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11800 | https://arxiv.org/html/2504.05496v1 | A Survey on Hypothesis Generation for Scientific Discovery in the Era of Large Language Models
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Table of Contents
Abstract
1 Introduction
2 Paper Collection Methodology
2.1 Search Strategy
2.2 Inclusion Criteria
2.3 Review Process
3 Methods for Scientific Hypothesis Generation
3.1 Human-Centric
3.2 Literature-based Discovery
Supervised Learning
3.3 LLM-Driven
Direct and Adversarial Prompting
Fine-tuning
Knowledge Integration
Multi-Agent System
4 Evaluation Methodologies
4.1 Human Expert Evaluation
Expert Assessment Protocols
Blind Review and Pairwise Comparison
Multi-Rater Reliability
4.2 Automated Evaluation
Text-based Relevance
Model-Based Metrics
Novelty Assessment
Domain-Specific Evaluation
5 Challenges and Future Research Directions
6 Conclusion
References
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License: CC BY 4.0
arXiv:2504.05496v1 [cs.CL] 07 Apr 2025
A Survey on Hypothesis Generation for Scientific Discovery in the Era of Large Language Models
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Atilla Kaan Alkan1,11, Shashwat Sourav2,11, Maja Jablonska3,11, Simone Astarita4,11, \AndRishabh Chakrabarty5,11, Nikhil Garuda6,11, Pranav Khetarpal7,11, Maciej Pióro8,11, \AndDimitrios Tanoglidis9,11, Kartheik G. Iyer10,11, Mugdha S. Polimera1,11, Michael J. Smith11, \AndTirthankar Ghosal12,11, Marc Huertas-Company13,11, Sandor Kruk14,11, \AndKevin Schawinski15,11 & Ioana Ciucă16,11
1 Center for Astrophysics, Harvard & Smithsonian, Cambridge, MA, USA
2 Washington University in St. Louis
3 Australian National University
4 European Commission, Joint Research Centre (JRC)
5 Intelligent Internet Inc.
6 University of Arizona
7 Indian Institute of Technology, Delhi
8 Institute of Fundamental Technological Research, Polish Academy of Sciences
9 Walgreens Boots Alliance AI Lab
10 Columbia University
11 UniverseTBD
12 Oak Ridge National Laboratory
13 Instituto de Astrofísica de Canarias
14 European Space Agency
15 Modulos AG
16 Stanford University
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Abstract
Report issue for preceding elementHypothesis generation is a fundamental step in scientific discovery, yet it is increasingly challenged by information overload and disciplinary fragmentation. Recent advances in Large Language Models (LLMs) have sparked growing interest in their potential to enhance and automate this process. This paper presents a comprehensive survey of hypothesis generation with LLMs by (i) reviewing existing methods, from simple prompting techniques to more complex frameworks, and proposing a taxonomy that categorizes these approaches; (ii) analyzing techniques for improving hypothesis quality, such as novelty boosting and structured reasoning; (iii) providing an overview of evaluation strategies; and (iv) discussing key challenges and future directions, including multimodal integration and human-AI collaboration. Our survey aims to serve as a reference for researchers exploring LLMs for hypothesis generation.
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1 Introduction
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Hypothesis generation is a fundamental component of scientific discovery, enabling researchers to formulate testable predictions and uncover new insights. This process has traditionally relied on human intuition, experience, and domain expertise. However, as the volume of scientific literature grows drastically, researchers face challenges in assimilating relevant knowledge across disciplines. This information saturation creates bottlenecks that hinder the discovery of new insights.
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From a philosophy of science perspective, a hypothesis can be defined as a tentative explanation or prediction about a phenomenon, formulated in a way that allows for empirical testing and potential falsification (Popper, 1959). Despite its crucial role in the scientific method, hypothesis generation remains constrained by disciplinary silos and cognitive overload. Traditional approaches struggle to integrate knowledge across fields, limiting researchers’ ability to identify interdisciplinary connections that may lead to groundbreaking discoveries.
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In this context, generative Large Language Models (LLMs) such as GPT (Radford & Narasimhan, 2018), PaLM (Chowdhery et al., 2022), LLaMA (Touvron et al., 2023), and Mistral (Jiang et al., 2023) have emerged as promising systems to overcome these barriers. By leveraging vast repositories of scientific texts, LLMs can process, synthesize, and generate novel hypotheses, assisting human expertise and facilitating interdisciplinary research. Since the introduction of LLMs, there has been a growing research interest in hypothesis generation using these models, as illustrated in. The number of research papers on this topic has significantly risen, highlighting the increasing recognition of LLMs’ potential in scientific exploration.
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Despite their promise, LLM-driven hypothesis generation presents several challenges. Evaluating generated hypotheses remains a complex issue, requiring novelty, relevance, feasibility, significance, and clarity assessment. A major concern is ensuring that LLMs generate innovative hypotheses rather than paraphrasing existing knowledge. Furthermore, the quality and diversity of training data play a crucial role in the effectiveness of these models. Biases present in the datasets can influence the generated hypotheses, potentially reinforcing existing perspectives while overlooking unconventional or groundbreaking ideas. Furthermore, integrating LLMs into the scientific process requires addressing issues related to interpretability, reliability, and validation of machine-generated hypotheses.
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In this paper, we aim to provide a comprehensive overview of the state of hypothesis generation using LLMs. We examine the current methodologies, categorize existing approaches into a structured taxonomy, and discuss the challenges and limitations inherent in this emerging field. By addressing these aspects, we seek to outline key research directions for the community on the generation of LLM-based hypotheses.
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2 Paper Collection Methodology
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We employed a systematic literature retrieval strategy combining keyword-based search and manual curation to construct a comprehensive and historically grounded survey of computational approaches to scientific hypothesis generation. The primary objective was to capture studies spanning both pre-LLM methods (e.g., literature-based discovery (LBD) and early NLP techniques) and more recent approaches involving LLMs.
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2.1 Search Strategy
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We queried the arXiv API to retrieve relevant publications for this survey paper. No explicit time range was imposed on the queries; the retrieved papers span a publication range from 2005 to 2025. To ensure domain relevance, we restricted the search to papers categorized under Computer Science, specifically within the cs.CL (Computation and Language) category. We used a curated list of search terms designed to reflect the thematic scope of our study, covering core concepts, traditional approaches, and recent methods involving natural language processing and large language models. The full list of search terms is presented in Table 1.
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| Core Concepts | Recent Techniques | Traditional Techniques |
---
| hypothesis generation | NLP for hypothesis generation | Swanson hypothesis discovery |
| scientific hypothesis generation | language models scientific discovery | open discovery system |
| scientific discovery | large language models hypothesis generation | ABC model literature discovery |
| automated scientific discovery | knowledge graph hypothesis generation | semantic predications scientific discovery |
| | question generation scientific research | semantic indexing hypothesis generation |
| | natural language processing scientific discovery | literature based discovery |
| | machine learning hypothesis generation | |
| | automated reasoning for discovery | |
| | discovery using LLMs | |
Table 1: Search terms used for systematic retrieval, grouped by theme.
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2.2 Inclusion Criteria
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Since our search was conducted exclusively via the arXiv API, the articles retrieved are primarily pre-prints. However, many of these works may have subsequently appeared in peer-reviewed journals, conference proceedings, or workshop venues. We included all papers that satisfied at least one of the following criteria, regardless of their publication status at the time of retrieval:
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•
The paper proposes or evaluates an automated or semi-automated method for scientific hypothesis generation;
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•
The work addresses scientific discovery through natural language processing, knowledge graph mining, or large language models;
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•
The paper contributes theoretical insights or historical perspectives on scientific discovery, particularly regarding the role of AI in hypothesis formulation.
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2.3 Review Process
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After initial filtering based on titles and abstracts, the remaining set of papers was manually screened for relevance and categorized according to methodological paradigm (e.g., LBD, NLP, LLMs, hybrid systems), scientific domain (e.g., biomedicine, astrophysics, chemistry), and hypothesis representation. This classification enabled us to trace the evolution of techniques and the shifts in hypothesis formalization over time.
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3 Methods for Scientific Hypothesis Generation
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In this section, we review the existing methods for generating scientific hypotheses. As illustrated in Figure 1, we present methods ranging from early approaches such as literature-based discovery (LBD), text mining, and statistical learning methods to more recent techniques, including graph-based models and LLM.
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Figure 1: Taxonomy of Methods for Scientific Hypothesis Generation (SHG).
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3.1 Human-Centric
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Human-centric methods are based on researchers’ expertise, intuition, and theoretical and practical knowledge. In this approach, individual insights gained through years of research and practical exposure play an important role in forming new hypotheses. Researchers engage in brainstorming sessions and discussions, where their knowledge and observational skills help identify new trends and anomalies (Swanson, 1986a; Nonaka, 2009). This method takes into account the ability of an individual to grasp context-specific nuances and reinterpret existing information. However, this method can have cognitive biases and the risk of overreliance on conventional paradigms, which may limit the exploration of less familiar or interdisciplinary ideas.
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3.2 Literature-based Discovery
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Literature-based discovery (LBD) leverages computational tools to mine the vast scientific literature for implicit or previously overlooked connections between concepts not directly linked in published research. The foundational idea of LBD introduced by Swanson (1986b), relies on the notion of ”undiscovered public knowledge”—information that exists in the literature but remains unconnected due to disciplinary silos or the overwhelming volume of publications. The author’s seminal study, which identified a potential link between fish oil and Raynaud’s syndrome by connecting disparate literature, remains a landmark example of this approach. Over time, LBD has evolved to incorporate advances in text mining, natural language processing (NLP), and semantic analysis, enabling the automated discovery of hidden relationships at scale (Smalheiser, 2017). Several systems have been developed to operationalize LBD. One of the earliest tools, ARROWSMITH (Smalheiser & Swanson, 1998), identifies intermediate terms (B) that link two disjoint sets of articles (A and C), thereby suggesting novel hypotheses. More recent systems have expanded the methodological toolkit. MOLIERE (Sybrandt et al., 2017), for example, builds semantic networks from MEDLINE and other biomedical data using topic modelling techniques such as Latent Dirichlet Allocation (Blei et al., 2003) and phrase mining to uncover latent themes and suggest short conceptual paths between topics. KnIT (Spangler et al., 2014) is another system that extracts factual statements from the literature, represents them in a queryable network and applies information diffusion algorithms to generate hypotheses, such as discovering novel kinases that phosphorylate p53. Other approaches rely on structured vocabularies like the Medical Subject Headings (MeSH, (Lipscomb, 2000)) to build profile-based representations of concepts and identify indirect links between them. Systems like DiseaseConnect (Liu et al., 2014) and BrainSCANr (Voytek & Voytek, 2012) also apply text mining to biomedical abstracts to uncover latent semantic features and generate novel associations. LBD remains especially valuable in information-rich domains, where human researchers may miss non-obvious connections across disciplines. By systematically surfacing these links, LBD tools support cross-domain discovery and can help accelerate hypothesis formulation in complex scientific landscapes.
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Supervised Learning
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Statistical computing methods are robust and data-driven for the hypothesis generation task. This set of methods uses statistical tools such as regression, clustering and Bayesian inference. On the other hand, computational approaches, including machine learning algorithms and network analysis, extend these tools by handling higher dimensional data and investigating complex, multivariate relations that may exist in the data. Some of the recent works (Breiman, 2003).
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3.3 LLM-Driven
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Large Language Models (LLMs) have recently emerged as powerful tools for scientific hypothesis generation. Figure 2 summarizes the pipeline and the common techniques used to generate hypothesis Their capacity to process vast corpora of scientific texts and synthesize information makes them particularly well-suited to this task. The literature reveals many methods that leverage LLMs to assist in or fully automate the generation of scientific hypotheses. These methods can be categorized as follows:
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Direct and Adversarial Prompting
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Direct prompting involves formulating clear and concise instructions to perform hypothesis generation directly using an LLM. In this method, users design a prompt explicitly asking the model to propose potential explanations or predictions based on a given context. One can ask the LLM a question or give it direct instruction about a topic, and it responds with possible ideas or explanations. This approach benefits from its simplicity and ease of implementation, allowing researchers to quickly gauge the model’s capacity for innovative reasoning (Radford & Narasimhan, 2018). However, the output quality highly depends on the prompt’s clarity and the model’s inherent understanding of the subject matter. The adversarial prompting approach is designed to make the LLM go beyond its standard response patterns by introducing counterfactual or challenging scenarios. By deliberately framing prompts in a way that exposes the model to unconventional perspectives, researchers can encourage the generation of hypotheses that diverge from common assumptions. This method can involve contrasting ideas or setting up dilemmas that force the model to explore unexplored paths of reasoning. Adversarial prompting not only tests the robustness of the LLM but also helps in detecting biases inherent in its training data, ultimately leading to more diversified and potentially groundbreaking insights (Chowdhery et al., 2023).
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Fine-tuning
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Rather than relying solely on prompting, this method involves fine-tuning an LLM on domain-specific datasets containing both foundational knowledge and corresponding hypotheses extracted from the literature. This tuning process enables the model to learn the patterns and context typically associated with hypothesis formulation. In one study, a temporally split biomedical dataset was used to test the model’s capacity to generate plausible hypotheses after fine-tuning.
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Knowledge Integration
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Some methods incorporate structured knowledge from scientific knowledge graphs to improve relevance and reduce hallucinations. These graphs encode entities and their relationships, which serve as grounding information for the LLM. A representative approach, KG-CoI (Knowledge-Grounded Chain of Ideas), uses graphs for context retrieval, chain-of-thought generation, and hallucination detection, improving the reliability of generated hypotheses (Xiong et al., 2024). Integrating knowledge in this form can help overcome some shortcomings of classical text-based RAG (Retrieval-Augumented Generation), such as the potential omission of rare but crucial information. Graph structures can also capture causal relationships between concepts and prove useful when generating hypotheses bridging selected concepts, as demonstrated by Tong et al. (2024).
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Multi-Agent System
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This approach introduces multiple LLM agents with different roles -such as analyst, scientist, or critic - that interact to collaboratively generate and evaluate hypotheses. Through dialogue and feedback between agents, this framework aims to produce more innovative and better-grounded scientific ideas.
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Figure 2: Pipeline of LLM-Driven Hypothesis Generation. The process begins with a research problem, which is processed by the LLM core. Various methodological branches (e.g., Direct & Adversarial Prompting, Fine-Tuning, and Knowledge Integration) contribute to a multi-agent framework that converges to generate hypotheses.
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4 Evaluation Methodologies
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Evaluating systems for scientific hypothesis generation is a complex task. Unlike traditional NLP evaluation, hypothesis generation aims to produce novel, plausible, and testable scientific ideas—often in domains where ground truth is incomplete or non-existent. This open-endedness renders standard evaluation metrics insufficient and necessitates a multi-faceted approach combining human expertise, automated metrics, multi-modal integration, and domain-specific validation. In this section, we first review established methodologies before outlining promising directions for future research.
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4.1 Human Expert Evaluation
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Expert Assessment Protocols
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Evaluations conducted by domain experts remain the most reliable method for assessing the relevance, originality, and scientific merit of machine-generated hypotheses. Over time, these assessments have become more structured and methodologically rigorous. Recent protocols have involved large panels of experts from diverse academic backgrounds to evaluate hypotheses along dimensions such as clarity, innovation potential, and expected impact. Comparative studies have shown that, when supported by LLMs, researchers can generate more compelling and diverse ideas than with traditional search-based workflows. Such findings suggest that expert-in-the-loop systems not only support hypothesis refinement but can also enhance ideation itself.
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In highly specialized fields such as biomedicine, structured evaluations have been designed to focus on clinical relevance and biological plausibility. Frameworks developed for this purpose often involve expert reviews centred on real-world applicability and potential translational impact. Some benchmark efforts have incorporated expert assessments across multiple research tasks, offering a broader view of how LLMs contribute to domain-specific scientific workflows.
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Blind Review and Pairwise Comparison
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To reduce bias and ensure fair evaluation, blind review protocols are increasingly employed. In these settings, experts are unaware whether a human or an AI system has generated a hypothesis. This approach has revealed that, in many cases, AI-generated hypotheses can be as highly rated—or even surpass—those written by human researchers regarding novelty and scientific interest. Building on this principle, some recent evaluation strategies employ direct pairwise comparisons in tournament-style formats, where hypotheses compete against each other and are ranked based on expert preference. These structured comparison schemes offer a scalable and interpretable method for evaluating generative systems.
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Multi-Rater Reliability
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One of the persistent challenges in expert-based evaluation is achieving consistency across annotators. Scientific hypothesis assessment often involves subjective judgment, leading to variability in ratings. Earlier studies have highlighted relatively low agreement levels among reviewers, emphasizing the complexity of the task. However, newer frameworks are addressing this by introducing more formalized scoring rubrics, multiple rounds of review, and collaborative assessment protocols. These improvements have contributed to more stable and reproducible evaluation outcomes, reflecting a growing understanding of effectively integrating human judgment into validating AI-generated scientific content.
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4.2 Automated Evaluation
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Text-based Relevance
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Initial efforts to evaluate LLM outputs relied heavily on surface-level metrics such as BLEU and ROUGE, which measure word overlap between generated and reference hypotheses. However, such metrics often fall short of capturing the semantic depth and scientific value of an idea. As a result, more sophisticated evaluation tools have been developed that incorporate semantic precision and recall, as well as hybrid scores that combine symbolic and neural representations. These allow for a more meaningful assessment of whether a hypothesis is contextually appropriate and scientifically relevant. Additionally, some benchmarks now include domain-specific metrics tailored to the complexity and requirements of particular research tasks, such as code execution or model reproducibility.
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Model-Based Metrics
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Recent evaluation frameworks have increasingly turned to large language models as evaluators of generated hypotheses. When fine-tuned or provided with structured prompts, these models can approximate human-level assessments across dimensions such as plausibility, novelty, and relevance. Some systems now rely on LLMs to score hypotheses using composite metrics that account not only for internal coherence but also for broader scientific context. For instance, measures have been developed to quantify how dissimilar a proposed idea is from past knowledge and how closely it aligns with emerging literature trends, thus reflecting historical uniqueness and prospective impact.
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Novelty Assessment
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Measuring novelty remains one of the central goals in hypothesis evaluation. Automated approaches have evolved to estimate the originality of ideas by analyzing their semantic distance from existing publications. This often involves embedding-based comparisons using pre-trained scientific language models combined with ranking strategies that assess the rarity or innovation of proposed connections. Some systems build structured citation graphs or ideation chains to contextualize a hypothesis within a broader intellectual lineage, enabling more informed judgments about its uniqueness.
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Domain-Specific Evaluation
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Evaluation strategies tailored to specific scientific fields are increasingly recognized as essential due to the varied standards of evidence, feasibility, and validation across disciplines. In biomedical research, hypothesis evaluation often relies on alignment with curated clinical databases or known gene-disease associations, enabling automated cross-referencing against structured biomedical knowledge. In the chemical sciences, evaluation protocols typically focus on structural validity and chemical plausibility, incorporating techniques such as molecular simulation or synthesis pathway prediction. Astronomy and astrophysics present unique challenges, where hypothesis evaluation may involve the integration of large-scale observational datasets or comparing generated hypotheses with complex knowledge graphs. Social science domains, on the other hand, prioritize theoretical grounding and temporal context, often requiring evaluation of whether a hypothesis is consistent with existing paradigms or predictive of future trends. These domain-specific practices underscore the importance of aligning evaluation methodologies with disciplinary norms, highlighting the need for adaptable frameworks that can accommodate the epistemological diversity of modern science.
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5 Challenges and Future Research Directions
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Despite substantial progress in developing LLM-based systems for scientific hypothesis generation, several critical challenges remain unresolved. One of the most pressing concerns is the issue of factual accuracy. LLMs are known to produce outputs that, while syntactically coherent and contextually plausible, can include erroneous or fabricated claims. This phenomenon, often referred to as hallucination, poses significant risks in scientific settings. Closely related is the challenge of interpretability. Most LLMs function as black-box systems, making it difficult to understand or trace the rationale behind specific hypotheses. This lack of transparency undermines trust and complicates the validation process, especially when hypotheses are intended to serve as the foundation for empirical research.
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Bias is another persistent issue. Given that LLMs are trained on large, heterogeneous corpora, they tend to reproduce—and occasionally amplify—preexisting societal biases. These biases can influence the direction and framing of generated hypotheses, potentially skewing research priorities and excluding underrepresented perspectives. At the same time, the computational cost of training and deploying these models remains prohibitive for many institutions. The high energy and hardware requirements not only limit accessibility but also raise concerns about environmental sustainability. Domain adaptation poses additional hurdles. While fine-tuning on specialized datasets can enhance performance in specific fields, it often introduces the risk of overfitting, compromising the model’s ability to generalize across topics. Furthermore, the ethical implications of AI-generated hypotheses—from questions of authorship and accountability to the potential misuse of misleading hypotheses—remain largely unaddressed, necessitating the development of robust governance mechanisms.
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To overcome these limitations, new methodological directions are emerging. Retrieval-augmented generation, which integrates LLMs with external scientific databases, offers a promising approach to grounding outputs in verifiable knowledge and reducing hallucinations. Another direction involves incorporating chain-of-thought reasoning or rationale tracing mechanisms, enabling models to generate not only hypotheses but also the reasoning pathways that led to their formulation. This increased transparency can help researchers evaluate the internal coherence and plausibility of generated ideas. Multi-agent collaborative frameworks are also gaining traction. Inspired by the collaborative nature of scientific inquiry, these systems simulate peer review or debate among virtual agents to refine and evaluate hypotheses dynamically. In the realm of fine-tuning, meta-learning and cross-domain transfer techniques are being explored to better balance specialization and generalization, allowing models to adapt flexibly to a variety of scientific domains without sacrificing rigour.
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From a computational perspective, advances in model compression and energy-efficient architectures are expected to democratize access to LLM-based tools, making them more practical for research institutions with limited resources. At the same time, methodological co-design with ethicists, legal scholars, and domain experts is increasingly recognized as essential to developing socially responsible AI tools. Future systems should embed ethical safeguards, including bias detection, provenance tracking, and clear attribution protocols, directly into their design.
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Complementing these methodological innovations, the field must adopt more sophisticated evaluation frameworks. Traditional metrics such as BLEU and ROUGE fall short of capturing hypotheses’ semantic depth and scientific merit. In response, several novel evaluation paradigms are being developed. Scientific verifiability benchmarks, for instance, aim to assess whether generated hypotheses can be empirically tested or verified in real-world research. Temporal evaluation methods propose tracking the evolution of ideas over time—through citations, modifications, or integrations into published work—to assess long-term scientific impact. Evaluating the diversity and redundancy of generated hypotheses has also become a key area of interest, as the capacity to propose a broad range of novel ideas is a fundamental indicator of exploratory potential.
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Another promising direction involves multi-modal evaluation, where hypotheses are assessed not just through text-based metrics but also through visualizations, structured knowledge graphs, or experimental data. Human-in-the-loop evaluation protocols are likewise gaining prominence. These frameworks involve iterative collaboration between researchers and models, enabling dynamic refinement and contextual validation of hypotheses. Finally, developing interdisciplinary evaluation standards is increasingly necessary as hypothesis-generation systems are deployed across various scientific domains. These standards must be flexible enough to accommodate domain-specific norms while preserving core principles such as novelty, relevance, verifiability, and scientific integrity.
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In summary, while LLM-based systems have demonstrated considerable potential in augmenting scientific discovery, their limitations call for caution and innovation. Addressing persistent challenges such as factual inaccuracy, opacity, and domain sensitivity will require a coordinated effort across AI, domain science, and ethics. At the same time, the emergence of new methodological and evaluation paradigms offers a promising path toward developing robust, transparent, and impactful hypothesis-generation tools that align with the evolving standards of scientific research.
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6 Conclusion
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While Large Language Models have revolutionized the domain of automated text generation, their application in scientific hypothesis generation is still in its nascent stages and filled with challenges. Issues such as factual inaccuracies, lack of interpretability, inherent biases, and high computational demands underscore the need for continued research and innovation. This paper has reviewed the state-of-the-art methods for LLM-driven hypothesis generation and critically examined the accompanying limitations. Future research must prioritize the development of more transparent, efficient, and ethically sound models that can reliably support scientific inquiry. By addressing these challenges through interdisciplinary collaboration and methodological advances, the scientific community can unlock the full potential of LLMs, ultimately paving the way for transformative breakthroughs in knowledge discovery.
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Voytek & Voytek (2012)↑ Jessica B. Voytek and Bradley Voytek. Automated cognome construction and semi-automated hypothesis generation. Journal of Neuroscience Methods, 208(1):92–100, 2012. ISSN 0165-0270. doi: URL
Xiong et al. (2024)↑ Guangzhi Xiong, Eric Xie, Amir Hassan Shariatmadari, Sikun Guo, Stefan Bekiranov, and Aidong Zhang. Improving scientific hypothesis generation with knowledge grounded large language models, 2024. URL
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Last updated: June 19, 2024
Summary
Embryogenesis is the process of embryonic development occurring in the first eight weeks after fertilization. After implantation of the blastocyst in the endometrium, the embryo consists of the embryoblast and the trophoblast. While the embryoblast further develops into different structures of the body, the trophoblast is mainly involved in the development of the placenta. The amniotic cavity, yolk sac, extraembryonic mesoderm, and the chorionic cavity develop during the second week. In weeks 3 and 4, the bilaminar disc differentiates into a trilaminar embryonic disc through the process of gastrulation. A number of structures develop from the three germ layers. The nervous system also develops during weeks 3 and 4 through the process of neurulation. Weeks 5–8 are mainly characterized by organogenesis and continued differentiation of embryonic tissue.
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Overview
Embryonic development
This table is based on time since fertilization (i.e., developmental age), instead of time since first day of last menstrual period (i.e., gestational age).
| Overview |
| Day/week | Characteristics | Possible disorders |
| Day 0 | Capacitation: maturation of the sperm in the female genital tract Fertilization (conception): usually occurs in the fallopian tubes (most commonly in the ampulla) within 1 day of ovulation + Consists of 2 phases: the acrosome reaction (dissolution of the spermatic cell membrane and the zona pellucida of the ovum) and impregnation (penetration of the sperm into the ovum) Conjugation (conception): fusion of the sperm and ovum to form the zygote (single cell) | Genopathies: injury to the gene locus Chromosomal aberrations (e.g., trisomy) Both occur before conception |
| Day 1–5 | The zygote travels down the fallopian tube towards the uterus Division of the zygote into the morula (16-cell mass) and then the blastocyst The blastocyst consists of 2 layers of cells: the outer trophoblasts and the inner embryoblasts. | Blastopathies + Congenital anomalies: complex congenital anomalies with double malformations or missing body parts + Occurrence: up to 14 days after conception + High risk of spontaneous miscarriage + Multiple pregnancies: conjoined twins especially |
| Day 6 | Implantation (embryology) of the blastocyst (most commonly into the anterior or posterior wall of the uterus) The trophoblast penetrates the endometrium. This may result in brief implantation bleeding: bloody vaginal discharge occurring during implantation (may be mistaken for menstruation) |
| Days 6–14 | Formation of the fetoplacental unit Trophoblast further divides into the syncytiotrophoblast, which is involved in the development of the placenta, and the cytotrophoblast, which forms the chorionic cavity, that surrounds the embryoblast. Syncytiotrophoblast begins secreting hCG (detectable in blood and urine one week and two weeks after conception, respectively) The maternal circulation becomes connected with the placental circulation The bilaminar embryonic disc develops (2 layers): epiblast and hypoblast + Epiblast → amniotic cavity + Hypoblast → yolk sac, lining of chorionic cavity |
| Weeks 3–4 | Gastrulation: Epiblast cells → primitive streak → trilaminar disc (endoderm, mesoderm, ectoderm) Notogenesis: axial mesoderm → notochord → induces development of neural plate from ectoderm Neurulation: ectoderm → neural plate → neural tube with neural pores, neural crest (closes by the end of week 4) Branchial apparatus → pharyngeal arches, pharyngeal pouches, pharyngeal grooves Aortic arches Development of upper and lower limb buds First heartbeat at week 4 of embryonic development (visible on transvaginal ultrasound) Primitive circulation with a tubular heart and initial intraembryonic blood vessels Morphogenesis Differentiation of the embryonic disc | Embryopathies: complex anomalies of individual organs during a time when the developing embryo is particularly susceptible to injury Etiology + Infections (e.g., rubella, CMV) + Drug toxicity (e.g., warfarin) |
| Week 5 | Weeks 5–8 of embryogenesis mainly involves organogenesis. + Rapid head growth through development of the brain and facial structures + The mesonephros, which is formed between weeks 3 and 5, bulges in the urogenital ridge. + Development and differentiation of additional pharyngeal arches |
| Week 6 | Digital ray development Auricular hillock development, which later becomes the auricle Eye is clearly recognizable through retinal pigment development Back starts to straighten Formation of a physiological umbilical hernia |
| Week 7 | The proximal bones of the upper limbs start to ossify. |
| Week 8 | Organogenesis is complete by the end of week 8. Recognizable human form Fingers are initially separated from one another by skin flaps, followed by complete separation. The proximal bones of the lower limbs start to ossify. Fetal movements begin. |
| Week 9–12 | Fetogenesis: the fetus matures and grows Genitals have sex-specific characteristics; sex can be determined on ultrasound at week 12. | Fetopathies: structural abnormalities that occur during the maturation and growth of the fetus once organ formation is complete Etiology + Infections (e.g., toxoplasmosis, syphilis) + Antibodies (e.g., Rh incompatibility) + Metabolic (e.g., fetal macrosomia) |
The embryo is extremely susceptible to teratogens from week 3 to week 8, when the process of organogenesis occurs.
The developmental events from week 2 to week 4 can be remembered as follows: At 2 weeks there are 2 layers (bilaminar disc), 3 weeks there are 3 layers (trilaminar disc) and 4 weeks there are 4 limb buds and 4 heart chambers present.
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Embryoblast and trophoblast development
Embryoblast
In the 2nd week of embryonic development (days 8–14), the embryoblast differentiates into two layers (epiblast and hypoblast), termed the bilaminar disc. After formation of the amniotic cavity and yolk sac, the bilaminar disc is sandwiched between them.
Embryoblast → bilaminar disc
Epiblast: columnar cells adjacent to amnioblasts → form the amniotic cavity → forms the embryo (begins the 3rd week of embryonic development)
Hypoblast
Cuboidal cells adjacent to the blastocyst cavity (blastocoel)
Forms yolk sac, lining of chorionic cavity
Fusion of epiblast and hypoblast → prechordal plate → mouth
All three germ layers (ectoderm, mesoderm, and endoderm), as well as the amniotic cavity and therefore the entire embryonic tissue, arise from the epiblast. The extraembryonic mesoderm and the yolk sac are derived from the hypoblast.
The bilaminar disc forms the dividing layer between the yolk sac and amniotic cavity.
Trophoblast
The trophoblast is the layer of cells that surrounds the blastocyst.
During week 2, the trophoblast divides into two layers, the cytotrophoblast and the syncytiotrophoblast.
They form the embryonic component of the placenta.
Cytotrophoblast
The inner layer of the chorionic villi; mitotically active
Function: cells migrate outward and fuse to become the syncytiotrophoblast
Cytotrophobasts make “cytos” (cells); Syncytiotrophoblasts synthesize hormones.
Syncytiotrophoblast
The outer layer of the chorionic villi; mitotically inactive
Function: invades the endometrium to create lacunae (spaces that fill with maternal blood)
Secretes beta-hCG to maintain the corpus luteum (which increases progesterone secretion by corpus luteum during the first trimester) and the decidua (starts when implantation of the blastocyst occurs)
Invades the endometrium to create lacunae (spaces that fill with maternal blood)
Proteolytic enzymes secreted by the syncytiotrophoblast break down the extracellular matrix
Formation of vacuoles within the syncytiotrophoblast, fusion of vacuoles to form lacunae (lacunar stage)
Invasion of syncytiotrophoblast into maternal capillaries (sinusoids)
Maternal sinusoids and lacunae fuse; maternal blood flows through the trophoblast → uteroplacental circulation
Does not express MHC class I on the surface of its cell and is, therefore, less susceptible to the mother's immune cells attack.
Formation of the chorionic cavity
Starting point: The amniotic cavity and primary yolk sac are surrounded by trophoblast cells.
Sequence of events
The trophoblast grows significantly faster than the surrounding structures → a gap forms between it and the amniotic cavity.
The gap is lined by the extraembryonic mesoderm.
Renewed gap formation in the extraembryonic mesoderm results in the formation of a new cavity
The extraembryonic mesoderm splits into two layers (somatopleuric and splanchnopleuric mesoderm) but remains attached at a connecting stalk.
Somatopleuric (parietal) mesoderm: lines the trophoblast
Splanchnopleuric (visceral) mesoderm: surrounds the amniotic cavity and yolk sac with an intermediate embryonic disc
Extraembryonic coelom (chorionic cavity)
Cavity that is formed between the somatopleuric and splanchnopleuric layers → precursor of the gestational sac
Lined by the chorion (outermost cell layer surrounding the embryo)
Connecting stalk: A derivative of extraembryonic mesoderm that connects the amniotic cavity and primitive yolk sac to the chorion; precursor of the umbilical cord
The extraembryonic coelom is also called the chorionic cavity, which is lined by the chorion.
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Gastrulation
Gastrulation is the formation of the trilaminar embryonic disc or gastrula through the migration of epiblast cells. Epiblast cells migrate through the primitive streak between the epiblast and hypoblast layers and form an intermediate cell layer called the intraembryonic mesoderm. The hypoblast is replaced by epiblast cells, from which the endoderm arises. The original epiblast becomes the ectoderm.
Origin: primitive streak, an elongated accumulation of epiblasts
Result: formation of the trilaminar embryonic disc (mesoderm, endoderm, and ectoderm)
Process: occurs in weeks 3–4
Mesoderm development (intraembryonic mesoderm)
Epiblast cells detach from the cellular complex below the primitive streak → formation of the primitive pit
Mesoderm cells migrate through primitive pit
Proliferation of cells on both sides of the primitive pit and formation of the mesoderm, a new layer between the epiblast and the hypoblast
Endoderm development
Epiblast cells migrate from the primitive streak and the primitive node to the hypoblast layer.
Hypoblast cells are invaded by epiblast cells and form an additional germ layer called the endoderm.
Ectoderm development: differentiation of remaining epiblast cells into ectoderm after migration of cells for endoderm, mesoderm, and notochord
All embryonic tissue originates from the epiblast.
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Notogenesis
Definition: development of the notochord, a rodlike structure between the ectoderm and endoderm that is essential for the development of the nervous system and primitive skeletal structures
Location: the notochord migrates along the primitive streak (the future craniocaudal axis), and ends at the prechordal plate. It is part of the axial mesoderm
Process: occurs weeks 3–4 (at the same time as development of the trilaminar embryonic disc)
Epiblast cells migrate from the primitive nodes and extend cranially to form the notochordal process.
Notochordal process margins converge until they merge into the rod-shaped notochord at the end of week 4
Derivatives
The notochord degenerates during the course of embryogenesis.
The nuclei pulposi of the intervertebral discs are remnants of the notochord.
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Neurulation
Neurulation is the formation of the neural tube and neural crests, which are the precursors to the central and peripheral nervous systems. During this process, the surface ectoderm is also formed, which gives rise to the epidermis.
Definition: formation of the neural tube and neural crests via neural plate folding
Process
The notochord secretes signaling molecules that trigger differentiation of the medial ectoderm (located directly above the notochord) into neural cells.
This region is termed the neural plate.
Sequence
The central area of the neural plate decreases (invagination) and becomes the neural groove.
The neural folds are found on both longitudinal sides of the neural groove.
The neural folds converge (week 3) and convert the neural groove into the neural tube.
The dorsal-most part of the neural tube forms the specialized neural crest.
The neural tube communicates with the amniotic cavity through cranial and caudal openings (neuropore).
Anterior neuropore: closes on ∼ days 24–25
Posterior neuropore: closes on ∼ days 26–27
Fusion of the opposite sides of the neural folds results in the formation of the surface ectoderm.
Neural fold fusion leads to several cells dissociating from the epithelial cluster, which deposit between the neural tube and surface ectoderm as the neural crest cells.
Derivatives
Neural tube derivatives
Central nervous system (brain and spinal cord), including neurons and glial cells (except microglia)
Retina
Neural crest derivatives
Peripheral nervous system neurons (in cranial, dorsal root, and autonomic ganglia) and glia (satellite cells and Schwann cells)
Melanocytes
Adrenal medulla
Leptomeninges (pia and arachnoid)
Odontoblasts
Enterochromaffin cells
Aorticopulmonary septum (spiral membrane)
Endocardial cushions (partially derived also from mesoderm)
Cranial bones
Surface ectoderm: forms the epidermis, adenohypophysis, olfactory epithelial cells and sensory organs of the ear
Notochord: differentiates into nucleus pulposus of the intervertebral disk
The entire nervous system develops from the ectoderm.
Neural tube defects are one of the most common CNS malformations and develop as a result of incomplete closure of the neural tube (e.g., spina bifida, anencephaly).
Neural crest derivatives (Cranial bones, Adrenal medulla, Leptomeninges, Melanocytes, Enterochromaffin cells, Tracheal cartilage, PNS ganglia, Odontoblasts, Spiral membrane, and Endocardial cushions): Embryos have CALMEST POSE.
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Branchial apparatus
The branchial apparatus is an embryological structure with five paired arches composed of mesodermal and neural crest cells bound externally by an ectodermal cleft and internally by an endodermal pouch, which differentiate into various head and neck structures. The branchial apparatus is externally visible below the developing brain of a 4-week-old embryo. The fifth arch regresses in utero and does not contribute to the development of any head and neck structures.
Pharyngeal groove: ectodermal origin
Pharyngeal arch
Central mesenchymal core, which contains a nerve, an artery, and cartilage
Derived from the neural crest and mesoderm
Pharyngeal pouch: endodermal origin
Derivatives from the outer to the inner layer (Groove: ectoderm; Arch: mesoderm and neural crest; Pouch: endoderm): GAP
Pharyngeal arches
| Pharyngeal arch derivatives |
| Pharyngeal arches (the fifth pharyngeal arch only exists transiently in human embryos) | Nerves (innervate the structures derived from the arches; do not originate from them) | Arteries | Muscles | Skeletal structures | Related pathologies |
| First pharyngeal arch (mandibular arch) | CN V3 | Maxillary artery | Muscles of mastication (masseter, temporalis, medial and lateral pterygoid muscles) Mylohyoid muscle Digastric muscle, anterior belly Tensor tympani muscle Tensor veli palatini muscle Anterior two-thirds of the tongue | Mandibular prominence + Meckel cartilage: malleus, incus + Mandible Maxillary prominence + Maxilla + Zygomatic bone + Palatine bone + Squamous temporal bone + Vomer Malleus and incus Sphenomandibular ligament | Pierre Robin sequence Treacher Collins syndrome Auricular atresia Isolated micrognathia Cleft lip Cleft palate |
| Second pharyngeal arch (hyoid arch) | CN VII | Obliterates completely | Muscles of facial expression (see development of the face) Digastric muscle, posterior belly Stylohyoid muscle Stapedius muscle Platysma Posterior one-third of the tongue | All the osseous structures originating from the second pharyngeal arch are derived from Reichert cartilage Stapes Styloid process Stylohyoid ligament Hyoid bones: lesser horn and body |
| Third pharyngeal arch | CN IX | Common carotid artery Proximal internal carotid artery | Stylopharyngeus muscle Posterior one-third of the tongue | Hyoid bones: greater horn and body | N/A |
| Fourth pharyngeal arch | CN X Superior laryngeal nerve | Left: aortic arch Right: subclavian artery | Middle and inferior pharyngeal constrictor muscles Cricothyroid muscle Palatopharyngeus muscle Levator veli palatini Posterior one-third of the tongue | Upper part of thyroid cartilage | N/A |
| Sixth pharyngeal arch | CN X Recurrent laryngeal nerve | Left: ductus arteriosus Truncus pulmonalis, left pulmonary artery Right pulmonary artery | Internal larynx muscles (other than cricothyroid) | Lower part of thyroid cartilage Arytenoid cartilage Corniculate cartilage Cuneiform cartilage | N/A |
Sensory and motor nerves do not derive from pharyngeal arches, but from the neural crest and neuroectoderm respectively.
Pharyngeal pouch
Derivatives
First pharyngeal pouch
Eustachian tube
Tympanic cavity
Mastoid air cells (ear structures covered by endoderm)
Second pharyngeal pouch: epithelial crypts of palatine tonsil, supratonsillar fossa
Third pharyngeal pouch
Dorsal wings: inferior parathyroid glands
Ventral wings: thymus
Fourth pharyngeal pouch
Dorsal wings: superior parathyroid glands
Ventral wings: ultimobranchial body, parafollicular cells (C cells) of the thyroid
The inferior parathyroid glands arise cranially (3rd pouch) but migrate caudally (lower poles of the thyroid gland). The superior parathyroid glands arise caudally (4th pouch) but migrate cranially (superior poles of the thyroid gland).
Pharyngeal grooves
Derivatives
The first pharyngeal groove develops into the external auditory meatus, the auditory canal, and the external aspect of the tympanic membrane.
The second to fourth pharyngeal grooves are obliterated in utero by the rapid growth of the second pharyngeal arch, and they do not differentiate into any of the head and neck structures.
Cervical sinus
The second pharyngeal arch; develops faster than the others → overlap and merging with the third and fourth arches; → grooves lose their connection with the amnion → formation of cervical sinus.
The cervical sinus is obliterated during later development due to the proliferation of the 2nd arch mesenchyme
Failure of the cervical sinus to obliterate completely results in branchial cleft sinus, which can lead to lateral cervical fistula formation
Located anterior to the sternocleidomastoid muscles
In contrast to a thyroglossal duct cyst, the swelling caused by the branchial cleft sinus does not move with swallowing.
A lateral cervical fistula is prone to infection and is a clear indication for operative treatment.
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Aortic arches
The aortic arches are blood vessels that run in between the pharyngeal pouches and form the major head and neck arteries. The arches develop in craniocaudal order, with the first two arches obliterating early and the fifth either never developing or also obliterating without giving rise to a vessel.
| Overview of aortic arch derivatives |
| Aortic arches | Derivatives |
| First | Maxillary artery: remnant of obliterated first arch |
| Second | Most parts of the second aortic arch obliterate early + Hyoid artery: remnant of the largely obliterated second arch + Stapedial artery: transient branch of the hyoid artery (typically regresses around week 10 of development) |
| Third | Common carotid Proximal internal carotid |
| Fourth | Right: proximal right subclavian artery Left: aortic arch (from left common carotid to left subclavian) |
| Sixth | Right: right pulmonary artery Left: left pulmonary artery, ductus arteriosus |
1st is maximal (maxillary artery); Second for Stapedial; C is the 3rd letter of the alphabet (Common Carotid, proximal internal Carotid); 4th arch and 4 limbs (systemic arch).
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Morphogenesis
Morphogenesis is the process by which the shape of an organism is generated. The embryo undergoes folding, resulting in transformation of the flat, embryonic disc into an embryo that approaches the human form during the course of the pregnancy. During the folding processes, the abdominal cavity, the abdominal wall, and the gut tube are formed. At the cranial and caudal embryonic poles, there is an area devoid of the mesoderm where the endoderm and ectoderm come into direct contact with one another, called the buccopharyngeal or cloacal membrane. The mouth and anus will later form in these areas.
Body axis determination
Dorsoventral body axis: formed as the bilaminar disc develops
Dorsal body axis: epiblast
Ventral body axis: hypoblast
Craniocaudal body axis
Formed by the extension of the primitive streak
Cranial pole: primitive nodes
Right-left differentiation: The positioning of unpaired chest and abdominal organs is determined by asymmetric expression of signaling molecules in the lateral plate mesoderm.
Situs inversus is a very rare congenital condition in which the chest and abdominal organs are reversed or mirrored. It is usually considered a benign condition, but can also present as part of a syndrome, e.g., Kartagener syndrome.
Craniocaudal folding
Process: The cranial and caudal embryonic poles curl, resulting in curving of the embryonic disc.
Result
The embryo develops a C form.
Caudal end: tail fold
Cranial end: head fold
Constriction of the yolk sac
Lateral folding
Process
Ventral outgrowth of the lateral margins of the embryonic disc
The ectoderm and the overlying lateral plate mesoderm on both sides converge.
Adhesion of both sides results in the formation of a lateral and anterior body wall as well as a cavity, the intraembryonic coelom
The endoderm and the overlying lateral plate mesoderm on both sides converge → gut tube formation
Result
Intraembryonic coelom
Pericardial cavity
Pleural cavity
Peritoneal cavity
Gut tube
Anterior end of the foregut
Foregut
Midgut
Hindgut
Further constriction of the yolk sac
The midgut stays connected to yolk sac remnants via the vitelline duct (omphalomesenteric duct). This duct is obliterated during the course of embryogenesis. Persistence of this duct most commonly results in Meckel diverticulum but could also cause retroumbilical cysts and fistulae.
Buccopharyngeal and cloacal membrane formation
Process
Buccopharyngeal membrane: region without mesoderm at the cranial embryonic pole
Stomodeum formation as a result of deep depression by the buccopharyngeal membrane due to expansion of the brain
The buccopharyngeal membrane ruptures shortly after folding → development of a link between the stomodeum (and therefore also the amniotic cavity) and the gut tube
Cloacal membrane: region without mesoderm at the caudal embryonic pole
Proctodeum formation as a result of inward folding of the ectoderm
The cloacal membrane tears significantly later than the buccopharyngeal membrane → a link develops between the amniotic cavity and urogenital tract/anal canal
Results
Formation of future body orifices
The embryo is connected to its external environment (e.g., amniotic fluid) via the body orifices.
Abnormalities of morphogenesis
| Overview |
| Process | Definition | Characteristics | Example |
| Agenesis | The organ is not present. | Primordial tissue from which the organ is derived is absent. | Agenesis of the corpus callosum Renal agenesis Müllerian agenesis |
| Aplasia | The organ is not present. | Primordial tissue from which the organ is derived is present. | Aplasia cutis congenita Thymic aplasia |
| Hypoplasia | Underdevelopment of an organ | Primordial tissue from which the organ is derived is present. | Hypoplastic right heart syndrome |
| Disruption | Breakdown of a previously normal structure or tissue | N/A | Amniotic band syndrome + Entrapment of fetal parts in fibrous amniotic bands + Clinical features - Constriction rings or bands - Autoamputation of digits or limbs - Syndactyly - Abdominal wall defects, visceral protrusions - Craniofacial anomalies |
| Deformation | Interruption of the normal development of an organ due to an extrinsic force (e.g., in case of multiple gestations or large leiomyomas) | Occurs after week 8 | Congenital talipes equinovarus Congenital torticollis Developmental dysplasia of the hip |
| Malformation | Interruption of the normal development of an organ due to an intrinsic process | Occurs between week 3 and week 8 | Arnold-Chiari malformation Arteriovenous malformation (an abnormal, high-flow connection between arteries and veins bypassing capillaries which develops due to disrupted angiogenesis) |
| Sequence | A single event that causes multiple abnormalities | N/A | Pierre Robin sequence Potter sequence |
| Syndrome | A set of clinical features that consistently occur together | Commonly caused by genetic mutations (e.g., trisomy 21) | Down syndrome Patau syndrome Edward syndrome |
| Field defect | Development of malformations in specific, contiguous area | N/A | Holoprosencephaly |
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Differentiation of the embryonic disc
Differentiation of the mesoderm
Axial mesoderm
Location: central region of the mesoderm
Components: notochord and prechordal mesoderm
Paraxial mesoderm
Location: tube-shaped area of the mesoderm surrounding the notochord
Components: somites
Process
The paraxial mesoderm becomes divided into segmented round cell clusters (somites) along the neural tube (week 4)
Up to the beginning of week 5, an initial 42–44 somite pairs are formed in a craniocaudal direction.
Degeneration of several somite pairs, so that 35–37 somite pairs remain
As primitive segments, somites determine body segmentation
Somite segments
Sclerotome: medial migration of cells toward the notochord and fusion of both sclerotomes of a somite pair
Dermomyotome
Dermatome: migration toward the surface ectoderm
Myotome: subdivision into the dorsal epaxial and ventral hypaxial skeletal muscles in the lateral and anterior regions of the thorax and abdomen
Intermediate mesoderm
Location: lateral to the paraxial mesoderm
Components: urogenital fold, consisting of the nephrogenic ridge and genital ridge
Lateral plate mesoderm
Location: lateral to the intermediate mesoderm
Sequence of events
In the second week of development, small folds develop in lateral aspect of the mesoderm.
These folds split horizontally to form two components.
Somatic (parietal) mesoderm: the dorsal layer that underlies the ectoderm and differentiates into the lining of the pleural, pericardial, and peritoneal cavities.
Splanchnic (visceral) mesoderm: the ventral layer that overlies the endoderm and differentiates into the visceral lining of internal organs.
The space between these components is called the coelom
The two coeloms from either side fuse at the end of the lateral folding of the embryo to form one large cavity, the intraembryonic coelom, which will differentiate into the thoracic and abdominal cavities (see “Morphogenesis” above).
Mesenchyme ≠ mesoderm: The mesoderm is one of the three germinal layers that differentiates into different tissues. The mesenchyme is embryonic connective tissue that develops from the mesoderm and other germ layers.
Fate mapping
A fate map is used to determine the origin of a cell lineage, e.g., a germ layer. The following table provides an overview of the various tissue types and structures that arise from the three germ layers.
| Overview |
| Germ layer | Germ layer structures | Differentiated tissue/organs |
| Ectoderm | Neuroectoderm (neural tube) | Central nervous system (except the anterior pituitary) + CNS neurons + Glia (astrocytes, oligodendrocytes, ependymal cells) + Pineal gland + Posterior pituitary (neurohypophysis) Retina |
| Neural crest | Nervous tissue + Neurons and glia of the peripheral nervous system - Dorsal root ganglion cells - Autonomic ganglion cells - Schwann cells - Satellite glial cells + Enteric nervous system Cranial and head structures + Dermis and subcutaneous tissue in the head region + Bones, cartilage, and connective tissue of the head region + Arachnoid mater and pia mater + Odontoblasts (dentin) and dental cement Other cell types + Chromaffin cells of the adrenal medulla + Skin melanocytes + Carotid body + Cardiac septum in the outflow tract Cardiac tissue + Endocardial cushions (form from the cells that undergo epithelial-mesenchymal transition) + Aortopulmonary septum |
| Surface ectodermal placodes | Olfactory epithelium Inner ear Lens Cranial nerve ganglia (partial) |
| Surface ectoderm | Epithelium + Oral cavity + Nasal cavity + Paranasal sinuses + Lacrimal ducts, + Ear canal, + Anal canal below the dentate line Epidermis and skin appendages: including hair, nails, glands (including sweat and mammary glands) Other cell types + Ameloblasts (enamel organ) of the teeth + Anterior pituitary (derived from Rathke pouch) + Parotid gland + Inner ear sensory organs |
| Mesoderm (intraembryonic mesoderm) | Axial | Prechordal | Bones that make up the base of the skull (partially) Extraocular muscles |
| Notochord | Nucleus pulposus of the intervertebral discs |
| Paraxial | Sclerotome | Vertebral bodies Annulus fibrosus of the intervertebral discs Ribs |
| Dermatome | Spinal meninges Dermis and cutis of the back and, partially, the head |
| Myotome | Skeletal muscles + Neck (hypomere) + Walls of the lateral and ventral trunk (hypomere) + Intrinsic back muscles (epimere) + Limbs (hypomere) |
| Intermediate | Kidneys (pronephros, mesonephros, metanephros) Gonads Renal and genital excretory ducts |
| Lateral plate mesoderm | Splanchnic (splanchnopleuric) mesoderm | Cardiovascular system + Heart + Blood vessels + Stem cells of erythroid, lymphoid, and myeloid origin (including microglia) + Lymph nodes and vessels Other cell types + Adrenal cortex + Spleen + Visceral layers of the pericardium, pleura, and peritoneum + Connective tissue and smooth muscle of the intestinal wall |
| Somatic (somatopleuric) mesoderm | Connective tissue of the anterior and lateral trunk wall Parietal layers of the pericardium, pleura, and peritoneum Sternum Limbs (all tissues except the muscles) |
| Endoderm | Epithelium + Pharynx + Eustachian tube + Tonsils + Thymus Epithelium and glands + Esophagus and gastrointestinal tract (except the parotid gland and the anal canal below the dentate line) + Liver, gallbladder, pancreas + Lungs + Bladder, prostate, urethra, and distal vagina + Follicular and parafollicular cells (C cells) in the thyroid + Parathyroid gland |
Ectoderm: ec-sternal layer; Mesoderm: middle layer; Endoderm: en-ternal layer.
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11802 | https://www.sjsu.edu/faculty/watkins/cardinals.htm | | San José State University |
| applet-magic.com Thayer Watkins Silicon Valley & Tornado Alley USA |
| |
| The Arithmetic of Cardinal Numbers |
Abstract: This is an explanation of the concept of cardinal numbers from the viewpoint
of naive set theory. The concept of equal number of elements in sets can be
extended to infinite sets. This leads to the fact that there are different
levels of infinity. The concepts of addition, multiplication and exponentiation
of cardinal numbers are introduced by means of the operations of set union,
Cartesian product and function. Some simple calculations are carried out
for addition, multiplication and exponentiation of infinite cardinals. Finally
some of the paradoxes of set theory are examined to show the need for an
axiomatic set theory.
Two sets are said to be equipolent if a one-to-one
correspondence between their elements can be established. There is a one-to-one
correspondence
between set A and set B if and only if there is function f:A→B such that
for any element a of A f(a) is defined and for any element b of B there is one
and only one element of A such that f(a)=b.
We can imagine all sets being grouped into classes in which all sets in
a class are equipolent. We say the relation of being equipolent
is an
equivalence relation. A relation R is an equivalence relation if three
conditions hold:
R is reflexive; i.e., XRX
R is symmetric; i.e., if XRY then YRX
R is transitive; i.e., if XRY and YRZ then XRZ
The set of sets that are equipolent are equivalence classes. Cardinal
numbers are simply the equivalence classes under the relation of equipolency.
One establishes the cardinality of a finite set by putting it into a one-to-one
correspondence with one of the sets, {1}, {1, 2}, {1, 2, 3}, ... In other
words one counts the elements of the set.
The Addition of Cardinal Numbers
Let n1 and n2 be cardinal numbers. There are then sets
A1 and A2 such that #(A1) = n1
and #(A2) = n2. It is essential that
A1 and A2 be disjoint. From A1 and A2
we can create A'1=
{(a1,1)|a1∈A1}=A1×{1} and
A'2={(a2,2)|a2∈A2}=A2×{2}.
The sum of cardinals is the cardinality of the union of the representative
sets for the cardinals; i.e.,
n1 + n2 = #(A'1) + #(A'2) = #(A'1∪A'2)
To be rigorous we must establish that addition of cardinals is well defined.
This means that we must show that if
#(A'1) = #(A"1) and #(A'2) = #(A"2) then #(A'1∪A'2) = #(A"1∪A"2)
To do this consider an element x from #(A'1∪A'2).
Let f1 be the one-to-one function from A'1 to
A"1) and f2 be the one-to-one function from
A'2 to A"2. The one-to-one onto function from
A'1∪A'2 to
A"1∪A"2 is then merely
f1 on A'1 and f2 on A'2.
Infinite Cardinal Numbers
If a set can be put into a one-to-one correspondence with the natural numbers
it is said to be countably infinite or denumerable. The cardinality of the natural numbers {0, 1, 2, 3, ... } is denoted as
aleph null, 0. If a set is finite or denumerable
it is said to be countable.
The original definition of infinite is "not finite," meaning "cannot be put into a one-to-one
correspondence with any finite set. Another criteria of infiniteness is
whether a set contains a proper subset that can be put into a one-to-one
correspondence with the set itself. It is easy to establish that if a set
is equipotent to a proper subset of itself then it cannot be finite. The
proof is by contradiction. Suppose a set S has one-to-one correspondences
with a proper subset T and with a finite set {1,2,...,n}. Then there
is a one-to-one correspondence between the proper subset T and the
finite set. But since T is a proper subset of S there exists an element of
S which does not belong to T, call it u. This would mean there would be
a one-to-one correspondence between T∪{u} and {1,2,..,n,n+1} and
hence between {1,2,..,n} and {1,2,..,n+1}. Therefore S cannot be finite.
The other proposition of interest is that if a set is not finite there
exists a one-to-one correspondence between the set and one of its
proper subsets.
This property of infinite sets provide the basis for some amusing illustrations.
N. Ya. Vilenkin in his Stories About Sets hypothesizes a hotel with
an infinite number of rooms. When a hotel with a finite number of rooms is
filled any new customers have to be turned away. For an infinite hotel that
is filled new customers still can be accomodated. The occupant in the first
room is moved to the second room. The second room's occupant is moved to
the third room and so on. The new customer can then be put in the first room.
If n new customers show the present occupant of room i can be moved to
room n+i for all values of i, thus freeing up space for the n new customers. If an infinite
number of new customers arrive they also can be accomodated by moving the
present occupant of any room, say room i, is moved into room 2i thus freeing
up all of the odd numbered rooms for the infinity of new customers.
The Ordering of Cardinal Numbers
If a set A can be put into a one-to-one correspondence with a subset of set B
then the cardinality of A is less than or equal to the cardinality of set B;
(A)≤#(B).
If set B can also be put into a one-to-one correspondence with a subset of set
A then it can be shown that set A and set B have the same cardinality
(The Schroeder-Bernstein Theorem).
For sets there is:
The Law of Trichotomy:
For any two sets A and B exactly one of the following three possibilities
is true:
(A) < #(B)
(A) = #(B)
(A) > #(B).
Cantor showed (Cantor's Theorem) that
for any set A, the cardinality of the set of all subsets of A is strictly greater than
the cardinality of A. The set of all subsets of A is called the Powerset of A, which is
denoed as P(A). Thus Cantor's Theorem is that #(A) < #(P(A)).
Thus for any cardinal number we
can find a larger cardinal number.
The Addition of Infinite Cardinal Numbers
Since addition of cardinal numbers is well defined we can find the sum of
any two cardinal numbers, n1 and n2, by finding any
two disjoint sets
A1 and A2 such that #(A1)
= n1 and #(A2) = n2
and finding the cardinality
of their union. For example, if n1 is a finite number n and
n2 = 0 then we find their sum
by choosing two disjoint sets such that the cardinality of one set is
equal to n and the cardinality of the other set is equal to
0 and then find the cardinality of the union
of the two sets. The following demonstrates that the sum of n and
0 is equal to
0.
#({1,2,3,...,n})=n and #({n+1,n+2,...}) = 0 #({1,2,3,...,n}∪{n+1,n+2,...}) = #({1,2,3,...})=0
We find that 0 +
0 = 0 because
we can use the set of odd numbers to represent the first
0 on the left and the set of even numbers
to represent the second 0 on the left.
The union of the odd numbers and even numbers gives the set of positive
integers which has the cardinality of 0.
#({1,3,5,...}) = 0 #({2,4,6,...}) = 0 so 0 + 0 = #({1,3,5,...})∪ {2,4,6,..}) = #({1,2,3,...}) = 0
Note that cancellation does not apply in the matter of infinite cardinal numbers; i.e.
a+c = b+c does not imply that a=b.
The sum of two transfinite cardinal numbers is equal to the larger of the
two. The cardinality of the real numbers is greater than 0
so the sum of the cardinality of the real number, usually denoted by a script
c and 0 is equal to the cardinality of the
real numbers.
Multiplication of Cardinal Numbers
Multiplication of two cardinal numbers is defined as the cardinality of the
Cartesian product of two sets having cardinalities equal to the two cardinals.
Let n1 and n1 be two cardinals and let
A1 and A1 such that #(A1)
= n1 and #(A2) = n2. There is no necessity
that A1 and A2 be disjoint. The product of
n1 and n2 is defined as:
n1n2 = #(A1)#(A2) = #(A1×A2)
The Cartesian product of the set of positive integers may be
enumerated in a systematic manner as indicated below. This establishes
that the set of ordered pairs of positive integers has the same cardinality
as the set of positive integers, 0.
The indicated pattern for the enumeration of the ordered pairs of integers
establishes a one-to-one correspondence between the set positive integers
P and the set of pairs of positive integers, P×P. Thus #(P×P)=#(P) and
hence 00=
0. This obviously implies that for any
finite cardinal n
0n = 0.
Exponentiation of Cardinal Numbers
Raising one cardinal number to a power equal to that of another cardinal
number can defined defined in terms of the set of functions mapping from
one set into another. Let the notation BA denoted the set of
all functions from a domain of set A to a range of set B. For finite sets
the number of different functions from a set A of n elements into a set B
of m elements is mn.
The raising of one cardinal number to a power of another cardinal number is
defined as follows. Let m and n be cardinal numbers and the sets A and B
such that #(A)=n and #(B)=m. Then:
mn = #(BA).
As with addition and multiplication we must establish that exponentiation is
well defined; i.e., if there are one-to-one correspondences between A and
A' and B and B', respectively, then there is a one-to-one correspondence
between BA and B'A'. Suppose f:A→A' and
g:B→B' are the one-to-one correspondences between A and A' and B and B'.
Let h be any function from A to B, h:A→B. Consider the function
h' ∈ B'A' defined by
h'(a') = b' = g(h(f-1(a'))).
This function takes any element a' belonging to A' and first obtains the
corresponding element a of A, then obtains the element b of B given by
h(a) and from b obtains the corresponding element of B', b'=g(b). Thus for
any function in BA there is one and only one function
corresponding to it in B'A'. The crucial question is whether
given any function in B'A' there is one and only one function in
BA corresponding to it. The corresponding function is determined
as follows. For any function h':A' → B' the corresponding function h
in BA is given by:
h(a) = g-1(h'(f(a))).
Thus there is a one-to-one correspondence between the set of functions
BA and the set of functions B'A'. There the
exponentiation of cardinal numbers defined previously is well defined.
There are two important properties of the powers of finite cardinal numbers:
(mn)p) = mnp mn+p = mnmp.
Below are proofs that these properties apply to all cardinal numbers.
Let A, B, and C be sets such that #(A)=n, #(B)=m and #(C)=p.
The first property requires that we show that:
#((BA)C) = #(BA×C).
The right hand side involves functions of the form f(a,c)=b, where a, b and
c are from the sets A, B and C, respectively. If we hold c fixed we get a
function from A to B, which might be expressed as fc(a)=b.
For different values of c we get the different functions of BA.
This is just a function of the form C→BA. Thus the two sides
describe the same set and thus the cardinalities are the same.
To prove the second property the sets A and C must be disjoint. To prove
the second property we must establish that
#(BA∪C) = #(BA×BC).
The set involved on the right side of the above equation is the set of
ordered pairs of functions of the form (f(a), g(c)) where a belongs to A and
c belongs to C and the values of both functions belong to B.
The set on the left hand side is the set of functions from
A∪C to B. But, because A and C are disjoint, the functions
from A∪C to B can be composed by taking any function
from A to B for the A-part of A∪ and any function from
C to B for the C-part of A∪C to B. Thus the left side
involves the same set of functions as the right side. Therefore the
cardinalities are the same.
Paradoxes of Set Theory Involving Cardinal Numbers
In formulating the notion of a cardinal number as an equivalence class of
sets are equipotent; i.e., sets that can be put into a one-to-one correspondence
there was the presumption that we could envision a set of all sets and also
a set of all sets that are equipotent. These presumptions are not valid and
this fact shows the need for a more sophisticated version of set theory,
an axiomatic set theory.
Consider the assumption that there is a set of all sets. This like assuming
that there a largest integer. Given any candidate for a largest integer
one can construct a larger integer; i.e., that integer plus one. Thus the concept
of a largest integer is contradictory.
Suppose
there were a set of all sets, S. From S we can construct the power set of
S, which by Cantor's Theorem is strictly larger than S.
Suppose now that C is the set of all sets which are equipotent. Consider
the power set of C and any function from P(C) to C×P(C), say f:P(C)→C×P(C).
Then f(P(C)) is a subset of C because for each c∈C,
f(c) is a set with the cardinality of C. Thus the cardinality of
f(P(C)) is less than or equal to the cardinality of C. Since the
sets in f(P(C)) are disjoint the cardinality of f(P(C)) is
the same as the cardinality of P(C). By Cantor's Theorem then
(P(C)) is strictly greater than #(C). Thus there is a contradiction
and therefore the concept of the set of all equipolent sets is invalid; i.e., contradictory.
Thus great care must be exercised in making set theory rigorous when dealing with sets
that are not finite.
| |
| HOME PAGE OF applet-magic HOME PAGE OF Thayer Watkins | |
11803 | https://www.geeksforgeeks.org/dsa/program-check-three-points-collinear/ | Program to check if three points are collinear
Last Updated :
03 Jun, 2022
Suggest changes
13 Likes
Given three points, check whether they lie on a straight (collinear) or not
Examples :
```
Input : (1, 1), (1, 4), (1, 5)
Output : Yes
The points lie on a straight line
Input : (1, 5), (2, 5), (4, 6)
Output : No
The points do not lie on a straight line
```
First approach
Three points lie on the straight line if the area formed by the triangle of these three points is zero. So we will check if the area formed by the triangle is zero or not
```
Formula for area of triangle is :
0.5 [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)]
The formula is basically half of determinant
value of following.
x1 x2 x3
y1 y2 y3
1 1 1
The above formula is derived from shoelace formula.
If this equals zero then points lie on a straight line
```
C++
````
// C++ program to check if three
// points are collinear or not
// using area of triangle.
include
include
include
using namespace std;
// function to check if point
// collinear or not
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
// Calculation the area of
// triangle. We have skipped
// multiplication with 0.5
// to avoid floating point
// computations
int a = x1 (y2 - y3) +
x2 (y3 - y1) +
x3 (y1 - y2);
if (a == 0)
cout << "Yes";
else
cout << "No";
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
// This code is contributed
// by Akanksha Rai(Abby_akku)
````
// C++ program to check if three
// C++ program to check if three
// points are collinear or not
// points are collinear or not
// using area of triangle.
// using area of triangle.
```
include #include
```
```
include #include
```
```
include #include
```
using namespace std; using namespace std
// function to check if point
// function to check if point
// collinear or not
// collinear or not
void collinear(int x1, int y1, int x2, void collinear int x1 int y1 int x2
int y2, int x3, int y3) int y2 int x3 int y3
{
// Calculation the area of
// Calculation the area of
// triangle. We have skipped
// triangle. We have skipped
// multiplication with 0.5
// multiplication with 0.5
// to avoid floating point
// to avoid floating point
// computations
// computations
int a = x1 (y2 - y3) + int a = x1 y2 - y3 +
x2 (y3 - y1) + x2 y3 - y1 +
x3 (y1 - y2); x3 y1 - y2
if (a == 0) if a == 0
cout << "Yes"; cout<< "Yes"
else else
cout << "No"; cout<< "No"
}
// Driver Code
// Driver Code
int main() int main
{
int x1 = 1, x2 = 1, x3 = 1, int x1 = 1 x2 = 1 x3 = 1
y1 = 1, y2 = 4, y3 = 5; y1 = 1 y2 = 4 y3 = 5
collinear(x1, y1, x2, y2, x3, y3); collinear x1 y1 x2 y2 x3 y3
return 0; return 0
}
// This code is contributed
// This code is contributed
// by Akanksha Rai(Abby_akku)
// by Akanksha Rai(Abby_akku)
C
````
// C program to check if three
// points are collinear or not
// using area of triangle.
include
include
include
// function to check if point
// collinear or not
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
// Calculation the area of
// triangle. We have skipped
// multiplication with 0.5
// to avoid floating point
// computations
int a = x1 (y2 - y3) +
x2 (y3 - y1) +
x3 (y1 - y2);
if (a == 0)
printf("Yes");
else
printf("No");
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
````
Java
````
// Java program to check if
// three points are collinear
// or not using area of triangle.
class GFG
{
// function to check if
// point collinear or not
static void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
/ Calculation the area of
triangle. We have skipped
multiplication with 0.5
to avoid floating point
computations /
int a = x1 (y2 - y3) +
x2 (y3 - y1) +
x3 (y1 - y2);
if (a == 0)
System.out.println("Yes");
else
System.out.println("No");
}
// Driver Code
public static void main(String args[])
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by Sam007.
````
Python
````
Python program to check
if three points are collinear
or not using area of triangle.
function to check if
point collinear or not
def collinear(x1, y1, x2, y2, x3, y3):
""" Calculation the area of
triangle. We have skipped
multiplication with 0.5 to
avoid floating point computations """
a = x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)
if (a == 0):
print "Yes"
else:
print "No"
Driver Code
x1, x2, x3, y1, y2, y3 = 1, 1, 1, 1, 4, 5
collinear(x1, y1, x2, y2, x3, y3)
This code is contributed
by Sachin Bisht
````
C#
````
// C# program to check if
// three points are collinear
// or not using area of triangle.
using System;
class GFG
{
/ function to check if
point collinear or not /
static void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
/ Calculation the area of
triangle. We have skipped
multiplication with 0.5 to
avoid floating point computations /
int a = x1 (y2 - y3) +
x2 (y3 - y1) +
x3 (y1 - y2);
if (a == 0)
Console.Write("Yes");
else
Console.Write("No");
}
// Driver code
public static void Main ()
{
int x1 = 1, x2 = 1, x3 = 1,
y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
}
}
// This code is contributed by Sam007.
````
PHP
````
php
// PHP or not using area of triangle.
/ function to check if
point collinear or not /
function collinear($x1, $y1, $x2,
$y2, $x3, $y3)
{
/ Calculation the area of
triangle. We have skipped
multiplication with 0.5 to
avoid floating point computations /
$a = $x1 ($y2 - $y3) +
$x2 ($y3 - $y1) +
$x3 ($y1 - $y2);
if ($a == 0)
printf("Yes");
else
printf("No");
}
// Driver Code
$x1 = 1; $x2 = 1; $x3 = 1;
$y1 = 1; $y2 = 4; $y3 = 5;
collinear($x1, $y1, $x2, $y2, $x3, $y3);
// This code is contributed by Sam007.
?
````
JavaScript
````
// Javascript program to check if three
// points are collinear or not
// using area of triangle.
function collinear(x1, y1, x2, y2, x3, y3)
{
// Calculation the area of
// triangle. We have skipped
// multiplication with 0.5
// to avoid floating point
// computations
var a = x1 (y2 - y3) +
x2 (y3 - y1) +
x3 (y1 - y2);
if (a == 0)
document.write("Yes");
else
document.write( "No");
}
var x1 = 1, x2 = 1, x3 = 1,y1 = 1, y2 = 4, y3 = 5;
collinear(x1, y1, x2, y2, x3, y3);
// This code is contributed by akshitsaxenaa09.
````
Output :
Yes
Time Complexity: O(1)
Auxiliary Space: O(1)
Second approach
```
For three points, slope of any pair of points
must be same as other pair.
For example, slope of line joining (x2, y2)
and (x3, y3), and line joining (x1, y1) and
(x2, y2) must be same.
(y3 - y2)/(x3 - x2) = (y2 - y1)/(x2 - x1)
In other words,
(y3 - y2)(x2 - x1) = (y2 - y1)(x3 - x2)
```
C++
````
// A C++ program
// Slope based solution to check
// if three points are collinear.
include
using namespace std;
/ function to check if
point collinear or not/
void collinear(int x1, int y1, int x2, int y2, int x3,
int y3)
{
if ((y3 - y2) (x2 - x1) == (y2 - y1) (x3 - x2))
cout << "Yes" << endl;
else
cout << "No" << endl;
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 0, y1 = 1, y2 = 6, y3 = 9;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
// The code is contributed by Gautam goel (gautamgoel962)
````
// A C++ program
// A C++ program
// Slope based solution to check
// Slope based solution to check
// if three points are collinear.
// if three points are collinear.
```
include #include
```
using namespace std; using namespace std
/ function to check if
/ function to check if
point collinear or not/
point collinear or not/
void collinear(int x1, int y1, int x2, int y2, int x3, void collinear int x1 int y1 int x2 int y2 int x3
int y3) int y3
{
if ((y3 - y2) (x2 - x1) == (y2 - y1) (x3 - x2)) if y3 - y2 x2 - x1 == y2 - y1 x3 - x2
cout << "Yes" << endl; cout<< "Yes"<< endl
else else
cout << "No" << endl; cout<< "No"<< endl
}
// Driver Code
// Driver Code
int main() int main
{
int x1 = 1, x2 = 1, x3 = 0, y1 = 1, y2 = 6, y3 = 9; int x1 = 1 x2 = 1 x3 = 0 y1 = 1 y2 = 6 y3 = 9
collinear(x1, y1, x2, y2, x3, y3); collinear x1 y1 x2 y2 x3 y3
return 0; return 0
}
// The code is contributed by Gautam goel (gautamgoel962)
// The code is contributed by Gautam goel (gautamgoel962)
C
````
// Slope based solution to check
// if three points are collinear.
include
include
/ function to check if
point collinear or not/
void collinear(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) (x2 - x1) ==
(y2 - y1) (x3 - x2))
printf("Yes");
else
printf("No");
}
// Driver Code
int main()
{
int x1 = 1, x2 = 1, x3 = 0,
y1 = 1, y2 = 6, y3 = 9;
collinear(x1, y1, x2, y2, x3, y3);
return 0;
}
````
Java
````
// Slope based solution to check
// if three points are collinear.
import java.io.;
class GFG {
/ function to check if
point collinear or not/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) (x2 - x1) ==
(y2 - y1) (x3 - x2))
System.out.println("Yes");
else
System.out.println("No");
}
// Driver Code
public static void main (String[] args) {
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
//This Code is Contributed by ajit
````
Python
````
Slope based solution to check if three
points are collinear.
function to check if
point collinear or not
def collinear(x1, y1, x2, y2, x3, y3):
if ((y3 - y2)(x2 - x1) == (y2 - y1)(x3 - x2)):
print ("Yes")
else:
print ("No")
Driver Code
x1, x2, x3, y1, y2, y3 = 1, 1, 0, 1, 6, 9
collinear(x1, y1, x2, y2, x3, y3);
This code is contributed
by Sachin Bisht
````
C#
````
// Slope based solution to check
// if three points are collinear.
using System;
class GFG
{
/ function to check if
point collinear or not/
static void cool_line(int x1, int y1, int x2,
int y2, int x3, int y3)
{
if ((y3 - y2) (x2 - x1) ==
(y2 - y1) (x3 - x2))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
// Driver Code
static public void Main ()
{
int a1 = 1, a2 = 1, a3 = 0,
b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
}
}
// This code is contributed by ajit
````
PHP
````
php
// Slope based solution to check
// if three points are collinear.
/ function to check if
point collinear or not/
function collinear($x1, $y1, $x2,
$y2, $x3, $y3)
{
if (($y3 - $y2) ($x2 - $x1) ==
($y2 - $y1) ($x3 - $x2))
echo ("Yes");
else
echo ("No");
}
// Driver Code
$x1 = 1;
$x2 = 1;
$x3 = 0;
$y1 = 1;
$y2 = 6;
$y3 = 9;
collinear($x1, $y1, $x2,
$y2, $x3, $y3);
// This code is contributed by ajit
?
````
JavaScript
````
// Slope based solution to check
// if three points are collinear.
/
function to check if point collinear or not
/
function cool_line(x1 , y1 , x2 , y2 , x3 , y3)
{
if ((y3 - y2) (x2 - x1) == (y2 - y1) (x3 - x2))
document.write("Yes");
else
document.write("No");
}
// Driver Code
var a1 = 1, a2 = 1, a3 = 0, b1 = 1, b2 = 6, b3 = 9;
cool_line(a1, b1, a2, b2, a3, b3);
// This code is contributed by aashish1995
````
Output :
No
Time Complexity: O(1)
Auxiliary Space: O(1)
P
pranav gupta
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11804 | https://www.organicchemistrytutor.com/topic/epoxide-opening/ | Skip to content
Epoxide Opening
Opening of epoxides is a seemingly simple reaction that can catch you by surprise if you’re not paying attention to the conditions and the reagents you’re working with. In this tutorial on epoxide opening reactions, we’ll focus on the regioselective opening of epoxides under acidic and basic conditions. Epoxides, known for their reactive three-membered ring structures, demonstrate distinct behaviors depending on the environment, leading to varied outcomes in synthesis. We’ll delve into the nuances of these reactions, highlighting the differences between opening epoxides from more or less substituted sides, and uncover the principles behind these pivotal organic reactions.
Epoxide Opening Overview
As I’ve mentioned a moment ago, epoxides can be opened in both acidic and basic conditions. The main difference between the two approaches is going to be the regioselectivity of the reaction. And by the regioselectivity here I mean which side we’re going to open our epoxide ring from.
So, if we look at an epoxide, it’s a three-membered ring with an oxygen atom in the ring. We also call epoxides oxidants, which is the IUPAC name for this functional group. If the molecule is not symmetrical, we can distinguish between the more and less substituted sides of the ring.
Now, depending on the nature of our reagents and conditions, the ring can open from either the more substituted side or from the less substituted side. TLDR; version is the acidic conditions open epoxides from the more substituted side, while the basic conditions open the epoxides from the less substituted side.
Now, let’s talk about the details of why that happens and how.
Epoxide Opening in Acidic Media
I’ll start by looking at the epoxide opening in acidic media. In acidic condition we’ll have a strong acid like H2SO4 present as a catalyst, or the reagent itself will be a strong acid, like, for instance HBr.
So, if I took my example molecule, 2,2-dimethyloxirane here, and treated it with HBr, I’d get the product, in which the oxygen will be still attached to the less substituted carbon of what used to be an epoxide ring, and the bromine will be connected to the more substituted carbon. Mechanistically speaking, this reaction starts by protonating the oxygen of our epoxide, followed by the ring opening by the nucleophilic bromide anion.
Interestingly enough, although this is a tertiary atom, the reaction seems to follow the SN2 kinetics. And, if any stereochemistry is present, we’ll see the inversion of stereochemistry here as well. Occasionally, we’ll see the formation of a carbocation in reactions like that, but those are not all too common. And even though there’s a possibility of a carbocation formation, we’ll treat this reaction as a “special case” of SN2 which works for tertiary carbons. We’ll also attribute this unusual reactivity to the ring strain.
Now, I’ve mentioned that we’ll also see the inversion of stereochemistry here. Indeed, if I took this molecule and treated it with HBr, the resulting product would have a carbon with the opposite stereoconfiguration. You can confirm that we do see the inversion here by assigning the stereodescriptors to our chiral atoms. The molecule on the left is the (S)-stereoisomer, while the product has the (R) stereoconfiguration. And by the way, using this opportunity I’m going to remind you that the inversion of stereoconfiguration does not always coincide with the change in the stereo descriptor. It works in most cases, but it depends on the nature of the leaving group and the nucleophile. And occasionally, you’ll end up with the same stereodescriptor for both starting material and the product because the CIP priorities can get all switched around with the new group in the molecule.
Epoxide Opening with Neutral Nucleophiles with Acid Catalysts
What if my reagent is not a strong acid on its own? Well, in this case, we’ll need to bring acid as a catalyst. Since we don’t want any competition between our conjugate base and the nucleophile, we’ll use non-nucleophilic acids like H2SO4 for our catalyst.
For instance, if I go back to my 2,2-dimethyloxirane and treat it with methanol, nothing will happen as methanol is not nearly acidic enough to protonate my epoxide and make it more reactive. This is where the sulfuric acid becomes important. It provides the acidic protons and enables this reaction to happen. So, we’ll protonate the epoxide first, and then a poor nucleophile that is our methanol can attack it. It happens from the more substituted side like in the previous examples, giving us the corresponding ether as the final product.
Epoxide Opening in Basic Conditions
Moving on to the opening of our epoxide in basic conditions, I wanna mention right the way, that it’s not always actually the basic conditions. I suppose, it’s called this way to drive the the point and emphasize attention on the differences in the conditions. And it’s easy to remember acid vs base. But it should’ve been called more like “epoxide opening with strong nucleophiles” instead. Yes, it doesn’t sound nearly as good, but it’s a more correct version. The thing is, while some nucleophiles are indeed bases, many are not.
For instance, if I treat my 2,2-dimethyloxirane with potassium cyanide, the cyanide is indeed a nucleophile. Is it a base though? Not at all!
Another important distinction here is that the strong nucleophiles typically open our epoxides from the less substituted side. So, mechanistically speaking, the reaction is your typical SN2 process. The nucleophile, the cyanide anion in this case, attacks the carbon breaking the carbon-oxygen bond and making an alkoxide intermediate, which quickly grabs the proton from the solvent giving us the final product.
And yes, our nucleophile can be a base as well. Like for instance, if I were to treat our good friend 2,2-dimethyloxirane here with a solution of sodium methoxide in methanol, I’ll get the methoxide attack the epoxide from the less substituted side as well. And as we know from all our previous organic chemistry studies, alkoxides are fairly basic.
Does it matter that the methoxide a base though? Nope, not really, unless, of course, you have some other acidic functional groups in your molecule that can participate in a proton transfer reaction. What does matter though, is that methoxide is a good nucleophile. And since the world of nucleophiles is really diverse, we have a plethora of possible products that we can make by opening epoxides with different nucleophiles. And what’s really cool is that they all will open epoxides from the less substituted side making this reaction a powerful tool in our synthesis arsenal!
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11805 | https://www.geeksforgeeks.org/maths/uses-of-percentage-in-daily-life/ | Applications of Percentage
Last Updated :
03 Sep, 2025
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A Percentage is a mathematical concept that represents a fraction of a whole expressed as a hundredth. It is widely used in various aspects of our daily lives, from financial transactions to scientific measurements.
Percentage has numerous applications in daily life, including:
1 / 3
Real-Life
Percentages are used in many everyday activities. Here are some important examples:
Calculating Interest - Percentages help calculate interest on loans and savings.
Getting Discount - Percentages are used to calculate price reductions.
Calculating Sales Tax - Percentages help determine taxes on goods and services.
Grading Assignment - Percentages are used to express marks and scores.
Finance and Accounting
Percentages are extensively used to calculate interest rates, profit margins, returns on investment, depreciation, taxes, and financial ratios.
Interest rates on loans and deposits are expressed as percentages to standardize calculations.
Profit margins and returns on investment (ROI) help investors and businesses compare performance across projects.
Percentages help compute depreciation of assets and assess tax liabilities accurately.
Sales and Marketing
Percentages are key in evaluating sales performance, pricing strategies, and customer behavior.
Discounts and markups are expressed in percentages to set prices and promotions.
Sales growth rates and conversion rates help assess business performance over time.
Market share percentages indicate a company’s position relative to competitors.
Economics
Percentages are vital in measuring economic changes and trends.
Inflation rates and interest rates are expressed in percentages for consistency.
Unemployment rates and GDP growth help policymakers analyze economic health.
Percentages simplify comparisons of economic indicators across countries or time periods.
Statistics
Percentages represent proportions and probabilities in data analysis.
They are used in frequency distributions, margins of error, and confidence intervals.
Percentages make complex data more interpretable, such as survey results or population statistics.
They help compare trends across different datasets efficiently.
Education
Percentages standardize evaluation and academic performance.
Grades, scores, and assessments are expressed as percentages to maintain consistency.
Percentages calculate attendance rates, graduation rates, and pass percentages.
They provide a clear measure of student performance and in
Healthcare:They help evaluate medical test results, mortality rates, disease prevalence, treatment success rates, and other healthcare statistics.
Healthcare
Percentages quantify medical outcomes and public health metrics.
Mortality rates, disease prevalence, and treatment success rates are reported in percentages.
Percentages help compare patient recovery rates and the effectiveness of interventions.
They assist policymakers in planning healthcare services and assessing population health
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11806 | https://www.cureus.com/articles/171092-utility-of-fluorescent-microscopy-for-rapid-intraoperative-diagnosis-of-rhinocerebral-mucormycosis-experience-from-a-tertiary-care-center-in-central-india | Utility of Fluorescent Microscopy for Rapid Intraoperative Diagnosis of Rhinocerebral Mucormycosis: Experience from a Tertiary Care Center in Central India | Cureus
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Utility of Fluorescent Microscopy for Rapid Intraoperative Diagnosis of Rhinocerebral Mucormycosis: Experience from a Tertiary Care Center in Central India
Nisha Meshram • Sachin Chaudhari • Milind Bhatkule • Rasika Gadkari
Nisha Meshram • Sachin Chaudhari • Milind Bhatkule • Rasika Gadkari
Published: March 31, 2025
DOI: 10.7759/cureus.81537
Open Access
Peer-Reviewed
Cite this article as: Meshram N, Chaudhari S, Bhatkule M, et al. (March 31, 2025) Utility of Fluorescent Microscopy for Rapid Intraoperative Diagnosis of Rhinocerebral Mucormycosis: Experience from a Tertiary Care Center in Central India. Cureus 17(3): e81537. doi:10.7759/cureus.81537
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Abstract
Rhino-orbital-cerebral mucormycosis is an uncommon life-threatening infection caused by the angioinvasive fungus Mucorales and is associated with high morbidity and mortality. In India, the pandemic of COVID-19 was associated with another deadly disease, rhinocerebral mucormycosis, which further complicated the course of the disease, necessitating an accurate and rapid diagnosis. Conventional methods of diagnosis, like fungal culture, histopathology, Gomori methenamine silver (GMS), and periodic acid-Schiff (PAS) stain, are not feasible for intraoperative diagnosis. We studied the use of fluorescent brightener calcofluor white (CFW) for rapid intraoperative diagnosis of mucormycosis and compared it with intraoperative crush smear cytology and frozen section.A total of 37 intraoperative samples were studied, of which 11 were positive for the fungus. Calcofluor white detected fungus in seven samples, while frozen section detected fungus in eight samples. Calcofluor white stain showed less sensitivity than the frozen section but had high specificity.In the presence of marked necrosis, suspicious fragments on frozen sections could be quickly confirmed by fluorescent stain. Thus, CFW direct microscopy is a useful adjunct for the rapid diagnosis of mucormycosis.
Introduction
From December 2019 to early 2023, the world experienced a deadly pandemic of COVID-19 with associated mortality and morbidity. India experienced a deadly second wave from around March 2021 to June 2021, which presented with more severe disease, resulting in increased use of invasive support systems and immunomodulation [1-4]. But in all this mayhem, no one could guess a new danger lurking around the corner, the epidemic of rhinocerebral mucormycosis . Globally, the highest number of cases were reported from India, amounting to more than 4,000 people with COVID-19-associated mucormycosis (CAM) [5-7].
Mucormycosis most commonly involves the rhinocerebral, pulmonary, gastrointestinal, and subcutaneous tissues . This disease is angio-invasive and causes rapid progression with tissue destruction, facial cellulitis, gangrenous mucosal changes in the nose and paranasal sinuses, with orbital and intracranial extension and death .The success of therapy depends on early detection, systemic antifungal therapy with liposomal amphotericin B, along with aggressive surgical debridement of affected tissue causing cosmetic disfigurement . Among the survivors, vision loss followed by facial deformity is a frequently encountered consequence of mucormycosis . Thus, the earliest diagnosis and immediate treatment are of utmost importance. Song et al. have suggested an algorithm for the early diagnosis and management of common invasive fungal infections (Aspergillus, candidiasis, cryptococcosis, and mucormycosis) .
Various methods for laboratory diagnosis of mucor include direct microscopy, cytology, histopathological examination, culture, and molecular methods like polymerase chain reaction (PCR)-based kits, fluorescent in-situ hybridization (FISH), or immunohistochemistry (IHC) on formalin-fixed paraffin-embedded tissue blocks [11-13]. Frozen section as a diagnostic method can be used in patients with rhino-cerebral mucormycosis in two scenarios: (1) to perform a rapid diagnosis allowing immediate administration of antifungal therapy, and (2) to guide complete eradication of the infected tissue during the surgical debridement . Routine stains for fungus, like Gomori methenamine silver (GMS) and periodic acid-Schiff (PAS), are time-consuming and technique-sensitive and hence cannot be used for intraoperative decision-making.
Calcofluor white (CFW), also called a fluorescent brightener, optical brightener, or “whitening agent,” is routinely used in the paper, textile, and related industries as an agent to whiten the papers and fabrics .Calcofluor white binds specifically to the beta (β) 1‑3, β 1‑4 glycoside chain of the chitin in the fungal cell wall, which almost immediately fluoresces under ultraviolet (UV) light with an intense apple green or bluish color against the white background, depending on the filter used. The addition of potassium hydroxide (KOH) in the CFW stain increases the visibility of the fungal elements in clinical samples .
The use of CFW stain for the detection of fungus was first described by Hageage and Harrington . Since then, it has been used by various researchers for the demonstration of fungi in different clinical samples [18-21]. McDermott et al. used calcofluor-stained tissue as a rapid technique for intraoperative diagnosis of fungus and for assessment of margin clearance during frozen sections [12, 18].Thus, the present study was undertaken to compare the efficacy of intraoperative CFW stain direct microscopy with intraoperative crush smear cytology and frozen sections for rapid diagnosis of rhinocerebral mucormycosis and other fungal infections in COVID-19-positive and post-COVID patients.
Part of this article was previously presented as a meeting abstract at the 50 th Annual Conference of the Indian Academy of Cytologists,held virtually from 19 to 21 November 2021.
Materials & Methods
Study design
This was a prospective, diagnostic test evaluation study performed over a period of three months, from May 2021 to July 2021, at the Department of Pathology of All India Institute of Medical Sciences, a tertiary care institute in Nagpur, India.
Study population
Inclusion Criteria
Intraoperative samples from suspected cases of COVID-19-associated rhinocerebral mucormycosis for frozen section diagnosis were included.
Exclusion Criteria
All other samples, like smears from nasal secretions, nasal swabs, sputum, or tracheal secretions, etc., were not included in the study.
Sample Size
A total of 37 samples from suspected cases of mucormycosis were included in the study; 22 were from male patients and 15 were from female patients.
Methodology
Endoscopic resection specimens or biopsies received for intraoperative diagnosis of fungal infection were divided into three parts. i) A small portion of the sample, preferably from the necrotic or blackish area, was used for the preparation of crush smears on a clean, grease-free slide, which was used for the preparation of calcofluor white wet mount. ii) Another small portion of the sample was used to prepare a crush smear, which was stained with hematoxylin and eosin stain (H&E) by the rapid method. These smears were called smears for intraoperative crush smear cytology. iii) The rest of the tissue was submitted for frozen section evaluation and stained with H&E by the rapid method.
Once the diagnosis on the frozen section was made & conveyed to the surgeon, the frozen tissue was submitted for routine histological processing for making formalin-fixed paraffin-embedded (FFPE) tissue blocks. Sections were cut and stained with PAS and H&E stain, and the sections were studied under a microscope for the presence of fungal elements. Histopathology diagnosis was considered the gold standard.
Calcofluor white stain wet mount preparation (direct microscopy)
One drop of ready-to-use CFW (Sigma Aldrich, St. Louis, MO), comprising 1 g/L CFW M2R, was added to the crush smear slide, followed by one drop of 10% KOH, and covered with a cover slip. The staining procedure was carried out in a darkroom setting. The CFW-KOH wet mount preparation was then examined under a fluorescence microscope (Leica DM750, Leica Microsystems, Wetzlar, Germany) using a blue light excitation filter, which has a 300-400 nm emission wavelength with excitation at around 355nm .
The procedure for the rapid staining method of H&E is outlined in Table 1.
Step no.Procedure Time
1.Immediately fix frozen sections in 95% ethyl alcohol 15 seconds
2.Transfer to formalin 10%, phosphate buffered 10 dips
3.Rinse well in distilled water 10 dips
4.Stain with hematoxylin stain, Harris modified 30 seconds
5.Wash well in two changes of distilled water 10 dips each
6.Place in 95% ethyl alcohol 10 dips
7.Counterstain in eosin Y working solution.15 sec
8.Dehydrate in two changes of 95% ethyl alcohol 10 dips each
9 Dehydrate in two changes of 100% ethyl alcohol 10 dips each
10.Clear in two changes of xylene 10 dips each
11.Coverslip with compatible mounting medium
Table 1: Procedure for the rapid staining method of hematoxylin and eosin stain
Interpretation
Calcofluor White Stain Wet Mount Preparation (Direct Microscopy)
The hyphae of Mucor stained blue against the black background. The fungal hyphae were sharply delineated and could be identified even on the scanner view. Brighter staining was observed when the calcofluor stain was allowed to set for a longer duration(Figure 1).
Figure 1: Calcofluor white stain under fluorescent light showing irregular, non-septate hyphae (original magnification 40x).
Intraoperative Crush Smear Cytology
The cytology smears showed broad, ribbon-like aseptate fungal hyphae with right-angled branching at irregular intervals.Many twisted and wrinkled forms were also seen (Figure 2).
Figure 2: Intraoperative crush smear cytology showing mucor hyphae, which are broad, ribbon-like and non-septate (H&E original magnification 400X).
Frozen Sections
On frozen tissue, Mucor was seen as broad, aseptate, ribbon-like, thin-walled hyphae with right-angled, irregular branching. Twisted, collapsed, wrinkled, and single hyphal forms were also seen (Figure 3).
Figure 3: Endoscopic biopsy tissue on frozen section showing invasion by mucor hyphae. Many twisted and bizarre forms are also seen in the background of necrotic material (H&E original magnification 400X).
Histopathology
The hyphae stained a faint blue with PAS stain.
Results
A total of 37 samples from cases suspected of mucormycosis were included in the study; 22 were from male patients and 15 from female patients. Of the 37 samples, 11 (30%) were positive for fungus while 26 (70%) were negative. Of the 11 samples positive for fungus, 10 (90%) showed the presence of mucormycosis, while one (10%) was positive for Aspergillus.
Of 11 positive samples, seven (63.6%) were positive with direct microscopy by CFW stain, while only four (36.3%) samples detected fungus on intraoperative crush smear cytology. Eight(72.7%) samples showed fungus on frozen tissue.
One sample was positive for fungus on the CFW stain wet mount, but did not show any fungus on frozen tissue. However, the routine paraffin sections from the same sample showed the presence of fungus. Two samples were negative by both CFW stain wet mount and crush smear cytology but showed fungus on frozen tissue and were reported as positive.
Three samples were negative for fungus on intraoperative tissue by all three methods, i.e., CFW stain wet mount, crush smear cytology, and frozen section, but showed fungus on routine histopathology sections.
Of the 26 negative samples on frozen tissue, separate (more) tissue was received in 25 cases, of which eight showed the presence of fungus. Seven samples showed Mucor, while one showed a mixed fungal infection of Mucor and Aspergillus. This disparity could be due to sampling error during freezing.
The sensitivity of the CFW stain wet mount was 63.6%, while specificity was 100% for the detection of fungus (Table 2). The positive predictive value (PPV) was 100%, while the negative predictive value (NPV) was 86.7%. The sensitivity of the frozen section was 72.7%, while the specificity was 100%. The PPV was 100% and the NPV was 89.66%. However, the sensitivity of intraoperative crush smear cytology was only 26.67%, but the specificity was 100%.
Test Positive samples Sensitivity Specificity
Calcofluor white stain 7 63.6%100%
Intraoperative crush smear cytology 4 26.6%100%
Frozen section 8 72.7%96.3%
Table 2: Comparison of calcofluor white direct microscopy, intraoperative crush smear cytology and frozen section as test methods as compared to the gold standard (histopathology; n=11)
Discussion
Rhinocerebral mucormycosis is reported to be exclusively seen in immunocompromised hosts, including uncontrolled diabetes mellitus, hematologic cancers, and solid organ or hematopoietic stem cell transplants . A study on 929 patients, done over 60 years, found poorly controlled diabetes mellitus as the most common risk factor (36%), followed by hematologic cancers (17%) and hematopoietic stem cell or solid organ transplant (12%) .
The development of CAM, uncontrolled diabetes, and systemic corticosteroid (over)treatment were the major risk factors in India and other countries . Nonjudicious use of antifungal agents, especially voriconazole as a prophylactic agent, further enhanced the possibility of mucormycosis. COVID‑19 infection itself is a predisposing risk factor as it causes lymphopenia and endothelitis, favoring the fungal invasion [3, 13]. While disease manifestation and incidence vary considerably between different geographical regions , rhino-orbital-cerebral and pulmonary diseases are the most frequent clinical presentations of CAM in patients with COVID-19.
To the best of our knowledge, this is probably the first study using CFW stain wet mount on intraoperative crush smears for the detection of rhinocerebral mucormycosis. Other authors have studied its use for the detection of fungal agents in different tissues and samples.
It was Darken, who in 1961 reported the uptake of CFW by growing yeast and higher fungi . Hageage and Harrington outlined the application of CFW to demonstrate hyphae and yeasts in clinical specimens in 1984 . Monheit et al. used it on lung and soft tissues for the intraoperative diagnosis of fungal infection. In 1986, Monheit et al. recommended the incorporation of CFW with routine Papanicolaou stain on cytology for the identification of the fungus. They reported that it increases both the sensitivity and specificity of the stain . Also, it doesn’t interfere with the diagnostic value of the Papanicolaou stain, and CFW does not distort or destroy cytopathologic features as done by GMS and PAS stains . As CFW fluoresces under UV light, it is easily visible in tissue samples. Also, it can be used on all types of samples, such as fresh, fixed, and frozen tissue and paraffin-embedded tissue blocks . Evans blue dye can be used as a counterstain with CFW, as it reduces the background fluorescence.
The hyphae of Mucor are variable in size, averaging 15-20 μm in diameter, and are aseptate.However, the creases and folds in the distorted hyphae resemble septa. When the tissue is handled or cut, the fluid contents of the hyphae escape, and they appear as hollow tubes or as twisted ribbons because of the folding of the empty tubes. The hyphae of Aspergillus are thin, slender, and have acute-angled branching .
In our study, out of 37 samples, 11 (30%) were positive for fungus while 26 (70%) were negative. Of the 11 cases positive for fungus, 10 (90%) showed the presence of mucormycosis, while one case (10%) was positive for Aspergillus.Of 11 positive samples, seven (63.6%) were positive with direct microscopy by CFW stain wet mount.When a CFW wet mount is observed under a brightfield microscope, the presence of cellular debris and necrotic material may obscure the fungus and cause a delay in identification, but under the fluorescent microscope, the fungus fluoresces brightly and is sharply delineated (Figures 4, 5).
Figure 4: Calcofluor white stain wet mount under light microscopy showing irregular non-septate hyphae in the background of debris (original magnification 400X).
Figure 5: Calcofluor white stain wet mount under light microscopy showing irregular non-septate hyphae in the background of debris (original magnification 400X).
Intraoperative crush smear cytology detected fungus in only four (36.3%) samples. The presence of necrotic cellular material, causing difficulty in the visualization of hyphae, could be one of the reasons for the lower detection rate. Also, due to time constraints, the smear was not observed as thoroughly as needed.
Eight (72.7%) samples showed fungus on frozen tissue. One sample was positive with the CFW stain wet mount, but did not show any fungus on frozen tissue. However, the routine paraffin sections from the same sample showed the presence of hyphae. This could be due to the fact that hyphae were hidden in the deeper part of the tissue, which was not exposed during sections taken for frozen sections, while for CFW stain, we selected preferably the necrotic, blackish tissue. Two samples were negative by both CFW stain wet mount and intraoperative crush smear cytology but showed fungus on frozen tissue and were reported as positive. This could be due to selective sampling for the wet mount.On frozen sections, mucor stains faint blue, and many twisted or bizarre forms are seen.Oedematous nasal mucosa combined with water artifacts in frozen can be confused with fungus when mixed with inflammatory cells and debris. Hence, an experienced pathologist is needed for the interpretation of frozen sections.
In the present study, the sensitivity of the CFW stain wet mount was 63.6%, while the specificity was 100% for the detection of fungus. The PPV was 100%, while the NPV was 86.7%. Four cases that were missed could be due to sampling error, as only a part of the tissue received was observed.The sensitivity of the frozen section was 72.7%, while the specificity was 100%. The PPV was 100% and the NPV was 89.66%. However, a prospective study of mucormycosis in north India by Bala et al. mentions that the detection rate of mucormycosis by CFW stain (84%) was better than histopathology (58%) and culture (61%) .
Hofman et al. demonstrated that the frozen section is a specific and sensitive method for early diagnosis of rhino-cerebral mucormycosis and improves the prognosis of this disease. Cytological imprints performed on surgical biopsies were less effective than the frozen section for the immediate diagnosis of mucormycosis because of the presence of hemorrhages and necrotic tissue. Inexperienced pathologists may miss the diagnosis of rhinocerebral mucormycosis during the frozen section when the Mucorales show some bizarre forms and are twisted .
Sharma et al. studied CFW stain for the diagnosis of corneal keratitis and found CFW stain to be superior over KOH with a positivity rate of 100% and 87.5%, respectively . Lackner et al. also recommended the use of optical brighteners, blankophor, and CFW in clinical specimens for a rapid diagnosis of mucormycosis .Shakthi et al. in their comparative study of KOH, CFW stain, and fungal culture in cases of onychomycosis found that the sensitivity and specificity of KOH wet mount was 83.05%, whereas CFW microscopy showed a sensitivity of 92.3%. They found CFW has high sensitivity and thus is the best method to detect fungal agents from clinically suspected onychomycosis cases .
Dass et al. in their study on 150 patients with onychomycosis found that direct microscopy with KOH mount, calcofluor mount, and mycological culture showed positive results in 84 (56%), 95 (63.33%), and 59 (39.33%), respectively. Mycological culture was the least sensitive method, and CFW was the most sensitive method among the three methods.
Punjabi et al. in their study to evaluate the staining efficacy of CFW and acridine orange for the detection of Candida species found that there is a statistically significant association between CFW fluorescent microscopy and the swab culture technique. The sensitivity and specificity of CFW were 85% and 50%, respectively . Jahanshahi et al. also found similar results in their study of Candida infection in cases of oral squamous cell carcinoma .
As CFW also stains other tissue elements like collagen, keratin, and elastin, it needs to be differentiated from threads, mucus strands, epithelial cells, and crystals. The type and intensity of fluorescence help to differentiate them. The advantages of CFW stain are that it is a simple and rapid technique. Light microscopy of a crush smear takes a minimum of four to five minutes for screening each slide, while under fluorescent light, calcofluor white stain preparation can be viewed or screened easily on a low power objective (10X), and it can be completed within one minute. Background or debris material can also be easily differentiated based on fluorescent properties.Calcofluor white is safe to use in the laboratory and does not have any health hazards. The disadvantage is that it is expensive due to the need for a fluorescent microscope. Also, every setup may not have the cryostat for the frozen section and fluorescent microscope in the same laboratory. The tissue sections need to be evaluated as soon as they are stained, as with time, the amount of fluorescence reduces .
The limitation of our study is the small sample size (as thankfully the second wave receded). However, the work is ongoing and being used for diagnosis in bronchoalveolar lavage (BAL) and sputum samples too.
Conclusions
Calcofluor white stain is a time-saving, specific, and sensitive technique for rapid detection of rhinocerebral mucormycosis. It can be used as an adjunct for intraoperative diagnosis of fungal infections, but not as a primary method of diagnosis. This stain is versatile and can be used with tissue sections, direct smears, and other clinical specimens. When the fungus is detected by CFW stain, a rapid diagnosis can be given even in the absence of classical forms on frozen sections. Rhinocerebral mucormycosis is a potentially life-threatening disease with severe morbidity. Hence, the use of all available modalities for rapid and early diagnosis is recommended.
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QUARTILE.EXC Function
QUARTILE.EXC Function
An Excel Statistical Function that excludes the 0 and the 100th percentile from the result, i.e., the minimum and maximum values returning only the three quartile values.
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Last Updated:November 14, 2024
In This Article
What Is The QUARTILE.EXC Function?
QUARTILE.EXC Function Formula
How To Use The QUARTILE.EXC Function In Excel?
QUARTILE.INC Vs. QUARTILE.EXC Function
QUARTILE Vs. QUARTILE.INC Function
What Is The QUARTILE.EXC Function?
What Is The QUARTILE.EXC Function?
QUARTILE.EXC Function Formula
How To Use The QUARTILE.EXC Function In Excel?
QUARTILE.INC Vs. QUARTILE.EXC Function
QUARTILE Vs. QUARTILE.INC Function
What is the QUARTILE.EXC Function?
The QUARTILE.EXC function in Excel excludes the 0 and the 100th percentile from the result, i.e., the minimum and maximum values returning only the three quartile values.
Wait, what?
What do we mean by a quartile, though? Well, quartiles are the numerical values that group a range of values into three quartiles.
The concept of Quartile is often compared with the percentile, which divides a dataset into 100 equal parts. Percentile has no universal definition. However, it can be interpreted as the number below which the ‘x’ percentage of data falls.
The three different quartiles that the function returns are:
1st Quartile(Q1)—Q1 represents the 25th percentile, implying 25% of the values in the data fall within the numerical value returned.
2nd Quartile(Q2)— Q2 represents the 50th percentile or the median/middle value below which 50% of the values fall.
3rd Quartile(Q3)—Q3 represents the 75th percentile, implying that 75% of the data values fall within the numerical value returned.
Apart from these three values, there are also the maximum and the minimum values, which are the 100th percentile and 0th percentile values, respectively.
However, the function does not return those values and does not even consider them in evaluating those numbers in the dataset.
Perhaps we should see an example to understand how the function works. But first, let’s see the process and its syntax.
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The QUARTILE.EXC function is an Excel Statistical Function used to calculate the exclusive quartile of a dataset, which divides the data into four equal parts.
Users provide two arguments to the QUARTILE.EXC function: the array of data points and the quartile number (1 to 4). It returns the value corresponding to the specified quartile.
Potential errors that users might encounter when using the QUARTILE.EXC function, such as providing non-numeric values or referencing cells with incorrect data, and how to address them effectively.
QUARTILE.EXC function finds applications in finance, economics, and other fields where analyzing data distributions and identifying percentile values are important decision-making, risk management, and forecasting.
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QUARTILE.EXC function Formula
The QUARTILE.EXC function is categorized as a Statistical function that returns the three quartiles for a range of values.
Previously, Excel offered only the QUARTILE function, but it was replaced by QUARTILE.EXC and QUARTILE.INC to provide more control to the users over what the function can do.
Both functions were introduced in the Excel 2010 version and have been present in all subsequent versions.
The syntax for the function is:
=QUARTILE.EXC(array,quart)
Where
array- (required) reference to the range of value in the dataset.
quart- (required) refers to the quartile value we intend to return.
The quart argument will only accept three values, as illustrated below:
Values| Value | Quart |
--- |
| 1 | Q1 - 25th percentile of values lie below this number |
| 2 | Q2 - 50th percentile of values lies below this number |
| 3 | Q3 - 75th percentile of values lies below this number |
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How to interpret the Quartile value?
Suppose you have the range of values as 1, 3, 5, 8, 9, 10, 11, 13, 15, and 18. We will try to find all three quartile values using the function.
Since the numbers are already arranged in ascending order, we know the minimum and maximum values, also called the 0th and the 100th percentile values.
However, we do not have to be concerned about those since the function does not consider those two values in the data range to evaluate the three-quartile values.
The tricky part is the interpretation of the quartile values. We get the Q1 as 4.5, which means that 25% of the values are below this number. The Q2 is 9.5 while the Q3 is 13.5, representing 50% and 75% of values in the dataset below those numbers.
We have eight numbers in the dataset apart from minimum and maximum values, i.e., excluding those.
Q2 is the median value from the dataset. Since we do not have the odd number of observations in the dataset, the result gets interpolated between 9 and 10, which are the two middle values. Thus, the resultant quartile value is equal to 9.5.
Since we know that Percentiles and Quartiles are closely related, assume that the remaining range of values is a total percentage of 100.
As there are eight observations, we will divide 100 by 8, which gives us each number weightage as 12.5% in the given dataset.
Now, coming to the Q1, which is equal to 4.5, has only 3 as the number below. Since a single number constitutes 12.5%, if there were two numbers, the weightage would have been equal to 25%, which defies the quart logic of values below the 25th percentile.
The same reasoning applies for Q3 from the bottom, where just one value is above the 75th percentile, equal to 15, while the rest fall below the Q3 or the 75th percentile.
It is important to remember that the values eventually get interpolated if there are many observations after excluding the minimum and maximum values from the dataset.
How to use the QUARTILE.EXC Function in Excel?
Suppose that you have the test scores for the students, as illustrated below:
We will use the formula=QUARTILE to get the first quartile value. For example, EXC($C$3:$C$12,1)in cell F5 gives us the result of 39.25.
To get the other values, we will copy the formula and change the quart argument, which provides us with the result:
How do we interpret the result?
Firstly, we will reorder the test scores in ascending order. This step is unnecessary; however, it will help you understand whether or not the quartile values returned are correct.
Select the range C3:C12 and press the keyboard keys Alt + H + S + S, which will give you the dialog box, as illustrated below:
Select 'Expand the selection' and then click on 'Sort.' The test scores sorted in ascending order would look like this:
First, we will cancel out the dataset's maximum and minimum values since the function does not include them in the quartiles' calculation.
Thus, we have eight values in the dataset, which means the quartile values would ultimately be interpolated since we have an even number of observations in the dataset.
The Q1, i.e., 39.25, represents the values below the 25th percentile. Dividing the observations by 100, we find that each number contributes 12.5% to the total. Thus, the test score of 37, forming 12.5%, is the only value under Q1.
Q2 is 62 while Q3 is 82.5, meaning that the 50th and 75th percentile values fall below that number(not even equal to that number).
Let's see another example where we might end up with an odd number of observations after the minimum and maximum values cancel out. Suppose you remove the dataset's test scores for 'Jenna Walton.'
The quartile values immediately change to different values. Here, we find that Q2 does not interpolate but is the median value, which is equal to 61 scored by High Larson.
However, Q1 and Q3 have values that result from interpolation between the two numbers. For example, Q1 is equal to 38.5 while Q3 is equal to 83.
Note
Using the function, you don't need to sort the data in ascending order. However, in our example, the data is arranged in ascending order to better understand what happens behind the scene when you use the formula in Excel.
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QUARTILE.INC vs. QUARTILE.EXC Function
The QUARTILE.INC was another function that Microsoft added as a newer version of Quartile in Excel 2010.
If the QUARTILE.EXC function excludes the minimum and maximum values from the dataset. Then, you can already imagine the QUARTILE.INC can do that.
Yes, you are right. It will 'include' the minimum and maximum values in our dataset and find the quartile values based on the entire range referenced in the formula.
The syntax for the function is
=QUARTILE.INC(array, quart)
where
array- (required) reference to the range of values in the dataset.
quart- (required) refers to the quartile value we intend to return.
Here, the quart argument can accept five different values, as illustrated below:
Values| Value | Quart |
--- |
| 0 | Minimum Value |
| 1 | Q1 - 25th percentile of values lie below this number |
| 2 | Q2 - 50th percentile of values lies below this number |
| 3 | Q3 - 75th percentile of values lies below this number |
| 4 | Maximum Value |
Let's see an example to see whether both functions return the same results. Suppose you have the data as illustrated below:
We will use the formula=QUARTILE.INC($C$3:$C$12, E5)in cell G5 and drag it down to cell G9. Similarly, we will use the formula =QUARTILE.EXC($C$3:$C$12, E5) in cell H5 and drag it to the H9 cell to give the result:
Firstly, we get the NUM! Error for QUARTILE.EXC since it does not return the minimum and maximum value or identify the quart arguments zero and four.
Suppose we arrange the salaries in ascending order. In that case, we get:
We see that Q2 equals $71,151.50 in both cases, implying 50 percent of the data falls below that number, which is fitting, as shown below:
The Q1 equals $64,366.75, where roughly 25 percent of the values from the dataset fall under that number.
The Q3 value says that 75 percent of the values from the dataset fall under this number, equal to $78,065.50.
The above interpretation is valid only for the QUARTILE.INC function and would change for its counterpart.
QUARTILE vs. QUARTILE.INC Function
The QUARTILE was the first native function used to calculate the quartile values in Excel. However, later on, in Excel 2010, it was replaced by two of its successors, i.e., QUARTILE.EXC and QUARTILE.INC
The function works similarly to QUARTILE.INC by returning the three quartiles and the minimum and maximum values.
Even though the function is replaced by its two successors, you can still find it under the compatibility section forbackward compatibility.
The syntax for the function is
=QUARTILE(array, quart)
Where
array- (required) reference to the range of values in a dataset.
quart- (required) refers to the quartile value we intend to return.
Again, the quart accepts only five different values, as illustrated below:
Values| Value | Quart |
--- |
| 0 | Minimum Value |
| 1 | Q1 - 25th percentile of values lie below this number |
| 2 | Q2 - 50th percentile of values lies below this number |
| 3 | Q3 - 75th percentile of values lies below this number |
| 4 | Maximum Value |
Suppose you have the test scores for students, as shown below:
Here, we will use the formula=QUARTILE($C$3:$C$12, E5)in cell G5 and drag it down to cell G9. Similarly, we will use the formula=QUARTILE.INC($C$3:$C$12, E5)in cell H5 to H9, which not surprisingly gives similar results as:
Since the QUARTILE function is the shadow of the newer version, both parts yield identical results.
However, it is still advisable to use the newer versions of the function as we might never know when Excel will discontinue the usage of the compatibility version.
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Binomial Theorem
The Binomial Theorem states that for real or complex, , and non-negativeinteger,
where is a binomial coefficient. In other words, the coefficients when is expanded and like terms are collected are the same as the entries in the th row of Pascal's Triangle.
For example, , with coefficients , , , etc.
Contents
[hide]
1 Proof
1.1 Proof via Induction
1.2 Proof using calculus
2 Generalizations
2.1 Proof
3 Usage
4 See also
Proof
There are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:
We can write . Repeatedly using the distributive property, we see that for a term , we must choose of the terms to contribute an to the term, and then each of the other terms of the product must contribute a . Thus, the coefficient of is the number of ways to choose objects from a set of size , or . Extending this to all possible values of from to , we see that , as claimed.
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Proof via Induction
Given the constants are all natural numbers, it's clear to see that . Assuming that , Therefore, if the theorem holds under , it must be valid. (Note that for )
Proof using calculus
The Taylor series for is for all .
Since , and power series for the same function are termwise equal, the series at is the convolution of the series at and . Examining the degree- term of each, which simplifies to for all natural numbers.
Generalizations
The Binomial Theorem was generalized by Isaac Newton, who used an infiniteseries to allow for complex exponents: For any real or complex, , and ,
.
Proof
Consider the function for constants . It is easy to see that . Then, we have . So, the Taylor series for centered at is
Usage
Many factorizations involve complicated polynomials with binomial coefficients. For example, if a contest problem involved the polynomial , one could factor it as such: . It is a good idea to be familiar with binomial expansions, including knowing the first few binomial coefficients.
See also
Combinatorics
Multinomial Theorem
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Lesson 9. The Complete Guide to Time in English
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Time in English
Knowing how to tell the time in the correct way will make you look like a native when you are visiting an English-speaking country. Recommended background: The Complete Guide of Numbers in English. Let’s learn about the time in English here!
Asking the Time in English
The most basic way to ask the time in English is:
What time is it?
Telling the Time in English
Hour
We use ‘of the clock’ (o’clock) to express the full hour. In simple, it is used when there is no minute available. Examples:
One o’clock.
1:00.
Three o’clock.
3:00.
Twelve o’clock.
12:00.
To answer in a full sentence, use the dummy subject: it is (shortened as it’s).
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It is seven o’clock.
It is 7:00.
Minutes
If there are minutes after the hour, do not use o’clock. There are two common ways to tell time in English: an easy and a general way.
Easy Way
The hours come first and followed by the minutes.
Structure: Number (Hours) + Number (Minutes)
Examples:
Six forty-five.
6:45.
Eight fifty-eight.
8:58.
For minutes from :01 until :09, pronounce ‘the number zero’ as ‘the letter o’.
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Three o two.
3:02.
Ten o five.
10:05.
General Way
In a general way, people use the word ‘past’ or ‘to’ between minutes and hours. The word ‘past’ is used for the minutes between 01 and 30, while ‘to’ is for minutes between 31 and 59.
Structure: Number (Minutes) + Past/To + Number (Hours)
Examples:
Ten past eleven.
11:10.
Eleven past ten.
10:11.
Since the minutes are mentioned earlier, we do not need to use the letter o for minutes from :01 to :09.
Five past three.
3:05.
Nine past four.
4:09.
