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Research:How we read Wikipedia and reader navigation - Meta
Research:How we read Wikipedia and reader navigation
Duration: 2020-January – 2020-December
In contrast to Wikipedia’s editor population, little is known about its readers; in large parts due to the challenges and restrictions when dealing with privacy-sensitive data. Only recently have we started to characterize wikipedia’s readership. For example, recent studies (Singer, Lemmerich, et al. 2017; Lemmerich et al. 2019) approached the question why we read wikipedia in order to identify the motivation, information need, and prior knowledge of different users. Here, we investigate whether and to what degree this is reflected in how we use wikipedia. That is, instead of looking at page views as isolated events, we consider user’s full reading session in order to characterize patterns of navigation within and across wikimedia projects, and as a result, better understand usage of wikipedia, in particular how readers use wikipedia to learn about specific topics.
There are systematic differences in the length of reading sessions depending on access-method and topic in which session starts
There are 2 distinct phases in reading sessions with respect to topical exploration (first increasing focus, broader exploration towards end)
Reader surveys suggest differences in topical exploration depending on information need and motivation (disclaimer: small and with given data not clear if statistically significant)
Established methods for defining reader sessions (1-hour cutoff between 2 consecutive pageviews) need to be revisited as particular choice not obvious
1 Background/Motivation/Goals
2.1 Defining reader sessions
2.2 Topic analysis
2.3 Navigation in topic space
2.3.1 Building continuous topic space using word embeddings
2.3.2 Example reading sessions
2.3.3 Evolution of topical spread during a reading session
2.3.4 Relation to motivation and information need
Background/Motivation/GoalsEdit
This project aims to answer questions such as: What are the strategies used for satisfying a user’s information needs and how do they vary, e.g., across different languages and projects? What are the reasons for when they fail, e.g. due to lack of accessibility; and how could we assist readers in finding information they are looking for? Answering these questions is pertinent to at least two of the five priorities laid out in the Wikimedia Foundation Medium-term plan 2019: worldwide readership (increase global readership) and platform evolution (ease-of-use and low barrier of entry). Thus, by providing a characterization of readership behavior and patterns, this project will be contributing towards the goal of addressing knowledge gaps in readership. Following the idea of the pipeline of online participation (Shaw & Halfaker 2018), this will also provide crucial insights into possible causes into knowledge gaps in contributorship and content. In the longer-term, we also hope to set the foundation for approaching questions on how users learn on Wikipedia (possibly over longer times), and how to support initiatives to train critical usage of wikipedia in order to increase resilience with respect to disinformation.
Starting exploratory research to empirically characterize navigation session in Wikipedia, e.g. quantifying different across Wikipedia editions, geographical locations, mobile access, topical content, etc.
Defining reader sessionsEdit
Histogram of interevent times in consecutive pageviews (log-binning)
Using webrequest logs, we identify reading sessions by grouping all pageviews for the same hash concatenating client_ip+user_agent+access_mode (removing automated traffic, only considering pageivews to main namespace, no redirects, etc).
One typical approach for identifying reading sessions is to break reading sessions if the time-difference between two pageviews is larger than a given threshold. Halfaker et al [1] suggested a cut-off of 1 hour for Wikipedia reading sessions. They convincingly showed that the distribution of interevent times (time between 2 consecutive pageviews) is bimodal distribution (with peaks at ~1min and ~1day) and identify a separation point at ~1hour (see Figure 1 bottom 3 panels). Following these insights, the choice of 1 hour has become the standard choice when constructing reader sessions in Wikipedia [2][3][4] .
Using a sample of ~500M sessions from 1 week of requests to enwiki, we found a slightly more subtle picture suggesting that a 1-hour cutoff might not be as obvious as suggested by the previous analysis. Looking at our data, we similarly observe 2 peaks at around 1 minute and 1 day. However, for the range in between there is no clear identifiable minimum which would suggest a cutoff at 1 hour. In fact, one could hypothesize whether there is a third peak at 1hour. In any case, any cutoff in the range 15min to 4 hours seems equally good and the distribution of interevent times seems insufficient to provide a clear answer. More research is needed to understand and solve this problem in a satisfactory way. While we will still choose 1 hour as a pragmatic choice for the cutoff, we suggest to check whether results obtained from reading sessions are robust when varying this cutoff (e.g. [15min, 1 hour, 4 hours]).
Topic analysisEdit
How do reading sessions vary across different topics in Wikipedia?
To answer this question, we assign each article a topic label from the WikiProjects-derived topic taxonomy (e.g. Culture.Sports or STEM.Physics). Specifically, we use the Wikidata topic model and choose the topic with the highest probability (the advantage of this model with respect to ORES' topic model is that the former can be applied to articles from any language using only the statements from the corresponding Wikidata-item and not the text of the article thus becoming language-independent.
First, we record the number of pageviews (across all sessions) for each topic and separate according to access method (desktop, mobile web, mobile app). Since there are many more pageviews to mobile web than to mobile app, for each access_method we calculate the fraction of pageviews across topics. According to our expectation, some topics are much more popular than others; for example 'Person' receives more than 10% of all pageviews whereas 'Culture.Crafts_and_Hobbies' receives less than 1/1000 of all pageviews (note that we do not account for the different number of articles in each topic). However, it is surprising that the distribution of pageviews across topics is very similar for the different access_methods. There are some notable differences, e.g. 'STEM.Technology' receives a higher fraction from desktop-readers, while 'Person' receives a higher fraction from web-users (mobile and app).
Second, we record the length of the reading session depending on the topic of the initial pageview. Specifically, we calculate the average (and standard error of the mean) length of sessions starting in a given topic conditioned on the different access methods. Across all topics, the average for mobile web is one pageview smaller than desktop and mobile app. The average varies substantially: 'Culture.Arts' has an average of 4 pageviews for desktop, while 'STEM.Technology' is around 2. This already suggests strong differences in the way Wikipedia is used depending on the topic of interest.
Navigation in topic spaceEdit
In this section, we measure navigation in topic space.
The discrete labels provide only a very coarse-grained view. Instead, we would like to describe navigation of a reader in a similar way we describe navigation when driving a car. Therefore, we first construct a continuous topic space (where similar articles are closer to each other) and then measure the topical spread of reading sessions in this space.
Building continuous topic space using word embeddingsEdit
Using the text of the articles, we use word-emebddings to project each article into a 2-dimensional topic space. The detailed workflow consists of the following steps: - extract all word tokens of the first section of an article using Wikitext Preprocessor from mediawiki-utilities/python-mwtext. this parses wikitext into text and performs tokenization using regex. note that there are more complex parser, such as mwparserfromhell, or tokenizers, such as spacy, but our approach is much faster. - use the fasttext word vectors (here for english) to obtain a 300-dimensional vector from the article text. specifically, we use the get_sentence_vector function which maps the string of a text (can be several) words to a 300-D vector. - use umap to project the 300-D topic space into a 2-D topic space.
We obtain a 2D-map of all articles in enwiki. We can see that similar articles will be closer to each other. For example, articles with a given topic label (obtained from the wikidata topic model) will be strongly localized in a particular region of the map. This suggests that our topical space captures the similarity of articles in terms of their content.
Example reading sessionsEdit
Once we have created the topical space, we can map each reading session from a sequence of page_ids to a sequence of 2D-coordinates. As an illustrative example, we show two reading sessions
a very focused reading session (red) with 34 pageviews
a very broad reading session (blue) with 38 pageviews.
We can quantify the spread of each session by calculating the pooled standard deviation:
{\displaystyle \sigma ={\sqrt {\sigma _{x}^{2}+\sigma _{y}^{2}}}}
. Large values would indicate a large topical spread whereas smaller values would indicate higher topical focus in the reading session. Accordingly, in the example we obtain values 0.54 (red) and 5.24 (blue).
Evolution of topical spread during a reading sessionEdit
We assess if and how the topical spread changes during a reading session. Previous research in 'laboratory settings' (i.e. where readers were tasked to find a navigation path from a source to a target page) found two distinct phases [5].
We calculate the spread
{\displaystyle \sigma }
for a window of 4 consecutive pageviews (other window sizes yield similar results) and move this window along the reading session from beginning to end. We further group reading sessions according to their length and show the mean and the standard error of the mean over the corresponding reading sessions.
In order to test whether the observations from the original data (blue) are genuine and not just an artifact of, e.g., the network topology of links, we compare with two null models in which we generate a synthetic set of reading sessions (each with the same starting page and the same length) assuming navigation as a random walk along the link network:
random walk: all outgoing-links have the same probability. for each page we get all outgoing links from pagelinks table and randomly choose one as the next pageview in the reading session.
clickstream: outgoing-links are weighted according to the number of times the link was used. for each page we get the number of times an outgoing link was clicked using the clickstream data and randomly choose (weighted with probability) one as the next pageview in the reading session.
there is a characteristic U-shape in the topical spread: in the first half/third there is an increase in topical focus (smaller <math> \sigma <\math>), towards the end the topical focus becomes much larger. Most importantly, this does not happen when just navigating randomly on the link network.
This suggests that there are two distinct phases in the navigation of reading sessions
This also clearly indicates that reading sessions are contextual and navigation depends on the previously visited pages (and not just the currently visited one). Compare previous studies critically assessing this aspect in navigation of web users [6] and findings that navigation in Wikigames can be memory-less[7]
longer sessions start with a higher topical spread from the very beginning
topical spread is smaller than for the null models in most cases (this is expected when readers do not just click links randomly); however, towards the end of sessions, the topical spread becomes larger suggesting that readers navigate content that is outside the reach of the link network.
Relation to motivation and information needEdit
Using ~5000 reading sessions from the Reader survey, we check whether and how the topical spread differs depending on the information need and the motivation.
Separating sessions of different length, we calculate the topical spread for the entire session conditioned on the response given in the survey with respect to information need and motivation.
Differences in spread across need and motivation (for same session length) are small and most likely not statistically significant; however, there seem to be interesting systematic shifts
information need: In-depth sessions have smaller average topical spread
motivation: intrinsic and work/school have smaller average topical spread
The topical spread increases with increasing session length (expected)
↑ Halfaker, Aaron; Keyes, Oliver; Kluver, Daniel; Thebault-Spieker, Jacob; Nguyen, Tien; Shores, Kenneth; Uduwage, Anuradha; Warncke-Wang, Morten (2015). "User Session Identification Based on Strong Regularities in Inter-activity Time". pp. 410–418. doi:10.1145/2736277.2741117.
↑ Singer, Philipp; Lemmerich, Florian; West, Robert; Zia, Leila; Wulczyn, Ellery; Strohmaier, Markus; Leskovec, Jure (2017). "Why We Read Wikipedia". pp. 1591–1600. doi:10.1145/3038912.3052716.
↑ Lemmerich, Florian; Sáez-Trumper, Diego; West, Robert; Zia, Leila (2019). "Why the World Reads Wikipedia". pp. 618–626. doi:10.1145/3289600.3291021.
↑ Paranjape, Ashwin; West, Robert; Zia, Leila; Leskovec, Jure (2016). "Improving Website Hyperlink Structure Using Server Logs". pp. 615–624. doi:10.1145/2835776.2835832.
↑ West, Robert; Leskovec, Jure (2012). "Human wayfinding in information networks". pp. 619–628. doi:10.1145/2187836.2187920.
↑ Chierichetti, Flavio; Kumar, Ravi; Raghavan, Prabhakar; Sarlos, Tamas (2012). "Are web users really Markovian?". pp. 609–618. doi:10.1145/2187836.2187919.
↑ Chen, Zhongxue; Singer, Philipp; Helic, Denis; Taraghi, Behnam; Strohmaier, Markus (2014). "Detecting Memory and Structure in Human Navigation Patterns Using Markov Chain Models of Varying Order". PLoS ONE 9 (7): e102070. ISSN 1932-6203. doi:10.1371/journal.pone.0102070.
