->Alice: I am good thanks!
And flow charts like this:
st=>start: Start
e=>end
op=>operation: My Operation
cond=>condition: Yes or No?
st->op->cond
cond(yes)->e
cond(no)->op | 2016-12-08 15:58:11 | {"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": 0.19618752598762512, "perplexity": 1732.9597937914302}, "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-2016-50/segments/1480698541864.24/warc/CC-MAIN-20161202170901-00369-ip-10-31-129-80.ec2.internal.warc.gz"} |
https://math.stackexchange.com/questions/2949318/how-to-find-formula-for-nth-partial-sum-of-telescopic-series | # How to find formula for nth partial sum of Telescopic Series
I have to find the formula for the nth partial sum of this series in order to determine if the series converges or diverges. Here's the equation:
$$\sum_{n=1}^\infty(\frac{1}{n}-\frac{1}{n+1})$$
What I did so far:
I made it into a single fraction:
$$\sum_{n=1}^\infty(\frac{(n+1)-n}{n(n+1)})$$
Then I split it up into partial fractions:
$$\frac{A}{n}-\frac{B}{n+1}=\frac{(n+1)-n}{n(n+1)}$$
$$\frac{An+A-Bn}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}$$
$$\frac{n(A-B)+A}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}$$
and so I got the two equations:
$$n(A+B)=-n$$ $$A=n+1$$
Does this look right so far? I know I'd have to use the Elimination method in order to find the sum.
• what is the partial sum for $n$ up to $2?$ What about up to $3?$ – Will Jagy Oct 10 '18 at 0:06
• Please don't make it into one fraction. Just add it as two fractions. Then it's pretty! – Don Thousand Oct 10 '18 at 0:07
• it's called "telescopic" for a reason. I have a feeling that a lot of fractions will cancel each other out and you'll only have the first and the last one left – Vasya Oct 10 '18 at 0:12
• Take a step back from the board: obviously $A=B=1$, that's where you started from. – Arnaud Mortier Oct 10 '18 at 0:17
$$S_1=\frac{1}{1}-\frac{1}{1+1}=\frac{1}{2}$$
$$S_2=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}=\frac{1}{1}-\frac{1}{3}$$
$$S_3=\frac{1}{1}-\frac{1}{1+1}+\frac{1}{2}-\frac{1}{2+1}+\frac{1}{3}-\frac{1}{3+1}=\frac{1}{1}-\frac{1}{3+1}$$
$$.$$
$$.$$
$$.$$
$$S_n=\frac{1}{1}-\frac{1}{n+1}$$ | 2019-09-22 18:40:08 | {"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 14, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7290040254592896, "perplexity": 228.9904782583437}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575627.91/warc/CC-MAIN-20190922180536-20190922202536-00022.warc.gz"} |
https://math.stackexchange.com/questions/937179/if-we-know-nullspace-of-matrix-how-to-find-reduced-row-echelon-form-of-that-mat | # If we know nullspace of matrix, how to find reduced row echelon form of that matrix?
vectors u = [4 1 0 0] and v = [1 0 2 1] form a base of nullspace of matrix $$A\in M_{5,4}(R)$$ Find a reduced row echelon form of Matrix A.
Since $n-r = dimN(A)$ we know we got two base variables and two free ones. And reduced row echelon form will look like this: $$\begin{bmatrix} 1 & 0 & a & b \\ 0 & 1 & c & d \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
I don't know where to go from here. I am also confused about something, if I plug in [4 1 0 0] instead of x1 x2 x3 and x4 in my "potential" reduced row echelon form I would get this:
1*4 + 0*1 + a*0 + b*0 = 0
0*4 + 1*1 + c*0 + d*0 = 0
and I get 4=0 and 1=0 from that. How is that possible? (I know I am making a mistake somewhere, just don't know where.)
