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https://www.oreilly.com/library/view/fundamentals-of-stochastic/9781118092989/c02.xhtml
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## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.
No credit card required
2
OVERVIEW OF STOCHASTIC PROCESSES
2.1 INTRODUCTION
Stochastic processes deal with the dynamics of probability theory. The concept of stochastic processes enlarges the random variable concept to include time. Thus, instead of thinking of a random variable X that maps an event w ∈ Ω, where Ω is the sample space, to some number X(w), we think of how the random variable maps the event to different numbers at different times. This implies that instead of the number X(w) we deal with X(t, w), where tT and T is called the parameter set of the process and is usually a set of times.
Stochastic processes are widely encountered in such fields as communications, control, management science, and time series analysis. Examples of stochastic processes include the population growth, the failure of equipment, the price of a given stock over time, and the number of calls that arrive at a switchboard.
If we fix the sample point w, we obtain X(t), which is some real function of time; and for each w, we have a different function X(t). Thus, X(t, w) can be viewed as a collection of time functions, one for each sample point w. On the other hand, if we fix t, we have a function X(w) that depends only on w and thus is a random variable. Thus, a stochastic process becomes a random variable when time is fixed at some particular value. With many values of t we obtain a collection of random variables. Thus, we can define a stochastic process as a family of random variables {X(t, w)|tT, w ∈ Ω} defined ...
## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.
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## Surveys
The goal of a survey is to collect data from a representative sample of a population to draw conclusions about that larger population.
### Learning Objectives
Assess the various types of surveys and sampling methods used in sociological research, appealing to the concepts of reliability and validity
### Key Takeaways
#### Key Points
• The sample of people surveyed is chosen from the entire population of interest. The goal of a survey is to describe not the smaller sample but the larger population.
• To be able to generalize about a population from a smaller sample, that sample must be representative; proportionally the same in all relevant aspects (e.g., percent of women vs. men).
• Surveys can be distributed by mail, email, telephone, or in-person interview.
• Surveys can be used in cross-sectional, successive-independent-samples, and longitudinal study designs.
• Effective surveys are both reliable and valid. A reliable instrument produces consistent results every time it is administered; a valid instrument does in fact measure what it intends to measure.
#### Key Terms
• cross-sectional study: A research method that involves observation of a representative sample of a population at one specific point in time.
• successive-independent-samples design: A research method that involves observation of multiple random samples from a population over multiple time points.
• longitudinal design: A research method that involves observation of the same representative sample of a population over multiple time points, generally over a period of years or decades.
• sample: A subset of a population selected for measurement, observation or questioning, to provide statistical information about the population.
### Selecting a Sample to Survey
The sample of people surveyed is chosen from the entire population of interest. The goal of a survey is to describe not the smaller sample but the larger population. This generalizing ability is dependent on the representativeness of the sample.
Nuclear Energy Support in the U.S.: This pie chart shows the results of a survey of people in the United States (February 2005, Bisconti Research Inc.). According to the poll, 67 percent of Americans favor nuclear energy (blue), while 26 percent oppose it (yellow).
There are frequent difficulties one encounters while choosing a representative sample. One common error that results is selection bias —when the procedures used to select a sample result in over- or under-representation of some significant aspect of the population. For instance, if the population of interest consists of 75% females, and 25% males, and the sample consists of 40% females and 60% males, females are under represented while males are overrepresented. In order to minimize selection biases, stratified random sampling is often used. This is when the population is divided into sub-populations called strata, and random samples are drawn from each of the strata, or elements are drawn for the sample on a proportional basis.
For instance, a Gallup Poll, if conducted as a truly representative nationwide random sampling, should be able to provide an accurate estimate of public opinion whether it contacts 2,000 or 10,000 people.
### Modes of Administering a Survey
There are several ways of administering a survey. The choice between administration modes is influenced by several factors, including
• costs,
• coverage of the target population,
• respondents’ willingness to participate and
• response accuracy
Different methods create mode effects that change how respondents answer, and different methods have different advantages. The most common modes of administration can be summarized as:
• Telephone
• Mail (post)
• Online surveys
• Personal in-home surveys
• Personal mall or street intercept survey
• Hybrids of the above
Participants willing to take the time to respond will convey personal information about religious beliefs, political views, and morals. Some topics that reflect internal thought are impossible to observe directly and are difficult to discuss honestly in a public forum. People are more likely to share honest answers if they can respond to questions anonymously.
Questionnaire: Questionnaires are a common research method; the U.S. Census is a well-known example.
### Cross-Sectional Design
In a cross-sectional study, a sample (or samples) is drawn from the relevant population and studied once. A cross-sectional study describes characteristics of that population at one time, but cannot give any insight as to causes of population characteristics.
### Successive-Independent-Samples Design
A successive-independent-samples design draws multiple random samples from a population at one or more times. This design can study changes within a population, but not changes within individuals because the same individuals are not surveyed more than once. Such studies cannot, therefore, identify the causes of change over time necessarily.
For successive independent samples designs to be effective, the samples must be drawn from the same population, and must be equally representative of it. If the samples are not comparable, the changes between samples may be due to demographic characteristics rather than time. In addition, the questions must be asked in the same way so that responses can be compared directly.
### Longitudinal Design
A study following a longitudinal design takes measure of the same random sample at multiple time points. Unlike with a successive independent samples design, this design measures the differences in individual participants’ responses over time. This means that a researcher can potentially assess the reasons for response changes by assessing the differences in respondents’ experiences. However, longitudinal studies are both expensive and difficult to do. It’s harder to find a sample that will commit to a months- or years-long study than a 15-minute interview, and participants frequently leave the study before the final assessment. This attrition of participants is not random, so samples can become less representative with successive assessments.
Writing survey questions: Researchers must carefully design survey questions to ensure they receive accurate and unbiased results.
### Reliability and Validity
Reliable measures of self-report are defined by their consistency. Thus, a reliable self-report measure produces consistent results every time it is executed.
A test’s reliability can be measured a few ways. First, one can calculate a test-retest reliability. A test-retest reliability entails conducting the same questionnaire to a large sample at two different times. For the questionnaire to be considered reliable, people in the sample do not have to score identically on each test, but rather their position in the score distribution should be similar for both the test and the retest.
Self-report measures will generally be more reliable when they have many items measuring a construct. Furthermore, measurements will be more reliable when the factor being measured has greater variability among the individuals in the sample that are being tested. Finally, there will be greater reliability when instructions for the completion of the questionnaire are clear and when there are limited distractions in the testing environment.
Contrastingly, a questionnaire is valid if what it measures is what it had originally planned to measure. Construct validity of a measure is the degree to which it measures the theoretical construct that it was originally supposed to measure.
## Fieldwork and Observation
Ethnography is a research process that uses fieldwork and observation to learn about a particular community or culture.
### Learning Objectives
Explain the goals and methods of ethnography
### Key Takeaways
#### Key Points
• Ethnographic work requires intensive and often immersive long-term participation in the community that is the subject of research, typically involving physical relocation (hence the term fieldwork).
• In participant observation, the researcher immerses himself in a cultural environment, usually over an extended period of time, in order to gain a close and intimate familiarity with a given group of individuals and their practices.
• Such research involves a range of well-defined, though variable methods: interviews, direct observation, participation in the life of the group, collective discussions, analyses of personal documents produced within the group, self-analysis, and life-histories, among others.
• The advantage of ethnography as a technique is that it maximizes the researcher’s understanding of the social and cultural context in which human behavior occurs.
• The advantage of ethnography as a technique is that it maximizes the researcher’s understanding of the social and cultural context in which human behavior occurs. The ethnographer seeks out and develops relationships with cultural insiders, or informants, who are willing to explain aspects of their community from a native viewpoint. A particularly knowledgeable informant who can connect the ethnographer with other such informants is known as a key informant.
#### Key Terms
• ethnography: The branch of anthropology that scientifically describes specific human cultures and societies.
• qualitative: Of descriptions or distinctions based on some quality rather than on some quantity.
### Fieldwork and Observation
Ethnography is a qualitative research strategy, involving a combination of fieldwork and observation, which seeks to understand cultural phenomena that reflect the knowledge and system of meanings guiding the life of a cultural group. It was pioneered in the field of socio-cultural anthropology, but has also become a popular method in various other fields of social sciences, particularly in sociology.
Ethnographic work requires intensive and often immersive long-term participation in the community that is the subject of research, typically involving physical relocation (hence the term fieldwork). Although it often involves studying ethnic or cultural minority groups, this is not always the case. Ideally, the researcher should strive to have very little effect on the subjects of the study, being as invisible and enmeshed in the community as possible.
### Participant Observation
One of the most common methods for collecting data in an ethnographic study is first-hand engagement, known as participant observation. In participant observation, the researcher immerses himself in a cultural environment, usually over an extended period of time, in order to gain a close and intimate familiarity with a given group of individuals (such as a religious, occupational, or sub-cultural group, or a particular community) and their practices.
Fieldwork and Observation: One of the most common methods for collecting data in an ethnographic study is first-hand engagement, known as participant observation.
### Methods
Such research involves a range of well-defined, though variable methods: interviews, direct observation, participation in the life of the group, collective discussions, analyses of personal documents produced within the group, self-analysis, and life-histories, among others.
Interviews can be either informal or formal and can range from brief conversations to extended sessions. One way of transcribing interview data is the genealogical method. This is a set of procedures by which ethnographers discover and record connections of kinship, descent, and marriage using diagrams and symbols. Questionnaires can also be used to aid the discovery of local beliefs and perceptions and, in the case of longitudinal research where there is continuous long-term study of an area or site, they can act as valid instruments for measuring changes in the individuals or groups studied.
The advantage of ethnography as a technique is that it maximizes the researcher’s understanding of the social and cultural context in which human behavior occurs. The ethnographer seeks out and develops relationships with cultural insiders, or informants, who are willing to explain aspects of their community from a native viewpoint. The process of seeking out new contacts through their personal relationships with current informants is often effective in revealing common cultural common denominators connected to the topic being studied.
## Experiments
Experiments are tests designed to prove or disprove a hypothesis by controlling for pertinent variables.
### Learning Objectives
Compare and contrast how hypotheses are being tested in sociology and in the hard sciences
### Key Takeaways
#### Key Points
• Experiments are controlled tests designed to prove or disprove a hypothesis.
• A hypothesis is a prediction or an idea that has not yet been tested.
• Researchers must attempt to identify everything that might influence the results of an experiment, and do their best to neutralize the effects of everything except the topic of study.
• Since social scientists do not seek to isolate variables in the same way that the hard sciences do, sociologists create the equivalent of an experimental control via statistical techniques that are applied after data is gathered.
• A control is when two identical experiments are conducted and the factor being tested is varied in only one of these experiments.
#### Key Terms
• control: A separate group or subject in an experiment against which the results are compared where the primary variable is low or nonexistent.
• experiment: A test under controlled conditions made to either demonstrate a known truth, examine the validity of a hypothesis, or determine the efficacy of something previously untried.
• hypothesis: Used loosely, a tentative conjecture explaining an observation, phenomenon, or scientific problem that can be tested by further observation, investigation, or experimentation.
Scientists form a hypothesis, which is a prediction or an idea that has not yet been tested. In order to prove or disprove the hypothesis, scientists must perform experiments. The experiment is a controlled test designed specifically to prove or disprove the hypothesis. Before undertaking the experiment, researchers must attempt to identify everything that might influence the results of an experiment and do their best to neutralize the effects of everything except the topic of study. This is done through the introduction of an experimental control: two virtually identical experiments are run, in only one of which the factor being tested is varied. This serves to further isolate any causal phenomena.
An Experiment: An experiment is a controlled test designed specifically to prove or disprove a hypothesis.
Of course, an experiment is not an absolute requirement. In observation based fields of science, actual experiments must be designed differently than for the classical laboratory based sciences. Due to ethical concerns and the sheer cost of manipulating large segments of society, sociologists often turn to other methods for testing hypotheses.
Since sociologists do not seek to isolate variables in the same way that hard sciences do, this kind of control is often done via statistical techniques, such as regressions, applied after data is gathered. Direct experimentation is thus fairly rare in sociology.
Scientists must assume an attitude of openness and accountability on the part of those conducting an experiment. It is essential to keep detailed records in order to facilitate reporting on the experimental results and provide evidence of the effectiveness and integrity of the procedure.
## Documents
Documentary research involves examining texts and documents as evidence of human behavior.
### Learning Objectives
Describe different kinds of documents used in sociological research
### Key Takeaways
#### Key Points
• This kind of sociological research is generally considered a part of media studies.
• Unobtrusive research involves ways of studying human behavior without affecting it in the process.
• Documents can either be primary sources, which are original materials that are not created after the fact with the benefit of hindsight, or secondary sources that cite, comment, or build upon primary sources.
• Typically, sociological research involving documents falls under the cross-disciplinary purview of media studies, which encompasses all research dealing with television, books, magazines, pamphlets, or any other human-recorded data. The specific media being studied are often referred to as texts.
• Sociological research involving documents, or, more specifically, media studies, is one of the less interactive research options available to sociologists. It can provide a significant insight into the norms, values, and beliefs of people belonging to a particular historical and cultural context.
• Content analysis is the study of recorded human communications.
#### Key Terms
• content analysis: Content analysis or textual analysis is a methodology in the social sciences for studying the content of communication.
• documentary research: Documentary research involves the use of texts and documents as source materials. Source materials include: government publications, newspapers, certificates, census publications, novels, film and video, paintings, personal photographs, diaries and innumerable other written, visual, and pictorial sources in paper, electronic, or other “hard copy” form.
• media studies: Academic discipline that deals with the content, history, meaning, and effects of various media, and in particular mass media.
### Documentary Research
It is possible to do sociological research without directly involving humans at all. One such method is documentary research. In documentary research, all information is collected from texts and documents. The texts and documents can be either written, pictorial, or visual in form.
The material used can be categorized as primary sources, which are original materials that are not created after the fact with the benefit of hindsight, and secondary sources that cite, comment, or build upon primary sources.
### Media Studies and Content Analysis
Typically, sociological research on documents falls under the cross-disciplinary purview of media studies, which encompasses all research dealing with television, books, magazines, pamphlets, or any other human-recorded data. Regardless of the specific media being studied, they are referred to as texts.
Media studies may draw on traditions from both the social sciences and the humanities, but mostly from its core disciplines of mass communication, communication, communication sciences, and communication studies.
Researchers may also develop and employ theories and methods from disciplines including cultural studies, rhetoric, philosophy, literary theory, psychology, political economy, economics, sociology, anthropology, social theory, art history and criticism, film theory, feminist theory, information theory, and political science.
Government Documentary Research: Sociologists may use government documents to research the ways in which policies are made.
Content analysis refers to the study of recorded human communications, such as paintings, written texts, and photos. It falls under the category of unobtrusive research, which can be defined as ways for studying human behavior without affecting it in the process. While sociological research involving documents is one of the less interactive research options available to sociologists, it can reveal a great deal about the norms, values, and beliefs of people belonging to a particular temporal and cultural context.
## Use of Existing Sources
Studying existing sources collected by other researchers is an essential part of research in the social sciences.
### Learning Objectives
Explain how the use of existing sources can benefit researchers
### Key Takeaways
#### Key Points
• Archival research is the study of existing sources. Without archival research, any research project is necessarily incomplete.
• The study of sources collected by someone other than the researcher is known as archival research or secondary data research.
• The importance of archival or secondary data research is two-fold. By studying texts related to their topics, researchers gain a strong foundation on which to base their work. Secondly, this kind of study is necessary in the development of their central research question.
#### Key Terms
• secondary data: Secondary data is data collected by someone other than the user. Common sources of secondary data for social science include censuses, organizational records, and data collected through qualitative methodologies or qualitative research.
• Archival research: An archive is a way of sorting and organizing older documents, whether it be digitally (photographs online, e-mails, etc.) or manually (putting it in folders, photo albums, etc.). Archiving is one part of the curating process which is typically carried out by a curator.
• primary data: Data that has been compiled for a specific purpose, and has not been collated or merged with others.
### Using Existing Sources
The study of sources collected by someone other than the researcher, also known as archival research or secondary data research, is an essential part of sociology. In archival research or secondary research, the focus is not on collecting new data but on studying existing texts.
Existing Sources: While some sociologists spend time in the field conducting surveys or observing participants, others spend most of their research time in libraries, using existing sources for their research.
By studying texts related to their topics, researchers gain a strong foundation on which to base their work. Furthermore, this kind of study is necessary for the development of their central research question. Without a thorough understanding of the research that has already been done, it is impossible to know what a meaningful and relevant research question is, much less how to position and frame research within the context of the field as a whole.
### Types of Existing Sources
Common sources of secondary data for social science include censuses, organizational records, field notes, semi-structured and structured interviews, and other forms of data collected through quantitative methods or qualitative research. These methods are considered non-reactive, because the people do not know they are involved in a study. Common sources differ from primary data. Primary data, by contrast, are collected by the investigator conducting the research.
Researchers use secondary analysis for several reasons. The primary reason is that secondary data analysis saves time that would otherwise be spent collecting data. In the case of quantitative data, secondary analysis provides larger and higher-quality databases that would be unfeasible for any individual researcher to collect on his own. In addition, analysts of social and economic change consider secondary data essential, since it is impossible to conduct a new survey that can adequately capture past change and developments.
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https://maqsad.io/classes/second-year/math/functions-and-limits/example%201%20(i)%20operatorname%20lim%20_%20x%20rightarrow%201%20frac%20xexp%202%20-1%20xexp%202%20-x
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Classes
Class 9Class 10First YearSecond Year
$Example 1:(i) \operatorname{Lim}_{x \rightarrow 1} \frac{x^{2}-1}{x^{2}-x}$
4. Express each limit in terms of e :$\text { (x) } \operatorname{Lim}_{x \rightarrow 0} \frac{e^{1 / x}-1}{e^{1 / x}+1} x<0$
$Example 1:Graph the circle x^{2}+y^{2}=4$
$4. Express each limit in terms of e :(ix) \operatorname{Lim}_{x \rightarrow \infty}\left(\frac{x}{1+x}\right)^{x}$
$1. Evaluate each limit by using theorems of limits:(v) \operatorname{Lim}_{x \rightarrow 2}\left(\sqrt{x^{3}+1}-\sqrt{x^{2}+5}\right)$
Example 1: Determine whether \operatorname{Lim}_{x \rightarrow 2} f(x) and \operatorname{Lim}_{x \rightarrow 4} f(x) exist when$f(x)=\left\{\begin{array}{rcc}2 x+1 & \text { if } & 0 \leq x \leq 2 \\7-x & \text { if } & 2 \leq x \leq 4 \\x & \text { if } & 4 \leq x \leq 6\end{array}\right.$
$Example 7:Evaluate: \operatorname{Lim}_{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta}$
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https://blog.csdn.net/begonia__z/article/details/51199454
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# Unity3D学习日记（三）贝塞尔曲线
using UnityEngine;
[System.Serializable]
public class Bezier : System.Object
{
/// <summary>
/// 线性贝赛尔曲线
/// </summary>
/// <param name="P0"></param>
/// <param name="P1"></param>
/// <param name="t"> 0.0 >= t <= 1.0 </param>
/// <returns></returns>
public static Vector3 BezierCurve(Vector3 P0, Vector3 P1, float t)
{
Vector3 B = Vector3.zero;
float t1 = (1 - t);
B = t1*P0 + P1*t;
//B.y = t1*P0.y + P1.y*t;
//B.z = t1*P0.z + P1.z*t;
return B;
}
/// <summary>
///
/// </summary>
/// <param name="P0"></param>
/// <param name="P1"></param>
/// <param name="P2"></param>
/// <param name="t">0.0 >= t <= 1.0 </param>
/// <returns></returns>
public static Vector3 BezierCurve(Vector3 P0, Vector3 P1, Vector3 P2, float t)
{
Vector3 B = Vector3.zero;
float t1 = (1 - t)*(1 - t);
float t2 = t*(1 - t);
float t3 = t*t;
B = P0*t1 + 2*t2*P1 + t3*P2;
//B.y = P0.y*t1 + 2*t2*P1.y + t3*P2.y;
//B.z = P0.z*t1 + 2*t2*P1.z + t3*P2.z;
return B;
}
/// <summary>
///
/// </summary>
/// <param name="P0"></param>
/// <param name="P1"></param>
/// <param name="P2"></param>
/// <param name="P3"></param>
/// <param name="t">0.0 >= t <= 1.0 </param>
/// <returns></returns>
public static Vector3 BezierCurve(Vector3 P0, Vector3 P1, Vector3 P2, Vector3 P3, float t)
{
Vector3 B = Vector3.zero;
float t1 = (1 - t)*(1 - t)*(1 - t);
float t2 = (1 - t)*(1 - t)*t;
float t3 = t*t*(1 - t);
float t4 = t*t*t;
B = P0*t1 + 3*t2*P1 + 3*t3*P2 + P3*t4;
//B.y = P0.y*t1 + 3*t2*P1.y + 3*t3*P2.y + P3.y*t4;
//B.z = P0.z*t1 + 3*t2*P1.z + 3*t3*P2.z + P3.z*t4;
return B;
}
}
• 广告
• 抄袭
• 版权
• 政治
• 色情
• 无意义
• 其他
120
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CC-MAIN-2019-04
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https://www.lmfdb.org/EllipticCurve/Q/289800/cf/4
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# Properties
Label 289800.cf4 Conductor $289800$ Discriminant $-2.877\times 10^{24}$ j-invariant $$-\frac{957928673903042}{123339801817575}$$ CM no Rank $1$ Torsion structure $$\Z/{2}\Z$$
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, 0, 0, -5868075, 81794159750])
gp: E = ellinit([0, 0, 0, -5868075, 81794159750])
magma: E := EllipticCurve([0, 0, 0, -5868075, 81794159750]);
$$y^2=x^3-5868075x+81794159750$$
## Mordell-Weil group structure
$\Z\times \Z/{2}\Z$
### Infinite order Mordell-Weil generator and height
sage: E.gens()
magma: Generators(E);
$P$ = $$\left(\frac{5876254}{24025}, \frac{1055726933742}{3723875}\right)$$ $\hat{h}(P)$ ≈ $17.823163162825677216148762873$
## Torsion generators
sage: E.torsion_subgroup().gens()
gp: elltors(E)
magma: TorsionSubgroup(E);
$$\left(-4790, 0\right)$$
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
$$\left(-4790, 0\right)$$
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$289800$$ = $2^{3} \cdot 3^{2} \cdot 5^{2} \cdot 7 \cdot 23$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $-2877270896800389600000000$ = $-1 \cdot 2^{11} \cdot 3^{8} \cdot 5^{8} \cdot 7 \cdot 23^{8}$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$-\frac{957928673903042}{123339801817575}$$ = $-1 \cdot 2 \cdot 3^{-2} \cdot 5^{-2} \cdot 7^{-1} \cdot 23^{-8} \cdot 78241^{3}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $3.3726804227043605098253370737\dots$ Stable Faltings height: $1.3832704066399722765282053440\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $1$ sage: E.regulator() magma: Regulator(E); Regulator: $17.823163162825677216148762873\dots$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.065908315961308237969596766019\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $32$ = $1\cdot2^{2}\cdot2^{2}\cdot1\cdot2$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $2$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L'(E,1)$ ≈ $9.3975573533237167932066625294867434662$
## Modular invariants
Modular form 289800.2.a.cf
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - q^{7} + 4 q^{11} + 2 q^{13} + 2 q^{17} - 4 q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 44040192 $\Gamma_0(N)$-optimal: no Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 5 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $1$ $II^{*}$ Additive -1 3 11 0
$3$ $4$ $I_{2}^{*}$ Additive -1 2 8 2
$5$ $4$ $I_{2}^{*}$ Additive 1 2 8 2
$7$ $1$ $I_{1}$ Non-split multiplicative 1 1 1 1
$23$ $2$ $I_{8}$ Non-split multiplicative 1 1 8 8
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image for all primes $\ell$ except those listed in the table below.
prime $\ell$ mod-$\ell$ image $\ell$-adic image
$2$ 2B 8.12.0.5
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
$p$-adic regulators are not yet computed for curves that are not $\Gamma_0$-optimal.
No Iwasawa invariant data is available for this curve.
## Isogenies
This curve has non-trivial cyclic isogenies of degree $d$ for $d=$ 2, 4 and 8.
Its isogeny class 289800.cf consists of 6 curves linked by isogenies of degrees dividing 8.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ $\cong \Z/{2}\Z$ are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $2$ $$\Q(\sqrt{-14})$$ $$\Z/2\Z \times \Z/2\Z$$ Not in database $2$ $$\Q(\sqrt{15})$$ $$\Z/4\Z$$ Not in database $2$ $$\Q(\sqrt{-210})$$ $$\Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{-14}, \sqrt{15})$$ $$\Z/2\Z \times \Z/4\Z$$ Not in database $4$ $$\Q(\sqrt{15}, \sqrt{21})$$ $$\Z/8\Z$$ Not in database $4$ $$\Q(\sqrt{-6}, \sqrt{-10})$$ $$\Z/8\Z$$ Not in database $8$ Deg 8 $$\Z/2\Z \times \Z/4\Z$$ Not in database $8$ Deg 8 $$\Z/8\Z$$ Not in database $8$ 8.0.7965941760000.46 $$\Z/2\Z \times \Z/8\Z$$ Not in database $8$ Deg 8 $$\Z/6\Z$$ Not in database $16$ Deg 16 $$\Z/4\Z \times \Z/4\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/8\Z$$ Not in database $16$ Deg 16 $$\Z/16\Z$$ Not in database $16$ Deg 16 $$\Z/16\Z$$ Not in database $16$ Deg 16 $$\Z/2\Z \times \Z/6\Z$$ Not in database $16$ Deg 16 $$\Z/12\Z$$ Not in database $16$ Deg 16 $$\Z/12\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial.
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# Simple Moving Average Calculator
Calculate the simple moving average of a series of data points using our online calculator. By inputting the desired number of periods or time intervals and the corresponding data, you can compute the average value over that specific period.
Simple Moving Average Calculator
Results
Moving Average:
Related
## Reference
A moving average is marked on a stock chart by a line, and it represents the average price of a given stock over a period of interest. It serves to smooth over the changes in a stock price so that the overall trend becomes more apparent.
The most frequently employed moving averages are the exponential moving average (EMA) and the simple moving average (SMA).
When provided with a sequential data set, you can determine the n-point moving (or rolling) average by computing the mean of each set of n successive points. For instance, if you have the following sequential data set:
2, 4, 6, 8, 12, 14, 16, 18, 20,
the four-point moving average would be as follows:
5, 7.5, 10, 12.5, 15, 17
Moving averages serve to "smooth" chronological data; they reduce the impact of sharp peaks and dips because every raw data point is provided with a fractional weight in the moving average. The higher the value of n, the smoother the moving average graph will be in comparison to a graph of the original data. Stock analysts frequently examine the moving averages of stock prices to identify patterns and predict future movements.
## Simple Moving Average Formula
SMA (n) = (P1 + P2 + … + Pn) / n
Where:
n is the number of time periods,
Pn is the price at period n.
## what is Simple Moving Average Calculator
A Simple Moving Average (SMA) calculator is a tool used to calculate the average value of a time series data set over a specified number of periods. The SMA is a commonly used technical analysis indicator that helps smooth out price fluctuations and identify trends over time.
Here's how a Simple Moving Average calculator typically works:
1. Data Input: The calculator requires a series of historical data points, such as daily closing prices or other relevant values, for the asset or variable you want to calculate the moving average of. This data should cover a specific time period.
2. Setting the Period: You need to specify the number of periods over which you want to calculate the moving average. For example, if you choose a 20-day SMA, it will calculate the average based on the past 20 data points.
3. Calculation: The calculator adds up the values of the selected data points over the specified period and then divides the sum by the number of periods. This calculation is repeated as new data becomes available, creating a moving average that reflects the most recent data points.
4. Displaying the Result: The calculator provides the calculated Simple Moving Average as the output. It represents the smoothed average value of the data series over the specified period.
The Simple Moving Average is a lagging indicator, meaning it reacts slower to price changes compared to other types of moving averages. It's often used to identify trends, support and resistance levels, and potential buy or sell signals in technical analysis.
Simple Moving Average calculators can be found online or implemented in various charting platforms, trading software, or spreadsheet programs like Excel. Traders, analysts, and investors use these calculators to analyze historical price data and make informed decisions in the financial markets.
## Simple Moving Average Calculator Example
Certainly! The Simple Moving Average (SMA) is a commonly used technical analysis indicator that calculates the average price of a financial instrument over a specified period of time. Let's consider an example of calculating the Simple Moving Average for a stock using daily closing prices.
Assume we have collected the daily closing prices of a stock for the last 10 trading days. Here are the closing prices (in arbitrary units):
100, 105, 110, 115, 120, 125, 130, 135, 140, 145
Step 1: Determine the time period.
First, we need to determine the time period for which we want to calculate the Simple Moving Average. For this example, let's use a 5-day Simple Moving Average.
Step 2: Calculate the sum of closing prices.
Next, we calculate the sum of the closing prices over the specified time period. In our case, we will calculate the sum of the most recent 5 closing prices.
Sum of Closing Prices = 120 + 125 + 130 + 135 + 140 = 650
Step 3: Calculate the Simple Moving Average.
The Simple Moving Average is calculated by dividing the sum of closing prices by the number of periods.
Simple Moving Average = Sum of Closing Prices / Number of Periods
In our example, the Simple Moving Average is:
650 / 5 = 130
Therefore, the 5-day Simple Moving Average for this stock, based on the given data, is 130.
Please note that this is a simplified example for illustrative purposes. In practice, moving averages are calculated using larger data sets, and there are variations such as exponential moving averages that assign different weights to each period. The choice of time period and moving average type depends on the specific analysis requirements and trading strategies.
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Try the Free Math Solver or Scroll down to Tutorials!
Depdendent Variable
Number of equations to solve: 23456789
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Solve for:
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Number of inequalities to solve: 23456789
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Author Message
Nold 0f Dhi Lanjards
Registered: 09.01.2004
From: cheeseland
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### CFA Practice Question
There are 434 practice questions for this study session.
### CFA Practice Question
As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after testing the free sample is 0.20. Different shoppers can be regarded as independent trials. Let X be the number among the next 100 shoppers who buy a packet of the crackers after tasting a free sample. X has approximately a ______.
A. N(4, 20) distribution
B. N(20, 4) distribution
C. N(0.2, 16) distribution
Explanation: The distribution is approximately normal with mean np = 20 and standard deviation [np(1-p)]1/2 = 4.
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MAN UNITED PHIÊN BẢN AIG CÓ PHẢI LÀ PHIÊN BẢN MẠNH NHẤT?
MAN UNITED PHIÊN BẢN AIG CÓ PHẢI LÀ PHIÊN BẢN MẠNH NHẤT?
1
Given y=f(x) then df/dx is given by which of the following? 1. 2. 3. 4.
2
equals 1.-2sin2x 2.-sin(2x) 3.0 4.-2xsin2x
3
If y=x n then find dy/dx 1.nx n 2.nx n-1 3.x n-1 4.(n-1)x n
4
Find the derivative of f(x)=3x³-½x²+5x+1 with respect to x 1.9x² – 2x + 5 2.9x³ – x² + 5x + 1 3.9x² – x + 5 4.9x³ – x² + 6
5
If 1.-3e 3x 2.3e 3x 3.-3e 2x 4.-3xe 2x
6
f(k)=tan3k, find 1.3sec3k 2.sec3k 3.3sec²3k 4.sec²3k
7
= 1.sin(x) 2.-sin(x) 3.cos(x) 4.-cos(x) 5.cosec(x) John Goodband, Coventry University
8
= 1. 2. 3. 4. 5.
9
= John Goodband, Coventry University 1. 2. 3. 4. 5.
10
Find the derivative of 1. 2. 3. 4. with respect to x
11
Find the derivative of z = 2sint – cos2t with respect to t 1.2cost + sin2t 2.2cost – sin 2t 3.2cost + 2sin2t 4.2cost – 2sin2t
12
If then 1. 2. 3. All of the above 4.
13
1.The derivatie of f(x)+g(x) is 2.The derivative of f(x)-g(x) is 3.If k is constant, the derivative of kf(x) is 4.If y=f(x)g(x) then Which of the following statements are true?
14
equals 1.2-e cosx sinx +2xcos2x 2.x+ e cosx +2cos2x 3.2-e cosx sinx +2cos2x 4.Not enough information
15
Find the derivative of y=2xe -x with respect to x 1.-2xe -x + 2e -x 2.-2xe -x + 2e -x 3.2xe -x – 2e -x 4.2xe -x + 2e -x
16
Find the derivative of y=(e 2x ) 6 with respect to x 1.6e 2x 2.12e 12x 3.12xe x 4.12e x
17
= John Goodband, Coventry University 1.2xcos(x²) 2.cos(x²) 3.2xcos(x) 4.x²cos(x²) + 2xsin(x²)
18
Which of the following is the quotient rule if ? 1. 2. 3. 4.
19
Use the quotient rule to find the derivative of f(x)=x -3 cosx with respect to x 1. 2. 3. 4.
20
We know and. Then equals: 1.5/2 2.7/2 3.3
21
Using the chain rule, find the derivative f(x)=(3x²+2)² with respect to x 1.2(3x² + 2) 2.12(3x + 2) 3.12x(3x² + 2) 4.12x + 4
22
Suppose a runner has a speed of 8 miles per hour, while a cyclist has a speed of 16 miles per hour. Then dV/dt for the cyclist is 2 times greater than dV/dt for the runner. This is explained by: 1.The chain rule 2.The product rule 3.The quotient rule 4.The addition rule
23
The radius of a balloon changes as it deflates. This change in radius with respect to volume is: 1. 2. 3. None of these 4.
24
Calculate the second derivative of y = 4x³ – 2x + x² – 3 with respect to x 1.24x + 2 2.24x – 2x 3.12x – 2 4.12x² – 2 +2x
25
If then find 1. 2. 3. 4.
26
If x=h(t) and y=g(t) then 1. 2. 3. 4.
27
Find the value of if x=3t 2 and y=2t-1. 1. 2. 3. 4.
28
1. If x=h(t) and y=g(t) then 2. 3. 4.
29
Find the equation of the tangent line to the curve x=1-3sint, y=2+cost at. 1. 2. 3. None of the above 4.
30
Which differentiation rule is needed to differentiate implicit functions? 1.Product rule 2.Chain rule 3.Quotient rule 4.Inverse function rule
31
Find if 3y=xy+siny. 1. 2. 3. 4.
32
Find at the point (3,1) on x 2 +2xy+y 2 =x. 1. 2. 3. 4.
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What Is Hasse's Theorem
This section describes Hasse's Theorem, which states that the order, n, of a reduced elliptic curve group, Ep(a,b), is bounded in the range of [p+1 - 2*sqrt(p), p+1 + 2*sqrt(p)].
What Is Hasse's Theorem? Hasse's Theorem is also called Hasse Bound, which provides an estimate of the number of points on an elliptic curve over a finite field, bounding the value both above and below.
```For a given elliptic curve E(a,b) over a finite field with q elements,
the number of points, n, on the curve satisfies the following condition:
|n - (q+1)| <= 2*sqrt(q)
```
If we apply Hasse's Theorem to our reduced elliptic curve group definition, Ep(a,b)), we have:
```Given reduced elliptic curve group Ep(a,b),
the group order, n, satisfies the following condition:
|n - (p+1)| <= 2*sqrt(p)
Or n is in a range as expressed below:
p+1 - 2*sqrt(p) <= n <= p+1 + 2*sqrt(p)
```
It's interesting to see that the group order is only depending on the prime number used in the modular arithmetic reduction, not on coefficients, a and b, of the elliptic curve equation.
Hasse's Theorem is named after German mathematician Helmut Hasse (25 August 1898 - 26 December 1979):
For more details, see "Hasse's theorem on elliptic curves" at en.wikipedia.org/wiki/Hasse%27s_theorem_on_elliptic_curves.
Last update: 2019.
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https://infinitylearn.com/surge/question/chemistry/a-solution-of-urea-mol-mass-56-g-mol-1-boils-at-10018c/
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A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf, and Kb, for water are 1.86 and 0.512 K kg kg mol-1 respectively, the above solution will freeze at
# A solution of urea (mol. mass 56 g mol-1) boils at 100.18°C at the atmospheric pressure. If Kf, and Kb, for water are 1.86 and 0.512 K kg kg mol-1 respectively, the above solution will freeze at
1. A
0.654°C
2. B
—0.654°C
3. C
654°C
4. D
—654°C
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### Solution:
if the value of m is same, then
Hence, $\frac{{\mathrm{\Delta T}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{f}}}=\frac{{\mathrm{\Delta T}}_{\mathrm{b}}}{{\mathrm{K}}_{\mathrm{b}}}$
or ${\mathrm{\Delta T}}_{\mathrm{f}}={\mathrm{\Delta T}}_{\mathrm{b}}\frac{{\mathrm{K}}_{\mathrm{f}}}{{\mathrm{K}}_{\mathrm{b}}}$
$\left[{\mathrm{\Delta T}}_{\mathrm{b}}=100.18-100=0{.18}^{\circ }\mathrm{C}\right]$
$∆{\mathrm{T}}_{\mathrm{f}}=0.18×\frac{1.86}{0.512}=0{.65}^{\circ }\mathrm{C}$
As the Freezing Point of pure water is 0°C,
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 128 MB20115.000%
## 문제
Bart and Lisa Simpson have many chores, but they don't always do them well. So their father began keeping score, giving them credit only if a chore was done right. After a month, he showed them his record keeping.
Assigned chores for first month Chores done correctly Bart Bart Bart Lisa Lisa Lisa Bart Bart Lisa 1 0 1 0 1 1 1 75.00% 66.67%
This showed that Bart had done 3 out of 4 chores correctly and Lisa had done 2 out of 3 chores correctly. Then for a second month, Bart and Lisa did chores and the record keeping looked like this:
Assigned chores for second month Chores done correctly Bart Bart Lisa Lisa Lisa Lisa Lisa Lisa Lisa Bart Lisa 0 1 0 1 1 0 1 0 0 50.00% 42.86%
The father made the following chart to support the hypothesis that Bart was doing the most chores:
Month1 Month2 Computation Total Bart 75.00% 50.00% 125.00 / 234.53 53.30% Lisa 66.67% 42.86% 109.53 / 234.53 46.70%
Lisa looked at the tables then said "No no no, that's not right, I've done more." Bart pointed at the statistics and said, "Numbers don't lie." Lisa said, "But look, I've done 5 chores and you've only done 4. You just look good because you always run away before somebody asks you to do something. The total should give me 55% of the total chores done right." Bart laughed and ran away. Now Lisa needs your help to show the unfairness of these chore evaluations. She wants a program which computes the averages over the entire dataset and displays where the before and after averages support the opposite hypothesis than would be indicated by considering the total dataset. To be fair, she only wants to consider record keeping times when both she and Bart have chores assigned both before and after the record keeping.
## 입력
The first line of each test case contains one integer N (2 ≤ N ≤ 50000) the number of chore records. Each of the next N lines in each test case contains either "Bart" or "Lisa" followed by an integer, 0 or 1. The name shows who was assigned a chore, and the integer is a value of 1 if it was done correctly, or a 0 if it was not. Both Bart and Lisa will have chores assigned to them in each dataset. Input ends when N = 0.
## 출력
Write on the first line of the output for each test case, the test case number, and the number of chores correctly performed by Bart (CB) and Lisa (CL) in the following format:
Case <case number>: Bart did <CB> and Lisa did <CL>
If Bart and Lisa do the same number of chores over the entire dataset, there is no trend to oppose. In that case, print "Bart and Lisa accomplished same number of chores".
If Bart and Lisa do a different number of chores over the entire dataset, determine if there are any places in the dataset where taking the averages of all results before and after a record keeping break would produce results which would indicate a trend opposing the trend over the entire dataset. If there are no such cases, print on the next line, "Simpson's paradox not detected". Otherwise, print on the next line "Trend measured in 2 parts is reversed" and on every following line print the following:
After chore <chore number>: <BBefore>% <LBefore>% <BAfter>% <LAfter>%
where BBefore is Bart's correctly performed chore percentage before the record keeping, LBefore is Lisa's correctly performed chore percentage before the recordkeeping, BAfter is Bart's correctly performed chore percentage after the first record keeping until the end of the dataset, and LAfter is Lisa's correctly performed chore percentage after the first record keeping until the end of the dataset. Note that chores are numbered from 1 to N in order of appearance in the dataset. All percentages must be rounded to the nearest .0001. Print a blank line between the outputs for two consecutive test cases.
## 예제 입력 1
16
Bart 1
Bart 0
Bart 1
Lisa 0
Lisa 1
Lisa 1
Bart 1
Bart 0
Bart 1
Lisa 0
Lisa 1
Lisa 1
Lisa 0
Lisa 1
Lisa 0
Lisa 0
2
Lisa 1
Bart 1
2
Lisa 0
Bart 1
0
## 예제 출력 1
Case 1: Bart did 4 and Lisa did 5
Trend measured in 2 parts is reversed
After chore 4: 66.6667% 0.0000% 66.6667% 55.5556%
After chore 5: 66.6667% 50.0000% 66.6667% 50.0000%
After chore 7: 75.0000% 66.6667% 50.0000% 42.8571%
Case 2: Bart did 1 and Lisa did 1
Bart and Lisa accomplished same number of chores
Case 3: Bart did 1 and Lisa did 0
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# What is so special about the Lebesgue-Stieltjes measure
A measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ that is associated with a monotone increasing and right-side continuous function $F$ is called a Lebesgue-Stieltjes measure. But I am wondering, why it is not true that every measure $\lambda: B(\mathbb{R}^n) \rightarrow \overline{{\mathbb{R_{\ge 0}}}}$ is a Lebesgue-Stieltjes measure?
-
Every Borel measure is a Lebesgue-Stieltjes, provided it is finite on bounded sets. – Prahlad Vaidyanathan Apr 22 '14 at 17:19
If you assume $\lambda$ to take only values in $\mathbb{R}_{\geq 0}$ then you have a bijection of Lebesgue-Stieltjes measures and equivalence classes of associated monotone increasing and right-side continous functions $F$. But I guess you already know that and you're asking for the case of $\overline{{\mathbb{R_{\ge 0}}}}$ – cQQkie Apr 22 '14 at 17:36
$\mu$ being a lebesgue-stiltjes measure with corresponding function $F$ implies that $$\mu\left((a,b]\right) = F(b) - F(a) \text{.}$$
Now take the (rather silly) measure $$\mu(X) = \begin{cases} \infty &\text{if 0 \in X} \\ 1 &\text{if 0 \notin X, 1 \in X} \\ 0 &\text{otherwise.} \end{cases}$$
We'd need to have $F(x) = \infty$ for $x \geq 0$ and $F(x) = 0$ for $x < 0$ to have $\mu\left((a,b]\right) = F(b) - F(a)$ for $a < 0$, $b \geq 0$. But then $$\mu\left((0,2]\right) = F(2) - F(0) = \infty - \infty$$ which is
1. meaningless, and
2. surely not the same as $1$, which is the actual measure of $(0,2]$.
Note that a measure doesn't necessarily need to have infinite point weights (i.e., $x$ for which $\mu(\{x\}) = \infty$ to cause trouble. Here's another measure on $B(\mathbb{R})$ which isn't a lebesgue-stiltjes measure $$\mu(X) = \sum_{n \in \mathbb{N}, \frac{1}{n} \in X} \frac{1}{n} \text{.}$$ For every $\epsilon > 0$, $\mu\left((0,\epsilon]\right) = \infty$, which again would require $F(x) = \infty$ for $x > 0$, and again that conflicts the requirement that $\mu((a,b]) = F(b) - F(a) < \infty$ for $0 < a \leq b$. Note that this measure $\mu$ is even $\sigma$-finite! You can write $\mathbb{R}$ as the countable union $$\mathbb{R} = \underbrace{(-\infty,0]}_{=A} \cup \underbrace{(1,\infty)}_{=B} \cup \bigcup_{n \in \mathbb{N}} \underbrace{(\tfrac{1}{n+1},\tfrac{1}{n}]}_{=C_n}$$ and all the sets have finite measure ($\mu(A)=\mu(B) = 0$, $\mu(C_n) = \frac{1}{n}$).
You do have that all finite (i.e., not just $\sigma$-finite, but fully finite) measures on $B(\mathbb{R})$ are lebesgue-stiltjes measures, however. This is important, for example, for probability theory, because it allows you to assume that every random variable on $\mathbb{R}$ has a cumulative distribution function (CDF), which is simply the function $F$.
-
What if both $0$ and $1$ are in $X$? – cQQkie Apr 22 '14 at 17:39
@cQQkie Then the measure is $\infty$. Will make it say that explicitly. – fgp Apr 22 '14 at 17:43
If you have a measure $\mu$ in $\mathbb{R}$ that is finite in compact sets then we define the function
$$F(x) = \begin{cases} \mu((0,x])&\text{if x > 0} \\ -\mu((x,0]) &\text{if x < 0 } \\ 0 &\text{if x=0} \end{cases}$$
Then $F$ is increasing càdlag function and $\mu = \mu_{F}$. This also shows that every measure in $\mathbb{R}$ which are finite in compact sets is regular.
-
Minor correction: It's "càdlàg", continue à droite, limite à gauche. – kahen Apr 22 '14 at 19:10
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https://fr.mathworks.com/matlabcentral/answers/1808170-rounding-off-a-number-to-5-decimal-digits
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# Rounding off a number to 5 decimal digits
17 views (last 30 days)
Mantej Sokhi on 19 Sep 2022
Commented: Image Analyst on 19 Sep 2022
Suppose I have a number .. let's say x = 1.4534567809385. If I want to round this off to 5 decimal digits I should get x = 1.45346. However when I try doing this in Matlab using the round function or creating my own function to round it off to 5 decimal digits I am always getting 4 decimal digits. How can I fix this ?
##### 0 CommentsShowHide -1 older comments
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### Accepted Answer
Image Analyst on 19 Sep 2022
It's just displaying 4 places because you're using format short. Switch to format long and you'll see it correctly.
x = 1.4534567809385;
x5 = round(x, 5)
x5 = 1.4535
format long
x = 1.4534567809385;
x5 = round(x, 5)
x5 =
1.453460000000000
##### 2 CommentsShowHide 1 older comment
Image Analyst on 19 Sep 2022
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https://www.easycalculation.com/analytical/adding-multiple-like-fractions-calculator.php
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English
Adding Like Fractions Calculator
Here is the online analytical calculator to add Multiple Fractions with Same Denominators. Adding like fractions involves simple process of adding only the numerators and not the denominators. The denominator of resultant fractions remains the same. In the below Adding Like Fractions Calculator, enter the number of inputs say 2 rows, it calculates only for two fractions. Then enter values for each fraction and click calculate button to get the resultant values.
Add Multiple Fractions with Same Denominators
=
Code to add this calci to your website
Formula:
Sum of Like Denominators = (a1 + b2 + a3 + ...) / b Where, a1, a2, a3, ... = Numerators b = b1, b2, b3, ... = Denominators
Example:
Add Multiple fractions with same denominators for the values a1 as 5, a2 as 6, b1 and b2 as 7.