Retrieved from "https://meta.wikimedia.org/w/index.php?title=Research:How_we_read_Wikipedia_and_reader_navigation&oldid=22242709" |
Farina, Alberto ; Sciunzi, Berardino ; Valdinoci, Enrico
We use a Poincaré type formula and level set analysis to detect one-dimensional symmetry of stable solutions of possibly degenerate or singular elliptic equations of the form
\phantom{\rule{0.166667em}{0ex}}\mathrm{div}\phantom{\rule{0.166667em}{0ex}}\left(a\left(|\nabla u\left(x\right)|\right)\nabla u\left(x\right)\right)+f\left(u\left(x\right)\right)=0.
Our setting is very general and, as particular cases, we obtain new proofs of a conjecture of De Giorgi for phase transitions in
{ℝ}^{2}
{ℝ}^{3}
and of the Bernstein problem on the flatness of minimal area graphs in
{ℝ}^{3}
. A one-dimensional symmetry result in the half-space is also obtained as a byproduct of our analysis. Our approach is also flexible to very degenerate operators: as an application, we prove one-dimensional symmetry for
1
-Laplacian type operators.
Classification : 32H02, 30C45
author = {Farina, Alberto and Sciunzi, Berardino and Valdinoci, Enrico},
title = {Bernstein and {De} {Giorgi} type problems: new results via a geometric approach},
AU - Farina, Alberto
AU - Sciunzi, Berardino
AU - Valdinoci, Enrico
TI - Bernstein and De Giorgi type problems: new results via a geometric approach
Farina, Alberto; Sciunzi, Berardino; Valdinoci, Enrico. Bernstein and De Giorgi type problems: new results via a geometric approach. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Série 5, Tome 7 (2008) no. 4, pp. 741-791. http://archive.numdam.org/item/ASNSP_2008_5_7_4_741_0/
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[2] L. Ambrosio and X. Cabré, Entire solutions of semilinear elliptic equations in
{ℝ}^{3}
and a conjecture of De Giorgi, J. Amer. Math. Soc. 13 (2000) (electronic), 725-739. | MR 1775735 | Zbl 0968.35041
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[6] L. Damascelli and B. Sciunzi, Regularity, monotonicity and symmetry of positive solutions of
m
-Laplace equations, J. Differential Equations 206 (2004), 483-515. | MR 2096703 | Zbl 1108.35069
[7] D. Danielli and N. Garofalo, Properties of entire solutions of non-uniformly elliptic equations arising in geometry and in phase transitions, Calc. Var. Partial Differential Equations 15 (2002), 451-491. | MR 1942128 | Zbl 1043.49018
[8] E. De Giorgi, Convergence problems for functionals and operators, In: “Proceedings of the International Meeting on Recent Methods in Nonlinear Analysis (Rome, 1978)”, Bologna, 1979, 131-188, Pitagora Editrice. | MR 533166 | Zbl 0405.49001
{C}^{1+\alpha }
local regularity of weak solutions of degenerate elliptic equations, Nonlinear Anal. 7 (1983), 827-850. | MR 709038 | Zbl 0539.35027
[10] L. C. Evans and R. F. Gariepy, “Measure theory and fine properties of functions”, Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992. | MR 1158660 | Zbl 0804.28001
[11] A. Farina, “Propriétés qualitatives de solutions d'équations et systèmes d'équations non-linéaires”, Habilitation à diriger des recherches, Paris VI, 2002.
[12] A. Farina, One-dimensional symmetry for solutions of quasilinear equations in
{ℝ}^{2}
, Boll. Unione Mat. Ital. Sez. B Artic. Ric. Mat. 6 (2003), 685-692. | MR 2014827 | Zbl 1115.35045
[13] A. Farina, Liouville-type theorems for elliptic problems, In: “Handbook of Differential Equations: Stationary Partial Differential Equations” M. Chipot (ed.), Vol. IV Elsevier B. V., Amsterdam, 2007, 61-116. | MR 2569331 | Zbl 1191.35128
[14] A. Farina, B. Sciunzi and E. Valdinoci, Bernstein and De Giorgi type problems: new results via a geometric approach, Preliminary version of this paper, available at http://cvgmt.sns.it/papers/, 2007. | MR 2483642 | Zbl 1180.35251
[15] A. Farina and E. Valdinoci,
1
D symmetry for solutions of semilinear and quasilinear elliptic equations, Preprint, 2008. | MR 2413100 | Zbl 1228.35105
[16] D. Fischer-Colbrie and R. Schoen, The structure of complete stable minimal surfaces in
3
-manifolds of nonnegative scalar curvature, Comm. Pure Appl. Math. 33 (1980), 199-211. | MR 562550 | Zbl 0439.53060
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Lepton - New World Encyclopedia
Previous (Leptis Magna)
Next (Les Paul)
Not-to-scale schematic of the classical view of a helium atom, showing two protons (red), two neutrons (green), and two electrons (yellow). Each electron belongs to the lepton group of elementary particles.
In particle physics, a lepton is one of the elementary (or fundamental) particles that are the building blocks of matter. Elementary particles are classified as fermions and bosons, and fermions are subdivided into leptons and quarks. A lepton is a fermion that does not experience the strong interaction (or strong nuclear force), which involves coupling with the bosons known as gluons. In other words, leptons are those fermions that "ignore" gluons. By comparison, quarks are fermions that couple with gluons to form composite particles such as protons and neutrons.
2 Properties of leptons
2.1 Quantum spin
3 Table of the leptons
As is the case for all fundamental particles, the lepton has properties of both a wave and a particle—it exhibits what is known as "wave-particle duality." The usual convention is to refer to such unified wave-particle fundamental entities as just "particles." The particle aspect is point-like even at scales thousands of times smaller than the proton size.
In the quantum view, there is a quantum probability wave and the particles jitter about in it over time.
According to the Oxford English Dictionary, the name "lepton" (from Greek leptos) was first used by physicist Léon Rosenfeld in 1948:
Following a suggestion of Prof. C. Møller, I adopt—as a pendant to "nucleon"—the denomination "lepton" (from λεπτός, small, thin, delicate) to denote a particle of small mass.[1] The name originated prior to the discovery in the 1970s of the heavy tau lepton, which is nearly twice the mass of a proton.
Properties of leptons
As is the case for all fundamental particles, the lepton is a unified entity of wave and particle—the wave-particle duality of quantum physics. The wave "tells" the particle what to do over time, while the interactions of the particle "tell" the wave how to develop and resonate. The particle aspect is point-like even at scales thousands of times smaller than the proton size. The usual convention is to refer to such unified wave-particle fundamental entities as just 'particles'.
There are three known flavors of lepton: the electron, the muon, and the tau. Each flavor is represented by a pair of particles called a weak doublet. One is a massive charged particle that bears the same name as its flavor (like the electron). The other is a nearly massless neutral particle called a neutrino (such as the electron neutrino). All six of these particles have corresponding antiparticles (such as the positron or the electron antineutrino). All known charged leptons have a single unit of negative or positive electric charge (depending on whether they are particles or antiparticles) and all of the neutrinos and antineutrinos have zero electric charge. The charged leptons have two possible spin states, while only one helicity is observed for the neutrinos (all the neutrinos are left-handed, and all the antineutrinos are right-handed).
The masses of the leptons also obey a simple relation, known as the Koide formula, but at present this relationship cannot be explained.
When particles interact, generally the number of leptons of the same type (electrons and electron neutrinos, muons and muon neutrinos, tau leptons and tau neutrinos) remains the same. This principle is known as conservation of lepton number. Conservation of the number of leptons of different flavors (for example, electron number or muon number) may sometimes be violated (as in neutrino oscillation). A much stronger conservation law is the total number of leptons of all flavors, which is violated by a tiny amount in the Standard Model by the so-called chiral anomaly.
The couplings of the leptons to gauge bosons are flavor-independent. This property is called lepton universality and has been tested in measurements of the tau and muon lifetimes and of Z-boson partial decay widths, particularly at the SLC and LEP experiments.
Fermions and bosons are distinguished by their quantum spin and the type of quantum probability statistics they obey: Fermi-Dirac probability or Bose-Einstein probability, neither of which is like classical probability. (This is a rough illustration of the difference: (one) The probability of two classical coins coming up the same side—HH or TT—is 50 percent. (two) For two boson coins, the probability of such a pair is 100 percent. (three) For two fermion coins, the probability of a pair is exactly zero percent, it is forbidden, and you always get HT. Fermions are said to have quantum spin -½, giving them the odd property of having to be rotated 720° in order to get back to where you started. (A familiar example of this sort of behavior is the Moebius Strip.) Bosons have quantum spin -1, and take the usual 360° to rotate back to where they started.
Table of the leptons
Charged lepton / antiparticle Neutrino / antineutrino
Electric charge (e)
Electron / Positron
{\displaystyle e^{-}\,/\,e^{+}}
−1 / +1 0.511 Electron neutrino / Electron antineutrino
{\displaystyle \nu _{e}\,/\,{\overline {\nu }}_{e}}
{\displaystyle \mu ^{-}\,/\,\mu ^{+}}
−1 / +1 105.7 Muon neutrino / Muon antineutrino
{\displaystyle \nu _{\mu }\,/\,{\overline {\nu }}_{\mu }}
{\displaystyle \tau ^{-}\,/\,\tau ^{+}}
{\displaystyle \nu _{\tau }\,/\,{\overline {\nu }}_{\tau }}
Note that the neutrino masses are known to be non-zero because of neutrino oscillation, but their masses are sufficiently light that they have not been measured directly as of 2007. The names "mu" and "tau" seem to have been selected due to their places in the Greek alphabet; mu is seven letters after epsilon (electron), whereas tau is seven letter after mu.
↑ Rosenfeld, Léon (1948). Nuclear Forces. Interscience Publishers, New York, xvii.
↑ 2.0 2.1 2.2 Laboratory measurements and limits for neutrino properties.
Griffiths, David J. 1987. Introduction to Elementary Particles. New York: Wiley. ISBN 0-471-60386-4
Povh, Bogdan. 1995. Particles and Nuclei: An Introduction to the Physical Concepts. Berlin: Springer-Verlag. ISBN 0-387-59439-6
Leptons Georgia State University.
History of "Lepton"
Retrieved from https://www.newworldencyclopedia.org/p/index.php?title=Lepton&oldid=1012709 |
Bijection, Injection and Surjection Problem Solving Practice Problems Online | Brilliant
10
dice are rolled. Each die is a regular
6
-sided die with numbers
1
6
labelled on the sides. How many different distinct sums of all 10 numbers are possible?
The sums
5 = 1+4
5 = 2 + 3
are the same sum (namely 5), and should only be counted once.
A=\{1, 2, \ldots, 6\}
B=\{b_1, b_2, \ldots , {b}_{5}\},
b_i
are all distinct. How many distinct functions from
A
B
are possible?
In New York city, all the streets are arranged in a grid. A hospital is located 5 blocks east and 6 blocks north of an accident. How many ways are there to get there, if we only go 1 block north or 1 block east at each intersection?
A product day is a day where the numerical value of the date, multiplied by the numerical value of the month, is equal to the numerical value of the year.
Starting from 1st of January, 1AD, how many product days are there?
Assume that the current convention of days and months extend back to 1st of January, 1 AD. You should include the 1st of January, 1AD into your count.
25th of December, 300 AD is a product day, because
25 \times 12 = 300
6
courses: Chemistry, Physics, Math, English, French and Biology. She has one textbook for each course and wants to place them on a shelf. How many ways can she arrange the textbooks so that the English textbook is placed before the French textbook?
If the books are ordered from left to right, then the "English textbook is placed before the French textbook" means that the English textbook was placed to the left of the French textbook. They do not need to be immediately beside each other. |
This problem is a checkpoint for simplifying expressions. It will be referred to as Checkpoint 7A .