Seconds
For expressing the full time (hours, minutes, and seconds included), use the following:
Structure: Number (Hours) + Number (Minutes) + Number (Seconds) + Seconds
Examples:
Eleven fifty-nine fifty-nine seconds.
11:59:59.
Three fifteen thirty-five seconds.
03:15:35.
a.m. and p.m.
To make the time distinction clear, English generally use the 12-hour clock system. We use ante meridiem (a.m.) to refer time between midnight 00:00 and midday 12:00 (before midday), while post meridiem (p.m.) refers to the time between midday 12:00 and midnight 12:00 (after midday). Therefore:
8.00 a.m. = 8 o’clock in the morning (08:00).
8.00 p.m. = 8 o’clock in the evening (20:00).
12.00 a.m. = 12 o’clock in the night (midnight).
12.00 p.m. = 12 o’clock in the afternoon (midday).
Note that we use the colon (:) to separate the 24-hour clock system. On the other hand, dot (.) is used to separate the 12-hour clock system.
Quarter and Half
Another way to express time is by using quarter and half. Quarter means 15 minutes, while half means 30 minutes in term of time. Thus:
6:15can be expressed by:
a quarter past six
six fifteen
fifteen past six
6:30can be expressed by:
half past six
six thirty
thirty past six
6:45can be expressed by:
a quarter to seven
six forty-five
forty-five past six
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11812 | https://en.wikipedia.org/wiki/Divergence_(computer_science) | Divergence (computer science) - Wikipedia
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1 DefinitionsToggle Definitions subsection
1.1 Rewriting
1.2 Denotational semantics
1.3 Concurrency theory
2 See also
3 Notes
4 References
Divergence (computer science)
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From Wikipedia, the free encyclopedia
Computation which does not terminate or terminates in an exceptional state
"Terminating" redirects here. For other uses, see Termination.
In computer science, a computation is said to diverge if it does not terminate or terminates in an exceptional state.: 377 Otherwise it is said to converge.[citation needed] In domains where computations are expected to be infinite, such as process calculi, a computation is said to diverge if it fails to be productive (i.e. to continue producing an action within a finite amount of time).
Definitions
[edit]
Various subfields of computer science use varying, but mathematically precise, definitions of what it means for a computation to converge or diverge.
Rewriting
[edit]
In abstract rewriting, an abstract rewriting system is called convergent if it is both confluent and terminating.
The notation t ↓ n means that t reduces to normal form n in zero or more reductions, t↓ means t reduces to some normal form in zero or more reductions, and t↑ means t does not reduce to a normal form; the latter is impossible in a terminating rewriting system.
In the lambda calculus an expression is divergent if it has no normal form.
Denotational semantics
[edit]
In denotational semantics an object functionf: A → B can be modelled as a mathematical functionf:A∪{⊥}→B∪{⊥}{\displaystyle f:A\cup {\perp }\rightarrow B\cup {\perp }} where ⊥ (bottom) indicates that the object function or its argument diverges.
Concurrency theory
[edit]
See also: Communicating sequential processes §Failures/divergences model
In the calculus of communicating sequential processes (CSP), divergence occurs when a process performs an endless series of hidden actions. For example, consider the following process, defined by CSP notation: C l o c k=t i c k→C l o c k{\displaystyle Clock=tick\rightarrow Clock} The traces of this process are defined as: traces(C l o c k)={⟨⟩,⟨t i c k⟩,⟨t i c k,t i c k⟩,…}={t i c k}∗{\displaystyle \operatorname {traces} (Clock)={\langle \rangle ,\langle tick\rangle ,\langle tick,tick\rangle ,\ldots }={tick}^{}} Now, consider the following process, which hides the tick event of the Clock process: P=C l o c k∖t i c k{\displaystyle P=Clock\setminus tick} As P{\displaystyle P} cannot do anything other than perform hidden actions forever, it is equivalent to the process that does nothing but diverge, denoted d i v{\displaystyle \mathbf {div} }. One semantic model of CSP is the failures-divergences model, which refines the stable failures model by distinguishing processes based on the sets of traces after which they can diverge.
See also
[edit]
Infinite loop
Termination analysis
Notes
[edit]
^C.A.R. Hoare (Oct 1969). "An Axiomatic Basis for Computer Programming"(PDF). Communications of the ACM. 12 (10): 576–583. doi:10.1145/363235.363259. S2CID207726175.
^Baader & Nipkow 1998, p.9.
^Pierce 2002, p.65.
^Roscoe, A.W. (2010). Understanding Concurrent Systems. Texts in Computer Science. doi:10.1007/978-1-84882-258-0. ISBN978-1-84882-257-3.
References
[edit]
Baader, Franz; Nipkow, Tobias (1998). Term Rewriting and All That. Cambridge University Press. ISBN9780521779203.
Pierce, Benjamin C. (2002). Types and Programming Languages. MIT Press.
J. M. R. Martin and S. A. Jassim (1997). "How to Design Deadlock-Free Networks Using CSP and Verification Tools: A Tutorial Introduction" in Proceedings of the WoTUG-20.
This computer science article is a stub. You can help Wikipedia by expanding it.
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11813 | https://arxiv.org/pdf/1304.3154 | arXiv:1304.3154v4 [math.CO] 24 Dec 2014
Journal of Combinatorics
Volume 0, Number 0, 1, 2014
An infinite cardinal version of Gallai’s Theorem for colorings of the plane
Jeremy F. Alm
We generalize a result of Tibor Gallai as follows: for any finite set of points S in the plane, if the plane is colored in finitely many colors, then there exist 2 ℵ0 monochromatic subsets of the plane homothetic to S. Furthermore, we prove an even stronger result for n-dimensional Euclidean space.
Introduction
1.1. History
In the late 1930s, Tibor Gallai proved the following result, which is a gener-alization to multiple dimensions of Van der Waerden’s celebrated theorem on arithmetic progressions in the integers:
Theorem 1.1 (Gallai’s Theorem on Zn). Let S be a finite subset of Zn.Then any finite coloring of Zn contains a monochromatic subset homothetic to S, i.e. a monochromatic subset of the form a + bS for some a ∈ Zn and
b ∈ Z.
Gallai himself did not publish the result; a proof first appeared in the literature in [ 4]. (For an interesting history, see Chapter 42 of Soifer’s book .) A derivation of Theorem 1.1 from the Hales-Jewett Theorem can be found in [ 3]. A Euclidean-space-version of Gallai’s Theorem is as follows:
Theorem 1.2 (Gallai’s Theorem on En). Let S be any finite subset of En.Then any finite coloring of En contains a monochromatic subset homothetic to S, i.e. a monochromatic subset of the form a + bS for a ∈ En and b ∈ R.
These two theorems are fundamental in Euclidean Ramsey theory. It seems natural, however, to suspect that Theorem 1.2 can be strengthened.
arXiv:
Keywords: Gallai’s Theorem, homothety, infinite cardinal, combinatorial geom-etry 2010 MSC codes 05D10, 52C10, 05C50
12 J. F. Alm Surely, we can find more than one monochromatic homothetic copy, no? In this paper, we generalize Theorem 1.2 by showing that there must in fact be uncountably many monochromatic homothetic copies. To the best of our knowledge, the first uncountability result of this type appeared in [ 2]:
Theorem 1.3. For every 2-coloring of E2, there exist uncountably many
r ∈ R+ such that there exists a monochromatic equilateral triangle of side length r.
The author’s first attempt to strengthen Gallai’s Theorem resulted in the following two theorems from [ 1], the second of which strengthens Theorem
1.3 :
Theorem 1.4. Let S be a finite configuration of points in the integer lattice
Z2. In any finite coloring of the plane E2, there exist 2ℵ0 monochromatic homothetic copies of S.
Theorem 1.5. For any finite coloring of E2 there exist 2ℵ0 monochromatic equilateral triangles, i.e., subsets homothetic to {(0 , 0) , (1 , 0) , (1 /2, √3/2) }.
In fact, the argument given in [ 1] for Theorem 1.5 will work for any 3-point configuration in E2. The problem for an arbitrary finite subset S ⊂ E2
was left open. The main result of the present paper is a solution to that problem:
Theorem 1.6. Let S be any finite set of points in E2. If the points of
E2 are colored in finitely many colors, then there exist 2ℵ0 pairwise-disjoint monochromatic subsets of E2 homothetic to S.
1.2. Notation
For a positive integer m, we let [ m] denote {1, 2, 3, . . . , m }. We let En denote
n-dimensional Euclidean space, and R the real line. For n > 1, elements of
En will be denoted with boldface, as in y, and 0 will denote the origin. Subsets of R and En will be denoted with capital script (“mathcal”) letters. For X ⊆ En, a ∈ R, and d ∈ En, let aX + d = {ax + d : x ∈ X } .Recall that for all n ∈ Z+, the cardinality of En is 2 ℵ0 .
Main Results
We will prove the following two results, whose conjunction is equivalent to Theorem 1.6 .Infinite cardinal Gallai 3
Theorem 2.1. Let S be any finite set of points in E2, not all on a line. If the points of E2 are colored in finitely many colors, then there exist 2ℵ0
pairwise-disjoint monochromatic subsets of E2 homothetic to S.
Theorem 2.2. Let S be any finite set of points in R. If the points of R
are colored in finitely many colors, then there exist 2ℵ0 pairwise-disjoint monochromatic subsets of R homothetic to S.Proof of Theorem 2.1 . Let S ⊂ E2, with |S| = n. Assume without loss of generality that S contains the origin 0, so that S = {0, y1, . . . , yn−1}. Let
A be a 2-by-( n − 1) matrix whose jth column is yj (considered as a 2-by-1 column vector):
A := [ y1y2 · · · yn−1].
Let T : En−1 → E2 be given by T (v) = Av. Note that since the points of
S do not lie all on a line, the set {y1, . . . , yn−1} spans E2 and hence T is surjective. Let c ∈ Z+, and let χ : E2 → [c] be a coloring of the plane with c
colors. Define χ′ : En−1 → [c] by χ′(v) = χ(T (v)). Let {u1, . . . , un−1} be the standard basis for En−1, and let U = {0, u1, . . . , un−1}.We must now partition En−1 into 2 ℵ0 parts. Consider the collection of cosets En−1/Zn−1. Each coset C ∈ En−1/Zn−1 is of the form Zn−1 + e for some e ∈ [0 , 1) n−1. Hence C is a translation of Zn−1. It will prove useful later in the proof to index the cosets by translation vectors from [0 , 1) n−1,as in Ce. Note that each e ∈ [0 , 1) n−1 uniquely identifies a coset. Each coset Ce is colored by χ′, so by Gallai’s Theorem on Zn we may conclude that there exists some monochromatic subset of Ce homothetic to
U. This subset will have the form aU + d + e, where a ∈ Z+ and d ∈ Zn−1.Now define
Se = T (aU + d + e) = {T (v) : v ∈ aU + d + e}.
It is clear that Se is monochromatic (under χ). To see that Se is homothetic to S, we recall that T was defined by T (v) = Av, where the columns of A
were the yi’s, so T sends the unit basis vectors in U to the yi’s in S. Hence
T (U) = S, and we have
T (aU + d + e) = aT (U) + T (d + e) = aS + T (d + e).
Therefore Se is homothetic to S. Since there are 2 ℵ0 cosets, we have 2 ℵ0
monochromatic homothetic copics of S.4 J. F. Alm It remains to show that, while many of these Se may overlap, it is possible to find 2 ℵ0 of them that are pairwise-disjoint. Since A has rank 2, the image under T of [0 , 1) n−1 contains a parallel-ogram P. Let Y = {m(yi − yj ) : m ∈ Z, 1 ≤ i ≤ j < n }, and define an equivalence relation ∼ on P as follows:
p ∼ q if and only if p − q ∈ Y .
Let P0 be a subset of P formed by choosing exactly one representative from each equivalence class. Since each equivalence class is at most countable,
|P 0| = 2 ℵ0 . For each p ∈ P 0, choose ep ∈ [0 , 1) n−1 such that T (ep) = p, and let E = {ep : p ∈ P 0}. Note that T is 1-1 on E, and that |E| = |P 0| = 2 ℵ0 .Now for each ep ∈ E there is a coset Cep that contains a monochromatic set homothetic to U of the form aU + d + ep, where a ∈ Z+ and d ∈ Zn−1.Since Z+ and Zn−1 are countable, there must be some a0 ∈ Z+ and d0 ∈
Zn−1 such that for 2 ℵ0 of the cosets Cep , the monochromatic subset of Cep
homothetic to U has the form a0U + d0 + ep.Let E′ be the subset of E consisting of all ep such that a0U + d0 + ep is monochromatic; note that |E ′| = 2 ℵ0 . We show that for any two distinct ep,
eq ∈ E ′, Sep and Seq are disjoint. Now
Sep = T (a0U + d0 + ep) = a0S + T (d0) + p
and
Seq = T (a0U + d0 + eq) = a0S + T (d0) + q.
Suppose there is some point common to both sets. Then there are yi, yj ∈ S
such that
a0yi + T (d0) + p = a0yj + T (d0) + q.
Rearranging yields
a0(yi − yj ) = q − p.
Hence p ∼ q, a contradiction. Therefore the sets Sep , for ep ∈ E ′, are pairwise disjoint. Since |E ′| = 2 ℵ0 ,the proof is complete.
The next proof will run in parallel to the previous one.
Proof of Theorem 2.2 . Let S ⊂ R, with |S| = n. Assume without loss of generality that 0 is the smallest number in S, so that S contains 0 < x 1 <x2 < . . . < x n−1. Let A be a 1-by-( n − 1) row vector
A := [ x1x2 · · · xn−1].Infinite cardinal Gallai 5Let T : En−1 → R be given by T (v) = Av.Let c ∈ Z+, and let χ : R → [c] be a coloring of the real line with c
colors. Define χ′ : En−1 → [c] by χ′(v) = χ(T (v)). Let {u1, . . . , un−1} be the standard basis for En−1, and let U = {0, u1, . . . , un−1}.As in the proof of Theorem 2.1 , we have for each e ∈ [0 , 1) n−1 a monochro-matic subset of aU + d + e and its image Se under T . The image of [0 , 1) n−1
under T contains an interval ( a, b ). Let Y = {m(xi − xj ) : m ∈ Z, 1 ≤ i ≤
j < n }, and define an equivalence relation ∼ on ( a, b ) as follows:
p ∼ q if and only if p − q ∈ Y .
Let ( a, b )⋆ be a subset of ( a, b ) containing exactly one representative from each equivalence class. Define ep, E, and E′ as in the proof of Theorem 2.1 .Then by the same argument given at the end of that proof, we have that for distinct ep and eq ∈ E ′, Sep and Seq are disjoint.
It is not hard to see how to extend this argument to k-dimensional Euclidean space. We have the following:
Theorem 2.3. Let n, k ∈ Z+, with n > k . Let S be an n-element subset of
Ek, whose points are not all contained in any (k−1) -dimensional hyperplane. If the points of Ek are colored in finitely many colors, then there exist 2ℵ0
pairwise-disjoint monochromatic subsets homothetic to S.
Instead of proving Theorem 2.3 , we will state and prove an even stronger version. Note that in the proof of Theorem 2.1 , the sets Sep all had the form
a0S + d + e, where a0 was a positive integer. Let us call the multiplicative constant a0 the dilation factor . We can use the fact that these dilation factors are integer-valued to get a stronger result.
Theorem 2.4. Let n, k ∈ Z+, with n > k . Let S be an n-element subset of
Ek, whose points are not all contained in any (k−1) -dimensional hyperplane. If the points of Ek are colored in finitely many colors, then there exist 2ℵ0
positive numbers r such that for each such r, there exist 2ℵ0 pairwise-disjoint monochromatic subsets homothetic to S with dilation factor r.Proof of Theorem 2.4 . Let S ⊂ Ek, with |S| = n. Assume without loss of generality that S contains the origin 0, so that S = {0, y1, . . . , yn−1}. Let
A be a k-by-( n − 1) matrix whose jth column is yj (considered as a k-by-1 column vector):
A := [ y1y2 · · · yn−1].6 J. F. Alm Let T : En−1 → Ek be given by T (v) = Av. Note that since the points of
S do not lie on any ( k − 1)-dimensional hyperplane, the set {y1, . . . , yn−1}
spans Ek and hence T is surjective. Let c ∈ Z+, and let χ : Ek → [c] be a coloring of Ek with c colors. Define
χ′ : En−1 → [c] by χ′(v) = χ(T (v)). Let {u1, . . . , un−1} be the standard basis for En−1, and let U = {0, u1, . . . , un−1}.Now we partition En−1 into cosets. Let r ∈ R+, and rZ := {rn : n ∈ Z}.We consider En−1/(rZ)n−1. Each coset C is a translation of ( rZ)n−1 by a vector e ∈ [0 , r )n−1. Again, we index the cosets by the vectors e.Each coset Ce is colored by χ′, so by Gallai’s Theorem on Zn we may conclude that there exists some monochromatic subset of Ce homothetic to rU. This subset will have the form ra U + d + e, where a ∈ Z+ and
d ∈ (rZ)n−1. Now define
Se = T (ra U + d + e) = {T (v) : v ∈ ra U + d + e}.
Again, Se is a monochromatic. We have
Se = T (ra U + d + e) = ra S + T (d + e),
so Se is homothetic to S with dilation factor ra .It remains to show that, while many of these Se may overlap, it is possible to find 2 ℵ0 of them that are pairwise disjoint. Since A has rank k, the image under T of [0 , r )n−1 contains a k-dimensional parallelotope P. Let Y = {m(yi − yj ) : m ∈ Z, 1 ≤ i ≤ j < n }, and define an equivalence relation ∼ on P as follows:
p ∼ q if and only if p − q ∈ Y .
Let P0 be a subset of P formed by choosing exactly one representative from each equivalence class. As before, for each p ∈ P 0 we choose ep ∈ [0 , r )n−1
such that T (ep) = p, and let E = {ep : p ∈ P 0}.Now for each ep ∈ E there is a coset Cep which contains a monochromatic set homothetic to U of the form ra U + d + e, where a ∈ Z+ and d ∈
(rZ)n−1. Again, there are a0 ∈ Z+ and d0 ∈ (rZ)n−1 such that for 2 ℵ0 of the cosets Cep , the monochromatic subset of Cep homothetic to U has the form
ra 0U + d0 + ep.Let E′ be the subset of E consisting of all ep such that ra 0U + d0 + ep is monochromatic. Then by the same argument given at the end of the proof of Theorem 2.1 , we have that Sep and Seq are disjoint for distinct ep and
eq.Infinite cardinal Gallai 7Now let R be a maximal subset of R+ that is linearly independent over
Z. For each r ∈ R there is some a0 such that there are 2 ℵ0 pairwise-disjoint subsets Sep with dilation factor ra 0. Since R is linearly independent over Z,no two dilation factors ra 0 coincide. Since R is maximal, |R| = 2 ℵ0 . This concludes the proof.
Finally, we note that the method of using linear transformations affords us an easy derivation of Gallai’s Theorem on En from Gallai’s Theorem on
Zn. We leave this derivation to the reader, since the necessary details are given above.
Acknowledgments
The author wishes to thank Jacob Manske for very many productive con-versations, and for critiquing an earlier draft of this paper. This paper is dedicated to my son Thomas, who was born three days after final accep-tance.
References
J. Alm, Toward an uncountable analogue of Gallai’s Theorem for color-ings of the plane , Geombinatorics 22 (2012), no. 3, 45–50. N. Brown, N. Dunfield, and G. Perry, Colorings of the plane I , Geombi-natorics 3 (1993), no. 2, 24–31. MR1241304 (94j:05052)
R. Graham, B. Rothschild, and J. Spencer, Ramsey theory , second ed., Wiley-Interscience Series in Discrete Mathematics and Optimization, John Wiley & Sons Inc., New York, 1990, A Wiley-Interscience Pub-lication. MR1044995 (90m:05003)
R. Rado, Note on combinatorial analysis , Proc. London Math. Soc. (2)
48 (1943), 122–160. MR0009007 (5,87a)
A. Soifer, The mathematical coloring book, Mathematics of color-ing and the colorful life of its creators , Springer, New York, 2009,
MR2458293 (2010a:05005) |
11814 | https://www.pbslearningmedia.org/resource/muen-math-ns-absvalue/absolute-value/ | Absolute Value | PBS LearningMedia
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This animated Math Shorts video explains absolute value, as demonstrated on the number line and through a real-life example. In the accompanying classroom activity, students watch the video and then play a game in which they move a penny along a number line in positive and negative directions. As they play, they use absolute value to track the total distance that the penny moves. To get the most from the lesson, students should be comfortable determining distance between positive and negative numbers on a number line.
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Common Core State Standards (7)
Grades 3-5
CCSS.Math.Content.5.NBT.A.3.b See anchor statement
Compare two decimals to thousandths based on meanings of the digits in each place, using >, =, and < symbols to record the results of comparisons.All “CCSS.Math.Content.5.NBT.A.3.b” resources
See all (2) Grades 3-5 standards
Grades 6-8
CCSS.Math.Content.6.NS.C.7.c See anchor statement
Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation.All “CCSS.Math.Content.6.NS.C.7.c” resources
See all (5) Grades 6-8 standards
College and Career Readiness Standards for Adult Education (8)
Grades 3-5
NS.A.3.c See anchor statement
Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation.All “NS.A.3.c” resources
See all (8) Grades 3-5 standards
Grades 6-8
NS.A.3.c See anchor statement
Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation.All “NS.A.3.c” resources
See all (8) Grades 6-8 standards
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Teaching Tips | Absolute Value
Learning Outcomes
Students will be able to
use absolute value to calculate the total distance moved in positive and negative directions on a number line
define the mathematical terms below
Common Core State Standards: 6.NS.C.7.c
Vocabulary: Absolute value, number line, negative numbers, absolute value notation
Materials: Per pair: pencils, paper, a red die and a blue die, a penny; 10 Steps Game handout, Number Line handout
Procedure:
Introduction and Video (5–10 minutes, whole group)
Prompt to find out what students know about absolute value, listening for connections to the number line, negative numbers, and absolute value notation.
If needed, give a brief introduction in preparation for the video.
Show students the video, pausing as follows:
At 1:02, ask students, How many blocks did Lucy walk forward? How many did she walk back? How do you know? Give students a moment to talk with a partner before resuming the video to reveal the answer.
10 Steps Game (15 minutes, pairs)
Distribute the materials and review the game instructions. Talk through the sample game as students use the penny to “act out” the game on their boards. When the game is over, ask students:
Where did the penny end up?
Which player, blue or red, moved the penny more? How do you know?
If students do not bring it up, point out that one approach is to add the moves each player made. Another is to note that the penny ends up on a negative number, so red must have moved more.
As students play a few games, circulate to check that they are distinguishing between total distance traveled and distance traveled in a particular direction.
Pose the following to students ready for a greater challenge: Suppose two people play the 10 Steps Game. Blue and red each roll three times. The penny ends up on 0 at the end of the game. Make up three possible rolls for each player.
Conclusion (5 minutes, whole group)
Engage students in reflecting on absolute value as the total distance moved by comparing it with displacement from the origin—that is, where the penny ends up. Begin by asking them about the games they played:
Did anyone play a game in which each player moved the penny the same amount?
Where did the penny end up?
Then, ask about a hypothetical situation:
I played a game in which the penny moved 12 steps. Can you tell whether red or blue rolled more? Can you tell if the penny ended up on a positive or negative number? Why or why not?
Suppose I moved 12 steps and the penny ended up at 2. Now can you tell whether blue or red moved more? Why or why not?
Activity Extension: Play a life-sized game of 10 Steps, with students taking turns standing in for the penny. Play outdoors with the number line marked in chalk or in the gym or classroom with the number line marked with tape on the floor.
This activity is based on work developed at TERC. |
11815 | https://www.chemteam.info/GasLaw/Gas-Boyle-Problems-only.html | ChemTeam: Gas Law Problems - Boyle's Law
Boyle's Law
Examples and problems only
Problems 1-15Ten examples
Problems 16-30KMT & Gas Laws Menu
Ten examples
Example #1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard pressure?
Example #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is 10.0 L?
Example #3: 9.48 L of a gas was at an unknown pressure. However, at standard pressure, its volume was measured to be 8.00 L. What was the unknown pressure?
Example #4: If we have 6.00 cm 3 of gas at a pressure of 10.0 N/cm 2 and we increase the pressure to 20.0 N/cm 2, what volume will the gas occupy?
Example #5: What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a cylinder whose volume is 26.0 liters?
Example #6: The volume of a gas is 6.10 L, measured at 1.00 atm. What is the pressure of the gas in mmHg if the volume is changed to 9.74 L?
Example #7: At 46.0 °C a sample of ammonia gas exerts a pressure of 5.30 atm. What is the pressure when the volume of the gas is reduced to one-eighth (0.125) of the original value at the same temperature?
Example #8: In order to measure the volume of a piece of apparatus, a chemist filled a 750. mL flask with 46.65 kPa pressure of gas, then expanded it into the apparatus. The final pressure was 14.95 kPa. Calculate the total volume occupied by the gas.
Example #9: Boyle's Law deals with the relationship between two of the variables (of four) that describe gas behavior. Which two variables are held constant in Boyle's Law problems?
(a) pressure/moles
(b) temperature/volume
(c) pressure/volume
(d) temperature/moles
(e) volume/moles
Example #10: A balloon contains 7.20 L of He. The pressure is reduced to 2.00 atm and the balloon expands to occupy a volume of 25.2 L. What was the initial pressure exerted on the balloon?
Bonus Example #1: An evacuated flask A, which has a volume of 30 mL, is attached to a second flask B containing an ideal gas at a pressure of 5 atm. When the two flasks are connected the pressure in the system drops to 2 atm. Calculate the volume of flask B.
Bonus Example #2: Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 39.0 mL. The first bulb has a volume of 56.0 mL and contains 5.92 atm of argon, the second bulb has a volume of 250.0 mL and contains 1.28 atm of neon, and the third bulb has a volume of 37.0 mL and contains 8.50 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm?
Problems 1-15
Problem #1: A gas occupies 12.3 liters at a pressure of 40.0 mmHg. What is the volume when the pressure is increased to 60.0 mmHg?
Problem #2: If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 2.50 atm?
Problem #3: To what pressure must a gas be compressed in order to get into a 3.00 cubic foot tank the entire weight of a gas that occupies 400.0 cu. ft. at standard pressure?
Problem #4: A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure becomes 3.00 atm?
Problem #5: A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes 15.0 L?
Problem #6: 500.0 mL of a gas is collected at 745.0 mmHg. What will the volume be at standard pressure?
Problem #7: Convert 350.0 mL at 740.0 mmHg to its new volume at standard pressure.
Problem #8: Convert 338 L at 63.0 atm to its new volume at standard pressure.
Problem #9: Convert 273.15 mL at 166.0 kPa to its new volume at standard pressure.
Problem #10: Convert 77.0 L at 18.0 mmHg to its new volume at standard pressure.
Problem #11: When the pressure on a gas increases, will the volume increase or decrease?
Problem #12: If the pressure on a gas is decreased by one-half, how large will the volume change be?
Problem #13: A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the pressure is increased to 1.25 atm.
Problem #14: 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00 atm?
Problem #15: 400.0 mL of a gas are under a pressure of 800.0 torr. What would the volume of the gas be at a pressure of 1000.0 torr?
Bonus Problem: A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 liters. If the balloon is filled with 2.0 liters of helium at sea level (101.3 kPa), and rises to an altitude at which the boiling temperature of water is only 88 degrees Celsius, will the balloon burst?
Problems 16-30
Problem #16: 4.00 L of a gas are under a pressure of 6.00 atm. What is the volume of the gas at 2.00 atm?
Problem #17: A gas occupies 25.3 mL at a pressure of 790.5 mmHg. Determine the volume if the pressure is reduced to 0.804 atm.
Problem #18: A sample of gas has a volume of 12.0 L and a pressure of 1.00 atm. If the pressure of gas is increased to 2.00 atm, what is the new volume of the gas?
Problem #19: A container of oxygen has a volume of 30.0 mL and a pressure of 4.00 atm. If the pressure of the oxygen gas is reduced to 2.00 atm and the temperature is kept constant, what is the new volume of the oxygen gas?
Problem #20: A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mmHg. Find the volume of the nitrogen when its pressure is changed to 400.0 mmHg while the temperature is held constant.
Problem #21: A 40.0 L tank of ammonia has a pressure of 8.00 atm. Calculate the volume of the ammonia if its pressure is changed to 12.0 atm while its temperature remains constant.
Problem #22: Two hundred liters of helium at 2.00 atm and 28.0 °C is placed into a tank with an internal pressure of 600.0 kPa. Find the volume of the helium after it is compressed into the tank when the temperature of the tank remains 28.0 °C.
Problem #23: You are now wearing scuba gear and swimming under water at a depth of 66.0 ft. You are breathing air at 3.00 atm and your lung volume is 10.0 L. Your scuba gauge indicates that your air supply is low so, to conserve air, you make a terrible and fatal mistake: you hold your breath while you surface. What happens to your lungs? Why?
Problem #24: Solve Boyle's Law equation for V 2.
Problem #25: Boyle's Law deals with what quantities?
(a) pressure/temperature
(b) pressure/volume
(c) volume/temperature
(d) moles/pressure
(e) none of these
Problem #26: A 1.5 liter flask is filled with nitrogen at a pressure of 12 atmospheres. What size flask would be required to hold this gas at a pressure of 2.0 atmospheres?
Problem #27: 300 mL of O 2 are collected at a pressure of 645 mm of mercury. What volume will this gas have at one atmosphere pressure?
Problem #28: How many cubic feet of air at standard conditions (1.00 atm.) are required to inflate a bicycle tire of 0.50 cu. ft. to a pressure of 3.00 atmospheres?
Problem #29: How much will the volume of 75.0 mL of neon change if the pressure is lowered from 50.0 torr to 8.00 torr?
Problem #30: A tank of helium has a volume of 50.0 liters and is under a pressure of 2000.0 p.s.i.. This gas is allowed to flow into a blimp until the pressure in the tank drops to 40.00 p.s.i. and the pressure in the blimp is 30.00 p.s.i.. What will be the volume of the blimp?
Bonus Problem: A spherical weather balloon is constructed so that the gas inside can expand as the balloon ascends to higher altitudes where the pressure is lower. If the radius of the spherical balloon is 2.5 m at sea level where the pressure is 1.004 x 10 5 Pa, what will be the radius at an altitude of about 10 km where the pressure of the gas is 2.799 x 10 4 Pa? For simplicity, assume the temperature has not changed.
Problems 1-15Ten examples
Problems 16-30KMT & Gas Laws Menu |
11816 | https://artofproblemsolving.com/community/c2437h1033423s3_polynomial_game?srsltid=AfmBOopDZs6qG7Uv29ZIyTthaMUp1cNLTdKiJkK1kuLn6qsYuU6YoeYF | An Excursion through Mathematical problems : Polynomial game
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Polynomial game
by boywholived, Oct 6, 2012, 12:02 PM
A polynomial has three undetermined coefficients denoted by stars. The players A and B move alternately, replacing a star by a real number until all stars are replaced. A wins if all zeros of the polynomial are complex.B wins if at least one zero is real. Show that B can win in spite of his only second move.
This problem is not yet unsolved(on AOPS) as on Oct 06 2012
The above is the link where I have posted. Check it to find any recent improvement
algebraPolynomialsTheory of equations
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11817 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8749206/ | Type 2 leprosy reaction presenting as a monoarthritis post multidrug therapy - PMC
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. 2022 Jan 6;27:e01386. doi: 10.1016/j.idcr.2022.e01386
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Type 2 leprosy reaction presenting as a monoarthritis post multidrug therapy
Isabela Maria Bernardes Goulart
Isabela Maria Bernardes Goulart
a School of Medicine, Federal University of Uberlândia (UFU), 1720, Pará avenue, Uberlândia, MG, Brazil
b Postgraduate Program in Health Sciences, School of Medicine, Federal University of Uberlândia (UFU), 94, Capricórnio Street, Uberlândia, MG, Brazil
c National Reference Center for Sanitary Dermatology and Leprosy, School of Medicine, Federal University of Uberlândia (UFU), 94, Capricórnio Street, Uberlândia, MG, Brazil
Find articles by Isabela Maria Bernardes Goulart
a,b,c, Marcela Araujo de Oliveira Santana
Marcela Araujo de Oliveira Santana
c National Reference Center for Sanitary Dermatology and Leprosy, School of Medicine, Federal University of Uberlândia (UFU), 94, Capricórnio Street, Uberlândia, MG, Brazil
Find articles by Marcela Araujo de Oliveira Santana
c,⁎, Willian Vargas Tenório da Costa
Willian Vargas Tenório da Costa
a School of Medicine, Federal University of Uberlândia (UFU), 1720, Pará avenue, Uberlândia, MG, Brazil
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a, Matthew Martin Pavelka
Matthew Martin Pavelka
d Indiana University School of Medicine, 620 N. Chestnut Street Holmstedt Hall 135, Terre Haute, IN 47809, USA
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d, Bruno de Carvalho Dornelas
Bruno de Carvalho Dornelas
a School of Medicine, Federal University of Uberlândia (UFU), 1720, Pará avenue, Uberlândia, MG, Brazil
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a School of Medicine, Federal University of Uberlândia (UFU), 1720, Pará avenue, Uberlândia, MG, Brazil
b Postgraduate Program in Health Sciences, School of Medicine, Federal University of Uberlândia (UFU), 94, Capricórnio Street, Uberlândia, MG, Brazil
c National Reference Center for Sanitary Dermatology and Leprosy, School of Medicine, Federal University of Uberlândia (UFU), 94, Capricórnio Street, Uberlândia, MG, Brazil
d Indiana University School of Medicine, 620 N. Chestnut Street Holmstedt Hall 135, Terre Haute, IN 47809, USA
⁎
Corresponding author. marcela20santana@ufu.br marcela20santana@gmail.com
Received 2021 Dec 14; Accepted 2022 Jan 5; Collection date 2022.
© 2022 The Authors
This is an open access article under the CC BY-NC-ND license (
PMC Copyright notice
PMCID: PMC8749206 PMID: 35036324
Abstract
Type 2 leprosy reaction, or erythema nodosum leprosum (ENL), involves a complex interaction between the host's immune system and Mycobacterium leprae. It may occur before, during, or after treatment and have a variable clinical presentation involving different body systems, such as skin, osteoarticular, kidneys, and others. Thus, the differential diagnosis, depending on its clinical presentation, can be broad and challenging. The authors report a case of a severe monoarthritis during a type 2 reaction after the multidrug therapy (MDT) was discharged and the investigation of the differential diagnoses.
Keywords: Leprosy, Mycobacterium leprae, Erythema nodosum, Arthritis
Introduction
Type 2 leprosy reaction, or erythema nodosum leprosum (ENL), is previously described in the literature as an immune response mediated by immune complexes that cause neutrophilic vasculitis , . Recent studies discussing the role of immune complexes in the pathogenesis of the disease reaction are still uncertain . However, there is evidence of in situ increase of interleukin-6 (IL-6), interleukin-8 (IL-8), and transforming growth factor beta 1 (TGF-β 1). Peripheral increases of tumor necrosis factor-alpha (TNF-α), interleukin 1-beta (IL-1β), and TGF-β are also observed, showing that there is a series of complex interactions in the immune system related to the pathogenesis of this reaction , , , , .
Leprosy reactions can occur before, during, or after leprosy treatment with multidrug therapy (MDT) . Among the risk factors related to its development are lepromatous leprosy and multibacillary disease, pregnancy, and lactation . Some authors also relate an increased risk of developing reactions with the presence of other infections after vaccinations, some medications, and other factors , . The possibility of a Sars-cov-2 infection during the pandemic is an additional concern for leprosy centers, especially due to the possible increased risk for a type 2 leprosy reaction . In general, studies show that reactions are more common in the first year of treatment, when there is a great bacillary destruction by an intense neutrophilic inflammatory infiltrate involving the regressing of the granuloma .
Clinical manifestations may include fever, asthenia, cutaneous and subcutaneous erythematous nodules, arthritis, synovitis, lymphadenopathy, neuritis, iridocyclitis, epididymo-orchitis, hepatitis, glomerulonephritis, among others , .
The authors report a case of severe monoarthritis due to a type 2 reaction after the end of MDT for leprosy.
Case presentation
A 57-year-old male was diagnosed with lepromatous leprosy and treated with MDT. Treatment ended 4 months before the arthritis onset. The patient was closely monitored, during and after the treatment, due to a type 2 reaction state, presenting with erythematous nodules, tibial nerve neuritis, and previous necrotizing erythema nodosum scars on the skin surface.
In the first type 2 reaction episode, thalidomide (400 mg/day), prednisone (60 mg/day), and clofazimine (at an anti-inflammatory dose of 300 mg/day) were prescribed. A decrease in the number and severity of reaction episodes was observed after a year. Pentoxifylline (800 mg/day) was added to the therapy concomitantly with a gradual decrease in the use of corticosteroids.
At the end of the first year, there was a recrudescence of the type 2 reaction, with mild erythema nodosum, right knee arthralgia, and tibial nerve neuritis. Prednisone, thalidomide, and pentoxifylline were prescribed at the same previous doses. In some weeks, it was possible to observe an improvement of the cutaneous condition. After 2 months there was a worsening of pain and edema in the right knee joint, local inflammatory signs, and limitation to active and passive movement on the physical exam. Radiograph exam showed periarticular soft tissue edema, decreased medial joint space, mild osteopenia, and calcification of the tibial artery.
Three arthrocentesis were performed, with an average removal of 60–80 ml of purulent secretion. The material was sent for biochemical examination, which revealed the presence of group III synovial fluid (very inflammatory,> 20,000 cells). Gram, cultures for bacteria, fungi, and Mycobacterium tuberculosis were requested, and all results were negative. The patient underwent arthroscopy of the right knee, which visualized synovial membrane hypertrophy, chondromalacia with the destruction of cartilage tissue, and purulent effusion. Biopsy and continuous local irrigation were performed. The biopsy showed synoviocyte hyperplasia, increased subintimal cellularity, presence of polymorphonuclear cells, specific Virchowian infiltration with some granular bacilli in its interior, obliterative and necrotizing vasculitis, and intracavitary exudate, all characteristic of a synovitis of erythema nodosum leprosum.
Discussion
Arthritis associated with a type 2 reaction may precede, be concomitant with, or appear after ENL. It commonly manifests as polyarthritis or, oligo and monoarthritis in rare cases. This arthritis has an abrupt onset, intense pain, morning stiffness, and functional limitation. The most affected joints are knees, proximal interphalangeal joints, metacarpophalangeal joints, wrists, ankles, metatarsophalangeal joints, and elbows , .
Treatment can be done with thalidomide (100–400 mg/per day), an immunomodulator derived from glutamic acid, and a potent selective inhibitor of TNF-α, in association with corticosteroids (1 mg/kg/day) for cases of neuritis, epididymo-orchitis, iridocyclitis, glomerulonephritis. Pentoxifylline (800–1200 mg/day) and clofazimine (300 mg/day, an anti-inflammatory dose) have been shown to help stabilize refractory reactions , .
Joint involvement in leprosy can be very variable. Possible presentations are the occurrence of arthritis due to direct Mycobacterium leprae infection unrelated to the reaction, swollen hands, and feet syndrome, Charcot neuroarthropathy, tenosynovitis, enthesitis, cryoglobulinemic vasculitis, in addition to possible other infections when the patient is using immunosuppressive therapy , . The presentation may be similar to other connective tissue diseases, making differential diagnosis difficult, which may delay diagnosis and adequate therapy , .
A high index of suspicion is required for cases of arthritis related to leprosy and leprosy reactions. In many cases, the differential diagnosis is difficult and extensive, especially when arthritis is included in the initial presentation of the disease . The mycobacterium can be difficult to isolate, and some diagnostic techniques, such as PCR, are not always available. In such cases, extensive investigation to exclude other causes and histopathological analysis may be necessary. Tests, such as rheumatoid factor and antinuclear antibodies, can often be positive in leprosy and make the diagnosis difficult .
In the case reported, the patient presented with improvement in the erythema nodosum and worsening of joint symptoms, an atypical evolution. There was also a concern for possible immunosuppression due to the chronic use of prednisone. All those factors led to the diagnostic hypothesis of septic arthritis, which was less likely after the negative results. Exams for possible rheumatoid arthritis and acute gout were also ordered and came back negative. The diagnosis of arthritis related to a type 2 reaction was confirmed by studying the synovial fluid and histopathology of the synovial membrane. As the destruction of the synovia was intense, there was only clinical improvement after an invasive local approach with continuous irrigation and maintenance of the instituted systemic therapy.
The occurrence of monoarthritis in patients with type 2 leprosy reaction after the end of MDT is not well documented in the literature. Studies on leprosy reactional arthritis are related to the presentation of cases during an ENL outbreak in patients during leprosy treatment , , .
Some papers have reported the study of the synovial membrane in arthritis secondary to leprosy reaction , and have verified that the synovial fluid of arthritis can be from group I (inflammatory –< 2000 cells), group II (inflammatory – 2000–20,000 cells) or group III (very inflammatory –> 20,000 cells). Mycobacterium leprae may be present in approximately 1/3 of the analyzed fluids. It has been shown that patients present group I fluids in the first episode and group II or III fluids in later episodes, or vice versa , . In the reported case, there was a predominance of group III synovial fluid and specific Virchowian infiltrates with granular bacilli in its interior, probably because it was a case of recurrent ENL in a patient discharged from polychemotherapy more than 4 months ago when the episode of arthritis occurred. These findings corroborate with previous studies regarding the synovial membrane, which show more than half of the cases were mild reactional synovitis. The most constant acute change was intracavitary exudation with a strong neutrophilic component, often with vacuoles and positive bacilloscopy in its interior , .
These studies have allowed us to explain the pathophysiology of arthritis in the type 2 reaction as a process identical to that of ENL of the skin. The synovial membrane of these patients is invaded by foci of Virchowian cells, and in these places there would be an antigen-antibody-complement reaction with the formation of insoluble complexes, triggering the inflammation in the synovia , . The intensity of arthritis would depend on the quantity and quality of specific infiltrate foci in the synovial membrane and would explain the different types of synovial fluid and histological aspects of the synovia , .
Reactional arthritis in leprosy patients can simulate osteoarthritis, rheumatoid and rheumatic arthritis, acute gout, septic arthritis, and other rheumatic diseases . Therefore, a careful differential diagnosis should be investigated to rule out the possibility of association with other rheumatic and infectious diseases. The presence of inflammation during an episode of ENL can affect the synovium of a large joint leading to severe reactional arthritis, even after drug discharge from MDT. It is important to have high diagnostic suspicion and prompt investigation to start drug treatment or invasive approach to avoid greater morbidity from this disease, which is already related to a high prevalence of disabling sequelae and great stigma.
CRediT authorship contribution statement
Isabela Maria Bernardes Goulart: Data collection, Writing – original draft, Writing – review & editing. Marcela Araujo de Oliveira Santana: Data collection, Writing – original draft, Writing – review & editing. Willian Vargas Tenório da Costa: Data collection, Writing – original draft, Writing – review & editing. Matthew Martin Pavelka: Writing – original draft, Writing – review & editing. Bruno de Carvalho Dornelas: Data collection, Writing – original draft, Writing – review & editing.
Ethical approval
None.
Consent
Written informed consent was obtained from the patient for publication of this case report and accompanying images. A copy of the written consent is available for review by the Editor-in-Chief of this journal on request.
Funding Statement
This work was supported by: Projeto fortalecimento da rede de atenção integral e integrada aos pacientes com hanseníase da região do Triângulo Mineiro e Alto Paranaíba, TED: 157/2019 – FN.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Contributor Information
Isabela Maria Bernardes Goulart, Email: imbgoulart@gmail.com.
Marcela Araujo de Oliveira Santana, Email: marcela20santana@ufu.br, marcela20santana@gmail.com.
Willian Vargas Tenório da Costa, Email: william.da@ufu.br.
Matthew Martin Pavelka, Email: matpavel@iu.edu.
Bruno de Carvalho Dornelas, Email: bruno.dornelas@ufu.br.
Data Availability
The data used in this study can be accessed upon contact with the corresponding author.
References
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Associated Data
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Data Availability Statement
The data used in this study can be accessed upon contact with the corresponding author.
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11818 | https://blog.csdn.net/u013066730/article/details/123377627 | 直线的极坐标方程-CSDN博客
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直线的极坐标方程
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本文介绍了如何在极坐标系中表示直线，通过举例说明不同类型的直线方程，包括与极轴平行、垂直的直线，以及一般直线的极坐标形式。利用直角三角形的几何关系和正弦定理推导出极坐标方程，并强调了在直线上任取点进行验证的重要性。
转自：直线的极坐标方程 - 知乎
简单回顾一下，在直角坐标系里，水平的叫x 轴，竖直的叫y 轴，两轴交点叫原点。我们描述一个点，会用它到两轴的距离 x、y 来表示①。
极坐标系②呢，只有一条射线，左端点叫极点，这条射线叫极轴。描述点的时候，会用这个点到极点的距离 ρ，以及该点和极点的连线 l，与极轴的夹角 θ 来表示。
虽然你现在不喜欢极坐标系，但是你不了解一个人的时候，也很难特别喜欢他。
今天先拿直线来说，你可能不喜欢，但要熬过认识它的第一阶段。
从最简单的开始：
01
如果我们把它放在原点，那你会说它的斜率是
这条直线的方程就是 y=x。
发现没有，你习惯性的表示 x 和 y 之间的关系，从而得到直线方程。
那如果把它放在极坐标系里呢，一样的思路，去找 ρ 和 θ 的关系就好。
这条直线上所有的点，到极轴的夹角都是
而这些点到极点的距离是任意的，所以直线方程就是
，吗？
觉不觉的有问题？我们随便取点验证一下。
所以其实，这条直线的方程的是由两条射线构成的，完整写出来就是：
我们发现，随便取点，是一个非常重要的方法！
用“在直线上任取点 A(ρ,θ)，观察图形，找 ρ 和 θ 的关系”这种方法，我们试试其他直线的方程怎么写：
02
一条过点 P(2,0)，且垂直于极轴的直线方程？
在直线 l 上任取一点 A(ρ,θ)，在图上找到 A 点的极径 ρ 和极角 θ。
从直角△OPA 里面可以发现：
，
这就是 ρ 和 θ 的关系，简单整理一下就好了：
所以构造直角三角形，是找几何关系的重要方法。同类型的特殊直线，还有平行于坐标轴的，你试一下：
03
在直线 l 上任取一点 A(ρ,θ)，在图上找到 A 点的极径 ρ 和极角 θ。
构造直角三角形，找 ρ 和 θ 的关系：
整理一下：
好，说了这么多特殊情况，一条长相随意的直线方程呢？
04
设直线 l 过点 P(r,α)，与极轴的夹角为 β，求直线 l 的极坐标方程。
题目的图
标上几何量的图
找到任意点 A(ρ,θ)的图
这次没有直角三角形了，三角形里还有哪些边角关系？——正弦定理！
找到 ρ 和 θ 的关系的图
在△OPA 中，由正弦定理得：
05
其实，如果你愿意抛开自己的求知欲去看前面的几个问题，你也可以把它们全都转化成直角坐标系里的问题来解，再用公式④转化成极坐标方程：
2.一条过点 P(2,0)，且垂直于极轴的直线方程？
4.设直线 l 过点 P(r,α)，与极轴的夹角为 β，求直线 l 的极坐标方程。
ps
①直角坐标系里，坐标有正负你知道的，我喜欢把符号理解成方向。
②关于极坐标系的概念，在这篇推文里：极坐标系。
③有的时候，为了表示方便，也可以像直角坐标系里那样，给 ρ 加上符号表示方向，这样的话 ρ∈R 就行了，直线写成：
.
④详情推荐两篇推文：直角坐标与极坐标的转化，直角坐标方程与极坐标方程的转化。
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参数 方程 和 极坐标
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27.参数 方程 和 极坐标 坐标 xxx 和 yyy 不直接相关，而是通过一个公共参数相联系 27.1 参数 方程 圆 方程 x2+y2=9 x^2+y^2=9 x2+y2=9 圆的参数化 方程 {x=3cos(t)其中0≤t≤2πy=3sin(t)\left{ \begin{aligned} x & = & 3\cos(t) &\ &&&\quad其中0\leq t \leq2\pi\ y & = & 3\sin(t) \ \end{ali
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用户可能需要一些基本例子来理解 极坐标 方程 的形式，比如圆、直线 或螺线在 极坐标 中的表示。例如，r = a表示圆，θ = constant表示射线，这些例子能帮助用户直观理解。此外，阿基米德螺线和对数螺线作为常见例子，可能...
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11819 | https://brainly.com/question/11472038 | [FREE] Simplify \sqrt{-40}. - brainly.com
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Simplify −40.
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Answer: 2 i 10
Step-by-step explanation: To simplify the square root of -40, think of the square root of -40 as the square root of -1 × the square root of 40.
The square root of -1 can be represented by the imaginary number i. So we have i × the square root of 40 or i 40.
Next, root 40 breaks down breaks down to 2 root 10. We can see this by setting up a factor tree for 40. 40 is 2 x 20, 20 is 2 x 10, and 10 is 2 x 5. We have a pair of 2's in our factor tree and we have 2 x 5 at the bottom.
So we have i x 2 root 10 or 2 i 10.
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To simplify −40, we recognize it as i⋅40, which simplifies to 2 i 10. This uses the concept of imaginary numbers, where i=−1.
Explanation
To simplify −40, we start by recognizing that the square root of a negative number can be expressed using the imaginary unit i, where i=−1. We can rewrite −40 as follows:
Factor the negative part: (\sqrt{-40} = \sqrt{-1 \cdot 40} = \sqrt{-1} \cdot \sqrt{40} = i \cdot \sqrt{40}.
Simplify the square root of 40: Next, we need to simplify 40. We can factor 40 to make its square root simpler:
The prime factorization of 40 is 40=4⋅10=2 2⋅10.
Therefore, we have (\sqrt{40} = \sqrt{4 \cdot 10} = \sqrt{4} \cdot \sqrt{10} = 2 \cdot \sqrt{10}.
Combine the results: Now we substitute back into our expression:
−40=i⋅40=i⋅(2⋅10)=2 i 10.
Thus, the simplified form of −40 is 2 i 10.
Examples & Evidence
An example of using this concept is simplifying −9 to 3 i because −9=−1⋅9=i⋅3=3 i.
The properties of square roots, particularly for negative numbers, can be derived from the definition of complex numbers, where the imaginary unit i is defined as −1. All steps and simplifications abide by established mathematical rules.
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if you know complex numbers:
sqrt (-40)
sqrt(-1) sqrt(40)
sqrt(-1) sqrt(4) sqrt(10)
i 2 sqrt(10)
2i sqrt(10)
If you do not know complex numbers:
it cannot be simplified because it is negative
no real solution
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### Free Mathematics solutions and answers
Community Answer simplify (5 square root 40)
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Community Answer 5.0 1 simplify to radical form the square root of 40 and then divide by 6
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11820 | https://www.quora.com/Why-are-the-integrals-of-odd-functions-equal-to-zero | Something went wrong. Wait a moment and try again.
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Why are the integrals of odd functions equal to zero?
Reynaldo Loya
B.S. in Mathematics & Computer Science, Texas A&M University - Kingsville (Graduated 2021)
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Upvoted by , 45 years as a university mathematics professor/ Ph.D. in mathematics · Author has 335 answers and 535.7K answer views
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6y
An odd function is defined such that f(−x)=−f(x). With that in mind, we can take the integral of an arbitrary odd function between one interval to its negative version: ∫a−af(x)dx
I=∫a−af(x)dx=∫0−af(x)dx+∫a0f(x)dxu=−x⟹du=−dx⟹u1=a⟹u2=0=−∫0af(−u)du+∫a0f(x)dx=−∫a0f(u)du+∫a0f(x)dx=−∫a0f(x)dx+∫a0f(x)dx=0
Note: So long as our integral has bounds and we evaluate it, we can use any variable; th
An odd function is defined such that f(−x)=−f(x). With that in mind, we can take the integral of an arbitrary odd function between one interval to its negative version: ∫a−af(x)dx
I=∫a−af(x)dx=∫0−af(x)dx+∫a0f(x)dxu=−x⟹du=−dx⟹u1=a⟹u2=0=−∫0af(−u)du+∫a0f(x)dx=−∫a0f(u)du+∫a0f(x)dx=−∫a0f(x)dx+∫a0f(x)dx=0
Note: So long as our integral has bounds and we evaluate it, we can use any variable; that’s why I switch back to x at the end.
Related questions
If the integral of a function is zero, what does it tells us about the function?
What is the integral of odd function?
Why is the integral of an even function over a closed interval always equal to zero?
Does the Integral of an even function always have to be an odd function?
If an integral is zero, is the integrand necessarily odd?
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
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Author has 6.8K answers and 52.8M answer views
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3y
Your question is too generalised!
Sometimes it is true, sometimes not.
This will explain it clearly in a nice diagram…
Look at this is a function y = f(x)
.
Your question is too generalised!
Sometimes it is true, sometimes not.
This will explain it clearly in a nice diagram…
Look at this is a function y = f(x)
.
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1y
The integral of an odd function over a symmetric interval around the origin is equal to zero due to the properties of odd functions and the concept of symmetry.
Definitions
Odd Function: A function f(x) is called odd if it satisfies the condition:
f(−x)=−f(x)for all x.
Symmetric Interval: An interval [−a,a] is symmetric about the origin.
Proof
To understand why the integral of an odd function over a symmetric interval is zero, consider the integral of an odd function f(x) over the interval [−a,a]:
∫a−af(x)dx.
Using the property of odd functions, we can transform th
The integral of an odd function over a symmetric interval around the origin is equal to zero due to the properties of odd functions and the concept of symmetry.
Definitions
Odd Function: A function f(x) is called odd if it satisfies the condition:
f(−x)=−f(x)for all x.
Symmetric Interval: An interval [−a,a] is symmetric about the origin.
Proof
To understand why the integral of an odd function over a symmetric interval is zero, consider the integral of an odd function f(x) over the interval [−a,a]:
∫a−af(x)dx.
Using the property of odd functions, we can transform the integral:
Split the integral into two parts:
∫a−af(x)dx=∫0−af(x)dx+∫a0f(x)dx.
Change the variable in the first integral. Let u=−x, then du=−dx, and when x=−a, u=a, and when x=0, u=0. Thus:
∫0−af(x)dx=∫0af(−u)(−du)=∫a0f(−u)du.
Since f(x) is odd, we have f(−u)=−f(u), so:
∫a0f(−u)du=∫a0−f(u)du=−∫a0f(u)du.
Now substituting back into the original integral:
∫a−af(x)dx=−∫a0f(u)du+∫a0f(u)du=0.
Conclusion
Thus, the integral of an odd function over a symmetric interval [−a,a] is zero:
∫a−af(x)dx=0.
This property highlights the cancellation of the areas under the curve of the function in the positive and negative halves of the interval.
Muhammad Abul Fazal
B.Sc. (Hons.) in Mathematical Physics, University of Alberta (Graduated 2023)
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Author has 101 answers and 381.2K answer views
·
7y
In explanatory words, in odd functions the area under interval [-a,a] can be divided into two equally opposite sign of integral i.e. area under increasing curve is equal to area under decreasing curve.
Here is an example of an even function curve i.e. sinx u can clearly see that area in interval [-pi/2,pi/2] is equally divided at origin and the curve from [-pi/2,0] is increasing and also curve in [0,pi/2] is also increasing so area under them add up to double.
Now contrast this with cosx which is an odd function.
You can see that curve from [-pi/2,0] increases while in [0,pi/2] decreases. The int
In explanatory words, in odd functions the area under interval [-a,a] can be divided into two equally opposite sign of integral i.e. area under increasing curve is equal to area under decreasing curve.
Here is an example of an even function curve i.e. sinx u can clearly see that area in interval [-pi/2,pi/2] is equally divided at origin and the curve from [-pi/2,0] is increasing and also curve in [0,pi/2] is also increasing so area under them add up to double.
Now contrast this with cosx which is an odd function.
You can see that curve from [-pi/2,0] increases while in [0,pi/2] decreases. The interval [-pi/2,pi/2] being divided into half by origin. Their area cancel ups.
But remember this is when you consider signs. If you want just plain area then in both cases halves are added up to double.
Now lets see its formal proof:
Hope that helped!
Related questions
How do I prove that improper integrals for odd functions like sin and 1/x are divergent (from negative infinity to infinity) and not equal to zero?
Why does an odd function in an integral = 0?
If the integral at any given point equals 0, shouldn't the integral of any function always equal 0?
Can an integral of a function be equal to zero in calculus? How would you prove it?
How can we prove that if the integral of a product of two functions is zero, then one function will be a constant multiple of the other?
Arjun Prasad
Studied at Kendriya Vidyalaya
·
5y
Originally Answered: Why is the integral of an odd function zero?
·
The integral of an odd function results for a 0 for integrating over a symmetrical interval.
And odd function is like-
f(x)+f(-x)=0
Some odd functions are -x^3,x^5,sin x.
Say you want to integrate sin x( which is an odd function) over the SYMMETRICAL INTERVAL [-π,π] then your integral results in 0. A symmetrical interval is like of the form [-A,A] where A is real.
This is the graph of sine function. The shaded area is the interval [-π,π]. When you are integrating over this interval, practically you are finding the area of this shaded portion. Since one shaded area amounts to the negative of the oth
The integral of an odd function results for a 0 for integrating over a symmetrical interval.
And odd function is like-
f(x)+f(-x)=0
Some odd functions are -x^3,x^5,sin x.
Say you want to integrate sin x( which is an odd function) over the SYMMETRICAL INTERVAL [-π,π] then your integral results in 0. A symmetrical interval is like of the form [-A,A] where A is real.
This is the graph of sine function. The shaded area is the interval [-π,π]. When you are integrating over this interval, practically you are finding the area of this shaded portion. Since one shaded area amounts to the negative of the other their sum results in 0.
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Muhammad Abdullah
Bs-Mathametics from Shah Abdul Latif University, Khairpur (Graduated 2022)
·
6y
univariate function is said to be oddprovided that . Geometrically, suchfunctions are symmetric about the origin. ... If an even function is differentiable, then its derivative is anodd function; what's more, if an odd function is integrable, then its integralover a symmetric interval , , is identically zero
Tanu Shyam Majumder
Former Ex Guest Lecturer of Shibpur Dinobondhu College (2014–2015)
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Author has 1.1K answers and 1.1M answer views
·
6y
Question is incomplete it should be definite integral and it is not valid for indefinite integral. Now integral 0ver (-a,a) f(x)dx =Integral over (-a,0) f(x)dx + Integral over (0,a) f(x)dx= Integral over (0,a)[f(-x)+f(x)]dx=Integral over (0,a) [-f(x)+f(x)]dx =Integral over (0,a)dx[ as f(x) is odd function so f(-x)=-f(x) and change the limit of 1st integral we obtain this result]=0(proved)
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Hilmar Zonneveld
Computer programmer since 1986
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Author has 58.5K answers and 19.4M answer views
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3y
Originally Answered: Why does an odd function in an integral = 0?
·
Well, it doesn’t necessarily.
It will only do so if you integrate from some constant -b to +b. In this case, let’s suppose - without loss of generality - that the integral from 0 to b is positive, and the integral from -b to 0 is negative. In this case, the two cancel out exactly if you add them - precisely because the function is odd.
Mike Hirschhorn
Honorary Associate Professor of Mathematics at UNSW
·
Author has 8.1K answers and 2.7M answer views
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3y
Originally Answered: Why does an odd function in an integral = 0?
·
If the integral over a balanced interval exists, then it is equal to zero. This follows from the definition of the integral as the limit of a sum.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist.
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Upvoted by
Terry Moore
, M.Sc. Mathematics, University of Southampton (1968) and
José Ilhano Silva
, M. S. Mathematics & Differential Geometry, Federal University of Ceará (2017) · Author has 6.8K answers and 52.8M answer views
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2y
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If the definite integral of a function equals zero, then what does that mean?
Definite integrals are about finding the area between a function and the x axis.
I can think of two cases…
If we integrate y = f(x) = 0 from any two points for instance from x = 1 to 2
we obviously get zero because the function is just the x axis so the area under it is zero
But the usual case is when the function has equal parts above and below the x axis such as y = f(x) = sin(x).
Areas under the x axis turn out to be negative.
In cases like this we usually want the area between the function and the x axis
which would need to be done in two parts…
Definite integrals are about finding the area between a function and the x axis.
I can think of two cases…
If we integrate y = f(x) = 0 from any two points for instance from x = 1 to 2
we obviously get zero because the function is just the x axis so the area under it is zero
But the usual case is when the function has equal parts above and below the x axis such as y = f(x) = sin(x).
Areas under the x axis turn out to be negative.
In cases like this we usually want the area between the function and the x axis
which would need to be done in two parts…
B. S. Thomson
Lived in Vancouver, BC
·
Author has 1.2K answers and 2.9M answer views
·
1y
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If the definite integral of a function equals zero, then what does that mean?
Not sure the other [current] answers say a lot here.
Q1. Suppose that is a continuous function and that [math]\int_a^b f(x)\,dx =0 [/math]. Is there anything I can conclude from that?
There are a few things:
(i) It is pretty easy to conclude that there is one point in math[/math] at least where [math]f(x)=0[/math]. Use the mean-value theorem for integrals, or use the intermediate value property of continuous functions. Or even write [math]F(x)= \int_a^x f(t)\,dt[/math] on [math][a,b] [/math] and use the fact that [math]F [/math] has a maximum or minimum somewhere in math[/math].
(ii) Another thing you can conclude is a bit trickier. A naive first though
Not sure the other [current] answers say a lot here.
Q1. Suppose that [math]f:[a,b]\to\mathbb R[/math] is a continuous function and that [math]\int_a^b f(x)\,dx =0 [/math]. Is there anything I can conclude from that?
There are a few things:
(i) It is pretty easy to conclude that there is one point in math[/math] at least where [math]f(x)=0[/math]. Use the mean-value theorem for integrals, or use the intermediate value property of continuous functions. Or even write [math]F(x)= \int_a^x f(t)\,dt[/math] on [math][a,b] [/math] and use the fact that [math]F [/math] has a maximum or minimum somewhere in math[/math].
(ii) Another thing you can conclude is a bit trickier. A naive first thought is that the integral on the intervals where [math] f[/math] is positive should balance with the integral on the intervals where [math]f[/math] is negative so that sum vanishes.
Define [math] f^+(x)= \max {f(x),0} [/math] and[math] f^-(x) = \max {-f(x),0}.[/math] Then [math]f = f^+ - f^- [/math]and all three functions are continuous. Hence
[math]0 = \int_a^b f(x)\,dx = \int_a^b f^+(x)\,dx - \int_a^b f^-(x)\,dx\tag1[/math]
and so
[math]\int_a^b f^+(x)\,dx = \int_a^b f^-(x)\,dx.\tag2[/math]
(iii) Or try this. Define [math]P = {x\in (a,b): f(x)>0} [/math] and [math]N = {x\in (a,b): f(x)<0}.[/math]
These are open sets. Open sets can be written as a union of open intervals, a finite union or maybe a sequence. So suppose [math]P = \bigcup_{i} (a_i,b_i)[/math] and [math]N= \bigcup_{i} (c_i,d_i) .[/math]
You can conclude that
[math]\sum_i \int_{a_i}^{b_i} f(x)\,dx= - \sum_i \int_{c_i}^{d_i} f(x)\,dx.\tag2[/math]
The sums might be finite but could also be infinite series. Not sure if many calculus students could show that, but with advanced tools it is easy enough.
Q2. Suppose that [math]f:[a,b]\to\mathbb R [/math] is a Riemann integrable function, and that [math]\int_a^b f(x)\,dx =0 . [/math] Is there anything I can conclude from that?
Not a whole lot. Such a Riemann integrable function can be badly discontinuous and so may have no zeros. The set of points where it is positive (negative) need not be open so the previous discussion does not apply.
You can show that
[math]0= \int_a^b f(x)\,dx = \int_P f(x)\,dx + \int_N f(x)\,dx \tag3[/math]
using the Lebesgue (not Riemann) integral.
You can refine this too: let [math]P'[/math] be the subset of [math]P[/math] containing just points at which [math]f[/math] is continuous. Similarly let [math]N '[/math] be the subset of [math]N [/math]containing just points at which [math]f [/math] is continuous. Then (3) can be written instead as
[math]0= \int_a^b f(x)\,dx = \int_{P'} f(x)\,dx + \int_{N'} f(x)\,dx. \tag4[/math]
Not sure if this is interesting to anyone else but it is a response to the question.
Exercise 1. Suppose that [math]f [/math] is a nonnegative continuous function on [math][a,b] [/math]and that [math] \int_a^bf(x)\,dx=0.[/math] What can you conclude?
Exercise 2. Suppose that [math]f[/math] is a nonnegative Riemann integrable function on [math][a,b][/math] and that [math]\int_a^bf(x)\,dx=0[/math]. What can you conclude?
Robert Chin
Phd in Physics - but trying to be an amateur mathematician.
·
Author has 163 answers and 396.9K answer views
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8y
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Why is integral x e^ (-\alpha x^2) an odd function and equal to 0?
Let us start with what is the definition of an odd function
[math]f(-x) = -f(x)[/math]
This is an odd function, similarly an even function is given by
[math]f(-x) = f(x)[/math]
How is an odd function useful, typically you have an integral with symmetric bounds i.e. [math][-L,L][/math] where [math]L>0[/math]
[math]\int_{-L}^L f(x)dx = \int_{-L}^0 f(x)dx + \int_0^L f(x)dx[/math]
now focusing on the first integral
[math]\int_{-L}^0 f(x)dx = \int_{L}^0f(-x)d(-x) = -\int_L^0f(-x)dx = \int_0^Lf(-x)dx [/math]
Taking the definition of an odd function we can re-write as
[math]\int_0^Lf(-x)dx = \int_0^L-f(x)dx =-\int_0^Lf(x)dx [/math]
Putting it into the original integral
[math]\int_{-L}^L f(x)dx = -\int_[/math]
Let us start with what is the definition of an odd function
[math]f(-x) = -f(x)[/math]
This is an odd function, similarly an even function is given by
[math]f(-x) = f(x)[/math]
How is an odd function useful, typically you have an integral with symmetric bounds i.e. [math][-L,L][/math] where [math]L>0[/math]
then
[math]\int_{-L}^L f(x)dx = \int_{-L}^0 f(x)dx + \int_0^L f(x)dx[/math]
now focusing on the first integral
[math]\int_{-L}^0 f(x)dx = \int_{L}^0f(-x)d(-x) = -\int_L^0f(-x)dx = \int_0^Lf(-x)dx [/math]
Taking the definition of an odd function we can re-write as
[math]\int_0^Lf(-x)dx = \int_0^L-f(x)dx =-\int_0^Lf(x)dx [/math]
Putting it into the original integral
[math]\int_{-L}^L f(x)dx = -\int_0^Lf(x)dx + \int_0^L f(x)dx[/math]
We can instantly see that the integrals on the right hand side balance and cancel
So an Odd function leads to
[math]\int_{-L}^L f(x)dx = 0[/math]
whilst even (playing the same game) leads to
[math]\int_{-L}^L f(x)dx = 2\int_0^Lf(x)dx[/math]
For your integral you have an odd function
[math]f(x) = x\mathrm{e}^{-ax^2}\implies f(-x) = -x\mathrm{e}^{-ax^2} = -f(x)[/math]
so if we integrate over symmetric limits then we have an integral that equates to zero.
David Joyce
Professor Emeritus of Mathematics at Clark University
·
Upvoted by
Justin Rising
, PhD in statistics · Author has 9.9K answers and 68.4M answer views
·
Updated 9y
Related
If the integral at any given point equals 0, shouldn't the integral of any function always equal 0?
The intention of the integral [math]\int_a^b f(x)\,dx[/math] of a positive function f is that it's supposed to represent the area under the curve y = f(x), above the x-axis, and between the vertical lines x = a and x = b.
When a = b, the region has no width—it's just a vertical line—and so the area under the curve is 0.
You ask how the integral at each point can be 0 while the integral over an interval can be greater than 0. Each vertical section a plane region is a line segment with area 0, but the whole plane region will have a positive area.
It is simply not the case that the area of a plane region
The intention of the integral [math]\int_a^b f(x)\,dx[/math] of a positive function f is that it's supposed to represent the area under the curve y = f(x), above the x-axis, and between the vertical lines x = a and x = b.
When a = b, the region has no width—it's just a vertical line—and so the area under the curve is 0.
You ask how the integral at each point can be 0 while the integral over an interval can be greater than 0. Each vertical section a plane region is a line segment with area 0, but the whole plane region will have a positive area.
It is simply not the case that the area of a plane region is equal to the sum of infinitely many 0's, each 0 representing the zero area of a vertical section of that plane region.
This seeming paradox was discussed by the ancient Greeks. See the article Continuity and Infinitesimals in the Stanford Encyclopedia of Philosophy. Aristotle, in particular, proposed that a line was not simply composed of its points, and a plane region was not simply composed of its linear sections.
The modern measure-theoretic interpretation is not the same as Aristotle's. Now we expect areas to be countably additive, but not uncountably additive. If you partition a plane region into either finitely many subregions, or countably infinitely many subregions, then the area of the whole region will be the sum of the areas of the subregions. On the other hand, if you partition the region into uncountably many subregions (as you do when you intersect it with uncountably many vertical lines), then the sum of areas of those subregions need not be the area of the whole region.
Related questions
If the integral of a function is zero, what does it tells us about the function?
What is the integral of odd function?
Why is the integral of an even function over a closed interval always equal to zero?
Does the Integral of an even function always have to be an odd function?
If an integral is zero, is the integrand necessarily odd?
How do I prove that improper integrals for odd functions like sin and 1/x are divergent (from negative infinity to infinity) and not equal to zero?
Why does an odd function in an integral = 0?
If the integral at any given point equals 0, shouldn't the integral of any function always equal 0?
Can an integral of a function be equal to zero in calculus? How would you prove it?
How can we prove that if the integral of a product of two functions is zero, then one function will be a constant multiple of the other?
Why is the "integral of odd functions rule" only valid for c>=0 (c being the interval [-c,c] of the integral)?
If the limit of an integral is zero, does that mean the function it represents is always zero?
What are some examples of functions that have zero derivative everywhere, but non-zero integral over any interval taken on its domain?
Is the integral and derivative of an even function always odd and vice versa?
How do I prove that the integral of an odd function is within a range?
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11822 | https://www.youtube.com/watch?v=0eJmak30Lpk | How to use reverse calculations to check your answers
Sum It Up Maths
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Posted: 16 Feb 2020
In this video will teach you how to properly use reverse calculations and how they can help you double check your answers
If you have any questions please ask in the comments section or our Instagram at sumitupmaths
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Transcript:
[Music] so what's up Matt okay so I'm going to show you how we would do a reverse calculation so what happens is we get to the end of our song and we want to check it's correct so we do the opposite in order to check that we get to our original start of the song so we're going backwards to check that everything works out the same as it should do so let me just show you how that might work so the opposite of AD is takeaway and the opposite of times is divided so if we're in I exam and we have a son which we add I'm going to use the easy numbers a so we just could focus on the concept so if we have three add two equals five our reverse calculation will be the opposite about it will be the takeaway and instead of ending with five we're going to start with five so five take away now this one two spaces they take away two equals and we can see the best three so the calculation has worked both ways so we know it's correct let's have another look at one maybe we would do a multiplication so if we had three times two equals six okay we've used times so the opposite of that is divided okay instead of ending with six we're going to start with six two stays where it is and we should six divided by two end up with our original number okay so that is the way that we do our reverse calculations whatever some you've got whether it's add takeaway times or divide so whichever of those more operations you will be able to use the opposite one to check that you can get back to your original amount okay the way of checking your calculations is to do an estimate so it may be that we want to just check a rough idea perhaps we're not good at decimal places or sort of thing so let's just say we have the number three point nine six add four point two one and we're going to add those together and we're not just sure about where we put the decimal afterwards okay so what we'll do is we'll round the three up to four okay this is a fullface that will go to four and we'll add it to four because this is less than 5 so we'll round it down to 4 so we know that our answer should equal roughly 8 okay and if we were to get 18 then we know we've done it wrong for going to get point 8 we know that we've done it wrong it is more useful I guess if we're doing x or divide because with that one there is war and chance of error for where to put the decimal spot okay there's more place I should say you |
11823 | https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems?srsltid=AfmBOooogXW1PaW0B072txeHZoHkCgBw2hHFGdyWLJIrCZw1PPpeO6qq | Art of Problem Solving
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2020 AMC 10B Problems
2020 AMC 10B (Answer Key)
Printable versions: Wiki • AoPS Resources • PDF
Instructions
This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator).
Figures are not necessarily drawn to scale.
You will have 75 minutes working time to complete the test.
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
Contents
1 Problem 1
2 Problem 2
3 Problem 3
4 Problem 4
5 Problem 5
6 Problem 6
7 Problem 7
8 Problem 8
9 Problem 9
10 Problem 10
11 Problem 11
12 Problem 12
13 Problem 13
14 Problem 14
15 Problem 15
16 Problem 16
17 Problem 17
18 Problem 18
19 Problem 19
20 Problem 20
21 Problem 21
22 Problem 22
23 Problem 23
24 Problem 24
25 Problem 25
26 See also
Problem 1
What is the value of
Solution
Problem 2
Carl has cubes each having side length , and Kate has cubes each having side length . What is the total volume of these cubes?
Solution
Problem 3
The ratio of to is , the ratio of to is , and the ratio of to is . What is the ratio of to
Solution
Problem 4
The acute angles of a right triangle are and , where and both and are prime numbers. What is the least possible value of ?
Solution
Problem 5
How many distinguishable arrangements are there of brown tile, purple tile, green tiles, and yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)
Solution
Problem 6
Driving along a highway, Megan noticed that her odometer showed (miles). This number is a palindrome-it reads the same forward and backward. Then hours later, the odometer displayed the next higher palindrome. What was her average speed, in miles per hour, during this -hour period?
Solution
Problem 7
How many positive even multiples of less than are perfect squares?
Solution
Problem 8
Points and lie in a plane with . How many locations for point in this plane are there such that the triangle with vertices , , and is a right triangle with area square units?
Solution
Problem 9
How many ordered pairs of integers satisfy the equation
Solution
Problem 10
A three-quarter sector of a circle of radius inches together with its interior can be rolled up to form the lateral surface area of a right circular cone by taping together along the two radii shown. What is the volume of the cone in cubic inches?
Solution
Problem 11
Ms. Carr asks her students to read any of the books on a reading list. Harold randomly selects books from this list, and Betty does the same. What is the probability that there are exactly books that they both select?
Solution
Problem 12
The decimal representation of consists of a string of zeros after the decimal point, followed by a and then several more digits. How many zeros are in that initial string of zeros after the decimal point?
Solution
Problem 13
Andy the Ant lives on a coordinate plane and is currently at facing east (that is, in the positive -direction). Andy moves unit and then turns left. From there, Andy moves units (north) and then turns left. He then moves units (west) and again turns left. Andy continues his progress, increasing his distance each time by unit and always turning left. What is the location of the point which Andy makes the left turn?
Solution
Problem 14
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?
Solution
Problem 15
Steve wrote the digits , , , , and in order repeatedly from left to right, forming a list of digits, beginning He then erased every third digit from his list (that is, the rd, th, th, digits from the left), then erased every fourth digit from the resulting list (that is, the th, th, th, digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions ?
Solution
Problem 16
Bela and Jenn play the following game on the closed interval of the real number line, where is a fixed integer greater than . They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval . Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?
Solution
Problem 17
There are people standing equally spaced around a circle. Each person knows exactly of the other people: the people standing next to him or her, as well as the person directly across the circle. How many ways are there for the people to split up into pairs so that the members of each pair know each other?
Solution
Problem 18
An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
Solution
Problem 19
In a certain card game, a player is dealt a hand of cards from a deck of distinct cards. The number of distinct (unordered) hands that can be dealt to the player can be written as . What is the digit ?
Solution
Problem 20
Let be a right rectangular prism (box) with edges lengths and , together with its interior. For real , let be the set of points in -dimensional space that lie within a distance of some point in . The volume of can be expressed as , where and are positive real numbers. What is
Solution
Problem 21
In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral , quadrilateral , and pentagon each has area What is ?
Solution
Problem 22
What is the remainder when is divided by ?
Solution
Problem 23
Square in the coordinate plane has vertices at the points and Consider the following four transformations:
a rotation of counterclockwise around the origin;
a rotation of clockwise around the origin;
a reflection across the -axis; and
a reflection across the -axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying and then would send the vertex at to and would send the vertex at to itself. How many sequences of transformations chosen from will send all of the labeled vertices back to their original positions? (For example, is one sequence of transformations that will send the vertices back to their original positions.)
Solution
Problem 24
How many positive integers satisfy(Recall that is the greatest integer not exceeding .)
Solution
Problem 25
Let denote the number of ways of writing the positive integer as a product
where , the are integers strictly greater than , and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counted as distinct). For example, the number can be written as , , and , so . What is ?
Solution
See also
2020 AMC 10B (Problems • Answer Key • Resources)
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Why a homogeneous magnetic field will not deflect neutrons? [closed]
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According to the comments below the answer of the question about Experiments with Alpha particles in a homogeneous magnetic field an inhomogeneous magnetic field will deflect neutrons. Why a homogeneous magnetic field will not deflect neutrons?
The use of an inhomogeneous magnetic field in the Stern-Gerlach-experiment is to separate particles in two regions and not to get a smear. But IF an inhomogeneous magnetic field will deflect neutrons THAN a homogeneous field has to deflect too?
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edited Apr 13, 2017 at 12:39
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asked Jul 20, 2016 at 19:51
HolgerFiedlerHolgerFiedler
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3 I'm voting to close this question as off-topic because of insufficient research effort.CuriousOne –CuriousOne 2016-07-20 20:35:25 +00:00 Commented Jul 20, 2016 at 20:35
It is not clear at all why you think that things at are deflected in inhomogeneous fields should also be deflected by homogeneous fields. It is not difficult to find out that the force on a magnetic moment depends on the spatial derivative of the magnetic field, which vanishes in the homogeneous case.ACuriousMind –ACuriousMind♦ 2016-07-21 13:16:14 +00:00 Commented Jul 21, 2016 at 13:16
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Lets treat this classically, for simplicity. Neutron has no charge. However, it has a intrinsic magnetic dipole moment μ μ. The force of the magnetic field applied to this dipole:
F=∇(μ⋅B)F=∇(μ⋅B)
Since you are assuming the magnetic field does not depend on space (its homogenous), then its "gradient" will be zero. That is:
F i=∑k∂∂x i(μ k B k)=∑k μ k∂B k∂x i=0,since∂B k∂x i=0 F i=∑k∂∂x i(μ k B k)=∑k μ k∂B k∂x i=0,since∂B k∂x i=0
Then, force is zero. Thus, no deflection.
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answered Jul 20, 2016 at 20:00
Physicist137Physicist137
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The gradient is made in the direction perpendicular to the trajectory of the particle?HolgerFiedler –HolgerFiedler 2016-07-20 20:06:34 +00:00 Commented Jul 20, 2016 at 20:06
3 @HolgerFiedler No. This is not F=q v×B F=q v×B. There is no velocity there, so, force is independent on trajectory (velocity). Force only depends on its position in space (ie, of the magnetic field in that position). The gradient is taken of the dot product of magnetic moment and external magnetic field, as its shown in answer.Physicist137 –Physicist137 2016-07-20 20:10:07 +00:00 Commented Jul 20, 2016 at 20:10
So independent from a homogeneous or an inhomogeneous field in the direction of the particles trajectory the gradient is zero.HolgerFiedler –HolgerFiedler 2016-07-20 20:19:08 +00:00 Commented Jul 20, 2016 at 20:19
3 @HolgerFiedler What? Why? Well, if field is inhomogeneous, then B is not constant spatially. Thus ∂B k∂x i≠0∂B k∂x i≠0. Thus, force will not be zero.Physicist137 –Physicist137 2016-07-20 20:32:02 +00:00 Commented Jul 20, 2016 at 20:32
1 I was going to write an answer almost identical to yours. It is necessary to have a nonconstant magnetic field.Lawrence B. Crowell –Lawrence B. Crowell 2016-07-20 21:49:11 +00:00 Commented Jul 20, 2016 at 21:49
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11825 | https://www.geeksforgeeks.org/maths/convert-decimal-to-fraction/ | Convert Decimal to Fraction - GeeksforGeeks
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Convert Decimal to Fraction
Last Updated : 23 Jul, 2025
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Converting a decimal to a fraction is the process of changing a number in decimal form into a fraction, which represents the same value in terms of parts of a whole number. Converting decimals into fractions is a key skill that helps clarify values and is useful in various practical situations.
Decimal: A decimal represents a number with a whole part and a fractional part, separated by a decimal point.Example:0.619, 50.5, etc.
Fraction: A fraction represents a part of a whole, expressed with a numerator (top) showing parts taken and a denominator (bottom) showing total parts.Example: 619/1000, etc.
Table of Content
Steps for Decimal to Fraction Conversion
Handling Repeating Decimals
Converting Negative Decimals to Fractions
Solved Example Problems
Steps for Decimal to Fraction Conversion
Here are the steps to convert a decimal number into a fraction:
Decimal to Fraction
Step 1: First, divide the decimal number by 1.
Example: Decimal number 0.565
On applying step 1:
0.565/1
Step 2:For every decimal point in the numerator multiply 10 for both numerator and denominator.
In this example, 2 numbers are given after the decimal point, so multiply with 10 two times for both numerator and denominator.
(0.565 × 10 ×10 x 10)/(1 ×10 ×10 x 10)
= 565/1000
Step 3:Perform simplification on the fraction formed from Step 2 until further simplification is not possible.
565/1000
= 113/200
These are the 3 steps that one must follow while converting from decimal to fraction.
Handling Repeating Decimals
Repeating decimals, like 0.333… (where 3 repeats infinitely), are a bit trickier to convert into fractions.
Here's how you can handle them:
Let x = 0.333….
Multiply both sides by 10 to shift the decimal point:
10x = 3.333…
3. Subtract the original equation from this new equation:
10x - x = 3.333… - 0.333…
This simplifies to:
9x = 3
4. Solve for x by dividing both sides by 9:
x = 3/9, which simplifies to 1/3.
Thus, 0.333… = 1/3. For more complex repeating decimals, the process is similar, but the multiplication factor changes depending on the number of repeating digits.
Converting Negative Decimals to Fractions
The process of converting negative decimals to fractions is almost identical to the steps we've discussed. The only difference is the negative sign. For example, to convert -0.625:
Write it as -0.625/1.
Multiply both top and bottom by 1000 (since there are three digits after the decimal).
Simplify the fraction -625/1000 by dividing both by 125, giving -5/8.
Negative decimal to fraction conversion requires keeping the negative sign intact throughout the process.
Read more:
Convert Fractions to Decimals
Convert Binary Fractions to Decimal
Solved Examples on Decimals to Fraction Conversion
Question 1: Convert a decimal 0.1 to a fraction
Solution:
Step - 1 Divide decimal with 1
= 0.1/1
Step - 2 As there is only 1 number after point so multiply 10 one time to both numerator and denominator.
= 0.1 ×10/1 ×10 = 1/10
Step - 3 The above generated can't be simplified further so we consider the above fraction 1/10 as final result.
So fraction of 0.1 = 1/10
Question 2: Convert a decimal 6.25 to a fraction
Solution:
Step - 1Divide decimal with 1
= 6.25/1
Step - 2As there are 2 numbers after point so multiply 10 two times with both numerator and denominator.
= 6.25 ×10 ×10/1 ×10 ×10 = 625/100
Step - 3This 625/100 fraction can be simplified to,
625/100 = 125/20 = 25/4
So fraction of 6.25 = 25/4
Question 3: Convert a decimal 6.25 into a mixed fraction.
Solution:
When a whole number is present before point in decimal number then separate that whole number from decimal number and follow the 3 steps for conversion.
6.25 = 6 + 0.25
Step - 1 Consider the digits that are after the decimal point i.e., 0.25 and Divide decimal with 1
= 0.25/1
Step - 2 As there are 2 numbers after point so multiply 10 two times with both numerator and denominator.
= 0.25 ×10 ×10/1 ×10 ×10 = 25/100
Step - 3 This 25/100 fraction can be simplified to
25/100 = 5/20 = 1/4
Add the separated digit 6 (done before step-1) to formed fraction.
So fraction of 6.25 =6 1 4 6\frac{1}{4}6 4 1
Question 4: Convert a decimal 4.372 into a mixed fraction.
Solution:
For conversion into mixed fraction, separate the whole number part before the decimal point from decimal value and follow the above specified 3 steps on numbers after decimal points.
4.372 = 4 + 0.372
Step - 1 Consider the digits that are after the decimal point i.e., 0.372 and Divide decimal with 1
= 0.372/1
Step - 2 As there are 3 numbers after point so multiply 10 three times with both numerator and denominator.
= 0.372 ×10 ×10 ×10/1 ×10 ×10 ×10 = 372/1000
Step - 3 This 372/1000 fraction can be simplified to,
372/1000 = 93/250
Add the separated digit 4 (done before step-1) to formed fraction
So fraction of 4.372 = 4 93 250 4\frac{93}{250}4 250 93
Question 5: Convert a decimal 0.33 to a fraction
Solution:
Step - 1 Divide decimal with 1
= 0.33/1
Step - 2 As there are 2 numbers after point so multiply 10 two times with both numerator and denominator.
= 0.33 ×10 ×10/1 ×10 ×10 = 33/100
Step - 3 This 33/100 fraction can't be simplified so it can be leaved as it is
So fraction of 0.33 = 33/100
Question 6: Convert a decimal 0.3333... to a fraction
Solution:
Step - 1 Divide decimal with 1
= 0.3333..../1
Note:For this kind of recurrence number i.e., 3 is recurring for infinite times we can't multiply 10 for each decimal point. In this case we multiply numerator and denominator with 3.
Step - 2 As this is a recurrence number we multiply 3 with numerator and denominator.
0.3333.... ×3/1 ×3 = 0.9999..../3
Step - 3Simply the above fraction.
As 0.9999... is close to 1, round up the numerator to 1.
0.9999..../3 can be simplified to 1/3.
So fraction of 0.3333... = 1/3
Question 7: Convert a negative decimal -0.75 to a fraction
Solution:
Step 1: Divide the negative decimal by 1.
−0.75/1
Step 2: Multiply both the numerator and the denominator by 100 to eliminate the decimal points (since there are two decimal places).−0.75×100/1×100=−75/100
Step 3: Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 25.
−75/100=−3/4
The fraction form of -0.75 is −3/4.
Question 8: Convert a repeating decimal 0.666... to a fraction
Solution:
Step 1: Let x=0.666...
Step 2: To eliminate the repeating decimal, multiply x by 10 (since the repeating part has one digit).
10x=6.666
Step 3: Subtract the original x from this new equation to isolate the repeating part.
10x − x = 6.666 − 0.666
9x = 6
Step 4: Solve for x by dividing both sides by 9.
x = 6/9
= 2/3.
Decimals - Simplification, Conversion and Operations
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11826 | https://pubmed.ncbi.nlm.nih.gov/1320514/ | Amantadine as N-methyl-D-aspartic acid receptor antagonist: new possibilities for therapeutic applications? - PubMed
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Clin Neurol Neurosurg
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. 1992:94 Suppl:S4-6.
doi: 10.1016/0303-8467(92)90006-o.
Amantadine as N-methyl-D-aspartic acid receptor antagonist: new possibilities for therapeutic applications?
J C Stoof1,J Booij,B Drukarch
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1 Department of Neurology, Medical Faculty, Free University, Amsterdam, The Netherlands.
PMID: 1320514
DOI: 10.1016/0303-8467(92)90006-o
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Amantadine as N-methyl-D-aspartic acid receptor antagonist: new possibilities for therapeutic applications?
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doi: 10.1016/0303-8467(92)90006-o.
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DOI: 10.1016/0303-8467(92)90006-o
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The N-methyl-D-aspartic acid (NMDA) receptor is an intriguing target for the development of drugs with anti-Parkinsonian activity as well as with protective actions against degenerative processes induced by brain ischemia. Amantadine is used in the treatment of Parkinson's disease without a well established mechanism of action. We show here that amantadine inhibits, in a non-competitive way, the NMDA receptor-mediated stimulation of acetylcholine release from rat neostriatum in vitro in "therapeutic" (i.e., low micromolar) concentrations. This indicates that amantadine might exert its anti-Parkinsonian effect via blockade of NMDA receptors. Sustained stimulation of NMDA receptors induces so-called excitotoxicity. Recently, it was demonstrated that amantadine is able to inhibit NMDA induced cell death in a neuronal culture. On the basis of these findings it seems worth investigating if amantadine is also able to protect against neurodegenerative processes caused by brain ischemia in vivo.
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[Ketamine inhibits n-methyl-d-aspartate (NMDA) receptor mediated acetylcholine release from rabbit caudate nucleus slices].Feuerstein TJ.Feuerstein TJ.Anaesthesist. 1994 Nov;43 Suppl 2:S48-51.Anaesthesist. 1994.PMID: 7840414 German.
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Glutamatergic drugs in Parkinson's disease.Lange KW, Riederer P.Lange KW, et al.Life Sci. 1994;55(25-26):2067-75. doi: 10.1016/0024-3205(94)00387-4.Life Sci. 1994.PMID: 7997066 Review.
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11827 | https://testbook.com/question-answer/the-difference-between-the-greatest-and-least-prim--5ad852c69db78831bbf42549 | [Solved] The difference between the greatest and least prime numbers
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Quantitative Aptitude
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The difference between the greatest and least prime numbers which are less than 100 is
97
94
96
95
Answer (Detailed Solution Below)
Option 4 : 95
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The greatest prime number less than 100 is 97. The least prime number is 2.
Difference = 97 - 2 = 95
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More Integers Questions
Q1.Let a, b, c, d, and e be positive integers, such that a, b, and c are even and d, e are odd. Consider the following statements: 1. (a + d + c) (b - e) is odd. 2. (a2 + b2 + c2) (d × e) is even. 3. (a - b - c) (d - e) is even. Which of the statements given above are correct?
Q2.A Question is given below followed by two Statements I and II. Question: mn = k, where m and n are prime numbers and k is an even number. What is the value of mn - n + 1? Statement - I: m > n Statement - II: One of the numbers from amongst m and n is 2.
Q3.What is the least fraction that must be added 1 1 3÷1 1 2÷1 1 9 to make the result an integer?
Q4.If 1 is added to each odd digit and 2 is subtracted from each even digit in the number 3816427, how many digits will appear more than once in the new number thus formed?
Q5.If 2 is added to each even digit and 1 is subtracted from each odd digit in the number 6435127, what will be the sum of the second digit from the left and the third digit from the right in the new number thus formed?
Q6.How many factors of 180 are perfect squares?
Q7.Which of the following is the example of composite number?
Q8.How many factors of 500 are perfect squares?
Q9.The sum of all odd numbers from 1 to 41 is:
Q10.If you write down the first 100 natural numbers, how many times will 8 be repeated?
More Number System Questions
Q1.If a shopkeeper gives a candy for Rs. 5 and also a candy for 2 wrappers of the same candy, then for Rs. 40 how many candies can a child eat?
Q2.In a series 1, 2, 3, 4, 5......, multiples of 2, 3, 4 and 5 are deleted from the series. In the remaining series, what will be the first number greater than 50?
Q3.The Lowest Common Multiple (LCM) of two numbers is 48. The numbers are in the ratio of 3 : 2. What is the sum of the numbers?
Q4.What is the number that when divided by 7, 9 and 12 leaves the same remainder 1 in each case ?
Q5.A set of natural numbers has to be formed that consists of six numbers, and the last number is 30. The First number of the set is the highest root of the given equation, and the second number of the set is (n + 4) - 2n, where 'n' is the difference between the roots of the given equation. The third number of the set is (n + 1)2- (n + 1). The fourth number of the set is (n + 2)2- (n + 2). The fifth number of the set is (n + 1)3– (n + 5). Note: x2- 14x + 48 = 0 Which of the statement/s is/are correct. I. The product of second and fifth number is perfect square. II. The average of last four number is 17. III. If the seventh number of the set is sixth number + n3, then the resultant number is a factor of 114.
Q6.When a number divided by 114 it leaves a remainder 14. What will be the remainder, when the same number divided by 19.
Q7.LCM and GCD of two numbers are 63 and 21 respectively. If one of them is 21, then find the second number?
Q8.The sum of two numbers is 384 HCF of the numbers is 48 the difference of the number is
Q9.If the 4-digit number 48ab is divisible by 2, 5 and 7, then what is the value of (10a - b) ?
Q10.The LCM of two numbers is 48. The numbers are in the ratio of 2∶ 3. Find the sum of the number.
More Quantitative Aptitude Questions
Q1.Solve the given series below and answer the following question. Series: 125r, M, 120q, (100p + 4), (30q + 42), (8r+ 4.8) Note: A) p and q are consecutive prime numbers where q > p. B) r = p + q C) r is a factor of 60 and greater than 10. Which of the series follows the same pattern as the above series? I) 2000, 1600, 1120, 672, 336, 134.4 II) 1000, 400, 240, 192, 192, 230.4 III) 1200, 1300, 1500, 1800, 2200, 2700
Q2.There have two equations given: Equation I. p2− 10Mp + 6N = 0 Equation II. q2+ Mq − 5N=0 Where M + N = 39, 3 ≤ M ≤ 5, and N is a multiple of 3. Find the relationship between p and q.
Q3.Two trains, Train A and Train B, have lengths of x meters and (x + 50) meters respectively. The ratio of the speed of Train A to Train B is 5 : 4. Train A crosses an electric pole in 12 seconds. When moving in the same direction, Train A completely overtakes Train B in _ seconds. If Train B were to cross a platform of 200 meters, it would take _ seconds. Which of the following values can fill the blanks in the same order? A: 130 and 35 B: 140 and 40 C: 150 and 45
Q4.A rectangular field has dimensions where one side is longer than the other. A person decides to walk directly across the field along the diagonal instead of walking along the two adjacent sides. By doing this, the person covers a distance that is shorter than the total walking path by exactly one-third of the length of the longer side. Determine the ratio of the length of the shorter side to the longer side of the field.
Q5.In a Monty Hall game with 3 doors, the host does not know where the car is and opens a door at random. By chance, it reveals a goat. You chose a door initially. What is your probability of winning if you stick with your original choice?
Q6.For the dataset: 10, 15, 20, 25, 30, 35, 40, 45, what is the value of the first quartile (Q1)?
Q7.For the dataset: 5, 8, 12, 15, 18, 22, 25, 30, what is the value of the 3rd decile (D3)?
Q8.What is the percentile rank of a score of 40 in a dataset where 60% of the data lies below 40?
Q9.What does Karl Pearson’s Coefficient of Skewness measure?
Q10.If the mean of a dataset is 50, the median is 48 and the standard deviation is 10, what is Karl Pearson’s coefficient of skewness?
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11828 | https://www.storyofmathematics.com/3-4-5-triangle/ | Home
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3 4 5 Right Triangles Explanation & Examples
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What is a 3-4-5 Right Triangle?
How to Solve a 3-4-5 Triangle
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3 4 5 Right Triangles Explanation & Examples
Right triangles are very useful in our daily life. The simpler the dimensions of a right triangle, the simpler is its use.
The ability to recognize special right triangles is the shortcut to solving problems involving right triangles. Instead of using the Pythagorean theorem, you can use special right triangle ratios to calculate the missing lengths.
They may have different dimensions, but the most common of them is the 3-4-5 right triangle. This article will discuss what a 3-4-5 right triangle is and how to solve problems involving the 3-4-5 right triangle.
A triangle is a two-dimensional polygon with three corners, three vertices, and three angles joined together, forming a closed diagram in geometry. There are different types of triangles depending on the side lengths and magnitude of their interior angles. For more details on triangles, you can go through the previous articles.
What is a 3-4-5 Right Triangle?
A 3-4-5 right triangle is a triangle whose side lengths are in the ratio of 3:4:5. In other words, a 3-4-5 triangle has the ratio of the sides in whole numbers called Pythagorean Triples.
This ratio can be given as:
Side 1: Side 2: Hypotenuse = 3n: 4n: 5n = 3: 4: 5
We can prove this by using the Pythagorean Theorem as follows:
a2 + b2 = c2
32 + 42 = 52
9 + 16 = 25
25 = 25
A 3-4-5 right triangle has the three internal angles as 36.87 °, 53.13 °, and 90 °. Therefore, a 3 4 5 right triangle can be classified as a scalene triangle because all its three sides lengths and internal angles are different
Remember that a 3-4-5 triangle does not mean that the ratios are exactly 3: 4: 5; it can be any common factor of these numbers. For example, a 3-4-5 triangle can also take the following forms:
6-8-10
9-12-15
12-16-20
15-20-25
How to Solve a 3-4-5 Triangle
Solving a 3-4-5 right triangle is the process of finding the missing side lengths of the triangle. The ratio of 3: 4: 5 allows us to quickly calculate various lengths in geometric problems without resorting to methods such as tables or the Pythagoras theorem.
Example 1
Find the length of one side of a right triangle in which hypotenuse and the other side measures 30 cm and 24 cm, respectively.
Solution
Test the ratio to see if it fits the 3n: 4n: 5n
?: 24: 30 = ?: 4(6): 5(6)
This must be a 3-4-5 right triangle, so we have;
n = 6
Hence the length of the other side is;
3n = 3(6) = 18 cm
Example 2
The longest edge and bottom edge of a sailboat’s triangular sail are 15 yards and 12 yards, respectively. How tall is the sail?
Solution
Test the ratio
?: 12: 15 = ? : 4(3) : 5(3)
Therefore, the value of n = 3
Substitute.
3n = 3(3) = 9
Hence, the height of the sail is 9 yards.
Example 3
Identify the 3-4-5 right triangle from the following list of triangles.
Triangle A 8, 8, 25
Triangle B 9, 12, 15
Triangle C 23, 27, 31
Triangle D 12, 16, 20
Triangle E 6, 8, 10
Solution
Test the ratio of each triangle.
A 8: 8: 25
B 9: 12: 15 (divide each term by 3)
= 3: 4: 5
C 23: 27: 31
D 12: 16: 20 (divide each term by 4)
= 3: 4: 5
E 6: 8: 10 (divide through by 2)
= 3: 4: 5
Therefore, triangles B, D, and E are 3-4-5 right triangles.
Example 4
Find the value of x in the figure shown below. Assume the triangle is a 3-4-5 right triangle.
Solution
Look for the factor n in a 3-4-5 right triangle.
?: 80: 100 = ?: 4(20): 5(20)
Hence, n = 20
Substitute in 3n: 4n: 5n.
3n = 3(20) = 60
Therefore, x = 60 m
Example 5
Calculate the length of the diagonal of a right triangle with side lengths of 6 inches and 8 inches.
Solution
Check the ratio if it fits the 3n: 4n: 5n ratio.
6: 8: ? = 3(2): 4(2):?
n =2
Substitute n= 2 in 5n.
5n = 5(2) = 10.
Therefore, the length of the diagonal is 10 inches.
Practice Questions
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In geometry, points, lines, and planes are fundamental concepts that form the basis for constructing and understanding more complex geometric figures and relationships.
Points:
A point is a fundamental geometric object with no size or dimension. It is represented by a dot and is considered to have only a location in space, without any length, width, or height. Points are often labeled with capital letters.
Example:
A on a coordinate plane represents a point with coordinates (3, 4).
Lines:
A line is a straight path that extends infinitely in both directions. It is defined by an infinite set of points and can be named by any two distinct points on the line. Lines have length but no width.
Example:
The line AB, where A(-1, 2) and B(3, 5), is the straight path passing through points A and B.
Planes:
A plane is a flat, two-dimensional surface that extends infinitely in all directions. It is defined by an infinite set of points and is typically named by three non-collinear points within the plane. A plane has length and width but no thickness.
Example:
The plane defined by points A(1, 0, 2), B(0, 1, 3), and C(2, 3, 1) is a flat surface extending infinitely.
Understanding their interrelationships:
Collinear Points:
Points that lie on the same straight line are called collinear points.
Example:
Points A, B, and C are collinear if they lie on the same line.
Coplanar Points:
Points that lie in the same plane are called coplanar points.
Example:
Points A, B, C, and D are coplanar if they lie in the same plane.
Intersection of Lines and Planes:
Lines and planes can intersect or be parallel to each other.
Example:
The line AB can intersect the plane defined by points C, D, and E.
Midpoint of a Line Segment:
The midpoint of a line segment is a point that divides the segment into two equal parts.
Example:
If A(1, 2) and B(5, 6) are the endpoints of a line segment, the midpoint is M(3, 4).
These basic geometric elements provide the foundation for constructing more complex figures and understanding the spatial relationships between them. The study of points, lines, and planes is fundamental to Euclidean geometry and is applicable across various branches of mathematics and science.
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11830 | https://squareonehealth.com/blog/the-risser-sign-what-it-is-and-how-its-used-in-scoliosis/ | Risser Sign In The General Diagnosis Of Scoliosis - Square ONE
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HomeBlogScoliosisRisser Sign In The General Diagnosis Of Scoliosis
Risser Sign In The General Diagnosis Of Scoliosis
When diagnosing scoliosis, specialists use a combination of physical examination and imaging tests (such as MRIs and x-rays).
First, they will evaluate the severity of the curve using the Cobb angle to measure the vertebral rotation. They will then use a degree of curvature chart (a chart that compares the Cobb angle with severity levels) to help assess the condition. If a curve is present and it measures greater than 10 degrees, and there is rotation in the spine a diagnosis of scoliosis is made.
However, when treating the condition, the Cobb angle of the patient’s spine isn’t the only factor to consider. A scoliosis specialist will also use the Risser Sign to form the patient’s overall diagnosis and treatment plan.
What Is The Risser Scale? An Introduction
The Risser scale is a system used to measure the maturity of a patient’s skeleton. More specifically, it measures the ossification of the iliac crest (the top of the pelvic bone) to help indicate how much more growth a patient’s spine will likely have. Ossification refers to the process in which cartilage is turned into bone.
By identifying the stage of ossification, the Risser scale can indicate how much more growth a patient might experience. This information can help a specialist determine whether or not immediate treatment is required and how the treatment plan should be adjusted over time. It can also help a specialist predict how likely it is that scoliosis will develop in a patient that has a mild spinal curve.
What Are The Risser Scale Stages?
The Risser scale stages are determined by looking at X-ray images of the pelvis and assessing how much cartilage growth is present along the iliac crest. The stages range from 0-5, with Risser 0 indicating no ossification of the iliac crest and Risser 5 indicating complete ossification.
Younger patients (generally under ten years old) are likely to be at a Risser 0 stage as their spine is still developing and growing, while patients who have already gone through puberty are likely to be at a Risser 4 or 5 stage as their spine has already finished growing.
There is a sub-scale within the Risser 0 grading of Risser 0- and 0+. Risser 0- is graded when the triradiate cartilage has not yet ossified. Risser 0+ indicates that the triradiate cartilage has ossified.
The patient’s Risser stage can then be used as a guide to estimate how much more growth is left in their skeleton. This is important for scoliosis patients, as it helps determine the best treatment course. Generally speaking, those with a lower Risser stage (0-2) are at a higher risk of scoliosis progression, so specialists may need to initiate a more-involved treatment plan.
On the other hand, patients with a higher Risser stage (3-5) can typically be monitored over time, as there is less risk of scoliosis progression. Of course, the severity of the curve also plays a vital role in treatment, and the scoliosis specialist will consider all of these factors before developing a treatment plan.
Risser scale progression
With all of that in mind, the following is a brief breakdown of each stage of the Risser scale:
Risser 0-: There has been no ossification of the iliac crest yet and the triradiate cartilage is open.
Risser 0+: There is no ossification of the iliac crest and the triradiate cartilage is closed.
Risser 1: Under 25 percent of the iliac crest has ossified. Patients at this stage tend to be at the beginning of puberty.
Risser 2: Between 25 and 50 percent of the iliac crest has ossified.
Risser 3: Between 50 and 75 percent of the iliac crest has ossified. Patients at this stage are usually in the middle of puberty.
Risser 4: Over 75 percent of the iliac crest has ossified.
Risser 5: The iliac crest is fully ossified. Patients at this stage are typically full-grown adults.
How Is The Risser Scale Used In Diagnosing And Treating Scoliosis?
By looking at the stage of ossification of the iliac crest, specialists can estimate how much more growth is left in the patient’s spine. Patients who are still growing (Risser 0-2) are at a higher risk of scoliosis progression and should be monitored more closely.
Those who have already reached full maturity (Risser 4-5) are less likely to experience further progression and may be able to wait before initiating a treatment plan, depending on the severity of their spinal curvature and the symptoms that they are experiencing.
Using Risser To Evaluate Factors That Develop Scoliosis
The Risser scale helps scoliosis specialists evaluate multiple factors that can contribute to the development of scoliosis. For example, since the Risser scale assesses the maturity of a patient’s skeleton, it can be used to gauge if growth spurts may have caused scoliosis.
Additionally, the Risser scale can help determine how much more growth may be left in the patient’s spine and if that could potentially cause the spinal curve to worsen.
Using Risser To Monitor Scoliosis Treatment Progress
The Risser scale can also be used to gauge the effectiveness of scoliosis treatments. In many cases, non-surgical treatment methods such as bracing and physical therapy may be used to help control the progression of the spinal curve.
For example, if the Cobb angle of the spine is no longer increasing and the patient’s Risser stage shows that they have achieved full skeletal maturity, this could indicate that the treatment has been effective at stopping progression.
What Can Influence The Risser Score?
The ossification of the iliac crest occurs at a different rate for everybody. For example, girls tend to experience puberty and a corresponding growth spurt earlier than boys, so their iliac crest ossification will usually be further along. Additionally, some people’s bones may mature faster than others, which can lead to a higher Risser score.
Genetics can also play a role in the ossification of the iliac crest. Certain genetic conditions can cause bones to mature faster or slower than usual. As such, specialists must evaluate a patient’s Risser score when assessing their scoliosis, as it can provide valuable information regarding the maturity of their bones and how much more growth they may experience.
What Are The Key Considerations When Using Risser?
When using the Risser scale, it’s important to remember that it doesn’t always accurately reflect a patient’s skeletal maturity. There are other factors that can influence the ossification of a patient’s iliac crest, including genetics and lifestyle choices.
While the Risser score can be beneficial, other vital considerations must also be taken into account when evaluating a patient’s scoliosis, including:
Patient Age: The Risser sign estimates the maturity of a patient’s bones, so age should be considered when assessing the Risser score.
Curvature Severity: The severity of a patient’s spinal curvature should also be considered when interpreting the Risser sign. Two curves measuring at the same Cobb angle may progress at different rates based on the Risser score.
Overall Health Status: It is essential to evaluate a patient’s overall health status when assessing the Risser sign as this can impact treatment options. For example, patients with certain medical conditions, such as those affecting bone density, may not respond as well to certain treatments.
Other Conditions: Any other conditions that may be influencing the patient’s scoliosis should also be considered when assessing the Risser sign. For example, Marfan’s syndrome or Ehler-Danlos Syndrome may cause scoliosis and should be considered in addition to the Risser score.
The Risser sign is a reliable method for assessing the maturity of a patient’s bones and determining if there may be further growth in their spine. It’s also helpful in monitoring the progression of scoliosis and gauging the effectiveness of treatment methods.
However, it is essential to note that the Risser sign does not measure spinal curvature or progression directly. Instead, it is used to indicate bone maturity, which can help inform treatment decisions.
What Are The Recommended Treatments For Scoliosis Based On Risser?
The Risser scale helps determine an effective treatment plan; however, Risser scores must be considered in conjunction with other factors, including the Cobb angle of the spine. For example, a child at a Risser 0 stage with a slightly moderate scoliosis curve (as defined by the Cobb angle measurement) is more at risk of significant progression.
In such a case, a brace might be recommended. However, someone with the same curve at a Risser 5 may not be prescribed a brace right away. Because their spine is already fully formed, their curve isn’t likely to progress nearly as fast, and a more conservative treatment plan may be recommended.
Get A Risser Assessment For Your Scoliosis
When it comes to diagnosing and treating scoliosis, a reputable scoliosis specialist will make a risk of progression assessment by taking into account the size of the curvature and the Risser sign via an X-ray evaluation. An experienced specialist can interpret the results and make recommendations based on the patient’s specific needs and goals.
At Square ONE Health, our scoliosis specialists provide comprehensive assessments, including Risser sign evaluations and curve size measurements, to determine the best treatment solution for each patient. In addition, we prescribe scoliosis rehabilitation, bracing, and exercise using several proven treatment modalities.
The methods include the Schroth Method, the Scientific Exercise Approach to Scoliosis (SEAS) method, and Chiropractic Biophysics (CBP), as well as bracing to help manage scoliosis and improve spinal health. Call our office today for more information about how we tailor our scoliosis treatment plans to meet each patient’s individual diagnosis.
At Square One Health, we will help you discover the path to a better quality of life. Schedule your consultation today!
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11831 | https://wisc.pb.unizin.org/madchem/chapter/17-7-electrolysis/ | 17.7 Electrolysis – Chemistry
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Contents
Preface
Chapter 1. Essential Ideas
Introduction
1.1 Chemistry in Context
1.2 Phases and Classification of Matter
1.3 Physical and Chemical Properties
1.4 Measurements
1.5 Measurement Uncertainty, Accuracy, and Precision
1.6 Mathematical Treatment of Measurement Results
Chapter 2. Atoms, Molecules, and Ions
Introduction
2.1 Early Ideas in Atomic Theory
2.2 Evolution of Atomic Theory
2.3 Atomic Structure and Symbolism
2.4 Chemical Formulas
2.5 The Periodic Table
2.6 Molecular and Ionic Compounds
2.7 Chemical Nomenclature
Chapter 3. Composition of Substances and Solutions
Introduction
3.1 Formula Mass and the Mole Concept
3.2 Determining Empirical and Molecular Formulas
3.3 Molarity
3.4 Other Units for Solution Concentrations
Chapter 4. Stoichiometry of Chemical Reactions
Introduction
4.1 Writing and Balancing Chemical Equations
4.2 Classifying Chemical Reactions
4.3 Reaction Stoichiometry
4.4 Reaction Yields
4.5 Quantitative Chemical Analysis
Chapter 5. Thermochemistry
Introduction
5.1 Energy Basics
5.2 Calorimetry
5.3 Enthalpy
Chapter 7. Chemical Bonding and Molecular Geometry
Introduction
7.1 Ionic Bonding
7.2 Covalent Bonding
7.3 Lewis Symbols and Structures
7.4 Formal Charges and Resonance
7.5 Strengths of Ionic and Covalent Bonds
7.6 Molecular Structure and Polarity
Chapter 6. Electronic Structure and Periodic Properties of Elements
Introduction
6.1 Electromagnetic Energy
6.2 The Bohr Model
6.3 Development of Quantum Theory
6.4 Electronic Structure of Atoms (Electron Configurations)
6.5 Periodic Variations in Element Properties
Chapter 8. Advanced Theories of Covalent Bonding
Introduction
8.1 Valence Bond Theory
8.2 Hybrid Atomic Orbitals
8.3 Multiple Bonds
8.4 Molecular Orbital Theory
Chapter 9. Gases
Introduction
9.1 Gas Pressure
9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law
9.3 Stoichiometry of Gaseous Substances, Mixtures, and Reactions
9.4 Effusion and Diffusion of Gases
9.5 The Kinetic-Molecular Theory
9.6 Non-Ideal Gas Behavior
Chapter 10. Liquids and Solids
Introduction
10.1 Intermolecular Forces
10.2 Properties of Liquids
10.3 Phase Transitions
10.4 Phase Diagrams
10.5 The Solid State of Matter
10.6 Lattice Structures in Crystalline Solids
Chapter 11. Solutions and Colloids
Introduction
11.1 The Dissolution Process
11.2 Electrolytes
11.3 Solubility
11.4 Colligative Properties
11.5 Colloids
Chapter 12. Kinetics
Introduction
12.1 Chemical Reaction Rates
12.2 Factors Affecting Reaction Rates
12.3 Rate Laws
12.4 Integrated Rate Laws
12.5 Collision Theory
12.6 Reaction Mechanisms
12.7 Catalysis
Chapter 13. Fundamental Equilibrium Concepts
Introduction
13.1 Chemical Equilibria
13.2 Equilibrium Constants
13.3 Shifting Equilibria: Le Châtelier’s Principle
13.4 Equilibrium Calculations
Chapter 14. Acid-Base Equilibria
Introduction
14.1 Brønsted-Lowry Acids and Bases
14.2 pH and pOH
14.3 Relative Strengths of Acids and Bases
14.4 Hydrolysis of Salt Solutions
14.5 Polyprotic Acids
14.6 Buffers
14.7 Acid-Base Titrations
Chapter 15. Equilibria of Other Reaction Classes
Introduction
15.1 Precipitation and Dissolution
15.2 Lewis Acids and Bases
15.3 Multiple Equilibria
Chapter 16. Thermodynamics
Introduction
16.1 Spontaneity
16.2 Entropy
16.3 The Second and Third Laws of Thermodynamics
16.4 Free Energy
Chapter 17. Electrochemistry
Introduction
17.1 Balancing Oxidation-Reduction Reactions
17.2 Galvanic Cells
17.3 Standard Reduction Potentials
17.4 The Nernst Equation
17.5 Batteries and Fuel Cells
17.6 Corrosion
17.7 Electrolysis
Chapter 18. Representative Metals, Metalloids, and Nonmetals
Introduction
18.1 Periodicity
18.2 Occurrence and Preparation of the Representative Metals
18.3 Structure and General Properties of the Metalloids
18.4 Structure and General Properties of the Nonmetals
18.5 Occurrence, Preparation, and Compounds of Hydrogen
18.6 Occurrence, Preparation, and Properties of Carbonates
18.7 Occurrence, Preparation, and Properties of Nitrogen
18.8 Occurrence, Preparation, and Properties of Phosphorus
18.9 Occurrence, Preparation, and Compounds of Oxygen
18.10 Occurrence, Preparation, and Properties of Sulfur
18.11 Occurrence, Preparation, and Properties of Halogens
18.12 Occurrence, Preparation, and Properties of the Noble Gases
Chapter 19. Transition Metals and Coordination Chemistry
Introduction
19.1 Occurrence, Preparation, and Properties of Transition Metals and Their Compounds
19.2 Coordination Chemistry of Transition Metals
19.3 Spectroscopic and Magnetic Properties of Coordination Compounds
Chapter 20. Organic Chemistry
Introduction
20.1 Hydrocarbons
20.2 Alcohols and Ethers
20.3 Aldehydes, Ketones, Carboxylic Acids, and Esters
20.4 Amines and Amides
Chapter 21. Nuclear Chemistry
Introduction
21.1 Nuclear Structure and Stability
21.2 Nuclear Equations
21.3 Radioactive Decay
21.4 Transmutation and Nuclear Energy
21.5 Uses of Radioisotopes
21.6 Biological Effects of Radiation
Appendix A: The Periodic Table
Appendix B: Essential Mathematics
Appendix C: Units and Conversion Factors
Appendix D: Fundamental Physical Constants
Appendix E: Water Properties
Appendix F: Composition of Commercial Acids and Bases
Appendix G: Standard Thermodynamic Properties for Selected Substances
Appendix H: Ionization Constants of Weak Acids
Appendix I: Ionization Constants of Weak Bases
Appendix J: Solubility Products
Appendix K: Formation Constants for Complex Ions
Appendix L: Standard Electrode (Half-Cell) Potentials
Appendix M: Half-Lives for Several Radioactive Isotopes
Chemistry
Chapter 17. Electrochemistry
17.7 Electrolysis
Learning Objectives
By the end of this section, you will be able to:
Describe electrolytic cells and their relationship to galvanic cells
Perform various calculations related to electrolysis
In galvanic cells, chemical energy is converted into electrical energy. The opposite is true for electrolytic cells. In electrolytic cells, electrical energy causes nonspontaneous reactions to occur in a process known as electrolysis. The charging electric car pictured in the Chapter 18 Introduction at the beginning of this chapter shows one such process. Electrical energy is converted into the chemical energy in the battery as it is charged. Once charged, the battery can be used to power the automobile.