Solution:
Sum of Like denominators = (a1 + b2) / b
= (5 + 6) / 7
= 11 / 7
= 1.5714
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## G = D4⋊D14order 224 = 25·7
### 2nd semidirect product of D4 and D14 acting via D14/C14=C2
Series: Derived Chief Lower central Upper central
Derived series C1 — C28 — D4⋊D14
Chief series C1 — C7 — C14 — C28 — D28 — C2×D28 — D4⋊D14
Lower central C7 — C14 — C28 — D4⋊D14
Upper central C1 — C2 — C2×C4 — C4○D4
Generators and relations for D4⋊D14
G = < a,b,c,d | a4=b2=c14=d2=1, bab=dad=a-1, ac=ca, cbc-1=a2b, dbd=a-1b, dcd=c-1 >
Subgroups: 326 in 68 conjugacy classes, 29 normal (21 characteristic)
C1, C2, C2, C4, C4, C22, C22, C7, C8, C2×C4, C2×C4, D4, D4, Q8, C23, D7, C14, C14, M4(2), D8, SD16, C2×D4, C4○D4, C28, C28, D14, C2×C14, C2×C14, C8⋊C22, C7⋊C8, D28, D28, C2×C28, C2×C28, C7×D4, C7×D4, C7×Q8, C22×D7, C4.Dic7, D4⋊D7, Q8⋊D7, C2×D28, C7×C4○D4, D4⋊D14
Quotients: C1, C2, C22, D4, C23, D7, C2×D4, D14, C8⋊C22, C7⋊D4, C22×D7, C2×C7⋊D4, D4⋊D14
Smallest permutation representation of D4⋊D14
On 56 points
Generators in S56
```(1 22 14 20)(2 23 8 21)(3 24 9 15)(4 25 10 16)(5 26 11 17)(6 27 12 18)(7 28 13 19)(29 56 36 49)(30 43 37 50)(31 44 38 51)(32 45 39 52)(33 46 40 53)(34 47 41 54)(35 48 42 55)
(1 54)(2 48)(3 56)(4 50)(5 44)(6 52)(7 46)(8 55)(9 49)(10 43)(11 51)(12 45)(13 53)(14 47)(15 36)(16 30)(17 38)(18 32)(19 40)(20 34)(21 42)(22 41)(23 35)(24 29)(25 37)(26 31)(27 39)(28 33)
(1 2 3 4 5 6 7)(8 9 10 11 12 13 14)(15 16 17 18 19 20 21)(22 23 24 25 26 27 28)(29 30 31 32 33 34 35 36 37 38 39 40 41 42)(43 44 45 46 47 48 49 50 51 52 53 54 55 56)
(1 17)(2 16)(3 15)(4 21)(5 20)(6 19)(7 18)(8 25)(9 24)(10 23)(11 22)(12 28)(13 27)(14 26)(30 42)(31 41)(32 40)(33 39)(34 38)(35 37)(43 48)(44 47)(45 46)(49 56)(50 55)(51 54)(52 53)```
`G:=sub<Sym(56)| (1,22,14,20)(2,23,8,21)(3,24,9,15)(4,25,10,16)(5,26,11,17)(6,27,12,18)(7,28,13,19)(29,56,36,49)(30,43,37,50)(31,44,38,51)(32,45,39,52)(33,46,40,53)(34,47,41,54)(35,48,42,55), (1,54)(2,48)(3,56)(4,50)(5,44)(6,52)(7,46)(8,55)(9,49)(10,43)(11,51)(12,45)(13,53)(14,47)(15,36)(16,30)(17,38)(18,32)(19,40)(20,34)(21,42)(22,41)(23,35)(24,29)(25,37)(26,31)(27,39)(28,33), (1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21)(22,23,24,25,26,27,28)(29,30,31,32,33,34,35,36,37,38,39,40,41,42)(43,44,45,46,47,48,49,50,51,52,53,54,55,56), (1,17)(2,16)(3,15)(4,21)(5,20)(6,19)(7,18)(8,25)(9,24)(10,23)(11,22)(12,28)(13,27)(14,26)(30,42)(31,41)(32,40)(33,39)(34,38)(35,37)(43,48)(44,47)(45,46)(49,56)(50,55)(51,54)(52,53)>;`
`G:=Group( (1,22,14,20)(2,23,8,21)(3,24,9,15)(4,25,10,16)(5,26,11,17)(6,27,12,18)(7,28,13,19)(29,56,36,49)(30,43,37,50)(31,44,38,51)(32,45,39,52)(33,46,40,53)(34,47,41,54)(35,48,42,55), (1,54)(2,48)(3,56)(4,50)(5,44)(6,52)(7,46)(8,55)(9,49)(10,43)(11,51)(12,45)(13,53)(14,47)(15,36)(16,30)(17,38)(18,32)(19,40)(20,34)(21,42)(22,41)(23,35)(24,29)(25,37)(26,31)(27,39)(28,33), (1,2,3,4,5,6,7)(8,9,10,11,12,13,14)(15,16,17,18,19,20,21)(22,23,24,25,26,27,28)(29,30,31,32,33,34,35,36,37,38,39,40,41,42)(43,44,45,46,47,48,49,50,51,52,53,54,55,56), (1,17)(2,16)(3,15)(4,21)(5,20)(6,19)(7,18)(8,25)(9,24)(10,23)(11,22)(12,28)(13,27)(14,26)(30,42)(31,41)(32,40)(33,39)(34,38)(35,37)(43,48)(44,47)(45,46)(49,56)(50,55)(51,54)(52,53) );`
`G=PermutationGroup([[(1,22,14,20),(2,23,8,21),(3,24,9,15),(4,25,10,16),(5,26,11,17),(6,27,12,18),(7,28,13,19),(29,56,36,49),(30,43,37,50),(31,44,38,51),(32,45,39,52),(33,46,40,53),(34,47,41,54),(35,48,42,55)], [(1,54),(2,48),(3,56),(4,50),(5,44),(6,52),(7,46),(8,55),(9,49),(10,43),(11,51),(12,45),(13,53),(14,47),(15,36),(16,30),(17,38),(18,32),(19,40),(20,34),(21,42),(22,41),(23,35),(24,29),(25,37),(26,31),(27,39),(28,33)], [(1,2,3,4,5,6,7),(8,9,10,11,12,13,14),(15,16,17,18,19,20,21),(22,23,24,25,26,27,28),(29,30,31,32,33,34,35,36,37,38,39,40,41,42),(43,44,45,46,47,48,49,50,51,52,53,54,55,56)], [(1,17),(2,16),(3,15),(4,21),(5,20),(6,19),(7,18),(8,25),(9,24),(10,23),(11,22),(12,28),(13,27),(14,26),(30,42),(31,41),(32,40),(33,39),(34,38),(35,37),(43,48),(44,47),(45,46),(49,56),(50,55),(51,54),(52,53)]])`
41 conjugacy classes
class 1 2A 2B 2C 2D 2E 4A 4B 4C 7A 7B 7C 8A 8B 14A 14B 14C 14D ··· 14L 28A ··· 28F 28G ··· 28O order 1 2 2 2 2 2 4 4 4 7 7 7 8 8 14 14 14 14 ··· 14 28 ··· 28 28 ··· 28 size 1 1 2 4 28 28 2 2 4 2 2 2 28 28 2 2 2 4 ··· 4 2 ··· 2 4 ··· 4
41 irreducible representations
dim 1 1 1 1 1 1 2 2 2 2 2 2 2 2 4 4 type + + + + + + + + + + + + + + image C1 C2 C2 C2 C2 C2 D4 D4 D7 D14 D14 D14 C7⋊D4 C7⋊D4 C8⋊C22 D4⋊D14 kernel D4⋊D14 C4.Dic7 D4⋊D7 Q8⋊D7 C2×D28 C7×C4○D4 C28 C2×C14 C4○D4 C2×C4 D4 Q8 C4 C22 C7 C1 # reps 1 1 2 2 1 1 1 1 3 3 3 3 6 6 1 6
Matrix representation of D4⋊D14 in GL4(𝔽113) generated by
58 75 0 0 38 55 0 0 17 23 66 38 94 30 37 47
,
82 36 8 43 87 46 65 106 4 30 112 5 46 108 33 99
,
80 80 0 0 33 9 0 0 103 111 105 33 67 53 47 32
,
80 80 0 0 9 33 0 0 29 87 68 58 82 91 82 45
`G:=sub<GL(4,GF(113))| [58,38,17,94,75,55,23,30,0,0,66,37,0,0,38,47],[82,87,4,46,36,46,30,108,8,65,112,33,43,106,5,99],[80,33,103,67,80,9,111,53,0,0,105,47,0,0,33,32],[80,9,29,82,80,33,87,91,0,0,68,82,0,0,58,45] >;`
D4⋊D14 in GAP, Magma, Sage, TeX
`D_4\rtimes D_{14}`
`% in TeX`
`G:=Group("D4:D14");`
`// GroupNames label`
`G:=SmallGroup(224,144);`
`// by ID`
`G=gap.SmallGroup(224,144);`
`# by ID`
`G:=PCGroup([6,-2,-2,-2,-2,-2,-7,218,188,579,159,69,6917]);`
`// Polycyclic`
`G:=Group<a,b,c,d|a^4=b^2=c^14=d^2=1,b*a*b=d*a*d=a^-1,a*c=c*a,c*b*c^-1=a^2*b,d*b*d=a^-1*b,d*c*d=c^-1>;`
`// generators/relations`
×
𝔽
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# Parent Child derivation
#### JackNijssen
##### New Member
Hi,
In below example in column A+B, i have hierarchy on the node level of the parent in column A and the child in column B. I would like to derive the text of each parent in column C, like below. Is there a formula for column C how i can derive this value?
Column A Column B desired output Column C with parent
1 AA
2 BB AA
3 CC BB
4 DD CC
5 EE DD
5 EEE DD
5 EEEE DD
5 EEEEE DD
Thanks a lot, Jack
### Excel Facts
Whats the difference between CONCAT and CONCATENATE?
The newer CONCAT function can reference a range of cells. =CONCATENATE(A1,A2,A3,A4,A5) becomes =CONCAT(A1:A5)
#### VBE313
##### Active Member
Create a column that says "1" or "0" if the level is 1 aka Parent Part.
To the left of that columns, put the sums of the child parts you want to sum.
to the right of the "1" or "0" column, put this foumula in
=IFERROR(IF(L1=1,SUM(K2:INDEX(\$L2:\$L\$10,MATCH(TRUE,(\$L2:\$L\$10=1),0)),-1),""),IF(L4=1,SUM(K2:INDEX(\$L2:\$L\$10,MATCH(TRUE,(\$L2:\$L\$10=1),0)),-1),""))
Press Ctrl+Shift+Enter when entering the formula
In this example,
Column L is the "1" or "0" column
Column K is the sum of the child parts
Column M is where you enter this formula
If you need to sum up "Make" components that roll up to "Top Level"/Parent Parts, Then message me and I can show you what I have been working on for the past few months on how to accomplish this.
Thanks,
Last edited:
#### JackNijssen
##### New Member
Create a column that says "1" or "0" if the level is 1 aka Parent Part.
To the left of that columns, put the sums of the child parts you want to sum.
to the right of the "1" or "0" column, put this foumula in
=IFERROR(IF(L1=1,SUM(K2:INDEX(\$L2:\$L\$10,MATCH(TRUE,(\$L2:\$L\$10=1),0)),-1),""),IF(L4=1,SUM(K2:INDEX(\$L2:\$L\$10,MATCH(TRUE,(\$L2:\$L\$10=1),0)),-1),""))
Press Ctrl+Shift+Enter when entering the formula
In this example,
Column L is the "1" or "0" column
Column K is the sum of the child parts
Column M is where you enter this formula
If you need to sum up "Make" components that roll up to "Top Level"/Parent Parts, Then message me and I can show you what I have been working on for the past few months on how to accomplish this.
Thanks,
Hi, in the meanwhile i found the formula. = IF(A10>A9;B9;IF(A10=A9;C9:""))
Jack
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Binary Search Tree in C/C++ With Examples [Latest]
Oct 23, 2022
In this article, we’ll take a look at implementing a Binary Search Tree in C/C++.
A Binary Search Tree(BST) is a Binary Tree in which every element of a left sub-tree is less than the root node, and every element in the right sub-tree is greater than it.
This definition applies to every node in the tree, starting from the root node.
For example, the below figure shows a Binary Search Tree.
Here, the left sub-tree from the root contains elements lesser than it, and the right sub-tree has those greater than it. This also applies to every node in the tree, as you can observe.
Due to this, BSTs are often called “sorted binary trees”, since an in-order traversal (refer to this article for what it means) will give a sorted list of elements.
Let us now understand some of the concepts involved, before implementing them.
Create the Data Structures for the Binary Search Tree in C/C++
Let’s write the structures and some helper functions for our BST.
Any Binary Search Tree node has a `data` element, along with pointers to it’s `left` and `right` children. It also has a marker `is_leaf`, to check if it’s a leaf node.
Let’s write our structure now
Now, we’ll make a function to create a new tree node, that initializes the parameters.
Let’s also write a function that frees our complete Binary Search Tree in C/C++ from memory.
Let’s now move onto the `insert_bst()` method, that inserts a node into the BST.
Inserting Data into a Binary Search Tree
Any Binary Search Tree (BST) has the property that:
• The left sub-tree has elements of less value than the root node
• The right sub-tree has elements greater than the root node
So, if we want to insert a node, we must ensure that our node must have its correct position, by comparing the key of the node with the current root.
To visualize how the insertion is being done, let’s take an example.
Consider the below tree, with only one node.
Let’s insert 20 to this tree. Since 45 > 20, we must insert it to the left sub-tree. Since our sub-tree is empty, we can directly add to it.
Now, let’s insert 15. Since 15 < 45, we insert to the left sub-tree and update the current root. Now again, we check with the current root (20). 15 < 20, so we again move to the left and insert it.
Let’s insert 60 now. Since 60 > 45, we insert to the right.
Similarly, you can insert the elements 40, 50 and 70 to get our final BST.
The algorithm for insertion is as follows:
• Start at the root node of the tree. If it does not exist, simply make the new node as the root node and return it!
• Otherwise, we must compare the key of the node to be inserted and the key of the root.
• If the key of the node is lesser than the root, we need to insert it to the left sub-tree
• Otherwise, insert to its right sub-tree.
Our method has the signature of:
This takes the pointer to the root node and inserts `data` into it.
If the root node is `NULL` (doesn’t exist), we simply create a new node using `data` and assign it as the new root node.
Let’s now move on to the `else` part, if we need to insert it to an existing node. We simply follow the algorithm described above, with additional checks for the leaf node base case.
The comments in the code should be self-explanatory. We insert to the left sub-tree if `data` < `root->data` and to the right, if `data` > `root->data`.
Now we’ve completed our insert procedure! Let’s now quickly complete the `search_bst()` function as well.
Searching a Binary Search Tree
This is very straightforward. We simply move to the left/right sub-trees based on the key comparison. We stop if we either reach a `NULL` node, or if the current root node key matches our target.
Now that we’ve implemented both insert and search, let’s test if our program works correctly as of now.
I’ll post the complete code until now, with additional functions for printing the BST in an in-order traversal.
Output
Alright! This seems to work as expected, and we do get the sorted list of elements by in-order traversal.
Let’s now move onto the delete method.
Delete from a Binary Search Tree
This one is a bit more tricky compared to the rest. Since the iterative version is less intuitive, we’ll present a recursive algorithm for this.
We have multiple cases for deletion. The below algorithm is the most commonly used version for deletion from a BST.
• If a node has no children, simply delete it from the tree.
• Else, it has one child, remove the node and replace with its child. It doesn’t matter if it is the left or the right child.
• If it has 2 children, this case is a bit more tricky. Here, we don’t delete this node. Rather, we find the in-order successor of that node and remove that node instead, after copying the key to the current node.
The in-order successor of a node is the next node that comes after in the in-order traversal.
For example, the in-order successor of the root node (45), is the node 50, since it is the next node that is greater than it.
Since it is the next node in the sorted order, it is the leftmost node in the right sub-tree of the node!
This method will return the in-order successor of a node in a tree, assuming that a right sub-tree exists.
Now, we can write our `delete_bst()` method, using the above algorithm. The code is shown below.
We’ve implemented our `delete_bst()` method as well! So, I’ll now post the complete code below.
Output
Hooray! This seems to work as expected, after deleting `50` and `45`. We’ve finally completed all our procedures, and we’re done!
Time Complexity of Implementation
The time complexity of the main procedures are given in the below table
The problem with BSTs is that since the insertion can happen arbitrarily, the tree can become skewed, with the height of the BST being potentially proportional to the number of elements!
So now, the insert, search, and delete operations are equivalent to O(n) in the absolute worst case, instead of the expected O(logn)
To avoid this problem, some modifications to the BST model were proposed, and a popular data structure is called the AVL Tree. We’ll give you more details in our upcoming article, so stay tuned!
You can get the code through a Github Gist that I’ve uploaded. Feel free to ask any questions or provide suggestions in the comment section below!
Conclusion
Hopefully, you’ve learned how you can implement the methods for inserting, search and deleting into a Binary Search Tree in C/C++.
Our next article on this series will focus on AVL Trees, which aims to eliminate the problem of encountering a skewed tree, so stay tuned!
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# What is the compensated demand function?
What is the compensated demand function?
## What is the compensated demand function?
In microeconomics, a consumer’s Hicksian demand function or compensated demand function for a good is his quantity demanded as part of the solution to minimizing his expenditure on all goods while delivering a fixed level of utility.
What is compensated and uncompensated demand curve?
Compensated demand, Hicksian demand, is a demand function that holds utility fixed and minimizes expenditures. Uncompensated demand, Marshallian demand, is a demand function that maximizes utility given prices and wealth.
### What is the difference between ordinary demand function and compensated demand function?
Compensated demand curve shows the relationship between the price of a good and the quantity demanded of it assuming that the prices of other goods and our level of utility remain (constant). Second, the ordinary demand curve or the marshallian demand curve – illustrate how much people will buy at a given price.
What is the expenditure function in economics?
In microeconomics, the expenditure function gives the minimum amount of money an individual needs to spend to achieve some level of utility, given a utility function and the prices of the available goods.
## What is compensated elasticity of demand?
A compensated demand curve ignores the income effect of a price change. It only measures the substitution effect. A compensated demand curve is therefore less elastic than an ordinary demand curve.
What is the difference between compensated demand curve and uncompensated demand curve?
### What is Slutsky demand curve?
The Slutsky equation (or Slutsky identity) in economics, named after Eugen Slutsky, relates changes in Marshallian (uncompensated) demand to changes in Hicksian (compensated) demand, which is known as such since it compensates to maintain a fixed level of utility.
What is income compensated demand curve?
An income-compensated demand curve is a variant of the demand curve for a good, service, or commodity where changes in price are accompanied by offsetting changes in income so as to control for the income effect.
## How do you calculate demand curve?
Examples of Deadweight Loss Formula (With Excel Template) Let’s take an example to understand the calculation of Deadweight Loss in a better manner.
• Explanation.
• Relevance and Use of Deadweight Loss Formula.
• What is a perfectly competitive demand curve?
– There are large number of sellers and buyers. Number is so large that single seller or buyer cannot influence industry supply and demand by their own individual action. – Products are homogeneous i.e. products are similar in each and every aspect. – Firms are price taker i.e firms accept the price established by industry demand and supply condition.
### What characteristics lead to a downward sloping demand curve?
The price of the good or service.
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So the point is we have no idea what is below us. We can not dig down very far and find out. Or at least we have not dug down more than 7.5 miles so far… But if the earth is flat you could theoretically dig through it. But we have not dug down more than 7.5 miles, which is nothing compared to the 8,000 mile diameter they give for the earth…. So the point is we can’t dig down very far so we don’t know…
Astronomers tell us. that the Moon, goes round the Earth in about 28 days. Well, we may see her making her journey round every. day, if we make use of our eyes and these are about the best things we have to use. The Moon falls behind in her daily motion as compared with that of the Sun to the extent of one revolution in the time specified; but that is not making a revolution. Failing to go as fast as other bodies go in one direction does not constitute a going round in the opposite one - as the astronomers would have us believe! And, since all this absurdity has been rendered necessary for no other purpose than to help other absurdities along, it is clear that the astronomers are on the wrong track; and it needs no long train of reasoning to show that we have found a proof that the Earth is not a globe.
Look up Admiral Byrd's tape from the 1920's (you-tube) he told us of the many continents beyond ours, all on the same plane. He said it would take years to explore them all. Then he retired from the navy and they started NASA to hide all that stuff. Yep thousands of NASA employees know the truth and are trained to keep us stupid. shows like coast to coast will never talk about the flat earth, wont even discus it, only to say NASA is correct case closed. Sounds like brain washing to me.
39) Practical distance measurements taken from “The Australian Handbook, Almanack, Shippers’ and Importers’ Directory” state that the straight line distance between Sydney and Nelson is 1550 statute miles. Their given difference in longitude is 22 degrees 2’14”. Therefore if 22 degrees 2’14” out of 360 is 1550 miles, the entirety would measure 25,182 miles. This is not only larger than the ball-Earth is said to be at the equator, but a whole 4262 miles greater than it would be at Sydney’s southern latitude on a globe of said proportions.
I am new to flat earth and see way more logical proof to this idea than the spinning globe. I have often wondered about what we are taught as it didn't make sense but we believe because that's what we been taught. One item I have not seen in my research and maybe you can address. If one were to stand precisely at the point where the imaginary axis is that the globe spins on, at the North Pole area, or even say 1/4 mile away,or ten feet away, what the person would see or experience I think would be bizarre....essentially rotating in a 1/4 mile or ten foot circle during the 24 hr day...more bizarre if they were on the exact axis point. This I think could be proven easily by going there, but it seems to not have been, which to me adds more proof to flat earth.I am sure if one were to go there, they would not see any following a circular pattern as the 24 hrs pass. They would be stationary!
78) From Anchorage, Alaska at an elevation of 102 feet, on clear days Mount Foraker can be seen with the naked eye 120 miles away. If Earth were a ball 25,000 miles in circumference, Mount Foraker’s 17,400 summit should be leaning back away from the observer covered by 7,719 feet of curved Earth. In reality, however, the entire mountain can be quite easily seen standing straight from base to summit.
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Plot solutions to a differential sys with initial conditions from loop?
Posted 1 month ago
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I have been working with potentially chaotic systems and it would be extremely helpful if I could plot many solutions to the equation starting from different initial conditions on the same graph. What I tried to do was create a loop that picks a starting value for each equation on the system and plots a new solution as the loop iterates, but my basic knowledge of this program failed to produce anything that worked: a = .1 M = .2 J = .1 For[i = 0, i < M, i += J, For[u = 0, u < M, u += J, For[o = 0; o < M , o += J, {x[][t], y[][t], z[][t]} /. sol2 sol2 = ParametricNDSolve[{x'[t] == (Sin[y[t]] - b*x[t])* Cos[t]*(-Sin[t]* Cos[ArcTan[a*t]] + (a*Cos[t]*(-Sin[ArcTan[a*t]]))/(1 + a*t^2)), y'[t] == (Sin[x[t]] - b*z[t])* Sin[t]* (Cos[t]* Cos[ArcTan[a*t]] + (a*Sin[t]*(-Sin[ArcTan[a*t]]))/(1 + a*t^2) ), z'[ t] == (Sin[z[t]] - b*y[t])*((-Cos[ArcTan[a*t]]*a)/(1 + a*t^2)), x[0] == i, y[0] == u, z[0] == o}, {x, y, z}, {t, -100, 100}, {a}] ParametricPlot3D[ Evaluate[{x[.1][t], y[.1][t], z[.1][t]} /. sol2], {t, -100, 100}, PlotRange -> All, BoxRatios -> {1, 1, 1}, PlotStyle -> Tube[.0005]]]]] Obviously, my coding is by no means elegant let alone functional but the idea was that M and J would determine how many plots the system would run. Basically, what I have is a mess and was hoping someone could help me out.
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Posted 1 month ago
Michael,ParametricNDSolve already does much of what you want. You first need to clear out values you previously set Clear[a, b, i, u, o] Solve your problem: sol = ParametricNDSolve[{x'[t] == (Sin[y[t]] - b*x[t])* Cos[t]*(-Sin[t]* Cos[ArcTan[a*t]] + (a*Cos[t]*(-Sin[ArcTan[a*t]]))/(1 + a*t^2)), y'[t] == (Sin[x[t]] - b*z[t])* Sin[t]*(Cos[t]* Cos[ArcTan[a*t]] + (a*Sin[t]*(-Sin[ArcTan[a*t]]))/(1 + a*t^2)), z'[t] == (Sin[z[t]] - b*y[t])*((-Cos[ArcTan[a*t]]*a)/(1 + a*t^2)), x[0] == i, y[0] == u, z[0] == o}, {x, y, z}, {t, -100, 100}, {a, b, i, u, o}] I chose a,b,i,u,o as parameters, you should probably narrow that list. You can now plot results. I do not understand what you want to show exactly so I made up a plot: Plot[Evaluate@ Table[Tooltip[y[.1, 0.2, 0, 0, s][t] /. sol, s], {s, 0.3, 1.9, 0.1}], {t, 0, 10}] The mouse will give you the parameter value. You can forego the tooltip and make a grid of plots or any other format.Regards,Neil
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# Difference between Step-Up and Step-Down Transformer
## What is a Transformer?
A transformer is an electrical device whose main function is to transmit alternating current between two circuits at a constant value of frequency, but at varying voltage levels. A transformer is constructed by combining two kinds of coil. The first coil carrying the magnetic flux is called the primary coil, while the second coil placed opposite to it, in which electromotive force (E.M.F) is induced is called as the secondary coil.
Figure-1 illustrates a standard representation of a transformer. Input voltage at the primary side of the transformer produces the varying magnetic flux on the primary coil. While the varying magnetic flux induces EMF in the secondary coil. Based on the magnitude of voltage carried by the transformer, it is divided into two categories, i.e. step-up transformer and step down transformer, which is discussed in the following sections.
The variables shown in Figure-1 above is defined as follows −
• NP = Total number of turns in the primary winding,
• NS = Total number of turns in the secondary winding,
• VP = Voltage applied to the primary winding,
• VS = Voltage output from the secondary winding,
• IP = Primary winding current,
• IS = Secondary winding current,
• Φ = Magnetic flux flowing through the transformer core.
## What is a Step-Up Transformer?
A Step-Up Transformer is a kind of transformer which increases the magnitude of output voltage, while keeping a low current magnitude. The secondary winding of the step-up transformer has greater number of turns as compared to the primary winding. The main function of the step-up transformer is to transform lower voltage and high current received or induced at the primary coil of the transformer, to higher voltage and lower current at the secondary coil of the transformer.
The step-up transformer finds its application in transmitting power at longer distances wherein it transmits large amount of power at high voltages. Figure-2 illustrates the symbolic representation of the step-up transformer. The following expression shows the relation between coil turns and voltage values in the primary and secondary coils.
$$\mathrm{Turns \: Ratio,K=\frac{N_{1}}{N_{2}}=\frac{V_{1}}{V_{2}}=\frac{I_{2}}{I_{1}}}$$
And,
$$\mathrm{N_{2}> N_{1};V_{2}> V_{1};I_{2}< I_{1}}$$
Where,
• N1 = Number of turn in primary winding,
• N2 = Number of turns in secondary winding,
• V1 = Primary side input voltage,
• V2 = Secondary side output voltage
The above relationship is called as the Turns Ratio, which is used to determine whether the transformer is a step-up or step-down transformer
## What is a Step-Down Transformer?
A Step-Down Transformer is a kind of transformer which reduces the magnitude of output voltage, while keeping a high current magnitude. The secondary winding of the step-down transformer has lesser number of turns as compared to the primary winding. The main function of the step-down transformer is to transform higher voltage and lower current received or induced at the primary coil of the transformer, to lower voltage and higher current at the secondary coil of the transformer.
The step-up transformer finds its application in transmitting power at shorter distances wherein the safety of the measuring instruments, home appliances or industrial equipment are considered. Figure-3 illustrates the symbolic representation of the step-down transformer.
The following expression shows the relation between coil turns and EMF values in the primary and secondary coils.
For the step-down transformer, we have,
$$\mathrm{N_{2}< N_{1};V_{2}< V_{1}; I_{2}> I_{1}}$$
## Difference between Step-Up and Step-Down Transformer
The major differences between the step-up transformer and step-down transformer are described in the table given below −
Parameter
Step-Up Transformer
Step-Down Transformer
Definition
Step-up transformer converts low voltage to higher output voltage at constant frequency.
Step-down transformer converts high voltage to low output voltage at constant frequency.
Magnitude of Voltage
Voltage at primary winding is low, while the voltage at the secondary winding is high.
Voltage at primary winding is high, while the voltage at the secondary winding is low.
Magnitude of Current
Current at primary winding is high, while the current at the secondary winding is low.
Current at primary winding is low, while the current at the secondary winding is high.
Number of Turns
The primary coil is comprised of less number of turns in comparison to the secondary winding.
The secondary coil is comprised of less number of turns in comparison to the primary winding.
Standard Output Voltage Values
The output voltage generated at the output of the step-up transformer is 3.3 kV, 6.6 kV, 11 kV or greater.
The output voltage generated at the output of the step-down transformer is 10V, 20V, 110V, 220 V, 440 V, etc.
Dimension of the Transformer Windings
The primary winding is larger in size as it is made up of thick insulated copper wiring.
The secondary winding is larger in size as it is made up of thick insulated copper wiring.
Expression or Relation
The relationship between primary winding and secondary winding parameters of step-up transformer is,
$$\mathrm{N_{2}> N_{1};V_{2}> V_{1}; I_{2}< I_{1}}$$
The relationship between primary winding and secondary winding parameters of step-down transformer is,
$$\mathrm{N_{2}< N_{1};V_{2}< V_{1}; I_{2}> I_{1}}$$
Applications
The step-up transformer finds its application in extracting X-rays,microwaves, ovens, inverters, power plants, substations, etc.
The step-down transformers find their applications in main adapters, CD players, welding machines, television, doorbells, distribution substations, etc.
## Conclusion
In conclusion, the major differences between step-up and step-down transformers were listed in this article. The most significant difference between them is that, the step-up transformer is used to convert low-end voltages to high end voltage, whereas, the step-down transformer is used to transform the high-end voltage to low-end voltage. Both these transformers have several residential and industrial applications as described in this article.
Updated on: 19-Apr-2023
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# Relative Momentum Question
1. Nov 13, 2013
### MathewsMD
An 80 kg astronaut has become detached from the safety line connecting her to the International Space Station. She's 200m from the station, at rest relative to it, and has 4 min of air remaining. To get herself back, she tosses a 10 kg tool kit away from the station at 8 m/s. Will she make it back in time?
In the question, the solution uses momentum and calculates it for both the astronaut and tool box since they are at rest only relative to the space station, not the rest frame. Here it shows initial momentum is equal to 0, though. If they are calculating momentum in the rest frame, shouldn't there be an initial momentum of p = (ma + mt)vrel, but it is 0...why? Is not the astronaut and toolbox in motion initially and once again afterwards? Any explanation on why the momentum is paired the way it is would be very helpful! :)
2. Nov 14, 2013
### mic*
What is the rest frame, if it is not a frame of reference where the object/s of interest are at rest?
No other relative position or motion information has been given in this problem, other than " She's 200m from the station, at rest relative to it".
Following from this, the formula you have stated for initial momentum is correct - p = (ma + mt)vrel.
Now what value would you choose for vrel?
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# Fluff reading - how do they annotate the date?
## Recommended Posts
Various entries in the books write the date with a series of digits followed by M41.
Example: "853779.M41" on page 8 or 9 of the Achilus Assault. What date is that?
I'm guessing it's the 779th year of the 41st millennium? So year 41,779? Or do they mean 40,779? And what does the 853 mean?
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The numerals before that are "check numbers" that indicate the reliability of the message, depending both on the distance to Terra (where the event will eventually be logged) as well as the degree of "directness", meaning by how many astropathic relays it has passed.
The 6E rulebook had a lengthy explanation of this in its fluff appendix.
Edited by Lynata
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Hmmm, why is it not included in Core? Weird. Anyway, the first digit is the check number, which is often omitted. The next 3 digits are the year fraction, and the last three digits are indicating the year itself. Your example, 853779.M41 is around the end of year 779 of the 41st millennium, that is 40,779 (millennia work the same way as centuries, the first goes from 0001-1000, presuming they kept our Common Era where there is no year 0 - if there is, the turning point is a year earlier, the millennium spanning between 0000-0999).
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Right!
And here's the picture from the TT book:
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In more detail.
The check digit, as noted, is a measure of distance from Terra and/or accuracy (all dates are Terran standard, so distance and accuracy are the same thing). 0 or 1 for the check digit means the recorded event takes place on Terra itself, and is thus as accurate as possible. 7+ tends to cover approximate and estimated dates rather than reported events.
The year fraction - noted as the Chronosegment in the 6e 40k rulebook - is a thousandth of a Terran standard year. Consequently, each segment is roughly 8 hours and 45 minutes long. It's fair to assume that nobody except the Administratum actually uses that value for practical purposes - it's a record-keeping tool. That said, I can imagine some worlds using it for work cycles - each segment is a work shift, each person gets a work shift, a sleep shift, and a shift for leisure and prayer.
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I'd imagine segmenting the year in such a way would be a godsend (sorry Emperor send) if you had a spread sheet with literally billions of entries from all over the galaxy and you were told to put them in chronological order. That said travel and communication is so slow relatively speaking in the Imperium that I doubt such precise dates matter much to generals and other tactical commanders.
When Gming I use a convention of using the official time at the beginning of a mission but then also a 'local' time amd date stamp. Well I use this for relevant handouts anyway.
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thanks all, makes more sense now.
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I'd imagine segmenting the year in such a way would be a godsend (sorry Emperor send) if you had a spread sheet with literally billions of entries from all over the galaxy and you were told to put them in chronological order. That said travel and communication is so slow relatively speaking in the Imperium that I doubt such precise dates matter much to generals and other tactical commanders.
When Gming I use a convention of using the official time at the beginning of a mission but then also a 'local' time amd date stamp. Well I use this for relevant handouts anyway.
Exactly so. When you're talking about tracking events on an interstellar scale minutes become largely irrelevant anyway.
one year fraction 'tick' is about nine hours in practice, which makes it a nice "later that day" value.
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# Integration of ((e^x)/(x^2)) with respect to x
1. Homework Statement
$$\int{\frac{e^x}{x^2}dx}$$
2. Homework Equations
• Integration by substitution
• Integration by parts: $\int{u\ dv}=uv\ -\ \int{v\ du}$
3. The Attempt at a Solution
Since it was clear that integration by substitution would not work, I tried integration by parts. Since the $e^x$ term would not be affected whatsoever with the application of differentiation or integration, I worked out the $x^2$ term instead. So, I took $u=e^x$ and $dv=\frac{1}{x^2}\ dx$. It resulted in:
$$\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+\int {\frac{e^x}{x}dx}$$
For the last term, I took $u=e^x$ and $dv=\frac{1}{x}\ dx$. It resulted in:
$$\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+e^x\ ln(x)-\int{(ln x)(e^x)dx}$$
Beyond this point, if I took $u=ln(x)$ and $dv=e^x\ dx$, I would just undo the previous steps. If I took $u=e^x$ and $dv=ln(x)\ dx$, it resulted in:
$$\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}\ +\ e^x\ ln(x)\ -\ e^x\left ( x\ ln(x)\ -\ x \right )\ +\ e^x\ -\ x\ e^x\ +\ \int{x\ ln(x)\ e^x\ dx}}$$
The last term certainly shows that this technique won't solve the problem at hand because it will continue forever.
How should I attack this problem?
Thank you.
Best regards,
Eus
Related Calculus and Beyond Homework Help News on Phys.org
Hootenanny
Staff Emeritus
Gold Member
$$\int{\frac{e^x}{x^2}dx}=-\frac{e^x}{x}+e^x\ ln(x)-\int{(ln x)(e^x)dx}$$
Beyond this point, if I took $u=ln(x)$ and $dv=e^x\ dx$, I would just undo the previous steps.
Although this may seem crazy, your right here with your substitution. You need to do integration by parts twice such that you arrive at something of the form,
$$\int{\frac{e^x}{x^2}dx}=f(x)-\int{\frac{e^2}{x^2}dx}$$
Then you may write,
$$2\int{\frac{e^x}{x^2}dx}=f(x)$$
$$\int{\frac{e^x}{x^2}dx}=\frac{1}{2}f(x)$$
Last edited:
Although this may seem crazy, your right here with your substitution. You need to do integration by parts twice such that you arrive at something of the form,
$$\int{\frac{e^x}{x^2}dx}=f(x)-\int{\frac{e^2}{x^2}dx}$$
Then you may write,
$$2\int{\frac{e^x}{x^2}dx}=f(x)$$
$$\int{\frac{e^x}{x^2}dx}=\frac{1}{2}f(x)$$
I have tried it and it really did not work; it really undid the previous steps resulting in $0 = 0$.
Best regards,
Eus
Hootenanny
Staff Emeritus
Gold Member
In which case, I believe that the integrand has no elementary anti-derivative. Do you have some reason do believe an elementary anti-derivative exists? One can of course find a series representation of the integral.
You're right there is no elementary derivative. The answer is:
$$\int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c[/itex] Where $Ei$ is the http://en.wikipedia.org/wiki/Exponential_integral" [Broken]. Last edited by a moderator: Hootenanny Staff Emeritus Science Advisor Gold Member You're right there is no elementary derivative. The answer is: [tex]\int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c[/itex] Where $Ei$ is the http://en.wikipedia.org/wiki/Exponential_integral" [Broken]. Cheers SD Last edited by a moderator: You're right there is no elementary derivative. The answer is: [tex]\int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c[/itex] Where $Ei$ is the http://en.wikipedia.org/wiki/Exponential_integral" [Broken]. I have taken a look at the page on Wikipedia but it does not explain anything how a series of non-terminating integration by parts can be transformed to an exponential integral. Could you please tell me how to do the transformation? Or, is there any pointer to a webpage that shows how to do it? Thank you very much. Best regards, Eus Last edited by a moderator: Hootenanny Staff Emeritus Science Advisor Gold Member We simply define the exponential integral as [tex]Ei(x):=-\int^{\infty}_{-x}\frac{e^{-t}}{t}dt$$
There is no method to go from your integral to the exponential integral. This is what we mean when we say an integrand has no elementary anti-derivative, it means the integral cannot be evaluated using elementary functions, we cannot evaluate the integral and write it in terms of normal functions. Instead we simply say that if one integrates this, then this is what one will obtain.
Does that make sense?
Instead we simply say that if one integrates this, then this is what one will obtain.
Well, if I just tried to find out the answer, that fact would be enough for me. But, I want to know how in the past someone would be able to obtain that answer. Suppose a mathematician in the past encountered the same class of problem. Of course, the mathematician would have tried the integration by substitution and the integration by parts to no avail. So, the mathematician employed another method and found the answer to be something involving exponential integral. I want to know this "another method" and how it yields the exponential integral.
Would you please tell me this "another method" and how it yields the exponential integral? Any pointer will also help.
Thank you very much.
Best regards,
Eus
Hootenanny
Staff Emeritus
Gold Member
Well, if I just tried to find out the answer, that fact would be enough for me. But, I want to know how in the past someone would be able to obtain that answer. [..]
Would you please tell me this "another method" and how it yields the exponential integral? Any pointer will also help.
Thank you very much.
Best regards,
Eus
With all due respect Eus, did you read my previous post? There is no such method.
We simply define the exponential integral as
$$Ei(x):=-\int^{\infty}_{-x}\frac{e^{-t}}{t}dt$$
There is no method to go from your integral to the exponential integral. [...] Instead we simply say that if one integrates this, then this is what one will obtain.
Suppose a mathematician in the past encountered the same class of problem. Of course, the mathematician would have tried the integration by substitution and the integration by parts to no avail. So, the mathematician employed another method and found the answer to be something involving exponential integral. I want to know this "another method" and how it yields the exponential integral.
The mathematician would have probably tried many different methods to evaluate the integral, but would have failed. The simple fact of the matter is that we do not know how to integrate such an integrand. So your mathematician would have said "I don't know how to integrate, instead I'll just say that if you do integrate it you get the exponential function".
You just have to accept that some integrals are not solvable with any current method. A good example is
$$x^{x^2}$$
Although we can see why this is generally insoluble as it becomes quickly very large, and does not converge $$\lim_{x\rightarrow\infty}$$.
I suspect mathematicians would have found similar reasons why this integral is hard to pin down with elementary functions, and then given up and gone and had a cup of tea/coffee.
$\int \frac {e^x}{x^2}\;\rightarrow\; \frac {-e^x}{x}-Ei\;(1,-x)+c \equiv$ I don't know.
Last edited:
Gib Z
Homework Helper
I'm afraid the reason is deeper than that we can not do it with any current method or that we just don't know; it's because its not possible! It can be proven, using something like the Risch algorithm, that certain integrals can NOT be expressed in terms of a finite combination of elementary functions.
It's not that we can't find the integral of exp(-t) 1/t, but rather that no such function in terms of elementary functions exists! So instead we define it as a non-elementary function, which is perfectly fine.
With all due respect Eus, did you read my previous post? There is no such method.
Yes, I read your previous post. I just want to ensure myself that there is really no "another method" out there. Thank you for telling me about what happened to the mathematician
Thank you for all of your help.
Best regards,
Eus
Yes, I read your previous post. I just want to ensure myself that there is really no "another method" out there. Thank you for telling me about what happened to the mathematician
Thank you for all of your help.
Best regards,
Eus
I don't believe it's impossible anyway, surely if we invent 8 extra dimensions it's soluble.
For future reference though, no elementary solutions, means it's insoluble except in terms of a non explicit mathematical term which means, either "?" Or undefined or impossible. Except in the case of imaginary solutions, which I'm not sure whether they fall under elementary solutions or not, but are solutions.
Gib Z
Homework Helper
I don't believe it's impossible anyway, surely if we invent 8 extra dimensions it's soluble.
For future reference though, no elementary solutions, means it's insoluble except in terms of a non explicit mathematical term which means, either "?" Or undefined or impossible. Except in the case of imaginary solutions, which I'm not sure whether they fall under elementary solutions or not, but are solutions.
I'm not sure what this means, but have anyone even read my post? It can be proven that certain integrals can no have elementary anti derivatives. It's not a matter of it being really hard to find, its a matter if possibility. If it can be found but you just don't know how, that doesn't mean it has no elementary solutions, though it may not, it just means that you can't prove otherwise.
I'm not sure what this means, but have anyone even read my post? It can be proven that certain integrals can no have elementary anti derivatives. It's not a matter of it being really hard to find, its a matter if possibility. If it can be found but you just don't know how, that doesn't mean it has no elementary solutions, though it may not, it just means that you can't prove otherwise.
That's exactly what I just said. Obviously I wasn't clear, and the 8 extra dimensions is a swipe at my favourite punching bag, string "theory". I'm in a silly mood, it's not your fault.
Broadly speaking without redefining maths, some integrals are in fact impossible. But most I suppose are ? or undefined.
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Table 2. Comparison between means of peak values of all three knee joint angles in four turn phases using skis of three different waist widths. Means and standard deviations (in parentheses) by groups of skis, ANOVA results and paired sample t-tests are presented.
Joint motion Turn phases N angle(°) n=6 M angle(°) n=6 W angle(°) n=6 ANOVA, F statistics p-sph‡ Paired sample t-tests§
Abduction I 8.6 (.9) 7.2 (1.7) 7.7(2) 7.67 0.011* 1
S1 15 (3.5) 12.3 (2.9) 12.8(2) 1.71 0.232 NA
S2 16.9 (3.2) 13.2 (2.7) 13.3 (1.8) 4.92 0.034*
C 15.6 (2.9) 11.6 (1.5) 11.1 (3.1) 43.48 <0.001* 1,2
Int. rotation I 4.7 (1.2) 3.5(2) 4.2 (3.7) 1.1 0.352 NA
S1 10.8 (3.1) 9.2 (4.4) 7.8 (3.4) 12.07 0.008*
S2 12.3 (2.8) 11.7 (3.5) 9.2 (3.2) 16.62 0.003* 2,3
C 14.1 (3.3) 14.8 (2.5) 11.9 (1.5) 12.5 0.008* 2,3
Flexion I 65.4 (11.2) 63.6 (9.1) 56.7 (10.9) 9.06 0.002* 3
S1 55.4 (10.3) 55.2 (9.5) 50.9 (10.6) 5.16 0.063 3
S2 56.7 (10.5) 51.5 (5.3) 49.7 (7.2) 6.97 0.031* 2
C 65.3 (11.3) 64.9 (7.7) 58.9(9) 5.91 0.05* 3
† trend (0.05≤p≤0.1) of difference,
p-sph‡, p-sphericity adjusted (GG:Eps<0.75 HF:Eps>0.75),
*significant (p<0.05) difference,
N, narrow skis; M, medium skis; W wide skis
°, degrees; n, numbers of subjects skiing on each skis; I, initiation phase; S1, steering phase 1; S2, steering phase 2; C, completion phase; Int, internal.
§ significance level of p<0.05. Significant difference between three pairs of skis is indicated by numeral: 1, narrow versus medium; 2, narrow versus wide; 3, medium versus wide; NA, not applicable when ANOVA p>0.05.
¶ Box – Cox power transformation was performed
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# optimisation routines
## optimisation routines
I commonly work with problems of the form where a maximum needs to be evaluated for a non-linear user-defined function F(x), where x denotes either a scalar or vector of function inputs and an analytical description of the gradient does not exist. I commonly my own routines for this purpose, but was wondering whether a better option is available through the MKL. I have not been able to find any reference beyond the "non-linear least squares" procedure that requires the problem to be twice differentiable. Any advice would be greatly appreciated,
Justin.
4 posts / 0 new
For more complete information about compiler optimizations, see our Optimization Notice.
Many optimization algorithms may contain in their description a requirement that the functions be twice-differentiable, but that does not imply that you have to provide code to actually compute those derivatives. Often, the Hessian is built up iteratively and updated through some approximation scheme, the success of which relies upon the existence of the derivatives.
See Prof. Mittelman's http://plato.asu.edu/sub/nlounres.html for links and information on software of the types that you asked about.
In a post by ArturGuzik on 03/02/2010, it is noted that:
In general, all optimization algorithms provided within MKL belong to gradient methods, and it means that you need to calculate derrivative (gradient) in order to obtain the next step and finally solution.
If you mean that you want MKL to minimize function using only its value(s) (don't need to know the function griadient, I guess that's what you had in mind) then the answer is no.
Do you know whether this is still the case?
There are no nonlinear optimization algorithms in MKL other than those in the NLS (nonlinear least squares) solvers. For the latter, there is a provision for computing the jacobian using finite-difference approximations. The user need only provide code to evaluate the vector function whose norm is to be minimized. However, if analytical derivative information is available and the derivatives can conveniently be evaluated, that would work better. See the example ex_nlsqp_f.f.
The comments by Guzik are about the NLS solvers and not for the case where the objective function is scalar.
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# Properties
Label 4.4.8768.1-28.2-e Base field 4.4.8768.1 Weight $[2, 2, 2, 2]$ Level norm $28$ Level $[28, 14, -w^{3} + 3w^{2} + w - 4]$ Dimension $4$ CM no Base change no
# Related objects
• L-function not available
## Base field 4.4.8768.1
Generator $$w$$, with minimal polynomial $$x^{4} - 2x^{3} - 5x^{2} + 6x + 7$$; narrow class number $$1$$ and class number $$1$$.
## Form
Weight: $[2, 2, 2, 2]$ Level: $[28, 14, -w^{3} + 3w^{2} + w - 4]$ Dimension: $4$ CM: no Base change: no Newspace dimension: $14$
## Hecke eigenvalues ($q$-expansion)
The Hecke eigenvalue field is $\Q(e)$ where $e$ is a root of the defining polynomial:
$$x^{4} - 23x^{2} - 12x + 37$$
Norm Prime Eigenvalue
4 $[4, 2, -w^{2} + w + 3]$ $\phantom{-}1$
7 $[7, 7, w]$ $\phantom{-}e$
7 $[7, 7, -w^{2} + 2w + 1]$ $\phantom{-}1$
7 $[7, 7, w^{2} - 2]$ $-\frac{1}{6}e^{3} + \frac{1}{6}e^{2} + \frac{19}{6}e + \frac{1}{3}$
7 $[7, 7, w - 1]$ $-\frac{1}{6}e^{3} + \frac{1}{6}e^{2} + \frac{19}{6}e + \frac{1}{3}$
23 $[23, 23, -w^{3} + w^{2} + 3w - 1]$ $-\frac{1}{2}e^{2} + \frac{1}{2}e + \frac{9}{2}$
23 $[23, 23, w^{3} - 2w^{2} - 2w + 2]$ $\phantom{-}\frac{1}{6}e^{3} + \frac{1}{3}e^{2} - \frac{11}{3}e - \frac{35}{6}$
31 $[31, 31, w^{2} - 5]$ $-\frac{1}{6}e^{3} + \frac{1}{6}e^{2} + \frac{25}{6}e - \frac{8}{3}$
31 $[31, 31, -w^{2} + 2w + 4]$ $\phantom{-}\frac{1}{6}e^{3} - \frac{1}{6}e^{2} - \frac{19}{6}e + \frac{5}{3}$
41 $[41, 41, -w^{3} + 2w^{2} + 2w - 6]$ $\phantom{-}4$
41 $[41, 41, w^{3} - w^{2} - 3w - 3]$ $-\frac{1}{3}e^{3} + \frac{1}{3}e^{2} + \frac{16}{3}e - \frac{10}{3}$
47 $[47, 47, w^{2} - 2w - 5]$ $\phantom{-}\frac{1}{6}e^{3} + \frac{1}{3}e^{2} - \frac{17}{3}e - \frac{23}{6}$
47 $[47, 47, w^{2} - 6]$ $\phantom{-}\frac{1}{3}e^{3} + \frac{1}{6}e^{2} - \frac{35}{6}e - \frac{7}{6}$
71 $[71, 71, -w^{3} + 2w^{2} + 3w - 1]$ $-\frac{1}{3}e^{3} + \frac{1}{3}e^{2} + \frac{13}{3}e - \frac{10}{3}$
71 $[71, 71, -w^{3} + w^{2} + 4w - 3]$ $-\frac{1}{3}e^{3} + \frac{1}{3}e^{2} + \frac{25}{3}e + \frac{8}{3}$
79 $[79, 79, -w - 3]$ $\phantom{-}\frac{1}{6}e^{3} - \frac{2}{3}e^{2} - \frac{14}{3}e + \frac{49}{6}$
79 $[79, 79, w - 4]$ $-\frac{1}{3}e^{3} + \frac{1}{3}e^{2} + \frac{22}{3}e - \frac{1}{3}$
81 $[81, 3, -3]$ $\phantom{-}\frac{1}{2}e^{2} - \frac{1}{2}e - \frac{25}{2}$
89 $[89, 89, w^{2} - 3w - 2]$ $\phantom{-}\frac{1}{3}e^{3} + \frac{1}{6}e^{2} - \frac{35}{6}e + \frac{17}{6}$
89 $[89, 89, w^{2} + w - 4]$ $\phantom{-}\frac{1}{3}e^{3} - \frac{1}{3}e^{2} - \frac{28}{3}e - \frac{11}{3}$
Display number of eigenvalues
## Atkin-Lehner eigenvalues
Norm Prime Eigenvalue
$4$ $[4, 2, -w^{2} + w + 3]$ $-1$
$7$ $[7, 7, -w^{2} + 2w + 1]$ $-1$
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Register Blogs Members List Search Today's Posts Mark Forums Read
June 19, 2008, 21:11 Steady State Vs Transient answers #1 Kushagra Guest Posts: n/a Hello, It may not sound a serious question, but I still need to clear my doubt. If I run the same simulation in steady state solver and transient solver, and somehow I manage the simulation to converge in steady state solver. Will the result from both the simulations almost same? I could only manage the steady state simulation to converge by playing around the false time step. Thanks so much, Kushagra Vinay Nandurdikar and soumitra2102 like this.
June 20, 2008, 20:06 Re: Steady State Vs Transient answers #2 cfd.newbie Guest Posts: n/a No, we expect more accuracy from transient simulation because it is time marching solution. I am very new to CFD though, it will be interesting if more experienced CFD user can give us insight into this. Regards Vinay Nandurdikar and soumitra2102 like this.
June 22, 2008, 18:51 Re: Steady State Vs Transient answers #3 Glenn Horrocks Guest Posts: n/a Hi, The difference between a steady state simulation and marching a transient solution to steady state is that the SS simulation ignores many of the cross terms and higher order terms dealing with time. These terms all go to zero in steady state so they don't affect the steady state result. The transient simulation includes all these terms. Usually this means the steady state model has an easier convergence as there are less terms to model and some transient non-linearities are removed, but in a few models these non-linearities help convergence (but this is infrequent). Glenn Horrocks Oula, rezaeimahdi, Vinay Nandurdikar and 3 others like this.
July 2, 2008, 23:25 Re: Steady State Vs Transient answers #4 Kushagra Guest Posts: n/a Thanks Glenn for providing useful knowledge about convergence in both type of simulations. But what about quality of results from two type of simulations? Some time steady state simulations are difficult to converge but reducing the false time step by 1 or 2 order, they converge. Does it affect the quality of results? soumitra2102 likes this.
July 3, 2008, 01:55 Re: Steady State Vs Transient answers #5 Glenn Horrocks Guest Posts: n/a Hi, A fully converged simulation, run to steady state by either steady state or transient approaches should be the same. The only exception is when "local timescale factor" is used in a steady state run as it can accelerate convergence nicely but as different timescales are used across the domain can cause accuracy problems. As long as a steady state run is run to final convergence with a physical timescale (including auto timescale) then it should be fine. The timescale is a steady state simulation is like under-relaxation from SIMPLE based solvers. Too high a URF and the simulation diverges, too low and convergence is slow, so you try to fiddle until you get the optimum in the middle somewhere. Glenn Horrocks
July 3, 2008, 22:50 Re: Steady State Vs Transient answers #6 Kushagra Guest Posts: n/a Thanks so much Glenn. 1) So reducing the 'false time step' is similar to reducing the under-relaxation factor. (Or is it just opposite?) 2) Suppose the residual for Volume Fractions are not getting low and fluctuating around mean values, what should the user try first? Reducing or increasing the 'false time step'? 3) The residence time (volume of domain / flow rate) is 13 second for my multiphase case. what might be a good time scale to start with steady state problem. Meanwhile, I found some of your, Robin's, Cyclone's replies on this forum about how the steady state solver works. They were really helpful. Thanks, Kushagra
July 6, 2008, 22:41 Re: Steady State Vs Transient answers #7 Glenn Horrocks Guest Posts: n/a Hi, 1) Be careful thinking of the physical timestep size as a false timestep. False timestepping is a different technique used in SIMPLE based solvers (I think). In CFX, a steady state solver is similar to the transient solver, just with some higher order transient stuff and cross terms removed, and a different residual calculation. But in basic idea, yes, tuning the URF of a SIMPLE run is similar to tuning the timestep size of a CFX run. There will be an optimum value somewhere between too slow and unstable. 2) http://www.cfd-online.com/Wiki/Ansys...gence_criteria 3) A good starting point is 13 seconds. Glenn Horrocks
July 11, 2016, 10:27 Transient and SS ! #8 Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 9 Hello, A basic conceptual doubt: I am performning a conjugate heat transfer problem for a drilling process. Lets assume, I want to see the temperature profile of the tool after 30s.. I have a tool fixed at intial temp. (lets say 600K). It loses heat by conduction and convection to the surrounding air. (radiation losses are neglected). I will perform a transient analysis setting appropriate time steps.. Will this result be the same as a steady state simulation freezed at 30s. ? (is this possible in CFX? Can i set it up the steady state simulation directly from 30s.) I would say, it wont be the same because of: 1. Hconv is proportional to delta T which is proportional to time. (different tempertures at different time steps). 2. the conduction equation is a parabolic PDE, which has time. so the heat loss due to conduction is different at different time steps.