4x^2+3x-7+(-2x^2)-2x+(-3)
-3x^2-2x+5+4x^2-7x+6
Answers and extra practice for the Checkpoint problems are located in the back of your printed textbook or in the Reference Tab of your eBook. If you have an eBook for Core Connections, Course 1, login and then click the following link: Checkpoint 7A: Simplifying Expressions |
Determine the following definite integrals: \int_{-1}^{3}|2x-x^{2}|dx
Determine the following definite integrals:
{\int }_{-1}^{3}|2x-{x}^{2}|dx
Evaluate the integrals:
f\left(x\right)={\int }_{-1}^{3}|2x-{x}^{2}|dx
Eliminate the absolutes,
-1\le x\le 0\text{ }f\left(x\right)=\left(-2x+{x}^{2}\right)
0\le x\le 2\text{ }f\left(x\right)=\left(2x-{x}^{2}\right)
2\le x\le 3\text{ }f\left(x\right)=\left(-2x+{x}^{2}\right)
f\left(x\right)={\int }_{-1}^{0}\left(-2x+{x}^{2}\right)dx+{\int }_{0}^{2}\left(2x-{x}^{2}\right)dx+{\int }_{2}^{3}\left(-2x+{x}^{2}\right)dx
={\left[-2\left(\frac{{x}^{2}}{2}\right)+\frac{{x}^{3}}{3}\right]}_{-1}^{0}+{\left[2\left(\frac{{x}^{2}}{2}\right)-\frac{{x}^{3}}{3}\right]}_{0}^{2}+{\left[-2\left(\frac{{x}^{2}}{2}\right)+\frac{{x}^{3}}{3}\right]}_{2}^{3}
={\left[-{x}^{2}+\frac{{x}^{3}}{3}\right]}_{-1}^{0}+{\left[{x}^{2}-\frac{{x}^{3}}{3}\right]}_{0}^{2}+{\left[-{x}^{2}+\frac{{x}^{3}}{3}\right]}_{2}^{3}
=\left[0-\left(-{\left(-1\right)}^{2}+\frac{{\left(-1\right)}^{3}}{3}\right)\right]+\left[{2}^{2}-\frac{{2}^{3}}{3}-0\right]+\left[\left(-{3}^{2}+\frac{{3}^{3}}{3}\right)-\left(-{2}^{2}+\frac{{2}^{3}}{3}\right)\right]
Evaulating the trigonometric integral
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx
To do this, I let
x=\mathrm{tan}u
dx={\mathrm{sec}}^{2}udu
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx=\int \frac{{\mathrm{sec}}^{2}\left\{u\right\}du}{{\left({\mathrm{tan}}^{2}\left\{u\right\}+1\right)}^{2}}
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx=\int \frac{1}{{\mathrm{sec}}^{2}\left\{u\right\}}du=\int {\mathrm{cos}}^{2}\left\{u\right\}du
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx=\int \frac{\mathrm{cos}\left\{\left(2u\right)\right\}+1}{2}du=\frac{\mathrm{sin}\left(u\right)}{4}+\frac{u}{2}
\int \frac{1}{{\left({x}^{2}+1\right)}^{2}}dx=\frac{\sqrt{1-{\mathrm{cos}}^{2}\left\{u\right\}}}{4}+\frac{u}{2}
Now, I think I am right so far but I do not know have to get rid of the u in the
{\mathrm{cos}}^{2}\left(u\right)
term. Please help.
Integral of
\int \left\{\frac{{\mathrm{cos}}^{4}x+{\mathrm{sin}}^{4}x}{\sqrt{1+\mathrm{cos}4x}}dx\right\}
60{m}^{4}-120{m}^{3}n+50{m}^{2}{n}^{2}
{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}15{\mathrm{sin}}^{4}3x\mathrm{cos}3xdx
\int \frac{dx}{\mathrm{sin}3x\mathrm{tan}3x}
{\int }_{0}^{x}\left(2x-y\right)dy
Evaluate the given integral.
\int \frac{x}{{x}^{2}+1}dx |
Simulation and Prediction in the App - MATLAB & Simulink - MathWorks Nordic
How to Plot Simulated and Predicted Model Output
Definition: Confidence Interval
To create a model output plot for parametric linear and nonlinear models in the System Identification app, select the Model output check box in the Model Views area. By default, this operation estimates the initial states from the data and plots the output of selected models for comparison.
To learn how to interpret the model output plot, see Interpret the Model Output Plot.
To change plot settings, see Change Model Output Plot Settings.
For general information about creating and working with plots, see Working with Plots.
The following figure shows a sample Model Output plot, created in the System Identification app.
The model output plot shows different information depending on the domain of the input-output validation data, as follows:
For time-domain validation data, the plot shows simulated or predicted model output.
For frequency-domain data, the plot shows the amplitude of the model response to the frequency-domain input signal. The model response is equal to the product of the Fourier transform of the input and the model frequency function.
For frequency-response data, the plot shows the amplitude of the model frequency response.
For linear models, you can estimate a model using time-domain data, and then validate the model using frequency domain data. For nonlinear models, you can only use time-domain data for both estimation and validation.
The right side of the plot displays the percentage of the output that the model reproduces (Best Fit), computed using the following equation:
\text{Best Fit}=\left(1-\frac{|y-\stackrel{^}{y}|}{|y-\overline{y}|}\right)×100
In this equation, y is the measured output,
\stackrel{^}{y}
is the simulated or predicted model output, and
\overline{y}
is the mean of y. 100% corresponds to a perfect fit, and 0% indicates that the fit is no better than guessing the output to be a constant (
\stackrel{^}{y}=\overline{y}
Because of the definition of Best Fit, it is possible for this value to be negative. A negative best fit is worse than 0% and can occur for the following reasons:
The estimation algorithm failed to converge.
The model was not estimated by minimizing
|y-\stackrel{^}{y}|
. Best Fit can be negative when you minimized 1-step-ahead prediction during the estimation, but validate using the simulated output
\stackrel{^}{y}
The validation data set was not preprocessed in the same way as the estimation data set.
The following table summarizes the Model Output plot settings.
Model Output Plot Settings
Display confidence intervals.
Confidence intervals are only available for simulated model output of linear models. Confidence internal are not available for nonlinear ARX and Hammerstein-Wiener models.
See Definition: Confidence Interval.
Change between simulated output or predicted output.
Prediction is only available for time-domain validation data.
Select Options > Simulated output or Options > k step ahead predicted output.
To change the prediction horizon, select Options > Set prediction horizon, and select the number of samples.
To enter your own prediction horizon, select Options > Set prediction horizon > Other. Enter the value in terms of the number of samples.
Display the actual output values (Signal plot), or the difference between model output and measured output (Error plot).
Select Options > Signal plot or Options > Error plot.
(Time-domain validation data only)
Set the time range for model output and the time interval for which the Best Fit value is computed.
Select Options > Customized time span for fit and enter the minimum and maximum time values. For example:
Select a different output.
The confidence interval corresponds to the range of output values with a specific probability of being the actual output of the system. The toolbox uses the estimated uncertainty in the model parameters to calculate confidence intervals and assumes the estimates have a Gaussian distribution.
For example, for a 95% confidence interval, the region around the nominal curve represents the range of values that have a 95% probability of being the true system response. You can specify the confidence interval as a probability (between 0 and 1) or as the number of standard deviations of a Gaussian distribution. For example, a probability of 0.99 (99%) corresponds to 2.58 standard deviations.
In the app, you can display a confidence interval on the plot to gain insight into the quality of a linear model. To learn how to show or hide confidence interval, see Change Model Output Plot Settings. |
networks(deprecated)/daughter - Maple Help
Home : Support : Online Help : networks(deprecated)/daughter
find daughters in a directed tree
daughter(v, G)
daughter(G)
Given a vertex, v, this routine reports the set of known daughters of v in the graph G. Such relationships are not always present in a given graph but are explicitly established by routines such as spantree() and shortpathtree(). Routines such as path() rely on this information when looking for paths and will return FAIL if it is not present.
If only the graph is mentioned then the actual daughter table indexed by vertices and specifying all known daughters in G is returned. Modifications to this table affect the actual graph.
In the two argument case, the first argument v may also be a set of vertices in which case the result is the set of daughters of the subgraph induced by v in G (ie. (the union of the daughters of each vertex in v minus the vertices in v).
This routine is normally loaded using the command with(networks) but may also be referenced using the full name networks[daughter](...).
\mathrm{with}\left(\mathrm{networks}\right):
G≔\mathrm{petersen}\left(\right):
\mathrm{daughter}\left(1,G\right)
\textcolor[rgb]{0,0,1}{\varnothing }
T≔\mathrm{shortpathtree}\left(G,1\right):
\mathrm{daughter}\left(8,T\right)
\textcolor[rgb]{0,0,1}{\varnothing }
\mathrm{daughter}\left(1,T\right)
{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}}
\mathrm{tbl}≔\mathrm{daughter}\left(T\right)
\textcolor[rgb]{0,0,1}{\mathrm{tbl}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{table}\textcolor[rgb]{0,0,1}{}\left([\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{5}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{6}\textcolor[rgb]{0,0,1}{=}{\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{7}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{8}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{10}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\varnothing }]\right)
\mathrm{path}\left([9,1],T\right)
\textcolor[rgb]{0,0,1}{\mathrm{FAIL}} |
Support vector machine (SVM) for one-class and binary classification - MATLAB - MathWorks Australia
f\left(x\right)=\left(x/s\right)\prime \beta +b.
\left\{\begin{array}{l}{\alpha }_{j}\left[{y}_{j}f\left({x}_{j}\right)-1+{\xi }_{j}\right]=0\\ {\xi }_{j}\left(C-{\alpha }_{j}\right)=0\end{array}
f\left({x}_{j}\right)=\varphi \left({x}_{j}\right)\prime \beta +b,
0.5\sum _{jk}{\alpha }_{j}{\alpha }_{k}G\left({x}_{j},{x}_{k}\right)
{\alpha }_{1},...,{\alpha }_{n}
\sum {\alpha }_{j}=n\nu
0\le {\alpha }_{j}\le 1
f\left(x\right)=x\prime \beta +b,
2/‖\beta ‖.
‖\beta ‖
0.5{‖\beta ‖}^{2}+C\sum {\xi }_{j}
{y}_{j}f\left({x}_{j}\right)\ge 1-{\xi }_{j}
{\xi }_{j}\ge 0
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}{x}_{j}\prime {x}_{k}-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}x\prime {x}_{j}+\stackrel{^}{b}.
\stackrel{^}{b}
{\stackrel{^}{\alpha }}_{j}
\stackrel{^}{\alpha }
\text{sign}\left(\stackrel{^}{f}\left(z\right)\right).
0.5\sum _{j=1}^{n}\sum _{k=1}^{n}{\alpha }_{j}{\alpha }_{k}{y}_{j}{y}_{k}G\left({x}_{j},{x}_{k}\right)-\sum _{j=1}^{n}{\alpha }_{j}
\sum {\alpha }_{j}{y}_{j}=0
0\le {\alpha }_{j}\le C
\stackrel{^}{f}\left(x\right)=\sum _{j=1}^{n}{\stackrel{^}{\alpha }}_{j}{y}_{j}G\left(x,{x}_{j}\right)+\stackrel{^}{b}.