The same principles are involved in electrolytic cells as in galvanic cells. We will look at three electrolytic cells and the quantitative aspects of electrolysis.
The Electrolysis of Molten Sodium Chloride
In molten sodium chloride, the ions are free to migrate to the electrodes of an electrolytic cell. A simplified diagram of the cell commercially used to produce sodium metal and chlorine gas is shown in Figure 1. Sodium is a strong reducing agent and chlorine is used to purify water, and is used in antiseptics and in paper production. The reactions are
anode:2 Cl−(l)Cl 2(g)+2 e−E Cl 2/Cl−∘=+1.3 V cathode:Na+(l)+e−Na(l)E Na+/Na∘=−2.7 V overall:2 Na+(l)+2 Cl−(l)2 Na(l)+Cl 2(g)E cell∘=−4.0 V
The power supply (battery) must supply a minimum of 4 V, but, in practice, the applied voltages are typically higher because of inefficiencies in the process itself.
Figure 1. Passing an electric current through molten sodium chloride decomposes the material into sodium metal and chlorine gas. Care must be taken to keep the products separated to prevent the spontaneous formation of sodium chloride.
The Electrolysis of Water
It is possible to split water into hydrogen and oxygen gas by electrolysis. Acids are typically added to increase the concentration of hydrogen ion in solution (Figure 2). The reactions are
anode:2 H 2 O(l)O 2(g)+4 H+(a q)+4 e−E anode∘=+1.229 V cathode:2 H+(a q)+2 e−H 2(g)E cathode∘=0 V overall:2 H 2 O(l)2 H 2(g)+O 2(g)E cell∘=−1.229 V
Note that the sulfuric acid is not consumed and that the volume of hydrogen gas produced is twice the volume of oxygen gas produced. The minimum applied voltage is 1.229 V.
Figure 2. Water decomposes into oxygen and hydrogen gas during electrolysis. Sulfuric acid was added to increase the concentration of hydrogen ions and the total number of ions in solution, but does not take part in the reaction. The volume of hydrogen gas collected is twice the volume of oxygen gas collected, due to the stoichiometry of the reaction.
The Electrolysis of Aqueous Sodium Chloride
The electrolysis of aqueous sodium chloride is the more common example of electrolysis because more than one species can be oxidized and reduced. Considering the anode first, the possible reactions are
(i)2 Cl−(a q)Cl 2(g)+2 e−E anode∘=+1.35827 V(ii)2 H 2 O(l)O 2(g)+4 H+(a q)+4 e−E anode∘=+1.229 V
These values suggest that water should be oxidized at the anode because a smaller potential would be needed—using reaction (ii) for the oxidation would give a less-negative cell potential. When the experiment is run, it turns out chlorine, not oxygen, is produced at the anode. The unexpected process is so common in electrochemistry that it has been given the name overpotential. The overpotential is the difference between the theoretical cell voltage and the actual voltage that is necessary to cause electrolysis. It turns out that the overpotential for oxygen is rather high and effectively makes the reduction potential more positive. As a result, under normal conditions, chlorine gas is what actually forms at the anode.
Now consider the cathode. Three reductions could occur:
(iii)2 H+(a q)+2 e−H 2(g)E cathode∘=0 V(iv)2 H 2 O(l)+2 e−H 2(g)+2 OH−(a q)E cathode∘=−0.8277 V(v)Na+(a q)+e−Na(s)E cathode∘=−2.71 V
Reaction (v) is ruled out because it has such a negative reduction potential. Under standard state conditions, reaction (iii) would be preferred to reaction (iv). However, the pH of a sodium chloride solution is 7, so the concentration of hydrogen ions is only 1× 10−7 M. At such low concentrations, reaction (iii) is unlikely and reaction (iv) occurs. The overall reaction is then
overall:2 H 2 O(l)+2 Cl−(a q)⟶H 2(g)+Cl 2(g)+2 OH−(a q)E cell∘=−2.186 V
As the reaction proceeds, hydroxide ions replace chloride ions in solution. Thus, sodium hydroxide can be obtained by evaporating the water after the electrolysis is complete. Sodium hydroxide is valuable in its own right and is used for things like oven cleaner, drain opener, and in the production of paper, fabrics, and soap.
Electroplating
An important use for electrolytic cells is in electroplating. Electroplating results in a thin coating of one metal on top of a conducting surface. Reasons for electroplating include making the object more corrosion resistant, strengthening the surface, producing a more attractive finish, or for purifying metal. The metals commonly used in electroplating include cadmium, chromium, copper, gold, nickel, silver, and tin. Common consumer products include silver-plated or gold-plated tableware, chrome-plated automobile parts, and jewelry. We can get an idea of how this works by investigating how silver-plated tableware is produced (Figure 3).
Figure 3. The spoon, which is made of an inexpensive metal, is connected to the negative terminal of the voltage source and acts as the cathode. The anode is a silver electrode. Both electrodes are immersed in a silver nitrate solution. When a steady current is passed through the solution, the net result is that silver metal is removed from the anode and deposited on the cathode.
In the figure, the anode consists of a silver electrode, shown on the left. The cathode is located on the right and is the spoon, which is made from inexpensive metal. Both electrodes are immersed in a solution of silver nitrate. As the potential is increased, current flows. Silver metal is lost at the anode as it goes into solution.
anode:Ag(s)⟶Ag+(a q)+e−
The mass of the cathode increases as silver ions from the solution are deposited onto the spoon
cathode:Ag+(a q)+e−⟶Ag(s)
The net result is the transfer of silver metal from the anode to the cathode. The quality of the object is usually determined by the thickness of the deposited silver and the rate of deposition.
Quantitative Aspects of Electrolysis
The amount of current that is allowed to flow in an electrolytic cell is related to the number of moles of electrons. The number of moles of electrons can be related to the reactants and products using stoichiometry. Recall that the SI unit for current (I) is the ampere (A), which is the equivalent of 1 coulomb per second (1 A = 1 C s). The total charge (Q, in coulombs) is given by
Q=I×t=n×F
Where t is the time in seconds, n the number of moles of electrons, and F is the Faraday constant.
Moles of electrons can be used in stoichiometry problems. The time required to deposit a specified amount of metal might also be requested, as in the second of the following examples.
Example 1
Converting Current to Moles of Electrons
In one process used for electroplating silver, a current of 10.23 A was passed through an electrolytic cell for exactly 1 hour. How many moles of electrons passed through the cell? What mass of silver was deposited at the cathode from the silver nitrate solution?
Solution
Faraday’s constant can be used to convert the charge (Q) into moles of electrons (n). The charge is the current (I) multiplied by the time
n=Q F=10.23 C s×1 hr×60 min hr×60 s min 96,485 C/mol e−=36,830 C 96,485 C/mol e−=0.3817 mol e−
From the problem, the solution contains AgNO 3, so the reaction at the cathode involves 1 mole of electrons for each mole of silver
cathode:Ag+(a q)+e−⟶Ag(s)
The atomic mass of silver is 107.9 g/mol, so
mass Ag=0.3817 mol e−×1 mol Ag 1 mol e−×107.9 g Ag 1 mol Ag=41.19 g Ag
Check your answer: From the stoichiometry, 1 mole of electrons would produce 1 mole of silver. Less than one-half a mole of electrons was involved and less than one-half a mole of silver was produced.
Check Your Learning
Aluminum metal can be made from aluminum ions by electrolysis. What is the half-reaction at the cathode? What mass of aluminum metal would be recovered if a current of 2.50 × 10 3 A passed through the solution for 15.0 minutes? Assume the yield is 100%.
Answer:
Al 3+(a q)+3 e−⟶Al(s); 7.77 mol Al = 210.0 g Al.
Example 2
Time Required for Deposition
In one application, a 0.010-mm layer of chromium must be deposited on a part with a total surface area of 3.3 m 2 from a solution of containing chromium(III) ions. How long would it take to deposit the layer of chromium if the current was 33.46 A? The density of chromium (metal) is 7.19 g/cm 3.
Solution
This problem brings in a number of topics covered earlier. An outline of what needs to be done is:
If the total charge can be determined, the time required is just the charge divided by the current
The total charge can be obtained from the amount of Cr needed and the stoichiometry
The amount of Cr can be obtained using the density and the volume Cr required
The volume Cr required is the thickness times the area
Solving in steps, and taking care with the units, the volume of Cr required is
volume=(0.010 mm×1 cm 10 mm)×(3.3 m 2×(10,000 cm 2 1 m 2))=33 cm 3
Cubic centimeters were used because they match the volume unit used for the density. The amount of Cr is then
mass=volume×density=33 cm 3×7.19 g cm 3=237 g Cr
mol Cr=237 g Cr×1 mol Cr 52.00 g Cr=4.56 mol Cr
Since the solution contains chromium(III) ions, 3 moles of electrons are required per mole of Cr. The total charge is then
Q=4.56 mol Cr×3 mol e−1 mol Cr×96485 C mol e−=1.32×10 6 C
The time required is then
t=Q I=1.32×10 6 C 33.46 C/s=3.95×10 4 s=11.0 hr
Check your answer: In a long problem like this, a single check is probably not enough. Each of the steps gives a reasonable number, so things are probably correct. Pay careful attention to unit conversions and the stoichiometry.
Check Your Learning
What mass of zinc is required to galvanize the top of a 3.00 m × 5.50 m sheet of iron to a thickness of 0.100 mm of zinc? If the zinc comes from a solution of Zn(NO 3)2 and the current is 25.5 A, how long will it take to galvanize the top of the iron? The density of zinc is 7.140 g/cm 3.
Answer:
231 g Zn required 446 minutes.
Key Concepts and Summary
Using electricity to force a nonspontaneous process to occur is electrolysis. Electrolytic cells are electrochemical cells with negative cell potentials (meaning a positive Gibbs free energy), and so are nonspontaneous. Electrolysis can occur in electrolytic cells by introducing a power supply, which supplies the energy to force the electrons to flow in the nonspontaneous direction. Electrolysis is done in solutions, which contain enough ions so current can flow. If the solution contains only one material, like the electrolysis of molten sodium chloride, it is a simple matter to determine what is oxidized and what is reduced. In more complicated systems, like the electrolysis of aqueous sodium chloride, more than one species can be oxidized or reduced and the standard reduction potentials are used to determine the most likely oxidation (the half-reaction with the largest [most positive] standard reduction potential) and reduction (the half-reaction with the smallest [least positive] standard reduction potential). Sometimes unexpected half-reactions occur because of overpotential. Overpotential is the difference between the theoretical half-reaction reduction potential and the actual voltage required. When present, the applied potential must be increased, making it possible for a different reaction to occur in the electrolytic cell. The total charge, Q, that passes through an electrolytic cell can be expressed as the current (I) multiplied by time (Q = It) or as the moles of electrons (n) multiplied by Faraday’s constant (Q = nF). These relationships can be used to determine things like the amount of material used or generated during electrolysis, how long the reaction must proceed, or what value of the current is required.
Key Equations
Q = I×t = n×F
Chemistry End of Chapter Exercises
Identify the reaction at the anode, reaction at the cathode, the overall reaction, and the approximate potential required for the electrolysis of the following molten salts. Assume standard states and that the standard reduction potentials in Appendix L are the same as those at each of the melting points. Assume the efficiency is 100%.
(a) CaCl 2
(b) LiH
(c) AlCl 3
(d) CrBr 3
What mass of each product is produced in each of the electrolytic cells of the previous problem if a total charge of 3.33 × 10 5 C passes through each cell? Assume the voltage is sufficient to perform the reduction.
How long would it take to reduce 1 mole of each of the following ions using the current indicated? Assume the voltage is sufficient to perform the reduction.
(a) Al 3+, 1.234 A
(b) Ca 2+, 22.2 A
(c) Cr 5+, 37.45 A
(d) Au 3+, 3.57 A
A current of 2.345 A passes through the cell shown in Figure 2 for 45 minutes. What is the volume of the hydrogen collected at room temperature if the pressure is exactly 1 atm? Assume the voltage is sufficient to perform the reduction. (Hint: Is hydrogen the only gas present above the water?)
An irregularly shaped metal part made from a particular alloy was galvanized with zinc using a Zn(NO 3)2 solution. When a current of 2.599 A was used, it took exactly 1 hour to deposit a 0.01123-mm layer of zinc on the part. What was the total surface area of the part? The density of zinc is 7.140 g/cm 3. Assume the efficiency is 100%.
Glossary
electrolysis process using electrical energy to cause a nonspontaneous process to occur electrolytic cell electrochemical cell in which electrolysis is used; electrochemical cell with negative cell potentials electroplating depositing a thin layer of one metal on top of a conducting surface overpotential difference between the theoretical potential and actual potential in an electrolytic cell; the “extra” voltage required to make some nonspontaneous electrochemical reaction to occur
Solutions
Answers to Chemistry End of Chapter Exercises
(a) mass Ca 69.1 g mass Cl 2 122 g; (b) mass Li 23.9 g mass H 2 3.48 g; (c) mass Al 31.0 g mass Cl 2 122 g; (d) mass Cr 59.8 g mass Br 2 276 g
0.79 L
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Kinematic Equations: Explanation, Review, and Examples
The Albert Team
Last Updated On:
Now that you’ve learned about displacement, velocity, and acceleration, you’re well on your way to being able to describe just about any motion you could observe around you with physics. All that’s left is to learn how these values really play into each other. We know a few ways to move between them, but they’re all pretty limited. What happens if you need to find displacement, but only know acceleration and time? We don’t have a way to combine all of those values yet. Enter the four kinematic equations.
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What We Review
The Kinematic Equations
The following four kinematic equations come up throughout physics from the earliest high school class to the highest level college course:
| |
| Kinematic Equationsv=v_{0}+at\Delta x=\dfrac{v+v_{0}}{2} t\Delta x=v_{0}t+\frac{1}{2}at^{2}v^{2}=v_{0}^{2}+2a\Delta x |
Don’t let all of these numbers and symbols intimidate you. We’ll talk through each one – what they mean and when we use them. By the end of this post, you’ll be a master of understanding and implementing each of these physics equations. Let’s start with defining what all of those symbols mean.
The First Kinematic Equation
| |
| v=v_{0}+at |
This physics equation would be read as “the final velocity is equal to the initial velocity plus acceleration times time”. All it means is that if you have constant acceleration for some amount of time, you can find the final velocity. You’ll use this one whenever you’re looking at changing velocities with a constant acceleration.
The Second Kinematic Equation
| |
| \Delta x=\dfrac{v+v_{0}}{2} t |
This one is read as “displacement equals final velocity plus initial velocity divided by two times time”. You’ll use this one whenever you don’t have an acceleration to work with but you need to relate a changing velocity to a displacement.
The Third Kinematic Equation
| |
| \Delta x=v_{0}t+\frac{1}{2}at^{2} |
This one may look a bit scarier as it is longer than the others, but it is read as “displacement equals initial velocity times time plus one half acceleration times time squared”. All it means is that our displacement can be related to our initial velocity and a constant acceleration without having to find the final velocity. You’ll use this one when final velocity is the only value you don’t know yet.
It is worth noting that this kinematic equation has another popular form: x=x_{0}+v_{0}t+\frac{1}{2}at^{2}. While that may seem even more intimidating, it’s actually exactly the same. The only difference here is that we have split up \Delta x into x-x_{0} and then solved to get x on its own. This version can be particularly helpful if you’re looking specifically for a final or initial position rather than just an overall displacement.
The Fourth Kinematic Equation
| |
| v^{2}=v_{0}^{2}+2a\Delta x |
Our last kinematic equation is read as “final velocity squared equals initial velocity squared plus two times acceleration times displacement”. It’s worth noting that this is the only kinematic equation without time in it. Many starting physicists have been stumped by reaching a problem without a value for time. While staring at an equation sheet riddled with letters and numbers can be overwhelming, remembering you have this one equation without time will come up again and again throughout your physics career.
It may be worth noting that all of these are kinematic equations for constant acceleration. While this may seem like a limitation, we learned before that high school physics courses generally utilize constant acceleration so we don’t need to worry about it changing yet. If you do find yourself in a more advanced course, new physics equations will be introduced at the appropriate times.
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How to Approach a Kinematics Problem
So now that we have all of these different kinematic equations, how do we know when to use them? How can we look at a physics word problem and know which of these equations to apply? You must use problem-solving steps. Follow these few steps when trying to solve any complex problems, and you won’t have a problem.
Step 1: Identify What You Know
This one probably seems obvious, but skipping it can be disastrous to any problem-solving endeavor. In physics problems, this just means pulling out values and directions. If you can add the symbol to go with the value (writing t=5\text{ s} instead of just 5\text{ s}, for example), even better. It’ll save time and make future steps even easier.
Step 2: Identify the Goal
In physics, this means figuring out what question you’re actually being asked. Does the question want you to find the displacement? The acceleration? How long did the movement take? Figure out what you’re being asked to do and then write down the symbol of the value you’re solving for with a question mark next to it (t=\text{?}, for example). Again, this feels obvious, but it’s also a vital step.
Step 3: Gather Your Tools
Generally, this means a calculator and an equation. You’ll want to look at all of the symbols you wrote down and pick the physics equation for all of them, including the unknown value. Writing everything down beforehand will make it easier to pull a relevant equation than having to remember what values you need while searching for the right equation. You can use the latter method, but you’re far more likely to make a mistake and feel frustrated that way.
Step 4: Put it all Together
Plug your values into your equation and solve for the unknown value. This will usually be your last step, though you may find yourself having to repeat it a few times for exceptionally complex problems. That probably won’t come up for quite a while, though. After you’ve found your answer, it’s generally a good idea to circle it to make it obvious. That way, whoever is grading you can find it easily and you can easily keep track of which problems you’ve already completed while flipping through your work.
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Kinematic Equation 1: Review and Examples
To learn how to solve problems with these new, longer equations, we’ll start with v=v_{0}+at. This kinematic equation shows a relationship between final velocity, initial velocity, constant acceleration, and time. We will explore this equation as it relates to physics word problems. This equation is set up to solve for velocity, but it can be rearranged to solve for any of the values it contains. For this physics equation and the ones following, we will look at one example finding the variable that has already been isolated and one where a new variable needs to be isolated using the steps we just outlined. So, let’s jump into applying this kinematic equation to a real-world problem.
Example 1
A car sits at rest waiting to merge onto a highway. When they have a chance, they accelerate at 4\text{ m/s}^2 for 7\text{ s}. What is the car’s final velocity?
Step 1: Identify What You Know
We have a clearly stated acceleration and time, but there’s no clearly defined initial velocity here. Instead, we have to take this from context. We know that the car “sits at rest” before it starts moving. This means that our initial velocity in this situation is zero. Other context clues for an object starting at rest is if it is “dropped” or if it “falls”. Our other known values will be even easier to pull as we were actually given numerical values. Now it’s time to put everything into a list.
v_{0}=0\text{ m/s}
a=4\text{ m/s}^2
t=7\text{ s}
Step 2: Identify the Goal
Our goal here was clearly stated: find the final velocity. We’ll still want to list that out so we can see exactly what symbols we have to work with on this problem.
v=\text{?}
Step 3: Gather Your Tools
We already know which of the kinematic equations we’re using, but if we didn’t, this would be where we search our equation sheet for the right one. Regardless, we’ll want to write that down too.
v=v_{0}+at
Step 4: Put it All Together
At this point, we’ll plug all of our values into our kinematic equation. If you’re working on paper, there’s no need to repeat anything we’ve put above. That being said, for the purposes of digital organization and so you can see the full problem in one spot, we will be rewriting things here.
v_{0}=0\text{ m/s}
a=4\text{ m/s}^2
t=7\text{ s}
v=\text{?}
v=v_{0}+at
v=0\text{ m/s}+4\text{ m/s}^2\cdot 7\text{ s}
v=28\text{ m/s}
Example 2
Now let’s get a bit trickier with a problem that will require us to rearrange our kinematic equation.
A ball rolls toward a hill at 3\text{ m/s}. It rolls down the hill for 5\text{ s} and has a final velocity of 18\text{ m/s}. What was the ball’s acceleration as it rolled down the hill?
Step 1: Identify What You Know
Just like before, we’ll make a list of our known values:
v_{0}=3\text{ m/s}
t=5\text{ s}
v=18\text{ m/s}
Step 2: Identify the Goal
Again, our goal was clearly stated, so let’s add it to our list:
a=\text{?}
Step 3: Gather Your Tools
We already know which equation we’re using, but let’s pretend we didn’t. We know that we need to solve for acceleration, but if you look at our original list of kinematic equations, there isn’t one that’s set up to solve for acceleration:
| |
| Kinematic Equationsv=v_{0}+at\Delta x=\dfrac{v+v_{0}}{2} t\Delta x=v_{0}t+\frac{1}{2}at^{2}v^{2}=v_{0}^{2}+2a\Delta x |
This begs the question, how to find acceleration (or any value) that hasn’t already been solved for? The answer is to rearrange an equation. First, though, we need to pick the right one. We start by getting rid of the second equation in this list as it doesn’t contain acceleration at all. Our options are now:
v=v_{0}+at
\Delta x=v_{0}t+\dfrac{1}{2}at^{2}
v^{2}=v_{0}^{2}+2a\Delta x
Now we’ll need to look at the first list we made of what we know. We know the initial velocity, time, and final velocity. There’s only one equation that has all the values we’re looking for and all of the values we know with none that we don’t. This is the first kinematic equation:
v=v_{0}+at
In this case, we knew the kinematic equation coming in so this process of elimination wasn’t necessary, but that won’t often be the case in the future. You’ll likely have to find the correct equation far more often than you’ll have it handed to you. It’s best to practice finding it now while we only have a few equations to work with.
Step 4: Put it All Together
Like before, we’ll be rewriting all of our relevant information below, but you won’t need to if you’re working on paper.
v_{0}=3\text{ m/s}
t=5\text{ s}
v=18\text{ m/s}
a=\text{?}
v=v_{0}+at
Although you can plug in values before rearranging the equation, in physics, you’ll usually see the equation be rearranged before values are added. This is mainly done to help keep units where they’re supposed to be and to avoid any mistakes that could come from moving numbers and units rather than just a variable. We’ll be taking the latter approach here. Follow the standard PEMDAS rules for rearranging the equation and then write it with the variable we’ve isolated on the left. While that last part isn’t necessary, it is a helpful organizational practice:
v-v_{0}=at
a=\dfrac{v-v_{0}}{t}
For a review of solving literal equations, visit this post! Now we can plug in those known values and solve:
a=\dfrac{18\text{ m/s}-3\text{ m/s}}{5\text{ s}}
a=3\text{ m/s}^2
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Kinematic Equation 2: Review and Examples
Next up in our four kinematics equations is \Delta x=\dfrac{v+v_{0}}{2} t. This one relates an object’s displacement to its average velocity and time. The right-hand side shows the final velocity plus the initial velocity divided by two – the sum of some values divided by the number of values, or the average. Although this equation doesn’t directly show a constant acceleration, it still assumes it. Applying this equation when acceleration isn’t constant can result in some error so best not to apply it if a changing acceleration is mentioned.
Example 1
A car starts out moving at 10\text{ m/s} and accelerates to a velocity of 24\text{ m/s}. What displacement does the car cover during this velocity change if it occurs over 10\text{ s}?
Step 1: Identify What You Know
v_{0}=10\text{ m/s}
v=24\text{ m/s}
t=10\text{ s}
Step 2: Identify the Goal
\Delta x=\text{?}
Step 3: Gather Your Tools
\Delta x=\dfrac{v+v_{0}}{2} t
Step 4: Put it All Together
This time around we won’t repeat everything here. Instead, We’ll jump straight into plugging in our values and solving our problem:
\Delta x=\dfrac{v+v_{0}}{2} t
\Delta x=\dfrac{24\text{ m/s}+10\text{ m/s}}{2} \cdot 10\text{ s}
\Delta x=\dfrac{34\text{ m/s}}{2} \cdot 10\text{ s}
\Delta x=17\text{ m/s} \cdot 10\text{ s}
\Delta x=170\text{ m}
Example 2
A ball slows down from 15\text{ m/s} to 3\text{ m/s} over a distance of 36\text{ m}. How long did this take?
Step 1: Identify What You Know
v_{0}=15\text{ m/s}
v=3\text{ m/s}
\Delta x=36\text{ m}
Step 2: Identify the Goal
t=\text{?}
Step 3: Gather Your Tools
We don’t have a kinematic equation for time specifically, but we learned before that we can rearrange certain equations to solve for different variables. So, we’ll pull the equation that has all of the values we need and isolate the variable we want later:
\Delta x=\dfrac{v+v_{0}}{2} t
Step 4: Put it All Together
Again, we won’t be rewriting anything, but we will begin by rearranging our equation to solve for time:
\Delta x=\dfrac{v+v_{0}}{2} t
2\Delta x=(v+v_{0})t
t=\dfrac{2\Delta x}{v+v_{0}}
Now we can plug in our known values and solve for time.
t=\dfrac{2\cdot 36\text{ m}}{3\text{ m/s}+15\text{ m/s}}
t=\dfrac{72\text{ m}}{18\text{ m/s}}
t=4\text{ s}
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Kinematic Equation 3: Review and Examples
Our next kinematic equation is \Delta x=v_{0}t+\frac{1}{2}at^{2}. This time we are relating our displacement to our initial velocity, time, and acceleration. The only odd thing you may notice is that it doesn’t include our final velocity, only the initial. This equation will come in handy when you don’t have a final velocity that was stated either directly as a number or by a phrase indicating the object came to rest. Just like before, we’ll use this equation first to find a displacement, and then we’ll rearrange it to find a different value.
Example 1
A rocket is cruising through space with a velocity of 50\text{ m/s} and burns some fuel to create a constant acceleration of 10\text{ m/s}^2. How far will it have traveled after 5\text{ s}?
Step 1: Identify What You Know
v_{0}=50\text{ m/s}
a=10\text{ m/s}^2
t=5\text{ s}
Step 2: Identify the Goal
\Delta x=\text{?}
Step 3: Gather Your Tools
\Delta x=v_{0}t+\frac{1}{2}at^{2}
Step 4: Put it All Together
\Delta x=v_{0}t+\frac{1}{2}at^2
\Delta x=(50\text{ m/s})(5\text{ s})+\frac{1}{2}(10\text{ m/s}^2)(5\text{ s})^2
\Delta x=250\text{ m}+\frac{1}{2}(10\text{ m/s}^2)(25\text{ s}^2)
\Delta x=250\text{ m}+\frac{1}{2}(250\text{ m})
\Delta x=250\text{ m}+125\text{ m}
\Delta x=375\text{ m}
At this point, it appears that these problems seem to be quite long and take several steps. While that is an inherent part of physics in many ways, it will start to seem simpler as time goes on. This problem presents the perfect example. While it may have been easy to combine lines 4 and 5 mathematically, they were shown separately here to make sure the process was as clear as possible. While you should always show all of the major steps of your problem-solving process, you may find that you are able to combine some of the smaller steps after some time of working with these kinematic equations.
Example 2
Later in its journey, the rocket is moving along at 20\text{ m/s} when it has to fire its thrusters again. This time it covers a distance of 500\text{ m} in 10\text{ s}. What was the rocket’s acceleration during this thruster burn?
Step 1: Identify What You Know
v_{0}=20\text{ m/s}
\Delta x=500\text{ m}
t=10\text{ s}
Step 2: Identify the Goal
a=\text{?}
Step 3: Gather Your Tools
\Delta x=v_{0}t+\frac{1}{2}at^{2}
Step 4: Put it All Together
As usual, we’ll begin by rearranging the equation, this time to solve for acceleration.
\Delta x=v_{0}t+\frac{1}{2}at^2
\Delta x-v_{0}t=\frac{1}{2}at^2
2(\Delta x-v_{0}t)=at^2
a=2\dfrac{\Delta x-v_{0}t}{t^2}
Now we can plug in our known values to find the value of our acceleration.
a=2\dfrac{500\text{ m}-20\text{ m/s}\cdot 10\text{ s}}{(10\text{ s})^2}
a=2\dfrac{500\text{ m}-200\text{ m}}{(10\text{ s})^2}
a=2\dfrac{300\text{ m}}{(10\text{ s})^2}
a=2\dfrac{300\text{ m}}{100\text{ s}^2}
a=2\cdot 3\text{ m/s}^2
a=6\text{ m/s}^2
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Kinematic Equation 4: Review and Examples
The last of the kinematic equations that we will look at is v^{2}=v_{0}^{2}+2a\Delta x. This one is generally the most complicated looking, but it’s also incredibly important as it is our only kinematic equation that does not involve time. It relates final velocity, initial velocity, acceleration, and displacement without needing a time over which a given motion occurred. For this equation, as with the others, let’s solve it as is and then rearrange it to solve for a different variable.
Example 1
A car exiting the highway begins with a speed of 25\text{ m/s} and travels down a 100\text{ m} long exit ramp with a deceleration (negative acceleration) of 3\text{ m/s}^2. What is the car’s velocity at the end of the exit ramp?
Step 1: Identify What You Know
v_{0}=25\text{ m/s}
\Delta x=100\text{ m}
a=-3\text{ m/s}^2
Note that our acceleration here is a negative value. That is because our problem statement gave us a deceleration instead of an acceleration. Whenever you have a deceleration, you’ll make the value negative to use it as an acceleration in your problem-solving. This also tells us that our final velocity should be less than our initial velocity so we can add that to the list of what we know as well.
Final velocity will be less than initial.
Being able to know something to help check your answer at the end is what makes this subject a bit easier than mathematics for some students.
Step 2: Identify the Goal
v=\text{?}
Step 3: Gather Your Tools
v^{2}=v_{0}^{2}+2a\Delta x
Step 4: Put it All Together
While we generally try to not have any operations going on for the isolated variable, sometimes it’s actually easier that way. Having your isolated variable raised to a power is generally a time to solve before simplifying. This may seem like an arbitrary rule, and in some ways it is, but as you continue through your physics journey you’ll come up with your own practices for making problem-solving easier.
v^{2}=v_{0}^{2}+2a\Delta x
v^{2}=(25\text{ m/s})^{2}+2\cdot (-3\text{ m/s}^{2})\cdot 100\text{ m}
v^{2}=625\text{ m}^{2}/\text{ s}^{2}+2\cdot (-3\text{ m/s}^{2})\cdot 100\text{ m}
v^{2}=625\text{ m}^{2}/\text{ s}^{2}+2\cdot (-300\text{ m}^{2}/\text{ s}^{2})
v^{2}=625\text{ m}^{2}/\text{ s}^{2}-600\text{ m}^{2}/\text{ s}^{2}
v^{2}=25\text{ m}^{2}/\text{ s}^{2}
Now that we have both sides simplified, we’ll take the square root to eliminate the exponent on the left-hand side:
v=5\text{ m/s}
If we remember back at the beginning, we said that our final velocity would have to be less than our initial velocity because the problem statement told us that we were decelerating. Our initial velocity was 25\text{ m/s} which is, indeed, greater than 5\text{ m/s} so our answer checks out.
Example 2
A ghost is sliding a wrench across a table to terrify the mortal onlooker. The wrench starts with a velocity of 2\text{ m/s} and accelerates to a velocity of 5\text{ m/s} over a distance of 7\text{ m}. What acceleration did the ghost move the wrench with?
Step 1: Identify What You Know
v_{0}=2\text{ m/s}
v=5\text{ m/s}
\Delta x=7\text{ m}
We can also make an inference about our acceleration here – that it will be positive. Not every problem will tell you clearly the direction of the acceleration, but if your final velocity is greater than your initial velocity, you can be sure that your acceleration will be positive.
Positive acceleration
You’ll get better at picking up on subtle hints like this as you continue your physics journey and your brain starts naturally picking up on some patterns. You’ll likely find this skill more and more helpful as it develops and as problems get more difficult.
Step 2: Identify the Goal
a=\text{?}
Step 3: Gather Your Tools
v^{2}=v_{0}^{2}+2a\Delta x
Step 4: Put it All Together
We’ll start by rearranging our equation to solve for acceleration.
v^{2}=v_{0}^{2}+2a\Delta x
v^{2}-v_{0}^{2}=2a\Delta x
a\Delta x=\frac{1}{2}(v^{2}-v_{0}^{2})
a=\dfrac{1}{2}\dfrac{v^{2}-v_{0}^{2}}{\Delta x}
As usual, now that we’ve rearranged our equation, we can plug in our values.
a=\dfrac{1}{2}\dfrac{(5\text{ m/s})^{2}-(2\text{ m/s})^{2}}{7\text{ m}}
a=\dfrac{1}{2}\dfrac{25\text{ m}^{2}/\text{s}^{2}-(2\text{ m/s})^{2}}{7\text{ m}}
a=\dfrac{1}{2}\dfrac{25\text{ m}^{2}/\text{s}^{2}-4\text{ m}^{2}/\text{s}^{2}}{7\text{ m}}
a=\dfrac{1}{2}\dfrac{21\text{ m}^{2}/\text{s}^{2}}{7\text{ m}}
a=\frac{1}{2}3\text{ m/s}^{2}
a=1.5\text{ m/s}^2
Again, we can go back to the beginning when we said our acceleration would be a positive number and confirm that it is.
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Problem-Solving Strategies
At this point, you’re likely getting the sense that physics will be a lot of complex problem-solving. If so, your senses are correct. In many ways, physics is the science of explaining nature with mathematical equations. There’s a lot that goes into developing and applying these equations, but at this point in your physics career, you’ll find that the majority of your time will likely be spent on applying equations to word problems. If you feel that your problem-solving skills could still use some honing, check out more examples and strategies from this post by the Physics Classroom or through this video-guided tutorial from Khan Academy.
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Conclusion
That was a lot of equations and examples to take in. Eventually, whether you’re figuring out how to find a constant acceleration or how to solve velocity when you don’t have a value for time, you’ll know exactly which of the four kinematic equations to apply and how. Just keep the problem-solving steps we’ve used here in mind, and you’ll be able to get through your physics course without any unsolvable problems.
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To prove the given inequality
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Question:-
If a,b,c a,b,c are positive real numbers which are in H.P. show that
a+b 2 a−b+c+b 2 c−b≥4 a+b 2 a−b+c+b 2 c−b≥4
Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
a+b 2 a−b+c+b 2 c−b 2≥a+b 2 a−b⋅c+b 2 c−b−−−−−−−−−−−−−√a+b 2 a−b+c+b 2 c−b 2≥a+b 2 a−b⋅c+b 2 c−b
Evaluating the right hand side of the inequality,
(a+b 2 a−b⋅c+b 2 c−b)=(b 2+(a+c)b+a c b 2−2 b(a+c)+4 a c)(a+b 2 a−b⋅c+b 2 c−b)=(b 2+(a+c)b+a c b 2−2 b(a+c)+4 a c)
Now, as a,b,c a,b,c are in H.P., hence we get the relation
b=2 a c a+c b=2 a c a+c
On substituting this result in the equation, we get
10 a 2 c 2+3 a c(a 2+c 2)4 a 2 c 2=5 2+3(a 2+c 2)4 a c 10 a 2 c 2+3 a c(a 2+c 2)4 a 2 c 2=5 2+3(a 2+c 2)4 a c
Now from the above calculations, what I observed is that, that for the proof to be valid, the condition 3(a 2+c 2)4 a c=3 2 3(a 2+c 2)4 a c=3 2 should be satisfied. The final conclusion to which I arrived was a=c a=c. Now although the numbers would still be in H.P. but it doesn't prove the result for the numbers which are in H.P. and are not equal.
Any other approach to the proof is invited too.
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asked Jun 30, 2016 at 13:35
user350331user350331
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Well, isn't it Harmonic Progression user350331 –user350331 2016-06-30 13:38:29 +00:00 Commented Jun 30, 2016 at 13:38
yes, given that b = (2ac)/(a+c), this name makes sense amakelov –amakelov 2016-06-30 13:39:26 +00:00 Commented Jun 30, 2016 at 13:39
@DietrichBurde Well...is it a joke?Soham –Soham 2016-06-30 13:39:39 +00:00 Commented Jun 30, 2016 at 13:39
@DietrichBurde Don't we commonly use the abbreviations-AP,GP and HP.....(I do)...Soham –Soham 2016-06-30 13:40:39 +00:00 Commented Jun 30, 2016 at 13:40
@user350331: aren't you almost done? You just need to observe that (a^2+c^2)/(2ac) \geq 1 (because in fact it suffices 3(a^2+c^2)/(4ac)\geq 3/2)amakelov –amakelov 2016-06-30 13:42:07 +00:00 Commented Jun 30, 2016 at 13:42
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So, why don't we rewrite the HP condition in a nicer way, as
2 b=1 a+1 c 2 b=1 a+1 c
and let's multiply the numerator and denominator of the first fraction by 2/b 2/b. We get
a+b 2 a−b=a(1/a+1/c)+2 2 a(1/a+1/c)−2=3+a/c 2 a/c=3 c+a 2 a a+b 2 a−b=a(1/a+1/c)+2 2 a(1/a+1/c)−2=3+a/c 2 a/c=3 c+a 2 a
Similarly, the other one becomes 3 a+c 2 c 3 a+c 2 c, so what we want to show reduces to
3 c+a 2 a+3 a+c 2 c≥4 3 c+a 2 a+3 a+c 2 c≥4
which reduces to
c a+a c≥2 c a+a c≥2
which follows directly from AM-GM.
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answered Jun 30, 2016 at 14:03
amakelovamakelov
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The theorem you are asked to prove is not true in general. The additional condition needed is that a≥b a≥b. And then you can do the proof without AM-GM.
Assuming (a,b,c)(a,b,c) are in H.P. in the order
a=1 x,b=1 x+d,c=1 x+2 d a=1 x,b=1 x+d,c=1 x+2 d
then the LHS is
4 x 2+8 d x+6 d 2 x 2+2 d x 4 x 2+8 d x+6 d 2 x 2+2 d x
Consider the case of x 2+2 d x>0 x 2+2 d x>0:
4 x 2+8 d x+6 d 2 x 2+2 d x−4=1 x 2+2 d[4 x 2+8 d x+6 d 2−4(x 2+2 d x)]=6 d 2 x 2+2 d≥=0 4 x 2+8 d x+6 d 2 x 2+2 d x−4=1 x 2+2 d[4 x 2+8 d x+6 d 2−4(x 2+2 d x)]=6 d 2 x 2+2 d≥=0
Now consider the case where x 2+2 d x<0 x 2+2 d x<0 and d≠0 d≠0:
4 x 2+8 d x+6 d 2 x 2+2 d x−4=6 d 2 x 2+2 d<0 4 x 2+8 d x+6 d 2 x 2+2 d x−4=6 d 2 x 2+2 d<0
So a good counterexample would be x=3,d=−1 x=3,d=−1, giving a=1 3,b=1 2,c=1 a=1 3,b=1 2,c=1.
If you add the condition that a≥b a≥b then you find that forces x 2+2 d x≥0 x 2+2 d x≥0 and the statement then is true.
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answered Jun 30, 2016 at 14:07
Mark FischlerMark Fischler
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1 I think the question has already forced the condition x 2+2 d x>0 x 2+2 d x>0, because if so is not the case then the number c c comes out as negative which can't be accepted as per the question. Also, from the condition a≥b a≥b, we get d≥0 d≥0, which doesn't seem to be a necessary condition as illustrated in your countew example. So, I think the question has no loopholes.user350331 –user350331 2016-07-04 08:35:44 +00:00 Commented Jul 4, 2016 at 8:35
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11834 | https://astronomy.stackexchange.com/questions/748/how-does-a-neutron-star-collapse-into-a-black-hole | Stack Exchange Network
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How does a neutron star collapse into a black hole?
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$\begingroup$
We know the spectacular explosions of supernovae, that when massive enough, form black holes. The explosive emission of both electromagnetic radiation and massive amounts of matter is clearly observable and studied quite thoroughly. If the star was massive enough, the remnant will be a black hole. If it wasn't massive enough, it will be a neutron star.
Now there's another mode of creation of black holes: the neutron star captures enough matter, or two neutron stars collide, and their combined mass creates enough gravity force to cause another collapse - into a black hole.
What effects are associated with this? Is there an explosive release of some kind of radiation or particles? Is it observable? What physical processes occur in the neutrons as they are subjected to the critical increase of pressure? What is the mass of the new black hole, comparing to its neutron star of origin?
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edited Oct 1, 2023 at 13:29
ProfRob
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$\begingroup$ There have been a few measurements of BH masses right in the middle of the mass gap. See, e.g. Zdziarski et al. 2013 (adsabs.harvard.edu/abs/2013MNRAS.429L.104Z) and Neustroev et al. 2014 (adsabs.harvard.edu/abs/2014MNRAS.445.2424N). $\endgroup$
user11330
– user11330
2016-03-25 13:39:56 +00:00
Commented Mar 25, 2016 at 13:39
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$\begingroup$ Interesting. But the uncertainties on the masses still allow them to be 4 solar masses or more in both cases. It is clearly a subject with a lot more work to be done and these two papers both provide an interesting discussion along the lines of what I present in my answer. $\endgroup$
ProfRob
– ProfRob
2016-03-25 23:26:26 +00:00
Commented Mar 25, 2016 at 23:26
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A neutron star must have a minimum mass of at least 1.4x solar masses (that is, 1.4x mass of our Sun) in order to become a neutron star in the first place. See Chandrasekhar limit on wikipedia for details.
A neutron star is formed during a supernova, an explosion of a star that is at least 8 solar masses.
The maximum mass of a neutron star is 3 solar masses. If it gets more massive than that, then it will collapse into a quark star, and then into a black hole.
We know that 1 electron + 1 proton = 1 neutron;
1 neutron = 3 quarks = up quark + down quark + down quark;
1 proton = 3 quarks = up quark + up quark + down quark;
A supernova results in either a neutron star (between 1.4 and 3 solar masses), a quark star(about 3 solar masses), or a black hole(greater than 3 solar masses), which is the remaining collapsed core of the star.
During a supernova, most of the stellar mass is blown off into space, forming elements heavier than iron which cannot be generated through stellar nucleosynthesis, because beyond iron, the star requires more energy to fuse the atoms than it gets back.
During the supernova collapse, the atoms in the core break up into electrons, protons and neutrons.
In the case that the supernova results in a neutron star core, the electrons and protons in the core are merged to become neutrons, so the newly born 20-km-diameter neutron star containing between 1.4 and 3 solar masses is like a giant atomic nucleus containing only neutrons.
If the neutron star's mass is then increased, neutrons become degenerate, breaking up into their constituent quarks, thus the star becomes a quark star; a further increase in mass results in a black hole.
The upper/lower mass limit for a quark star is not known (or at least I couldn't find it), in any case, it is a narrow band around 3 solar masses, which is the minimum stable mass of a black hole.
When you talk about a black hole with a stable mass (at least 3 solar masses), it is good to consider that they come in 4 flavors: rotating-charged, rotating-uncharged, non-rotating-charged, non-rotating-uncharged.
What we would see visually during the transformation would be a hard radiation flash. This is because during the collapse, the particles on/near the surface have time to emit hard radiation as they break up before going into the event horizon; so this could be one of the causes of gamma ray bursts (GRBs).
We know that atoms break up into protons, neutrons, electrons under pressure.
Under more pressure, protons and electrons combine into neutrons.
Under even more pressure, neutrons break down into quarks.
Under still more pressure, perhaps quarks break down into still smaller particles.
Ultimately the smallest particle is a string: open or closed loop, and has a Planck length, which is many orders of magnitude smaller than a quark. if a string is magnified so it is 1 millimeter in length, then a proton would have a diameter that would fit snugly between the Sun and Epsilon Eridani, 10.5 light years away; that's how big a proton is compared to a string, so you can imagine there are perhaps quite a few intermediate things between quarks and strings.
Currently it looks like several more decades will be needed to figure out all the math in string theory, and if there is anything smaller than strings then a new theory will be required, but so far string theory looks good; see the book Elegant Universe by Brian Greene.
A string is pure energy and Einstein said mass is just a form of energy, so the collapse into a black hole really breaks down the structure of energy that gives the appearance of mass/matter/baryonic particles, and leaves the mass in its most simple form, open or closed strings, that is, pure energy bound by gravity.
We know that black holes (which are not really holes or singularities, as they do have mass, radius, rotation, charge and hence density, which varies with radius) can evaporate, giving up their entire mass in the form of radiation, thus proving they are actually energy. Evaporation of a black hole occurs if its mass is below the minimum mass of a stable black hole, which is 3 solar masses; the Schwarzschild radius equation even tells you what the radius of a black hole is given its mass, and vice versa.
So you could transform anything you want, such as your pencil, into a black hole if you wanted to, and could compress it into the required size for it to become a black hole; it is just that it would immediately transform itself (evaporate) completely into a flash of hard radiation, because a pencil is less than the stable black hole mass (3 solar masses).
This is why the CERN experiment could never have created a black hole to swallow the Earth - a subatomic black hole, even one with the mass of the entire Earth, or the Sun, would evaporate before swallowing anything; there is not enough mass in our solar system to make a stable (3 solar mass) black hole.
A simple way for a neutron star to become more massive in order to be able to turn into a black hole is to be part of a binary system, where it is close enough to another star that the neutron star and its binary pair orbit each other, and the neutron star siphons off gas from the other star, thus gaining mass.
Here is a nice drawing showing exactly that.
Matter falling into a black hole is accelerated toward light speed. As it is accelerated, the matter breaks down into subatomic particles and hard radiation, that is, X-rays and gamma rays. A black hole itself is not visible, but the light from infalling matter that is accelerated and broken up into particles is visible. Black holes can also cause a gravitational lens effect on the light of background stars/galaxies.
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$\begingroup$ I'll just list the inaccuracies of this answer: (i) Neutrons stars must be more massive than 1.4Msun. Not true and several are known not to be. The Chandrasekhar mass depends on composition - supernovae cores are not made of carbon (for which 1.4 Msun is appropriate). (ii) The maximum mass of a neutron star is at least 2Msun (the highest measured). We do not know how much higher, but general relativity places an upper bound of about 3Msun. (iii) Nobody knows if quark stars exist. (iv) Neutron stars are not only made of neutrons. (v) The neutrons in a neutrons star are already degenerate. $\endgroup$
ProfRob
– ProfRob
2014-12-17 19:23:02 +00:00
Commented Dec 17, 2014 at 19:23
13
$\begingroup$ (vi) Black holes appear observationally to have a minimum mass of about 4-5Msun (Ozel at al. 2012). (vii) The minimum stable mass for a black hole is definitely not 3Msun. (viii) GRBs are not caused by matter falling into black holes (or provide a reference for any work that says so). (ix) Black hole evaporation may be relevant for micro blackholes it is utterly irrelevant for stellar-sized black holes. (x) The paragraph about the pencil vanishing in a flash is nonsense. $\endgroup$
ProfRob
– ProfRob
2014-12-17 19:32:09 +00:00
Commented Dec 17, 2014 at 19:32
2
$\begingroup$ Might I just mention two things: First off yes maybe his answer had some figures that included exceptions but I don't see why this requires in depth listing. Secondly I must mention that you are supporting string theory as if it is incontrovertible, which I am afraid is not true in the slightest. It is, of coarse, a legitimate theory, however you really must mention that. $\endgroup$
trevorKirkby
– trevorKirkby
2014-12-21 22:20:47 +00:00
Commented Dec 21, 2014 at 22:20
3
$\begingroup$ @userLTK The largest measured neutron star mass is 2 solar masses. The gap you comment on is dealt with in my answer and there are at least two classes of explanation for it. Neutron stars have a GR instability that makes them collapse well before they get near their Schwarzschild radii. $\endgroup$
ProfRob
– ProfRob
2015-06-02 06:20:21 +00:00
Commented Jun 2, 2015 at 6:20
3
$\begingroup$ I don't know why you speculate about possible components of quarks. Quarks (and leptons) are fundamental in the Standard Model, there's no evidence that they are composite particles. And even in string theory, a quark isn't made of strings, it is a string in a particular vibrational mode. $\endgroup$
PM 2Ring
– PM 2Ring
2017-04-29 14:28:43 +00:00
Commented Apr 29, 2017 at 14:28
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Just to focus on one part of your question. Whilst it might be possible for a neutron star to accrete material, or for two neutron stars to collide, in order to form black holes, this kind of event must be quite rare (although see below).
The distribution of measured neutron star and black holes masses can be fitted with an estimated true distribution. Here it is, from Ozel et al. (2012). You can see there is a distinct gap between the highest mass neutron stars (currently the record holder has a mass of about $2M_{\odot}$ and the smallest black holes (about $5M_{\odot}$). This confirms slightly earlier work by Farr et al. (2011).
The merging of neutron stars must happen though. The obvious example is the Hulse-Taylor binary neutron star system, where the two object are spiralling together, presumably by the emission of gravitational waves, and will merge in about 300 million years. The combined mass of the 2 neutron stars is $2.83M_{\odot}$, but the mass of any black hole they create would be lower, with the difference radiated away as neutrinos and gravitational waves.
Merging neutron stars (or merging neutron star + black hole binaries) are thought to be the progenitors of short duration gamma ray bursts or so-called Kilonova events that are generally seen in high redshift galaxies. These typically last a second or less, but involve an energy release of about $\sim 10^{44}$ J. They may produce a black hole, or perhaps a more massive neutron star. There will also be a gravitational wave signature (a "chirp") that could be detected by the next generation of gravitational wave experiments (now a reality). These black holes may be isolated and hence not represented in the mass distribution above. A further observational signature of these events may be in the form of the current levels of a number of heavy r-process elements, like Iridium and Gold, that may mostly be produced in these events.
As for accretion onto an existing neutron star - well it looks quite rare because there may be a large gap between the highest masses at which neutron stars are produced in supernovae (maybe $1.5M_{\odot}$) and the maximum mass of a neutron star. We know that the latter is at least $2M_{\odot}$, but it could be higher, perhaps $3M_{\odot}$, the maximum allowed by General Relativity. As to the outcome of this hypothetical event, well sticking to non-speculative physics, the most likely thing to happen would be the production of massive hyperons in the neutron star core at sufficiently high densities ($>10^{18}$ kg/m$^3$), which would lead to an instability (due to the removal of degenerate neutrons that are providing the majority of support); the neutron star may then slip inside its event horizon (about 6km for a $2M_{\odot}$ neutron star) and become a black hole. Some sort of explosion seems unlikely, though a gravitational wave signature might be possible.
EDIT: An update on the NS/BH mass distribution above. I saw a talk recently at a conference - the explanation of the distribution has two broad thrusts; either the black holes are not produced in this mass range because of the physics of the progenitors, or there is a strong observational bias against seeing them. An example of the former explanation can be found in Kochanek (2014), who proposes that there is a class of "failed supernovae" between 16 and 25$M_{\odot}$ that do manage to eject their envelopes in weak transient events, but leave behind their helium cores to form the lowest mass 5-8$M_{\odot}$ black holes. Lower mass progenitors are then responsible for the neutron stars.
The observational bias is that the companions to the lowest mass black holes in binary systems may be always overflowing their Roche lobes. The resultant accretion signature swamps the companion spectrum and prevents a dynamical mass estimate (e.g. Fryer 1999). The Chandra Galactic Bulge Survey is attempting to find examples of quiescent, relatively low X-ray luminosity, eclipsing compact binaries, with which to measure a more unbiased black hole mass distribution.
Further Edit: There continue to be challenges and claims that there are "low-mass" black holes that could be formed via accretion-induced collapse of a neutron star (pointed out by Alexandra Veledina). For instance Cygnus-X3 has a claimed mass of $2.4^{+2.1}_{-1.1}\ M_{\odot}$ according to Zdziarski et al. 2013, but these observations lack the precision to be really sure yet.
Addition 30/6/20:
A compact object has now been found with a precise mass measurement between $2.5-2.67 M_{\odot}$. It was detected through a gravitational wave signature during its merger with a much more massive black hole. This object was either the most massive neutron star found yet, or the least massive black hole (Abbott et al. 2020).
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11835 | https://www.youtube.com/watch?v=0wREydE3sJw | Long Division: Dividing by a 2-Digit Number | A Step-By-Step Review | Math with Mr. J
Math with Mr. J
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Welcome to Long Division: Dividing by a 2-Digit Number (A Step-By-Step Review) with Mr. J! Need a refresher on dividing by 2-digit divisors (aka dividing by 2-digit numbers)? You're in the right place!
Everyone needs a refresher from time to time, so this video is for you if you're looking for help with how to divide. Mr. J will go through an example of dividing a 3-digit number by a 2-digit number, an example of dividing a 4-digit number by a 2-digit number, and explain the steps of long division.
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[Music] welcome to math with mr j [Music] in this video i'm going to go through a quick review of division and specifically we're going to be working with two digit divisors that just means we're dividing by two digit numbers now this is going to be the exact same process as when we divide by a one digit number if it's been a while and you need a quick refresher this should be helpful whether you're in middle school high school college continuing your education as an adult or maybe you just learned this recently really no matter what your situation is here are a couple of examples to help you out everyone needs a refresher from time to time let's jump into number one where we have 962 divided by 20 and the first thing that we're going to do we're going to set this up now 962 is the number we are dividing it's called the dividend so it goes under the division bar so we have 962 divided by and then 20 is our divisor it's the number we're dividing by now we're ready to go through the division process these steps divide multiply subtract bring down repeat and we start with divide so we have 9 divided by 20. how many whole groups of 20 are in 9 well we can't do that so we need to go to the next digit over and use that 6 so we have the 9 and the 6 so we have 96 96 divided by 20. how many whole groups of 20 are in 96 well four groups of 20 gets us to 80 and then five groups of 20 gets us to a hundred so five is too many it's going to be four so we need to put a four above the 96 don't put it above the nine put it above the six because we did 96 divided by 20. then we come around and multiply so four times twenty that gives us eighty then we subtract six minus zero six and then nine minus eight one after we subtract we bring down so let's bring down this 2 and then we repeat so we go back to divide so now we have 162 divided by 20. how many whole groups of 20 are in 162 well eight groups of 20 gets us to 160 and that's as close as we are going to get so 162 divided by 20. eight whole groups of 20 in 162 then we come around and multiply so 8 times 20 is 160 subtract 2 minus zero is two six minus six is zero and one minus one is zero so we have two now after subtracting we bring down but we don't have anything to bring down we went all the way over to the ones place so that 2 is going to be our remainder so 48 remainder 2 is our answer let's move on to number 2 where we have 6 800 divided by 73 so we are dividing 6865 that's our dividend it goes under our division bar and we are dividing by 73 73 is our divisor now we go through our steps so we start with divide we have 6 divided by 73 how many whole groups of 73 are in six we can't do that so we go to the next digit we have an eight so we take a look at 68. how many whole groups of 73 are in 68 we can't do that either so we need to go to the next digit over which is another six so we have 686 divided by 73 so we need to figure out how many whole groups of 73 are in 686 now in order to figure out how many whole groups of 73 are in 686 we need to estimate and check now i always like to use something i know as a reference point something to go off of in order to make better estimates so for example i always like to start with ten i like to think about ten so seventy three times ten is seven hundred thirty now we can go off of that because we have 686 which is kind of close to 730 so our estimate should be close to 10 groups of 73 so let's try 9 so i'm going to come to the side and do 73 times 9 to see where 9 groups of 73 gets us so 9 times 3 27 then we have 9 times 7 which is 63 plus 2 657 and that's as close as we are going to get so nine whole groups of 73 and that 9 needs to go above the 686 that last 6 we used now we multiply i'm going to do this problem without drawing those arrows so 9 times 73 is six hundred fifty-seven then we subtract so six minus seven we need to borrow sixteen minus seven is nine seven minus five is two and then six minus six is zero after subtracting we bring down so let's bring down the five and we have 295 after we bring down that 5 we repeat so we go back to divide so we have 295 divided by 73. we need to figure out how many whole groups of 73 are in 295 so let's use nine groups of 73 as a reference point nine groups of 73 or 73 times 9 gave us 657 295 is about half of that so we need to scale that back let's try 73 times five and see how close we get and then we can make adjustments if need be let's come to the left here where i have some room so 73 times five five times three is fifteen five times seven is thirty five plus one is thirty six so we get 365 which is too high we don't have enough for five whole groups of 73 but that's still useful because we know we need to scale back so let's try four so 73 times four four times three is twelve carry the one four times seven is 28 plus one is twenty nine so 292 it's going to be four whole groups of 73 in 295 so let's put our 4 up here and then we multiply 4 times 73 is 292 subtract 5 minus two is three and then we have nine minus nine is zero and two minus two is zero after subtracting we bring down we went all the way over to the ones place we do not have anything else to bring down so that three is our remainder so we get 94 remainder 3 for our final answer now one more thing i do want to mention is the difference between numbers 1 and 2. although we divided by a two-digit number in both of these problems number two took more work and more time but that's perfectly okay in number one we divided by 20. that's a much easier number to work with than the 73 in number two so that's something to keep in mind as you go through these types of problems some numbers are easier to work with than others and some problems take more work than others so there you have it there's a quick review of division and specifically two digit divisors so we divided by two digit numbers i hope that helped thanks so much for watching until next time peace you |
11836 | https://pmc.ncbi.nlm.nih.gov/articles/PMC4543024/ | Kallmann syndrome in a female adolescent: a new mutation in the FGFR1 gene - PMC
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BMJ Case Rep
. 2012 Jun 29;2012:bcr1220115380. doi: 10.1136/bcr-12-2011-5380
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Kallmann syndrome in a female adolescent: a new mutation in the FGFR1 gene
Ana Novo
Ana Novo
1 Department of Paediatrics, Centro Hospitalar do Porto, Porto, Portugal
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1, Isabel Couto Guerra
Isabel Couto Guerra
1 Department of Paediatrics, Centro Hospitalar do Porto, Porto, Portugal
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1, Felisbela Rocha
Felisbela Rocha
2 Department of Paediatrics, Centro Hospitalar do Médio Ave, Unidade de Famalicão, Famalicão, Portugal
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2, Susana Gama-de-Sousa
Susana Gama-de-Sousa
2 Department of Paediatrics, Centro Hospitalar do Médio Ave, Unidade de Famalicão, Famalicão, Portugal
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2, Teresa Borges
Teresa Borges
3 Department of Paediatric Endocrinology, Centro Hospitalar do Porto, Porto, Portugal
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3, Rita Cerqueira
Rita Cerqueira
4 Department of Clinical Genetics, CGC Genetics/Centro de Genética Clínica, Porto, Portugal
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4, Purificação Tavares
Purificação Tavares
4 Department of Clinical Genetics, CGC Genetics/Centro de Genética Clínica, Porto, Portugal
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4, Paula Fonseca
Paula Fonseca
2 Department of Paediatrics, Centro Hospitalar do Médio Ave, Unidade de Famalicão, Famalicão, Portugal
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2
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1 Department of Paediatrics, Centro Hospitalar do Porto, Porto, Portugal
2 Department of Paediatrics, Centro Hospitalar do Médio Ave, Unidade de Famalicão, Famalicão, Portugal
3 Department of Paediatric Endocrinology, Centro Hospitalar do Porto, Porto, Portugal
4 Department of Clinical Genetics, CGC Genetics/Centro de Genética Clínica, Porto, Portugal
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Correspondence to Dr Ana Novo, anocas.novo@gmail.com
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Collection date 2012.
2012 BMJ Publishing Group Ltd
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PMCID: PMC4543024 PMID: 22751423
Abstract
The Kallmann syndrome is characterised by the association of hypogonadotropic hypogonadism and hypo/anosmia. It represents a phenotypically and genotypically heterogeneous clinical entity, with six genes identified so far in the literature—KAL1, FGFR1, PROKR2, PROK2, CHD7 and FGF8. Mutations in the FGFR1 gene can be found in approximately 10% of the patients. The authors present the case of a female adolescent with hypogonadotropic hypogonadism and impaired olfactory acuity in the presence of hypoplasia of the nasal sulcus and agenesis of the olfactory bulbs. The molecular analysis of the fibroblast growth factor receptor 1 identified a heterozygous mutation c.1377_78insA (p.V460SfsX3) in exon 10 of FGFR1 gene. This mutation has not yet been reported in the literature. A theoretical review of clinical features and therapeutic approach of this syndrome is also presented.
Background
The Kallmann syndrome (KS) is a rare clinical entity, with an estimated prevalence of 1:8000 in males and 1:40 000 in females,12 characterised by the association of hypogonadotropic hypogonadism and hypo/anosmia.13
The first description of this syndrome was attributed to the Spanish anatomist Aureliano Maestre de San Juan, who, in 1856, reported the absence of cerebral olfactory structures in a patient with small testicules, micropenis, absence of pubic air and anosmia. However, it was not until the next century (1944) that this syndrome was recognised as a clinical entity by the American geneticist Franz Josef Kallmann, who, while studying the presence of hypogonadism and anosmia in three distinct families, proved its inherited trait.1–4
Considering the pathophysiological point of view, KS results from an embryogenic defect. The impaired olfactory acuity occurs as a consequence of a unilateral or bilateral hypoplasia/agenesis of the olfactory system. Embryologically, the gonadotropin-releasing hormone (GnRH) neurons derive from the olfactory placodium and migrate to the hypothalamus through the olfactory beam. The normal development of the olfactory bulb is therefore an absolute condition to an adequate migration of the GnRH neurons to the hypothalamus, which explains the involvement of the gonadotropic axis observed in this syndrome.35
The KS may occur in both sporadic and inherited forms. As the majority of cases are sporadic, a high frequency of de novo mutations in the genes involved is suggested.
In the inherited form, the transmission is autosomal recessive, autosomal dominant with incomplete penetration or X linked.12
Until present, six causal genes have been identified—KAL1, FGFR1, PROKR2, PROK2, CHD7 and FGF8—and KS is classified according to this into six different subtypes, from KAL1 to KAL6, respectively.126
Until now, the finding of KAL1 mutations in male patients with KS has demonstrated that this gene is responsible for the X linked recessive form of the disease.2
Mutations in these six genes are observed in only 30% of all patients with KS, suggesting a possible involvement of other genes in its aetiology.12
Case presentation
The authors report the case of a Caucasian female adolescent, referred to the Paediatric Outpatient Clinic at 11 years of age because of short stature.
She was the only child of young, healthy and non-consanguineous parents. No family history of pubertal delay or known inherited familial diseases was reported.
Pregnancy, birth and neonatal period had occurred uneventfully. Anthropometry at birth was adequate for gestational age. From the second year of life, a linear growth (both weight and stature) below percentile 3 was observed. She also presented adequate neurodevelopment skills.
At first evaluation (11 years of age) she had a good general appearance, weighed 26 kg and was 127 cm tall (SD score (SDS) −2.56; percentile <1); her body mass index was of 16.1 kg/m 2 (SDS −0.82; percentile 20.5) and she was prepubertal. Physical examination was otherwise normal, with no dysmorphic features or malformations.
The target height was 154.5 cm (SDS −1.25; percentile 10.5) and growth velocity during the previous year had been normal.
Investigations
The initial laboratory workup revealed a normal hormonal basal study according to age, sex and Tanner stage (follicle-stimulating hormone (FSH), luteinising hormone (LH) and thyroid function), IGF 1 within normal range value (262 μg/l), a female karyotype (46, XX), negative antitransglutaminase IgA antibodies and a bone age of 10 years.
Blood count, extended biochemistry, serum IgA levels, erythrocyte sedimentation rate and urinalysis were all within the normal range; microbiological examination of the urine and parasitological examination of faeces were negative.
Follow-up was carried out in an outpatient care department and 2 years later, at the age of 13 years and 6 months, she was still prepubertal. Hormonal study was repeated at this point revealing low values of FSH, LH and oestradiol for sex and chronological age.
An luteinizing hormone-releasing hormone (LHRH) stimulation test was performed demonstrating a prepubertal LH response. The remaining pituitary function was normal. The abdominal and pelvic ultrasonography also revealed prepubertal uterine dimensions, with no anomalies detected.
The review of the anamnesis was able to identify anosmia.
Considering the clinical association of hypogonadotropic hypogonadism and anosmia, an encephalic and pituitary MRI was performed, which showed a hypoplasia of the nasal sulcus and bilateral agenesis of the olfactory bulbs.
The molecular analysis of the fibroblast growth factor receptor 1 (FGFR1), by sequence analysis of all codifying region (exons 2–18), and including the adjacent intronic regions, allowed the identification of the mutation c.1377_78insA (p.V460SfsX3) in heterozygous on exon 10.
Treatment
At the age of 14 , treatment with transdermal oestradiol was started, beginning with an initial dose of 6.25 μg/day, and gradually increasing to a maximum dose of 25 μg/day. Three months after this therapeutic approach she began her pubertal development.
Outcome and follow-up
At the age of 17, her height was 149.5 cm (SDS −2.11; percentile 1.77), her body mass index was of 20.27 kg/m 2 (SDS −0.29; percentile 38.64) and she was on a Tanner stage 3 with normal uterine dimensions for chronological age. A progestative was introduced and, 1 month later, a withdrawal haemorrhage occurred.
At present, at the age of 18, her height is 151 cm (SDS −1.84; percentile 3.3) with a body mass index of 20.74 kg/m 2 (percentile 43.59).
Discussion
In the case reported, the presence of hypogonadotropic hypogonadism in association with olfactory impairment and hypoplasia/agenesis of the encephalic olfactory structures, allowed the clinical diagnosis of KS. The molecular identification of a new mutation on the FGFR1 gene corroborated it.
Clinically, the KS is mainly characterised by absence or incomplete puberty in the presence of hypogonadotropic hypogonadism with hypo/anosmia. Patients rarely value the latter sign during the clinical history, which emphasises the importance of its exclusion when a hypogonadotropic hypogonadism is suspected.2
This also happened in our patient who did not spontaneously mention the presence of an olfactory impairment.
KS is a heterogenic entity. The degree of olfactory and hypogonadism can substantially vary, among unrelated patients and within the same family or even between monozygotic twins.1
A subgroup of patients with KS can present other clinical manifestations, such as: bimanual synkinesis, abnormal ocular movements, congenital ptosis, coloboma of the iris, nasal cartilage agenesis, malformation of the external ear, cleft lip or palate, unilateral or bilateral renal agenesis, agenesis of the corpus callosum, hypodontia, skeletal anomalies of the feet and hands, obesity and sleep disorders.125
In the reported case, none of these associated signs and symptoms was present.
The sequencing analysis of the FGFR1 gene allowed the identification of a c.1377_78insA (p.V460SfsX3) mutation in heterozygous state located on exon 10. This c.1377_78insA mutation causes a frameshift starting with codon Val460 changing it to serine and creates a premature stop codon at position 3 of the new reading frame, denoted (p.V460SfsX3).
Although the mutation identified in this case has not yet been published in the literature or inserted on the Human Gene Mutation Database Professional, it can be regarded as pathogenic since protein function is compromised, being consistent with the diagnosis of KS 2 (OMIM #147950).
According to Dodé and Hardelin,1 mutations in the FGFR1 gene are present in about 10% of all cases of KS, corresponding, in 30% of those cases, to de novo mutations.
Our patient seems to represent one of such cases, since, from the clinical point of view, the family history was negative. Nevertheless, considering the wide intrafamilial phenotypic variability, only an FGFR1 gene molecular analysis performed on each family member would allow a definitive conclusion on the supposed occurrence of a de novo mutation in this case.
The therapeutic approach of the KS is based on the management of hypogonadism, promoting the development of secondary sexual characteristics as a primary goal and, secondarily, fertility.12
Similar to other causes of hypogonadism, a chemical induction of puberty is required, through the administration of sexual steroids.12
On the female sex, the hormone replacement therapy with oestrogens should be gradual, beginning with low doses of oestrogens, to mimic, as much as possible, physiological puberty. Concerning the route of oestrogen administration, transdermal application seems to be a more physiological alternative, when compared with to oral administration.
Initial treatment includes the administration of a low dose of oestrogens, approximately 6.25 µg/day if transdermal, with a progressive dose increase during a period of 2 years. The progestative should be associated only when withdrawal haemorrhage occurs or otherwise not until 2 years after the introduction of oestrogens, in a way to provide adequate breast and uterine development.7 This last approach was the one chosen for our patient.
To induce ovulation in females and enhance the production of spermatozoids in men, gonadotropins may be used, as well as a pulsatile administration of GnRH. Both approaches are effective in the vast majority of patients.12
Although previously thought to be a lifelong disorder, some cases of reversible KS may occur, characterised by the existence of fertility in the absence of treatment with GnRH or gonadotropins and/or by the improvement in testosterone secretion after discontinuation of treatment with gonadotropins, GnRH or testosterone replacement.3
Those cases were all reported in males,38 and therefore it is not expected that such a situation would occur in our female patient.
Learning points.
In patients with suspected hypogonadotropic hypogonadism it is important to value the presence of other clinical signs such as hypo/anosmia.
Genetic testing strategy in Kallmann syndrome is based on patient gender and family history (if any), the supposed mode of disease inheritance, as well as the presence of additional clinical anomalies.
Since up to 30% of the mutations found in the FGFR1 gene might be de novo mutations, this possibility should be considered when assessing recurrence risk of this genetic form in a family.
Footnotes
Competing interests: None.
Patient consent: Obtained.
References
1.Dodé C, Hardelin JP. Kallmann syndrome. Eur J Hum Genet 2009;17:139–46. [DOI] [PMC free article] [PubMed] [Google Scholar]
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3.Ribeiro RS, Vieira TC, Abucham J. Reversible Kallmann syndrome: report of the first case with a KAL1 mutation and literature review. Eur J Endocrinol 2007;156:285–90. [DOI] [PubMed] [Google Scholar]
4.Cadman SM, Kim SH, Hu Y, et al. Molecular pathogenesis of Kallmann's syndrome. Horm Res 2007;67:231–42. [DOI] [PubMed] [Google Scholar]
5.Roux N. Isolated gonadotropic deficiency with and without anosmia: a developmental defect or a neuroendocrine regulation abnormality of gonadotropic axis. Horm Res 2005;64(Suppl 2):48–55. [DOI] [PubMed] [Google Scholar]
6.Kim HG, Kurth I, Lan F, et al. Mutations in CHD7, encoding a chromatin-remodeling protein, cause idiopathic hypogonadotropic hypogonadism and Kallmann syndrome. Am J Hum Genet 2008;83:511–19. [DOI] [PMC free article] [PubMed] [Google Scholar]
7.Bondy CA, Turner Syndrome Study Group. Care of girls and women with Turner syndrome: a guideline of the turner syndrome study group. J Clin Endocrinol Metab 2007;92:10–25. [DOI] [PubMed] [Google Scholar]
8.Sinisi AA, Asco R, Bellastella G, et al. Homozygous mutation in the prokineticin-receptor2 gene (Val274Asp) presenting as reversible Kallmann syndrome and persistent oligozoospermia: case report. Hum Reprod 2008;23:2380–4. [DOI] [PubMed] [Google Scholar]
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11838 | https://www.scirp.org/journal/paperinformation?paperid=38785 | Modeling a General Equation for Pool Boiling Heat Transfer
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Advances in Chemical Engineering and Science>Vol.3 No.4, October 2013
Modeling a General Equation for Pool Boiling Heat Transfer ()
Mohammed Salah Hameed, Abdul Rahman Khan, A. A. Mahdi
DOI:10.4236/aces.2013.34037PDFHTML 10,118 Downloads 13,581 ViewsCitations
Abstract
It is recognized that the nucleate pool boiling data available in literature are mainly related to four known correlations, each differs from the other by a varying magnitude of constant coefficients, depending on restrictive experimental conditions. The present work is concerned in developing an empirically generalized correlation, which covers the entire range of nucleate boiling with a minimum possible deviation from experimental data. The least squares multiple regression technique is used to evaluate the best coefficient value used in the correlations. An empirical correlation that fits a broader scope of available data has been developed by a non-linear solution technique leading to the following equation: where the coefficients R 1 and R 3 both represent the effect of surface-liquid combination. They are assessed independently for the used surface material and liquid.
Keywords
Pool Boiling; Nucleate Boiling; Linear and Non-Linear Technique Methods; Heat Transfer
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Hameed, M. , Khan, A. and Mahdi, A. (2013) Modeling a General Equation for Pool Boiling Heat Transfer. Advances in Chemical Engineering and Science, 3, 294-303. doi: 10.4236/aces.2013.34037.
1. Introduction
Boiling is a complex process and an intensive work is needed for its understanding. Within the last decades, several nucleate boiling models were formulated and could be grouped in two main categories: a) Bubble Agitation Models and b) Macro/Micro Layer Evaporation Models.
The bubble agitation models are based on the principle of agitating the liquid, but they carry away little heat. The heat transfer is considered within the turbulent forced convection. The obtained empirical pool boiling heat transfer models employ dimensionless groups based on both fluid and solid properties while the main constant in the model is found to depend on the geometry of the heater. The models found in literature are useful within the range of database used in developing their derivation.
Bubble agitation mechanism together with Helmholtz-instability mechanism can be used either to explain the heat transfer at the low heat flux regime or to explain CHF (Critical Heat Flux). They cannot account for the continuity of the pool-boiling curve. On the other hand, the macro/micro layer evaporation reproduces the poolboiling curve from the nuclear boiling to transition boiling. The macro/micro layer models play an important role in high heat flux region. The liquid layer includes the micro layer underneath the bubble and the macro layer on the base of coalescence and dries out periodically . Heramura & Katto assume the liquid-vapor interface is stationary and the entire surface heat flux contributes to macro layer evaporation.
Several numerical models were proposed based on the macro layer theory among that of Maruyama et al. . Zhao et al. put forward a model for transient pool boiling heat transfer. The model employed is too high heating rate to be realized in practical experiments for a horizontal surface. He et al. concluded that the macro layer model is more suitable for the high heat flux regime. Dhir confirmed that numerical simulations are not a substitute for detailed experiments. The experimental results are needed to validate the simulations. Numerical simulations provide additional insights into the boiling phenomena.
Within the late decade, many researchers worked on viewing the pool boiling in microgravity (in the absent of buoyancy) to understand the lower limit of forced convection. Several workers are Lee , Herman , Wan & Zhao , and Kubota et al. .
Ji et al. enhanced the pool boiling heat transfer in microgravity by using porous coating heating surface at atmospheric pressure and slightly moderate superheats.
Other researchers [11,12] enhanced the pool boiling by using nanofluid (water mixed with extremely small amount of nanosized particles). They concluded the enhancement of the thermal conductivity and convection heat transfer capability of the suspended particles of nanometer in size for many volume fractions of nanofluids.
The results of workers [6-12] can be used as a guidance in formulating proper equations that can be used in design. The aim of the present work is to use bubble agitation models to obtain a generalized empirical correlation that gives the best possible representation of collected data. Pool of data is collected from literature for various liquids effects with different plain test surfaces. For this purpose, linear and non-linear programming techniques were used in the evaluation of the proposed correlation.
2. Theoretical Analysis of Bubble Agitation Models
The primary requirement for nucleation to occur or for a nucleus to subsist in a liquid is that the liquid should be superheated. There are two types of nuclei. One type is formed in a pure liquid; it can be either a high energy molecular group resulting from thermal fluctuations of liquid molecules, or a cavity resulting from a local pressure reduction such as that occurs in accelerated flow. The other type, formed on a foreign object can be either a cavity on the heating wall or suspended foreign material with a non-wetted surface.
Rohsenow assumes that the movement of bubbles at the instant of breaking away from the heating surface is of prime importance and obtained Equation (1) for heat transfer in the region of nucleation pool boiling.
(1)
The recommended variation of r is within 0.8 to 2.0. Evaluation of C sf from experimental results of many workers prove to be a parameter which does not pick out only the nucleation ability of heating surface but contains the effect of physical properties of liquid.
Rohsenow proposed the surface factor C sf to prescribe the condition of heating surface in nucleate boiling. Various investigators utilized this factor in their determination of empirical expressions. The surface factor is defined by Equation (1), generally known as Rohsenow empirical correlation.
Forster and Zuber indicated that small bubbles grow rapidly and large ones slowly, but the degree of agitation in the surrounding liquid due to bubble growth remains the same. They derived the following empirical correlation:
(2)
Equation (2) predicts the same heat transfer coefficient for a liquid boiling on any hot surface (for all heterogeneous cases only) or boiling in bulk (for all homogeneous cases only). Rohsenow’s Equation (1) was developed and applied to the heterogeneous case only.
Forster and Greif suggested a different approach by considering that the mechanism of high heat transfer rate, during nucleate boiling, is mainly due to the liquidvapor exchange. They obtained a dimensional empirical correlation, for the pool boiling heat flux q in water at 100 - 4763 kN/m 2, as shown in Equation (3).
(3)
This correlation is not as widely verified as that of Rohsenow.
Gupta and Varshney obtained experimental data for boiling heat transfer, using distilled water, benzene and toluene as liquids over a heated horizontal cylinder made of stainless steel. Their data was correlated by the following dimensionless empirical correlation:
(4)
where Nu B and Pe B are the Nusselt and Peclet number of boiling respectively. Or it can be written as:
(5)
In order to derive a general correlation based on bubble agitation phenomena to be more versatile than the correlations existing in literature, a search was made through published work in literature and found that the following four well known empirical correlations referred to in most publications:
a) Rohsenow correlation (Equation(1))
b) Forster and Zuber correlation (Equation (2))
c) Forster and Greif correlation (Equation (3))
d) Gupta and Varshney correlation (Equation (5)).
For the sake of analysis, experimental data collected from many literatures [19-24] tabulated as heat flux (q), surface temperature (T s), heat transfer coefficient (h), and coefficient h (equal to h/q 0.7). Moreover, physical and thermodynamic properties collected at the reported experimental conditions from literature [25-29] to be used in the analysis. The properties include, liquid thermal conductivity (K L), liquid heat capacity (C L), density of vapor (r V), surface tension of liquid (s), saturation temperature (T sat), density of liquid (r L), latent heat of vaporization (H fg), and viscosity of liquid (m L).
The above stated correlations are of dimensionless form with the exception of the Forster and Greif correlation (Equation (4)). These equations can be represented by general equation as shown in Appendix. Many modifications to linear correlations have been tried to minimize the sum of squares of errors and to conclude some general correlations.
3. Results and Discussion
Boiling heat transfer studied earlier indicated that several variables are important in nucleate boiling such as pressure, fluid properties, surface condition, boiling temperature, kind and relative amount of impurities. The practical data showed that changes in magnitude of these properties and conditions could significantly affect pool boiling heat transfer.
A graphical analyses for 56 sets of literature data was used in studying the effect of heat flux, (q), and operating pressure (P), on boiling heat transfer coefficient, (h). Figures 1-3 show the variation of heat transfer coefficient with heat transfer flux, (q). The lines in the figures are the best-fit lines of the reported data. Figure 1 plotted for various liquids at different operating pressure and test surface. Figure 2 corresponds to various liquid-surface combinations at constant atmospheric pressure. Figure 3 reflects the behavior of various metal surfaces and operating pressures for the same liquid.
All the data can be represented by the empirical, Equation (6), with an average percentage error ranging from 0.012 to 11.8
(6)
The proportionality constant h is proved to be a function of pressure and liquid-surface combination. Cichelli
Figure 1. Heat transfer coefficient (h) versus heat flux (q) at various pressures.
Figure 2. Heat transfer coefficient (h) versus heat flux (q) at atmospheric pressure.
Figure 3. Heat transfer coefficient (h) versus heat flux (q) for various metal surfaces.
and Bonilla confirmed that the coefficient of heat transfer increases with absolute operating pressure in the nucleate boiling zone. They reported the following correlation:
(7)
Figure 4 shows the variation of the proportionality constant (h) as function of pressure (P) for a definite liquid-surface combination. Equation (8) represents the relationship between h and P, that was obtained from best data fit of Figure 4.
(8)
The overall dependence of (h) on operating pressure (P) and heat flux (q) for different liquid-surface combinations is shown in Figure 5.
3.1. Linear Programming Analysis of Empirical Correlations
Equation (A.4) in Appendix used to test the validity of the published correlations. It is used to formulate a correlation that shows the best fit of the experimental data. All the data cited in the literature from [19-24] classified as eight liquids (Water, Benzene, Methanol, Carbon Tetrachloride, n-Butanol, Isopropanol, n-Amyl Alcohol, and n-Heptane) and four surfaces (Brass, Copper, Nichrome, and Stainless Steel) at different operating pressures grouped in 56 data sets.
The applicability of the four empirical correlations, Equations (1)-(3) and (5), in representing the data was test by using linear-programming; that by fixing some of
Figure 4. Proportionality constant (h) versus operating pressure (P).
Figure 5. Heat transfer coefficient (h) dependence on operating pressure and heat flux (q).
the coefficients and evaluating the others and the average percentage error is used as test criteria for comparison purposes.
3.1.1. Rohsenow’s Correlation
Equation (9) is a general expression for Rohsenow’s correlation while the exact expression stated as in Equation (1).
(9)
The coefficient C sf reported, in the literature, to vary with each liquid-surface combination and it is independent of pressure . The validity of Equation (9) was tested for the entire collected data by applying the leastsquares method.
In the initial analysis of data, the pressure was assumed constant and the obtained results showed inconsistency in the calculated values of constants for various experimental conditions tested. The inconsistency in values of constants is most likely due to pressure effect, which was not considered as variable during the initial analysis of data. On the next try, a pressure parameter was introduced in an attempt to narrow the variation in the values of constants for different systems and to conclude general correlation. Four different expressions of pressure parameter cited from literature , and stated in Equation (10), was used and expected to have an effect on heat transfer in the region of pool boiling nucleation:
(10)
Each of these pressure expressions selected to replace the coefficient C sf in Equation (9) and the obtained results were compared and checked. The analysis found that the pressure parameter gives the lowest minimum percentage error between the others forms and selected to replace C sf and modify Equation (9) to the form showing in Equation (11).
(11)
To make the present work more general, the analysis was repeated for various liquids at certain test surface and different operating pressures, that by calculating all the coefficients (R 1, R 2, R 3, and R 4) of Equation (11) for each set of data. The values of coefficients were selected, by the help of Equation (A.4), on the basis of using Equation (11) with the lowest average percentage error.
3.1.2. Gupta and Varshney Correlation
Equation (12) is a general expression for Gupta and Varshney empirical correlation, while Equation (5) represent the exact form.
(12)
The pressure term is not included in the correlation of Gupta and Varshney as was the case of Rohsenow’s empirical correlation, Equation (1). Equation (12) differs from Rohensow’s Equation (9) by including the density ratio term and classifying other dimensionless term in well-known groups. A similar data analysis used for Equation (12) as it was with the case of Rohsenow’s Correlation. By substituting the various pressure forms of Equation (10) in place of coefficient R 3 in Equation (12), it is found that pressure expression gave the minimum average percentage error. This term is then chosen as the best fit expression for pressure and used in Equation (12) to obtain Equation (13).
(13)
In order to make the present work more general, the analysis repeated for various liquids at specific test surface and different operating pressures then follow the same procedure as in case of Rohsenow’s correlation. The analysis concluded that the modified Equation (13) provides a better data fit than that of Gupta and Varshney Equation (5).
3.1.3. Forster and Zuber Correlation
Equation (14) is a general expression for Forster and Zuber empirical correlation, while the exact form is given in Equation (2).
(14)
The pressure term is taken care of in Forster and Zubers correlation, Equation (2). By applying a similar procedure as in previous cases, it was found that the validity of the above equation is restricted to specific experimental data near critical temperature difference. By checking the above results for Forster and Zuber empirical correlation, Equation (14), it is found that the overall average percentage errors is very high in predicting the published data under consideration.
3.1.4. Forster and Greif Empirical Correlation
Equation (15) is a general expression for Forster and Greif dimensional empirical correlation, while the exact form is given in Equation (3).
(15)
Using the same sets of data analyzed previously gave a higher average percentage error as compared with the dimensionless empirical correlations as shown in Table 1.
The variation of some thermodynamic properties as function of pressure and temperature is not reported in literature and these properties were considered constant during the calculations, which may be the cause of the large average percentage error. Moreover, this empirical correlation dealt with fluid side effect of the problem and ignored the surface side effect on the nucleate boiling behavior. This might have added more error to the validity of this correlation.
The linear programming analysis recommended the use of the modified correlations of Rohsenow and Gupta & Varshney as they give closer prediction to the experimental data than in case of using Forster & Zuber and Forster & Greif modified correlations as showing in Table 1. Hence, the last two correlations excluded from any further analysis.
Table 1. Linear programming results for Equation (11) to Equation (15).
Where APE = Average Percentage Errors; CC = Correlation Coefficient.
3.2. Non-Linear Programming Analysis of Empirical Correlations
The data were re-analyzed by non-linear programming methods by using the modified correlations of Rohsenow, Equation (11), and Gupta & Varshney Equation (13) in an attempt to improve the correlations for lower average percentage error.
By assume that: in Equation (4) is equal to X and in Equation (13) is equal to Y, then use either of the following equations:
Binominal Expression:
(16)
or non-linear Expression:
(17)
to replace the in Equation (11) or in Equation (13) respectively. It is found from data fittings that a better representation can be obtained by using the binominal expression, Equation (16), with in place of in Equation (11) and using the binominal expression, Equation (16), with in place of in Equation (13).
Various forms of expressions for the constant R 7 was tried for both Rohsenow and Gupta & Varshney modified correlations. It was found that keeping R 7 as a constant and independent of other parameters yielded a better data representation. It was also concluded that the average percentage error would not be improved by using the non linear equations instead of the linear equation as showing in Table 2.
General Empirical Correlation
By using the best data fit for Equation (11) for different surfaces (Nichrome, Copper, Brass, and Stainless Steel), the variation of the powers of pressure expression term and Prandtl number in the equation were found to be approximately equal to 0.08 and 1.0 respectively leading to the following generalized Equation (18).
(18)
The coefficients R 1 and R 3 represent the effect of surface-liquid combination. They are assessed independently for each surface by the least-squares linear regression method and the results are stated in Table 3.
A similar analysis tried for Equation (13) and concluded that the powers of pressure term, Peclet number (Pe B), and density ratio term (r V/r L) were relatively independent of surface—liquid combination as compared with the coefficient R 3 and the power of Prandtl number, R 4. The best form of Equation (13) was tested for different data sets and concluded Equation (19).
(19)
A similar way was followed for Equation (19), to that of Equation (18), in finding the power R 4 and the coefficient R 3 and their best values are given in Table 3. The result of analysis of Equation (18) and Equation (19) listed in Table 3 suggested the use of Equation (19) in preference to Equation (18).
The applicability of Equaiton (19) was examined for different surfaces as showing in Figures 6-9. The equation found to fit well for all the data with the exception of Brass. The deviation in the results for Brass is due to
Table 2. Comparison of linear, binominal, and non linear expressions versions of Equations (11) and (13).
Table 3. Linear programming results for Equations (18) and (19).
Figure 6. Experimental data predictions using Equation (19) for Nichrome surface.
Figure 7. Experimental data predictions using Equation (19) for Copper surface.
Figure 8. Experimental data predictions using Equation (19) for Brass surface.
Figure 9. Experimental data prediction using Equation (19) for Stainless Steel surface.
limited available data at very low pressure. Equation (20) represents the dimensionless form of Equation (19).
(20)
The analysis concluded that Equation (20) is valid for the entire available data and represent a more generalized correlation than the correlations found in literature.
4. Conclusions
A graphical analysis concluded that the empirical Equation (6) is showing the effect of heat flux (q) and operating pressure (P) on the boiling heat transfer coefficient (h).
(6)
where h is a function of pressure and for different liquid-surface combinations, it is found to vary with the pressure as follows:
(8)
56 sets of literature data were tested on each of the four known correlations, Rohsenow, Forster & Zuber, Forster & Greif, and Gupta & Varshney, by using the linear and non-linear programming solution. The concluded results show that any of these correlations does not fit the entire data satisfactory. To improve their predictions, the correlations were modified, including additional parameters in an attempt to close up the deviation in the values of calculated parameters. The modified correlations of Rohsenow and Gupta & Varshney responded better to the applied modification than that of Foster & Zuber and Foster & Grief and they were considered for further analysis.
The least squares multiple regression technique [30,31] is used to evaluate the best possible values of the constant coefficients in the correlation. The cumulative error squares were minimized by using an ordinary optimum seeking technique. Linear, binominal & non-linear correlations were tested in concluding the final correlation.
The use of non-linear solution technique did not improve correlations 11 and 13 that were concluded by the linear technique and hence Equation (20) gives the best representation of the entire tested data.
Appendix
Sum of Squares of Errors
Equations (1) to (4) can be represented by a general equation:
where
(A.1)
This equation represents a general form of those correlations and simplified by taking logarithms of both sides.
or
(A.2)
For N number of data readings there will be N number of linear equations, while for the determination of k coefficients only k equations are required. A least squares multiple regression technique [30,31] was used to evaluate the best possible coefficient from raw data readings. The cumulative error squares minimized by an ordinary optimum seeking technique resulting into k number of equations to provide k number of coefficients for the entire data. These equations mathematically represented in the form:
(A.3)
The sum of squares of errors is expressed as:
(A.4)
Nomenclature
C L—Heat capacity of liquid, J/kg×˚C C s f—Surface factor F m—Nucleation factor g—Acceleration of gravity, m/s 2
g c—Conversion ratio, kg m/kg·s 2
h—Heat transfer coefficient W/m 2·˚C h—Proportionality constant H fg—Latent heat of vaporation, J/kg K L—Thermal conductivity, W/m ˚C Nu B—Nusselt number for boiling =
P—Operating pressure (kN/m 2)
Pe B—Peolet number for boiling =
Pr—Prandtl number q—Heat Flux, W/m 2 s
—coefficients Re—Reynolds number Re b—Reynolds number for bubbles =
Sr—Superheat ratio =
T s—Test surface temperature, ˚C T sat—Saturation temperature, ˚C
Greek Letters
a L Thermal diffusivity, m 2/s,
DP Pressure difference corresponds to (T s–T sat), kN/m 2
m L Viscosity of liquid, kg/m×s r V Density of vapor, kg/m 3
r L Density of liquid, kg/m 3
s Surface tension, kg/s 2
Subscripts
b Refers to bubble property
B Refers to boiling condition
L Refers to liquid condition
sf Refers to surface factor
s Refers to surface condition
sat Refers to saturation condition
v Refers to vapor condition
NOTES
Conflicts of Interest
The authors declare no conflicts of interest.
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Andrew Go Lee, MD, Guy V. Jirawuthiworavong, M.D., M.A., Joseph Pecha, Nita Bhat, MBBS, MS, Shruthi Harish Bindiganavile, MBBS, MS
All contributors:
Guy V. Jirawuthiworavong, M.D., M.A., Aaron M. Miller, MD, MBA, FAAP, Nagham Al-Zubidi, MD, Michael S. Vaphiades, DO, Aroucha Vickers, DO, Shruthi Harish Bindiganavile, MBBS, MS, M. Tariq Bhatti, MD, Alvaro Mejia-Vergara, MD
Assigned editor:
Aroucha Vickers, DO
Review:
Assigned status Up to Date
by Aroucha Vickers, DO on June 2, 2025.
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Contents
1 Introduction
2 Disease Entity
2.1 Disease
2.2 Etiology
2.3 Risk Factors
2.4 General Pathology
2.5 Pathophysiology
3 Diagnosis
3.1 History
3.2 Symptoms
3.3 Signs
3.4 Physical examination
3.5 Clinical Diagnosis
3.6 Diagnostic Procedures
3.7 Laboratory Tests
3.8 Differential diagnosis
3.8.1 Differential Diagnosis for Retrobulbar Optic Neuritis (normal appearance of optic nerve and vision loss)
3.8.2 With a Relative Afferent Pupillary Defect
3.8.3 No Relative Afferent Pupillary Defect
3.8.4 Differential Diagnosis of Optic Neuritis (unilateral optic disc edema, vision loss)
4 Management
4.1 ONTT Outcomes
4.2 Medical therapy
4.3 Medical follow-up
4.4 Complications
4.5 Prognosis
5 Additional Resources
6 References
Introduction
Demyelinating Optic Neuritis
ICD-10
H46
ICD-9
377.3
Demyelinating optic neuritis (ON) in an adult is one of the most common reasons for a young patient to seek consultation with a neuro-ophthalmologist. This brief guide will help the general ophthalmologist to understand the following:
ON is a clinical diagnosis, thus magnetic resonance imaging (MRI) of the brain and orbit with gadolinium is helpful to document optic nerve enhancement as well as demyelinating white matter lesions in the brain. Although in typical cases, serologic testing for inflammatory, infiltrative, or infectious etiologies are not necessary, in atypical cases these tests may help to differentiate demyelinating or idiopathic ON from alternative etiologies. Likewise, cerebrospinal (CSF) analysis is not required for the diagnosis, but can be considered for cases which are atypical (e.g., marked visual loss at onset, bilateral ON, cases which are progressive or do not recover).
Typical (i.e., acute, unilateral vision loss, relative afferent pupillary defect (RAPD), pain with eye movement, and normal fundus exam in a young patient) ON will usually recover with good visual prognosis by a few weeks and normal or near normal visual function by 6-12 months. Lack of visual recovery or progressive visual loss is atypical for ON.
The final amount of vision recovery is independent of treatment with or without corticosteroids, however in the ON Treatment Trial (ONTT), intravenous (IV) steroids hastened the speed of recovery but not the final visual outcome compared with either oral prednisone (1 mg/kg/day) or placebo. Thus, although the natural history and prognosis of typical ON is very favorable with or without steroid treatment, steroids can be considered for speeding the rate of recovery.
Oral prednisone alone in standard doses (1 mg/kg/day) however is not recommended for ON because in the ONTT there was a higher rate of recurrence compared to either placebo or intravenous (IV) steroids followed by oral prednisone. Higher dose (1250 mg/day) oral prednisone however could be considered on a case by case basis.
MRI of the brain is helpful after an initial episode of ON because the number of lesions seen can stratify the patient’s risk for developing clinically definite (CD) multiple sclerosis (MS) after a clinically isolated syndrome (CIS). MRI of brain can detect radiographic dissemination in time and space as described by the McDonald criteria.
The most common neurologic disorder associated with ON is MS and ophthalmologist can play a vital role to decrease morbidity from MS by understanding the role of early diagnosis and referral to neurology for consideration for MS treatment with disease modifying drugs (DMD).
This article will focus on adults, 18 years and older who present with ON.
Disease Entity
Optic neuritis ICD 377.3
Retrobulbar neuritis (acute) ICD 377.32
Optic neuritis unspecified, ICD 377.30
Disease
ON is characterized by acute, unilateral decreased vision, optic nerve dysfunction (decreased central and/or peripheral vision, decreased color vision, decreased contrast/brightness sense, RAPD and tends to be associated with periorbital pain worse with eye movement. Periocular pain (62 to 92%) may precede the visual loss by days and is characteristically precipitated or aggravated by eye movement (87%). When the clinical history and examination suggest ON and the optic disc appears normal, the term retrobulbar ON is used. It may be the initial presentation of MS with obvious implications for future diagnosis, prognosis and treatment. Prevalence roughly estimated to be 1-5/100,000 depending upon geography and ethnicity.
Etiology
Believed to be an autoimmune, inflammatory, demyelination process, many different pathophysiologic mechanisms have been proposed and remain under investigation.
Risk Factors
3:2-female:male ratio
Young age (20-45 years old)
A prodromal flu like illness commonly accompanies the event but does not always occur
Patients with MS often have ON and up to 75% of patients with CDMS will have at least one episode of ON in their lifetime. In addition, autopsies of MS patients show up to 90% had some optic nerve involvement.
General Pathology
Immune-mediated inflammatory demyelination of the optic nerve. The myelin undergoes destruction causing axons to poorly conduct impulses and the axons themselves become damaged, depending on the degree of inflammation within the lesion. Thus, retinal ganglion cell axons often become damaged in ON.
Pathophysiology
After myelin destruction occurs, the retinal ganglion cell axons start to degenerate. Monocytes localize along blood vessels and macrophages follow to remove myelin. Astrocytes then proliferate with deposition of glial tissue where axons may have been present before. These gliotic (sclerotic) areas can be multiple in number throughout the brain and spinal cord, hence the term multiple sclerosis.
Many different theories have been proposed for MS but none are proven. One theory is that a viral trigger may be responsible for the autoimmune inflammatory response. One recent study found viral DNA in the CSF of some ON patients indicating that there might be an associated role for varicella zoster virus and herpes simplex virus in the development of this disease.
Diagnosis
ON is a clinical diagnosis that is made when a patient presents with:
Unilateral (but may be bilateral) decreased visual acuity
Pain with eye movement and/or periorbital discomfort
RAPD if unilateral or bilateral and asymmetric
Visual field defect (nerve fiber layer or central loss)
Normal-appearing fundus (retrobulbar ON) although optic nerve swelling can be seen in up to 35% of cases.
History
The importance of the optic nerve has been known since the time of the Greeks. Galen recognized it as a conduit from the brain to the eye, but not until the Renaissance period was the optic nerve and the retina recognized to be the actual organ of vision instead of the lens. Despite misunderstandings about the exact function of the optic nerve, early medieval physicians including Hunayn ibn Ishaq in the 800s recognized that a poorly reactive dilated pupil could indicate damage or obstruction at the level of the optic nerve.
After the introduction of the ophthalmoscope in the late 1800s, a clearer understanding of ocular diseases with respect to etiology was worked out. Thomas Allbutt first suggested that ON might be caused by toxins and infectious agents and made the point that it was most often associated with other neurological abnormalities.
Interestingly, nearly 2000 years of medical history recognized some association between menstruation cessation and optic nerve disease, but not until Harvey Cushing in the early 1900s was the etiology shown to be pituitary tumors compressing the optic chiasm.
Symptoms
Prodromal viral illness, may or may not be present
Acute vision loss that tends to worsen over days and nadirs around 2 weeks prior to improvement. Eye discomfort or pain particularly with eye movement (mild to severe)
Phosphenes may occur (flashes of lights)
Vision may become blurry when body temperature rises during exercise or bathing in warm to hot water (Uhthoff phenomenon)
"Washed out" color vision
Altered perception of motion (Pulfrich phenomenon)
Signs
Decreased visual acuity
Decreased color vision (red desaturation) and/or decreased contrast/brightness sense
RAPD unless both eyes are affected or RAPD was present prior in contralateral eye
Other efferent lesions may be present in MS (e.g., ocular dysmetria or internuclear ophthalmoplegia)
Any nerve fiber layer or central scotoma on Amsler grid and/or confrontation visual fields or formal visual field testing
Typically normal fundus but optic disc edema can be seen in up to 35%. Thus lack of optic disc edema is the rule rather than the exception for ON.
Retinal vascular sheathing, pars planitis (periphlebitis occurs in 5-10% of MS patients)
Physical examination
Snellen visual acuity with best correction for distance and/or near vision
Color vision and/or Contrast measurement (e.g., Pelli-Robson or VisTech)
Swinging flashlight test for RAPD
Evaluation of extraocular movement
Confrontation and automated visual fields (although automated perimetry has proven extremely useful in establishing the progression and recovery of visual field loss in ON, the patterns of visual field defects are not particularly helpful in distinguishing ON from any other form of optic nerve disease (eg, ischemic, compressive)).
Amsler grid with best correction central visual field defects (be aware that this test may also be abnormal in macular disease)
Biomicroscopy/direct ophthalmoscopy of the optic nerve and retina
Clinical Diagnosis
ON is diagnosed clinically by symptoms of acute unilateral decrease in vision, eye pain (especially with movement), RAPD (in unilateral or bilateral asymmetric cases), decreased color vision/contrast/brightness sense and a visual field defect.
Less commonly, patients may present with absence of eye pain, absence of orbital discomfort, bilateral, marked vision loss, and severe optic disc swelling with hemorrhages or retinal exudates. These cases should be considered for further work up to determine if an underlying inflammatory or infectious process can be found.
An MRI depicting enhancement of the optic nerve after administration of gadolinium is helpful but not required to make the clinical diagnosis. The main indication of MRI is derived from its prognostic value for the presence of or future development of MS. MRI is far superior to computed tomography (CT) in the diagnosis of ON. In a retrospective observational series of patients with acute ON who had gadolinium-enhanced fat-suppressed cranial MRI scans within 20 days of the visual loss, enhancement of the orbital optic nerve occurred in (94%) of patients. Optic nerve enhancement is not specific for ON and a similar appearance on MRI can be seen in many other inflammatory optic neuropathies.
Diagnostic Procedures
Optical coherence tomography (OCT) of the peripapillary retinal nerve fiber layer (pRNFL) is helpful as an adjunctive measurement of nerve function in patients with ON. OCT can quantify the onset of optic disc pallor that ensues 6-8 weeks post onset of ON. Optic atrophy can be subtle on biomicroscopy, thus OCT is useful for detection and quantification of the optic atrophy. OCT of the pRNFL also serves in patient education and documentation. Figure 1 below shows an example of an OCT image from a patient’s right eye 2.5 months after and left eye 3.5 months after presentation. In retrobulbar ON however, the OCT may initially be normal and only show pRNFL or macular ganglion cell loss downstream after recovery from acute ON.
It is known that ON demonstrates a delayed response on pattern visual evoked potentials (P-VEPs), and there has been recent interest in using trans-orbital sonography to enhance diagnostic accuracy for ON by measuring differences in optic nerve sheath diameter. One study showed a 68% sensitivity and 88% specificity for trans-orbital sonography but it was not found to be superior to P-VEP.
In order to determine whether ON represents a CIS or related to MS, the McDonald criteria are applied. These criteria aim to diagnose MS when a patient presents with CIS likely to be associated with MS. Recent recommendations are to obtain an MRI in all patients for whom the clinician has a high suspicion of MS. Spinal cord MRI in these cases is not mandatory but should be considered in patients with spinal involvement, patients with progressive symptoms, and patients who are from atypical populations (e.g. non-white or elderly individuals). Figure 2 below summarizes the 2017 McDonald criteria. Changes from the 2010 criteria include a stipulation that in patients with a clinically isolated syndrome and clinical or imaging evidence of dissemination in space who don’t fulfill the 2010 criteria for dissemination in time according to MRI, CSF analysis showing oligoclonal bands provides sufficient support for a confident diagnosis of MS. Importantly, lesions of the optic nerve seen on imaging should not be considered in determination of dissemination in space when applying the McDonald criteria due to insufficient evidence.
The criteria also state that on MRI, one or more T2 hyperintense lesions in the periventricular, cortical or juxtacortical, infratentorial, or spinal cord regions is demonstrative of dissemination in space. There is no need to differentiate between symptomatic and asymptomatic lesions for diagnostic purposes in the most updated guidelines. Dissemination in time is demonstrated by gadolinium-enhanced and non-enhanced lesions appearing simultaneously, or new lesions appearing on recent imaging compared to older imaging.
The clinician should keep in mind that periventricular lesions, which can be classic for MS, are also nonspecific, and can be seen as non-specific findings in the elderly.
When approaching a patient with ON, an MRI of brain and orbits should be obtained. Imaging can show some broad differences between etiologies of ON whether myelin oligodendrocyte glycoprotein (MOG)-ON, neuromyelitis optica spectrum disorder (NMOSD) -ON, or MS-ON. Bilateral involvement of the optic nerves is more common in NMOSD-ON and MOG-ON. Retrobulbar optic nerve involvement is seen more often in MOG-ON, and intracranial involvement is seen more often in NMOSD-ON. MOG-ON may be associated with optic nerve sheath and surrounding orbital fat enhancement. Figure 3 summarizes the imaging characteristics of ON.
Figure 1
Figure 2: 2017 McDonald criteria for diagnosing multiple sclerosis in patients with an attack at onset
| | | |
---
| | Number of lesions with objective clinical evidence | Additional data needed for diagnosing multiple sclerosis |
| ≥2 clinical attacks | ≥2 | None |
| ≥2 clinical attacks | 1 (plus clear historical evidence of a previous attack with a lesion in a distinct anatomical location) | None |
| ≥2 clinical attacks | 1 | Dissemination in space demonstrated by an additional clinical attack implicating a different CNS site or by MRI |
| 1 clinical attack | ≥2 | Dissemination in time demonstrated by an additional clinical attack or by MRI OR demonstration of CSF-specific oligoclonal bands |
| 1 clinical attack | 1 | Dissemination in space demonstrated by an additional clinical attack implicating a different CNS site or by MRI AND Dissemination in time demonstrated by an additional clinical attack or by MRI OR demonstration of CSF-specific oligoclonal bands |
Figure 3: Diagnostic Imaging Clues in ON
| | | | |
--- --- |
| | NMOSD-ON | MOG-ON | MS-ON |
| Distribution of ON lesions | Bilateral | Bilateral | Unilateral |
| Segment involvement | Intracranial, chiasmal, optic tract | Retrobulbar | Retrobulbar and canalicular |
| Length of lesions | Longitudinally extensive | Longitudinally extensive | Short segment/focal |
| Degree of ON swelling | Mild | Severe | Mild |
| Location of postcontrast enhancement | Optic nerve | Optic nerve and perineural | Optic nerve |
| Presence of brain MRI lesions | Commonly observed | Infrequently observed | Frequently observed |
| Location/characteristics of brain lesions | Hypothalamic lesions more common than MOG-ON and MS-ON; posterior fossa and periaqueductal gray | Large, tumefactive lesions; cortical and subcortical lesions | Periventricular, ovoid lesions; subcortical and juxtacortical lesions |
A lumbar puncture is not required for the diagnosis of ON. A lumbar puncture can also help to rule out elevated intracranial pressure, and in cases of atypical bilateral ON with bilateral anterior disc edema. The CSF can also be examined for infectious and inflammatory conditions as well in atypical cases of optic neuritis.
Laboratory Tests
Bilateral ON with poor visual recovery is atypical and further testing with blood tests for NMOSD, MOG and other infectious and inflammatory diseases and neuroimaging of the brain and spine may be considered.
If vision loss has occurred in a young male with a significant family history of maternally-related males with bilateral vision loss, genetic consulting and testing should be entertained for Leber hereditary optic neuropathy (LHON). Patients with LHON are much less likely to recover vision, and the contralateral eye is often affected within weeks to months of the first eye.
Differential diagnosis
Differential Diagnosis for Retrobulbar Optic Neuritis (normal appearance of optic nerve and vision loss)
Of note, the conditions noted below tend to be painless in nature, whereas 92% of patients with demyelinating ON present with some form of eye pain and/or eye pain with movement.
With a Relative Afferent Pupillary Defect
Compressive lesions like meningioma or sphenoid sinus mucocele
Central serous chorioretinopathy (OCT of the macula can help to rule out macular etiology)
Incipient central retinal artery occlusion
Posterior ischemic optic neuropathy
Toxic/metabolic or drug-induced (ethambutol, linezolid, imatinib – these present early without optic atrophy and tend to be more symmetric)
Paracentral acute middle maculopathy
No Relative Afferent Pupillary Defect
Visual field defects from lesions beyond the lateral geniculate body.
Retinal degeneration such as retinitis pigmentosa
Macular disease (OCT of the macula helpful)
Age-related macular degeneration
Macular edema (post-cataract surgery, diabetic)
Central serous chorioretinopathy
Macular hole (traumatic, idiopathic)
Epiretinal membrane/surface wrinkling maculopathy
Acute macular neuroretinopathy (AMNR)
Differential Diagnosis of Optic Neuritis (unilateral optic disc edema, vision loss)
NMOSD (Devic disease)
MOG immunoglobulin G-associated disorder (MOGAD)
Glial fibrillary acidic protein (GFAP) meningoencephalomyelitis (often visually asymptomatic with bilateral disc edema and imaging showing linear periventricular radial enhancement)
Anterior ischemic optic neuropathy (tends to be painless in nature)
Neuroretinitis (early before the macular star formation)
Chronic relapsing inflammatory optic neuropathy (CRION) (painful)
Pending central retinal vein occlusion (painless)
Optic papillitis in patient with uveitis (e.g. Vogt-Koyanagi-Harada (VKH) syndrome, sarcoidosis)
Diabetic papillopathy (painless)
Compressive lesion along the anterior pathway of the optic nerve (up to and including optic chiasm)
Optic disc drusen (pseudodisc edema, visual field defects common due to crowding of the disc, painless)
Infectious ON (e.g. syphilis, Lyme, Herpes virus (HSV, VZV); Bartonella—often with macular star in neuroretinitis)
Vasculitis (e.g. systemic lupus erythematosus, granulomatosis with polyangiitis)
Malignant hypertension (tends to be bilateral)
Neoplastic (e.g. CNS leukemia, CNS lymphoma, metastatic lesion)
Posterior scleritis (pain with eye movement, decreased vision, B-scan with thickened choroid and fluid-"T-sign")
LHON (painless, telangiectatic disc vessels)
Vascular lesion (e.g. juxtapapillary hemangioblastoma, combined hamartoma of the retina and retinal pigment epithelium)
Radiation induced optic neuropathy
Multiple evanescent white dot syndrome (MEWDS) (painless, "wreath sign" on fluorescein angiography)
Management
The ONTT was a landmark multi-center, randomized, prospective, controlled clinical trial designed to evaluate the efficacy and safety of oral prednisone vs. intravenous methylprednisolone followed by oral prednisone as compared with oral placebo for the treatment of acute ON. Patients were recruited from July 1988 to June 1991. 457 patients (85% Caucasian, range of 18-60 years old) at 15 different medical centers with presentation of symptoms of decreased vision at 8 days or less without any prior episodes of ON, not currently on any prednisone, and without any significant systemic diseases were randomized to either: 1) oral placebo, 2) 250mg IV solumedrol q6hrs for 3 days then 1mg/kg of prednisone for 11 days, or 3) 1mg/kg of PO prednisone alone for 14 days. Primary outcome measures were contrast sensitivity and visual field and secondary measures were vision and color. The patients were followed for a minimum of 6 months.
ONTT Outcomes
IV steroid treated patients recovered faster within the first 4-6 weeks post onset. However at 6 months to 10 years out, there was no statistical difference in final visual outcome between IV-steroid treated patients and placebo.
Oral steroids in conventional doses alone are contraindicated for acute ON because of higher rate of recurrences of ON
Practically speaking, physicians should consider the various causes of ON and have a high suspicion for ON as a clinically isolated syndrome associated with MS.
Treatment involves 1000 mg IV methylprednisolone daily for 3 days followed by an 11 day oral prednisone taper. This treatment has been shown to accelerate visual recovery, but not improve long-term outcomes. Oral formulations which approximate the bioavailability of the IV formulation also exist.Adrenocorticotropic hormone has also been approved in subcutaneous or intramuscular injections for treatment. Depending on the etiology (NMOSD, MOG, MS, etc.) and response to therapy of ON there may be a role for plasma exchange (PLEX) and IV immunoglobulin therapy depending on clinician judgement. Further study determining timing of treatment and benefits of these modalities is needed.
Medical therapy
Side effects of IV steroids should be fully discussed, especially in diabetics.
Low dose (60 mg/day) oral steroids are contraindicated in patients who have acute ON especially in patients that carry a diagnosis of MS
Some MS specialists and neuro-ophthalmologists are using a methylprednisolone (Solumedrol) smoothie. Instead of administering the 1 gram of Solumedrol IV, they dilute the powder in a smoothie drink and the patient drinks one dose daily for 3 days. The smoothie drink camouflages the bitter taste. It is generally not available at local pharmacies in this form. Also, repository corticotropin (intramuscular/subcutaneous) injection at 80 units/1 ml has been used to treat acute ON.
Medical follow-up
Visual acuity testing (color vision and/or contrast can elucidate subtle changes not seen on Snellen acuity testing).
If neuroretinitis is unable to be ruled out as the cause, a dilated exam 4 weeks later should show the macular star if neuroretinitis was indeed the cause. A macular star figure would not be expected in MS however and suggests an atypical ON.
A fluorescein angiogram may be performed to rule out true optic nerve edema if LHON or other non-MS etiologies are suspected.
Visual field testing upon onset if possible (Humphrey or Goldmann) and repeat at follow up visual fields at 3, 6, and 12 months.
OCT of the pRNFL at onset, at 6-8 week visit, and at 6 month follow up for quantitation of optic atrophy.
All patients with suspected ON should have a fat-suppressed contrast-enhanced MRI at the time of presentation for diagnostic and prognostic purposes.
Complications
Visual prognosis is excellent with normal to near normal recovery. Patients may still complain of decreased brightness sense, contrast deficit, and loss of stereopsis. It is important to forewarn them at the beginning that even though visual recovery is the norm, they still have the possibility of:
Permanent visual loss either mild (20/30) to severe (20/200 or worse)
Permanent scotomas that may limit driving
Recurrences
IV steroids at high dosages for the treatment of ON according to the ONTT can cause insomnia, mood changes, dyspepsia, weight gain, flushing, nausea, vomiting, and elevated blood pressure. Patients should be advised to take a proton pump inhibitor such as omeprazole and may need to seek medical attention for anxiety and insomnia. Diabetics may be at risk for episodes of hyperglycemia and also for diabetic ketoacidosis. Co-management with the patient's primary physician, internist or endocrinologist is a must. Patients tend to have more side effects from the oral prednisone taper. Some clinicians abbreviate the taper or prescribe no taper at all, others prescribe a prolonged taper over one month. It is advisable to prescribe calcium and vitamin D while taking the corticosteroids.
Prognosis
The vast majority (94%) of patients recover vision to 20/40 or better at 5 years. Only 3% of patients had 20/200 or worse visual outcomes at 5 years (based on the ONTT).
Visual recovery tends to occur by 1 month after onset and the majority recover within 1-3 months. At 6 months, patients tend to have similar visual outcomes no matter if they were treated with IV steroids or placebo. Vision improvement can take up to one year. Prolonged pain with eye movement, lack of recovery, recurrence within 2 months would alert the physician to re-evaluate for atypical causes of ON such as sarcoidosis, syphilis or an idiopathic autoimmune steroid-responsive ON .
As stated above, up to 28% of patients developed recurrent ON associated with oral prednisone. As a result, oral steroids at a dose of 1 mg/kg/day are contraindicated in the management of acute ON.
Since ON is common among patients who have MS, (up to 75% have at least one episode of ON in their lifetime), these patients are at risk for developing CDMS. The ONTT showed that even without any brain lesions present on MRI, there still was a 16% chance of developing MS in 5 years, 22% in 10 years and 25% in 15 years. One recent study in South Korea showed that the 7 year conversion rate to MS was 10.6%.
At 10 years, the overall risk of MS was 38% after an initial episode of ON. That risk is 56% if the MRI had one or more brain lesions. The risk is lower if the patient is male, disc swelling is present, pain is absent, the patient has no visual light perception, severe optic disc swelling with hemorrhages are seen, or macular exudates such as a retinal star pattern occur. In patients with CDMS, the median time to diagnosis was 3 years and 34% of the diagnoses were made in the first 2 years whereas 72% were made within 5 years.
At 15 years, the overall risk of MS was 50% after an initial episode of ON and strongly related to presence of lesions on a baseline non-contrast-enhanced MRI of the brain. Twenty-five percent of patients with no lesions on baseline brain MRI developed MS during follow-up compared with 72% of patients with 1 or more lesions. Baseline factors associated with a lower risk for MS included male sex, sever optic disc swelling, and atypical features of ON.
Patients with signs and symptoms of a CIS such as ON require prompt MRI brain imaging for lesions to determine if patients are at high risk for the development of CDMS. The role of the initiation of DMDs for MS at the time of the first ON episode should be considered depending on the MRI brain findings as well as the neurological examination.
Additional Resources
Patient Information Brochure. North American Neuro Ophthalmology Society (NANOS). Accessed March 3, 2025
Boyd K, DeAngelis KD. What Is Optic Neuritis? American Academy of Ophthalmology. EyeSmart/Eye health. Accessed March 23, 2023.
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Sellebjerg F1, Nielsen HS, Frederiksen JL, Olesen J. A randomized, controlled trial of oral high-dose methylprednisolone in acute optic neuritis. Neurology. 1999 ;52(7):1479-84.
Kupersmith MJ, Alban T, Zeiffer B, Lefton D: Contrast-enhanced MRI in acute optic neuritis: relationship to visual performance. Brain. 2002;125:812-22.
Optic Neuritis Study Group. Multiple sclerosis risk after optic neuritis: final optic neuritis treatment trial follow-up. Arch Neurol. 2008;65(6):727-32.
Chen JJ, Pittock SJ, Flanagan EP, Lennon VA, Bhatti MT. ON in the era of biomarkers. Surv Ophthalmol. 2020;65(1):12-17.
Iorga RE, Moraru A, Ozturk MR, Costin D. The role of optical coherence tomography in optic neuropathies. Romanian journal of ophthalmology. 2018;62(1):3-14.
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Introduction to MathCounts video.
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Alabama Statewide High School Mathematics Contesthomepage.