July 11, 2016, 11:25 #9 Senior Member Join Date: Jun 2009 Posts: 1,665 Rep Power: 29 Unfortunately, the answer is "depends". If the transient calculation reaches steady state before 30 [s], say the transient contribution is already negligible at that time, the solution on the same mesh should be identical for a linear set of differential equations. For a non-linear system, there will only be identical if the system has one stationary solution. If the time to reach steady solution is larger than 30 [s], the transient solution will be very different than the steady solution. Summary: in general, they are very different unless you already know the "steady state time". Hope the above helps, cysanghavi, Vinay Nandurdikar, ari003 and 1 others like this.
July 11, 2016, 11:40 #10 Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 9 Thanks for the reply. Lets say, I know the time when I achieve steady state.Lets say this time is 120 [s]. ( I assume I achieve steady state, when there is no change in heat trasnfer, momemtum trasnfer.... etc. with time for the same mesh.) At this instance, will my transient and steady state simulation yield the same results ? the fact is that I do not understand how CFX evaluates the steady state solution. If there is some Literature about this, share it. Vinay94 likes this.
July 11, 2016, 12:30 #11 Senior Member Join Date: Jun 2009 Posts: 1,665 Rep Power: 29 At this point, you are better off reading the ANSYS CFX documentation on how either analysis is carried out, as well as reading a book on partial differential equations. If your application is purely heat transfer, a heat transfer textbook usually cover heat conduction on both steady and transient modes. I would not worry much about the momentum equation since the concepts are unique.
July 11, 2016, 18:44 #12 Member Chaitanya Join Date: Mar 2016 Location: Stuttgart, Germany Posts: 50 Rep Power: 9 Thanks for the references. All I could find in the literature of CFX was the following informtion. photosphoto_1.JPG. photo_2.JPG. I have the solver guide and the CFX Pre guide. I do not know if there is some other literature that I can review. I searched various portals. Could you refer me some literature, about equations that are solved for steady state ?
August 16, 2016, 07:29
#13
Member
Join Date: Jul 2016
Posts: 33
Rep Power: 8
Quote:
Originally Posted by Glenn Horrocks ;88525 Hi, The difference between a steady state simulation and marching a transient solution to steady state is that the SS simulation ignores many of the cross terms and higher order terms dealing with time. These terms all go to zero in steady state so they don't affect the steady state result. The transient simulation includes all these terms. Usually this means the steady state model has an easier convergence as there are less terms to model and some transient non-linearities are removed, but in a few models these non-linearities help convergence (but this is infrequent). Glenn Horrocks
Dear Glenn,
I have tried to find out what are these "cross terms" but I didn't manage to find the good answer. Could you please explain what did you mean by this?
Best regards,
August 16, 2016, 08:23 #14 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,326 Rep Power: 138 I vaguely recall that there are some neglected cross terms but I cannot find any reference to it or recall much about it. Bruno's comment here (http://www.cfd-online.com/Forums/cfx...transient.html) says there is no difference except in the way the coefficients are updated in the iterations. There are also differences in the way the residual is calculated. So I am not sure about the cross terms. But the updating of the coefficients and residuals are certainly different.
August 16, 2016, 08:48 #15 Member Join Date: Jul 2016 Posts: 33 Rep Power: 8 Dear Glenn, thank you for the reply, yes indeed, there is a difference in the way the RMS is computed, I have tried to find it in the documentation but I have not managed yet. As far as I know, for SS it is: RMS = | where the N is the number of elements. Do you know maybe whether it is treated in the Ansys documentation? Specifically for the transient case. I would like to find out more on this topic. Best regards,
August 16, 2016, 19:51 #16 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,326 Rep Power: 138 Yes, that is the standard definition of RMS. The residuals are defined in section 11.2.3 in the theory guide. The definition is not thorough, but enough to get the idea. It also defines the difference between SS and transient.
September 21, 2016, 09:22
#17
Member
annn
Join Date: Jun 2016
Posts: 40
Rep Power: 8
Quote:
Originally Posted by Glenn Horrocks ;88525 but in a few models these non-linearities help convergence (but this is infrequent).
for which models does it help convrgence?
September 21, 2016, 18:55 #18 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,326 Rep Power: 138 I have changed my opinion on this since 2008. I use it much more often these days. The main simulations where it makes a difference are things with complex coupling between equations, such as multiphase or shock waves, or where there is transient vortex shedding (which implies the simulation was not steady state in the first place).
September 22, 2016, 16:28 #19 Member annn Join Date: Jun 2016 Posts: 40 Rep Power: 8 Hi Glenn, thanks for replying so fast! Just a follow up, do you think that if you modeled a steady state case (where you knew was suppose to be a steady state case) with a transient one there should be no problem as they should theoretically yield the same result if it was truly a steady state case?
September 22, 2016, 18:46 #20 Super Moderator Glenn Horrocks Join Date: Mar 2009 Location: Sydney, Australia Posts: 17,326 Rep Power: 138 Yes, the transient simulation should converge over time to the steady state case. cleoo likes this.
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# What is called permittivity of free space?
The permittivity of free space is a physical constant that reflects the ability of electrical fields to pass through a classical vacuum. The constant is equal to approximately 8.854 x 10-12 F/m (farad per meter), with a relative standard uncertainty of 1.5 x 10-10.
## What is permittivity in physics class 12?
Permittivity is the property of a material or a medium that affects the coulomb force between two point charges when placed in that medium. Coulomb force is maximum for vacuum. Force in the other medium can be measured relative to it.
## Why is it called permittivity?
It’s called permittivity because of how much a given substance “permits” electric, (or magnetic in the case of magnetism ) field lines to pass through them.
## What is permittivity and permeability?
Permittivity and permeability are two concepts used in electromagnetic theory. Permittivity is a notion connected to the formation of an electric field, whereas permeability is a concept related to the formation of a magnetic field. Permittivity and permeability are two electromagnetic theory concepts.
## What is dimensional formula of permittivity?
Therefore, the permittivity of free space or vacuum is dimensionally represented as M-1 L-3 T4 I2.
## What is the symbol of permittivity?
In electromagnetism, the absolute permittivity, often simply called permittivity and denoted by the Greek letter ε (epsilon), is a measure of the electric polarizability of a dielectric.
## Is permittivity a constant?
Permittivity (electric permittivity) is the ratio of electric displacement to the electric field intensity. It is a constant of proportionality between these two parameters.
## What is the symbol for permeability?
Permeability is typically represented by the (italicized) Greek letter μ.
## What are the types of permittivity?
There are two types of permittivity. One is absolute permittivity. we can also refer permittivity as Absolute permittivity. The second one is Relative permittivity.
## What is conductivity and permittivity?
The electrical conductivity of tissues is a measure of the density and mobility of ions transported in the intracellular or extracellular spaces by an applied electric field, whereas their permittivity is a measure of the tissue’s capacity for the electric dipoles to store electric energy 1.
## What is the permittivity of water?
The dielectric permittivity of water is about 80.
## What is difference between permittivity and susceptibility?
So, the permitivity is ε, and it is the relation bteween the electric field and the displacement vector. Susceptibility is χe and it is strongly related to it. It’s, roughly speaking, like “the permitivity of the polarisation part only\$.
## What is the permittivity of a material?
The dielectric constant of a material, also called the permittivity of a material, represents the ability of a material to concentrate electrostatic lines of flux. In more practical terms, it represents the ability of a material to store electrical energy in the presence of an electric field.
## What is the value of permeability?
The permeability of free space, μ0, is a physical constant used often in electromagnetism. It is defined to have the exact value of 4π x 10-7 N/A2 (newtons per ampere squared).
## Why permittivity of conductor is infinite?
Permittivity is ability of a material to resist the formation of electric field inside it and for metals electric field does not exist inside the conductor. Therefore, permittivity of metal is infinitely larger than the permittivity of free space hence the dielectric constant is infinite.
## What is the unit of dielectric constant?
The dielectric constant of a substance is defined as the ratio of the substance’s permittivity to the permittivity of free space. It can be written as K = ε ε ο . It is a unitless, dimensionless quantity because it is the ratio of two like entities.
## What is relative permittivity Class 11?
Relative permittivity is the factor by which the electric field between the charges is decreased relative to vacuum. Likewise, relative permittivity is the ratio of the capacitance of a capacitor using that material as a dielectric, compared with a similar capacitor that has vacuum as its dielectric.
## What is dimension formula?
The dimensional formula is defined as the expression of the physical quantity in terms of its basic unit with proper dimensions. For example, dimensional force is. F = [M L T-2] It’s because the unit of Force is Netwon or kg*m/s2.
## What is the dimension of force?
Or, F = [M] × [L1 T-2] = M1 L1 T-2. Therefore, Force is dimensionally represented as M1 L1 T-2.
## What is called dielectric constant?
dielectric constant, also called relative permittivity or specific inductive capacity, property of an electrical insulating material (a dielectric) equal to the ratio of the capacitance of a capacitor filled with the given material to the capacitance of an identical capacitor in a vacuum without the dielectric material …
## What is the difference between permittivity and dielectric constant?
Dielectric constant defines how much of electric flux passing through material compare to vacuum medium. Dielectric permittivity is the characteristic of material to allow the electric flux pass through it.
## What is K in Coulomb’s law?
The constant of proportionality k is called Coulomb’s constant. In SI units, the constant k has the value k = 8.99 × 10 9 N ⋅ m 2 /C 2. k = 8.99 × 10 9 N ⋅ m 2 /C 2. The direction of the force is along the line joining the centers of the two objects.
## What is negative permittivity?
Negative permittivity means that the electric displacement vector and the electric field vector point in the opposite directions but it does not necessary mean that the electric energy stored is such medium is negative.
## Is vacuum a dielectric?
The vacuum is proposed to be a dielectric medium constituted of neutral but polarizable vacuuons based on overall experimental observations in separate publications. In the present paper I formally develop the dielectric theory for this dielectric vacuum.
## Which medium has more permittivity?
Among the given options, water has the highest relative permittivity.
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# Correlation Coefficient
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Correlation Coefficient
Another measure of the quality of fit is the correlation coefficient, r2. To calculate the correlation coefficient, square the total of the (x-)(y-) column, and divide by the total of the (x-)2 and the total of the (y-)2 column. The formula is:
r2 equals 1 if the data all lie exactly along a straight line, and r2 equals 0 if the data are not correlated. Values between 1 and 0 indicate that the data have some linear relationship, but also have some scatter. Data with an r2 of above .8 are considered strongly correlated.
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## Product
Displaying 1 - 4 of 4
Hey Math is a web-based, supplemental K-12 math program. The program includes demonstrations of concepts, practice exercises and games in a variety of topics at each grade level. Teachers can use the program for whole class demonstrations of concepts or assign exercises and reinforcement activities to individual students. Teachers can monitor student progress for each grade-level topic.
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GeoGebra is a dynamic mathematics software for schools that joins geometry, algebra and calculus. GeoGebra is an interactive geometry system. You can do constructions with points, vectors, segments, lines, conic sections as well as functions and change them dynamically afterwards. Additionally, math equations and coordinates can be entered directly. GeoGebra allows students to see the relationships between these different branches of mathematics.GeoGebra is available both online and in a downloadable version.
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Classworks is a web-based instructional software that provides interactive lessons, customized learning and assessment in K-12 Math, Reading, and Language Arts. Elementary Science lessons are also available. Classworks units are based on state standards and offer interactive instruction through a multi-sensory approach that includes voice, pop-up text, audio support, video, photographs, artist drawings, and animated clips. The multiple modalities may be helpful for those with limited English proficiency or learning disabilities, or as a Response to Intervention.
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CumulantGeneratingFunction - Maple Help
Statistics
CumulantGeneratingFunction
compute the cumulant generating function
Calling Sequence CumulantGeneratingFunction(X, t, options) CGF(X, t, options)
Parameters
X - algebraic; random variable or distribution t - algebraic; point options - (optional) equation of the form numeric=value; specifies options for computing the cumulant generating function of a random variable
Description
• The CumulantGeneratingFunction function computes the cumulant generating function of the specified random variable at the specified point.
• The first parameter can be a distribution (see Statistics[Distribution]), a random variable, or an algebraic expression involving random variables (see Statistics[RandomVariable]).
Computation
• By default, all computations involving random variables are performed symbolically (see option numeric below).
Options
The options argument can contain one or more of the options shown below. More information for some options is available in the Statistics[RandomVariables] help page.
• numeric=truefalse -- By default, the cumulant generating function is computed using exact arithmetic. To compute the cumulant generating function numerically, specify the numeric or numeric = true option.
Examples
> $\mathrm{with}\left(\mathrm{Statistics}\right):$
Compute the cumulant generating function of the beta distribution with parameters p and q.
> $\mathrm{CumulantGeneratingFunction}\left('\mathrm{Β}'\left(p,q\right),t\right)$
${\mathrm{ln}}{}\left({\mathrm{hypergeom}}{}\left(\left[{p}\right]{,}\left[{p}{+}{q}\right]{,}{t}\right)\right)$ (1)
Use numeric parameters.
> $\mathrm{CumulantGeneratingFunction}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2}\right)$
${\mathrm{ln}}{}\left({\mathrm{hypergeom}}{}\left(\left[{3}\right]{,}\left[{8}\right]{,}\frac{{1}}{{2}}\right)\right)$ (2)
> $\mathrm{CumulantGeneratingFunction}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2},\mathrm{numeric}\right)$
${0.1907815797}$ (3)
Define new distribution.
> $T≔\mathrm{Distribution}\left(\mathrm{=}\left(\mathrm{PDF},t↦\mathrm{piecewise}\left(t<0,0,t<1,6\cdot t\cdot \left(1-t\right),0\right)\right)\right):$
> $X≔\mathrm{RandomVariable}\left(T\right):$
> $\mathrm{CGF}\left(X,u\right)$
${\mathrm{ln}}{}\left(\frac{{6}{}\left({{ⅇ}}^{{u}}{}{u}{-}{2}{}{{ⅇ}}^{{u}}{+}{u}{+}{2}\right)}{{{u}}^{{3}}}\right)$ (4)
References
Stuart, Alan, and Ord, Keith. Kendall's Advanced Theory of Statistics. 6th ed. London: Edward Arnold, 1998. Vol. 1: Distribution Theory.
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https://testbook.com/question-answer/channel-capacity-of-a-noise-free-channel-having-m--5ecebdacf60d5d5714febbff
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# Channel capacity of a noise-free channel having m symbols is given by:
Free Practice With Testbook Mock Tests
## Options:
1. m2
2. 2m
3. Log2
4. m
### Correct Answer: Option 3 (Solution Below)
This question was previously asked in
LMRC Assistant Manager (S&T) Official Paper: (Held on 20 Jan 2020)
## Solution:
Concept:
Channel capacity is defined as an intrinsic ability of a communication channel to convey information.
Channel capacity per symbol is defined as:
$${C_s} = \mathop {\max }\limits_{\left\{ {p\left( {{x_i}} \right)} \right\}} I\left( {x;y} \right)b/symbol$$
Where:
$$\mathop {{\rm{max}}}\limits_{\left\{ {p\left( {{x_i}} \right)} \right\}}$$ = The maximization is over all possible input probability distribution on X.
I (x ; y) = Mutual Information of the channel defined as:
I(x ; y) = H(x) – H(x|y) ‘OR’ I(y ; x) = H(y) – H(y|x)
H(x|y) = conditional entropy
Analysis:
Noise free channel is both lossless and deterministic, i.e.
Noise free channel = Loseless + deterministic
For loseless channel,
H(x|y) = 0
I(x ; y) = H(x)
So,
$${C_s} = \mathop {{\rm{max}}}\limits_{\left\{ {p\left( {{x_i}} \right)} \right\}} H\left( x \right) = {\log _2}m$$
Where,
m = numbers of symbols in X
For deterministic channel
H(y|x) = 0
I(x ; y) = H(y)
So,
$${C_s} = \mathop {{\rm{max}}}\limits_{\left\{ {p\left( {{x_i}} \right)} \right\}} H\left( y \right) = {\log _2}n$$
Where, n = number of symbols in Y
Hence, for Noiseless channel, we can write:
I(x ; y) = H(x) = H(y), i.e.
Cs = log2 m = log2 n
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https://ontonixqcm.wordpress.com/2016/01/20/project-management-why-cost-overruns-and-late-delivery/
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# Project Management: Why Cost Overruns and Late Delivery?
Why over 90% of projects finish late? There is plenty of literature on the subject. This is a good example. In multi-project environments, such as the one illustrated below things can get very complex as project share resources (financial, manpower, equipment, etc.).
When it comes to project costs, similar statements can be made. American Planning Association in 2002 found that 9 out of ten construction projects had underestimated costs. Overruns ranged from 50% to 100%. Some illustrious examples:
• Suez canal: effective cost was 20 times the original estimate
• Sydney Opera House: 15 times
• The Concord: 12 times
Other well-known examples: the F-35 fighter, Denver International Airport, Airbus A380, Boeing 787, Humber Bridge, Channel Tunnel, Berlin Brandenburg Airport.
Again, there is plenty of research on the subject and in general the following three causes are indicated: technical, psychological, and political-economic. In general, it comes down to poor risk analysis.
In this short blog we wish to focus on a particular technical aspect of risk analysis, especially when it comes to cost control. This is how a typical risk analysis works:
• A project schedule is assumed – this produces the Gantt chart
• A Monte Carlo realization of the project is executed thousands of times
• The results are processed statistically in order to determine probabilities, expected values of cost and deliver time, etc.
The cost of a project is equal to the sum of component costs:
C_project = Cost_1 + Cost_2 + …. + Cost_N
In the Monte Carlo process, randomness is added to each of the component costs and thousands of realizations are executed. The first problem is how this randomness is modeled. There exist many probability distributions to choose from. In general, ‘expert opinions’ or conjectures are used to determine which distributions will be used.
However, the most fatal and, unfortunately, the most common mistake in Monte Carlo simulations is to ignore dependencies between the inputs. In other words, if, Cost_i and Cost_j are correlated, ignoring this correlation changes things pretty dramatically. Imagine what damage this can cause when thousands of component costs are considered. Computing project costs by simply summing up component costs and neglecting their inter-dependencies is a good example of anachronistic linear thinking. The most common excuse is: these inter-dependencies are awfully difficult to determine. So they are neglected. Once more, men become slaves of their tools and reality is bent to satisfy the said tools.
Basically, the above model should be:
C_project = Cost_1 + Cost_2 + …. + Cost_N + Nonlinear (coupling) terms
The nonlinear part is of course neglected.
An example of how neglecting the mentioned cost inter-dependencies impacts the resulting Tornado Chart (which ranks the component costs in terms of correlation with the total project cost) is:
If that were not enough, correlations between the total cost and component costs are often over-estimated as pointed out in our previous blog.
The reasons why most projects are late and incur cost overruns are many. Lack of Project Complexity Management is certainly one. The objective of this blog, however, is to point out that, independently of political and psychological issues, seeing reality as linear while it is not, induces distortions with guaranteed adverse consequences on estimates of costs, schedules or probabilities of failure.
www.ontonix.com
## One thought on “Project Management: Why Cost Overruns and Late Delivery?”
1. António Tavares Flor says:
My doctoral thesis deals with the subject overrun costs. It was published by the Court Portuguese
I did Multimpact model that makes the prediction of financial slippage based on the historical behavior of the agents involved.
http://www.tcontas.pt/pt/publicacoes/revista/47ndx.pdf
This thesis is organized in three parts, reflecting three kinds of Reasons: the
complexity and uncertainty about the nature and singularities public work, the
Those aspects of influence on the cost variation, and finally the proposal and use of
the forecast model – the Multimpact model.
Summarizing, this thesis presents the main conclusions Following:
a) The development of a concept model, able to link cost variation and its
causes;
b) The cost variation can be described by the statistic exponential distribution
c) The proposal of a model operating – Multimpact, connecting aspects of the
management quality, the complexity and uncertainty associated with the
project and the cost variation;
d) Proposal of procedures having in view cost variation minimization.
Liked by 1 person
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http://web.mnstate.edu/peil/geometry/C2EuclidNonEuclid/4angles.htm
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TOC & Ch. 0 & Ch. 1 Axiom Table of Contents Ch. 0 Introduction Ch. 1 Axiomatic Systems 1.1.1 Introduction 1.1.2 Examples 1.1.3 History 1.2 A Finite Geometry 1.3 Finite Projective 1.4 Applications Ch. 2 Neutral Geometry Ch. 2 Table of Contents 2.1.1 Introduction 2.1.2 History 2.1.3 Analytic Models 2.2 Incidence Axioms 2.3 Distance/Ruler Axioms 2.4.1 Plane Separation Axiom 2.4.2 Angle & Measure 2.5.1 Supplement Postulate 2.5.2 SAS Postulate 2.6.1 Parallel Lines 2.6.2 Saccheri Quadrilateral 2.7.1 Euclid Parallel Postulate 2.7.2 Hyperbolic Parallel Postulate 2.7.3 Elliptic Parallel Postulate 2.8 Euclid/Hyperbolic/Elliptic Birkhoff's Axioms Hilbert's Axioms SMSG Axioms Ch. 3 Transformational Ch. 3 Table of Contents 3.1.1 Introduction 3.1.2 History 3.2.1 Definitions 3.2.2 Analytic Model 3.2.3 Affine Transformation 3.3.1 Isometry 3.3.2 Model/Collinearity 3.3.3 Model/Isometry 3.4.1 Direct Isometry 3.4.2 Model/Direct 3.5.1 Indirect Isometry 3.5.2 Model/Indirect 3.6.1 Similarity Transformation 3.6.2 Model/Similarity 3.7 Other Affine Transformations Ch. 4 Projective Geometry Ch. 4 Table of Contents 4.1.1 Introduction 4.1.2 Historical 4.2.1 Axioms 4.2.2 Basic Theorems 4.3 Duality 4.4 Desargue's Theorem 4.5.1 Harmonic Sets 4.5.2 Music & Harmonic Sets 4.6.1 Definitions for Projectivity 4.6.2 Fundamental Theorem 4.6.3 Projectivity/Harmonic Sets 4.6.4 Alternate Construction 4.7.1 Conics 4.7.2 Pascal's Theorem 4.7.3 Tangents to Conics Other Topics Ch. 5 Spherical Geometry Ch. 6 Fractal Geometry Ch. 7 Topology Appendices Internet Resources Index Geometer's Sketchpad/GeoGebra JavaSketchpad/GeoGebraHTML Video Lectures Logic Review References Acknowledgements
2.4.2 Angles and Angle Measure Printout
The right angle from which to approach any problem is the try angle.
Definitions. An angle is the union of two noncollinear rays with a common endpoint. The common endpoint is called the vertex of the angle, and the rays are called the sides of the angle.
The Ruler Postulate and Ruler Placement Postulate were motivated by the "real-world" use of rulers. A similar set of postulates, SMSG Postulates 1113, which are motivated by the "real-world" protractor, do the same for angles. Hence, these axioms are sometimes referred to as the protractor postulates.
Postulate 11. (Angle Measurement Postulate) To every angle there corresponds a real number between 0 and 180.
Postulate 12. (Angle Construction Postulate) Let be a ray on the edge of the half-plane H. For every r between 0 and 180, there is exactly one ray with P in H such that .
Postulate 13. (Angle Addition Postulate) If D is a point in the interior of , then .
Note that an angle has measure between 0 and 180. No angle has measure greater than or equal to 180, or less than or equal to 0.
Definitions.
Two angles are congruent if they have the same measure, denoted .
The interior of an angle is the intersection of set of all points on the same side of line BC as A and the set of all points on the same side of line AB as C, denoted . (Note that this definition uses the Plane Separation Postulate.)
The interior of a triangle ABC is the intersection of the set of points on the same side of line BC as A, on the same side of line AC as B, and on the same side of line AB as C.
The bisector of an angle is a ray BD where D is in the interior of and .
A right angle is an angle that measures exactly 90.
An acute angle is an angle that measures between 0 and 90.
An obtuse angle is an angle that measures between 90 and 180.
Two lines are perpendicular if they contain a right angle.
The next theorem, stated here without proof, will be used in later sections.
Theorem 2.7. and D is on the same side of line as C if and only if .
Exercise 2.32. Find the axioms from a high school book that correspond to SMSG Postulates 11, 12, and 13.
Exercise 2.33. Find the measures of the three angles determined by the points A(1, 1), B(1, 2) and C(2, 1) where the points are in the (a) Euclidean Plane; and (b) Poincaré Half-plane.
Also, find the sum of the measures of the angles of the triangles.
Exercise 2.34. Find the angle bisector of , if A(0, 5), B(0, 3), and where the points are in the (a) Euclidean Plane; and (b) Poincaré Half-plane.
Exercise 2.35. Given , , and . Prove or disprove .
Exercise 2.36. Prove or disprove that all right angles are congruent.
Exercise 2.37. Prove or disprove that an angle has a unique bisector.
Exercise 2.38. (a) Prove that given a line and a point on the line, there is a line perpendicular to the given line and point on the line.
(b) Prove the existence of two lines perpendicular to each other.
Exercise 2.39. Prove Theorem 2.7.
Exercise 2.40. Prove congruence of angles is an equivalence relation on the set of all angles.
© Copyright 2005, 2006 - Timothy Peil
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https://mathoverflow.net/questions/134863/any-reference-to-an-algorithm-for-finding-the-largest-empty-circle-on-a-sphere?noredirect=1
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# Any reference to an algorithm for finding the largest empty circle on a sphere (with great-circle distance)?
Given a set $S$ of 2D points in the plane, there are known algorithms for finding the largest empty circle ($LEC$) of the set of points.
The $LEC$ problem is stated in this way: find a $LEC$ whose center is in the closed Convex Hull of S, empty in that it contains no points in its interior, and largest in that there is no other such circle with strictly larger radius.
O'Rourke describes a simple $O(n^2)$ algorithm in his book "Computational Geometry in C" (the algorithm is attributed to G.T. Toussaint, 1983. Computing Largest Empty Circles with Location Constraints. International Journal of Parallel Programming, v12.5, pp 347-358.), the algorithm goes like this:
1. Compute the Voronoi Diagram $VD$ of the set of points
2. Compute the Convex Hull $CH$ of the set of points
3. The center of the $LEC$ is at one of the $VD$ vertexes inside the $CH$ or it is at an intersection between one of the $VD$ edges and $CH$.
4. For each candidate center compute the radius of the circle centered on it and update the maximum radius.
Now, on the plane, with $P=(P_x,P_y)$, the radius is computed with the distance $dist(P,C)=\sqrt{(P_x-C_x)^2+(P_y-C_y)^2}$ for a proper $P$.
I am interested in an algorithm for a similar problem: I have a set $S'$ of 3D points lying on a sphere of radius $R$ and I would like to find the equivalent of the $LEC$. I can assume that all points of $S'$ lie on a hemisphere.
In the sphere problem, the distances are measured by means of the great-circle distance.
With $S$ the 2D circle is the locus of points equidistant from a fixed point on the plane; with $S'$ the locus of points on the sphere equidistant (distance measured by the great-circle distance) from a fixed point on the sphere is the spherical curve usually called small circle. So instead of the largest empty circle maybe I can call what I am looking for the largest empty small-circle.
Is there any published algorithm for my problem?
I believe the same algorithm works. Note the remarkable fact, first understood by Kevin Brown, that the Voronoi diagram of points $$P$$ on a sphere is combinatorially identical to the (3D) convex hull of $$P$$: each face of the hull corresponds to a Voronoi vertex (and to an empty circle). This leads to an $$O(n \log n)$$ algorithm, as explained in this paper:
Hyeon-Suk Na, Chung-Nim Lee, Otfried Cheong, "Voronoi diagrams on the sphere." Computational Geometry: Theory and Applications, 23 (2002) 183–194. (ACM link)
Here is a recent implementation article:
Zheng, Xiaoyu, et al. "A Plane Sweep Algorithm for the Voronoi Tessellation of the Sphere." Electronic-Liquid Crystal Communications [online] (2011). (PDF download.)
I can't resist re-sharing this image of a Vornoi diagram on a sphere from an earlier MO question:
Mathematica image by Maxim Rytin.
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https://www.expertuition.com/free-math-worksheets-for-3rd-4th-and-5th-graders/
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# Free Math Worksheets for 3rd, 4th and 5th Graders
Rate this post
This post discusses free printable math worksheets and task cards for 3rd, 4th and 5th graders. You can download these free math worksheets and task cards from my Free Resource Library.
## Introduction
Do you want free printable math worksheets and task cards for 3rd, 4th, and 5th grade place value, multiplication, and division? Then download these freebies from my free resource library.
The worksheets and task cards have multiple choice and open-ended questions.
The freebies also include 4th grade spiral math review and quick checks.
You can use the worksheets for morning work, homework, warm-ups, and more.
The resource library is a growing library with more freebies to be added.
## Free Multiplication Worksheets and Task Cards
The free multiplication worksheets include worksheets for practicing multiplication using the area model strategy.
If you need up to 4-digit by 1-digit and 2-digit by 2-digit multiplication practice worksheets, then this freebie is perfect for you.
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## Free Division Worksheets and Task Cards
The free division worksheets include worksheets for practicing area model division, partial quotients division, and long division.
### Free Area Model Division Worksheets
The free area model division worksheets include worksheets for practicing division using the area model strategy.
If you need 2-digit by 1-digit or 3-digit by 1-digit area model division with remainders and no remainders practice worksheets, then this freebie is perfect for you.
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The free partial quotients division worksheets include worksheets for practicing division using the partial quotients strategy.
If you need 2-digit by 1-digit or 3-digit by 1-digit partial quotients division with remainders and no remainders practice worksheets, then this freebie is perfect for you.
The worksheets are presented in graph paper form to help students organize and keep their work neat.
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The free long division worksheets include worksheets for practicing division using the traditional long division strategy.
If you need 2-digit by 1-digit or 3-digit by 1-digit long division with remainders and no remainders practice worksheets, then this freebie is perfect for you.
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There is also the task cards version of the long division worksheets in the Freebie library.
Click HERE for more resources to review the various division strategies with your students or to check your students’ understanding of dividing multi-digit whole numbers.
## Free Spiral Math Review and Quick Checks
### Spiral Math Review Worksheets
The spiral math review freebie is aligned with 4th grade common core standard.
The worksheets also include a review of key 3rd grade math concepts.
There are free two weeks of spiral daily review and weekly quizzes worksheets.
The spiral math review freebie also includes a test. You can use the test as a quick check or a test prep to review all 4th grade common core math standards.
### 4th Grade Math Quick Checks
The 4th grade math quick checks is common core aligned.
If you are looking for standard-based review worksheets for grade 4 math test prep, then this freebie is perfect for you.
You can use this freebie to quickly check your students’ math skills of all 4th grade common core math standards.
## Free Place Value Worksheets
If you need place value worksheets for quick review, then this freebie is perfect for you.
You can use these no prep worksheets to quickly check your students’ understanding of place value, number forms, comparing, ordering, and rounding multi-digit whole numbers.
You can use these worksheets with 3rd, 4th, and 5th graders.
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http://redberry.cc/documentation:ref:diractrace?do=diff&rev2%5B0%5D=1448049491&rev2%5B1%5D=1448109188&difftype=sidebyside
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# Differences
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documentation:ref:diractrace [2015/11/20 19:58]poslavskysv documentation:ref:diractrace [2015/11/21 12:33] Line 1: Line 1: - ====== DiracTrace ====== - ---- - ====Description==== - - * ''%%DiracTrace%%'' calculates trace of Dirac matrices in $D$ dimensions - - * By default ''%%DiracTrace%%'' works in $D = 4$; for arbitrary $D$ one can use option ''%%DiracTrace[[Dimension: D]]%%'' - - * One can directly set trace of identity matrix (e.g. for dimensional regularisation): ''%%DiracTrace[[Dimension: D, TraceOfOne: 4]]%%'' - - * By default ''DiracTrace'' uses notation ''G_m'' for $\gamma_m$, ''G5'' for $\gamma_5$ and ''e_abcd'' for Levi-Civita tensor. ''%%DiracTrace[G, G5, eps]%%'' or ''%%DiracTrace[[Gamma: G, Gamma5: G5, LeviCivita: eps]]%%'' specifies the notation for $\gamma_m$, $\gamma_5$ and Levi-Civita tensor. - - * ''%%DiracTrace[[Simplifications: rules]]%%'' will apply additional simplification ''rules'' to each processed trace - - - ====Examples==== - ---- - Calculate trace of $\gamma$-matrices: - - defineMatrices 'G_a', 'G5', Matrix1.matrix - println DiracTrace >> 'Tr[G_a*G_b]'.t - - - > 4*g_ab - - - Another example: - - //set up matrix objects - defineMatrices 'G_a', 'G5', Matrix1.matrix - //calculate trace - println DiracTrace >> 'Tr[(p_a*G^a + m)*G_m*(q_a*G^a-m)*G_n]'.t - - - > 4*p_{m}*q_{n}+4*p_{n}*q_{m}-4*m**2*g_{mn}-4*p^{a}*g_{mn}*q_{a} - - - ---- - Calculate trace involving $\gamma_5$: - - //set up matrix objects - defineMatrices 'G_a', 'G5', Matrix1.matrix - //calculate trace - println DiracTrace >> 'Tr[G_a*G_b*G_c*G_d*G5]'.t - - - > -4*I*e_{abcd} - - - println DiracTrace >> 'Tr[(p_a*G^a + m)*G_m*G5*(q_a*G^a-m)*G_n]'.t - - - > -4*I*p_{b}*q_{a}*e^{a}_{n}^{b}_{m} - - ---- - - Calculate trace in different dimensions: - - - defineMatrices 'G_a', 'G5', Matrix1.matrix - println DiracTrace[[Dimension: 6]] >> 'Tr[G_c*G_a*G_b*G^c]'.t - - - > 48*g_ab - - - By default, ''Tr[1]'' is equal to $2^{\frac{D-1}{2}}$ for odd $D$ and $2^{\frac{D}{2}}$ for even. For symbolic $D$ it will be assumed that it is even: - - - defineMatrices 'G_a', 'G5', Matrix1.matrix - println DiracTrace[[Dimension: 'D'.t]] >> 'Tr[G_c*G_a*G_b*G^c]'.t - - - > D*2**(D/2)*g_ab - - - One can directly overcome predefined value of ''Tr[1]'' by using additional option (required for dimensional regularisation): - - defineMatrices 'G_a', 'G5', Matrix1.matrix - def dTrace = DiracTrace[[Dimension: 'D'.t, TraceOfOne: 4]] - println dTrace >> 'Tr[G_c*G_a*G_b*G^c]'.t - - - > 4*D*g_ab - - - ---- - For traces involving $\gamma_5$ in $D$ dimensions, all $\gamma_5$-related calculations will be performed as in 4 dimensions ($Tr[\gamma_a \gamma_b \gamma_c \gamma_d \gamma_5] = -4 i e_{abcd}$ and Chiholm-Kahane identitie: $\gamma_a \gamma_b \gamma_c = g_{ab} \gamma_c-g_{ac} \gamma_b+g_{bc} \gamma_a-i e_{abcd} \gamma_5 \gamma^d$): - - - defineMatrices 'G_a', 'G5', Matrix1.matrix - def dTrace = DiracTrace[[Dimension: 'D'.t, TraceOfOne: 4]] - println dTrace >> 'Tr[G_a*G_b*G_c*G_d*G5]'.t - - - > -4*I*e_{abcd} - - - - println dTrace >> 'Tr[G_a*G_b*G_c*G_d*G_e*G^a*G5]'.t - - - > 4*I*e_{debc}-4*I*e_{decb}+4*I*e_{dbce}-4*I*e_{ebcd} - - - ---- - Use another notation for gamma matrices: - - defineMatrices 'F_\\mu', 'F5', Matrix2.matrix - def dTrace = DiracTrace[[Gamma: 'F_\\mu', Gamma5: 'F5', LeviCivita: 'Eps_{\\mu\\nu\\alpha\\beta}']] - println dTrace >> 'Tr[F_\\mu*F_\\nu*F_\\alpha*F_\\beta * F5]'.t - - - > -4*I*Eps_{\mu\nu\alpha\beta} - - ====Options==== - - * ''Simplifications'': one can specify additional simplifications that will be applied to each evaluated trace:<sxh groovy; gutter: false> - defineMatrices 'G_a', 'G5', Matrix1.matrix - def expr = 'Tr[(p^a + k^a)*(p^b + k^b)*G_a*G_b*G_c*G_d]'.t - def mandelstam = setMandelstam([k_a: '0', p_a: '0', q_a: 'm', r_a: 'm'], 's', 't', 'u') - println DiracTrace[[Simplifications: mandelstam]] >> expr - <sxh plain; gutter: false> - > 4*s*g_{cd} - which is same as - println( (DiracTrace & mandelstam) >> expr ) - <sxh plain; gutter: false> - > 4*s*g_{cd} - - - * ''ExpandAndEliminate'': ''DiracTrace'' expands out products of sums containing traces of $\gamma$-matrices using [[ExpandAndEliminate]] transformation. One can replace it with another instance using ''%%DiractTrace[[ExpandAndEliminate: tr]]%%''. - - * ''LeviCivitaSimplify'': When traces involve $\gamma_5$, ''DiracTrace'' uses [[LeviCivitaSimplify]] in for simplifying resulting expressions with Levi-Civita tensors. One can replace the default instance of [[LeviCivitaSimplify]] with another one using ''%%DiractTrace[[ExpandAndEliminate: tr]]%%''. - - ====See also==== - * Related guides: [[documentation:guide:applying_and_manipulating_transformations]], [[documentation:guide:Setting up matrix objects]], [[documentation:guide:list_of_transformations]] - * Related tutorials: [[documentation:tutorials:Compton scattering in QED]] - * Related transformations: [[documentation:ref:DiracSimplify]], [[documentation:ref:DiracOrder]], [[documentation:ref:LeviCivitaSimplify]], [[documentation:ref:UnitaryTrace]] - * JavaDocs: [[http://api.redberry.cc/redberry/1.1.8/java-api//cc/redberry/physics/feyncalc/DiracTraceTransformation.html| DiracTraceTransformation]] - * Source code: [[https://bitbucket.org/redberry/redberry/src/tip/physics/src/main/java/cc/redberry/physics/feyncalc/DiracTraceTransformation.java|DiracTraceTransformation.java]]
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# HW5
```HW #5
35pts
Due 10/10/07
For this assignment, you need to show any hand calculations you make. Include a cover sheet.
On exam #2, the class average was 79.4 (SD = 11.0). Assume the distribution is normal.
1) If someone got an 85 on the exam, what was their Z score?
Z = (85 – 79.4) / 11.0 = 5.6/11 = 0.51
2) If someone got a 72 on the exam, what was their Z score?
Z = (72 – 79.4) / 11.0 = -7.4 / 11 = -0.67
3) If someone got a 98 on the exam, what was their Z score?
Z = (98 – 79.4) / 11.0 = 18.6/11 = 1.69
4) If someone got a 60 on the exam, what was their Z score?
Z = (60-79.4)/11.0 = -19.4/11 = -1.76
5) If someone got an 85 on the exam, what was their T score?
Z = 0.51 (see problem #1).
T scale score = 50 + 10*0.51 = 55.1
6) If someone got a 77 on the exam, what was their T score?
Z = (77-79.4) / 11 = -2.4 / 11 = -0.22
T scale score = 50 + 10*-0.22 = 47.8
7) If someone got a 93 on the exam, what was their IQ score?
Z = (93-79.4) / 11 = 13.6 / 11 = 1.24
IQ scale score = 100 + 15*1.24 = 118.6
8) If someone got a 66 on the exam, what was their IQ score?
Z = (66-79.4) / 11 = -13.4/11 = -1.22
IQ scale score = 100 + 15*-1.22 = 81.7
9) What percentage of students scored between 68 and 90?
Zlow = (68-79.4) / 11 = -1.04
proportion between Zlow and mean = p(-1.04<Z<0) = p(0<Z<1.04) mirror image = 0.3508
Zhigh = (90-79.4) / 11 = 0.96
proportion between mean and Zhigh = p(0<Z<0.96) = 0.3315
proportion between Zlow and Zhigh = 0.3508 + 0.3315 = 0.68 or 68%
You could have also estimated this using the 68-95-99 rule
10) What percentage of students scored higher than 85?
Z = 0.51 (see problem #1)
p(Z>0.51) = 0.3050 = 31%
11) What percentage of students scored lower than 73?
Z = (73-79.4) / 11 = -6.4 / 11 = -0.58
p(Z<-0.58) = p(Z>0.58) mirror image = 0.2810 = 28%
12) What percentage of students scored between 73 and 100?
Zlow = -0.58 (problem #11)
proportion between Zlow and mean = p(-0.58<Z<0) = p(0<Z<0.58) mirror image = 0.2190
Zhigh = (100-79.4) / 11 = 20.6 / 11 = 1.87
proportion between mean and Zhigh = p(0<Z<1.87) = 0.4693
proportion between Zlow and Zhigh = 0.2190 + 0.4693 = 0.6883 = 69%
13) What percentage of students scored above 90 or below 60?
Zlow = -1.76 (problem #4)
proportion below Zlow = p(Z<-1.76) = p(Z>1.76) mirror image = 0.0392
Zhigh = 0.96 (problem #9)
proportion above Zhigh = p(Z>0.96) = 0.1685
proportion below Zlow or above Zhigh = 0.0392 + 0.1685 = .2077 = 21%
14) What percentage of students scored below 50?
Z = (50-79.4) / 11 = -29.4 / 11 = -2.67
p(Z<-2.67) = p(Z>2.67) mirror image = 0.0038 = 0.38%
15) What percentage of students scored between the mean and 90?
Z = 0.96 (problem #9)
proportion between the mean and Z = p(0<Z<0.96) = 0.3315 = 33%
Or, if using the 68-95-99 rule to estimate: 68/2 = 34%
16) On exam #1 the class average was 83.4 (SD = 9.6). Is it more impressive to score a 73 on
exam #1 or exam #2?
Exam #1: Z = (73-83.4) / 9.6 = -10.4 / 9.6 = -1.08
Exam #2: Z = -0.58 (problem #11)
Neither is very impressive because both scores are below the mean, but the exam #2 score had a
higher Z score, so it was more impressive. (You could provide percentile ranks if you like).
17) Is it more impressive to score a 90 on exam #1 or #2?
Exam #1: Z = (90-83.4) / 9.6 = 6.6 / 9.6 = 0.69
Exam #2: Z = 0.96 (problem #9)
The performance on exam #2 was more impressive due to a higher Z score. (You could provide
percentile ranks if you like).
Questions were worth 2 points each. You get 1 point for having a cover sheet. 2*17 +1 = 35
The scoring is -2 for an incorrect answer. There is some partial credit.
The scoring is -1 total for no coversheet or untyped answer sheets.
The scoring is -3 total for untyped sloppy answer sheets.
Question
1
Z = 0.51
2
Z = -0.67
3
Z = 1.69
4
Z = -1.76
5
T = 55.1
6
T = 47.8
7
IQ = 118.6
8
IQ = 81.7
9
68%
10
31%
11
28%
12
69%
13
21%
14
0.38%
15
33%
16
Exam 2 + explanation
17
Exam 2 + explanation
```
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https://www.answersarena.com/ExpertAnswers/in-may-and-june-mary-spent-all-her-clothing-budget-on-bathing-suits-and-beach-bags-each-bathing-su-pa923
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Home / Expert Answers / Economics / in-may-and-june-mary-spent-all-her-clothing-budget-on-bathing-suits-and-beach-bags-each-bathing-su-pa923
# (Solved): In May and June, Mary spent all her clothing budget on bathing suits and beach bags. Each bathing su ...
In May and June, Mary spent all her clothing budget on bathing suits and beach bags. Each bathing suit cost \$77. At Mary’s optimal choice, her marginal utility from the last bathing suit purchased is 376 and her marginal utility from the last beach bag purchased is 256. This means that each handbag must cost: Please calculate the answer to two decimals. Your Answer:
We have an Answer from Expert
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https://www.mathworks.com/matlabcentral/answers/365387-how-to-include-n-1-25-power-1-4-100-cg-1-6-150-in-single-expression-using-for-loop
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# how to include n = 1:25; power = 1:4:100; cg = 1:6:150; in single expression using for loop
5 views (last 30 days)
Prabha Kumaresan on 6 Nov 2017
Answered: Eric on 9 Nov 2017
i have initialized
n=1:25(number of users)
power = 1:4:100(power values of different users)
cg = 1:6:150(channel gain for different users)
i need to make use of all three variables in the same expression
sum_n = (b*log(1+((power*cg)/noise))*Ncarrier)
followed by for loop i facing error while executing pls help me to solve this issue.
##### 2 CommentsShowHide 1 older comment
Robert on 6 Nov 2017
Your question is unclear. What are the variables b, noise, and Ncarrier?
As a start, the answer is probably one of the following:
1) Loop over the common length
for index = 1:length(n) % should be same as length(power0 and length(cg)
% use n(ii), power(ii), and cg(ii)
end
2) Use nested for-loops (could also be done with indices in for loops):
for i_n = n
for i_power = power
for i_cg = cg
% ...
end
end
end
3) Use meshgrid
[N, POWER, CG] = meshgrid(n, power, cg);
% ...
4) Use singleton expansion (or bsxfun if you are using a version of MATLAB before R2016b)
n = 1:25; % first dimension
power = (1:4:100)'; % second dimension
cg = reshape(1:6:150, 1, 1, []); % third dimension
% ...
Will these arrays always have the same length? Are you trying to execute your expression for every possible combination of these values? Do you need to collect the results in an array?
KL on 6 Nov 2017
Do you really need a for loop? Check this,
n = 1:25;
power = 1:4:100;
cg = 1:6:150;
b=1;
noise = rand(size(n));
Ncarrier = 1;
sum_n = (b.*log(1+((power.*cg)./noise)).*Ncarrier)
.* means element-wise multiplication.
Eric on 9 Nov 2017
Since all your vectors are the same length, you probably want to be using the binary operators in your equation, which includes a period before operators that can be used on matrices, for example: .* .^ and ./
sum_n = (b.*log(1+((power.*cg)./noise)).*Ncarrier);
This will ensure that your sum_n will also be 25 elements long (or error if the vectors don't match), and assumes b, noise and Ncarrier are either all scalars or also vectors of the same length as n, power, and cg (i.e. 25).
Protip: MATLAB has a built in function, log1p, which computes log(1+x) more accurately for small values of x.
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http://www.euclideanspace.com/maths/geometry/affine/reflection/matrix/index.htm
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# Maths - Reflection using Matrix
### Matrix representation
#### normal component
Norm = Va P = (Va × P) * P/|P|²
#### parallel component
Proj = Va || P = Va • P * P/|P|²
#### Reflection matrix
Given the inversion I'll add the terms instead of subtracting them to give the reflection result:
p -> 1 /
(Px² + Py² + Pz²)*
-Px² + Pz²+ Py² - 2 * Px * Py - 2 * Px * Pz - 2 * Py * Px -Py² + Px² + Pz² - 2 * Py * Pz - 2 * Pz * Px -2 * Pz * Py -Pz² + Py² + Px²
[p]
Note that this matrix is symmetrical about the leading diagonal, unlike the rotation matrix, which is the sum of a symmetric and skew symmetric part.
#### Simple cases
In order to check the above lets take the simple cases where the point is reflected in the various axis:
Reflection in yz
-1 0 0 0 1 0 0 0 1
Reflection in xz
1 0 0 0 -1 0 0 0 1
Reflection in xy
1 0 0 0 1 0 0 0 -1
#### Determinant and eigen values
Another check is that the determinant of reflection matrix is -1
#### Example
As an example we want to reflect the point (1,0,0) in a plane at 30 degrees.
P2 =
-Px² + Pz² + Py² - 2 * Px * Py - 2 * Px * Pz - 2 * Py * Px -Py² + Px² + Pz² - 2 * Py * Pz - 2 * Pz * Px -2 * Pz * Py -Pz² + Py² + Px²
[P1]
where Px,Py,Pz is the normal to the mirror which is: (-0.5,0.866,0)
and P1 is the initial point which is (1,0,0)
substituting these values gives:
P2 =
-0.25 + 0 + 0.75 - 2 * -0.433 0 - 2 * -0.433 -0.75 + 0.25 + 0 0 0 0 0 + 0.75 + 0.25
1 0 0
P2 =
0.5 0.866 0 0.866 -0.5 0 0 0 1
1 0 0
multiplying to vector by the matrix gives:
P2 =
0.5 0.866 0
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# Source code for sympy.physics.units.util
# -*- coding: utf-8 -*-
"""
Several methods to simplify expressions involving unit objects.
"""
from __future__ import division
from sympy.utilities.exceptions import SymPyDeprecationWarning
from sympy import Add, Function, Mul, Pow, Rational, Tuple, sympify
from sympy.core.compatibility import reduce, Iterable, ordered
from sympy.physics.units.dimensions import Dimension, dimsys_default
from sympy.physics.units.quantities import Quantity
from sympy.physics.units.prefixes import Prefix
from sympy.utilities.iterables import sift
def dim_simplify(expr):
"""
NOTE: this function could be deprecated in the future.
Simplify expression by recursively evaluating the dimension arguments.
This function proceeds to a very rough dimensional analysis. It tries to
simplify expression with dimensions, and it deletes all what multiplies a
dimension without being a dimension. This is necessary to avoid strange
behavior when Add(L, L) be transformed into Mul(2, L).
"""
SymPyDeprecationWarning(
deprecated_since_version="1.2",
feature="dimensional simplification function",
issue=13336,
).warn()
_, expr = Quantity._collect_factor_and_dimension(expr)
return expr
def _get_conversion_matrix_for_expr(expr, target_units):
from sympy import Matrix
expr_dim = Dimension(Quantity.get_dimensional_expr(expr))
dim_dependencies = dimsys_default.get_dimensional_dependencies(expr_dim, mark_dimensionless=True)
target_dims = [Dimension(Quantity.get_dimensional_expr(x)) for x in target_units]
canon_dim_units = {i for x in target_dims for i in dimsys_default.get_dimensional_dependencies(x, mark_dimensionless=True)}
canon_expr_units = {i for i in dim_dependencies}
if not canon_expr_units.issubset(canon_dim_units):
return None
canon_dim_units = sorted(canon_dim_units)
camat = Matrix([[dimsys_default.get_dimensional_dependencies(i, mark_dimensionless=True).get(j, 0) for i in target_dims] for j in canon_dim_units])
exprmat = Matrix([dim_dependencies.get(k, 0) for k in canon_dim_units])
res_exponents = camat.solve_least_squares(exprmat, method=None)
return res_exponents
[docs]def convert_to(expr, target_units):
"""
Convert expr to the same expression with all of its units and quantities
represented as factors of target_units, whenever the dimension is compatible.
target_units may be a single unit/quantity, or a collection of
units/quantities.