The reduced row echelon form of a matrix is unique. The fact that the two given vectors form a basis of the null space means that the reduced form of the homogeneous linear system associated to the matrix is $$\begin{cases} x_1=4x_2+x_4\\ x_3=2x_4 \end{cases}$$ because, for $x_2=1$ and $x_4=0$ we get the first vector and with $x_2=0$ and $x_4=1$ we get the second vector. So the reduced system can be written $$\begin{cases} x_1-4x_2-x_4=0\\ x_3-2x_4=0 \end{cases}$$ which corresponds to the matrix $$\begin{bmatrix} 1 & -4 & 0 & -1 \\ 0 & 0 & 1 & -2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$
• @Jay That's not a subspace either; if $A$ has $U$ as null space, then also $-A$ has, but $A+(-A)$ has a different null space. I think this warrants a fresh question on the site, but you probably mean "The set of matrices whose null space contains a fixed subspace” (this is indeed a subspace of the vector space of $5\times4$ matrices). – egreg Sep 19 '14 at 15:08
• @Jay If the kernel of a linear map $f\colon V\to W$ contains a subspace $U$, then it induces a linear map $\bar{f}\colon V/U\to W$. Conversely a linear map $g\colon V/U\to W$ can be composed with the canonical projection. So your subspace has the same dimension as $\operatorname{Hom}(V/U,W)$. In the present case, $(4-2)\cdot 5=10$. – egreg Sep 19 '14 at 15:21 | 2019-08-18 19:06:11 | {"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": 0.8400630950927734, "perplexity": 91.91436283959654}, "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/1566027313996.39/warc/CC-MAIN-20190818185421-20190818211421-00268.warc.gz"} |
https://www.transtutors.com/questions/santinis-new-contract-for-2016-indicates-the-following-compensation-and-benefits-ben-2570136.htm | # Santinis new contract for 2016 indicates the following compensation and benefits: Benefit Descrip...
PLEASE SHOW WORK AND USE 2016 TAX LAWS.
Santinis new contract for 2016 indicates the following compensation and benefits: Benefit Description Amount Salary $130,000 Health insurance 9,000 Restricted stock grant 2,500 Bonus 5,000 Hawaii trip 4,000 Group-term life insurance 1,600 Parking ($285 per month) 3,420 Santini is 54 years old at the end of 2016. He is single and has no dependents. Assume that the employer matches $1 for$1 for the first $6,000 that the employee contributes to his 401(k) during the year. The 100 lsos each allow the purchase of 10 shares of stock at a strike price of$5 (also the market price on the date of grant). The isos vest in two years when the stock price is expected to be $15 and Santini expects to sell the shares in three years when the market price is$20. The restricted stock grant is 500 shares granted when the market price was $5 per share. Assume that the stock vests on December 31, 2016, and that the market price on that date is$7.50 per share. Also assume that Santini is willing to make any elections to reduce equity-based compensation taxes. The Hawaii trip was given to him as the outstanding sales person for 2015. The group-term life policy gives him \$150,000 of coverage. Assume that Santini does not itemize deductions for the year. Assume that Santini makes a section 83(b election. Determine Santini's taxable income and income tax liability for 2016. Use Tax rate schedules and Exhibit 12-10. Round your answer to the nearest whole dollar amount. Leave no answer blank. Enter zero if applicable.) Amount Description Taxable Benefits Salary Restricted stock grant Bonus Hawaii trip Life Insurance (taxable portion) Parking ISO AGI Standard Deduction Personal Exemption Taxable Income Income Tax Liability | 2019-01-23 20:23:04 | {"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": 0.2847675383090973, "perplexity": 7352.274050727439}, "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-2019-04/segments/1547584350539.86/warc/CC-MAIN-20190123193004-20190123215004-00244.warc.gz"} |
https://math.stackexchange.com/questions/2949416/use-standard-form-of-line-equation-to-find-line-using-two-points | # Use Standard Form of line equation to find line using two points?
I can find the equation of a line using the slope-intercept form if I have two points on the line.
However I was trying to do the same with the standard form of the equation of a line, $$ax+by=c$$, and I can't make it happen.
The way it looks, there are three unknowns and only two equations in our system, where we would have $$ax_1+by_1=c$$$$ax_2+by_2=c$$ I'm not sure what $$c$$ is in this context or how to get a value for it with just two points as total information.
If it turns out it's actually not possible to find the equation of a line using the standard form, why is that the case?
• Divide the whole equation by $b$. $\frac ab$ and $\frac cb$ are just as arbitrary as $a$ and $c$. Voila, you have only unknowns. – Don Thousand Oct 10 '18 at 1:53
• @RushabhMehta, Thank you that works, but I still don't understand what $c$ represents geometrically I guess. – jeremy radcliff Oct 10 '18 at 2:20
• $\frac cb$ represents the $y$ intercept – Don Thousand Oct 10 '18 at 2:23
• @RushabhMehta, wouldn't it be $\frac{c}{b}$? Or is it the same? – jeremy radcliff Oct 10 '18 at 2:25
• You are correct. That's a goof on my part. – Don Thousand Oct 10 '18 at 2:25
As you’ve noted, the system of equations in the unknown coefficients $$a$$, $$b$$ and $$c$$ is underdetermined. This means that there’s an infinite number of solutions to the system. (We know that the system has a solution since there is a line through the two points.) This shouldn’t really come as a surprise: if $$ax+by+c=0$$ is an equation of the line, you can multiply it by any nonzero scalar and get another equation of the same line. So, any of the solutions to the system of equations will give you a valid equation for the line. If you want to pick one in particular, a convenient choice is to choose $$a$$ and $$b$$ so that $$a^2+b^2=1$$ and $$a\ge0$$. With this choice, the normal vector $$(a,b)$$ is a unit vector and $$|c|$$ is then the distance of the line from the origin.