{C}_{j}=n{C}_{0}{w}_{j}^{\ast },
{x}_{j}^{\ast }=\frac{{x}_{j}-{\mu }_{j}^{\ast }}{{\sigma }_{j}^{\ast }},
\begin{array}{c}{\mu }_{j}^{\ast }=\frac{1}{\sum _{k}{w}_{k}^{*}}\sum _{k}{w}_{k}^{*}{x}_{jk},\\ {\left({\sigma }_{j}^{\ast }\right)}^{2}=\frac{{v}_{1}}{{v}_{1}^{2}-{v}_{2}}\sum _{k}{w}_{k}^{*}{\left({x}_{jk}-{\mu }_{j}^{\ast }\right)}^{2},\\ {v}_{1}=\sum _{j}{w}_{j}^{*},\\ {v}_{2}=\sum _{j}{\left({w}_{j}^{*}\right)}^{2}.\end{array}
\sum _{j=1}^{n}{\alpha }_{j}=n\nu . |
Diversification of Portfolios - MATLAB & Simulink Example - MathWorks í•œêµ
{\mathrm{μ}}_{0}
\begin{array}{l}\mathrm{min}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}\\ \mathrm{st}.\text{ââââ}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathit{x}}_{\mathit{i}}=1\\ \text{ââââââââââ}{\mathit{x}}^{\mathit{T}}\mathrm{μ}â¥{\mathrm{μ}}_{0}\\ \text{ââââââââââ}\mathit{x}â¥0\end{array}
\begin{array}{l}\mathrm{min}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}\\ \mathrm{st}.\text{ââââ}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathit{x}}_{\mathit{i}}=1\\ \text{ââââââââââ}\mathit{x}â¥0\end{array}
\mathit{X}
\mathit{X}=\left\{\mathit{x}\text{â}|\text{â}\mathit{x}â¥0,\text{â}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathit{x}}_{\mathit{i}}=1\right\}
\underset{\mathit{x}â\mathit{X}}{\mathrm{min}}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}
{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\text{â}\mathit{x}
\mathrm{HH}\left(\mathit{x}\right)=\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathit{x}}_{\mathit{i}}^{2}
\underset{\mathit{x}â\mathit{X}}{\mathrm{min}}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathit{x}
\mathit{X}
\mathit{X}
\underset{\mathit{x}â\mathit{X}}{\mathrm{min}}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}+{\mathrm{λ}}_{\mathrm{HH}}{\mathit{x}}^{\mathit{T}}\mathit{x}
{\mathrm{λ}}_{\mathrm{HH}}=0
{\mathrm{λ}}_{\mathrm{HH}}â\infty ,
\mathrm{MDP}\left(\mathit{x}\right)=-\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathrm{Ï}}_{\mathit{i}}{\mathit{x}}_{\mathit{i}}
{\mathrm{Ï}}_{\mathit{i}}
\mathit{i}
\mathrm{Ï}\left(\mathit{x}\right)=\text{â}\frac{{\mathit{x}}^{\mathit{T}}\mathrm{Ï}}{\sqrt{{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}}}
\mathrm{Ï}\left(\mathit{x}\right)
\mathrm{Ï}\left(\mathit{x}\right)>1
\mathrm{Ï}\left(\mathit{x}\right)â1
\mathrm{Ï}\left(\mathit{x}\right)
\underset{\mathit{x}â\mathit{X}}{\mathrm{max}}\text{âââ}\frac{{\mathrm{Ï}}^{\mathit{T}}\mathit{x}}{\sqrt{{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}}}
Instead of solving the nonlinear, nonconvex optimization problem above, the problem is rewritten as an equivalent convex quadratic problem. The reformulation follows the same idea used to maximize the Sharpe ratio (Cornuejols & Tütüncü [3]). The equivalent quadratic problem results in the following:
\begin{array}{l}\underset{\mathit{y}â¥0,\text{â}\mathrm{Ï}â¥0}{\mathrm{min}}\text{â}{\text{â}\text{â}\mathit{y}}^{\mathit{T}}\mathrm{Σ}\mathit{y}\\ \text{â}\text{â}\text{â}\text{â}\mathit{s}.\mathit{t}{\text{âââââ}\mathrm{Ï}}^{\mathit{T}}\mathit{y}=1\\ \text{âââââââââââââââ}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}{\mathit{y}}_{\mathit{i}}=\mathrm{Ï}\end{array}
\frac{\mathit{y}}{\mathrm{Ï}}
{\stackrel{Ë}{\mathrm{λ}}}_{\mathrm{MDP}}>0
\underset{\mathit{x}â\mathit{X}}{\text{â}\mathrm{min}}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}-{\mathrm{λ}}_{\mathrm{MDP}}\text{â}{\mathrm{Ï}}^{\mathit{T}}\mathit{x}
{\mathrm{λ}}_{\mathrm{MDP}}=0
{\mathrm{λ}}_{\mathrm{MDP}}â\left[0,{\stackrel{Ë}{\mathrm{λ}}}_{\mathrm{MDP}}\right]
\mathrm{ERC}\left(\mathit{x}\right)=-\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}\mathrm{ln}\left({\mathit{x}}_{\mathit{i}}\right)
\begin{array}{l}\underset{\mathit{y}â¥0}{\mathrm{min}}\text{â}{\mathit{y}}^{\mathit{T}}\mathrm{Σ}\text{â}\mathit{y}\\ \mathrm{st}.\text{ââ}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}\mathrm{ln}\left({\mathit{x}}_{\mathit{i}}\right)â¥\mathit{c}\end{array}
\mathit{x}
\mathit{x}=\frac{\mathit{y}}{{â}_{\mathit{i}}{\mathit{y}}_{\mathit{i}}}
\mathit{c}>0
\underset{\mathit{x}â\mathit{X}}{\mathrm{min}}\text{âââ}{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}-{\mathrm{λ}}_{\mathrm{ERC}}\text{â}\underset{\mathit{i}=1}{\overset{\mathit{n}}{â}}\mathrm{ln}\left({\mathit{x}}_{\mathit{i}}\right)
{\stackrel{Ë}{\mathrm{λ}}}_{\mathrm{ERC}}
{\mathrm{λ}}_{\mathrm{ERC}}=0
{\mathrm{λ}}_{\mathrm{ERC}}â\left[0,{\stackrel{Ë}{\mathrm{λ}}}_{\mathrm{ERC}}\right]
TnonZero=1×7 table
{\left(\mathrm{PRC}\right)}_{\mathit{i}}=\frac{{\mathit{x}}_{\mathit{i}}{\left(\mathrm{Σ}\mathit{x}\right)}_{\mathit{i}}}{{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}}
\mathit{i}
{\mathrm{Ï}}_{\mathrm{iP}}=\frac{{\left(\mathrm{Σ}\mathit{x}\right)}_{\mathit{i}}}{{\mathrm{Ï}}_{\mathit{i}}\sqrt{{\mathit{x}}^{\mathit{T}}\mathrm{Σ}\mathit{x}}}
Maillard, S., Roncalli, T., & Teïletche, J. "The Properties of Equally Weighted Risk Contribution Portfolios." The Journal of Portfolio Management, 36(4), 2010, pp. 60-70.
Tütüncü, R., Peña, J., Cornuéjols, G. Optimization Methods in Finance. United Kingdom: Cambridge University Press, 2018. |
I_st=stim_amplitudeiftime≥stim_start∧time≦stim_end∧time-stim_start-⌊time-stim_startstim_period⌋stim_period≦stim_duration0otherwiseddtimeV=dVdtdVdt=-i_Na+i_Ca_L+i_Ca_T+i_Kr+i_Ks+i_K_ATP+i_to+i_K1+i_Kp+i_p_Ca+i_Na_b+i_Ca_b+i_NaK+i_NaCa+i_ns_Ca+I_st
E_Na=RTFlnNaoNaii_Na=g_Nam3hjV-E_Na
E0_m=V+47.13alpha_m=0.32E0_m1-ⅇ-0.1E0_mif|E0_m|≥delta_m3.2otherwisebeta_m=0.08ⅇ-V11ddtimem=alpha_m1-m-beta_mm
alpha_h=0.135ⅇ80+V-6.8ifV<-400otherwisebeta_h=3.56ⅇ0.079V+310000ⅇ0.35VifV<-4010.131+ⅇV+10.66-11.1otherwiseddtimeh=alpha_h1-h-beta_hh
Component: fast_sodium_current_j_gate
alpha_j=-127140ⅇ0.2444V+3.474 e -5ⅇ-0.04391VV+37.781+ⅇ0.311V+79.23ifV<-400otherwisebeta_j=0.1212ⅇ-0.01052V1+ⅇ-0.1378V+40.14ifV<-400.3ⅇ-2.535 e -7V1+ⅇ-0.1V+32otherwiseddtimej=alpha_j1-j-beta_jj
I_CaCa=P_Ca22VF2RTgamma_CaiCaiⅇ2VFRT-gamma_CaoCaoⅇ2VFRT-1I_CaNa=P_Na12VF2RTgamma_NaiNaiⅇ1VFRT-gamma_NaoNaoⅇ1VFRT-1I_CaK=P_K12VF2RTgamma_KiKiⅇ1VFRT-gamma_KoKoⅇ1VFRT-1i_CaCa=dff_CaI_CaCai_CaNa=dff_CaI_CaNai_CaK=dff_CaI_CaKi_Ca_L=i_CaCa+i_CaK+i_CaNa
E0_d=V+10d_infinity=11+ⅇ-E0_d6.24tau_d=10.0356.24if|E0_d|<1 e -5d_infinity1-ⅇ-E0_d6.240.035E0_dotherwisealpha_d=d_infinitytau_dbeta_d=1-d_infinitytau_dddtimed=alpha_d1-d-beta_dd
f_infinity=11+ⅇV+328+0.61+ⅇ50-V20tau_f=10.0197ⅇ-0.0337V+102+0.02alpha_f=f_infinitytau_fbeta_f=1-f_infinitytau_fddtimef=alpha_f1-f-beta_ff
Component: L_type_Ca_channel_f_Ca_gate
f_Ca=11+CaiKm_Ca
i_Ca_T=g_CaTbbgV-E_Ca
Component: T_type_Ca_channel_b_gate
b_inf=11+ⅇ-V+1410.8tau_b=3.7+6.11+ⅇV+254.5ddtimeb=b_inf-btau_b
Component: T_type_Ca_channel_g_gate
g_inf=11+ⅇV+605.6tau_g=-0.875V+12ifV≦012otherwiseddtimeg=g_inf-gtau_g
g_Kr=0.02614Ko5.4Rect=11+ⅇV+922.4i_Kr=g_KrxrRectV-E_K
Component: rapid_delayed_rectifier_potassium_current_xr_gate
xr_infinity=11+ⅇ-V+21.57.5tau_xr=10.00138V+14.21-ⅇ-0.123V+14.2+0.00061V+38.9ⅇ0.145V+38.9-1ddtimexr=xr_infinity-xrtau_xr
E_Ks=RTFlnKo+PNaKNaoKi+PNaKNaig_Ks=0.1251+0.61+3.8 e -5Cai1.4i_Ks=g_Ksxs1xs2V-E_Ks
Component: slow_delayed_rectifier_potassium_current_xs1_gate
xs1_infinity=11+ⅇ-V-1.516.7tau_xs1=17.19 e -5V+301-ⅇ-0.148V+30+0.000131V+30ⅇ0.0687V+30-1ddtimexs1=xs1_infinity-xs1tau_xs1
xs2_infinity=11+ⅇ-V-1.516.7tau_xs2=47.19 e -5V+301-ⅇ-0.148V+30+0.000131V+30ⅇ0.0687V+30-1ddtimexs2=xs2_infinity-xs2tau_xs2
g_K1=0.75Ko5.4E_K=RTFlnKoKii_K1=g_K1K1_infinityV-E_K
Component: time_independent_potassium_current_K1_gate
alpha_K1=1.021+ⅇ0.2385V-E_K-59.215beta_K1=10.49124ⅇ0.08032V-E_K+5.476+ⅇ0.06175V-E_K-594.311+ⅇ-0.5143V-E_K+4.753K1_infinity=alpha_K1alpha_K1+beta_K1
Component: plateau_potassium_current
Kp=11+ⅇ7.488-V5.98i_Kp=g_KpKpV-E_K
Component: ATP_sensitive_potassium_current
g_K_ATP=0.000193nicholsareapATP=11+ATPikATPhATPGKbaraATP=g_K_ATPpATPKo4nATPi_K_ATP=GKbaraATPV-E_K
g_to=00.5rvdv=ⅇV100i_to=g_tozdv3ydvrvdvV-E_K
Component: transient_outward_current_zdv_gate
alpha_zdv=10ⅇV-40251+ⅇV-4025beta_zdv=10ⅇ-V+90251+ⅇ-V+9025tau_zdv=1alpha_zdv+beta_zdvzdv_ss=alpha_zdvalpha_zdv+beta_zdvddtimezdv=zdv_ss-zdvtau_zdv
Component: transient_outward_current_ydv_gate
alpha_ydv=0.0151+ⅇV+605beta_ydv=0.1ⅇV+2551+ⅇV+255tau_ydv=1alpha_ydv+beta_ydvydv_ss=alpha_ydvalpha_ydv+beta_ydvddtimeydv=ydv_ss-ydvtau_ydv
Component: sarcolemmal_calcium_pump
i_p_Ca=I_pCaCaiK_mpCa+Cai
i_Na_b=g_NabV-E_Na
E_Ca=RT2FlnCaoCaii_Ca_b=g_CabV-E_Ca
sigma=17ⅇNao67.3-1f_NaK=11+0.1245ⅇ-0.1VFRT+0.0365sigmaⅇ-VFRTi_NaK=I_NaKf_NaK11+K_mNaiNai2KoKo+K_mKo
Component: non_specific_calcium_activated_current
P_ns_Ca=01.75 e -7I_ns_Na=P_ns_Ca12VF2RTgamma_NaiNaiⅇ1VFRT-gamma_NaoNaoⅇ1VFRT-1I_ns_K=P_ns_Ca12VF2RTgamma_KiKiⅇ1VFRT-gamma_KoKoⅇ1VFRT-1i_ns_Na=I_ns_Na11+K_m_ns_CaCai3i_ns_K=I_ns_K11+K_m_ns_CaCai3i_ns_Ca=i_ns_Na+i_ns_K
i_NaCa=K_NaCaⅇgamman_NaCa-2VFRTNain_NaCaCao-ⅇgamma-1n_NaCa-2VFRTNaon_NaCaCai1+d_NaCaCaiNaon_NaCa+CaoNain_NaCa1+Cai0.0069
Component: calcium_dynamics
V_JSR=0.00480.68V_myoV_NSR=0.05520.68V_myoddtimeAPtrack=1001-APtrack-0.5APtrackifdVdt>150-0.5APtrackotherwiseddtimeAPtrack2=1001-APtrack2-0.5APtrack2ifAPtrack<0.2∧APtrack>0.18-0.5APtrack2otherwiseddtimeAPtrack3=1001-APtrack3-0.5APtrack3ifAPtrack<0.2∧APtrack>0.18-0.01APtrack3otherwiseddtimeCainfluxtrack=-1A_capi_CaCa+i_Ca_T-i_NaCa+i_p_Ca+i_Ca_b2V_myoFifAPtrack>0.20ifAPtrack2>0.01∧APtrack≦0.2-0.5CainfluxtrackotherwiseddtimeOVRLDtrack=501-OVRLDtrackif11+K_mCSQNCa_JSR>CSQNthresh∧OVRLDtrack3<0.37∧APtrack3<0.37-0.5OVRLDtrackotherwiseddtimeOVRLDtrack2=501-OVRLDtrack2ifOVRLDtrack>Logicthresh∧OVRLDtrack2<Logicthresh-0.5OVRLDtrack2otherwiseddtimeOVRLDtrack3=501-OVRLDtrack3ifOVRLDtrack>Logicthresh∧OVRLDtrack3<Logicthresh-0.01OVRLDtrack3otherwiseG_rel=G_rel_maxCainfluxtrack-delta_Ca_ithK_mrel+Cainfluxtrack-delta_Ca_ith1-APtrack2APtrack2ifCainfluxtrack>delta_Ca_ithG_rel_overload1-OVRLDtrack2OVRLDtrack2ifCainfluxtrack≦delta_Ca_ith∧OVRLDtrack2>00otherwisei_rel=G_relCa_JSR-Caii_up=I_upCaiCai+K_mupK_leak=I_upCa_NSR_maxi_leak=K_leakCa_NSRi_tr=Ca_NSR-Ca_JSRtau_trddtimeCa_JSR=0.0011+CSQN_maxK_mCSQNK_mCSQN+Ca_JSR2i_tr-i_relddtimeCa_NSR=0.001-i_trV_JSRV_NSR-i_leak+i_upddtimeCai=0.0011+CMDN_maxK_mCMDNK_mCMDN+Cai2+Tn_maxK_mTnK_mTn+Cai2-1A_capi_CaCa+i_Ca_T-i_NaCa+i_p_Ca+i_Ca_b2V_myoF+i_relV_JSRV_myo+i_leak-i_upV_NSRV_myo