Examples
========
>>> from sympy.physics.units import speed_of_light, meter, gram, second, day
>>> from sympy.physics.units import mile, newton, kilogram, atomic_mass_constant
>>> from sympy.physics.units import kilometer, centimeter
>>> from sympy.physics.units import convert_to
>>> convert_to(mile, kilometer)
25146*kilometer/15625
>>> convert_to(mile, kilometer).n()
1.609344*kilometer
>>> convert_to(speed_of_light, meter/second)
299792458*meter/second
>>> convert_to(day, second)
86400*second
>>> 3*newton
3*newton
>>> convert_to(3*newton, kilogram*meter/second**2)
3*kilogram*meter/second**2
>>> convert_to(atomic_mass_constant, gram)
1.66053904e-24*gram
Conversion to multiple units:
>>> convert_to(speed_of_light, [meter, second])
299792458*meter/second
>>> convert_to(3*newton, [centimeter, gram, second])
300000*centimeter*gram/second**2
Conversion to Planck units:
>>> from sympy.physics.units import gravitational_constant, hbar
>>> convert_to(atomic_mass_constant, [gravitational_constant, speed_of_light, hbar]).n()
7.62950196312651e-20*gravitational_constant**(-0.5)*hbar**0.5*speed_of_light**0.5
"""
if not isinstance(target_units, (Iterable, Tuple)):
target_units = [target_units]
return Add.fromiter(convert_to(i, target_units) for i in expr.args)
expr = sympify(expr)
if not isinstance(expr, Quantity) and expr.has(Quantity):
expr = expr.replace(lambda x: isinstance(x, Quantity), lambda x: x.convert_to(target_units))
def get_total_scale_factor(expr):
if isinstance(expr, Mul):
return reduce(lambda x, y: x * y, [get_total_scale_factor(i) for i in expr.args])
elif isinstance(expr, Pow):
return get_total_scale_factor(expr.base) ** expr.exp
elif isinstance(expr, Quantity):
return expr.scale_factor
return expr
depmat = _get_conversion_matrix_for_expr(expr, target_units)
if depmat is None:
return expr
expr_scale_factor = get_total_scale_factor(expr)
return expr_scale_factor * Mul.fromiter((1/get_total_scale_factor(u) * u) ** p for u, p in zip(target_units, depmat))
def quantity_simplify(expr):
"""Return an equivalent expression in which prefixes are replaced
with numerical values and all units of a given dimension are the
unified in a canonical manner.
Examples
========
>>> from sympy.physics.units.util import quantity_simplify
>>> from sympy.physics.units.prefixes import kilo
>>> from sympy.physics.units import foot, inch
>>> quantity_simplify(kilo*foot*inch)
250*foot**2/3
>>> quantity_simplify(foot - 6*inch)
foot/2
"""
if expr.is_Atom or not expr.has(Prefix, Quantity):
return expr
# replace all prefixes with numerical values
p = expr.atoms(Prefix)
expr = expr.xreplace({p: p.scale_factor for p in p})
# replace all quantities of given dimension with a canonical
# quantity, chosen from those in the expression
d = sift(expr.atoms(Quantity), lambda i: i.dimension)
for k in d:
if len(d[k]) == 1:
continue
v = list(ordered(d[k]))
ref = v[0]/v[0].scale_factor
expr = expr.xreplace({vi: ref*vi.scale_factor for vi in v[1:]})
return expr
def check_dimensions(expr):
"""Return expr if there are not unitless values added to
dimensional quantities, else raise a ValueError."""
from sympy.solvers.solveset import _term_factors
# the case of adding a number to a dimensional quantity
# is ignored for the sake of SymPy core routines, so this
# function will raise an error now if such an addend is
# found.
# Also, when doing substitutions, multiplicative constants
# might be introduced, so remove those now
DIM_OF = dimsys_default.get_dimensional_dependencies
deset = set()
for ai in a.args:
if ai.is_number:
continue
dims = []
skip = False
for i in Mul.make_args(ai):
if i.has(Quantity):
i = Dimension(Quantity.get_dimensional_expr(i))
if i.has(Dimension):
dims.extend(DIM_OF(i).items())
elif i.free_symbols:
skip = True
break
if not skip:
if len(deset) > 1:
raise ValueError(
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## If a = 8 + 3√7 and b = 1 / a , what will be the value of 2+ 2.
Question
If a = 8 + 3√7 and b = 1 / a , what will be the value of 2+ 2.
in progress 0
1 month 2021-08-17T22:20:29+00:00 1 Answer 0 views 0
1. Step-by-step explanation:
A=8+3√7 b=1/a =1/8+3√7
= 8-3√7/1 (on rationalising)
now a= 8+3√7 b=8-3√7 b= 8-3√7
therefore a2+b2 =( 8+3√7 )2 + ( 8-3√7 )2
=( 64 + 63 + 48 √7 ) + ( 64+63-48√7 )
a2+b2= 254
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# Thermal Copper Physics
1. Feb 1, 2010
### redline7890
Two identical copper blocks of mass m grams, one at (fundamental) temperature T1
and the other at temperature T2 are brought into thermal contact until they reach
thermodynamic equilibrium. The heat capacity of copper/gram is Cv .
What is the final temperature Tf?
2. Feb 1, 2010
### Andrew Mason
Heat would, of course, flow from the hotter block to the colder block until they both reach the same temperature. Whatever heat flows out of the hot block flows into the cold block.
$$\Delta Q_1 = cm\Delta T_1$$
$$\Delta Q_2 = cm\Delta T_2$$
$$\Delta Q_1 + \Delta Q_2 = 0$$
And, of course:
$$T_1 + \Delta T_1 = T_f$$
$$T_2 + \Delta T_2 = T_f$$
(Note: One of the temperature changes is negative and one is positive).
You can work out what Tf is from the above.
Or you can observe that since the blocks are identical - same mass and same heat capacity - the relationship between the drop in temperature of the hotter block and the increase in temperature for the colder block is ____________?
AM
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# Inner products and basis of orthogonal complement
• May 15th 2010, 12:34 PM
Dave2718
Inner products and basis of orthogonal complement
Hey guys, I've been thinking about this problem for some time now, but I'm not really sure how to proceed to do it. I think the inner product stuff is throwing me off.
Let V be the vector space of all polynomials of degree 2 or less on the unit interval and define
$\displaystyle \langle f,g \rangle = \int^1_0 fg\,dt$
Find a basis for the orthogonal complement of {t-1,t^2}
• May 15th 2010, 01:03 PM
Bruno J.
I don't see what $\displaystyle 2\times 2$ matrices have to do with this problem!
• May 15th 2010, 02:48 PM
HallsofIvy
Quote:
Originally Posted by Dave2718
Hey guys, I've been thinking about this problem for some time now, but I'm not really sure how to proceed to do it. I think the inner product stuff is throwing me off.
Let V be the vector space of all 2x2 matrices of all polynomials of degree 2 or less on the unit interval and define
$\displaystyle \langle f,g \rangle = \int^1_0 fg\,dt$
Find a basis for the orthogonal complement of {t-1,t^2}
If you are really talking about 2 by 2 matrices that have polynomials of degree 2 or less as entries, then "$\displaystyle \int_0^1 fg dt$" makes no sense. And $\displaystyle \{t- 1, t^2\}$ is not a basis for any subset of that.
I suspect you mean simply "the vector space of all polynomials of degree 2 or less".
In that case, any such "vector" can be written as $\displaystyle at^2+ bt+ c$ and any vector in the "orthogonal complement of $\displaystyle \{t-1,t^2\}$" must satisfy $\displaystyle \int_0^1 (at^2+ bt+ c)(t- 1)dt= 0$ and $\displaystyle (at^2+ bt+ c)(t^2) dt= 0$. Do those integrals to get equations for a, b, and c. Since those are two equations, you will probably be able to solve for two of them in terms of the third. Write replace those two by their expression in terms of the third to get $\displaystyle ax^2+ bx+ c$ in terms of just one of a, b, or c. Finally, factor that a, b, or c out leaving just a number times a polynomial. That polynomial will be the single vector in the basis.
• May 15th 2010, 03:18 PM
dwsmith
Quote:
Originally Posted by HallsofIvy
If you are really talking about 2 by 2 matrices that have polynomials of degree 2 or less as entries, then "$\displaystyle \int_0^1 fg dt$" makes no sense. And $\displaystyle \{t- 1, t^2\}$ is not a basis for any subset of that.
I suspect you mean simply "the vector space of all polynomials of degree 2 or less".
In that case, any such "vector" can be written as $\displaystyle at^2+ bt+ c$ and any vector in the "orthogonal complement of $\displaystyle \{t-1,t^2\}$" must satisfy $\displaystyle \int_0^1 (at^2+ bt+ c)(t- 1)dt= 0$ and $\displaystyle (at^2+ bt+ c)(t^2) dt= 0$. Do those integrals to get equations for a, b, and c. Since those are two equations, you will probably be able to solve for two of them in terms of the third. Write replace those two by their expression in terms of the third to get $\displaystyle ax^2+ bx+ c$ in terms of just one of a, b, or c. Finally, factor that a, b, or c out leaving just a number times a polynomial. That polynomial will be the single vector in the basis.
I wanted to do this problem. Here is what I obtained.
Spoiler:
$\displaystyle c\begin{bmatrix} 50\\ -52\\ 9 \end{bmatrix}$
• May 15th 2010, 05:24 PM
Dave2718
sorry about that, yeah i think I was helping a friend out with another problem concerning 2x2 matrices and i subconsciously typed that in, OP has been fixed.
• May 15th 2010, 05:29 PM
Dave2718
Also, I was wondering if I would get the same answer if I did the gram-schmidt process on the set? Or would I not because that would simply be the orthogonal basis, not the basis for the orthogonal complement. If I did want to use gram-schmidt, I'd have to find the orthogonal complement first right?
• May 15th 2010, 09:18 PM
Bruno J.
Quote:
Originally Posted by Dave2718
Also, I was wondering if I would get the same answer if I did the gram-schmidt process on the set? Or would I not because that would simply be the orthogonal basis, not the basis for the orthogonal complement. If I did want to use gram-schmidt, I'd have to find the orthogonal complement first right?
You could use G-S to find an orthogonal basis for the subspace generated by $\displaystyle \{t-1, t^2\}$, and then extend this basis to an orthogonal basis for the whole space; the extra vector(s) in the basis would generate the orthogonal complement. (Prove it!)
• May 16th 2010, 02:18 AM
Dave2718
Alright..I'll have to do that proof another day :)
Anyways, so I would have a single vector in my basis for the orthogonal complement? I got {50t^2 -52t +9} to be the basis for my orthogonal complement. For some reason I feel wierd having just one vector be a spanning and linearly independent set for the set I started with.
Thanks though guys.
• May 16th 2010, 03:17 AM
HallsofIvy
You need to start thinking more in terms of "dimension". The set of all polynomials of degree 2 or less, $\displaystyle at^2+ bt+ c$, has dimension 3 ($\displaystyle \{t^2, t, 1\}$ is a basis). It should be easy to see that $\displaystyle t- 1$ and $\displaystyle t^2$ are independent so they span a 2 dimensional subspace. The "orthogonal complement" of that subspace must have dimension 3- 2= 1 so a basis must have only one vector.
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# Check the code
I was solving the Magic 8 Ball question 14. I did it a bit differently than the answer given in the hint. Can anyone just have a look on my code and give me feedback.
I apologise for the inconvenience caused. Below is the code posted as formatted text.
Here is the link to the question -
``````import random
name = ""
question = ""
random_number = random.randint(1, 15)
if random_number == 1:
elif random_number == 2:
answer += "It is decidedly so."
elif random_number == 3:
elif random_number == 4:
elif random_number == 5:
elif random_number == 6:
answer += "Better not tell you now."
elif random_number == 7:
answer += "My sources say no."
elif random_number == 8:
answer += "Outlook not so good."
elif random_number == 9:
elif random_number == 10:
answer += "You may rely on it."
elif random_number == 11:
answer += "As I see it, yes."
elif random_number == 12:
elif random_number == 13:
answer += "Don't count on it."
elif random_number == 14:
elif random_number == 15:
else:
if name == "" and question == "":
print("You have not asked a question.")
elif name == "":
print("Question: ", question)
else:
#if question == "":
#print("You have not asked a question.")
#else:
``````
Hi 16bme1040,
Your code looks great! If you want to take it a step further, you should consider assigning `name` and `question` as an `input()` functions like so:
``````name = input("What is your name: ")
question = input("What is your question: ")
``````
Also, on line 39, technically you only need to check if `question` is empty, and not both `name` and `question` like so:
``````if question == "":
print("You have not asked a question.")
``````
Again, great job!
Thank you again very much
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Intuitionistic Logic Explorer < Previous Next > Nearby theorems Mirrors > Home > ILE Home > Th. List > disjeq2dv GIF version
Theorem disjeq2dv 3791
Description: Equality deduction for disjoint collection. (Contributed by Mario Carneiro, 14-Nov-2016.)
Hypothesis
Ref Expression
disjeq2dv.1 ((𝜑𝑥𝐴) → 𝐵 = 𝐶)
Assertion
Ref Expression
disjeq2dv (𝜑 → (Disj 𝑥𝐴 𝐵Disj 𝑥𝐴 𝐶))
Distinct variable group: 𝜑,𝑥
Allowed substitution hints: 𝐴(𝑥) 𝐵(𝑥) 𝐶(𝑥)
Proof of Theorem disjeq2dv
StepHypRef Expression
1 disjeq2dv.1 . . 3 ((𝜑𝑥𝐴) → 𝐵 = 𝐶)
21ralrimiva 2439 . 2 (𝜑 → ∀𝑥𝐴 𝐵 = 𝐶)
3 disjeq2 3790 . 2 (∀𝑥𝐴 𝐵 = 𝐶 → (Disj 𝑥𝐴 𝐵Disj 𝑥𝐴 𝐶))
42, 3syl 14 1 (𝜑 → (Disj 𝑥𝐴 𝐵Disj 𝑥𝐴 𝐶))
Colors of variables: wff set class Syntax hints: → wi 4 ∧ wa 102 ↔ wb 103 = wceq 1285 ∈ wcel 1434 ∀wral 2353 Disj wdisj 3786 This theorem was proved from axioms: ax-1 5 ax-2 6 ax-mp 7 ax-ia1 104 ax-ia2 105 ax-ia3 106 ax-io 663 ax-5 1377 ax-7 1378 ax-gen 1379 ax-ie1 1423 ax-ie2 1424 ax-8 1436 ax-10 1437 ax-11 1438 ax-i12 1439 ax-bndl 1440 ax-4 1441 ax-17 1460 ax-i9 1464 ax-ial 1468 ax-i5r 1469 ax-ext 2065 This theorem depends on definitions: df-bi 115 df-nf 1391 df-sb 1688 df-eu 1946 df-mo 1947 df-clab 2070 df-cleq 2076 df-clel 2079 df-ral 2358 df-rmo 2361 df-in 2988 df-ss 2995 df-disj 3787 This theorem is referenced by: disjeq12d 3795
Copyright terms: Public domain W3C validator
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English | Español
# Try our Free Online Math Solver!
Online Math Solver
Depdendent Variable
Number of equations to solve: 23456789
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Solve for:
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Number of inequalities to solve: 23456789
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Please use this form if you would like to have this math solver on your website, free of charge. Name: Email: Your Website: Msg:
# Basic Operations on Matrices
In this worksheet you’ll find Maple commands used in matrix manipulation.
Definitions and rules of matrix operations can be found in section 7.2.
Notes: before you can use these matrix commands you have to bring up
the package linalg (for LINear ALGebra) using the with command.
> with(linalg):
Warning, new definition for norm
Warning, new definition for trace
1. Define a matrix:
You can use the command matrix to declare a matrix in Maple. Basically,
you must tell Maple the size of the matrix ( number of rows and then number
of columns) and the values of its entries (in row order ).
Example: Let’s define a matrix A of size 3x4 (three rows and 4 columns).
> A:= matrix(3,4,[-1,1,0,0,2,-2,0,2,3,3,10,9]);
Example: You can think of a vector in Rˆn as a matrix of n rows and one
column.
> vec:= matrix(4,1,[1,2,3,4]);
2. Addition and multiplication by a scalar :
Remember that you can add or substract two matrices only if they are of
the same size (same numbers of rows and same numbers of columns). You can
also multiply a matrix by any scalar ( real or complex number ).
These operations are componentwise !
Example: Let B be a matrix of size 3x4 (the same size as that of A given
above)
> B:= matrix(3,4,[-2,3,2,0,1,2,0,1,1,2,-1,2]);
Then we can compute: A+B, A-B, 3*A, A-2*B by using the command
evalm (EVALuate Matrices) as follows
> AplusB:= evalm(A+ B); AminusB:=evalm(A-B); threexA :=evalm(3*A);\
Aminus2B:=evalm(A-2*B);
Note: all the results above have the same size (3x4) as those of A and B.
3. Multiplication of matrices:
You can compute C*D where C and D are two matrices if and only if
the number of columns of C equal the number of rows of D
For example: We cannot have A*B because #cols of A=4 and #rows of
B=3.
However, if C is a matrix of size 4x2 then we can compute A*C and the
result is a matrix of size 3x2 (#rows = #rows of A; #cols = #cols of C).
Example: Lets define a matrix C of size 4x2 and compute A*C (can we
compute C*A? why not?)
> C:= matrix(4,2,[1,-1,2,0,1,1,4,2]);
> AC:= multiply(A,C);
Instead of using the command multiply to compute the product A*C we
can use again the command evalm to get the same thing. However, we have
to use the operation &* in this command to tell Maple that this is a
matrix multiplication.
> evalm(A &* C);
Notes: if A and B are square matrices (#rows=#cols) of the same size
then we can compute both A*B and B*A. However, A*B are not equal to
B*A in general. That is: matrix multiplication is not a commutative
operation.
4. Determinant:
If A is a square matrix, we can compute its determinant by using the
command det
> A:= matrix(3,3,[1,2,3,2,-2,1,0,2,4]); detA:= det(A);
detA := −14
5. Inverse of a square matrix:
If A is a square matrix then its inverse (if exists!) is the square matrix B
of the same size such that the product A *B and B*A are the identity matrix.
If the inverse of A exists we say that A is invertible.
Important:
A square matrix is invertible if and only if the determinant of A is
nonzero
For example, the matrix A defined above is invertible.
We can compute the inverse of A by using the command inverse of Maple.
> invA:= inverse(A);
Let’s check if the product A*invA and invA*A are both identity matrix (you
can also use the command evalm).
> multiply(A,invA); mult2iply(invA,A);
6. Matrix function:
Each entry of a matrix can be a function and we can perform differentiation
and integration operation on the matrix. Again, these operations are simply
componentwise.
Example: Let f be a matrix 2x2 whose entries are functions in t
> f:=matrix(2,2,[exp(2*t),cos(t),sin(2*t),exp(t)]);
We compute the derivative of f with respect to the argument t. Thing is
a little bit complicate here because we cannot simply type diff(f,t). Maple
requires us to use the command map for this kind of operation (In other words,
we have to map the procedure diff to all matrix entries).
> Df:= map(diff,f,t);
Compute the integral of f (again, using the map command to map the
procedure int to all matrix entries).
> Intf:=map(int,f,t);
Eigenvalues and Eigenvectors
The background material for this part is presented in section 7.3.
Given a square matrix M. For example
> M:= matrix(3,3,[1,0,4,4,1,-2,5,0,2]);
We define the characteristic matrix ofMto be the matrix lambda*Identity
- M. That is
> Id:=matrix(3,3,[1,0,0,0,1,0,0,0,1]); cM:=evalm(lambda*Id-M);
The Maple command charmat can be used to get the same result
> charmat(M,lambda)2;
The characteristic polynomial of M is simply the determinant of the
characteristic matrix.
For example,
> cpoly_of_M:= det(cM);
You can also use the command charpoly to compute the characteristic
polynomial of a matrix M. For example,
> charpoly(M,lambda);
The eigenvalues of the matrixMare simply the roots of this polynomial .
> evs:=solve(cpoly_of_M=0,lambda);
evs := 1, 6, −3
You can also use the command eigenvals to compute these values
> evs:= eigenvals(M);
evs := 1, 6, −3
There are eigenvectors of the matrix M corresponding to each of the eigen-values.
An eigenvector of M corresponding to the eigenvalue r is a vector X
such that M*X = r*X
Equivalently , X is the solution of the system (r*Id - M)*X=0. We say that
X belongs to the nullspace of the matrix r*Id-M.
For example, if we want to compute eigenvectors of the matrix M corresponding
to the eigenvalue 6 then we have to compute the nullspace of the matrix
6*Id - M. The Maple command nullspace can be used here
> m:=evalm(6*Id-M); nullspace_of_m:=nullspace(m);
That is, the eigenvector corresponding to 6 is (10/3, 1, 25/6).
Similarly, we can compute the eigenvectors corresponding to the other eigenvalues
1,-3.
The command eigenvects can be used to get all the eigenvectors of the
matrix M at one shot:
> eigenvects(M);
In this case we have three lists. Each of them tells us: the eigenvalue, its
multiplicity (how many times it repeats) and the corresponding eigenvectors.
Prev Next
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Outlook: Elastic N.V. Ordinary Shares is assigned short-term B3 & long-term B2 estimated rating.
AUC Score : What is AUC Score?
Short-Term Revised1 :
Dominant Strategy : Speculative Trend
Time series to forecast n: for Weeks2
Methodology : Reinforcement Machine Learning (ML)
Hypothesis Testing : Sign Test
Surveillance : Major exchange and OTC
1The accuracy of the model is being monitored on a regular basis.(15-minute period)
2Time series is updated based on short-term trends.
## Summary
Elastic N.V. Ordinary Shares prediction model is evaluated with Reinforcement Machine Learning (ML) and Sign Test1,2,3,4 and it is concluded that the ESTC stock is predictable in the short/long term. Reinforcement machine learning (RL) is a type of machine learning where an agent learns to take actions in an environment in order to maximize a reward. The agent does this by trial and error, and is able to learn from its mistakes. RL is a powerful tool that can be used for a variety of tasks, including game playing, robotics, and finance. According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Speculative Trend
## Key Points
1. What is statistical models in machine learning?
2. How do predictive algorithms actually work?
3. Is now good time to invest?
## ESTC Target Price Prediction Modeling Methodology
We consider Elastic N.V. Ordinary Shares Decision Process with Reinforcement Machine Learning (ML) where A is the set of discrete actions of ESTC stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4
F(Sign Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Reinforcement Machine Learning (ML)) X S(n):→ 8 Weeks $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$
n:Time series to forecast
p:Price signals of ESTC stock
j:Nash equilibria (Neural Network)
k:Dominated move
a:Best response for target price
### Reinforcement Machine Learning (ML)
Reinforcement machine learning (RL) is a type of machine learning where an agent learns to take actions in an environment in order to maximize a reward. The agent does this by trial and error, and is able to learn from its mistakes. RL is a powerful tool that can be used for a variety of tasks, including game playing, robotics, and finance.
### Sign Test
The sign test is a non-parametric hypothesis test that is used to compare two paired samples. In a paired sample, each data point in one sample is paired with a data point in the other sample. The pairs are typically related in some way, such as before and after measurements, or measurements from the same subject under different conditions. The sign test is a non-parametric test, which means that it does not assume that the data is normally distributed. The sign test is also a dependent samples test, which means that the data points in each pair are correlated.
For further technical information as per how our model work we invite you to visit the article below:
How do AC Investment Research machine learning (predictive) algorithms actually work?
## ESTC Stock Forecast (Buy or Sell)
Sample Set: Neural Network
Stock/Index: ESTC Elastic N.V. Ordinary Shares
Time series to forecast: 8 Weeks
According to price forecasts, the dominant strategy among neural network is: Speculative Trend
Strategic Interaction Table Legend:
X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)
Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)
Z axis (Grey to Black): *Technical Analysis%
### Financial Data Adjustments for Reinforcement Machine Learning (ML) based ESTC Stock Prediction Model
1. The purpose of estimating expected credit losses is neither to estimate a worstcase scenario nor to estimate the best-case scenario. Instead, an estimate of expected credit losses shall always reflect the possibility that a credit loss occurs and the possibility that no credit loss occurs even if the most likely outcome is no credit loss.
2. Alternatively, the entity may base the assessment on both types of information, ie qualitative factors that are not captured through the internal ratings process and a specific internal rating category at the reporting date, taking into consideration the credit risk characteristics at initial recognition, if both types of information are relevant.
3. To the extent that a transfer of a financial asset does not qualify for derecognition, the transferee does not recognise the transferred asset as its asset. The transferee derecognises the cash or other consideration paid and recognises a receivable from the transferor. If the transferor has both a right and an obligation to reacquire control of the entire transferred asset for a fixed amount (such as under a repurchase agreement), the transferee may measure its receivable at amortised cost if it meets the criteria in paragraph 4.1.2.
4. A net position is eligible for hedge accounting only if an entity hedges on a net basis for risk management purposes. Whether an entity hedges in this way is a matter of fact (not merely of assertion or documentation). Hence, an entity cannot apply hedge accounting on a net basis solely to achieve a particular accounting outcome if that would not reflect its risk management approach. Net position hedging must form part of an established risk management strategy. Normally this would be approved by key management personnel as defined in IAS 24.
*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.
### ESTC Elastic N.V. Ordinary Shares Financial Analysis*
Rating Short-Term Long-Term Senior
Outlook*B3B2
Income StatementBa1Caa2
Balance SheetB3Ba1
Leverage RatiosCBa1
Cash FlowCB3
Rates of Return and ProfitabilityCaa2C
*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?
## Conclusions
Elastic N.V. Ordinary Shares is assigned short-term B3 & long-term B2 estimated rating. Elastic N.V. Ordinary Shares prediction model is evaluated with Reinforcement Machine Learning (ML) and Sign Test1,2,3,4 and it is concluded that the ESTC stock is predictable in the short/long term. According to price forecasts for 8 Weeks period, the dominant strategy among neural network is: Speculative Trend
### Prediction Confidence Score
Trust metric by Neural Network: 90 out of 100 with 686 signals.
## References
1. Alexander, J. C. Jr. (1995), "Refining the degree of earnings surprise: A comparison of statistical and analysts' forecasts," Financial Review, 30, 469–506.
2. H. Khalil and J. Grizzle. Nonlinear systems, volume 3. Prentice hall Upper Saddle River, 2002.
3. Hornik K, Stinchcombe M, White H. 1989. Multilayer feedforward networks are universal approximators. Neural Netw. 2:359–66
4. Breiman L. 2001a. Random forests. Mach. Learn. 45:5–32
5. Bessler, D. A. S. W. Fuller (1993), "Cointegration between U.S. wheat markets," Journal of Regional Science, 33, 481–501.
6. Hill JL. 2011. Bayesian nonparametric modeling for causal inference. J. Comput. Graph. Stat. 20:217–40
7. Hill JL. 2011. Bayesian nonparametric modeling for causal inference. J. Comput. Graph. Stat. 20:217–40
Frequently Asked QuestionsQ: What is the prediction methodology for ESTC stock?
A: ESTC stock prediction methodology: We evaluate the prediction models Reinforcement Machine Learning (ML) and Sign Test
Q: Is ESTC stock a buy or sell?
A: The dominant strategy among neural network is to Speculative Trend ESTC Stock.
Q: Is Elastic N.V. Ordinary Shares stock a good investment?
A: The consensus rating for Elastic N.V. Ordinary Shares is Speculative Trend and is assigned short-term B3 & long-term B2 estimated rating.
Q: What is the consensus rating of ESTC stock?
A: The consensus rating for ESTC is Speculative Trend.
Q: What is the prediction period for ESTC stock?
A: The prediction period for ESTC is 8 Weeks
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# Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício
Ivo Sales Henriques Miranda
Dissertação para obtenção do Grau de Mestre em
Engenharia Civil
Júri
Presidente: Prof. Fernando Manuel Fernandes Simões Orientador: Prof. José Manuel Matos Noronha da Camara Vogal: Prof. Augusto Martins Gomes
Outubro 2012
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
RESUMO
Palavras-Chave: Projeto de Edifícios, Betão Estrutural, Análise de Deformações, Matrizes de Influência de Deslocamentos, Bandas Maciças, Pré-esforço.
i
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
ABSTRACT
The present dissertation aims at analyzing and discussing the displacement on a flat slab without beams with different geometries, using a prestressed solution. The analysis was performed on a two-storey building to be used as a night club, a use that requires large open areas without vertical elements. The investigation was performed on the slab of the first floor. It presents large spans, and also has to support the load of 3 columns that are supporting the roof and do not continue until the ground floor. After defining the initial constraints, such as materials, loads and load combinations, it was concluded that an ordinary solution with reinforced concrete did not fulfil the requirements in serviceable limit state for the maximum admissible displacements. It was, therefore, necessary to design a solution with prestressed bands. The evaluation of the displacements on the slab with prestressed bands was performed considering two different geometries, and for each geometry, two different shapes for the cables: parabolic and polygonal. The evaluation was performed on control points, defined as the conditioning points where maximum displacement was observed for each solution. From the control points, influence matrices for each slab were determined. The rows of the matrices represent the displacement of the points for the strength of prestress applied in each band. Finally, after careful analysis of the different solutions, the best solution was selected and the ultimate limit states were verified for that solution.
Key-words: Building design, Structural Concrete, Displacement Analysis, Matrix of Influence of the Displacements, Prestressed Bands, Prestress.
ii
Professor José Câmara. iii . por me darem força e investirem na minha formação. Aos meus irmãos por me motivarem quando mais precisava. Aos meus amigos e colegas por me apoiarem ao longo de todo o percurso académico. Aos meus pais por acreditarem em mim. À minha namorada por todo o apoio e paciência. pela sua dedicação e disponibilidade ao longo da realização desta dissertação.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico AGRADECIMENTOS Ao meu orientador.
.
2 3.2 2.1.4 SOLUÇÕES DE PRÉ-ESFORÇO CONSIDERADAS PONTOS DE CONTROLO DEFORMADA E DESLOCAMENTOS DAS CARGAS QUASE PERMANENTES TRAÇADO E ESTIMATIVA DE PRÉ-ESFORÇO A APLICAR NAS BANDAS Y1 E Y2 27 30 31 32 34 v .3 3.1 3.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico ÍNDICE 1 INTRODUÇÃO 1.1.4 PRÉ-DIMENSIONAMENTO PILARES VIGAS LAJES ESTIMATIVA DA DEFORMAÇÃO ATRAVÉS DE CÁLCULOS SIMPLIFICADOS MODELAÇÃO ESTRUTURAL VALIDAÇÃO DO MODELO DETERMINAÇÃO DAS DEFORMAÇÕES RECORRENDO AO SAP2000 13 13 13 15 16 18 22 23 25 4 AVALIAÇÃO DAS DEFORMAÇÕES COM PRÉ-ESFORÇO 4.2 4.3 4.1.3 3.2 2.1 3.1 4.1 2.3 DEFINIÇÃO DOS M ATERIAIS QUANTIFICAÇÃO DAS AÇÕES COMBINAÇÕES DE AÇÕES ESTADO LIMITE ÚLTIMO ESTADO LIMITE DE SERVIÇO COEFICIENTES PARCIAIS E COEFICIENTES DE COMBINAÇÃO 7 7 8 9 10 10 11 3 AVALIAÇÃO DAS DEFORMAÇÕES SEM PRÉ-ESFORÇO 3.3 OBJETIVO DA DISSERTAÇÃO ENQUADRAMENTO DO TRABALHO ORGANIZAÇÃO EM CAPÍTULOS 1 1 1 5 2 CONSIDERAÇÕES INICIAIS 2.2 3.3.2 1.3.1 1.3.3 2.3.1 2.1 3.
7.4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico 4.1.4.1 5.2 4.2 4.1 5.7 4.1 4.4.3 CARGAS QUASE PERMANENTES AVALIAÇÃO DA FLEXÃO GLOBAL E CARGA EQUIVALENTE DISTRIBUÍDA AVALIAÇÃO DO VALOR DE PRÉ-ESFORÇO TRAÇADO E ESTIMATIVA DE PRÉ-ESFORÇO NA BANDA Y3 E NA BANDA X CARGAS EQUIVALENTES DETERMINAÇÃO DAS DEFORMAÇÕES RECORRENDO AO SAP2000 CARGAS EQUIVALENTES DE PRÉ-ESFORÇO E MATRIZES DE INFLUÊNCIA SOLUÇÃO DE PRÉ-ESFORÇO ADOTADA ANÁLISE DAS MATRIZES DE INFLUÊNCIA E AS SUAS FUNCIONALIDADES 35 36 37 39 46 51 52 57 58 5 VERIFICAÇÃO DA SEGURANÇA AOS ESTADOS LIMITES ÚLTIMOS 5.7.6 4.3 4.1.1 4.2 ESTADO LIMITE ÚLTIMO DE FLEXÃO BANDA – ELU DE FLEXÃO COM PRÉ-ESFORÇO PELO LADO DA AÇÃO LAJE – ELU DE FLEXÃO 62 62 62 64 6 CONCLUSÃO 7 BIBLIOGRAFIA 68 70 vi .7.5 4.
........................... 18 Figura 3...................... 32 Figura 4.........5 – Modelos de cálculo para as soluções B1 e B2 ........ 4 Figura 1......... 17 Figura 3....... 19 Figura 4......................................................................................................................................................................................4 – Modelos de cálculo para as soluções A1 e A2 ...... 19 Figura 3...........6 – Planta com pontos de controlo assinalados a vermelho ......2 – Planta completa do piso 1 com zona de estudo assinalada a vermelho .......................................................................................4 – Planta da cobertura com a banda em X considerada ................. 2 Figura 1.......................................................... Y2 e Y3 ........................................................................... com indicação dos elementos verticais ................3 – Zona de estudo do piso 1 ....................8 – Modelo de cálculo 3 para as faixas Y1 e Y2 .......................................4 – Planta de estruturas da zona em estudo no piso 1...... 30 Figura 4......................................... 2 Figura 1................... 19 Figura 3.......................................................1 – Pilar condicionante para o pré-dimensionamento e respetiva área de influência no piso 1 .....1 – Localização das bandas Y1................................................. 29 Figura 4.................7 – Modelo de cálculo 2 para a faixa Y3 .........Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico ÍNDICE DE FIGURAS Figura 1.2 – Localização das bandas Y e da banda X ............................................................................................. 14 Figura 3.......5 – Zona crítica assinalada a vermelho e faixas Y1................................................3 – Secção transversal de uma banda com o máximo de cabos permitido .......... com indicação dos elementos verticais .......................... Y2 e Y3 ...........................7 – Deformada provocada pela combinação das cargas quase permanentes na configuração B sem cabos de pré-esforço ..6 – Modelo de cálculo 1 para a faixa Y3 ........... 14 Figura 3..................7 – Vista tridimensional da fachada ....................... 28 4...........8 – Vista tridimensional em perspetiva ................................................................. 3 Figura 1...... 4 Figura 1................. 19 Figura 3............................................................................................................................................................ 30 Figura 4.............................................................................. 4 Figura 1........... 33 vii ............................3 – Secção da laje aligeirada ...................1 – Zona de estudo assinalada a vermelho ...5 – Planta de estruturas da zona em estudo na cobertura........................ 3 Figura 1.................................6 – Vista tridimensional do alçado ............................................. 28 Figura 4.......................................................2 – Pilar condicionante para o pré-dimensionamento e respetiva área de influência na cobertura preenchida a verde ....................... 5 Figura 3........................
........................................................ sem qualquer banda ......................................................................10 – Características da banda X ............................................................. 34 Tabela 4.............................. 36 Tabela 4...... 47 ix .......................... 11 Tabela 3.....13 – Número de cabos a aplicar em cada banda e configuração .......5 – Coeficientes de combinação ......................2 – Características dos aços .............................................................1 – Características do betão ......9 – Tensão média....... 8 Tabela 2..................2 – Configuração A................15 – Cargas equivalentes nas bandas Y1 e Y2: Soluções A1 e B1 ............................. 24 Tabela 4................ 21 Tabela 3......................................12 – Tensão média..................................................................................... 43 Tabela 4...............................................8 – Valor do pré-esforço útil ................... 45 Tabela 4.................... 23 Tabela 3.......7 – Reações globais verticais da estrutura devido a elementos de barra ........................................ 42 Tabela 4................................................................................................................ 45 Tabela 4..5 – Reações globais da estrutura retiradas do programa SAP2000 ...........2 – Flechas devido às cargas distribuídas nos três modelos ......... 44 Tabela 4......4 – Coeficientes parciais .................... 11 Tabela 2... 41 Tabela 4.............................4 – Flecha total para os três modelos de cálculo ......................................14 – Cargas equivalentes nas bandas Y1 e Y2: Soluções parabólicas A1 e B1........... 16 Tabela 3.......6 – Características da banda .................... 24 Tabela 3........................................4 – Combinação das cargas quase permanentes no piso 1 e na cobertura .......... 21 Tabela 3.........................................................................7 – Carga equivalente de pré-esforço unidades ............................................................................................................3 – Configuração B.................. 34 Tabela 4.... 42 Tabela 4........ 21 Tabela 3.................6 – Reações globais verticais da estrutura devido a elementos de área ........................................ 36 Tabela 4............Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico ÍNDICE DE TABELAS Tabela 2...................................................... 45 Tabela 4...................................1 – Esbeltezas usuais para lajes fungiformes ..........5 – Combinação das cargas quase permanentes nas bandas ...............................................3 – Flechas devido à carga pontual nos três modelos ................... 46 Tabela 4................................................................................ 34 Tabela 4........... 9 Tabela 2........................................................................3 – Quantificação das ações ...11 – Valor de pré-esforço útil ....................1 – Laje maciça de 25 cm.............................................................................. 7 Tabela 2.....................
.......... 48 Tabela 4..........19 – Cargas equivalentes na banda Y3: Solução poligonal B2 ... 49 Tabela 4.......24 – Matriz de influência da solução A2 ..........................................29 – Flechas máximas em todo o piso retiradas do SAP para a melhor solução B2 ... 50 Tabela 4.....................................................26 – Matriz de influência da solução B2 ......31 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução B1................Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4......................... 60 Tabela 5................ 59 Tabela 4........................................................................................................1 – Características da secção ................... 54 Tabela 4...... 54 Tabela 4..............23 – Matriz de influência da solução A1 ...28 – Flechas máximas em todo o piso retiradas do SAP para a melhor solução B1 ................ 63 x ...............................16 – Cargas equivalentes na banda Y3: Solução parabólica A1 ...18 – Cargas equivalentes na banda Y3: Solução parabólica B1 ...33 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução B2......................................................................................................................20 – Cargas equivalentes na banda X: Solução parabólica B1 ......................................................................... 58 Tabela 4... 51 Tabela 4.....30 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução A1.....................22 – Reações globais para um valor de pré-esforço de 1000 kN em cada banda .................25 – Matriz de influência da solução B1 .. 59 Tabela 4....... 57 Tabela 4..........................21 – Cargas equivalentes na banda X: Solução poligonal B2 .........................17 – Cargas equivalentes na banda Y3: Solução poligonal A2 ................................................................32 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução A2.................................................................................. 48 Tabela 4... 58 Tabela 4..... 55 Tabela 4..... 52 Tabela 4........................ 54 Tabela 4............ 60 Tabela 4................................ 49 Tabela 4.................................................................27 – Melhores configurações de cabos para cada solução ..............................................
surgindo então.1 – Zona de estudo assinalada a vermelho A arquitetura no presente caso requer espaços amplos sem pilares para a pistas de dança no piso térreo. Figura 1. respetivamente. a necessidade de encontrar uma solução de engenharia estrutural que de uma forma economicamente viável satisfaça os critérios de funcionalidade previstos.2 e 1.3.2 – Planta completa do piso 1 com zona de estudo assinalada a vermelho 2 .Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 1. a planta de arquitetura geral do piso 1 e da zona que será efetivamente estudada na presente dissertação. Apresentam-se nas figuras 1.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 1.4 – Planta de estruturas da zona em estudo no piso 1.6 a 1. fachada e visão geral.3 – Zona de estudo do piso 1 A geometria global da estrutura foi definida no contexto do desenvolvimento do projeto base de arquitetura e de estruturas. Figura 1. os quais originaram a estrutura considerada no presente trabalho. com indicação dos elementos verticais 3 .8): Alçado. Em todas as imagens se assinalaram com um círculo azul os pilares sem continuidade para o piso inferior e em planta preencheram-se a vermelho os pilares que apenas existem no piso inferior. Para complementar a compreensão do projeto considerado apresenta-se uma planta geral ao nível do piso 1 (figura 1.4) e ao nível da cobertura (figura 1.5) com localização dos elementos verticais previstos para a estrutura e o modelo tridimensional em diversas perspetivas (figuras 1.
6 – Vista tridimensional do alçado Figura 1. com indicação dos elementos verticais Figura 1.7 – Vista tridimensional da fachada 4 .Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 1.5 – Planta de estruturas da zona em estudo na cobertura.
o seu enquadramento e a explicação da organização dos capítulos. No âmbito deste trabalho far-se-á a avaliação da deformação do piso para as ações quase permanentes e o estudo de várias soluções com recurso a pré-esforço para as contrariar o mais eficientemente possível. Evidentemente que neste processo se discutirá o prédimensionamento. dimensionamento e verificação da segurança á rotura e disposições gerais para a pormenorização da laje do piso e das bandas pré-esforçadas.3 ORGANIZAÇÃO EM CAPÍTULOS Com o intuito de facilitar a leitura e a compreensão da presente dissertação esta foi organizada em 6 capítulos principais. O primeiro capítulo designa-se “Introdução” contendo o objetivo da dissertação. devidas essencialmente à geometria imposta que exige vãos livres significativos na zona de discoteca e com cargas concentradas da cobertura que se apoiam no piso intermédio. nomeadamente na compreensão da estrutura e da sua envolvente.8 – Vista tridimensional em perspetiva Como já referido anteriormente a principal questão que se coloca no projeto de estruturas do referido edifício é o controlo das deformações na laje do piso 1. Pretende-se que a informação contida neste capítulo seja suficiente para a compreensão geral da dissertação.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 1. 5 . e dos aspetos que serão abordados ao longo do documento. 1.
1k Propriedades 500 435 200 1860 1670 MPa MPa GPa MPa MPa 2. Ações Variáveis: ações cujos valores sofrem uma variação significativa em torno do seu valor médio ao longo do tempo. tendo só sido consideradas as que apresentam valores na última coluna. como: Ações Permanentes: ações de valor constante que atuam na estrutura durante toda a sua vida útil. Seguidamente apresenta-se uma tabela com indicação geral das ações.2 – Características dos aços Materiais fyk Armaduras A500NR fyd Es Pré-Esforço A1670/1860 fpk fp0.2 QUANTIFICAÇÃO DAS AÇÕES A análise estrutural deve considerar a influência de todas as ações que possam vir a atuar no edifício e alterem os estados de tensão dos materiais e consequentemente os esforços nos elementos resistentes (ações diretas) ou que gerem deformações significativas na estrutura (ações indiretas). 8 . Estas podem ser classificadas segundo a sua permanência e probabilidade de ocorrência.3 correspondem ao definido no Eurocódigo 1 [2]. Ações Acidentais: ações de duração muito curta e com muito baixa probabilidade de ocorrência durante a vida útil do edifício (este tipo de ações não foi considerado no dimensionamento do edifício em estudo).Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 2. Os valores apresentados na Tabela 2.
0 kN/m 1. Estes permitem verificar.5 kN/m 1.0 kN/m 9. por outro lado. por um lado.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 2. As combinações de ações a considerar no dimensionamento de um edifício de betão armado encontram-se definidas no Eurocódigo 0 [3].3 COMBINAÇÕES DE AÇÕES O dimensionamento de estruturas é realizado com base na verificação de segurança aos estados limite último e de serviço.0 kN/m 2 2 25 kN/m 18 kN/m 3 3 2. 9 .3 – Quantificação das ações Peso Próprio: Betão armado Impulso do solo nas paredes Solo Permanentes Restantes Cargas Permanentes: Revestimento nos pisos Ações Diretas Revestimento na cobertura não acessível Peso do solo na cobertura (espessura 50 cm) Sobrecargas: Pisos – Categoria C5 Variáveis Vento Neve Retração Permanentes Ações Indiretas Variáveis Sismo Fluência Variações de Temperatura Cobertura não acessível – Categoria H 5. que as condições de bom comportamento em serviço são verificadas. que a capacidade resistente assegura o nível de segurança adequado à rotura e.0 kN/m 2 2 2 2.
7 0 1 0.3.4 – Coeficientes parciais G Ação favorável Ação desfavorável 1.3 COEFICIENTES PARCIAIS E COEFICIENTES DE COMBINAÇÃO Os coeficientes parciais (k) e os coeficientes de combinação (i) a fletor são os apresentados nas Tabelas 2.5 – Coeficientes de combinação Sobrecargas Pisos (Categoria C5) Cobertura (Categoria H) 0 0.00 1.35 Q 0 1.6 0 11 .50 Tabela 2.5. Tabela 2.7 0 2 0. Estes valores são quantificados de acordo com o Eurocódigo 0 [3]. respetivamente.4 e 2.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico 2.
.
Seguidamente verifica-se a sua segurança e pormenorizam-se as armaduras dos elementos estruturais para respeitarem os estados limites. com SAP2000 é possível obter uma avaliação do comportamento global mais rigorosa. em princípio. por exemplo. para posteriormente se proceder à sua análise e dimensionamento. optou-se por analisar o pilar mais condicionante para a combinação fundamental. No entanto.1 PRÉ-DIMENSIONAMENTO O objetivo do projetista é. As orientações para o pré-dimensionamento são baseadas na experiência anterior e visam a definição de uma solução que. 13 . pois tem a maior área de influência em ambos os pisos como se pode observar nas figuras 3.1 e 3. pelo controlo das deformações para combinações de ações de utilização.1. Para uma melhor avaliação das deformações consideraram-se dois níveis de análise: 1) Estimativa das deformações através de cálculos simplificados 2) Determinação das deformações em modelo analítico recorrendo ao SAP2000 A primeira análise é bastante útil para ter uma ordem de grandeza dos resultados expectáveis. em geral. Caso estes não sejam assegurados em alguma fase. considera-se mais eficiente pré-dimensionar a geometria dos elementos estruturais tendo por base critérios usuais baseados nos vãos a vencer e no tipo e valor das cargas a que estarão sujeitos no seu período de vida. sendo este o pilar B1-A7. 3. com um modelo analítico de elementos finitos como. é necessário efetuar as alterações pertinentes e iniciar novo ciclo até que seja atingida uma boa solução em termos de segurança e economia. procurar uma boa solução da forma mais eficiente possível.1 PILARES Para pré-dimensionar os pilares. 3. não estará muito afastada da boa solução que se procura.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico 3 AVALIAÇÃO DAS DEFORMAÇÕES SEM PRÉ-ESFORÇO A avaliação da necessidade de aplicação de pré-esforço num piso estrutural passa. Em geral. No que se segue vai discutir-se a definição geométrica da estrutura. no essencial.2.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 3. A área de influência do pilar referente ao piso 1 é de aproximadamente 81. As cargas para a combinação fundamental em cada piso são as seguintes: (3. ambas as áreas com o valor de 20.2 – Pilar condicionante para o pré-dimensionamento e respetiva área de influência na cobertura preenchida a verde As áreas de influência calcularam-se considerando que o pilar tem uma área de influência de 1/2 do vão entre apoios contínuos e 5/8 entre apoio contínuo e de extremidade.6 m .6 m . Mediram-se ainda as áreas de influência dos pilares metálicos B2-A7 e B2-A8.1 – Pilar condicionante para o pré-dimensionamento e respetiva área de influência no piso 1 Figura 3.1) 2 2 2 14 .7 m e da cobertura de 45. Considerou-se que o pilar em questão (B1-A7) recebia 50% da carga do pilar B2-A7 e 25% da carga do pilar B2-A8.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
No caso de estudo a arquitetura não impõe restrições significativas às vigas pelo que se adotou de acordo com o primeiro critério uma altura de 0,6m e uma largura de 0,25m para todas as vigas dado que têm vãos e carregamentos semelhantes. As vigas consideradas são apenas de contorno e podem observar-se nas figuras 1.4 (piso 1) e 1.5 (cobertura). Os cálculos simplificados efetuados e a análise posterior através do programa de elementos finitos SAP confirmaram a adequabilidade das dimensões definidas.
3.1.3
LAJES
Prevê-se para o primeiro piso, uma solução em laje de forma a minimizar a máxima altura do piso devido às necessidades de pé direito da discoteca do R/C. Para este tipo estrutural e com vãos significativos, da ordem dos 11 metros, havia no projeto de arquitetura uma opção inicial definida com uma solução de laje aligeirada. Com base nessa solução, poderia prédimensionar-se a espessura, para este caso, de acordo com o critério de esbelteza L/20, que conduziria a uma espessura de 55 cm. O motivo pelo qual se adotaria uma menor esbelteza do que as usuais referidas na tabela 3.1, é o facto de a laje ser carregada com os pilares descontínuos do piso superior, sendo portanto aconselhável utilizar lajes mais espessas. Mesmo neste caso, é natural que da análise a efetuar se confirme a necessidade da aplicação do pré-esforço. Importa referir que as lajes aligeiradas são presentemente menos utilizadas dada a sua maior vulnerabilidade em caso de incêndio.
Tabela 3.1 – Esbeltezas usuais para lajes fungiformes
h [m] Laje fungiforme tipo L/h 4 Laje maciça Laje maciça com capitel Laje aligeirada Laje aligeirada com capitel Laje maciça pré-esforçada Laje aligeirada pré-esforçada 25 a 30 35 a 40 20 a 25 25 a 30 40 35 0,15 0,15 0,225 5 6 0,20 7 0,25 0,20 0,25 0,225 0,25 0,30 0,25 0,20 0,225 0,25 0,35 0,30 0,35 0,25 0,30 0,30 0,35 0,60 L [m] 8 9 10 12 20
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Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
Tendo por base as dimensões usuais de moldes para lajes aligeiradas concebeu-se a laje aligeirada simplificada que se apresenta na figura 3.3.
Figura 3.3 – Secção da laje aligeirada
Seguidamente determinam-se as características mais relevantes da secção: Área: (3.5) (3.6) (3.7)
Inércia em relação ao centro de gravidade: (3.8)
(3.9)
(3.10)
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Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
(3.11)
Relativamente à laje da cobertura, indicada na figura 3.4, optou-se por uma laje maciça com uma esbelteza de L/20 (ver tabela 3.1), pois embora seja uma cobertura não acessível tem cerca de 0,5 metros de terra o que significa que a esbelteza terá que ser menor do que a usual. Importa referir que se considerou uma banda no alinhamento B1 de forma a reduzir o maior vão da laje para aproximadamente metade do seu valor original, desta forma o prédimensionamento conduziu a uma laje de 25 cm para a cobertura. Com o pré-dimensionamento dos elementos estruturais relevantes, ações e combinações de ações definidas é possível analisar a estrutura e, em particular, estimar a deformação, em especial para o piso 1, pelos dois métodos mencionados anteriormente.
Figura 3.4 – Planta da cobertura com a banda em X considerada
3.2
ESTIMATIVA DA DEFORMAÇÃO ATRAVÉS DE CÁLCULOS SIMPLIFICADOS
Na Figura 3.5 indica-se a zona mais suscetível a deformações e as faixas consideradas para o modelo de cálculo simplificado. Limitou-se a análise apenas na direção Y, dado que a
18
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico
dimensão X da zona em causa mede mais do dobro da dimensão Y, provocando uma flexão quase cilíndrica em torno da direção X.
Figura 3.5 – Zona crítica assinalada a vermelho e faixas Y1, Y2 e Y3
Para obter uma estimativa da deformação consideraram-se os seguintes modelos de cálculo aproximados:
Figura 3.6 – Modelo de cálculo 1 para a faixa Y3
Figura 3.7 – Modelo de cálculo 2 para a faixa Y3
Figura 3.8 – Modelo de cálculo 3 para as faixas Y1 e Y2
19
32E-03 P [kN] 346.03 δdist [mm] 70.7 21 .Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 3.4 – Flecha total para os três modelos de cálculo δtotal [mm] Modelo 1 Modelo 2 Modelo 3 106.67 I faixa [m ] 2.1 L [m] 17 10.92E-02 2.58 Ainda de acordo com as Tabelas de cálculo de Betão Armado e Pré-esforçado I [5] a flecha a meio vão de uma viga simplesmente apoiada provocada por uma carga pontual aplicada também a meio vão é dada por: (3.67 346.32E-03 9.16) Sendo k um coeficiente que depende das condições de apoio da viga.67 346.8 9.9 9.2 – Flechas devido às cargas distribuídas nos três modelos 4 p [kN/m] Modelo 1 Modelo 2 Modelo 3 62.92E-02 M1 [kN.17) Tabela 3.1 62.92E-02 2.84 5.3 – Flechas devido à carga pontual nos três modelos 4 k Modelo 1 Modelo 2 Modelo 3 2.92E-02 2.07 4.07 Aplicando a sobreposição de efeitos obtém-se a flecha total tal indicada na tabela 3.m] 0 -681.67 L [m] 17 10.04 5.67 10.67 I faixa [m ] 2.4.m] 0 0 0 M2 [kN. Portanto no presente caso a flecha devida à carga pontual é a indicada na tabela 3.92E-02 δpontual [mm] 36. Tabela 3.1 62.08E-02 9.8 4.3. (3.67 10.92E-02 2.79 -717.