• Thanks, this is the kind of explanation I was looking for and it makes sense. I changed the accepted answer to yours since it answers my question better. – jeremy radcliff Oct 10 '18 at 9:54
If you have two points start with $$y=ax+b$$
where you only have to find $$a$$ and $$b$$.
At this point you either plug your points in and get
$$ax_1+b = y_1$$
$$ax_2+b = y_2$$ to solve for $$a$$ and $$b$$, and then you change your equation $$y-ax=b$$ or $$ax-y = -b$$ which is in standard form.
The other way is to find the slope intercept form $$y=mx+b$$ and change it to the standard form by moving $$mx$$ to the left side of you equation.
• Thank you, but I already know how to solve in slope-intercept form. I was looking to solve directly in standard form $ax+by=c$ without turning it into s-i form. What Rushabh Metha mentioned about dividing by $c$ makes sense, but I still don't really understand where $c$ comes from and what it really means geometrically to divide by $c$ – jeremy radcliff Oct 10 '18 at 2:19
• With $c$ you have one degree of freedom. you can pick $c=1$ and solve $ax+by=1$ with the given two points to find $a$ and $b$ from there. – Mohammad Riazi-Kermani Oct 10 '18 at 2:33
• What if the line is vertical? – amd Oct 10 '18 at 4:57
• Then the form $x=C$ is the standard form. – Mohammad Riazi-Kermani Oct 10 '18 at 5:31
• My point is that you can’t get to that from $y=ax+b$. – amd Oct 10 '18 at 17:56 | 2019-06-26 00:35:29 | {"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": 25, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6877769231796265, "perplexity": 146.9049694531726}, "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-2019-26/segments/1560627999964.8/warc/CC-MAIN-20190625233231-20190626015231-00320.warc.gz"} |
https://math.stackexchange.com/questions/2412192/given-a-ring-homomorphism-and-a-unital-module-make-another-module | Given a ring homomorphism and a unital module, make another module
This question is from Abstract Algebra book by Pierre Grillet (2nd edition).
Let $\phi:R\to S$ be a homomorphism of rings with identity and let $A$ be a unital left $S$-module. Make $A$ a unital left $R$-module.
Aim is to find a map $g:R \times A \to A$ that satisfies module conditions.
We have $\phi(1_R)=1_S$ and since $A$ is a unital left $S$-module, we have a function $f: S \times A \to A$ with $s \in S, a\in A, f(s,a)=sa$, and since it is unital, we also have $f(1_S,a)=a$ for all $a\in A$.
Can I define the function $g$ as $g:f\circ\phi$, the composition?
If it is possible, we have $(1_R,a)\to (1_S,a)\to a$, which gives unitarity.
Also commutativity is done, since for $r_1,r_2\in R, a\in A$, $R$ action is like, $(r_1+r_2)a=\phi(r_1+r_2)a=[\phi(r_1)+\phi(r_2)]a=\phi(r_1)a+\phi(r_2)a$ which is equal to $r_1a+r_2a$
But $r_1(r_2a)=(r_1r_2)a$ doesn't seem to hold because $\phi$ is a ring homomorphism and we can't have $\phi(r1r2)=r_1\phi(r_2)$.
I might be totally wrong on my work, if so, can someone help me find the function?
• We have $(r_1r_2)a=\phi(r_1r_2)a=(\phi(r_1)\phi(r_2))a=\phi(r_1)(\phi(r_2)a)=r_1(r_2a)$. – Aweygan Aug 31 '17 at 12:57
• You're making it too complicated. First think about the case where $R$ is a subring of $S$ (so $\phi$ is the inclusion map). Then obviously every $S$-module is an $R$-module. All you do is restrict the scalar multiplication to $R$. The general case is straightforward from there. – D_S Aug 31 '17 at 13:58
It may be simpler. Given $r \in R$ and $a \in A$, define your scaling as $$r \cdot a= \phi(r) \cdot a.$$ Now check that everything is kosher because $\phi$ is a 1-preserving ring map.
My point being you don't need to necessarily write the structure as a composition (which in your case of $f \circ \phi$ isn't defined).
Giving an $S$-module action on a abelian group $A$ is equivalent to giving a ring homomorphism $\alpha: S \to \mbox{End}(A)$.