volume=πpreplengthradius2V_myo=0.68volumeddtimeNai=-0.001i_Na+i_CaNa+i_Na_b+i_ns_Na+i_NaCa3+i_NaK3A_capV_myoFddtimeKi=-0.001i_CaK+i_Kr+i_Ks+i_K1+i_Kp+i_K_ATP+i_to+i_ns_K+-i_NaK2A_capV_myoF |
Positive Periodic Solution for Second-Order Singular Semipositone Differential Equations
Xiumei Xing, "Positive Periodic Solution for Second-Order Singular Semipositone Differential Equations", Abstract and Applied Analysis, vol. 2013, Article ID 310469, 7 pages, 2013. https://doi.org/10.1155/2013/310469
Xiumei Xing1
1School of Mathematics and Statistics, Yili Normal University, Yining City 835000, China
We study the existence of a positive periodic solution for second-order singular semipositone differential equation by a nonlinear alternative principle of Leray-Schauder. Truncation plays an important role in the analysis of the uniform positive lower bound for all the solutions of the equation. Recent results in the literature (Chu et al., 2010) are generalized.
In this paper, we study the existence of positive -periodic solutions for the following singular semipositone differential equation: where and the nonlinearity satisfies for some . In particular, the nonlinearity may have a repulsive singularity at , which means that Electrostatic or gravitational forces are the most important examples of singular interactions.
During the last two decades, the study of the existence of periodic solutions for singular differential equations has attracted the attention of many researchers [1–4]. Some strong force conditions introduced by Gordon [5] are standard in the related earlier works [6, 7]. Compared with the case of a strong singularity, the study of the existence of periodic solutions under the presence of a weak singularity is more recent [2, 8, 9], but has also attracted many researchers. Some classical tools have been used to study singular differential equations in the literature, including the method of upper and lower solutions [10], degree theory [11], some fixed point theorem in cones for completely continuous operators [12], Schauder’s fixed point theorem [8, 9, 13], and a nonlinear Leray-Schauder alternative principle [2, 3, 14, 15].
However the singular differential equations, in which there is the damping term, that is, the nonlinearity is dependent on the derivative, has not attracted much attention in the literature. Several existence results can be found in [14, 16, 17].
The aim of this paper is to further show that the nonlinear Leray-Schauder alternative principle can be applied to (1) in the semipositone cases, that is, for some .
The remainder of the paper is organized as follows. In Section 2, we state some known results. In Section 3, the main results of this paper are stated and proved. To illustrate our result, we select the following system: where , , , is a positive parameter, is a -periodic function.
In this paper, let us fix some notations to be used in the following: given , we write if for almost everywhere and it is positive in a set of positive measure. The usual -norm is denoted by . and the essential supremum and infinum of a given function , if they exist.
We say that associated to the periodic boundary conditions is nonresonant when its unique solutions is the trival one. When (4)-(5) is nonresonant, as a consequence of Fredholm’s alternative, the nonhomogeneous equation admits a unique -periodic solution, which can be written as where is the Green’s function of problem (4)-(5). Throughout this paper, we assume that the following standing hypothesis is satisfied.(A) The Green function , associated with (4)-(5), is positive for all .
In other words, the strict antimaximum principle holds for (4)-(5).
Definition 1. We say that (4) admits the antimaximum principle if (6) has a unique -periodic solution for any and the unique -periodic solution for all if .
Under hypothesis (A), we denote Thus and . We also use to denote the unique periodic solution of (6) with under condition (5), that is, . In particular, .
With the help of [18, 19], the authors give a sufficient condition to ensure that (4) admits the antimaximum principle in [14]. In order to state this result, let us define the functions
Lemma 2 (see [14, Corollary 2.6]). Assume that and the following two inequalities are satisfied: where . Then the Green’s function , associated with (5), is positive for all .
Next, recall a well-known nonlinear alternative principle of Leray-Schauder, which can be found in [20] and has been used by Meehan and O’Regan in [4].
Lemma 3. Assume is an open subset of a convex set in a normed linear space and . Let be a compact and continuous map. Then one of the following two conclusions holds:(I) has at least one fixed point in .(II)There exists and such that .
In applications below, we take with the norm and define .
In this section, we prove a new existence result of (1).
Theorem 4. Suppose that (4) satisfies (A) and Furthermore, assume that there exist three constants such that:(H1) for all .(H2) for , where the nonincreasing continuous function satisfies and .(H3), for all , where is nonincreasing in and , are nondecreasing in . (H4) where Then (1) has at least one positive periodic solution with .
Proof. For convinence, let us write , , where . Let First we show that has a solution satisfying (5), and for . If this is true, it is easy to see that will be a positive solution of (1)–(5) with .
Choose such that , and then let .
Consider the family of equations where , , and .
A -periodic solution of (17) is just a fixed of the operator equation where and is a completely continuous operator defined by where we have used the fact
We claim that for any -periodic solution of (17) satisfies Note that the solution of (17) is also satisfies the following equivalent equation Integrating (22) from to , we obtain
By the periodic boundary conditions, we have for some . Therefore,where we have used the assumption (11) and . Therefore, which implies that (21) holds. In particular, let in (17), we have
Choose such that , and then let . The following lemma holds.
Lemma 5. There exists an integer large enough such that, for all ,
Proof. The lower bound in (27) is established by using the strong force condition of . By condition (H2), there exists and a continuous function such that for all , where satisfies also the strong force condition like in (H2).
For , let , .
If , due to , (27) holds.
If , we claim that, for all , Otherwise, suppose that for some . Then it is easy to verify In fact, if , we obtain from (28) and, if , we have Integrating (22) (with ) from to , we deduce that where estimation (30) and the fact are used. This is a contradiction. Hence (29) holds.
Due to , that is, for some . By (29), there exists (without loss of generality, we assume .) such that and for .
It can be checked that where is defined by (15).
In fact, if is such that , we have and, if is such that , we have So (34) holds.
Using (17) (with ) for and the estimation (34), we have, for As , for all , so is strictly increasing on . We use to denote the inverse function of restricted to .
Suppose that (27) does not hold, that is, for some , . Then there would exist such that and Multiplying (17) (with ) by and integrating from to , we obtain
By the facts , , and the definition of , we can obtain , together with , implies that the second term and the third term are bounded. The first term is which is also bounded. As a consequence, there exists a such that On the other hand, by (H2), we can choose large enough such that for all . So (27) holds.
Furthermore, we can prove has a uniform positive lower bound .
Lemma 6. There exist a constant such that, for all ,
Proof. Multiplying (17) (with ) by and integrating from to , we obtain
In the same way as in the proof of (41), one way readily prove that the right-hand side of the above equality is bounded. On the other hand, if , by (H2), if . Thus we know that there exists a constant such that . Hence (43) holds.
Next, we will prove (17) has periodic solution .
For , we can choose such that , which together with (H4) imply Let . For , consider (17).
Next we claim that any fixed point of (18) for any must satisfy . So, by using the Leray-Schauder alternative principle, (17) (with ) has a periodic solution . Otherwise, assume that is a fixed point of (18) for some such that . Note that For , we have By (27) and assumption (H3), for all and , we have Therefore, This is a contradiction to the choice of and the claim is proved.
The fact and show that is a bounded and equicontinuous family on . Now Arzela-Ascoli Theorem guarantees that has a subsequence , converging uniformly on to a function . From the fact and , satisfies for all . Moreover, satisfies the integral equation Letting , we arrive at where the uniform continuity of on is used. Therefore, is a positive periodic solution of (16) and . Thus we complete the prove of Theorem 4.
Corollary 7. Let the nonlinearity in (1) be where , , , is a positive parameter, is a -periodic function. (i)If , then (1) has at least one positive periodic solution for each .(ii)If , then (1) has at least one positive periodic solution for each , where is some positive constant.
Proof. We will apply Theorem 4 with and , , . Then condition (H1)–(H3) are satisfied and existence condition (H4) becomes So (1) has at least one positive periodic solution for Note that if and if . We have the desired results.