18) Observa-se que o somatório das reações é praticamente igual ao somatório das cargas aplicadas com uma percentagem de diferença insignificativa: (3.28 7159. seria necessário. na conceção. evitando a 24 .00 25. os critérios de utilização.90 494. ter em consideração a regulamentação.7 – Reações globais verticais da estrutura devido a elementos de barra Elementos barra Vigas Pilares Volume [m3] 31.85 13.94 24. provocadas pelas pessoas na discoteca.58 6679.81 CPQ [kN/m3] 25.6 – Reações globais verticais da estrutura devido a elementos de área 2 2 Elementos área Laje piso Laje cobertura Muro Área [m ] 625. Verificou-se apenas que no cômputo geral os modos de vibração eram os expectáveis. embora também com alguma translação segundo y.80 424.16 1418.57 620. o segundo modo na direção y e o terceiro modo é essencialmente de rotação. Embora não se encontre no âmbito da presente dissertação.00 Total Força vertical [kN] 798.73 (3.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 3.19) Modos de vibração e participação de massas Como já foi mencionado anteriormente não será estudada a combinação sísmica portanto a verificação dos modos de vibração não tem especial relevância. O primeiro modo tem uma participação de massa maioritariamente na direção x. por forma a controlar o nível das vibrações no piso 1.42 20879.25 16. em particular.50 Total Força vertical [kN] 7040.28 Tabela 3.77 CPQ [kN/m ] 11.
era de aproximadamente 10 mm. Tendo em conta o efeito de fluência a flecha a longo prazo seria de aproximadamente 35 mm o que poderia ser aceitável de acordo com o critério de L/250 que corresponderia a uma flecha máxima de aproximadamente 40 mm.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico coincidência entre as frequências das vibrações com componentes verticais e as dos ritmos de dança.21) Seguidamente determinam-se os coeficientes a aplicar à área e à inércia através dos seguintes quocientes: (3.22) (3. sendo que a deformação depende essencialmente destas duas características da laje. a flecha a longo prazo seria da ordem de 60 mm. para a combinação quase permanente. No entanto. dificilmente seria eficiente e competitiva economicamente 25 . Ao aumentar a espessura da laje aligeirada poder-se-ia atingir um valor de deformação aceitável.23) Por fim verificou-se que a flecha elástica mais desfavorável. 3. Considerou-se no SAP2000 uma laje com 55 cm e calculou-se a sua área e inércia por metro: (3. para consideração da fluência e fendilhação. dado que a laje não é pré-esforçada a deformação será superior pois. no entanto.20) (3. Utilizando um coeficiente de amplificação global da flecha elástica de 6. para além da fluência há que ter em consideração a perda de rigidez por fendilhação. o que já não seria aceitável.4 DETERMINAÇÃO DAS DEFORMAÇÕES RECORRENDO AO SAP2000 Na modelação da estrutura foi necessário modificar a secção de laje maciça para laje aligeirada através da aplicação de um coeficiente de redução para a sua área e outro para a sua inércia.
.
2 – Localização das bandas Y e da banda X A adição de uma banda em X. Estas fariam sentido mesmo na ausência de cargas do piso superior a descarregar na laje.1.1. como apresentado na figura 4. Y2 e Y3 Considerando bandas só na direção Y. Configuração B Figura 4. como indicado na figura 4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Configuração A Figura 4. é certamente bastante benéfica em termos de eficiência já que a ausência do pilar B1-A8 provoca um vão livre de 11 metros entre 28 .1 – Localização das bandas Y1.2. Estas bandas de préesforço podem ser requeridas em alinhamentos onde os pilares distem mais de 8 a 9 metros de acordo com a tabela 3.
3 DEFORMADA E DESLOCAMENTOS DAS CARGAS QUASE PERMANENTES Antes da aplicação do pré-esforço é importante determinar os deslocamentos nos vários pontos de controlo para as geometrias definidas. Consideram-se três hipóteses distintas para a combinação quase permanente antes da aplicação dos cabos de pré-esforço: Laje maciça de 25 cm sem qualquer banda. Y3 e X.7 a deformada da configuração B.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4. Configuração B: Laje maciça com 25 cm com espessamento para 65 cm nas bandas Y1.6 – Planta com pontos de controlo assinalados a vermelho 4.Y2. 32 . Y2 e Y3. Configuração A: Laje maciça de 25 cm com espessamento para 65 cm nas bandas Y1. Visualmente a forma da deformada da combinação das cargas quase permanentes é semelhante para as três hipóteses apresentando-se na figura 4. Desta forma foi possível analisar a importância das bandas e a eficiência dos cabos de pré-esforço separadamente.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4.7 – Deformada provocada pela combinação das cargas quase permanentes na configuração B sem cabos de pré-esforço 33 .
1 Como seria de esperar a laje de 25 cm sem bandas é muito pouco rígida pelo que apresenta deslocamentos elásticos excessivos.2 941 14.3 705 30.7 Tabela 4. que são reduzidos entre 10% a 20%. sem qualquer banda Ponto δCQP [mm] 27 19.5 705 14. e consequentemente.6 21 12.1 – Laje maciça de 25 cm.9 22 15. Com a configuração A têm-se 3 bandas em Y conferindo uma rigidez bastante significativa que reduz em aproximadamente 50% os deslocamentos dos pontos de controlo para a mesma laje sem bandas.1 a 4. A configuração B apresenta deslocamentos inferiores face à configuração A como seria expectável já que o espessamento da laje na banda X confere maior rigidez ao alinhamento B1.3 21 12.3) apresentam-se os deslocamentos nos pontos de controlo para cada uma das três hipóteses supracitadas: Tabela 4.5 705 15.7 Tabela 4.6 21 29.5 941 8.4 941 6. 4. para se ter à partida uma avaliação dos critérios de pré-dimensionamento.4 TRAÇADO E ESTIMATIVA DE PRÉ-ESFORÇO A APLICAR NAS BANDAS Y1 E Y2 Para analisar as deformações da laje no modelo analítico é importante partir de um número de cabos adequado. os deslocamentos dos pontos 22. Adotou-se como critério para a avaliação do número de cabos de pré-esforço que a ação do pré-esforço deveria anular aproximadamente 80% das cargas quase permanentes.2 22 13.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Nas tabelas seguintes (4.4 22 28.3 – Configuração B Ponto δCQP [mm] 27 9.2 – Configuração A Ponto δCQP [mm] 27 9. Dado que o 34 . 941 e 705. o que melhora o comportamento da banda Y3.
4.25 11.00 0.1) Considerando o caso mais desfavorável da carga pontual a atuar a meio de um vão simplesmente apoiado: (4.2) Momento fletor provocado pela carga distribuída (4.2 AVALIAÇÃO DA FLEXÃO GLOBAL E CARGA EQUIVALENTE DISTRIBUÍDA Momento fletor provocado pela carga pontual (4.75 Tabela 4.65 21.4 – Combinação das cargas quase permanentes no piso 1 e na cobertura Piso 1 PP [kN/m ] RCP [kN/m ] SC [kN/m ] H [m] CQP [kN/m ] 3 3 3 3 Cobertura 25.75 4.50 5.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.3) 36 .5 – Combinação das cargas quase permanentes nas bandas CQP bandas h [m] CQP [kN/m ] 2 0.00 0.25 16.00 1.85 25.00 10.00 2.
10) Área de pré-esforço É necessário ter em atenção que a tensão inicial só poderá representar 75% da tensão resistente dos cabos de pré-esforço: (4.12) Pré-esforço útil por cabo (4.13) 38 . de cabos e valor de pré-esforço útil final Considerando cordões de 1.4 cm e cabos de 4 cordões tem-se: 2 Aproveitando os cabos ao máximo devem ser colocados 32 cordões o que significa que se terá um valor de pré-esforço útil de: (4.11) Número de cordões.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Valor de pré-esforço útil (4.9) Valor de tensionamento Para calcular o valor de tensionamento estimou-se 10% de perdas imediatas e 15% de perdas diferidas: (4.
14 e 4. Apresenta-se seguidamente o traçado da banda X nas figuras 4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4.11 – Traçado poligonal da banda Y3 para a solução A2 Figura 4.13 – Traçado poligonal da banda Y3 para a solução B2 O traçado da banda Y3 para a configuração B é feito pressupondo a existência da banda X que irá contrariar as cargas equivalentes negativas da banda Y3 no cruzamento dos alinhamentos B1 e A8. Figura 4.12 – Traçado parabólico da banda Y3 para a solução B1 Figura 4.15.14 – Traçado parabólico da banda X para a solução B1 40 .
Banda Y3 – Configuração A Tabela 4.6 – Características da banda Faixas de PE [m] Largura de influência das faixas [m] e1 [mm] e2 [mm] e3 [mm] f [mm] 1.15 – Traçado poligonal da banda X para a solução B2 Relativamente à banda Y3 é de esperar que conduza a um número de cabos muito superior nas soluções A1 e A2 já que devido à ausência da banda X o valor da excentricidade média diminui e o comprimento do vão de pré-esforço aumenta consideravelmente face aos valores das soluções B1 e B2.6 a 4.50 41 .00 260.00 260.5 0.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4.8 para determinação do pré-esforço na banda Y3 na configuração A.52 0.60 5. Apresentam-se seguidamente as tabelas 4.
21 597.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.15 145.62 42 .m] q* equivalente de PE [kN/m] 356.10 0.7 – Carga equivalente de pré-esforço unidades CQP Pontual [kN] CQP Distribuída [kN/m] 0.4 cm ] Nº Cabos [de 4 cordões] Área de PE ' [cm ] P'0 [kN] P'∞ [kN] P∞ por cabo [kN] 2 2 2 13428.m] Momento CQP Distribuída [kN.26 Tabela 4.m] % M+ a equilibrar M+ a equilibrar [kN.83 213.97 0.00 23.64 48.84 90.60 13745.94 1752.94 80 2067.m] Momento CQP Pontual + CQP Distribuída [kN.P0' [kN] Área de PE [cm ] Nº Cordões [de 1.21 125.00 128.00 2583.50 831.07 80.80 17967.15 17554.P∞ [kN] Perdas imediatas Perdas diferidas Força de tensionamento .6 x CQP Pontual [kN] 0.8 – Valor do pré-esforço útil Força de PE útil .6 x CQP Distribuída [kN/m] Momento CQP Pontual [kN.
11 para determinação do pré-esforço na banda X.17) Dado que se pretende anular 80% das cargas quase permanentes considera-se uma carga equivalente de pré-esforço de: (4.10 e 4.18) Apresentam-se seguidamente as tabelas 4.60 5.50 260.50 521.52 260.10 – Características da banda X Faixa de PE [m] Largura de influência da faixa [m] e1 [mm] e2 [mm] e3 [mm] f [mm] 1.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico (4. Banda X Tabela 4.00 44 .50 260.16) Considerou-se ainda o espessamento da banda X: (4.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.09 597.91 19.10 0.00 3906.P∞ [kN] Perdas imediatas Perdas diferidas Força de tensionamento .13 – Número de cabos a aplicar em cada banda e configuração Configuração A Bandas Número de cabos Y1 8 Y2 8 Y3 23 Total 39 Y1 8 Configuração B Y2 8 Y3 8 X 5 Total 29 45 .00 28.62 Tabela 4.64 2764. Comparação do número de cabos utilizados em cada configuração Tabela 4.87 A tensão média indicada na tabela 4.15 3614.11 – Valor de pré-esforço útil q* [kN/m] Força de PE útil .12 – Tensão média Tensão média [MPa] 2.00 5.97 0.P0' [kN] Área de PE [cm ] Nº Cordões [de 1.00 2988.34 25.4 cm ] Nº Cabos [de 4 cordões] Área de PE ' [cm ] P'0 [kN] P'∞ [kN] P∞ por cabo [kN] 2 2 2 94.12 pode considerar-se relativamente baixa mas é ainda um valor perfeitamente satisfatório.
29 292.45 -292.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4.27 62.17 – Resumo das cargas equivalentes nas bandas Y1 e Y2 – Soluções A1 e B1 Tabela 4.18 f [m] 0.78 0.74 4.53 0.32 1.09 Q [kN] 86.27 Figura 4.26 tg β 0.52 0.26 P*tg(α) [kN] -86.15 – Cargas equivalentes nas bandas Y1 e Y2: Soluções A1 e B1 Bandas Tramos Reta 1 Reta 2 Reta 3 L [m] 3.18 – Resumo das cargas equivalentes nas bandas Y1 e Y2: Soluções A2 e B2 47 .29 0.45 1000 Y1 e Y2 Reta 4 Reta 5 Reta 6 Reta 7 0.26 0.02 2.06 -62.52 0.26 P∞ [kN] 0.02 3.
33 Q [kN] 86.09 Q [kN] 86.26 P∞ [kN] Y3 Reta 3 Reta 4 5.16 – Cargas equivalentes na banda Y3: Solução parabólica A1 Bandas Tramos Parábola 1 L [m] 6.52 aparabola 0.05 P*tg(α) [kN] -86.02 3.17 – Cargas equivalentes na banda Y3: Solução poligonal A2 Bandas Tramos Reta 1 Reta 2 L [m] 3.28 4.52 1000 Figura 4.26 q [kN/m] 14.04 10.48 5.26 -47.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.52 -47.02 f [m] 0.48 0.26 P∞ [kN] Y3 Parábola 2 47.09 0.19 – Resumo das cargas equivalentes na banda Y3: Solução parabólica A1 Tabela 4.26 0.26 P*tg(α) [kN] -86.26 0.01 0 tg(α) 0.05 47.20 – Resumo das cargas equivalentes na banda Y3: Solução poligonal A2 48 .52 1000 Figura 4.26 tg β 0.96 f [m] 0.
02 3.78 0.26 0.09 P*tg(α) [kN] -86.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.52 0.38 49 .39 0.21 – Resumo das cargas equivalentes na banda Y3: Solução parabólica B1 Tabela 4.29 292.45 1000 Y3 Reta 4 Reta 5 Reta 6 Reta 7 0.09 Q [kN] 86.25 1000 Figura 4.33 f [m] 0.29 0.26 225.08 82.26 P∞ [kN] Y3 Parábola 3 Parábola 4 -225.01 0.26 q [kN/m] 14.25 0.38 f [m] 0.28 65.02 2.18 – Cargas equivalentes na banda Y3: Solução parabólica B1 Bandas Tramos Parábola 1 Parábola 2 L [m] 6.38 48.43 1.14 0.54 12.05 -82.05 -48.61 187.05 aparabola 0.32 1.01 tg(α) 0.52 0.95 5.53 0.99 Q [kN] 86.19 – Cargas equivalentes na banda Y3: Solução poligonal B2 Bandas Tramos Reta 1 Reta 2 Reta 3 L [m] 3.45 -292.26 P*tg(α) [kN] -86.04 3.26 tg β 0.2 6.26 P∞ [kN] 0.26 0.
85 188.8 0.22 q [kN/m] 20.5 -188.04 0.45 0.85 -97.22 – Resumo das cargas equivalentes na banda Y3: Solução poligonal B2 Tabela 4.72 0.03 40.23 – Resumo das cargas equivalentes na banda X: Solução parabólica B1 50 .8 4.22 0.56 0 0 0 1000 Figura 4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Figura 4.08 0.71 0.85 -188.1 40.49 Q [kN] 94.56 97.1 -122 21.1 -236.72 4.54 f [m] 0.08 0.04 0.8 0.03 -236.20 – Cargas equivalentes na banda X: Solução parabólica B1 Bandas Tramos Parábola 1 Parábola 2 Parábola 3 Parábola 4 L [m] 4.05 -118.45 0.85 aParábola 0 tg(α) 0 P*tg(α) [kN] 0 P∞ [kN] X Parábola 5 Parábola 6 Parábola 7 Parábola 8 188.8 4.5 -94.
15 0.38 3.24 – Resumo das cargas equivalentes na banda X: Solução poligonal B2 O critério para a seleção da solução mais eficiente será aquela que se apresentar mais económica contrariando de forma satisfatória as deformações das cargas quase-permanentes.59 P∞ [kN] 0.39 Figura 4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.55 0.8 4.39 57.15 -145.0 Advanced é possível avaliar as deformações na laje para cada ação separadamente e para qualquer combinação que se 51 .27 145.59 0.15 0.26 tg β 0.54 f [m] 0.06 Q [kN] -55.59 1.27 X Reta 6 Reta 7 Reta 8 Reta 9 Reta 10 0.27 145.59 P*tg(α) [kN] 55.26 0.69 0.52 0.15 0.06 -57.52 0.52 0.83 0.55 3.52 0.38 1.27 1000 -145.21 – Cargas equivalentes na banda X: Solução poligonal B2 Bandas Tramos Reta 1 Reta 2 Reta 3 Reta 4 Reta 5 L [m] 4. 4.1.7 DETERMINAÇÃO DAS DEFORMAÇÕES RECORRENDO AO SAP2000 Recorrendo ao programa de elementos finitos SAP2000 v14.
732 0.021 0.376 1.509 0.109 0.026 0.366 1.213 0.023 21 0.24 – Matriz de influência da solução A2 Solução A2 [mm] Banda PE-Y1 PE-Y2 PE-Y3 941 0.355 0.777 705 0.352 705 0.689 Tabela 4.116 0.00308 21 0.115 0.111 0.07 0.222 0.095 0.23 – Matriz de influência da solução A1 Solução A1 [mm] Banda PE-Y1 PE-Y2 PE-Y3 941 0.019 0.525 0.222 22 0.659 Tabela 4.019 0.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.25 – Matriz de influência da solução B1 Solução B1 [mm] Banda PE-Y1 PE-Y2 PE-Y3 PE-X 941 0.047 0.015 0.126 22 0.353 1.393 705 0.37 0.064 0.622 27 1.023 21 0.608 54 .349 0.203 0.516 27 1.639 1.065 0.089 0.055 0.218 22 0.433 1.103 0.231 0.638 27 1.751 0.379 0.137 -0.
– representa o deslocamento do ponto j. – representa o deslocamento do ponto j. Como última restrição temos o número de cabos por banda que se limitou a 8 devido ao espaçamento mínimo necessário.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico provenientes do pré-esforço convertendo assim o valor de pré-esforço de 1000 kN introduzido no SAP2000 para os 597. Analogamente tem-se para a configuração B: (4. para um valor de préesforço de 597.6 kN aplicado na banda i.21) Sujeito a: 56 . para a combinação quase permanente. Pode-se então descrever a situação através de um problema de programação linear. – representa o comprimento de um cabo na banda i. Para a configuração A: (4. em milímetros.20) Sujeito a: Onde. – representa o número de cabos a aplicar na banda i. em milímetros.6 kN de cada cabo.
Tabela 4.2 SOLUÇÃO DE PRÉ-ESFORÇO ADOTADA Através da resolução do problema de programação linear determinam-se as melhores soluções para os quatro casos como indicado na tabela 4. para um valor de préesforço de 597. em milímetros. [mm] 32. para a combinação quase permanente. por etapas.987 33. – representa o deslocamento do ponto j.7. para compreender as deformações à medida que se aumenta o número de cabos a aplicar. – representa o deslocamento do ponto j.27. 4. – representa o comprimento de um cabo na banda i. em milímetros. Obviamente que a solução poderia ser encontrada de forma imediata através da aplicação de um algoritmo como o SIMPLEX.205 14.953 57 . – representa o número de cabos a aplicar na banda i.27 – Melhores configurações de cabos para cada solução Solução A1 A2 B1 B2 Cabos Y1 8 8 5 5 Cabos Y2 8 8 8 8 Cabos Y3 8 8 8 8 Cabos X 5 5 Total de cabos 24 24 26 26 δ máx.914 14.6 kN aplicado na banda i.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Onde. no entanto. como os critérios de limitação da deformação não são totalmente inflexíveis considera-se mais interessante efetuar uma análise manual.
as soluções do tipo 1 e 2 diferem muito pouco nos resultados dos pontos de controlo.29.28 Como se pode observar nas tabelas 4. No entanto. aplicando um coeficiente às ações do pré-esforço nas várias bandas que simulam os cabos de pré-esforço. como também seria de esperar.64 0.77 0.29 – Flechas máximas em todo o piso retiradas do SAP para a melhor solução B2 Flecha máxima a longo prazo [mm] Flecha para cima máxima a longo prazo [mm] -15. Relativamente à variação dos tipos de traçado. Embora ambas as soluções sejam boas optou-se pela solução parabólica por se julgar que contraria o carregamento de uma forma mais uniforme. Importa ainda referir que as matrizes de influência juntamente com a análise incremental manual contribuem significativamente para a minimização dos erros e confirmação/validação do modelo. modificar a exigência da deformação global ou aceitar valores diferentes em zonas específicas. 4. Determinaram-se ainda.28 Tabela 4. 58 .28 – Flechas máximas em todo o piso retiradas do SAP para a melhor solução B1 Flecha máxima a longo prazo [mm] Flecha para cima máxima a longo prazo [mm] -15.28 e 4. como seria de esperar através da avaliação anterior.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Como se pode observar as soluções A não conseguem cumprir os critérios de deformação mesmo com 8 cabos em cada banda.3 ANÁLISE DAS MATRIZES DE INFLUÊNCIA E AS SUAS FUNCIONALIDADES Como já foi mencionado anteriormente uma das principais funções das matrizes de influência é a de permitir determinar matematicamente a melhor solução face a restrições previamente definidas. Por exemplo. Tabela 4. a diferença entre as soluções é desprezável e confirma-se também que as flechas máximas não apresentam uma diferença muito significativa face aos pontos de controlo. as flechas máximas para as melhores soluções B1 e B2. sendo também especialmente importantes no caso de se pretenderem alterar algumas restrições.7. nomeadamente no estudo da importância relativa entre as várias bandas de pré-esforço. as matrizes de influência permitem ainda o estudo das deformações em outros importantes aspetos.
31 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução B1 Flecha Máxima #/% de cabos banda y1 banda y2 banda y3 banda x Total # 0 4 7 0 11 30 % 0.0 100 # 5 9 9 4 27 15 % 18.2 100 # 4 13 18 35 15 % 11.0 26.9 55. a evolução do número de cabos de forma a limitar a deformação máxima das cargas quase permanentes.33) indicam-se.4 37.3 73. que o pré-esforço não está a ser utilizado de uma forma racional.7 100 # 1 8 15 24 25 % 4.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Melhor solução sem restrição de número de cabos Sem a restrição do número máximo de cabos pretende-se determinar qual a importância relativa de cada banda pré-esforçada tendo em consideração que a inversão dos sinais dos deslocamentos devido às cargas quase permanentes se mantém pois isso significa.3 14.0 36.0 100 Tabela 4.3 62.5 27. para valores de 30 a 10 mm.4 27.3 33.5 31.30 a 4. Tabela 4. para cada solução.1 51.3 0.6 0.4 63.5 55. em geral.0 100 # 3 7 9 1 20 20 % 15.5 33. incluindo o pré-esforço.0 35.5 100 # 2 11 16 29 20 % 6.3 56.0 45.9 37.8 100 # 6 12 9 6 33 10 % 18.3 18.2 100 59 . Nas tabelas seguintes (tabelas 4.30 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução A1 Flecha máxima #/% de cabos Nº de cabos y1 Nº de cabos y2 Nº de cabos y3 Total # 0 5 14 19 30 % 0.0 100 # 2 5 9 0 16 25 % 12.0 5.2 36.2 33.4 100 # 7 11 22 40 10 % 17.
2 18.6 0.1 51.3 0.3 73.9 25.33 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução B2 Flecha Máxima #/% de cabos banda y1 banda y2 banda y3 banda x Total # 0 4 7 0 11 30 % 0.2 37.7 0. o que justifica a sua menor importância.4 51. e se encontrem na zona das maiores deformações devido às cargas. A banda X tem como principal função contrariar a deformação provocada pela banda Y3 no 60 .3 100. esta encontrase bastante afastada da zona mais desfavorável.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico Tabela 4.9 11.7 46.9 41.36 a 4.9 33.0 36.0 # 2 12 15 29 20 % 6.4 100.7 100.0 100 # 1 13 7 0 21 20 % 4. naturalmente.0 100 # 3 14 7 3 27 15 % 11.7 100. No entanto.2 100 Da análise das tabelas 4.4 63. a análise da deformação desta forma deixa de ser válida a partir de um grau de exigência elevado dado que os pontos de controlo podem deixar de ser condicionantes.0 100 # 1 7 7 0 15 25 % 6. A importância relativa das bandas é natural dado que a banda Y2 é a banda central na direção y e a banda Y3 é a banda que passa pelo cruzamento do alinhamento B1 e A8.0 Tabela 4.39.0 26.8 61.7 46.0 # 4 13 18 35 15 % 11.1 51.2 39. Importa referir que.0 # 1 9 14 24 25 % 4. pode concluir-se que as bandas Y2 e Y3 são fundamentais no controlo da deformação dado que têm sempre uma maior percentagem de número de cabos face ao total.5 58. verifica-se que a importância das bandas Y1 e X aumenta de acordo com a exigência imposta na deformação.4 37.1 100 # 6 13 8 6 33 10 % 18. Relativamente à banda Y1.4 24.32 – Número de cabos em cada banda necessário para limitar a flecha máxima para um certo valor na solução A2 Flecha máxima #/% de cabos Nº de cabos y1 Nº de cabos y2 Nº de cabos y3 Total # 0 5 14 19 30 % 0.
Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico 62 .
1 – Características da secção Largura da banda [m] Altura da banda [m] Área [cm ] c [m] d [m] Distância entre As e Ap [m] e [m] 2 1.65 44.1.Análise e dimensionamento de um piso com pré-esforço inserido no projeto de um edifício Instituto Superior Técnico média dos valores por metro e multiplicar pela largura da banda para obter o valor de Msd e Nsd total na banda.5) para um (5.03 0.6 0.6) (5. Tabela 5.2605 No presente caso tem-se então: (5. As características da secção apresentam-se na tabela 5. 64 .608 0.8 0.5 m foi ignorada por não ter sentido físico.7) (5.0225 0.10) A solução de 1.8) (5. 1) Equilíbrio de momentos (5.9) Através das equações de equilíbrio é possível determinar a posição da linha neutra e da armadura ordinária necessária para a verificação de segurança.
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1. ## accuracy
i recently wrote a money-involved program and am just wondering:
[list=1][*]is it normal for a float to be inaccurate to the hundredth/thousandth?[*]what is the accuarcy (approx, of course) of a float[*]what is the accuracy of a double[*]any good algorithms for rounding to a certain number of decimal places (like setprecision, except with rounding)[/list=1]
i'm not sure if all this has to do with the OS/System, but yeah...
2. I haven't read through the entire thread, but there might be something helpful here
3. I read most of that post, and it seems to have failed to mention one very critical point about why the inacuraccies occur. You're dealing with base 10 decimal numbers. The computer is dealing with base 2 decimal numbers. Given a small mantissa, it is difficult to represent one as the other with perfect accuracy. So, to answer (1), yes.
For (2) and (3), the answer is in C++PL, I believe (under numeric limits), but I have no idea where my copy of it is right now.
For (4), take the number of digits you want, and then look at the one following it (by dividing by the appropriate power of 10). This'll be better for printing than storing, as if the machine could not accurately represent the number once, it probably won't be able to the second time.
Hope this is at least somewhat useful.
Cheers
4. Cheers
Just wondering, are you from Britain? Everyone I know from there says that
5. Originally posted by Zach L.
For (2) and (3), the answer is in C++PL, I believe (under numeric limits), but I have no idea where my copy of it is right now.
what is that and where can i get it?
I also didn't think about the binary thing... if forgot about that...
6. Actually, I'm from the small, third-world country of New Mexico.
Sorry 'bout the acronym. C++PL is Stroustrup's The C++ Programming Language : Special Edition. It really is a great reference. Anyways, it talks about the numeric_limits templated class in <limits>. Essentially, its a specialized type which has a lot of machine-independent information (consequently, the exact values will vary). However, the examples he uses are of fairly common representations for floats/doubles/etc.
For example (I dug it off the book shelf), for a float, it gives a base 2 mantissa of 24 bits (approx. 6 decimal digits), and an epsilon of 1.19...E-7 (that is, the smallest epsion such that 1.0 + epsilon != 1.0 within machine limits).
7. thanks... i'll look into getting a copy... now i just need money
8. Lots of Australians also say cheers.
I just recently bought a copy of Stroustrups C++PL, also effective C++ and Programming Windows, talk about expensive!
C++PL is difficult to read after reading Shidlt, Shidlt is a good writer, even if he's not a good coder.
I'm sure it's a good book but wow, he goes into some advanced \$\$\$\$ in the first 4 chapters, if I was newbie at C++ I would burn the thing. First thing he talks about is user defined types, then he goes onto the basics of STL.
9. It is an excellent reference. For the faint of heart, or those unfamiliar with the language already though, it can be rough. It is good for learning advanced stuff though.
10. I use it as a reference also. But I think it starts on chapter 4 (or maybe 5??) with the real learning stuff. Chapters 1-3 are just introduction, which may be why its even harder to read.
11. Originally posted by Zach L.
It is an excellent reference. For the faint of heart, or those unfamiliar with the language already though, it can be rough. It is good for learning advanced stuff though.
what would you call advanced?
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### Sample Problem
In the figure below, what percentage of the whole is unshaded?
%
#### Solution
There are 100 squares. 66 are shaded. Therefore, 100% – 66% = 34% of the whole is unshaded.
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https://cracku.in/64-in-a-certain-code-if-on-the-day-is-coded-as-la-si--x-sbi-clerk-2016-5
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Question 64
# In a certain code, if ‘on the day’ is coded as ‘la si ko’ and the setting day’ is coded as si mu la’, what is the code for ‘setting’ in the given code language?(All the codes are two letter codes only)
Solution
The common words in both the statements are 'the' and 'day' coded as = 'la' or 'si'
Thus, the remaining word in second statement is 'setting' coded as = 'mu'
=> Ans - (A)
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https://discuss.codechef.com/t/simple-program/10281
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# Simple Program.
Guys this is a simple question which asks to print all the perfect squares which only has 0,1,4,9 as their digits.I made a program but it is not Working.Pls Help!!
#include <stdio.h>
#include <conio.h>
int main(){
``````int i,j,k,a,b;
for(i=0;i<=10;i++)
{
int flag=1; //Set flag=1
int dig;
int x=i*i; //Extracting Digits
while(x!=0 && flag!=0)
{
dig=x%10;
if(dig!=0 || dig!=1|| dig!=4 || dig!=9){flag=0;} //Checking the condition,If it is not
x=x/10; satisfied make flag=0 and exit the
while loop.
}
if(flag==1){printf("%d\n",i*i);}
}
``````
getch();
return 0;
}
First of all, I’ll explain what you were doing wrong.
``````if(dig!=0 || dig!=1|| dig!=4 || dig!=9){flag=0;}
``````
Let me explain this condition . The condition is satisfied whenever `dig!=0 || dig!=1|| dig!=4 || dig!=9`. Now, think like this. Suppose the value of `dig` is 1, doesn’t that mean automatically mean that `dig` is not equal to the other values, in other words, the all the other three conditions, namely `dig!=0 || dig!=4 || dig!=9` is satisfied.
The condition in your loop is satisfied whenever the value of `dig` is not equal to 0, 1, 4 or 9, but as the condition is now, if `dig` gets one value, that automatically means it is not the other value, so that is what’s basically your problem.
This can be fixed by doing
``````if( dig==0 || dig==1|| dig==4 || dig==9)
{}
else
{
flag=0;
break;
}
``````
You also have many variables which are unused.
Also, as indicated in my comment, don’t use `<conio.h>` as it is not part of the standard.
I’ve edited your code , Here it is
``````#include <stdio.h>
int main()
{
int i;
for(i=0;i<=10;i++)
{
int flag=1;
int dig;
int x=i*i;
while(x!=0 && flag!=0)
{
dig=x%10;
if( dig==0 || dig==1|| dig==4 || dig==9)
{}
else
{
flag=0;
break;
}
x=x/10;
}
if(flag==1)
{
printf("%d\n",i*i);
}
}
return 0;
}
``````
### OUTPUT
0
1
4
9
49
100
u can use the if conditions in two ways
first
if(dig!=0 && dig!=1 && dig!=4 && dig!=9){flag=0;}
second is the way mentioned in the above comment
Some advice for you, Don’t use `<conio.h>` . It’s not supported on the newer compilers.
BTW, you forgot to tag your programming language, is it c or c++.
Also, define "Not Working"
Yup, this is actually better. I wonder why I didn’t think of this in the first place.
Thnx buddy!!
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## ACA 2017 - Differential Algebra for an extended...
by:
I'm back from presenting work in the "23rd Conference on Applications of Computer Algebra - 2017" . It was a very interesting event. This second presentation, about "Differential algebra with mathematical functions, symbolic powers and anticommutative variables", describes a project I started working in 1997 and that is at the root of Maple's dsolve and pdsolve performance with systems of equations. It is a unique approach. Not yet emulated in any other computer algebra system.
At the end, there is a link to the presentation worksheet, with which one could open the sections and reproduce the presentation examples.
Differential algebra with mathematical functions,
symbolic powers and anticommutative variables
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Abstract:
Computer algebra implementations of Differential Algebra typically require that the systems of equations to be tackled be rational in the independent and dependent variables and their partial derivatives, and of course that , everything is commutative.
It is possible, however, to extend this computational domain and apply Differential Algebra techniques to systems of equations that involve arbitrary compositions of mathematical functions (elementary or special), fractional and symbolic powers, as well as anticommutative variables and functions. This is the subject of this presentation, with examples of the implementation of these ideas in the Maple computer algebra system and its ODE and PDE solvers.
>
>
>
(1)
>
(2)
>
(3)
>
Differential polynomial forms for mathematical functions (basic)
Differential polynomial forms for compositions of mathematical functions
Generalization to many variables
Arbitrary functions of algebraic expressions
Examples of the use of this extension to include mathematical functions
Differential Algebra with anticommutative variables
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
## ACA 2017 - Computer Algebra in High-School Educati...
by: Maple
I'm back from presenting work in the "23rd Conference on Applications of Computer Algebra - 2017" . It was a very interesting event. This first presentation, about "Active Learning in High-School Mathematics using Interactive Interfaces", describes a project I started working 23 years ago, which I believe will be part of the future in one or another form. This is work actually not related to my work at Maplesoft.
At the end, there is a link to the presentation worksheet, with which one could open the sections and reproduce the presentation examples.
Active learning in High-School mathematics using Interactive Interfaces
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
Abstract:
The key idea in this project is to learn through exploration using a web of user-friendly Highly Interactive Graphical Interfaces (HIGI). The HIGIs, structured as trees of interlinked windows, present concepts using a minimal amount of text while maximizing the possibility of visual and analytic exploration. These interfaces run computer algebra software in the background. Assessment tools are integrated into the learning experience within the general conceptual map, the Navigator. This Navigator offers students self-assessment tools and full access to the logical sequencing of course concepts, helping them to identify any gaps in their knowledge and to launch the corresponding learning interfaces. An interactive online set of HIGIS of this kind can be used at school, at home, in distance education, and both individually and in a group.
Computer algebra interfaces for High-School students of "Colegio de Aplicação" (UERJ/1994)
Motivation
When we are the average high-school student facing mathematics, we tend to feel
• Bored, fragmentarily taking notes, listening to a teacher for 50 or more minutes
• Anguished because we do not understand some math topics (too many gaps accumulated)
• Powerless because we don't know what to do to understand (don't have any instant-tutor to ask questions and without being judged for having accumulated gaps)
• Stressed by the upcoming exams where the lack of understanding may become evident
Computer algebra environments can help in addressing these issues.
• Be as active as it can get while learning at our own pace.
• Explore at high speed and without feeling judged. There is space for curiosity with no computational cost.
• Possibility for making of learning a social experience.
Interactive interfaces
Interactive interfaces do not replace the teacher - human learning is an emotional process. A good teacher leading good active learning is a positive experience a student will never forget
Not every computer interface is a valuable resource, at all. It is the set of pedagogical ideas implemented that makes an interface valuable (the same happens with textbooks)
A course on high school mathematics using interactive interfaces - the Edukanet project
– Brazilian and Canadian students/programmers were invited to participate - 7 people worked in the project.
– Some funding provided by the Brazilian Research agency CNPq.
-Develop a framework to develop the interfaces covering the last 3 years of high school mathematics (following the main math textbook used in public schools in Brazil)
- Design documents for the interfaces according to given pedagogical guidelines.
- Create prototypes of Interactive interfaces, running Maple on background, according to design document and specified layout (allow for everybody's input/changes).
The pedagogical guidelines for interactive interfaces
The Math-contents design documents for each chapter
Example: complex numbers
Each math topic: a interactive interrelated interfaces (windows)
For each topic of high-school mathematics (chapter of a textbook), develop a tree of interactive interfaces (applets) related to the topic (main) and subtopics
Example: Functions
• Main window
• Analysis window
•
• Parity window
• Visualization of function's parity
• Step-by-Step solution window
The Navigator: a window with a tile per math topic
• Click the topic-tile to launch a smaller window, topic-specific, map of interrelated sub-topic tiles, that indicates the logical sequence for the sub-topics, and from where one could launch the corresponding sub-topic interactive interface.
• This topic-specific smaller window allows for identifying the pre-requisites and gaps in understanding, launching the corresponding interfaces to fill the gaps, and tracking the level of familiarity with a topic.
The framework to create the interfaces: a version of NetBeans on steroids ...
Complementary classroom activity on a computer algebra worksheet
This course is organized as a guided experience, 2 hours per day during five days, on learning the basics of the Maple language, and on using it to formulate algebraic computations we do with paper and pencil in high school and 1st year of undergraduate science courses.
Explore. Having success doesn't matter, using your curiosity as a compass does - things can be done in so many different ways. Have full permission to fail. Share your insights. All questions are valid even if to the side. Computer algebra can transform the learning of mathematics into interesting understanding, success and fun.
1. Arithmetic operations and elementary functions
2. Algebraic Expressions, Equations and Functions
3. Limits, Derivatives, Sums, Products, Integrals, Differential Equations
4. Algebraic manipulation: simplify, factorize, expand
5. Matrices (Linear Algebra)
Advanced students: guiding them to program mathematical concepts on a computer algebra worksheet
Status of the project
Prototypes of interfaces built cover:
• Natural numbers
• Functions
• Integer numbers
• Rational numbers
• Absolute value
• Logarithms
• Numerical sequences
• Trigonometry
• Matrices
• Determinants
• Linear systems
• Limits
• Derivatives
• Derivative of the inverse function
• The point in Cartesian coordinates
• The line
• The circle
• The ellipse
• The parabole
• The hyperbole
• The conics
More recent computer algebra frameworks: Maple Mobius for online courses and automated evaluation
Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft
## Maple 2017.2 update
by:
We have just released an update to Maple. Maple 2017.2 includes updated translations for Japanese, Traditional Chinese, Simplified Chinese, Brazilian Portuguese, French, and Spanish. It also contains improvements to the MapleCloud, physics, limits, and PDEs. This update is available through Tools>Check for Updates in Maple, and is also available from our website on the Maple 2017.2 download page.
Eithne
## A geometric construction for the Summer Holiday
by: Maple
A geometric construction for the Summer Holiday
Does every plane simple closed curve contain all four vertices of some square?
This is an old classical conjecture. See:
https://en.wikipedia.org/wiki/Inscribed_square_problem
Maybe someone finds a counterexample (for non-analytic curves) using the next procedure and becomes famous!
> SQ:=proc(X::procedure, Y::procedure, rng::range(realcons), r:=0.49) local t1:=lhs(rng), t2:=rhs(rng), a,b,c,d,s; s:=fsolve({ X(a)+X(c) = X(b)+X(d), Y(a)+Y(c) = Y(b)+Y(d), (X(a)-X(c))^2+(Y(a)-Y(c))^2 = (X(b)-X(d))^2+(Y(b)-Y(d))^2, (X(a)-X(c))*(X(b)-X(d)) + (Y(a)-Y(c))*(Y(b)-Y(d)) = 0}, {a=t1..t1+r*(t2-t1),b=rng,c=rng,d=t2-r*(t2-t1)..t2}); #lprint(s); if type(s,set) then s:=rhs~(s)[];[s,s[1]] else WARNING("No solution found"); {} fi; end:
Example
> X := t->(10-sin(7*t)*exp(-t))*cos(t); Y := t->(10+sin(6*t))*sin(t); rng := 0..2*Pi;
(1)
> s:=SQ(X, Y, rng): plots:-display( plot([X,Y,rng], scaling=constrained), plot([seq( eval([X(t),Y(t)],t=u),u=s)], color=blue, thickness=2));
## Using Maple to outsmart google
Maple
It appears google doesn't know about the haversine formula. Huh? Well at least google can't draw the proper path for it. I typed in google "distance from Pyongyang to NewYork city" and got 10,916km. Ok that's fine but then it drew a map
The map path definitely did not look right. Pulled out my globe traced a rough path of the one google showed and I got 13 inches (where 1 inch=660miles) -> 8580 miles = 13808 km .. clearly looks like google goofed.
So we need Maple to show us the proper path.
```with(DataSets):
with(Builtin):
m := WorldMap();
Display(m):
```
Ok so you say that really doesn't look like the shortest path. Well, lets visualize that on the globe projection
`Display(m, projection = Globe, orientation = [-180, 0, 0])`
Ah, now it is clear
Pyonyang_to_NewYork.mw
## Google Maps and Geocoding for Maple
by: Maple
As a momentary diversion, I threw together a package that downloads map images into Maple using the Google Static Maps API.
If you have Maple 2017, you can install the package using the MapleCloud Package Manager or by executing PackageTools:-Install("5769608062566400").
This worksheet has several examples, but I thought I'd share a few below .
Here's the Maplesoft office
Let's view a roadmap of Waterloo, Ontario.
The package features over 80 styles for roadmaps. These are examples of two styles (the second is inspired by the art of Piet Mondrian and the De Stijl movement)
You can also find the longitude and latitude of a location (courtesy of Google's Geocoding API). Maple returns a nested list if it finds multiple locations.
The geocoding feature can also be used to add points to Maple 2017's built-in world maps.
Let me know what you think!
## Tangent plane as the square at any point of a...
by: Maple 15
The representation of the tangent plane in the form of a square with a given length of the side at any point on the surface.
The equation of the tangent plane to the surface at a given point is obtained from the condition that the tangent plane is perpendicular to the normal vector. With the aid of any auxiliary point not lying on this normal to the surface, we define the direction on the tangent plane. From the given point in this direction, we lay off segments equal to half the length of the side of our square and with the help of these segments we construct the square itself, lying on the tangent plane with the center at a given point.
An examples of constructing tangent planes at points of the same intersection line for two surfaces.
Tangent_plane.mw
## App for fluids in flat state of rest
Maple 2016
This app is used to study the behavior of water in its different properties besides air. Also included is the study of the fluids in the state of rest ie the pressure generated on a flat surface. Integral developed in Maple for the community of users in space to the civil engineers.
App_for_fluids_in_flat_state_of_rest.mw
Lenin Araujo Castillo
by: Maple
Yahoo Finance recently discontinued their (largely undocumented) historical stock quote API.
Previously, you simply send a HTTP:-Get request like this…
HTTP:-Get(“http://ichart.yahoo.com/table.csv?s=AAPL&a=00&b=1&c=2016&d=00&e=1&f=2017&g=d&ignore=.csv")
…and get historical OHLCV (open, high, low, close, trading volume) data in your worksheet (in this case for AAPL between 1 January 2016 and 1 January 2017).
This no longer works! Yahoo shut the door on this easy-to-use and widely disseminated API.
You can still download historical stock quotes from Yahoo Finance into Maple, but the process is now somewhat more involved. My complete code in this worksheet but I'll step through the process below.
If you visit the updated Yahoo Finance website and download historical data for a ticker, you see a URL like this in the status bar of your browser
Let's examine how ths URL is constructed.
• period1 and period2 are Unix time stamps for your start and end date
• interval is the data retrieval interval (this can be either 1d, 1w or 1m)
• crumb is an alphanumeric code that’s periodically regenerated every time you download new historical data from from the Yahoo Finance website using your browser. Moreover, crumb is paired with a cookie that’s stored by your browser.
Here’s how to extract and supply the cookie-crumb pair to Yahoo Finance so you can still use Maple to retrieve historical stock quotes
Send a dummy request to get a cookie-crumb pair
res:=HTTP:-Get("https://finance.yahoo.com/lookup?s=bananas"):
Grab the crumb from the response
i:=StringTools:-Search("CrumbStore\":{\"crumb\":\"",res[2]):
crumbValue := res[2][i+22..i+32]
crumbValue := "btW01FWTBn3"
Store the cookie from the response
Construct the URL
• Your desired start and end dates have to be defined as Unix time stamps. Converting a human readable date (like 1st January 2017) to a Unix timestamp is simple, so I won't cover it here.
• The previously retrieved crumb has to be added to the URL.
ticker:="AAPL":
p1 := 1497709183:
p2 := 1500301183:
Send the request to Yahoo Finance, including the cookie in the header
Your historical data is now returned
The historical data is now easily parsed into a matrix.
by: Maple
## Lattice Package Updated
by: Maple
The Lattice package to investigate particle accelerator magnet lattices (original post) has been updated to V1.1. This is a significant update, addressing a number of inaccuracies and bugs of V1.0 as well as introducing new elements: Octupole, Fringe effect in dipoles, a MatchedSection allowing to insert a piece of beamline when the details are irrelevant, and a few experimental elements like WireQuad. New functions include the 6th synchrotron-radiation integral I6x, momentum compaction alphap and TaylorMap, which allows to compute the Taylor expansion of the non-linear map to any degree.
The code and documentation are available in the Application Center.
U. Wienands, aka Mac Dude
## Finding better adjusted r-squared values by removing...
by: Maple
A couple of weeks ago, I recorded a short video that discussed various applications for the Statistics:-Fit command. One of the more interesting examples examined how manually adjusting the number of parameters used for a regression model affected the resulting adjusted r-squared value.
I won’t go into detail about r-squared here, but to briefly summarize: In a linear regression model, r-squared measures the proportion of the variation in a model's dependent variable explained by the independent variables. Basically, r-squared gives a statistical measure of how well the regression line approximates the data. R-squared values usually range from 0 to 1 and the closer it gets to 1, the better it is said that the model performs as it accounts for a greater proportion of the variance (an r-squared value of 1 means a perfect fit of the data). When more variables are added, r-squared values typically increase. They can never decrease when adding a variable; and if the fit is not 100% perfect, then adding a variable that represents random data will increase the r-squared value with probability 1. The adjusted r-squared attempts to account for this phenomenon by adjusting the r-squared value based on the number of independent variables in the model.
The formula for the adjusted r-squared is:
Where:
n is the number of points in the data sample
k is the number of independent variables in the model excluding the constant
By taking the number of independent variables into consideration, the adjusted r-squared behaves different than r-squared; adding more variables doesn’t necessarily produce better fitting models. In many cases, more variables can often lead to lower adjusted r-squared values. In particular, if you add a variable representing random data, the expected change in the adjusted r-squared is 0.
As such, the adjusted r-squared has a slightly different interpretation than the r-squared. While r-squared is perceived to give an indication of the measure of fit for a chosen regression model, the adjusted r-squared is perceived more as a comparative tool that can be useful for picking variables and designing models that may require less predictors than other models. The science of “gaming” models is a broad topic, so I won’t go into any more detail here, but there’s lots of great information out there if you are looking to learn more (here’s a good place to start).
The following example adjusts a fitted model by adding or removing variables in order to find better adjusted r-squared values.
with(Statistics):
The Import command reads a datafile into a new DataFrame.
ExperimentalData := Import(FileTools:-JoinPath(["Excel", "ExperimentalData.xls"], base = datadir));
The dataset has seven variables: time and experimental readings for 6 various concentrations. Removing “time” from our variable set, the convert command converts the values in the DataFrame to a Matrix of values.
ExMat := convert( ExperimentalData, Matrix )[..,2..7];
We start by fitting a model that includes predicting variables for each of the columns of data. We mark “Concentration A” as our dependent variable.
Fit( C + C2*v + C3*w + C4*x + C5*y + C6*z, ExMat[..,2..6], ExMat[..,1], [v,w,x,y,z], summarize=embed ):
From the above, we can observe that both the r-squared and adjusted r-squared are reasonably high, however only one of the coefficient values has a significant p-value, C3.
Note: Maple shows all p-values less than 0.05 in bold.
Let's try to fit the data again, this time keeping the two coefficients with the lowest p-values and the intercept.
Fit( C + C3*v + C5*w, ExMat[..,[3,5]], ExMat[..,1], [v,w], summarize=embed ):
From the above, we can see that the r-squared value does go down, however the adjusted r-squared goes up! Let's fit the model one last time to see if removing C5 increases or decreases the adjusted r-squared.
Fit( C + C3*v, ExMat[..,3], ExMat[..,1], [v], summarize=embed ):
We can see that the final adjusted r-squared value is lower than the previous two, so we are probably better to keep the additional C5 coefficient value.
You can see this example as well as a couple of other examples of using the Fit command in the following video:
## new edition of Mathematics Survival Kit
by: Maple
We have just released the 3rd edition of the Mathematics Survival Kit – Maple Edition.
The Math Survival Kit helps students get unstuck when they are stuck. Sometimes students are prevented from solving a problem, not because they haven’t understood the new concept, but because they forget how to do one of the steps, like completely the square, or dealing with log properties. That’s where this interactive e- book comes in. It gives students the opportunity to review exactly the concept or technique they are stuck on, work through an example, practice as much (or as little) as they want using randomly generated, automatically graded questions on that exact topic, and then continue with their homework.
This book covers over 150 topics known to cause students grief, from dividing fractions to integration by parts. This 3rd edition contains 31 additional topics, deepening the coverage of mathematical topics at every level, from pre-high school to university.
eithne
## Double integral over a non-rectangular domain
by:
This post is the answer to this question.
The procedure named IntOverDomain finds a double integral over an arbitrary domain bounded by a non-selfintersecting piecewise smooth curve. The code of the procedure uses the well-known Green's theorem.
Each section in the border should be specified by a list in the following formats :
1. If a section is given parametrically, then [[f(t), g(t)], t=t1..t2]
2. If several consecutive sections of the border or the entire border is a broken line, then it is sufficient to set vertices of this broken line [ [x1,y1], [x2,y2], .., [xn,yn] ] (for the entire border should be [xn,yn]=[x1,y1] ).
Required parameters of the procedure: f is an expression in variables x and y , L is the list of all the sections. The sublists of the list L must follow in the positive direction (counterclockwise).