Therefore, an $R$-module structure on $A$ is given by the composite ring homomorphism $\alpha \circ \phi : R \to S \to \mbox{End}(A)$. | 2019-10-14 06:24:49 | {"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": 0.9800615310668945, "perplexity": 164.6636216034676}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "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/1570986649232.14/warc/CC-MAIN-20191014052140-20191014075140-00391.warc.gz"} |
http://www.co-pylit.org/courses/cosc2325/attiny85sim/02-test-driven-development.html | # Testing, Testing, 123¶
Read time: 27 minutes (6982 words)
There is an old saying in software development that goes like this:
If you did not test it, it does not work!
We are about to set off on an adventure to build a simulator for a complete computer system. Our model is the attiny85 chip, but we will only build a part of this machine.
We will be building a bunch of parts, each one representing a digital component inside a real machine. We will wire those components up in the same way real digital components are interconnected in a chip.
So, how will we know if anything in this setup works, without waiting for the complete application to come together?
The answer is simple: we will test each part in isolation!
Huh?
Most of your work as a developer will involve only part of some bigger project. The entire project may be building a huge application and the team may involve hundreds of developers, eah building only one part of the overall system.
Since you will not have access to the entire code base, you need a way to test just your part, so you can gain confidence that your part is “production ready”.
But you have bit been trained to do testing. You are still learning how to write simple programs. We need a simple test scheme.
## Testing a Function¶
Suppose you need to write a magical square root routine that is much faster than the canned on available on every system. You need to use the same interface as the “real” square root routine, so you build a program that looks like this as a start:
main.cpp
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 #include #include float sqrt2(float x) { return sqrt(x); } int main(void) { float x = 2.0; std::cout << "Testing sqrt2" << std::endl; std::cout << "\tthe square root of " << x << " is " << sqrt2(x) << std::endl; }
Note
Believe it or not, I have given an assignment in a beginning programming class to calculate the square root using an iteration scheme, and had students submit exactly this code. It ain’t about the answer, it is about the method!
$g++ -o test main.cpp $ ./test
Testing sqrt2
the square root of 2 is 1.41421
Does this constitute a good test? Not really! All it proves is that the code runs, and produces some answer. You still need to decide if this answer if correct, and that is left to you to figure out.
Even worse, this is just one test, and that is definitely not enough to prove that the code is working.
Let’s see is we can do better.
## Breaking out a Unit of code¶
We are going to test a chunk of this code, not the complete application. The only part we are really concerned with is the sqrt function. Let’s start off by breaking that part out into its own file:
main.cpp
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 #include float sqrt2(float x); int main(void) { float x = 2.0; std::cout << "Testing sqrt2" << std::endl; std::cout << "\tthe square root of " << x << " is " << sqrt2(x) << std::endl; }
sqrt2.cpp
1 2 3 4 5 #include float sqrt2(float x) { return sqrt(x); }
We can still run this by doing this:
$g++ -o test main.cpp sqrt2.cpp $ ./test
Testing sqrt2
the square root of 2 is 1.41421
Now the program is configured for testing. We can compile the function alone and build a test application that exercises it. We will do that using the catch.hpp framework.
## Catch Testing¶
Basically, all we need to do to test this function separately from any application that will use it (like our main.cpp here) is to set up a separate directory where all test files will live. I usually set up a tests directory in the project directory for this purpose.
All we need to do is download the simple header file that makes up the catch.hpp test framework, and either land it in the project include directory, or place it in the tests folder directly.
Next, we need to build a single file in the tests directory named test-main.cpp. This file will cause catch.hpp to build a main program that will be used in the test application we will build.
This file is very simple:
tests/test_main.cpp
1 2 3 #define CATCH_CONFIG_MAIN #include "catch.hpp"
Not much there. The single define in this file is what triggers the C++ preprocessor to add code to this file that is your main function. We do not need to worry about the magic happening here.
To get started with this testing framework, let;s make sure the test system works. Here is a sanity check that proves things are working properly:
tests/test_sanity.cpp
1 2 3 4 5 #include "catch.hpp" TEST_CASE( "sanity test", "sanity" ){ REQUIRE(true); }
Look at the code here. The include brings in the catch.hpp magic. The TEST_CASE definition is in the catch.hpp header, and it creates a function that the main test logic will call. The two parameters in this function give a name to the test, and a group name you can use to only run tests associated with the named “group” We will ignore that for now.
Inside this magic test function, you see a single line that constitutes the “test we want to run. In this example, all we want to do is test that the test passes. The stuff inside the parentheses is a logical expression (in this case, just the value true). If that expression evaluates to true, catch.hpp will assume that the test passed. If it produced false, the test failed.