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Math and CS Research is a math and computer science publication with articles and problem sets on a wide range of topics.
Past USAMTS problems can be found at the USAMTS homepage.
The Kalva site is one of the best resources for math problems on the planet. (Currently offline - but a few mirrors are available, e.g here.)
Nick's Mathematical Puzzles -- Challenging problems with hints and solutions.
Canadian Mathematical Olympiad are hosted here by the Canadian Mathematical Society.
Problems of the All-Soviet-Union math competitions 1961-1986 - Many problems, no solutions. [Site no longer exists. Site has been replaced by a web capture]
Past International Mathematical Talent Search (IMTS) problems can be found here
Olympiad Math Madness - Stacks of challenging problems, no solutions. [Site no longer exists and has been replaced by this web capture]
Articles
Time Management
Pros and Cons of Math Competitions by Richard Rusczyk.
Establishing a Positive Culture of Expectation in Math Education by Sister Scholastica Award winner Darryl Hill.
Stop Making Stupid Mistakes by Richard Rusczyk.
What Questions Really Are the Stupid Questions? by Richard Rusczyk.
Learning Through Teaching
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Inequalities by Dr. Kiran Kedlaya
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Number Theory by Naoki Sato
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11841 | https://www.mathsisfun.com/numbers/scientific-notation.html | Powers of 10
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Scientific Notation
Scientific Notation (also called Standard Form in Britain) is a special way of writing numbers:
| | | |
---
| Like this: | | |
| | | |
| Or this: | | |
It makes it easy to use very large or very small values.
OK, How Does it Work?
Example: 700
Why is 700 written as 7 × 102 in Scientific Notation ?
700 = 7 × 100
and 100 = 102 (see powers of 10)
so 700 = 7 × 102
Both 700 and 7 × 102 have the same value, just shown in different ways.
Example: 4,900,000,000
1,000,000,000 = 109 ,
so 4,900,000,000 = 4.9 × 109 in Scientific Notation
Example: It is about 385,000,000 m from the Earth to the Moon
100,000,000 = 108 ,
385,000,0000 = 3.85 × 108 in Scientific Notation
The number is written in two parts:
Just the digits, with the decimal point placed after the first digit, followed by
× 10 to a power that puts the decimal point where it should be
(i.e. it shows how many places to move the decimal point).
In this example, 5326.6 is written as 5.3266 × 103,
because 5326.6 = 5.3266 × 1000 = 5.3266 × 103
Try It Yourself
Enter a number and see it in Scientific Notation:
images/sci-notation.js
Now try to use Scientific Notation yourself:
images/sci-notation-num.js
Other Ways of Writing It
3.1 × 10^8
We can use the ^ symbol (above the 6 on a keyboard), as it is easy to type.
Example: 3 × 10^4 is the same as 3 × 104
3 × 10^4 = 3 × 10 × 10 × 10 × 10 = 30,000
Calculators often use "E" or "e" like this:
Example: 6E+5 is the same as 6 × 105
6E+5 = 6 × 10 × 10 × 10 × 10 × 10 = 600,000
Example: 3.12E4 is the same as 3.12 × 104
3.12E4 = 3.12 × 10 × 10 × 10 × 10 = 31,200
How to Do it
To figure out the power of 10, think "how many places do I move the decimal point?"
| | |
--- |
| | When the number is 10 or greater, the decimal point has to move to the left, and the power of 10 is positive. |
| | |
| | When the number is smaller than 1, the decimal point has to move to the right, so the power of 10 is negative. |
Example: 0.0055 is written 5.5 × 10-3
Because 0.0055 = 5.5 × 0.001 = 5.5 × 10-3
Example: 3.2 is written 3.2 × 100
We didn't have to move the decimal point at all, so the power is 100
But it is now in Scientific Notation
Check!
After putting the number in Scientific Notation, just check that:
The "digits" part is between 1 and 10 (it can be 1, but never 10)
The "power" part shows exactly how many places to move the decimal point
Why Use It?
Because it makes it easier when dealing with very big or very small numbers, which are common in Scientific and Engineering work.
Example: it is easier to write (and read) 1.3 × 10-9 than 0.0000000013
It can also make calculations easier, as in this example:
Example: a tiny space inside a computer chip has been measured to be 0.00000256m wide, 0.00000014m long and 0.000275m high.
What is its volume?
Let's first convert the three lengths into scientific notation:
width: 0.000 002 56m = 2.56×10-6
length: 0.000 000 14m = 1.4×10-7
height: 0.000 275m = 2.75×10-4
Then multiply the digits together (ignoring the ×10s):
2.56 × 1.4 × 2.75 = 9.856
Last, multiply the ×10s:
10-6 × 10-7 × 10-4 = 10-17 (easier than it looks, just add −6, −4 and −7 together)
The result is 9.856×10-17 m3
It is used a lot in Science:
Example: Suns, Moons and Planets
Our Sun has a Mass of 1.988 × 1030 kg.
Easier than writing 1,988,000,000,000,000,000,000,000,000,000 kg
(and that number gives a false sense of many digits of accuracy.)
| | |
--- |
| Play With It! Use Scientific Notation in Gravity Freeplay | |
It can also save space! Here is what happens when you double on each square of a chess board:
Values are rounded off, so 53,6870,912 is shown as just 5×108
That last value, shown as 9×1018 is actually 9,223,372,036,854,775,808
Engineering Notation
Engineering Notation is like Scientific Notation, except that we only use powers of ten that are multiples of 3 (such as 103, 10-3, 1012 etc).
Examples:
2,700 is written 2.7 × 103
27,000 is written 27 × 103
270,000 is written 270 × 103
2,700,000 is written 2.7 × 106
Example: 0.00012 is written 120 × 10-6
Notice that the "digits" part can now be between 1 and 1,000 (it can be 1, but never 1,000).
The advantage is that we can replace the ×10s with Metric Numbers. So we can use standard words (such as thousand or million), prefixes (such as kilo, mega) or the symbol (k, M, etc)
Powers of 3 fit our metric number system neatly:
106 is mega, 103 is kilo, 10-3 is milli, etc.
Example: 19,300 meters is written 19.3 × 103 m, or 19.3 km
Example: 0.00012 seconds is written 120 × 10-6 s, or 120 microseconds
Copyright © 2025 Rod Pierce |
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Table of Contents
Abstract
Introduction
Historical Traceability and Structural Analysis
Molecular Biosynthesis
Bioenrichment of STX: Pathways, Influencing Factors, and Ecological Impacts
Toxicity Mechanisms of STX
Self-Resistance Mechanisms to STX Enlighten Treatment Development
Detection
Conclusions
Author Contributions
Funding
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Acknowledgments
Conflicts of Interest
References
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Open AccessReview
Saxitoxin: A Comprehensive Review of Its History, Structure, Toxicology, Biosynthesis, Detection, and Preventive Implications
by
Huiyun Deng
Huiyun Deng
SciProfilesScilitPreprints.orgGoogle Scholar
1,2,†,
Xinrui Shang
Xinrui Shang
SciProfilesScilitPreprints.orgGoogle Scholar
1,2,†,
Hu Zhu
Hu Zhu
SciProfilesScilitPreprints.orgGoogle Scholar
3,
Ning Huang
Ning Huang
SciProfilesScilitPreprints.orgGoogle Scholar
2,4,,
Lianghua Wang
Lianghua Wang
SciProfilesScilitPreprints.orgGoogle Scholar
2,4, and
Mingjuan Sun
Mingjuan Sun
SciProfilesScilitPreprints.orgGoogle Scholar
2,4,
1
Department of Student team, College of Basic Medical Sciences, Naval Medical University, Shanghai 200433, China
2
Department of Biochemistry and Molecular Biology, College of Basic Medical Sciences, Naval Medical University, Shanghai 200433, China
3
College of Chemistry and Materials Science, Fujian Normal University, Fuzhou 350007, China
4
Key Laboratory of Biosafety Defense (Naval Medical University), Ministry of Education, Shanghai 200433, China
Authors to whom correspondence should be addressed.
†
These authors contributed equally to this work.
Mar. Drugs 2025, 23(7), 277;
Submission received: 20 May 2025
/
Revised: 28 June 2025
/
Accepted: 30 June 2025
/
Published: 2 July 2025
(This article belongs to the Special Issue Marine Biotoxins 3.0)
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Abstract
Saxitoxin (STX) is a potent toxin produced by marine dinoflagellates and freshwater or brackish water cyanobacteria, and is a member of the paralytic shellfish toxins (PSTs). As a highly specific blocker of voltage-gated sodium channels (NaVs), STX blocks sodium ion influx, thereby inhibiting nerve impulse transmission and leading to systemic physiological dysfunctions in the nervous, respiratory, cardiovascular, and digestive systems. Severe exposure can lead to paralysis, respiratory failure, and mortality. STX primarily enters the human body through the consumption of contaminated shellfish, posing a significant public health risk as the causative agent of paralytic shellfish poisoning (PSP). Beyond its acute toxicity, STX exerts cascading impacts on food safety, marine ecosystem integrity, and economic stability, particularly in regions affected by harmful algal blooms (HABs). Moreover, the complex molecular structure of STX—tricyclic skeleton and biguanide group—and its diverse analogs (more than 50 derivatives) have made it the focus of research on natural toxins. In this review, we traced the discovery history, chemical structure, molecular biosynthesis, biological enrichment mechanisms, and toxicological actions of STX. Moreover, we highlighted recent advancements in the potential for detection and treatment strategies of STX. By integrating multidisciplinary insights, this review aims to provide a holistic understanding of STX and to guide future research directions for its prevention, management, and potential applications.
Keywords:
saxitoxin (STX); paralytic shellfish toxins; voltage-gated sodium channels (NaVs); mechanism; treatment; detection
1. Introduction
The ocean, as a rich reservoir of natural products, exhibits immense potential for drug research. Saxitoxin (STX) is a remarkable yet perilous neurotoxin, and as a pivotal component of the paralytic shellfish toxins (PSTs), STX is also one of the primary initiators of paralytic shellfish poisoning (PSP) [1,2,3], posing a substantial threat to both human health and the marine ecosystem.
STX emanates from a plethora of sources. Research indicates that STX can be produced both by dinoflagellates in marine environments and by cyanobacteria in freshwater environments. The former includes Alexandrium (comprising 10 species), Pyrodinium bahamense, Gymnodinium catenatum, and Centrodinium punctatum, which have long been considered the primary contributors to PSP [4,5,6,7,8]. Freshwater cyanobacteria include Raphidiopsis brookii, Anabaena circinalis, Aphanizomenon sp., Raphidiopsis raciborskii, and Microseira wollei, among others. It has been confirmed that 15 species of freshwater cyanobacteria possess the capability for STX synthesis [9,10,11,12,13]. Through the intricate food chain, STX accumulates in organisms such as shellfish and fish. Once humans consume these contaminated seafood items, they are at risk of intoxication. The symptoms of STX poisoning are diverse and severe, affecting the nervous system and including nausea, vomiting, oral paresthesia, visual disturbances, muscle paralysis, and myalgia. These symptoms may appear within 0.5 to 2 h after ingestion, and in severe cases, it can lead to respiratory paralysis and death [14,15]. Annually, there are approximately 2000 PSP cases worldwide, with a mortality rate as high as 15% in a comprehensive review , highlighting the urgent necessity for in-depth research on STX.
Internationally, the World Health Organization (WHO) and the European Food Safety Authority (EFSA) have set the limit for STX in 100 g of shellfish soft tissue at 80 μg . The United States Environmental Protection Agency (EPA) has placed it on the candidate pollutant list 4 (CCL4) and list 5 (CCL5), and has established health advisories for drinking water regarding microcystin and cylindrospermopsin. Additionally, due to its high toxicity, stability, and potential use as a biological warfare agent, STX has received considerable attention in fields such as food safety, neuroscience, and ecotoxicology [18,19,20,21,22], and is also listed as a controlled substance under the Chemical Weapons Convention; related experiments are subject to strict safety approval.
Although numerous studies have demonstrated that the toxicity of STX is fundamentally attributed to its potent blockade of voltage-gated sodium channels (NaVs) as a specific site-1 sodium channel blocker (S1SCB) acting on the α-subunit [23,24], recent investigations have uncovered that β subunits may also exert a specific modulatory role in mediating the interaction between STX and sodium channels [25,26]. Moreover, different subtypes of NaVs in the human body exhibit varying binding capacities for STX [27,28,29]. Notably, apart from its primary action on NaVs, STX may also affect non-NaV targets to exert more complex toxic effects [30,31]. These aspects collectively constitute the intricate toxicity mechanism of STX. STX has a minimum lethal dose of 1–4 mg/kg and a median lethal dose (LD50) of merely 10 μg/kg (intraperitoneal injection in mice) [32,33], rendering it extremely toxic.
Over the years, extensive studies on STX have been undertaken. By systematically integrating the full-chain evidence spanning from “discovery” to “mechanism” and “application”, the present review innovatively constructs an interdisciplinary research framework, which comprehensively covers STX’s basic toxicology, biosynthesis pathways, and advanced detection technologies. It provides a systematic theoretical foundation for STX prevention and control, precise toxin monitoring, and rational drug design, thereby bridging the gap between fundamental research and practical applications in the field of marine toxins.
2. Historical Traceability and Structural Analysis
2.1. First Detection and Process
The discovery of STX can be traced back to 1957, when Hermann Sommer’s team first isolated the toxin from the Alaska butter clam (Saxidomus gigantea) and named it after its host . In the early stages, research primarily focused on its toxic characteristics and found that it is closely related to PSP. The structural analysis of STX has been hindered due to the lack of high-purity samples and advanced characterization techniques.
In 1975, its chemical structure was completely analyzed; Schantz et al. further purified it and revealed its three-dimensional chemical structure through X-ray diffraction and nuclear magnetic resonance (NMR) techniques, confirming it as a tetrahydropurine derivative and unveiling its unique biguanide and tricyclic skeletal characteristics, while also affirming its role as the primary toxic component of PSP and its unique biguanide, tricyclic skeleton and hydrated ketone structure, laying the foundation for subsequent toxicity mechanism and synthesis research [35,36,37].
In 1977, the Tanino H. team successfully completed the total synthesis of STX racemate (d,l-STX) for the first time, marking the beginning of the artificial synthesis of STX . Subsequently, the researchers developed various synthetic strategies for its complex triazole guanidine structure. For example, the Fleming team reported asymmetric synthesis methods in the 2000s, where natural STX possessed a specific optical purity (+)-configuration; the team achieved the stereoselective synthesis of (+)-STX through 17 reaction steps [39,40]. Iwamoto et al. optimized the synthesis route using the cyanation intermediate, which improved the yield [41,42]. These synthetic breakthroughs not only advanced the development of STX analogs (such as neoSTX and deoxynivalenol), but also provided tools for researching their structure–activity relationships.
2.2. Molecular Structure Characteristics
STX belongs to the class of pyrrolidine and purine alkaloids characterized by high polarity and hydrophilicity. Its molecular formula is C10H17N7O4 (molecular weight = 299.286 g/mol) and it has a CAS number of 35523-89-8. It belongs to pyrrolopurine alkaloids and tetrahydropurine derivatives, with its molecular framework consisting of a tricyclic fused system.
The active sites are concentrated in the guanidine groups at positions 7, 8, and 9, as well as in the C12 hydroxyl group. Additionally, while the formyl side chain is not a core active group, its spatial orientation can influence the binding specificity of the toxin to different subtypes of sodium channels (such as the differences between skeletal muscle and cardiac channels) [43,44,45]. Through reductive preparation, two stereoisomers can be derived: (12R)-dihydrosaxitoxin (12α-OH) and (12S)-dihydrosaxitoxin (12β-OH). The inactive dihydrosaxitoxin features a 12α-OH configuration, with its side-chain conformation resembling that of the ketone form. In contrast, the side-chain conformation of the 12β-OH isomer is similar to the hydrated form of the original toxin. Notably, the stereoelectronic effects of the 12β-OH group exert a significant influence on the side-chain conformation, which is critical for toxicity . Notably, it has been shown that zetekitoxin AB (ZTX), as the only macrocyclic member of the STX family, has a C6-C11 isoxazolidine macrocyclic structure that is speculated to be the key to enhanced toxicity (LD50 is lower than STX). The macrocyclic backbone of ZTX enhances the binding affinity with NaVs by limiting the hydration conformation of C12 (ΔGhγ = 4.29 kcal/mol). This finding challenges the conventional wisdom that C12-hydrated ketones are necessary for the high toxicity of STX .
The STX pure product is a white crystalline solid, readily soluble in water, slightly soluble in methanol and ethanol, and insoluble in non-polar solvents. Analysis through the X-ray single-crystal diffraction of its aqueous solution crystallization reveals that the molecules form a strong hydrogen bond network through amino and sulfonic acid groups, and that the crystal structure contains ordered layers of water molecules [35,36]. It is readily soluble in water, sparingly soluble in methanol and ethanol, and insoluble in non-polar solvents. This solubility profile arises from the high polarity of the guanidino groups in the STX molecule, which facilitate the formation of extensive hydrogen-bonding networks, particularly in protic solvents; STX is thermally stable and exhibits high stability under acidic conditions (pH < 4) and low temperatures (−20 °C). It can be stored long-term in diluted hydrochloric acid without losing activity. Carbamate hydrolysis occurs only in highly concentrated acid solutions. Due to its heat and acid resistance, conventional cooking and processing cannot destroy STX, posing a significant threat to food hygiene and safety. However, it is prone to oxidation and inactivation in alkaline environments, with its toxicity rapidly diminishing [48,49]. Specifically, at a neutral pH, STX molecules exist in equilibrium in a divalent cationic hydrate, monovalent cationic hydrate, and monovalent ketone form. With the increase in pH, the C8 guanidine group donates hydronium ions, leading to a decrease in charge value, Additionally, the equilibrium between the ketone and hydrate forms affects the charge state of the molecule, potentially altering its interaction with sodium channels [46,50]. Saxitoxin with a deprotonated C8 guanidino group is susceptible to oxidation, so STX exhibits a more degradable characteristic when the pH value is >8 . Furthermore, STX can be stored stably for a long time under light-proof conditions, while light exposure may accelerate its oxidative degradation. Solid STX remains stable in dry air, but is prone to oxidation when it absorbs moisture or is in solution. Therefore, it should be stored in a sealed and light-proof container .
The STX family includes various natural analogs. To date, 57 types of STX analogs have been described, all sharing a tetrahydropurine ring skeleton that can be substituted at positions C11, N1, and C13. Their structural differences primarily arise from the substitution or modification of functional groups. The saxitoxins themselves are further divided into classes such as C toxins, G toxins, and LW toxins depending on their chemical structures and substitutions . They can be classified into several categories, such as non-sulfated, mono-sulfated, di-sulfated, and hydrophobic analogs, each with varying levels of toxicity [53,54]. The new stonehouse clam toxin (neo-STX), based on STX, has an added hydroxyl group at the N1 position, enhancing the hydrogen bond interaction with the sodium channel Tyr401, resulting in a higher affinity than STX . In decarbamoylsaxitoxin (dc-STX), the C12 trancarbamyl group is missing, and the toxicity is reduced, but some channel inhibitory activity is retained . Gonyautoxin (GTX) is generated through mono-sulfation at the N21 or O22 position of STX, which can then be di-sulfated to produce C-toxins . The research on STX analogs is crucial for reducing PST toxicity and chemical or enzymatic degradation, in order to develop detoxification methods, while also exploring the unique medicinal potential of these toxins .
3. Molecular Biosynthesis
3.1. Biosynthetic Gene Clusters and Synthetic Pathways
The biosynthetic gene cluster of STX (Sxt gene cluster) was first identified in cyanobacteria, and genomic research showed that this gene cluster contained approximately 36 genes that encoded various functional enzymes, such as methyltransferases and aminotransferases .
The biosynthesis of STX is a highly complex process (Figure 1), commencing with the polyketide synthase-like enzyme SxtA, and this enzyme contains multiple catalytic domains (sxtA1-A4). Among these, the sxtA4 domain (functions as condensation domain) is considered a critical marker for toxin synthesis. SxtA is responsible for initiating the synthesis of the STX core structure using arginine and acetyl-coenzyme A (acetyl-CoA) as starting substrates. This step generates the intermediate product 4-amino-3-oxoadipate, which provides the structural framework for subsequent reactions. The subsequent step involves the participation of SxtG and SxtB in the guanylation reaction, catalyzed by guanylyltransferase, which introduces the guanidine derived from arginine into the molecular framework, forming the characteristic guanidine ring structure of the toxin. SxtS is a 2-oxoglutarate-dependent (2OG) dioxygenase that executes the consecutive epoxidation of new double bonds and opens the epoxide to form an aldehyde, accompanied by bicyclization. The dehydrogenase SxtU then reduces the terminal aldehyde of the STX precursor. Following this, SxtH and SxtT (the terminal monooxygenase subunit and the associated ring-hydroxylating dioxygenase encoded by the respective genes) catalyze the consecutive hydroxylation at C-12, resulting in dcSTX. Subsequently, SxtI combines with SxtJ and SxtK to catalyze the formation of STX, with the entire process encompassing approximately 30 biosynthetic steps [1,59] (Figure 1).
Figure 1.
Biosynthesis pathway of STX [1,59]: Steps 1–3, Claisen condensation: Acetyl-CoA transfers an acetate group to the SxtA complex (subunits: MTF, ACP, ACTP, and AONS); methyl donation from SAM generates propionyl-SxtA, which undergoes Claisen condensation with arginine. Step 4, amidino group transfer: SxtG transfers an amidino group from arginine to the SxtA-derived intermediate, forming 4,7-diguanidino-3-oxoheptane. Steps 5–7, cyclization and double bond formation: Step 5, SxtB mediates cyclization of the amidino intermediate, forming the first heterocycle; Step 6, SxtD introduces a C1–C5 double bond; Step 7, SxtS catalyzes epoxidation of the new double bond, and epoxide ring opening generates aldehydes, triggering bicyclization. Steps 8–9, reduction and hydroxylation: Step 8, SxtU reduces the terminal aldehyde to an alcohol; Step9, SxtH/SxtT catalyze C12 dihydroxylation, with O2 and succinate as co-substrates. Step 10, carbamoyl transfer: the SxtI/SxtJ/SxtK complex transfers a carbamoyl group to the dihydroxylated intermediate (dsSTX), yielding saxitoxin (STX).
On the basis of STX, STX analogs are regenerated through specific reactions, and this process and the transport of toxins are involved in the transport of sxtL, sxtN, sxtO, sxtR, sxtX, sxtW, sxtZ, sxtPER, and other proteins encoded by the sxtACT gene [1,60,61].
The conservation analysis of the gene clusters indicates that the Sxt gene clusters of different toxin-producing cyanobacteria (such as Raphidiopsis brookii and Cylindrospermopsis raciborskii) have high homology, but that there are species-specific modifying enzymes that lead to the diversity of toxin analogs [62,63,64].
Additionally, research utilizing the Alexandrium minutum strain indicates a significant correlation between the toxin content of the strain and the expression of the fourth domain in the sxtA gene (sxtA4), as well as the presence of genetic subtypes. The quantitative polymerase chain reaction (qPCR) determination based on sxtA4 has been applied in the study of toxic algal blooms associated with PST in marine environments, demonstrating broad application prospects [60,65]. There are also newly designed sxtB TaqMan probes for the targeted detection of the STX’s biosynthetic gene sxtB from Alexandrium catenella and Alexandrium Pacificum . Notably, dinoflagellates, as the primary producers of STX, have posed challenges in identifying their biosynthetic genes. A recent comparative transcriptomic study of Gymnodinium catenatum identified over 1000 homologs of STX biosynthesis genes in this species. Among these, sxtA and sxtG exhibited significantly higher expression levels than other genes, suggesting their central roles in the toxin biosynthesis pathway. Intriguingly, the study revealed the absence of sxtB. This finding implies that Gymnodinium catenatum may employ alternative metabolic routes for STX biosynthesis or rely on functionally redundant genes to compensate for the loss of sxtB .
3.2. Environmental Regulation of Toxin Synthesis
The biosynthesis of STX is coordinately regulated by various environmental factors, including nutritional conditions, light, temperature, salinity, and others [68,69,70].
Different toxin-producing algae have their own suitable temperature ranges for growth and STX synthesis. Within the appropriate temperature range, enzymatic activity is high, and reactions related to the STX synthesis pathway can proceed smoothly. Extreme temperatures, whether too high or too low, can affect enzyme activity and thus inhibit STX synthesis. Additionally, temperature can influence the expression of core genes sxtA and sxtG, thereby affecting STX content, particularly evidenced by the increased expression of sxtA and sxtG under low-temperature stress .
The intensity and quality of light can affect the photosynthesis and related metabolic pathways of toxin-producing algae, thereby influencing the synthesis of STX [72,73]. Appropriate light intensity can promote the growth and metabolism of algae, facilitating the synthesis of STX, while excessive or insufficient light can inhibit STX production. Research indicated that an increase in temperature combined with adequate light can enhance the biomass of Raphidiopsis raciborskii. However, excessively high temperatures can lead to a decrease in its toxicity .
Toxic cyanobacteria (such as Raphidiopsis raciborskii) and dinoflagellates (such as the genus Alexandrium) perceive environmental changes and dynamically adjust the expression of the Sxt gene cluster to adapt to ecological pressures and maintain competitive advantages [75,76]. Specifically, changes in the nitrogen and phosphorus content in nutrients lead to nutritional stress, with nitrogen stress inhibiting the growth of Alexandrium minutum and reducing the biosynthesis of STX, while phosphorus deficiency can lead to an increase in the biosynthesis of saxitoxin [77,78]. The NO3− concentration also affects the expression of sxt core genes and affects the production of STXs, suggesting that N is directly involved in the biosynthetic pathway of STXs .
Apart from the natural environment, the biological environment can also affect the content of STX. For instance, under the chemical induction of the zooplankton Daphnia gessneri, the sxt genes of Raphidiopsis raciborskii (Cyanobacteria) are upregulated . Other studies have shown that Jannaschia cystaugens (representing Alexandrium cyst formation-promoting bacteria, Alex-CFPB) significantly influences the life cycle and toxin production of Alexandrium pacificum. Co-culturing with Jannaschia cystaugens induces stress that promotes the synthesis of highly toxic PST, leading to increased intracellular toxin levels, specifically involving oxidative stress, nutrient stress, and quorum sensing signals, all of which contribute to an increase in toxin synthesis .
3.3. Latest Research Developments
Although significant achievements have been made in the biosynthesis research of STX, many details still require refinement.
In exploring the details of the STX biosynthetic pathway, multi-omics technologies such as metabolomics and proteomics have played a crucial role. Through metabolomics, it is possible to comprehensively analyze the metabolite profiles of toxic algae at different growth stages and under various environmental conditions, and by utilizing proteomics, it is possible to identify differentially expressed proteins in toxic algae under varying environments, thus providing a pathway for the in-depth exploration of the dynamic changes in the STX biosynthetic pathway under environmental influences .
In addition, biotechnological techniques provide new strategies for the efficient production of STX, and researchers are attempting to reconstruct the STX biosynthetic pathway in heterologous hosts (such as Escherichia coli and Synechocystis) to achieve the extracellular synthesis of PST [82,83]. Jiao Y et al. established an asymmetric synthesis strategy featuring an intramolecular [2 + 2] photocycloaddition of an alkenylboronate ester equipped with a new chiral auxiliary, leveraging chiral additives to stabilize the transition state through hydrogen bonding, which achieved high stereoselectivity in constructing tricyclic frameworks and offered a novel paradigm for synthesizing complex toxins . Furthermore, Greenhough H et al. scaled up the cultivation of Alexandrium pacificum using a 1250 L photobioreactor (PBR), thereby surpassing the production limitations of traditional natural sources for GTX-1,4. This advancement provides a gram-scale supply of toxins for preclinical research and offers a groundbreaking solution for the production of STX and its analogs .
4. Bioenrichment of STX: Pathways, Influencing Factors, and Ecological Impacts
The traditional route by which STX causes human poisoning involves the production of toxic dinoflagellates (such as the genus Alexandrium), their concentration through filter-feeding bivalves (mussels, oysters, and clams), and subsequent ingestion by humans, representing a common pathway in reported poisoning incidents. This pathway is closely associated with the species composition, geographic distribution, and toxin-producing capacity of the involved dinoflagellate populations. These factors are further influenced by environmental conditions, including salinity, temperature, and eutrophication levels [86,87,88].
Different traditional media have variations in the speed of toxin removal, retention time, and the retention of toxic components. The toxicity produced by dinoflagellate species also varies, resulting in different toxic effects on humans. Many countries have implemented monitoring programs for STX in toxic microalgae and shellfish, which help minimize public health risks .
Non-traditional media include non-bivalve invertebrates (gastropods and cephalopods), crustaceans, fish, annelid worms such as Eudistylia sp., and echinoderms like Asterias amurensis, Astropecten scoparius, Astropecten polyacanthus, and Pisaster ochraceus. Among these, although gastropods and cephalopods accumulate STX, they do not cause significant negative effects, while the yellow crab is the most commonly known non-filter-feeding non-molluscan species containing STXs. The mechanism by which STX is found in fish involves the transport of STXs through the food chain and the dissemination and accumulation of toxins through zooplankton [90,91]. Research indicates that the internal organs are the primary sites for PST accumulation. Simultaneously, these organs exhibit a higher STX clearance rate and excrete via the kidneys. Among STX and its analogs, those with higher solubility have a greater excretion rate [77,92]. The regulation and control of these non-traditional media is of significant importance for preventing STX poisoning incidents.
In the field of ecotoxicology, the accumulation of STX at the top of the food chain can lead to population cascading effects. For instance, high concentrations of STX in shellfish can cause foodborne poisoning, including PSP [93,94]. Toxins’ metabolic transformation significantly affects their toxicity level. The latest studies show that when mussels are exposed to dynamic algal cell densities (simulating natural blooms), the concentration of PSTs (paralytic shellfish toxins) in their bodies increases by 1.8 times compared with exposure to fixed densities. Moreover, low-toxicity sulfocarbamoyl toxins (such as GTX5) are converted into highly toxic derivatives (e.g., dcSTX) through decarbamoylation reactions, a process associated with the inhibition of sulfotransferase activity in the hepatopancreas . The consumption of contaminated fish may lead to the mass death of marine predators that feed on fish (such as seabirds and cetaceans) [96,97,98]. Exposure to STX during the development of zebrafish can lead to changes in the expression of genes related to axon growth and affect the functionality of NaVs . Studies on Saxidomus gigantea (Alaska butter clams) have shown that when humans consume 200 g of tissue containing 900 μg of saxitoxin (STX), the probability of severe symptoms in average males reaches 11%, and the mortality risk is 0.27%. This confirms the potential threats of STX to the ecological chain and human health .
The acute guideline value for STXs proposed by the WHO is 0.003 mg/L (3 μg/L) for the total STXs (including congeners, free forms, and cell-bound forms). Exposure routes include drinking contaminated surface water, consuming contaminated marine shellfish, and having contact with high-concentration water bodies during recreational activities. In terms of protection, avoiding contact and accidental ingestion is fundamental. So far, there is no specifically approved antidote, and treatment after poisoning mainly focuses on symptomatic and supportive care [101,102]. Health risks should be addressed through water source protection and water treatment process control, while paying attention to their impacts on the sensory quality of drinking water. Controlling water eutrophication can reduce the risk of cyanobacterial blooms. Additionally, water contaminated with STXs can be treated to mitigate poisoning risks. Extracellular STXs can be physically removed via activated carbon adsorption or membrane filtration . Alternatively, chemical inactivation methods such as potassium permanganate, ozonation, and chlorination, etc., can disrupt the toxin structure. However, the risks of disinfection by-products should be noted [104,105]. In addition, H2O2 may not be able to effectively degrade saxitoxin . Biological activity-based removal or degradation methods, such as adsorption by lactic acid bacteria and degradation by bacteria isolated from the digestive tract of blue mussels (Mytilus edulis), have been reported . To remove intracellular toxins, complete cyanobacterial cells need to be eliminated, which can be achieved through traditional coagulation and sand filtration processes . During the removal process, care must be taken to prevent cell lysis and the release of toxins, as the review showed.
5. Toxicity Mechanisms of STX
STX, as a guanidine alkaloid, features a core structure with two guanidine groups, both belonging to the guanidinium toxin family with TTX. These guanidine groups exhibit an extremely high affinity with and ion flux blocking ability towards NaVs [25,110], which endows STX with high biological activity and extreme toxicity.
5.1. Blocking Mechanism of STX on NaVs
Toxins can be roughly divided into two categories: pore blockers that physically close the pore of the channel and “gating modifier toxins” (GMTs) that act on the voltage-sensing domain of the channel to cause abnormal activation (opening) or inactivation (closing) of the sodium ion channel.
Mammalian NaVs consist of a pore-forming α subunit and one or more β subunits. The α subunit is the main part that forms the ion channel pore and has voltage-sensing and gating functions [111,112]. The α subunit is folded from a single polypeptide chain containing four homologous repeat sequences. Each repeat sequence contains six transmembrane helix segments (S1–S6). These segments are connected by extracellular and intracellular loops. S1–S4 form the voltage sensor, and S5 (outer helix), S6 (inner helix), and the P loop that re-enters the extracellular membrane between S5 and S6 together form the pore domain [113,114]. Site 1 on the α subunit is the common receptor-binding site for these types of toxins, such as STX, tetrodotoxin (TTX), and polypeptide toxins like μ-conotoxin [25,115]. The active sites of STX are the guanidyl groups at positions 7, 8, and 9. They have a high affinity with the amino acid residues at site 1 of the voltage-gated Na⁺ channel on the excitable cell membrane, thereby physically blocking the pore . Through the analysis of the structure of the sodium channel–STX complex via X-ray crystallography and cryo-electron microscopy, it was observed that after STX binding, the selective filter region of the channel contracted, narrowing the ion permeation path and preventing sodium ions from passing through smoothly .
STX blocks the influx of sodium ions, inhibiting the establishment and collapse of the transmembrane electrochemical charge gradient in animal nerve, muscle, and endocrine cells. This ion imbalance hinders the initiation of action potentials [117,118], leading to serious consequences. The symptoms of STX poisoning usually appear within 30 min after ingestion, including oral numbness, limb paralysis, and breathing difficulties, etc. In severe cases, it can cause death within a few hours [119,120] (Figure 2).
Figure 2.
Schematic of STX Blockade Mechanism on NaVs: (A) Structural organization of NaV complex: The NaV complex is composed of a pore-forming α subunit and one or more β subunits . The α subunit, a single polypeptide chain with four homologous repeat domains (I–IV), constitutes the ion-conducting pore and voltage-sensing apparatus. Each domain contains six transmembrane helices (S1–S6)—S1–S4 form the voltage sensor, while S5–S6 form the pore domain ; (B) STX binding and functional disruption: Upon binding to the outer vestibule of the α-subunit pore via site 1, STX induces conformational contraction of the selective filter, restricting Na⁺ influx . This α-subunit-mediated disruption abrogates neuronal action potential generation (Vmax reduction ≥ 80%) and propagation, leading to clinical conditions: facial paresthesia, limb paralysis, and respiratory arrest, etc. [23,30]. (Some drawing elements are from www.figdraw.com).
5.2. Association of β Subunits in the Action of STX
Furthermore, the process of STX blocking sodium channels also involves interactions with β subunits. The β subunit is not just a simple accessory to the pore-forming α subunit. Under normal circumstances, β subunits play an important role in the localization of sodium channels, function regulation, and coupling with other cell signaling pathways [115,121], allowing for the precise control of excitability in a cell type-specific manner. At the same time, the regulatory effect of the β subunit on voltage-gated potassium channels affects neuronal excitability. Abnormal neuronal excitability is related to many nervous system diseases. Therefore, this regulatory effect may play a role in the occurrence and development of nervous system-related diseases .
Messner, D.J. and Catterall, W.A. detected the functional integrity of the resulting protein complexes by selectively removing β1 or β2 subunits from the detergent-solubilized channels in rat brains. They found that selectively removing the β1 subunit from the detergent-solubilized channels in rat brains led to a complete loss of 3H-labeled STX binding activity in sodium channels, confirming that the presence of the β1 subunit is crucial for maintaining the high-affinity conformation of the STX binding site on sodium channels .
Nonetheless, currently, there are very few studies on the specific role of β subunits in the process of STX toxicity, and further exploration is needed.
5.3. Affinity Difference in Different Subtypes of NaVs to STX
In the human body, there are nine subtypes of sodium channels, such as Nav1.1–Nav1.9. They are specifically distributed in different tissues. The amino acid sequences, spatial conformations, and accessory subunits of different subtypes of sodium channels vary [27,28,123], resulting in different binding abilities of STX to each subtype.
However, STX binds to the amino acid residues at site 1, which is located on the outer surface of the excitable cell membrane and closer to the outer opening of the sodium channel [29,43]. Sodium channels in nerve tissues have a high affinity for STX. Ritchie and Rogart conducted binding experiments using radioactively labeled STX with rabbit vagus nerves, lobster walking leg nerves, and gar olfactory nerve fibers and found that STX can tightly bind to the NaVs on nerve cell membranes and effectively block nerve impulse conduction at extremely low concentrations .
Nav1.4 is mainly distributed in skeletal muscle. The high affinity of Nav1.4 in skeletal muscle for STX may be determined by the charge complementarity of acidic residues in the pore region, the cooperative effect of domains, and subtype-specific conformations [113,125], which provides a key molecular target for the development of toxin antagonists targeting skeletal muscle sodium channels. In contrast, the sodium channels in cardiac muscle tissue have a relatively low affinity for STX . Catterall and Coopersmith used radioactive ligand binding experiments to detect STX receptor sites in vertebrate heart tissues. However, the binding ability of heart tissues to STX is weaker than that of nerve tissues, and a higher concentration of STX is required to produce a significant inhibitory effect .
5.4. Potential Toxicity Targets of STX
Interestingly, although the toxicity of STX mainly stems from the inhibition of specific NaVs, STX can also affect other channels, enzymes, and proteins, thereby exerting toxic effects . Studies have found that STX also affects L-type calcium channels and HERG potassium channels , interfering with the ion balance and electrophysiological activities within cells. In addition, STX may also act on neuronal nitric oxide synthase, STX-metabolizing enzymes, a transferrin-like family of proteins, and a unique protein found in the blood of pufferfish. Although the specific mechanisms and degrees of influence remain unclear, these additional effects may further exacerbate the toxic effects of STX on organisms [120,128].
5.5. Synergistic Effect of STX with TTX
In addition to structural analogs, the combined exposure effects of STX with other toxins have garnered increasing attention in recent years. Despite their pronounced chemical, structural, and affinity disparities, both STX and TTX are categorized as S1SCBs, and their neurotoxicity may exhibit additive effects [129,130].
Furthermore, investigations into the combined toxic effects of STX and TTX have made significant strides recently. Concurrently, Andrea Boente-Juncal et al. validated through chronic toxicity experiments that combined exposure to TTX and STX elicits additive toxic effects, particularly when the STX dosage exceeds the maximum exposure threshold recommended by the EFSA, leading to significantly elevated mortality and biochemical derangements. They demonstrated via intragastric gavage that co-exposure to STX and TTX in mice induced significantly more aberrant changes in blood biochemical parameters and higher mortality rates compared to single-toxin exposure groups, thereby confirming their combined toxic effects . Nevertheless, the intragastric gavage methodology harbors methodological limitations, as it may lead to inadvertent pulmonary aspiration of the administered toxins—potentially causing rapid lethality—and mechanical reflux induced by gastric tube insertion [132,133]. Nevertheless, Finch et al. systematically demonstrated the additive toxicity of STX and TTX through LD50 determinations via multiple routes (intraperitoneal injection and feeding), a comparison of predictive models with experimental data, and consistency analysis of action mechanisms and clinical symptoms . Despite the lack of in-depth explanations of the mechanisms of the combined effects of toxins, these findings provide a scientific basis for incorporating TTX into the regulatory system of PSTs.
6. Self-Resistance Mechanisms to STX Enlighten Treatment Development
Currently, there is no specific antidote for STX poisoning. Treatment mainly involves symptomatic supportive therapies such as artificial ventilation and gastric lavage, as the review summarized . However, research on the strategies of organisms in nature to cope with STX exposure has inspired the research on STX treatment [136,137]. In recent years, relevant research achievements in exploring STX antagonists and optimizing toxin delivery systems have also provided clues for the development of drug treatments.
6.1. Target Protein Resistance Mutations
Target protein resistance mutations play a crucial role in the research on the mechanisms of biological toxicity resistance and have been investigated in depth and systematically [137,138]. Soft-shell clams (Mya arenaria) from areas affected by “red tides” have been found to be more resistant to PSTs. This is due to a natural mutation of a single amino acid residue, which reduces the affinity of the STX binding site in the sodium channel pore of resistant soft-shell clams (but not sensitive ones) to one-thousandth of the original level . When cysteine (Cys) at position 374 of the cardiac sodium channel α subunit (RHI) is replaced by tyrosine (Tyr) (similar to the situation of the μl subtype), the sensitivity of RHI to STX changes significantly , which may significantly affect the STX resistance of organisms. When single-gene mutations were studied in the SS2 and adjacent regions of the four internal repeat sequences in sodium channel II, it was found that mutations involving a negatively charged residue clustered at specific positions significantly changed the STX sensitivity, and that these mutations may affect the extracellular mouth and/or pore wall structure of the sodium channel . All these mutations have an impact on the STX resistance of organisms.
6.2. Toxin Sequestration
Soluble STX-binding proteins found in some organisms such as frogs, pufferfish, cockles, and crabs constitute a special mechanism for organisms to cope with toxins—toxin sequestration . Among the strategies of organisms to cope with toxins, the resistance mechanism of frogs to STX has attracted much attention . It has been found that saxiphilin (Sxph) in bullfrogs can bind to STX with a high affinity (Kd is approximately 0.2 nM) [143,144,145]. In the study of the binding site of STX to NaVs, it was found that Sxph and Navs use similar molecular logic to recognize STX. Both use side-chain carboxylates to interact with the five-membered and six-membered guanidyl groups of STX. Moreover, Sxph can bind STX through a specific binding pocket in the C1 domain, thus acting as a high-affinity “toxin sponge” protein to prevent STX poisoning. In addition, a similar sequence of Sxph, SxphNP, was found in the Himalayan frog (Nanorana parkeri), which also has the key characteristics of recognizing STX [128,146]. These findings expand the research direction of developing effective STX antidotes based on biological resistance.
6.3. Biotransformation and Metabolism
Some reports have indicated that the digestive glands of bivalve molluscs possess a high capacity for PST transformation [147,148], suggesting the presence of microorganisms and/or toxin-transforming enzymes within them, as demonstrated in the review . Vasama Mshou et al. were the first to confirm that Lactobacillus rhamnosus GG and LC-705 strains (both in viable and non-viable forms) can remove STX and neoSTX from neutral solutions with a pH of 7.3 and acidic solutions with a pH of 2, achieving removal rates of as high as 77–97.2% . Pseudoalteromonas bacteria are capable of degrading PSTs, and STX is one of them. These bacteria have the potential to remove STX. Such bacteria are relatively common in the marine environment. They may degrade toxins through special enzymes secreted by themselves or metabolic pathways. Marine bacterial isolates cultured from the digestive tracts of blue mussels (Mytilus edulis) contaminated with PST can reduce the overall toxicity of algal extracts to less than 10% of the original within 3 days . These bacteria and similar ones may assist marine bivalves in the natural metabolism and elimination of PSTs and other marine biotoxins, and also promote research on the biological treatment of STX poisoning.
Meanwhile, some newly reported C-11 hydroxyl analogs of PSTs in shellfish are considered novel derivatives of PSTs and are called “M-toxins” [152,153,154,155]. The research team led by Li Aifeng confirmed that the lower the pH and temperature under experimental conditions, the more stable the M-toxins. Compared with common components of PSTs, the stability of M-toxins is poor. They also confirmed the chemical transformation pathways of “M-toxins” as M1→M3→M5, M2→M4→M6, and NEO→M10 . This is of great significance for understanding the detoxification mechanism of shellfish and achieving rapid detoxification by converting highly toxic PSTs into weakly toxic M-toxins. It also provides new ideas for studying the detoxification mechanism of organisms to STX.
7. Detection
In the context of the high risk of STX, countries around the world have listed it as a mandatory item for aquatic product safety inspection, and the detection technology of STX and its derivatives is the key to its bioaccumulation and toxicological mechanism study. At present, there are various detection methods, mainly including biological detection methods and instrument detection methods, etc.
Biological detection methods, based on the biological activity of toxins, were the main means for early STX detection. The Mouse Bioassay (MBA) is a classic method for early STX detection. It involves the intraperitoneal injection of sample extracts and semi-quantitative analysis of STX toxicity based on mouse death time. Since it was confirmed as the statutory detection method by the Association of Official Analytical Chemists (AOAC) in 1958 (AOAC 959.08), it has been used as the reference standard for other detection techniques . However, it has limitations such as an inability to distinguish toxin analogs, low sensitivity, and animal ethics controversies . Additionally, the Daphnia acute toxicity test utilizes the sensitivity of Daphnia pulex (limit of detection (LOD): 0.3–10 ng equivalent to saxitoxin (Eq.STX)/L) and Moina micrura (LOD: 2.7–10.7 ng Eq.STX/L) to STXs, serving as a rapid screening tool for environmental samples . Similarly, the speckled cockroach (Nauphoeta cinerea) bioassay can act as a supplementary field screening tool for STXs, with its core value lying in cost and speed advantages, though specificity remains low .
Cell-level detection, achieved through toxicity effects mediated by NaVs, can replace traditional mouse bioassays. For example, the Neuro-2a cell-based assay operates on the principle that veratridine, in the presence of ouabain, increases sodium influx in the mouse neuroblastoma cell line Neuro-2a (ATCC, CCL131), leading to cell swelling and death . STX specifically blocks NaVs, antagonizing the toxic effect of veratridine. The toxicity of STX is indirectly reflected by detecting functional changes in intracellular ion channels or cytotoxic effects (such as MTT metabolism and lactate dehydrogenase release). Similarly, this method can also be used for the detection of sodium channel-blocking toxins such as TTX [161,162]. However, this method requires a long analysis time. The analysis time can be shortened by using modified forms of short toxins or the calcium agonist maitotoxin [163,164]. In addition, it also has high requirements for reagents, professional knowledge, and equipment.
With the acquisition of anti-STX antibodies, immunoassay techniques have been developed. The enzyme-linked immunosorbent assay (ELISA) method is the most commonly used, relying on the color reaction between enzyme-labeled antibodies and substrates for detection, including direct and indirect ELISA [165,166]. Immunoassay can achieve the rapid screening of PSP, but its cross-reactivity to different toxins is low, making it difficult to fully reflect the composition of toxic components in samples. Therefore, when detecting STX and its analogs, additional approaches are required to identify and quantify specific toxins, which highly depends on laboratory equipment . The subsequently developed MIST Alert™ rapid detection method can be used for the determination of the total PSP toxicity and related toxin toxicity in acidic crude toxin extracts. It achieves a 100% detection rate for toxic extracts containing at least 80 μg Eq.STX/100 g, with a LOD of approximately 40 μg Eq.STX/100 g (varying with toxin profiles). This method features high sensitivity, rapid detection, and does not require specialized tissue culture equipment or expertise. However, it is unable to distinguish individual toxin components [167,168]. Rabbit monoclonal antibodies (r-mAbs) developed by Li et al. achieved the ultra-sensitive recognition of STX through a single-cell sorting and cross-screening strategy. The sensitivity is 100 times higher than that of traditional mouse antibodies, which solves the problem of the poor stability of traditional antibodies in complex matrices .
The receptor binding assay (AOAC 2011.27) utilizes the principle that PSP toxins can competitively bind to NaVs with [3H]-STX diHCl. This method applies isotope labeling to detect PSP toxins, featuring rapidity, reliability, and accuracy . The substitution of Na+ channel proteins with saxiphilin, which specifically binds to STX, has been studied to be specific for STX identification and can be distinguished from TTX identification [171,172].
In chemical instrumental analysis, high-performance liquid chromatography (HPLC) was the first instrumental analysis technique applied to the detection of PSTs. It offers advantages such as high sensitivity, strong specificity, a low LOD (0.1 μg/kg), and fast analysis speed. HPLC can simultaneously provide qualitative and quantitative information for multiple toxins, making it the core method for toxin detection currently. Due to the weak chromophoric group of STX, pre-column oxidation (AOAC 2005.06) or post-column oxidation (AOAC 2011.02) is required before detection to generate fluorescent chromophores (e.g., dimethylaminobenzoic acid derivatives), which are then detected using a fluorescence detector. Among them, post-column oxidation HPLC is noted for its operational simplicity, strong matrix tolerance, and high stability [173,174].
Subsequently, with the advancement of mass spectrometry technology, the combination of liquid chromatography and mass spectrometry (LC/MS) has emerged as one of the preferred analytical methods for the highly selective and sensitive quantification of toxins. In a study, dansyl chloride was used for the chemical derivatization of STX, followed by separation via ultra-high-performance liquid chromatography (UHPLC) coupled with heated electrospray ionization (HESI), and quantification using a Q-Exactive mass spectrometer. This approach achieved a LOD as low as 0.01 μg/L [175,176]. LC-MS/MS serves as the gold standard for detecting guanidine toxins. Its chromatographic separation module is compatible with both HPLC and UHPLC systems . Liquid chromatography separates STX via a chromatographic column by leveraging the partitioning difference between the stationary phase and mobile phase. Meanwhile, mass spectrometry ionizes STX through techniques such as electrospray ionization (ESI), generates characteristic product ions via collision-induced dissociation (CID), and enables the simultaneous qualitative and quantitative analysis of STX and more than 50 of its analogs using the multiple reaction monitoring (MRM) mode . Subsequent improvements in the separation and extraction methods for STX have been made. For example, the combination of isotope dilution liquid chromatography tandem mass spectrometry (HILIC) and the isotope dilution method has enabled the simultaneous detection of multiple toxins and improved sensitivity [179,180]. The extraction and purification of STX are performed using immunoaffinity column (IAC) combined with solid-phase extraction (SPE) technology. STX is extracted with phosphate-buffered saline (PBS), separated on a TSK-GEL Amide column, and detected via LC-MS/MS. This method completely eliminates the matrix effects and does not require matrix matching, meeting the requirements for the detection of trace STX in bivalve aquatic products . Additionally, extraction is performed by acidifying plasma samples and precipitating with acetonitrile, followed by LC-MS-MS analysis. The entire process takes ≤30 min, with a LOD of 2.8 ng/mL and a lower limit of quantitation (LOQ) of 5.0 ng/mL . This technique utilizes the characteristic fragmentation of the guanidino group in STX molecules, combined with the high sensitivity of electrospray ionization (ESI) to distinguish structurally similar toxins (such as neoSTX and dcSTX) and achieve accurate confirmation in complex matrices (e.g., biological tissues and seawater). For instance, Dahlmann et al. used LC-ESI-MS/MS to simultaneously detect STX and microcystins in phytoplankton, confirming the tricyclic skeleton structure of STX through MS/MS fragment analysis . LC-MS/MS offers significant advantages in STX detection, enabling simultaneous multi-component analysis without the need for derivatization, thus ensuring high operational efficiency. However, it also has limitations, such as expensive instrumentation and complex maintenance.
Thin-layer chromatography (TLC) is one of the traditional methods for detecting PSP toxins, which is currently used less frequently . Research has been conducted on the chromatographic column for the separation of STX and its congeners using thin-layer chromatography in a rod shape, with detection by a flame thermionic detector (FTID). The detection limit for STX was 5 ng, while the detection limits for the other congeners also reached the ng level . In addition, traditional techniques for detecting STX toxins include gas chromatography, ion-exchange column chromatography, and electrophoresis [184,186,187,188]. However, these traditional methods are currently less used due to their low resolution and poor sensitivity.
In recent years, new detection technologies have been continuously emerging. For example, some research is dedicated to developing detection methods based on biosensors, utilizing the specific interactions between biological recognition elements and STX, achieving the rapid and sensitive detection of STX through signal transduction, thereby enhancing selectivity and sensitivity [189,190]. These methods hold promise for overcoming some of the limitations of the traditional techniques, such as expensive instrumentation and complex operation, featuring advantages of portability, rapidity, high sensitivity, and low cost. They provide new possibilities for on-site detection and real-time monitoring. However, they also suffer from limitations, including complicated fabrication, short operational lifespan, complex manipulation, high professional requirements, and significant susceptibility to matrix effects [191,192,193]. In addition, microfluidic chip technology has also begun to be applied in the field of STX detection. The microfluidic chip can integrate multiple steps such as sample processing, reaction, and detection onto a small chip, offering advantages such as fast analysis speed, the low consumption of samples and reagents, and the ability to achieve high-throughput detection, thereby providing direction for the miniaturization, automation, and integration of STX detection. However, its fabrication is complicated, and each detection causes the consumption of the detection chip, resulting in a short operational lifespan [194,195].
Based on the existing research methods and their characteristics, combined with the scenario requirements for STX detection (on-site screening, laboratory confirmation, and research-grade analysis), the following feasible detection system of “rapid screening–precise quantification–mechanism verification” is proposed. By integrating the sensitivity, specificity, and operational complexity of different techniques, this system enables efficient detection. Quality control is essential before and during the detection process. The first stage involves on-site rapid screening, applicable to critical scenarios including aquatic product harvesting sites, the emergency monitoring of environmental water bodies, and initial food safety screening. The Daphnia acute toxicity bioassay and MIST Alert™ rapid test are employed here, leveraging their combined advantages of operational simplicity, cost-effectiveness, and rapid response. These techniques enable the preliminary triage of non-toxic samples, thereby streamlining the subsequent analytical workload. The second stage is the laboratory confirmation and quantification stage (applicable scenarios: regulatory sample re-inspection, market access testing, and acquisition of basic data for toxicological research). Firstly, the indirect ELISA method is adopted, which uses the specific binding of enzyme-labeled antibodies to STX for color development, rapidly quantifies the total STX equivalents in samples, and screens positive samples. The second step is precise analysis via HPLC (post-column oxidation method) or LC-MS/MS, with LC-MS/MS being the preferred choice. The third stage is the research-grade analysis and mechanism verification stage (applicable scenarios: STX toxicological mechanism research, identification of novel toxin analogs, and drug development target screening), involving the utilization of cell-based assays (Neuro-2a cell-based assay) and receptor-binding assays. By integrating LC-MS/MS for structural characterization with cell function analysis, this stage deepens STX research from the multi-dimensional perspectives of “toxicity effect–molecular mechanism–structural characterization” (Table 1).
Table 1.
Established detection system for saxitoxin (STX).
8. Conclusions
STX, as a potent neurotoxin derived from marine dinoflagellates and freshwater or brackish water cyanobacteria, has been the subject of extensive research in recent years. This review has comprehensively summarized the current state of knowledge regarding STX, encompassing its discovery history, chemical structure, molecular biosynthesis, bioenrichment, toxicity mechanisms, self-resistance mechanisms, and detection methods.
The discovery and structural elucidation of STX were pivotal milestones in understanding its properties and biological activities. The identification of its unique tricyclic skeleton and biguanide group provided a foundation for subsequent studies on its interactions with biological targets, particularly NaVs.
In the field of toxicity mechanism research, the traditional mechanism of STX as a specific S1SCB of NaVs has been systematically elucidated. However, recent studies have continuously broken through the boundaries of cognition, not only discovering potential novel action targets of STX beyond NaVs, but also achieving key advancements in the exploration of biological natural resistance mechanisms. These innovative discoveries provide entirely new ideas and theoretical support for the development of STX-related drugs. Scholars such as Mingxuan Kai et al. have developed Neuron-MOF/SxtN-NPs (neuronal membrane-coated MOF nanoparticles loaded with SxtN enzyme), a dual-mode cellular nanoparticle formulation. They enable broad-spectrum neurotoxin neutralization via membrane mimicry and continuous detoxification through enzymatic activity. In STX-intoxicated mice, the nanoparticles show enhanced survival, no acute toxicity, and validate a biomimetic nanomedicine platform for neurotoxin countermeasures . Dali Wang et al. have pioneered the design of an aptamer-based small-molecule therapeutic sustained-release system, applying STX to nerve blocks , which has inspired the progress of “seeking drugs from the ocean”.
As a highly toxic marine neurotoxin, STX as a single component has posed a threat to the marine ecological balance and human health. In natural environments, STX often forms complex contamination systems with multiple toxicants, and more than 50 natural structural analogs of STX have been identified. In this context, research on the combined action of toxins is gradually becoming a frontier in the field of STX toxicology. This research not only helps to reveal the real toxic risks of natural contamination systems, but also provides key scientific evidence for establishing the safety thresholds of marine toxins based on mixture effects and developing multi-target detoxification strategies.
In terms of detection technology, although numerous methods have been developed and applied to STX detection, and LC-MS/MS has become the most reliable method for detecting STX, systematically comparing the advantages and disadvantages of the existing detection methods and scientifically classifying their applicable scenarios still holds significant practical significance for achieving the rapid and precise detection of STX.
Author Contributions
Writing—original draft preparation, H.D. and X.S.; writing—review and editing, M.S. and N.H.; supervision, H.Z.; investigation, L.W. All authors have read and agreed to the published version of the manuscript.
Funding
This work was supported by the National Natural Science Foundation of China (NSFC) under Grant No. 82173732 and the National Key Research and Development Program of China under Grant No. 2019YFC0312603. Lianghua Wang received financial support from the above grants and acknowledges their contribution.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
No new data were generated in this review.
Acknowledgments
The contributions of all reviewers and editors are deeply valued by the authors. The authors extend sincere gratitude to the peer reviewers for their meticulous and constructive suggestions on the sections of toxicological mechanisms, biosynthesis process, and detection methods significantly enhanced the scientific rigor and clarity of the draft. Special appreciation is also due to the editors for their continuous guidance and reminders throughout the review process, which ensured the manuscript’s compliance with academic standards and publication requirements. The authors extend heartfelt gratitude to Andrew Turner for his generous provision of references about the detection of paralytic shellfish toxins.
Conflicts of Interest
The authors declare no conflicts of interest.
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Figure 1.
Biosynthesis pathway of STX [1,59]: Steps 1–3, Claisen condensation: Acetyl-CoA transfers an acetate group to the SxtA complex (subunits: MTF, ACP, ACTP, and AONS); methyl donation from SAM generates propionyl-SxtA, which undergoes Claisen condensation with arginine. Step 4, amidino group transfer: SxtG transfers an amidino group from arginine to the SxtA-derived intermediate, forming 4,7-diguanidino-3-oxoheptane. Steps 5–7, cyclization and double bond formation: Step 5, SxtB mediates cyclization of the amidino intermediate, forming the first heterocycle; Step 6, SxtD introduces a C1–C5 double bond; Step 7, SxtS catalyzes epoxidation of the new double bond, and epoxide ring opening generates aldehydes, triggering bicyclization. Steps 8–9, reduction and hydroxylation: Step 8, SxtU reduces the terminal aldehyde to an alcohol; Step9, SxtH/SxtT catalyze C12 dihydroxylation, with O2 and succinate as co-substrates. Step 10, carbamoyl transfer: the SxtI/SxtJ/SxtK complex transfers a carbamoyl group to the dihydroxylated intermediate (dsSTX), yielding saxitoxin (STX).
Figure 2.
Schematic of STX Blockade Mechanism on NaVs: (A) Structural organization of NaV complex: The NaV complex is composed of a pore-forming α subunit and one or more β subunits . The α subunit, a single polypeptide chain with four homologous repeat domains (I–IV), constitutes the ion-conducting pore and voltage-sensing apparatus. Each domain contains six transmembrane helices (S1–S6)—S1–S4 form the voltage sensor, while S5–S6 form the pore domain ; (B) STX binding and functional disruption: Upon binding to the outer vestibule of the α-subunit pore via site 1, STX induces conformational contraction of the selective filter, restricting Na⁺ influx . This α-subunit-mediated disruption abrogates neuronal action potential generation (Vmax reduction ≥ 80%) and propagation, leading to clinical conditions: facial paresthesia, limb paralysis, and respiratory arrest, etc. [23,30]. (Some drawing elements are from www.figdraw.com).
Table 1.
Established detection system for saxitoxin (STX).
| Stage | Technical Method | LOD | Application Scenarios | Key Features | Ref |
--- --- --- |
| On-site rapid screening | Daphnia acute toxicity bioassay | 0.3–10.7 ng Eq.STX/L | Aquatic product preliminary screening | Low cost and simple operation | |
| MIST Alert™ rapid test | 40 μg Eq.STX/100 g | Aquatic product preliminary screening | Equipment-free and adapted for on-site rapid turnaround | |
| Laboratory confirmation and quantification | ELISA | 0.015 ng/mL | Positive sample screening and preliminary quantification | High stability and applicable for rapid triage of large-scale samples | |
| HPLC (post-column oxidation method) | 0.1 μg/kg | Routine sample quantification | High specificity | |
| LC-MS/MS | 0.1 μg/kg | Complex matrices and multitoxin simultaneous analysis | High specificity and capable of structural confirmation | |
| Research-grade analysis and mechanism verification | Neuro-2a cell-based assay | 2 ng/L | Toxicological mechanisms and low-dose effect studies | Direct correlation between STX ion channel blockade toxicity and biological effect simulation | |
| Receptor-binding assay | 2 ug Eq.STX/L | Toxin–receptor interaction mechanism | Real-time reflection of binding kinetics and high sensitivity | |
: The limit of detection (LOD) may be affected by various factors such as experimental conditions, instrument performance, and sample matrix, which may vary in actual detection.
| | |
--- |
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Deng, H.; Shang, X.; Zhu, H.; Huang, N.; Wang, L.; Sun, M.
Saxitoxin: A Comprehensive Review of Its History, Structure, Toxicology, Biosynthesis, Detection, and Preventive Implications. Mar. Drugs 2025, 23, 277.
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Saxitoxin: A Comprehensive Review of Its History, Structure, Toxicology, Biosynthesis, Detection, and Preventive Implications. Marine Drugs. 2025; 23(7):277.
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(2025). Saxitoxin: A Comprehensive Review of Its History, Structure, Toxicology, Biosynthesis, Detection, and Preventive Implications. Marine Drugs, 23(7), 277.
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11843 | https://pmc.ncbi.nlm.nih.gov/articles/PMC9331643/ | Primary Amenorrhea with Apparently Absent Uterus: A Report of Three Cases - PMC
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. 2022 Jul 25;11(15):4305. doi: 10.3390/jcm11154305
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Primary Amenorrhea with Apparently Absent Uterus: A Report of Three Cases
Eva Porsius
Eva Porsius
1 Department of Pediatric Endocrine Disease, Amalia Children’s Hospital, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; hedi.claahsen@radboudumc.nl
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1,, Marian Spath
Marian Spath
2 Department of Obstetrics and Gynaecology, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; marian.spath@radboudumc.nl (M.S.); kirsten.kluivers@radboudumc.nl (K.K.)
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2, Kirsten Kluivers
Kirsten Kluivers
2 Department of Obstetrics and Gynaecology, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; marian.spath@radboudumc.nl (M.S.); kirsten.kluivers@radboudumc.nl (K.K.)
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2, Willemijn Klein
Willemijn Klein
3 Department of Medical Imaging, Radiology, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; willemijn.klein@radboudumc.nl
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3, Hedi Claahsen-van der Grinten
Hedi Claahsen-van der Grinten
1 Department of Pediatric Endocrine Disease, Amalia Children’s Hospital, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; hedi.claahsen@radboudumc.nl
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1 Department of Pediatric Endocrine Disease, Amalia Children’s Hospital, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; hedi.claahsen@radboudumc.nl
2 Department of Obstetrics and Gynaecology, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; marian.spath@radboudumc.nl (M.S.); kirsten.kluivers@radboudumc.nl (K.K.)
3 Department of Medical Imaging, Radiology, Radboud University Medical Centre, P.O. Box 9101, 6500 HB Nijmegen, The Netherlands; willemijn.klein@radboudumc.nl
Correspondence: evaporsius@gmail.com
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K Katharina Rall: Academic Editor
Received 2022 Jun 20; Accepted 2022 Jul 20; Collection date 2022 Aug.
© 2022 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC9331643 PMID: 35893396
Abstract
Background: The apparent absence of a uterus upon imaging women with primary amenorrhea appears to lead to a high risk of misdiagnosis, which will lead to significant mental distress in patients. Case: Three young females with primary amenorrhea were referred with a diagnosis of Mayer–Rokitansky–Kuster–Hauser syndrome based on radiological findings of an apparently absent uterus. In two patients, the absence of the uterus could be confirmed, but with various diagnoses. The other patient had a normal but unstimulated uterus due to her hypoestrogenic state. Summary and Conclusion: The presented cases illustrate the broad differential diagnoses and the specific pitfalls of primary amenorrhea with an apparently absent uterus upon imaging. A well-established diagnosis was only possible through a thorough correlation of imaging findings with clinical history, biochemical findings and physical examination.
Keywords: female genital malformations, primary amenorrhea, absent uterus, Müllerian agenesis, DSD
1. Introduction
Primary amenorrhea is a common problem that leads to severe distress in adolescents. Normal puberty starts with breast development at the age of eight to thirteen years. Menarche usually occurs two years after the first signs of puberty, at age eleven to fifteen years. Primary amenorrhea is defined as the absence of menarche by age fifteen or more than three years after the start of breast development . The cause of primary amenorrhea can be due to anatomical abnormalities of the uterus or its outflow tract, and endocrine disorders of the gonads, pituitary gland or hypothalamus [2,3]. The presence or absence of a uterus can help to categorize primary amenorrhea, as illustrated in Figure 1.
Figure 1.
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Differential diagnoses of primary amenorrhea with apparently absent uterus on imaging in women with normal external genitalia. LH: luteinizing hormone; FSH: follicle-stimulating hormone; MRKHS: Mayer–Rokitansky–Küster–Hauser syndrome; DSD: differences in sex development; CAIS: complete androgen insensitivity syndrome.
Uterus absence in otherwise healthy females with normal external genitalia can be caused by different conditions within differences in sex development (DSD) conditions. Therefore, a DSD condition, such as Müllerian agenesis, testosterone synthesis defect or complete androgen insensitivity syndrome (CAIS), should be considered as the underlying cause of primary amenorrhea.
Pelvic ultrasound and magnetic resonance imaging (MRI) are good methods to visualize Müllerian structures [3,4]. When the uterus cannot be detected by ultrasound, MRI is advised for detailed imaging of Müllerian structures, ovaries, the vagina and kidneys . It has to be considered that, especially in prepubertal state, a present uterus may remain undetected, since in the hypoestrogenic state the uterus is small and not easy to visualize. The interpretation of imaging findings by an experienced radiologist with knowledge of the normal appearance of the uterus and ovaries during different phases of growth and development is therefore preferred .
In this paper, we present three cases of young women who were referred to our hospital presenting with primary amenorrhea and an apparently absent uterus. Our aim is to point out the importance of correlating imaging findings with clinical history, biochemical findings and physical examination, and provide guidance for distinguishing the various diagnoses of primary amenorrhea with an absent uterus upon imaging. This is important because these cases are rare, but various medical practitioners could encounter them and misdiagnosis will lead to significant mental distress in patients.
2. Case 1
A seventeen-year-old girl was referred to the local gynecologist because of primary amenorrhea. Puberty had started with breast development and pubarche at the age of ten. She had used the oral contraceptive pill, but no withdrawal bleeding ever occurred. According to the patient, penetration during sexual intercourse was possible without pain. However, from a physical examination and vaginal ultrasound, a short and blind-ending vagina was found. The MRI suggested an absent uterus, normally located ovaries and normal kidneys (Figure 2). With the suspicion of Mayer–Rokitansky–Küster–Hauser syndrome (MRKHS), the patient was referred to our expert center. An MRI demonstrated no vaginal, myometrial or cervical tissue, with some fibrous and lipomatous tissue at this location. The ovaries had normal cystic appearance and were situated in the broad ligament. Karyotype was 46,XX. A hormonal examination showed normal endocrine parameters (Table 1). With these results, a diagnosis of MRKHS was confirmed and the patient received information about the etiology and consequences of this diagnosis.
Figure 2.
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Case 1. (a) Coronal T2-weighted image of the abdomen, showing two normal kidneys (white arrows) as well as two normal ovaries (blue arrows). (b) Mid-sagittal T2-weighted image of the pelvis, demonstrating a triangular shape between bladder and rectum, which is the fibro-adipose remnant tissue of the aplastic uterus, without any endometrial or myometrial organization (white arrow).
Table 1.
Overview of important findings in three cases with primary amenorrhea, based on Mayer–Rokitansky–Küster–Hauser syndrome (MRKHS), complete androgen insensitivity syndrome (CAIS) and hypogonadotropic hypogonadism of unknown origin.
| | Case 1 | Case 2 | Case 3 |
:---: :---: |
| Diagnosis | MRKHS | CAIS | Hypogonadotropic hypogonadism e.c.i. |
| History | Normal start of breast development and pubarche | Normal start of breast development, no pubarche | Delayed and incomplete breast development, normal pubarche |
| Physical examination | Tanner M4 P4 Female external genitalia with no virilization Short blind-ending vagina | Tanner M5 P1 Female external genitalia with no virilization. Tall stature. No axillary hair | Tanner M1-2 P2-3 Female external genitalia with no virilization BMI 15.8 kg/m 2 (−2.2 SD) Height 1.56 m (−1.78 SD) |
| Biochemical analysis | reference values | | | |
| LH | 2.4–12.6 E/L (follicular), 14.0–95.6 (ovular), 1.0–11.4 (luteal) | 38.8 | 24 | 0.5 |
| FSH | 3.5–12.5 E/L (follicular); 4.7–21.5 (ovular); 1.7–7.7 (luteal) | 6.6 | 2.8 | 1.2 |
| Estradiol | 45–854 pmol/L (follicular); 151–1461 (ovular); 82–1251 (luteal) | 470 | 78 | <18 |
| AMH | 0.52–12.01 ug/L | 2.5 | 1292 | 10.1 |
| Testosterone | 0.52–2.0 nmol/L | | 13 | |
| Progesterone | <1.3 (follicular); 1.3–12 (mid cycle); 19–120 (luteal) nmol/L | 1.7 | <0.25 | |
| Prolactin | 100–760 mE/L | | | 250 |
| TSH | 0.27–4.20 mE/L | | | 2.5 |
| Karyotype | 46,XX | 46,XY | 46,XX |
| Imaging findings | ultrasound | No visualization of uterus or ovaries | No visualization of uterus or ovaries | No visualization of uterus or ovaries |
| | MRI report on primary investigation | Absent uterus, impression of hypoplastic vagina. Normal multicystic ovaries | Absent uterus and presence of a short vagina. Normal ovaries | Normal multicystic ovaries and a small uterus of 2.3 × 1.3 cm, described as rudimentary uterus |
| | MRI after referral | Lipofibromatous tissue at location of the uterus, no endo- myometrial or cervical structures. Normal multicystic ovaries | Underdeveloped Müllerian structures. Gonads with (ovo)testicular aspect | Small uterus with normal endo- and myometrial tissue, normal multicystic ovaries |
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BMI: body mass index; SD: standard deviation; MRI: magnetic resonance imaging; LH: luteinizing hormone; FSH: follicle-stimulating hormone; AMH: anti-Müllerian hormone; TSH: thyroid-stimulating hormone.
3. Case 2
A seventeen-year-old girl presented with primary amenorrhea after a normal start to puberty. Breast development started at the age of twelve years and she had a growth spurt two years later. The physical examination showed the normal anatomy of the vulva and shaved pubic hair, and the vaginal depth upon digital palpation was 3–4 cm (reference 7–15 cm ). The uterus and ovaries could not be visualized by pelvic ultrasound (Figure 3a). MRI performed in the local hospital confirmed the absence of a normal uterus and presence of a short vagina. In this MRI report, normal ovaries were described. These results lead to the assumption of an MRKHS diagnosis. This was explained to the patient and vaginal dilation therapy was suggested.
Figure 3.
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Case 2. (a) Ultrasound of the pelvis, demonstrating the rectum (blue arrow) direct dorsal to the bladder (white arrow), without uterus in between. (b) Coronal T2-weighted MR image of the pelvis, with the white arrows indicating the gonads with partially solid tissue and partially cystic, a suspected aspect of ovotestes.
She was referred to our expert center with suspicion of MRKHS. In the physical exam, we noted her tall stature (height +1.17 SDS) and very sparse axillary and pubic hair. After the specialized radiologist reviewed the MRI, the previously described ovaries actually appeared as atypical gonadal structures with a large solid part, suspected of being (ovo)testicular tissue (Figure 3b). A hormonal examination showed high levels of anti-Müllerian hormone (AMH) and testosterone (Table 1). Genetic diagnostics confirmed the suspicion of a 46,XY karyotype and an androgen receptor gene mutation (hemizygous variant NM_000044.2(AR):c.2594A>T, p.(Asp865Val)), confirming the diagnosis of CAIS. Follow up by our DSD team is continued to guide her during vaginal dilation therapy, offer psychological support and screening for a possible malignant transformation of the abdominal testes.
4. Case 3
A sixteen-year-old girl presented with delayed puberty and primary amenorrhea. Breast development started at age fourteen, but no further progression occurred after tanner stages 1–2. Axillary and pubic hair also developed up to tanner stage 2. There were no signs of virilization. She had a short stature and low weight (Table 1). Bone age was delayed 3 years according to Gruelich and Pyle using the automatic BoneXpert method . There were no signs of eating disorders, malabsorption or other chronic illness. The mother had normal puberty with menarche at twelve years old. The father experienced late but spontaneous puberty. There was no history of medication use, except sporadic corticosteroid ointment for eczema. Hormonal serum analysis showed low FSH, LH and estradiol levels, normal thyroid function and prolactin levels (Table 1). The karyotype was 46,XX. Abdominal ultrasound performed at the local hospital showed no uterus or ovaries. Subsequent MRI showed normal ovaries and a small uterus (2.3 × 1.3 cm), described as a rudimentary uterus suspected for MRKHS.
She was referred to our expert center with a diagnosis of MRKHS. Ultrasound and MRI revision by our radiologist showed a small uterus with visible myo- and endometrial tissue, which was interpreted as an unstimulated hypoestrogenic uterus (Figure 4). A further diagnostic evaluation of the etiology of hypogonadotropic hypogonadism was performed. Cerebral MRI showed no pathology in the hypothalamus or pituitary gland. Exome sequencing known genes connected to hypogonadotropic hypogonadism showed no mutations. Pubertal induction with estrogen therapy was started, which resulted in breast development, endometrial thickening and menarche.
Figure 4.
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Case 3. (a) Ultrasound of the pelvis. The white arrow indicates a small uterus behind the bladder. (b). Mid-sagittal T2-weighted MR image, demonstrating a small uterus (white arrow). (c) The oblique-coronal T2-weighted images demonstrate clearly that there is an architecture of endometrial and myometrial tissue (white arrow).
5. Discussion
This study presents three patients with primary amenorrhea, who were referred to our expert clinic with a (mis)diagnosis of MRKHS because of radiological findings of an apparently absent uterus. In two patients, the absence of the uterus could be confirmed, but with various diagnoses (MRKHS resp. CAIS). The other patient had a normal but unstimulated uterus. Our case report shows that the apparent absence of a uterus upon imaging in women with primary amenorrhea appears to lead to a high risk of misdiagnosis. The presented cases illustrate the broad range of diagnoses of primary amenorrhea and specific pitfalls. A well-established diagnosis was only possible through considering the combination of clinical history, physical examination, biochemical evaluation and assessment of imaging by an experienced radiologist.
During the prenatal development of 46,XX individuals, the Müllerian ducts fuse in absence of AMH to form the uterovaginal canal. In syndromes with Müllerian agenesis, such as MRKHS, this process fails, resulting in the absence of the uterus and upper part of the vagina . In 46,XY individuals, AMH stimulates the regression of Müllerian ducts, resulting in absence of the uterus. Testosterone and dihydrotestosterone are responsible for the virilization of external genitalia. The effect of these androgens is compromised in androgen receptor defects and testosterone synthesis or metabolizing defects. A 46,XY female with CAIS but normal functioning AMH will thereby have external female features in the absence of a uterus and will have testes instead of ovaries .
The first step in the evaluation of primary amenorrhea is carefully looking through clinical history and physical examination. Women with MRKHS experience a normal start of puberty with age-appropriate breast development and pubarche, but primary amenorrhea as an isolated symptom . In CAIS, there is a normal onset of breast development, but no or very sparse axillary and pubic hair. A Tanner stage examination is important since patients will often unknowingly misinterpret pubarche. Height is taller than general. Acne and sweat odor are absent. Breasts and female adiposity can develop through the action of estradiol deriving from the peripheral aromatization of testosterone . Delayed pubertal development in 46,XX individuals with an absence of breast development and menarche is caused by estrogen absence, due to ovarial or pituitary/hypothalamic causes.
Thus, assessing patient history and physical findings carefully an accurate diagnosis can be made. However, as the impact of these diagnoses is great, biochemical analysis, imaging or karyotype analysis is conducted to support this. A careful consideration of necessary diagnostics will prevent misinterpretation and misdiagnosis.
LH, FSH, progesterone, estradiol, TSH and prolactin can guide in finding possible endocrine causes of primary amenorrhea . In response to our cases, we suggest adding AMH and testosterone to give direction in determining possible 46,XY DSD conditions. Karyotype analysis should always be performed when 46,XY DSD conditions are considered.
In women with normal secondary sexual characteristics but primary amenorrhea, pelvic ultrasound is advised to search for possible anatomical causes [2,3]. In case of an apparently absent uterus on ultrasound, we advise performing MRI and consulting a specialized radiologist. In CAIS patients, no Müllerian structures will be found and gonadal tissue might be testicular or maldifferentiated. In women with an absent uterus because of MRKHS, absence might not always be a full absence. Wang et al. described bilateral Müllerian rudiments seen on MRI in 95% of MRKHS women and normally located ovaries in 68.7%. The Müllerian rudiments are generally small with only one-layer differentiation. There is also a high incidence of renal (~30%) and skeletal (~10–40%) anomalies .
In women with delayed breast development, uterus imaging is not advised, since mostly estrogen deficiency will be the cause of primary amenorrhea. An important consideration is that, in the hypoestrogenic state, the uterus is small and unstimulated and can therefore be difficult to visualize. Berglund et al. described the association of an apparently absent uterus with primary ovarian insufficiency in their review of the literature. Estrogen deficiency has previously resulted in misdiagnosis as MRKHS in 22/25 patients (88%) with primary ovarian insufficiency. Michala et al. demonstrates this as well in their description of ten patients with misdiagnoses of uterine agenesis based on imaging and laparoscopic findings. They suggest that imaging should be undertaken by clinicians with experience in management of this age group and in some girls it may be necessary to delay final diagnosis until after puberty or the administration of estrogen therapy.
In case of absent uterus upon imaging, a progestin challenge test will probably not lead to vaginal bleeding. In modern practice, there seems to be no additional value for this test, but a high risk of further delay of correct diagnosis.
The diagnosis of an absent uterus and vagina leads to a negative impact on the affected women’s level of psychological distress and self-esteem . Additionally, these women are confronted with sexuality and fertility issues at a young age. In the case of the misdiagnosis of MRKHS instead of CAIS, the adolescent will be confronted with confusing information: first the absence of uterus and vagina and subsequent information on afunctional gonads. This has even further fertility consequences and, therefore, it is important to make an early correct diagnosis. Confirmation by a DSD team is advised . This specialized team consists of a gynecologist, endocrinologist, radiologist, psychologist, geneticist, urologist and sexologist.
6. Conclusions
The presented cases illustrate the broad differential diagnosis and the specific pitfalls of primary amenorrhea with an apparently absent uterus upon imaging. It is important not to draw premature conclusions from findings of an apparently absent uterus from imaging alone. A well-established diagnosis was only possible with a thorough correlation of imaging findings with clinical history, biochemical findings and physical examination.
Acknowledgments
We would like to thank our patients for their permission to use their history and clinical data.
Author Contributions
E.P. and H.C.-v.d.G. participated in the design and planning of the article. E.P. performed data collection and drafted the manuscript. K.K., M.S., W.K. and H.C.-v.d.G. revised the manuscript critically. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
Ethical review and approval were waived for this study due to the rules of our hospitals Research Ethics Committee. Retrospective research with patient files, where the patients are not physically involved in the research and the data were collected primarily for patients care and not primarily for research purposes, is not considered to be subject to the Medical Research Involving Human Subjects Act.
Informed Consent Statement
Written permission to report these cases was obtained from all patients. No ethics committee or institutional review board approval was required, according to the local Research Ethics Committee.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research received no external funding.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
References
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1. Introduction
2. Case 1
3. Case 2
4. Case 3
5. Discussion
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How To Calculate The Power Factor Of Load Bank？
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How To Calculate The Power Factor Of Load Bank？
How To Calculate The Power Factor Of Load Bank？
2022-06-06
Friends who are often exposed to dummy load testing know that AC circuits contain terms such as resistance, capacitance or resistance, inductance (active power and reactive power).So in order for us to calculate the total power consumed, we need to know the phase difference between the sinusoidal waveforms of the voltage and current.
In AC circuits, the voltage and current waveforms are sine waves, so their amplitudes change over time.Since we know that power is voltage times current(P = VI), maximum power occurs when the two voltage and current waveforms are aligned with each other.That is, their peaks and zero-crossings occur at the same time.When this happens, the two waveforms are said to be "in phase".
By defining the total impedance of the circuit, the three main elements in an AC circuit that can affect the relationship between the voltage and current waveforms and their phase difference are resistors, capacitors, and inductors.
The impedance(Z) of an AC circuit is equivalent to the resistance calculated in a DC circuit, and the impedance is measured in ohms.For AC circuits, impedance is usually defined as the ratio of the voltage phasor to the current phasor produced by the circuit elements.A phasor is a straight line drawn in such a way that the magnitude of the voltage or current is represented by its length, and its phase difference relative to other phasor lines is represented by its angular position relative to the other phasor lines.
AC circuits contain resistance and reactance that combine to provide a total impedance(Z) that limits the flow of current around the circuit.But the impedance of an AC circuit is not equal to the algebraic sum of the ohmic values of resistance and reactance, because pure resistance and pure reactance are 90o out of phase with each other.But we can use this 90o phase difference as the sides of a right triangle, called an impedance triangle, where impedance is the hypotenuse determined by the Pythagorean theorem.
This geometric relationship between resistance, reactance, and impedance can be visually represented by using the impedance triangle as shown.
Note that impedance is the vector sum of resistance and reactance, and it has not only a magnitude(Z), but also a phase angle(Φ), which represents the phase difference between resistance and reactance.Also note that as the frequency changes, the triangle changes shape due to the change in reactance(X) .Of course, the resistance(R)will always remain the same.
We can take this idea a step further by transforming the impedance triangle into a power triangle that represents the three elements of power in an AC circuit.Ohm's Law tells us that in a DC circuit, the power(P) in watts is equal to the square of the current(I 2)times the resistance(R).So we can multiply the three sides of the impedance triangle above by I 2 to get the corresponding power triangle as:
Active Power P = I 2 R Watts,(W)
Reactive power Q = I 2 X reactive volt-ampere,(VAr)
Apparent power S = I 2 Z volt-ampere,(VA)
real power in an AC circuit
Active Power(P), also known as Active Power or Active Power, performs "real work" in a circuit.Real power (in watts) defines the power dissipated by the resistive portion of the circuit.Then the actual power(P)in the AC circuit is the same as the power P in the DC circuit.So just like a DC circuit, it is always calculated as I 2R, where R is the total resistive component of the circuit.
Since the resistance does not create any phasor difference (phase shift) between the voltage and current waveforms, all useful power is transferred directly to the resistance and converted to heat, light and work.Then the power dissipated by the resistor is the real power, basically the average power of the circuit.
To find the corresponding active power value, the voltage and current rms values are multiplied by the cosine of the phase angle.
Active Power P = I 2 R = V I cos (Φ)Watts,(W)
But since they have no phase difference between the voltage and current in a resistive circuit, the phase shift between the two waveforms will be zero(0).Then:
Actual power(P)is in watts, voltage(V)is in rms volts, and current(I)is in rms amps.
The actual power is then the I2R resistive element measured in watts, which is what you read on your utility meter in watts(W), kilowatts(kW),and megawatts(MW).Note that the real power P is always positive.
Reactive power in AC circuits
Reactive power(Q), (sometimes called reactive power) is the power dissipated in an AC circuit that does no useful work but has a large effect on the phase shift between the voltage and current waveforms.Reactive power is related to the reactance created by inductors and capacitors, which can counteract the effects of active power.There is no reactive power in a DC circuit.
Unlike active power(P) , which does all the work, reactive power(Q)takes power away from the circuit due to the creation and reduction of induced magnetic fields and capacitive electrostatic fields, making it harder to supply active power directly to a circuit or load .
The power stored by an inductor in its magnetic field attempts to control the current flow, while the power stored by the capacitor's electrostatic field attempts to control the voltage.The result is that the capacitor"produces" reactive power and the inductor "consumes" reactive power.This means that they both consume power and return power to the source, so they don't consume any real power.
To find the reactive power, the voltage and current rms values are multiplied by the sine of the phase angle.
Reactive power Q = I 2 X = VIsin(Φ)reactive volt-ampere,(VAr's)
Since there is a 90o phase difference between the voltage and current waveforms in pure reactance (inductive or capacitive),multiplying VI by sin(Φ)yields a vertical component that is 90 out of phase with each reactance oOther, so:
where reactive power(Q)is in reactive volt-amperes, voltage(V)is in rms volts, and current(I)is in rms amperes.
Then reactive power represents the product of volts and amperes,90 degrees out of phase with each other, but in general, there can be any phase angle Φ between voltage and current.
Therefore, reactive power is an I 2 X reactive element, and its units are volt-ampere reactive(VAr), kilovolt-ampere reactive(kVAr),and megavolt-ampere reactive(MVAr).
Apparent Power in AC Circuits
We have seen above that the active power is dissipated by the resistance and the reactive power is supplied to the reactance.Therefore, the current and voltage waveforms are not in phase due to the difference between the circuit resistive and reactive components.
Then there is a mathematical relationship between active power (P) and reactive power (Q ), called complex power.The product of the rms voltage V applied to an AC circuit and the rms current I flowing into that circuit is called the "volt-ampere product" (VA), symbol S, and its magnitude is often referred to as apparent power.
This complex power is not equal to the algebraic sum of the active and reactive powers added together, but rather the vector sum of P and Q given in volt- amperes(VA) .It is a complex power represented by a power triangle.The rms value of the volt-ampere product is often referred to as apparent power, because "obviously" this is the total power dissipated by the circuit, even though the actual power doing work is much less.
Since apparent power consists of two components, resistive power is in-phase power or active power in watts, and reactive power is out-of-phase power in volt-amperes, we can show the vector sum of these two power components in terms of power triangles form appears.A power triangle has four parts:P,Q,S and θ.
The three elements that make up a power source in an AC circuit can be represented graphically by the three sides of a right triangle, roughly the same as the impedance triangle above.As shown, the horizontal (adjacent) side of the power triangle represents the circuit active power (P), the vertical (opposite) side represents the circuit reactive power (Q), and the hypotenuse represents the apparent power produced (S).
P is the I 2 R or real power to perform the work, in watts,W
Q is I 2X or reactive power in volt-amperes reactive,VAr
S is I2Z or apparent power in VA,VA
Φ is the phase angle in degrees.The larger the phase angle, the greater the reactive power
Cos(Φ) = P/S = W/VA =power factor,pf
Sin(Φ) = Q/S = VAr/VA
Tan(Φ) = Q/P = VAr/W
Power factor is calculated as the ratio of real power to apparent power, since this ratio is equal to cos(Φ).
The power factor cos(Φ)is an important part of the AC circuit, and it can also be expressed by circuit impedance or circuit power.Power factor is defined as the ratio of real power(P)to apparent power(S), usually expressed as a decimal value such as 0.95, or as a percentage:95%.
The power factor defines the phase angle between the current and voltage waveforms, where I and V are the magnitudes of the rms values of the current and voltage .Note that it doesn't matter whether the phase angle is the difference between current and voltage or whether the phase angle is the difference between voltage and current.The mathematical relationship is as follows:
We said earlier that in a purely resistive circuit, the current and voltage waveforms are in phase with each other, so when the phase difference is zero(0 o), the actual power dissipated is the same as the apparent power.So the power factor is:
Power factor,pf = cos 0 o= 1.0
That is, the watts consumed are the same as the volt-amps consumed, resulting in a power factor of 1.0 or 100% .In this case it is called unity power factor.
We also said above that in a pure reactive circuit, the current and voltage waveforms are 90o out of phase with each other.Since the phase difference is ninety degrees(90 o), the power factor will be:
Power factor,pf = cos 90 o= 0
That is, the wattage consumed is zero, but there is still voltage and current supplying the reactive load.Obviously, reducing the reactive VAr component of the power triangle will result in a decrease in θ, thereby increasing the power factor to 1, ie unity.It is also desirable to have a high power factor, as this makes the most efficient use of the circuit carrying current to the load.
We can then write the relationship between active power, apparent power, and circuit power factor as:
An inductive circuit whose current"lags" the voltage(ELI)is said to have a lagging power factor, while a capacitive circuit whose current "leads" the voltage(ICE)is said to have a leading power factor.
A wire wound coil with an inductance of 180mH and a resistance of 35Ω was connected to a 100V 50Hz power supply.Calculate:a)the impedance of the coil,b) the current,c) the power factor, and d)the apparent power dissipated.
Also draw the resulting power triangle for the coil above.
Data given:R = 35 Ω,L = 180mH,V = 100V and ƒ= 50Hz.
At a power factor of 0.5263 or 52.63%, the coil requires 150 VA of power to produce 79 watts of useful work.In other words, at 52.63%power factor, the coil needs 89%more current to do the same job, which is a lot of wasted current.
Adding a power factor correction capacitor (32.3uF in this case) across the coil to increase the power factor above 0.95 or 95%will greatly reduce the reactive power consumed by the coil as these capacitors act as reactive current generation machine, thereby reducing the total amount of current consumed.
Power Triangle and Power Factor Summary
We have seen here that the three elements of electrical power in an AC circuit, namely active power,reactive power and apparent power , can be represented by the three sides of a triangle called the power triangle .Since these three elements are represented by a"right triangle", their relationship can be defined as:S 2= P 2+ Q 2, where:P is the active power in watts (W) and Q is the active power in watts(W)Reactive power in volt-ampere reactive(VAr),S is the apparent power in volt-ampere(VA) .
We also saw that in an AC circuit, the quantity cos(Φ) is called the power factor.The power factor of an AC circuit is defined as the ratio of the active power(W) consumed by the circuit to the apparent power(VA)consumed by the same circuit.So this gives us: Power Factor=Real Power/Apparent Power, or pf = W/VA.
Then the cosine of the resulting angle between the current and the voltage is the power factor.Usually power factor is expressed as a percentage, such as 95%, but it can also be expressed as a decimal value, such as 0.95.
When the power factor is equal to 1.0(units) or 100%, i.e. when the actual power dissipated is equal to the apparent power of the circuit, the phase angle between the current and the voltage is 0 o, because:cos-1(1.0) = 0 o.When the power factor is zero(0), the phase angle between the current and voltage will be 90 degrees because:cos-1 ( 0) = 90 degrees.In this case, the actual power dissipated by the AC circuit is zero, regardless of the circuit current.
In a real AC circuit, the power factor can be between 0 and 1.0, depending on the passive components in the connected load.For resistive loads or circuits (the most common case), the power factor will"lag".In a capacitive-resistive circuit, the power factor will "lead".AC circuits can then be defined as having unity, lagging or leading power factor.
A poor power factor with a value close to zero(0) will dissipate wasted power and thus reduce the efficiency of the circuit, while a circuit or load with a power factor close to one(1.0)or unity(100%)will be more efficient.This is because a circuit or load with a low power factor requires more current than the same circuit or load with a power factor close to 1.0(units).
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11845 | https://www.tsfx.edu.au/resources/N4%20-%20Pages%20from%20Sullivan_AlgTrig_Ch6-3-EXP%20MODELS%2050650.pdf?srsltid=AfmBOop9mPsseGs2YqIo851IUdXOZAK8PXESLV1g9hXBJCRVepJk7_eM | 478 CHAPTER 6 Exponential and Logarithmic Functions
Find Equations of Populations That Obey the Law of Uninhibited Growth
Many natural phenomena have been found to follow the law that an amount A
varies with time t according to the function
1
6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models OBJECTIVES 1 Find Equations of Populations That Obey the Law of Uninhibited Growth (p. 478)
2 Find Equations of Populations That Obey the Law of Decay (p. 480)
3 Use Newton’s Law of Cooling (p. 481)
4 Use Logistic Models (p. 483)
A 1t2 = A0 e kt (1)
Here A0 is the original amount 1t = 02 and k 0 is a constant. If k 7 0, then equation (1) states that the amount A is increasing over time; if k 6 0, the amount A is decreasing over time. In either case, when an amount A
varies over time according to equation (1), it is said to follow the exponential law , or the law of uninhibited growth 1k 7 02 or decay 1k 6 02. See Figure 41. For example, in Section 6.7, continuously compounded interest was shown to follow the law of uninhibited growth. In this section we shall look at some additional phenomena that follow the exponential law. Cell division is the growth process of many living organisms, such as amoebas, plants, and human skin cells. Based on an ideal situation in which no cells die and no by-products are produced, the number of cells present at a given time follows the law of uninhibited growth. Actually, however, after enough time has passed, growth at an exponential rate will cease as a consequence of factors such as lack of living space and dwindling food supply. The law of uninhibited growth accurately models only the early stages of the cell division process. The cell division process begins with a culture containing N0 cells. Each cell in the culture grows for a certain period of time and then divides into two identical cells. Assume that the time needed for each cell to divide in two is constant and does not change as the number of cells increases. These new cells then grow, and eventually each divides in two, and so on.
Uninhibited Growth of Cells
A model that gives the number N of cells in a culture after a time t has passed (in the early stages of growth) is
N 1t2 = N0 e kt k 7 0 (2)
where N0 is the initial number of cells and k is a positive constant that represents the growth rate of the cells. Using formula (2) to model the growth of cells employs a function that yields positive real numbers, even though the number of cells being counted must be an integer. This is a common practice in many applications.
A
(a) A(t) A0ek t ,k 0
tA0
tAA0
(b) A(t) A0ekt ,k 0
Figure 41
Exponential growth Exponential decay
k 0
ttAA0
(b) A(t) A0ekt ,k 0SECTION 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models
479
Bacterial Growth
A colony of bacteria that grows according to the law of uninhibited growth is modeled by the function N 1t2 = 100 e0.045 t, where N is measured in grams and t is measured in days. (a) Determine the initial amount of bacteria. (b) What is the growth rate of the bacteria? (c) What is the population after 5 days? (d) How long will it take for the population to reach 140 grams? (e) What is the doubling time for the population? (a) The initial amount of bacteria, N0 , is obtained when t = 0, so
N0 = N 102 = 100 e0.045 102 = 100 grams (b) Compare N 1t2 = 100 e0.045 t to N 1t2 = N0 e kt . The value of k, 0.045, indicates a growth rate of 4.5%. (c) The population after 5 days is N 152 = 100 e0.045 152 ≈ 125.2 grams. (d) To find how long it takes for the population to reach 140 grams, solve the equation N 1t2 = 140. 100 e0.045 t = 140
e0.045 t = 1.4 0.045 t = ln 1.4
t = ln 1.4
0.045
≈ 7.5 days The population reaches 140 grams in about 7.5 days. (e) The population doubles when N 1t2 = 200 grams, so the doubling time is found by solving the equation 200 = 100 e0.045 t for t.200 = 100 e0.045 t
2 = e0.045 t
ln 2 = 0.045 t
t = ln 2
0.045
≈ 15.4 days The population doubles approximately every 15.4 days.
Now Work P R O B L E M 1
Bacterial Growth
A colony of bacteria increases according to the law of uninhibited growth. (a) If N is the number of cells and t is the time in hours, express N as a function of t.(b) If the number of bacteria doubles in 3 hours, find the function that gives the number of cells in the culture. (c) How long will it take for the size of the colony to triple? (d) How long will it take for the population to double a second time (that is, to increase four times)? (a) Using formula (2), the number N of cells at time t is
N 1t2 = N0 e kt
where N0 is the initial number of bacteria present and k is a positive number.
EXAMPL E 1
Solution
Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045. Divide both sides of the equation by 100. Rewrite as a logarithm. Divide both sides of the equation by 0.045.
•
EXAMPL E 2
Solution 480 CHAPTER 6 Exponential and Logarithmic Functions
(b) To find the growth rate k, note that the number of cells doubles in 3 hours, so
N 132 = 2N0But N 132 = N0 e k132, so
N0 e k132 = 2N0
e3k = 2
3k = ln 2
k = 1
3 ln 2 ≈ 0.23105 The function that models this growth process is therefore
N(t) = N0 e0.23105 t
(c) The time t needed for the size of the colony to triple requires that N = 3N0 .Substitute 3 N0 for N to get 3N0 = N0 e0.23105 t
3 = e0.23105 t
0.23105 t = ln 3
t = ln 3
0.23105 ≈ 4.755 hours It will take about 4.755 hours, or 4 hours and 45 minutes, for the size of the colony to triple. (d) If a population doubles in 3 hours, it will double a second time in 3 more hours, for a total time of 6 hours.
Find Equations of Populations That Obey the Law of Decay
Radioactive materials follow the law of uninhibited decay.
Divide both sides by N0.Write the exponential equation as a logarithm.
•
2Uninhibited Radioactive Decay
The amount A of a radioactive material present at time t is given by
A 1t2 = A0 e kt k 6 0 (3)
where A0 is the original amount of radioactive material and k is a negative number that represents the rate of decay. All radioactive substances have a specific half-life, which is the time required for half of the radioactive substance to decay. Carbon dating uses the fact that all living organisms contain two kinds of carbon, carbon-12 (a stable carbon) and carbon-14 (a radioactive carbon with a half-life of 5730 years). While an organism is living, the ratio of carbon-12 to carbon-14 is constant. But when an organism dies, the original amount of carbon-12 present remains unchanged, whereas the amount of carbon-14 begins to decrease. This change in the amount of carbon-14 present relative to the amount of carbon-12 present makes it possible to calculate when the organism died.
Estimating the Age of Ancient Tools
Traces of burned wood along with ancient stone tools in an archeological dig in Chile were found to contain approximately 1.67% of the original amount of carbon-14. If the half-life of carbon-14 is 5730 years, approximately when was the tree cut and burned?
EXAMPL E 3SECTION 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models 481
Using formula (3), the amount A of carbon-14 present at time t is
A 1t2 = A0 e kt
where A0 is the original amount of carbon-14 present and k is a negative number. We first seek the number k. To find it, we use the fact that after 5730 years, half of the original amount of carbon-14 remains, so A 15730 2 = 1
2 A0 . Then 12 A0 = A0 e k15730 2
12 = e5730 k
5730 k = ln 12
k = 1
5730 ln 12 ≈ - 0.000120968 Formula (3) therefore becomes
A(t) = A0 e - 0.000120968 t
If the amount A of carbon-14 now present is 1.67% of the original amount, it follows that 0.0167 A0 = A0 e - 0.000120968 t
0.0167 = e - 0.000120968 t
0.000120968 t = ln 0.0167
t = ln 0.0167
0.000120968 ≈ 33,830 years The tree was cut and burned about 33,830 years ago. Some archeologists use this conclusion to argue that humans lived in the Americas nearly 34,000 years ago, much earlier than is generally accepted. •
Now Work P R O B L E M 3
Use Newton’s Law of Cooling
Newton’s Law of Cooling states that the temperature of a heated object decreases exponentially over time toward the temperature of the surrounding medium.
Solution
Divide both sides of the equation by A 0.Rewrite as a logarithm. Divide both sides of the equation by A 0.Rewrite as a logarithm.
3
Named after Sir Isaac Newton (1643–1727), one of the cofounders of calculus.
Newton’s Law of Cooling
The temperature u of a heated object at a given time t can be modeled by the following function:
u1t2 = T + 1u0 - T2e kt k 6 0 (4)
where T is the constant temperature of the surrounding medium, u0 is the initial temperature of the heated object, and k is a negative constant.
Using Newton’s Law of Cooling
An object is heated to 100°C (degrees Celsius) and is then allowed to cool in a room whose air temperature is 30°C. (a) If the temperature of the object is 80°C after 5 minutes, when will its temperature be 50°C? (b) Determine the elapsed time before the temperature of the object is 35°C. (c) What do you notice about the temperature as time passes?
EXAMPL E 4482 CHAPTER 6 Exponential and Logarithmic Functions
(a) Using formula (4) with T = 30 and u0 = 100, the temperature u(t) (in degrees Celsius) of the object at time t (in minutes) is
u1t2 = 30 + 1100 - 30 2e kt = 30 + 70 e kt (5)
where k is a negative constant. To find k, use the fact that u = 80 when t = 5. Then
u1t2 = 30 + 70 e kt
80 = 30 + 70 e k152
50 = 70 e5k
e5k = 50
70
5k = ln 57
k = 1
5 ln 57 ≈ - 0.0673 Formula (5) therefore becomes
u(t) = 30 + 70 e - 0.0673 t (6)
To find t when u = 50 °C, solve the equation 50 = 30 + 70 e - 0.0673 t
20 = 70 e - 0.0673 t
e - 0.0673 t = 20
70
0.0673 t = ln 27
t =
ln 27
0.0673 ≈ 18.6 minutes The temperature of the object will be 50°C after about 18.6 minutes, or 18 minutes, 36 seconds. (b) Use equation (6) to find t when u = 35 °C. 35 = 30 + 70 e - 0.0673 t
5 = 70 e - 0.0673 t
e - 0.0673 t = 5
70
0.0673 t = ln 570
t =
ln 570
0.0673 ≈ 39.2 minutes The object will reach a temperature of 35°C after about 39.2 minutes. (c) Look at equation (6). As t increases, the exponent - 0.0673 t becomes unbounded in the negative direction. As a result, the value of e - 0.0673 t approaches zero, so the value of u, the temperature of the object, approaches 30°C, the air temperature of the room. •
Now Work P R O B L E M 1 3
Solution
u(5) = 80
Simplify. Solve for e5k.Rewrite as a logarithm. Solve for k.Simplify. Rewrite as a logarithm. Solve for t.Simplify. Rewrite as a logarithm. Solve for t.SECTION 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models 483 Use Logistic Models
The exponential growth model A 1t2 = A0 e kt , k 7 0, assumes uninhibited growth, meaning that the value of the function grows without limit. Recall that cell division could be modeled using this function, assuming that no cells die and no by-products are produced. However, cell division eventually is limited by factors such as living space and food supply. The logistic model , given next, can describe situations where the growth or decay of the dependent variable is limited.
4Logistic Model
In a logistic model, the population P after time t is given by the function
P1t2 = c
1 + ae -bt (7)
where a, b, and c are constants with a 7 0 and c 7 0. The model is a growth model if b 7 0; the model is a decay model if b 6 0. The number c is called the carrying capacity (for growth models) because the value P1t2 approaches c as t approaches infinity; that is, lim
tSq
P1t2 = c. The number
0 b 0 is the growth rate for b 7 0 and the decay rate for b 6 0. Figure 42(a) shows the graph of a typical logistic growth function, and Figure 42(b) shows the graph of a typical logistic decay function.
Figure 42
(0, P(0))
tP(t)y c
c1–2Inflection point
(a) P(t)5,b.0c
11ae 2bt
Logistic growth
(0, P(0))
c1–2
tP(t)y c
Inflection point Logistic decay
(b) P(t)5,b,0c
11ae 2bt
Properties of the Logistic Model, Equation (7)
The domain is the set of all real numbers. The range is the interval 10, c2,where c is the carrying capacity.
There are no x-intercepts; the y-intercept is P102.
There are two horizontal asymptotes: y = 0 and y = c.
P1t2 is an increasing function if b 7 0 and a decreasing function if b 6 0.
There is an inflection point where P1t2 equals 12 of the carrying capacity. The inflection point is the point on the graph where the graph changes from being curved upward to being curved downward for growth functions, and the point where the graph changes from being curved downward to being curved upward for decay functions.
The graph is smooth and continuous, with no corners or gaps. Based on the figures, the following properties of logistic functions emerge. 484 CHAPTER 6 Exponential and Logarithmic Functions
Figure 43
0
250 250 25
Y25115
Y15230
1156.5 e20.37 x
Fruit Fly Population
Fruit flies are placed in a half-pint milk bottle with a banana (for food) and yeast plants (for food and to provide a stimulus to lay eggs). Suppose that the fruit fly population after t days is given by
P1t2 = 230
1 + 56.5 e-0.37 t
(a) State the carrying capacity and the growth rate. (b) Determine the initial population. (c) What is the population after 5 days? (d) How long does it take for the population to reach 180? (e) Use a graphing utility to determine how long it takes for the population to reach one-half of the carrying capacity. (a) As t S q , e-0.37 t S 0 and P1t2 S 230
1 . The carrying capacity of the half-pint bottle is 230 fruit flies. The growth rate is 0 b 0 = 0 0.37 0 = 37 , per day. (b) To find the initial number of fruit flies in the half-pint bottle, evaluate P102.
P102 = 230 1 + 56.5 e-0.37 102 = 230 1 + 56.5 = 4
So, initially, there were 4 fruit flies in the half-pint bottle. (c) After 5 days the number of fruit flies in the half-pint bottle is
P152 = 230
1 + 56.5 e-0.37 152 ≈ 23 fruit flies After 5 days, there are approximately 23 fruit flies in the bottle. (d) To determine when the population of fruit flies will be 180, solve the equation
P1t2 = 180. 230 1 + 56.5 e-0.37 t = 180 230 = 180 11 + 56.5 e-0.37 t2
1.2778 = 1 + 56.5 e-0.37 t
0.2778 = 56.5 e-0.37 t
0.0049 = e-0.37 t
ln 10.0049 2 = - 0.37 t
t ≈ 14.4 days
It will take approximately 14.4 days (14 days, 10 hours) for the population to reach 180 fruit flies. (e) One-half of the carrying capacity is 115 fruit flies. Solve P1t2 = 115 by graphing
Y1 = 230 1 + 56.5 e-0.37 t and Y2 = 115 and using INTERSECT. See Figure 43. The population will reach one-half of the carrying capacity in about 10.9 days (10 days, 22 hours). •
Look at Figure 43. Notice the point where the graph reaches 115 fruit flies (one-half of the carrying capacity): The graph changes from being curved upward to
EXAMPL E 5
Solution
Divide both sides by 180. Subtract 1 from both sides. Divide both sides by 56.5. Rewrite as a logarithmic expression. Divide both sides by −0.37. SECTION 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models 485
being curved downward. Using the language of calculus, we say the graph changes from increasing at an increasing rate to increasing at a decreasing rate. For any logistic growth function, when the population reaches one-half the carrying capacity, the population growth starts to slow down.
Now Work P R O B L E M 2 3
Exploration
On the same viewing rectangle, graph
Y1=500
1+24 e-0.03 tand Y2=500 1+24 e-0.08 t
What effect does the growth rate
0 b 0 have on the logistic growth function?
Wood Products
The EFISCEN wood product model classifies wood products according to their life-span. There are four classifications: short (1 year), medium short (4 years), medium long (16 years), and long (50 years). Based on data obtained from the European Forest Institute, the percentage of remaining wood products after t years for wood products with long life-spans (such as those used in the building industry) is given by
P1t2 = 100.3952 1 + 0.0316 e0.0581 t
(a) What is the decay rate? (b) What is the percentage of remaining wood products after 10 years? (c) How long does it take for the percentage of remaining wood products to reach 50%? (d) Explain why the numerator given in the model is reasonable. (a) The decay rate is 0 b 0 = 0 - 0.0581 0 = 5.81 , per year. (b) Evaluate P110 2.
P110 2 = 100.3952
1 + 0.0316 e0.0581 110 2 ≈ 95.0 So 95% of long-life-span wood products remain after 10 years. (c) Solve the equation P1t2 = 50. 100.3952 1 + 0.0316 e0.0581 t = 50 100.3952 = 50 11 + 0.0316 e0.0581 t2
2.0079 = 1 + 0.0316 e0.0581 t
1.0079 = 0.0316 e0.0581 t
31.8956 = e0.0581 t
ln 131.8956 2 = 0.0581 t
t ≈ 59.6 years
It will take approximately 59.6 years for the percentage of long-life-span wood products remaining to reach 50%. (d) The numerator of 100.3952 is reasonable because the maximum percentage of wood products remaining that is possible is 100%. •
EXAMPL E 6
Solution
Divide both sides by 50. Subtract 1 from both sides. Divide both sides by 0.0316. Rewrite as a logarithmic expression. Divide both sides by 0.0581. 486 CHAPTER 6 Exponential and Logarithmic Functions
Applications and Extensions
Growth of an Insect Population The size P of a certain insect population at time t (in days) obeys the law of uninhibited growth P1t2 = 500 e0.02 t.(a) Determine the number of insects at t = 0 days. (b) What is the growth rate of the insect population? (c) What is the population after 10 days? (d) When will the insect population reach 800? (e) When will the insect population double?
Growth of Bacteria The number N of bacteria present in a culture at time t (in hours) obeys the law of uninhibited growth N1t2 = 1000 e0.01 t.(a) Determine the number of bacteria at t = 0 hours. (b) What is the growth rate of the bacteria? (c) What is the population after 4 hours? (d) When will the number of bacteria reach 1700? (e) When will the number of bacteria double?
Radioactive Decay Strontium-90 is a radioactive material that decays according to the function A1t2 = A0 e-0.0244 t,where A0 is the initial amount present and A is the amount present at time t (in years). Assume that a scientist has a sample of 500 grams of strontium-90. (a) What is the decay rate of strontium-90? (b) How much strontium-90 is left after 10 years? (c) When will 400 grams of strontium-90 be left? (d) What is the half-life of strontium-90?
Radioactive Decay Iodine-131 is a radioactive material that decays according to the function A1t2 = A0 e-0.087 t, where
A0 is the initial amount present and A is the amount present at time t (in days). Assume that a scientist has a sample of 100 grams of iodine-131. (a) What is the decay rate of iodine-131? (b) How much iodine-131 is left after 9 days? (c) When will 70 grams of iodine-131 be left? (d) What is the half-life of iodine-131?
Growth of a Colony of Mosquitoes The population of a colony of mosquitoes obeys the law of uninhibited growth. (a) If N is the population of the colony and t is the time in days, express N as a function of t.(b) If there are 1000 mosquitoes initially and there are 1800 after 1 day, what is the size of the colony after 3 days? (c) How long is it until there are 10,000 mosquitoes?
Bacterial Growth A culture of bacteria obeys the law of uninhibited growth. (a) If N is the number of bacteria in the culture and t is the time in hours, express N as a function of t.(b) If 500 bacteria are present initially and there are 800 after 1 hour, how many will be present in the culture after 5 hours? (c) How long is it until there are 20,000 bacteria?
Population Growth The population of a southern city follows the exponential law. (a) If N is the population of the city and t is the time in years, express N as a function of t.(b) If the population doubled in size over an 18-month period and the current population is 10,000, what will the population be 2 years from now?
Population Decline The population of a midwestern city follows the exponential law. (a) If N is the population of the city and t is the time in years, express N as a function of t.(b) If the population decreased from 900,000 to 800,000 from 2008 to 2010, what will the population be in 2012?
Radioactive Decay The half-life of radium is 1690 years. If 10 grams is present now, how much will be present in 50 years?
Radioactive Decay The half-life of radioactive potassium is 1.3 billion years. If 10 grams is present now, how much will be present in 100 years? In 1000 years?
Estimating the Age of a Tree A piece of charcoal is found to contain 30% of the carbon-14 that it originally had. When did the tree die from which the charcoal came? Use 5730 years as the half-life of carbon-14.
Estimating the Age of a Fossil A fossilized leaf contains 70% of its normal amount of carbon-14. How old is the fossil?
Cooling Time of a Pizza Pan A pizza pan is removed at 5:00 from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5 minutes, the temperature of the pan is 300°F. (a) At what time is the temperature of the pan 135°F? (b) Determine the time that needs to elapse before the temperature of the pan is 160°F. (c) What do you notice about the temperature as time passes?
6.8 Assess Your Understanding
Newton’s Law of Cooling A thermometer reading 72°F is placed in a refrigerator where the temperature is a constant 38°F. (a) If the thermometer reads 60°F after 2 minutes, what will it read after 7 minutes? (b) How long will it take before the thermometer reads 39°F? (c) Determine the time that must elapse before the thermometer reads 45°F. (d) What do you notice about the temperature as time passes?
Newton’s Law of Heating A thermometer reading 8°C is brought into a room with a constant temperature of 35°C. If the thermometer reads 15°C after 3 minutes, what will it read after being in the room for 5 minutes? For 10 minutes? [Hint: You need to construct a formula similar to equation (4).] SECTION 6.8 Exponential Growth and Decay Models; Newton’s Law; Logistic Growth and Decay Models 487
Author’s Note: Surprisingly, the chemical formulas for glucose and fructose are the same: This is not a typo.
Warming Time of a Beer Stein A beer stein has a temperature of 28°F. It is placed in a room with a constant temperature of 70°F. After 10 minutes, the temperature of the stein has risen to 35°F. What will the temperature of the stein be after 30 minutes? How long will it take the stein to reach a temperature of 45°F? (See the hint given for Problem 15.)
Decomposition of Chlorine in a Pool Under certain water conditions, the free chlorine (hypochlorous acid, HOCl) in a swimming pool decomposes according to the law of uninhibited decay. After shocking his pool, Ben tested the water and found the amount of free chlorine to be 2.5 parts per million (ppm). Twenty-four hours later, Ben tested the water again and found the amount of free chlorine to be 2.2 ppm. What will be the reading after 3 days (that is, 72 hours)? When the chlorine level reaches 1.0 ppm, Ben must shock the pool again. How long can Ben go before he must shock the pool again?
Decomposition of Dinitrogen Pentoxide At 45°C, dinitrogen pentoxide (N 2O 5 ) decomposes into nitrous dioxide (NO 2 )and oxygen (O 2 ) according to the law of uninhibited decay. An initial amount of 0.25 M N 2O 5 (M is a measure of concentration known as molarity) decomposes to 0.15 M N2O 5 in 17 minutes. What concentration of N 2O 5 will remain after 30 minutes? How long will it take until only 0.01 M N2O 5 remains?
Decomposition of Sucrose Reacting with water in an acidic solution at 35°C, sucrose (C 12 H 22 O 11 ) decomposes into glucose (C 6H 12 O 6 ) and fructose (C 6H 12 O 6 ) according to the law of uninhibited decay. An initial concentration of 0.40 M of sucrose decomposes to 0.36 M sucrose in 30 minutes. What concentration of sucrose will remain after 2 hours? How long will it take until only 0.10 M sucrose remains?
Decomposition of Salt in Water Salt (NaCl) decomposes in water into sodium (Na +) and chloride (Cl -) ions according to the law of uninhibited decay. If the initial amount of salt is 25 kilograms and, after 10 hours, 15 kilograms of salt is left, how much salt is left after 1 day? How long does it take until 12 kilogram of salt is left?
Radioactivity from Chernobyl After the release of radioactive material into the atmosphere from a nuclear power plant at Chernobyl (Ukraine) in 1986, the hay in Austria was contaminated by iodine 131 (half-life 8 days). If it is safe to feed the hay to cows when 10% of the iodine 131 remains, how long did the farmers need to wait to use this hay?
Word Users According to a survey by Olsten Staffing Services, the percentage of companies reporting usage of Microsoft Word t years since 1984 is given by
P1t2 = 99.744 1 + 3.014 e-0.799 t
(a) What is the growth rate in the percentage of Microsoft Word users? (b) Use a graphing utility to graph P = P1t2.(c) What was the percentage of Microsoft Word users in 1990? (d) During what year did the percentage of Microsoft Word users reach 90%? (e) Explain why the numerator given in the model is reasonable. What does it imply?
Home Computers The logistic model
P1t2 = 95.4993
1 + 0.0405 e0.1968 t
represents the percentage of households that do not own a personal computer t years since 1984. (a) Evaluate and interpret P102.(b) Use a graphing utility to graph P = P1t2.(c) What percentage of households did not own a personal computer in 1995? (d) In what year did the percentage of households that do not own a personal computer reach 10%?