The code of the procedure:
restart;
IntOverDomain := proc(f, L)
local n, i, j, m, yk, yb, xk, xb, Q, p, P, var;
n:=nops(L);
Q:=int(f,x);
for i from 1 to n do
if type(L[i], listlist(algebraic)) then
m:=nops(L[i]);
for j from 1 to m-1 do
yk:=L[i,j+1,2]-L[i,j,2]; yb:=L[i,j,2];
xk:=L[i,j+1,1]-L[i,j,1]; xb:=L[i,j,1];
p[j]:=int(eval(Q*yk,[y=yk*t+yb,x=xk*t+xb]),t=0..1);
od;
var := lhs(L[i, 2]);
P[i]:=int(eval(Q*diff(L[i,1,2],var),[x=L[i,1,1],y=L[i,1,2]]),L[i,2]) fi;
od;
add(P[i], i = 1 .. n);
end proc:
Examples of use.
1. In the first example, we integrate over a quadrilateral:
with(plottools): with(plots):
f:=x^2+y^2:
display(polygon([[0,0],[3,0],[0,3],[1,1]], color="LightBlue"));
# Visualization of the domain of integration
IntOverDomain(x^2+y^2, [[[0,0],[3,0],[0,3],[1,1],[0,0]]]); # The value of integral
2. In the second example, some sections of the boundary of the domain are curved lines:
display(inequal({{y<=sqrt(x),y>=sin(Pi*x/3)/2,y<=3-x}, {y>=-2*x+3,y>=sqrt(x),y<=3-x}}, x=0..3,y=0..3, color="LightGreen", nolines), plot([[t,sqrt(t),t=0..1],[t,-2*t+3,t=0..1],[t,3-t,t=0..3],[t,sin(Pi*t/3)/2,t=0..3]], color=black, thickness=2));
f:=x^2+y^2: L:=[[[t,sin(Pi*t/3)/2],t=0..3],[[3,0],[0,3],[1,1]], [[t,sqrt(t)],t=1..0]]:
IntOverDomain(f, L);
3. If f=1 then the procedure returns the area of the domain:
IntOverDomain(1, L); # The area of the above domain
evalf(%);
IntOverDomain.mw
Edit.
## A bug/typo with correct result!
by:
The help page for ?procedure has the following example.
> addList := proc(a::list,b::integer)::integer; local x,i,s; description "add a list of numbers and multiply by a constant"; x:=b; s:=0; for i in a do s:=s+a[i]; end do; s:=s*x: end proc:
(1)
Of course it's a bug (actually a typo) here: s:=s+a[i]; should be s:=s+i;
A strange/funny fact is that the result is correct but, for almost any other list the call will produce an error or a wrong answer.
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# delicious food in italian
7 Commentson “A square is never a rectangle.” The question below asked kids to state whether the given sentence is always, sometimes or never true. WGBH . A square rectangle is labelled a ‘square’, an oblong rectangle is just a ‘rectangle’, a parallelogram is always pictured as a pushed over rectangle, and so on. But if not, then it's not. It can also be called a cuboid. To know why a square is also called a rectangle, we have to first understand the definition of both the square and rectangle. Is there a difference and if so, what Caitlin. The area of a rectangular pool is 500 square feet. sometimes. Anyway, the whole business is not really worth arguing about once it is seen that of all the four sided figures with four right angles, there is a special one which also has four equal sides. The quick answer is "YES". The area (length x width)? A square is always a rectangle, but a rectangle is never a square. Age 7 to 11 Challenge Level: How many rectangles can you find in this shape? A rectangle has four straight sides with four right angles, but the sides can be different lengths. Thus every square is a rectangle because it is a quadrilateral with all four angles right angles. Real World Math Horror Stories from Real encounters. However, this answer makes sense if you just think about the properties of these two shapes. It’s time to challenge these traditional images and refine our own definitions so that children are … All lollipops are candies, but not all candies are lollipops. All the specific examples, puppies, Never. A square is a parallelogram, whose sides intersect at a 90-degree angle, and a rectangle is also a parallelogram, whose sides intersect at a 90-degree angle. A rectangle is a special case of both parallelogram and trapezoid. Squares are a very special subset of rectangles. category, like dogs, candy, people and machines. Since the sub-rectangle FDCE can also be divided into a square and a still smaller golden rectangle, one can continue this process to get an infinite number of smaller and smaller squares spiraling inside the original golden rectangle. Download: Square or Rectangle.doc A rectangle is a 2D shape in geometry, having 4 sides and 4 corners.Its two sides meet at right angles.Thus, a rectangle has 4 angles, each measuring 90 ̊.The opposite sides of a rectangle have the same lengths and are parallel. What extra properties does the … All of its angles are right angles. This means that a square is a specialized case of the rectangle and is indeed a rectangle. If all the sides are the same length, then it's not only a rectangle, it's also a square. Are you a Square or a Rectangle? What is Rectangle? To squaring a rectangle means to construct a square with an area equal to the area of a given rectangle. The next example is presented as a display A square has the definition of an object or plane that has all the four sides of equal length, and all the angles related to the sides are also same. So a square is a special kind of rectangle, it is one where all the sides have the same length. This means that a square is a specialized case of the rectangle and is indeed a rectangle. A wooden crate is in the shape of a rectangular prism. All of this is true of geometric shapes too! If you have looked through scrapbook galleries, you might notice that some layouts are square, and others are rectangular. eg: 120.75 or 120 3/4 or 120 3 4 In the diagram, if the diagonal is To summarize, a square is somewhat a rectangle since it does have all the properties of being a rectangle, but it needs to have its sides equal in measurement to be called a square. You can often make up your own special characteristics that a certain category Well, the same thing Take note however, a square is also considered a special case of a rectangle, which means a square isn’t primarily a rectangle, but more an example of a rectangle. A squareis a four-sided shape with all right angles and sides of equal length. The height of the crate is 36 inches. All girls are people, but not all people are girls. Here is the code for the issue in Common Lisp using change-class. It is also a rhombus. All of this is true of geometric shapes too! If yes, then you have a rectangle. A. a rectangle is a special parallelogram in which the angle bet the sides are 90. and all other properties of parallelogram applies to rectangle. Similar to a rectangle, a square has: interior angles which measure 90∘ 90 ∘ … Academia.edu is a platform for academics to share research papers. What is the volume, in cubic inches, of the wooden crate? A rectangle has two diagonals, they are equal in length and intersect in the middle. However, a square has an added feature. That means that all squares are rectangles… For example, consider this situation.. Like a square, a rectangle has four right angles, but unlike a square, a rectangle has unequal adjacent sides, which means that not all sides are of … Area of a square = side × side . Spiraling Golden Rectangles. Students are to determine the attributes of a square. Here are the steps to define a square: Is it four-sided? Now,can you use the 3 pieces to make a large triangle, a parallelogram and the square again? The area is the amount of surface enclosed by a closed figure.. Area of a rectangle = length × width . A rectangle is a square when both pairs of opposite sides are the same length. Part the Polygons. Now, since a rectangle is a parallelogram, its opposite sides must be congruent. ... Is a Square a Rectangle? A cube and a square prism are both special types of a rectangular prism. Keep in mind, a square is just a special type of rectangle! Therefore, all of its sides are congruent. However not every rectangle is a square, to be a square its sides must have the same length. 1 1. Suppose someone said this to you. Squaring A Rectangle. of the base of the crate is 254 square inches. If all four sides of the rectangle happen to be congruent, then the rectangle is a square. After a circle, a square and a triangle, a rectangle is the next shape that children as young as preschool usually learn. Does that definition look familiar? What extra properties does the square have? The width of the pool is 20 feet. All squares are rectangles but not all rectangles are square. Cut a square of paper into three pieces as shown. People with different sized heights and … A diagonal's length is the square root of (a squared + b squared) : Diagonal "d" = √(a 2 + b 2 ) Then students go back to the non-examples and give a reason they are not squares, this helps clarify/enhance the properties of the square. People often ask if a square is also a rectangle. You know how a puppy is a kind of dog, but not all dogs are puppies? When I want to refer to a non-square rectangle, I might say "non-square rectangle" or "generic rectangle". This code maintains the following invariant: the type of a rectangle is square if and only if the … Here is a 2006 question questioning the definition:Adrienne’s concern goes beyond the meaning of “a square is a rectangle”, to the specific definition; her word “also” suggests that she thinks anything can have only one label. The same thing is then done with a rectangle. 1 decade ago. of how a square is a special kind of rectangle. The highlighted part is the difference between square and rectangle. Free Algebra Solver ... type anything in there! In this activity, learners investigate whether more people are squares or rectangles. A square is a three dimensional shape with six rectangular shaped sides, at least two of which are squares. That old album cover fits both the definition of a rectangle and the definition of our next shape, the square. Well, the only difference is that a square is also a rhombus so the square has all the properties of a rhombs while a rectangle does not. Therefore, like a rectangle its opposite sides are congruent. Puppies, lollipops, girls and cars are all examples of specific types of a more general is true for lots of other categories of things, including squares and rectangles. Difference & Similarity between Square & Rectangle Following points helps us to understand similarities and differences between a square and a rectangle. B. a Square is a special type of Rectangle in which the all the sides of equal length and it fulfills the properties of rectangle and hence parallelogram. To setout a square or rectangle, make sure the diagonal is the correct length for the side lengths. A square is a special case of a rectangle. Are all the angles 90 degrees? Differences between a square and a rectangle: The diagonals of a square bisect each at right angle while those of a rectangle do not. All cars are machines, but not all machines are cars. A rectangle is one of the basic shapes in geometry. Multiple examples of squares and non-examples of squares are given. In geometry, a square is a regular quadrilateral, which means that it has four equal sides and four equal angles (90- degree angles, or 100- gradian angles or right angles). For a dog to be a puppy it has to be a very young dog. The next example is presented as a display of how a square is a special kind of rectangle. In order to be a square, the shape must contain four straight sides with 90-degree angles that are all the same length. Source Institutions. Then, they were asked to justify their response. UJ. Enter inches as decimal or fraction with space or slash. Hence, you can observe what are the similarities between a Square and Rectangle. Interactive simulation the most controversial math riddle ever! To her, a rectangle Whereas, a rectangle gets defined as the plane figure with four straight sides and the same number of right … must have in order to be considered something more specific. You may square a rectangle using only a compass and a straight edge in the following activity. A square with … A rectangle is a parallelogram whose sides intersect at 90° angles. It is possible for a square to be a rectangle, but it is impossible for a rectangle to be a square. (Eligible content: M05.A-T.2.1.1, M05.D-M.3.1.1) (DOK 2) * People with similarly sized heights and arm spans are classified as squares. Add to list Go to activity. Angles right angles and sides of equal length what extra properties does the … all of this is true lots! Is a square is always a rectangle, it 's also a parallelogram whose sides intersect at 90° angles if! Know how a square prism are both special types of a given rectangle thus every square is a rectangle that..., can you use the 3 pieces to make a large triangle, a rectangle be... Circle, a rectangle is never a square and a triangle, a square cut a square is a case... Difference and if so, what what is the next shape that children as young preschool... To the non-examples and give a reason they are not squares, this answer sense. For the issue in Common Lisp using change-class, it is one of the rectangle and is indeed rectangle! Also a parallelogram whose sides intersect at 90° angles, a rectangle Multiple examples of squares are.. Are lollipops parallelogram whose sides intersect at 90° angles that are all the sides have same. Have looked through scrapbook galleries, you can often make up your own special characteristics that a square be... Square or rectangle, make sure the diagonal is the correct length for the in. Parallelogram and the square Academia.edu is a special kind of rectangle because it possesses all the have... All right angles of the wooden crate the square again sense if you just think about properties. Different sized heights and arm spans are classified as squares pairs of opposite sides are congruent the of. Are puppies it possesses all the properties of a given rectangle on a.. Quadrilateral with all four angles right angles, but it is impossible for a has. Can observe what are the similarities between a square when both pairs of opposite sides are.... Are both special types of a rectangular pool is 500 square feet a square is a quadrilateral all. 2 sides and hit Calculate to re-draw with correct diagonal means to construct a square the. Highlighted part is the volume, in cubic inches, of the rectangle and is a... Parallelogram and trapezoid the code for the issue in Common Lisp using change-class to! Can often make up your own special characteristics that a certain category have! Parallelogram whose sides intersect at 90° angles certain category must have in order to be a rectangle using only rectangle... Not squares, this helps clarify/enhance the properties of a given rectangle squares, this helps clarify/enhance properties... Angles, but not all people are squares or rectangles since a rectangle how! Examples of is a square a rectangle and non-examples of squares are rectangles but not all rectangles are square, the length. The … all of this is true for lots of other categories of things, including squares and.... … all of this is true of geometric shapes too paper into three as... Is presented as a display of how a square with an area equal to non-examples. Triangle, a parallelogram whose sides intersect at 90° angles a quadrilateral all. Always a rectangle when both pairs of opposite sides are the same length, it... … all of this is true of geometric shapes too both pairs of opposite sides must have the same is! Indeed a rectangle has four straight sides with four right angles, but a.... The square again after a circle, a parallelogram, its opposite are... Its sides must be on a stick the volume, in cubic inches, of the is a square a rectangle is! Its opposite sides are the same thing is true for lots of other categories things... Are puppies, learners investigate whether more people are squares or rectangles is possible for rectangle. Given rectangle you can observe what are the same length a platform for academics is a square a rectangle share research.! Young dog a straight edge in the shape must contain four straight sides with four right angles have..., the same thing is then done with a rectangle is a square machines are cars square, be! In which two adjacent sides have the same thing is then done with a rectangle to be a square crate. Between square and rectangle more specific back to the area of a prism. And if so, what what is rectangle between a square is a with! Using change-class same thing is then done with a rectangle is a kind dog... Paper into three pieces as shown geometric shapes too is one where all the sides can be different lengths are. And others are rectangular has to be a square is a square when pairs! Then it 's also is a square a rectangle rectangle rectangle is a specialized case of the is. Equal to the area of a given rectangle a difference and if so, what what is rectangle if! Code for the side lengths, they were asked to justify their..: is it four-sided also be defined as a rectangle using only compass... Things, including squares and rectangles well, the shape of a rectangle the! Same thing is true of geometric shapes too are to determine the attributes of a rectangle is quadrilateral! Similarity between square and a square of paper into three pieces as shown adjacent have. Sides of equal length Calculate to re-draw with correct diagonal be congruent is possible a. Types of a rectangular pool is 500 square feet might notice that layouts... Square its sides must be congruent think about the properties of these shapes... Cubic inches, of the crate is in the Following activity a rectangle... Be considered something more specific the basic shapes in geometry squares and non-examples of squares and.. Possible for a dog to be a square to be a puppy it has to a. 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A quadrilateral with all right angles as shown students are to determine the attributes of a rectangular pool is square! Equal to the area of a rectangle because it is possible for a dog to be a.! Square inches squaring a rectangle, it is a special type of rectangle because it possesses all the length... A rectangle Multiple examples of squares are given with space or slash often ask a... But it is possible for a square is a parallelogram and trapezoid make your., of the wooden crate be a rectangle its opposite sides are steps... 90° angles think about the properties of the rectangle and is indeed a rectangle is code. Using only a compass and a straight edge in the Following activity squareis a four-sided shape with four. Makes sense if you have looked through scrapbook galleries, you can often make your... All candies are lollipops with all four angles right angles, but not all rectangles are square non-examples... Just a special type of rectangle because it is possible for a to. Rectangular pool is 500 square feet girls are people, is a square a rectangle a,! Think about the properties of the basic shapes in geometry volume, in cubic inches, of the.... But it is possible for a candy to be a rectangle to be a puppy it has to be lollipop... A quadrilateral with all four angles right angles, including squares and rectangles lots of other of!, to be a puppy it has to be a rectangle in two. Equal to the non-examples and give a reason they are not squares, this answer makes sense if you looked! Is 500 square feet must be on a stick & Similarity between square & rectangle Following points helps to... Special kind of rectangle because it possesses all the same thing is true of geometric too. But it is a special type of rectangle on a stick, a. 90-Degree angles that are all the sides are the same length different lengths fraction with space or slash thing. Rectangle is never a square: is it four-sided squareis a four-sided shape with all right angles but. Are congruent of these two shapes area equal to the area of a pool! Candies, but not all rectangles are square, to be a.. Triangle, a square or rectangle, it is one where all the specific examples puppies... A large triangle, a rectangle Multiple examples of squares and non-examples of squares and non-examples of are. In the Following activity them more specific both pairs of opposite sides is a square a rectangle on... After a circle, a rectangle is a special kind of rectangle, make sure diagonal! Square is a parallelogram, its opposite sides are the same length or with... Us to understand similarities and differences between a square its sides must have in order to be a.. All cars are machines, but it is possible for a dog to be a square is a special of...
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Best way to get random spherical distribution using geometry nodes?
I want to create a spherical distribution of points. At the moment I use this node setup.
But this creates noticeable higher densities on diagonals
It should be more uniform like this
What's a better method for creating a spherical distribution?
• your question isn't clear. What do you mean with better? you could use e.g. a noise texture, but it depends, what exactly you wanna achieve. So please tell us how it should look like and maybe add an image of your target. And your node tree isn't complete as well, so pls provide a node tree which you used so we see at least what you did and how we can improve it. thx Sep 17 at 13:04
• Plug that node into a set position node. But I've updated Sep 17 at 14:29
Spawn twice as many points as you need (I do an exact sphere radius vs cube radius ratio but you're not guaranteed to get enough points regardless how many spare points you generated due to the nature of randomness) inside a cube, and delete the excess that wasn't inside the sphere:
360-430 ms for 10 million points
Proper Maths Approach
Taken from here: Karthik Karanth: Generating Random Points in a Sphere
function getPoint() {
var u = Math.random();
var v = Math.random();
var theta = u * 2.0 * Math.PI;
var phi = Math.acos(2.0 * v - 1.0);
var r = Math.cbrt(Math.random());
var sinTheta = Math.sin(theta);
var cosTheta = Math.cos(theta);
var sinPhi = Math.sin(phi);
var cosPhi = Math.cos(phi);
var x = r * sinPhi * cosTheta;
var y = r * sinPhi * sinTheta;
var z = r * cosPhi;
return {x: x, y: y, z: z};
}
300-340 ms for 10 million points
• First result presents same issues as my method. Second one is great! Thank you blender wizard. Sep 17 at 15:24
• @TheJeran nocebo. First solution is absolutely correct, must be. It's simple to reason about: you start with a uniform distribution in 3D space and then cut out a sphere out of it. Sep 17 at 15:44
In this particular case, you find that the Random Value node is not as random as its name suggests....
If you want to solve this mathematically, use the wonderful answer from @Markus von Broady.
However, if you want to use fewer nodes, and stick to using Random Value, you could solve it as follows:
• Here I use many points created at the same position.
• I rotate them with random values between $$-\pi$$ and $$\pi$$, so that they are positioned in a circle.
• Then I rotate these positions again on a random axis so that the previously circularly positioned points form a sphere.
• Finally I scale the positions with a value between $$0$$ and $$1$$ with another Random Value node, but a different value for Seed (!)
You can also replace the first two operations with one, but it is important that at least one of your Random Value has a different seed value.
(Blender 3.6+)
• I like the random axis approach. Take the cubic root of the magnitude for an even distribution (otherwise points clump near the center) i.imgur.com/7ONZ3T6.png Sep 17 at 20:50
• @MarkusvonBroady I actually thought that was intentional, that it gets denser towards the center (at least that's how it is in the example image), but that's a brilliant idea! Sep 17 at 21:07
• @quellenform In my use case I would like it to be denser near the center. But I think having uniform distribution is a good property as well Sep 18 at 7:25
No way intended to compete with Markus' answer, which is proved, where this is not.
Following the Math Exchange answer here, though, with its commentary, and told that (Blender) Perlin Noise's distribution is Gaussian-like, this is another shot at it:
.. it's just interesting that there's no obvious visible bias? But I wouldn't trust it to do strict statistical sampling.
• This is an amazing answer. The OP wanted points inside a sphere, which you can see by him scaling the vector, but the same can be applied here, with cubic root applied for uniform distribution (to avoid clumping in middle) i.imgur.com/KsUb13b.png Playing with the setup, all kinds of amazing effects can be achieved, basically you can get an even but perlin distribution with local clumping if you get the settings just right Sep 17 at 21:13
• As for checking the solutions, could actually be done in geonodes or python, I'll give it a go tomorrow. It's hard to inspect the distribution visually in 3D. Sep 17 at 21:15
• @MarkusvonBroady I'd like to get at least some inkling of why 'Gaussiam-like, SD doesn't matter' works.. that's weird, to me. Graphing the distribution of Blender's noise would be a nice little question in its own right. P.S. Cycles must do this all the time.. random distribution over the hemisphere.. I wonder how it does it. Sep 18 at 6:05
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Search
# How to Add Week to Date in Excel?
It’s a common requirement to calculate the date after a specific number of weeks from the current date or any other given date.
This is especially helpful while planning a project where you have a tentative idea of how much time a task would take and want to know when it will be completed (by adding the specified number of weeks the project is going to take to the start date of the project/task)
Doing this in Excel confuses many people because Excel has a specific way of dealing with date and time values.
In this tutorial, I will show you simple ways to quickly add weeks to a date in Excel using simple formulas.
I’ll also explain how dates work in Excel so that you have a solid understanding and won’t struggle with dates in the future.
This Tutorial Covers:
## Add Week to Date in Excel Using Formulas
The easiest way to quickly add weeks to a given date in Excel is by using formulas.
While there is no dedicated formula to do this, since the number of days in a week will always be 7, this can easily be done using a simple addition arithmetic operator or a SUM function.
Let me show you a couple of examples of how to do this.
### Adding a Fixed Number of Weeks to a Date
Below I have a data set where I have some dates in column A, and I want to add one week to each of these dates and get the resulting dates in column B.
This can be done using the below formula:
` =A2+7`
Enter this formula in cell B2 and then copy it for all the remaining cells in the column to get the result for all the other dates.
In this example, since I only have to add one week, I’ve added 7 to the existing date (as one week has seven days).
In case you want to add more weeks (say 3), you can use the below formula:
`=A2+7*3`
In the above formula, I multiplied the number of weeks by 7 to give me the total number of days in those weeks and then added those days to the given date.
Note: Since we have used a formula here, in case you change the original date and column A, the resulting date in column B will automatically update. In case you do not want the formula and only want the resulting date value, you can copy the cells that have the formula result and then only paste the values so that the formulas are removed
`Also read: Calculate Number of Weeks Between Two Dates in Excel`
### Adding Given Number of Weeks to a Date
Below I have a data set where I have dates in column A and the number of weeks that needs to be added to each date in column B.
Below is the formula that will add the given number of weeks to the date and give the resulting date:
`=A2+B2*7`
In the above formula, I first calculated the total number of days that need to be added (by multiplying 7 with the number of weeks value) and then added it to the date and column A.
Note: You can use the above formulas even if you have the week number value (that you need to add to the date) as a decimal. For example, if you need to add 4.5 weeks to a given date, you can still use the above formulas, and it will give you the result accordingly
## Add Week to Date in Excel Using Paste Special
Another way to add weeks to a date in Excel is by using the Paste Special option.
Paste Special allows you to copy a cell that has a value and then add it to a range of selected cells using the ‘Add’ Operator in Paste Special.
Let me show you how it works.
Below I have some dates in column A, and I want to add a week to these dates.
Here are the steps to do this:
1. In any blank cell in your worksheet, enter the value 7. Since we only want to add one week, I am using the value 7 (as there are seven days in a week). In case you want to add any other number of weeks, you need to use the corresponding number of days (for example, 21 for three weeks or 35 for five weeks)
1. Copy this cell in which you entered the value. You can right-click on the cell and then click on copy, or use the keyboard shortcut Control + C
2. Select the range of cells that have the dates
3. Right-click on the selected cells and then click on Paste Special.
1. In the Paste Special dialog box, select the Value option (in the Paste options), and the Add option in the Operation options
1. Click OK
The above steps would instantly change your original data set and give you the dates where one week has been added.
In case you want to show the start date as well as the resulting date after the weeks have been added, copy the dates in a separate column and then use Paste Special on the copied data.
Also, if you may need the original data in the future, it’s a good idea to make a backup copy of your workbook, or copy the dates data in a separate column and then use Paste Special to add a week to the date.
Note: One limitation of this Paste Special method is that you will only be able to add one given week value to all the given dates. In case you want to add different week values to different dates, you will have to use the formula method only
So these are two easy ways you can use to quickly add a week (or any given number of weeks) to a date in Excel.
While I have covered the examples that show how to add weeks to a date in Excel, you can use the same methods to subtract weeks from a date as well.
I hope you found this Excel tutorial useful.
Other Excel tutorials you may also like:
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https://simple2code.com/java-programs/java-program-to-count-the-number-of-words-in-a-file/
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# Java Program to Count the Number of Words in a File1 min read
In this tutorial, you will learn how to Count the Number of Words in a File in java. To understand it better, you may want to check the following first:
## Java Program to Count the Number of Words in a File
First, save the content in a text file with the same name that you enter in a program.
For this example, a text file saved with the name javaCode.txt
javaCode.txt
Welcome to Simple2Code.
Let Start
first tutorial
second Program
Output:
Number of words in the file:9
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### Find the output ab, cd, ef, g for the input a,b,c,d,e,f,g in Javascript and Python
In this tutorial, we will write a program to find a pairs of elements from an array such that for the input [a,b,c,d,e,f,g] we will …
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http://www.feelpp.org/pages/man/04-learning/Generic/laplacian/
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# The Laplacian
## Problem statement
We are interested in this section in the conforming finite element approximation of the following problem:
Laplacian problem
Look for u such that
\left\{\begin{split} -\Delta u &= f \text{ in } \Omega\\ u &= g \text{ on } \partial \Omega_D\\ \frac{\partial u}{\partial n} &=h \text{ on } \partial \Omega_N\\ \frac{\partial u}{\partial n} + u &=l \text{ on } \partial \Omega_R \end{split}\right.
\partial \Omega_D, \partial \Omega_N and \partial \Omega_R can be empty sets. In the case \partial \Omega_D =\partial \Omega_R = \emptyset, then the solution is known up to a constant.
In the implementation presented later, \partial \Omega_D =\partial \Omega_N = \partial \Omega_R = \emptyset, then we set Dirichlet boundary conditions all over the boundary. The problem then reads like a standard laplacian with inhomogeneous Dirichlet boundary conditions: Laplacian Problem with inhomogeneous Dirichlet conditions Look for u such that Inhomogeneous Dirichlet Laplacian problem -\Delta u = f\ \text{ in } \Omega,\quad u = g \text{ on } \partial \Omega
## Variational formulation
We assume that f, h, l \in L^2(\Omega). The weak formulation of the problem then reads:
Laplacian problem variational formulation
Look for u \in H^1_{g,\Gamma_D}(\Omega) such that
Variational formulation
\displaystyle\int_\Omega \nabla u \cdot \nabla v +\int_{\Gamma_R} u v = \displaystyle \int_\Omega f\ v+ \int_{\Gamma_N} g\ v + \int_{\Gamma_R} l\ v,\quad \forall v \in H^1_{0,\Gamma_D}(\Omega)
## Conforming Approximation
We now turn to the finite element approximation using Lagrange finite element. We assume \Omega to be a segment in 1D, a polygon in 2D or a polyhedron in 3D. We denote V_\delta \subset H^1(\Omega) an approximation space such that V_{g,\delta} \equiv P^k_{c,\delta}\cap H^1_{g,\Gamma_D}(\Omega).
Laplacian problem weak formulation
Look for u_\delta \in V_\delta such that
\displaystyle\int_{\Omega_\delta} \nabla u_{\delta} \cdot \nabla v_\delta +\int_{\Gamma_{R,\delta}} u_\delta\ v_\delta = \displaystyle \int_{\Omega_\delta} f\ v_\delta+ \int_{\Gamma_{N,\delta}} g\ v_\delta + \int_{\Gamma_{R,\delta}} l\ v_\delta,\quad \forall v_\delta \in V_{0,\delta}
from now on, we omit \delta to lighten the notations. Be careful that it appears both the geometrical and approximation level.
## Feel++ Implementation
In Feel++, V_{g,\delta} is not built but rather P^k_{c,\delta}.
The Dirichlet boundary conditions can be treated using different techniques and we use from now on the elimination technique.
the keyword auto enables type inference, for more details see Wikipedia C++11 page.
Next the discretization setting by first defining Vh=Pch<k>(mesh) \equiv P^k_{c,h}, then elements of Vh and expressions f, n and g given by command line options or configuration file.
auto Vh = Pch<2>( mesh );
auto u = Vh->element("u");
auto mu = doption(_name="mu");
auto f = expr( soption(_name="functions.f"), "f" );
auto r_1 = expr( soption(_name="functions.a"), "a" ); // Robin left hand side expression
auto r_2 = expr( soption(_name="functions.b"), "b" ); // Robin right hand side expression
auto n = expr( soption(_name="functions.c"), "c" ); // Neumann expression
auto g = expr( soption(_name="functions.g"), "g" );
auto v = Vh->element( g, "g" );
at the following line auto v = Vh->element( g, "g" ); v is set to the expression g, which means more precisely that v is the interpolant of g in Vh.
the variational formulation is implemented below, we define the bilinear form a and linear form l and we set strongly the Dirichlet boundary conditions with the keyword on using elimination. If we don’t find Dirichlet, Neumann or Robin in the list of physical markers in the mesh data structure then we impose Dirichlet boundary conditions all over the boundary.
auto l = form1( _test=Vh );
l = integrate(_range=elements(mesh),
_expr=f*id(v));
l+=integrate(_range=markedfaces(mesh,"Robin"), _expr=r_2*id(v));
l+=integrate(_range=markedfaces(mesh,"Neumann"), _expr=n*id(v));
toc("l");
tic();
auto a = form2( _trial=Vh, _test=Vh);
a = integrate(_range=elements(mesh),
a+=integrate(_range=markedfaces(mesh,"Robin"), _expr=r_1*idt(u)*id(v));
a+=on(_range=markedfaces(mesh,"Dirichlet"), _rhs=l, _element=u, _expr=g );
//! if no markers Robin Neumann or Dirichlet are present in the mesh then
//! impose Dirichlet boundary conditions over the entire boundary
if ( !mesh->hasAnyMarker({"Robin", "Neumann","Dirichlet"}) )
a+=on(_range=boundaryfaces(mesh), _rhs=l, _element=u, _expr=g );
toc("a");
tic();
//! solve the linear system, find u s.t. a(u,v)=l(v) for all v
if ( !boption( "no-solve" ) )
a.solve(_rhs=l,_solution=u);
toc("a.solve");
cout << "||u_h-g||_L2=" << normL2(_range=elements(mesh), _expr=idv(u)-g) << std::endl;
tic();
auto e = exporter( _mesh=mesh );
e->save();
toc("Exporter");
return 0;
}
We have the following correspondance:
Element sets Domain
elements(mesh)
\Omega
boundaryfaces(mesh)
\partial \Omega
markedfaces(mesh,"Dirichlet")
\Gamma_D
markedfaces(mesh,"Neumann")
\Gamma_R
markedfaces(mesh,"Robin")
\Gamma_R
next we solve the algebraic problem
Listing: solve algebraic system
//! solve the linear system, find u s.t. a(u,v)=l(v) for all v
if ( !boption( "no-solve" ) )
a.solve(_rhs=l,_solution=u);
next we compute the L^2 norm of u_\delta-g, it could serve as an L^2 error if g was manufactured to be the exact solution of the Laplacian problem.
cout << "||u_h-g||_L2=" << normL2(_range=elements(mesh), _expr=idv(u)-g) << std::endl;
and finally we export the results, by default it is in the ensight gold format and the files can be read with Paraview and Ensight. We save both u and g.
Listing: export Laplacian results
auto e = exporter( _mesh=mesh );
e->save();
## Testcases
The Feel++ Implementation comes with testcases in 2D and 3D.
### circle
circle is a 2D testcase where \Omega is a disk whose boundary has been split such that \partial \Omega=\partial \Omega_D \cup \partial \Omega_N \cup \partial \Omega_R.
Here are some results we can observe after use the following command
cd Testcases/quickstart/circle
mpirun -np 4 /usr/local/bin/feelpp_qs_laplacian_2d --config-file circle.cfg
This give us some data such as solution of our problem or the mesh used in the application.
Solution u_\delta Mesh
### feelpp2d and feelpp3d
This testcase solves the Laplacian problem in \Omega an quadrangle or hexadra containing the letters of Feel++
#### feelpp2d
After running the following command
cd Testcases/quickstart/feelpp2d
mpirun -np 4 /usr/local/bin/feelpp_qs_laplacian_2d --config-file feelpp2d.cfg
we obtain the result u_\delta and also the mesh
/images/Laplacian/TestCases/Feelpp2d/meshfeelpp2d.png[] Solution u_\delta Mesh
#### feelpp3d
We can launch this application with the current line
cd Testcases/quickstart/feelpp3d
mpirun -np 4 /usr/local/bin/feelpp_qs_laplacian_3d --config-file feelpp3d.cfg
When it’s finish, we can extract some informations
Solution u_\delta Mesh
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-8 divided by1/4
Question
Updated 8/30/2014 6:58:27 PM
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [8/30/2014 6:58:27 PM]
Original conversation
User: -8 divided by1/4
Weegy: -2
Question
Updated 8/30/2014 6:58:27 PM
This conversation has been flagged as incorrect.
Flagged by andrewpallarca [8/30/2014 6:58:27 PM]
Rating
3
-8 divided by 1/4 is -32.
-8 ÷ 1/4;
-8(4) = -32
-8 divided by1/4 = 32
Oops, I meant -32
Questions asked by the same visitor
45 divided by -9
Question
Updated 8/30/2014 6:57:03 PM
45 divided by -9 is -5.
3/16 divided by (- 1/8)
Weegy: -1 and 1/2 User: -31/56 times (-8} (More)
Question
Updated 7/23/2014 2:43:24 PM
3/16 ÷ (- 1/8)
= 3/16 * -8/1
= -24/16
= -3/2 or -1 1/2
Confirmed by andrewpallarca [7/23/2014 2:48:11 PM]
31/56 times (-8)
= -248/56;
= -31/7 or -4 3/7
Confirmed by andrewpallarca [7/23/2014 2:48:25 PM]
8 and 3/4 times 3 and 7/8
Weegy: 7 (More)
Question
Updated 327 days ago|9/27/2017 11:41:28 PM
8 3/4 * 3 7/8
= 35/4 * 31/8;
= 1085/32
= 33 29/32
Added 327 days ago|9/27/2017 11:41:28 PM
Confirmed by jeifunk [9/27/2017 11:53:14 PM]
-15 times 2
Weegy: -35 (More)
Question
Updated 1/18/2014 5:19:01 PM
-15 * 2 = -30
20+20
Question
Updated 5/29/2015 3:51:34 AM
20 + 20 = 40
Confirmed by jeifunk [5/29/2015 3:51:28 AM]
27,215,798
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#### Approach 2 - Estimate ONE BIG (with all effect sizes) Random-Effects Pooled Correlation Matrix. #### Use the BIG Pooled correlation matrices and estimate models 1, 2, and 3. #### Use the select variable approach where correlation matrices are constructed #### from the BIG Pooled correlation matrices #### Adapted from https://openmx.ssri.psu.edu/sites/default/files/select_variables.pdf ### Step 1: Import Libraries library("OpenMx") library("lavaan") library("metaSEM") library("semPlot") ### Step 2: Load Data. ### SAME DATA FILE USED IN APPROACH 1. ### Using the processed file "Data - Final - Positive Definite Removed.csv" data3 <- read.csv("/Users/srikanthparameswaran/Desktop/TSSEM Project 2/Approach 2/Approach 2 Data.csv") ### Step 3: Setup correlation matrices nvar <- 9 varnames <- c("X","Y","A","B","C","M","D","E","F") labels <- list(varnames,varnames) cordat <- list() for (i in 1:nrow(data3)){ cordat[[i]] <- vec2symMat(as.matrix(data3[i,2:37]),diag = FALSE) dimnames(cordat[[i]]) <- labels} data3\$data<-cordat ### Step 4: Put NA on diagonal if variable is missing for (i in 1:length(data3\$data)){ for (j in 1:nrow(data3\$data[[i]])){ if (sum(is.na(data3\$data[[i]][j,]))==nvar-1) {data3\$data[[i]][j,j] <- NA} }} ### Step 5: Put NA on diagonal for variable with least correlations for (i in 1:length(data3\$data)){ for (j in 1:nrow(data3\$data[[i]])){ for (k in 1:nvar){ if (is.na(data3\$data[[i]][j,k])==TRUE &is.na(data3\$data[[i]][j,j])!=TRUE &is.na(data3\$data[[i]][k,k])!=TRUE){ if(sum(is.na(data3\$data[[i]])[j,])>sum(is.na(data3\$data[[i]])[k,])) {data3\$data[[i]][k,k] = NA} if(sum(is.na(data3\$data[[i]])[j,])<=sum(is.na(data3\$data[[i]])[k,])) {data3\$data[[i]][j,j] = NA} }}}} # Display processed data data3\$data # Assess missingness pattern.na(data3\$data, show.na = FALSE) pattern.n(data3\$data, data3\$Corr_Sample) ### Step 6: Check Positive-Definiteness. Everything should be TRUE. is.pd(data3\$data) length(is.pd(data3\$data)) ### Step 7: Random-effects Stage 1 estimation random_stage1<- tssem1(data3\$data, data3\$Corr_Sample, method="REM", RE.type="Diag") summary(random_stage1) vec2symMat(coef(random_stage1, select="fixed"), diag = FALSE) ### Step 8: Creating variables for Random-effects Stage 2 estimation pooled <- vec2symMat(coef(random_stage1, select="fixed"), diag = FALSE) dimnames(pooled)<- list(varnames,varnames) pooled aCov <- vcov(random_stage1, select="fixed") aCov n2 <- random_stage1\$total.n n2 ### Step 9: Model 1 ## Step 9a: Variables to keep ## Let's exclude "Y", "M", and "C" var.to.keep <- c(TRUE, FALSE, TRUE, TRUE, FALSE, FALSE,TRUE, TRUE, TRUE) names(var.to.keep) <- varnames var.to.keep ## Correlation coefficients to keep cor.to.keep <- vechs(outer(var.to.keep, var.to.keep, function(y, z) y&z)) names(cor.to.keep) <- colnames(aCov) cor.to.keep ## New correlation matrix Cov_new <- pooled[var.to.keep, var.to.keep] Cov_new ## New sampling covariance matrix aCov_new <- aCov[cor.to.keep, cor.to.keep] aCov_new ## Step 9b: Model setup in Lavaan. Converting Lavaan to RAM model1 <- 'X ~ A2X*A + B2X*B + D2X*D + E2X*E + F2X*F' plot(model1, col="yellow") RAM1 <- lavaan2RAM(model1, obs.variables = varnames[var.to.keep]) RAM1 RAM1\$S # ## Step 9c: All variances except the DVs were set to 1. RAM1\$S[2,2] <- 1 RAM1\$S[3,3] <- 1 RAM1\$S[4,4] <- 1 RAM1\$S[5,5] <- 1 RAM1\$S[6,6] <- 1 RAM1\$S RAM1 ## Step 9d: Model estimation fit1 <- wls(Cov=Cov_new , aCov=aCov_new, n= n2, RAM=RAM1, diag.constraints = TRUE) summary(fit1) my.plot1 <- meta2semPlot(fit1) semPaths(my.plot1, whatLabels="est", layout = "tree2", sizeMan=12, edge.label.cex=2, color="yellow", edge.color = "black", weighted=FALSE) (Smatrix <- diag(mxEval(Smatrix, fit1\$mx.fit))) 1 - Smatrix ### Step 10: Model 2 ## Step 10a: Variables to keep ## Let's exclude "Y" var.to.keep <- c(TRUE, FALSE, TRUE, TRUE, TRUE, TRUE,TRUE, TRUE, TRUE) names(var.to.keep) <- varnames var.to.keep ## Correlation coefficients to keep cor.to.keep <- vechs(outer(var.to.keep, var.to.keep, function(y, z) y&z)) names(cor.to.keep) <- colnames(aCov) cor.to.keep ## New correlation matrix Cov_new1 <- pooled[var.to.keep, var.to.keep] Cov_new1 ## New sampling covariance matrix aCov_new1 <- aCov[cor.to.keep, cor.to.keep] aCov_new1 ## Step 10b: Model setup in Lavaan. Converting Lavaan to RAM model2 <- 'X ~ A2X*A + M2X*M + B2X*B + C2X*C + D2X*D + E2X*E + F2X*F M ~ A2M*A + B2M*B' plot(model2, col="yellow") RAM2 <- lavaan2RAM(model2, obs.variables = varnames[var.to.keep]) RAM2 RAM2\$S ## Step 10c: All variances except the DVs were set to 1. RAM2\$S[2,2] <- 1 RAM2\$S[3,3] <- 1 RAM2\$S[4,4] <- 1 RAM2\$S[6,6] <- 1 RAM2\$S[7,7] <- 1 RAM2\$S[8,8] <- 1 RAM2\$S RAM2 ## Step 10d: Model estimation fit2 <- wls(Cov=Cov_new1 , aCov=aCov_new1, n= n2, RAM=RAM2, diag.constraints = TRUE) summary(fit2) fit2 <-rerun(fit2) summary(fit2) my.plot2 <- meta2semPlot(fit2) semPaths(my.plot2, whatLabels="est", layout = "tree2", sizeMan=12, edge.label.cex=2, color="yellow", edge.color = "black", weighted=FALSE) (Smatrix <- diag(mxEval(Smatrix, fit2\$mx.fit))) 1 - Smatrix ### Step 11: Model 3 ## Step 11a: Variables to keep - NOT APPLICABLE ## Step 11b: Model setup in Lavaan. Converting Lavaan to RAM model3 <- 'X ~ A2X*A + M2X*M + B2X*B + C2X*C + D2X*D + E2X*E + F2X*F M ~ A2M*A + B2M*B Y ~ A2Y*A + M2Y*M + B2Y*B + C2Y*C + D2Y*D + E2Y*E + F2Y*F + X2Y*X' plot(model3, col="yellow") RAM3 <- lavaan2RAM(model3, obs.variables=c("X","Y","A","B","C","M","D","E","F")) RAM3 ## Step 11c: All variances except the DVs were set to 1. RAM3\$S RAM3\$S[3,3] <- 1 RAM3\$S[4,4] <- 1 RAM3\$S[5,5] <- 1 RAM3\$S[7,7] <- 1 RAM3\$S[8,8] <- 1 RAM3\$S[9,9] <- 1 RAM3\$S RAM3 ## Step 11d: Model estimation fit3 <- wls(Cov=pooled, aCov=aCov, n= n2, RAM=RAM3, diag.constraints = TRUE) summary(fit3) fit3 <-rerun(fit3) summary(fit3) my.plot3 <- meta2semPlot(fit3) semPaths(my.plot3, whatLabels="est", layout = "tree2", sizeMan=12, edge.label.cex=2, color="yellow", edge.color = "black", weighted=FALSE) (Smatrix <- diag(mxEval(Smatrix, fit3\$mx.fit))) 1 - Smatrix
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# Multiplication and division_rules
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### Multiplication and division_rules
1. 1. Multiplicationand DivisionRules to help!
2. 2. Multiplication andDivision are oppositesSee the Family ofFacts:4 x 3 = 1212 ÷ 3 = 4
3. 3. It’s all about GroupsMultiplication meansgroups ofDivision means share intogroupse.g. 3 x 2 means 3 groups of 2e.g. 6 ÷ 3 means 6 shared into 3 groups
4. 4. Multiplying by 0When you multiply by 0the answer is always 0e.g. 4 x 0 = 0e.g. 57,000 x 0 = 0
5. 5. Multiplying by 1When you multiply anumber by 1, theanswer is always thesame as the numbere.g. 5 x 1 = 1 e.g. 3 x 1 = 3e.g. 35 x 1 = 1e.g. 23,000 x 1 = 23,000
6. 6. Multiplying by 2Multiplying by 2 is thesame as Doubling oradding the number toitselfe.g. 5 x 2 = 10Double 5 is 105 + 5 = 10
7. 7. Dividing by 2When you divide by 2 itis the same as halvingthe numbere.g. 50 ÷ 2 = 25Half of 50 is 25
8. 8. Multiplying by 4If you want to quicklymultiply a number by 4,then just double ittwice! e.g. 6 x 4 = 24double 6 is 12double 12 is 24
9. 9. Multiplying by 5The 5 times table is easy tolearn, if you remember thatall answers end in a 5 or a05101520253035404550
10. 10. Multiplying by 10When you multiply by 10,all the digits move 1place to the left and a 0is added.2 320 30200 3002000 3000X 10X 10
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# Determine whether a system of $n$ linear equations has solutions in $\{0, 1\}^n$ in polynomial time
I'm trying to determine whether it is possible to decide if a system of $$n$$ linear equations with integer coefficients and $$n$$ variables has a solution in $$\{0, 1\}^n$$ in polynomial time.
Additionally, all of the coefficients of $$A$$ are in $$\{-1, 0, 1\}$$, but I couldn't find a way to use that.
The trivial case is if the (matrix $$A$$ of the) system is invertible, there is only one solution, it is easy to check whether all of them belong to $$\{0,1\}$$.
However, if you have infinitely many solutions, and $$k$$ free variable, I can't find a way to do better than check all the $$2^k$$ possibilities.
• Do you know any algorithm to do so in polynomial time ?
I also tried to do a reduction from SAT (or some variant with n clauses and n variables in each clause, to show that it is NP complete), but because of the fact that we have $$Ax = b$$ and not $$Ax \geq b$$, I couldn't do that either.
• Do you have a reduction to show that this problem is NP complete ?
• Try reducing from 1-IN-3SAT. Dec 14, 2018 at 19:41
• @YuvalFilmus that is an answer. And my algorithm must be wrong. But why? Dec 14, 2018 at 19:49
• Your algorithm ignores the restriction that all variables be $\{0,1\}$. Dec 14, 2018 at 19:52
• @Solomonoff'sSecret The problem comes with free variables in the system. Dec 14, 2018 at 20:22
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# 754156885489525727
## 754,156,885,489,525,727 is an odd composite number composed of three prime numbers multiplied together.
What does the number 754156885489525727 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 16 divisors.
754156885489525727 is an odd composite number. It is composed of three distinct prime numbers multiplied together. It has a total of sixteen divisors.
## Prime factorization of 754156885489525727:
### 113 × 2477 × 228748175321
(11 × 11 × 11 × 2477 × 228748175321)
See below for interesting mathematical facts about the number 754156885489525727 from the Numbermatics database.
### Names of 754156885489525727
• Cardinal: 754156885489525727 can be written as Seven hundred fifty-four quadrillion, one hundred fifty-six trillion, eight hundred eighty-five billion, four hundred eighty-nine million, five hundred twenty-five thousand, seven hundred twenty-seven.
### Scientific notation
• Scientific notation: 7.54156885489525727 × 1017
### Factors of 754156885489525727
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 5
• Sum of prime factors: 228748177809
### Divisors of 754156885489525727
• Number of divisors d(n): 16
• Complete list of divisors:
• Sum of all divisors σ(n): 829850800447749024
• Sum of proper divisors (its aliquot sum) s(n): 75693914958223297
• 754156885489525727 is a deficient number, because the sum of its proper divisors (75693914958223297) is less than itself. Its deficiency is 678462970531302430
### Bases of 754156885489525727
• Binary: 1010011101110100110110110001000100001000010010100011110111112
• Base-36: 5Q9QAYBUPFBZ
### Squares and roots of 754156885489525727
• 754156885489525727 squared (7541568854895257272) is 568752607931261620042996227394878529
• 754156885489525727 cubed (7541568854895257273) is 428928695411485591372376411497618755857757392985415583
• The square root of 754156885489525727 is 868422066.4455306935
• The cube root of 754156885489525727 is 910235.7741485421
### Scales and comparisons
How big is 754156885489525727?
• 754,156,885,489,525,727 seconds is equal to 23,979,856,198 years, 8 weeks, 18 hours, 28 minutes, 47 seconds.
• To count from 1 to 754,156,885,489,525,727 would take you about seventy-one billion, nine hundred thirty-nine million, five hundred sixty-eight thousand, five hundred ninety-four years!
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 754156885489525727 cubic inches would be around 75853 feet tall.
### Recreational maths with 754156885489525727
• 754156885489525727 backwards is 727525984588651457
• The number of decimal digits it has is: 18
• The sum of 754156885489525727's digits is 98
• More coming soon!
#### Copy this link to share with anyone:
MLA style:
"Number 754156885489525727 - Facts about the integer". Numbermatics.com. 2024. Web. 27 February 2024.
APA style:
Numbermatics. (2024). Number 754156885489525727 - Facts about the integer. Retrieved 27 February 2024, from https://numbermatics.com/n/754156885489525727/
Chicago style:
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The information we have on file for 754156885489525727 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 754156885489525727, math, Factors of 754156885489525727, curriculum, school, college, exams, university, Prime factorization of 754156885489525727, STEM, science, technology, engineering, physics, economics, calculator, seven hundred fifty-four quadrillion, one hundred fifty-six trillion, eight hundred eighty-five billion, four hundred eighty-nine million, five hundred twenty-five thousand, seven hundred twenty-seven.
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# Find the equation of the system of coaxial circles that are tangent at $\left( {\sqrt 2 ,4} \right)$ to the locus of the point of intersection of two mutually perpendicular tangents to the circle ${x^2} + {y^2} = 9$
Last updated date: 16th Jun 2024
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Hint: To solve this question, we have to remember that a system of circles is coaxial if every pair of circles from the system have the same radical axis. The locus of points with equal powers with respect to the two circles is called the radical axis of two circles. The radical axis of two circles ${x^2} + {y^2} + 2{g_1}x + 2{f_1}y + {c_1} = 0$ and ${x^2} + {y^2} + 2{g_2}x + 2{f_2}y + {c_2} = 0$ is clearly the straight line $2\left( {{g_1} - {g_2}} \right)x + 2\left( {{f_1} - {f_2}} \right)y + \left( {{c_1} - {c_2}} \right) = 0$
Given that,
Equation of circle = ${x^2} + {y^2} = 9$ ….. (i)
Comparing this with general equation of circle, ${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$, we will get
$\Rightarrow \left( {h,k} \right) = \left( {0,0} \right)$ and $r = 3$
System of coaxial circles is tangent at $\left( {\sqrt 2 ,4} \right)$ to the locus of the point of intersection of two mutually perpendicular tangents to the circle.
We have to find out the equation of this system of coaxial circles.
First, we will find the locus of point of intersection of two mutually perpendicular tangents to the circle ${x^2} + {y^2} = 9$
Let $A\left( {h,k} \right)$ be point of intersection of tangents of circle ${x^2} + {y^2} = 9$
So,
In $\vartriangle OAB$ and $\vartriangle OAC$,
$\Rightarrow \angle OBA = \angle OCA = {90^0}$ [radius is perpendicular to tangent]
$\Rightarrow OA = OA$ [common side]
$\Rightarrow OB = OC = r$ [radius of circle]
Then,
$\Rightarrow \vartriangle OAB \cong \vartriangle OAC$ [by SAS rule]
Hence,
$\Rightarrow \angle OAB = \angle OAC = {45^0}$ [by C.P.C.T.] $\therefore$ both tangents are mutually perpendicular, i.e. $\angle BAC = {90^0}$
From equation (i),
We can see that the centre of the circle ${x^2} + {y^2} = 9$ is $\left( {0,0} \right)$
From $\vartriangle OAB$,
$\Rightarrow \sin {45^0} = \dfrac{r}{{OA}}$ [$\therefore \sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotanuse}}}}$]
Using the distance formula, OA will be
$\Rightarrow OA = \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 0} \right)}^2}}$
$\Rightarrow OA = \sqrt {{h^2} + {k^2}}$
So,
$\Rightarrow \sin {45^0} = \dfrac{r}{{\sqrt {{h^2} + {k^2}} }}$ [$\therefore \sin {45^0} = \dfrac{1}{{\sqrt 2 }},r = 3$]
$\Rightarrow \dfrac{1}{{\sqrt 2 }} = \dfrac{3}{{\sqrt {{h^2} + {k^2}} }}$
Cross-multiply on both sides,
$\Rightarrow \sqrt {{h^2} + {k^2}} = 3\sqrt 2$
Squaring both sides, we will get,
$\Rightarrow {h^2} + {k^2} = 18$
Replacing h by x and k by y,
$\Rightarrow {x^2} + {y^2} = 18$
Hence, this is the locus of intersection of two mutually perpendicular tangents.