To make testing easy, we need to modify our Makefile so it will build our test application, and run the tests. Since the test code is just another C++ program, all we need to do it teach make where our test files live, and then set things up to build the test application.
Here is a basic Makefile that does this work. We will add the appropriate lines to our project Makefile later.
Makefile
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 include mk/util.mak .DEFAULT_GOAL := all TARGET := attiny85sim TSTTGT := testapp .PHONY: all all: $(TSTTGT) ## build simulator app .PHONY: test test:$(TSTTGT) ./$(TSTTGT) .PHONY: run run:$(TARGET) ./$(TARGET) machine.hdl # program build rules$(TARGET): $(OBJS)$(CXX) -o $@$^ $(shell ./scripts/inc_version.sh build)$(TSTTGT): $(TOBJS)$(LOBJS) $(CXX) -o$@ $^ I am betting this test will pass. Let’s see: $ make clean
rm -f testapp tests/test_main.o tests/test_sanity.o
$make g++ -c -std=c++11 -Iinclude tests/test_main.cpp -o tests/test_main.o g++ -c -std=c++11 -Iinclude tests/test_sanity.cpp -o tests/test_sanity.o g++ -o testapp tests/test_main.o tests/test_sanity.o $ make test
./testapp
===============================================================================
All tests passed (1 assertion in 1 test case)
That says we are running properly
Now that we know that the testing system works, we are ready to really test our code.
I leave the sanity test in place, as it does not hurt anything. Create a new test file that looks like this:
tests/test_sqrt2.cpp
1 2 3 4 5 6 7 8 9 10 #include "catch.hpp" #include "sqrt2.h" #include TEST_CASE( "basic sqrt test", "sqrt2" ){ float x = 2.0; float result = sqrt2(x); float checkval = sqrt(x); REQUIRE(result == checkval); }
This time, we set up the test code to access the function we want to test. I added a header file for this, since I should have set that up in splitting the code into two files. That header just has the prototype for the sqrt2 function.
include/sqrt2.h
1 2 float sqrt2(float x);
Now, we need to compile the new test and regenerate the testapp program.
$make clean rm -f testapp tests/test_main.o tests/test_sanity.o tests/test_sqrt2.o $ make
g++ -c -std=c++11 -Iinclude tests/test_main.cpp -o tests/test_main.o
g++ -c -std=c++11 -Iinclude tests/test_sanity.cpp -o tests/test_sanity.o
g++ -c -std=c++11 -Iinclude tests/test_sqrt2.cpp -o tests/test_sqrt2.o
g++ -o testapp tests/test_main.o tests/test_sanity.o tests/test_sqrt2.o lib/sqrt2.o
The project Makefile we are using in this class compiles any code in the lib directory. Since our test function is in that directory, this compiles sqrt2.cpp and creates the object file we need to link with our test code. Notice that we add this object file to the link step when we built the testapp program.
Run the new test:
$make test ./testapp =============================================================================== All tests passed (2 assertions in 2 test cases) It looks like this function is working. Well it should work, all we are doing it verifying that the real sqrt method returns the same value as our new sqtr2 function. Obviously, in a real development project, we would write our own code to calculate the square root, and then this test is going to tell you something useful! ## Improving the test¶ Just running one test is not enough to gain confidence in your code. You should run several tests to see how it performs. Here is a better version of the test code for our function: tests/test_sqrt2.cpp 1 2 3 4 5 6 7 8 9 10 11 12 13 14 #include "catch.hpp" #include "sqrt2.h" #include float test_set[] = {2.0,3.0,4.0,5.0,6.0}; TEST_CASE( "basic sqrt test", "sqrt2" ){ for(int i=0;i<5;i++) { float x = test_set[i]; float result = sqrt2(x); float checkval = sqrt(x); REQUIRE(result == checkval); } } Now we have $ make clean
rm -f testapp tests/test_main.o tests/test_sanity.o tests/test_sqrt2.o
$make g++ -c -std=c++11 -Iinclude tests/test_main.cpp -o tests/test_main.o g++ -c -std=c++11 -Iinclude tests/test_sanity.cpp -o tests/test_sanity.o g++ -c -std=c++11 -Iinclude tests/test_sqrt2.cpp -o tests/test_sqrt2.o g++ -o testapp tests/test_main.o tests/test_sanity.o tests/test_sqrt2.o lib/sqrt2.o $ make test
./testapp
===============================================================================
All tests passed (6 assertions in 2 test cases)
The number of passing tests is going up. In a real project, you may have hundreds of tests cases you run. We never throw away tests. As the project evolves, new code could break old code that used to work. Running all the tests over and over is called regression testing. (We want to make sure we are moving forward, mot backward in our work!)