Source: U.S. Department of Commerce
Farmers The logistic model
W 1t2 = 14,656,248 1 + 0.059 e0.057 t
represents the number of farm workers in the United States
t years after 1910. (a) Evaluate and interpret W 102.(b) Use a graphing utility to graph W = W 1t2.(c) How many farm workers were there in the United States in 2010? (d) When did the number of farm workers in the United States reach 10,000,000? (e) According to this model, what happens to the number of farm workers in the United States as t approaches q ?Based on this result, do you think that it is reasonable to use this model to predict the number of farm workers in the United States in 2060? Why?
Source: U.S. Department of Agriculture
Birthdays The logistic model
P1n2 = 113.3198
1 + 0.115 e0.0912 n
models the probability that, in a room of n people, no two people share the same birthday. (a) Use a graphing utility to graph P = P1n2.(b) In a room of n = 15 people, what is the probability that no two share the same birthday? (c) How many people must be in a room before the probability that no two people share the same birthday falls below 10%? (d) What happens to the probability as n increases? Explain what this result means.
Population of an Endangered Species Environmentalists often capture an endangered species and transport the species to a controlled environment where the species can produce offspring and regenerate its population. Suppose that six American bald eagles are captured, transported to Montana, and set free. Based on experience, the environmentalists expect the population to grow according to the model
P1t2 = 500 1 + 83.33 e-0.162 t
where t is measured in years. (continued on the next page) 488 CHAPTER 6 Exponential and Logarithmic Functions
(a) Determine the carrying capacity of the environment. (b) What is the growth rate of the bald eagle? (c) What is the population after 3 years? (d) When will the population be 300 eagles? (e) How long does it take for the population to reach one-half of the carrying capacity?
The Challenger Disaster After the Challenger disaster in 1986, a study was made of the 23 launches that preceded the fatal flight. A mathematical model was developed involving the relationship between the Fahrenheit temperature x around the O-rings and the number y of eroded or leaky primary O-rings. The model stated that
y = 6
1 + e-15.085 - 0.1156 x2
where the number 6 indicates the 6 primary O-rings on the spacecraft. (a) What is the predicted number of eroded or leaky primary O-rings at a temperature of 100°F? (b) What is the predicted number of eroded or leaky primary O-rings at a temperature of 60°F? (c) What is the predicted number of eroded or leaky primary O-rings at a temperature of 30°F? (d) Graph the equation using a graphing utility. At what temperature is the predicted number of eroded or leaky O-rings 1? 3? 5?
Source: Linda Tappin, “Analyzing Data Relating to the
Challenger Disaster,” Mathematics Teacher, Vol. 87 ,No. 6, September 1994, pp. 423–426. Problems 28 and 29 use the following discussion: Uninhibited growth can be modeled by exponential functions other than A(t) = A0e kt . For example, if an initial population P0
requires n units of time to double, then the function P (t) = P0 # 2t>n
models the size of the population at time t. Likewise, a population requiring n units of time to triple can be modeled by P (t) = P0 # 3t>n.
Growth of a Human Population The population of a town is growing exponentially. (a) If its population doubled in size over an 8-year period and the current population is 25,000, write an exponential function of the form P(t) = P0 # 2t>n that models the population. (b) What will the population be in 3 years? (c) When will the population reach 80,000? (d) Express the model from part (a) in the form A(t) = A0e kt .
Growth of an Insect Population An insect population grows exponentially. (a) If the population triples in 20 days, and 50 insects are present initially, write an exponential function of the form P(t) = P0 # 3t>n that models the population. (b) What will the population be in 47 days? (c) When will the population reach 700? (d) Express the model from part (a) in the form A(t) = A0e kt .
Retain Your Knowledge
Problems 30–33 are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam.
Find the equation of the linear function f that passes through the points (4, 1) and (8, - 5).
Determine whether the graphs of the linear functions
f (x) = 5x - 1 and g(x) = 1
5x + 1 are parallel, perpendicular, or neither.
Write the logarithmic expression ln ax2 1yz b as the sum and /or difference of logarithms. Express powers as factors.
Rationalize the denominator of 10
23 25 . |
11846 | https://www.hotshotsecret.com/diesel-engine-knocking/?srsltid=AfmBOorr6cLEJNsPV_77q3rSnpT6KvlVuz-x-rbejTNF1JgAHPr2LIjV | Diesel Engine Knocking | What is Knocking In Your Engine?
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Diesel Engine Knocking Prevention & Solutions
January 28, 2024
An unexplained sound coming from your engine often signifies an internal problem. While a humming sound could be normal, a knocking, rattling or thumping noise demands immediate action.
Engine knocking could be a sign of a bigger problem, either with your injector timing, fuel or the vehicle’s mechanics.Ignoring this noise could reduce fuel efficiency and lead to permanent engine damage, so it’s important that you know what to do when you start hearing knocking sounds in your vehicle.
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What Is Engine Knocking?
Engine knocking is a sound that occurs when your vehicle’s cylinders don’t burn fuel evenly. Ideally, the mix of fuel and air in each cylinder will be balanced so that small pockets of fuel and air burn in small amounts. When the mix becomes unbalanced, multiple fuel and air pockets burn at once, producing a knocking noise.
The knocking noise in the engine may start out quietly, becoming louder as you continue to drive — or on later trips if the issue isn’t treated. Left unresolved, engine knocking can cause damage to the pistons and cylinder walls of the engine.
Diesel Knock Causes
Engine knocking can occur in many ways. It can happen when a part of the engine gets worn out or due to poor-quality fuel. It may also result from ignition troubles or mechanical issues.
Fuel-Related Reasons
Your choice of fuel is critical to engine performance and the vehicle as a whole. Engine knock can be attributed to these fuel issues:
Poor-Quality Diesel Fuel:Low-quality orcontaminated fuel, can disrupt the combustion process, leading to premature or uneven ignition.
Wrong cetane rating:The cetane number determines diesel fuel ignition. A higher cetane rating means faster fuel ignition and more complete combustion. A low cetane rating, meanwhile, means longer ignition and more time to complete combustion. Typically, the higher the cetane rating, the better for diesel engines, with the ideal range between 50 and 55.
Unevenfuel-to-air ratio:When there’s too much air and not enough fuel in the cylinders, your engine needs multiple combustions to operate. Frequent multiple combustions could lead to engine knock.
Badfuel additives:Using additives not optimized for diesel engines or injecting more than the appropriate amount can clog fuel lines, cause incomplete fuel combustion and damage electrical components, all contributing to the possibility of engine knock.
Ignition-Related Reasons
Fuel isn’t the only possible cause of engine knocking. Sometimes, these or other problems with the vehicle’s ignition system may be to blame:
Ignition timing issues: Timing is crucial to the engine. When the air-fuel mixture ignites prematurely, it pushes against the pistons, producing loud pinging sounds from the engine.
Loose timing chain:Over time, the timing chain can become loose or broken, causing a rattling sound in the engine. This sound is noticeable when the engine heats up and, if left unresolved, may lead to significant damage. Repairing the chain or lubricating the assembly can help correct the problem and prevent further damage to your vehicle’s engine.
Faulty fuel pressure regulator:A bad fuel pressure regulator affects fuel delivery and may allow more air in the air-fuel mixture. The correct balance of air and fuel is critical to the engine’s overall function, so a malfunctioning fuel pressure regulator can be a sign of trouble.
Mechanical Reasons
Mechanical issues with the engine occasionally lead to a knocking noise. Some of the most common mechanical causes of diesel engine knock include:
Excessive carbonbuildup:Carbon deposits are inevitable. They can develop in various parts of the engine system, such as the cylinders, chamber walls and fuel injectors, accumulating over time. When excessive carbon deposits build up in the combustion chambers and other components, it can create a loud, thumping noise, reduce fuel efficiency and cause overheating.
Compression problems:A problem with the compression system, such as leaky valves, damaged cylinders or worn-out piston rings, may cause your vehicle to make a knocking sound.
Worn-out parts:Engine knock may also be caused by worn-out mechanical components such as the timing belt, bearings and crankshaft. Regular maintenance helps you spot these problems immediately so you can make the necessary fixes.
Overheating:Low oil or coolant volume due to hot weather or a problem with the cylinders can cause your vehicle to overheat. Overheating leads to an increase in cylinder pressure, which causes incomplete combustion and the possibility of engine knocking.Along with the pinging noise, you may notice other signs of overheating, such as smoke from the engine, a burning smell and the temperature gauge in the red zone.
Other engine components, such as the piston rings and valve seals, can impact engine performance and cause engine pinging or detonation. Piston rings play a significant role in maintaining compression and preventing combustion gases from escaping by sealing the piston and the cylinder wall. When they get worn out, gases may escape to the crankcase, creating a lean air-fuel mixture.
Valve seals have a similar role to piston rings. They prevent oil from leaking through the valve guides and into the combustion chamber. Damaged valve seals allow more oil to enter the chamber, leading to an oil-rich air-fuel mixture. In both scenarios, an engine knock is possible.
Common Symptoms of Diesel Knock
Pay attention to these signs if you suspect that your diesel engine is developing problems:
Audible knocking sounds:Listen for a high-pitched or rattling noise coming from the engine when driving under load or accelerating.
Vibrations or rough engine performance:Excessive vibration and overheating are common indicators of engine knock.
Loss of power or reduced fuel efficiency: Engine knock is characterized by a noticeable loss of power when accelerating or driving uphill. In severe cases, you may feel some hesitation in your engine and an increase in fuel consumption.
Engine Knock Prevention and Solutions
Hot Shot’s Secret offers a range of high-performance additives and treatments specifically designed to enhance diesel engine performance and address common issues like engine knock. Here are some of their top products to consider:
Diesel Extreme
•Purpose: Deep cleans your entire fuel system.
• How it Helps: Diesel Extreme removes carbon deposits, cleans injectors, and eliminates water contamination in the fuel system. By restoring fuel system efficiency, it helps prevent uneven combustion and reduces the risk of knock.
Everyday Diesel Treatment (EDT)
• Purpose: Enhances fuel quality with every fill-up.
• How it Helps:EDT increases the cetane number of diesel fuel, ensuring a more complete and efficient burn. It also protects against fuel system wear and improves lubricity, reducing the likelihood of injector-related issues.
Stiction Eliminator
• Purpose: Cleans and protects your engine’s internal components, anything the oil touches.
• How it Helps: Stiction Eliminator removes sludge and varnish buildup, ensuring smoother operation of critical engine components like rings and lifters. This can help eliminate the mechanical causes of knock.
FR3 Friction Reducer
• Purpose: Reduces engine wear and improves performance.
• How it Helps: By minimizing friction in the engine, FR3 enhances overall efficiency and reduces stress on internal components, helping prevent knock caused by mechanical wear.
Reduce Engine Knock With Hot Shot’s Secrets
Explore the Hot Shot’s Secret range of fuel additivesto keep your engine running smoothly and help prevent engine knock. Using Diesel Extreme and EDT prevent many issues that can cause engine knocking. Browse our selection today andcontact us with any questionsor for more product information.
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11847 | https://www.youtube.com/watch?v=u3O0QKk33B0 | How to balance CO(g) + O2(g) → CO2(g)
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Balance the following equation:
CO(g) + O2(g) → CO2(g)
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BalancingEquations #BalancingChemistry #Chemistry
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Transcript:
today we're going to balance the following equation we have carbon monoxide so co plus oxygen gas which is o2 and that will produce carbon dioxide co2 so i already wrote down our chart and the elements that i see on both of the reactants and the products now all we have to do is we just have to count how many we have i see that i have one carbon right so i have one of these but then when it comes to oxygen i see that i have oxygen here and here on my left side so i have to count both of them there's one oxygen here plus two oxygens here so one plus two will get me a total of three oxygens and then coming to my product side i see that i have one carbon so one and i only have this two right as the subscript for the oxygen so i have two oxygens now by looking at this chart i see that the carbons are balanced but the oxygens are not now i do see a problem here i don't like odd numbers so if i can change an odd number into an even number i'm going to do that because it just makes the balancing so much easier now when we did the calculation to get the oxygens on the left-hand side remember there was one oxygen plus two this was the problem so if i just turn this number into a even number i'll get an even number for my reactant side so this is just an educated guess right and that's what balancing is i'm going to just put a 2 in front of here but whatever number that you put in front of a compound you have to distribute you have to be fair so this two actually is going to change the carbon and the oxygen number so now i have two carbons so no more one anymore it turns into a two but now i have two oxygens plus two other oxygens so two plus two is now four and looky there now it makes it much easier so whether i want to balance the carbon or the oxygen is fine with me let's just balance the carbon right so it looks like i'm going to put a number here right what times 1 will get me to 2. yeah i'll put a 2 in front of here but since i put a 2 in front of the compound i have to play fair so it's going to change the carbon and the oxygen number so now i have 2 carbons and now two times two is four oxygens and now if i look i balanced all of my elements so this is the final answer so guys hopefully this helped right we have two cos plus one o2 which will yield us two co2s thank you so much for tuning in if this helped please click the like button and subscribe to the channel if you want to thank you so much and have a great day bye |
11848 | https://gubner.ece.wisc.edu/notes/MagnitudeAndPhaseOfComplexNumbers.pdf | MagnitudeAndPhaseOfComplexNumbers.tex 7/6/17, 7/12/17, March 30, 2021 Magnitude and Phase of Complex Numbers John A. Gubner Department of Electrical and Computer Engineering University of Wisconsin–Madison Abstract Every nonzero complex number can be expressed in terms of its magnitude and angle. This angle is sometimes called the phase or argument of the complex number.
Although formulas for the angle of a complex number are a bit complicated, the angle has some properties that are simple to describe. In particular, when the complex number is a function of frequency, we derive a simple formula for the derivative of the argument.
If you find this writeup useful, or if you find typos or mistakes, please let me know at John.Gubner@wisc.edu Properties of the Angle of a Complex Number Recall that every nonzero complex number z = x+ jy can be written in the form rejθ, where r := |z| := p x2 +y2 is the magnitude of z, and θ is the phase, angle, or argument of z. Common notations for θ include ∠z and argz. With this notation, we can write z = |z|ejargz = |z|∠z. For each z ̸= 0, there are infinitely many possible values for argz, which all differ from each other by an integer multiple of 2π. For this reason, it is sometimes convenient to use the principal angle or principal argument of z, which is the unique value of θ ∈(−π,π] for which z = |z|ejθ. The principal argument is denoted by Argz with an uppercase “A”. We show below that Arg(x+ jy) = tan−1(y/x), x > 0, right half-plane, tan−1(y/x)+π, x < 0,y ≥0, upper left-half-plane, tan−1(y/x)−π, x < 0,y < 0, lower left-half-plane, π/2, x = 0,y > 0, + j-axis, −π/2, x = 0,y < 0, −j-axis, undefined, x = 0,y = 0, origin, (1) 1 MagnitudeAndPhaseOfComplexNumbers.tex 7/6/17, 7/12/17, March 30, 2021 and that Arg(x+ jy) = π/2−tan−1(x/y), y > 0, upper half-plane, −π/2−tan−1(x/y), y < 0, lower half-plane, 0, y = 0,x > 0, positive real axis, π, y = 0,x < 0, negative real axis, undefined, x = 0,y = 0, origin.
(2) To obtain Argz in MATLAB, use angle(z), and note that angle(0) returns 0.
Derivation of (1) and (2). Consider a complex number z = x + jy in the first quad-rant, as shown in Figure 1. When x and y are both positive, θ ∈(0,π/2), and the the-0 θ z = x+ jy x jy Figure 1. A complex number z = x+ jy in the first quadrant of the complex plane.
ory of right triangles tells us that tanθ = y/x. Similarly, tan(π/2−θ) = x/y. Hence, we have two formulas to choose from: θ = tan−1(y/x) and θ = π/2 −tan−1(x/y).
The first formula holds even for x > 0 and y = 0; i.e., for z on the positive real axis, the argument is zero. The second formula holds even for x = 0 and y > 0; i.e., for z on the +j-axis, the argument is π/2.
Now consider z = x + jy in the second quadrant, as shown in Figure 2. Since 0 x θ z = x+ jy jy Figure 2. A complex number z = x+ jy in the second quadrant of the complex plane.
θ > π/2, we apply the theory of right triangles to the the supplementary angle π −θ.
Since x is negative, the length of the horizontal side is −x. Thus, tan(π −θ) = y/(−x), and, since tan−1 is an odd function, it follows that θ = π +tan−1(y/x). Since the complementary angle of π −θ is π/2 −(π −θ) = θ −π/2, we have tan(θ − π/2) = −x/y, and it follows that θ = π/2−tan−1(x/y). Notice this formula is the same as the one derived in the first quadrant! Hence, for any z in the strict upper half-plane, its argument is π/2−tan−1(x/y).
2 MagnitudeAndPhaseOfComplexNumbers.tex 7/6/17, 7/12/17, March 30, 2021 For the third quadrant, it can similarly be shown that θ = tan−1(y/x) −π and θ = −π/2−tan−1(x/y).
For the fourth quadrant, θ = tan−1(y/x) and θ = −π/2 −tan−1(x/y). Hence, the argument of any z in the strict lower half-plane is −π/2 −tan−1(x/y), and the argument of any z in the strict right half-plane is tan−1(y/x).
Functions of Frequency. Suppose z( f) = x( f) + jy( f), where x( f) and y( f) are differentiable functions of f. Put θ( f) := arg(x( f)+ jy(f)). Our goal is to compute θ ′( f) for all f with x(f)2 +y( f)2 > 0. Note that differentiation removes any multiple of 2π that distinguishes different versions of arg. Hence, we can use any version that is convenient. There are four (overlapping) cases to consider: x( f) > 0, x( f) < 0, y( f) > 0, and y(f) < 0.
For x( f) > 0, we use the first formula in (1) to write tan(θ(f)) = tan(tan−1(y( f)/x( f)) = y( f)/x( f).
Differentiating tan(θ( f)) = y(f)/x( f) with respect to f yields sec2(θ( f))θ ′(f) = x( f)y′( f)−y(f)x′( f) x( f)2 .
Since sec2 = 1+tan2, and since tan2(θ( f)) = y( f)2/x( f)2, we can write 1+y( f)2/x( f)2 θ ′(f) = x( f)y′( f)−y( f)x′(f) x( f)2 .
Solving for θ ′( f), we obtain ∂ ∂f arg(x( f)+ jy( f)) = x( f)y ′( f)−y( f)x ′( f) x( f)2 +y( f)2 .
(3) For y( f) < 0, we use the second formula in (2), the identity tan(π/2 −α) = 1/tanα, and the fact that the tangent is odd. It follows that tan(θ( f)) = tan(−π/2−tan−1(x( f)/y( f))) = −tan(π/2+tan−1(x( f)/y( f))) = −1/tan(−tan−1(x( f)/y( f))) = y( f)/x(f).
Arguing as above, we see that (3) holds for y( f) < 0. It can similarly be shown that (3) holds for y( f) > 0.
To handle the case x( f) < 0, we cannot use Arg because it is discontinuous across the negative real axis. However, if we add 2π to the third formula in (1), the result will be identical to the second formula in (1). In other words, for z = x + jy in the 3 MagnitudeAndPhaseOfComplexNumbers.tex 7/6/17, 7/12/17, March 30, 2021 strict left half-plane, we can write arg(x + jy) = tan−1(y/x) + π. Since the tangent function has period π, we have tan(θ( f)) = tan(arg(x( f)+ jy( f))) = tan(tan−1(y( f)/x( f))+π) = y( f)/x( f), and (3) holds for x( f) < 0.
Putting everything together, we have ∂ ∂f arg(x( f)+ jy(f)) = x( f)y ′(f)−y( f)x ′( f) x( f)2 +y( f)2 , x( f)2 +y( f)2 > 0.
(4) 4 |
11849 | https://math.stackexchange.com/questions/1912007/domain-and-range-of-a-square-root | functions - Domain and range of a square root - Mathematics Stack Exchange
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Domain and range of a square root
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I got a question given by my lecturer to find the domain and range of this function:
y=2−x 2−−−−−√y=2−x 2
For the domain, we start by restricting the square root to be non-negative. After factorising, it becomes (x+2–√)(x−2–√)≤0(x+2)(x−2)≤0. This is my explanation.
But the lecturer at this point jumps to the conclusion that y:[−2–√,2–√]↦[0,2]y:[−2,2]↦[0,2] without any explanation.
I tried to substitute 2 as the value of y but it is wrong. Can anyone care to explain how he derived the domain from the factorisation and range?
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edited Sep 2, 2016 at 11:55
Parcly Taxel
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asked Sep 2, 2016 at 11:47
BryanBryan
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The correct factorization is
2−x 2=(2–√+x)(2–√−x)2−x 2=(2+x)(2−x)
where the left factor is nonnegative for x≥−2–√x≥−2 and the right factor is nonnegative for x≤2–√x≤2. Thus the function is defined on [−2–√,2–√][−2,2]. The factors can´t be both negative since the conditions x<−2–√x<−2 and 2–√<x 2<x exclude each other. Thus [−2–√,2–√][−2,2] is the maximal set of real numbers where the function is defined. Obviously the minimal value 0 0 is taken for x∈{−2–√,2–√}x∈{−2,2} and because of the monotonicity of the squareroot the maximal value is taken when the radicand is maximal thus for x=0 x=0 and this maximal value is 2–√2. Thus the mapping is a mapping [−2–√,2–√]→[0,2–√][−2,2]→[0,2] The image of f f is I m(f)=[0,2–√]I m(f)=[0,2].
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edited Sep 2, 2016 at 12:09
answered Sep 2, 2016 at 12:03
Peter MelechPeter Melech
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I'm still confused with how range is being identified. Isn't it found by just subbing the individual highest / lowest domain into the equation to find the range (y)?Bryan –Bryan 2016-09-02 17:22:23 +00:00 Commented Sep 2, 2016 at 17:22
I'm a little confused over the fact that radicand is maximal thus for x=0. Isn't the maximum value of the domain(x) is √2 like how it is in the domain [−√2, √2]?Bryan –Bryan 2016-09-02 18:04:35 +00:00 Commented Sep 2, 2016 at 18:04
The minimum is 0 0 since the square root is non-negative and the maximum is taken for the maximal value of the radicand because of the monotonicity of the square root and 2−x 2 2−x 2 is maximal for x=0 x=0 The maximum that is taken is 2–√2 Peter Melech –Peter Melech 2016-09-04 09:55:17 +00:00 Commented Sep 4, 2016 at 9:55
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Your lecturer appears to have made a typo.
it should have been [−2–√,2–√]→[0,2–√][−2,2]→[0,2]
You have made the correct start by saying that the square root must be non-negative. Using that information you can then change your equation from this:
y=2−x 2−−−−−√y=2−x 2
to this:
2−x 2≥0 2−x 2≥0
a bit of rearranging and out pops the domain limits.
2≥x 2 2≥x 2
−2–√≤x≤2–√−2≤x≤2
so from this we can see the domain limits are [−2–√,2–√][−2,2]
Then to get the upper bound on the range, maximise the radicand (the part under the square root). x=0→y=2–√x=0→y=2
So you end up with [−2–√,2–√]→[0,2–√][−2,2]→[0,2]
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answered Sep 2, 2016 at 12:28
Ben CrossleyBen Crossley
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I'm still confused with how range is being identified. Isn't it found by just subbing the individual highest / lowest domain into the equation to find the range (y)?Bryan –Bryan 2016-09-02 17:22:25 +00:00 Commented Sep 2, 2016 at 17:22
I'm curious how does maximizing the radicand (the part under the square root) gives x = 0. However, the maximum value of the domain is √2Bryan –Bryan 2016-09-02 18:01:24 +00:00 Commented Sep 2, 2016 at 18:01
@TeoChuenWeiBryan It may help to look at a graph of it. wolframalpha.com/input/… It isn't necessarily the limits of the domain that create the limiting values of the range. You basically set yourself a challenge. Find the maximum value you can create by varying x (the upper limit) and find the lowest value you can make (lower limit)Ben Crossley –Ben Crossley 2016-09-02 20:36:15 +00:00 Commented Sep 2, 2016 at 20:36
To find the upper limit of (√2−x 2)(2−x 2) you need to make 2−x 2 2−x 2 as big as possible, and so x x as small as possible. Making x x as small as possible within your domain is the same as saying x=0 x=0 To find the lower limit of (√2−x 2)(2−x 2) you need to make 2−x 2 2−x 2 as small as possible, and so x x as big as possible. This is the same as taking either limit of your domain.Ben Crossley –Ben Crossley 2016-09-02 20:36:19 +00:00 Commented Sep 2, 2016 at 20:36
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Anyway, it's much simpler to use absolute values:
2−x 2≥0⟺x 2≤2⟺x 2−−√=|x|≤2–√⟺−2–√≤x≤2–√.2−x 2≥0⟺x 2≤2⟺x 2=|x|≤2⟺−2≤x≤2.
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answered Sep 2, 2016 at 12:32
BernardBernard
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11850 | https://www.fq.math.ca/Scanned/36-1/drmota.pdf | THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Michael Drmota and Johannes Gajdoslk Department of Discrete Mathematics, Technical University of Vienna Wiedner HaupstraBe 8-10, A-1040, Vienna (Submitted June 1996) 1. INTRODUCTION Let G = (Gw) be a strictly increasing sequence of positive integers with Gl = l. Then every nonnegative integer n has a digital expansion i>\ with respect to basis G, where the digits si - si{ji) > 0 are integers. This digital expansion is unique, when one assumes that the digits et are chosen in such a way that the digital sum Z7>! si is as small as possible; in this case, we will call the digital expansion a proper digital expansion. It is easy to see that the following algorithm provides this expansion. 1. For n = 0, we have £,-(«) = 0 for every i > 1. 2. If Gj <n< Gj+l and n' = n-Gj has the proper expansion n' = Zz>isffjt, then the expan-sion of n = Sz->i etGt is given by si = s\ for i & j and by Sj =e'. + l. The most prominent digital expansions are related to linear recurring sequences G = (Gn), e.g., the binary (resp. the #-ary) expansion relies on Gn = 2n~l (resp. on G„ = qn~l). If Gn are the Fibo-nacci numbers, i.e., Gn = Fn+l, then we obtain the Zeckendorf expansion. For each digital expansion with respect to a basis G, we can define a partial order in a quite natural way. We will say a<Gh if and only if et(a) < st{b) for every i > 1. It is well known that for every partial order there is a Mobius function (see , ). Let sG(n) denote the sum of digits of n. Then it will turn out that the Mobius function juG of a digital expansion to a basis G is given by juG(n) = (~l)%(w) if maXy^j si{n) < 1 and by juG(n) — 0 otherwise. If G is a proper linear recurring sequence and if the initial conditions of G are properly chosen (see Section 3), then n=0 is either bounded or MG(N) = Sa(N):=Z(-iyaW, which we will see from calculating the Mobius function in Section 2. (We always define empty sums to be zero, i.e., MG(N) = SG(N) := 0 for N < 0.) This work was supported by the Austrian" Science Foundation, grant P10187-PHY. This paper, presented at the Seventh International Research Conference held in Graz, Austria, in July 1996, was scheduled to appear in the Conference Proceedings. However, due to limitations placed by the publisher on the number of pages allowed for the Proceedings, we are publishing the article in this issue of The Fibonacci Quarterly to assure its presentation to the widest possible number of readers in the mathematics community. 1998] 3 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS In Section 3 we will formulate conditions for G, under which we will be able to derive for-mulas for SG(N). We will also obtain a recursive formula for the generating function of SG(Gn), which we will analyze in Section 4 in order to obtain asymptotic information about SG(N). Our main interest lies in the distribution of the SG(N) (resp. MG(N)) when 0<N <m for large m. This means that we count the number of times SG(N) takes a certain value k when 0 < N < m: let dm(k): = | {0 < N < m: SG(N) = k}\ be this number and let Xm be a random variable with probability distribution f(Xm = k) = dm(k)/m. Then we are interested in the asymptotic dis-tribution of Xm for m-> oo. Depending on the nature of the recurrence relation for G, we will observe significantly different behavior of Xm. First, we distinguish two cases: 1. either SG(Gn) is bounded for all initial conditions of G (Section 4.1), or 2. there are initial conditions of G such that SG(Gn) is unbounded (Section 4.2). Since we can establish a linear recurrence relation for the SG(Gn), the first case is equivalent to the assumption that the characteristic polynomial of this recursion is a product of some zr~v (r-v>0) and certain different cyclotomic polynomials. In this case, we can derive asymptotic formulas for EXW and VXm, provided that the sequence G satisfies a certain technical condition. Our main result (Theorem 2) says that, in the case of unbounded variance, Xm satisfies a central limit theorem. (Note that there are sequences G for which VXm is bounded, e.g., Gn = 2W-1.) 2. THE MOBIUS FUNCTION OF A DIGITAL EXPANSION Let G = (G„) be a strictly increasing sequence of integers with Gx = 1. As mentioned above, every nonnegative integer n has a digital expansion n - Z/^i^/Gy with nonnegative integral digits sr It is called proper digital expansion for n if the digital sum Zz>i st is as small as possible. Lemma 1: LetTi^Z^^G, be a proper digital expansion for n. Then any sum of the form 2/>i^z'Q with integral digits e\, i > 1, satisfying 0< s\ < st is a proper digital representation for some«'<«. Proof: First, note that it follows from the algorithm stated in the Introduction that any digital expansion of the form nj = S/=i £iGi < n is a proper one. Next, we will use induction on the digital sum 5' = Z/>i^J, where 0<8f ii e\Gt has digital sum s'. There exists j > 1 such that s'j > 0 and s\ - 0 for i> j . Then Gj <n' <nj <Gj+l. Therefore, n" = n'-Gj can be represented by n" = Z/=i£,"Gz with s'j-s'j-I and s'(-s\ for ij. Since ()<£,"<£; and its digital sum satisfies E/^i^r= s'-li£-G7 is a proper expan-sion for n'. D Now we introduce the Mobius functions ju(x, y) of a locally finite partial order < on a set X, i.e., all intervals [x, y] = {u e X: x < u < y) are finite (see , ). Any function f:X2-J»C that satisfies f(x, y) = 0 for x & y will be called an arithmetical function. The convolution / g of two arithmetical functions/, g is given by x<u<y 4 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Obviously 8, defined by S(x, y) = 1 for x = y and S(x, y) = 0 otherwise, is the unit element of . Furthermore, if f(x, x)^0 for every XGX, then there always exists an inverse arithmetical function f~l satisfying f~lf = S. The Mobius function JU is defined as the inverse function of C, given by g(x, y) = 1 if x < y and by £(x, y)=0 otherwise. Especially, if g = £ f, then/can be recovered by / = jug. (We intend to use this Mobius function in future work for sieve methods in connection with specific problems of digital expansions.) Theorem 1: Let i s\ Gt and /2 = Z/>i£"Q are proper digital expansions of nonnegative integers m, n with m<G n, i.e., e\ < e" for all /'. Then ( 0 if there is an / with e"- e\ > 1, (-ifixto-'i) otherwise. Proof: Since there is a natural bijection between [m, n] = {d eN0|jw <G d <G n} and [0, n-m\ we have ju(m, n) = ju(0, n-m) if m 0. 0<Gd<Gn Assume for a moment that € "< 1 for all i. We show that //(0, ZyljQ,) = (~lf DY induction on the digital sum s = k. Clearly, for s = 0, we have ju(09 0) = 1 = (-1)°. Now assume that s>\ and that //(0, Z%1 Gt.) = (-1) for all k < s. Then 0 - X M(09d) -(//(o,o))+(//(o,a0)+//(o,a1)+---+Mo,aJ_1)) ( ( s~i ^ + ( / I ( 0 , G , 0 + G O H V K O , G ^ =i+(;>-i> l+(s) (" 1)2+••• +G- Oc-1)-1-^^ 2 ^ ) Because of S}=0(/)(-l); = 0, it follows that ju(0, SJIQQ,) = (-1)', which proves the theorem in this special case. Nov/ suppose that kGt with i > 1 and k > 1 is a proper digital expansion. Then 0 = ju(0,0) + ju(0, G,) + • • • +//(0, kGt). Notice that /i(0,0) +ju(0, Gf) = 0. Hence, it follows that ju(0,2Gt) = ju(0,3Gi) = - = ju(0,kGi) = 0. Next, we show by induction on the digital sum s(ri) = Z;>i £" that /i(0, w) = 0 whenever there is an i with £'/> 1. We must start with s(n) = 2 because e"> 1 cannot be satisfied when s(ri) < 2. Suppose that s(n) = 2 and that there is some i with e"> 1. Then m = 2G,- and /i(0, m) = 0. Now assume the assertion holds for all natural numbers / with s(l) . Then 1998] 5 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS -M(0,ri)= £ ju(0,d)= £ M(0,d)+ X fM 0<Gd<Gn Q^QdKQn^i-.s^d)^! 0< Gd< G«,3i:/(<0>l I M(P,d). Q<Gd<Gn,^i\Ei{d)<\ Define nx := Zz>imin(£", l)Gr Because of the existence of/with £">l, we have 0<nx <n and 0<G^<Gw,V/:£'/(c?)<l Q<Gd2Gn_l for all n>. Then MG(N) is bounded by 1. On the other hand, if Gn < 2Gn__x for all n > 1, then MG(N) = SG(N):=Nf(-iyc(»\ where sG(n) denotes the digital sum sG(n) = Z,>i ex of the proper digital representation i>\ Proof: Due to Theorem 1, only those n with expansion coefficients 0 or 1 enter the sum. If Gn > 2Gn_l for all n > 1, then all the digital expansions Sz>i stGt with si G {0,1} are proper ones. Hence, MG(N) attains only the same values as in the binary case in which the corresponding sum is 0 or ±1. If Gn < 2Gn_l for all n > 1, then in all the proper digital expansions only the digits 0 and 1 can occur, and the assertion follows from Theorem 1 with m - 0. D Remark 1: We will see later that for all G considered here, (al + l)Gn_l >Gn > a^jn_Y holds for n > r; therefore, Gn < 2Gn_l for all n > 1 is equivalent to ax - 2 and r = 1 or ax = 1 when the initial conditions of G are properly chosen. But if aY > 2 or ax = 2 and r > 1, and if Gn > 2Gn_x holds for the initial values, then Proposition 1 applies and MG{N) is bounded. Because of this, we will investigate the function SG(N) rather than MG{N), keeping in mind that, in most cases, when MG(N) is of interest, both are the same. Remark 2: If Gn = 2"~\ then tn = (-l)^(w) is the Thue-Morse sequence . Since t2n + t2n+l = 0, we have £G(2ft + l) = t2n = tn9 and we also have SG(2ri) = 0. Thus, it is not really interesting to study SG(N) in this case. 6 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS 3. DIGITAL EXPANSIONS AND GENERATING FUNCTIONS From this point on we will consider only integral linear recurring sequences G = (Gn)n>x that satisfy assumptions 1-5 below (in Section 4.1 we will also need assumption 6): 1. Gx = l and G„+x>Gn for n>. 2. Gn = S[=1 ciiGn_i for n > r with some integers at > 0. 3. Gn_j > EJ=y+i @jGn-i for w > r and 1 < j < r. 4. G satisfies no linear recursion with constant integer coefficients with a smaller degree. 5. The characteristic polynomial zr - E[=1 atzr~l = Il[=1(^ - at) (of the above recursion) has only one real, positive, and simple root ax of maximal modulus. 6. Let bt = (af mod2)(-l)ai+'"+a'-1 (a, mod2 = 0 if a;, is even and at mod2 = 1 otherwise). Then is a product of zr~v (r - v > 0) and different cyclotomic polynomials Q>kh(z) (kx<k2<--< kk,\ all of them dividing zp -1 with some fixed p>r. Furthermore, none of the at and no quotient at I otj (i ^ j) is a p^ root of unity. Assumptions 1, 2, and 4 are natural. Therefore, only conditions 3, 5, and 6 need to be motivated. Assumption 3 is necessary to show that S{Gn) satisfies a linear recurrence; especially, it im-plies (6) in Proposition 2. From assumption 5, we obtain Gn = (ial~l + 0{{axy)n) with some /? > 0 and 0 < y < 1. Note that assumptions 2 and 3 imply (ax + 1)G„_X ^Gn> afjn_x for n > r, which gives ax < ax < ax +1. Similarly, we get ax > at for all i. The first part of assumption 6 (concerning the cyclotomic factors) ensures that S(Gn) is bounded. The assumption that at and af I ay are not /7th roots of unity is frequently used in problems concerning digital expansions with respect to linear recurring sequences and avoids technical difficulties (see Lemma 2). Usually, assumptions 3 and 5 are replaced by the stronger condition ax > a2 > • • • > ar and certain assumptions on the initial values of G (see, e.g., ; in this case, the second part of assumption 6 is also satisfied). However, there are other interesting examples, e.g., ax =ar = 1, a2 = • •. = ar_x - 0, that satisfy the above assumptions and are not of the form ax > a2 > • • • > ar. From here on, let G = (Gn) be a fixed linear recurring sequence with assumptions 1-5. For notational convenience, we will omit the index G in the sequel. Proposition 2: Let ht = (a, mod 2)(-l)ai+'""K-1 {at mod 2 = 0 if at is even and at mod 2 = 1 other-wise). Then S(Gn) = SG(Gn) satisfies the linear recurrence S(G„) = X W ? „ - , ) for»>r. (2) 1=1 Furthermore, if n has the proper digital expansion n = Hl J=l SJGJ, then ( i \ i ^ Z ^ =2(^mod2X-l)^1+-"+tf|5,(Gy). (3) V/=1 J /=! 1998] 7 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Proof: We will first establish a set identity that holds for all nonnegative integers sjy regard-less of whether Zy=i SJGJ is a proper digital expansion or not: O^aKpfiLpL i / j=\ \h=j+i } i /-i r / 0 ^ < ^ y = U U ZAGA+« J y=l i=0 IA=y+1 iGj<a<(i + l)Gj\ (4) ,-i f / = U U Z^Gh\ + iGJ+a 0<a where each union is disjoint. (Again, empty sums are set at zero.) Now set l = n-l, Sj =an-j for n-r r, after interchanging i and 7 and shifting /'->/?-/', h-^n-h, {a n-\ at-\ 0<a< I^Gy=UU I ^ + ^ + f l 0 r, 3,-1 r a,-l G ^ - l S(G„) = Z(-ir ( a ) = Z Z Z {-\y^^ahG"-h+jG"-i+a) a=0 /'=! ;=0 a=0 =t£lY1(-^+/^"B=Z(-i)a:ti,,i^(GJ1_,)5;l(-i)/ = X(«, mod2X-l)(zfc,'fl)5(GfM) = I W ^ , ) / = 1 J = l with £ > := (a, mod 2)(-l)ai+'"+a'1. Note that assumption 3 from above ensures that n-i \ i-i V/P=I y A=I You only have to start with m = lL1h=\ahGn_h+jGn_i+a and apply the algorithms stated in the Introduction to deduce that sn_h{m) = ah, \<h<i and sn_^m) - j . (Of course, this procedure is standard in the study of such digital sequences (cf. , ). This proves equation (2). The proof of (3) is quite similar. If we set Zy=i epj =:m + slGl in (4), we get {a\0<a<m + £lGl}={J{iGl+a\0<a<Gl}u{elGl+a\0 r, the relation DG (z) = Y y ^ t i ^ C ^ ^ (z{-i) a> +'" a>- +J\ ( m 7=1 j=Q Proof: Suppose n > r. An iterated use of (7) gives, for 1 < i < r, y < at, and m < G„„;, + jG„_i+m) = (aj mod 2)<S,(G„_1) + (-1)°' (a2 mod 2),S'(G„_2) + • • • + (-l)-.+-"^-2(a-_1mod2)5(G^+1) + (-l)fll+""f<'wC7mod2)5(GllW) = 2 w ^ ) + (-l)fl,+ "^'-,0"mod2)S(G^^^^ Note that, for / = 1, we just obtain S(jGn_l +m) = (Jmod2)S(Gn_l) + (-iyS(m). Hence, by using (5) and (8), we get G„-l r q-lG^-l JJ / z \ _ y z % ) = y y y£zS(alG^.l + ---+al_lGn_M+jGn_i+m) m=0 j=\ y=o m=0 i=l J=0 m=0 i=i y=o 4. ASYMPTOTIC ANALYSIS We distinguish two cases: either S(Gn) is bounded for all suitable initial conditions of G or it is not. The first case will be of special interest. It turns out that in this case the distribution of the values of S(N) approximates a normal distribution for all suitable initial conditions of G (see Theorem 2). 4.1 Bounded S(Gm) Proposition 3: Suppose that S{Gn) is bounded. Then S(Gn) satisfies a linear recursion for n > N with some J S T , whose characteristic polynomial is a product of different cyclotomic polynomials. Remark: This motivates the first part of assumption 6 in Section 3. Proof: We know that every S(m) is an integer and, therefore, can only attain a finite number of distinct values. So we see from (2) that S(Gn) must be periodic (in n) for n> N. Let/? > r be 1998] 9 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS some period of S(Gn) and assume n > N. Then S(Gn+p) - S(Gn) = 0, which implies that S(Gn) is a linear combination of powers of /7th roots of unity. Let m(z) be the product of all cyclotomic polynomials corresponding to those roots of unity which appear in the representation of S(Gn). Then S(Gn) satisfies the linear recurrence related to m(z), D Proposition 4: Suppose that S{Gn) is bounded. Then DG„(Z) (defined in (8)) and Don(z~l) satisfy, for n > T V , a homogeneous linear recurrence with (in n) constant coefficients af(z) that are analytic around z = 1 and satisfy at(z) = at(z~l). Proof: Let/? > r be a period of S(Gn). Then, by splitting (9) into four parts, we get / = max(0,fc-r) k-\ i=k+p-r p-1 / = max(0,fc-r) i-k+p-r with , / T \ - U SZh=\ ' 6/»%-/»-(ai+ - +ak-i-i mod 2)m,) /k,i\Z)~nk-JZ „, /-A _ |T Jsfci"1 hf»k-h+(oi+ ••• +«-/-imod2M-) Yk,p+i\Z)-nk-iZ bkJ\Z)-nk+p-iZ £k,P+i(z) = hk+p-i^ , where /^, := £(G7), 0 < A : < p and 0 </ < p and [ | { 0 < 7 < a / | j ^ a 1 + ..-+a/1(2)}| f o r l < / < r , 4 = 5 = 0 otherwise, |{0^y<flil7sa 1 + .-+fli.1 + l(2)}| f o r l < / < r , 0 otherwise. In the case 1 < i < r, we can calculate A = a,+l (10) (11) for a, = ^ + • • •+a,! (2) = 1 (2), otherwise, / - $ = & , : Furthermore, we define yp+Kp+i(z~l) = ykJ(z), yp+Ki{z-l) = yk>p+i{z), CP+k,P+i(z~x) = Ck,i{z)> ( (DG(z),DG(z),...,DGn(z)y ) Hi% JQ+sp y v ^O+sp^ Jl+spy -Jl+sp \ (»o^-^DG(z-^...,DG(z->)y Jp-+sp •Jp-l+sp y 10 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS r(z) = ( ru00 riAzA( (TkAzy)o<kj<p (rk,p+i(zy)o<k,i<P \Y p+k, p+i \z))o<k, i<p 7 M = (ziM zu(z)l=((CkAz))o<k,i<P (Ck,P+i(z))o1(z) z2j2(z)J {(Cp+kAz))o,k,«p (Cp+Kp+Mo^pf Then the identities d 2 » = du(2-'), r„(z) = ru(-»), r21(z) = r ^ z " 1 ) , Z„(z) = Z u(z- 1), and Z2 j(z) = Z u(z _ 1) hold and (10) becomes ds(z) = r(z)ds(z)+Z(z)ds_1(z), or, fortnally, di(z)-((I-r(z))-1Z(z))dJ_1(z). Since the quadratic matrix T(l) consists of four quadratic p x p -blocks that are lower triangle matrices with zero diagonal, it is an easy exercise to show that I-T(l) is invertible. Hence, (I - T(z)) is invertible in a neighborhood of z = 1. Call P(Z)(/) •= det(/I- 0(z)) the characteristic polynomial of the matrix ®(z):=(I-r(z))-lZ(z). Then, by the theorem of Cayley-Hamilton, P(Z)(0(z))=O. From this, we see that the sequence (DGi+ (z))s>o satisfies a linear homogeneous recursion. Finally, it follows from the definition of FandZ that P(z)(/) = P(z_i)(/), from which we see ihatai(z) = ai(z-1). D Let 4(ZX 1^/^2/7, denote the roots of the polynomial P(2)(/), where z varies in a suffi-ciently small neighborhood of z = 1. Since ai(z-l) = ai(z), they satisfy 4(z_1) = 4( z)- Further-more, there exist functions Bkfi(z, s) that are polynomials in s such that A%+„() = 2X/(vM()'. (12) 7 Since-Dbk+ (1) = Gfc+Jp ~/^af-1(af ), it might be expected that (locally around z-X) there exists a unique root Ax(z) (satisfying ^(1) = a{) of maximal modulus which is simple. The following lemma shows that this is true if assumption 6 in Section 3 holds. Lemma- 2: Suppose that assumptions 1-6 in Section 3 hold and let v:= max{l</ - ^0}. Then, with the above notation, the 2p roots of P(i)(/) are af, 1 < / < r, where au 1 < / < r, denote the roots of zr - YI^\ajZr~J, 0 with multiplicity 2p-r-v, and 1 with multiplicity v. Proof: From A?^(1) = G+J, = I , / ? , ^ we see that af surely are roots of P(i>(/)-Since I-T(l) is invertible, the multiplicity of 0 is 2p minus the rank of Z(l). Z(l) has a simple block structure. It is an easy exercise to show that its rank equals r + v. (Recall that ^ + hj = at and ht - ht - bt.) Similarly, the multiplicity of 1 is 2p minus the rank of 1998] 11 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS K = ( K U KL K;; K^i-rcD-zd). Observe that r k | K U K u^j_ ^ fK u+K 1 } Kl2 K u and that K u + K12 (resp. K u - K 1 2 ) are cyclic matrices with entries 1, -al9 ...,-ar,0,...,0 (resp. 1, -bx,..., -br, 0,..., 0). By [3, Lemma 3], the rank of Kx?1 + KX 2 is p (resp. the rank of K x J - K J 2 is p-v), v being equal to the number of different /? roots of unity that are roots of zr - Sy=i bjZr~J. Thus, rk K = 2p - v, which completes the proof of the lemma. D Let us define discrete random variables Xm by J>(Xm = k) = ^ . (13) (Recall that dm(k): = |{0n\A. n> n-A.r>'n\2 From here on, we will assume 1-6 in Section 3. Lemma 3: Let Ax(z) be the unique root of maximal modulus of P(/)(z). Then we have A"(l) ^ 0, A"(X\ uG :=EXG =0(1) and a% :=VXG = « £ + 0(l) ^^k+sp Uk+sp V ' °ifc+p Uk+sp A (I) as 5 -> oo. Furthermore, if A"(l) 0, then Eexp ( Xa. ~Ua. \ k+sp ^k+sp lt~ \ ^k+sp t2[, .„( 1 = c x p | ~ l l + 0 O T as s—> oo. This means that Xam is asymptotically Gaussian with mean juGm and variance aGm. Proof: Let A(z) = ^(z) and Bk(z) = Bkl(z, s) in (12) (where the s-degree of the polynomial Bkl(z9 s) is zero). Since A'(I) = 0, we obtain from (12) by differentiation, DGkj) = Bk(i)A(iy+o((A()rn D'G^ (1) = Bk()A(\y Jg + O {{A{)YY), D^Jl) = Bk(l)A(iy(s^+^yO((A(l)yy), with some 0 < y < 1 properly chosen. From Z)Gjfc+j (1) = Gk+sp9 we get 12 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS W), Both Dbk+sp(l) andDgk+sp(l) are real, and because of Bk(l) = B^f'1 eR+, B'k() is real. Further-more, A"{) and 2%'(1) are real, too. From this, we obtain that EXo^=^(l+°(rs)) = o(i), VXr. Jk+sp f sA"()^Bll(l) (W)^ \ U (i+0(rs)) = s^-+o(i), A{) Bk{) {Bk(l) from which it is clear that A"() > 0. Using A'() = ^'"0) = °, we get A(e'y = Aiiyexp^^&y + Oist)). Now suppose ^4"(1) > 0, then we have where the 0-constants are independent of A : . For any fixed t, we get r4 + l Eexp it-Jk+sp ' ^k+sp Jk+sp J Jk+sp v G, exp k+sp -it- ^ k+sp Or Jk+sp J —"-Tjl,+0l7?JJ-Thus, by Levi's theorem (see ), the normalized random variables (Xom - HGJI &Gm converge weakly to normal distribution. Remark: The use of generating functions for the proof of asymptotic normality probably started with Bender's paper . Further references can be found in . Now we will turn our attention to Xm9 where m need not be an element of the basis G. Theorem 2: Suppose that G = (G„) satisfies a linear recursion with restrictions 1-6 of Section 3. Then, with the above notation, we have EXm = 0(l) and VXm=t^ + 0(t), P AQ) Xm being defined as in (13) and / being the length of the digital expansion of m. If A"(X)> 0, then Xm is asymptotically Gaussian with mean value EXm and variance ¥Xm -clogm for some constant c>0, i.e., 2K J-CO 1998] 13 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Remark: The special case of G„ = Fn+l (which leads to the original Zeckendorf representation) was discussed in . There are also recent contributions to similar questions, e.g., Dumont and Thomas prove asymptotic normality for substitution sequences by a different method, and Barat and Grabner show the existence of a limiting distribution of G-additive functions. Proof: Let m = S/=i ^G,- be the digital expansion of m. Iterated use of equation (7) yields, for 1 < j < /, i < ej9 and a < Gjy ( i (. i-^hGh+iGJ+a = (,mod2)S(Gt) + (-l)"5j 2>A+'G>+" = (sl mod 2)S{Gl) + (-1)' (s M mod 2)5(GM) + • • • + (-1)'+'"+^2 {sj+l mod 2)S(Gj+l) + (-l)£'+':-+s^ (i mod 2)S(Gj) + (-iyi+'"+€ j+l+i S(a) = X (-l)€ l+'"+£p+l (ep mod 2)S(Gp) + ( - l ) ^ + - + ^ (i mod 2)S(Gj) + (-if+'"+8'+l+i S(a), and from (4) we see that I dm(k) = 0<a<^siGk\S(a) = k I Bj-\ = 1 1 y=i /=o \0<a<G, Y^e^ + iGj+aUk] VW+i ) J -EI ;=1 ;=0 0 / + •••+,+,+/ Y| k - Z(-1)"""""'1 S(Gp)- (-l)/+'"+^+1 (i mod 2)S(Gj) P=j+i „-l(2) /J / / , - l y=i /=o f £, + -+SJ+l+i w k- i(-l)€ l+'"+€ p+lmp-(-rfl+-+€ J+l(imod2)mj P=j+\ and AreZ y=l /=0 keZ £l+-+£j+l+i\ \ \ k - £(-l)/+""+Vl^(G,) - (-l)'+"+^(i mod 2)S(Gj) v «p-i(2) JJ ;=1 |=o fceZ (-l)'+-+;+'+,fc+ 2(-l)'+""+ir|'+,^+(-l)ff'+""y+1(/mod2)»i>/ 14 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS =1; £(-ir'+"~''« y^" 1^' +£^'('mod2)'"-')JD (z((-i)"+""M'M+')) in which = 1 1 ^ 1 Vi ;=1 7=0 6(7,0 = i;.(-l) e , +"^ ,i» p + (-l)" +"^ ,(/mod2)« y, .«1(2) (14) c(7,o=(-ir/+"^+,+'-Differentiation of (14) yields £()=I i(6a,'yu-0AJ,(^a-0)+^')^(^/'0K/.»K(y,')). y=i <=o zj-(zD>m(z)) = Z X(C/.i)2zbO'0DG(zc^) + 2b(j,i)zb^ DG(z«^)c<J, tyV <% ,=1 ,=0 + z bu- '\zcU- °DG (zcU- °)+z2c°- > D g (zc(A °))). It is an easy exercise to show Hl J=l(l-j + l)kGj^CkGt. Because the ntj are bounded, we get b(j, i) = 0(1- /' + l) (uniformly in i) and ^ 0 ) = Z £ (bU,i)DGa) + c(j,i)DGj(l)) j=\ 1=0 7 / O I(/-7 + l)G y 0(G,) = 0(rn) and < f c woo) « ^ s,-i = II(O-,0 2AJ/I)+2C/,I-K/,0^(I)+^0)+^ /(I)) lz=l y=l ,'=0 / A"() l A"() P A() P A() + 0{m) mj-m+o(m)-1998] 15 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Thus, we have EXm = 0(l) and VXm = -L^-+0(l). Furthermore, by using (14), we obtain ( A.(^-)=S«p ;=i \ i V ^=1(2) x E e x p j ^ ( - l ) ^ it r n f /+...+o 1+;-and for any fixed t, A ^ P ^ - i ) ' ' "«')) = A , exp ( A = Gj exp (M())) 1 + 0 ' l \Jf; -i-iy , Gj = G/-'2/2exp ftlzl + a-L. 2 I ,yfj, and |U(^(-l)--+-Omod2H)/£(l + o(^) •=<j\1+°\ziT l h ^ e x p f o r 1 II V7 where the (9-constants do not depend on / ory. Thus we get, for 0 < & < y, 77 r (-;»<;<! I W U •\ylJJ) i<y</-/» = i _ I j A e x p ( O ( / » - ) + O ( ^ f ) ] + 0 ( G l w , ) = Z W l + 0(('-)j + 0(oi-',)= 2«A+0("""" 1)+0(ai-") /-/9<;</ /-/d<;</ and, finally (for any fixed t), 16 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Eexpl itXm Mm) = Dm{elt/CTm)Qxp(-it^ 1 o-m m I <j„ |Vl + 0(/H)) ( I and Xm is asymptotically Gaussian with mean iim and variance a2 m. D The condition that zr-Z|=1&^r-/ (where v = max{l<i <r\bt ^ 0}) is a product of zr~v and different cyclotomic polynomials is rather restrictive in the case in which Gn < 2Gn_x for n > 1. Proposition 5: Suppose that G = (Gw) satisfies a linear recursion with restrictions 1-5 of Section 3 such that Gn < 2Gn_x for n > 1. Then zr - J?i=l b^1"1 is a product of zr~v and different cyclo-tomic polynomials, where v = max{l <i 1 we are in the first case. So let us assume ax = 1, then it follows that ay e {0,1}, ar = 1, and therefore v = r. From this, we see that zr - E[=i btzr~l must be a symmetric polynomial that yields at - ar_t for all l<i'<r. Now suppose ax =--- = a/1 = l = ar =-«- = ar/+1 and ax=0 = ar_f forsomel<i< r-i. Then, by assumption 3 in Section 3, we have that Gn_rM > YIj=r_iJ>rlapn_j = EJ=r_/+1 Gn_j for n>r or, equivalently, that Gn >Sy=iGn_j for n>i. Because Gn - Zy=i<pn-j for « > r, it follows that TJj^+idjG^j >Gn_j forn>r. On the other hand we have, again by assumption 3, that G„, > YJj=i+iafGrJ-j f°r n> r> fr°m which we see that Gn = Zy=\ai+JGn_j for n> r-i, a contradiction to assumption 4. Now let r = 1 andax = 2, then v = 0 and B(z)-z. Finally, suppose ax = a2 = ••• = ar = 1. Then 6/ = (-iy+1 and ,=o z + l is of the desired type. • 4,2 Unbounded S(Gn) Proposition 6: If S(Gn) is unbounded, then there exists some a with \l, real numbers \(0'a)") 1=1 for some y e(0,1). Proof: Since S(Gn) satisfies the linear recurrence of Proposition 2, this representation fol-lows immediately. D 1998] 17 THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Theorem 3: Suppose that G = (G„) satisfies a linear recurrence as above such that S(Gn) is unbounded. Then l i m s u p i o g ( l ^ ) l ) = i o g g m^oo logm logal Proof: First, it follows from Proposition 6 that r log(\S(m)\) ^ v log(\S(Gn)\) log a hm sup , v /u > lim sup , =-r-g—. w^oo logm m_>^ logG„ loga^ The upper bound follows from the second part of Proposition 2 and again by an application of Proposition 6: Let m = Zy=i £jGj be the proper digital expansion of m and let C, K > 0 be large enough so that |/?,(w) + /?/(w)| < Cw^ for all w, i. Then we have, for / -» oo, log(l^)l) ^ lQgK=il^(G/)l) ^ log(/a'(C/D + Cy)) logw logO/G,) log^+logG, 2Gn for all n > 1, then the answer is very easy (see Proposition 1). Therefore, we restricted ourselves to the case Gn+l < 2Gn for all n > 1. Here juG(n) = (~l)SG(n). Thus, fdG{n) 0 for all n > 0 and MG(N) = SG(N) is exactly the difference between the number of n < N with juG{ri) = 1 and the number of n< N with juG(ri) = - 1 . In the case of linear recurring sequences G = (Gn) (satisfying certain natural conditions), we proved that in any case MG(N)~o(N), i.e., -1,+ 1 are asymp-totically equidistributed. More precisely, we discussed the distribution of values of SG(N) (which can also be con-sidered in the case G„+1 > 2G„). It turns out that there are two essentially different cases, the case of bounded SG(G„) and the case of unbounded SG(Gn). If SG(G„) k unbounded, then SG(N) has an almost fractal structure (see Theorem 3 and the Remark following it). However, if SG(Gn) is bounded for all suitable initial conditions of G, then the values SG(N) admit a Gaussian limit law in the following sense: If Xn is a random variable defined by J>(XN=k) = jj{r,<N\SG(n) = k}\ then XN is asymptotically Gaussian with bounded mean value and variance VXN -clogN, pro-vided that c 0 (Theorem 2). 18 [FEB. THE PARITY OF THE SUM-OF-DIGITS-FUNCTION OF GENERALIZED ZECKENDORF REPRESENTATIONS Since SG(Gn) satisfies the linear recurrence (2), it follows that SG(Gn) Is periodic (for suffi-ciently large ri) if it is bounded. This can only occur for all suitable initial conditions of G if and only If the roots of the characteristic polynomial B(z) = zr-'Zr JsslbjZr~J of (2) are 0 or roots of unity. Therefore, the assumption on B(z) in Theorem 2, this is assumption 6 in Section 3, is quite natural. Finally, we want to recall that the only recurring sequences G = G(n) satisfying assumptions 1-5 such that ax = 1 (I.e., Gn+l <2Gn) and that B(z) is the product of zr~v and cyclotomic poly-nomials are generalized Fibonacci numbers (Proposition 5). They satisfy a recursion of the form Gn = Gn_x + '-+Gn_r. Here Theorem 2 applies. Hence, the values of MG(N) with respect to generalized Zeckendorf representations satisfy a central limit law. REFERENCES 1. G. Barat & P. J. Grabner. "Distribution Properties of G-Additive Functions." J. Number Th. 60(1996):103-23. 2. E. A. Bender. "Central and Local Limit Theorems Applied to Asymptotic Enumeration." J. Combin. Theory Ser. A 15 (1973):91-111. 3. M. Drmota & M. Skalba. "Relations between Polynomial Roots." ActaArith 71 (1995):65-77. 4. M. Drmota & M. Skalba. "The Parity of the Zeckendorf Sum-of-DIglts-Function." Manu-script, 1995. 5. M. Drmota & M. Sorla. "Marking In Combinatorial Constructions: Generating Functions and Limiting Distributions." Theor. Comput. Set 144 (1995):67-99. 6. J. M. Dumont & A. Thomas. "Gaussian Asymptotic Properties of the Sum-of-Digits Func-tion." J. Number Th. 62 (1997): 19-38. 7. W. Feller. An Introduction to Probability Theory and Its Applications. Vols. I and II. New York: Wiley, 1966. 8. P. J. Grabner & R. F. Tlchy. "Contributions to Digit Expansions with Respect to Linear Recurrences." J. Number Th.. 36 (1990): 160-69. 9. P. J. Grabner & R. F. Tichy. "a-Expansions, Linear Recurrences, and the Sum-of-Digits Function." ManuscriptaMath. 70 (1991):311-24. 10. J. H. Van Lint & R. M. Wilson. A Course in Combinatorics. Cambridge: Cambridge Uni-versity Press, 1992. 11. H. M. Morse. "Recurrent Geodesies on a Surface of Negative Curvature." Trans. Amer. Math. Soc. 22 (1921):84-100. 12. A. Petho & R. F. Tichy. "On Digit Expansions with Respect to Linear Recurrences." J. Number Th. 33 (1989):243-56. 13. R. P. Stanley. Enumerative Combinatorics. Vol. I. Monterey, CA: Wadsworth & Brooks/-Cole Advanced Books & Software, 1986. AMS Classification Numbers: 11A63, 11B39 1998] 19 |
11851 | https://www.mathadventures.org/curriculum/The_Fibonacci_Sequence_and_the_Golden_Ratio_Worksheet.pdf | Name: ___ Math Adventures Week 4: The Fibonacci Sequence and the Golden Ratio In the 1200s, an Italian mathematician named Leonardo Fibonacci asked the question, “If we start with 2 newborn rabbits in a pen, how many rabbits will there be after 1 year?” Rabbit Rules: ● Start with 1 pair of newborn rabbits.
● When the rabbits are 1 month old, they are teenagers. Teenagers cannot have children.
● When the rabbits are 2 months old, they are adults. A pair of rabbits will give birth to 1 pair of newborn rabbits the same month they become adults.
● A pair of adult rabbits gives birth to 1 pair of newborn rabbits each month.
● Rabbits never die. Once a rabbit becomes an adult, it remains an adult forever.
1. Suppose there are 5 pairs of newborn rabbits in a given month.
a.
How many pairs of adults must there be in that same month?
b. How many pairs of teenagers will there be in the next month?
2. Suppose there is 1 pair of newborn rabbits in December (Month 0).
a.
How many pairs of each type of rabbit will there be in January (Month 1)?
i.
Newborn: ii.
Teenager: iii.
Adult: b. What about February (Month 2)?
i.
Newborn: ii.
Teenager: iii.
Adult: 1 MathAdventures.org 3. Starting with 1 pair of newborn rabbits born in Month 0, fill out the chart below.
Month 0 1 2 3 4 5 6 7 8 9 10 11 12 Newborn 1 Teenager 0 Adult 0 Total 1 The sequence of numbers in the bottom row that represents the total number of pairs of rabbits is called the Fibonacci sequence!
4. How is the total number of pairs of rabbits in Month 2 related to the total number of pairs in Month 0 and Month 1?
5. How is the total number of pairs in Month 5 related to the total number of pairs in Month 3 and Month 4?
6. How is the total number of pairs in any month related to the total number of pairs in the previous months?
The Fibonacci sequence is also seen in other instances of nature! For example, the spiral of the nautilus seashell is related to the Fibonacci sequence.
2 MathAdventures.org 7. In the diagram below, assume that the side lengths of the smallest, inside squares (A and B) are 1 cm long. Find the side lengths of the rest of the squares by starting from the center and working outward. Fill out the table below with the side lengths.
A B C D E F G H 1 1 8. What do you notice about the side lengths?
9. Calculate the ratios of successive terms in the Fibonacci sequence using the table above. Round to the nearest hundredth.
_ _____ _ _ 𝐵 𝐴= 𝐶 𝐵= 𝐷 𝐶= 𝐸 𝐷= _ _ _ 𝐹 𝐸= 𝐺 𝐹= 𝐻 𝐺= 10. What number do your ratios seem to approach?
3 MathAdventures.org 11. In the pentagram (5-pointed star) below, measure the lengths of A, B, C, and D in centimeters to the nearest tenth of a centimeter.
A: _ cm B: _ cm C: _ cm D: _ cm 12. Calculate the following ratios using the lengths you found above. Round to the nearest tenth.
Then, find the average of your three ratios.
_ _ _ Average = _ 𝐵 𝐴= 𝐶 𝐵= 𝐷 𝐶= The ratios found in the nautilus, the pentagram, and many other shapes and patterns in nature are all a number called phi. The ratios of successive terms in the Fibonacci sequence approach phi as well. Phi is represented by the Greek symbol Φ and is also known as the golden ratio.
Rounded to the nearest thousandth, Φ = _.
13. Draw the most beautifully proportioned rectangle you can in the space below.
4 MathAdventures.org The Ancient Greeks thought the following rectangle was the most beautiful rectangle of all, and used it frequently in their architecture. What makes it so special? Let’s find out!
14. Measure the length and width of the rectangle to the nearest tenth of a centimeter.
Length: _ cm Width: _ cm 15. Calculate the ratio of the length to the width. Round to the nearest tenth.
= __ 𝑙𝑒𝑛𝑔𝑡ℎ 𝑤𝑖𝑑𝑡ℎ The golden rectangle is a rectangle whose sides are in the golden ratio ( ). If we cut a 𝑙𝑒𝑛𝑔𝑡ℎ 𝑤𝑖𝑑𝑡ℎ= Φ square from one side of a golden rectangle, we get another golden rectangle.
5 MathAdventures.org Lesson Summary The Fibonacci sequence was developed by Leonardo Fibonacci in trying to answer the question, “If we start with 2 newborn rabbits in a pen, how many rabbits will there be after 1 year?” ● There will be 233 rabbits after 1 year.
Each term in the Fibonacci sequence is the sum of the previous two terms.
The Fibonacci sequence is also seen in other instances in nature: ● The lengths in the spiral of the nautilus seashell ● Plant growth ○ Seeds ○ Flower petals ○ Pinecones The ratios found in the nautilus, the pentagram, and many other shapes and patterns in nature are all a number called phi. The ratios of successive terms in the Fibonacci sequence approach phi as well. Phi is represented by the Greek symbol Φ and is also known as the golden ratio.
Rounded to the nearest thousandth, Φ = 1.618.
The golden rectangle is a rectangle whose sides are in the golden ratio ( ). The Ancient 𝑙𝑒𝑛𝑔𝑡ℎ 𝑤𝑖𝑑𝑡ℎ= Φ Greeks thought the golden rectangle was the most beautiful rectangle of all, and used it frequently in their architecture. If we cut a square from one side of a golden rectangle, we get another golden rectangle.
References: Olga Radko Endowed Math Circle archive, TeachEngineering 6 MathAdventures.org |
11852 | https://people.richland.edu/james/lecture/m116/logs/logs.html | 4.2 - Logarithmic Functions and Their Graphs
Inverse of Exponential Functions
We stated in the section on exponential functions, that
exponential functions were one-to-one. One-to-one
functions had the special property that they have inverses
that are also functions. And, as many of you said in class,
and I'm so glad you remember, one-to-one functions can
be applied to both sides of an equation. They also pass a
horizontal line test.
This section is about the inverse of the exponential
function. The inverse of an exponential function is a
logarithmic function. Remember that the inverse of a
function is obtained by switching the x and y coordinates. This reflects the
graph about the line y=x. As you can tell from the graph to the right, the
logarithmic curve is a reflection of the exponential curve.
The table below demonstrates how the x and y values of the points on the expontential
curve can be switched to find the coordinates of the points on the logarithmic
curve.
| Point on exponential curve | Corresponding point on logarithmic curve |
--- |
| (-3, 1/8) | (1/8, -3) |
| (-2, 1/4) | (1/4, -2) |
| (-1, 1/2) | (1/2, -1) |
| (0, 1) | (1, 0) |
| (1, 2) | (2, 1) |
| (2, 4) | (4, 2) |
| (3, 8) | (8, 3) |
Comparison of Exponential and Logarithmic Functions
Let's look at some of the properties
of the two functions.
The standard form for a logarithmic function is: y = loga x
Note, if the "a" in the expression above is not a subscript (lower than
the "log"),
then you need to update your
web browser.
| | Exponential | Logarithmic |
---
| Function | y=ax, a>0, a≠1 | y=loga x, a>0, a≠1 |
| Domain | all reals | x > 0 |
| Range | y > 0 | all reals |
| intercept | y = 1 | x = 1 |
| increasing | when a > 1 | when a > 1 |
| decreasing | when 0 < a < 1 | when 0 < a < 1 |
| asymptote | y-axis | x-axis |
| continuous | yes | yes |
| smooth | yes | yes |
Working Definition of Logarithm
In the exponential function, the x was the exponent. The purpose of the inverse of a function is
to tell you what x value was used when you already know the y value. So, the purpose of the
logarithm is to tell you the exponent.
Thus, our simple definition of a logarithm is that it is an exponent.
Another way of looking at the expression "loga x" is "to what power (exponent) must a be raised
to get x?"
Equivalent Forms
The logarithmic form of the equation y=loga x is equivalent to the exponential form x=ay.
To rewrite one form in the other, keep the base the same, and switch sides with the other two
values.
Properties of Logarithms
loga 1 = 0 because a0 = 1
: No matter what the base is, as long as it is legal, the log of 1 is always 0.
That's because
logarithmic curves always pass through (1,0)
loga a = 1 because a1 = a
: Any value raised to the first power is that
same value.
loga ax = x
: The log base a of x and a to the x power are inverse functions. Whenever inverse
functions are applied to each other, they inverse out, and you're left with
the
argument, in this case, x.
loga x = loga y implies that x = y
: If two logs with the same base are equal, then
the arguments must be equal.
loga x = logb x implies that a = b
: If two logarithms with the same argument are
equal, then the bases must be equal.
Common Logs and Natural Logs
There are two logarithm buttons on your calculator. One is marked "log" and
the other is marked
"ln". Neither one of these has the base written in. The base can be determined,
however, by looking at the inverse function, which is written above the key and
accessed by the 2nd key.
Common Logarithm (base 10)
When you see "log" written, with no base, assume the base is 10.
That is: log x = log10 x.
Some of the applications that use common logarithms are in pH (to measure acidity), decibels
(sound intensity), the Richter scale (earthquakes).
An interesting (possibly) side note about pH. "Chapter 50: Sewers" of the Village
of Forsyth Code requires forbids the discharge of waste with a pH of less than
5.5 or higher than 10.5
(section 50.07).
Common logs also serve another purpose. Every increase of 1 in a common logarithm
is the result of 10 times the argument. That is, an earthquake of 6.3 has 10
times the magnitude of a
5.3 earthquake. The decibel level of loud rock music or a chainsaw (115 decibels
= 11.5 bels) is 10 times
louder than chickens inside a building (105 decibels = 10.5 bels)
Natural Logarithms (base e)
Remember that number e, that we had from the previous section? You know, the one that was
approximately 2.718281828 (but doesn't repeat or terminate). That is the base for the natural
logarithm.
When you see "ln" written, the base is e.
That is: ln x = loge x
Exponential growth and decay models are one application that use natural logarithms. This
includes continuous compounding, radioactive decay (half-life), population growth. Typically
applications where a process is continually happening. Now, these applications were first
mentioned in the exponential section, but you will be able to solve for the other variables
involved (after section 4) using logarithms.
In higher level mathematics, the natural logarithm is the logarithm of choice. There are several
special properties of the natural logarithm function, and it's inverse function, that make life much
easier in calculus.
Since "ln x" and "ex" are inverse functions of each other, any time an "ln" and "e" appear right
next to each other, with absolutely nothing in between them (that is, when they
are composed with each other), then they inverse out, and you're left with
the argument. |
11853 | https://faculty.etsu.edu/gardnerr/Set-Theory-Intro/Notes-Halmos/Halmos-Section-1.pdf | Section 1. The Axiom of Extension 1 Section 1. The Axiom of Extension Note. In this section, we define the equality of two sets in the Axiom of Exten-sion. We consider subset inclusion and briefly mention the properties of reflexive, symmetric/antisymmetric, and transitive.
Note. As with the axiomatic development of geometry, we leave certain terms undefined (they are “primitives”) and instead give their relationships to other un-defined objects. The properties of these objects are given by the axioms and result-ing theorems. For more on this approach, see my online notes for ”Introduction to Modern Geometry” (MATH 4157/5157) on Section 1.3. Axiomatic Systems. Our fundamental undefined terms are set and element.
Note. Informally, we use the term set to indicate a “collection’ or “family” of objects called elements or “members.” For example, in geometry a line is a set of points. In addition, a plane is a set of lines (so that a plane it a set of sets).
Symbolically, if element x belongs (or is contained in) set A, we write x ∈A.
Note. We want to put a relation of equality on certain sets. If sets A and B are equal, then we write A = B. If they are not equal, we write A ̸= B. More formally, equality is defined in our first axiom.
Section 1. The Axiom of Extension 2 Axiom of Extension.
Two sets are equal if and only if they have the same elements.
Note. Halmos illustrates the Axiom of Extension by considering an analogous setting where the Axiom of Extension does not hold. Suppose we consider people instead of sets. For x and A people, we write x ∈A whenever x is an ancestor of A. The analogous claim to the Axiom of Extension would be that two people are equal if an only if they have the same ancestor. This need not be true, however.
When “two” people are equal then they have the same ancestors (the “only if” part), but having the same ancestors does not imply the two people are necessarily equal (i.e., the same; they could be siblings). That is, the “if” part does not hold.
Definition. If A and B are sets and if every element of A is an element of B, the A is a subset of B, or B includes A, denoted A ⊂B or B ⊃A. If A ⊂B and A ̸= B then A is a proper subset of B.
Note. We do not have a notational way to distinguish between subsets and proper subsets. In other sources, “A is a subset of B” may be denoted A ⊆B and “A is a proper subset of B” may be denoted A ⊊B. More confusingly, “A is a proper subset of B” may be denoted A ⊂B (the same notation we use for “subset”), so be careful when reading different texts as to which notation they use.
Section 1. The Axiom of Extension 3 Note. Notice that for every set, A ⊂A; that is, subset inclusion is reflexive. If A, B, and C are sets such that A ⊂B and B ⊂C, then A ⊂C; that is, subset inclusion is transitive. Notice that set equality is also reflexive and transitive.
Note/Theorem 1.A. If A and B are sets such that A ⊂B and B ⊂A, then A and B have the same elements and therefore, by the Axiom of Extension, A = B.
Note. The result of Theorem 1.A shows that subset inclusion is antisymmetric (as is ≤on the set of real numbers, for example). Set equality is symmetric, since A = B implies that B = A.
Note. We can reword the Axiom of Extension in terms of subsets as: If A and B are sets, then a necessary and sufficient condition that A = B is that both A ⊂B and B ⊂A. You will repeatedly use this in your mathematical career when showing the equality of two sets.
Revised: 1/2/2023 |
11854 | https://bicycleinfrastructuremanuals.com/manuals7/ECF-Geometric-Design-Parameters-for-Cycling-Infrastructure-2022.pdf | Geometric design parameters for cycling infrastructure ECF gratefully acknowledges financial support from the cycling industry via Cycling Industries Europe ECF gratefully acknowledges financial support from the LIFE Programme of the European Union www.ecf.com Publishing credits Author Aleksander Buczyński, Policy Officer – Infrastructure, a.buczynski@ecf.com Editors Cian Delaney Thomas Delrive Design Omer Malak © European Cyclists’ Federation, February 2022 This report is also available online at www.ecf.com EU CYCLE The project has been co-financed by the European Union European Regional Development Fund (ERDF) and made possible by the INTERREG EUROPE Programme. The information in this document only reflects the author's views. The Interreg Europe programme authorities are not liable for any use that may be made of the information contained therein. ECF | Geometric design parameters for cycling infrastructure 3 Contents Introduction .......................................................................................................................................... 4 1. Design speed ................................................................................................................................ 5 2. Minimum horizontal curve radius ................................................................................................... 7 3. Minimum vertical curve radius ......................................................................................................11 4. Minimum sight distances ..............................................................................................................14 5. Recommendations .......................................................................................................................17 6. Analysed standards and guidelines ..............................................................................................19 6.1 Austria ..................................................................................................................................21 6.2 Belgium.................................................................................................................................22 6.3 Bulgaria ................................................................................................................................23 6.4 Croatia ..................................................................................................................................24 6.5 Czechia.................................................................................................................................25 6.6 Denmark ...............................................................................................................................25 6.7 Finland ..................................................................................................................................27 6.8 Germany ...............................................................................................................................29 6.9 Greece ..................................................................................................................................30 6.10 Ireland ..................................................................................................................................30 6.11 Netherlands ..........................................................................................................................32 6.12 Poland ..................................................................................................................................33 6.13 Slovakia ................................................................................................................................34 6.14 Spain (Catalonia) ..................................................................................................................35 6.15 UK ........................................................................................................................................37 7. Final remarks ...............................................................................................................................38 ECF | Geometric design parameters for cycling infrastructure 4 Introduction The geometry of cycling infrastructure is a critical aspect of its safety and comfort. Bicycles need to maintain a certain speed to keep their balance. They travel in curves and cannot change direction at straight angles. Sharp turns make it difficult for cyclists to stay on track or maintain stability, and difficult for other road users to predict the bicycle trajectory. This can lead to falls, “run-off-the-road” accidents and collisions between cyclists, other vehicles or pedestrians. Unsuitable geometry can also exclude some user groups, in particular those who need dedicated cycling infrastructure the most, such as elderly cyclists and parents with children. Figure 1. Bicycles travel in curves and lean into the curves. In section 1 of this factsheet we look at the design speeds of cycling infrastructure adopted across different national or regional standards and guidelines, as well as factors affecting the choice of the design speed. Sections 2, 3 and 4 look at the key parameters determined by the design speed (horizontal/vertical curve radii and sight distances1) and then compare specific values given by different documents. Section 5 provides key recommendations that can be used as a reference if a national standard is missing or does not cover the specific parameter. For each of the countries and regions included in the comparisons, the relevant sections of the analysed documents are summarised in section 6. The focus of the factsheet is on the most common requirements in already existing national and regional regulations and guidelines. It should however be noted that there are also more in-depth, non-normative analytical models which consider more parameters.2 1 Visibility splays on crossings are also determined on basis of design speed, but they have been omitted in this comparison. We reckon it makes more sense to discuss these together with other factors influencing crossing design. 2 For example, “Analytical Geometric Design of Bicycle Paths” (Zain Ul-Abdin, Sarmad Zaman Rajper, Ken Schotte, Pieter De Winne, and Hans De Backer, 2020, considers also ratio of curvature for upcoming and previous road segments, and transition curves. ECF | Geometric design parameters for cycling infrastructure 5 1. Design speed Design speed defines how fast cyclists can travel along the route section without endangering their safety. High design speed means shorter travel times, and therefore increases the competitiveness of cycling. Consistent design speed reduces the need of braking and accelerating. Good view of the route ahead gives advance time to make decisions. National and regional standards vary design speed depending on the role of the route in the network and its location. Typically, it means 20 km/h for local and 30 km/h for main cycle routes, with even higher values (35-45 km/h) for cycle highways or outside built-up areas. Regardless of the route category, design speed needs to be increased on inclines/declines. On the other hand, some standards allow somewhat reduced design speed in the intersection area. 10-12 km/h is usually used as the lowest possible threshold; below that speed a standard two-wheeled cycle becomes unstable Table 1. Comparison of design speeds in national and regional standards and guidelines. Guidelines/ standard Lowest possible Side routes Main routes Inclines Increased for other reason Austria 20-30 km/h 30 km/h (3%) 40 km/h (6%) Belgium (Flanders) 20 km/h 30 km/h Bulgaria 20 km/h 30 km/h Croatia Czechia 10 km/h 20-25 km/h 30 km/h (3%) 40 km/h (6%) Denmark 30 km/h 35 km/h 36 km/h (3%) 40 km/h (5%) Finland 25-30 km/h 40 km/h +10 km/h 45 km/h base for cycle highways or if mopeds allowed ECF | Geometric design parameters for cycling infrastructure 6 Guidelines/ standard Lowest possible Side routes Main routes Inclines Increased for other reason Germany3 10 km/h 20 km/h 30 km/h 40 km/h Greece Not defined Italy 20-25 km/h 40 km/h for gradients >5% Ireland 10-12 km/h 30 km/h 40 km/h 50 km/h for slopes >5% longer than 150 m Netherlands 12 km/h 20 km/h 30 km/h 35-40 km/h 40 km/h outside built-up areas Poland 12 km/h 20 km/h 30 km/h Slovakia 10 km/h 25 km/h 40 km/h Spain (Catalonia) 20 km/h 30 km/h 50 km/h +2 km/h per 1% incline UK 20 km/h 30 km/h 40 km/h for gradients >3% 3 In ERA 2010 the speeds assigned to route categories are given as ranges of travel speeds, not design speeds. In the table, higher end of the travel speed range is given as an equivalent of the design speed; lower end of the lowest category is in the “lowest possible column”. ECF | Geometric design parameters for cycling infrastructure 7 2. Minimum horizontal curve radius In all design standards and guidelines, except for Denmark, minimum horizontal curve radius is clearly tied to design speed. Germany and Catalonia also consider the type of surface, with Catalonia additionally considering the cant (elevation of one side) of the curve. Nearly all standards agree that a curve radius between 20 and 25 m is required for a design speed of 30 km/h. Requirements vary more both for lower and higher speeds: between 8 and 15 m for 20 km/h and between 25 and 47 m for 40 km/h. Table 2. Comparison of requirements for horizontal curve radii in national and regional standards and guidelines. Guidelines/ standard Lowest possible (10-12 km/h or intersection) 20 km/h 30 km/h 40 km/h Notes Austria 4 m 8 m 22 m Belgium 3-4 m (Brussels) 15 m (Flanders) 35 m (Flanders) Bulgaria 4 m 15 m 25 m 45 m Up to 80 m for 50 km/h Czechia 2.5 m 8 m 22 m Width of the track must be increased by minimum: 0.25 m for R < 22 m, 0.5 m for R < 8 m Croatia 5 m (1 m) 10 m 17.5 m 25 m Denmark 16-105 m, not clearly tied to design speed Recommended tracing 210-360 m; for horizontal radii of less than 50 m a transverse inclination toward the bend’s centre may be necessary ECF | Geometric design parameters for cycling infrastructure 8 Guidelines/ standard Lowest possible (10-12 km/h or intersection) 20 km/h 30 km/h 40 km/h Notes Finland 10-20 m 25 m (25 km/h) 30 m 55 m Up to 130 m for 55 km/h (downhill sections of cycle highways) Germany 10 m 20 m 30 m 50-130% higher radii required for unpaved surfaces Greece Not defined Italy 3-5 m Not defined Curve radii should be suitable for design speed, but no formula/values given Ireland 3.6-4 m 8.4-10 m 16-25 m 20-25 m Up to 94 m for 50 km/h Netherlands 5 m 10 m 20 m 25 m (estimate) Increase of width by 0.5 m advised in bends Poland 5 m 10 m 20 m Formula for other design speeds given. Slovakia 2.5 m 8 m 22 m Spain (Catalonia) 10 m 24 m 47 m 70-80% higher radii required for unpaved surfaces UK 4 m 15 m 25 m 40 m In some documents, the growth of the required curve radius with design speed is given as linear or nearly linear (Germany, Netherlands, UK), in some it is quadratic (Catalonia, probably also Austria and Czechia, though no explicit formula is given in the latter cases). The difference is negligible for lower speeds but becomes significant above 25 km/h. Given the growing popularity of electrically power assisted cycles and the potential of speed pedelecs (reaching speeds up to 45 km/h) in rural areas, additional analysis or research could be desirable. ECF | Geometric design parameters for cycling infrastructure 9 Figure 2. Comparison of requirements for horizontal curve radii on asphalted cycle tracks in national and regional standards and guidelines. In the Catalonian formula we assumed p = 0.02 (transverse inclination of 2% towards the bend’s centre). To allow the full width of a cycle track to be effectively used, horizontal curve radius should be measured to the inner edge of the track. Figure 3. Lack of curves or insufficient curve radius can cause head-on collisions, or single-vehicle accidents as a result of collision avoidance manoeuvre. Several documents require or advise increased width of cycle tracks in bends. This is to accommodate the fact that cyclists lean into the curves (at higher speeds) or require additional space for balancing on the bike (at lower speeds). Unless stated otherwise, the radii are calculated for clean asphalt surfaces. Non-asphalted or poorly maintained surfaces require roughly 1.5-2 times higher curve radii because of lower friction coefficient. 0 5 10 15 20 25 30 35 40 45 50 55 60 10 15 20 25 30 35 40 Horizontal curve radius Design speed [km/h] Bulgaria Catalonia Croatia Czechia & Slovakia Finland Germany & UK Netherlands & Poland ECF | Geometric design parameters for cycling infrastructure 10 Figure 4. Bicycles need curves also to change direction at junctions. Not taking this into account results in cyclists travelling outside allocated space, potentially surprising other road users. Figure 5. Because of insufficient curve radius at the connection of cycle tracks, the cyclist joining from the crossing will enter the part of the cycle track meant for traffic in the opposite direction. ECF | Geometric design parameters for cycling infrastructure 11 3. Minimum vertical curve radius Vertical curves are applied to avoid sudden changes of gradient. They need to ensure comfort and stability of riding, so the bicycle wheel does not jump on concave “bump”, or crash into the opposite wall of a convex “hole”. Vertical curves, especially concave, have also an impact on sight distance. Figure 6. Vertical curves radii . Table 3. Comparison of requirements for vertical curve radii in national and regional standards and guidelines. Country 20 km/h 30 km/h 40 km/h Austria 20 m concave 10 m convex 40 m concave 20 m convex 65 m concave 40 m convex Belgium Not specified Bulgaria Not specified Croatia 40 m concave 25 m convex 80 m concave 50 m convex 150 m concave 100 m convex Czechia 20 m concave 10 m convex 40 m concave 20 m convex 65 m concave 40 m convex Denmark 175-580 m, not clearly tied to design speed ECF | Geometric design parameters for cycling infrastructure 12 Country 20 km/h 30 km/h 40 km/h Finland 225 m concave 50 m convex (25 km/h) 385 m concave 70 m convex 940 m concave 125 m convex Germany 40 m concave 25 m convex 80 m concave 50 m convex 150 m concave 100 m convex Greece Not specified Italy Not specified Ireland Not specified Netherlands Not specified Poland Not specified Slovakia 20 m concave 10 m convex 30 m concave 20 m convex 40 m Spain (Catalonia) 10 m 20 m 40 m UK 300 m concave 250 m convex Vertical curve radius represented the biggest variety in the analysed parameters. Half of analysed documents do not mention it at all, some set the minimum requirements relatively low (10-40 m), other go as high as 250-300 m (up to 1370 m in Finland). Wherever there is a distinction between requirements for concave and convex curves, higher minimum curve radius is required for concave. 4 This can be justified by the impact of concave curves on visibility of the route ahead. 4 Literal translation of some regulations or guidelines can suggest that it is not always the case (for example in Croatia), but the terms concave and convex are used differently in traffic engineering across European countries. The comparison takes these differences into account and normalises the terminology to the interpretation shown on the figure in the beginning of the section. ECF | Geometric design parameters for cycling infrastructure 13 Figure 7. This obstacle, hidden behind a vertical curve for the cyclists coming from the opposite direction, was the cause of a serious accident. Photo credit: Agnieszka Fabiańczyk. ECF | Geometric design parameters for cycling infrastructure 14 4. Minimum sight distances A cyclist should be able to see the road ahead of them enough to have time to react to obstacles or other users. The required sight distance calculations involve the reaction time and the time needed to decelerate. Table 4. Comparison of requirements for sight distances in national and regional standards and guidelines Country 20 km/h 30 km/h 40 km/h Notes Austria 15 m 25 m 40 m Belgium Not specified Bulgaria 25 m 35-45 m Croatia 30 m 40 m Czechia 15 m 25 m 40 m Denmark Not specified5 Meeting the curve radii criteria should ensure sufficient visibility Finland 26-34 m (25 km/h) 34-45 m 53-73 m Higher values required on downhill section Double the values for encounter sight distance Germany 15 m 25 m 40 m Greece Not specified Italy 10-18 m 18-31 m 29-46 m Values calculated for the maximum allowed friction coefficient of 0.35 5 A more general Handbook on Basic Design of Traffic Areas lists minimum stopping sight distances between 18 and 49 m, depending on the incline. ECF | Geometric design parameters for cycling infrastructure 15 Country 20 km/h 30 km/h 40 km/h Notes Ireland 15 m (10 km/h) 35 m 60 m (50 km/h) Should be increased by 50% on loose surfaces Netherlands 21-60 m 35-84 m 42-112 m Distinguishes between “sight distance in motion” and “stopping sight distance” Poland 21 m 45 m 40 m 75 m Required minimum Recommended minimum Slovakia 15 m 25 m 30 m Double values required on downhills >5% and unpaved sections Spain (Catalonia) 20-25 m 35-45 m 55-70 m Higher values required on downhill sections; add values for both directions on bidirectional tracks UK 17 m 31 m 47 m Table 5. Comparison of underlying assumptions for calculating the stopping sight distance (where explicitly given) Reaction time Deceleration Eye level Obstacle level Denmark 2 s 2 m/s2 Finland 2 s 2 m/s2 1.5 m 0.0 m Italy 1-2.5 s up to 3.4 m/s2 Ireland 2 s 1.5 m/s2 1.0-2.2 m 0.0-2.2 m Netherlands 2 s 1.5 m/s2 Spain (Catalonia) 2.5 s 2.5 m/s2 1.4 m 0.0 m UK 0.8-2.2 m 0.0-2.4 m ECF | Geometric design parameters for cycling infrastructure 16 Figure 8. Example of a curve with radius sufficient for design speed of 30 km/h, but no satisfactory sight distance. The opaque obstacle inside the curve limits forward visibility below of 35 m (minimum value from the Dutch manual) When evaluating the sight distance, a range of eye heights should be considered to accommodate different cyclists (children, adults) and different cycles (including recumbent). Specific values (0.8-2.2 m) are given for example by the UK guidelines. On bidirectional cycle tracks the visibility should be ensured for double stopping sight distance (or sum of stopping distances for cyclists coming from the opposite directions). If it is not possible, the directions of traffic should be clearly separated. Figure 9. A bidirectional cycle track split into two unidirectional to ensure sufficient curve radius and sight distance without cutting the tree. ECF | Geometric design parameters for cycling infrastructure 17 5. Recommendations The recommended quality criteria are grouped into three levels, following the classification of users developed in the frame of the EuroVelo European Certification Standard:6 • Essential: minimum requirements covering the most basic user needs. Can be applied if space is limited and the expected usage is not high. If the cycle infrastructure does not meet the Essential criteria, it is probably not safe and it definitely should not be obligatory to use. On this level, design speeds of 20 km/h and exceptionally 10 km/h are considered. When translating this design speed into minimum geometric parameters, we used the first quartiles from the analysed documents (which means that 75% of standards require higher parameters for the same design speeds). • Important: addresses the needs of most regular and occasional users and is sufficient for typical cycle routes. On this level, a design speed of 30 km/h is assumed and median values from the analysed guidance documents are given as recommended. • Additional: covers the widest range of cycle users, including those travelling on road bikes, electrically assisted pedal cycles, velomobiles, handbikes, tandems and bikes with trailers. It is an equivalent of a cycle highway in most guidelines, and can be considered the aspiration level for main, future-proof functional cycling connections. When translating the design speed into geometric parameters for this level, the third quartiles from the analysed documents are listed as recommended. Figure 10. Additional criteria accommodate widest range of users and can be considered the aspiration level for main, future-proof cycling connections 6 ECF | Geometric design parameters for cycling infrastructure 18 Table 6. Summary recommendations for geometric design parameters. Criteria level When to apply Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance Essential Exceptional cases: when changing direction of travel, for example on an intersection or at a junction of cycle tracks 10 km/h 3 m 20 m concave 10 m convex 15 m Local cycle routes with low expected usage 20 km/h 10 m Important Most typical cycle routes 30 km/h 22 m 60 m concave 35 m convex 35 m Additional Main, future-proof functional cycle routes Downhill sections on all routes 40 km/h 45 m 190 m concave 110 m convex 57 m ECF | Geometric design parameters for cycling infrastructure 19 6. Analysed standards and guidelines The analysed documents include standards and guidelines with varying level of detail and scope. They come from countries and regions with varying climates, level of cycle use, population density and topography. In the following subsections we provide short reviews of each of them, pointing out their unique features. Table 7. Standards and guidelines included in the comparison. Country Document Publication year Link Austria RVS 03.02.13 Radverkehr (Guidelines and regulations for roads: cycle traffic) 2011 Available here (in German) Planungsleitfaden Radverkehr (Planning guide cycle traffic) 2007 Available here (in German) Belgium Vademecum Fietsvoorzieningen (Handbook cycle facilities) 2017 Available here (in Dutch) Vadémécum vélo en Région de Bruxelles-Capitale: Aménagements cyclables séparés de la chaussée partie 17 / Fietsvademecum Brussels Hoofdstedelijk Gewest: Van de rijbaan afgescheiden fietsinfrastructuur deel 18 (Handbook cycling: cycle infrastructure separated from the carriageway part 1) 2018 Available here (in French) Available here (in Dutch) Bulgaria Наредба № рд-02-20-2 от 20 декември 2017 г. за планиране и проектиране на комуникационно-транспортната система на урбанизираните територии (Ordinance on planning and design of the communication and transport system in urban areas) 2017 Available here (in Bulgarian) 7 8 ECF | Geometric design parameters for cycling infrastructure 20 Country Document Publication year Link Czechia Navrhování komunikací pro cyklisty (Design of routes for cyclists) 2017 Available here (in Czech) Denmark Collection of cycle concepts 2012 Available here (in English) Håndbog i cykeltrafik. En samling af de danske vejregler på cykelområdet (Handbook cycling traffic. A collection of the Danish road rules regarding cycling) 2014 Available here (in Danish) Håndbog supercykelstier. Anlæg og planlægning (Handbook supercyclehighways. Construction and planning) 2016 Available here (in Danish) Finland Pyöräliikenteen Suunnittelu (Cycling Design) 2020 Available here (in Finnish) France No relevant document identified Germany Empfehlungen für Radverkehrsanlagen (Recommendations for cycling facilities) 2010 Available here (in German) Greece Τεχνικές οδηγίες για υποδομές ποδηλάτων (Technical instructions for bicycle infrastructure) 2016 Available here (in Greek) Italy Regolamento recante norme per la definizione delle caratteristiche tecniche delle piste ciclabili (Regulation laying down rules for the definition of the technical characteristics of cycle paths) 1999 Available here (in Italian) Ireland National Cycle Manual 2011 Available here (in English) Rural Cycleway Design 2017 Available here (in English) Malta No relevant document identified ECF | Geometric design parameters for cycling infrastructure 21 Country Document Publication year Link Netherlands Design Manual on Bicycle Traffic 2016 Available here (in English) Poland Wytyczne organizacji bezpiecznego ruchu rowerowego (Guidelines for organisation of safe bicycle traffic) 2019 Available here (in Polish) Slovakia Technické podmienky. Navrhovanie cyklistickej infraštruktúry (Technical requirements. Cycle infrastructure design) 2019 Available here (in Slovakian) Spain (Catalonia) Manual for the design of cyclepaths in Catalonia 2008 Available here (in English) UK Cycle infrastructure design (LTN 1/20) 2020 Available here (in English) 6.1 Austria Documents: • RVS 03.02.13 Radverkehr (Guidelines and regulations for roads: cycle traffic), 2011 (federal)9 • Planungsleitfaden Radverkehr (Planning guide for cycle traffic), 2007 (regional/Tirol)10 The regional guidance recommends design speeds between 20 and 30 km/h within built-up areas (page 14). These design speeds are translated into minimum horizontal curve radii between 8 and 22 m, with 4 m permitted at intersection areas (page 15). The document stresses the safety aspect of sufficient curve radii, noting that tight bends force the cyclist to concentrate on the driving technique, and thus divert attention away from the road traffic. The table with minimum stopping sight distances includes additionally a design speed of 40 km/h. While in the regional document it is not explicitly stated where this increased design speed should be applied, the federal document clarifies that 40 km/h should be applied on inclines with 6% gradient. The federal document also lists requirements for vertical curves, not copied to the regional guide. For convex curves, the minimum curve radii vary between 10 and 40 m; for concave, between 20 m and 65 m, depending on the design speed. 9 10 ECF | Geometric design parameters for cycling infrastructure 22 Table 8. Geometric design parameters for cycling infrastructure in Austria. Route Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance In the intersection area 4 m Within built-up areas 20 km/h 8 m 20 m concave, 10 m convex 15 m 30 km/h 22 m 40 m concave, 20 m convex 25 m General, 3% gradient 6% gradient 40 km/h 65 m concave, 40 m convex 40 m 6.2 Belgium Documents: • Flanders: Vademecum Fietsvoorzieningen (Handbook cycle facilities), 201711 • Brussels: Vadémécum vélo en Région de Bruxelles-Capitale: Aménagements cyclables séparés de la chaussée partie 112 / Fietsvademecum Brussels Hoofdstedelijk Gewest: Van de rijbaan afgescheiden fietsinfrastructuur deel 113 (Handbook cycling: cycle infrastructure separated from the carriageway part 1), 2018 In Belgium, separate handbooks for cycling infrastructure exist for two out of the three federated regions (Flanders and Brussels Capital Region). No similar document has been identified for the third region, Wallonia, or at the federal level. Although the Flemish handbook for cycle facilities is a comprehensive document with more than 200 pages, geometric design parameters are only briefly mentioned in chapter 2. Minimum horizontal curve radius is given for design speeds of 20 and 30 km/h. No specific requirements regarding vertical curves or sight distances are provided. 11 chapter 2: 12 13 ECF | Geometric design parameters for cycling infrastructure 23 Table 9. Geometric design parameters for cycling infrastructure in Flanders. Route Design speed Minimum horizontal curve radius Connecting routes 20 km/h 15 m Cycle highways 30 km/h 35 m The Brussels handbook is even more brief, with curves discussed in section 6.3. Minimum horizontal curve radius is set at 10 m, always measured to the internal edge of the curve. For connections of cycle tracks, 4 m is the recommended and 3 m the minimum horizontal curve radius. The document stresses (also in other parts) that cyclists are unable to make a 90 degree turn in place. Design speeds, vertical curves and stopping distances are not mentioned. 6.3 Bulgaria Document: Наредба № рд-02-20-2 от 20 декември 2017 г. за планиране и проектиране на комуникационно-транспортната система на урбанизираните територии (Ordinance on planning and design of the communication and transport system in urban areas), 201714 Bicycle infrastructure design is regulated in chapter 3, section 2 of the regulation (articles 61-70). Key parameters are gathered in tables in appendix 8 to article 62. Table 10. Geometric design parameters for cycling infrastructure in Bulgaria. Route Design speed Minimum horizontal curve radius Minimum visibility distance Minimum stopping sight distance ? 10 km/h 4 m Connecting routes 20 km/h 15 m 25 m 25 m Through routes 30 km/h 25 m 45 m 35 m ? 40 km/h 45 m ? 50 km/h 80 m Table 1 in the appendix sets design speed to 30 km/h for through cycle routes, and to 20 km/h for connecting routes. This translates to minimum visibility distances of respectively 45 m and 25 m, and stopping sight distances 35 m and 25 m. Table 3 in the same appendix provides minimum values of horizontal curve radii for a wider range of design speeds, from 10 km/h to 50 km/h. It is not however explained when designs speeds other than 20 or 30 km/h should be used. Vertical curve radii are not discussed for cycling infrastructure. 14 ECF | Geometric design parameters for cycling infrastructure 24 6.4 Croatia Document: Pravilnik o biciklističkoj infrastrukturi (Regulation on bicycle infrastructure), 201615 Geometric parameters are provided in chapter V.II of the regulation, articles 16 (horizontal curves), 19 (vertical curves) and 20 (stopping sight distance). No specific design speeds seem to be required. Different ranges of design speed are used for different parameters: • 12-20 km/h for horizontal curve radii in table 1, • 12-40 km/h for horizontal curve radii on figure 1, • 20-50 km/h for vertical curve radii in table 3, • 20-30 km/h for stopping sight distances in table 4. Article 16(2) allows to use an exceptionally low minimum horizontal curve radius of 1.0 m for cycle tracks at intersections. In horizontal curves, the cross slope is generally required to be directed towards the centre of the curve. Vertical curves are only obligatory if the change in gradient exceeds 5%. Table 11. Geometric design parameters for cycling infrastructure in Croatia. Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance concave convex 12 km/h 5 m 16 km/h 8 m 20 km/h 10 m 40 m 25 m 30 m 30 km/h 17.5 m16 80 m 50 m 40 m 40 km/h 25 m17 150 m 100 m 50 km/h 300 m 200 m 15 16 Estimated from the graph. 17 Estimated from the graph. ECF | Geometric design parameters for cycling infrastructure 25 6.5 Czechia Document: Navrhování komunikací pro cyklisty (Design of routes for cyclists), 201718 Chapter 3.1.3 “Návrhová rychlost” of the Czech standard assumes basic design speed of 20-25 km/h, with higher values needed on descents, and 10 km/h allowed in exceptional cases. This has been translated to horizontal curve radii between 2.5 and 22 m (chapter 3.1.5 “Směrové vedení a rozšíření ve směrovém oblouku”), with the required widening of the track of minimum 0.25 m for radii below 22 m, and minimum 0.5 m for radii below 8 m. Stopping sight distances for different design speeds are listed in chapter “3.1.4 Délka rozhledu pro zastavení”. These values are calculated for asphalt surface. For unpaved roads required minimums need to be increased by 50%. The same increase is recommended for inclines exceeding 5%.19 Table 12. Geometric design parameters for cycling infrastructure in Czechia. Route Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance In the intersection area (exceptionally, in justified cases) 10 km/h 2.5 m 9 m ? 15 km/h 4.5 m Basic 20 km/h 8 m 20 m concave 10 m convex 15 m 25 km/h 14 m More than 3% descent 30 km/h 22 m 40 m concave 20 m convex 25 m More than 6% descent 40 km/h 65 m concave 40 m convex 40 m + 50% 6.6 Denmark Documents: • Collection of cycle concepts, 201220 18 19 This might seem to be superfluous considering that slopes already require higher design speeds, but actually slopes have double effect on stopping distance: the initial speeds are higher, and the deceleration is lower. 20 ECF | Geometric design parameters for cycling infrastructure 26 • Håndbog i cykeltrafik. En samling af de danske vejregler på cykelområdet (Handbook on cycling traffic. A collection of the Danish road rules regarding cycling”), 201421 • Håndbog supercykelstier. Anlæg og planlægning (Handbook on supercyclehighways. Construction and planning), 201622 In the Danish “Collection of cycle concepts,” the section on “Horizontal and vertical radii” explains that curves on cycle tracks should be passable at 30 km/h for bicycles and mopeds. It does not provide specific dimensions, noting only that in horizontal radii of less than 50 m a transverse inclination toward the bend’s centre may be necessary. Further on, in the section on gradients, the manual notes that cyclists’ safety drops if a steep incline is combined with a sharp horizontal curve. At a gradient of 5%, cycle paths should be dimensioned for 40 km/h and for 36 km/h at a gradient of 3% (the increase is equal to the 2 km/h per 1% gradient given in the Catalonian manual, see section 6.14. In the handbook for cycle superhighways, 35 km/h is assumed as a basis for cycle highways. Table 13. Design speeds for cycling infrastructure in Denmark. Route Design speed General 30 km/h Cycle highways 35 km/h Gradient 3% 36 km/h Gradient 5% 40 km/h Another overview publication, “Handbook on cycling traffic,” differentiates curve radii depending on whether the cycle track is designated for bicycles only or also for moped. Table 14. Geometric design parameters for cycling infrastructure in Denmark Route Horizontal curve radius Vertical curve radius Exceptional minimum Minimum Minimum normal Recommended tracing Minimum radius Recommended minimum radius Pedal bicycles only 16 m 40 m 60 m 210 m 175 m 340 m Mopeds allowed 20 m 70 m 105 m 360 m 300 m 580 m 21 22 ECF | Geometric design parameters for cycling infrastructure 27 The manual does not specify sight distances but notes that observing the given curve radii requirements ensures sufficient visibility on the cycle track. The more general “Handbook on the basis of traffic area design. Construction and planning”23 calculates the minimum stopping sight distances for cyclists on basis of the following assumptions: reaction time 2 seconds, deceleration 2 m/s2 on flat sections. Initial speed depends on gradient only and varies between 20 km/h (5% uphill) and 34 km/h (5% downhill). This translates to stopping sight distances between 18 and 49 m, depending on the incline. As the speeds assumed are not aligned with the recommendations of the cycle-specific guidelines, we do not include these values in the summary table. 6.7 Finland Document: Pyöräliikenteen Suunnittelu (Cycling Design), 202024 The guidelines are published by the Finnish Transport Infrastructure Agency Design and are obligatory for cycling infrastructure managed on the national level. Municipalities often also apply the same parameters but, in some cases, have their own standards.25 Design speeds are defined in table 22 in section 4.9.1. of the document. Section 4.9.2. and table 23 cover the sight distances, section 4.9.3. and table 24 cover horizontal curve radii, while section 4.9.4 with tables 25 and 26 cover vertical curve radii. Table 15. Design speeds for cycle routes in Finland. Route Design speed Local 25 km/h Regional 30 km/h Main 40 km/h (30 km/h in tightly built environment) Cycle highways 45 km/h (30 km/h in tightly built environment) Regional and main routes with mopeds allowed 45 km/h Design speeds vary from 25 km/h to 45 km/h, with the highest applied to cycle highways and sections of other routes where mopeds are allowed to use the cycling infrastructure. For establishing the minimum horizontal curve radii, on long descends the design speed needs to be additionally increased by 10 km/h. 23 Håndbog grundlag for udformning af trafikarealer. Anlæg og planlægning: 24 25 For example, Helsinki: ECF | Geometric design parameters for cycling infrastructure 28 Table 16. Minimum horizontal curve radius in Finland. Modified design speed Minimum horizontal curve radius 25 km/h 25 m 30 km/h 30 m 40 km/h 55 m 45 km/h 75 m 50 km/h 100 m 55 km/h 130 m Reduced curve radii (10-20 m) are allowed in built-up areas in the vicinity of an intersection. On the other hand, if the change in direction exceeds 90 degrees, higher curve radius should be applied. For sight distances and vertical curve radii, unmodified design speed is used as a base, but gradient is taken into account as a separate parameter for sight distance. Table 17. Minimum vertical curve radii and stopping sight distances for cycling infrastructure in Finland. Design speed Minimum vertical curve radius Minimum stopping sight distance Concave Convex Gradient 0% Gradient 5% Gradient 8% 25 km/h 225 m 50 m 26 m 30 m 34 m 30 km/h 385 m 70 m 34 m 40 m 45 m 40 km/h 940 m 125 m 53 m 63 m 73 m 45 km/h 1370 m 160 m 64 m 77 m 89 m Sight distance in motion (encounter sight distance) is calculated as double of the stopping sight distance. Eye height is assumed to be 1.5 m. For stopping sight distance on concave vertical curves obstacles on surface level (0.0 m) need to be visible. ECF | Geometric design parameters for cycling infrastructure 29 6.8 Germany Document: Empfehlungen für Radverkehrsanlagen (Recommendations for cycling facilities), 201026 German federal recommendations for cycling facilities present the desired speeds for different categories of cycle routes in table 2 on page 10. However, the values given are travel speeds for the whole route (taking into account time lost at crossings and waiting on traffic lights, for example), not design speeds for route elements. Table 18. Desired travel speed for different route categories in Germany. Route category Travel speed AR II – AR IV (interregional, regional and connections) 20-30 km/h IR II (fast local cycle routes) 15-25 km/h IR III – IR V (other local cycle routes) 10-20 km/h Further on, table 6 on page 16 gives geometric parameters for specific design speeds between 20 and 40 km/h. No clear relation with route categories or table 2 is established, only a remark that speeds 20 and 30 km/h are suitable for flat sections, while 40 km/h might be necessary on longer downhill stretches. For horizontal curves, the table differentiates between asphalt and unbound surfaces; for vertical curves – between concave and convex. Table 19. Geometric design parameters for cycling infrastructure in Germany. Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance 20 km/h 10 m asphalt / concrete 15 m unbound surfaces 40 m concave 25 m convex 15 m 30 km/h 20 m asphalt / concrete 35 m unbound surfaces 80 m concave 50 m convex 25 m 40 km/h 30 m asphalt / concrete 70 m unbound surfaces 150 m concave 100 m convex 40 m The recommendations clarify that the given geometric parameters are applicable also at intersections for the route going straight ahead. 26 ECF | Geometric design parameters for cycling infrastructure 30 In addition to the federal recommendations, there are more detailed standards in different federate states. However, the two documents we examined: • Radnetz Hessen: Qualitätsstandards und Musterlösungen27 (Hessen) • Qualitätsstandards für Radschnellverbindungen28 (Baden-Württemberg) with regards to the geometric design parameters simply refer to the federal recommendation, copying the parameters for asphalted tracks with a design speed of 30 km/h. 6.9 Greece Document: Τεχνικές οδηγίες για υποδομές ποδηλάτων (Technical instructions for bicycle infrastructure), 201629 The Greek regulation, while quite extensive (232 pages) does not include guidance or quality requirements on design speed, sight distances, horizontal or vertical curve radii. 6.10 Ireland Documents: • National Cycle Manual, 201130 • Rural Cycleway Design, 201731 The National Cycle Manual does not contain a dedicated section on geometric design parameters. Some information on curve radii and sight distances is scattered around section 4.10, on transitions between different types of cycling infrastructure, and section 5.3, on public lighting. In the Rural Cycleway Design (RCD) the information is better organised (in sections 5 and 6.1), but the title implies that the scope of the document is limited. In the National Cycle Manual, the graph in section 4.10.1 implies that cycling32 speeds between 12 and 40 km/h are considered, providing a formula for horizontal curve radius:33 Radius = 0.6 speed – 3.62 In the same section, a curve radius of 16 m is recommended to accommodate cyclists travelling at 30 km/h as a general principle, with tighter curves allowed near junctions. Section 4.10.3 lists values of 10-25 m for cycling speeds between 20 and 40 km/h, giving downhill sections as an example of where 27 28 29 30 PDF available at: 31 32 In section 4 the National Cycle Manual the term “cycling speed” is used. 33 Similar to the formula provided by the Dutch manual, but with lower coefficient. ECF | Geometric design parameters for cycling infrastructure 31 40 km/h should be applied. Stopping sight distance is mentioned in section 5.3.3.2 (on public lighting), as directly proportional to design speed. Minimum cycling sight distance between 20 and 40 m are required. In the Rural Cycleway Design, design speeds between 10 and 50 km/h are listed (section 5.1), and translated to minimum horizontal curve radii between 4 and 94 m. For sight distances, a distinction between dynamic and stopping sight distances is made. In the first case a visibility envelope between heights of 1.0 and 2.2 m is considered (figure 5.1), in the second – eye height between 1.0 and 2.2 m and obstacles between 0.0 and 2.2 m (figure 5.2). A reaction time of 2 seconds and a deceleration rate of 0.15 g are assumed. The document specified also that minimum stopping sight distances should be increased by 50% on loose surface tracks. Table 20. Cycling/design speed for cycling infrastructure in Ireland Type of the route/context National Cycle Manual Rural Cycleway Design Minimum 12 km/h (threshold of good stability) 10 km/h (on approach to obstacles) Standard 30 km/h 30 km/h Downhill 40 km/h 50 km/h Table 21. Horizontal curve radii in Ireland Cycling/design speed National Cycle Manual Rural Cycleway Design Section 4.10.1 Section 4.10.3 10 km/h 4 m 12 km/h 3.6 m 20 km/h 8.4 m 10 m 30 km/h 16 (14) m 20 m 25 m 40 km/h 20 m 25 m 50 km/h 94 m Table 22. Minimum sight distances in Ireland Cycling/design speed National Cycle Manual Rural Cycleway Design Dynamic sight distance Stopping sight distance 10 km/h 20-40 m (proportional to design speed, but no exact formula given) 15 m 15 m 30 km/h 65 m 35 m 50 km/h 110 m 60 m ECF | Geometric design parameters for cycling infrastructure 32 6.11 Netherlands Document: Design Manual on Bicycle Traffic, 201634 The Dutch design manual, probably the most-often quoted set of guidelines for cycling infrastructure, discusses design speeds in chapters 3.2 and 3.3 (see page 46). The manual notes that for typical two-wheeled cycles, speeds of around 15 km/h are needed to cycle stably. With lower speeds cyclists need additional space, up to 0.8 m, to maintain balance. Speeds below 12 km/h are particularly problematic for elderly cyclists. The consequences for curve radii and sight distances are elaborated further on in chapter 3.4 “Bends and view” (page 50-52). The manual quotes research demonstrating a linear dependency between the design speed and curve radius, with a concrete formula provided in an earlier edition of the manual35: R = 0.68 V – 3.62 where R – curve radius in meters, V – design speed in km/h The manual also quotes research showing the following: • With straight angles or insufficient curve radii cyclists need to travel outside the designed space, for example taking the part of a bidirectional cycle track designated for the opposite direction.36 • Sharp bends or insufficient sight distance can cause single-vehicle bicycle accidents, as well as accidents involving multiple bicycles.37 A more specialised “Design guide for bridges for slow traffic”38 also warns that sharp bends on downhill slopes or just after them might cause single-vehicle bicycle accidents, and bicycle-bicycle accidents in case of two-directional cycle tracks. Because cyclists lean into curves, increasing the width on bends by around 0.5 m is advised. Regarding sight distances, the manual distinguishes sight distance in motion and stopping sight distance. The first one bases on the fact that the cyclist needs to be able to see the road ahead of them, on a distance equivalent to covered in 4-5 seconds (minimum) or 8-10 seconds (recommended). Stopping distances are calculated with the assumption of 2 seconds reaction time and deceleration of 1.5 m/s2. 34 35 1993 edition. The following article is referenced as a source for the formula: “Het verband tussen de snelheid van fietsers en hun vrij gekozen baan bij haakse bochten”; A. Dijkstra. In: Verkeerskunde 33, nr. 9, 1982. 36 37 38 Ontwerpwijzer bruggen voor langzaam verkeer: ECF | Geometric design parameters for cycling infrastructure 33 Table 23. Geometric design parameters for cycling infrastructure in Netherlands Route Design speed Minimum horizontal curve radius Minimum sight distance in motion Minimum stopping sight distance Lower limit 12 km/h 5 m Basic network 20 km/h 10 m 22-30 m (recommended: 44-60 m) 21 m (Main) cycle route 30 km/h 20 m 35-42 m (recommended: 70-84 m) 40 m Slopes 35-40 km/h Outside built-up areas 40 km/h Not specified, can be estimated to ~25 m basing on the graph Not specified, can be estimated to 44-56 m (recommended: 88-112 m) Not specified, can be estimated to 42 m 6.12 Poland Document: Wytyczne organizacji bezpiecznego ruchu rowerowego (Guidelines for organisation of safe bicycle traffic), 201939 The Polish guidelines define the horizontal curve radii in section 6.1.1., and sight distances in section 6.1.3. Vertical curve radii are not mentioned. Horizontal curve radius should be measured to the inner side of the curve. Table 24. Geometric design parameters for cycling infrastructure in Poland. Route Design speed Minimum horizontal curve radius Recommended minimum sight distance in motion Minimum stopping sight distance Minimum values 12 km/h 5 m Local and distributor cycle route 20 km/h 10 m 45 m 21 m Main cycle route 30 km/h 20 m 70 m 40 m 39 ECF | Geometric design parameters for cycling infrastructure 34 Given values seem to be based on the Dutch manual (discussed in section 6.11), with the following modifications: • No design speeds higher than 30 km/h are mentioned. • Formula for calculating horizontal curve radii is explicitly given. • For recommended minimum sights distance in motion, the lower borders of the ranges are used. • Additional minimum value for horizontal curve radii of 2 m is specified (allowed in situations where the cyclist needs to stop in order to give way). • The requirements on slopes are not increased, there is only a mention in section 6.1.2 that for gradients above 5% sharp curves should be avoided. 6.13 Slovakia Document: Technické podmienky. Navrhovanie cyklistickej infraštruktúry (Technical requirements. Cycle infrastructure design), 201940 Design parameters are discussed in chapter 4 of the requirements. Section 4.8 “Návrhová rýchlosť” sets the default design speed to 25 km/h. In the areas of intersections or pedestrian crossings, the design speed can be reduced to 10 km/h. On the other hand, if gradient exceed 3%, 40 km/h design speed is required. In section 4.4 “Smerové oblúky”, table 8 translates design speeds between 10 and 30 km/h to horizontal curve radii between 2.5 and 22 m (so 40 km/h is not covered). In radii below 22 m, the track should be widened by at least 0.25 m, and in radii below 14 m – by at least 0.5 m. Additional widening is required on slopes exceeding 3%. In section 4.5 “Výškové vedenie” tables 9 and 10 provide minimum and recommended vertical curve radii for design speeds between 20 and 40 km/h (but not for the default design speed of 25 km/h). Finally, stopping sight distances for different design speeds are listed in section 4.6 “Rozhľadové pomery”, tables 11 & 12. On flat and asphalted sections, a cyclist should be able to see at least 15-30 m of the route ahead. Double stopping sight distances are required for downhills exceeding 5% and unpaved roads. If the required distances cannot be achieved, traffic mirrors can be used to provide a better view of the route ahead. For overtaking, uniform sight distance of 100 m is required. If it is not available, it should be indicated by horizontal markings. Note: although the Czech document “Navrhování komunikací pro cyklisty” (see section 6.5) is referenced many times in the chapter, some of the Slovakian design parameters are different. 40 ECF | Geometric design parameters for cycling infrastructure 35 Table 25. Geometric design parameters for cycling infrastructure in Slovakia. Route Design speed Minimum horizontal curve radius Vertical curve radius Stopping sight distance minimum recom-mended minimum gradients >5% & unpaved Intersection areas 10 km/h 2.5 m ? 15 km/h 4.5 m 20 km/h 8 m 20 m concave 10 m convex 40 m concave 25 m convex 15 m 30 m Default 25 km/h 14 m 20 m 40 m ? 30 km/h 22 m 30 m concave 20 m convex 80 m concave 50 m convex 25 m 50 m Gradient >3% 40 km/h 40 m concave 40 m convex 150 m concave 100 m convex 30 m 60 m 6.14 Spain (Catalonia) Document: Manual for the design of cycle paths in Catalonia, 200841 The Catalonian guidelines offer one of the highest ranges of design speeds, providing geometric parameters for cycling infrastructure up to 50 km/h (to accommodate sport cyclists or steep slopes?) The highest speeds are assumed for greenways with separations from pedestrians. Slopes increase the required design speed by 2 km/h per 1% incline. The manual consistently differentiates geometric design parameters depending on the surface material. It observes correctly that unpaved surfaces (for example, stabilised gravel) are characterised by significantly lower friction coefficient. Therefore, they require larger curve radii and longer stopping sight distances. 41 ECF | Geometric design parameters for cycling infrastructure 36 Table 26. Geometric design parameters for cycling infrastructure in Catalonia. Route Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance Greenway minimum 20 km/h 10 m paved 17 m unpaved 10 m 20 m on flat 25 m on 10% descent Greenway general, bicycle lane minimum 30 km/h 24 m paved 44 m unpaved 20 m 35 m flat 45 m on 10% descent 40 km/h 47 m paved 84 m unpaved 40 m 55 m flat 70 m on 10% descent Greenway with separation for pedestrians, bicycle lane 50 km/h 86 m paved 151 m unpaved 70 m 75 m flat 100 m on 10% descent Effect of slopes +2 km/h per 1% incline See formula The manual also provides a formula for individual calculation of horizontal curve radii, depending on design speed, cant of the curve and transversal friction coefficient. In contrast to linear dependency listed in the Dutch manual, the Catalonian document recommends curve radii proportional to square of the design speed. R = V^2/127 (p + f) Where: • R = minimum radius of a curve [m] • V = design speed [km/h] • p = cant of the curve • f = transversal friction coefficient For stopping sight distances, the required lateral visibility should be equivalent to the sum of stopping distances for cyclists coming from the opposite directions. ECF | Geometric design parameters for cycling infrastructure 37 6.15 UK Document: Cycle infrastructure design (LTN 1/20), 202042 Chapter 5.6 of the guidance lists design speeds from 20 to 40 km/h, with 30 km/h being the default speed, 20 km/h absolute minimum and 40 km/h required on downhills with gradients exceeding 3%. The manual notes that the buffer space needed to maintain balance when cycling rises from 0.2 to 0.8 m when the speed drops below 7 mph (~11 km/h). The table defining minimum horizontal radii in chapter 5.9 includes additional design speed of 10 km/h, below the “absolute minimum” listed in 5.6. It is not clear in what (exceptional?) situations this speed and curve radii can be used. Table 27. Geometric design parameters for cycling infrastructure in the UK. Route Design speed Minimum horizontal curve radius Minimum vertical curve radius Minimum stopping sight distance ? 10 km/h 4 m Crest (concave): 300 m Sag (convex): 250 m Absolute minimum 20 km/h 15 m 17 m General cycle tracks 30 km/h 25 m 31 m Downhill gradients >3% 40 km/h 40 m 47 m The guidance notes that introducing barriers or bends to slow cyclists is likely to increase the potential for user conflict and may prevent access for larger cycles and disabled people and so should not be used. For designing vertical curves, the manual varies the curve length (not radius) depending on the difference between the gradients on neighbouring sections. The given method of calculation translates to nearly constant values of minimum radii: around 250 m for convex curves and 300 m for concave curves. For stopping sight distances (chapter 5.7), the guidance recommends between 17-47 m. Obstacles should be visible from an eye height in the range of 0.8-2.2 m, to accommodate a range of cyclists including recumbent users, children and adults 42 ECF | Geometric design parameters for cycling infrastructure 38 7. Final remarks The comparison was created with the help of ECF member organisations, with contributions from our project partners and participants of the workshops organised in the frame of Safer Cycling Advocacy Programme.43 We would like to thank in particular: • Luc Goffinet, GRACQ (Belgium) • Ross Goorden, Fietsersbond (Netherlands) • Matti Koistinen, Finnish Cyclists' Federation • Niccolo Minotti, ECF • Spiros Papageorgiou, Cities for Cycling (Greece) • Philipp Schober, Radlobby Wien • Antonios Sifakis, Hellenic Urban Cycling Federation • James Wightman, Rota (Malta) • Matej Zganec, ECF A considerable effort has been made to ensure that the information presented is current and accurate. If outdated or incorrect information is brought to our attention, ECF will correct or remove it. Please also let us know if you would like to see other standards or guidelines added to the comparison or if you know about other relevant research on cycling infrastructure geometry that should be mentioned in the document. 43 ECF | Geometric design parameters for cycling infrastructure 39 ECF | Geometric design parameters for cycling infrastructure 1 European Cyclists’ Federation Mundo Madou Avenue des Arts 7-8 B-1210 Brussels +32 2 329 03 80 office@ecf.com |
11855 | https://www.concord.edu/wp-content/uploads/NUMBERS-IN-BASE-5.pdf | 1 NUMBERS IN BASE 5 Base 5 Number System There’s nothing magical about base 10; it’s just what we’re the most familiar with. We’ll be using base 5 as our primary example (Think about why base 10 is natural. Why do you think base 5 might make sense?) We’ll be using base 5 to illustrate the structure behind place value and operations. You’ve been using base 10 your whole life; switching bases helps you see the places where students might run into trouble because you struggle in base 5. In base 5, we can write any number, as big or small as we want, using only 5 digits: 0, 1, 2, 3, 4. We can tell the size of the number by the position it’s in. Consider 14302 (base 5). The 4 doesn’t stand for just 4, it’s really 4 x 125 because of its position. 1 4 3 0 2 54 53 52 51 50 625 125 25 5 1 Stack of Cubes Cubes Flats Longs Units Keep in mind I can keep adding powers of five and get as big as I want, or I can add a decimal point and go down to fifths (5-1), twenty-fifths (5-2), one twenty-fifths (5-3), etc., which are just smaller powers of 5. As you go up each place value you multiply by five (as you go down, you divide by five). 2 Base 5 Blocks We can model numbers in base 5 with Base 5 Blocks. They look a lot like base 10 blocks; they’re just set up in groups of 5 instead of 10. There are blocks for ones, fives, twenty-fives, and one twenty-fives. The idea can be extended, but those are the most common ones used. Ones: commonly called units. Really just a single cube. Figure 1: Unit CC0 1.0 Fives: commonly called longs. Five units connected. Figure 2: Long Twenty-fives: commonly called flats. Five longs connected. Figure 3: Flat One twenty-fives: commonly called blocks or cubes. Some resources make a flat printable version called a long flat (5 flats connected in a line). Five flats stacked. Figure 4: Cube You can extend the idea: the next size up would be five blocks stacked together (block long?), then five of those (block block?), etc. 3 Realistically, it’s easier to simplify and just draw 2 dimensional versions. You do need to know what base you’re working with, though—the simplified images are the same regardless of the base. Figure 5: Simplified Base Blocks To represent a number using base 5 blocks, you simply include the same number of each type as block as the digit associated with that place value. Figure 6: 14302 (base 5) in Base 10 blocks |
11856 | https://brilliant.org/wiki/sat-direct-and-inverse-variation/ | SAT Direct and Inverse Variation | Brilliant Math & Science Wiki
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Tatiana Georgieva contributed
To solve problems about direct and inverse variation on the SAT, you need to know:
how to work with fractions
the rules of exponents
how to translate words into math
the definition of direct variation
If y y y varies directly with x x x, then there is a constant k k k for which
y=k⋅x,k≠0 y=k\cdot x, \quad k\neq0 y=k⋅x,k=0
k k k is called the constant of proportionality, or the constant of variation.