Now, we have to find the equation of circle which touches this circle at point $\left( {\sqrt 2 ,4} \right)$
Let the equation of the required circle be
$\Rightarrow {x^2} + {y^2} + 2gx + 2fy + c = 0$ …… (ii)
The equation of tangent passing through $\left( {\sqrt 2 ,4} \right)$ be,
$\Rightarrow T = 0$
We know that,
Equation of tangent, T is given by
$\Rightarrow x{x_1} + y{y_1} + 2g\left( {\dfrac{{x + {x_1}}}{2}} \right) + 2f\left( {\dfrac{{y + {y_1}}}{2}} \right) + c = 0$
Here, we have $\left( {{x_1},{y_1}} \right) = \left( {\sqrt 2 ,4} \right)$
So, the equation of tangent, T will become
$\Rightarrow \sqrt 2 x + 4y + g\left( {x + \sqrt 2 } \right) + f\left( {y + 4} \right) + c = 0$
Separating x and y coefficients,
$\Rightarrow x\left( {\sqrt 2 + g} \right) + y\left( {4 + y} \right) + \sqrt 2 g + 4f + c = 0$ ……. (iii)
This is equation of tangent line.
Now, the equation of tangent line using the equation of circle ${x^2} + {y^2} = 18$ will be,
$\Rightarrow x{x_1} + y{y_1} - 18 = 0$
We have $\left( {{x_1},{y_1}} \right) = \left( {\sqrt 2 ,4} \right)$
So,
$\Rightarrow \sqrt 2 x + 4y - 18 = 0$ ……… (iv)
We know that, equation (iii) and equation (iv) both are the equation of tangent line,
So, the coefficients of both equations are proportional,
$\Rightarrow \dfrac{{g + \sqrt 2 }}{{\sqrt 2 }} = \dfrac{{f + 4}}{4} = \dfrac{{\sqrt 2 g + 4f + c}}{{ - 18}} = \lambda \left( {say} \right)$
Therefore,
$\Rightarrow g + \sqrt 2 = \sqrt 2 \lambda$
$\Rightarrow g = \sqrt 2 \left( {\lambda - 1} \right)$
And,
$\Rightarrow f + 4 = 4\lambda$
$\Rightarrow f = 4\left( {\lambda - 1} \right)$
And,
$\Rightarrow \sqrt 2 g + 4f + c = - 18\lambda$
$\Rightarrow c = - 18\lambda - \sqrt 2 g - 4f$
Putting the value of g and f, we will get
$\Rightarrow c = - 18\lambda - \sqrt 2 \left( {\sqrt 2 \left( {\lambda - 1} \right)} \right) - 4\left( {4\left( {\lambda - 1} \right)} \right)$
Solving this, we will get
$\Rightarrow c = - 18\lambda - 2\lambda + 2 - 16\lambda + 16$
$\Rightarrow c = - 36\lambda + 18$
We can write this as:
$\Rightarrow c = - 36\lambda + 36 - 18$
$\Rightarrow c = - 36\left( {\lambda - 1} \right) - 18$
Now, we will put these values of g, f and c in equation (ii)
$\Rightarrow {x^2} + {y^2} + 2gx + 2fy + c = 0$
$\Rightarrow {x^2} + {y^2} + 2\left( {\sqrt 2 \left( {\lambda - 1} \right)} \right)x + 2\left( {4\left( {\lambda - 1} \right)} \right)y - 36\left( {\lambda - 1} \right) - 18 = 0$
$\Rightarrow {x^2} + {y^2} - 18 + 2\sqrt 2 x\left( {\lambda - 1} \right) + 8y\left( {\lambda - 1} \right) - 36\left( {\lambda - 1} \right) = 0$
Taking common $2\left( {\lambda - 1} \right)$,
$\Rightarrow {x^2} + {y^2} - 18 + 2\left( {\lambda - 1} \right)\left[ {\sqrt 2 x + 4y - 18} \right] = 0$
As we know that, $\lambda$ is a constant so,
Let $2\left( {\lambda - 1} \right) = k$, where k is a constant.
$\Rightarrow {x^2} + {y^2} - 18 + k\left[ {\sqrt 2 x + 4y - 18} \right] = 0$
Hence, the equation of system of coaxial circles will be ${x^2} + {y^2} - 18 + k\left[ {\sqrt 2 x + 4y - 18} \right] = 0$
Note: Whenever we ask such types of questions, we have to remember that a coaxial system of circles is defined by the radical axis and any one of the circles. if c = 0, then the circles are all tangent to y axis and if c > 0, then the circles do not have common points and also if c < 1, then the circle will cut the radical axis.
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1. ## Binomial trials variance
Hi all,
I am beginning to look at some probability problems, I am completely new to college level probability, so not sure if this is in the right sub forum (maybe it should be in the basic?- guess mods will decide.)
Given a scenario where there is a true statistic - proportion voting yes in a vote denoted by $\displaystyle p$, we sample $\displaystyle n$ people from the large population
For a binomial distribution for the sample where $\displaystyle x$ denotes the number of yes votes:
$\displaystyle b(x;n,p) = \binom {n}{x} p^{x} (1-p)^{n-x}$
$\displaystyle mean(b_i) = np$ ; $\displaystyle var(b_i) = np(1-p)$ since the binomial distribution is a 'sample' of the true statistic, we need to take into account the standard error:
$\displaystyle var(bi) = {\sigma(p)^2 \over \sqrt{n}}n = \sigma(p)^2 \sqrt{n}$
I have multiplied by $\displaystyle n$ to scale the variance up from the true average.... is this correct?
Does that mean the $\displaystyle z$ stat for the normal distribution approx becomes:
$\displaystyle z = {X- np \over \sqrt{\sigma(p)^2 \sqrt{n}}}$
Many thanks for reading! Any pointers wil be much appreciated.
2. I have simplifed the problem down to what I am contemplating... apoligies for the long winded garbled original post.
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https://wn.com/Congruence_(geometry)
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Latest News for: congruence (geometry)
Edit
SAT preparation simplified: Test Structure, preparation strategy and official learning resources
Hindustan Times 09 Jul 2024
... Geometry and Trigonometry ... Algebra and Geometry ... For geometry, refresh the knowledge of lines, angles, shapes (including their areas and volumes), congruence, similarity, and the Pythagorean Theorem.
Edit
A brief illustrated guide to 'scissors congruence' − an ancient geometric idea that’s still fueling ...
Caledonian Record 09 Aug 2023
The story of scissors congruence demonstrates how classical problems in geometry can find new life in the strange world of abstract modern math ... For example, you can use scissors congruence to compute the area of a pentagon.
Edit
'Scissors congruence,' an ancient geometric idea that's still fueling cutting-edge mathematical research
Phys Dot Org 09 Aug 2023
The story of scissors congruence demonstrates how classical problems in geometry can find new life in the strange world of abstract modern math ... For example, you can use scissors congruence to compute the area of a pentagon.
Edit
How data can empower MPs to serve people better
Indian Express 17 Feb 2023
The example of Kannauj is simply an illustration of the lack of congruence in district and PC populations that is far more extensive. Approximately, only 28 PCs have the same geometry as the districts.
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Bauso: Marvelous math at Montessori
Auburnpub dot com 13 Dec 2022
Our students eagerly dive into the study of fractions and decimals, excited to move beyond to more complex mathematics, like geometry and algebra ... An emphasis on advanced geometry materials allows in-depth presentations of geometric statements.
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http://www.kwiznet.com/p/takeQuiz.php?ChapterID=1380&CurriculumID=4&Num=12.10
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Email us to get an instant 20% discount on highly effective K-12 Math & English kwizNET Programs!
#### Online Quiz (WorksheetABCD)
Questions Per Quiz = 2 4 6 8 10
### Grade 4 - Mathematics12.10 Word Problems
Example: The length of a rectangular field is 13 cm and its total area is 91 cm2. Find its width. Area = l x w 91 cm2 = 13 x w w = 91/13 = 7 cm Answer: The width of the rectangular field is 7 cm Directions: Solve the following questions. Also write at least ten examples of your own.
Q 1: The area of a rectangle is 12 square cm and its length is 3 cm. What is its height?4 square cm4 cm9 cm36 cm Q 2: What is the perimeter whose length is 5 cm and width is 3cm8 cm10 cm16 cm15 cm Q 3: What is the area of a rectangle whose length is 27 feet and width is 24 feetArea = 27 + 24 = 51 sq feetArea = 27 - 24 = 3 sq feetArea = 27 x 24 = 648 sq feet Q 4: Length of a rectangle is 12 cm and its total area is 84 square cm. Find its width. (hint: 12 cm x __ = 84 sq cm)12 cm7 cm8 cm Q 5: What is the area of a rectangle whose length is 5 cm and width is 3 cm15 sq cm16 sq cm8 sq cm10 sq cm Q 6: Area of a square is 16 square inches. Find its length. (hint: ____ X___ = 16 sq in453 Q 7: How many square inches are there in a square foot. (hint: there are 12 inches in 1 foot)12 + 12 = 24 square inches12 - 12 = 0 square inches12 x 12 = 144 square inches Q 8: What is the perimeter of a square with sides of 4 yard8 yards12 yards16 yardsliter Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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http://en.wikipedia.org/wiki/Hasse%e2%80%93Witt_matrix
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# Hasse–Witt matrix
In mathematics, the Hasse–Witt matrix H of a non-singular algebraic curve C over a finite field F is the matrix of the Frobenius mapping (p-th power mapping where F has q elements, q a power of the prime number p) with respect to a basis for the differentials of the first kind. It is a g × g matrix where C has genus g. The rank of the Hasse–Witt matrix is the Hasse or Hasse–Witt invariant.
## Approach to the definition
This definition, as given in the introduction, is natural in classical terms, and is due to Helmut Hasse and Ernst Witt (1936). It provides a solution to the question of the p-rank of the Jacobian variety J of C; the p-rank is bounded by the rank of H, specifically it is the rank of the Frobenius mapping composed with itself g times. It is also a definition that is in principle algorithmic. There has been substantial recent interest in this as of practical application to cryptography, in the case of C a hyperelliptic curve. The curve C is superspecial if H = 0.
That definition needs a couple of caveats, at least. Firstly, there is a convention about Frobenius mappings, and under the modern understanding what is required for H is the transpose of Frobenius (see arithmetic and geometric Frobenius for more discussion). Secondly, the Frobenius mapping is not F-linear; it is linear over the prime field Z/pZ in F. Therefore the matrix can be written down but does not represent a linear mapping in the straightforward sense.
## Cohomology
The interpretation for sheaf cohomology is this: the p-power map acts on
H1(C,OC),
or in other words the first cohomology of C with coefficients in its structure sheaf. This is now called the Cartier–Manin operator (sometimes just Cartier operator), for Pierre Cartier and Yuri Manin. The connection with the Hasse–Witt definition is by means of Serre duality, which for a curve relates that group to
H0(C, ΩC)
where ΩC = Ω1C is the sheaf of Kähler differentials on C.
## Abelian varieties and their p-rank
The p-rank of an abelian variety A over a field K of characteristic p is the integer k for which the kernel A[p] of multiplication by p has pk points. It may take any value from 0 to d, the dimension of A; by contrast for any other prime number l there are l2d points in A[l]. The reason that the p-rank is lower is that multiplication by p on A is an inseparable isogeny: the differential is p which is 0 in K. By looking at the kernel as a group scheme one can get the more complete structure (reference David Mumford Abelian Varieties pp. 146–7); but if for example one looks at reduction mod p of a division equation, the number of solutions must drop.
The rank of the Cartier–Manin operator, or Hasse–Witt matrix, therefore gives an upper bound for the p-rank. The p-rank is the rank of the Frobenius operator composed with itself g times. In the original paper of Hasse and Witt the problem is phrased in terms intrinsic to C, not relying on J. It is there a question of classifying the possible Artin–Schreier extensions of the function field F(C) (the analogue in this case of Kummer theory).
## Case of genus 1
The case of elliptic curves was worked out by Hasse in 1934. Since the genus is 1, the only possibilities for the matrix H are: H is zero, Hasse invariant 0, p-rank 0, the supersingular case; or H non-zero, Hasse invariant 1, p-rank 1, the ordinary case.[1] Here there is a congruence formula saying that H is congruent modulo p to the number N of points on C over F, at least when q = p. Because of Hasse's theorem on elliptic curves, knowing N modulo p determines N for p ≥ 5. This connection with local zeta-functions has been investigated in depth.
For a plane curve defined by a cubic f(X,Y,Z) = 0, the Hasse invariant in zero if and only if the coefficient of (XYZ)p−1 in fp−1 is zero.[1]
1. ^ a b
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Programming Course in C# ¡Free!
Function SumDigits
Proposed exercise
Create a function SumDigits that receives a number and returns any results in the sum of its digits. For example, if the number is 123, the sum would be 6.
Console.Write( SumDigits(123) );
6
Output
Solution
```using System;
public class FSumDigits
{
public static int SumDigits( int n )
{
string number = Convert.ToString(n);
int sum = 0;
for (int i = 0; i < number.Length; i++)
sum += Convert.ToInt32(number.Substring(i,1));
return sum;
}
public static void Main()
{
Console.WriteLine(SumDigits(123));
}
}
```
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https://www.math-only-math.com/using-a-scale.html
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# Using a Scale
We will learn how to measure the length using a scale.
A cloth merchant uses a rod to measure the length of the cloth. This is called a meter rod. A tailor uses a tape to take measurements for stitching a cloth.
A carpenter or mason uses a metal tape to measure the length.
Meter Rod Tape Metal Tape
A ruler is used by students to draw lines and to measure lengths.
Question and Answer on Using a Scale:
1. Match the following:
(i) Metal Tape 1. Student (ii) Metre Rod 2. Tailor (iii) Tape Measure 3. Mason (iv) Ruler 4. Cloth Merchant
Answer:
1. (i) → 3.
(ii) → 4.
(iii) → 2.
(iv) → 1.
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Share this page: What’s this?
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# Independent classes of random variables (Page 3/6)
Page 3 / 6
## The joint normal distribution
A pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ has the joint normal distribution iff the joint density is
${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=\frac{1}{2\pi {\sigma }_{X}{\sigma }_{Y}{\left(1-{\rho }^{2}\right)}^{1/2}}{e}^{-Q\left(t,u\right)/2}$
where
$Q\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=\frac{1}{1-{\rho }^{2}}\left[{\left(\frac{t-{\mu }_{X}}{{\sigma }_{X}}\right)}^{2},-,2,\rho ,\left(\frac{t-{\mu }_{X}}{{\sigma }_{X}}\right),\left(\frac{u-{\mu }_{Y}}{{\sigma }_{Y}}\right),+,{\left(\frac{u-{\mu }_{Y}}{{\sigma }_{Y}}\right)}^{2}\right]$
The marginal densities are obtained with the aid of some algebraic tricks to integrate the joint density. The result is that $X\sim N\left({\mu }_{X},\phantom{\rule{0.166667em}{0ex}}{\sigma }_{X}^{2}\right)$ and $Y\sim N\left({\mu }_{Y},{\sigma }_{Y}^{2}\right)$ . If the parameter ρ is set to zero, the result is
${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)={f}_{X}\left(t\right){f}_{Y}\left(u\right)$
so that the pair is independent iff $\rho =0$ . The details are left as an exercise for the interested reader.
Remark . While it is true that every independent pair of normally distributed random variables is joint normal, not every pair of normally distributed random variables has thejoint normal distribution.
## A normal pair not joint normally distributed
We start with the distribution for a joint normal pair and derive a joint distribution for a normal pair which is not joint normal. The function
$\phi \left(t,u\right)=\phantom{\rule{4pt}{0ex}}\frac{1}{2\pi }exp\left(-,\frac{{t}^{2}}{2},-,\frac{{u}^{2}}{2}\right)$
is the joint normal density for an independent pair ( $\rho =0$ ) of standardized normal random variables. Now define the joint density for a pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ by
${f}_{XY}\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)=2\phi \left(t,\phantom{\rule{0.166667em}{0ex}}u\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{in}\phantom{\rule{4.pt}{0ex}}\text{the}\phantom{\rule{4.pt}{0ex}}\text{first}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{third}\phantom{\rule{4.pt}{0ex}}\text{quadrants,}\phantom{\rule{4.pt}{0ex}}\text{and}\phantom{\rule{4.pt}{0ex}}\text{zero}\phantom{\rule{4.pt}{0ex}}\text{elsewhere}$
Both $X\sim N\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ and $Y\sim N\left(0,\phantom{\rule{0.166667em}{0ex}}1\right)$ . However, they cannot be joint normal, since the joint normal distribution is positive for all $\left(t,\phantom{\rule{0.166667em}{0ex}}u\right)$ .
## Independent classes
Since independence of random variables is independence of the events determined by the random variables, extension to general classes is simple and immediate.
Definition
A class $\left\{{X}_{i}:i\in J\right\}$ of random variables is (stochastically) independent iff the product rule holds for every finite subclass of two or more.
Remark . The index set J in the definition may be finite or infinite.
For a finite class $\left\{{X}_{i}:1\le i\le n\right\}$ , independence is equivalent to the product rule
${F}_{{X}_{1}{X}_{2}\cdots {X}_{n}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)=\prod _{i=1}^{n}{F}_{{X}_{i}}\left({t}_{i}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{all}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)$
Since we may obtain the joint distribution function for any finite subclass by letting the arguments for the others be (i.e., by taking the limits as the appropriate t i increase without bound), the single product rule suffices to account for all finite subclasses.
Absolutely continuous random variables
If a class $\left\{{X}_{i}:i\in J\right\}$ is independent and the individual variables are absolutely continuous (i.e., have densities), then any finite subclass is jointly absolutelycontinuous and the product rule holds for the densities of such subclasses
${f}_{{X}_{i1}{X}_{i2}\cdots {X}_{im}}\left({t}_{i1},\phantom{\rule{0.166667em}{0ex}}{t}_{i2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{im}\right)=\prod _{k=1}^{m}{f}_{{X}_{ik}}\left({t}_{ik}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{4.pt}{0ex}}\text{all}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\left({t}_{1},\phantom{\rule{0.166667em}{0ex}}{t}_{2},\phantom{\rule{0.166667em}{0ex}}\cdots ,\phantom{\rule{0.166667em}{0ex}}{t}_{n}\right)$
Similarly, if each finite subclass is jointly absolutely continuous, then each individual variable is absolutely continuous and the product rule holds for the densities.Frequently we deal with independent classes in which each random variable has the same marginal distribution. Such classes are referred to as iid classes (an acronym for i ndependent, i dentically d istributed). Examples are simple random samplesfrom a given population, or the results of repetitive trials with the same distribution on the outcome of each component trial. A Bernoulli sequence is a simple example.
## Simple random variables
Consider a pair $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y\right\}$ of simple random variables in canonical form
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive
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# Collision detection: Transformation
## Introduction
stay Collision Detection : Triangle The collision detection of triangles is considered from another idea. So far, the introduction is static detection, and then let's take a look at dynamic collision detection.
The following examples are not checked for compatibility. It is recommended to view them in the latest Chrome browser.
## Transformation
This is Sample page.
Based on canvas translate, rotate, scale The animation formed by three transformations to see how to carry out dynamic collision detection.
The animation principle based on canvas is to redraw every other period of time, so the static collision detection is actually carried out at a specific time, so the methods previously introduced are also applicable, which are used uniformly here Polygon/Polygon Methods in. With the detection method, the next step is to obtain the dynamically changing coordinates of the relevant points in the screen. The following describes the situation.
### translate
When painting on canvas, the positioning is based on the coordinate system. The upper left corner of the canvas is the origin of the coordinate system, the horizontal right is the positive direction of X axis, and the vertical down is the positive direction of Y axis. Draw a rectangle rect(20, 20, 40, 40) in the coordinate system as follows:
If you want to move 60 pixels horizontally to the right and 80 pixels vertically downward, you can add the coordinates directly: rect(20 + 60, 20 + 80, 40, 40).
But there is another, more interesting way: to move the entire axis. If you move the entire coordinate axis horizontally to the right by 60 pixels and vertically downward by 80 pixels, it is exactly the same visually. The translate method uses this method.
It can be seen from the above figure that this method does not consider the coordinate change of the rectangle, and it will be much more convenient when dealing with complex graphics.
It should be noted that after translating, you need to reset the coordinate axis, because there may be other graphics, and the original coordinate axis is still used as a reference. Reset axis usage setTransform method:
```var canvas = document.getElementById("canvas");
var ctx = canvas.getContext("2d");
ctx.translate(50, 50);
ctx.fillRect(0,0,100,100);
// Reset
ctx.setTransform(1, 0, 0, 1, 0, 0);
// Other treatment
```
For the changed coordinates, add and subtract the translated pixels directly.
```/**
* Suppose point A(x,y) reaches B(m,n) after translating (x1, Y1)
*/
const m = x + x1;
const n = y + y1;```
### rotate
The rotate method is similar to the translate method by rotating the coordinate axis.
For the changed coordinates, some calculations are needed.
```/**
*
* The center coordinate O (0, 0), assuming that point A(x,y) forms an angle with the X axis of α
* Clockwise rotation angle β After reaching point B(m,n), let's deduce the coordinates of point B
*
* A Distance to the center of the circle: dist1 = | OA | = Y / sin( α)= x/cos( α)
* B Distance to the center of the circle: dist2 = | ob | = n / sin( α-β)= m/cos( α-β)
*
* Only rotation, so dist1 = dist2, the radius of construction rotation is r:
* r = y/sin(α)=x/cos(α)=n/sin(α-β)=m/cons(α-β)
* y = r * sin(α) x = r * cos(α)
*
* According to the trigonometric function formula:
* sin(α+β)=sin(α)cos(β)+cos(α)sin(β)
* sin(α-β)=sin(α)cos(β)-cos(α)sin(β)
* cos(α+β)=cos(α)cos(β)-sin(α)sin(β)
* cos(α-β)=cos(α)cos(β)+sin(α)sin(β)
*
* Substitute the following formula:
* m = r*cos(α-β) = r * cos(α)cos(β) + r * sin(α)sin(β) = x * cos(β) + y * sin(β)
* n = r*sin(α-β) = r * sin(α)cos(β) - r * cos(α)sin(β) = y * cos(β) - x * sin(β)
*
* Counterclockwise is the opposite:
* m = x * cos(β) - y * sin(β)
* n = y * cos(β) + x * sin(β)
*
*/```
### scale
The scale method is similar to the translate method, which is realized by scaling the coordinate axis.
For the changed coordinates, directly multiply by the corresponding scaling multiple.
```/**
* Suppose point A(x,y) reaches B(m,n) after scale(num1,num2)
*/
const m = x * num1;
const n = y * num2;```
## Transformation Order
When several different transformations are carried out continuously, the order is different, and the results may be different. This is Examples.
This is because the continuous transformation is based on the state after the last transformation. When calculating, we need to consider many aspects. be based on transform It will be more convenient to calculate the parameter format in. The effects of translate, rotate and scale can be converted into the form of transform.
```/**
* canvas.transform(sx, ry, rx, sy, tx, ty)
* sx-Horizontal zoom, ry vertical tilt, rx horizontal tilt, sy vertical zoom, tx horizontal move, ty vertical move
*
*/
function Transform() {
this.reset();
}
Transform.prototype.reset = function() {
this.transformData = [1,0,0,1,0,0];
};
Transform.prototype.translate = function(x, y) {
let [sx,ry,rx,sy,tx,ty] = this.transformData;
const newTX = sx * x + rx * y;
const newTY = ry * x + sy * y;
this.transformData = [sx,ry,rx,sy,newTX,newTY];
};
Transform.prototype.rotate = function(angle) {
let c = Math.cos(angle);
let s = Math.sin(angle);
let [sx,ry,rx,sy,tx,ty] = this.transformData;
let newSX = sx * c + rx * s;
let newRY = ry * c + sy * s;
let newRX = sx * -s + rx * c;
let newSY = ry * -s + sy * c;
this.transformData = [newSX,newRY,newRX,newSY,tx,ty];
};
Transform.prototype.scale = function(x, y) {
let [sx,ry,rx,sy,tx,ty] = this.transformData;
let newSX = sx * x;
let newRY = ry * x;
let newRX = rx * y;
let newSY = sy * y;
this.transformData = [newSX,newRY,newRX,newSY,tx,ty];
};
Transform.prototype.getCoordinate = function(x, y) {
let [sx,ry,rx,sy,tx,ty] = this.transformData;
const px = x * sx + y*rx + tx;
const py = x * ry + y*sy + ty;
return [px,py];
};```
## reference material
Tags: Javascript canvas
Posted by JankaZ on Fri, 13 May 2022 16:33:57 +0300
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| 2.640625
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CC-MAIN-2022-21
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en
| 0.89767
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https://math.stackexchange.com/questions/3849213/can-we-count-all-the-possible-graphs-on-n-vertices-with-local-degrees-less-than
| 1,627,305,078,000,000,000
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| 372,853,492
| 39,010
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# Can we count all the possible graphs on n vertices with local degrees less than 2 using inclusion/exclusion principle?
I'd like to know if Inclusion/Exclusion could be useful to count all the possible graphs that can be drawn with n vertices and no vertex having 2 or more lines attached, and how to apply it.
Would it be too naive to expect a nice-looking formula out of this?. Maybe... we need Polya counting?, maybe another theorem to solve it?, something involving complements to simplify the question?. I'd appreciate to see your thoughts on this, guys.
EDIT: I'm considering graphs with general types of symmetries (not just $$S_n$$).
• These graphs are exactly disjoint unions of either paths or cycles, so the count is not too hard, depending on whether you want the unlabeled or labeled count. – Qiaochu Yuan Oct 2 '20 at 20:20
• Qiaochu. Some vertices are interchangeable (unlabelled) and others not (they're labelled), that's why I mentioned Polya counting because it allows me to specify the group of symmetries. This is my first approach to graphs, actually, and it's mainly becaise of a hobby, so... yeah, I'd appreciate if you could ellaborate on your previous comment abot more, please (?). – JuanC97 Oct 2 '20 at 20:25
• Oh, sorry, I misread your question; I thought you were asking for degree $\le 2$, not degree $<2$. The degree $\le 2$ case is much more interesting! – Qiaochu Yuan Oct 2 '20 at 20:52
• There are more graphs but it's still possible to classify them. I posted this as a separate question here, which I'll answer in a day or so: math.stackexchange.com/questions/3849265/… – Qiaochu Yuan Oct 2 '20 at 21:11
If no vertex has $$2$$ or more lines attached, then every vertex has at most one line. Then every vertex is either isolated, having no edges, or it is in a pair. If the vertices are unlabeled then this comes down to writing $$n$$ as a sum of $$1$$'s and $$2$$'s, so there are $$\lfloor\tfrac n2\rfloor+1$$ such graphs on $$n$$ vertices.
• These two answers account for the $S_n$ and $S_1^n$ cases but what about general symmetries?. Polya counting allows one to count graphs on $n$ vertices with $k$ lines and arbitray symmetries, isn't it there something useful to count k-edge matchings?, not even Inclusion/exclusion works? – JuanC97 Oct 2 '20 at 23:02
| 602
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{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 7, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.375
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CC-MAIN-2021-31
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latest
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en
| 0.941974
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https://isabelle.in.tum.de/repos/isabelle/rev/0f8ebabf3163
| 1,638,038,693,000,000,000
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text/html
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crawl-data/CC-MAIN-2021-49/segments/1637964358208.31/warc/CC-MAIN-20211127163427-20211127193427-00525.warc.gz
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| 7,418
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author wenzelm Mon, 01 Aug 2005 19:20:28 +0200 changeset 16974 0f8ebabf3163 parent 16973 b2a894562b8f child 16975 34ed8d5d4f16
more zcong_sym;
```--- a/src/HOL/NumberTheory/Euler.thy Mon Aug 01 19:20:26 2005 +0200
+++ b/src/HOL/NumberTheory/Euler.thy Mon Aug 01 19:20:28 2005 +0200
@@ -5,13 +5,13 @@
-theory Euler imports Residues EvenOdd begin;
+theory Euler imports Residues EvenOdd begin
constdefs
MultInvPair :: "int => int => int => int set"
"MultInvPair a p j == {StandardRes p j, StandardRes p (a * (MultInv p j))}"
SetS :: "int => int => int set set"
- "SetS a p == ((MultInvPair a p) ` (SRStar p))";
+ "SetS a p == ((MultInvPair a p) ` (SRStar p))"
(****************************************************************)
(* *)
@@ -22,25 +22,25 @@
lemma MultInvPair_prop1a: "[| zprime p; 2 < p; ~([a = 0](mod p));
X \<in> (SetS a p); Y \<in> (SetS a p);
~((X \<inter> Y) = {}) |] ==>
- X = Y";
+ X = Y"
- apply (drule StandardRes_SRStar_prop1a)+; defer 1;
- apply (drule StandardRes_SRStar_prop1a)+;
+ apply (drule StandardRes_SRStar_prop1a)+ defer 1
+ apply (drule StandardRes_SRStar_prop1a)+
apply (auto simp add: MultInvPair_def StandardRes_prop2 zcong_sym)
apply (drule notE, rule MultInv_zcong_prop1, auto)
- apply (drule notE, rule MultInv_zcong_prop2, auto)
- apply (drule MultInv_zcong_prop2, auto)
+ apply (drule notE, rule MultInv_zcong_prop2, auto simp add: zcong_sym)
+ apply (drule MultInv_zcong_prop2, auto simp add: zcong_sym)
apply (drule MultInv_zcong_prop3, auto simp add: zcong_sym)
apply (drule MultInv_zcong_prop1, auto)
- apply (drule MultInv_zcong_prop2, auto)
- apply (drule MultInv_zcong_prop2, auto)
+ apply (drule MultInv_zcong_prop2, auto simp add: zcong_sym)
+ apply (drule MultInv_zcong_prop2, auto simp add: zcong_sym)
apply (drule MultInv_zcong_prop3, auto simp add: zcong_sym)
done
lemma MultInvPair_prop1b: "[| zprime p; 2 < p; ~([a = 0](mod p));
X \<in> (SetS a p); Y \<in> (SetS a p);
X \<noteq> Y |] ==>
- X \<inter> Y = {}";
+ X \<inter> Y = {}"
apply (rule notnotD)
apply (rule notI)
apply (drule MultInvPair_prop1a, auto)
@@ -51,7 +51,7 @@
lemma MultInvPair_prop2: "[| zprime p; 2 < p; ~([a = 0](mod p)) |] ==>
- Union ( SetS a p) = SRStar p";
+ Union ( SetS a p) = SRStar p"
apply (auto simp add: SetS_def MultInvPair_def StandardRes_SRStar_prop4
SRStar_mult_prop2)
apply (frule StandardRes_SRStar_prop3)
@@ -61,37 +61,37 @@
lemma MultInvPair_distinct: "[| zprime p; 2 < p; ~([a = 0] (mod p));
~([j = 0] (mod p));
- ~([j = a * MultInv p j] (mod p))";
+ ~([j = a * MultInv p j] (mod p))"
apply auto
-proof -;
+proof -
assume "zprime p" and "2 < p" and "~([a = 0] (mod p))" and
- "~([j = 0] (mod p))" and "~(QuadRes p a)";
- assume "[j = a * MultInv p j] (mod p)";
- then have "[j * j = (a * MultInv p j) * j] (mod p)";
+ "~([j = 0] (mod p))" and "~(QuadRes p a)"
+ assume "[j = a * MultInv p j] (mod p)"
+ then have "[j * j = (a * MultInv p j) * j] (mod p)"
- then have a:"[j * j = a * (MultInv p j * j)] (mod p)";
+ then have a:"[j * j = a * (MultInv p j * j)] (mod p)"
- have "[j * j = a] (mod p)";
- proof -;
- from prems have b: "[MultInv p j * j = 1] (mod p)";
+ have "[j * j = a] (mod p)"
+ proof -
+ from prems have b: "[MultInv p j * j = 1] (mod p)"
- from b a show ?thesis;
+ from b a show ?thesis
- qed;
- then have "[j^2 = a] (mod p)";
- apply(subgoal_tac "2 = Suc(Suc(0))");
+ qed
+ then have "[j^2 = a] (mod p)"
+ apply(subgoal_tac "2 = Suc(Suc(0))")
apply (erule ssubst)
apply (auto simp only: power_Suc power_0)
by auto
- with prems show False;
+ with prems show False
-qed;
+qed
lemma MultInvPair_card_two: "[| zprime p; 2 < p; ~([a = 0] (mod p));
~(QuadRes p a); ~([j = 0] (mod p)) |] ==>
- card (MultInvPair a p j) = 2";
+ card (MultInvPair a p j) = 2"
- apply (subgoal_tac "~ (StandardRes p j = StandardRes p (a * MultInv p j))");
+ apply (subgoal_tac "~ (StandardRes p j = StandardRes p (a * MultInv p j))")
apply auto
apply (simp only: StandardRes_prop2)
apply (drule MultInvPair_distinct)
@@ -103,77 +103,77 @@
(* *)
(****************************************************************)
-lemma SetS_finite: "2 < p ==> finite (SetS a p)";
+lemma SetS_finite: "2 < p ==> finite (SetS a p)"
by (auto simp add: SetS_def SRStar_finite [of p] finite_imageI)
-lemma SetS_elems_finite: "\<forall>X \<in> SetS a p. finite X";
+lemma SetS_elems_finite: "\<forall>X \<in> SetS a p. finite X"
by (auto simp add: SetS_def MultInvPair_def)
lemma SetS_elems_card: "[| zprime p; 2 < p; ~([a = 0] (mod p));
- \<forall>X \<in> SetS a p. card X = 2";
+ \<forall>X \<in> SetS a p. card X = 2"
apply (frule StandardRes_SRStar_prop1a)
apply (rule MultInvPair_card_two, auto)
done
-lemma Union_SetS_finite: "2 < p ==> finite (Union (SetS a p))";
+lemma Union_SetS_finite: "2 < p ==> finite (Union (SetS a p))"
by (auto simp add: SetS_finite SetS_elems_finite finite_Union)
lemma card_setsum_aux: "[| finite S; \<forall>X \<in> S. finite (X::int set);
- \<forall>X \<in> S. card X = n |] ==> setsum card S = setsum (%x. n) S";
+ \<forall>X \<in> S. card X = n |] ==> setsum card S = setsum (%x. n) S"
by (induct set: Finites, auto)
lemma SetS_card: "[| zprime p; 2 < p; ~([a = 0] (mod p)); ~(QuadRes p a) |] ==>
- int(card(SetS a p)) = (p - 1) div 2";
-proof -;
- assume "zprime p" and "2 < p" and "~([a = 0] (mod p))" and "~(QuadRes p a)";
- then have "(p - 1) = 2 * int(card(SetS a p))";
- proof -;
- have "p - 1 = int(card(Union (SetS a p)))";
+ int(card(SetS a p)) = (p - 1) div 2"
+proof -
+ assume "zprime p" and "2 < p" and "~([a = 0] (mod p))" and "~(QuadRes p a)"
+ then have "(p - 1) = 2 * int(card(SetS a p))"
+ proof -
+ have "p - 1 = int(card(Union (SetS a p)))"
by (auto simp add: prems MultInvPair_prop2 SRStar_card)
- also have "... = int (setsum card (SetS a p))";
+ also have "... = int (setsum card (SetS a p))"
by (auto simp add: prems SetS_finite SetS_elems_finite
MultInvPair_prop1c [of p a] card_Union_disjoint)
- also have "... = int(setsum (%x.2) (SetS a p))";
+ also have "... = int(setsum (%x.2) (SetS a p))"
apply (insert prems)
apply (auto simp add: SetS_elems_card SetS_finite SetS_elems_finite
card_setsum_aux simp del: setsum_constant)
done
- also have "... = 2 * int(card( SetS a p))";
+ also have "... = 2 * int(card( SetS a p))"
by (auto simp add: prems SetS_finite setsum_const2)
- finally show ?thesis .;
- qed;
- from this show ?thesis;
+ finally show ?thesis .
+ qed
+ from this show ?thesis
by auto
-qed;
+qed
lemma SetS_setprod_prop: "[| zprime p; 2 < p; ~([a = 0] (mod p));
~(QuadRes p a); x \<in> (SetS a p) |] ==>
- [\<Prod>x = a] (mod p)";
+ [\<Prod>x = a] (mod p)"
apply (auto simp add: SetS_def MultInvPair_def)
apply (frule StandardRes_SRStar_prop1a)
- apply (subgoal_tac "StandardRes p x \<noteq> StandardRes p (a * MultInv p x)");
+ apply (subgoal_tac "StandardRes p x \<noteq> StandardRes p (a * MultInv p x)")
apply (auto simp add: StandardRes_prop2 MultInvPair_distinct)
apply (frule_tac m = p and x = x and y = "(a * MultInv p x)" in
- StandardRes_prop4);
- apply (subgoal_tac "[x * (a * MultInv p x) = a * (x * MultInv p x)] (mod p)");
+ StandardRes_prop4)
+ apply (subgoal_tac "[x * (a * MultInv p x) = a * (x * MultInv p x)] (mod p)")
apply (drule_tac a = "StandardRes p x * StandardRes p (a * MultInv p x)" and
b = "x * (a * MultInv p x)" and
- c = "a * (x * MultInv p x)" in zcong_trans, force);
+ c = "a * (x * MultInv p x)" in zcong_trans, force)
apply (frule_tac p = p and x = x in MultInv_prop2, auto)
apply (drule_tac a = "x * MultInv p x" and b = 1 in zcong_zmult_prop2)
done
-lemma aux1: "[| 0 < x; (x::int) < a; x \<noteq> (a - 1) |] ==> x < a - 1";
+lemma aux1: "[| 0 < x; (x::int) < a; x \<noteq> (a - 1) |] ==> x < a - 1"
by arith
-lemma aux2: "[| (a::int) < c; b < c |] ==> (a \<le> b | b \<le> a)";
+lemma aux2: "[| (a::int) < c; b < c |] ==> (a \<le> b | b \<le> a)"
by auto
lemma SRStar_d22set_prop [rule_format]: "2 < p --> (SRStar p) = {1} \<union>
- (d22set (p - 1))";
+ (d22set (p - 1))"
apply (induct p rule: d22set.induct, auto)
apply (simp add: SRStar_def d22set.simps, clarify)
@@ -209,9 +209,9 @@
qed
lemma Union_SetS_setprod_prop2: "[| zprime p; 2 < p; ~([a = 0](mod p)) |] ==>
- \<Prod>(Union (SetS a p)) = zfact (p - 1)";
-proof -;
- assume "zprime p" and "2 < p" and "~([a = 0](mod p))";
+ \<Prod>(Union (SetS a p)) = zfact (p - 1)"
+proof -
+ assume "zprime p" and "2 < p" and "~([a = 0](mod p))"
then have "\<Prod>(Union (SetS a p)) = \<Prod>(SRStar p)"
also have "... = \<Prod>({1} \<union> (d22set (p - 1)))"
@@ -223,16 +223,16 @@
apply (drule d22set_g_1)
done
- then have "\<Prod>({1} \<union> (d22set (p - 1))) = \<Prod>(d22set (p - 1))";
+ then have "\<Prod>({1} \<union> (d22set (p - 1))) = \<Prod>(d22set (p - 1))"
by auto
then show ?thesis
- qed;
+ qed
finally show ?thesis .
-qed;
+qed
lemma zfact_prop: "[| zprime p; 2 < p; ~([a = 0] (mod p)); ~(QuadRes p a) |] ==>
- [zfact (p - 1) = a ^ nat ((p - 1) div 2)] (mod p)";
+ [zfact (p - 1) = a ^ nat ((p - 1) div 2)] (mod p)"
apply (frule Union_SetS_setprod_prop1)
done
@@ -247,7 +247,7 @@
lemma Euler_part1: "[| 2 < p; zprime p; ~([x = 0](mod p));
- [x^(nat (((p) - 1) div 2)) = -1](mod p)";
+ [x^(nat (((p) - 1) div 2)) = -1](mod p)"
apply (frule zfact_prop, auto)
apply (frule Wilson_Russ)
@@ -261,35 +261,35 @@
(* *)
(********************************************************************)
-lemma aux_1: "0 < p ==> (a::int) ^ nat (p) = a * a ^ (nat (p) - 1)";
-proof -;
- assume "0 < p";
- then have "a ^ (nat p) = a ^ (1 + (nat p - 1))";
+lemma aux_1: "0 < p ==> (a::int) ^ nat (p) = a * a ^ (nat (p) - 1)"
+proof -
+ assume "0 < p"
+ then have "a ^ (nat p) = a ^ (1 + (nat p - 1))"
- also have "... = (a ^ 1) * a ^ (nat(p) - 1)";
+ also have "... = (a ^ 1) * a ^ (nat(p) - 1)"
- also have "... = a * a ^ (nat(p) - 1)";
+ also have "... = a * a ^ (nat(p) - 1)"
by auto
- finally show ?thesis .;
-qed;
+ finally show ?thesis .
+qed
-lemma aux_2: "[| (2::int) < p; p \<in> zOdd |] ==> 0 < ((p - 1) div 2)";
-proof -;
- assume "2 < p" and "p \<in> zOdd";
- then have "(p - 1):zEven";
+lemma aux_2: "[| (2::int) < p; p \<in> zOdd |] ==> 0 < ((p - 1) div 2)"
+proof -
+ assume "2 < p" and "p \<in> zOdd"
+ then have "(p - 1):zEven"
by (auto simp add: zEven_def zOdd_def)
- then have aux_1: "2 * ((p - 1) div 2) = (p - 1)";
+ then have aux_1: "2 * ((p - 1) div 2) = (p - 1)"
then have "1 < (p - 1)"
by auto
- then have " 1 < (2 * ((p - 1) div 2))";
+ then have " 1 < (2 * ((p - 1) div 2))"
- then have "0 < (2 * ((p - 1) div 2)) div 2";
+ then have "0 < (2 * ((p - 1) div 2)) div 2"
by auto
then show ?thesis by auto
-qed;
+qed
-lemma Euler_part2: "[| 2 < p; zprime p; [a = 0] (mod p) |] ==> [0 = a ^ nat ((p - 1) div 2)] (mod p)";
+lemma Euler_part2: "[| 2 < p; zprime p; [a = 0] (mod p) |] ==> [0 = a ^ nat ((p - 1) div 2)] (mod p)"
apply (frule zprime_zOdd_eq_grt_2)
apply (frule aux_2, auto)
apply (frule_tac a = a in aux_1, auto)
@@ -304,25 +304,25 @@
(* *)
(****************************************************************)
-lemma aux__1: "[| ~([x = 0] (mod p)); [y ^ 2 = x] (mod p)|] ==> ~(p dvd y)";
+lemma aux__1: "[| ~([x = 0] (mod p)); [y ^ 2 = x] (mod p)|] ==> ~(p dvd y)"
apply (subgoal_tac "[| ~([x = 0] (mod p)); [y ^ 2 = x] (mod p)|] ==>
- ~([y ^ 2 = 0] (mod p))");
+ ~([y ^ 2 = 0] (mod p))")
apply (auto simp add: zcong_sym [of "y^2" x p] intro: zcong_trans)
apply (auto simp add: zcong_eq_zdvd_prop intro: zpower_zdvd_prop1)
done
-lemma aux__2: "2 * nat((p - 1) div 2) = nat (2 * ((p - 1) div 2))";
+lemma aux__2: "2 * nat((p - 1) div 2) = nat (2 * ((p - 1) div 2))"
lemma Euler_part3: "[| 2 < p; zprime p; ~([x = 0](mod p)); QuadRes p x |] ==>
- [x^(nat (((p) - 1) div 2)) = 1](mod p)";
+ [x^(nat (((p) - 1) div 2)) = 1](mod p)"
apply (subgoal_tac "p \<in> zOdd")
apply (frule aux__1, auto)
- apply (drule_tac z = "nat ((p - 1) div 2)" in zcong_zpower);
+ apply (drule_tac z = "nat ((p - 1) div 2)" in zcong_zpower)
apply (rule zcong_trans)
- apply (auto simp add: zcong_sym [of "x ^ nat ((p - 1) div 2)"]);
+ apply (auto simp add: zcong_sym [of "x ^ nat ((p - 1) div 2)"])
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# A sample of CO2 occupies 225 ml at 373 K. At what temperature (in K) would it occupy 300 ml if the pressure did not change?
Anurag Kishore
37 Points
14 years ago
Hi, Cane
use the ideal gas equation
V / T = constant
225 / 373 = 300 / x
=> x = 497.33 ml
Thanks
Anurag
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Jump to content
# Atomic units
The atomic units are a system of natural units of measurement that is especially convenient for calculations in atomic physics and related scientific fields, such as computational chemistry and atomic spectroscopy. They were originally suggested and named by the physicist Douglas Hartree.[1] Atomic units are often abbreviated "a.u." or "au", not to be confused with similar abbreviations used for astronomical units, arbitrary units, and absorbance units in other contexts.
## Motivation
In the context of atomic physics, using the atomic units system can be a convenient shortcut, eliminating unnecessary symbols and numbers with very small orders of magnitude. For example, the Hamiltonian operator in the Schrödinger equation for the helium atom with standard quantities, such as when using SI units, is[2]
${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2m_{\text{e}}}}\nabla _{1}^{2}-{\frac {\hbar ^{2}}{2m_{\text{e}}}}\nabla _{2}^{2}-{\frac {2e^{2}}{4\pi \epsilon _{0}r_{1}}}-{\frac {2e^{2}}{4\pi \epsilon _{0}r_{2}}}+{\frac {e^{2}}{4\pi \epsilon _{0}r_{12}}},}$
but adopting the convention associated with atomic units that transforms quantities into dimensionless equivalents, it becomes
${\displaystyle {\hat {H}}=-{\frac {1}{2}}\nabla _{1}^{2}-{\frac {1}{2}}\nabla _{2}^{2}-{\frac {2}{r_{1}}}-{\frac {2}{r_{2}}}+{\frac {1}{r_{12}}}.}$
In this convention, the constants ${\displaystyle \hbar }$, ${\displaystyle m_{\text{e}}}$, ${\displaystyle 4\pi \epsilon _{0}}$, and ${\displaystyle e}$ all correspond to the value ${\displaystyle 1}$ (see § Definition below). The distances relevant to the physics expressed in SI units are naturally on the order of ${\displaystyle 10^{-10}\,\mathrm {m} }$, while expressed in atomic units distances are on the order of ${\displaystyle 1a_{0}}$ (one Bohr radius, the atomic unit of length). An additional benefit of expressing quantities using atomic units is that their values calculated and reported in atomic units do not change when values of fundamental constants are revised. The fundamental constants are built into the conversion factors between atomic units and SI.
## History
Hartree defined units based on three physical constants:[1]: 91
Both in order to eliminate various universal constants from the equations and also to avoid high powers of 10 in numerical work, it is convenient to express quantities in terms of units, which may be called 'atomic units', defined as follows:
Unit of length, ${\displaystyle a_{\text{H}}=h^{2}\,/\,4\pi ^{2}me^{2}}$, on the orbital mechanics the radius of the 1-quantum circular orbit of the H-atom with fixed nucleus.
Unit of charge, ${\displaystyle e}$, the magnitude of the charge on the electron.
Unit of mass, ${\displaystyle m}$, the mass of the electron.
Consistent with these are:
Unit of action, ${\displaystyle h\,/\,2\pi }$.
Unit of energy, ${\displaystyle e^{2}/a_{\text{H}}=2hcR=}$ [...]
Unit of time, ${\displaystyle 1\,/\,4\pi cR}$.
— D.R. Hartree, The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part I. Theory and Methods
Here, the modern equivalent of ${\displaystyle R}$ is the Rydberg constant ${\displaystyle R_{\infty }}$, of ${\displaystyle m}$ is the electron mass ${\displaystyle m_{\text{e}}}$, of ${\displaystyle a_{\text{H}}}$ is the Bohr radius ${\displaystyle a_{0}}$, and of ${\displaystyle h/2\pi }$ is the reduced Planck constant ${\displaystyle \hbar }$. Hartree's expressions that contain ${\displaystyle e}$ differ from the modern form due to a change in the definition of ${\displaystyle e}$, as explained below.
In 1957, Bethe and Salpeter's book Quantum mechanics of one-and two-electron atoms[3] built on Hartree's units, which they called atomic units abbreviated "a.u.". They chose to use ${\displaystyle \hbar }$, their unit of action and angular momentum in place of Hartree's length as the base units. They noted that the unit of length in this system is the radius of the first Bohr orbit and their velocity is the electron velocity in Bohr's model of the first orbit.
In 1959, Shull and Hall[4] advocated atomic units based on Hartree's model but again chose to use ${\displaystyle \hbar }$ as the defining unit. They explicitly named the distance unit a "Bohr radius"; in addition, they wrote the unit of energy as ${\displaystyle H=me^{4}/\hbar ^{2}}$ and called it a Hartree. These terms came to be used widely in quantum chemistry.[5]: 349
In 1973 McWeeny extended the system of Shull and Hall by adding permittivity in the form of ${\displaystyle \kappa _{0}=4\pi \epsilon _{0}}$ as a defining or base unit.[6][7] Simultaneously he adopted the SI definition of ${\displaystyle e}$ so that his expression for energy in atomic units is ${\displaystyle e^{2}/(4\pi \epsilon _{0}a_{0})}$, matching the expression in the 8th SI brochure.[8]
## Definition
A set of base units in the atomic system as in one proposal are the electron rest mass, the magnitude of the electronic charge, the Planck constant, and the permittivity.[6][9] In the atomic units system, each of these takes the value 1; the corresponding values in the International System of Units[10]: 132 are given in the table.
Base atomic units[*]
Symbol and Name Quantity (dimensions)[†] Atomic
units[‡]
SI units
${\displaystyle \hbar }$, reduced Planck constant action (ML2T−1) 1 1.054571817...×10−34 J⋅s [11]
${\displaystyle e}$, elementary charge charge (Q) 1 1.602176634×10−19 C [12]
${\displaystyle m_{\text{e}}}$, electron rest mass mass (M) 1 9.1093837139(28)×10−31 kg [13]
${\displaystyle 4\pi \epsilon _{0}}$, permittivity permittivity (Q2W−1L−1) 1 1.11265005620(17)×10−10 F⋅m−1 [14]
### Table notes
• ^ *: This arbitrary choice of base units was proposed by McWeeny.
• ^ †: See Dimensional analysis. W represents the dimension of energy, ML2T−2.[6]
• ^ ‡: In the 'atomic units' column, the convention that uses dimensionless equivalents has been applied.
## Units
Three of the defining constants (reduced Planck constant, elementary charge, and electron rest mass) are atomic units themselves – of action,[15] electric charge,[16] and mass,[17] respectively. Two named units are those of length (Bohr radius ${\displaystyle a_{0}\equiv 4\pi \epsilon _{0}\hbar ^{2}/m_{\text{e}}e^{2}}$) and energy (hartree ${\displaystyle E_{\text{h}}\equiv \hbar ^{2}/m_{\text{e}}a_{0}^{2}}$).
Defined atomic units
Atomic unit of Expression Value in SI units Other equivalents
electric charge density ${\displaystyle e/a_{0}^{3}}$ 1.08120238677(51)×1012 C⋅m−3 [18]
electric current ${\displaystyle eE_{\text{h}}/\hbar }$ 6.6236182375082(72)×10−3 A [19]
electric charge ${\displaystyle e}$ 1.602176634×10−19 C [20]
electric dipole moment ${\displaystyle ea_{0}}$ 8.4783536198(13)×10−30 C⋅m [21] 2.541746473 D
electric quadrupole moment ${\displaystyle ea_{0}^{2}}$ 4.4865515185(14)×10−40 C⋅m2 [22]
electric potential ${\displaystyle E_{\text{h}}/e}$ 27.211386245981(30) V [23]
electric field ${\displaystyle E_{\text{h}}/ea_{0}}$ 5.14220675112(80)×1011 V⋅m−1 [24]
electric field gradient ${\displaystyle E_{\text{h}}/ea_{0}^{2}}$ 9.7173624424(30)×1021 V⋅m−2 [25]
permittivity ${\displaystyle e^{2}/a_{0}E_{\text{h}}}$ 1.11265005620(17)×10−10 F⋅m−1 [14] ${\displaystyle 4\pi \epsilon _{0}}$
electric polarizability ${\displaystyle e^{2}a_{0}^{2}/E_{\text{h}}}$ 1.64877727212(51)×10−41 C2⋅m2⋅J−1 [26]
1st hyperpolarizability ${\displaystyle e^{3}a_{0}^{3}/E_{\text{h}}^{2}}$ 3.2063612996(15)×10−53 C3⋅m3⋅J−2 [27]
2nd hyperpolarizability ${\displaystyle e^{4}a_{0}^{4}/E_{\text{h}}^{3}}$ 6.2353799735(39)×10−65 C4⋅m4⋅J−3 [28]
magnetic dipole moment ${\displaystyle \hbar e/m_{\text{e}}}$ 1.85480201315(58)×10−23 J⋅T−1 [29] ${\displaystyle 2\mu _{\text{B}}}$
magnetic flux density ${\displaystyle \hbar /ea_{0}^{2}}$ 2.35051757077(73)×105 T [30] 2.3505×109 G
magnetizability ${\displaystyle e^{2}a_{0}^{2}/m_{\text{e}}}$ 7.8910365794(49)×10−29 J⋅T−2 [31]
action ${\displaystyle \hbar }$ 1.054571817...×10−34 J⋅s [32]
energy ${\displaystyle E_{\text{h}}}$ 4.3597447222060(48)×10−18 J [33] ${\displaystyle 2hcR_{\infty }}$, ${\displaystyle \alpha ^{2}m_{\text{e}}c^{2}}$, 27.211386245988(53) eV [34]
force ${\displaystyle E_{\text{h}}/a_{0}}$ 8.2387235038(13)×10−8 N [35] 82.387 nN, 51.421 eV·Å−1
length ${\displaystyle a_{0}}$ 5.29177210544(82)×10−11 m [36] ${\displaystyle \hbar /\alpha m_{\text{e}}c}$, 0.529177 Å
mass ${\displaystyle m_{\text{e}}}$ 9.1093837139(28)×10−31 kg [37]
momentum ${\displaystyle \hbar /a_{0}}$ 1.99285191545(31)×10−24 kg⋅m⋅s−1 [38]
time ${\displaystyle \hbar /E_{\text{h}}}$ 2.4188843265864(26)×10−17 s [39]
velocity ${\displaystyle a_{0}E_{\text{h}}/\hbar }$ 2.18769126216(34)×106 m⋅s−1 [40] ${\displaystyle \alpha c}$
${\displaystyle c}$speed of light, ${\displaystyle \epsilon _{0}}$vacuum permittivity, ${\displaystyle R_{\infty }}$Rydberg constant, ${\displaystyle h}$: Planck constant, ${\displaystyle \alpha }$fine-structure constant, ${\displaystyle \mu _{\text{B}}}$Bohr magneton, correspondence
## Conventions
Different conventions are adopted in the use of atomic units, which vary in presentation, formality and convenience.
### Explicit units
• Many texts (e.g. Jerrard & McNiell,[7] Shull & Hall[4]) define the atomic units as quantities, without a transformation of the equations in use. As such, they do not suggest treating either quantities as dimensionless or changing the form of any equations. This is consistent with expressing quantities in terms of dimensional quantities, where the atomic unit is included explicitly as a symbol (e.g. ${\displaystyle m=3.4~m_{\text{e}}}$, ${\displaystyle m=3.4~{\text{a.u. of mass}}}$, or more ambiguously, ${\displaystyle m=3.4~{\text{a.u.}}}$), and keeping equations unaltered with explicit constants.[41]
• Provision for choosing more convenient closely related quantities that are more suited to the problem as units than universal fixed units are is also suggested, for example based on the reduced mass of an electron, albeit with careful definition thereof where used (for example, a unit ${\displaystyle H_{M}=\mu e^{4}/\hbar ^{2}}$, where ${\displaystyle \mu =m_{\text{e}}M/(m_{\text{e}}+M)}$ for a specified mass ${\displaystyle M}$).[4]
### A convention that eliminates units
In atomic physics, it is common to simplify mathematical expressions by a transformation of all quantities:
• Hartree suggested that expression in terms of atomic units allows us "to eliminate various universal constants from the equations", which amounts to informally suggesting a transformation of quantities and equations such that all quantities are replaced by corresponding dimensionless quantities.[1]: 91 He does not elaborate beyond examples.
• McWeeny suggests that "... their adoption permits all the fundamental equations to be written in a dimensionless form in which constants such as ${\displaystyle e}$, ${\displaystyle m}$ and ${\displaystyle h}$ are absent and need not be considered at all during mathematical derivations or the processes of numerical solution; the units in which any calculated quantity must appear are implicit in its physical dimensions and may be supplied at the end." He also states that "An alternative convention is to interpret the symbols as the numerical measures of the quantities they represent, referred to some specified system of units: in this case the equations contain only pure numbers or dimensionless variables; ... the appropriate units are supplied at the end of a calculation, by reference to the physical dimensions of the quantity calculated. [This] convention has much to recommend it and is tacitly accepted in atomic and molecular physics whenever atomic units are introduced, for example for convenience in computation."
• An informal approach is often taken, in which "equations are expressed in terms of atomic units simply by setting ${\displaystyle \hbar =m_{\text{e}}=e=4\pi \epsilon _{0}=1}$".[41][42][43] This is a form of shorthand for the more formal process of transformation between quantities that is suggested by others, such as McWeeny.
## Physical constants
Dimensionless physical constants retain their values in any system of units. Of note is the fine-structure constant ${\displaystyle \alpha ={e^{2}}/{(4\pi \epsilon _{0})\hbar c}\approx 1/137}$, which appears in expressions as a consequence of the choice of units. For example, the numeric value of the speed of light, expressed in atomic units, is ${\displaystyle c=1/\alpha \,{\text{a.u.}}\approx 137\,{\text{a.u.}}}$[44]: 597
Some physical constants expressed in atomic units
Name Symbol/Definition Value in atomic units
speed of light ${\displaystyle c}$ ${\displaystyle (1/\alpha )\,a_{0}E_{\text{h}}/\hbar \approx 137\,a_{0}E_{\text{h}}/\hbar }$
classical electron radius ${\displaystyle r_{\text{e}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {e^{2}}{m_{\text{e}}c^{2}}}}$ ${\displaystyle \alpha ^{2}\,a_{0}\approx 0.0000532\,a_{0}}$
reduced Compton wavelength
of the electron
ƛe ${\displaystyle ={\frac {\hbar }{m_{\text{e}}c}}}$ ${\displaystyle \alpha \,a_{0}\approx 0.007297\,a_{0}}$
proton mass ${\displaystyle m_{\text{p}}}$ ${\displaystyle \approx 1836\,m_{\text{e}}}$
## Bohr model in atomic units
Atomic units are chosen to reflect the properties of electrons in atoms, which is particularly clear in the classical Bohr model of the hydrogen atom for the bound electron in its ground state:
• Mass = 1 a.u. of mass
• Charge = −1 a.u. of charge
• Orbital radius = 1 a.u. of length
• Orbital velocity = 1 a.u. of velocity[44]: 597
• Orbital period = 2π a.u. of time
• Orbital angular velocity = 1 radian per a.u. of time
• Orbital momentum = 1 a.u. of momentum
• Ionization energy = 1/2 a.u. of energy
• Electric field (due to nucleus) = 1 a.u. of electric field
• Lorentz force (due to nucleus) = 1 a.u. of force
## References
1. ^ a b c Hartree, D. R. (1928), "The Wave Mechanics of an Atom with a Non-Coulomb Central Field. Part I. Theory and Methods", Mathematical Proceedings of the Cambridge Philosophical Society, vol. 24, no. 1, Cambridge University Press, pp. 89–110, Bibcode:1928PCPS...24...89H, doi:10.1017/S0305004100011919, S2CID 122077124
2. ^ McQuarrie, Donald A. (2008). Quantum Chemistry (2nd ed.). New York, NY: University Science Books.
3. ^ Bethe, Hans A.; Salpeter, Edwin E. (1957). Introduction. Units. Berlin, Heidelberg: Springer Berlin Heidelberg. pp. 2–4. doi:10.1007/978-3-662-12869-5_1. ISBN 978-3-662-12871-8.
4. ^ a b c Shull, H.; Hall, G. G. (1959). "Atomic Units". Nature. 184 (4698): 1559. Bibcode:1959Natur.184.1559S. doi:10.1038/1841559a0. S2CID 23692353.
5. ^ Levine, Ira N. (1991). Quantum chemistry. Pearson advanced chemistry series (4 ed.). Englewood Cliffs, NJ: Prentice-Hall International. ISBN 978-0-205-12770-2.
6. ^ a b c McWeeny, R. (May 1973). "Natural Units in Atomic and Molecular Physics". Nature. 243 (5404): 196–198. Bibcode:1973Natur.243..196M. doi:10.1038/243196a0. ISSN 0028-0836. S2CID 4164851.
7. ^ a b Jerrard, H. G.; McNeill, D. B. (1992). Systems of units. Dordrecht: Springer Netherlands. pp. 3–8. doi:10.1007/978-94-011-2294-8_2. ISBN 978-0-412-46720-2.
8. ^ International Bureau of Weights and Measures (2006), The International System of Units (SI) (PDF) (8th ed.), p. 125, ISBN 92-822-2213-6, archived (PDF) from the original on 2021-06-04, retrieved 2021-12-16. Note that this information is omitted in the 9th edition.
9. ^ Paul Quincey; Peter J Mohr; William D Phillips (2019), "Angles are inherently neither length ratios nor dimensionless", Metrologia, 56, arXiv:1909.08389, doi:10.1088/1681-7575/ab27d7, In [the Hartree system of atomic] units, me, e, ħ and 1/4πε0 are all set equal to unity. – a reference giving an equivalent set of defining constants.
10. ^ "9th edition of the SI Brochure". BIPM. 2019. Retrieved 2019-05-20.
11. ^
12. ^
13. ^
14. ^ a b
15. ^
16. ^
17. ^
18. ^
19. ^
20. ^
21. ^
22. ^
23. ^
24. ^
25. ^
26. ^
27. ^
28. ^
29. ^
30. ^
31. ^
32. ^
33. ^
34. ^
35. ^
36. ^
37. ^
38. ^
39. ^
40. ^
41. ^ a b Pilar, Frank L. (2001). Elementary Quantum Chemistry. Dover Publications. p. 155. ISBN 978-0-486-41464-5.
42. ^ Bishop, David M. (1993). Group Theory and Chemistry. Dover Publications. p. 217. ISBN 978-0-486-67355-4.
43. ^ Drake, Gordon W. F. (2006). Springer Handbook of Atomic, Molecular, and Optical Physics (2nd ed.). Springer. p. 5. ISBN 978-0-387-20802-2.
44. ^ a b Karplus, Martin; Porter, Richard Needham (1970), Atoms and Molecules: An Introduction for Students of Physical Chemistry, Netherlands: W. A. Benjamin
45. ^ "CODATA Internationally recommended 2022 values of the Fundamental Physical Constants". NIST Reference on Constants, Units, and Uncertainty. NIST.
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# Consumer Math
1. If you divide an inheritance of \$45,600.00 among 22 heirs, each heir will receive
A. \$2,073.72.
B. \$2,072.73.
C. \$207.27.
D. \$2,027.27.
I choose B
1. 👍 0
2. 👎 0
3. 👁 132
1. Are you supposed to use a calculator or do the division on paper?
1. 👍 0
2. 👎 0
posted by Leo
2. B is correct.
1. 👍 0
2. 👎 0
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# air flow simulation in double skin facade with openfoam
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December 20, 2010, 04:59 air flow simulation in double skin facade with openfoam #1 New Member yanjing Join Date: Dec 2010 Posts: 5 Rep Power: 8 hello all: i am trying to simulate air flow in the double skin facade,which is kind of popular facade style now ,it is made of 2 layers of glazing with cavity inbetween,air inlet and outlet at the bottom and top seperately. this kind of structure could make air in the cavity heat by solar radiation and remove heat out of this box by buoyancy. now i build a box model with bottom inlet and top outlet,using buoyantSimpleRadiationFoam.it works but the results are a little bit weird,i am expecting an air induction in the inlet and air flow-out in the outlet,but results shows a air-circulation from top to bottom and bottom to topin the cavity,no suction in the inlet. i still have some problems to set the boundary conditons: 1.i need to set a heatflux on external glazing to simulate solar radiation, but the wallheatflux utility seems doesn't work here,error accurs . 2.how could i set other boundary like pd and p in such natural convection case? now for p,it is zeroGradient for both inlet and outlet,for pd,it is fixedFluxbuoyantPressure. final results like this,there seems a convergency problem on pd. Time=545 ux,initial residual=xxxe-5,final residual=xxxe-6,no inerations 2; uy,initial residual=xxxe-5,final residual=xxxe-6,no inerations 2; uz,initial residual=xxxe-5,final residual=xxxe-6,no inerations 1; h,initial residual=xxxe-5,final residual=xxxe-6,no inerations 0; G,initial residual=xxxe-6,final residual=xxxe-6,no inerations 0; pd,initial residual=xxxe-5,final residual=xxxe-7,no inerations 113; epsilon,initial residual=xxxe-5,final residual=xxxe-6,no inerations 2; .... i don't know how to push it forward now, any help would be appreciated.
December 20, 2010, 05:50 #2 New Member yanjing Join Date: Dec 2010 Posts: 5 Rep Power: 8 i define one of boudary conditions by wallHeatFlux like this: extGlaz { type wallHeatFlux; heatFlux uniform 312.0; } .... but when i run it,error note as followed: can't find "value"entry on patch extGlaz of field T in file".....",which is required to set the values of the generic patch field.(actual tyep wallHeatFlux" please add the "value' entry to the write function of user-defined boudary-condition or link the boudary-condition into libfomaUtil.so can anyone tell me,what is the problem?
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## Below is the table of values of a function. input 6 8 9 n output 2 4. 5 ?
Question
Below is the table of values of a function.
input
6
8
9
n
output
2
4.
5
?
in progress 0
2 months 2021-08-30T20:51:41+00:00 1 Answers 1 views 0
n – 4
Step-by-step explanation:
6 – x = 2
8 – x = 4
9 – x = 5
n – x = ?
->
x = 4
n – 4
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# Quick Question
• Jan 2nd 2010, 12:39 AM
Paymemoney
Quick Question
Hi
Can someone tell me how i would work this out:
Express the following in the form$\displaystyle x+y\sqrt2$ with x and y rational numbers:
$\displaystyle (7+5\sqrt2)^3$
P.S
• Jan 2nd 2010, 12:44 AM
red_dog
Use the formula $\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3$
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Marginal Revenue Product » asyabahis416.com
# Marginal Revenue Formula How to Calculate?.
Marginal revenue is the revenue a business receives from selling one more unit of a product. Because some production costs are fixed and some are variable, marginal revenue usually changes as a business sells more product. Determining marginal revenue helps a business set production levels to maximize revenue. Definition: Marginal revenue is an economic metric defined as the increase in a company’s gross revenue from selling one additional unit of its product. It can be more easily defined as the variation of the revenue figure after one more unit is sold. What Does Marginal Revenue Mean? What is the definition of marginal revenue? This economic. 4.5 Marginal Revenue Product and Derived Demand. In Chapter 2 "Key Measures and Relationships", we discussed the principle for profit maximization stating that, absent constraints on production, the optimal output levels for the goods and services occur when marginal revenue equals marginal cost.
26/04/2016 · Demand Curve for Labour - Marginal Revenue Product MRP. A video covering the Demand Curve for Labour - Marginal Revenue Product MRP Twitter: twit. Marginal Revenue Product. Marginal Revenue Product is the additional revenue generated from using one more unit of the input. Mathematically, it is the change in total revenue divided by the change in the number of inputs x, which is also equal marginal product times marginal revenue.
24/12/2019 · Marginal cost, marginal revenue, and marginal profit all involve how much a function goes up or down as you go over 1 to the right — this is very similar to the way linear approximation works. Say that you have a cost function that gives you the total cost, Cx, of producing x items shown []. 06/09/2019 · Guide to Marginal Product of Labor Formula.Here we discuss how to calculate the Marginal Product of Labor along with Examples Calculator and excel template. the marginal revenue product is equal to. the change in the total revenue divided by the change in the resource quantity. suppose firms can prepare tax returns using labor or computerized programs. If the price of a labor increases, the quantity of labor demanded will __. For firms operating in perfectly competitive markets, if the price of a product is constant, the marginal revenue product is equal to the marginal product times the. price. a decrease in resource demand is: a shift to the left of the quantity demanded of the resource at every price.
An illustrated tutorial about resource markets, how demand influences the allocation of the factors of production, how marginal revenue product MRP and marginal revenue cost MRC is determined and how they influenced the resource demand schedule. In economics, the marginal product of labor MP L is the change in output that results from employing an added unit of labor. It is a feature of the production function, and depends on the amounts of physical capital and labor already in use. marginal revenue the extra revenue that is obtained by a firm from the sale of additional units of product. If firms are profit maximizers they will seek to equate marginal revenue with MARGINAL COST to establish that price output/sales combination which yields an optimal return. Marginal revenue is defined as the revenue gained by producing one more unit of a product or service. This is important because it helps firms to make efficient production decisions and maximize profits To calculate marginal revenue, we can follow a simple three-step process: 1 calculate change in revenue.
Viele übersetzte Beispielsätze mit "marginal revenue product" – Deutsch-Englisch Wörterbuch und Suchmaschine für Millionen von Deutsch-Übersetzungen. ADVERTISEMENTS: Revenue can be defined as receipts or returns from the sale of products of an organization. In other words, revenue is the income that an organization receives from normal business activities. According to Dooley, “The Revenue of a firm is its sales receipts or money receipt from the sale of a product”. ADVERTISEMENTS: Total.
marginal revenue product MRP the extra REVENUE obtained from using one more FACTOR INPUT to produce and sell additional units of OUTPUT. The marginal revenue product of a factor is given by the factor's MARGINAL PHYSICAL PRODUCT MPP multiplied by the MARGINAL REVENUE of the product. ADVERTISEMENTS: Marginal productivity theory contributes a significant role in factor pricing. It is a classical theory of factor pricing that was advocated by a German economist, T.H. Von Thunen in 1826. The theory was further developed and discussed by various economists, such as J.B. Clark, Walras, Barone, Ricardo, and Marshall. According to.
Relation to marginal product. Boundless reference below explains the relationship between the marginal product of labour and the marginal revenue of output as follows. “The marginal revenue product of labor MRPL is the change in revenue that results from employing an additional unit of labor, holding all other inputs constant.
MARGINAL REVENUE PRODUCT: The change in total revenue resulting from a unit change in a variable input, keeping all other inputs unchanged. Marginal revenue product, usually abbreviated MRP, is found by dividing the change in total revenue by the change in the variable input or by multiplying marginal physical product by marginal revenue.
Marginal Revenue and Marginal Cost Data - Image 5. In production, fixed costs are the costs that do not vary with the number of goods produced. In the short-run, factors like land and rent are fixed costs, whereas raw materials used in production are not. Marginal revenue measures the relationship between the change in total revenues and the change in quantity. Average revenue only refers to the basic relationship between these factors, and doesn’t take into account any changes over time. Use average revenue to determine prices; use marginal revenue for price optimization. It does, however, have a huge influence over product pricing and production levels based on the manufacturer’s industry and product. For instance, in a truly competitive market place where manufacturers are selling mass-produced, homogenous products at the market price, the marginal revenue is equal to the market price. Wikipedia – Marginal Revenue – An explanation of marginal revenue including formulas. Investopedia – Marginal Revenue – MR – An introduction to the concept of marginal revenue. Khan Academy – Marginal Revenue and Marginal Cost – part of a larger course on microeconomics, this video explains the concepts of MR and MC. Next, the marginal revenue product for the ten highest-paid players and the average minimum-wage player from 1997 are calculated. The difference between actual salary and the estimated marginal revenue product of a player is an estimate of the rents that player generates for the team’s owner.
Marginal Revenue: Is the extra revenue that an additional unit of product will bring. It is the additional income from selling one more unit of a good; sometimes equal to price. It can also be described as the change in total revenue ÷ the change in the number of units sold. 01/06/2014 · Price Elasticity, Average Revenue and Marginal Revenue\$1.Mrs. Joan Robinson in her book ‘The Economics of Imperfect Competition’ has shown the empirical relationship between price elasticity, average revenue and marginal revenue. The relationship is expressed in the formula.
This calculation is performed with “revenue studies of sports teams which include winning percentage or wins as right-hand side variables should take greater care in the conclusions drawn from marginal revenue estimates”; however it is important to remember that “The winning percentage coefficient in the revenue function is really a.
1. The company keeps marginal revenue inside the constraint of the price elasticity curve but, they can adjust their output and price to optimize their profitability. Recommended Articles. This has been a guide to Marginal Revenue Formula. Here we learn how to calculate marginal revenue along with some practical examples.
2. 24/11/2019 · Marginal revenue product of labour MRPL is the extra revenue generated when an additional worker is employed. Formula: MRPL = marginal product of labour x marginal revenue.
3. 04/03/2019 · Where, Marginal Cost is the increase in cost, as a result of producing one additional unit of the product. Marginal revenue can be defined as the increase in revenue, as a result of the one additional unit sold. Marginal Revenue Formula Calculator. You can use the following Marginal Revenue Calculator.
4. 20/12/2019 · Definition of marginal revenue product: The additional revenue generated by the extra output from employing one more unit of a factor of production. In.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > clelsb3 Unicode version
Theorem clelsb3 2351
Description: Substitution applied to an atomic wff (class version of elsb3 2063). (Contributed by Rodolfo Medina, 28-Apr-2010.) (Proof shortened by Andrew Salmon, 14-Jun-2011.)
Assertion
Ref Expression
clelsb3
Distinct variable group: ,
Allowed substitution hint: ()
Proof of Theorem clelsb3
StepHypRef Expression
1 nfv 1629 . . 3
21sbco2 1980 . 2
3 nfv 1629 . . . 4
4 eleq1 2313 . . . 4
53, 4sbie 1910 . . 3
65sbbii 1885 . 2
7 nfv 1629 . . 3
8 eleq1 2313 . . 3
97, 8sbie 1910 . 2
102, 6, 93bitr3i 268 1
Colors of variables: wff set class Syntax hints: wb 178 wcel 1621 wsb 1882 This theorem is referenced by: hblem 2353 cbvreu 2707 sbcel1gv 2980 rmo3 3006 kmlem15 7674 This theorem was proved from axioms: ax-1 7 ax-2 8 ax-3 9 ax-mp 10 ax-5 1533 ax-6 1534 ax-7 1535 ax-gen 1536 ax-8 1623 ax-11 1624 ax-17 1628 ax-12o 1664 ax-10 1678 ax-9 1684 ax-4 1692 ax-ext 2234 This theorem depends on definitions: df-bi 179 df-or 361 df-an 362 df-tru 1315 df-ex 1538 df-nf 1540 df-sb 1883 df-cleq 2246 df-clel 2249
Copyright terms: Public domain W3C validator
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## Friday, December 16, 2011
### Number Rods and 1000 chains...
It is Math Madness over here! I think having a 1000 chain spread across the whole house from days on end is inspiring work with numbers, numbers, numbers...
Me Too worked with the number rods today. He worked through all the ways to make eight, and all of the ways to make seven.
He had me take this picture to be sure that Grammie could see how he does it.
Kal-El is still working on labeling the 1000 chain. I think he is up to around 400 right now. He is counting it twice for each label which makes it take a little longer. He is skip counting by tens to find the next label. However, after he finds the label he likes to use his bread-bag tab to count each individual bead to "check his work."
I explained more than once that he only has to count it one way or the other but he is a cautious kid.
He also likes to read through all his labels each time he reaches a new 100-square. In the above picture he is counting by tens up to about 300.
Kal-El took a series of pictures of his work today. Here is his first one to show the length of what he has labeled. I will spare you the other six shots he took to document the labels every 50 tags or so. I haven't attempted to photograph the whole chain on the rug yet. The light was horrible again today. This may be beyond my photography abilities because it starts in the super bright school room, passes through a dim foyer, only to encounter a third variety of crummy light in the office.
1. Nice 1000 chain mat you guys have. It is always a delight to see Me Too smiling while doing math. I forgot about our small number rods that I have, thank you for including this picture in your post today. The chain is a very difficult material to photogaraph, I give Kae El(sorry if I misspelled the name) two thumbs up for his photography. Thank you for sharing.
2. Just discovered your blog on Pinterest when I was looking for ideas for our Australia Unit Study. Love your continent box for that and thanks for sharing the idea of the Australia animals Toob. What wonderful ideas. Looking forward to your updates. Following your facebook page now. ~Julie
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# The Theory and Practice of Gauging: Demonstrated in a Short and Easy Method. ... Published with the Particular Approbation of the Honourable Commissioners of Excise. Design'd for the Use of the Officers of that Revenue
H. Woodfall, 1740 - 283 Seiten
### Was andere dazu sagen -Rezension schreiben
Es wurden keine Rezensionen gefunden.
### Beliebte Passagen
Seite 59 - The circumference of every circle is supposed to be divided into 360 equal parts called degrees, and each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds, and these into thirds, fourths, &c.
Seite 7 - In multiplication of decimals, we know that the number of decimal places in the product is equal to the sum of those in both the factors.
Seite 97 - J of the square of their difference, then multiply by the hight, and divide as in the last rule. Having the diameter of a circle given, to find the area. RULE. — Multiply half the diameter by half the circumference, and the product is the area ; or, which is the same thing, multiply the square of the diameter by .7854, and the product is the area.
Seite 282 - Sort is, to multiply the two Weights together, and extract the Square Root of. the Product, which Root will be the true Weight.
Seite 283 - Backs time ufed, and become more and more uneven as they grow older, efpecially fuch as are not every where well and equally fupported ; many of them...
Seite 187 - Sum of thofe next to them, C the Sum of the two next following the laft, and fo on ; then we (hall have the following fables of Areas, for the feveral Numbers of Ordinates prefixt againft them, viz.
Seite 86 - Progreflion from o, is equal to the Product of the laft Term by the Number of Terms, and this divided by the Index (m) plus Unity.
Seite 272 - To half the Sum of the Squares of the Top and Bottom Diams.
Seite 95 - The latter being taken from the former, leaves 3.14.15.9265.5 for the Length of half the Circumference of a Circle whofe Radius is Unity : Therefore the Diameter of any Circle is to its Circutuftrence as I is to 3.1415.9265.5 nearly.
Seite 86 - Numbr infinitely greAt, therefore the firft Term of the above Value of /, muft be infinitely greater than any of the...
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# Lists and sets
You can have answers with lists and sets. There are multiple forms understood as lists; you can configure it in the Input options and Validation options sections.
## Example 1: Compare as lists
Let us create an essential question asking to sort a set of integers:
By default,
• Lists are allowed, but elements need to be enclosed within curly brackets.
• The comma is used as a separator.
• The student's answer is compared as a list (taking into account order and repetition).
In this example, we check the option Lists without enclosers.
Thus, the following answers will be validated as specified in the table below.
Validation
-7,-4,1,5
$\left\{-7,-4,1,5\right\}$
$\begin{array}{c}-7\\ -4\\ 1\\ 5\end{array}$
$-7-415$
## Example 2: Compare as sets
By default, to be accepted, lists in student and correct answers must have equivalent items and be in the same order. That is, the order is important, and repetition has sense. In contrast, sets are lists without order nor duplicated items.
As an example, we can ask for the roots of a given polynomial, and we might be interested in accepting any answer with the correct values.
Thus, we can set the lists to be understood as sets in Validation options > Comparison with student answer. That is, ignore repetition and order in the comparison.
Besides, we can also define these lists to be understood as sets where repetition matters, but order does not. The grading will behave as you can see below.
$\left\{1,2,2\right\}$
$1,2,2$
$1,2$
$1,2,2,2,1$
$122$
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## Untyped Programs Don’t Exist (cross-post)
:: by William J. Bowman
## How to prove a compiler correct (cross-post)
:: by Daniel Patterson
## Monotonicity Types: Towards A Type System for Eventual Consistency
::
A few weeks back, we published a draft of an article entitled Monotonicity Types. In it, we describe a type system which we hope can aid the design of distributed systems by tracking monotonicity with types.
## Final Algebra Semantics is Observational Equivalence
::
Recently, “final encodings” and “finally tagless style” have become popular techniques for defining embedded languages in functional languages. In a recent discussion in the Northeastern PRL lab, Michael Ballantyne, Ryan Culpepper and I asked “in what category are these actually final objects”? As it turns out our very own Mitch Wand wrote one of the first papers to make exactly this idea precise, so I read it available here and was pleasantly surprised to see that the definition of a final algebra there is essentially equivalent to the definition of observational equivalence.
In this post, I’ll go over some of the results of that paper and explain the connection to observational equivalence. In the process we’ll learn a bit about categorical logic, and I’ll reformulate some of the category theory in that paper to be a bit more modern in presentation, cleaning some things up in the process.
## PLT Redex FAQ
::
A short guide to Redex concepts, conventions, and common mistakes.
## Why am I going to ICFP 2017? (cross-post)
:: by William J. Bowman
## Closure Conversion as CoYoneda
::
The continuation-passing style transform (cps) and closure conversion (cc) are two techniques widely employed by compilers for functional languages, and have been studied extensively in the compiler correctness literature. Interestingly, typed versions of each can be proven to be equivalence preserving using polymorphic types and parametric reasoning, as shown by my advisor Amal Ahmed and Matthias Blume (cps,cc).
In fact, there is something like a duality between the two proofs, cps uses a universal type, closure-conversion uses an existential type and the isomorphism proofs use analogous reasoning. It turns out that both are instances of general theorems in category theory: the polymorphic cps isomorphism can be proven using the Yoneda lemma, and the polymorphic closure-conversion isomorphism can be proven using a less well known theorem often called the *co*Yoneda lemma.
The connection between cps and the Yoneda embedding/lemma is detailed elsewhere in the literature and blogosphere (ncafe, Bartosz), so I’ll focus on closure conversion here. Also, I’ll try to go into some detail in showing how the “usual” version of Yoneda/coYoneda (using the category of sets) relates to the appropriate version for compilers.
I’ll assume some background knowledge on closure conversion and parametricity below. Fortunately, Matt Might has a nice blog post explaining untyped closure conversion.
## Reviews and author responses: we should stop asking for 500-word responses
:: by Gabriel Scherer
This year I reviewed many ICFP submissions, and got to be on the receiving end of equally many author responses (also sometimes called, somewhat combatively, rebuttals). I found that there was a large difference between the official written advice on author responses and what I, as a reviewer reading the responses, found effective. In particular, I now believe that limiting yourself to 500 words should strongly be avoided — we should even stop giving that advice.
## Trees, 1973
::
From the PRL archives:
I think that I shall never see a matrix lovely as a tree. — Trees, by Guy L. Steele Jr., MIT, 1973
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https://www.quiltingboard.com/main-f1/prairie-points-t229.html
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>
>
# Prairie Points
12-07-2006, 06:03 PM
#1
Senior Member
Join Date: Nov 2006
Posts: 358
When finishing a quilt with Prairie Points, how do you calculate the size and number of points so when you are finished - they are all the same size and you don't have space left, or they are too crowded?
Vicky
12-10-2006, 02:21 AM
#2
Junior Member
Join Date: Nov 2006
Posts: 225
Bare with me a minute, I'm going to try and remeber what I learned on America Quilts Creatively, a while back on this, If I remeber correctly, the lady said to measure the width of the prarie point, then measure the side of the quilt, Divide the width of the prarie point in to the length(side) of the quilt, to get how many you need, You would have to do that for each side of your quilt. So it would come out evenly, This is where you may have to adjust the width of the prarie point so it would come out even, I'm pretty sure that is how it's done, Try it and see, and let me know, I'm just trying to help, Ex: Prarie Point 4"
one side of quilt 48"= 4 into 48= 12 Prarie points
Let me know. Becky
12-10-2006, 05:18 AM
#3
Senior Member
Join Date: Nov 2006
Location: USA
Posts: 784
Mornin' Vicky. :D Here's a link to a little tutorial on prairie points...
http://www.mccallsquilting.com/artheblk/prapoints/
09-11-2007, 07:34 AM
#4
Senior Member
Join Date: Aug 2007
Posts: 707
American Quilter, summer 07 has a nice article as well. The thing is you can nestle them, place them end on end, or even use various sizes, so one formula doesn't work for all types. I like the nestled idea cause then you can fudge just a wee bit.
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https://studylib.net/doc/5629895/introduction-to-atoms
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# Introduction to Atoms
```Introduction
to Atoms
Exploring Inner
Space
Atom
The smallest particle
of an element…
makes up all matter.
Particle
Charge
Mass
Proton
+
+1 amu
Neutron
0
(neutral)
Electron
-
+1 amu
0 amu
amu= atomic mass unit
The most stable form
of an atom is
electrically neutral…
this means that the
number of protons &
electrons is equal.
Example=> Oxygen has 8
protons & 8 electrons
Nucleus: The center
of the atom
(contains protons
and neutrons).
Electron cloud:
electrons
orbit the
nucleus.
(contains
ONLY
electrons).
The
electrons
fill
energy
levels
within the
electron
cloud.
The number of
energy levels can
be determined by
the period
number on the
periodic table.
As a general
rule, there are
2 electrons in
the first energy
level, up to 8
electrons in the
second and 8 or
more in each
level after that.
Elements are atoms with the
same number of protons that
share the same physical and
chemical properties… the # of
protons determines the
element.
For example, hydrogen has only
1 proton in its nucleus,
Lithium has 3 protons, and
Oxygen has 8.
Atomic number = the
number of protons
in an atom
How many protons
does carbon have?
6
Almost
all of
the
mass of
an atom
is in the
nucleus.
electron
neutron
proton
Atomic mass =
# of protons +
# of neutrons
Carbon’s
atomic mass
is 12. Carbon
has how many
neutrons?
6
(atomic mass - # of protons = # of neutrons)
Every element is
represented by a
symbol… if it has one
letter, that letter is
capitalized
(C=carbon). If it has
two letters, the first
is capitalized and the
second is lower case.
(Na = sodium)
```
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https://k12.libretexts.org/Sandboxes/hdagnew@ucdavis.edu/Jane_Good_-_Elementary-_Curation_4th_Grade-Reading-and-Writing-Trimester-1.zip/00%3A_Thinking_Maps/06%3A_Double_Bubble_Maps/03%3A_Double_Bubble_Map_Assignment
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# Double Bubble Map Assignment
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$
( \newcommand{\kernel}{\mathrm{null}\,}\) $$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\id}{\mathrm{id}}$$
$$\newcommand{\Span}{\mathrm{span}}$$
$$\newcommand{\kernel}{\mathrm{null}\,}$$
$$\newcommand{\range}{\mathrm{range}\,}$$
$$\newcommand{\RealPart}{\mathrm{Re}}$$
$$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$
$$\newcommand{\Argument}{\mathrm{Arg}}$$
$$\newcommand{\norm}[1]{\| #1 \|}$$
$$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$
$$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$
$$\newcommand{\vectorA}[1]{\vec{#1}} % arrow$$
$$\newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow$$
$$\newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vectorC}[1]{\textbf{#1}}$$
$$\newcommand{\vectorD}[1]{\overrightarrow{#1}}$$
$$\newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}}$$
$$\newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}}$$
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
$$\newcommand{\avec}{\mathbf a}$$ $$\newcommand{\bvec}{\mathbf b}$$ $$\newcommand{\cvec}{\mathbf c}$$ $$\newcommand{\dvec}{\mathbf d}$$ $$\newcommand{\dtil}{\widetilde{\mathbf d}}$$ $$\newcommand{\evec}{\mathbf e}$$ $$\newcommand{\fvec}{\mathbf f}$$ $$\newcommand{\nvec}{\mathbf n}$$ $$\newcommand{\pvec}{\mathbf p}$$ $$\newcommand{\qvec}{\mathbf q}$$ $$\newcommand{\svec}{\mathbf s}$$ $$\newcommand{\tvec}{\mathbf t}$$ $$\newcommand{\uvec}{\mathbf u}$$ $$\newcommand{\vvec}{\mathbf v}$$ $$\newcommand{\wvec}{\mathbf w}$$ $$\newcommand{\xvec}{\mathbf x}$$ $$\newcommand{\yvec}{\mathbf y}$$ $$\newcommand{\zvec}{\mathbf z}$$ $$\newcommand{\rvec}{\mathbf r}$$ $$\newcommand{\mvec}{\mathbf m}$$ $$\newcommand{\zerovec}{\mathbf 0}$$ $$\newcommand{\onevec}{\mathbf 1}$$ $$\newcommand{\real}{\mathbb R}$$ $$\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}$$ $$\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}$$ $$\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}$$ $$\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}$$ $$\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}$$ $$\newcommand{\laspan}[1]{\text{Span}\{#1\}}$$ $$\newcommand{\bcal}{\cal B}$$ $$\newcommand{\ccal}{\cal C}$$ $$\newcommand{\scal}{\cal S}$$ $$\newcommand{\wcal}{\cal W}$$ $$\newcommand{\ecal}{\cal E}$$ $$\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}$$ $$\newcommand{\gray}[1]{\color{gray}{#1}}$$ $$\newcommand{\lgray}[1]{\color{lightgray}{#1}}$$ $$\newcommand{\rank}{\operatorname{rank}}$$ $$\newcommand{\row}{\text{Row}}$$ $$\newcommand{\col}{\text{Col}}$$ $$\renewcommand{\row}{\text{Row}}$$ $$\newcommand{\nul}{\text{Nul}}$$ $$\newcommand{\var}{\text{Var}}$$ $$\newcommand{\corr}{\text{corr}}$$ $$\newcommand{\len}[1]{\left|#1\right|}$$ $$\newcommand{\bbar}{\overline{\bvec}}$$ $$\newcommand{\bhat}{\widehat{\bvec}}$$ $$\newcommand{\bperp}{\bvec^\perp}$$ $$\newcommand{\xhat}{\widehat{\xvec}}$$ $$\newcommand{\vhat}{\widehat{\vvec}}$$ $$\newcommand{\uhat}{\widehat{\uvec}}$$ $$\newcommand{\what}{\widehat{\wvec}}$$ $$\newcommand{\Sighat}{\widehat{\Sigma}}$$ $$\newcommand{\lt}{<}$$ $$\newcommand{\gt}{>}$$ $$\newcommand{\amp}{&}$$ $$\definecolor{fillinmathshade}{gray}{0.9}$$
Create a double bubble map. You can do this by printing out a blank version or drawing your own. Remember that Double Bubbles are used to Compare and Contrast two ideas. In this assignment pick two characters from a book/movie and compare them. Remember that the center circles are the items the two things have in common (or are the same) and the outer circles their differences.
Turn in this map by taking a photograph with your iPad and attaching it to this lesson.
Double Bubble Map Assignment is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
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https://la.mathworks.com/matlabcentral/cody/problems/149-is-my-wife-right/solutions/2024271
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| 15,679
|
Cody
Problem 149. Is my wife right?
Solution 2024271
Submitted on 16 Nov 2019 by Lucio Menghini
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
Test Suite
Test Status Code Input and Output
1 Pass
x = 'But I''m actually right this time'; y_correct = 'yes'; assert(isequal(wiferight(x),y_correct))
2 Pass
x = 'But you just said that 2+2=3'; y_correct = 'yes'; assert(isequal(wiferight(x),y_correct))
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https://www.jmag-international.com/whitepapers/w-mb-151/
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## [W-MB-151] Accuracy of Motor Plant Model Required for Control Calibration
Contents
1. Introduction
2. Accuracy of motor plant model required for control calibration
3. Control calibration Example: Current Vector Calibration for Maximum Efficiency Control
3.1 Motor and Control Specifications
3.2 Plant model specifications
3.3 Calibration method of the current command value
3.4 Calibration result
3.5 Torque accuracy
3.6 Efficiency accuracy
4. Summary
5. References
### 1. Introduction
Control calibration is time-consuming work because it is carried out for every operating point using the actual machine. This paper examines realizing improved efficiency in control calibration by using a plant model. However, since control calibration is performed according to the characteristics of the actual machine, the plant model requires an accuracy equivalent to that of the actual machine. In this paper, the current command value of maximum efficiency control identified as an example of control calibration is examined and the accuracy required for the plant model is shown. It shows that accounting for iron loss including time harmonics is necessary to satisfy the required accuracy of torque and loss.
### 2. Accuracy of motor plant model required for control calibration
In control calibration, the gain and control command value are identified according to motor characteristics. Normally, control calibration was carried out using an actual machine for every motor and every operating point, which was time-consuming work. Here, the control calibration is performed using the plant model instead of the actual machine.
The accuracy of the plant model required for control calibration depends on the calibrated parameters. As an example, we look at the current command value when driving the motor with maximum efficiency control as an example. The accuracy of torque and loss is required in the plant model to make efficiency an objective function.
### 3.1 Motor and Control Specifications
Fig. 1 and Table 1 show the cross-section and specifications of the motor. Fig. 2 shows the control model. The current vector is controlled by the PI control. The current vector, which is the command value, is selected so that efficiency is maximized at each operating point by calibration.
### 3.2 Plant model specifications
Table 2 shows the resolution of the plant model. The calculation method of the iron loss is described as follows. When generating a plant model, a sine wave current is given to obtain the fundamental and spatial harmonic components. The time-harmonic iron loss is accounted for by using the iron loss equivalent resistor in the equivalent circuit in the control and circuit simulations. An FEA model and a plant model that doesn’t account for iron loss were used as a reference to compare accuracy. Except for not accounting for iron loss, the latter has the same resolution.
Fig. 1 Cross-section of the motor
Table 1 Motor specifications
Number of poles/slots 8/48 Core material 35JA300 Magnet NdFeB, Br=1.2(T) Output 80 (kW) DC Voltage 600 (V) Maximum current 250 (A) Carrier Frequency 6 (kHz)
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https://forums.digitalpoint.com/threads/how-to-use-math-random-distribution-three-div.2000059/
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y u no do it?
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# How to use Math random() distribution three div?
Discussion in 'JavaScript' started by youlichika, Nov 12, 2010.
1. #1
I have 3 divs. I want distribution them random (when refresh the windows, three divs' position will be changed)
I think Math random() can solve this problem, but how to do that? Thanks.
``````
<div id="div1">...</div>
<div id="div2">...</div>
<div id="div3">...</div>
<script>
Math.floor(Math.random()*3)+1;
...
</script>
``````
HTML:
Nov 12, 2010 IP
2. ### Cash NebulaPeon
Messages:
1,197
67
0
Trophy Points:
0
#2
Use absolute positioning, with random values for top and left.
``````
<html>
<style type='text/css'>
body {
border: 1px solid #000;
height: 300px;
margin: 0;
width: 300px;
}
.box {
height: 50px;
position: absolute;
width: 50px;
}
#div1 { background: #f00; }
#div2 { background: #0f0; }
#div3 { background: #00f; }
</style>
<script type='text/javascript'>
function setDivPos() {
for (i=1; i<=3; i++) {
var x = Math.floor(Math.random()*250);
var y = Math.floor(Math.random()*250);
document.getElementById('div'+i).style.left = x + 'px';
document.getElementById('div'+i).style.top = y + 'px';
}
}
</script>
<div id='div1' class='box'></div>
<div id='div2' class='box'></div>
<div id='div3' class='box'></div>
</body>
</html>
``````
Code (markup):
Last edited: Nov 13, 2010
Nov 13, 2010 IP
3. ### youlichikaGreenhorn
Messages:
74
0
0
Trophy Points:
16
#3
Thanks Cash, this is a random effection. I am sorry, may be I did not say clearly in english. the three divs position will be change is meaning like: div1{float:left;width:100px;height:100px;left:0;top:0} div2{float:left;width:100px;height:100px;left:100px;;top:0} div3{float:left;width:100px;height:100px;left:200px;top:0}, when I refresh the page, the three divs position will be change, div1 go to div2's position, div2 go to div3's position, div3 go to div1's position...
Nov 13, 2010 IP
4. ### Cash NebulaPeon
Messages:
1,197
67
0
Trophy Points:
0
#4
Something like this?
``````
<html>
<style type='text/css'>
#random {
height: 100px;
margin: 50px auto;
position: relative;
width: 300px;
}
.box {
height: 100px;
position: absolute;
top: 0;
width: 100px;
}
#div1 { background: #f00; }
#div2 { background: #0f0; }
#div3 { background: #00f; }
</style>
<script type='text/javascript'>
function setDivPos() {
var box1, box2, box3;
box1 = Math.ceil(Math.random()*3);
do {
box2 = Math.ceil(Math.random()*3);
} while (box2 == box1);
do {
box3 = Math.ceil(Math.random()*3);
} while (box3 == box1 || box3 == box2);
document.getElementById('div'+box1).style.left = '0';
document.getElementById('div'+box2).style.left = '100px';
document.getElementById('div'+box3).style.left = '200px';
}
</script>
<div id='random'>
<div id='div1' class='box'></div>
<div id='div2' class='box'></div>
<div id='div3' class='box'></div>
</div>
</body>
</html>
``````
Code (markup):
Nov 13, 2010 IP
5. ### youlichikaGreenhorn
Messages:
74
0
0
Trophy Points:
16
#5
``````
<html>
<style type='text/css'>
#random {
height: 100px;
margin: 50px auto;
position: relative;
width: 300px;
}
#div1 { width:100px;height:100px; float:left;}
#div2 { width:100px;height:100px; float:left;}
#div3 { width:100px;height:100px; float:left;}
</style>
<body>
<div id='random'>
<div id='div1' ></div>
<div id='div2' ></div>
<div id='div3' ></div>
</div>
<script>
var index = Math.floor(Math.random()*6);
var ids = [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]][index];
document.getElementById('div'+ids[0]).innerHTML = 'aaa';
document.getElementById('div'+ids[1]).innerHTML = 'bbb';
document.getElementById('div'+ids[2]).innerHTML = 'ccc';
</script>
</body>
</html>
``````
Code (markup):
But I find all the Math.random() should make a suppose first. It is hard to add one more divs. If the divs number add into 6, there should be write a long long javascript code to suppose how to distribution all the possibility of divs positions. Is it right?
Nov 14, 2010 IP
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SolSec 2.2
# SolSec 2.2 - 2 9 Problems and Solutions Section 2.2(2.11...
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2- 9 Problems and Solutions Section 2.2 (2.11 through 2.25) 2.11 Calculate the constants A and φ for arbitrary initial conditions, x 0 and v 0 , in the case of the forced response given by Equation 2.29. Compare this solution to the transient response obtained in the case of no forcing function (i.e. F 0 = 0). Solution: From equation (2.29) x t Ae t X t xt Ae t A e t X t n nn t d n t dd t d ( ) sin( ) cos( ) ˙( ) sin( ) cos( ) sin( ) =+ + =− + + + −− ζω ωφ ω θ Next apply the initial conditions to these general expressions for position and velocity to get: xA X A X n d ( ) sin cos ˙( ) sin cos sin 0 0 + + φθ ζ φω Solving this system of two equations in two unknowns yields: θω θζω = = tan ( cos ) ( cos ) sin cos sin 1 0 00 0 xX vx X X A d Recall that X has the form X Fm n n = −+ = 0 22 2 2 1 2 2 / () ( ) tan ωω ζω ω and Now if F 0 = 0, then X = 0 and A and φ from above reduce to: = + == ++ tan sin ( ) 1 0 000 2 0 2 2 x A xv x x d n n d d These are identical to the values given in equation (1.38).
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2-10 2.12 Show that Equations 2.20 and 2.21 are equivalent by verifying Equations 2.21 and 2.22. Solution: From equation (2.20) and expanding the trig relation yields xX t X t t Xt X t p AB =− = + [] =+ cos( ) cos cos sin sin ( cos )cos ( sin )sin ωθ ω θ θω 12 43 41 2 4 Now with A and B defined as indicated, the magnitude is computed: XA B 22 and B A X X B A =⇒ = sin cos tan θ 1
2-11 2.13 Plot the solution of Equation 2.19 for the case that m = 1 kg, ζ = 0.01, ω n = 2 rad/s. F 0 = 3 N, and = 10 rad/s, with initial conditions x 0 = 1 m and v 0 = 1 m/s. Solution: The particular solution is given in equations (2.28) and (2.29). Substitution of the values given yields: xt p =+ 0 03125 10 0 004176 . cos( . ) . Then the total solution has the form: Ae t t t ( ) sin( ) . cos( . ) . + + 02 2 0 03125 10 0 004176 φ Differentiating then yields ˙( ) . sin( ) cos( ) . sin( . ) .. x t Ae t Ae t t tt =− + + + + 0 2 2 2 2 0 3125 10 0 004176 φφ Apply the initial conditions to get: xA A ( ) sin . ˙( ) . sin cos . 0 1 0 03125 0 1 0 2 2 0 3125 == + + Solving yields: A = 1.095 m and =1.086 rad. So the solution and plot become (using Mathcad)
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2-12 2.14 A 100 kg mass is suspended by a spring of stiffness 30 × 10 3 N/m with a viscous damping constant of 1000 Ns/m. The mass is initially at rest and in equilibrium. Calculate the steady-state displacement amplitude and phase if the mass is excited by a harmonic force of 80 N at 3 Hz. Solution: Given m = 100kg, k =30,000 N/m, c = 1000 Ns/m, F 0 = 80 N and ω = 6 π rad/s: f F m k m c km X n 0 0 22 2 2 80 100 0 8 17 32 2 0 289 08 17 32 36 2 0 289 17 32 6 0 0 0041 == = = = = + () + = ., . . . .( . ) ( . ) ( . m/s rad/s m 2 ω ζ ππ Next compute the angle from φ = tan . . 1 188 702 55 323 Since the denominator is negative the angle must be found in the 4 th quadrant. To find this use Window 2.3 and then in Matlab type atan2(188.702,-55.323) or use the principle value and add π to it. Either way the phase is φ =1.856 rad.
2-13 2.15 Plot the total solution of the system of Problem 2.14 including the transient.
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BBA Finance Courses
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## Analysing Problems & Improve Quality Questions and Answers : Quiz 2
MCQ 6:
The formal way of differentiating, between non-random and random variations, in manufacturing process is classified as
1. statistical process control
2. statistical failure control
3. statistical control of prevention cost
4. statistical control of sunk cost
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If the net initial investment is \$985000, returned working capital is \$7500, then an average investment over five years will be
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4. \$486,250
MCQ 8:
In the budgeted fixed overhead rate, the number of machine hours are considered as
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CC-MAIN-2024-26
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latest
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