This is still now a really good set of tests, but we are at least exercising the function more now. You should test “edge cases” and even test with bad data, and see what happens.
Your goal is to write a set of test cases that make you confident in this “unit” of code, and be willing to turn it loose on real users!
## Testing Classes¶
The strategy for testing a C++ class is identical to what we just did. You include the class header file in your test file, then create an object from the class. You should exercise that object, and use any accessor methods you have to make sure you are getting the results you expect.
Once you get the hang of this, adding testing to your development workflow is pretty easy to do, and then you will begin really testing your code. | 2019-01-20 19:55:18 | {"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": 0.3289285898208618, "perplexity": 1328.8875255965295}, "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-2019-04/segments/1547583730728.68/warc/CC-MAIN-20190120184253-20190120210253-00475.warc.gz"} |
http://server1.wikisky.org/starview?object_type=1&object_id=4889&object_name=HR+6341&locale=TR | WIKISKY.ORG
Home Baþlangýç Evrende yaþayabilmek için News@Sky Gökyüzü görüntüsü Koleksiyon Forum Blog New! SSS Basýn Giriþ
# HD 154228
Ýçindekiler
### Görüntüler
Resim Yükleyin
DSS Images Other Images
### Ýlgili Makaleler
Automated spectroscopic abundances of A and F-type stars using echelle spectrographs. II. Abundances of 140 A-F stars from ELODIEUsing the method presented in Erspamer & North (\cite{erspamer},hereafter Paper I), detailed abundances of 140 stars are presented. Theuncertainties characteristic of this method are presented and discussed.In particular, we show that for a S/N ratio higher than 200, the methodis applicable to stars with a rotational velocity as high as 200 kms-1. There is no correlation between abundances and Vsin i,except a spurious one for Sr, Sc and Na which we explain by the smallnumber of lines of these elements combined with a locally biasedcontinuum. Metallic giants (Hauck \cite{hauck}) show larger abundancesthan normal giants for at least 8 elements: Al, Ca, Ti, Cr, Mn, Fe, Niand Ba. The anticorrelation for Na, Mg, Si, Ca, Fe and Ni with Vsin isuggested by Varenne & Monier (\cite{varenne99}) is not confirmed.The predictions of the Montréal models (e.g. Richard et al.\cite{richard01}) are not fulfilled in general. However, a correlationbetween left [(Fe)/(H)right ] and log g is found for stars of 1.8 to 2.0M_sun. Various possible causes are discussed, but the physical realityof this correlation seems inescapable.Based on observations collected at the 1.93 m telescope at theObservatoire de Haute-Provence (St-Michel l'Observatoire, France) andCORALIE.Based on observations collected at the Swiss 1.2 m Leonard Eulertelescopes at the European Southern Observatory (La Silla, Chile).Tables 5 and 6 are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u.strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/398/1121 Rotational velocities of A-type stars in the northern hemisphere. II. Measurement of v sin iThis work is the second part of the set of measurements of v sin i forA-type stars, begun by Royer et al. (\cite{Ror_02a}). Spectra of 249 B8to F2-type stars brighter than V=7 have been collected at Observatoirede Haute-Provence (OHP). Fourier transforms of several line profiles inthe range 4200-4600 Å are used to derive v sin i from thefrequency of the first zero. Statistical analysis of the sampleindicates that measurement error mainly depends on v sin i and thisrelative error of the rotational velocity is found to be about 5% onaverage. The systematic shift with respect to standard values fromSlettebak et al. (\cite{Slk_75}), previously found in the first paper,is here confirmed. Comparisons with data from the literature agree withour findings: v sin i values from Slettebak et al. are underestimatedand the relation between both scales follows a linear law ensuremath vsin inew = 1.03 v sin iold+7.7. Finally, thesedata are combined with those from the previous paper (Royer et al.\cite{Ror_02a}), together with the catalogue of Abt & Morrell(\cite{AbtMol95}). The resulting sample includes some 2150 stars withhomogenized rotational velocities. Based on observations made atObservatoire de Haute Provence (CNRS), France. Tables \ref{results} and\ref{merging} are only available in electronic form at the CDS viaanonymous ftp to cdsarc.u-strasbg.fr (130.79.125.