Notice that as x x x increases (or decreases) by some factor, y y y increases (or decreases) by the same factor.
Sometimes the definition is written as follows to demonstrate that the ratio of y y y and x x x is constant:
y x=k\frac{y}{x}=k x y=k
the definition of inverse variation
If y y y varies inversely with x x x, then there is a constant k k k, such that:
y=k⋅1 x,k≠0 y=k\cdot\frac{1}{x}, \quad k\neq0 y=k⋅x 1,k=0
k k k is called the constant of proportionality, or the constant of variation.
Notice that as x x x increases (or decreases) by some factor, y y y (decreases or increases) by the same factor.
Sometimes the definition is written as follows to demonstrate that the product of x x x and y y y is constant:
x⋅y=k x\cdot y =k x⋅y=k
Contents
Examples
Review
SAT Tips for Direct Variation
Examples
p p p varies directly with n n n. Which of the following equations could represent the relationship between p p p and n n n?
(A) p=1 2 n\ \ p=\frac{1}{2n}p=2 n 1
(B) n=2 p\ \ n=\frac{2}{p}n=p 2
(C) p⋅n=1 2\ \ p\cdot n=\frac{1}{2}p⋅n=2 1
(D) p⋅n=2\ \ p\cdot n=2 p⋅n=2
(E) p=1 2⋅n\ \ p=\frac{1}{2}\cdot n p=2 1⋅n
Correct Answer: E
Solution 1:
By the definition of direct variation, for some constant k k k, p=k n p=kn p=kn. Here, k=1 2.k=\frac{1}{2}.k=2 1. You should be able to tell that this answer choice is correct at a glance.
Solution 2:
Tip: Replace variables with numbers.
In a direct variation, as one variable increases by a factor, the other one increases by the same factor; and as one variable decreases by a factor, the other one decreases by the same factor. We analyze each choice to see which one meets this expectation.
(A) For n=1,p=1 2⋅1=1 2.n=1, p=\frac{1}{2\cdot1}=\frac{1}{2}.n=1,p=2⋅1 1=2 1.
For n=2,p=1 2⋅2=1 4.n=2, p=\frac{1}{2\cdot2}=\frac{1}{4}.n=2,p=2⋅2 1=4 1.
Contrary to the definition of direct variation, as n n n doubles, p p p is divided by 4 4 4. Eliminate (A).
(B) For p=1,n=2 1=2.p=1, n=\frac{2}{1}=2.p=1,n=1 2=2.
For p=2,n=2 2=1.p=2, n=\frac{2}{2}=1.p=2,n=2 2=1.
Contrary to the definition of direct variation, as p p p doubles, n n n halves. Eliminate (B).
(C) Dividing both sides of this equation by n n n, we get:
p⋅n=1 2 p \cdot n=\frac{1}{2}p⋅n=2 1p=1 2 n p=\frac{1}{2n}p=2 n 1
This is identical to (A), which we already eliminated. Eliminate (C).
(D) Dividing both sides of this equation by n n n, we get:
p⋅n=2 p \cdot n=2 p⋅n=2 p=2 n p=\frac{2}{n}p=n 2
For n=1,p=2 1=2.n=1, p=\frac{2}{1}=2.n=1,p=1 2=2.
For n=2,p=2 2=1.n=2, p=\frac{2}{2}=1.n=2,p=2 2=1.
As n n n doubles, p p p halves. This contradicts the definition of direct variation. Eliminate (D).
(E) For n=1,p=1 2⋅1=1 2.n=1, p=\frac{1}{2} \cdot 1 = \frac{1}{2}.n=1,p=2 1⋅1=2 1.
For n=2,p=1 2⋅2=2 2=1.n=2, p=\frac{1}{2} \cdot 2 = \frac{2}{2}=1.n=2,p=2 1⋅2=2 2=1.
As n n n doubles, so does p p p. Therefore, this is the correct answer.
Incorrect Choices:
(A), (B), (C), and (D)
See Solution 2 for how to eliminate these answers.
A B C D E
Reveal the answer
a a a varies directly with b b b.
b b b varies indirectly with c c c.
If the above statements are true for nonzero a,b,a, b,a,b, and c c c, which of the following statements must be true?
(A) a\ \ a a varies directly with c c c and c c c varies directly with a a a.
(B) a\ \ a a varies directly with c c c and c c c varies indirectly with a a a.
(C) a\ \ a a varies indirectly with c c c and c c c varies directly with a a a.
(D) a\ \ a a varies indirectly with c c c and c c c varies indirectly with a a a.
(E) \ \ None of the above.
The correct answer is: D
If y y y is inversely proportional to x x x, and y=20 y=20 y=20 when x=4 x=4 x=4, what is the value of y y y when x=10 x=10 x=10?
(A) 4\ \ 4 4
(B) 5\ \ 5 5
(C) 8\ \ 8 8
(D) 20\ \ 20 20
(E) 50\ \ 50 50
Correct Answer: C
Solution 1:
If y y y is inversely proportional to x x x, then for some constant k≠0,y=k 1 x.k\neq 0, y=k\frac{1}{x}.k=0,y=k x 1.
We are told that when x=4,y=20 x=4, y=20 x=4,y=20. So,
y=k x definition of inverse variation(1)20=k 4 replace variables with given values(2)80=k multiply both sides by 4(3)\begin{array}{l c l l l} y&=&\frac{k}{x} &\quad \text{definition of inverse variation} &(1)\ 20&=&\frac{k}{4} &\quad \text{replace variables with given values} &(2)\ 80&=&k &\quad \text{multiply both sides by}\ 4 &(3)\ \end{array}y 20 80===x k4 kkdefinition of inverse variation replace variables with given values multiply both sides by 4(1)(2)(3)
We found the constant k k k. Therefore, y=80 x y=\frac{80}{x}y=x 80.
When x=10 x=10 x=10, y=80 10=8 y=\frac{80}{10}=8 y=10 80=8.
Solution 2:
When x=4 x=4 x=4, y=20 y=20 y=20. We are looking for y y y when x=10 x=10 x=10. By definition, when two quantities vary inversely, their product is constant. Here we create two products, set them equal to each other, and solve for the unknown.
20⋅4=y⋅10 set the two products equal to each other 80=y⋅10 20⋅4=80 8=y divide both sides by 10\begin{array}{l c l l l} 20\cdot 4&=&y \cdot 10 &\quad \text{set the two products equal to each other}\ 80 &=&y \cdot 10 &\quad 20\cdot 4=80\ 8 &=& y &\quad \text{divide both sides by}\ 10\ \end{array}20⋅4 80 8===y⋅10 y⋅10 yset the two products equal to each other 20⋅4=80 divide both sides by 10
Incorrect Choices:
(A)
Tip: Just because a number appears in the question doesn’t mean it is the answer.
(B)
This answer is meant to confuse you. It is the quotient of the numbers 20 20 20 and 4 4 4, which appear in the prompt.
(D)
Tip: Just because a number appears in the question doesn’t mean it is the answer.
(E)
If we vary y y y directly with x x x, instead of inversely, we will start with the equation y=k x y=kx y=k x. Using y=20 y=20 y=20 and x=4 x=4 x=4, we will find that 20=k⋅4 20=k\cdot 4 20=k⋅4. Dividing both sides by 4 4 4, we will get k=5 k=5 k=5. Then, for x=10,y=5⋅10=50 x=10, y=5\cdot 10 =50 x=10,y=5⋅10=50. But, this is wrong.
Review
If you thought these examples difficult and you need to review the material, these links will help:
Direct variation
Indirect variation
SAT Exponents
SAT Translating Word Problems
Fractions
Combining Like Terms
Distributive Property
a(b+c)=a b+a c a(b+c)=ab+ac a(b+c)=ab+a c
Simple Equations
SAT Tips for Direct Variation
Know the Rules of Exponents.
Follow order of operations.
SAT General Tips
Cite as: SAT Direct and Inverse Variation. Brilliant.org. Retrieved 11:05, September 29, 2025, from
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11857 | https://math.stackexchange.com/questions/1298668/proof-of-sum-in-an-inequality | calculus - Proof of sum in an inequality - Mathematics Stack Exchange
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Proof of sum in an inequality
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Asked 10 years, 4 months ago
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I was having hard time solving this one, any help will be greatly appreciated. prove that:
39 e 2≤∑n=1∞4 n 2−1 e n−3 e≤54 e 2 39 e 2≤∑n=1∞4 n 2−1 e n−3 e≤54 e 2
calculus
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asked May 25, 2015 at 21:12
Roy1191191Roy1191191
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Hint:
Let f(x)=4 x 2−1 e x f(x)=4 x 2−1 e x. Note that this function decreases for x≥3 x≥3.
Further, we can check that f(2)>∫3 2 f(x)d x f(2)>∫2 3 f(x)d x.
⟹∫3 2 f(x)d x+∫∞3 f(x)d x<f(2)+∑3∞f(n)<f(2)+∫∞2 f(x)d x⟹∫2 3 f(x)d x+∫3∞f(x)d x<f(2)+∑3∞f(n)<f(2)+∫2∞f(x)d x
⟹∫∞2 f(x)d x<∑2∞f(n)<f(2)+∫∞2 f(x)d x⟹∫2∞f(x)d x<∑2∞f(n)<f(2)+∫2∞f(x)d x
⟹39 e 2<∑2∞f(n)<54 e 2⟹39 e 2<∑2∞f(n)<54 e 2
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answered May 26, 2015 at 7:15
MacavityMacavity
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1 This is a nice argument!user84413 –user84413 2015-05-26 15:06:19 +00:00 Commented May 26, 2015 at 15:06
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We know ∑∞n=0 x n=1 1−x∑n=0∞x n=1 1−x. Taking derivatives ∑∞n=1 n x n−1=1(1−x)2∑n=1∞n x n−1=1(1−x)2. Multiply both sides by x x and take another derivative to get ∑∞n=1 n 2 x n−1=1−x 2(1−x)4∑n=1∞n 2 x n−1=1−x 2(1−x)4. Thus ∑∞n=1 4 n 2 x n=4 x(1−x 2)(1−x)4∑n=1∞4 n 2 x n=4 x(1−x 2)(1−x)4. Thus ∑∞n=1(4 n 2 x n−x n)=4 x(1−x 2)(1−x)4−∑∞n=1 x n=4 x(1−x 2)(1−x)4−x 1−x∑n=1∞(4 n 2 x n−x n)=4 x(1−x 2)(1−x)4−∑n=1∞x n=4 x(1−x 2)(1−x)4−x 1−x. Thus ∑∞n=1(4 n 2−1)x n=4 x(1−x 2)(1−x)4−x 1−x∑n=1∞(4 n 2−1)x n=4 x(1−x 2)(1−x)4−x 1−x. Plug in 1 e 1 e to get ∑∞n=1 4 n 2−1 e n=4 e(e+1)(e−1)3−1 e−1∑n=1∞4 n 2−1 e n=4 e(e+1)(e−1)3−1 e−1.
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edited May 25, 2015 at 22:47
answered May 25, 2015 at 22:37
Gregory GrantGregory Grant
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How do you finish proving the inequalities using this?user84413 –user84413 2015-05-25 23:56:11 +00:00 Commented May 25, 2015 at 23:56
@user84413 I was wondering that myself. I did cross multiply the first one and simplify and it was evident. I didn't check the other one. But I don't think this is the solution whoever asked the question has in mind, because otherwise why didn't they just ask for an exact value?Gregory Grant –Gregory Grant 2015-05-26 00:05:52 +00:00 Commented May 26, 2015 at 0:05
Thanks for your reply, and you're right that the first inequality can be checked using this. The other one doesn't seem quite so easy to verify, though, so I suspect you are right.user84413 –user84413 2015-05-26 00:23:44 +00:00 Commented May 26, 2015 at 0:23
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4∑n=1∞n 2(1 e)n 4∑n=1∞n 2(1 e)n is related to the infinite series: ∑n=1∞x n,|x|<1∑n=1∞x n,|x|<1 through twice differentiation. Can you continue?
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answered May 25, 2015 at 21:16
DeepSeaDeepSea
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not really can you elaborate?Roy1191191 –Roy1191191 2015-05-25 21:26:43 +00:00 Commented May 25, 2015 at 21:26
you can use this series to find the sum and take it from there.DeepSea –DeepSea 2015-05-25 21:27:16 +00:00 Commented May 25, 2015 at 21:27
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11858 | https://www.youtube.com/watch?v=7Qs7u-IV6Xs | Axial Stress Example 3 - Mechanics of Materials
Civil Engineering with Amir
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- Normal Stress; Axial Stress (Example 3)
In this video, we demonstrate the process of calculating the axial stress in two vertical rods supporting a horizontal rigid member. This question was selected from Chapter 2 (Stress and Strain; Introduction to Design) of Craig's Mechanics of Materials (3rd edition).
Mechanics_of_Materials #Solid_Mechanics #Strength_of_Materials #Structural_Mechanics #Material_Strength #Mechanics_of_Solids #Civil_Engineering #Mechanical_Engineering #Problem_Solving #Tutorial #Stress #Strain
Transcript:
in this question we have a rigid beam which is supported by two vertical rods number one and number two in the first part of the question it's asking if we apply this load P at a distance of one meter from point A and two meters from point B what would be the axial stresses in Rod number one and number two and in the second part it is asking at what distance X from a you must apply the load so we can get the same axial stresses in row number one and number two for the first part we can use this simplified diagram and use the equations of equilibrium to calculate F1 and F2 in the y direction we can write Sigma f y is equal to zero so F1 plus F2 minus 60 is equal to zero so F1 plus F2 is equal to 60. and if we take the moment about point a you can write Sigma M A is equal to zero this direction so 60 multiplied by 1 meter minus F2 multiplied by 3 meters is equal to zero so F2 is equal to 20 k n and now if we put this F2 in this equation we have F1 plus 20 is equal to 60 so F1 is equal to 40 km Now using this equation for axial stress which is force divided by cross-sectional area if you put these uh forces that we determined and use the cross-section area for each Rod we can calculate their axial stresses for Rod number one T1 is equal to 25 millimeters before 2 is 20 millimeters so we can write A1 is equal to pie 0.025 meter and a power of 2 divided by 4 which is approximately equal to 4.91 [Music] multiplied by 10 and the power of negative 4 square meter and for member two we have a two equal to pi multiplied by 0.020 meter and a power of two divided by four which is approximately equal to 3.14 multiplied by 10 in the power of minus 4 square meter now we're going to write down these values into the table so for F1 we have 40 for F2 we have 20 for A1 we have 4.91 and for A2 we have 3.14 so for Sigma 1 if we divide 40 by 4.91 multiplied by 10 and the power of minus 4 you'll get a stress equal to 81 .487 megapascal and for Sigma 2 if we divide this value to this value we will get stress equal to 63 point six six two megapascal so this was part one of the question in the part B we don't know this distance but we know that the stress in this rod and this Rod are the same so we can write Sigma 1 is equal to F1 divided by A1 and sigma 2 is F2 divided by A2 and we know Sigma 1 and sigma 2 are equal and we know the value of A1 and A2 we already calculated them here so F1 divided by 4.91 multiplied by 10 and a power of four this one is F2 divided by 3.14 multiplied by 10 in the power of minus 4 and we know Sigma 1 is equal to Sigma 2 so we can write F1 divided by 4.91 is equal to F2 divided by 3.14 so F2 is equal to 0.64 F1 and now here we can substitute F2 with this value if you write the equation of equilibrium in the y direction we have Sigma FY equal to zero so F1 plus 0.684 F1 is equal to 60 so F1 is equal to 36 point 59 km and F2 is equal to 36.59 multiplied by 0.60 4 so F2 would be equal to 23.41 kn and in order to determine X we can take the moments about point a again m a is equal to zero so 60 multiplied by X minus 23.41 multiplied by 3 is equal to zero so X is almost equal to 1.171 meter and now using this table again if we divide this value by this value we will get an axial stress equal to 74 point five three one Mega Pascal and we know if we divide this value over this value we will get the same answer so 74 point five three one Mega possible that's it thanks for watching this video please subscribe to our Channel if you haven't already and press the like button if you like this post feel free to write your comments below if you have any suggestions or comments and see you in the next video thanks |
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Contents
1 Physical properties
1.1 Atomic
1.2 Bulk
1.3 Isotopes
2 Chemistry
2.1 Inorganic compounds
2.1.1 Lead(II)
2.1.2 Lead(IV)
2.1.3 Other oxidation states
2.2 Organolead
3 Origin and occurrence
3.1 In space
3.2 On Earth
4 Etymology
5 History
5.1 Prehistory and early history
5.2 Classical era
5.2.1 Confusion with tin and antimony
5.3 Middle Ages and the Renaissance
5.4 Outside Europe and Asia
5.5 Industrial Revolution
5.6 Modern era
6 Production
6.1 Primary
6.1.1 Two-stage process
6.1.2 Direct process
6.1.3 Alternatives
6.2 Secondary
7 Applications
7.1 Elemental form
7.2 Batteries
7.3 Coating for cables
7.4 Compounds
8 Biological effects
8.1 Toxicity
8.2 Effects
8.3 Exposure sources
8.3.1 Lead in foods
8.3.2 Lead in plastic toys
8.4 Treatment
9 Environmental effects
10 Restriction and remediation
11 See also
12 Notes
13 References
14 Bibliography
15 Further reading
16 External links
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From Wikipedia, the free encyclopedia
This article is about the chemical element. For other uses, see Lead (disambiguation).
"Leads" redirects here; not to be confused with Leeds.
Chemical element with atomic number 82 (Pb)
Lead, 82Pb
| Lead |
| Pronunciation | /lɛd/ ⓘ (led) |
| Appearance | metallic gray |
| Standard atomic weight Ar°(Pb) |
| [206.14, 207.94] 207.2±1.1 (abridged) |
| |
| Lead in the periodic table |
| | | | | --- | Hydrogen | | Helium | | Lithium | Beryllium | | Boron | Carbon | Nitrogen | Oxygen | Fluorine | Neon | | Sodium | Magnesium | | Aluminium | Silicon | Phosphorus | Sulfur | Chlorine | Argon | | Potassium | Calcium | | Scandium | Titanium | Vanadium | Chromium | Manganese | Iron | Cobalt | Nickel | Copper | Zinc | Gallium | Germanium | Arsenic | Selenium | Bromine | Krypton | | Rubidium | Strontium | | | Yttrium | Zirconium | Niobium | Molybdenum | Technetium | Ruthenium | Rhodium | Palladium | Silver | Cadmium | Indium | Tin | Antimony | Tellurium | Iodine | Xenon | | Caesium | Barium | Lanthanum | Cerium | Praseodymium | Neodymium | Promethium | Samarium | Europium | Gadolinium | Terbium | Dysprosium | Holmium | Erbium | Thulium | Ytterbium | Lutetium | Hafnium | Tantalum | Tungsten | Rhenium | Osmium | Iridium | Platinum | Gold | Mercury (element) | Thallium | Lead | Bismuth | Polonium | Astatine | Radon | | Francium | Radium | Actinium | Thorium | Protactinium | Uranium | Neptunium | Plutonium | Americium | Curium | Berkelium | Californium | Einsteinium | Fermium | Mendelevium | Nobelium | Lawrencium | Rutherfordium | Dubnium | Seaborgium | Bohrium | Hassium | Meitnerium | Darmstadtium | Roentgenium | Copernicium | Nihonium | Flerovium | Moscovium | Livermorium | Tennessine | Oganesson | Sn↑Pb↓Fl thallium ← lead → bismuth |
| Atomic number (Z) | 82 |
| Group | group 14 (carbon group) |
| Period | period 6 |
| Block | p-block |
| Electron configuration | [Xe] 4f14 5d10 6s2 6p2 |
| Electrons per shell | 2, 8, 18, 32, 18, 4 |
| Physical properties |
| Phase at STP | solid |
| Melting point | 600.61 K (327.46 °C, 621.43 °F) |
| Boiling point | 2022 K (1749 °C, 3180 °F) |
| Density (at 20 °C) | 11.348 g/cm3 |
| when liquid (at m.p.) | 10.66 g/cm3 |
| Heat of fusion | 4.77 kJ/mol |
| Heat of vaporization | 179.5 kJ/mol |
| Molar heat capacity | 26.650 J/(mol·K) |
| Vapor pressure | P (Pa) | 1 | 10 | 100 | 1 k | 10 k | 100 k | --- --- --- | at T (K) | 978 | 1088 | 1229 | 1412 | 1660 | 2027 | |
| Atomic properties |
| Oxidation states | common: +2, +4 −4, −2,? −1,? 0, +1,? +3? |
| Electronegativity | Pauling scale: 2.33 (in +4), 1.87 (in +2) |
| Ionization energies | 1st: 715.6 kJ/mol 2nd: 1450.5 kJ/mol 3rd: 3081.5 kJ/mol |
| Atomic radius | empirical: 175 pm |
| Covalent radius | 146±5 pm |
| Van der Waals radius | 202 pm |
| Spectral lines of lead |
| Other properties |
| Natural occurrence | primordial |
| Crystal structure | face-centered cubic (fcc) (cF4) |
| Lattice constant | a = 494.99 pm (at 20 °C) |
| Thermal expansion | 28.73×10−6/K (at 20 °C) |
| Thermal conductivity | 35.3 W/(m⋅K) |
| Electrical resistivity | 208 nΩ⋅m (at 20 °C) |
| Magnetic ordering | diamagnetic |
| Molar magnetic susceptibility | −23.0×10−6 cm3/mol (at 298 K) |
| Young's modulus | 16 GPa |
| Shear modulus | 5.6 GPa |
| Bulk modulus | 46 GPa |
| Speed of sound thin rod | 1190 m/s (at r.t.) (annealed) |
| Poisson ratio | 0.44 |
| Mohs hardness | 1.5 |
| Brinell hardness | 38–50 MPa |
| CAS Number | 7439-92-1 |
| History |
| Naming | possibly from a PIE root meaning “to flow”, for its low melting point |
| Discovery | Middle East (7000 BCE) |
| Symbol | "Pb": from Latin plumbum |
| Isotopes of lead v e |
| |
| | Main isotopes | Decay | --- | | Isotope | abundance | half-life (t1/2) | mode | product | | 202Pb | synth | 5.25×104 y | ε | 202Tl | | 204Pb | 1.40% | stable | | 205Pb | synth | 1.70×107 y | ε | 205Tl | | 206Pb | 24.1% | stable | | 207Pb | 22.1% | stable | | 208Pb | 52.4% | stable | | 209Pb | trace | 3.235 h | β− | 209Bi | | 210Pb | trace | 22.2 y | β− | 210Bi | | α | 206Hg | | 211Pb | trace | 36.16 min | β− | 211Bi | | 212Pb | trace | 10.627 h | β− | 212Bi | | 214Pb | trace | 27.06 min | β− | 214Bi | Isotopic abundances vary greatly by sample |
| |
| Category: Lead view talk edit | references |
Lead (/lɛd/ ⓘ) is a chemical element with the symbol Pb (from the Latin plumbum) and atomic number 82. It is a heavy metal, denser than most common materials. Lead is soft, malleable, and has a relatively low melting point. When freshly cut, it appears shiny gray with a bluish tint, but tarnishes to dull gray on exposure to air. Lead has the highest atomic number of any stable element, and three of its isotopes are endpoints of major nuclear decay chains of heavier elements.
Lead is a relatively unreactive post-transition metal. Its weak metallic character is shown by its amphoteric behavior: lead and lead oxides react with both acids and bases, and it tends to form covalent bonds. Lead compounds usually occur in the +2 oxidation state rather than the +4 state common in lighter members of the carbon group, with exceptions mostly limited to organolead compounds. Like the lighter members of the group, lead can bond with itself, forming chains and polyhedral structures.
Easily extracted from its ores, lead was known to prehistoric peoples in the Near East. Galena is its principal ore and often contains silver, encouraging its widespread extraction and use in ancient Rome. Production declined after the fall of Rome and did not reach similar levels until the Industrial Revolution. Lead played a role in developing the printing press, as movable type could be readily cast from lead alloys. In 2014, annual global production was about ten million tonnes, over half from recycling. Lead's high density, low melting point, ductility, and resistance to oxidation, together with its abundance and low cost, supported its extensive use in construction, plumbing, batteries, ammunition, weights, solders, pewter, fusible alloys, lead paints, leaded gasoline, and radiation shielding.
Lead is a neurotoxin that accumulates in soft tissues and bones. It damages the nervous system, interferes with biological enzymes, and can cause neurological disorders ranging from behavioral problems to brain damage. It also affects cardiovascular and renal systems. Lead's toxicity was noted by ancient Greek and Roman writers, but became widely recognized in Europe in the late 19th century.
Physical properties
[edit]
Atomic
[edit]
A lead atom has 82 electrons, with the electron configuration [Xe]4f145d106s26p2. The combined first and second ionization energies—the total energy required to remove the two 6p electrons—are similar to those of tin, lead's immediate neighbor above in the carbon group. This is unusual, as ionization energies typically decrease down a group due to the outer electrons being farther from the nucleus and more shielded by inner orbitals. However, the sum of the first four ionization energies of lead is higher than that of tin, contrary to periodic trends. This anomaly is explained by relativistic effects, which become significant in heavier atoms. These effects contract the s and p orbitals, giving lead's 6s electrons greater binding energies than its 5s electrons. This leads to the inert pair effect, where the 6s electrons are less likely to participate in bonding. The result is stabilization of the +2 oxidation state and unusually long distances between nearest atoms in crystalline lead.
Lighter carbon-group congeners of lead form stable or metastable allotropes with the tetrahedrally coordinated, covalently bonded diamond cubic structure. In these elements, the s- and p-orbital energy levels are close enough to allow mixing into four hybrid sp3 orbitals. In lead, however, the inert pair effect increases the separation between s- and p-orbitals so much that the energy gain from hybridization is insufficient to overcome this gap. Instead of a diamond cubic arrangement, lead forms metallic bonds in which only the p-electrons are delocalized and shared among Pb2+ ions. Consequently, lead adopts a face-centered cubic structure, similar to the divalent metals calcium and strontium.[a][b][c]
Bulk
[edit]
Pure lead has a bright, shiny gray appearance with a faint blue tint. It tarnishes when exposed to moist air, developing a dull surface whose color depends on environmental conditions. Lead is characterized by high density, malleability, ductility, and resistance to corrosion due to passivation.
Its close-packed face-centered cubic structure and high atomic mass give lead a density of 11.34 g/cm3, greater than that of common metals such as iron (7.87 g/cm3), copper (8.93 g/cm3), and zinc (7.14 g/cm3). This high density is the origin of the idiom to go over like a lead balloon.[d] Some rarer metals are denser: tungsten and gold are both 19.3 g/cm3, while osmium—the densest known metal—has a density of 22.59 g/cm3, nearly twice that of lead.
Lead is soft, with a Mohs hardness of 1.5, and can be scratched with a fingernail. It is very malleable and moderately ductile.[e] Its bulk modulus—a measure of resistance to compression—is 45.8 GPa, compared with 75.2 GPa for aluminium, 137.8 GPa for copper, and 160–169 GPa for mild steel. Lead's tensile strength is low, at 12–17 MPa (around six times lower than aluminium, ten times lower than copper, and fifteen times lower than mild steel). Strength can be increased by alloying with small amounts of copper or antimony.
Lead melts at 327.5 °C (621.5 °F), a relatively low melting point compared to most metals,[f] and has a boiling point of 1749 °C (3180 °F), the lowest among the carbon-group elements. Its electrical resistivity at 20 °C is 192 nanoohm-meters, almost an order of magnitude higher than that of good conductors (copper: 15.43 nΩ·m; gold: 20.51 nΩ·m; aluminium: 24.15 nΩ·m). Lead becomes a superconductor below 7.19 K, which is the highest critical temperature among type-I superconductors and the third highest among the elemental superconductors.
Isotopes
[edit]
Main article: Isotopes of lead
Natural lead consists of four stable isotopes with mass numbers 204, 206, 207, and 208, along with traces of six short-lived radioisotopes with mass numbers 209–214. The relatively high number of isotopes is consistent with lead's even atomic number.[g] Lead has a magic number of protons (82), making its nucleus especially stable according to the nuclear shell model. Lead-208 also has 126 neutrons, another magic number, which may account for its exceptional stability.
With its high atomic number, lead is the heaviest element whose natural isotopes are considered stable; lead-208 is the heaviest stable nucleus known. This distinction previously belonged to bismuth (atomic number 83) until its sole primordial isotope, bismuth-209, was found in 2003 to decay extremely slowly.[h] Although the four stable isotopes of lead could theoretically undergo alpha decay to mercury isotopes with an energy release, no such decay has been observed; their predicted half-lives range from 1035 to 10189 years, at least 1025 times the current age of the universe.
Three of lead's stable isotopes—lead-206, lead-207, and lead-208—are the end products of the three major natural decay chains: the uranium chain (from uranium-238), the actinium chain (from uranium-235), and the thorium chain (from thorium-232), respectively. The isotopic composition of a rock sample depends on the presence of these parent isotopes; for example, lead-208 abundance can vary from about 52% in ordinary samples to as much as 90% in thorium ores. For this reason, the standard atomic weight of lead is reported to only one decimal place. Over time, the ratios of these isotopes to lead-204 increase as they are produced by radioactive decay. These variations allow for lead–lead and uranium–lead dating. Lead-207 exhibits nuclear magnetic resonance, a property used to study its compounds in both solution and solid states, including in biological systems such as the human body.
Chemistry
[edit]
When exposed to moist air, bulk lead develops a protective surface layer of variable composition. Lead(II) carbonate is a common constituent, and in urban or maritime environments, lead(II) sulfate or lead(II) chloride may also be present. This layer renders bulk lead effectively inert under atmospheric conditions. In contrast, finely powdered lead, like many metals, is pyrophoric and burns with a bluish-white flame.
Lead reacts with fluorine at room temperature to form lead(II) fluoride. Its reaction with chlorine is similar but requires heating, as the resulting chloride layer reduces further reactivity. Molten lead combines with the chalcogens to produce lead(II) chalcogenides.
The metal resists attack by sulfuric and phosphoric acids but not by hydrochloric or nitric acids; the difference arises from the insolubility and subsequent passivation of certain lead salts. Organic acids, such as acetic acid, dissolve lead in the presence of oxygen. Concentrated alkalis can also dissolve lead, producing plumbites.
Inorganic compounds
[edit]
See also: Lead compounds
Lead exhibits two principal oxidation states: +4 and +2. While the tetravalent state is characteristic of the carbon group, the divalent state is rare for carbon and silicon, less common for germanium, significant but not dominant for tin, and the most prevalent for lead. This predominance is linked to relativistic effects—specifically the inert pair effect—which occurs when there is a large electronegativity difference between lead and anions such as oxide, halide, or nitride. In such cases, lead develops a pronounced partial positive charge, causing a stronger contraction of the 6s orbital compared to the 6p orbital and rendering it relatively unreactive in ionic compounds. The inert pair effect is less pronounced in compounds where lead forms covalent bonds with elements of similar electronegativity, such as carbon in organolead compounds. In these, the 6s and 6p orbitals remain comparable in size, and sp3 hybridization remains energetically favorable, making lead predominantly tetravalent in such cases.
The electronegativity values further reflect this behavior: lead(II) has a value of 1.87, and lead(IV) has 2.33. This represents a reversal in the general trend of increasing stability of the +4 oxidation state down the carbon group; by comparison, tin has electronegativities of 1.80 (+2 state) and 1.96 (+4 state).
Lead(II)
[edit]
Lead(II) compounds are characteristic of the inorganic chemistry of lead. Even strong oxidizing agents like fluorine and chlorine react with lead to give only PbF2 and PbCl2. Lead(II) ions are usually colorless in solution, and partially hydrolyze to form Pb(OH)+ and finally [Pb4(OH)4]4+ (in which the hydroxyl ions act as bridging ligands), but are not reducing agents as tin(II) ions are. Techniques for identifying the presence of the Pb2+ ion in water generally rely on the precipitation of lead(II) chloride using dilute hydrochloric acid. As the chloride salt is sparingly soluble in water, in very dilute solutions the precipitation of lead(II) sulfide is instead achieved by bubbling hydrogen sulfide through the solution.
Lead monoxide exists in two polymorphs, litharge α-PbO (red) and massicot β-PbO (yellow), the latter being stable only above around 488 °C. Litharge is the most commonly used inorganic compound of lead. There is no lead(II) hydroxide; increasing the pH of solutions of lead(II) salts leads to hydrolysis and condensation. Lead commonly reacts with heavier chalcogens. Lead sulfide is a semiconductor, a photoconductor, and an extremely sensitive infrared radiation detector. The other two chalcogenides, lead selenide and lead telluride, are likewise photoconducting. They are unusual in that their color becomes lighter going down the group.
Lead dihalides are well-characterized; this includes the diastatide and mixed halides, such as PbFCl. The relative insolubility of the latter forms a useful basis for the gravimetric determination of fluorine. The difluoride was the first solid ionically conducting compound to be discovered (in 1834, by Michael Faraday). The other dihalides decompose on exposure to ultraviolet or visible light, especially the diiodide. Many lead(II) pseudohalides are known, such as the cyanide, cyanate, and thiocyanate. Lead(II) forms an extensive variety of halide coordination complexes, such as [PbCl4]2−, [PbCl6]4−, and the [Pb2Cl9]n5n− chain anion.
Lead(II) sulfate is insoluble in water, like the sulfates of other heavy divalent cations. Lead(II) nitrate and lead(II) acetate are very soluble, and this is exploited in the synthesis of other lead compounds.
Lead(IV)
[edit]
Few inorganic lead(IV) compounds are known. They are only formed in highly oxidizing solutions and do not normally exist under standard conditions. Lead(II) oxide gives a mixed oxide on further oxidation, Pb3O4. It is described as lead(II,IV) oxide, or structurally 2PbO·PbO2, and is the best-known mixed valence lead compound. Lead dioxide is a strong oxidizing agent, capable of oxidizing hydrochloric acid to chlorine gas. This is because the expected PbCl4 that would be produced is unstable and spontaneously decomposes to PbCl2 and Cl2. Analogously to lead monoxide, lead dioxide is capable of forming plumbate anions. Lead disulfide and lead diselenide are only stable at high pressures. Lead tetrafluoride, a yellow crystalline powder, is stable, but less so than the difluoride. Lead tetrachloride (a yellow oil) decomposes at room temperature, lead tetrabromide is less stable still, and the existence of lead tetraiodide is questionable.
Other oxidation states
[edit]
See also: Plumbide
Some lead compounds exist in formal oxidation states other than +4 or +2. Lead(III) may be obtained, as an intermediate between lead(II) and lead(IV), in larger organolead complexes; this oxidation state is not stable, as both the lead(III) ion and the larger complexes containing it are radicals. The same applies for lead(I), which can be found in such radical species.
Numerous mixed lead(II,IV) oxides are known. When PbO2 is heated in air, it becomes Pb12O19 at 293 °C, Pb12O17 at 351 °C, Pb3O4 at 374 °C, and finally PbO at 605 °C. A further sesquioxide, Pb2O3, can be obtained at high pressure, along with several non-stoichiometric phases. Many of them show defective fluorite structures in which some oxygen atoms are replaced by vacancies: PbO can be considered as having such a structure, with every alternate layer of oxygen atoms absent.
Negative oxidation states can occur as Zintl phases, as either free lead anions, as in Ba2Pb, with lead formally being lead(−IV), or in oxygen-sensitive ring-shaped or polyhedral cluster ions such as the trigonal bipyramidal Pb52− ion, where two lead atoms are lead(−I) and three are lead(0). In such anions, each atom is at a polyhedral vertex and contributes two electrons to each covalent bond along an edge from their sp3 hybrid orbitals, the other two being an external lone pair. They may be made in liquid ammonia via the reduction of lead by sodium.
Organolead
[edit]
Main article: Organolead compound
Lead can form multiply-bonded chains, a property it shares with its lighter homologs in the carbon group. Its capacity to do so is much less because the Pb–Pb bond energy is over three and a half times lower than that of the C–C bond. With itself, lead can build metal–metal bonds of an order up to three. With carbon, lead forms organolead compounds similar to, but generally less stable than, typical organic compounds (due to the Pb–C bond being rather weak). This makes the organometallic chemistry of lead far less wide-ranging than that of tin. Lead predominantly forms organolead(IV) compounds, even when starting with inorganic lead(II) reactants; very few organolead(II) compounds are known. The most well-characterized exceptions are Pb[CH(SiMe3)2]2 and plumbocene.
The lead analog of the simplest organic compound, methane, is plumbane. Plumbane may be obtained in a reaction between metallic lead and atomic hydrogen. Two simple derivatives, tetramethyllead and tetraethyllead, are the best-known organolead compounds. These compounds are relatively stable: tetraethyllead only starts to decompose if heated or if exposed to sunlight or ultraviolet light.[i] With sodium metal, lead readily forms an equimolar alloy that reacts with alkyl halides to form organometallic compounds such as tetraethyllead. The oxidizing nature of many organolead compounds is usefully exploited: lead tetraacetate is an important laboratory reagent for oxidation in organic synthesis. Tetraethyllead, once added to automotive gasoline, was produced in larger quantities than any other organometallic compound, and is still widely used in fuel for small aircraft. Other organolead compounds are less chemically stable. For many organic compounds, a lead analog does not exist.
Origin and occurrence
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Solar System abundances
| Atomicnumber | Element | Relativeamount |
| 42 | Molybdenum | 0.798 |
| 46 | Palladium | 0.440 |
| 50 | Tin | 1.146 |
| 78 | Platinum | 0.417 |
| 80 | Mercury | 0.127 |
| 82 | Lead | 1 |
| 90 | Thorium | 0.011 |
| 92 | Uranium | 0.003 |
In space
[edit]
Lead's per-particle abundance in the Solar System is 0.121 ppb (parts per billion).[j] This figure is two and a half times higher than that of platinum, eight times more than mercury, and seventeen times more than gold. The amount of lead in the universe is slowly increasing as most heavier atoms (all of which are unstable) gradually decay to lead. The abundance of lead in the Solar System since its formation 4.5 billion years ago has increased by about 0.75%. The Solar System abundances table shows that lead, despite its relatively high atomic number, is more prevalent than most other elements with atomic numbers greater than 40.
Primordial lead—which comprises the isotopes lead-204, lead-206, lead-207, and lead-208—was mostly created as a result of repetitive neutron capture processes occurring in stars. The two main modes of capture are the s- and r-processes.
In the s-process (s is for "slow"), captures are separated by years or decades, allowing less stable nuclei to undergo beta decay. A stable thallium-203 nucleus can capture a neutron and become thallium-204; this undergoes beta decay to give stable lead-204; on capturing another neutron, it becomes lead-205, which has a half-life of around 17 million years. Further captures result in lead-206, lead-207, and lead-208. On capturing another neutron, lead-208 becomes lead-209, which quickly decays into bismuth-209. On capturing another neutron, bismuth-209 becomes bismuth-210, and this beta decays to polonium-210, which alpha decays to lead-206. The cycle hence ends at lead-206, lead-207, lead-208, and bismuth-209.
In the r-process (r is for "rapid"), captures happen faster than nuclei can decay. This occurs in environments with a high neutron density, such as a supernova or the merger of two neutron stars. The neutron flux involved may be on the order of 1022 neutrons per square centimeter per second. The r-process does not form as much lead as the s-process. It tends to stop once neutron-rich nuclei reach 126 neutrons. At this point, the neutrons are arranged in complete shells in the atomic nucleus, and it becomes harder to energetically accommodate more of them. When the neutron flux subsides, these nuclei beta decay into stable isotopes of osmium, iridium, platinum.
On Earth
[edit]
Lead is classified as a chalcophile under the Goldschmidt classification, meaning it is generally found combined with sulfur. It rarely occurs in its native, metallic form. Many lead minerals are relatively light and, over the course of the Earth's history, have remained in the crust instead of sinking deeper into the Earth's interior. This accounts for lead's relatively high crustal abundance of 14 ppm; it is the 36th most abundant element in the crust.[k]
The main lead-bearing mineral is galena (PbS), which is mostly found with zinc ores. Most other lead minerals are related to galena in some way; boulangerite, Pb5Sb4S11, is a mixed sulfide derived from galena; anglesite, PbSO4, is a product of galena oxidation; and cerussite or white lead ore, PbCO3, is a decomposition product of galena. Arsenic, tin, antimony, silver, gold, copper, bismuth are common impurities in lead minerals.
World lead resources exceed two billion tons. Significant deposits are located in Australia, China, Ireland, Mexico, Peru, Portugal, Russia, United States. Global reserves—resources that are economically feasible to extract—totaled 88 million tons in 2016, of which Australia had 35 million, China 17 million, Russia 6.4 million.
Typical background concentrations of lead do not exceed 0.1 μg/m3 in the atmosphere; 100 mg/kg in soil; 4 mg/kg in vegetation, 5 μg/L in fresh water and seawater.
Etymology
[edit]
The modern English word lead is of Germanic origin; it comes from the Middle English leed and Old English lēad (with the macron above the "e" signifying that the vowel sound of that letter is long). The Old English word is derived from the hypothetical reconstructed Proto-Germanic lauda- ('lead'). According to linguistic theory, this word bore descendants in multiple Germanic languages of exactly the same meaning.
There is no consensus on the origin of the Proto-Germanic lauda-. One hypothesis suggests it is derived from Proto-Indo-European lAudh- ('lead'; capitalization of the vowel is equivalent to the macron). Another hypothesis suggests it is borrowed from Proto-Celtic ɸloud-io- ('lead'). This word is related to the Latin plumbum, which gave the element its chemical symbol Pb. The word ɸloud-io- is thought to be the origin of Proto-Germanic bliwa- (which also means 'lead'), from which stemmed the German Blei.
The name of the chemical element is not related to the verb of the same spelling, which is derived from Proto-Germanic laidijan- ('to lead').
History
[edit]
Prehistory and early history
[edit]
Metallic lead beads dating back to 7000–6500 BC have been found in Asia Minor and may represent the first example of metal smelting. At that time, lead had few (if any) applications due to its softness and dull appearance. The major reason for the spread of lead production was its association with silver, which may be obtained by burning galena (a common lead mineral). The Ancient Egyptians were the first to use lead minerals in cosmetics, an application that spread to Ancient Greece and beyond; the Egyptians had used lead for sinkers in fishing nets, glazes, glasses, enamels, ornaments. Various civilizations of the Fertile Crescent used lead as a writing material, as coins, and as a construction material. Lead was used by the ancient Chinese as a stimulant, as currency, as contraceptive, and in chopsticks. The Indus Valley civilization and the Mesoamericans used it for making amulets; and the eastern and southern Africans used lead in wire drawing.
Classical era
[edit]
Because silver was extensively used as a decorative material and an exchange medium, lead deposits came to be worked in Asia Minor from 3000 BC; later, lead deposits were developed in the Aegean and Laurion. These three regions collectively dominated production of mined lead until c. 1200 BC. Beginning c. 2000 BC, the Phoenicians worked deposits in the Iberian peninsula; by 1600 BC, lead mining existed in Cyprus, Greece, and Sardinia.
Rome's territorial expansion in Europe and across the Mediterranean, and its development of mining, led to it becoming the greatest producer of lead during the classical era, with an estimated annual output peaking at 80,000 tonnes. Like their predecessors, the Romans obtained lead mostly as a by-product of silver smelting. Lead mining occurred in central Europe, Britain, Balkans, Greece, Anatolia, Hispania, the latter accounting for 40% of world production.
Lead tablets were commonly used as a material for letters. Lead coffins, cast in flat sand forms and with interchangeable motifs to suit the faith of the deceased, were used in ancient Judea. Lead was used to make sling bullets from the 5th century BC. In Roman times, lead sling bullets were amply used, and were effective at a distance of between 100 and 150 meters. The Balearic slingers, used as mercenaries in Carthaginian and Roman armies, were famous for their shooting distance and accuracy.
Lead was used for making water pipes in the Roman Empire; the Latin word for the metal, plumbum, is the origin of the English word "plumbing". Its ease of working, its low melting point enabling the easy fabrication of completely waterproof welded joints, and its resistance to corrosion ensured its widespread use in other applications, including pharmaceuticals, roofing, currency, warfare. Writers of the time, such as Cato the Elder, Columella, and Pliny the Elder, recommended lead (and lead-coated) vessels for the preparation of sweeteners and preservatives added to wine and food. The lead conferred an agreeable taste due to the formation of "sugar of lead" (lead(II) acetate), whereas copper vessels imparted a bitter flavor through verdigris formation.
This metal was by far the most used material in classical antiquity, and it is appropriate to refer to the (Roman) Lead Age. Lead was to the Romans what plastic is to us.
Heinz Eschnauer and Markus Stoeppler"Wine—An enological specimen bank", 1992
The Roman author Vitruvius reported the health dangers of lead and modern writers have suggested that lead poisoning played a major role in the decline of the Roman Empire.[m] Other researchers have criticized such claims, pointing out, for instance, that not all abdominal pain is caused by lead poisoning. According to archaeological research, Roman lead pipes increased lead levels in tap water but such an effect was "unlikely to have been truly harmful". When lead poisoning did occur, victims were called "saturnine", dark and cynical, after the ghoulish father of the gods, Saturn. By association, lead was considered the father of all metals. Its status in Roman society was low as it was readily available and cheap.
Confusion with tin and antimony
[edit]
Since the Bronze Age, metallurgists and engineers have understood the difference between rare and valuable tin, essential for alloying with copper to produce tough and corrosion resistant bronze, and 'cheap and cheerful' lead. However, the nomenclature in some languages is similar. Romans called lead plumbum nigrum ("black lead"), and tin plumbum candidum ("bright lead"). The association of lead and tin can be seen in other languages: the word olovo in Czech translates to "lead", but in Russian, its cognate олово (olovo) means "tin". To add to the confusion, lead bore a close relation to antimony: both elements commonly occur as sulfides (galena and stibnite), often together. Pliny incorrectly wrote that stibnite would give lead on heating, instead of antimony. In countries such as Turkey and India, the originally Persian name surma (Persian: سرمه) came to refer to either antimony sulfide or lead sulfide, and in some languages, such as Russian, gave its name to antimony (сурьма).
Middle Ages and the Renaissance
[edit]
Lead mining in Western Europe declined after the fall of the Western Roman Empire, with Al-Andalus being the only region having a significant output. The largest production of lead occurred in South Asia and East Asia, especially China and India, where lead mining grew rapidly.
In Europe, lead production began to increase in the 11th and 12th centuries, when it was again used for roofing and piping. Starting in the 13th century, lead was used to create stained glass. In the European and Muslim traditions of alchemy, lead (symbol ♄ in the European tradition) was considered an impure base metal which, by the separation, purification and balancing of its constituent essences, could be transformed to pure and incorruptible gold. During the period, lead was used increasingly for adulterating wine. The use of such wine was forbidden for use in Christian rites by a papal bull in 1498, but it continued to be imbibed and resulted in mass poisonings up to the late 18th century. Lead was a key material in parts of the printing press, and lead dust was commonly inhaled by print workers, causing lead poisoning. Lead also became the chief material for making bullets for firearms: it was cheap, less damaging to iron gun barrels, had a higher density (which allowed for better retention of velocity), and its lower melting point made the production of bullets easier as they could be made using a wood fire. Lead, in the form of Venetian ceruse, was extensively used in cosmetics by Western European aristocracy as whitened faces were regarded as a sign of modesty. This practice later expanded to white wigs and eyeliners, and only faded out with the French Revolution in the late 18th century. A similar fashion appeared in Japan in the 18th century with the emergence of the geishas, a practice that continued long into the 20th century. The white faces of women "came to represent their feminine virtue as Japanese women", with lead commonly used in the whitener.
Outside Europe and Asia
[edit]
In the New World, lead production was recorded soon after the arrival of European settlers. The earliest record dates to 1621 in the English Colony of Virginia, fourteen years after its foundation. In Australia, the first mine opened by colonists on the continent was a lead mine, in 1841. In Africa, lead mining and smelting were known in the Benue Trough and the lower Congo Basin, where lead was used for trade with Europeans, and as a currency by the 17th century, well before the scramble for Africa.
Industrial Revolution
[edit]
In the second half of the 18th century, Britain, and later continental Europe and the United States, experienced the Industrial Revolution. This was the first time during which lead production rates exceeded those of Rome. Britain was the leading producer, losing this status by the mid-19th century with the depletion of its mines and the development of lead mining in Germany, Spain, and the United States. By 1900, the United States was the leader in global lead production, and other non-European nations—Canada, Mexico, and Australia—had begun significant production; production outside Europe exceeded that within. A great share of the demand for lead came from plumbing and painting—lead paints were in regular use. were linked; British physician Alfred Baring Garrod noted a third of his gout patients were plumbers and painters. The effects of chronic ingestion of lead, including mental disorders, were also studied in the 19th century. The first laws aimed at decreasing lead poisoning in factories were enacted during the 1870s and 1880s in the United Kingdom.
Modern era
[edit]
Further evidence of the threat that lead posed to humans was discovered in the late 19th and early 20th centuries. Mechanisms of harm were better understood, lead blindness was documented, and the element was phased out of public use in the United States and Europe. The United Kingdom introduced mandatory factory inspections in 1878 and appointed the first Medical Inspector of Factories in 1898; as a result, a 25-fold decrease in lead poisoning incidents from 1900 to 1944 was reported. Most European countries banned lead paint—commonly used because of its opacity and water resistance—for interiors by 1930.
The last major human exposure to lead was the addition of tetraethyllead to gasoline as an antiknock agent, a practice that originated in the United States in 1921. It was phased out in the United States and the European Union by 2000.
In the 1970s, the United States and Western European countries introduced legislation to reduce lead air pollution. The impact was significant: while a study conducted by the Centers for Disease Control and Prevention in the United States in 1976–1980 showed that 77.8% of the population had elevated blood lead levels, in 1991–1994, a study by the same institute showed the share of people with such high levels dropped to 2.2%. The main product made of lead by the end of the 20th century was the lead–acid battery.
From 1960 to 1990, lead output in the Western Bloc grew by about 31%. The share of the world's lead production by the Eastern Bloc increased from 10% to 30%, from 1950 to 1990, with the Soviet Union being the world's largest producer during the mid-1970s and the 1980s, and China starting major lead production in the late 20th century. Unlike the European communist countries, China was largely unindustrialized by the mid-20th century; in 2004, China surpassed Australia as the largest producer of lead. As was the case during European industrialization, lead has had a negative effect on health in China.
Production
[edit]
As of 2014, production of lead is increasing worldwide due to its use in lead–acid batteries. There are two major categories of production: primary from mined ores, and secondary from scrap. In 2014, 4.58 million metric tons came from primary production and 5.64 million from secondary production. The top three producers of mined lead concentrate in that year were China, Australia, and United States. The top three producers of refined lead were China, United States, and India. According to the Metal Stocks in Society report of 2010, the total amount of lead in use, stockpiled, discarded, or dissipated into the environment, on a global basis, is 8 kg per capita. Much of this is in more developed countries (20–150 kg per capita) rather than less developed ones (1–4 kg per capita).
The primary and secondary lead production processes are similar. Some primary production plants now supplement their operations with scrap lead, and this trend is likely to increase in the future. Given adequate techniques, lead obtained via secondary processes is indistinguishable from lead obtained via primary processes. Scrap lead from the building trade is usually fairly clean and is re-melted without the need for smelting, though refining is sometimes needed. Secondary lead production is therefore cheaper, in terms of energy requirements, than is primary production, often by 50% or more.
Primary
[edit]
Most lead ores contain a low percentage of lead (rich ores have a typical content of 3–8%) which must be concentrated for extraction. During initial processing, ores typically undergo crushing, dense-medium separation, grinding, froth flotation, drying. The resulting concentrate, which has a lead content of 30–80% by mass (regularly 50–60%), is then turned into (impure) lead metal.
There are two main ways of doing this: a two-stage process involving roasting followed by blast furnace extraction, carried out in separate vessels; or a direct process in which the extraction of the concentrate occurs in a single vessel. The latter has become the most common route, though the former is still significant.
World's largest mining countries of lead, 2016
| Country | Output(thousand tons) |
| China | 2,400 |
| Australia | 500 |
| United States | 335 |
| Peru | 310 |
| Mexico | 250 |
| Russia | 225 |
| India | 135 |
| Bolivia | 80 |
| Sweden | 76 |
| Turkey | 75 |
| Iran | 41 |
| Kazakhstan | 41 |
| Poland | 40 |
| South Africa | 40 |
| North Korea | 35 |
| Ireland | 33 |
| Macedonia | 33 |
| Other countries | 170 |
Two-stage process
[edit]
First, the sulfide concentrate is roasted in air to oxidize the lead sulfide:
: 2 PbS(s) + 3 O2(g) → 2 PbO(s) + 2 SO2(g)↑
As the original concentrate was not pure lead sulfide, roasting yields not only the desired lead(II) oxide, but a mixture of oxides, sulfates, and silicates of lead and of the other metals contained in the ore. This impure lead oxide is reduced in a coke-fired blast furnace to the (again, impure) metal:
: 2 PbO(s) + C(s) → 2 Pb(s) + CO2(g)↑
Impurities are mostly arsenic, antimony, bismuth, zinc, copper, silver, and gold. Typically they are removed in a series of pyrometallurgical processes. The melt is treated in a reverberatory furnace with air, steam, sulfur, which oxidizes the impurities except for silver, gold, bismuth. Oxidized contaminants float to the top of the melt and are skimmed off. Metallic silver and gold are removed and recovered economically by means of the Parkes process, in which zinc is added to lead. Zinc, which is immiscible in lead, dissolves the silver and gold. The zinc solution can be separated from the lead, and the silver and gold retrieved. De-silvered lead is freed of bismuth by the Betterton–Kroll process, treating it with metallic calcium and magnesium. The resulting bismuth dross can be skimmed off.
Alternatively to the pyrometallurgical processes, very pure lead can be obtained by processing smelted lead electrolytically using the Betts process. Anodes of impure lead and cathodes of pure lead are placed in an electrolyte of lead fluorosilicate (PbSiF6). Once electrical potential is applied, impure lead at the anode dissolves and plates onto the cathode, leaving the majority of the impurities in solution. This is a high-cost process and thus mostly reserved for refining bullion containing high percentages of impurities.
Direct process
[edit]
In this process, lead bullion and slag is obtained directly from lead concentrates. The lead sulfide concentrate is melted in a furnace and oxidized, forming lead monoxide. Carbon (as coke or coal gas[n]) is added to the molten charge along with fluxing agents. The lead monoxide is thereby reduced to metallic lead, in the midst of a slag rich in lead monoxide.
If the input is rich in lead, as much as 80% of the original lead can be obtained as bullion; the remaining 20% forms a slag rich in lead monoxide. For a low-grade feed, all of the lead can be oxidized to a high-lead slag. Metallic lead is further obtained from the high-lead (25–40%) slags via submerged fuel combustion or injection, reduction assisted by an electric furnace, or a combination of both.
Alternatives
[edit]
Research on a cleaner, less energy-intensive lead extraction process continues; a major drawback is that either too much lead is lost as waste, or the alternatives result in a high sulfur content in the resulting lead metal. Hydrometallurgical extraction, in which anodes of impure lead are immersed into an electrolyte and pure lead is deposited (electrowound) onto cathodes, is a technique that may have potential, but is not currently economical except in cases where electricity is very cheap.
Secondary
[edit]
Further information: Battery recycling § Lead–acid batteries
Smelting, which is an essential part of the primary production, is often skipped during secondary production. It is only performed when metallic lead has undergone significant oxidation. The process is similar to that of primary production in either a blast furnace or a rotary furnace, with the essential difference being the greater variability of yields: blast furnaces produce hard lead (10% antimony) while reverberatory and rotary kiln furnaces produce semisoft lead (3–4% antimony).
The ISASMELT process is a more recent smelting method that may act as an extension to primary production; battery paste from spent lead–acid batteries (containing lead sulfate and lead oxides) has its sulfate removed by treating it with alkali, and is then treated in a coal-fueled furnace in the presence of oxygen, which yields impure lead, with antimony the most common impurity. Refining of secondary lead is similar to that of primary lead; some refining processes may be skipped depending on the material recycled and its potential contamination.
Of the sources of lead for recycling, lead–acid batteries are the most important; lead pipe, sheet, and cable sheathing are also significant.
Applications
[edit]
Contrary to popular belief, pencil leads in wooden pencils have never been made from lead. When the pencil originated as a wrapped graphite writing tool, the particular type of graphite used was named plumbago (literally, lead mockup).
Elemental form
[edit]
Lead metal has several useful mechanical properties, including high density, low melting point, ductility, and relative inertness. Many metals are superior to lead in some of these aspects but are generally less common and more difficult to extract from parent ores. Lead's toxicity has led to its phasing out for some uses.
Lead was used to cover the ramparts protecting the ascent to the Alamut Castle in Persia, which could absorb attacks by siege engines.
Lead has been used for bullets since their invention in the Middle Ages. It is inexpensive; its low melting point means small arms ammunition and shotgun pellets can be cast with minimal technical equipment; and it is denser than other common metals, which allows for better retention of velocity. It remains the main material for bullets, alloyed with other metals as hardeners. Concerns have been raised that lead bullets used for hunting can damage the environment.[o] Shotgun cartridges used for waterfowl hunting must today be lead-free in the United States, Canada, and in Europe.
Lead's high density and resistance to corrosion have been exploited in a number of related applications. It is used as ballast in sailboat keels; its density allows it to take up a small volume and minimize water resistance, thus counterbalancing the heeling effect of wind on the sails. It is used in scuba diving weight belts to counteract the diver's buoyancy. In 1993, the base of the Leaning Tower of Pisa was stabilized with 600 tonnes of lead. Because of its corrosion resistance, lead is used as a protective sheath for underwater cables.
Lead has many uses in the construction industry; lead sheets are used as architectural metals in roofing material, cladding, flashing, gutters and gutter joints, roof parapets. Lead is still used in statues and sculptures,[p] including for armatures. In the past it was often used to balance the wheels of cars; for environmental reasons this use is being phased out in favor of other materials.
Lead is added to copper alloys, such as brass and bronze, to improve machinability and for its lubricating qualities. Being practically insoluble in copper, the lead forms solid globules in imperfections throughout the alloy, such as grain boundaries. In low concentrations, as well as acting as a lubricant, the globules hinder the formation of swarf as the alloy is worked, thereby improving machinability. Copper alloys with larger concentrations of lead are used in bearings. The lead provides lubrication, and the copper provides the load-bearing support.
Lead's high density, atomic number, and formability form the basis for use of lead as a barrier that absorbs sound, vibration, and radiation. Lead has no natural resonance frequencies; as a result, sheet-lead is used as a sound deadening layer in the walls, floors, and ceilings of sound studios. Organ pipes are often made from a lead alloy, mixed with various amounts of tin to control the tone of each pipe. Lead is an established shielding material from radiation in nuclear science and in X-ray rooms due to its denseness and high attenuation coefficient. Molten lead has been used as a coolant for lead-cooled fast reactors.
Batteries
[edit]
The largest use of lead in the early 21st century is in lead–acid batteries. The lead in batteries undergoes no direct contact with humans, so there are fewer toxicity concerns.[q] People who work in lead battery production or recycling plants may be exposed to lead dust and inhale it. The reactions in the battery between lead, lead dioxide, and sulfuric acid provide a reliable source of voltage.[r] Supercapacitors incorporating lead–acid batteries have been installed in kilowatt and megawatt scale applications in Australia, Japan, and the United States in frequency regulation, solar smoothing and shifting, wind smoothing, and other applications. These batteries have lower energy density and charge-discharge efficiency than lithium-ion batteries, but are significantly cheaper.
Coating for cables
[edit]
Lead is used in high voltage power cables as shell material to prevent water diffusion into insulation; this use is decreasing as lead is being phased out. Its use in solder for electronics is also being phased out by some countries to reduce the amount of environmentally hazardous waste. Lead is one of three metals used in the Oddy test for museum materials, helping detect organic acids, aldehydes, acidic gases.
Compounds
[edit]
In addition to being the main application for lead metal, lead–acid batteries are also the main consumer of lead compounds. The energy storage/release reaction used in these devices involves lead sulfate and lead dioxide:
: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
Other applications of lead compounds are very specialized and often fading. Lead-based coloring agents are used in ceramic glazes and glass, especially for red and yellow shades. While lead paints are phased out in Europe and North America, they remain in use in less developed countries such as China, India, or Indonesia. Lead tetraacetate and lead dioxide are used as oxidizing agents in organic chemistry. Lead is frequently used in the polyvinyl chloride coating of electrical cords. It can be used to treat candle wicks to ensure a longer, more even burn. Because of its toxicity, European and North American manufacturers use alternatives such as zinc. Lead glass is composed of 12–28% lead oxide, changing its optical characteristics and reducing the transmission of ionizing radiation, a property used in old TVs and computer monitors with cathode-ray tubes. Lead-based semiconductors such as lead telluride and lead selenide are used in photovoltaic cells and infrared detectors.
Biological effects
[edit]
Main article: Lead poisoning
Lead
| Hazards |
| GHS labelling: |
| Pictograms | |
| Signal word | Danger |
| Hazard statements | H302, H332, H351, H360Df, H373, H410 |
| Precautionary statements | P201, P261, P273, P304, P308, P312, P313, P340, P391 |
| NFPA 704 (fire diamond) | 2 0 0 |
Chemical compound
Lead has no confirmed biological role, and there is no confirmed safe level of lead exposure. A 2009 Canadian–American study concluded that even at levels that are considered to pose little to no risk, lead may cause "adverse mental health outcomes". Its prevalence in the human body—at an adult average of 120 mg[s]—is nevertheless exceeded only by zinc (2500 mg) and iron (4000 mg) among the heavy metals. Lead salts are very efficiently absorbed by the body. A small amount of lead (1%) is stored in bones; the rest is excreted in urine and feces within a few weeks of exposure. Only about a third of lead is excreted by a child. Continual exposure may result in the bioaccumulation of lead.
Toxicity
[edit]
Lead is a highly poisonous metal (whether inhaled or swallowed), affecting almost every organ and system in the human body. At airborne levels of 100 mg/m3, it is immediately dangerous to life and health. Most ingested lead is absorbed into the bloodstream. The primary cause of its toxicity is its predilection for interfering with the proper functioning of enzymes. It does so by binding to the sulfhydryl groups found on many enzymes, or mimicking and displacing other metals that act as cofactors in many enzymatic reactions. The essential metals that lead interacts with include calcium, iron, and zinc. High levels of calcium and iron tend to provide some protection from lead poisoning; low levels cause increased susceptibility.
Effects
[edit]
Lead can cause severe damage to the brain and kidneys and, ultimately, death. By mimicking calcium, lead can cross the blood–brain barrier. It degrades the myelin sheaths of neurons, reduces their numbers, interferes with neurotransmission routes, and decreases neuronal growth. In the human body, lead inhibits porphobilinogen synthase and ferrochelatase, preventing both porphobilinogen formation and the incorporation of iron into protoporphyrin IX, the final step in heme synthesis. This causes ineffective heme synthesis and microcytic anemia.
Symptoms of lead poisoning include nephropathy, colic-like abdominal pains, and possibly weakness in the fingers, wrists, or ankles. Small blood pressure increases, particularly in middle-aged and older people, may be apparent and can cause anemia. Several studies, mostly cross-sectional, found an association between increased lead exposure and decreased heart rate variability. In pregnant women, high levels of exposure to lead may cause miscarriage. Chronic, high-level exposure has been shown to reduce fertility in males.
In a child's developing brain, lead interferes with synapse formation in the cerebral cortex, neurochemical development (including that of neurotransmitters), and the organization of ion channels. Early childhood exposure has been linked with an increased risk of sleep disturbances and excessive daytime drowsiness in later childhood. High blood levels are associated with delayed puberty in girls. The rise and fall in exposure to airborne lead from the combustion of tetraethyl lead in gasoline during the 20th century has been linked with historical increases and decreases in crime levels.
Exposure sources
[edit]
Lead exposure is a global issue since lead mining and smelting, and battery manufacturing, disposal, and recycling, are common in many countries. Lead enters the body via inhalation, ingestion, or skin absorption. Almost all inhaled lead is absorbed into the body; for ingestion, the rate is 20–70%, with children absorbing a higher percentage than adults.
Poisoning typically results from ingestion of food or water contaminated with lead, and less commonly after accidental ingestion of contaminated soil, dust, or lead-based paint. Seawater products can contain lead if affected by nearby industrial waters. Fruit and vegetables can be contaminated by high levels of lead in the soils they were grown in. Soil can be contaminated through particulate accumulation from lead in pipes, lead paint, residual emissions from leaded gasoline.
The use of lead for water pipes is a problem in areas with soft or acidic water. Hard water forms insoluble protective layers on the inner surface of the pipes, whereas soft and acidic water dissolves the lead pipes. Dissolved carbon dioxide in the carried water may result in the formation of soluble lead bicarbonate; oxygenated water may similarly dissolve lead as lead(II) hydroxide. Drinking such water, over time, can cause health problems due to the toxicity of the dissolved lead. The harder the water the more calcium bicarbonate and sulfate it contains, and the more the inside of the pipes are coated with a protective layer of lead carbonate or lead sulfate.
Ingestion of applied lead-based paint is the major source of exposure for children: a direct source is chewing on old painted window sills. Additionally, as lead paint on a surface deteriorates, it peels and is pulverized into dust. The dust then enters the body through hand-to-mouth contact or contaminated food or drink. Ingesting certain home remedies may result in exposure to lead or its compounds.
Inhalation is the second major exposure pathway, affecting smokers and especially workers in lead-related occupations. Cigarette smoke contains, among other toxic substances, radioactive lead-210. "As a result of EPA's regulatory efforts, levels of lead in the air [in the United States] decreased by 86 percent between 2010 and 2020." The concentration of lead in the air in the United States fell below the national standard of 0.15 μg/m3 in 2014.
Skin exposure may be significant for people working with organic lead compounds. The rate of skin absorption is lower for inorganic lead.
Lead in foods
[edit]
Lead may be found in food when food is grown in soil that is high in lead, airborne lead contaminates the crops, animals eat lead in their diet, or lead enters the food either from what it was stored or cooked in. Ingestion of lead paint and batteries is also a route of exposure for livestock, which can subsequently affect humans. Milk produced by contaminated cattle can be diluted to a lower lead concentration and sold for consumption.
In Bangladesh, lead compounds have been added to turmeric to make it more yellow. This is believed to have started in the 1980s and continues as of 2019[update]. It is believed to be one of the main sources of high lead levels in the country. In Hong Kong the maximum allowed lead level in food is 6 parts per million in solids and 1 part per million in liquids.
Lead-containing dust can settle on drying cocoa beans when they are set outside near polluting industrial plants. In December 2022, Consumer Reports tested 28 dark chocolate brands and found that 23 of them contained potentially harmful levels of lead, cadmium or both. They have urged the chocolate makers to reduce the level of lead which could be harmful, especially to a developing fetus.
In March 2024, the US Food and Drug Administration recommended a voluntary recall on 6 brands of cinnamon due to contamination with lead, after 500 reports of child lead poisoning. The FDA determined that cinnamon was adulterated with lead chromate.
Lead in plastic toys
[edit]
According to the United States Center for Disease Control, the use of lead in plastics has not been banned as of 2024. Lead softens the plastic and makes it more flexible so that it can go back to its original shape. Habitual chewing on colored plastic insulation from stripped electrical wires was found to cause elevated lead levels in a 46-year-old man. Lead may be used in plastic toys to stabilize molecules from heat. Lead dust can be formed when plastic is exposed to sunlight, air, and detergents that break down the chemical bond between the lead and plastics.
Treatment
[edit]
See also: Chelation therapy
Treatment for lead poisoning normally involves the administration of dimercaprol and succimer. Acute cases may require the use of disodium calcium edetate, the calcium chelate, and the disodium salt of ethylenediaminetetraacetic acid (EDTA). It has a greater affinity for lead than calcium, with the result that lead chelate is formed by exchange and excreted in the urine, leaving behind harmless calcium.
Environmental effects
[edit]
The extraction, production, use, and disposal of lead and its products have caused significant contamination of the Earth's soils and waters. Atmospheric emissions of lead were at their peak during the Industrial Revolution, and the leaded gasoline period in the second half of the twentieth century.
Lead releases originate from natural sources (i.e., concentration of the naturally occurring lead), industrial production, incineration and recycling, and mobilization of previously buried lead. In particular, as lead has been phased out from other uses, in the Global South, lead recycling operations designed to extract cheap lead used for global manufacturing have become a well documented source of exposure. Elevated concentrations of lead persist in soils and sediments in post-industrial and urban areas; industrial emissions, including those arising from coal burning, continue in many parts of the world, particularly in the developing countries.
Lead can accumulate in soils, especially those with a high organic content, where it remains for hundreds to thousands of years. Environmental lead can compete with other metals found in and on plant surfaces potentially inhibiting photosynthesis and at high enough concentrations, negatively affecting plant growth and survival. Contamination of soils and plants can allow lead to ascend the food chain affecting microorganisms and animals. In animals, lead exhibits toxicity in many organs, damaging the nervous, renal, reproductive, hematopoietic, and cardiovascular systems after ingestion, inhalation, or skin absorption. Fish uptake lead from both water and sediment; bioaccumulation in the food chain poses a hazard to fish, birds, and sea mammals.
Anthropogenic lead includes lead from shot and sinkers. These are among the most potent sources of lead contamination along with lead production sites. Lead was banned for shot and sinkers in the United States in 2017, although that ban was only effective for a month, and a similar ban is being considered in the European Union.
Analytical methods for the determination of lead in the environment include spectrophotometry, X-ray fluorescence, atomic spectroscopy, and electrochemical methods. A specific ion-selective electrode has been developed based on the ionophore S,S'-methylenebis(N,N-diisobutyldithiocarbamate). An important biomarker assay for lead poisoning is δ-aminolevulinic acid levels in plasma, serum, and urine.
Restriction and remediation
[edit]
By the mid-1980s, there was significant decline in the use of lead in industry. In the United States, environmental regulations reduced or eliminated the use of lead in non-battery products, including gasoline, paints, solders, and water systems. Particulate control devices were installed in coal-fired power plants to capture lead emissions. In 1992, U.S. Congress required the Environmental Protection Agency to reduce the blood lead levels of the country's children. Lead use was further curtailed by the European Union's 2003 Restriction of Hazardous Substances Directive. A large drop in lead deposition occurred in the Netherlands after the 1993 national ban on use of lead shot for hunting and sport shooting: from 230 tonnes in 1990 to 47.5 tonnes in 1995. The usage of lead in Avgas 100LL for general aviation is allowed in the EU as of 2022.
In the United States, the permissible exposure limit for lead in the workplace, comprising metallic lead, inorganic lead compounds, and lead soaps, was set at 50 μg/m3 over an 8-hour workday, and the blood lead level limit at 5 μg per 100 g of blood in 2012. Lead may still be found in harmful quantities in stoneware, vinyl (such as that used for tubing and the insulation of electrical cords), and Chinese brass.[t] Old houses may still contain lead paint. White lead paint has been withdrawn from sale in industrialized countries, but specialized uses of other pigments such as yellow lead chromate remain, especially in road pavement marking paint. Stripping old paint by sanding produces dust which can be inhaled. Lead abatement programs have been mandated by some authorities in properties where young children live. The usage of lead in Avgas 100LL for general aviation is generally allowed in United States as of 2023.
Lead waste, depending on the jurisdiction and the nature of the waste, may be treated as household waste (to facilitate lead abatement activities), or potentially hazardous waste requiring specialized treatment or storage. Lead is released into the environment in shooting places and a number of lead management practices have been developed to counter the lead contamination. Lead migration can be enhanced in acidic soils; to counter that, it is advised soils be treated with lime to neutralize the soils and prevent leaching of lead.
Research has been conducted on how to remove lead from biosystems by biological means: Fish bones are being researched for their ability to bioremediate lead in contaminated soil. The fungus Aspergillus versicolor is effective at absorbing lead ions from industrial waste before being released to water bodies. Several bacteria have been researched for their ability to remove lead from the environment, including the sulfate-reducing bacteria Desulfovibrio and Desulfotomaculum, both of which are highly effective in aqueous solutions. Millet grass Urochloa ramosa has the ability to accumulate significant amounts of metals such as lead and zinc in its shoot and root tissues making it an important plant for remediation of contaminated soils.
See also
[edit]
Derek Bryce-Smith– one of the earliest campaigners against lead in petrol in the UK
Thomas Midgley Jr. – discovered that the addition of tetraethyllead to gasoline prevented "knocking" in internal combustion engines
Clair Cameron Patterson – instrumental in the banning of tetraethyllead in gasoline in the US and lead solder in food cans.
Robert A. Kehoe – foremost medical advocate for the use of tetraethyllead as an additive in gasoline.
Notes
[edit]
^ The tetrahedral allotrope of tin is called α- or gray tin and is stable only at or below 13.2 °C (55.8 °F). The stable form of tin above this temperature is called β- or white tin and has a distorted face centered cubic (tetragonal) structure which can be derived by compressing the tetrahedra of gray tin along their cubic axes. White tin effectively has a structure intermediate between the regular tetrahedral structure of gray tin, and the regular face centered cubic structure of lead, consistent with the general trend of increasing metallic character going down any representative group.
^ A quasicrystalline thin-film allotrope of lead, with pentagonal symmetry, was reported in 2013. The allotrope was obtained by depositing lead atoms on the surface of an icosahedral silver-indium-ytterbium quasicrystal. Its conductivity was not recorded.
^ Diamond cubic structures with lattice parameters around the lattice parameter of silicon exists both in thin lead and tin films, and in massive lead and tin, freshly solidified in vacuum of ~5 x 10−6 Torr. Experimental evidence for almost identical structures of at least three oxide types is presented, demonstrating that lead and tin behave like silicon not only in the initial stages of crystallization, but also in the initial stages of oxidation.
^ British English: to go down like a lead balloon.
^ Malleability describes how easily it deforms under compression, whereas ductility means its ability to stretch.
^ A (wet) finger can be dipped into molten lead without risk of a burning injury.
^ An even number of either protons or neutrons generally increases the nuclear stability of isotopes, compared to isotopes with odd numbers. No elements with odd atomic numbers have more than two stable isotopes; even-numbered elements have multiple stable isotopes, with tin (element 50) having the highest number of isotopes of all elements, ten. See Even and odd atomic nuclei for more details.
^ The half-life found in the experiment was 1.9×1019 years. A kilogram of natural bismuth would have an activity value of approximately 0.003 becquerels (decays per second). For comparison, the activity value of natural radiation in the human body is around 65 becquerels per kilogram of body weight (4500 becquerels on average).
^ Tetraphenyllead is even more thermally stable, decomposing at 270 °C.
^ Abundances in the source are listed relative to silicon rather than in per-particle notation. The sum of all elements per 106 parts of silicon is 2.6682×1010 parts; lead comprises 3.258 parts.
^ Elemental abundance figures are estimates and their details may vary from source to source.
^ The inscription reads: "Made when the Emperor Vespasian was consul for the ninth term and the Emperor Titus was consul for the seventh term, when Gnaeus Iulius Agricola was imperial governor (of Britain)."
^ The fact that Julius Caesar fathered only one child, as well as the alleged sterility of his successor, Caesar Augustus, have been attributed to lead poisoning.
^ Gaseous by-product of the coking process, containing carbon monoxide, hydrogen and methane; used as a fuel.
^ California began banning lead bullets for hunting on that basis in July 2015.
^ For example, a firm "...producing quality [lead] garden ornament from our studio in West London for over a century".
^ Potential injuries to regular users of such batteries are not related to lead's toxicity.
^ See for details on how a lead–acid battery works.
^ Rates vary greatly by country.
^ An alloy of brass (copper and zinc) with lead, iron, tin, and sometimes antimony.
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Further reading
[edit]
Astrid, S.; Helmut, S.; Sigel, R. K. O., eds. (2017). Lead: Its Effects on Environment and Health. Metal Ions in Life Sciences. Vol. 17. De Gruyter. ISBN 978-3-11-044107-9. Table of contents
Casas, J. S.; Sordo, J., eds. (2006). Lead Chemistry, Analytical Aspects. Environmental Impacts and Health Effects. Elsevier. ISBN 978-0-444-52945-9.
Hunt, A.; Abraham, J. L.; Judson, B.; Berry, C. L. (2003). "Toxicologic and epidemiologic clues from the characterization of the 1952 London smog fine particulate matter in archival autopsy lung tissues". Environmental Health Perspectives. 111 (9): 1209–1214. Bibcode:2003EnvHP.111.1209H. doi:10.1289/ehp.6114. PMC 1241576. PMID 12842775.
Ingalls, Walter Renton (1865-), Lead Smelting and Refining, With Some Notes on Lead Mining
Zhang, Hao; Wei, Kai; Zhang, Mengyu; Liu, Rutao; Chen, Yadong (2014). "Assessing the mechanism of DNA damage induced by lead through direct and indirect interactions". Journal of Photochemistry and Photobiology B: Biology. 136: 46–53. Bibcode:2014JPPB..136...46Z. doi:10.1016/j.jphotobiol.2014.04.020. PMID 24844619.
External links
[edit]
Wikisource has the text of the 1879 American Cyclopædia article Lead.
Look up lead in Wiktionary, the free dictionary.
Wikimedia Commons has media related to Lead.
The Toxicology of Heavy Metals: Getting the Lead Out, American Society for Clinical Pathology
COVID-19 to reduce global lead production by 5,2% in 2020 (with a figure showing global lead production, 2010–2024)
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| | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- | | | 1 | 2 | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | | 1 | H | | He | | 2 | Li | Be | | B | C | N | O | F | Ne | | 3 | Na | Mg | | Al | Si | P | S | Cl | Ar | | 4 | K | Ca | | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr | | 5 | Rb | Sr | | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe | | 6 | Cs | Ba | La | Ce | Pr | Nd | Pm | Sm | Eu | Gd | Tb | Dy | Ho | Er | Tm | Yb | Lu | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | Tl | Pb | Bi | Po | At | Rn | | 7 | Fr | Ra | Ac | Th | Pa | U | Np | Pu | Am | Cm | Bk | Cf | Es | Fm | Md | No | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | |
| | | | | | --- --- | | s-block | f-block | d-block | p-block | |
| v t e Lead compounds |
| Pb(II) | Pb(BiO3)2 PbBr2 Pb(C5H5)2 Pb(C2H3O2)2 PbC2O4 PbC32H16N8 PbCl2 Pb(ClO4)2 PbCO3 PbCrO4 PbF2 PbFBr PbFCl PbHAsO4 PbI2 Pb(C11H23COO)2 Pb(NO3)2 Pb(N3)2 PbO Pb(OH)2 PbPo PbP7 Pb3(PO4)2 PbS Pb(SCN)2 PbSe PbSO4 PbSeO4 PbTe PbTiO3 PbGeO3 C36H70PbO4 PbO2−2 PbC2 (hypothetical) |
| Pb(II,IV) | Pb3O4 |
| Pb(IV) | Pb(C2H3O2)4 PbCl4 PbF4 PbH4 PbO2 PbS2 plumbate |
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11860 | https://www.txdot.gov/manuals/des/hyd/chapter-4--hydrology/section-11--time-of-concentration/manning-s-roughness-coefficient-values.html | Published Time: Tue, 09 Sep 2025 16:49:30 GMT
Manning’s Roughness Coefficient Values
Hydraulic Design Manual
Manning’s Roughness Coefficient Values
Manning’s roughness coefficients are used to calculate flows using Manning’s equation. Values from
American Society of Civil Engineers
(ASCE) 1992,
FHWA
2001, and Chow 1959 are reproduced in Table 4-7, Table 4-8, and Table 4-9.
Table 4-7: Manning’s Roughness Coefficients for Open Channels
Type of channelManning’s n
A. Natural streams
Minor streams (top width at flood stage < 100 ft)
a. Clean, straight, full, no rifts or deep pools 0.025-0.033
b. Same as a, but more stones and weeds 0.030-0.040
c. Clean, winding, some pools and shoals 0.033-0.045
d. Same as c, but some weeds and stones 0.035-0.050
e. Same as d, lower stages, more ineffective 0.040-0.055
f. Same as d, more stones 0.045-0.060
g. Sluggish reaches, weedy, deep pools 0.050-0.080
h. Very weedy, heavy stand of timber and underbrush 0.075-0.150
i. Mountain streams with gravel and cobbles, few boulders on bottom 0.030-0.050
j. Mountain streams with cobbles and large boulders on bottom 0.040-0.070
Floodplains
a. Pasture, no brush, short grass 0.025-0.035
b. Pasture, no brush, high grass 0.030-0.050
c. Cultivated areas, no crop 0.020-0.040
d. Cultivated areas, mature row crops 0.025-0.045
e. Cultivated areas, mature field crops 0.030-0.050
f. Scattered brush, heavy weeds 0.035-0.070
g. Light brush and trees in winter 0.035-0.060
h. Light brush and trees in summer 0.040-0.080
i. Medium to dense brush in winter 0.045-0.110
j. Medium to dense brush in summer 0.070-0.160
k. Trees, dense willows summer, straight 0.110-0.200
l. Trees, cleared land with tree stumps, no sprouts 0.030-0.050
m. Trees, cleared land with tree stumps, with sprouts 0.050-0.080
n. Trees, heavy stand of timber, few down trees, flood stage below branches 0.080-0.120
o. Trees, heavy stand of timber, few down trees, flood stage reaching branches 0.100-0.160
Major streams (top width at flood stage > 100 ft)
a. Regular section with no boulders or brush 0.025-0.060
b. Irregular rough section 0.035-0.100
B. Excavated or dredged channels
Earth, straight and uniform
a. Clean, recently completed 0.016-0.020
b. Clean, after weathering 0.018-0.025
c. Gravel, uniform section, clean 0.022-0.030
d. With short grass, few weeds 0.022-0.033
Earth, winding and sluggish
a. No vegetation 0.023-0.030
b. Grass, some weeds 0.025-0.033
c. Deep weeds or aquatic plants in deep channels 0.030-0.040
d. Earth bottom and rubble sides 0.028-0.035
e. Stony bottom and weedy banks 0.025-0.040
f. Cobble bottom and clean sides 0.030-0.050
g. Winding, sluggish, stony bottom, weedy banks 0.025-0.040
h. Dense weeds as high as flow depth 0.050-0.120
Dragline-excavated or dredged
a. No vegetation 0.025-0.033
b. Light brush on banks 0.035-0.060
Rock cuts
a. Smooth and uniform 0.025-0.040
b. Jagged and irregular 0.035-0.050
Unmaintained channels
a. Dense weeds, high as flow depth 0.050-0.120
b. Clean bottom, brush on sides 0.040-0.080
c. Clean bottom, brush on sides, highest stage 0.045-0.110
d. Dense brush, high stage 0.080-0.140
C. Lined channels
Asphalt 0.013-0.016
Brick (in cement mortar)0.012-0.018
Concrete
a. Trowel finish 0.011-0.015
b. Float finish 0.013-0.016
c. Unfinished 0.014-0.020
d. Gunite, regular 0.016-0.023
e. Gunite, wavy 0.018-0.025
Riprap (n-value depends on rock size)0.020-0.035
Vegetal lining 0.030-0.500
Table 4-8: Manning’s Coefficients for Streets and Gutters
Type of gutter or pavementManning’s n
Concrete gutter, troweled finish 0.012
Asphalt pavement: smooth texture 0.013
Asphalt pavement: rough texture 0.016
Concrete gutter with asphalt pavement: smooth texture 0.013
Concrete gutter with asphalt pavement: rough texture 0.015
Concrete pavement: float finish 0.014
Concrete pavement: broom finish 0.016
Table 4-8 note: For gutters with small slope or where sediment may accumulate, increase n values by 0.02 (USDOT, FHWA 2001).
Table 4-9: Manning’s Roughness Coefficients for Closed Conduits (ASCE 1982, FHWA 2001)
MaterialManning’s n
Asbestos-cement pipe 0.011-0.015
Brick 0.013-0.017
Cast iron pipe
Cement-lined & seal coated 0.011-0.015
Concrete (monolithic)
Smooth forms 0.012-0.014
Rough forms 0.015-0.017
Concrete pipe 0.011-0.015
Box (smooth)0.012-0.015
Corrugated-metal pipe -- (2-1/2 in. x 1/2 in. corrugations)
Plain 0.022-0.026
Paved invert 0.018-0.022
Spun asphalt lined 0.011-0.015
Plastic pipe (smooth)0.011-0.015
Corrugated-metal pipe -- (2-2/3 in. by 1/2 in. annular)0.022-0.027
Corrugated-metal pipe -- (2-2/3 in. by 1/2 in. helical)0.011-0.023
Corrugated-metal pipe -- (6 in. by 1 in. helical)0.022-0.025
Corrugated-metal pipe -- (5 in. by 1 in. helical)0.025–0.026
Corrugated-metal pipe -- (3 in. by 1 in. helical)0.027–0.028
Corrugated-metal pipe -- (6 in. by 2 in. structural plate)0.033–0.035
Corrugated-metal pipe -- (9 in. by 2-1/2 in. structural plate)0.033–0.037
Corrugated polyethylene 0.010–0.013
Smooth 0.009-0.015
Corrugated 0.018–0.025
Spiral rib metal pipe (smooth)0.012-0.013
Vitrified clay
Pipes 0.011-0.015
Liner plates 0.013-0.017
Polyvinyl chloride (PVC) (smooth)0.009-0.011
Table 4-9 note: Manning’s n for corrugated pipes is a function of the corrugation size, pipe size, and whether the corrugations are annular or helical (see USGS 1993).
Section 11: Time of Concentration |
11861 | https://www.youtube.com/watch?v=-BywefmMuTc | Learn to Simplify Fractions to Lowest Terms | Step-by-Step
Math and Science
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Posted: 16 Apr 2023
Are you struggling to simplify fractions? Do you find yourself getting lost in the process and ending up with incorrect answers? Look no further, as we've got you covered! In this video, we'll show you step-by-step how to simplify fractions like a pro, so you can tackle even the toughest fractions problems with ease.
First, we'll start with the basics of what it means to simplify a fraction. We'll explain the concept of equivalent fractions, and how you can use them to simplify fractions by dividing the numerator and denominator by their greatest common factor. We'll also cover the different methods for finding the greatest common factor, including listing factors, prime factorization, and the Euclidean algorithm.
Next, we'll dive into some examples of simplifying fractions using each of these methods. We'll walk you through the steps of simplifying fractions with clear, concise explanations and plenty of examples to help reinforce your understanding. We'll also discuss common mistakes to watch out for and how to avoid them.
But that's not all – we'll also show you some tricks and tips for simplifying fractions even faster, so you can save time on exams and tests. We'll share shortcuts for simplifying fractions with numbers that end in zeros, and explain how to simplify mixed numbers and improper fractions.
By the end of this video, you'll be a master of simplifying fractions, and you'll be able to tackle even the toughest fractions problems with confidence. Whether you're a student struggling with math homework, or a teacher looking for new ways to teach fractions, this video is perfect for you.
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what about the fraction 3 6 and we want to do what we call simplifying that fraction now first of all what is 3 6 right let's take a look at what that actually is here's 1 6 2 6 3 6. this is 3 6 of a fraction now let me ask you something if I turn it maybe this way it might help C A little better what does that actually look like to you well three six one six two six three six here it is but what does it look like it looks exactly like half of a pizza in fact I can hold up this is the fraction one half I can hold up the fraction one half and they look and they are exactly the same thing so we can see from these magnets that the fraction one half of a pizza is exactly the same as 3 6. why is that by the way because if you have a pizza that you cut into six pieces but I take three of the pieces then I've taken half of the pizza so we have to have a way or we want to have a way to take any fraction that we have and simplify it into a simpler version of itself so let me move this down below right here and let's try to do that the way you do it is the following put a little equal sign here rewrite the fraction 3 6. now is there a number that you can think of where you could divide the top and the bottom and it would go evenly could you divide by two for instance well 6 divided by 2 is a is a goes a whole number of times but what is three divided by two three divided by 2 does not divide evenly so we can't divide by two because we can't do it to the top and to the bottom what about the number three what if we choose to divide the top of this fraction by the number three and also the bottom of the fraction by the number three what would happen well three divided by 3 on the top is just one and six divided by 3 on the bottom is just 2. so what we have said here is that the fraction one-half here which is represented here is exactly the same thing as the fraction 3 6 and of course you can see from these that this is true so let me rotate this just to kind of get them out of the way a little bit more and we'll leave it up on the board what you would say is that this is the simplified form or the lowest terms form of the original fraction it's important for you to understand that the fractions mean the same thing 3 6 really does mean exactly the same amount of stuff is one half but when we give answers and problems we usually want to show the simplified version of the fraction into lowest terms like in general when we solve problems we always want to simplify as the final step to get the fraction into lowest terms into simplest form all right so now that we have the idea down let's go conquer the next one let's say we have the fraction 4 12 and I'm going to ask you is this in lowest terms or can we simplify this into lowest terms well what you need to be asking yourself is can I divide the top and bottom by something anything I can think of and the answer is yes I notice that this is a whole number uh an even number I mean this is a four and this is a 12. so actually I can take this 4 12. right and I can because I know that they're both even numbers I can divide the top by two and I can divide the Bottom by two right what will I get I'll get 4 divided by 2 is 2. and 12 divided by 2 is 6. so I'll get a new fraction and we're saying that the fraction 2 6 is exactly the same thing as the fraction 4 12. but then I look at this and I say well wait a minute is this in simplest form and I say no it's not because I could still divide the Top Again by 2 and I could divide the bottom Again by two so let's do that 2 6. let's see what happens when I divide the top by 2 and the Bottom by two what do I get right 2 divided by 2 is on the top uh 1 and 6 divided by 2 on the bottom is 3. and so I get an answer of 1 3. so what we're saying and then we ask ourselves is this the final answer is this the simplified form 1 divided by three or one over three I cannot simplify this further because I cannot divide top and bottom by something to make it simpler I can't do it because even if I divide by 3 that would work down here but I can't divide this by three it doesn't go evenly so I'm done I stop at the fraction one-third all right so let's see first of all if this is even true let's take a look at the fraction 4 12 here here's 1 12. here's two twelfths here's three twelfths here's 4 12. so this is the actual amount of pizza notice the slices when you cut a pizza into 12 pieces pretty small 4 12. now let's go all the way to the answer we said the answer was one-third we're saying that these two fractions even though they look totally different actually represent the same thing one-third is exactly the same thing as four twelfths so that's why they're equivalent all of the numbers in the fraction look totally different but the actual fraction means the same thing now notice that first we divided by 2 to get this and then we simplified further in a second step to get this let's see what this is equal to what is 2 6 here so let me go over here 2 6. here's 1 6 Here's 2 6 because these are now cut into six larger slices there's two six and I think you can see that one-third is exactly the same as 2 6 which is also exactly the same as 4 12. so you see what's going on here when I divide this number a fraction by 2 on the top and two on the bottom I get this this fraction means exactly the same as this but it's not fully simplified yet but when we divide by two again we get to this which again means exactly the same amount of pizza but this is fully simplified because I cannot further divide this anymore so when you see a teacher or a test tell you simplify the fraction into lowest terms you want to keep simplifying as far or as much as you need to in order to get to a point where you can't do it anymore where you can't do it anymore now one more thing I want to show you before we move on let me uh let's see here yeah I guess I'll just leave it up here just like this let's do the problem again 4 12. all right what we did is we divided by two got an answer then we divided by 2 again and got an answer but if you notice carefully you can actually divide by two of course we did that but actually there's something else that divides into top and bottom here what if I divide by four divide this by 4 and also this by four divide Top by four divide Bottom by four what will I get 4 divided by 4 is 1 and 12 divided by 4 is 3. notice I get exactly the same answer one third is the final answer to this problem and I get the same answer no matter how I simplify the problem so here's the bottom line when you're simplifying fractions what you want to do is try to divide the top and bottom by the largest thing possible that you can do because if you do that then you can get to the simplified answer in only one step but a lot of times we don't often think of the best number to divide by but we might notice that there are even numbers and we can divide by two and if you divide by two then you get to an intermediate answer and then you divide by two again so I guess my point is I don't really care what you divide by as long as it's legal I'm okay with it some teachers are going to get mad at you if you take more than one step to get to the answer I'm not going to be like that I don't care I just want you to get to the right answer so if you happen to see that you can divide by four awesome do it if you're not sure about that but you know you can divide by two then just do it multiple times that's fine with me too to me as long as you get the right answer I'm happy all right let's take a look at the problem two-tenths how do we simplify that we rewrite the fraction as two over ten what can we divide top and bottom by well I can divide the top and the Bottom by two because both are divisible by two two divided by two is one ten divided by 2 is 5. so what we're claiming is that the fraction one-fifth is exactly the same thing as two tenths so what we want to do is see if that's true here's one-tenth here's two tenths and then we're asking ourselves is the fraction one-fifth here the same and of course we can pick it up and put it right on top you can see that they're not just close they're not just kind of sore to equal they're exactly the same thing two tenths is exactly the same thing as one-fifth so for the rest of the problems we're not going to use the magnets but I just want you to make sure that that you know that you can divide by whatever you want as long as it's legal and if you get to a place where you can still simplify it further then just do it again if it takes more than one step then all it means is there was probably something larger that you could have divided by but it's not like wrong if you take more than one step I'm just telling you there's more than one path to the answer in math I think I've been telling you over and over there's almost always more than one way to get to an answer so when I'm doing these problems for you you might see a different way in a different path to the answer and that is okay let's take a look at the problem 9 18. let's simplify 9 18. this is an excellent example of that all right what can we do well we can divide the top and bottom by three how do I know because I just know that this is divisible by 3 and this is divisible by three so let's do it divide Top by three divide Bottom by three what do I get 9 divided by 3 is 3 18 divided by 3 is 6. now I'm going to circle this answer but then I notice well wait a minute I can further simplify this further 3 6. I can divide it by three again okay I can divide it by 3 again so I get to an intermediate answer but this is not fully simplified I can divide it again by 3. 3 divided by three is one six divided by 3 is 2. and so the answer is finally one half because I cannot simplify or divide this any further to make it any simpler and it took kind of like two steps to get to here step number one and then for here to get to step number two but let's really quickly just do the whole thing again 9 18. notice that it took two steps so there must be something even bigger I can divide by if I happen to see it what about if I divide the top by 9 and the Bottom by 9. now you might not have seen that but if you do you know that 9 divided by 9 is 1. 18 divided by 9 is 2 because 2 times 9 is 18. and you can see you get exactly the same answer so I guess I'm going to stop solving all of these problems two ways what I want you to know is that I don't care if you see the perfect thing to divide by the best thing to divide by here would be 9 because it only gives one step but oftentimes I don't really see that either and it might take more than one step to get there so as we go forward if it takes me a different number of steps than you fine there's more than one way to do these problems all right so let's do the next problem what about four twentieths right so I'm going to write four twentieths I know right away I could divide by four and then divide by four because I know that they're both divisible but let's just say that I didn't notice that I know these are even numbers so I divide by two and by two four divided by two is two twenty divided by 2 is 10. but then I notice oh man I I can do it again because these are again even numbers so I write two tenths down again and then I say all right I'm going to divide the top by two and the Bottom by two two divided by two is one ten divided by 2 is 5. and the answer is one-fifth if I did not divide by 2 and then by 2 again what if I noticed and started off by just dividing by four well four divided by 4 is 1. 20 divided by 4 is 5. I would go straight from here to the answer in one step if I choose the best thing to divide by but to me as long as you get the right answer that's fine I'm totally fine with that all right enough of that let's take a look at 12 15. let's simplify that rewrite the fraction put the 12 and the 15th draw your longer fraction bar what can I divide this by well I know I can divide by three and by 3. so put up here I'm going to divide by 3 divide by 3. 12 divided by 3 is 4 because 4 times 3 is 12. 15 divided by 3 is 5 because 5 times 3 is 15. so the answer is four-fifths and I can't simplify this further because I can't divide top and bottom by something to make that any simpler all right what about the uh problem eight twentieths right now I know that there are probably other things I can divide by but let's just say I don't know and I just noticed that these are even numbers right so I'll just divide this by two because I know they're both even that's definitely going to work 8 divided by 2 is 4. 20 divided by 2 is 10. and then I say is this simplified nope because it's still even numbers so I just write the four tenths again and I do the same thing again I can divide both by 2. the top by two divided Bottom by 2 I get 4 divided by 2 is 2 and 10 divided by 2 is 5. and so the answer is two-fifths there is nothing I can divide further to make this simpler now uh I said I won't do this for every problem but I can't resist doing it here let's rework the problem again what else could I divide by what else is divisible here actually you can divide the top and bottom by four you may not see that right away but 8 divided by 4 is 2. 20 divided by 4 is what 5 because 5 times 4 is 20. so you see you get the same answer it's just if I pick the bigger number uh which is by the way called the greatest common factor that's why we learned what the greatest common factor was because if you always divide by the greatest common factor then you'll get the simplified fraction in the fewest number of steps so if you happen to see that you can divide by four then you're going to get the answer in one step but if you're like the rest of us who doesn't always see that just divide by 2 because it's even divide by two again maybe you have to do it twice but then you always get the right answer with no problems all right let's take a look at our final few problems we'll take a look at 14 18. what can I divide top and bottom by well I can see right away that these are even numbers so I'm going to divide the top and the Bottom by 2. Top by two Bottom by two what is 14 divided by 2. 7 because 7 times 2 is 14. what is 18 divided by 2 9 because 9 times 2 is 18 the answer is 7 9. I cannot divide top and bottom by anything further to make this any simpler so I am done all right only two more problems what about 26 28. right 26 28. again I see that they're both even numbers so write 26 on the top 28 on the bottom and I'm going to divide the top and the Bottom by two now these are a little tricky because they're larger numbers right but when you think about it I think you could convince yourself that 26 divided by 2 is 13 and 28 divided by 2 is 14. and if you're not sure about that you can go off to the side 26 divided by 2 and go through it all and you'll get 13 and do this one you'll get 14. and I know we've covered all that in the past so at some point I have to start assuming that you know how to divide numbers which we've done before you get 13 14 here and these are the final answers I cannot simplify that any further all right and then finally here's our last problem what about 24 and 96. so I write 24 again and 96 again now I don't really know exactly what the best number is to divide by but I know that they're both even so I'm going to divide the top and by 2 and I'll divide again the Bottom by two now you might need to go off to the side and take 24 divide by 2 and so I want to get the answer but I think you can convince yourself that this is 12 on the top because 12 times 2 is 24. and then 96 divided by 2 is actually going to work out to be 48 you take 96 divide by 2. we've done all that many times you get an answer of 48. but this is not perfectly simplified because these are again uh even numbers so what will I do I will divide the top and bottom by 2 again right what am I going to get 12 divided by 2 is 6 and 48 divided by 2 when you do the division and work that out you'll get 24. these again are even numbers so 6 24. here I'm going to be a little bit smarter I can divide by 2 again because there are even numbers but I realize now that I could just divide this by six I can divide this by 6 and this by 6 because they're both divisible by six six divided by six is one 24 divided by 6 is 4 because 4 times 6 is 24. so the answer is 1 4. so notice that we had like a multi-step thing going on here we divided one time to get this thing then we divided another time to get this thing and then I guess I'll do it like this another time to get this so we did one division two division three divisions and it always worked out but it was multiple steps so when you have multiple steps like that to simplify a fraction all it means is that there was a larger number I could divide top and bottom by to get exactly the same thing what do you think that is I think I could actually probably divide the top and bottom by 24. what I'm claiming is if I take the top and the bottom and divide by 24. and divide by 24. what will I get the answer is 1 4 because 4 times 24 is 96 let's check it 24 times 4. 4 times 4 16 carry the 1. 2 times 4 is 8 plus 1 is 9. so 4 times 24 is this and 1 times 24 is this so if I had looked at this problem and I was a human calculator which I'm not but if I was I could say oh I'll just divide top and bottom by 24 and I'll get the answer in one step but nobody can do that I mean I don't think anybody most people cannot do that so instead do it like the rest of us well I'll divide by 2 and get this now actually we divided by 2 again but here if you're clever you can see that you can divide top and bottom by 12 because 12 divided by 12 is 1. 48 divided by 12 is 4. remember for multiplication 12 times 4 is 48. so you might have been able to shortcut a couple steps by dividing top and bottom by 12 here but if you didn't think of it divide by 2 again and then here we divided by 6 and so on and get the final answer of 1 4. so simplifying fractions is really important because when we add or subtract or multiply or divide fractions the answer we get we'll always have to check if it's in lowest terms and if it's not we'll have to do this process to get it into lowest terms don't forget what's happening what we're saying is 2496 of a pizza is exactly the same amount of food as 12 48 of a pizza or 12 slices out of 48 which is exactly the same thing as six slices out of 24 which is exactly the same thing as one slice out of four they all represent the same thing but they're just written slightly different that's all that that means so I'd like you to solve all of these yourself follow me on to part two we'll wrap up our skills with simplifying fractions learn anything at mathandscience.com |
11862 | https://physics.stackexchange.com/questions/333449/bullet-interception | kinematics - Bullet interception? - Physics Stack Exchange
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Bullet interception?
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I am programming a space game, and I put this on the physics stack exchange because I need some help in the physics behind it. Let's say there is a moving enemy ship - referred to as "target", and we want to lead our ship's torpedo to a point where the bullet will intercept the moving target. Before we go on further, let's assume some constraints:
The motion is constant, there is no acceleration or net forces being applied.
We will not account for the angular velocity of our ship to meet the intersection point.
The problem is in two dimensions. X and Y plane. No Z component.
The origin is relative to the top left at (0,0). For this case however, I want to make all vectors relative to our ship.
We will assume the ship (torpedo) has no initial velocity right now.
The motion is constant in terms of the space game server I am running, I will factor gravitation later and angular velocity for the best result. For now, let's break down the variables we know:
Target :
Velocity of the moving target (including direction) as a vector
Displacement of the target to the ship as a vector.
Initial position of the ship as a point.
Ship (Torpedo to be specific) :
Speed of the bullet (constant) (magnitude of it's velocity)
Initial position of the bullet (in this case, the same as the position of the ship)
Now that I have covered that, let's state what we need to figure out:
I wan't to find the point of intersection so that if I shoot a torpedo in that direction, it will perfectly intercept the target.
I need to find the final position of the ship as a function of time. Then I can make a displacement vector from the final position of the ship to the initial position of the ship.
Here are the steps I am taking:
We have several vectors here - U, V and C. Our origin is our ship (the torpedo).
Vector U is the displacement vector from our ship to the target's initial position
Vector V is the displacement vector from the target's initial position to the target's final position as a function of time (need to solve for that)
Vector C is the resultant vector of U and V. Vector C will tell me the point of intersection
My program already has a Vector class, so there shouldn't be any worry in implementation. This is in Java btw. (Again, I put this in a physics forum to talk about the physics of what's going on).
In my program, I can easily calculate the distance between the ship and the target's initial position using a head minus tail rule. In this example, Vector U is <7,3>.
Vector V on the other hand is the troubling part. Vector V is a vector from it's predicted final position to it's initial position. The question is : how do we find this?
I have modeled the predicted final position with an equation. (Pt)f stands for final position. (Pt)i stands for initial position. Note t is the subscript for target. i means initial. Vt = initial target velocity. T means time.
(Pt)f = (Pt)i + Vt T.
(Pt)f is a vector, (Pt)i is a vector, Vt is a vector and time is a scalar. If you think about it, I'm doing a vector addition of two position values (You can think of Vttime as a position vector as a result of velocity scaled by time).
(Pt)f is going to be the vector sum of (Pt)i + (Vt T).
I am concerned with finding T though. Here's what I know:
The bullet needs to be at the same point. We can model the bullet by saying (Pb)f = VbT.
If they need to be at the same point, that means that (Pb)f = (Pt)f.
That means that VbT = (Pt)i + (Vt T). We can solve for T.
I did the algebra and concluded that T = (Pt)i /((Vb) - (Vt))
The problem is, this is Time, a scalar quantity. How do I find that? I can't just divide two vectors. Vb-Vt is just vector subtraction. How do I divide that from the vector (Pt)i?
Additionally, is this the correct approach? In the end, I can create vector V in my diagram by finding a vector between (Pt)f and (Pt)i. Then, once I have that, I can find the point of intersection by adding vector U and V. Then I have to convert that point to world space (remember origin is top left at (0,0)). I have a function that will rotate my ship towards that point and shoot in that point's direction.
Let me know if I have to be more clear, and thanks for the help!
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edited May 16, 2017 at 5:17
Qmechanic♦
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asked May 16, 2017 at 4:24
Maheer AeronMaheer Aeron
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You had the right approach up until you tried to divide by a vector to solve for t t. So, starting from equating the final position of the target and the bullet:
v⃗b t=p⃗t+v⃗t t.v→b t=p→t+v→t t.
Where v⃗b v→b is the velocity of the bullet, p⃗t p→t is the current position of the target, v⃗t v→t is the velocity of the target, and t t is time.
Ultimately, we want v⃗b v→b. We know the speed of the bullet, but not the direction. So, we can dot both sides of the equation with themselves to get the equation in non-vector form in terms of quantities we know.
(v⃗b t⋅v⃗b t)=(p⃗t+v⃗t t)⋅(p⃗t+v⃗t t).(v→b t⋅v→b t)=(p→t+v→t t)⋅(p→t+v→t t).
Expanding, we get
|v⃗b|2 t 2=|p⃗t|2+2(p⃗t⋅v⃗t)t+|v⃗t|2 t 2.|v→b|2 t 2=|p→t|2+2(p→t⋅v→t)t+|v→t|2 t 2.
0=|p⃗t|2+2(p⃗t⋅v⃗t)t+(|v⃗t|2−|v⃗b|2)t 2.0=|p→t|2+2(p→t⋅v→t)t+(|v→t|2−|v→b|2)t 2.
As a check on the correctness of this equation, set the speed of the target to zero, and you'll see that the time is the distance to the target divided by the speed of the bullet, as it should be. Now we have a quadratic equation in t t in terms of non-vector quantities we know. Solving for t t:
t=−2(p⃗t⋅v⃗t)±4(p⃗t⋅v⃗t)2−4(|v⃗t|2−|v⃗b|2)|p⃗t|2−−−−−−−−−−−−−−−−−−−−−−−−−√2(|v⃗t|2−|v⃗b|2)t=−2(p→t⋅v→t)±4(p→t⋅v→t)2−4(|v→t|2−|v→b|2)|p→t|2 2(|v→t|2−|v→b|2)
t=−(p⃗t⋅v⃗t)±(p⃗t⋅v⃗t)2−(|v⃗t|2−|v⃗b|2)|p⃗t|2−−−−−−−−−−−−−−−−−−−−−−−√|v⃗t|2−|v⃗b|2 t=−(p→t⋅v→t)±(p→t⋅v→t)2−(|v→t|2−|v→b|2)|p→t|2|v→t|2−|v→b|2
If the bullet is slower than the target, there will be one positive and one negative solution for t t. Choose the positive one since we want to hit the target in the future. Depending on the sign and magnitude of p⃗t⋅v⃗t,p→t⋅v→t, either the ++ or −− solution could end up positive, so calculate both and take the positive one.