The research of X. Xing is supported by the Fund of the Key Disciplines in the General Colleges and Universities of Xin Jiang Uygur Autonomous Region (Grant no. 2012ZDKK13). It is a pleasure for the author to thank Professor J. Chu for his encouragement and helpful suggestions.
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Copyright © 2013 Xiumei Xing. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. |
Boolean prime ideal theorem - formulasearchengine
In mathematics, a prime ideal theorem guarantees the existence of certain types of subsets in a given algebra. A common example is the Boolean prime ideal theorem, which states that ideals in a Boolean algebra can be extended to prime ideals. A variation of this statement for filters on sets is known as the ultrafilter lemma. Other theorems are obtained by considering different mathematical structures with appropriate notions of ideals, for example, rings and prime ideals (of ring theory), or distributive lattices and maximal ideals (of order theory). This article focuses on prime ideal theorems from order theory.
Although the various prime ideal theorems may appear simple and intuitive, they cannot be derived in general from the axioms of Zermelo–Fraenkel set theory without the axiom of choice (abbreviated ZF). Instead, some of the statements turn out to be equivalent to the axiom of choice (AC), while others—the Boolean prime ideal theorem, for instance—represent a property that is strictly weaker than AC. It is due to this intermediate status between ZF and ZF + AC (ZFC) that the Boolean prime ideal theorem is often taken as an axiom of set theory. The abbreviations BPI or PIT (for Boolean algebras) are sometimes used to refer to this additional axiom.
1 Prime ideal theorems
3 Further prime ideal theorems
4 The ultrafilter lemma
Prime ideal theorems
Recall that an order ideal is a (non-empty) directed lower set. If the considered poset has binary suprema (a.k.a. joins), as do the posets within this article, then this is equivalently characterized as a non-empty lower set I that is closed for binary suprema (i.e. x, y in I imply x
{\displaystyle \vee }
y in I). An ideal I is prime if its set-theoretic complement in the poset is a filter. Ideals are proper if they are not equal to the whole poset.
Historically, the first statement relating to later prime ideal theorems was in fact referring to filters—subsets that are ideals with respect to the dual order. The ultrafilter lemma states that every filter on a set is contained within some maximal (proper) filter—an ultrafilter. Recall that filters on sets are proper filters of the Boolean algebra of its powerset. In this special case, maximal filters (i.e. filters that are not strict subsets of any proper filter) and prime filters (i.e. filters that with each union of subsets X and Y contain also X or Y) coincide. The dual of this statement thus assures that every ideal of a powerset is contained in a prime ideal.
The above statement led to various generalized prime ideal theorems, each of which exists in a weak and in a strong form. Weak prime ideal theorems state that every non-trivial algebra of a certain class has at least one prime ideal. In contrast, strong prime ideal theorems require that every ideal that is disjoint from a given filter can be extended to a prime ideal that is still disjoint from that filter. In the case of algebras that are not posets, one uses different substructures instead of filters. Many forms of these theorems are actually known to be equivalent, so that the assertion that "PIT" holds is usually taken as the assertion that the corresponding statement for Boolean algebras (BPI) is valid.
Another variation of similar theorems is obtained by replacing each occurrence of prime ideal by maximal ideal. The corresponding maximal ideal theorems (MIT) are often—though not always—stronger than their PIT equivalents.
The Boolean prime ideal theorem is the strong prime ideal theorem for Boolean algebras. Thus the formal statement is:
Let B be a Boolean algebra, let I be an ideal and let F be a filter of B, such that I and F are disjoint. Then I is contained in some prime ideal of B that is disjoint from F.
The weak prime ideal theorem for Boolean algebras simply states:
Every Boolean algebra contains a prime ideal.
We refer to these statements as the weak and strong BPI. The two are equivalent, as the strong BPI clearly implies the weak BPI, and the reverse implication can be achieved by using the weak BPI to find prime ideals in the appropriate quotient algebra.
The BPI can be expressed in various ways. For this purpose, recall the following theorem:
For any ideal I of a Boolean algebra B, the following are equivalent:
I is a prime ideal.
I is a maximal ideal, i.e. for any proper ideal J, if I is contained in J then I = J.
For every element a of B, I contains exactly one of {a, ¬a}.
This theorem is a well-known fact for Boolean algebras. Its dual establishes the equivalence of prime filters and ultrafilters. Note that the last property is in fact self-dual—only the prior assumption that I is an ideal gives the full characterization. All of the implications within this theorem can be proven in ZF.
Thus the following (strong) maximal ideal theorem (MIT) for Boolean algebras is equivalent to BPI:
Let B be a Boolean algebra, let I be an ideal and let F be a filter of B, such that I and F are disjoint. Then I is contained in some maximal ideal of B that is disjoint from F.
Note that one requires "global" maximality, not just maximality with respect to being disjoint from F. Yet, this variation yields another equivalent characterization of BPI:
Let B be a Boolean algebra, let I be an ideal and let F be a filter of B, such that I and F are disjoint. Then I is contained in some ideal of B that is maximal among all ideals disjoint from F.
The fact that this statement is equivalent to BPI is easily established by noting the following theorem: For any distributive lattice L, if an ideal I is maximal among all ideals of L that are disjoint to a given filter F, then I is a prime ideal. The proof for this statement (which can again be carried out in ZF set theory) is included in the article on ideals. Since any Boolean algebra is a distributive lattice, this shows the desired implication.
All of the above statements are now easily seen to be equivalent. Going even further, one can exploit the fact the dual orders of Boolean algebras are exactly the Boolean algebras themselves. Hence, when taking the equivalent duals of all former statements, one ends up with a number of theorems that equally apply to Boolean algebras, but where every occurrence of ideal is replaced by filter. It is worth noting that for the special case where the Boolean algebra under consideration is a powerset with the subset ordering, the "maximal filter theorem" is called the ultrafilter lemma.
Summing up, for Boolean algebras, the weak and strong MIT, the weak and strong PIT, and these statements with filters in place of ideals are all equivalent. It is known that all of these statements are consequences of the Axiom of Choice, AC, (the easy proof makes use of Zorn's lemma), but cannot be proven in ZF (Zermelo-Fraenkel set theory without AC), if ZF is consistent. Yet, the BPI is strictly weaker than the axiom of choice, though the proof of this statement, due to J. D. Halpern and Azriel Lévy is rather non-trivial.
Further prime ideal theorems
The prototypical properties that were discussed for Boolean algebras in the above section can easily be modified to include more general lattices, such as distributive lattices or Heyting algebras. However, in these cases maximal ideals are different from prime ideals, and the relation between PITs and MITs is not obvious.
Indeed, it turns out that the MITs for distributive lattices and even for Heyting algebras are equivalent to the axiom of choice. On the other hand, it is known that the strong PIT for distributive lattices is equivalent to BPI (i.e. to the MIT and PIT for Boolean algebras). Hence this statement is strictly weaker than the axiom of choice. Furthermore, observe that Heyting algebras are not self dual, and thus using filters in place of ideals yields different theorems in this setting. Maybe surprisingly, the MIT for the duals of Heyting algebras is not stronger than BPI, which is in sharp contrast to the abovementioned MIT for Heyting algebras.
Finally, prime ideal theorems do also exist for other (not order-theoretical) abstract algebras. For example, the MIT for rings implies the axiom of choice. This situation requires to replace the order-theoretic term "filter" by other concepts—for rings a "multiplicatively closed subset" is appropriate.
The ultrafilter lemma
A filter on a set X is a nonempty collection of nonempty subsets of X that is closed under finite intersection and under superset. An ultrafilter is a maximal filter. The ultrafilter lemma states that every filter on a set X is a subset of some ultrafilter on X.[1] This lemma is most often used in the study of topology. An ultrafilter that does not contain finite sets is called non-principal. The ultrafilter lemma, and in particular the existence of non-principal ultrafilters (consider the filter of all sets with finite complements), follows easily from Zorn's lemma.
The ultrafilter lemma is equivalent to the Boolean prime ideal theorem, with the equivalence provable in ZF set theory without the axiom of choice. The idea behind the proof is that the subsets of any set form a Boolean algebra partially ordered by inclusion, and any Boolean algebra is representable as an algebra of sets by Stone's representation theorem.
Intuitively, the Boolean prime ideal theorem states that there are "enough" prime ideals in a Boolean algebra in the sense that we can extend every ideal to a maximal one. This is of practical importance for proving Stone's representation theorem for Boolean algebras, a special case of Stone duality, in which one equips the set of all prime ideals with a certain topology and can indeed regain the original Boolean algebra (up to isomorphism) from this data. Furthermore, it turns out that in applications one can freely choose either to work with prime ideals or with prime filters, because every ideal uniquely determines a filter: the set of all Boolean complements of its elements. Both approaches are found in the literature.
Many other theorems of general topology that are often said to rely on the axiom of choice are in fact equivalent to BPI. For example, the theorem that a product of compact Hausdorff spaces is compact is equivalent to it. If we leave out "Hausdorff" we get a theorem equivalent to the full axiom of choice.
A not too well known application of the Boolean prime ideal theorem is the existence of a non-measurable set[2] (the example usually given is the Vitali set, which requires the axiom of choice). From this and the fact that the BPI is strictly weaker than the axiom of choice, it follows that the existence of non-measurable sets is strictly weaker than the axiom of choice.
In linear algebra, the Boolean prime ideal theorem can be used to prove that any two bases of a given vector space have the same cardinality.
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Modify Properties of Conditional Mean Model Objects - MATLAB & Simulink - MathWorks 日本
A model created by arima has values assigned to all model properties. To change any of these property values, you do not need to reconstruct the whole model. You can modify property values of an existing model using dot notation. That is, type the model name, then the property name, separated by '.' (a period).
Modify the model to remove the constant term:
The updated constant term now appears in the model output.
Be aware that every model property has a data type. Any modifications you make to a property value must be consistent with the data type of the property. For example, AR, MA, SAR, and SMA are all cell vectors. This mean you must index them using cell array syntax.
To modify the property value of AR, assign AR a cell array. Here, assign known AR coefficient values:
Mdl.AR = {0.8,-0.4}
The updated model now has AR coefficients with the specified equality constraints.
Distribution = Mdl.Distribution
The degrees of freedom property of the model is updated. Note that the DoF field of Distribution is not directly assignable. For example, Mdl.Distribution.DoF = 8 is not a valid assignment. However, you can get the individual fields:
Mdl.Distribution.DoF
You can modify Mdl to include, for example, two coefficients
{\mathrm{β}}_{1}=0.2
{\mathrm{β}}_{2}=4
corresponding to two predictor series. Since Beta has not been specified yet, you have not seen it in the output. To include it, enter:
Mdl.Beta=[0.2 4]
Description: "ARIMAX(2,0,0) Model (t Distribution)"
Beta: [0.2 4]
P. This property updates automatically when any of p (degree of the nonseasonal AR operator),
{p}_{s}
(degree of the seasonal AR operator), D (degree of nonseasonal differencing), or s (degree of seasonal differencing) changes.
Q. This property updates automatically when either q (degree of the nonseasonal MA operator), or
{q}_{s}
(degree of the seasonal MA operator) changes.
Not all name-value pair arguments you can use for model creation are properties of the created model. Specifically, you can specify the arguments ARLags, MALags, SARLags, and SMALags during model creation. These are not, however, properties of arima models. This means you cannot retrieve or modify them in an existing model.
The nonseasonal and seasonal AR and MA lags update automatically if you add any elements to (or remove from) the coefficient cell arrays AR, MA, SAR, or SMA.
For example, specify an AR(2) model:
The model output shows nonzero AR coefficients at lags 1 and 2.
Add a new AR term at lag 12:
Mdl.AR{12} = NaN
Description: "ARIMA(12,0,0) Model (Gaussian Distribution)"
AR: {NaN NaN NaN} at lags [1 2 12]
The three nonzero coefficients at lags 1, 2, and 12 now display in the model output. However, the cell array assigned to AR returns twelve elements:
Mdl.AR
{[NaN]} {[NaN]} {[0]} {[0]} {[0]} {[0]} {[0]} {[0]}
AR has zero coefficients at all the interim lags to maintain consistency with traditional MATLAB® cell array indexing.
arima | struct |
Twenty-five percent of the students at Marcus Garvey Middle School bring their lunches from home.