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcat?J/A+A/393/897 A spectroscopic survey for lambda Bootis stars. II. The observational datalambda Bootis stars comprise only a small number of all A-type stars andare characterized as nonmagnetic, Population i, late B to early F-typedwarfs which show significant underabundances of metals whereas thelight elements (C, N, O and S) are almost normal abundant compared tothe Sun. In the second paper on a spectroscopic survey for lambda Bootisstars, we present the spectral classifications of all program starsobserved. These stars were selected on the basis of their Strömgrenuvbybeta colors as lambda Bootis candidates. In total, 708 objects insix open clusters, the Orion OB1 association and the Galactic field wereclassified. In addition, 9 serendipity non-candidates in the vicinity ofour program stars as well as 15 Guide Star Catalogue stars were observedresulting in a total of 732 classified stars. The 15 objects from theGuide Star Catalogue are part of a program for the classification ofapparent variable stars from the Fine Guidance Sensors of the HubbleSpace Telescope. A grid of 105 MK standard as well as pathological''stars guarantees a precise classification. A comparison of our spectralclassification with the extensive work of Abt & Morrell(\cite{Abt95}) shows no significant differences. The derived types are0.23 +/- 0.09 (rms error per measurement) subclasses later and 0.30 +/-0.08 luminosity classes more luminous than those of Abt & Morrell(\cite{Abt95}) based on a sample of 160 objects in common. The estimatederrors of the means are +/- 0.1 subclasses. The characteristics of oursample are discussed in respect to the distribution on the sky, apparentvisual magnitudes and Strömgren uvbybeta colors. Based onobservations from the Observatoire de Haute-Provence, OsservatorioAstronomico di Padova-Asiago, Observatório do Pico dosDias-LNA/CNPq/MCT, Chews Ridge Observatory (MIRA) and University ofToronto Southern Observatory (Las Campanas). Catalogue of Apparent Diameters and Absolute Radii of Stars (CADARS) - Third edition - Comments and statisticsThe Catalogue, available at the Centre de Données Stellaires deStrasbourg, consists of 13 573 records concerning the results obtainedfrom different methods for 7778 stars, reported in the literature. Thefollowing data are listed for each star: identifications, apparentmagnitude, spectral type, apparent diameter in arcsec, absolute radiusin solar units, method of determination, reference, remarks. Commentsand statistics obtained from CADARS are given. The Catalogue isavailable in electronic form at the CDS via anonymous ftp tocdsarc.u-strasbg.fr (130.79.128.5) or viahttp://cdsweb.u-strasbg.fr/cgi-bin/qcar?J/A+A/367/521 The photoelectric astrolabe catalogue of Yunnan Observatory (YPAC).The positions of 53 FK5, 70 FK5 Extension and 486 GC stars are given forthe equator and equinox J2000.0 and for the mean observation epoch ofeach star. They are determined with the photoelectric astrolabe ofYunnan Observatory. The internal mean errors in right ascension anddeclination are +/- 0.046" and +/- 0.059", respectively. The meanobservation epoch is 1989.51. The Relation between Rotational Velocities and Spectral Peculiarities among A-Type StarsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1995ApJS...99..135A&db_key=AST Vitesses radiales. Catalogue WEB: Wilson Evans Batten. Subtittle: Radial velocities: The Wilson-Evans-Batten catalogue.We give a common version of the two catalogues of Mean Radial Velocitiesby Wilson (1963) and Evans (1978) to which we have added the catalogueof spectroscopic binary systems (Batten et al. 1989). For each star,when possible, we give: 1) an acronym to enter SIMBAD (Set ofIdentifications Measurements and Bibliography for Astronomical Data) ofthe CDS (Centre de Donnees Astronomiques de Strasbourg). 2) the numberHIC of the HIPPARCOS catalogue (Turon 1992). 3) the CCDM number(Catalogue des Composantes des etoiles Doubles et Multiples) byDommanget & Nys (1994). For the cluster stars, a precise study hasbeen done, on the identificator numbers. Numerous remarks point out theproblems we have had to deal with. 'Normal' main sequence AO stars of low rotational velocityA set of v sin i values has been derived in a homogeneous way for allthe northern unevolved AO stars which are listed in the Bright StarCatalogue and which are known to rotate more slowly than 40 km/s. Themethod used for the v sin i measurement employs the FFT algorithm and isdesigned to make the best use possible of line profiles obtained fromphotographic spectrograms. It is interesting to notice that, from afirst glance analysis of the spectra obtained, the suspicion arises thatAO single stars with atmospheric solar composition are very rare or evennonexistent among slow rotators (v sin i of less than 15 km/s). ICCD speckle observations of binary stars. I - A survey for duplicity among the bright starsA survey of a sample of 672 stars from the Yale Bright Star Catalog(Hoffleit, 1982) has been carried out using speckle interferometry onthe 3.6-cm Canada-France-Hawaii Telescope in order to establish thebinary star frequency within the sample. This effort was motivated bythe need for a more observationally determined basis for predicting thefrequency of failure of the Hubble Space Telescope (HST) fine-guidancesensors to achieve guide-star lock due to duplicity. This survey of 426dwarfs and 246 evolved stars yielded measurements of 52 newly discoveredbinaries and 60 previously known binary systems. It is shown that thefrequency of close visual binaries in the separation range 0.04-0.25arcsec is 11 percent, or nearly 3.5 times that previously known. Research Note - Absolute UBV Photometry at the Zacatecas ObservatoryAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1985RMxAA..11...55S&db_key=AST Photometric investigations of the Ap-star HD 153882File measurements of the Ap-star HD 153882 showed no short-time lightvariation, although subsequent photoelectric investigations using allavailable data from five comparison stars indicated a probablevariability. A variation exposure is found with a period of 6.01 d and arelatively clear double ripple in the wavelength range from 322 nm to765 nm. Significant differences in amplitude and form are discussed, andfluctuations up to 0.02 m are explained by a fade in and mixing of theindicator, and short-time variations which are determined byquasi-periodic variations of the time base. The existence of secondaryvariation is also discovered, and is found to originate from thecomparison star HD 154228. Photoelectric K-line indices for 165 B, A and F stars.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1978A&AS...34..441P&db_key=AST Spectral classification from the ultraviolet line features of S2/68 spectra. III - Early A-type starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1978A&AS...33...15C&db_key=AST Space velocities and ages of nearby early-type starsPhotometric distances and space velocities have been calculated for 458B0-A0 stars with apparent magnitudes not exceeding 6.5. UsingStromgren's ubvy-beta photometry the effective temperature and theposition in bolometric magnitude over the zero-age main sequence of thestars were derived. These quantities were used to obtain age and massfor 423 of the stars by interpolation in the models of stellar evolutionfor the chemical composition (X Z) = (0.7, 0.03). A relation forderiving interstellar reddening for normal stars in the intermediategroup is given. Photoelectric measures of variable stars observed at MT. Kobau (1970-73).Not Available Investigations of the light variation of twelve AP stars in ten spectral regionsTwelve Ap stars of known period and large amplitude were photometricallyinvestigated with the seven-color system based on glass filtersaccording to Straizys (1971) supplemented by three Schott filtercombinations in the red (total range 3400 to 7600 A). The light curvesfor each spectral range are presented, and some conclusions regardingthe importance of the data for possible model representations of theatmospheres of magnetic stars are set forth. Absolute luminosity calibration of Stroemgren's 'intermediate group'A relation defining the luminosity index for Stroemgren's (1966)intermediate group (A0 to A3 stars) in terms of absolute magnitude iscalibrated using a method based on the principle of maximum likelihood.This relation is also calibrated for the case when the 'a' index iscorrected for reddening. For both relations, calculations are made ofthe magnitude dispersion, the mean velocity components and correspondingdispersion, and the precision of each parameter. The results are shownto be in fairly good agreement with Stroemgren's (1966) values, and arelation incorporating the corrected 'a' index is proposed formain-sequence stars. The absolute magnitudes obtained with a relation ofthe present type are compared with those derived from trigonometricparallaxes and with those obtained by Eggen (1972). Rotational Velocities of a0 StarsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1974ApJS...28..101D&db_key=AST Luminosities and motions of AO to A2 starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1972PASP...84..757E&db_key=AST Four-color and Hβ photometry for the brighter AO type starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1972A&AS....5..109C&db_key=AST Rotation and chemical abundances in the peculiar A stars. II.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970MNRAS.147...75D&db_key=AST A catalogue of proper motions for 437 A starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970A&AS....1..189F&db_key=AST Photoelectric observations of early A starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1970A&AS....1..165J&db_key=AST Photometric behavior of magnetic starsAbstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1968ApJ...154..945S&db_key=AST Observations spectrographiques d'etoiles A a spectre particulier et a raies metalliques.Not Available Light Variations of Periodic Magnetic Variable StarsNot Available Catalogue et bibliographie des étoiles A à spectre particulier Premier supplémentNot Available Relation between Light-variation and Magnetic Variation in Magnetic Alpha VariablesNot Available Catalogue et bibliographie des étoiles A à spectre particulierNot Available Spectral Classification of 533 B8-A2 Stars and the Mean Absolute Magnitude of a0 V Stars.Abstract image available at:http://adsabs.harvard.edu/cgi-bin/nph-bib_query?1959ApJ...130..159O&db_key=AST
Yeni bir Makale Öner | 2013-05-26 02:55:06 | {"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": 0.6883813142776489, "perplexity": 9391.758407381729}, "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-2013-20/segments/1368706499548/warc/CC-MAIN-20130516121459-00000-ip-10-60-113-184.ec2.internal.warc.gz"} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.