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edited Oct 20, 2022 at 4:51
answered May 16, 2017 at 6:22
Mark HMark H
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What benefit will the bullets velocity get me in your last equation? I had a function in my code that allows me to rotate towards a point. I can find the intersection point by finding the final positon of the target with the new time you solved. Then I can create vector V (displacement from targets inital to final positon) and add it to U (displacement of ship to target) to get vector C. Then I can convert C to a point and rotate towards it. If I find the angle of the velocity vector, would this just benefit me in efficiency such that I automatically know the angle to rotate to?Maheer Aeron –Maheer Aeron 2017-05-16 21:43:17 +00:00 Commented May 16, 2017 at 21:43
@MaheerAeron The method you describe will work. If you only need the direction to fire, then your vector C and my final vector v⃗b v→b are equivalent, since they are parallel. There won't be any difference in efficiency either.Mark H –Mark H 2017-05-16 22:17:59 +00:00 Commented May 16, 2017 at 22:17
What's the difference between solving for time like your method versus splitting it into x and y components. I'm assuming you'll find time for the x component and a separate time for the y component then? Then you'd take the timeX and multiply the x component of the targets velocity and you can take the timeY and multiply the y component of the targets velocity.Maheer Aeron –Maheer Aeron 2017-05-19 23:20:20 +00:00 Commented May 19, 2017 at 23:20
@MaheerAeron Time isn't a vector. The time found for the x components has to be the same as for the y components in order for the bullet to hit the target. Separating into components doesn't seem easier than my method. I'll edit my answer to put in the final formula for t.Mark H –Mark H 2017-05-20 00:05:16 +00:00 Commented May 20, 2017 at 0:05
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In problems like this it is usually very convenient to set the time as additional dimension of your vector space. In this case you make your 2D problem 3D with the z z-component actually being time. Solving your then 3D vector equations gives the time immediately as 3rd component.
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answered May 16, 2017 at 8:49
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AM-GM Inequality
In algebra, the AM-GM Inequality, also known formally as the Inequality of Arithmetic and Geometric Means or informally as AM-GM, is an inequality that states that any list of nonnegative reals' arithmetic mean is greater than or equal to its geometric mean. Furthermore, the two means are equal if and only if every number in the list is the same.
In symbols, the inequality states that for any real numbers , with equality if and only if .
The AM-GM Inequality is among the most famous inequalities in algebra and has cemented itself as ubiquitous across almost all competitions. Applications exist at introductory, intermediate, and olympiad level problems, with AM-GM being particularly crucial in proof-based contests.
Contents
[hide]
1 Proofs
2 Generalizations
2.1 Weighted AM-GM Inequality
2.2 Mean Inequality Chain
2.3 Power Mean Inequality
3 Problems
3.1 Introductory
3.2 Intermediate
3.3 Olympiad
4 See Also
Proofs
Main article: Proofs of AM-GM
All known proofs of AM-GM use induction or other, more advanced inequalities. Furthermore, they are all more complex than their usage in introductory and most intermediate competitions. AM-GM's most elementary proof utilizes Cauchy Induction, a variant of induction where one proves a result for , uses induction to extend this to all powers of , and then shows that assuming the result for implies it holds for .
Generalizations
The AM-GM Inequality has been generalized into several other inequalities. In addition to those listed, the Minkowski Inequality and Muirhead's Inequality are also generalizations of AM-GM.
Weighted AM-GM Inequality
The Weighted AM-GM Inequality relates the weighted arithmetic and geometric means. It states that for any list of weights such that , with equality if and only if . When , the weighted form is reduced to the AM-GM Inequality. Several proofs of the Weighted AM-GM Inequality can be found in the proofs of AM-GM article.
Mean Inequality Chain
Main article: Mean Inequality Chain
The Mean Inequality Chain, also called the RMS-AM-GM-HM Inequality, relates the root mean square, arithmetic mean, geometric mean, and harmonic mean of a list of nonnegative reals. In particular, it states that with equality if and only if . As with AM-GM, there also exists a weighted version of the Mean Inequality Chain.
Power Mean Inequality
Main article: Power Mean Inequality
The Power Mean Inequality relates all the different power means of a list of nonnegative reals. The power mean is defined as follows: The Power Mean inequality then states that if , then , with equality holding if and only if Plugging into this inequality reduces it to AM-GM, and gives the Mean Inequality Chain. As with AM-GM, there also exists a weighted version of the Power Mean Inequality.
Problems
Introductory
For nonnegative real numbers , demonstrate that if then . (Solution)
Find the maximum of for all positive . (Solution)
Intermediate
Find the minimum value of for .
(Source)
Olympiad
Let , , and be positive real numbers. Prove that
(Source)
See Also
Proofs of AM-GM
Mean Inequality Chain
Power Mean Inequality
Cauchy-Schwarz Inequality
Inequality
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in:Math exercises, High school statistics and probability exercises, Probability and statistics: Random variables and probability distributions
Constructing probability distributions
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The Constructing probability distributions exercise appears under the High school statistics and probability Math Mission. This exercise helps to explore creating a probability distribution (or random variable assignment) for a situation.
Types of Problems[]
There are two types of problems in this exercise:
Create distribution from experiment: This problem provides some empirical data from a survey or situation. The student is asked to fill the histogram with the appropriate probabilities.
Create distribution from experiment
Create distribution from theory: This problem describes a theoretical situation. The student is asked to fill the histogram with the appropriate probabilities.
Create distribution from theory
Strategies[]
This exercise is medium to get accuracy badges because the theory problems require knowledge of multiplication principles of probability. The speed badges are difficult because the current time tolerances are very tight considering the time it takes to find the probabilities, especially on the theory problems.
All probabilities add up to 100%, so if you can figure out all but one of the probabilities, you can find the last by subtraction.
A tree diagram is a great way to organize the information for the theory problems.
Real-life Applications[]
Probability, along with decimals, percents, and fractions are used to determine the probability of a basketball player making a shot.
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Types of Functions
Functional Equations
Functions (mathematics)
Basic Functions
Mathematical Sciences
Mathematical Functions
5
How do I find all functions that satisfy f(x) =f(1/x)?
Mark Taranto
PhD in Finance, University of California, Berkeley
·
Upvoted by
Clyde Monma
, B.S. Computer Science & Mathematics, Washington State University (1974) · Author has 37.3K answers and 22.9M answer views
·
2y
There are an uncountably infinite number of such functions. For example, it is true for every constant function. The fact that you don’t state that it is continuous or differentiable complicates it even more.
for example, these two functions meet your criteria:
f(X) = 0 if X is rational and X if it is irrational.
f(X) = max(X, 1/X). We could get much more complicated.
Related questions
Which of the following functions f is f (x) = f (1 – x) for all x?
What functions satisfy f(x) - f(x-1) = 10/x?
How do I find all the functions satisfying f(x) ×f(1/x) =f(x) +f(1/x)?
How can I find all functions which satisfy f (x+f(x)) =x+f(f(x))?
Which functions can satisfy f''(x) =f(x).f'(x) and f'(x) =f''(x).f(x)? (individual functions for each equation)
Clyde Monma
B.S. in Computer Science & Mathematics, Washington State University (Graduated 1974)
·
Author has 418 answers and 91.3K answer views
·
2y
An interesting example is
f(x) = | log |x| |
since log(1/x) = log 1 - log x = - log x
It is continuous & differentiable everywhere except x=0
Balázs Iván József
Master in Mathematics, Eötvös Loránd University (Graduated 1983)
·
Author has 5.2K answers and 1.8M answer views
·
2y
Pair the numbers in the domain with their reciprocals and simply assign f’s values otherwise arbitrarily to the members of these pairs.
Silvio Capobianco
Senior Researcher at Tallinn University of Technology (2009–present)
·
Author has 3.3K answers and 619.1K answer views
·
2y
Do you require that the function be continuous?
Because if you don’t, the characteristic function of the set of rational numbers satisfies f(x) = f(1/x).
But if you do, then I suspect that only constant functions will do
Related questions
Can I find a function that meets f(x) =f (1/x+1)?
What is an example of a function f(x) such that f(f(x)) = x?
Is f(x) =1/x a function?
What is a function which satisfies f(x+1) =f(x) and f(1) =1?
What is the definition of a function where f(x) =f(-x)?
Jan van Delden
MSc Math and still interested
·
Author has 4.8K answers and 6.5M answer views
·
2y
Related
Can I find a function that meets f(x) =f (1/x+1)?
I don’t know about you, but one could try.
Just to be clear the question is to find a function f(x) such that
f(x)=f(1x+1)
Since we may not divide by 0 we can’t define f(0), but it doesn’t stop there. Solve
0=1x+1
and we see that we can’t define f(−1) as well, since we would get
f(−1)=f(0)=f(10+1)
In fact, with g(x)=1/x+1, then any other value x such that
gn(x)!=(g∘g∘⋯∘g)n times(x)=0
will bring us into trouble. All those values x may be retrieved by defining the inverse of g(x):
h(x)=g^{-1
I don’t know about you, but one could try.
Just to be clear the question is to find a function f(x) such that
f(x)=f(1x+1)
Since we may not divide by 0 we can’t define f(0), but it doesn’t stop there. Solve
0=1x+1
and we see that we can’t define f(−1) as well, since we would get
f(−1)=f(0)=f(10+1)
In fact, with g(x)=1/x+1, then any other value x such that
gn(x)!=(g∘g∘⋯∘g)n times(x)=0
will bring us into trouble. All those values x may be retrieved by defining the inverse of g(x):
h(x)=g−1(x)=1x−1
and computing all values xn=hn(0), for which we have x0=0 and
xn=1xn−1−1
The first few of those values xn, for which we can’t define f(x) are in the set
S0={0,−1/1,−1/2,−2/3,−3/5,−5/8,…}
You may notice that the numerators and denominators are part of the Fibonacci sequence. In fact the xn converge to ψ=−1/φ, where φ=(1+√5)/2, the golden ratio.
Both numbers φ,ψ are roots to x2n−xn−1=0 an equation you may find if we simply set xn=xn−1.
Convergence of g(x) may be established by studying |g′(x)|. I’ll omit the details, but φ is an attracting fixed point of g(x) and the other root is repelling. All values x∈R∖S0 converge to φ. Similarly h(x) converges for all x∈R∖S0 to ψ (and φ is repelling).
The main point is that excluding one value of x, here x0=0, excludes infinitely many xn for which we can define f(x). However we may define f(x) for x∈{φ,ψ}.
How about x∈R∖S0, i.e. other values of x?
For every other value x we know that we may define a similar set Sx=hn(x)∪gn(x) all of which elements should have the same value for f(⋅). All its elements xn will eventually converge to φ, by g since these xn are in the basin of attraction of φ. and if we iterate by h to ψ. If the function is not to be continuous, we may define f for an uncountable set of sets Sx. As long as we make sure that we don’t use x∈S0.
If f(x) is to be continuous then f(x) needs to be constant for those x where f(⋅) is defined, since the elements xn=gn(x) converge to the same value φ for every x.
Continuity in a neighourhood of φ means that for positive y∈R sufficiently close to φ() that for every ϵ>0 there is a δ such that
|y−φ|<δ⟹|f(y)−f(φ)|<ϵ
We certainly can pick y equal to a value xn∈Sx arbitrary close to φ, since the xn converge to φ. But if we set f(xn)≠f(φ) then the right hand side is a fixed positive constant, which can only be smaller than ϵ for every positive ϵ if f(xn)=f(φ).
() The set S0 doesn’t contain positive values y.
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Vagabond Guy
Costal Cruiser Who Loves Sailing
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Updated May 1
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Airline Mistake Fare Alerts are my favorite travel hack for getting cheap flights all over the world for a fraction of the cost!
(Check out the screenshots below to show you the kinds of deals you can get)
👉 Instead of looking for cheap flights, let them find you!
Sign up for airline mistake fare and flash sales alerts and quit planning trips like it’s 1960—choosing dates and a destination first, then spending days (or weeks) trying to make those locked-in plans fit your budget. That’s like walking onto a car lot, announcing the exact model and purchase date, and hoping the price magically works out.
This makes zero sense! You would never agree to buy a certain model of car on a specific date, and then just be stuck with either paying whatever they wanted on that date, or worse you not have a car at all. There is no other thing in your life where you plan with such insanity as one does vacations. Why would you not just choose from a list of top destinations that are currently offering a huge discount on flights? You actually would, you just didn’t know there was options to be notified of every amazing deal and that’s the purpose of this post.
Mistake fare and flash sales alerts can save you 50–90% on every single flight you book! This one trick has allowed me to see more of the world than I thought I would ever be able to. Prior to planning my travel experiences in this manner, we always went on cruises because it was the only affordable way to do travel for me. Not anymore!
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There are a handful of websites that do this, here’s my favorite. They don’t sell airfare; they only find the best deals and alert you of those deals via email.
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Pablo Emanuel
MSc in Math and IMO medalist
·
Author has 846 answers and 5.4M answer views
·
4y
Related
How do I find all functions f satisfying f(x+1) +1=f(f(x) +1) for f: N -->N?
From now on I’ll assume 0∈N, but everything would be essentially the same if we assume N starting at 1 (left as an exercise for the reader).
First thing to test is whether f is injective. So assume f(x)=f(y), with y=x+k>x.
Then
f(x+1)+1=f(f(x)+1)=f(f(y)+1)=f(y+1)+1⟹f(x+1)=f(y+1)
By induction, that means that, for any m>0,f(x+m)=f(y+m). In particular, f(x+k)=f(y)=f(x+2k)=f(x). That means that, starting on x, f is periodic with period k. In particular the image set of f, f(N)={f(0),f(1),⋯,f(x+k−1)} is finite. So, in short,
Claim 1: Either
From now on I’ll assume 0∈N, but everything would be essentially the same if we assume N starting at 1 (left as an exercise for the reader).
First thing to test is whether f is injective. So assume f(x)=f(y), with y=x+k>x.
Then
f(x+1)+1=f(f(x)+1)=f(f(y)+1)=f(y+1)+1⟹f(x+1)=f(y+1)
By induction, that means that, for any m>0,f(x+m)=f(y+m). In particular, f(x+k)=f(y)=f(x+2k)=f(x). That means that, starting on x, f is periodic with period k. In particular the image set of f, f(N)={f(0),f(1),⋯,f(x+k−1)} is finite. So, in short,
Claim 1: Either f is injective, or f(N) is finite.
Now, assume that z∈f([1,∞)). Then there is x>0 such that f(x)=z. But that means that z+1=f(x−1+1)+1=f(f(x−1)+1), that is, z+1∈f([1,∞)). That implies
Claim 2: f([1,∞))=[c,∞) for some c∈N.
Corollary: f(N) is infinite, hence f is injective.
Now, let a=f(0). Then we have
f(1)+1=f(f(0)+1)=f(a+1)⟹f(1)=f(a+1)−1
f(2)+1=f(f(1)+1)=f(f(a+1))⟹f(2)=f2(a+1)−1
f(3)+1=f(f(2)+1)=f(f2(a+1))⟹f(3)=f3(a+1)−1
By induction, f(n)=fn(a+1)−1
Therefore
Claim 3: The orbit of a+1, O(a+1)={fn(a+1)|n∈N} is exactly [c+1,∞)∪{a+1}.
But now,
(f(O(a+1))=([c+1,∞)∪{a+1})∖{a+1}=[c,∞)∖{c,a+1}(1),
since f is injective, therefore the orbit starting on n = 1 will be exactly the orbit starting at n = 0 minus the initial point. But, on the other hand,
f(O(a+1))=f([c+1,∞))∪{f(a+1)}=(f([1,∞))∖{f(1),f(2),⋯f(c)})∪{f(a+1)}=([c,∞)∖{f(1),f(2),⋯f(c)})∪{f(a+1)}(2)
But, on expression 2, we’re excluding exactly c or c−1points from [c,∞), whereas, in expression 1, we’re excluding 1 or 2 points. So we have 3 cases:
Case 1: c = 1.
In this case f([1,∞))=[1,∞) which constrains a=f(0) to be 0. But that means that f(1)+1=f(1), contradiction. ■
Case 2: c = 2.
In this case f([1,∞))=[2,∞), and we’ve seen that f(0)≠0. Therefore, we need to have a=f(0)=1. Therefore, O(2)=[2,∞), and, (1) and (2) turn into
f(O(2))=[2,∞)∖{2}=[3,∞)=[2,∞)∖{f(1)}
which means f(1)=2. But we know that f(1)=f(2)−1, therefore, f(2)=3. But that means that f(3)=f2(2)=1+f(2)=4. By induction, we have f(n)=n+1∀n, and, in fact, this satisfies the functional equation. ■
Case 3: c = 3
That can only happen when, on expression (1) we’re excluding exactly 2 points, i.e., when c≠a+1, and on expression (2) we’re excluding exactly c−1 points, i.e. a+1∈{1,2,⋯,c}, which, since c = 3, means a≠2 and a∈{0,1,2}. So a=0 or a=1. But we already found a contradiction if a=f(0)=0. Therefore, we need to have a=1
So that means that the sets excluded in (1) and (2) are respectively {2,3} and {f(1),f(3)}, which have to be equal. But, by hypothesis, f([1,∞))=[3,∞), therefore neither f(1) nor f(3) can be 2, contradiction. ■
Therefore, the only function satisfying this equation is f(n)=n+1.
Jonathan Devor
PhD in Astronomy, Harvard University (Graduated 2008)
·
Author has 3.7K answers and 19.3M answer views
·
11mo
Related
How do I find all functions
f
:
R
→
R
,
such that
f
(
x
−
f
(
y
)
)
=
1
−
x
−
y
,
∀
x
,
y
∈
R
?
We’re given the functional equation: f(x−f(y))=1−x−y
Let’s define: c=f(1)
Now, setting y=1 gives us:
f(x−f(1))=−x⟹
f(x−c)=−x⟹
f(x)=−x−c
If we plug back this solution into the original equation, we get:
f(x+y+c)=1−x−y⟹
(−x−y−c)−c=1−x−y⟹
−2c=1⟹
c=−12
In conclusion, the only solution is:
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Pablo Emanuel
Masters in Mathematics, IMPA (Graduated 1996)
·
Upvoted by
Ermal Feleqi
, PhD Mathematics, University of Padua (2010) · Author has 846 answers and 5.4M answer views
·
1y
Related
What is the definition of a function f(x) such that f(f(x)) =f(x)?
Functions that have this property are called idempotent.
If [math]f:A \to B[/math] is such a function, then we need to have [math]f(A) \subset A[/math], so that [math]f(f(x))[/math] even makes sense, and also that [math]f[/math] restricted to [math]f(A)[/math] to be the identity function.
The identity function itself is a trivial example, with [math]f(A)=A[/math], but here are some non-trivial examples:
[math]f:\R^3 \to \R^3[/math] given by a linear projection onto a plane, e.g. [math]f(x,y,z)=(x,y,0)[/math],
[math]f:\R \to \R[/math], [math]f(x)=|x|[/math],
[math]f:\R \to \R[/math], [math]f(x)=\lfloor x \rfloor[/math],
The function that map the set of citizens of a republic to the president of their republic.
The mapping from a field to its algebraic cl
Functions that have this property are called idempotent.
If [math]f:A \to B[/math] is such a function, then we need to have [math]f(A) \subset A[/math], so that [math]f(f(x))[/math] even makes sense, and also that [math]f[/math] restricted to [math]f(A)[/math] to be the identity function.
The identity function itself is a trivial example, with [math]f(A)=A[/math], but here are some non-trivial examples:
[math]f:\R^3 \to \R^3[/math] given by a linear projection onto a plane, e.g. [math]f(x,y,z)=(x,y,0)[/math],
[math]f:\R \to \R[/math], [math]f(x)=|x|[/math],
[math]f:\R \to \R[/math], [math]f(x)=\lfloor x \rfloor[/math],
The function that map the set of citizens of a republic to the president of their republic.
The mapping from a field to its algebraic closure.
The mapping from a subset of the plane to its convex hull.
The map from a natural number to its remainder when divided by 42.
Any constant function.
Idempotent functions are important in computer science, because you don’t need to worry whether they’re invoked more than once - it’s especially important when messages are sent over unreliable channels, and we want to be able to retry without being sure if the original call will eventually be received, causing the function to be invoked twice.
Kumar Kishlay
PGP 2018-20, IIMA
·
Updated 8y
Related
How do I find all the functions satisfying f(x) ×f(1/x) =f(x) +f(1/x)?
I am assuming that you understand that the solution to the problem f(x).f(1/x)=1 is f(x)=
[math]x^n or-x^n[/math]
.
Thanks.
I am assuming that you understand that the solution to the problem f(x).f(1/x)=1 is f(x)=
[math]x^n or-x^n[/math]
.
Thanks.
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Jan van Delden
MSc Math and still interested
·
Author has 4.8K answers and 6.5M answer views
·
3y
Related
What is f(x) if f'(x) =f(x+1)?
If we try [math]f(x)=e^{\alpha x}[/math] we find that we need to solve:
[math]\alpha=e^{\alpha}[/math]
which may be recast to:
[math]-1 = -\alpha e^{-\alpha}[/math]
And we find with this trial function:
[math]\alpha_n =- W_n(-1), n\in \mathbb{Z}[/math]
where [math]W_n(\cdot)[/math] is the Lambert-W function, and [math]n[/math] indicates the branch. If we pick [math]n=0[/math] we specify the principal branch.
[math]\alpha_0 \approx 0.3181-1.337 i[/math]
[math]\alpha_{-1} \approx 0.3181+1.337 i[/math]
These two are complex conjugates. Since the differential equation is linear we may add any linear combination to try and find a real solution. Here a simple addition will work, we find that (where I choose to use the app
If we try [math]f(x)=e^{\alpha x}[/math] we find that we need to solve:
[math]\alpha=e^{\alpha}[/math]
which may be recast to:
[math]-1 = -\alpha e^{-\alpha}[/math]
And we find with this trial function:
[math]\alpha_n =- W_n(-1), n\in \mathbb{Z}[/math]
where [math]W_n(\cdot)[/math] is the Lambert-W function, and [math]n[/math] indicates the branch. If we pick [math]n=0[/math] we specify the principal branch.
[math]\alpha_0 \approx 0.3181-1.337 i[/math]
[math]\alpha_{-1} \approx 0.3181+1.337 i[/math]
These two are complex conjugates. Since the differential equation is linear we may add any linear combination to try and find a real solution. Here a simple addition will work, we find that (where I choose to use the approximate values)
[math]f(x)=c e^{0.3181 x} \cos(1.337 x)[/math]
is a solution, for any constant [math]c[/math]. Subtraction is fine as well:
[math]f(x)=c e^{0.3181 x} \sin(1.337 x)[/math]
Or addition of these two, for any constants:
[math]f(x)=e^{0.3181 x} ( c_1 \cos(1.337 x)+c_2 \sin(1.337 x))[/math]
If you dislike complex numbers, or are not that acquainted yet then you might better try:
[math]f(x)=e^{ax}\cos(bx)[/math]
substitute, expand [math]\cos(b(x+1))[/math] and you may retrieve:
[math]\left{\begin{array}{l} &a=e^a \cos(b)\ &b=e^a \sin(b)\end{array}\right.[/math]
from which:
[math]a=b \cot(b)[/math]
is easy to deduce. Substitute into any of the two given equations and use a numerical tool to retrieve the value for [math]b[/math].
Oliver Jennrich
Native speaker, born and raised in Germany.
·
Author has 766 answers and 1.2M answer views
·
8y
Related
Are there any functions such that f(g(x)) = x and f(x) = 1/g(x)?
Well, we can try [math]f(x) = x^a[/math] and [math]g(x) = x^b[/math].
Then[math] f(g(x)) = (x^b)^a = x^{ab}[/math]. If this is supposed to be equal to [math]x[/math], it means that [math]ab=1[/math]. The second requirement is [math]a=-b[/math], so [math]-b^2=1[/math] or [math]b=\sqrt{-1}=\mathrm{i}[/math] and [math]a=-\mathrm{i}[/math].
So there you go: [math]f(x)=x^{-\mathrm{i}}, g(x) = x^{\mathrm{i}}[/math]
Of course, there might be other solutions, including the rather trivial [math]f(x)=1, g(x)=1[/math]
W. Dale Hall
Ph.D in Mathematics & Physics, University of Wisconsin - Madison (Graduated 1978)
·
Author has 276 answers and 178.7K answer views
·
1y
Related
How do I Find f(x) =1/×?
I’m pretty confident that this isn’t the question you meant to ask, unless you’re hopelessly lost. After all, you’ve just displayed the very thing you’re asking us (that is, the Quora world) to find.
Perhaps you want to locate a particular fact about this function, or maybe your calculator has a key that computes [math]\frac {1}{x}[/math] for whatever value [math]x[/math] is currently displayed. If you were to ask something more specific, I think your responses would be more helpful.
Alexander Farrugia
Ph.D. in Mathematics, University of Malta (Graduated 2016)
·
Author has 3.2K answers and 27.4M answer views
·
6y
Related
What is the function where f'(x) =1/f(x), holds true?
We can alternatively write down the relation as
This reminds me of an application of the product rule, specifically the following:
Thus [math]f^\prime(x)f(x)=1[/math] means that
[math]\displaystyle\big(f(x)\big)^2=\int 2\,dx = 2x+K.[/math]
for any constant [math]K[/math].
Related questions
Which of the following functions f is f (x) = f (1 – x) for all x?
What functions satisfy f(x) - f(x-1) = 10/x?
How do I find all the functions satisfying f(x) ×f(1/x) =f(x) +f(1/x)?
How can I find all functions which satisfy f (x+f(x)) =x+f(f(x))?
Which functions can satisfy f''(x) =f(x).f'(x) and f'(x) =f''(x).f(x)? (individual functions for each equation)
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11869 | https://www.khanacademy.org/math/grade-6-math-tx/x3931b57772b927b3:equivalent-expressions/x3931b57772b927b3:greatest-common-factor/v/distributive-property-exercise-2 | Factor with the distributive property (video) | Khan Academy
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6th grade math (TX TEKS)
Course: 6th grade math (TX TEKS)>Unit 9
Lesson 1: Greatest common factor
Greatest common factor examples
Greatest common factor explained
Greatest common factor
Factor with the distributive property
Factor with the distributive property (no variables)
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6th grade math (TX TEKS)>
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Greatest common factor
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Factor with the distributive property
NJ.Math: 6.EE.A.3, 6.EE.A.4
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The distributive property helps us factor out the greatest common factor (GCF) in math problems. By finding the GCF of two numbers, we can rewrite the expression as a product, making calculations easier. Understanding this property is essential for simplifying expressions and solving equations.
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Sara Lynn Younes 6 years ago Posted 6 years ago. Direct link to Sara Lynn Younes's post “Why can’t we just add and...” more Why can’t we just add and get the answer? It’s easier. 35 plus 50= 85. I don’t want to go with that process every time I add.
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Ali W. 3 years ago Posted 3 years ago. Direct link to Ali W.'s post “This isn't to find the su...” more This isn't to find the sum, it's to find the greatest common factor which is useful in different ways. It's applied to later algebruh.
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sgh07 4 years ago Posted 4 years ago. Direct link to sgh07's post “no five hundred would be ...” more no five hundred would be 250,000. 200 times 50 would be 10,000.
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ayanverma30 2 years ago Posted 2 years ago. Direct link to ayanverma30's post “bro u have to learn to fa...” more bro u have to learn to factor with the distributive for later stuff
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AceSpace 5 years ago Posted 5 years ago. Direct link to AceSpace's post “I seem to understand how ...” more I seem to understand how to do this just fine but i notise that when Im doing prime factorizing I dont get all possible greatest factors, am I doing something wrong? Or is there a more accurate way of factorizing that will get me all the factors or actual largest factor? Otherwise factorizing doesnt work well in these cases and its better to just list all the factors the other way?
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Video transcript
[Voiceover] We're asked to apply the distributive property to factor out the greatest common factor, and we have 35 plus 50 is equal to, so let me get my scratch pad out. So we have 35 plus 50 is equal to, now what is the greatest common factor of 35 and 50. So what's the largest whole number that's divisible into both of these. Well I could write 35 as, let's see, I could write that as five times seven, and I could write 50 as five times ten, and so we see five is the greatest common factor. Seven and ten don't have any factors in common. So I could rewrite this, I could write 35 as equal to five times seven and I could rewrite 50 as equal to, get another color here, I could rewrite 50 as five times ten, and of course, I'm adding them, so I have plus right over here. If I want, I could put parentheses, but order of operations would make me do the multiplication first, anyway. But now I want to factor out that greatest common factor. I want to factor out the five. So what I'm really doing right over here is I'm unwinding the distributive property. So if I factor out a five, this is going to be equal to, this is going to be equal to let's factor out the five. Five times, so you do 35 divided by five, you're just left with the seven. You're just left with the seven over here. So you're just left with the seven after you've factored out the five, and over here, you're just left with the ten. So five, or seven plus ten, and we're done. 35 plus 50 is equal to five times seven plus ten. So let me now go and type that in. So this is the same thing as five, five times seven plus ten. And you know you've factored out the greatest common factor because seven and ten don't have any factors in common anymore. They're called relatively prime. They have no factors in common other than one. So we could now check that. Let's do a couple more of these. Apply the distributive property to factor out the greatest common factor. So let's see if we can just do this one a little bit faster. So let's see, the largest number it's divisible, in both 75 and 20, I don't know, let me try five. So if I say five times, so 75 divided by five, let's see is going to be, it's going to be 15. Is that right? Yeah, cause five times ten is 50, five times five is 25. Yeah, so it's 15, and I got that 15 by dividing 75 by five. So 15 plus, and then 20 divided by five would be, 20 divided by five would be four. 15 plus four. Let's see, did I do this right? 15 and four don't have any factors in common, and if I were to apply the distributive property here, I'd have five times 15 is 75. Five times four is 20. Yeah, I'm feeling good about that. We got it right.
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11870 | https://en.wikipedia.org/wiki/Stupor | Jump to content
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Contents
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1 Signs and symptoms
2 Causes
3 Management
4 See also
5 Notes
6 References
7 External links
Stupor
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From Wikipedia, the free encyclopedia
Medical condition
| Stupor |
| A patient in catatonic stupor |
| Specialty | Neurology, Psychiatry |
Stupor is the lack of critical mental function and a level of consciousness, in which an affected person is almost entirely unresponsive and responds only to intense stimuli such as pain. The word derives from the Latin stupor ("numbness, insensibility").
Signs and symptoms
[edit]
Stupor is characterized by impaired reaction to external stimuli. Those in a stuporous state are rigid, mute and only appear to be conscious, as the eyes are open and follow surrounding objects. If not stimulated externally, a patient with stupor will appear to be in a sleepy state most of the time. In some extreme cases of severe depressive disorders the patient can become motionless, lose their appetite and become mute. Short periods of restricted responsivity can be achieved by intense stimulation (e.g. pain, bright light, loud noise, shock).[citation needed]
Causes
[edit]
Stupor is associated with infectious diseases, complicated toxic states (e.g. heavy metals), severe hypothermia, mental illnesses (e.g. schizophrenia, major depressive disorder), epilepsy, vascular illnesses (e.g. hypertensive encephalopathy), acute stress reaction (shock), neoplasms (e.g. brain tumors), brain disorders (e.g. alzheimers, dementia, fatal insomnia), B12 deficiency, major trauma, alcohol poisoning, vitamin D excess, and other conditions.
Lesions of the ascending reticular activation system on height of the pons and metencephalon have been shown to cause stupor. The incidence is higher after left-sided lesions.[citation needed]
Management
[edit]
Because stupors are caused by another health condition, treatment focuses on uncovering and treating the cause. Doctors may administer IV antibiotics or fluids to treat infections and nutritional deficits, or conduct an MRI to check for lesions on the brain.[citation needed]
See also
[edit]
Torpor
Notes
[edit]
^ a b Gelder, M.; Mayou, R.; Geddes, J. (2005). Psychiatry (3rd ed.). New York: Oxford.
^ Berrios, G. E. (1981). "Stupor: A Conceptual History". Psychological Medicine. 11 (4): 677–688. doi:10.1017/S0033291700041179. PMID 7034030. S2CID 26932116.
^ Berrios, G. E. (1981). "Stupor Revisited". Comprehensive Psychiatry. 22 (5): 466–478. doi:10.1016/0010-440X(81)90035-3. PMID 7187867.
References
[edit]
C. Lafosse, Zakboek Neuropsychologische Symptomatologie, p. 37, ISBN 90-334-3995-6.
External links
[edit]
The dictionary definition of stupor at Wiktionary
| | |
--- |
| Classification | D ICD-10: R40.1 ICD-9-CM: 780.09 MeSH: D053608 |
| v t e Disorders of consciousness |
| Unconsciousness | Minimally conscious state Persistent vegetative state Obtundation Coma Brainstem death Stupor + Sopor + Sleep + Somnolence Cataplexy |
| Syncope | Heat syncope Vasovagal episode |
| Alteration ofconsciousness | Locked-in syndrome |
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11871 | https://rgmia.org/papers/v7n4/vpn.pdf | Equations and Inequalities Involving vp(n!) Mehdi Hassani Department of Mathematics Institute for Advanced Studies in Basic Sciences Zanjan, Iran mhassani@iasbs.ac.ir Abstract In this paper we study vp(n!), the greatest power of prime p in factorization of n!. We find some lower and upper bounds for vp(n!), and we show that vp(n!) = n p−1 + O(ln n).
By using above mentioned bounds, we study the equation vp(n!) = v for a fixed positive integer v. Also, we study the triangle inequality about vp(n!), and show that the inequality pvp(n!) > qvq(n!) holds for primes p < q and sufficiently large values of n.
2000 Mathematics Subject Classification: 05A10, 11A41, 26D15, 26D20.
Keywords: factorial function, prime number, inequality.
1 Introduction As we know, for every n ∈N, n! = 1 × 2 × 3 × · · · × n. Let vp(n!) be the highest power of prime p in factorization of n! to prime numbers. It is well-known that (see or ) vp(n!) = ∞ X k=1 n pk = [ ln n ln p ] X k=1 n pk , (1) in which [x] is the largest integer less than or equal to x. An elementary problem about n! is finding the number of zeros at the end of it, in which clearly its answer is v5(n!). The inverse of this problem is very nice; for example finding values of n in which n! terminates in 37 zeros , and generally finding values of n such that vp(n!) = v. We show that if vp(n!) = v has a solution then it has exactly p solutions.
For doing these, we need some properties of [x], such as [x] + [y] ≤[x + y] (x, y ∈R), (2) and hx n i = [x] n (x ∈R, n ∈N).
(3) 2 Estimating vp(n!) Theorem 1 For every n ∈N and prime p, such that p ≤n, we have: n −p p −1 −ln n ln p < vp(n!) ≤n −1 p −1.
(4) Proof: According to the relation (1), we have vp(n!) = Pm k=1[ n pk ] in which m = [ ln n ln p], and since x −1 < [x] ≤x, we obtain n m X k=1 1 pk −m < vp(n!) ≤n m X k=1 1 pk , considering Pm k=1 1 pk = 1− 1 pm p−1 , we yield that n p −1(1 −1 pm) −m < vp(n!) ≤ n p −1(1 −1 pm), and combining this inequality with ln n ln p −1 < m ≤ln n ln p completes the proof.
Corollary 1 For every n ∈N and prime p, such that p ≤n, we have: vp(n!) = n p −1 + O(ln n).
Proof: By using (4), we have 0 < n p−1 −vp(n!) ln n < 1 ln p, and this yields the result.
2 Note that the above corollary asserts that n! ends approximately in n 4 zeros .
Corollary 2 For every n ∈N and prime p, such that p ≤n, and for all a ∈(0, +∞) we have: n −p p −1 − 1 ln p n a + ln a −1 < vp(n!).
(5) Proof: Consider the function f(x) = ln x. Since, f ′′(x) = −1 x2, ln x is a concave function and so, for every a ∈(0, +∞) we have ln x ≤ln a + 1 a(x −a), combining this with the left hand side of (4) completes the proof.
3 Study of the Equation vp(n!) = v Suppose v ∈N is given.
We are interested to find the values of n such that in factorization of n!, the highest power of p, is equal to v. First, we find some lower and upper bounds for these n’s.
Lemma 1 Suppose v ∈N and p is a prime and vp(n!) = v, then we have 1 + (p −1)v ≤n < v + p p−1 + ln(1+(p−1)v) ln p − 1 ln p 1 p−1 − 1 (1+(p−1)v) ln p .
(6) Proof: For Proving the left hand side of (6), use right hand side of (4) with assump-tion vp(n!) = v, and for proving the right hand side of (6), use (5) with a = 1+(p−1)v.
The lemma 1 suggest an interval for the solution of vp(n!) = v. In the next lemma we show that it is sufficient one check only multiples of p in above interval.
Lemma 2 Suppose m ∈N and p is a prime, then we have vp((pm + p)!) −vp((pm)!) ≥1.
(7) Proof: By using (1) and (2) we have vp((pm + p)!) = ∞ X k=1 pm + p pk ≥ ∞ X k=1 pm pk + ∞ X k=1 p pk = 1 + vp((pm)!), and this completes the proof.
3 In the next lemma, we show that if vp(n!) = v has a solution, then it has exactly p solutions. In fact, the next lemma asserts that if vp((mp)!) = v holds, then for all 0 ≤r ≤p −1, vp((mp + r)!) = v also holds.
Lemma 3 Suppose m ∈N and p is a prime, then we have vp((m + 1)!) ≥vp(m!), (8) and vp((pm + p −1)!) = vp((pm)!).
(9) Proof: For proving (8), use (1) and (2) as follows vp((m + 1)!) = ∞ X k=1 m + 1 pk ≥ ∞ X k=1 m pk + ∞ X k=1 1 pk = ∞ X k=1 m pk = vp(m!).
For proving (9), it is enough to show that for all k ∈N, [pm+p−1 pk ] = [pm pk ] and we do this by induction on k; for k = 1, clearly [ pm+p−1 p ] = [ pm p ]. Now, by using (3) we have pm + p −1 pk+1 = " pm+p−1 pk p # = h pm+p−1 pk i p = h pm pk i p = " pm pk p # = pm pk+1 .
This completes the proof.
So, we have proved that Theorem 2 Suppose v ∈N and p is a prime. For solving the equation vp(n!) = v, it is sufficient to check the values n = mp, in which m ∈N and 1 + (p −1)v p ≤m ≤ " v + p p−1 + ln(1+(p−1)v) ln p − 1 ln p p p−1 − p (1+(p−1)v) ln p # .
(10) Also, if n = mp is a solution of vp(n!) = v, then it has exactly p solutions n = mp+r, in which 0 ≤r ≤p −1.
Note and Problem 1 As we see, there is no guarantee for existing a solution for vp(n!) = v. In fact we need to show that {vp(n!)|n ∈N} = N; however, computational observations suggest that n = p||1+(p−1)v p || usually is a solution, such that ||x|| is the nearest integer to x, but we can’t prove it.
Note and Problem 2 Other problems can lead us to other equations involving vp(n!); for example, suppose n, v ∈N given, find the value of prime p such that vp(n!) = v.
Or, suppose p and q are primes and f : N2 →N is a prime value function, for which n’s we have vp(n!) + vq(n!) = vf(p,q)(n!)? And many other problems!
4 4 Triangle Inequality Concerning vp(n!) In this section we are going to compare vp((m + n)!) and vp(m!) + vp(n!).
Theorem 3 For every m, n ∈N and prime p, such that p ≤min{m, n}, we have vp((m + n)!) ≥vp(m!) + vp(n!), (11) and vp((m + n)!) −vp(m!) −vp(n!) = O(ln(mn)).
(12) Proof: By using (1) and (2), we have vp((m + n)!) = ∞ X k=1 m + n pk ≥ ∞ X k=1 m pk + ∞ X k=1 n pk = vp(m!) + vp(n!).
Also, by using (4) and (11) we obtain 0 ≤vp((m + n)!) −vp(m!) −vp(n!) < 2p −1 p −1 + ln(mn) ln p ≤3 + ln(mn) ln 2 , this completes the proof.
More generally, if n1, n2, · · · , nt ∈N and p is a prime, in which p ≤min{n1, n2, · · · , nt}, by using an extension of (2), we obtain vp(( t X k=1 nk)!) ≥ t X k=1 vp(nk!), and by using this inequality and (4), we yield that 0 ≤vp(( t X k=1 nk)!) − t X k=1 vp(nk!) < kp −1 p −1 + ln(n1n2 · · · nt) ln p ≤2k −1 + ln(n1n2 · · · nt) ln p , and consequently we have vp(( t X k=1 nk)!) − t X k=1 vp(nk!) = O(ln(n1n2 · · · nt)).
Note and Problem 3 Suppose f : Nt →N is a function and p is a prime. For which n1, n2, · · · , nt ∈N, we have vp((f(n1, n2, · · · , nt)!) ≥f(vp(n1!), vp(n2!), · · · , vp(nt!))?
Also, we can consider the above question in other view points.
5 5 The Inequality pvp(n!) > qvq(n!) Suppose p and q are primes and p < q. Since vp(n!) ≥vq(n!), comparing pvp(n!) and qvq(n!) become a nice problem. In , by using elementary properties about [x], it is considered the inequality pvp(n!) > qvq(n!) in some special cases, beside it is shown that 2v2(n!) > 3v3(n!) holds for all n ≥4. In this section we study pvp(n!) > qvq(n!) in more general case and also reprove 2v2(n!) > 3v3(n!).
Lemma 4 Suppose p and q are primes and p < q, then pq−1 > qp−1.
Proof: Consider the function f(x) = x 1 x−1 (x ≥2).
A simple calculation yields that for x ≥2 we have f ′(x) = −x x−2 x−1(x ln x −x + 1) (x −1)2 < 0, so, f is strictly decreasing and f(p) > f(q). This completes the proof.
Theorem 4 Suppose p and q are primes and p < q, then for sufficiently large n’s we have pvp(n!) > qvq(n!).
(13) Proof: Since p < q, the lemma 4 yields that pq−1 qp−1 > 1 and so, there exits N ∈N such that for n > N we have pq−1 qp−1 n ≥pp(q−1) qp−1 n(p−1)(q−1).
Thus, pn(q−1) n(p−1)(q−1)pp(q−1) ≥qn(p−1) qp−1 , and therefor, p n p−1 np p p−1 ≥q n q−1 q 1 q−1 .
So, we obtain p n−p p−1 −ln n ln p ≥q n−1 q−1 , and considering this inequality with (4), completes the proof.
6 Corollary 3 For n = 2 and n ≥4 we have 2v2(n!) > 3v3(n!).
(14) Proof: It is easy to see that for n ≥30 we have (4 3)n ≥16 3 n2, and by theorem 4, we yield (14) for n ≥30. For n = 2 and 4 < n < 30 check it by a computer.
A Computational Note. In the theorem 4, the relation (13) holds for n > N (see its proof). We can check (13) for n ≤N at most by checking the following number of cases: R(N) := # {(p, q, n)| p, q ∈P, n = 3, 4, · · · , N, and p < q ≤N} , in which P is the set of all primes. If, π(x) = The number of primes≤x, then we have R(N) = N X n=3 # {(p, q)| p, q ∈P, and p < q ≤n} = 1 2 N X n=3 π(n)(π(n) −1).
But, clearly π(n) < n and this yields that R(N) < N 3 6 .
Of course, we have other bounds for π(n) sharper that n such as π(n) < n ln n 1 + 1 ln n + 2.25 ln2 n (n ≥355991), and by using this bound we can find sharper bounds for R(N).
Acknowledgement. I deem my duty to thank A. Abedin-Zade and Y. Rudghar-Amoli for their comments on the Note and Problem 1.
References Andrew Adler and John E. Coury, The Theory of Numbers, Bartlett Publishers, 1995.
7 Ion B˘ al˘ acenoiu, Remarkable inequalities, Proceedings of the First International Conference on Smarandache Type Notions in Number Theory (Craiova, 1997), 131135, Am. Res. Press, Lupton, AZ, 1997.
David M. Burton, Elementary Number Theory (Second Edition), Universal Book Stall, 1990.
P. Dusart, In´ egalit´ es explicites pour ψ(X), θ(X), π(X) et les nombres premiers, C. R. Math. Acad. Sci. Soc. R. Can. 21 (1999), no. 2, 53–59.
Melvyn B. Nathanson, Elementary Methods in Number Theory, Springer, 2000.
8 |
11872 | https://pangea.stanford.edu/~edunham/publications/Prochnow_etal_axisymmetric-waves_CF17.pdf | Computers and Fluids 149 (2017) 138–149 Contents lists available at ScienceDirect Computers and Fluids journal homepage: www.elsevier.com/locate/compfluid Treatment of the polar coordinate singularity in axisymmetric wave propagation using high-order summation-by-parts operators on a staggered grid Bo Prochnow a , Ossian O’Reilly a , b , ∗, Eric M. Dunham a , c , N. Anders Petersson d a Department of Geophysics, Stanford University, Stanford, CA, USA b Department of Mathematics, Division of Computational Mathematics Linköping University, SE-581 83 Linköping, Sweden c Institute for Computational and Mathematical Engineering, Stanford University, Stanford, CA, USA d Center for Applied Scientific Computing, Lawrence Livermore National Laboratory, PO Box 808, Livermore,CA 94551, United States a r t i c l e i n f o Article history: Received 29 July 2016 Revised 7 February 2017 Accepted 14 March 2017 Available online 16 March 2017 Keywords: Polar coordinate singularity Summation-by-parts High-order finite difference method Staggered grid Axisymmetric wave propagation a b s t r a c t We develop a high-order finite difference scheme for axisymmetric wave propagation in a cylindrical conduit filled with a viscous fluid. The scheme is provably stable, and overcomes the difficulty of the polar coordinate singularity in the radial component of the diffusion operator. The finite difference ap- proximation satisfies the principle of summation-by-parts (SBP), which is used to establish stability using the energy method. To treat the coordinate singularity without losing the SBP property of the scheme, a staggered grid is introduced and quadrature rules with weights set to zero at the endpoints are consid- ered. The accuracy of the scheme is studied both for a model problem with periodic boundary conditions at the ends of the conduit and its practical utility is demonstrated by modeling acoustic-gravity waves in a magmatic conduit. © 2017 Elsevier Ltd. All rights reserved. 1. Introduction Wave propagation in fluid-filled tubes or conduits arises in many systems, ranging from pulsatile flow in blood vessels [1,2] to acoustic-gravity waves in magmatic conduits beneath volcanoes . Fluid viscosity, a key source of dissipation, introduces diffusion terms to the governing equations. When expressed in cylindrical coordinates, the radial component of the diffusion operator (i.e., the Laplacian) takes the form 1 r ∂ ∂r r ∂v ∂r (1.1) for velocity field v along the axis of the conduit. Finite difference discretization of this operator must contend with the coordinate singularity at r = 0 , the center of the conduit. There have been several approaches to the treatment of the 1/ r coordinate singularity using finite difference methods. These methods are not necessarily limited to axisymmetric problems. As summarized in [4,5] , the main approaches to treating this ∗Corresponding author. E-mail addresses: bprochnow@alumni.stanford.edu (B. Prochnow), ooreilly@stanford.edu (O. O’Reilly), edunham@stanford.edu (E.M. Dunham), petersson1@llnl.gov (N.A. Petersson). pole have been to use analytic techniques or choose a specific discretization to treat the singularity, or to transform the domain or governing equations to avoid imposing a boundary condition at the pole. The former approach is used in [6–8] , which impose l’Hôpital’s rule near the pole to write any 1/ r terms in a form that is nonsingular near r = 0 . Mohseni and Colonius avoid the need for a special boundary closure (and the correspond- ing potential for reduced boundary accuracy) by remapping the computational domain such that no grid point is placed on the singularity. In , the incompressible Navier–Stokes equations in cylindrical coordinates are solved on a staggered grid. Terms like (1.1) are never evaluated at r = 0 ; similar terms with coordinate singularities involving the radial flux are handled by using r times the radial flux as a dependent variable instead of the radial flux. Summation-by-parts (SBP) finite difference methods, first in- troduced in and summarized in [12,13] , are well-suited to modeling wave propagation in fluid-filled conduits since they allow for the construction of higher-order spatial discretizations for which the system energy rate can be computed and used to establish time-stability. While we focus on a system with periodic boundary conditions on the top and bottom of the conduit, more general boundary conditions can be weakly enforced using the simultaneous-approximation-term (SAT) technique [13–15] . The implementation of SBP methods for wave propagation in cylindri- 0045-7930/© 2017 Elsevier Ltd. All rights reserved. B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 139 cal coordinates requires finding a way to treat the 1/ r singularity while preserving the SBP properties of the operators. In , a second order accurate discretization is constructed by using the ax- isymmetry of the system to impose a regularity condition at r = 0 . This approach uses standard second order centered difference op- erators (modified to satisfy the regularity condition), but is not di- rectly generalizable to higher orders of accuracy. A related method is used in to construct SBP operators that are second- and fourth-order accurate on interior points by manipulating the sten- cil of standard difference operators near the boundary using regu- larity conditions and l’Hôpital’s rule to eliminate singular terms. Here we present a novel discretization scheme utilizing SBP finite difference operators on a staggered grid. This approach permits construction of high-order-accurate operators with proper boundary treatment, thereby allowing stability proofs via energy estimates. We present these operators first in the 1-D (radial) context, then generalize to the 2-D problem of axisymmetric, long-wavelength waves propagating in a cylindrical conduit filled with a viscous fluid. This problem features narrow boundary layers near the conduit walls when the viscosity is sufficiently small. After establishing stability and accuracy of the method, we present results regarding optimal selection of grid spacings in the radial and axial directions based on consideration of the relative size of the wavelength and boundary layer thickness. We close by demonstrating the utility of these new operators for the more complex problem of acoustic-gravity waves in magmatic conduits. While the operators presented here are specifically designed for the coordinate singularity problem for the radial component of the Laplacian operator in cylindrical coordinates, it is likely that the approach could be extended to related problems in spherical coordinates, non-axisymmetric problems, or other coordinate systems having coordinate singularities. 2. Continuous formulation of the model problem 2.1. Governing equations We consider wave propagation and diffusion in a cylindrical conduit. The fluid pressure p = p(t, z) and the axial component of the velocity field v = v (t, z, r) are treated as being axisymmetric (no azimuthal dependence), where r and z are the dimensionless radial and axial coordinates, respectively. We define a two- dimensional rectangular domain as 0 ≤r ≤1 and 0 ≤z ≤2 π, where r = 0 is at the center of the conduit and r = 1 is at the wall, while z = 0 is the conduit bottom and z = 2 π is the top. By restricting attention to wavelengths much greater than the conduit radius, the radial momentum balance establishes, to a good ap- proximation, the uniformity of pressure in the radial direction . In this limit, a dimensionless and linearized statement of conser- vation of mass, together with linearized relations between density perturbations and pressure (and possibly also perturbations in cross-sectional area and local pressure), becomes ∂p ∂t + ∂u ∂z = 0 , (2.1) where u (t, z) = 2 1 0 v (t, z, r) r dr (2.2) is the cross-sectionally averaged fluid velocity. The factor of 2 is included such that if v is constant the cross-sectionally averaged velocity u will equal v exactly. The linearized, dimensionless statement of conservation of momentum includes both a pressure gradient term and a viscous term from shearing on axisymmetric surfaces of constant r : ∂v ∂t + ∂p ∂z = ϵ 1 r ∂ ∂r r ∂v ∂r . (2.3) Additional viscous terms in the linearized Navier–Stokes equation are negligible in comparison in the long wavelength limit of interest. The dimensionless parameter ϵ is defined in terms of the fluid viscosity μ, density ρ, and conduit radius R as ϵ = μ ρR 2 ω , (2.4) where ω = ck is the characteristic angular frequency, defined in terms of the acoustic wave speed c and the axial wavenumber k . For small values of ϵ, narrow boundary layers develop near the conduit walls, where a no-slip boundary condition is imposed: v (t, z, r = 1) = 0 . (2.5) To make the presentation more concise, we consider periodic boundary conditions in the axial direction: v (t, z = 2 π, r) = v (t, z = 0 , r) , p(t, z = 2 π) = p(t, z = 0) . (2.6) 2.2. Energy balance In this section, we derive an energy estimate for the problem. In cylindrical coordinates, we define the continuous L 2 -norm for a real-valued function ψ( t, z, r ) as ∥ ψ∥ 2 = 2 π 0 1 0 ψ 2 r dr dz, and define the energy E(t) = 1 2 ∥ v ∥ 2 + 1 2 ∥ p∥ 2 = 1 2 2 π 0 1 0 (v 2 + p 2 ) r dr dz. (2.7) The first term is the kinetic energy of the fluid, while the second term is the potential energy stored through compression/expansion of the fluid and/or elastic deformation of the conduit walls. Prior to the imposition of boundary conditions, the rate of change of energy is dE dt = −1 2 [ up ] 2 π z=0 + ϵ 2 π 0 v r ∂v ∂r 1 r=0 dz −ϵ 2 π 0 1 0 ∂v ∂r 2 r dr dz. (2.8) The first term in this expression, which corresponds to the work done by external forces at the top and bottom of the conduit ( z = 2 π and z = 0 ), equals zero when the periodic boundary condition (2.6) is imposed. The second term represents the change in energy due to external forces at the walls of the conduit ( r = 1 ). When the no-slip boundary condition (2.5) is imposed, this term is zero at r = 1 . With no sources or sinks of mass or singular forces along the axis, v and ∂ v / ∂ r are bounded at r = 0 , so the term is zero at r = 0 as well. The energy rate thus reduces to dE dt = −ϵ ∂v ∂r 2 . (2.9) The right hand side of this equation represents the rate of energy loss due to viscous dissipation in the fluid. Because ϵ ≥0, dE / dt ≤0. 3. Finite difference approximation of the radial diffusion operator The construction of a provably stable numerical scheme re- quires special treatment of the coordinate singularity at r = 0 . To overcome this challenge, we introduce a shifted grid with interior points offset by r /2 relative to an equidistant grid with grid spacing r . As shown in Section 2.2 , no boundary condi- tion should be specified at r = 0 . We proceed by constructing summation-by-parts operators that also possess this property. 140 B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 Fig. 1. The r −and r + grids used in the discretization. The difference operator D + acts on a grid function on the r −grid and approximates the first derivative on the r + grid. The difference operator D −is defined in a similar manner. 3.1. Definitions We define two grids in the radial direction: r + , which is an equidistant grid with grid spacing r = 1 /N, and r −, which is staggered relative to r + (see Fig. 1 ). We have r + = [ r 0 r 1 . . . r N ] T ∈ R N+1 , r −= r 0 r 1 / 2 r 3 / 2 . . . r N−1 / 2 r N T ∈ R N+2 , where r j = j r, for j ∈ [0 , N] , r j+ 1 / 2 = j + 1 2 r, for j ∈ [0 , N −1] . Note that the endpoints of both grids coincide and that the grid point r 0 is placed at r = 0 . We introduce grid functions φ+ and φ−that represent a continuous function φ( r ) on the grids r + and r −, respectively. Thus, φ+ = [ φ0 φ1 . . . φN ] T , φ−= φ0 φ1 / 2 . . . φN−1 / 2 φN T . (3.1) We define the difference operator D + to act on a grid function φ−and approximate the first derivative on the r + grid. Similarly, we define the difference operator D −to act on a grid function φ+ and approximate the derivative on the r −grid. Fig. 1 illustrates an example with second order accurate difference operators, ∂φ ∂r r j ≈(D (2) + φ−) j = φ j+ 1 / 2 −φ j−1 / 2 r , ∂φ ∂r r j−1 / 2 ≈(D (2) −φ+ ) j−1 / 2 = φ j −φ j−1 r . Difference operators that satisfy the principle of summation-by- parts are defined by combining central difference approximations in the interior of the domain and one-sided difference approxima- tions near the boundary. Definition 1. The pair of difference operators D + ∈ R (N+1) ×(N+2) and D −∈ R (N+2) ×(N+1) are first derivative staggered grid summation-by-parts operators if 1. The difference operators D + and D −are 2 p th order accurate in the interior and at least p th order accurate near the boundary. 2. There exists (semi-)definite diagonal matrices P + and P −that define the discrete scalar products, ψ T + P + φ+ := ( ψ + , φ+ ) + = r N j=0 w j ψ j φ j , (3.2) ψ T −P −φ−:= ( ψ −, φ−) −= r N−1 j=0 ˜ w j+ 1 / 2 ψ j+ 1 / 2 φ j+ 1 / 2 , (3.3) with weights w j > 0, ˜ w j+ 1 / 2 > 0 , and corresponding (semi-) norms ∥ φ+ ∥ 2 + = ( φ+ , φ+ ) + and ∥ φ−∥ 2 −= ( φ−, φ−) −. 3. The first and last diagonal elements of P −are zero. 4. The difference operators satisfy the summation-by-parts prop- erty ( ψ + , D + φ−) + + (D −ψ + , φ−) −= ψ N φN −ψ 0 φ0 , (3.4) for all real-valued grid functions ψ + and φ−. This property can be expressed in matrix form as P + D + + D T −P −= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −1 0 0 0 . . . . . . 0 0 0 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ =: B . (3.5) Remark 1. By construction, the scalar products (3.2) and (3.3) rep- resent quadrature rules and their order of accuracy is addressed in Appendix A . Remark 2. The requirement that the first and last diagonal ele- ments of P −are zero is essential for the treatment of the po- lar coordinate singularity. This implies that the first and last rows of D −are only determined by the accuracy constraint (1) of Definition 1 and can be chosen independently of all other rows in D −. As an example, we show a pair of finite difference operators that satisfy the conditions of Definition 1 for p = 1 : D + = 1 r ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −2 2 −1 1 . . . . . . −1 1 −2 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ , (3.6) D −= 1 r ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ −1 1 −1 1 . . . . . . −1 1 −1 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ . (3.7) The corresponding weights in the scalar products are P + = r diag ( 1 / 2 , 1 , . . . , 1 , 1 / 2 ) and P −= r diag (0 , 1 , . . . , 1 , 0) . 3.2. Construction of higher order staggered grid SBP operators Standard SBP approximations of first derivatives are commonly defined by D = P −1 Q , where Q + Q T = diag (−1 , 0 , . . . , 0 , 1) . Un- fortunately, this approach does not generalize to the staggered case because the matrix P −is singular, necessitating a different approach. Let m ( p ) ≥2 p be an integer, which equals the number of modified difference stencils near each boundary. The starting point of our construction is to choose a compact interior difference stencil of order 2 p . In the regular grid, the interior stencil will be used at the points r j , for j ∈ [ m, N −m ] . In the staggered grid, the interior stencil will be used at r j−1 / 2 , for j ∈ [ m, N −m + 1] . At all interior points, the weights in the scalar products are taken to be to w j = 1 and ˜ w j−1 / 2 = 1 . Let k ≥0 be an integer and let the grid functions r k + and r k −, be discretizations of the monomial function f(r) = r k , i.e., r k + = [ r k 0 , r k 1 , . . . , r k N ] T and r k −= [ r k 0 , r k 1 / 2 , . . . , r k N−1 / 2 , r k N ] T . We define a pair of staggered grid SBP operators to have p th order boundary accuracy if they satisfy Q + r k −= k P + r k −1 + , (3.8) Q −r k + = k P −r k −1 − . (3.9) for k = 0 , 1 , . . . , p. Here, Q + = P + D + and Q −= P −D −. Note that the first condition is equivalent to D + r k −= k r k −1 + , but the second condition can not be written on this form because P −is singular. Since the interior stencil is 2 p th order accurate and the weights in the scalar products equal one in the interior, we only need to B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 141 enforce (3.8) and (3.9) for the first and last m rows. Furthermore, symmetry considerations imply that the last m rows of Q + and Q − can be constructed from the first m rows of the same matrix. We therefore only need to consider the first m rows of (3.8) and (3.9) . Because the first and last diagonal elements of P −are zero, the first and last columns of Q T −= D T −P −are zero. The SBP property (3.5) implies that the only non-zero element in the first column of Q + = P + D + is (Q + ) 00 = −1 . By the same argument, the only non-zero element in its last column is (Q + ) N,N+1 = 1 . For example, the interior difference stencil for 4th order interior accuracy ( p = 2 ) correspond to Q + φ+ j = 1 24 φ j−3 / 2 −9 8 φ j−1 / 2 + 9 8 φ j+ 1 / 2 −1 24 φ j+ 3 / 2 . The above considerations imply that Q + must have the following structure: Q + = ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ −1 q 01 q 02 q 03 q 04 0 0 0 0 0 0 q 11 q 12 q 13 q 14 0 0 0 0 0 0 q 21 q 22 q 23 q 24 0 0 0 0 0 0 q 31 q 32 q 33 q 34 1 / 24 0 0 0 0 0 0 0 1 / 24 −9 / 8 9 / 8 −1 / 24 0 0 0 0 0 0 0 −1 / 24 −q 34 −q 33 −q 32 −q 31 0 0 0 0 0 0 −q 24 −q 23 −q 22 −q 21 0 0 0 0 0 0 −q 14 −q 13 −q 12 −q 11 0 0 0 0 0 0 −q 04 −q 03 −q 02 −q 01 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ (3.10) To conserve space, the matrix only has one row where the interior stencil is used. In practice, that stencil is used at all interior points, which gives the matrix a banded structure. For symmetry reasons, the 4 × 4 block of unknown coefficents q ij at the top of Q + also occurs at the bottom, but with opposite sign and reversed rows and columns. The structure of the matrices Q + and Q −is discussed in detail in Section 3.3 . The matrix (3.10) has 16 unknown coefficients q ij . There are also 4 unknown weights in P + , and 4 unknown weights of P −. These 24 unknowns are determined by satisfying the first 4 equations of (3.8) , the first 4 equations of (3.9) , and the SBP constraint (3.5) . This leads to an under-determined system where the solution is chosen to minimize the spectral radius of D + and D −, their truncation errors, or some combination thereof. We have derived SBP operators with 4 th , 6 th , and 8 th order interior accuracy. MATLAB implementations of these operators are available at: . 3.3. Enforcement of boundary conditions To demonstrate how to enforce boundary conditions, we con- sider the diffusion equation in one dimension as a motivating example. Here we focus on the diffusion operator in a Cartesian coordinate system. The key results from this section will be used to enforce the no-slip boundary condition. We contend with the polar coordinate singularity in the axisymmetric case in the subsequent section. Consider the diffusion equation ∂u ∂t = ∂ 2 u ∂x 2 , 0 ≤x ≤1 , t ≥0 , (3.11) where u = u (x, t) . The energy method is applied by multiplying (3.11) with u and integrating over the domain, and integrating by parts: 1 2 d∥ u ∥ 2 dt = u ∂u ∂x 1 x =0 − 1 x =0 ∂u ∂x 2 dx, (3.12) where ∥ u ∥ 2 = 1 0 u 2 dx . We obtain an energy estimate for this prob- lem by enforcing appropriate boundary conditions at each end- point. We restrict attention to homogeneous Dirichlet conditions: u (0 , t) = 0 and u (1 , t) = 0 , (3.13) or homogeneous Neumann conditions: ∂u ∂x (0 , t) = 0 , and ∂u ∂x (1 , t) = 0 . (3.14) Either choice of these conditions results in the energy rate 1 2 d∥ u ∥ 2 dt = − 1 x =0 ∂u ∂x 2 dx ≤0 . (3.15) Next, we construct a semi-discrete approximation of (3.11) us- ing the staggered SBP operators and enforce the boundary conditions by injection. To separate out the boundary points from the interior points we introduce the matrix S , S = ⎡ ⎢ ⎢ ⎣ 0 1 0 0 1 0 . . . . . . . . . 0 1 0 ⎤ ⎥ ⎥ ⎦ ∈ R N ×(N +2) . (3.16) Let v −= [ v 0 v 1 / 2 v 3 / 2 . . . v N−1 / 2 v N ] T ∈ R (N+2) denote a stag- gered grid function and let ¯ v = [ v 1 / 2 v 3 / 2 . . . v N−1 / 2 ] T ∈ R N be the corresponding interior grid function. We have ¯ v = Sv −, v −= v 0 ¯ v v N . More generally, SA removes the first and last rows from A ∈ R (N+2) ×n , and BS T removes the first and last columns from B ∈ R m ×(N+2) . Thus, the diagonal matrix ¯ P −= SP −S T = diag ( ˜ w 1 / 2 , ˜ w 3 / 2 , . . . , ˜ w N−1 / 2 ) ∈ R N×N , is positive definite. Because of the accuracy constraint (3.9) , the first and last rows of Q −are zero. We can therefore only define the difference approximation at the interior staggered points, i.e., ¯ u = ¯ D −v + , where ¯ D −= ¯ P −1 −¯ Q −∈ R N ×(N +1) , ¯ Q −= SQ −∈ R N ×(N +1) . (3.17) To separate out boundary and interior points when evaluating D + v −, we notice that Q + has the structure Q + = −e 0 ¯ Q + e N , ¯ Q + = Q + S T , (3.18) where e 0 = [1 0 0 . . . 0] T ∈ R (N+1) and e N = [0 0 . . . 0 1] T ∈ R (N+1) . Because D + = P −1 + Q + , D + v −= P −1 + −v 0 e 0 + v N e N + ¯ Q + ¯ v = : P −1 + ( −v 0 e 0 + v N e N ) + ¯ D + ¯ v , ¯ D + = P −1 + ¯ Q + . (3.19) Using the above definitions, the semi-discrete approximation of (3.11) becomes d ¯ v dt = ¯ D − P −1 + (v N e N −v 0 e 0 ) + ¯ D + ¯ v , (3.20) where v N and v 0 are determined from the boundary conditions. The semi-discrete approximation is stable because it satisfies an energy rate equation analogous to (3.15) , as is made precise by the following lemma. Lemma 3.1. The semi-discrete approximation (3.20) satisfies the en- ergy rate equation 1 2 d∥ ¯ v ∥ 2 − dt = −( ¯ D + ¯ v ) N v N + ( ¯ D + ¯ v ) 0 v 0 −( ¯ D + ¯ v ) T P + ( ¯ D + ¯ v ) . (3.21) 142 B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 Proof. Our proof is based on the energy method. Taking the time derivative of ∥ ¯ v ∥ 2 −, using (3.20) , and the above definition of ¯ D −, results in 1 2 d∥ ¯ v ∥ 2 − dt = ¯ v T ¯ Q −P −1 + (v N e N −v 0 e 0 ) + ¯ v T ¯ Q −¯ D + ¯ v . (3.22) Because of (3.18) , it follows from the SBP property (3.5) that ¯ Q + + ¯ Q T −= 0 . (3.23) Hence, the definition of ¯ D + gives ¯ v T ¯ Q −P −1 + = −( ¯ D + ¯ v ) T and we ar- rive at (3.21) . □ Corollary 3.2. (Dirichlet boundary conditions) If the initial conditions satisfy v 0 = 0 and v N = 0 , the semi-discrete approximation d ¯ v dt = ¯ D −¯ D + ¯ v , t ≥0 , (3.24) satisfies homogeneous Dirichlet boundary conditions for all t > 0, and is stable. Proof. The homogeneous Dirichlet condition (3.13) are discretized by v 0 = 0 and v N = 0 . (3.25) Inserting (3.25) into (3.20) , leads to (3.24) . Because the scheme does not change the boundary values of v −, the Dirichlet condi- tions are satisfied for all t > 0. By inserting (3.25) into (3.21) , it follows that this semi-discrete approximation satisfies the energy rate equation 1 2 d∥ ¯ v ∥ 2 − dt = −( ¯ D + ¯ v ) T P + ( ¯ D + ¯ v ) ≤0 , which proves energy stability. □ Corollary 3.3. (Neumann boundary conditions) The semi-discrete ap- proximation d ¯ v dt = ¯ D − −( ¯ D + ¯ v ) N e N −( ¯ D + ¯ v ) 0 e 0 + ¯ D + ¯ v , (3.26) is stable and enforces the discrete homogeneous Neumann conditions (D + v −) 0 = 0 and (D + v −) N = 0 . Proof. The discrete boundary conditions are equivalent to (Q + v −) 0 = 0 and (Q + v −) N = 0 . Because the matrix Q + has the decomposition (3.18) , these conditions are satisfied by taking −v 0 + ( ¯ Q + ¯ v ) 0 = 0 and v N + ( ¯ Q + ¯ v ) N = 0 . (3.27) Inserting (3.27) into (3.20) , leads to (3.26) . By inserting (3.27) into (3.21) , it follows that this semi-discrete approximation satisfies the energy rate 1 2 d∥ ¯ v ∥ 2 − dt = ( ¯ D + ¯ v ) 2 N (P + ) NN + ( ¯ D + ¯ v ) 2 0 (P + ) 00 −( ¯ D + ¯ v ) T P + ( ¯ D + ¯ v ) = − N−1 j=1 ( ¯ D + ¯ v ) 2 j (P + ) j j ≤0 . Thus the discretization is energy stable. □ 3.4. Construction of the radial diffusion operator We now return to the model problem in cylindrical coordi- nates. By focusing only on the radial and time dependence of v , the diffusive part of the momentum balance expression (2.3) can be expressed in the form ∂v ∂t = ϵ r ∂ ∂r r ∂v ∂r (3.28) for v = v (t, r) and with the no-slip boundary condition v (t, r = 1) = 0 . (3.29) We approximate (3.28) by d ¯ v dt = ϵD 2 v . The difference operator D 2 approximates the radial component of the Laplacian and is constructed along the lines of (3.20) : D 2 v −= ¯ R −1 −¯ D − P −1 + (r N v N e N −r 0 v 0 e 0 ) + R + ¯ D + ¯ v ∈ R N , where R + = diag ( r + ) ∈ R (N+1) ×(N+1) and ¯ R −= diag ( ¯ r −) ∈R N×N , ¯ r −= Sr −. This difference approximation avoids the computation of the numerical solution at the coordinate singularity r = 0 and also injects the no-slip boundary condition (3.29) by taking v N = 0 . Since r 0 = 0 , it is not possible to specify v 0 . After injecting the no-slip boundary condition, we get D 2 ≡¯ R −1 −¯ D −R + ¯ D + ∈ R N×N (3.30) and d ¯ v dt = ϵD 2 ¯ v . Hence, the dependent variables consist of the interior points in the staggered grid. In the second order case ( p = 1 ), the radial diffusion operator becomes D 2 = 1 r 2 ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ −r 1 r 1 / 2 r 1 r 1 / 2 r 1 r 3 / 2 −r 1 + r 2 r 3 / 2 r 2 r 3 / 2 . . . . . . . . . r N−2 r N−3 / 2 −r N−2 + r N−1 r N−3 / 2 r N−1 r N−3 / 2 r N−1 r N−1 / 2 −r N−1 +2 r N r N−1 / 2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ . 4. Semi-discrete approximation of the model problem We now implement the discrete operator (3.30) in the context of the model problem defined in Section 2 . This 2-D problem requires the construction of a grid in the z direction with cor- responding ∂ / ∂ z operators in addition to the radial grids and associated difference operators defined in Section 3 . We define circulant first derivative operators that act in the z direction and reflect the periodicity condition at the bottom ( z = 0 ) and top ( z = 2 π) of the domain. By discretizing the 2-D problem using SBP operators, we prove energy stability for the system. 4.1. Grid definitions To create a discretization of our 2-D model problem, we use the r + grid defined in Section 3 , as well as the truncated grid ¯ r −= Sr −. The number of grid points on ¯ r −is N r . The z direction can be discretized using either a single grid for both p and v or using staggered grids. Since the dispersion error is smaller for central first derivatives on staggered grids compared to collocated grids, we discretize the z direction using staggered grids as well. The grids are given by: z + = [ z 0 z 1 z 2 . . . z N z −1 ] T , z −= [ z 1 2 z 3 2 . . . z N z −1 2 ] T , (4.1) where z j = j z for 0 ≤j ≤N z −1 , z j+1 / 2 = j + 1 2 z for 0 ≤j ≤N z −1 for step size z = 2 π/N z , where N z is the number of grid points in the z direction. It is important to note that these grids do not have the same structure as the staggered grids in the radial direction (i.e., the + and - grids do not share the same endpoints). The point z N z is not stored, since by the periodicity condition it is equal to z 0 . B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 143 Fig. 2. The mesh used in the implementation of the model problem. p and u are 1D fields stored on the z + and z −grids respectively, while v is a 2-D field stored on the ¯ r −and z −grids. 4.2. Numerical scheme The velocity ¯ v (along with other 2-D fields) is stored in a vector using column-major ordering (along the radial direction first): ¯ v = [ ⃗ v 1 2 ⃗ v 3 2 . . . ⃗ v N z −1 2 ] T ∈ R (N r N z ) . (4.2) The vectors ⃗ v j containing the velocity values along the radial direction at some z j are ⃗ v j = [ v 1 2 ,j v 3 2 ,j . . . v N r −1 2 ,j ] ∈ R 1 ×N r , (4.3) where v ij = v (t, z j , r i ) . The cross-sectionally averaged velocity is stored on the z −grid and the pressure on the z + grid: p = [ p 0 p 1 p 2 . . . p N z −1 ] T ∈ R N z , u = [ u 1 / 2 u 3 / 2 . . . u N z −1 / 2 ] T ∈ R N z , where p j = p(t, z j ) and u j = u (t, z j ) ; neither depends on r . The grid points at which the fields are stored are shown in Fig. 2 . In the z direction, the derivatives of the continuous functions u ( t, z ) and p ( t, z ) are approximated using standard central differ- ence operators D z −and D z + on a staggered, periodic grid. The op- erator D z −acts on quantities stored on the z + grid, with the re- sulting product stored on the z −grid, while D z + acts on quantities stored on the z −grid, with the product stored on the z + grid. Due to periodicity in the z direction, the difference operators D z −and D z + satisfy the SBP property: D z + = −D T z −. (4.4) Using these operators, the z derivative of the continuous function p ( t, z ) is approximated in terms of the discrete function p (stored on the z + grid) as ∂p ∂z ≈D z −p . To apply these one-dimensional difference operators to a two- dimensional field (such as v ), we must apply them at each point in the radial direction. This is accomplished by using the Kronecker product, defined by A B = ⎡ ⎣ a 11 B . . . a 1 n B . . . . . . . . . a m 1 B . . . a mn B ⎤ ⎦ . We can then apply the operator D 2 , which acts in the radial direc- tion only, to the two dimensional velocity field v to approximate the viscous term in the governing Eq. (2.3) : 1 r ∂ ∂r r ∂v (t, z, r) ∂r ≈(I z D 2 ) ¯ v , (4.5) where I z ∈ R N z ×N z is an identity matrix. The semi-discrete form of the governing Eqs. (2.1) and (2.3) can thus be written as dp dt = −D z + u , (4.6) d ¯ v dt = −(D z −p e −) + ϵ(I z D 2 ) ¯ v , (4.7) where e −= [1 1 . . . 1] T ∈ R N r . As in the continuous case, we have defined the cross-sectionally averaged velocity u = (I z w T ) ¯ v ≈2 1 0 v (t, z, r) r dr, (4.8) where w is the cross-sectional averaging operator, constructed using the quadrature rule P −on the staggered radial grid: w T = 2 e T −¯ R −¯ P −∈ R 1 ×N r . (4.9) The discrete cross-sectionally averaged velocity u (4.8) approxi- mates the continuous cross-sectionally averaged velocity u ( t, z ) defined in (2.2) . 4.3. Energy balance To compute the discrete energy, we introduce a two- dimensional discrete grid function a −defined on the ¯ r −and z −grids in the same manner as v in (4.2) and (4.3) , as well as a + defined on the r + and z + grids. We then define discrete norms for two-dimensional fields on both radial grids: ∥ a −∥ 2 −= a T −P −a −, ∥ a + ∥ 2 + = a T + P + a + , where P −= P z ¯ R −¯ P − and P + = P z R + P + (4.10) are quadrature rules approximating the area of . Since the solution is periodic in the z direction, P z = z I z ∈ R N z ×N z . This quadrature rule is therefore the same for both grids z −and z + . Using these norms, the discrete energy is defined as E h = 1 2 ∥ ¯ v ∥ 2 −+ 1 2 ∥ p e −∥ 2 −= 1 2 ¯ v T (P z ¯ R −¯ P −) ¯ v + 1 4 p T P z p . (4.11) The result ∥ p e −∥ 2 −= p T P z p / 2 follows from the definition of the cross-sectionally averaged operator (4.9) : ∥ p e −∥ 2 −= (p T e T −)(P z ¯ R −¯ P −)(p e −) = (p T P z p e T −¯ R −¯ P −e −) = 1 2 (p T P z p w T e −) = 1 2 p T P z p , where we used the fact that w T e −= 2 r=1 r=0 rdr = 1 (see Appendix A ). Using the fact that the quadrature rules are symmetric, positive definite matrices, we compute the energy rate for the system by taking the time derivative of (4.11) and inserting (4.6) and (4.7) into the result, which yields dE h dt = −1 2 p T P z D z + u −¯ v T (P z D z −p ¯ R −¯ P −e −) + ϵ ¯ v T (P z ¯ P −¯ D −R + ¯ D + ) ¯ v . (4.12) From the definition of the cross-sectionally averaged operator (4.9) , we obtain ¯ v T (P z D z −p ¯ R −¯ P −e −) = p T D T z −P z (I z e T −¯ R −¯ P −) ¯ v 144 B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 Fig. 3. Radial profiles of the velocity v at the middle of the conduit ( z = π). Figure (a) shows velocity profiles for ϵ = 10 −2 after 0, 1, and 2 periods, showing amplitude decay due to viscous dissipation. Figure (b) shows velocity profiles (at t = 0 ) for different values of ϵ. As ϵ is increased, the viscous boundary layer becomes wider. = 1 2 p T D T z −P z (I z w T ) ¯ v = 1 2 p T D T z −P z u . Then (4.12) becomes dE h dt = −1 2 p T (P z D z + + D T z −P z ) u + ϵ ¯ v T (P z ¯ P −¯ D −R + ¯ D + ) ¯ v . (4.13) Due to the SBP property of the z operators given in (4.4) , the first two terms on the right-hand side of (4.13) cancel each other out. Applying the SBP property for the radial operators (3.5) , we substitute for ¯ D −to simplify (4.13) : dE h dt = −ϵ ¯ v T (P z ¯ D T + P + R + ¯ D + ) ¯ v + ϵ ¯ v T (P z ¯ B T R + ¯ D + ) ¯ v . The truncated boundary matrix ¯ B T is a zero matrix, so only the viscous dissipation term is left: dE h dt = −ϵ∥ (I z ¯ D + ) ¯ v ∥ 2 + . (4.14) The scheme is stable if the right-hand side is non-positive. Stability is addressed in the following proposition: Proposition 4.1. The discrete energy rate in (4.14) satisfies dE h dt = −ϵ∥ (I z ¯ D + ) ¯ v ∥ 2 + < 0 , for ϵ > 0 and for all ¯ v ̸ = 0 . Proof. The proof is given Appendix B for the second order accurate difference approximation. The proposition is numerically verified to hold for higher orders as well. □ 5. Convergence tests We investigate the accuracy of our numerical scheme by con- structing a smooth analytic solution with a harmonic pressure of the form p(t, z) = Re e i (z−ωt) , (5.1) for complex ω to be determined below; Re( ω) is the angular frequency of oscillation. We chose the nondimensional domain length in the axial direction to be exactly one wavelength, so the wavenumber k is equal to 1. Following , we solve the momentum balance to find the velocity v ( t, z, r ) satisfying the no-slip boundary condition v (t, z, r = 1) = 0 as v (t, z, r) = Re 1 ω 1 −J 0 (αr) J 0 (α) e i (z−ωt) , (5.2) where α = − iω/ϵ, and J 0 ( x ) is a Bessel function of the first kind. Fig. 3 a shows radial profiles of the velocity for ϵ = 10 −2 at a fixed depth z = π after 0, 1, and 2 periods T = 2 π/ Re (ω) . The velocity retains the same profile but decreases in amplitude with each cycle due to viscous dissipation. In Fig. 3 b, profiles at t = 0 are shown for different values of ϵ. As ϵ is increased, the width of the viscous boundary layer increases. In the limit of large ϵ the flow is fully developed (the boundary layers meet at the center of the conduit to form a parabolic velocity profile), while as ϵ approaches 0 the flow approaches plug flow behavior. Applying the width averaging (2.2) to the velocity given in (5.2) , we obtain the cross-sectionally averaged velocity: u (t, z) = Re 1 ω 1 −2 α J 1 (α) J 0 (α) e i (z−ωt) , (5.3) Upon satisfying the mass balance equation, we obtain the disper- sion relation ω 2 = 1 −2 α J 1 (α) J 0 (α) . (5.4) In general, solving the dispersion relation for ω yields complex values, whose real and imaginary parts correspond to the oscil- latory and decaying behavior of the wave, respectively. In the inviscid limit ( ϵ = 0 ), ω → ±1, and waves propagate without en- ergy loss. As ϵ is increased, the solutions to (5.4) acquire negative imaginary components, which correspond to dissipative behavior in the boundary layer. For large enough ϵ, the real components of the solutions approach zero, producing overdamped wave behavior in which the waves decay without oscillating. 5.1. Optimal discretization Given a desired error level, or tolerance, and value of ϵ, we determine the optimal discretization by comparing the error contributions due to discretization in the r and z directions. The norm of the error is computed using ∥ e ∥ 2 −= ∥ v −v ∗∥ 2 −+ ∥ ( p −p ∗) e −∥ 2 −, where v ∗and p ∗is the velocity and pressure given by the analytic solution. The numerical solution is integrated in time using a low- storage fourth-order Runge–Kutta method and the norm of the error is measured after one oscillation period ( t max = 2 π/ Re (ω) ). In Fig. 4 , the error contributions from the grid refinement in the z and r directions are visualized by fixing the grid refinement level in one direction and varying it in the other direction. In Fig. 4 a, the value of N z is held fixed while N r is varied, while the reverse is true in Fig. 4 b. The optimal discretization for a desired error level is where the error contributions from the r and z discretizations are approximately equal, corresponding to the location where the error curves begin to plateau in the grid refinement plots. Decreasing ϵ narrows the boundary layer in which viscous effects act. To resolve this narrower boundary layer B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 145 Fig. 4. Error contributions from r (a) and z (b) grid refinement using 2nd order interior operators, with ϵ = 10 −2 . Fig. 5. Comparison of error contributions from z grid refinement using 2nd order interior operators with ϵ = 10 −2 (a) and ϵ = 10 −3 (b). When ϵ is reduced, viscous effects are confined to a narrower boundary layer, so a higher resolution (larger N r ) is needed to achieve the same accuracy. to a desired accuracy, the grid resolution in the radial direction must be increased. A grid refinement for ϵ = 10 −3 is shown in Fig. 5 b. Comparing Fig. 5 a and b, we see that when ϵ is reduced by a factor of 10, the number of radial grid points N r needed for an optimal discretization for a given N z increases. Since the grid is uniform, the numerical solution in the center of the conduit becomes increasingly over resolved as the width of the boundary layer decreases. To avoid over resolving the numerical solution, one can adopt a coordinate transform in the radial direction that clusters most of the grid points in the boundary layer. 5.2. Convergence rate We compute the accuracy of the simulation for operators of 2nd, 4th, 6th, and 8th order interior accuracy, with the results shown in Fig. 6 . The order of accuracy of the diffusion difference operator is p at the boundary and 2 p in the interior. The expected global convergence rate is p + 1 [19,20] . By using factor of two refinements, we estimate the convergence rate using Rate = log 2 ∥ e j−1 ∥ − ∥ e j ∥ − , (5.5) where ∥ e j ∥ −denotes the error on a mesh j . For our convergence tests, we choose each successive mesh j to have twice the number of grid points in both the radial and axial directions as the previ- ous mesh j −1 . Following the optimal discretization for ϵ = 10 −2 shown in Fig. 4 , we choose the same number of grid points in the radial and axial directions. The errors and convergence rates at each grid refinement are shown in Table 1 . The error converges towards zero at the expected convergence rate ( p + 1 ). Fig. 6. Convergence rates for 2nd through 8th order interior operators, with ϵ = 10 −2 and equal number of grid points in the z and r directions. 6. Application to acoustic-gravity waves in magmatic conduits The problem of axisymmetric wave propagation with viscous dissipation in a conduit arises in a range of different contexts, in- cluding arterial flow , aeroacoustic applications including airflow in human phonation , and acoustic-gravity waves in magmatic conduits . Many of these application problems cannot be solved directly due to spatially variable material properties, time-varying boundary conditions, or multiphase fluids in the conduit. The problem of wave propagation in magmatic conduits is of particular interest to us, and we introduce it here as a more complex problem on which we can apply the results developed in the previous sections. In this problem, z is depth (positive up), and the magma properties are functions of depth. In Karlstrom and 146 B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 Table 1 Error magnitudes and rates for simulations using operators with 2nd through 8th order interior accuracy. N z N r 2nd order interior accuracy 4th order interior accuracy error magnitude error rate error magnitude error rate 16 4 . 18 × 10 −2 1 . 43 × 10 −2 32 1 . 11 × 10 −2 1 .91 1 . 40 × 10 −3 3 .32 64 2 . 80 × 10 −3 1 .98 1 . 14 × 10 −4 3 .65 128 7 . 09 × 10 −4 1 .99 8 . 26 × 10 −6 3 .79 N z N r 6th order interior accuracy 8th order interior accuracy error magnitude error rate error magnitude error rate 16 6 . 90 × 10 −3 4 . 10 × 10 −3 32 2 . 82 × 10 −4 4 .61 8 . 73 × 10 −5 5 .55 64 8 . 47 × 10 −6 5 .06 1 . 93 × 10 −6 5 .50 128 2 . 85 × 10 −7 4 .89 4 . 84 × 10 −8 5 .32 Dunham , a model is presented for acoustic-gravity waves in a cylindrical conduit filled with viscous, compressible magma in the presence of a gravitational field. Magma is a mixture of liquid melt and gas bubbles or volatiles that can be fully or partially dissolved in the melt according to pressure-dependent solubility laws. The model is motivated by observations at Kileaua Volcano, Hawaii, where an open lava lake sits atop a magma column extending down at least a few kilometers. Huge rockfalls into the lava lake trigger oscillations that are interpreted as acoustic-gravity wave eigenmodes of the system. The governing equations are derived by considering small perturbations about a rest state of thermodynamic and mechan- ical equilibrium (the magmastatic base state); the liquid and gas phases are assumed to move with the same velocity. The exchange of mass between liquid and gas phases is described in terms of non-equilibrium bubble growth and resorption (BGR) of volatiles, parametrized as a relaxation process over a characteristic timescale τ. The governing equations are obtained by linearizing the equations for mass and momentum balance in the magma, the BGR evolution equation involving the volatile solubility law, and the magma equation of state: ρ ∂v ∂t + ∂p ∂z + ρg K p −ρgan −μ r ∂ ∂r r ∂v ∂r = 0 , (6.1) 1 K ∂p ∂t + ∂u ∂z −ρg K u + a (n + bp) τ = 0 , (6.2) ∂n ∂t + ρgbu + n + bp τ = 0 , (6.3) u = 2 R 2 R 0 v r dr, (6.4) where p ( t, z ), v ( t, z, r ), and u ( t, z ) are, as before, perturbations in pressure, velocity, and cross-sectionally averaged velocity; and n ( t, z ) is the perturbation in the mass fraction of exsolved gas. In addition, ρ( z ) is the density in the magmastatic state, K ( z ) is the bulk modulus, g is acceleration due to gravity, μ is viscosity, R is the conduit radius, and a ( z ) ≥0 and b ( z ) ≥0 are coefficients quantifying processes related to the influence of volatiles on the magma density and the pressure-dependence of volatile solubility. Note that these equations, unlike those in the model problem, are dimensional. As in the model problem, we enforce a no-slip boundary condition: v (t, z, r = R ) = 0 . (6.5) The top and bottom boundary conditions for the conduit require more careful treatment than in the model problem. At the bottom of the conduit ( z = 0 ), choices of boundary condition include but are not limited to a no flow boundary ( v (t, z = 0 , r) = 0 ) or a zero Fig. 7. Radial profiles of the vertical magma velocity at z = 0 . 8 L (above the exsolu- tion depth) for magma viscosity μ = 10 Pa s. As in the model problem, the viscous effects manif est in boundary layers near the conduit wall. pressure perturbation boundary ( p(t, z = 0) = 0 ); we focus on the latter here. At the top of the conduit, the magma is in contact with the atmosphere, so pressure is set equal to atmospheric pressure (plus some perturbation p force ( t ) associated with the impulsive forcing from the rockfalls) on that moving surface located at L + h (t) , where L is the nominal conduit length and h ( t ) is the change in height. Linearization of this condition results in the top boundary condition p(t, z = L ) −ρ(L ) gh (t) = p force (t) , (6.6) where the height evolves according to dh dt = u (t, z = L ) . (6.7) Eqs. (6.2) and (6.3) are discretized using SBP operators on a regular grid with boundary conditions weakly enforced using simultaneous approximation terms (SAT) . This discretization is possible to perform because there is no r dependence in the variables ( p, n, u ). The SAT technique is implemented using the method and choice of SAT parameters described in , where the target boundary values are determined by preserving the char- acteristic variables carrying information out of the domain. The radial diffusion term in (6.1) is discretized using the same operator D 2 as in the model problem, i.e., by (4.5) . The integral in (6.4) is discretized using the ¯ P −operator, see (4.8) and (4.9) . We note that viscosity was neglected in the original study ; accounting for viscous dissipation appears to be essential to properly match observations, thus motivating this study. The simulation results for a conduit of length L = 2 km and radius R = 15 m containing magma with a viscosity of μ = 10 Pa s are shown in Figs. 7 –9 . In these figures, the system is forced by a B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 147 Fig. 8. Space-time plot of pressure perturbation in response to rockfall onto the surface of the magma column. Fig. 9. Space-time plot of vertical velocity along the vertical cross-section at r = R/ 2 in response to rockfall onto the surface of the magma column. Gaussian pressure pulse p force ( t ) at the surface ( z = L ) with an am- plitude of 50 kPa and 1 s duration. The increasing pressure with depth in the magmastatic base state, together with the pressure- dependence of volatile solubility, leads to segmentation of the conduit into two distinct sections. At depths below z = L/ 2 (the exsolution depth), all volatiles are dissolved (making the magma less compressible and more dense); above the exsolution depth, gas bubbles are present, making the magma more compressible and less dense. As in the model problem (see Fig. 3 a), the viscous effects manif est in narrow boundary layers near the wall of the conduit and the amplitude of the oscillations decays with each cycle, as seen in Fig. 7 . The response of the magma column to the Gaussian pressure impulse (i.e., rockfall) is seen in Figs. 8 and 9 , which show the time evolution of the pressure and fluid velocity at r = R/ 2 , respectively. The disturbance at the surface creates a wave that travels down into the conduit at a relatively constant speed until it reaches the exsolution depth, where it excites ringing behavior in the deeper conduit section. These oscillations are damped out by the fluid viscosity and non-equilibrium bubble growth and resorption over the course of several minutes. With the addition of viscosity, many features observed in the Kilauea data are reproduced in the model. 7. Conclusion We have developed a numerical model for the propagation of waves in a fluid-filled tube or conduit with dissipation due to fluid viscosity. To model viscous dissipation, we have developed a method for constructing a provably stable method introducing a summation-by-parts operator D 2 for approximating the radial component of the Laplacian in cylindrical coordinates. This op- erator is constructed by defining a staggered grid in the radial direction, on which we define first derivative SBP difference operators. This operator does not include the grid points at the endpoints r = 0 and r = 1 and exactly enforces the no-slip bound- ary condition at r = 1 . To prove stability, we derived an energy estimate for the semi-discrete approximation. As an example of an application problem which is not analyt- ically solvable, we applied the numerical method to the governing equations for acoustic-gravity waves propagating in a magmatic conduit. This problem introduces complexities not seen in the model problem, including a multiphase fluid and depth-dependent material properties, which make it solvable only by numerical techniques. Related application problems arise in other contexts, including arterial flow and human phonation , and could constitute a basis for further work. A potential future development is to reduce the stencil width of the difference approximation. Since the approximation of the radial component of the Laplacian operator was constructed by applying the first derivative twice, this results in a stencil width of 4 p 2 −1 , where 2 p is the interior accuracy. The optimal stencil width is 2 p + 1 . For p = 1 (second order interior accuracy), the stencil width is optimal. For higher orders of accuracy, it might be possible to construct a difference approximation of the Laplacian with optimal stencil width using the approaches taken in [22–25] . 148 B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 While the analysis presented in this work is restricted to cylindrical coordinates, it is likely that the staggered grid oper- ators could be used to handle similar coordinate singularities in spherical coordinates. Furthermore, our method of handling the 1/ r coordinate singularity does not make use of the symmetry condition at r = 0 that is specific to axisymmetric problems, which indicates that this method could be applied to non-axisymmetric problems in both cylindrical and spherical coordinates. Extending this method to fully three dimensional non-axisymmetric prob- lems would require the definition of derivative operators in the azimuthal direction, and careful treatment of the truncation step. 8. Funding This work was partially supported by a Chevron fellowship to O. O’Reilly. Acknowledgments This work was performed under the auspices of the U.S. De- partment of Energy by Lawrence Livermore National Laboratory under contract DE-AC52-07NA27344 (technical report number LLNL-JRNL-698320). Appendix A. Accuracy of quadrature rules Lemma A.1. Let the SBP operators D + and D −be pth-order-accurate on the boundary. Then the order of the quadrature rules P + and P − is at least max (2 , (2 p −2)) for p ≥1 . Proof. Let r j = a + jh, h = (b −a ) /N, and r + = [ r 0 , r 1 , . . . , r N ] T be the vector of grid point locations for the regular grid. The corresponding location vector for the staggered grid is r −= [ r 0 , r 1 / 2 , . . . , r N−1 / 2 , r N ] T . Denote a grid function on the regular grid by u + = [ u 0 , u 1 , . . . , u N ] T and on the staggered grid by u −= [ u 0 , u 1 / 2 , . . . , u N−1 / 2 , u N ] T . Recall the summation-by-parts property (3.5) for first derivatives. By applying this property to the sec- ond derivatives D −D + and D + D −, and taking the scalar products (3.2) and (3.3) it follows that (u + , D + D −v + ) + = −(D −u + , D −v + ) −+ u N (D −v + ) N −u 0 (D −v + ) 0 , (A.1) (u −, D −D + v −) −= −(D + u + , D + v −) + + u N (D + v −) N −u 0 (D −v −) 0 . (A.2) The SBP operators exactly satisfy ⎧ ⎪ ⎨ ⎪ ⎩ D + r s −= s r s −1 + , D −r s + = s r s −1 −, D −D + r s −= s (s −1) r s −2 −, D + D −r s + = s (s −1) r s −2 + , s ∈ [1 , p] , (A.3) where r s + = [ r s 0 , r s 1 , . . . , r s N ] T and r s −= [ r s 0 , r s 1 / 2 , . . . , r s N−1 / 2 , r s N ] T . Case I : p = 1 . In this case, P + is the trapezoidal rule in (3.7) and P −is the midpoint rule in (3.7) , which are second-order-accurate. Case II : Even degree polynomials. Taking u + = r s + , v + = r s + , u −= r s −, and v −= r s −in (A.1) and (A.2) , and using (A.3) yields (s −1)(e + , r 2 s −2 + ) + = −s (e −, r 2 s −2 − ) −+ r 2 s −1 N −r 2 s −1 0 , s ∈ [2 , p] , (A.4) (s −1)(e −, r 2 s −2 − ) −= −s (e + , r 2 s −2 + ) + + r 2 s −1 N −r 2 s −1 0 , s ∈ [2 , p] , (A.5) where e + = [1 , 1 , . . . , 1] T ∈ R N+1 and e −= [1 , 1 , . . . , 1] T ∈ R N+2 . By forming the difference between (A.5) and (A.4) (e + , r 2 s −2 + ) + = (e −, r 2 s −2 − ) −. (A.6) Inserting (A.6) into (A.4) and (A.5) leads to (e + , r 2 s −2 + ) + = (e −, r 2 s −2 − ) −= 1 2 s −1 b 2 s −1 −a 2 s −1 , s ∈ [2 , p] . The right-hand side is exactly equal to the result obtained by the integration b a r 2 s −2 dr = 1 2 s −1 b 2 s −1 −a 2 s −1 , which means that the quadrature rules are exact for all even degree polynomials up to degree 2 p −2 . Case III : Odd degree polynomials. Now taking u + = r s + and v + = r s −1 + and u −= r s −and v −= r s −1 − in (A.1) and (A.2) , and using (A.3) leads to (s −2)(e + , r 2 s −3 + ) + = −s (e −, r 2 s −3 − ) −+ r 2 s −2 N −r 2 s −2 0 , s ∈ [2 , p] , (A.7) (s −2)(e −, r 2 s −3 − ) −= −s (e + , r 2 s −3 + ) + + r 2 s −2 N −r 2 s −2 0 , s ∈ [2 , p] . (A.8) By forming the difference between (A.8) and (A.7) (e + , r 2 s −3 + ) + = (e −, r 2 s −3 − ) −. (A.9) Inserting (A.9) into (A.7) and (A.8) leads to (e + , r 2 s −3 + ) + = (e −, r 2 s −3 − ) −= 1 2 s −2 b 2 s −2 −a 2 s −2 , s ∈ [2 , p] , which means that the quadrature rules are exact for all odd degree polynomials up to degree 2 p −3 . □ Remark 3. For p > 1, there is numerical evidence that the quadra- ture rules are of order 2 p −1 . Appendix B. Proof of Proposition 4.1 (second order case) We need to show that the following holds: ¯ v T A ¯ v = ∥ (I z ¯ D + ) ¯ v ∥ 2 + > 0 , for all ¯ v ̸ = 0 , i.e. A > 0. From the definition of the norm ∥ · ∥ 2 + in (4.10) , we get A = P z ( ¯ D T + P + R + ¯ D + ) . Since P z = zI z > 0 , we only need to consider ¯ A = ¯ D T + P + R + ¯ D + ∈ R N×N . If ¯ A > 0 then the null space nul l ( ¯ A ) = { x ∈ R N ̸ = 0 | Ax = 0 } is empty. Since R + is zero on the diagonal in the first row and both P + and R + are diagonal and positive everywhere else, it is sufficient to show that there exists no vector ¯ x = ( ¯ x 0 ¯ x 1 . . . ¯ x N−1 ) T ∈ R N ̸ = 0 such that ¯ D + ¯ x = y , where y = (α, 0 , 0 , . . . , 0) T ∈ R (N+1) for α ̸ = 0. In the second order case, the matrix ¯ D + is given by ¯ D + = 1 r ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ 2 −1 1 −1 1 −1 1 . . . . . . −1 1 −2 ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ∈ R (N+1) ×N . Consider the equation system ¯ D + ¯ x = 0 . We use a proof by con- tradiction and assume that there exists x ̸ = 0. From the last equation we obtain ¯ x N−1 = 0 . Inserting this result into the second last equation yields ¯ x N−2 = 0 . Repeating the procedure until the second equation yields ¯ x 0 = 0 , which implies that ¯ x = 0 is the only solution. We have arrived at a contradiction and therefore conclude that ¯ A > 0 , which completes the proof. B. Prochnow et al. / Computers and Fluids 149 (2017) 138–149 149 Fig. B10. The minimum modulus eigenvalue of ¯ A / r as a function of the number of grid points. For higher order of accuracy, we numerically compute the minimum modulus eigenvalue λmin ≡ min λi ∈ λ( ¯ A / r) | λi | . We use 16 to 400 grid points. Fig. B.10 shows that λmin is nonzero. References Womersley JR. Method for the calculation of velocity, rate of flow and viscous drag in arteries when the pressure gradient is known. J Phys- iol 1955;127(3):553–63. doi: 10.1113/jphysiol.1955.sp005276 . 10.1113/jphysiol.1955.sp005276 . Atabek HB, Lew HS. Wave propagation through a viscous incompressible fluid contained in an initially stressed elastic tube. Biophys J 1966;6(4):481–503 . 19210972[pmid]; . Karlstrom L, Dunham EM. Excitation and resonance of acoustic- gravity waves in a column of stratified, bubbly magma. J Fluid Mech 2016;797:431–70. doi: 10.1017/jfm.2016.257 . article _ S0 0221120160 02573 . Tavakoli E, Lessani B, Hosseini R. High-order pole-treatment in cylindrical co- ordinates for incompressible flow simulations with finite-difference collocated schemes. J Comput Phys 2015;296:1–24. 042 . 0219991150 02995 . Oguic R, Viazzo S, Poncet S. A parallelized multidomain compact solver for incompressible turbulent flows in cylindrical geometries. J Comput Phys 2015;300:710–31. . sciencedirect.com/science/article/pii/S0 0219991150 05203 . Griffin MD, Jones E, Anderson JD. A computational fluid dynamic tech- nique valid at the centerline for non-axisymmetric problems in cylin- drical coordinates. J Comput Phys 1979;30(3):352–60. 1016/0021-9991(79)90120-7 . 0021999179901207 . Constantinescu G, Lele S. A highly accurate technique for the treatment of flow equations at the polar axis in cylindrical coordinates using series expansions. J Comput Phys 2002;183(1):165–86. 06/jcph.20 02.7187 . . Morinishi Y, Vasilyev OV, Ogi T. Fully conservative finite difference scheme in cylindrical coordinates for incompressible flow simulations. J Comput Phys 2004;197(2):686–710. . sciencedirect.com/science/article/pii/S0 0219991030 06594 . Mohseni K, Colonius T. Numerical treatment of polar coordinate singularities. J Comput Phys 20 0 0;157(2):787–95. . . Verzicco R, Orlandi P. A finite-difference scheme for three-dimensional in- compressible flows in cylindrical coordinates. J Comput Phys 1996;123(2):402– 14. 06/jcph.1996.0 033 . science/article/pii/S0021999196900339 . Kreiss H-O, Scherer G. Finite element and finite difference methods for hy- perbolic partial differential equations. In: Boor Cd, editor. Mathematical As- pects of Finite Elements in Partial Differential Equations. Academic Press; 1974. p. 195–212. ISBN 978-0-12-208350-1 . article/pii/B9780122083501500121 . Strand B. Summation by parts for finite difference approximations for d/dx. J Comput Phys 1994;110(1):47–67. . . Svärd M, Nordström J. Review of summation-by-parts schemes for initial- boundary-value problems. J Comput Phys 2014;268:17–38. org/10.1016/j.jcp.2014.02.031 . S0 0219991140 0151X . Carpenter MH, Gottlieb D, Abarbanel S. Time-stable boundary con- ditions for finite-difference schemes solving hyperbolic systems: methodology and application to high-order compact schemes. J Com- put Phys 1994;111(2):220–36. . . Fernández DCDR, Hicken JE, Zingg DW. Review of summation-by-parts opera- tors with simultaneous approximation terms for the numerical solution of par- tial differential equations. Comput Fluids 2014;95:171–96. 1016/j.compfluid.2014.02.016 . S0 0457930140 0 0796 . Calabrese G, Neilsen D. Spherical excision for moving black holes and summa- tion by parts for axisymmetric systems. Phys Rev D 2004;69:044020. doi: 10. 1103/PhysRevD.69.044020 . . Gundlach C, Martín-García JM, Garfinkle D. Summation by parts methods for spherical harmonic decompositions of the wave equation in any dimen- sions. Classical Quantum Gravity 2013;30(14):145003 . 0264-9381/30/i=14/a=145003 . Carpenter MH, Kennedy CA. Fourth-order 2n-storage runge-kutta schemes. Tech. Rep. NASA-TM-109112. NASA; 1994 . 199400284 4 4 . Gustafsson B. The convergence rate for difference approximations to mixed initial boundary value problems. Math Comput 1975;29(130):396–406 . http: //www.jstor.org/stable/2005559 . Svärd M, Nordström J. On the order of accuracy for difference approximations of initial-boundary value problems. J Comput Phys 2006;218(1):333–52. doi: 10. 1016/j.jcp.2006.02.014 . . Larsson M, Müller B. Numerical simulation of confined pulsating jets in hu- man phonation. Comput Fluids 2009;38(7):1375–83. j.compfluid.2008.01.033 . Special Issue Dedicated to Professor Alain Lerat on the Occasion of his 60th Birthday; pii/S0 0457930 080 0 0662 . Nilsson S, Petersson NA, Sjögreen B, Kreiss H-O. Stable difference approxi- mations for the elastic wave equation in second order formulation. SIAM J Numer Anal 2007;45(5):1902–36. doi: 10.1137/060663520 . 1137/060663520 . Mattsson K. Summation by parts operators for finite difference approx- imations of second-derivatives with variable coefficients. J Sci Comput 2012;51(3):650–82. doi: 10.1007/s10915- 011- 9525- z . s10915- 011- 9525- z . Sjögreen B, Petersson NA. A fourth order accurate finite difference scheme for the elastic wave equation in second order formulation. J Sci Comput 2012;52(1):17–48. doi: 10.1007/s10915- 011- 9531- 1 . s10915- 011- 9531- 1 . Fernández DCDR, Zingg DW. Generalized summation-by-parts operators for the second derivative. SIAM Journal on Scientific Computing 2015;37(6):A2840–64. doi: 10.1137/140992205 . . |
11873 | https://www.unitsconverters.com/en/Milligramperliter-To-Gramper100ml/Unittounit-4519-4540 | Milligram per Liter to Gram per 100mL
(mg/L to gram/dL)
| | | |
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| -10% Copy +10% | | -10% Copy +10% |
| | = | |
| Grain per Gallon(UK) [gr/gal(UK)] Grain per Gallon(US) [gr/gal(US)] Gram per 100mL [gram/dL] Gram per Cubic Centimeter [g/cm³] Gram per Cubic Meter [g/m³] Gram per Deciliter [g/dL] Gram per Liter [g/L] Gram per Milliliter [g/mL] Kilogram per Cubic Meter [kg/m³] Kilogram per Liter [kg/L] Milligram per Deciliter [mg/dL] Milligram per Liter [mg/L] Pound per Cubic Foot [lb/ft³] Pound per Gallon [lb/gal(UK)] Pound per Gallon(US) [lb/gal(US)] Pound per Million Gallon [lb/million gal(US)] Pound per Million Gallon (UK) [lb/million gal(UK)] | ⇄ | Grain per Gallon(UK) [gr/gal(UK)] Grain per Gallon(US) [gr/gal(US)] Gram per 100mL [gram/dL] Gram per Cubic Centimeter [g/cm³] Gram per Cubic Meter [g/m³] Gram per Deciliter [g/dL] Gram per Liter [g/L] Gram per Milliliter [g/mL] Kilogram per Cubic Meter [kg/m³] Kilogram per Liter [kg/L] Milligram per Deciliter [mg/dL] Milligram per Liter [mg/L] Pound per Cubic Foot [lb/ft³] Pound per Gallon [lb/gal(UK)] Pound per Gallon(US) [lb/gal(US)] Pound per Million Gallon [lb/million gal(US)] Pound per Million Gallon (UK) [lb/million gal(UK)] |
| | |
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| 2X of 1 ▶ | 1/2X of 1 ▶ |
| 5X of 1 ▶ | 1/5X of 1 ▶ |
| 8X of 1 ▶ | 1/8X of 1 ▶ |
Other Milligram per Liter Conversions
Milligram per Liters to Gram per 100mLs Conversion
mg/L stands for milligram per liters and gram/dL stands for gram per 100mls. The formula used in milligram per liters to gram per 100mls conversion is 1 Milligram per Liter = 0.0001 Gram per 100mL. In other words, 1 milligram per liter is 10000 times smaller than a gram per 100ml. To convert all types of measurement units, you can used this tool which is able to provide you conversions on a scale.
Convert Milligram per Liter to Gram per 100mL
How to convert milligram per liter to gram per 100ml? In the mass concentration measurement, first choose milligram per liter from the left dropdown and gram per 100ml from the right dropdown, enter the value you want to convert and click on 'convert'. Want a reverse calculation from gram per 100ml to milligram per liter? You can check our gram per 100ml to milligram per liter converter.
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Milligram per Liters to Gram per 100mLs Converter
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like mass concentration finds its use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps in the conversion of different units of measurement like mg/L to gram/dL through multiplicative conversion factors. When you are converting mass concentration, you need a Milligram per Liters to Gram per 100mLs converter that is elaborate and still easy to use. Converting Milligram per Liter to Gram per 100mL is easy, for you only have to select the units first and the value you want to convert. If you encounter any issues to convert, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in Milligram per Liter to Gram per 100mL conversion along with a table representing the entire conversion. |
11874 | https://fiveable.me/key-terms/ap-chem/boyles-law | Boyle's Law - (AP Chemistry) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Boyle's Law
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Definition
Boyle's Law is a basic principle in gas physics that states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure.
Related terms
Charles' Law:This law states that volume and temperature are directly proportional to each other as long as pressure remains constant.
Avogadro's Law:
It states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules.
Ideal Gas Law:
The equation PV=nRT where P is the pressure, V is the volume, n is number of moles, R is universal gas constant and T is temperature; combines Boyle’s law, Charles’ law and Avogadro’s law.
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11875 | https://stackoverflow.com/questions/2850572/is-a-line-formed-by-two-points-greater-than-45-degrees-off-of-the-horizontal | c# - Is a line formed by two points greater than 45 degrees off of the horizontal - Stack Overflow
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Is a line formed by two points greater than 45 degrees off of the horizontal
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Asked 15 years, 4 months ago
Modified6 years, 5 months ago
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I am trying to find out if a line defined by two points is greater than or equal to 90 degrees compared to the horizontal. Here is the code I used
bool moreThan90 = false;
double angle = Math.Atan((double)(EndingLocation.Y - Location.Y) / (double)(EndingLocation.X - Location.X));
if (angle >= Math.PI / 2.0 || angle <= -Math.PI / 2.0)
moreThan90 = true;
Did I do this correctly or is there a better built in function in .Net that will find this?
EDIT -- Actually I messed up my question I ment to say 45 off of horizontal not 90. however the answers got me to a point where I can figure it out (really I just needed to be pointed at Atan2).
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edited Apr 30, 2019 at 6:58
Bhargav Rao
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asked May 17, 2010 at 15:51
Scott ChamberlainScott Chamberlain
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1 Do you mean 'Does the line defined by points x and y intersect a horizontal line at an angle greater than 90 degrees ?' If you do, the answer is yes (draw a couple of examples to convince yourself of this) unless the line is vertical or horizontal. This should lead you to refining your question.High Performance Mark –High Performance Mark 2010-05-17 15:53:40 +00:00 Commented May 17, 2010 at 15:53
I know the math is correct, I was wonder if there is a better built in function that is more efferent (I hate doing floating point division if I don't have to).Scott Chamberlain –Scott Chamberlain 2010-05-17 15:55:37 +00:00 Commented May 17, 2010 at 15:55
@Scott Chamerlain take a look at my answer, this is do-able without any division if you are using the horizontal for the X-Axis every time.msarchet –msarchet 2010-05-17 16:02:27 +00:00 Commented May 17, 2010 at 16:02
@Slaks - As per my original (wrong) question Pheelicks is the most efficient answer.Scott Chamberlain –Scott Chamberlain 2010-05-17 16:11:34 +00:00 Commented May 17, 2010 at 16:11
@Scott Chamberlain - why specifically would you hate to do floating point division?Daniel Earwicker –Daniel Earwicker 2010-05-17 16:12:33 +00:00 Commented May 17, 2010 at 16:12
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A line that is more than 90 degrees from the horizontal will have its EndLocation.x at a smaller x value than Location.x.
So you don't need all the atan nonsense, this should be enough:
if (EndingLocation.X < Location.X)
moreThan90 = true;
EDIT:
Seems the OP meant 45 degrees not 90, which means that the above simplification no longer holds. For this it might be better to use atan2 (as Slaks pointed out) But in the spirit of not using tan:
if (Math.Abs(EndingLocation.X - Location.X) > Math.Abs(EndingLocation.Y - Location.Y) &&
EndingLocation.X < Location.X)
moreThan45 = true;
Note that you only need the 2nd check if you only want lines which point to the right
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edited May 17, 2010 at 16:14
answered May 17, 2010 at 15:55
pheelickspheelicks
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SLaks
SLaksOver a year ago
Wrong. This is more permissive than Atan
2010-05-17T15:57:16.587Z+00:00
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msarchet
msarchetOver a year ago
@SLaks, but it is correct. Assuming that your using the horizontal as the X-Axis, and @Scott Chamberlain said he would rather not do floating point division if he didn't have to.
2010-05-17T15:59:51.51Z+00:00
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pheelicks
pheelicksOver a year ago
Before downvoting, I suggest that you think - or draw a diagram. I am right
2010-05-17T16:00:48.323Z+00:00
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GrahamS
GrahamSOver a year ago
Seems reasonable to me given the question. (Perhaps the OP needs to reword it a bit)
2010-05-17T16:03:31.8Z+00:00
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Scott Chamberlain
Scott ChamberlainOver a year ago
I messed up my original question, I meant to say 45. you are correct for the 90 degree case. Edit your question so I can edit my vote
2010-05-17T16:04:18.157Z+00:00
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You should call Math.Atan2, like this:
```
double angle = Math.Atan2(EndingLocation.Y - Location.Y,
EndingLocation.X - Location.X);
if (Math.Abs(angle) >= Math.PI / 2.0)
moreThan90 = true;
```
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edited May 17, 2010 at 15:58
answered May 17, 2010 at 15:52
SLaksSLaks
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I wouldn't imagine that there is a library method for finding the angle between the two vectors, you doing this correctly (the math is right) and a quick glance around msdn and google didn't provide me with anything. I would use SLaks' version of calling the Math.Atan method.
An interesting thing to note since you are using the 'horizontal' as your plane to determine if the angle is greater than 90 degrees. If endingLocation.x < Location.X your angle will always be 'greater' than 90 degrees, if you are measuring from the positive X-Axis.
Edit: Original question was changed to 45 degree check.
The section below is a discussion of how to do this without doing floating point division per a comment that the OP made.
To find out if you have a 45 degree angle we know a few things without actually having to call ATan on the points.
first the slope of a 45 degree angle is 1. So if
Math.Abs((EndLocation.y - location.y)/(EndLocation.X - Location.X)) > 1
You have an angle that is > 45 degrees, however as permutations of a 45 degree angle occur 4 times in a circle. We need to check a few things.
If EndLocation.X < Location.X then the angle is greater than 45 degrees. This represents all angles that are left of the Y Axis (90 - 270). To determine if the angle is greater than 45 degrees we only need to know if the absolute value of the slope is greater than 1. This will always be true for the following.
Math.Abs(EndLocation.Y - Location.Y) > Math.Abs(EndLocation.X - Location.X).
So with a if statement following something like
If (EndLocation.X < Location.X) OrElse (Math.Abs(EndLocation.Y - Location.Y) > Math.Abs(EndLocation.X - Location.X) Then AngleGreaterThan45 = True.
We can determine if the angle is greater than 45 degrees without the need to perform any floating point calculations.
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edited May 17, 2010 at 16:26
answered May 17, 2010 at 15:58
msarchetmsarchet
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11876 | https://www.dictionary.com/browse/chimerical | Daily Crossword
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View synonyms for chimerical
chimerical
Also chi·mer·ic
[ki-mer-i-kuhl, -meer-, kahy-]
adjective
unreal; imaginary; visionary.
a chimerical terrestrial paradise.
Synonyms: fantastic, illusory
Antonyms: real
2. wildly fanciful; highly unrealistic.
a chimerical plan.
chimerical
/ kɪ-, kaɪˈmɛrɪkəl, kaɪˈmɛrɪk/
adjective
wildly fanciful; imaginary
given to or indulging in fantasies
Discover More
Other Word Forms
chimerically adverb
chimericalness noun
nonchimeric adjective
nonchimerical adjective
nonchimerically adverb
Discover More
Word History and Origins
Origin ofchimerical1
First recorded in 1630–40; chimer(a) + -ical
Discover More
Example Sentences
Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com.
Bessent’s statement about the millions of “illegal aliens” coming out of the workforce is especially chimerical.
FromLos Angeles Times
Projections of economic gains from major sporting events are typically optimistic, euphoric, chimerical or conjectural.
FromLos Angeles Times
Sharply vivid rather than suggestively chimerical, the scenes and dances had a trim, finely honed character.
FromNew York Times
Claims made for the abilities or perils of AI chatbots have often turned out to be mistaken or chimerical.
FromLos Angeles Times
Which is to say, this was a precious, anticipated, frankly anxious affair — the materialization of a beloved and mercurial performer moving from the chimerical to the literal.
FromNew York Times
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chimerechi meson |
11877 | https://arminstraub.com/downloads/teaching/linearalgebra-fall14/lecture15.pdf | Review • Vectors v1, , vp are linearly dependent if x1v1 + x2v2 + + xpvp = 0, and not all the coefficients are zero.