225
students do not bring their lunch. How many students attend the school? Draw and label a diagram to show the number and percent of each group of students.
25\%
of the students bring lunches from home, then
75\%
of the students must not bring their lunches to school.
225
75\%
of the students, how many total students attend the school and how many bring lunches from home?
After finding out each value, create a diagram to illustrate this ratio. |
Find the solution of limit displaystylelim_{{{left({x},{y}right)}rightarrow{0},{0}}}frac{{sqrt{{{x}^{2}+{y}^{2}}}}}{{{x}^{2}+{y}^{2}}} by using
Find the solution of limit \lim_{(x,y)->0,0} sqrt((x^2+y^2)/(x^2+y^2)) by using the polar coordinates system
Find the solution of limit
\underset{\left(x,y\right)\to 0,0}{lim}\frac{\sqrt{{x}^{2}+{y}^{2}}}{{x}^{2}+{y}^{2}}
by using the polar coordinates system.
\underset{\left(x,y\right)\to 0,0}{lim}\left(\sqrt{{x}^{2}+{y}^{2}}\left({x}^{2}+{y}^{2}\right)\right)=\underset{\left(x,y\right)\to 0,0}{lim}f\left(x,y\right)
f\left(x,y\right)=\left(\sqrt{{x}^{2}+{y}^{2}}\right).\text{Using polar form}\text{ }x=r\mathrm{cos}0,y=r\mathrm{sin}0
f\left(r,0\right)=\frac{r}{{r}^{2}\left({\mathrm{cos}}^{2}0+{\mathrm{sin}}^{2}0\right)}=\frac{r}{{r}^{2}}
\underset{r\to 0,0\to 0}{lim}f\left(r,0\right)=\underset{r\to 0}{lim}\frac{r}{{r}^{2}}=\mathrm{\infty }
\underset{\left(x,y\right)\to 0,0}{lim}\frac{{\sqrt{x}}^{2}+{y}^{2}}{{x}^{2}+{y}^{2}}=\mathrm{\infty }
Two snowcats tow a housing unit to a new location at McMurdo Base, Antarctica, as shown in the figure. The sum of the forces
{F}_{A}
{F}_{B}
exerted on the unit by the horizontal cables is parallel to the line L, and
{F}_{A}=4200
N. Determine
{F}_{B}
? Determine the magnitude of
{F}_{A}+{F}_{B}
A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 100 m/s. The rocket moves for 3.00 s a long its initial line of motion with an acceleration of 30.0 m/s/s. At this time, its engines fail and the rocket proceeds to move as a projectile. Find:
a) the maximum altitude reached by the rocket
b) its total time of flight
c) its horizontal range.
A 0.48 kg piece of wood floats in water but is found to sinkin alcohol (specific gravity= 0.79) in which it has an apparentmass 0.035 kg. What is the SG of the wood?
A car initially traveling eastward turns north by traveling in a circular path at uniform speed as in the figure below. The length of the arc ABC is 235 m, and the car completes the turn in 33.0 s. (Enter only the answers in the input boxes separately given.)
(a) What is the acceleration when the car is at B located at an angle of 35.0°? Express your answer in terms of the unit vectors
\stackrel{^}{i}
\stackrel{^}{j}
1. (Enter in box 1)
\frac{m}{{s}^{2}}\stackrel{^}{i}+\left(\text{ Enter in box 2 }\right)\frac{m}{{s}^{2}}\stackrel{^}{j}
(b) Determine the car's average speed.
3. ( Enter in box 3) m/s
(c) Determine its average acceleration during the 33.0-s interval.
4. ( Enter in box 4)
\frac{m}{{s}^{2}}\stackrel{^}{i}+
\frac{m}{{s}^{2}}\stackrel{^}{j}
y\left(x\right)=x{\mathrm{tanh}}^{-1}\left(x\right)+\mathrm{ln} \left(\sqrt{1-{x}^{2}}\right)
Two 2.1cm diameter disks face each other, 2.9mm apart. They are charged to
±10nC
a) what is the electric field strength between two disks?
b) a proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
A pump is used to transport water to a higher reservoir. If the water temperature is
{20}^{\circ }C
, determine the lowest pressure that can exist in the pump without cavitation. |
Permutations | Brilliant Math & Science Wiki
Mei Li, U Z, Ashish Menon, and
InkWhite 2000
In combinatorics, a permutation is an ordering of a list of objects. For example, arranging four people in a line is equivalent to finding permutations of four objects. More abstractly, each of the following is a permutation of the letters
a, b, c,
and
\begin{aligned} → \ & a, b, c, d\\ → \ & a, c, d, b\\ → \ & b, d, a, c\\ → \ & d, c, b, a\\ → \ & c, a, d, b. \end{aligned}
Note that all of the objects must appear in a permutation and two orderings are considered different if some object appears in a different place in the orderings.
All possible arrangements or permutations of a,b,c,d.
Permutations are important in a variety of counting problems (particularly those in which order is important), as well as various other areas of mathematics; for example, the determinant is often defined using permutations.
Permutations of a Set of Distinct Objects
Permutations of a Subset of Distinct Objects
The simplest example of a permutation is the case where all objects need to be arranged, as the introduction did for
a, b, c, d.
The question thus becomes the following:
Given a list of objects, how can all possible permutations be listed?
There are several algorithms for enumerating all permutations; one example is the following recursive algorithm:
If the list contains a single element, then return the single element.
If the list contains more than one element, loop through each element in the list, returning this element concatenated with all permutations of the remaining
n-1
Lisa has 5 different ornaments she wants to arrange in a line on her mantle. How many ways can she arrange the ornaments?
We can think of Lisa’s mantle as having five positions in a line. There are 5 ornaments, which gives 5 choices for which ornament goes into the first position. After placing the first ornament, there are 4 choices of which ornament to put into the second position. Repeating this argument, there are 3 choices for the third position, 2 choices for the fourth position, and 1 choice for the last position. By the rule of product, the total number of ways to place the ornaments is
5 \times 4 \times 3 \times 2 \times 1 = 120. \ _\square
n
distinct objects, how many different permutations of the objects are there?
Since each permutation is an ordering, start with an empty ordering which consists of
n
positions in a line to be filled by the
n
n
choices for which object to place in the first position. After the first object is placed, there are
n-1
n-1
n-2
n-3
choices for the fourth position, and so on. For the
n^\text{th}
position, the number of choices is
n - (n-1)= 1.
Then the rule of product implies the total number of orderings is
n \times (n-1) \times (n-2) \times (n-3) \times \cdots \times 1 = n!.\ _\square
n
n!
denotes the factorial of
n
and refers to the product of all positive integers from
1
n
0!
is the empty product and is defined to be
1
The argument above shows the following result:
n
n!
n.
Jack is playing with a standard deck of
52
playing cards. He shuffles the cards and then turns the top card over to show an ace of spades. If he continues to deal the cards from the top of the deck, how many different permutations are there for the remaining cards in the deck?
Since the first card is an ace of spades, there are
51
remaining distinct playing cards in the deck. Then there are
51!
different permutations of the remaining cards.
_\square
The first few values of
n!
\begin{array}{c}&1!= 1, &2! = 2, &3! = 6, &4! = 24, &5! =120, &6! = 720, &7! = 5040, &\ldots .\end{array}
As the above list shows,
n!
grows very quickly with
n
60!
, for example, is already larger than the number of atoms in the observable universe.
In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent??
The 4 men can be seated at the circular table such that there is a vacant seat between every pair of women.
The number of ways in which these 4 men can be seated at the circular table is
3! = 6.
Now the 4 vacant seats can be occupied by the women in
4! = 24
ways. Thus, the required number of ways is
6\times24 = 144
_\square
Suppose Ellie is choosing a secret passcode consisting of the digits
0,1,2, \ldots, k
k \leq 9
. She would like her passcode to use each digit at most once and because she is concerned about security, she would like to choose a value of
k
such that the number of possible permutations is at least 250,000. What is the smallest value of
k
Ellie can use?
8! = 40320
9! = 362880.
Therefore, Ellie needs at least
9
digits in her passcode. Since the digits start from
0
, the smallest value of
k
Ellie can choose is
k = 8.
_\square
More precisely, Stirling's formula states that
n!
n! \sim \sqrt{2 \pi n}\left(\frac{n}{e}\right)^n.
The precise bounds for the approximation are shown in the inequality
\sqrt{2\pi}\ n^{n+1/2}e^{-n} \le n! \le e\ n^{n+1/2}e^{-n}.
In the case that the objects are not distinct, we have a permutations with repetition problem; in the case that the permutation must satisfy certain constraints, we have a permutations with restriction problem.
How many 5-digit numbers without repetition of digits can be formed using the digits 0, 2, 4, 6, 8?
Lisa has 13 different ornaments and wants to put 4 ornaments on her mantle. In how many ways is this possible?
Using the product rule, Lisa has 13 choices for which ornament to put in the first position, 12 for the second position, 11 for the third position, and 10 for the fourth position. So the total number of choices she has is
13 \times 12 \times 11 \times 10
\frac{13!}{9!}
Using the same argument, we can proceed with the general case. If we have
n
objects and want to arrange
k
of them in a row, there are
\frac{n!}{(n-k)!}
ways to do this. This is also known as a
k
n
P_k ^n
How many 3-digit numbers can be formed without repetition of digits?
The hundredth place cannot contain
0
, but we can put any of the other numbers in the hundredths place. So, it can be filled in
9
ways. Now we cannot use the number we used in the hundredths place in the tenth place. But we can use
0
. So, we can fill the tenth place in
9
ways, too. The units digit can be filled in
8
places since two numbers are now unavailable. So, the total number of 3-digit numbers without repetition of digits that can be formed is
9×9×8 = 648.\ _\square
How many 4-letter passwords can be formed if the characters allowed to use without repetition are 0, 1, 2, 3, ..., 9 and A, B, C, ..., Z and a, b, c, ..., z?
26+26+10=62
choices to choose from. So we have to arrange 4 objects out of 62 available objects, the number of ways of doing which is equal to
\frac {62!}{(62-4)!}=62\times 61\times 60\times 59=13388280.
_\square
Suppose Lisa has 13 different ornaments and would like to place 4 of the ornaments in a row on her mantle, and Anna has 12 different ornaments and would like to place 5 of the ornaments in a row on her mantle. Does Lisa have more choices in the possible number of ways to place her ornaments or does Anna?
Since Lisa has 13 ornaments and would like to place 4 of them on her mantle, she has
\frac{13!}{(13-4)!} = \frac{13!}{9!} = 13 \times 12 \times 11 \times 10
choices in the possible number of ways to place her ornaments. Similarly, since Anna has 12 ornaments and would like to place 5 of them on her mantle, she has
\frac{12!}{(12-5)!} = \frac{12!}{7!} = 12 \times 11 \times 10 \times 9 \times 8
choices in the possible number of ways to place her ornaments. Now,
13 \times 12 \times 11 \times 10 < 12 \times 11 \times 10 \times 9 \times 8
since by canceling terms, we can see that
13 < 9 \times 8
. This shows that Anna has more choices in the possible ways to place her ornaments.
_\square
In general, if Anna has
12
different ornaments and would like to place
k
of them on the mantle
(
1 < k \leq 12)
and if Lisa has
13
k-1
of them on a mantle, for what values of
k
does Anna have more choices in the possible number of ways to place all of her ornaments?