• The columns of A are linearly independent each column of A contains a pivot.
• Are the vectors 1 1 1 , 1 2 3 , −1 1 3 independent?
1 1 −1 1 2 1 1 3 3 ⇝ 1 1 −1 0 1 2 0 2 4 ⇝ 1 1 −1 0 1 2 0 0 0 So: no, they are dependent! (Coeff’s x3 = 1, x2 = −2, x1 = 3) • Any set of 11 vectors in R10 is linearly dependent.
A basis of a vector space Definition 1. A set of vectors {v1, , vp} in V is a basis of V if • V = span{v1, , vp}, and • the vectors v1, , vp are linearly independent.
In other words, {v1, , vp} in V is a basis of V if and only if every vector w in V can be uniquely expressed as w = c1v1 + + cpvp.
Example 2. Let e1 = 1 0 0 , e2 = 0 1 0 , e3 = 0 0 1 .
Show that {e1, e2, e3} is a basis of R3.
It is called the standard basis.
Solution.
• Clearly, span{e1, e2, e3} = R3.
• {e1, e2, e3} are independent, because 1 0 0 0 1 0 0 0 1 has a pivot in each column.
Definition 3. V is said to have dimension p if it has a basis consisting of p vectors.
Armin Straub astraub@illinois.edu 1 This definition makes sense because if V has a basis of p vectors, then every basis of V has p vectors.
Why? (Think of V = R3.) A basis of R3 cannot have more than 3 vectors, because any set of 4 or more vectors in R3 is linearly dependent.
A basis of R3 cannot have less than 3 vectors, because 2 vectors span at most a plane (challenge: can you think of an argument that is more “rigorous”?).
Example 4. R3 has dimension 3.
Indeed, the standard basis 1 0 0 , 0 1 0 , 0 0 1 has three elements.
Likewise, Rn has dimension n.
Example 5. Not all vector spaces have a finite basis. For instance, the vector space of all polynomials has infinite dimension.
Its standard basis is 1, t, t2, t3, This is indeed a basis, because any polynomial can be written as a unique linear combination: p(t) = a0 + a1t + + antn for some n.
Recall that vectors in V form a basis of V if they span V and if they are linearly independent. If we know the dimension of V , we only need to check one of these two conditions: Theorem 6. Suppose that V has dimension d.
• A set of d vectors in V are a basis if they span V .
• A set of d vectors in V are a basis if they are linearly independent.
Why?
• If the d vectors were not independent, then d −1 of them would still span V . In the end, we would find a basis of less than d vectors.
• If the d vectors would not span V , then we could add another vector to the set and have d + 1 independent ones.
Example 7. Are the following sets a basis for R3?
(a) ( 1 2 0 , 0 1 1 ) No, the set has less than 3 elements.
(b) ( 1 2 0 , 0 1 1 , 1 0 3 , −1 2 0 ) No, the set has more than 3 elements.
Armin Straub astraub@illinois.edu 2 (c) ( 1 2 0 , 0 1 1 , 1 0 3 ) The set has 3 elements. Hence, it is a basis if and only if the vectors are independent.
1 0 1 2 1 0 0 1 3 ⇝ 1 0 1 0 1 −2 0 1 3 ⇝ 1 0 1 0 1 −2 0 0 5 Since each column contains a pivot, the three vectors are independent.
Hence, this is a basis of R3.
Example 8. Let P2 be the space of polynomials of degree at most 2.
• What is the dimension of P2?
• Is {t, 1 −t, 1 + t −t2} a basis of P2?
Solution.
• The standard basis for P2 is {1, t, t2}.
This is indeed a basis because every polynomial a0 + a1t + a2t2 can clearly be written as a linear combination of 1, t, t2 in a unique way.
Hence, P2 has dimension 3.
• The set {t, 1 −t, 1 + t −t2} has 3 elements. Hence, it is a basis if and only if the three polynomials are linearly independent.
We need to check whether x1t + x2(1 −t) + x3(1 + t −t2) (x2+x3)+(x1−x2+x3)t−x3t2 = 0 has only the trivial solution x1 = x2 = x3 = 0.
We get the equations x2 + x3 = 0 x1 −x2 + x3 = 0 −x3 = 0 which clearly only have the trivial solution. (If you don’t see it, solve the system!) Hence, {t, 1 −t, 1 + t −t2} is a basis of P2.
Armin Straub astraub@illinois.edu 3 Shrinking and expanding sets of vectors We can find a basis for V = span{v1, , vp} by discarding, if necessary, some of the vectors in the spanning set.
Example 9. Produce a basis of R2 from the vectors v1 = 1 2 , v2 = −2 −4 , v3 = 1 1 .
Solution. Three vectors in R2 have to be linearly dependent.
Here, we notice that v2 = −2v1.
The remaining vectors {v1, v3} are a basis of R2, because the two vectors are clearly independent.
Checking our understanding Example 10. Subspaces of R3 can have dimension 0, 1, 2, 3.
• The only 0-dimensional subspace is {0}.
• A 1-dimensional subspace is of the form span{v} where v 0.
These subspaces are lines through the origin.
• A 2-dimensional subspace is of the form span{v, w} where v and w are not multiples of each other.
These subspaces are planes through the origin.
• The only 3-dimensional subspace is R3 itself.
True or false?
• Suppose that V has dimension n. Then any set in V containing more than n vectors must be linearly dependent.
That’s correct.
• The space Pn of polynomials of degree at most n has dimension n + 1.
True, as well. A basis is {1, t, t2, , tn}.
• The vector space of functions f: R →R is infinite-dimensional.
Yes. A still-infinite-dimensional subspace are the polynomials.
• Consider V = span{v1, , vp}. If one of the vectors, say vk, in the spanning set is a linear combination of the remaining ones, then the remaining vectors still span V .
True, vk is not adding anything new.
Armin Straub astraub@illinois.edu 4 |
11878 | https://math.stackexchange.com/questions/1457490/prove-the-laplace-transform-of-sinhat | calculus - Prove the laplace transform of $\sinh(at)$? - Mathematics Stack Exchange
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Prove the laplace transform of sinh(a t)sinh(a t)?
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My problem today is as above. Here is what I have done:
Use integral definition of laplace transform to get
∫∞0 sinh(a t)exp(−s t)d t∫0∞sinh(a t)exp(−s t)d t
=lim b→∞∫b 0 sinh(a t)exp(−s t)d t=lim b→∞∫0 b sinh(a t)exp(−s t)d t
By the definition sinh(a t)=1 2(exp(a t)−exp(−a t))sinh(a t)=1 2(exp(a t)−exp(−a t))
We can write
=lim b→∞1 2∫b 0(exp(a t)−exp(−a t))exp(−s t)d t=lim b→∞1 2∫0 b(exp(a t)−exp(−a t))exp(−s t)d t
=lim b→∞1 2∫b 0 exp((a−s)t)−exp((−a−s)t)d t=lim b→∞1 2∫0 b exp((a−s)t)−exp((−a−s)t)d t
lim b→∞1 2[(1 a−s exp((a−s)b)−1−a−s exp((−a−s)b))−(1 a−s−1−a−s)]lim b→∞1 2[(1 a−s exp((a−s)b)−1−a−s exp((−a−s)b))−(1 a−s−1−a−s)]
But taking the limit as b→∞b→∞ yields infinity terms. What have I done wrong here?
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asked Sep 30, 2015 at 3:31
PythonNewbPythonNewb
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1 We do not usually worry about the LT existing for all s s it is sufficient (in this case) that it exist for s>|a|s>|a|.Conrad Turner –Conrad Turner 2015-09-30 03:37:03 +00:00 Commented Sep 30, 2015 at 3:37
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Hint. If s>|a|s>|a| then exp((a−s)b)exp((a−s)b) tends to 0 0, not ∞∞, if b→∞b→∞.
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answered Sep 30, 2015 at 3:35
DavidDavid
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I'm not quite sure if I did it right but I think I did. Just correct me if I'm wrong.
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edited Feb 27, 2017 at 16:25
Daniel Fischer
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answered Feb 27, 2017 at 15:19
R.YawnR.Yawn
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11879 | https://www.oxfordlearnersdictionaries.com/us/definition/american_english/braggadocio | Definition of braggadocio noun from the Oxford Advanced American Dictionary
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11880 | https://tmedweb.tulane.edu/pharmwiki/doku.php/cromolyn_sodium | cromolyn_sodium [TUSOM | Pharmwiki]
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Trace:•cromolyn_sodium
cromolyn_sodium
Cromolyn Sodium
Trade Names: generic (only a nebulizer formulation is available in the United States)
Drug Class: an inhaled anti-inflammatory agent; cromoglycate
Mechanism of Action:
inhibits both antigen- and exercise-induced asthma, and chronic use (four times daily) slightly reduces the overall level of bronchial reactivity.
Note: cromolyn and nedocromil sodium have no effect on airway smooth muscle tone, and are ineffective in reversing asthmatic bronchospasm.
The cellular mechanism is thought to be an alteration in the function of delayed chloride channels in the cell membrane that inhibits cellular activation in eosinophils & specific mast cells subtypes found in the lung.
Indications:
Prophylaxis for inhibiting asthma caused by allergens or exercise
ineffective when used after bronchoconstriction has developed, following the release of inflammatory mediators by mast cells or basophils
these agents can be administered 10-15 minutes before exposure to a known asthma trigger (Kemp, 2014)
cromolyn sodium & nedocromil are considered as less effective than inhaled glucocorticoids, but are devoid of systemic side effects, and represent an alternative or additive agent for nebulizer-based therapy.
cromolyn sodium is also available OTC for treatment of allergy and other mast cell disorders in the form of a nasal spray, and oral solution, and as an ophthalmic solution.
Contraindications:
known hypersensitivity
Pharmacokinetics:
a generic nebulizer solution is available (10 mg/mL in 2 mL sterile ampules)
the Intal Inhaler formulation is no longer available in the USA
Side Effects:
throat irritation
bronchospasm
Availability of Degranulation Inhibitors:
In the USA,neither cromolyn nor nedocromil is available in Metered Dose Inhalers (MDIs), which was their most popular use in the past. This is due to the discontinuation of the use of chlorofluorcarbon propellants that had previously been used for these inhalers, and the decision by their manufacturers to not invest in new hydrocarbon-free formulations.1)
The efficacy of these drugs as prophylactic agents has been considered as “limited”, and the discontinuation of the easy-to-use MDI formulations for these agents has almost completely eliminated their clinical use in the USA.
The only formulation of the two “cromoglycates” available to treat asthma are solutions of cromolyn (10 mg/mL) for nebulization
Nedocromil is no longer available in the USA (Kemp, 2014).
Pronunciation: CROW-mow-lin
References:
Bhakta NR, Choo E (2024): Drugs used in asthma (Chapter 20). In:Katzung's Basic and Clinical Pharmacology. 16th Edition. Vanderah TW (Editor). McGraw-Hill / Lange. (Access-Medicine).
Kemp JP (2014): The use of chromones (cromoglycates) in the treatment of asthma. In: UpToDate. Waltham, MA. Cited 11/4/15.
rxlist.com (Cromolyn Sodium Nebulizer Solution)
Keywords
cromolyn sodium
1)
Editorial comment: this was most likely a market driven decision - e.g. why invest in developing a product that very few people are using, and is of limited therapeutic benefit
cromolyn_sodium.txt · Last modified: 2024/08/23 17:46 by cclarks
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11881 | https://link.springer.com/rwe/10.1007/978-94-007-2739-7_415 | Log in
Transient Heat Conduction in Semi-infinite Solid with Surface Convection
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pp 6181–6186
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Encyclopedia of Thermal Stresses
Marcin Trojan3
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Overview
In this section, the heat conduction problem for the semi-infinite solid, with convection boundary condition, will be solved.
A semi-infinite solid is an idealized body that has a single plane surface and extends to infinity in all directions except one. If a sudden change is imposed to this surface, transient one-dimensional conduction will occur within the solid.
We can use the general heat equation, which has the following form :
$$ \frac{{{\partial^2}T}}{{\partial {x^2}}}=\frac{1}{\kappa}\frac{{\partial T}}{{\partial t}} $$
(1)
Equation (1) is a second order in displacement and first order in time; therefore, we need an initial condition and two boundary conditions in order to solve it.
Assumptions are the same, which is one dimensional heat transfer without heat generation. For the initial condition, we can specify that the temperature inside the solid is uniform at ( {T_i} ):
$$ T(x,0)={T_i} $$
(2)
The interior boundary condition is
$$ T(x\to \infty, t)={T_i} $$
...
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References
Carslaw HS, Jeager JC (2008) Conduction of heat in solid, 2nd edn. Oxford University Press, Oxford
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2. Thomson WJ (1997) Atlas for computing mathematical functions. Wiley, New York
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3. Ȍzişik MN (1993) Heat conduction, 2nd edn. Wiley, New York
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4. Taler J, Duda P (2006) Solving direct and inverse heat conduction problems. Springer, Berlin/Heidelberg/New York
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5. Tautz H (1971) Wärmeleitung und Temperaturausgleich. Verlag Chemie, Weinheim
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Author information
Authors and Affiliations
Institute of Thermal Power Engineering, Faculty of Mechanical Engineering, Cracow University of Technology, Cracow, Poland
Marcin Trojan
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Department of Mechanical Engineering, Rochester Institute of Technology, Rochester, NY, USA
Richard B. Hetnarski
2. Naples, FL, USA
Richard B. Hetnarski
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Trojan, M. (2014). Transient Heat Conduction in Semi-infinite Solid with Surface Convection. In: Hetnarski, R.B. (eds) Encyclopedia of Thermal Stresses. Springer, Dordrecht.
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11882 | https://mathworld.wolfram.com/SquareRootInequality.html | Square Root Inequality -- from Wolfram MathWorld
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11883 | https://math.stackexchange.com/questions/2620364/composite-of-two-rotations-with-different-centers | plane geometry - Composite of two rotations with different centers - Mathematics Stack Exchange
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Composite of two rotations with different centers
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Asked 7 years, 8 months ago
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What is the composite of two rotations having different centers? What type of transformation (translation, rotation, etc) is it? Or there is no rule for it?
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edited Jan 25, 2018 at 11:49
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Try out a few examples and see what happens. What about 90∘90∘ clockwise about the origin, and then 90∘90∘ counterclockwise about (0,1)(0,1)? What if the last rotation is clockwise? What if it's 45∘45∘? What if the first rotation is 45∘45∘? Does it look nicer if the second point isn't (0,1)(0,1) but instead the point that the first rotation moves (0,1)(0,1) to? What if you pick a different point for the second rotation? Some experimentation like this ought to lead you to concrete ideas that might possibly be provable.
This kind of experimentation is how new math is made, and while I'm certain that the answer to your problem is known, you shouldn't let an opportunity like this to make your own, genuine discovery go to waste.
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answered Jan 25, 2018 at 9:28
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Rotations are plane isometries and there are four types of plane isometries:
rotations;
translations;
reflections;
glide reflections.
The first two types preserve orientation, whereas the last two reverse it. Since the composition of plane isometries is again a plane isometry, and since rotations preserve orientation, the composition of two rotation can only be either a rotation or a translation. It turns out that it is a translation if and only if the sum of the rotation angles is a multiple of 360∘360∘.
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edited Jan 24, 2023 at 1:38
answered Jan 25, 2018 at 9:43
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11884 | https://www.sciencedirect.com/science/article/pii/S2666776225000730 | Quick buys for prevention and control of noncommunicable diseases - ScienceDirect
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Introduction
The 25 quick buys
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Appendix A. Supplementary data
References
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The Lancet Regional Health - Europe
Volume 52, May 2025, 101281
Health Policy
Quick buys for prevention and control of noncommunicable diseases
Author links open overlay panel Gauden Galea a, Allison Ekberg a, Angela Ciobanu a, Marilys Corbex a, Jill Farrington a, Carina Ferreira-Bores a, Daša Kokole a, María Lasierra Losada a, Maria Neufeld a, Ivo Rakovac a, Elena Tsoy a, Kremlin Wickramasinghe a, Julianne Williams a, Martin McKee b, David Stuckler c
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From inertia to impact: delivering real solutions for non-communicable diseases
The Lancet Regional Health - Europe, Volume 52, May 2025, Pages 101315
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From inertia to impact: delivering real solutions for non-communicable diseases
The Lancet Regional Health - Europe, Volume 52, May 2025, Pages 101315
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Summary
Despite their established effectiveness, uptake of the WHO best buys for tackling non-communicable diseases (NCDs) has been uneven and disappointing. Here we introduce the “quick buys”, an evidence-based set of cost-effective interventions with measurable public health impacts within five years. We reviewed 49 interventions previously established as cost-effective (<$I20,000 per disability-adjusted life-year averted) to identify the earliest possible detectable effect on high-level population health targets. Using a strict evidence hierarchy, including Cochrane and systematic reviews, we estimated the effects of each intervention against global targets agreed upon by countries. Quick buys were defined as those interventions that could exhibit measurable effects within 5 years, aligning with average electoral cycles in across the WHO European Region. Of the 49 interventions, 25 qualified as quick buys, including those relating to tobacco (n=5), alcohol (n=4), unhealthy diet (n=3), physical inactivity (n=1), cardiovascular disease (n=3), diabetes (n=4), chronic respiratory disease (n=1), and cancer (n=4). These findings not only offer guidance to policymakers deciding on interventions that align with short-term political cycles but also have the potential to accelerate progress to global health targets, particularly the 2030 Sustainable Development Goal of reducing premature NCD mortality by one-third.
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Keywords
NCD
Best buy
Quick buy
SDG
Search strategy and selection criteria
References for this Health Policy paper were identified through searches of Cochrane Reviews and PubMed from database inception through to November 1, 2024 for best buys using search terms based on keywords for each intervention taken from Appendix 3 of the WHO Global NCD Action Plan 2013–2030. For each candidate quick buy, we sought to ascertain the earliest possible detectable effect on population health defined as the identification of significant effect in either a meta-analysis or a study contained in a systematic review. Our search followed a search hierarchy, prioritising Cochrane Reviews, followed by systematic reviews and/or meta-analysis, narrative reviews and finally individual research articles. For more details, see the main text.
Introduction
In 2010, WHO developed a package of evidence-based, cost-effective interventions that could reduce the burden of non-communicable diseases (NCDs).1 Acting at the individual and population level, these interventions targeted the leading NCD risk factors (tobacco and alcohol use, unhealthy diet, and physical inactivity) and four disease groups (cardiovascular disease, diabetes, chronic respiratory disease, and cancer). The resulting list was known as the ‘NCD best buys’, and helped focus both NCD advocacy worldwide and policymaking at the country level.
However, a decade and a half later, implementation and enforcement of the best buys has been disappointing. Progress towards the nine global voluntary targets agreed to in the NCD Global Monitoring Framework is slow and uneven.2,3 It is estimated that without increased uptake of these effective interventions, half of all countries will miss the 2030 Sustainable Development Goal (SDG) Target 3.4 to reduce NCD-related premature mortality by one-third.4 The reasons for slow progress are multiple but may include the false perception that the benefits of NCD interventions take too long to realise and are thus misaligned with short-term political cycles.5,6 Politicians are likely to favour measures that yield results that they can take credit for, although empirical research shows that this varies with their career stage.7 In addition, the best buys need to be more relevant to higher-income countries, given that they were initially envisioned and promoted to apply to low-income and middle-income countries.
To address these misconceptions, we undertook a review of the best buys and other recommended interventions for preventing and controlling NCDs to determine when the soonest public health impact would become apparent and to construct clear and measurable timelines to support action by policymakers. Although countries should primarily focus on the epidemiological need and cost-effectiveness of interventions when deciding priorities, this paper adds the dimension of time to impact. We also recognise that policymakers will consider other factors, including whether a single intervention might bring multiple benefits over different timescales. With an eye on the 2030 SDG deadline, we sought to identify, from among the best buys and other recommended interventions, those that can be expected to demonstrate impact on population health within five years or less—and in some cases immediately; these are the quick buys.
The methods used to identify NCD best buys has been described in detail elsewhere, including updates in 2017 and 2022.8,9 Cost effectiveness of those best buys has been determined using the WHO-CHOICE methodology.10 All costs are expressed in International Dollars (Int$), a hypothetical unit of currency with the same purchasing power parity as the US Dollar in the US at the same time. These methods projected cost-effectiveness for low-middle, lower-middle, and upper middle-income countries, but not high-income countries, even though the underlying evidence primarily came from high-income countries.9,11 While it is beyond the scope of this paper to review how the best buys were selected, the challenges involved, and any critiques of them, we note that they do not include certain other WHO products, such as the menu of cost-effective interventions for mental health, the recommended interventions to address the impact of air pollution, and the menu of cost-effective interventions for oral health. Furthermore, we are limited to those interventions that have been evaluated using WHO-CHOICE, so the exclusion of other interventions does not mean that they are not cost-effective, affordable, or feasible.
Identifying quick buys
Our approach started with a systematic review of the canonical ‘best buys’ (<Int$ 100 per disability-adjusted life-year [DALY] averted) and interventions with a cost-effectiveness ratio between Int$100/DALY and Int$500/DALY (previously referred to as ‘good buys’) through to interventions up to $I20,000 per DALY averted, as taken from Appendix 3 of the WHO Global NCD Action Plan 2013–2030 as updated in December 2022.9 An intervention was included only if it met the best buy criteria in at least one country income stratum. This yielded 49 candidate interventions addressing tobacco (n=7), alcohol (n=5), unhealthy diet (n=7), physical inactivity (n=2), cardiovascular disease (CVD, n=6), diabetes (n=7), chronic obstructive pulmonary disease (COPD, n=4) and cancer (n=11).
For each candidate quick buy, we then sought to ascertain the earliest possible detectable effect on population health defined as the identification of significant effect in either a meta-analysis or a study contained in a systematic review.
Our search followed the standard evidence hierarchy but recognised that natural experiment designs are often more appropriate for population-level interventions.12 Where possible, we first sought a Cochrane Review, followed by a PubMed search for systematic reviews, with pre-registration if available. For interventions for which systematic reviews were unavailable, we searched for published peer-reviewed literature reviews. Thus, we only used the review that occupied the highest position in this evidence hierarchy. Taking alcohol as an example, for the five alcohol interventions, we identified two associated Cochrane Reviews pertaining to restrictions on alcohol advertising in both adults and adolescents as well as on brief alcohol interventions. A systematic review was used to identify the effect of price increases from excise taxes and for the effect of sobriety checkpoints (see Appendix 1 flow diagram for inclusion).
We then searched each review for the earliest possible significant effect on a UN or other established high-level, multi-country target for reducing risk factor exposure or disease outcomes. These targets included, where possible, the SDG 2030 targets relevant to NCDs13 (which covered premature mortality, alcohol and tobacco) or the WHO NCD Global Monitoring Framework 2025 targets14 (which were used for unhealthy diet, physical inactivity and CVD), the Global Strategy to Eliminate Cervical Cancer15 and the global initiatives on breast cancer16 and on childhood cancer17 (for cancer targets) or the Global Diabetes Targets.18 Of note, some of these targets focused on consumption- or treatment access metrics while others focused on prevalence or mortality targets, making them not directly comparable. However, these global targets were used for the analysis given the political commitments made by countries to achieve them. Finally, we categorised a quick buy as an intervention that could exert an effect within 5 years, whether at individual or population levels, chosen to reflect the timing of average electoral cycles in high-income, European countries and the time remaining before the deadline of the SDGs and Global Action Plan in 2030.
The 25 quick buys
Table 1, Table 2, Table 3, Table 4, Table 5, Table 6, Table 7, Table 8 present the findings for each of the 49 candidate interventions. Of these, 25 were identified that met the criterion for a quick buy of having a detectable effect within 5 years.
Table 1. Empirical evidence of the timing of best buys and recommended interventions' effects related to tobacco.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Review source(s) |
--- ---
| Increase excise taxes and prices on tobacco products | Yes | 4 months19 | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Wilson et al. 201220 |
| Implement large graphic health warnings on all tobacco packages, accompanied by plain/standardized packaging | Yes | 14 months21 | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | McNeill et al., 2017; Pang et al., 202122,23 |
| Enact and enforce comprehensive bans on tobacco advertising, promotion and sponsorship | Yes | 2 years24 | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Henriksen 201225,26 |
| Eliminate exposure to second-hand tobacco smoke in all indoor workplaces, public places, public transport | Yes | Immediate27 | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Frazer et al., 2016a and 2016b28,29 |
| Implement effective mass media campaigns that educate the public about the harms of smoking/tobacco use and secondhand smoke, and encourage behavior change | No | ≤7 years | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Bala et al., 201730,31 |
| Provision of cost-covered effective population-wide support (including brief advice, national toll-free quit line services and mCessation) for tobacco cessation to all tobacco users | No | >7 years | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Silagy et al., 2001; Lancaster et al. 201732,33 |
| Provision of cost-covered effective pharmacological interventions to all tobacco users who want to quit through the use of nicotine replacement therapy (NRT), Bupropion and Verenicline | Yes | 6 months34 | Strengthen the implementation of the World Health Organization Framework Convention on Tobacco Control in all countries, as appropriate. 3.a.1. Age-standardized prevalence of current tobacco use among persons aged 15 years and older | Bergen et al. 201435 |
Table 2. Empirical evidence of the timing of best buys and recommended interventions' effects related to alcohol.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Source(s) |
--- ---
| Increase excise taxes on alcoholic beverages | Yes | Immediate19,36 | SDG 3.5.2 Alcohol per capita consumption (aged 15 years and older) within a calendar year in litres of pure alcohol | Kilian et al. 202336 |
| Enact and enforce bans or comprehensive restrictions on exposure to alcohol advertising (across multiple types of media) | Yes | Immediate37 | SDG 3.5.2 Alcohol per capita consumption (aged 15 years and older) within a calendar year in litres of pure alcohol | Siegfried et al. 201437 |
| Enact and enforce restrictions on the physical availability of retailed alcohol (via reduced hours of sale) | Yes | Immediate36 | SDG 3.5.2 Alcohol per capita consumption (aged 15 years and older) within a calendar year in litres of pure alcohol | Kilian et al. 202336 |
| Enact and enforce drink-driving laws and blood alcohol concentration limits via sobriety checkpoints | No | n/a | SDG 3.5.2 Alcohol per capita consumption (aged 15 years and older) within a calendar year in litres of pure alcohol | Bergen et al. 201438 |
| Provide brief psychosocial intervention for persons with hazardous and harmful alcohol use | Yes | 12 months | SDG 3.5.2 Alcohol per capita consumption (aged 15 years and older) within a calendar year in litres of pure alcohol | Kaner et al. 201839 |
Table 3. Empirical evidence of the timing of best buys and recommended interventions' effects related to unhealthy diet.
| Intervention | Quick Buys | Evidence of Earliest Possible Effect on UN-linked targets | UN Target/Indicator | Source(s) |
--- ---
| Reformulation policies for healthier food and beverage products (e.g., elimination of trans-fatty acids and/or reduction of saturated fats, free sugars and/or sodium) | Yes | 1 year40,41 | WHO NCD 2025 Targets: A 30% relative reduction in mean population intake of salt/NCD Voluntary 2025 target: 30% reduction in salt/sodium intake | McLaren et al. 201640 |
| Front-of-pack labelling as part of comprehensive nutrition labelling policies for facilitating consumers' understanding and choice of food for healthy diets | Yes | Immediate42 | WHO NCD 2025 Targets: A 30% relative reduction in mean population intake of salt/NCD Voluntary 2025 target: 30% reduction in salt/sodium intake | An et al., 2021; Croker et al. 202043,44 |
| Public food procurement and service policies for healthy diets (e.g., to reduce the intake of free sugars, sodium, and unhealthy fats, and to increase the consumption of legumes, whole grains, fruits and vegetables) | No | n/a | WHO NCD 2025 Targets: A 30% relative reduction in mean population intake of salt/NCD Voluntary 2025 target: 30% reduction in salt/sodium intake | McLaren et al. 201640 |
| Behaviour change communication and mass media campaigns for healthy diets (e.g., to reduce the intake of energy, free sugars, sodium, and unhealthy fats, and to increase the consumption of legumes, whole grains, fruits and vegetables) | Yes | 3 years40,45 | WHO NCD 2025 Targets: A 30% relative reduction in mean population intake of salt/NCD Voluntary 2025 target: 30% reduction in salt/sodium intake | McLaren et al. 201640 |
| Policies to protect children from the harmful impact of food marketing on diet | No | Not identified | Not identified | Not identified |
| Protection, promotion and support of optimal breastfeeding practices | No | Not identified | Not identified | Not identified |
| Taxation on sugar-sweetened beverages as part of fiscal policies for healthy diets | No | Not identified | WHO NCD 2025 Targets: Halt the rise in diabetes and obesity | Pfinder et al. 202046 |
Table 4. Empirical evidence of the timing of best buys and recommended interventions' effects related to physical inactivity.
| Intervention | Quick buys | Evidence of earliest possible effect | UN target/indicator | Source(s) |
--- ---
| Brief counselling intervention on physical activity in primary health care | Yes | 4 weeks | WHO NCD Target 2025: a 10% relative reduction in the prevalence of insufficient physical activity | Lamming et al. 201747 |
| Physical activity public education and awareness campaign | No | Not identified | WHO NCD Target 2025: a 10% relative reduction in the prevalence of insufficient physical activity | Baker et al. 201548 |
Table 5. Empirical evidence of the timing of best buys and recommended interventions' effects on cardiovascular disease.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Review source(s) |
--- ---
| Pharmacological treatment of hypertension in adults using either of the following: thiazide and thiazide-like agents; angiotensin converting enzyme inhibitors (ACE-Is)/angiotensin-receptor blocker (ARBs); calcium channel blockers (CCBs) | Yes | 10 days49 | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Perez et al. 200949 |
| Drug therapy (treatment with an antihypertensive and statin) to control CVD risk using a total risk approach and counselling to individuals who have had a heart attack or stroke and to persons with high risk (≥20%) of a fatal and nonfatal cardiovascular event in the next 10 years using the updated WHO CVD risk charts Drug therapy (treatment with an antihypertensive) to control CVD risk using a total risk approach and counselling to individuals who have had a heart attack or stroke and to persons with high risk (≥10%) of a fatal and non-fatal cardiovascular event in the next 10 years using the updated WHO CVD risk charts | No | Not identified | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Bahiru et al., 2017; Wang et al. 202050,51 |
| Treatment new cases of acute myocardial infarction with acetylsalicylic acid initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate Treatment new cases of acute myocardial infarction with acetylsalicylic acid and thrombolysis, with patients initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate Treatment of new cases of acute myocardial infarction with acetylsalicylic acid, thrombolysis and clopidogrel, with patients initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate | Yes | Immediate | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Husted et al. 198952 |
| Treatment of acute ischemic stroke with intravenous thrombolytic therapy | Yes | 3 months | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Wardlaw et al. 201453 |
| Primary prevention of rheumatic fever and rheumatic heart diseases by increasing appropriate treatment of streptococcal pharyngitis at the primary care level | No | Not identified | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Not identified |
| Secondary prevention of rheumatic fever and rheumatic heart disease by developing a register of patients who receive regular prophylactic penicillin | No | Not identified | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Not identified |
Table 6. Empirical evidence of the timing of best buys and recommended interventions' effects on diabetes.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Source(s) |
--- ---
| Foot care to prevent amputation in people with diabetes (including educational programmes, access to appropriate footwear, multidisciplinary clinics) | No | Not identified | Not identified | Not identified |
| Diabetic retinopathy screening for all diabetes patients and laser photocoagulation for prevention of blindness | No | Not identified | Not identified | Not identified |
| Glycaemic control for people with diabetes, along with standard home glucose monitoring for people treated with insulin to reduce diabetes complications | Yes | Immediate | Global diabetes target 2: 80% of people with diagnosed diabetes have good control of glycaemia (fasting plasma glucose <9.9 mmol/l) | Hemmingsen et al. 201154 |
| Screening of people with diabetes for albuminuria and treatment with angiotensin-converting enzyme inhibitor for the prevention and delay of renal disease | Yes | <4.5 years | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | HOPE Investigators, 200055 |
| Control of blood pressure in people with diabetes | Yes | 12 months56 | Global diabetes target 3: 80% of people with diagnosed diabetes have good control of blood pressure (<140/90 mmHg) | Brunstrom et al. 201656 |
| Secondary prevention of rheumatic fever and rheumatic heart disease by developing a register of patients who receive regular prophylactic penicillin | No | Not identified | Not identified | Not identified |
| Statin use in people with diabetes >40 years old | Yes | Immediate | Global diabetes target 4: 60% of people with diabetes of 40 years or older receive statins | Yang et al. 202257 |
Table 7. Empirical evidence of the timing of best buys and recommended interventions' effects on chronic respiratory diseases.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Source(s) |
--- ---
| Acute treatment of asthma exacerbations with inhaled bronchodilators and oral steroids | No | Not identified | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | O'Byrne et al. 201958 |
| Acute treatment of COPD exacerbations with inhaled bronchodilators and oral steroids | Yes | 6 months | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Salpeter et al. 200659 |
| Long-term management of asthma with inhaled bronchodilator and low-dose beclometasone | No | Not identified | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | O'Byrne et al. 201958 |
| Long-term management of COPD with inhaled bronchodilator | No | >7 years | SDG Target 3.4: reduce premature mortality from NCDs by one-third by 2030 | Walters et al. 199660 |
Table 8. Empirical evidence of the timing of best buys and recommended interventions' effects on cancer.
| Intervention | Quick buys | Evidence of earliest possible effect on UN-linked targets | UN target/indicator | Source(s) |
--- ---
| Vaccination against human papillomavirus (1–2 doses) of 9–14 year old girls | Yes | Immediate | Global Strategy to Eliminate Cervical Cancer 90-70-90 Targets: 90% of girls fully vaccinated with the HPV vaccine by the age of 15 | Staley et al. 202161 |
| Cervical cancer: HPV DNA screening, starting at the age of 30 years with regular screening every 5–10 years (using a screen-and-treat approach or screen, triage and treat approach) | Yes | Immediate | Global Strategy to Eliminate Cervical Cancer 90-70-90 Targets: 70% of women screened using a high-performance test by the age of 35, and again by the age of 45 | Staley et al. 202161 |
| Cervical cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment | Yes | Immediate | Global Strategy to Eliminate Cervical Cancer 90-70-90 Targets: 90% of women with pre-cancer treated and 90% of women with invasive cancer managed. | Staley et al. 202161 |
| Breast cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment | Yes | Immediate | WHO Global Breast Cancer Initiative: reduce breast cancer mortality by 2.5% per year | n/a |
| Breast Cancer: screening with mammography (once every 2 years for women aged 50–69 years) linked with timely diagnostic work-up and comprehensive breast cancer treatment in setting where mammographic screening programme is recommended | No | >10 years | WHO Global Breast Cancer Initiative: reduce breast cancer mortality by 2.5% per year | Gotzsche et al. 201362 |
| Colorectal cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment | No | Not identified | Not identified | Not identified |
| Colorectal cancer screening: population based programme including through stool-based tests, as appropriate, at age >50 years, linked with timely treatment in settings where screening programme is recommended | No | Not identified | Not identified | Not identified |
| Prevention of liver cancer through hepatitis B immunization | No | Not identified | Not identified | Not identified |
| Childhood cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment, focusing on 6 index cancers of WHO Global Initiative for Childhood Cancer | No | Not identified | Not identified | Not identified |
| Prostate cancer: early diagnosis programmes linked with timely diagnostic work-up and comprehensive cancer treatment | No | Not identified | Not identified | Not identified |
| Early detection and comprehensive treatment of cancer for those living with HIV | No | Not identified | Not identified | Not identified |
We disaggregate these findings by NCD risk factor or diseases (tobacco, alcohol, unhealthy diet, physical activity, cardiovascular disease, diabetes, chronic respiratory diseases, and cancer) below, starting with tobacco.
Tobacco
Five out of seven tobacco interventions had evidence of impacts within 5 years in the included review articles (Table 1). The fastest effect was for eliminating exposure to second-hand smoke, which had immediately detectable effects.19,20 This was followed by increasing excise taxes and prices, which demonstrated a significant effect at 4 months21,22; followed by nicotine replacement therapy (6 months), graphic health warnings (14 months), and enacting and enforcing comprehensive bans on tobacco advertising, promotion or sponsorship (2 years).
Two interventions did not meet the 5-year threshold for inclusion. Implementing media campaigns to educate the public about tobacco-related harms had an earliest identified effect within 7 years.30,31 Similarly, providing population-wide support, such as brief advice, national toll-free quit lines and m-cessation (using messaging) services to all tobacco users, had an earliest potential effect estimated at beyond 7 years.32,33
Alcohol
All three interventions affecting alcohol price and availability had immediate effects (Table 2). Increasing excise taxes,19,36 enacting and enforcing bans or comprehensive restrictions on exposure to alcohol advertising,37 and enacting and enforcing restrictions on the physical availability of alcohol had immediate impacts on alcohol per capita consumption in persons aged 15 or older.36 Brief psychosocial interventions for persons with hazardous and harmful alcohol use also met our criterion, with detectable impacts under 12 months post-intervention.39 However, we could not identify the timing for an effect of the WHO-recommended intervention to enact and enforce drunk-driving laws at sobriety checkpoints on blood alcohol concentrations. Although we identified systematic reviews finding positive effects on alcohol-related car crashes and fatalities, studies were insufficient to support an effect on per capita alcohol consumption.
Unhealthy diet
Unhealthy diet interventions were, unless otherwise specified, linked to the WHO NCD Global Monitoring Framework targets of 30% relative reductions in the mean population salt intake. Three of the seven met the criteria for quick buys (Table 3). These were: reformulation policies for healthier food and beverage products (achieving impact within 1 year)40,41; front-of-pack labelling as part of comprehensive nutrition labelling policies for facilitating consumers’ choices (immediate)42, 43, 44; and mass media campaign and behavioural change communication for healthy diets (within 3 years).40,45
Three recommended interventions with an average cost-effectiveness ratio>Int$100 did not have identified effects on UN targets within the 5-year timeline. For policies to protect children from the harmful effects of food marketing and protection, promotion and support of optimal breastfeeding practices, the link to health outcomes was non-specific. In contrast, for taxes on sugar-sweetened beverages, the corresponding target from WHO NCD Global Monitoring Framework goals14 was to halt the rise in obesity. Although systematic reviews identified effects on sugar consumption, and this would plausibly translate into obesity reductions, the reviews included provided insufficient evidence of an impact on obesity prevalence within five years.46
Physical inactivity
Two interventions related to physical inactivity were examined (Table 4): public education and awareness campaigns and brief counselling interventions in primary care. We were unable to identify a detectable effect of public education and awareness within the 5-year timeline, consistent with other evidence that education campaigns, often advocated by producers of harmful products, have little effect. However, brief counselling interventions did meet our criteria for a quick buy, with evidence of a significant improvement in physical activity within 4 weeks.
Cardiovascular disease
Six interventions were evaluated for reducing premature CVD mortality (Table 5). Of these, three had evidence of impacts on UN targets within 5 years. Treating hypertension in adults reduced CVD mortality in as few as 10 days, and there was a virtually immediate reduction in mortality associated with administering acetylsalicylic acid to those experiencing a myocardial infarction. Finally, trials of treatment of acute ischaemic stroke with thrombolytic therapy typically use 3-month mortality as an early endpoint, finding significant benefits by then in the included review, given the mechanisms of action the effects can be expected to arise immediately.
Three of the interventions included did not achieve an effect on CVD outcomes within 5 years. These were primary prevention of rheumatic fever through treating streptococcal pharyngitis (likely because rheumatic heart valve damage is a late complication); and secondary prevention of rheumatic fever and rheumatic heart disease by developing a register of patients who receive regular prophylactic penicillin. Additionally, we were unable to find evidence for effects on CVD mortality within the 5-year timeline for one intervention: combined drug therapy and counselling to control CVD risk using a total risk approach for to individuals who either had a stroke or heart attack or were at high risk for a CVD event in the next decade.
Diabetes
We analysed seven diabetes interventions (Table 6), including four with an average cost-effectiveness ratio>Int$100 and one best buy. Four met criteria for quick buys. We found significant mortality reductions linked to blood pressure control in people with diabetes at 12 months. However, other interventions had results that were inevitable. Thus, statin use in people with diabetes >40 years old had immediate effects on the WHO diabetes statin use target of 60 and, glycaemic control for people with diabetes had immediate effects on achieving WHO targets for 80% of persons with diagnosed diabetes meeting targets for HbA1c (<8%). Screening people with diabetes for albuminuria and treatment with angiotensin-converting enzyme to delay and prevent renal disease was linked to mortality reductions at 4.5 years.
Three interventions did not have a corresponding UN target: foot care to prevent amputation; diabetic retinopathy screening and laser photocoagulation for preventing blindness; and secondary prevention of rheumatic fever by developing a register of patients who receive regular prophylactic penicillin.
Chronic respiratory diseases
The chronic respiratory disease candidate interventions involved acute and long-term management of asthma and chronic obstructive pulmonary disease (COPD) exacerbations with inhaled bronchodilators and oral steroids (Table 7). The four interventions were evaluated for their ability to achieve the NCD premature mortality target. None of the asthma interventions had a detectable effect on COPD mortality endpoints within 5 years. Although this may seem surprising given the immediate symptomatic benefits of treatment, trials and observational studies have produced conflicting results63 and few studies have evaluated models of care designed to increase uptake of these treatments.64 A further consideration is that asthma mortality is an uncommon event, making it difficult to identify in small trials of short duration.58 Alternatively, treating COPD with anticholinergic drugs was linked to a significant reduction in mortality rates at 6 months.59
Cancer
Interventions related to cancer (Table 8) focused on cervical cancer, breast cancer, colorectal cancer, liver cancer, childhood cancer, and cancers in people living with HIV. All interventions for cervical cancer showed immediate effects on the 90-70-90 targets (90% of girls fully vaccinated with HPV vaccine by age 15; 70% of women screened using a high-performance test by age 35 and again by age 45; 90% of women with pre-cancer treated and 90% of women with invasive cancer managed) set out in the Global Strategy to Eliminate Cervical Cancer. These included vaccinations against human papillomavirus (1–2 doses) for 9–14 year-old girls and screening for human papillomavirus DNA every 5–10 years from age 30. Mammography for breast cancer did not exhibit significant effects in the included review within the 5-year time limit for reducing breast cancer mortality by 2.5% per year. We were unable to identify UN-linked targets for programmes for early diagnosis of childhood cancer, programmes for early detection and comprehensive treatment of cancer for those living with HIV, population-based colorectal cancer or prostate cancer screening, and prevention of liver cancer through hepatitis B immunisation.
Towards faster progress
From a public health perspective, all interventions that are cost-effective should be implemented but, in reality, we need to convince often sceptical politicians faced with multiple demands for action. Recognising the incentives they face, such as the desire to be re-elected, as well as the time preferences that everyone incorporates in their decisions, it is intuitive that measures that achieve results faster will be more attractive, all else being equal, and we know that myriad factors are taken into account, many involving the commercial determinants of health.65 In making this case, it is also important to emphasise that interventions that exploit reductions of demand for hazardous products by raising taxes will generate revenue, although we caution against linking these funds for prevention or treatment as it can create perverse incentives to maintain this funding stream.66
Out of 49 potential interventions, we identified 25 showing an effect within 5 years Panel 1, and in some cases immediately, on a UN-linked target. These quick buys have ‘face validity’ as having plausible, rapid effects given the natural history of the NCDs in question. Yet, as with the WHO best buys, they have several important limitations.
Panel 1
Quick Buys
Tobacco
•Increase excise taxes and prices on tobacco products
•Implement large graphic health warnings on all tobacco packages, accompanied by plain/standardized packaging
•Enact and enforce comprehensive bans on tobacco advertising, promotion and sponsorship
•Eliminate exposure to second-hand tobacco smoke in all indoor workplaces, public places, public transport
•Provision of cost-covered effective pharmacological interventions to all tobacco users who want to quit through the use of nicotine replacement therapy (NRT), Bupropion and Verenicline
Alcohol
•Increase excise taxes on alcoholic beverages
•Enact and enforce bans or comprehensive restrictions on exposure to alcohol advertising (across multiple types of media)
•Enact and enforce restrictions on the physical availability of retailed alcohol (via reduced hours of sale)
•Provide brief psychosocial intervention for persons with hazardous and harmful alcohol use
Unhealthy diet
•Reformulation policies for healthier food and beverage products (e.g., elimination of trans-fatty acids and/or reduction of saturated fats, free sugars and/or sodium)
•Front-of-pack labelling as part of comprehensive nutrition labelling policies for facilitating consumers' understanding and choice of food for healthy diets
•Behaviour change communication and mass media campaigns for healthy diets (e.g., to reduce the intake of energy, free sugars, sodium, and unhealthy fats, and to increase the consumption of legumes, whole grains, fruits and vegetables)
Physical inactivity
•Brief counselling intervention on physical activity in primary health care
Cardiovascular disease
•Pharmacological treatment of hypertension in adults using either of the following: thiazide and thiazide-like agents; angiotensin converting enzyme inhibitors (ACE-Is)/angiotensin-receptor blocker (ARBs); calcium channel blockers (CCBs)
•Treatment new cases of acute myocardial infarction with acetylsalicylic acid initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate; Treatment new cases of acute myocardial infarction with acetylsalicylic acid and thrombolysis, with patients initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate; Treatment of new cases of acute myocardial infarction with acetylsalicylic acid, thrombolysis and clopidogrel, with patients initially treated in a hospital setting with follow up carried out through primary health care facilities at a 95% coverage rate
•Treatment of acute ischemic stroke with intravenous thrombolytic therapy
Diabetes
•Glycaemic control for people with diabetes, along with standard home glucose monitoring for people treated with insulin to reduce diabetes complications
•Screening of people with diabetes for albuminuria and treatment with angiotensin-converting enzyme inhibitor for the prevention and delay of renal disease
•Control of blood pressure in people with diabetes
•Statin use in people with diabetes >40 years old
COPD
•Acute treatment of COPD exacerbations with inhaled bronchodilators and oral steroids
Cancer
•Vaccination against human papillomavirus (1–2 doses) of 9–14 year old girls
•Cervical cancer: HPV DNA screening, starting at the age of 30 years with regular screening every 5–10 years (using a screen-and-treat approach or screen, triage and treat approach)
•Cervical cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment
•Breast cancer: early diagnosis programs linked with timely diagnostic work-up and comprehensive cancer treatment
First, consistent with the dominant paradigm in evaluating health interventions, we have used the conventional evidence hierarchy that privileges randomised controlled trials. Yet, when evaluating population-level interventions, which include many of those we have examined, this will rarely be the most appropriate approach, either because of feasibility or the ability to generalise from specific contexts.67 Rather, it will be more appropriate to take advantage of natural experiments, of which there are a growing number, employing interrupted time series analysis, synthetic control designs and others. However, these are still few in number. A further constraint is that, in some cases, the WHO recommends population interventions which, at the time of writing, lack foundation in systematic reviews, such as educational interventions to increase physical inactivity. Although these question the selection of the interventions which our study drew upon to identify quick buys, it is beyond the scope of this analysis to revisit the methodology employed to create the best buys. It is important to note that our failure to find evidence using this hierarchical approach does not indicate a lack of evidence. The NCD best buys are, in themselves, limited, excluding important conditions like fatty liver disease and chronic renal disease.68 We also lack evidence of the cumulative effect of interventions. Other research, for example on tobacco, shows that the addition of measures is non-linear, with the greatest benefits accruing from a comprehensive package.69 Another challenge is that costs of some interventions can change substantially, for example when patents expire on innovative medicines.70
More generally it is important to note that several effective interventions were not included for evaluation (e.g., surveillance, monitoring, governance, joint interventions). Some best buys, such as those on unhealthy diets, were not well specified. It is possible that they might have an effect on salt, but not fat consumption, and the policy would still qualify as a quick buy. Other best buys require certain prerequisites, such as the existence of infrastructure to deliver a model of care, which may not always be a realistic assumption.
As in most public health interventions, the “precautionary principle” applies. For example, even if there is no randomised controlled trial showing that restricting marketing of unhealthy products to children can reduce obesity levels in children, this should not preclude action. This is particularly important given how the manufacturers of these products exploit any uncertainty about evidence to undermine public health.71 These findings should not be misinterpreted or misused to argue against implementation of the best buys, but rather to support their increased uptake.
Our approach also sought to identify the earliest possible detectable effect based on evidence from meta-analyses or estimates in systematic reviews of a biologically plausible effect on NCD-related endpoints. This highlights plausibility and a best-case scenario, but does not address either robustness or the magnitude of the effect size. This requires further research, informed by an understanding of the biological mechanisms involved to ensure that findings are scientifically and epidemiologically plausible. Importantly, the risk function associated with exposure to certain substances is heterogenous and asymmetrical. For example, the effect of smoking on cardiovascular illness, mediated largely through endothelial dysfunction, can be measured in days whereas that on lung cancer involves a lag of many years. It takes several years of heavy drinking to acquire cirrhosis but withdrawal can reduce death rates rapidly. A further complication is that even if the biological effects are rapid, the process of implementing policies may be prolonged.
We also sought to link timing to global targets, for policy coherence. Yet these targets themselves vary and are debated.72 We have taken a technical approach to targets that are set as part of a political process. We hope this will help inform the next iteration of the global discussion on accountable targets: the 2025 Political Declaration of the forthcoming UN High Level Meeting on NCDs and Mental Health.
Conclusions and future directions
Notwithstanding these limitations, our attempt to demonstrate the potential for rapid impacts on health and progress toward UN targets marks an important advance and contribution to the global effort to tackle NCDs. Further, we note that within the WHO European Region, 10 countries (specifically, Belgium, Denmark, Estonia, Israel, Kazakhstan, Luxembourg, Netherlands, Norway, Sweden, and Switzerland) have already achieved the European Programme of Work target of a 25% reduction in premature mortality from NCDs ahead of schedule. They have implemented comprehensive policies, reducing multiple risk factors, reducing preventable and treatable mortality and CVD and cancer mortality.
We note that “quick buys” should not be conflated with “quick wins.” This latter concept is commonly used to denote interventions that are easy to implement, so-called ‘low-hanging fruit’. This set of quick buys may not be easy to implement. Yet they are cost-effective means of attaining rapid population benefit.
Future research should look to expand previous research on how countries achieved this success73 as well as assess the impact of combinations of interventions. This study also draws attention to the need for more natural experiments. Future research could also consider a wider range of outcomes at different stages along the causal pathways of disease.
Contributors
GG, DS and AE conceived of the paper and design. DS implemented the search protocol and drafted the paper. MM contributed to interpreting the data. GG, DS, AE and MM contributed to developing the search protocol, inclusion criteria, and drafting and editing the manuscript. AC, MC, JF, CF, DK, ML, MN, IR, ET, KW and JW contributed to review and editing of the manuscript.
Declaration of interests
DS received support from the WHO to conduct the research. All other authors have no conflicts of interest to declare.
Acknowledgements
This paper is part of WHO/EURO's RACE to the Finish, an effort to accelerate the uptake of effective policy interventions and support countries to improve the health and wellbeing of their populations as well as achieve their SDG commitment to reduce premature mortality from NCDs by one third by 2030. This work is financially supported by a grant from the governments of Denmark, Estonia, Finland, France, Iceland, Ireland, Latvia, Lithuania, Luxembourg, Norway, Slovenia and Sweden (Award number: 75511).
Appendix A. Supplementary data
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Appendix Figures.
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2025, International Journal of Drug Policy Show abstract Digital labelling through quick-response (QR) codes is increasingly proposed by the alcohol producers to inform consumers without significantly modifying their original labels. Current alcohol labelling policy discourse in Europe often revolves around the advantages and disadvantages of digital labelling, yet evidence on actual practices and the information provided through QR codes is lacking. This study aimed to assess the presence of QR codes on alcohol labels in 13 European countries, as well as the accessibility and content of the information behind those. The study methodology was developed with the EVID-ACTION Youth Network, with members sampling stores and products based on a co-developed mapping protocol. Thirty-four stores in 25 cities across 13 European countries were visited between April and August 2024. Descriptive analysis was conducted by beverage type and country. Of 1815 products examined, 31 % contained QR codes (23 % beers, 37 % wines, 30 % spirits). Most QR codes (84 %) were positioned on the back of containers, and 61 % had no explanatory text about their purpose. Most accessed websites were in local languages (75 %), with 36 % requiring age information to enter. Websites most commonly contained brand/drink information (46 %), followed by nutritional information (42 %), health information (42 %), and ingredient information (41 %). Almost a third of alcoholic beverages in 13 European countries contained QR code. However, most labels did not specify the purpose of the code. Furthermore, given the website content, there is concern that they serve as a promotional tool rather than providing access to nutritional and risk information.
### Healthy choices, healthy futures: promoting healthy nutrition to curb non-communicable diseases
2025, Trends in Endocrinology and Metabolism Show abstract Non-communicable diseases (NCDs) pose a serious challenge to global public health, compounded by unhealthy nutrition and obesogenic environments. This science and society article emphasizes several evidence-based behavior-change strategies, policy interventions, and meaningful youth engagement to promote healthy nutrition choices among children and adolescents.
© 2025 Published by Elsevier Ltd.
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How to Find the Focus, Vertex, and Directrix of a Parabola?
You can easily find the focus, vertex, and directrix from the standard form of a parabola.
A parabola consists of three parts: Vertex, Focus, and Directrix. The vertex of a parabola is the maximum or minimum of the parabola and the focus of a parabola is a fixed point that lies inside the parabola. The directrix is outside of the parabola and parallel to the axis of the parabola.
Related Topic
How to Write the Equation of Parabola
Step-by-Step Guide to Finding the Focus, Vertex, and Directrix of a Parabola
The standard form of Parabola when it opens up or down is (𝑥−ℎ)2=4 𝑝(𝑦−𝑘)(x−h)2=4 p(y−k), where the focus is ℎ,𝑘+𝑝 h,k+p and the directrix is 𝑦=𝑘−𝑝 y=k−p.
The standard form of Parabola when it opens right or left is (𝑦+𝑘)2=4 𝑝(𝑥−ℎ)(y+k)2=4 p(x−h), where the focus is ℎ+𝑝,𝑘 h+p,k and the directrix is 𝑥=ℎ−𝑝 x=h−p.
For a Parabola in the form 𝑦=𝑎 𝑥 2+𝑏 𝑥+𝑐 y=a x 2+b x+c:
Vertex: (−𝑏 2 𝑎,4 𝑎 𝑐−𝑏 2 4 𝑎)(−b 2 a,4 a c−b 2 4 a), Focus: (−𝑏 2 𝑎,4 𝑎 𝑐−𝑏 2+1 4 𝑎)(−b 2 a,4 a c−b 2+1 4 a), Direcrix: 𝑦=𝑐−(𝑏 2+1)4 𝑎 y=c−(b 2+1)4 a.
Finding the Focus, Vertex, and Directrix of a parabola – Example 1:
Find the vertex and focus of this parabola: 𝑦=3 𝑥 2+6 𝑥 y=3 x 2+6 x
Solution:
The Parabola given parameters are: 𝑎=3,𝑏=6,𝑐=0 a=3,b=6,c=0
Substitute the values in vertex formula: (−𝑏 2 𝑎,4 𝑎 𝑐−𝑏 2 4 𝑎)=(−6 2(3),4(3)(0)−6 2 4(3))(−b 2 a,4 a c−b 2 4 a)=(−6 2(3),4(3)(0)−6 2 4(3))
Therefore, the vertex of the parabola is (−1,3)(−1,3).
To find the focus of the parabola, substitute the values in the focus formula: (−𝑏 2 𝑎,4 𝑎 𝑐−𝑏 2+1 4 𝑎)=(−6 2(3),4(3)(0)−6 2+1 4(3))(−b 2 a,4 a c−b 2+1 4 a)=(−6 2(3),4(3)(0)−6 2+1 4(3))
Focus of parabola is (−1,−35 12)(−1,−35 12).
Exercises for Finding the Focus, Vertex, and Directrix of Parabola
Find the vertex and focus of each parabola.
(𝑦−2)2=3(𝑥−5)2(y−2)2=3(x−5)2
𝑦=4 𝑥 2+𝑥−1 y=4 x 2+x−1
𝑦=𝑥 2+2 𝑥+3 y=x 2+2 x+3
𝑥=𝑦 2−4 𝑦 x=y 2−4 y
𝑉 𝑒 𝑟 𝑡 𝑒 𝑥:(5,2),V e r t e x:(5,2),𝑓 𝑜 𝑐 𝑢 𝑠:(5,25 12)f o c u s:(5,25 12)
𝑉 𝑒 𝑟 𝑡 𝑒 𝑥:(−1 8,−17 16),𝑓 𝑜 𝑐 𝑢 𝑠:(−1 8,−1)V e r t e x:(−1 8,−17 16),f o c u s:(−1 8,−1)
𝑉 𝑒 𝑟 𝑡 𝑒 𝑥:(−1,2),𝑓 𝑜 𝑐 𝑢 𝑠:(−1,9 4)V e r t e x:(−1,2),f o c u s:(−1,9 4)
𝑉 𝑒 𝑟 𝑡 𝑒 𝑥:(−4,2),𝑓 𝑜 𝑐 𝑢 𝑠:(−15 4,2)V e r t e x:(−4,2),f o c u s:(−15 4,2)
by: Effortless Math Team about 4 years ago (category: Articles)
Effortless Math Team
3 weeks ago
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11886 | https://www.chegg.com/homework-help/questions-and-answers/part-water-outdoor-fountain-follows-path-shaped-like-parabola-arch-created-water-32-inches-q66174931 | Solved Part A The water from an outdoor fountain follows a | Chegg.com
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Part A The water from an outdoor fountain follows a path that is shaped like a parabola. The arch created by the water is 32 inches wide and 27 inches high Use the description of the fountain to label the three points on the graph, modeling the path of the water. y 25 20 15 10 00 5 10 15 20 25 Part B Find a quadratic function, W. that models the water
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Question: Part A The water from an outdoor fountain follows a path that is shaped like a parabola. The arch created by the water is 32 inches wide and 27 inches high Use the description of the fountain to label the three points on the graph, modeling the path of the water. y 25 20 15 10 00 5 10 15 20 25 Part B Find a quadratic function, W. that models the water
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To approach Part A, identify the three key points provided on the graph: the vertices at (0,0), (16,27), and (32,0). vertex form of the parabola is given by here vertex is ..................…
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Part A The water from an outdoor fountain follows a path that is shaped like a parabola. The arch created by the water is 32 inches wide and 27 inches high Use the description of the fountain to label the three points on the graph, modeling the path of the water. y 25 20 15 10 00 5 10 15 20 25 Part B Find a quadratic function, W. that models the water height 1 Part If the arch were 32 inches wide but 44 inches high, how could you modify your function IV to model the new archi? Explain or show your reasoning.
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11887 | https://khanacademy.fandom.com/wiki/Factor_polynomials_with_special_product_forms | Factor polynomials with special product forms | Khan Academy Wiki | Fandom
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in:Math exercises, Algebra II exercises, Algebra II: Polynomial expressions, equations, and functions, Mathematics III exercises
Factor polynomials with special product forms
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| Factor polynomials with special product forms |
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| Description |
| Exercise Name: | Factor polynomials with special product forms |
| Math Missions: | Algebra II Math Mission, Mathematics III Math Mission |
| Types of Problems: | 1 |
The Factor polynomials with special product forms exercise appears under the Algebra II Math Mission and Mathematics III Math Mission. This exercise practices factoring "advanced" polynomials (i.e. polynomials of various degrees and or with two variables) using special product factorization methods.
Types of Problems[]
There is one type of problem in this exercise:
Find the special product form, then find the equivalent expression: This problem provides a quadratic expression. The student is expected to find the special product form, then find the equivalent expression with the same special product form. Students should note that a{\displaystyle {a}} and b{\displaystyle {b}} represent either positive rational numbers or terms with positive coefficients.
Find the special product form, then find the equivalent expression
Strategies[]
Knowledge of any quadratic factoring technique can be used to ensure accuracy and efficiency on this exercise.
One such process for factoring a quadratic such as this is to call the coefficient to the x{\displaystyle {x}} the "adding" term and the product of the constant and the leading coefficient to be called the "multiplying" term. Then find two numbers that add to the "adding" number and multiply to the "multiplying" number.
Another possible technique is to use the quadratic formula to find the roots and use these to determine the factoring.
Real-life Applications[]
Polynomials can be used via power series to represent many complicated functions and calculus is simpler to perform on polynomials than other functions. Thus any comfort with polynomials should increase calculus ability.
Factoring polynomials is an important skill in solving inequalities and equations with polynomials.
Factoring polynomials can be used for solving equations with polynomials, such as those that appear in many business models.
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11889 | https://www.crashwhite.com/apphysics/materials/presentations/sem2-ch13/index.html | AP Physics - Unit 4 - Chapter 13
Chapter 13: Electromagnetic Induction
13.0. Overview
Up to this point, we've seen electric fields produced by electric charges:
... and magnetic fields produced by moving charges:
Here's a question: If currents produce magnetic fields, can magnetic fields produce currents?
Let's get started!
Faraday's Law of Induction
Applications
Motional EMF
Eddy currents
13.1. Faraday's Law of Induction
The English scientist and experimentalist Michael Faraday, in 1831, designed an experiment to demonstrate that a changing magnetic field produces a flow of charge (a current). Independently, the American scientist Joseph Henry discovered the same effect in 1832.
Michael Faraday's Experiment
The experimental setup shown here demonstrates the principal of electromagnetic induction:
On the left side of the diagram, when the switch is closed, a battery causes a current to flow in the circuit. The coil of wire wrapped around the left side of the torus didn't have any current in it at first, and thus had no magnetic field. After closing the switch, there is now a magnetic field in the coil wrapped around the torus.
A changing magnetic field in the torus results in a change in magnetic flux in the torus.
On the right side of the diagram, the coil of wire wrapped around the torus experiences a change in magnetic flux, which produces an EMF in the circuit on the right, as described by Faraday's Law of Electromagnetic Induction.
Note that the EMF is maintained only as long as the magnetic flux keeps changing.
Definition: Faraday's Law of Induction
The emf ε induced in a circuit is directly proportional to the time-rate of change of magnetic flux through the circuit.
If we want to induce a flow of current, we need to change the magnetic flux, which we can do by changing B, A or θ. For some conducting loops such as a coil of wire, N represents the number of coils in the loop, which has the effect of increasing the net area of the loop.
Changing magnetic field
A coil is wrapped with 200 turns of wire on the perimeter of a square frame of sides 18 cm. The total resistance of the coil is 2.0 ohms. A uniform magnetic field is turned on perpendicular to the plane of the coil.
If the field changes linearly from 0 to 0.50 Wb/m 2 in a time of 0.80s, find the magnitude of the induced emf in the coil while the field is changing.
Find the magnitude of the current induced in the coil while the field is changing.
Solution)
Area = 0.0324 m 2 , emf = 4.1 V
2.05 A
13.1.1. Lenz's Law
Faraday’s Law allows us to calculate the magnitude of an induced emf (and thus current), but to determine direction of current flow, we use Lenz's Law.
Definition: Lenz's Law
Lenz's Law states that the polarity of the induced emf is such that it produces a current that will create a magnetic flux to oppose the change in magnetic flux through the loop.
In the diagram here of a conducting hoop, what happens if the magnetic field, pointing to the right, starts to decrease?
There is a magnetic field through the conductor that is producing a flux through the area of the loop.
As the field decreases, so does the flux.
The changing flux produces an emf with a current, but in which direction?
Lenz's law predicts an induced current that has its own magnetic field, opposing the change in flux. What does that mean?
If flux is decreasing for the magnetic field to the right, the induced current is going to oppose that decreasing flux. It will do so if the induced current has a magnetic field in the same direction as B.
The direction of current that has a magnetic field in the same direction as B would be (according to the RHR) down in front and up in back. Therefore, that is the direction of the current caused by the induced emf.
Practicing Lenz's Law
In each of the following diagrams, determine the direction the current for the emf induced in the conducting loop.
Magnet stationary with magnetic field to the right, conducting loop (green) moving to the right
2. Magnet stationary with magnetic field to the right, conducting loop (green) moving to the left
3. Magnet stationary with magnetic field to the left, conducting loop (green) moving to the left
4. Rotating conducting loop, loop area in B field getting smaller
5. Rotating conducting loop, loop area in B field getting larger
Solutions)
Down in front, up in back
Up in front, down in back
Down in front, up in back
Down in front, up in back
Up in front, down in back
More Lenz's Law Practice
Examine the following diagram. Find the direction of the induced current in the loop:
at the instant the switch is closed
after the switch has been closed for several seconds
when the switch has been opened again
Solution)
When switch is closed, current begins to flow in loop, creating increasing magnetic field to left. To oppose this change in flux, current in the loop is induced in a "down in front, up in back" direction.
When switch has been closed for several seconds, no more current flows in the loop, because there is no more change in magnetic flux.
When the switch is opened again, the magnetic field to the left begins to decrease. Loop current opposes this decrease by having a direction that is in the same direction as the magnetic field, ie."up in front, down in back."
13.1.2. Induced EMF and Electric Field
We've determined that a changing magnetic flux induces an emf and a resulting current flow in a conducting loop, so... an electric field must be present in the conductor. We can't analyze that electric field using our electrostatic analysis because in those situations, the electric field E was conservative. Here, as charges move around in the circle, , so the induced electric field E is non-conservative.
Definition: Faraday's Law, Revised
A changing magnetic flux through a conductor produces an emf that is associated with a time-varying, non-conservative electric field E.
Time-varying electric field
For a conducting loop of radius r exposed to a time-changing magnetic field B oriented perpendicular to the plane of the loop, determine the magnitude of the changing electric field E in the conductor.
Solution)
13.2. Applications
Electromagnetic induction isn't just an esoteric physics topics—it is used dozens of applications, from your electric toothbrush to your phone charger, from the outlets in the walls of your house to electric guitars.
You wouldn't even be sitting in a well-lit room reading this on a computer without Faraday's Law of Induction providing you with the power to light your lights and operate your computer.
Here are some ways that law works in some devices around us.
13.2.1. Ground Fault Interrupter
A ground-fault interrupt (GFI) is a strategy for cutting off the current to a circuit in an emergency.
1. Ordinarily, there is no net current enclose by gray loop, so there is no magnetic field around the wires, and the green coil wrapped around the grey loop has no emf or current.
2. If there’s a ground fault (current going through the coil and into a person, for example, but not coming back through the circuit), Ampere's Law results in a magnetic field in the gray loop, increasing flux, and inducing an emf in the green coil that activates a circuit breaker, cutting the circuit.
Typically, a GFI will cut the power to a circuit within 25-40 milliseconds, preventing significant amounts of current from entering a person's body, where it risks interfering with the fibrillation activity of the heart.
13.2.2. Electric guitar pickups
An acoustic or classical guitar has vibrating strings that cause the hollow-body of the guitar to resonate as well, which physically moves air molecules in an area that are picked up as sound by ears.
Some acoustic/classical guitars have a small microphone that picks up the sound waves, converts those vibrations to electrical signals, and delivers them to an amplifier which moves a large speaker at the same rate as the guitar string, increasing the volume of the sound without changing its frequency.
An electric guitar, however, has "pickups" that convert the sound to electrical signals for transmission to the amplifier.
1. A small magnet in the pickup sits below each metal string, magnetizing the section of the string just above it.
2. When the magnetized string begins to vibrate, a nearby coil experiences a change in magnetic flux at the same frequency as the string. This changing flux produces an emf/current which is transmitted to the amplifier.
13.3. Motional EMF
Related to the idea of an emf produced by a changing magnetic flux, there is an emf produced by the motion of a conductor through a magnetic field, or motional emf.
Definition: Motional EMF
When we move a conductor through a magnetic field, motional emf occurs. Consider the conducting rod moving through a magnetic field as shown here. What happens to charges in the conductor as the rod moves to the right?
"Free charges" in the conductor have a velocity as a result of being moved through the magnetic field, so they have a velocity v, and thus experience a magnetic force:
By the Right-Hand Rule, this causes positive charges to be forced upwards in the rod, with negative charge downwards. This separation of charges causes there to be an electric field set up in the rod, away from positive charges (downward), and this field applies an electrical force to counter the magnetic force.
Charges will continue to migrate upwards due to the magnetic force until the magnetic force up and the electric force down are in dynamic equilibrium.
At this point there is an electric potential difference between the upper and lower ends of the rod that we can calculate:
This is an interesting result that doesn't require any knowledge of magnetic flux to understand. But let's see what happens to that same example if we set the conducting bar as a moving component of a circuit.
Moving conductor in a circuit, part 1
This conductor, moving to the right with a velocity v is sliding on conducting rails that are connected by a resistor, making a conducting loop. According to Faraday's Law of Induction, changing flux through that loop will induce an emf. Identify the emf in the circuit as a function of the quantities given.
Solution)
Notice that this result is the same that we arrived at in using motional emf in the previous example.
Moving conductor in a circuit, part 2 - Lenz's Law
What direction is the current flowing in this loop?
Solution)
Determine direction of magnetic field: "into the page"
Determine whether flux in the area is increasing or decreasing: "increasing" as the area gets larger
Use RHR, with thumb pointing in opposite direction of increase—"out of the page"—to determine direction of current flow: "clockwise", or "up" in the bar.
As we're moving that bar with a velocity v across the magnetic field, the emf causes a current to move through the bar... and a current-carrying conductor in a magnetic field experiences a magnetic force!
Let's try to calculate how much force, how much Power it takes to make it all happen, and how much Power gets dissipated by the circuit.
Moving conductor in a circuit, part 3
As we move the current-carrying bar through the magnetic field, there’s a magnetic force that acts on it. If we move the bar with a constant velocity v across conducting rails with negligible friction, calculate
the magnetic force acting on the bar
the power we're using to move the bar
the power dissipated by the circuit
Solution)
We've already determined that the emf induced in the circuit is Blv. With that in mind:
The magnetic force acting on the bar is
the power we're using to move the bar
the power dissipated by the circuit is
Note that the Power applied to the experiment and the Power dissipated by the circuit are the same! Conservation of Energy!
Let's see one more interesting application of this experimental set-up.
Give it a shove
A bar of mass m and length l is given an initial velocity v 0 and released so that it slides on two parallel frictionless rails in the presence of a magnetic field B as shown, with load resistance R attached. Although there is no friction force opposing the motion, there is a magnetic force in the opposite direction of v. Calculate v as a function of time t.
Solution)
The magnetic force acting on the current induced in the bar acts in the opposite (negative) direction of the velocity:
Manipulate the equations to isolate v on one side and t on the other, and then integrate with respect to those two variables:
or:
13.4. Eddy Currents
If changing magnetic flux produces emf and currents in a circuit, those same changes in magnetic flux can induce eddy currents (circulating free charges in a bulk metal moving through a magnetic field).
Eddy currents
In the example shown here, a flat conducting rectangle is attached to a pendulum arm, and is swinging down into a gravity field. As free charges in the metal enter the magnetic field, they are forced upwards in the conductor, resulting in a circular flow of charge in the rectangle. The current in the part of the conductor that is in the magnetic field is subjected to a magnetic force according to F = Il × B, in opposition to the motion of the pendulum's arm.
What happens as the flat conducting rectangle begins to leave the magnetic field on the opposite side?
Because eddy currents produce retarding forces, they can be used in as braking systems for mass transit systems. If eddy currents are not desired, then the bulk metal is often split into thin layers that are laminated together. |
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A comparison of two numbers by division. A quotient used to compare two or more quantities of the same units of measure. A statement of the relative size of two quantities (numbers, functions, and so on), expressed as a quotient.
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How can we show n⊂m→n∈m∨n=m n⊂m→n∈m∨n=m?
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I want to prove that for any natural numbers n,m n,m it holds that:
n⊂m↔n∈m∨n=m n⊂m↔n∈m∨n=m
"⇐""⇐": If n∈m∨n=m n∈m∨n=m, using the sentence: For any natural numbers m,n m,n it holds that n∈m→n⊂m n∈m→n⊂m, we conclude that n⊂m n⊂m.
"⇒""⇒":Could we justify it like that?
n⊂m→n=m∨n⊊m n⊂m→n=m∨n⊊m
When n=m n=m, the desired property is satisfied.
When n⊊m n⊊m How could we show that n=m∨n⊊m n=m∨n⊊m ? Could we show it using induction? If so, how could we do this?
elementary-set-theory
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edited Jan 4, 2015 at 9:39
evindaevinda
asked Dec 6, 2014 at 21:49
evindaevinda
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1 Careful here! m⊂m m⊂m but {m}∉m{m}∉m. You mean n⊊m→{n}∈m n⊊m→{n}∈m AlexR –AlexR 2014-12-06 21:52:08 +00:00 Commented Dec 6, 2014 at 21:52
@AlexR How else could we show the implication?evinda –evinda 2014-12-06 21:53:37 +00:00 Commented Dec 6, 2014 at 21:53
1 You're on the right track. Make a trivial case n=m n=m and a nontrivial case n⊊m n⊊m ;)AlexR –AlexR 2014-12-06 21:54:44 +00:00 Commented Dec 6, 2014 at 21:54
1 No we can't but we can conclude from {n}∈m{n}∈m that n∈m n∈m by construction of m m.AlexR –AlexR 2014-12-06 22:06:32 +00:00 Commented Dec 6, 2014 at 22:06
1 It follows inductively from the successor function S(n)={n,{n}}S(n)={n,{n}}AlexR –AlexR 2014-12-06 22:14:08 +00:00 Commented Dec 6, 2014 at 22:14
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Lemma 1: Every natural number is a transitive set.
Proof: Induction. 0=∅0=∅, so it's trivially transitive. Assume n n is transitive. Then S(n)=n∪{n}S(n)=n∪{n} is transitive: let k∈m∈S(n)k∈m∈S(n). Then either m∈n m∈n or m=n m=n. If m∈n m∈n, then since n n is transitive, k∈n⊊S(n)k∈n⊊S(n), so k∈S(n)k∈S(n). Otherwise, m=n m=n, in which case k∈n k∈n directly, and thus k∈S(n)k∈S(n). By induction, every natural number is transitive.
Note: Being transitive is equivalent to all elements also being subsets.
Lemma 2: No natural number is an element of itself.
Proof: Induction. 0=∅0=∅ and ∅∉∅∅∉∅. Now assume n∉n n∉n. S(n)=n∪{n}S(n)=n∪{n}. If S(n)∈S(n)S(n)∈S(n), then either S(n)=n S(n)=n or S(n)∈n S(n)∈n. If S(n)=n S(n)=n, then {n}⊆n⟹n∈n{n}⊆n⟹n∈n, contradicting the inductive hypothesis. If S(n)∈n S(n)∈n, then since n∈S(n)n∈S(n), and since n n is transitive, we have n∈n n∈n, again contradicting the inductive hypothesis. Thus, no natural number is an element of itself.
Lemma 3: ω ω is totally ordered by ∈∈.
Proof: By Lemma 1, ∈∈ gives a transitive relation on ω ω. By Lemma 2, this relation is irreflexive. Thus, we need only show that for any n,m∈ω n,m∈ω, at least one of m∈n,m=n,n∈m m∈n,m=n,n∈m is true (transitivity and irreflexivity prevent more than one from being true). We fix m m and induct on n n. Firstly, 0∈m 0∈m or 0=m 0=m. Next, if n∈m n∈m, then S(n)∈m S(n)∈m or (n)=m(n)=m. If n=m n=m, then m∈S(n)m∈S(n). Finally, if m∈n m∈n, then m∈n∈S(n)⟹m∈S(n)m∈n∈S(n)⟹m∈S(n), since S(n)S(n) is transitive. Thus, m m is comparable to any natural number. Since m m was arbitrary, any two natural numbers are comparable, so ω ω is totally ordered by ∈∈.
Theorem: If n,m∈ω n,m∈ω, then n⊊m⟹n∈m n⊊m⟹n∈m.
Assume n⊊m n⊊m. We seek to show that n∈m n∈m. Since ω ω is totally ordered by elementhood, we have either n∈m n∈m or m∈n m∈n. If m∈n m∈n, then since n n is transitive, all elements of n n, including m m, are subsets, so n⊊m⊆n⟹n⊊n n⊊m⊆n⟹n⊊n, a contradiction.
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answered Dec 7, 2014 at 3:42
NishantNishant
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This is more elegant, I think (+1).AlexR –AlexR 2014-12-07 13:02:11 +00:00 Commented Dec 7, 2014 at 13:02
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Proposition 1: {n}∈m→n∈m{n}∈m→n∈m
Proof: For m=∅m=∅ we are okay since there is no {n}∈m{n}∈m.
Now let {n}∈m→n∈m{n}∈m→n∈m. We must show that {n}∈S(m)→n∈S(m){n}∈S(m)→n∈S(m).
S(m)=m∪{m}S(m)=m∪{m}
Thus {n}∈S(m)→{n}∈m∨{n}={m}{n}∈S(m)→{n}∈m∨{n}={m}. In the first case we have by condition n∈m n∈m and since m⊂S(m)m⊂S(m), n∈S(m)n∈S(m). The second case {n}={m}→n=m→n∈S(m){n}={m}→n=m→n∈S(m), thus n∈S(m)n∈S(m). □◻
Proposition 2: n⊂m→n∈m∨n=m n⊂m→n∈m∨n=m
Proof: If n=m n=m then clearly n⊂m n⊂m, now assume n⊂m n⊂m and n≠m n≠m, i.e. n⊊m n⊊m. Now if n⊊m n⊊m then {n}∈m{n}∈m as you said. Finally by Proposition 1, n∈m n∈m.
Alternative last step without Proposition 1: Since n⊊m n⊊m, n∪{n}=S(n)⊂m n∪{n}=S(n)⊂m so especially n∈S(n)⊂m n∈S(n)⊂m, i.e. n∈m n∈m as claimed.
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edited Dec 7, 2014 at 12:58
answered Dec 6, 2014 at 22:25
AlexRAlexR
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Alex R Couldn't we also justify it like that: n⊂m↔n=m∨n⊊m n⊂m↔n=m∨n⊊m When n=m n=m, the desired property is satisfied. When n⊊m n⊊m, then {n}∈m→{n}⊂m→n∈m{n}∈m→{n}⊂m→n∈m. Or am I wrong?evinda –evinda 2014-12-06 22:51:12 +00:00 Commented Dec 6, 2014 at 22:51
1 @evinda That's what I did but with reasoning why those →→ hold.AlexR –AlexR 2014-12-06 22:52:41 +00:00 Commented Dec 6, 2014 at 22:52
Do these →→ have to be proven, or are they known?evinda –evinda 2014-12-06 22:54:34 +00:00 Commented Dec 6, 2014 at 22:54
1 @evinda The {n}∈m→{n}⊂m{n}∈m→{n}⊂m is nontrivial. It's exactly my proposition 1.AlexR –AlexR 2014-12-06 22:55:51 +00:00 Commented Dec 6, 2014 at 22:55
1 I think S(n):=n∪{n}S(n):=n∪{n}.Nishant –Nishant 2014-12-07 03:43:11 +00:00 Commented Dec 7, 2014 at 3:43
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11892 | https://encyclopediaofmath.org/wiki/Z-transform | Z-transform - Encyclopedia of Mathematics
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Z-transform
From Encyclopedia of Mathematics
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2020 Mathematics Subject Classification: Primary:05A15 [MSN][ZBL]
Z-transformation
This transform method may be traced back to A. De Moivre [a5] around the year 1730 when he introduced the concept of "generating functions" in probability theory. Closely related to generating functions is the Z-transform, which may be considered as the discrete analogue of the Laplace transform. The Z-transform is widely used in the analysis and design of digital control, and signal processing [a4], [a2], [a3], [a6].
The Z-transform of a sequence x(n)x(n), n∈Z n∈Z, that is identically zero for negative integers, is defined as
x~(z)=Z(x(n))=∑j=0∞x(j)z−j,(a1)(a1)x~(z)=Z(x(n))=∑j=0∞x(j)z−j,
where z z is a complex number.
By the root test, the series (a1) converges if |z|>R|z|>R, where R=limsup n→∞|x(n)|1/n R=limsup n→∞|x(n)|1/n. The number R R is called the radius of convergence of the series (a1).
[x]
Contents
1 Example 1.
2 Example 2.
3 Properties of the Z-transform.
4 Inverse Z-transform.
4.1 Example.
5 Pairs of Z-transforms.
5.1 References
Example 1.
The Z-transform of {a n}{a n} is given by
Z(a n)=∑j=0∞a j z−j=z z−a for|z|>1.Z(a n)=∑j=0∞a j z−j=z z−a for|z|>1.
Example 2.
The Z-transform of the Kronecker-delta sequence
δ k(n)={1 0 if n=k if n≠k,δ k(n)={1 if n=k 0 if n≠k,
is given by
Z(δ k(n))=∑j=0∞δ k(j)z−j=z−k for all z.Z(δ k(n))=∑j=0∞δ k(j)z−j=z−k for all z.
Properties of the Z-transform.
i) Linearity: Let R 1 R 1 and R 2 R 2 be the radii of convergence of the sequences x(n)x(n) and y(n)y(n). Then for any α,β∈C α,β∈C,
Z(α x(n)+β y(n))=α Z(x(n))+β Z(y(n)),Z(α x(n)+β y(n))=α Z(x(n))+β Z(y(n)),
for|z|>max{R 1,R 2}.for|z|>max{R 1,R 2}.
ii) Shifting: Let R R be the radius of convergence of Z(x(n))Z(x(n)). Then, for k∈Z+k∈Z+,
a) Right-shifting: Z[x(n−k)]=z−k Z(x(n))Z[x(n−k)]=z−k Z(x(n)), for |z|>R|z|>R;
b) Left-shifting: Z(x(n+k))=z k Z(x(n))−∑k−1 r=0 x(r)z k−r Z(x(n+k))=z k Z(x(n))−∑r=0 k−1 x(r)z k−r, for |z|>R|z|>R.
iii) Initial and final value.
a) Initial value theorem: lim|z|→∞x~(z)=x(0)lim|z|→∞x~(z)=x(0);
b) Final value theorem: x(∞)=lim n→∞x(n)=lim z→1(z−1)Z(x(n))x(∞)=lim n→∞x(n)=lim z→1(z−1)Z(x(n)).
iv) Convolution: The convolution of two sequences x(n)x(n) and y(n)y(n) is defined by
x(n)∗y(n)=∑j=0 n x(n−j)y(j)=∑j=0 n x(n)y(n−j)x(n)∗y(n)=∑j=0 n x(n−j)y(j)=∑j=0 n x(n)y(n−j)
and its Z-transform is given by
Z(x(n)∗y(n))=Z(x(n)).Z(y(n)).Z(x(n)∗y(n))=Z(x(n)).Z(y(n)).
Inverse Z-transform.
If x~(z)=Z(x(n))x~(z)=Z(x(n)), then the inverse Z-transform is defined as Z−1(x~(z))=x(n)Z−1(x~(z))=x(n). Notice that by Laurent's theorem [a1] (cf. also Laurent series), the inverse Z-transform is unique [a2]. Consider a circle c c centred at the origin of the z z-plane and enclosing all the poles of z n−1 x~(z)z n−1 x~(z). Then, by the Cauchy integral theorem[a1], the inversion formula is given by
x(n)=1 2 π i∮c x˜(z)z n−1 d z x(n)=1 2 π i∮c x~(z)z n−1 d z
and by the residue theorem (cf. also Residue of an analytic function) [a1], x(n)=∑(residues of z n−1 x~(z))x(n)=∑(residues of z n−1 x~(z)).
If x~(z)z n−1=h(z)/g(z)x~(z)z n−1=h(z)/g(z) in its reduced form, then the poles of x~(z)z n−1 x~(z)z n−1 are the zeros of g(z)g(z).
a) If g(z)g(z) has simple zeros, then the residue K i K i corresponding to the zero z i z i is given by
K i=lim z→z i[(z−z i)h(z)g(z)].K i=lim z→z i[(z−z i)h(z)g(z)].
b) If g(z)g(z) has multiple zeros, then the residue K i K i at the zero z i z i with multiplicity r r is given by
K i=1(r−1)!lim z→z i d n d z r−1[(z−z i)r h(z)g(z)].K i=1(r−1)!lim z→z id n d z r−1[(z−z i)r h(z)g(z)].
The most practical method of finding the inverse Z-transform is the use of partial-fractions techniques as illustrated by the following example.
Example.
See also [a2]. Suppose the problem is to solve the difference equation
x(n+4)+9 x(n+3)+30 x(n+2)+20 x(n+1)+24 x(n)=0,(1)(1)x(n+4)+9 x(n+3)+30 x(n+2)+20 x(n+1)+24 x(n)=0,
where x(0)=0 x(0)=0, x(1)=0 x(1)=0, x(2)=1 x(2)=1, x(3)=10 x(3)=10.
Taking the Z-transform yields
Z(x(n))=z(z−1)(z+2)3(z+3)=Z(x(n))=z(z−1)(z+2)3(z+3)=
=−4 z z+2+4 z(z+2)2−3 z(z+2)3+4 z z+3.=−4 z z+2+4 z(z+2)2−3 z(z+2)3+4 z z+3.
Taking the inverse Z-transform of both sides yields
x(n)=(3 4 n 2−11 4 n−4)(−2)n+4(−3)n.x(n)=(3 4 n 2−11 4 n−4)(−2)n+4(−3)n.
Pairs of Z-transforms.
x(n)x(n)Z(x(n))Z(x(n)) a n a n z/z−a z/z−a n k n k k!z/(z−1)k+1 k!z/(z−1)k+1 n k a n n k a n(−1)k D k(z/(z−1)(−1)k D k(z/(z−1); D=z d/d z D=z d/d z sin n w sinn w z sin w/(z 2−2 z cos w+1)z sinw/(z 2−2 z cosw+1) cos n w cosn w z(z−cos w)/(z 2−2 z cos w+1)z(z−cosw)/(z 2−2 z cosw+1) δ k(n)δ k(n)z−k z−k sinh n w sinhn w z sinh w/(z 2−2 z cosh w+1)z sinhw/(z 2−2 z coshw+1) cosh n w coshn w z(z−cosh w)/(z 2−2 z cosh w+1)z(z−coshw)/(z 2−2 z coshw+1).
References
[a1]R.V. Churchill, J.W. Brown, "Complex variables and applications" , McGraw-Hill (1990)
[a2]S. Elaydi, "An introduction to difference equations" , Springer (1999) (Edition: Second)
[a3]A.J. Jerri, "Linear difference equations with discrete transform methods" , Kluwer Acad. Publ. (1996)
[a4]E. Jury, "Theory and application of the z-transform method" , Robert E. Krieger (1964)
[a5]A. De Moivre, "Miscellanew, Analytica de Seriebus et Quatratoris" , London (1730)
[a6]A.D. Poularikas, "The transforms and applications" , CRC (1996)
How to Cite This Entry:
Z-transform. Encyclopedia of Mathematics. URL:
This article was adapted from an original article by S. Elaydi (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article
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The Posterior Belly of Digastric Muscle as the Landmark in Facial Nerve Anastomosis Surgery: Anatomical Study and Case Illustration
Affiliations
The Posterior Belly of Digastric Muscle as the Landmark in Facial Nerve Anastomosis Surgery: Anatomical Study and Case Illustration
Authors
Affiliations
Abstract
Objective:
To establish a new method for fast exposure of the facial nerve and hypoglossal nerve during facial nerve anastomosis surgery.
Methods:
Dissection of 12 formalin-fixed cadaveric specimens was performed to explore the positional relationship between the posterior belly of digastric muscle (PBD) and the facial nerve and hypoglossal nerve. We retrospectively reviewed patients who underwent facial nerve reconstruction surgery between 2015 and 2020 at Xuanwu Hospital, Capital Medical University, and the optimized surgical strategy based on the PBD was proposed.
Results:
The trunk of the hypoglossal nerve runs across the external carotid artery after giving off the descendens hypoglossi located within the 1-cm scope deep to the junction of the tendon and belly of the PBD. The mean depth difference between the hypoglossal nerve and the junction of the tendon and belly of the PBD was 5.48 ± 2.24 mm (range, 1.88-9.27 mm). The stylomastoid foramen segment of the facial nerve was revealed after the parotid gland was dissected within the angle between the anterior margin of the mastoid tip and the inferior margin of the cartilage of the external acoustic meatus.
Conclusions:
The facial nerve and hypoglossal nerve can be rapidly identified using the PBD as an anatomical landmark. The end-to-end facial-descendens hypoglossi anastomosis is a reliable facial nerve reconstruction method for patients whose facial nerve was damaged during operation.
Keywords:
Descendens hypoglossi; Facial nerve anastomosis; Facial paralysis; Hypoglossal nerve; Posterior belly of digastric muscle.
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| Metri (m) | Nanometri (nm) |
--- |
| 1 | 1000000000 |
| 2 | 2000000000 |
| 3 | 3000000000 |
| 4 | 4000000000 |
| 5 | 5000000000 |
| 6 | 6000000000 |
| 7 | 7000000000 |
| 8 | 8000000000 |
| 9 | 9000000000 |
| 10 | 10000000000 |
| Metri (m) | Nanometri (nm) |
--- |
| 11 | 11000000000 |
| 12 | 12000000000 |
| 13 | 13000000000 |
| 14 | 14000000000 |
| 15 | 15000000000 |
| 16 | 16000000000 |
| 17 | 17000000000 |
| 18 | 18000000000 |
| 19 | 19000000000 |
| 20 | 20000000000 |
Tabella di conversione in Metri (m) per tutte le unità di Lunghezza
| Convertire dametri (m) |
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| 1 m=0.00062137119223733 mi | Miglia |
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| 1 m=39.370078740158 in | Pollici |
| 1 m=0.00053995680345572 nmi | Miglia nautiche |
| 1 m=3.2407792896349E-17 pc | Parsec |
| 1 m=1.0570008340246E-16 ly | Anni luce |
| 1 m=6.6845871222685E-12 au | Unità astronomiche |
| Convertire inmetri (m) |
| Chilometri | 1 km=1000 m |
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| Nanometri | 1 nm=1E-09 m |
| Miglia | 1 mi=1609.344 m |
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| Piedi | 1 ft=0.3048 m |
| Pollici | 1 in=0.0254 m |
| Miglia nautiche | 1 nmi=1852 m |
| Parsec | 1 pc=3.08567758131E+16 m |
| Anni luce | 1 ly=9.4607304725808E+15 m |
| Unità astronomiche | 1 au=149597870700 m |
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2025 Langley's Adventitious Angles - Wikipedia
Langley's Adventitious Angles is a geometric puzzle proposed by Edward Mann Langley in 1922, where one must deduce an angle in an isosceles triangle from given angles. The problem has variou…
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Langley's Adventitious Angles
Langley's Adventitious Angles
Langley's Adventitious Angles
is a puzzle in which one must infer an angle in a geometric diagram from other given angles. It was posed by Edward Mann Langley in
The Mathematical Gazette
in 1922.
In its original form the problem was as follows:
is an isosceles triangle with at to cuts in at to cuts in Prove
The problem of calculating angle is a standard applicationof Hansen's resection. Such calculations can establish thatis within any desired precision of , but being of only finiteprecision, always leave doubt about the exact value.A direct proof using classical geometry was developed by James Mercer in 1923.
This solution involves drawing one additional line, and then making repeated use of the fact that the internal angles of a triangle add up to 180° to prove that several triangles drawn within the large triangle are all isosceles.
Draw at to intersecting at and draw (See figure on the lower right.)Since and then and triangle is isosceles with Since and then and triangle is isosceles with Since and then triangle is equilateral.Since and then and triangle is isosceles with
The problem Solution
2 0 2 5/3/6 1 4:2 8 L a n g l e y's A d v e n t i t i o u s A n g l e s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/L a n g l e y%2 7 s _ A d v e n t i t i o u s _ A n g l e s 1/3
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Solution to Langley's 80-80-20 triangle problem adventitious quadrangles problem
Therefore all the red lines in the figure are equal.Since triangle is isosceles with Therefore
Many other solutions are possible. Cut the Knot list twelve different solutions and several alternative problems with the same 80-80-20 triangle but different internal angles.
A quadrilateral such as BCEF is called an
adventitious quadrangle
when the angles between its diagonals and sides are all rational angles, angles that give rational numbers when measured in degrees or other units for which the whole circle is a rational number. Numerous adventitious quadrangles beyond the one appearing in Langley's puzzle have been constructed.They form several infinite families and an additional set of sporadic examples.
Classifying the adventitious quadrangles (which need not be convex) turns out to be equivalent to classifying all triple intersections of diagonals in regular polygons. This was solved by Gerrit Bol in 1936 (Beantwoording van prijsvraag # 17,Nieuw-Archief voor Wiskunde 18, pages 14–66). He in fact classified (though with a few errors) all multiple intersections of diagonals in regular polygons. His results (all done by hand)were confirmed with computer, and the errors corrected, by Bjorn Poonen and Michael Rubinstein in 1998.
The article contains a history of the problem and a picture featuring the regular triacontagon and its diagonals.In 2015, an anonymous Japanese woman using the pen name"aerile re" published the first known method (the method of 3 circumcenters) to construct a proof in elementary geometry for a special class of adventitious quadrangles problem.
This work solves the first of the three unsolved problems listed by Rigby in his 1978 paper.
Langley, E. M. (1922), "Problem 644",
The Mathematical Gazette
,
11
: 173.2. Darling, David (2004),
The Universal Book of Mathematics: From Abracadabra to Zeno's Paradoxes
( A180), John Wiley & Sons, p. 180, ISBN 978047127 0478.
Generalization References
2 0 2 5/3/6 1 4:2 8 L a n g l e y's A d v e n t i t i o u s A n g l e s - W i k i p e d i a h t t p s://e n.w i k i p e d i a.o r g/w i k i/L a n g l e y%2 7 s _ A d v e n t i t i o u s _ A n g l e s 2/3
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Tr ipp, Colin (1975), "Adventitious angles",
The Mathematical Gazette
,
59
(408): 98–106,doi:10.2307/361 6644 ( 7%2F361664 4), JSTO R 3616644 ( w.jstor.org/stable/3616644).4. Bogomolny, Alexander, "The 80-80-20 Triangle" (
www.cut-the-knot.org
, retrieved 2018-06-03 5. Rigby, J. F. (1978), "Adventitious quadrangles: a geometrical approach",
The Mathematical Gazette
,
62
(421): 183–191, doi:10.2307/3616687 ( 7%2F3616687), JSTOR 3616687 ( MR 0513855 (ht tps://mathscinet.ams.or g/mathscinet-getitem?mr=05 13855).6. Poonen, Bjorn; Rubinstein, Michael (1998), "The number of intersection points made by the diagonals of a regular polygon" ( th.mit.edu/~poonen/papers/ngon.pdf) (PDF),
SIAM Journal on Discrete Mathematics
,
11
(1): 135–156,doi:10.1137/S08 954801952 81246 ( 7%2FS089 548019528 1246),S2CID 8673508 ( 508).7. Saito, Hiroshi (2016), "The adventitious quadrangles was solved completely by the elementary solution" ( e=201602),
Gendaisūgaku (現代数学)
(in Japanese),
49
(590): 66–73, ISSN 2187-6495 ( aerile_re (2015-10-27),
The last challenge from Geometry the Great
( e.org/web/2016 041602543 6/ hoo.co.jp/detail/n365238) (in Japanese), archived from the original ( 8) on 2016-04-16.9. Saito, Hiroshi (2016-12-11),
Introducing "3 circumcenter method"
( umcenter_en.html) - English tra nslation of the article from
Gendaisūgaku
(現代数学).Angular Angst ( e/kmath277/km ath277.htm),MathPages
Retrieved from "
External links
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11896 | https://pubmed.ncbi.nlm.nih.gov/21132302/ | Corticosteroids and the risk of scleroderma renal crisis: a systematic review - PubMed
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Rheumatol Int
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. 2012 Mar;32(3):645-53.
doi: 10.1007/s00296-010-1697-6. Epub 2010 Dec 4.
Corticosteroids and the risk of scleroderma renal crisis: a systematic review
Gerald Trang1,Russell Steele,Murray Baron,Marie Hudson
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Affiliation
1 McGill University, Room A-216, 3755 Cote Ste Catherine Street, Montreal, QC, H3T 1E2, Canada.
PMID: 21132302
DOI: 10.1007/s00296-010-1697-6
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Corticosteroids and the risk of scleroderma renal crisis: a systematic review
Gerald Trang et al. Rheumatol Int.2012 Mar.
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. 2012 Mar;32(3):645-53.
doi: 10.1007/s00296-010-1697-6. Epub 2010 Dec 4.
Authors
Gerald Trang1,Russell Steele,Murray Baron,Marie Hudson
Affiliation
1 McGill University, Room A-216, 3755 Cote Ste Catherine Street, Montreal, QC, H3T 1E2, Canada.
PMID: 21132302
DOI: 10.1007/s00296-010-1697-6
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Scleroderma renal crisis (SRC) has been associated with the use of corticosteroids (CS) in retrospective studies. Using an evidence-based approach, we undertook a systematic review of the literature to identify prospective studies in which scleroderma patients were administered CS to ascertain the risk of SRC in those patients. A comprehensive search was conducted using Medline, EMBASE, the Cochrane Library, and Web of Science. All original prospective clinical studies were eligible if they enrolled SSc patients newly treated with CS. Selected studies were reviewed, and data extraction was systematically performed for the dose and duration of the CS intervention as well as the occurrence of SRC. Twenty-six studies with a total of 500 SSc patients commencing new CS therapy were included in the systematic review. Ten definite cases of SRC, equivalent to a rate of 2%, were identified. In the subset of early diffuse patients, the rate of SRC was 4%. All 10 definite cases of SRC occurred in patients who received medium- to high-dose CS therapy. Seven cases occurred in the setting of stem cell transplant. CS are associated with SRC, although this may be due to confounding by disease severity and/or co-intervention. Great caution must continue to be exerted when initiating such therapy, especially in high doses and in the early diffuse subset of SSc patients.
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References
Clin Rheumatol. 2006 Mar;25(2):205-12 - PubMed
Rheumatology (Oxford). 2008 Oct;47 Suppl 5:v54-6 - PubMed
Clin Rheumatol. 2003 Oct;22(4-5):289-94 - PubMed
Mod Rheumatol. 2005;15(2):134-8 - PubMed
Int J Dermatol. 1995 Oct;34(10):726-8 - PubMed
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11897 | https://www.math.cmu.edu/~mlavrov/arml/16-17/number-theory-09-11-16-solutions.pdf | Modular Arithmetic and Divisibility Number Theory Annie Xu and Emily Zhu September 11, 2016 1 Solutions Note: in these solutions, primes are assumed to be positive!
1. Using modular arithmetic, show that 3 divides n if and only if 3 divides the sum of the digits of n. Do the same for 9. Can you find something similar for 11?
We only provide the solution for the case of 11. (The case for 3 was done in class and the case for 9 is identical.) We claim that 11 divides n if and only if 11 divides the alternating sum of the digits. We can represent n = dk10k + dk−110k−1 + · · · + d1101 + d0. Now assume that 11 divides n or 0 ≡n (mod 11). Note that 10 ≡−1 (mod 11) so: n = dk10k+dk−110k−1+· · ·+d1101+d0 ≡dk(−1)k+dk−1(−1)k−1+· · ·+d1(−1)1+d0 (mod 11) Thus, 11 divides the alternating sum of digits as well. For the converse, assume that 11 divides dk(−1)k + dk−1(−1)k−1 + · · · + d1(−1)1 + d0, the alternating sum of digits. Then: dk(−1)k+dk−1(−1)k−1+· · ·+d1(−1)1+d0 ≡dk10k+dk−110k−1+· · ·+d1101+d0 = n (mod 11) Thus, 11 divides n as well. Note that this actually proves something stronger–n is actually always congruent to its alternating sum of digits (mod 11) regardless of if n ≡0 (mod 11).
2. Find gcd(221, 299) and gcd(2520, 399).
We calculate gcd(221, 299).
gcd(221, 299) = gcd(221, 299 −221) 299 = 221 + 78 gcd(221, 78) = gcd(78, 221 −2 · 78) 221 = 2 × 78 + 65 gcd(78, 65) = gcd(65, 78 −65) 78 = 65 + 13 gcd(65, 13) = gcd(13, 65 −5 · 13) 65 = 5 × 13 + 0 = gcd(13, 0) So gcd(221, 299) = 13.
We calculate gcd(2520, 399).
gcd(2520, 399) = gcd(399, 2520 −6 · 399) 2520 = 6 × 399 + 126 gcd(399, 126) = gcd(399, 399 −3 · 126) 399 = 3 × 126 + 21 gcd(126, 21) = gcd(21, 126 −6 · 21) 126 = 6 × 21 + 0 = gcd(21, 0) So gcd(2520, 399) = 21.
1 3. 333 + 999 ≡3 + 4 ≡7 ≡2 (mod 5).
3333 × 7777 = 3 × 2 ≡6 ≡1 (mod 5).
4. How many steps does it take the Euclidean Algorithm to reach (1, 0) when the input is (n + 1, n)?
We trace through the steps of the Euclidean Algorithm for gcd(n + 1, n): gcd(n + 1, n) = gcd(n, n + 1 −n) n + 1 = (n) + 1 gcd(n, 1) = gcd(1, n −1 · n) n = 1 × n + 0 = gcd(1, 0) = 1 So we see that this takes 2 steps.
5. Let n be a positive integer. Construct a set of n consecutive positive integers that are not prime.
Let n be a positive integer. Then take (n + 1)! + 2, (n + 1)! + 3, . . . , (n + 1)! + (n + 1), which are n consecutive positive integers. Note that all of these integers are composite (not prime): for each i = 2, . . . , n + 1, i divides (n + i)! + i. Then, note that since (n+i)!
i > 0, we have that (n+i)!+i i = (n+i)!
i + 1 > 1. As such, (n + i)! + i is the product of 2 integers where neither is 1, making it composite.
6. Find all positive integers n such that (n + 1) divides (n2 + 1).
Note that if gcd(n + 1, n2 + 1) = n + 1, then (n + 1) divides (n2 + 1). We (kind of) use the Euclidean Algorithm.
gcd(n2 + 1, n + 1) = gcd(n + 1, n2 + 1 −(n −1)(n + 1)) n2 + 1 = (n −1)(n + 1) + 2 = gcd(n + 1, 2) From, here, we see that gcd(n + 1, 2) = n + 1 means that n + 1 = 2 or n = 1 is the only such n.
Alternate. Substitute m = n+1 (and note m ≥2). Then n2+1 = (m−1)2+1 = m2−2m+2.
Since we want m to divide m2 −2m + 2 and m divides m2 −2m, we need m to divide 2. As m ≥2, we must have m = 2 or n = 1.
7. Find all primes in the form n3 −1.
Note that n3−1 = (n−1)(n2+n+1). Since n2+n+1 ≥0 for all integers n. (Convince yourself of this!), if n ≤1, then n3−1 ≤0 and cannot be prime. If n ≥3 then n3−1 = (n−1)(n2+n+1) with neither factor being 1. The only remaining case is n = 2, in which n3 −1 = 7, which is prime.
8. What is the largest positive integer n for which (n + 10) divides n3 + 100?
This uses the same idea as problem 6.
Let m = n + 10 and note that m ≥11.
Then n3 + 100 = (m −10)3 + 100 = m3 −30m2 + 300m −1000 + 100 = m3 −30m2 + 300m −900.
Since m divides the first 3 terms, it remains for m to divide 900. The largest such m is then 900, making the largest n = 890.
2 9. Show that 1 . . . 1 | {z } 91 ones is composite.
We claim that 1111111 divides 1 . . . 1 | {z } 91 ones . Note that 1 . . . 1 | {z } 91 ones = 1111111 × (100 + 107 + 1014 + · · · + 1084), which gives that 1 . . . 1 | {z } 91 ones is composite.
10. A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?
We first count the number of leap years between 1812, inclusive and 2012, exclusive. Among the 200 years, 50 are divisible by 4 of which 1900 is not a leap year but 2000 is. Thus, there are 49 leap years. Then, note that there are 365 days in a non-leap year, which is 1 (mod 7) (so every year, a date is shifted by 1 as a day in the week). Letting Sunday be 0 (mod 7), then Tuesday is 2 (mod 7). If n is the day of the week when Dickens was born, note that n + 200 · 1 + 49 ≡2 (mod 7) ⇒n + 4 ≡2 (mod 7) ⇒n ≡5 (mod 7), so Dickens was born on a Friday.
11. What is the largest prime factor of 7999488?
Note that this is 8000000 −512 = 512(15625 −1) = 512(15624) = 29 × 23 × 32 × 7 × 31, so 31 is the largest prime factor.
12. An n-digit number is cute if its n digits are an arrangement of the set {1, 2, . . . , n} and its first k digits form an integer that is divisible by k, for k = 1, 2, . . . , n. For example, 321 is a cute 3-digit integer because 1 divides 3, 2 divides 32 and 3 divides 321. How many cute 6-digit numbers are there?
We begin to construct the number abcdef. Note that e = 5 since 5 must divide abcde. Note that b, d, f are some permutation of 2, 4, 6 since ab, abcd, abcdef are also divisible by 2. As such a, c are some permutation of 1, 3. Note that then since 3 divides abc, as shown in problem 1, 3 must divide a + b + c = 1 + 3 + b = 4 + b. As such, b = 2 is necessary. Then note that since 4 must divide abcd, by divisibility rules, 4 must divide cd.
• If d = 4, then c must be 2, but c must be odd, so this is impossible.
• If d = 6, then c can be either 1 or 3, and then taking a to be the remaining odd number and f = 4 works.
Thus, the only cute numbers are 123654 and 321654, and so there are 2 cute 6-digit numbers.
13. An old receipt has faded.
It reads 88 chickens at the total of $x4.2y, where x and y are unreadable digits. How much did each chicken cost?
Note that we want x42y to be a 4-digit number divisible by 88. Since x42y is divisible by 8, we know that by divisibility rules, 42y is divisible by 8, so we must have that y = 4.
By divisibility rules for 11, note that −x + 4 −2 + y ≡0 (mod 11) ⇒−x + 2 + 4 ≡0 (mod 11) ⇒x ≡6 (mod 11). As such, we must have x = 6 so the total cost was $64.24 and the cost of one chicken was that $0.74.
14. Find the smallest positive integer such that n 2 is a square and n 3 is a cube.
3 Note that clearly n = 2a3b for some natural numbers a, b is necessary for the smallest such n.
By the conditions, √ 2a−13b must be an integer, so 2 must divide a −1 and b, and 3 √ 2a3b−1 must be an integer, so 3 must divide a and b −1. Then a = 3, b = 4 are the smallest possible values for a, b (check that smaller values fail). Thus, 648 is the smallest positive integer.
15. and 16. were challenge problems. :) 4 |
11898 | https://www.ck12.org/flexi/physical-science/Valence-Electrons/determine-the-atoms-of-two-chemical-elements-that-have-the-same-number-of-the-valence-electrons-in-a-ground-state-barium-sodium-chlorine-tellurium-manganese/ | Flexi answers - Determine the atoms of two chemical elements that have the same number of the valence electrons in a ground state: barium, sodium, chlorine, tellurium, manganese. | CK-12 Foundation
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Valence Electrons
Question
Determine the atoms of two chemical elements that have the same number of the valence electrons in a ground state: barium, sodium, chlorine, tellurium, manganese.
Flexi Says:
Barium (Ba) and Sodium (Na) both have 2 valence electrons in their ground state.
Barium, with atomic number 56, has an electron configuration of [Xe] 6s 2, which means it has 2 valence electrons in the 6s orbital.
Sodium, with atomic number 11, has an electron configuration of [Ne] 3s 1, which means it has 1 valence electron in the 3s orbital.
Chlorine (Cl) and Tellurium (Te) both have 7 valence electrons in their ground state.
Chlorine, with atomic number 17, has an electron configuration of [Ne] 3s 2 3p 5, which means it has 5 valence electrons in the 3p orbital and 2 in the 3s orbital, totaling 7.
Tellurium, with atomic number 52, has an electron configuration of [Kr] 5s 2 4d 10 5p 4, which means it has 4 valence electrons in the 5p orbital and 2 in the 5s orbital, totaling 6.
Manganese (Mn) has 2 valence electrons in its ground state. With atomic number 25, its electron configuration is [Ar] 4s 2 3d 5, which means it has 2 valence electrons in the 4s orbital.
So, Barium and Manganese have the same number of valence electrons (2), and Chlorine has the same number of valence electrons as Tellurium (7). Sodium does not share its number of valence electrons (1) with any of the other elements mentioned.
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11899 | https://www.zhihu.com/question/22629654 | 数字的中文读法和大写转换标准应该是怎样的？ - 知乎
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数字的中文读法和大写转换标准应该是怎样的？
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因为最近也想写读法生成器所以思考了一些细节问题，并没有得到什么特定的解，所以来增加一点混乱度。
题主提问的“标准读法”大部分人都知道，问题就在细节上，于是把这些细节问题列一列吧，有些细节其实还跟地域有关……
比如十和一十，粗分有三种模式，一个是程序输出普遍采用的【全部一十】；二是应该是大部分人用的【仅把整个数字的开头替换成十】；三似乎是我到魔都之后才发现似乎江浙地区偏向于【全部用十】，于是出现了〝三百十〞（本地人说普通话）这种让我卡一下的说法……
但是其实这个也不是绝对的，就算是我认为大部分人也有搞不清楚的地方……来读一个10'0010'0010，开头是〝十亿〞，最后是〝零一十〞，中间是〝十万〞还是〝一十万〞？
二和两也有很多规则。基本地说，避免〝两十〞，单个数不读〝两〞，就不会有原则性问题了。然而转换器精确定义是不可能的，最简单的例子，单个2，单独为二，后有量词则为两，量词为两又读二，作为一个转换器还得关注后面是不是量词（当然一定要搞也不是不行，同情一下日语，要关注量词进行尾字和量词变音，不少语音合成都实现了，比如 AquesTalk）。所以程序输出全部用二似乎不会有什么原则性问题也就这样了……虽然没有两很多时候还是会不习惯的吧……
零的问题，也就是最大的一个坑……也就是所谓连续只读一个这样的模糊定义，就算只用万和亿（不引入兆京再往上的东西了，再大亿字叠用）也会形成坑。因为问题就出在这个每一组内的千位上。
1000'1000，1000'1000'0000，1000'0000'1000，1000'0000'0000'1000，1000'1000'1000'1000，1000'0000'1000'0000，1000'0000'1000'0000'1000，1000'0000'0000'0000'1000'0000
这样所谓【某部分已经读完于是不加零】的说法就会面临一些挑战。
另外就是分级的问题，大数读法其实有多种规范的。有【上数、中数、万进、下数】四种。参见喂鸡中文数字（然而它也没说明以上这些细节问题，主要是历史方面的材料）。
以万进来说，【个十百千】为一级，【万亿兆京垓秭壤沟涧正载】为一级（高一级中间用穿插低一级来描述）；以中数（亿进）来说就变成【个十百千】一级，【万】自成一级，【亿兆京垓秭壤沟涧正载】是更高的一级；上数则是从万万为亿，亿亿为兆，兆兆为京，这样万以上所有各自自成一级。
这个也同样影响到刚才的读零问题，如果归纳为某一级读完了后面紧接就不读零的话……这个定义就不太好搞了。毕竟写程序生成就是把规则完善到每一个细处（虽然并无卵用）。
----以下是原题无关---
另外我还挺想支持喂鸡所说的《华严经》大数名呢，随便输入多大都能读了（只要是在一个你电脑装得下的字符串里面xsk）
——但是问题来了，它中间最小的级为7位，上面就是按上数，下面又多一个【洛叉】等于十万（喂鸡上没有但是其他地方考证有），然后一百洛叉等于一【倶胝】，这就是7个0的最小一级，于是一级里面，最大数会变成【九十九洛叉九万九千九百九十九】。这样一级内前面出0了的各种情况又怎么办呢【【【
---更新---
经过我长达两个月的思考和一晚上的奋战，搞出了一个读法生成器，同时也对一些细节有了更深入的理解。
注意这个目前是不分二两皆用二，十百千皆用一十一百一千的。主要的处理如题所说在0的读法上。
题主问题举例的答案：
text
二千零一十四
十万二千
二亿二千零二万二千二百二十二
二十二亿零二万二千二百
二千零一十京零二千零一十亿
最终应该搞明白了，读零与否的准确判法。当某处是否读零存疑：
取其前面部分，对于使得整个读法合法的任意后接部分，都能确定的那些数位的最低位，与接下来实际接上的那个数位的位置紧邻，则不读零。
举例：
10,1000：十万■一千
前面部分为 十万，因为后面不管接什么都知道他会是10xxxxxxxxxx……【后面可能还钻出个亿来】，但是因为万字已吐出，要让那个0改变必须前面是 十X万，所以这个0已经确定，接下来这个 一 所在的是千位，与之紧邻，所以不读零，十万一千。若最后是 百，则跟他隔着一位了，就得有零，十万零一百。（目前生成结果是一十万一千）
101：一百■一（这里暂不考虑 一百一 可以解为一百一十的简读法）
前面 一百 后接任意的话可以搞出1xxxxx……，而接下来的 一 表示的位并没有跟他相邻，所以读出零，一百零一。
1,0000,1000：一亿■一千
一亿 后面可以接几千万，能确定的最低位还是亿位，决定到了1,xxxx,xxxx。但是实际给出的这个就只有个 一千，没有挨着，所以读零，一亿零一千。
1,0000,1000：（Hard模式）
数字跟上面一样，但是只用万字读，八个零读作一万万（追随某大爷的连书万字XX个）。则是 一万万■一千。这里情况不同了，一万万 决定到了1,0000,xxxx，后面接的数再不可能影响到前一个 万 后面的数字。而接下来给的 一千 刚好贴着最高位，所以不读零，一万万一千。
以上讨论都是万进。||||||||行内分割线||||||||| 然而接下来贴一张采用“上数”内部测试图供感受：
大概还会添加各种读法配置项（日本中国），下标采用高精度计算，然后挑战华严经大数（
兴许还能包英语的两种系统【只是英语的千以下3位有点烦
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中国矿业大学（北京） 岩土工程硕士
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一亿亿以及以上的数字，日常生活中基本不会涉及，好像是没有见过什么规定的读法。不过题主其他的问题，应该都是小学老师都会教的。
中文数位有：个，十，百，千，万，十万，百万，千万，亿，十亿，百亿，千亿，万亿，十万亿，百万亿，千万亿，亿亿……
「兆」在大陆也有使用，不过含义不一，「百万、万亿、亿亿」都可能是「兆」。中国数位其实不像英语是三位等差排列的（10^3,10^6,10^9,10^12...），中文除了「千」以外，其他的基本的数位是乘方的关系，十乘以十是百，百乘以百是万，万乘以万是亿。「亿亿」我找不到一个特定的数位，暂以「兆」记，那亿乘以亿就是兆。所以中文基本的数位是10,10^2,10^4,10^8,10^16...
然后是「十」和「一十」的问题。10至19以及以这些数字开头的多位数，以「十」开头，如十五，十万，十亿等。两位数以上，在数字中部出现，则用「一十几」，如一百一十，一千零一十，一万零一十等。
「二」和「两」的问题。两亿，两万，两千，两百，都可以，但是20只能是二十，200用二百也更好。22,2222,2222是「二十二亿两千二百二十二万两千二百二十二」。
关于「零」和「〇」的问题，数字中一律用「零」，只有页码、年代等编号中数的空位才能用「〇」。数位中间无论多少个0，都读成一个「零」。2014是「两千零一十四」，20014是「二十万零一十四」，201400是「二十万零一千四百」。
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```python3
一万以内的用这个，一万以上的懒得想了，大概是递归的用这个？
所以这种面试题到底有什么好玩的呦,我现在写的出来，面试再出还是答不出来嘛，八股
def print_chinese(num):
base = ['', '十', '百', '千', '万']
ns = '零一二三四五六七八九'
l = []
while int(num) > 0:
l.append(int(num) % 10)
num = int(num / 10)
l.reverse()
i = 0
res = ''
while i != len(l):
if l[i] == 0:
while i != len(l) and l[i] == 0:
i = i + 1
if i != len(l):
res = res + '零'
continue
res = res + ns[l[i]]
res = res + base[len(l) - 1 - i]
i = i + 1
return res
```
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