Since Anna has
12
ornaments and would like to place
k
of them on her mantle, she has
\frac{12!}{(12-k)!} = 12 \times 11 \times \cdots \times (12 - k + 1)
choices in the possible number of ways to place her ornaments. Similarly, since Lisa has
13
k-1
\frac{13!}{\big(13-(k-1)\big)!} = 13 \times 12 \times \cdots \times (13 - k + 2)
choices in the possible number of ways to place her ornaments. Now, let's compare the two quantities by cancelling terms:
\begin{aligned} 13 \times 12 \times \cdots \times (13 - k+2) & \quad ? \quad 12 \times 11 \times \cdots \times (12 - k + 1)\\\\ 13 & \quad ? \quad (12 - k + 2) \times (12 - k + 1). \end{aligned}
1 < k \leq 12
, we can check that for
k = 10, 11, 12
13 > (12 - k + 2) \times (12 - k + 1).
k = 1, 2, \ldots ,9
13 < (12 - k + 2) \times (12 - k + 1).
This shows that Anna has more choices in the possible ways to place her ornaments for
k = 1,2, \ldots ,9
_\square
Here are some examples for permutations with repetition:
Main Article: Permutations with Repetition
How many different 6-digit numbers can be obtained by using all of the digits
5,\ 5,\ 7,\ 7,\ 7,\ 8?
We first count the total number of permutations of all six digits. This gives a total of
6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
permutations. Now, there are two 5's, so the repeated 5's can be permuted in
2!
ways and the six digit number will remain the same. Similarly, there are three 7's, so the repeated 7's can be permuted in
3!
ways and the six digit number will remain the same. This shows the number of different 6-digit numbers using all of the digits is
\frac{6!}{2! 3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1 }{2! 3!} = 5 \times 4 \times 3 = 20 \times 3 = 60.\ _\square
\text{"BRILLIANT"}?
\large \text{HUNG WOEI NEOH}
From the name above, how many different names (including the above one) can be created such that the relative order of consonants and vowels does not change.
"The relative order of consonants and vowels does not change" means that vowels and consonants should occupy the same places as they did in the original name, but they can be juggled with their places.
An example would be "GEHW NIOO HEUN".
Main Article: Permutations - With Restriction
We have to decide if we want to place the dog ornaments first, or the cat ornaments first, which gives us 2 possibilities. We can arrange the cat ornaments in
6!
ways and the dog ornaments in
4!
2 \times 6! \times 4! = 34560
_\square
How many ways are there to arrange 5 red, 5 blue, and 5 green balls in a row such that no two blue balls lie next to each other?
How many positive integers less than 10000 are there such that they each have only 3 and/or 7 as their digit(s)?
\text{ABHIRAM}
From the name above, how many words with or without meaning can be formed which do not end with
M?
4\times4
2\times2
4\times4
Given a permutation problem, how do we determine which category the problem falls under and which technique should be applied to solve the problem? It may be useful to first ask yourself a few questions:
Are the objects all distinct?
How many objects are there in total?
How many objects are we asked to place into an ordering?
Are there any restrictions on the objects or on the ordering?
Based on the answers to these questions, it may become easier to decide which technique should be applied. Working through many examples is one way to become better at recognizing whether a permutation problem should fall in the category of permutation with or without repetition, or permutation with or without restriction.
Cite as: Permutations. Brilliant.org. Retrieved from https://brilliant.org/wiki/permutations/ |
Estimate frequency response and spectrum using spectral analysis with frequency-dependent resolution - MATLAB spafdr
y\left(t\right)=G\left(q\right)u\left(t\right)+v\left(t\right)
where Φυ(ω) is the spectrum of υ(t). data contains the output-input data as an iddata object. The data can be complex valued, and either time or frequency domain. It can also be an idfrd object containing frequency-response data. g is an idfrd object with the estimate of
G\left({e}^{i\omega }\right)
at the frequencies ω specified by row vector w. g also includes information about the spectrum estimate of Φυ(ω) at the same frequencies. Both results are returned with estimated covariances, included in g. The normalization of the spectrum is the same as described in spa.
\frac{2\pi }{N{T}_{s}}
\frac{\pi }{{T}_{s}} |
EUDML | Foliations, groupoids and Baum-Connes conjecture. EuDML | Foliations, groupoids and Baum-Connes conjecture.
Foliations, groupoids and Baum-Connes conjecture.
Macho-Stadler, M.
Macho-Stadler, M.. "Foliations, groupoids and Baum-Connes conjecture.." Zapiski Nauchnykh Seminarov POMI 266 (2000): 169-187. <http://eudml.org/doc/232535>.
@article{Macho2000,
author = {Macho-Stadler, M.},
keywords = {foliated manifolds; Baum-Connes conjecture; Morita-equivalence of groupoids; Lie groupoids; groupoid -algebras; groupoid -algebras},
title = {Foliations, groupoids and Baum-Connes conjecture.},
AU - Macho-Stadler, M.
TI - Foliations, groupoids and Baum-Connes conjecture.
KW - foliated manifolds; Baum-Connes conjecture; Morita-equivalence of groupoids; Lie groupoids; groupoid -algebras; groupoid -algebras
foliated manifolds, Baum-Connes conjecture, Morita-equivalence of groupoids, Lie groupoids, groupoid
{C}^{*}
-algebras, groupoid
{C}^{*}
Topological groupoids (including differentiable and Lie groupoids)
Articles by Macho-Stadler |
Use the Laplace transform to solve the given integral equation (s)=2t^2+int_0^t
Use the Laplace transform to solve the given integral equation (s)=2t^2+int_0^tsinleft[2(t-tau)right]x(tau)d tau
Use the Laplace transform to solve the given integral equation
\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau
Corben Pittman
\left(s\right)=2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau
Taking Laplace transform on both sides of the equation
L\left\{x\left(t\right)\right\}=L\left\{2{t}^{2}+{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}
By using linearity property of Laplace transform
L\left[af\left(t\right)+bg\left(t\right)\right]=aL\left\{f\left(t\right)\right\}+bL\left\{f\left(t\right)\right\}
where a, b are constants.
L\left\{x\left(t\right)\right\}=2L\left\{{t}^{2}\right\}+L\left\{{\int }_{0}^{t}sin\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}
By using convolution theorem
L\left\{{\int }_{0}^{t}\mathrm{sin}\left[2\left(t-\tau \right)\right]x\left(\tau \right)d\tau \right\}=L\left\{\mathrm{sin}\left(2t\right)\right\}L\left\{x\left(t\right)\right\}
L\left\{x\left(t\right)\right\}=2L\left\{{t}^{2}\right\}+L\left\{\mathrm{sin}\left(2t\right)\right\}L\left\{x\left(t\right)\right\}
L\left\{{t}^{n}\right\}=\frac{n!}{{s}^{n+1}},L\left\{\mathrm{sin}at\right\}=\frac{a}{{s}^{2}+{a}^{2}}
x\left(s\right)=2\left(\frac{2!}{{s}^{2+1}}\right)+\left(\frac{2}{{s}^{2}+{2}^{2}}\right)x\left(s\right)
x\left(s\right)=\frac{4}{{s}^{3}}+\frac{2}{{s}^{2}+4}x\left(s\right)
Solve above equation for x(s)
x\left(s\right)-\frac{2}{{s}^{2}+4}x\left(s\right)=\frac{4}{{s}^{3}}
\left(1-\frac{2}{{s}^{2}+4}\right)x\left(s\right)=\frac{4}{{s}^{3}}
\left(\frac{{s}^{2}+2}{{s}^{2}+4}\right)x\left(s\right)=\frac{4}{{s}^{3}}
x\left(s\right)=\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}
By using partial fractions:
\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}=\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}
Taking inverse Laplace transform on both the sides.
{L}^{-1}\left\{\frac{4\left({s}^{2}+4\right)}{{s}^{3}\left({s}^{2}+2\right)}\right\}={L}^{-1}\left\{\frac{-2}{s}+\frac{8}{{s}^{3}}+\frac{2s}{{s}^{2}+2}\right\}
Using linearity property of Inverse Laplace transform,
{L}^{-1}\left\{x\left(s\right)\right\}=-2{L}^{-1}\left\{\frac{1}{s}\right\}+8{L}^{-1}\left\{\frac{1}{{s}^{3}}\right\}+2{L}^{-1}\left\{\frac{s}{{s}^{2}+2}\right\}
Use formulae for Inverse Laplace transform,
Solve the differential equations:
\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)dy={y}^{2}dx
4{t}^{2}+5{t}^{3}
How to prove that the inverse Laplace Transform of zero is zero itself?
{L}^{-1}\left\{0\right\}=0
I know that the inverse Laplace Transform of a constant is Dirac's Delta. But I think that that applies only to positive constants.
laplace transform of 2^[t] where [] is greatest integer function <=t
Use the Laplace transform to solve the heat equation
{u}_{t}={u}_{xx}0<x<1\text{ and }t>0
u\left(x,0\right)=\mathrm{sin}\left(\pi x\right)
u\left(0,t\right)=u\left(1,t\right)=0
\frac{dy}{dx}=-\frac{ax+by}{bx+cy} |
\frac{a}{b}=\left(\frac{a}{b}×100\right)%
\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
P{\left(1+\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}
P{\left(1-\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
A shopkeeper marks the price of the article in such a way that after allowing 28% discount, he wants a gain of 12%. If the marked price is ₹224. then the cost price of the article is:
A) ₹168 B) ₹144
C) ₹120 D) ₹196
A) ₹168
B) ₹144
C) ₹120
D) ₹196
Answer & Explanation Answer: B) ₹144
The price of sugar is increased by 24%. A person wants to increase his expenditure by 15% only. By what percentage, correct to one decimal place, should he reduce his consumption?
The price of sugar is increased by 18%. A person wants to increase the expenditure by 12% only. By what percent, correct to one decimalplace, should he decrease his consumption?
A) 5.6% B) 5.1%
C) 6% D) 5.3%
Answer & Explanation Answer: B) 5.1%
Each side of a rectangular field is increased by 10%. Then the percentage increase in the area of the field is:
The price of sugar is increased by 17%. A person wants to increase his expenditure by 7% only. By what percentage, correct to one decimal place, should he reduce his consumption?
C) 8.7% D) 8.5%
Answer & Explanation Answer: D) 8.5%
The price of sugar is increased by 17%. A person wants to increase his expenditure by 8% only. By what percent should he decrease his consumption, nearest to one decimal place?
The sum of the number of male and female students in an institute is 100. If the number of male students is x, then the number of female students becomes x% of the total number of students. Find the number of male students.
In a class of 50 students, 60% are boys. The average of marks of the boys is 62, and that of the girls is 68. What is the average marks of the whole class? |
Using calculus, it can be shown that the tangent function can be approximated by
Using calculus, it can be shown that the tangent function can be approximated by the polynomial tan x approx x + frac{2x^{3}}{3!} + frac{16x^{5}}{5!}
Using calculus, it can be shown that the tangent function can be approximated by the polynomial
\mathrm{tan}\text{ }x\text{ }\approx \text{ }x\text{ }+\text{ }\frac{2{x}^{3}}{3!}\text{ }+\text{ }\frac{16{x}^{5}}{5!}
where x is in radians. Use a graphing utility to graph the tangent function and its polynomial approximation in the same viewing window. How do the graphs.
odgovoreh
The red graph is tangent and the blue graph is approximation. The graphs aresimilar close to when
x=0
Graph the polynomial function.
f\left(x\right)=-{x}^{4}+8x
(a) find the Maclaurin polynomial
{P}_{3}\left(x\right)
for f(x), (b) complete the following
\begin{array}{|cccccccc|}\hline x:& -0.75& -0.50& -0.25& 0& 0.25& 0.50& 0.75\\ \hline\end{array}
for f(x) and
{P}_{3}\left(x\right)
and (c) sketch the graphs of f(x) and
{P}_{3}\left(x\right)
on the same set of coordinate axes.
f\left(x\right)=\mathrm{arctan}x
For the following exercise, for each polynomial
f\left(x\right)=\text{ }\frac{1}{2}{x}^{2}\text{ }-\text{ }1
: a) find the degree, b) find the zeros, if any, c) find the y-intercept(s), if any, d) use the leading coefficient to determine the graph’s end behavior, e) determine algebraically whether the polynomial is even, odd, or neither.
Graph each polynomial function. Factor first if the expression is not in factored form.
f\left(x\right)=\left(3x-1\right){\left(x+2\right)}^{2}
\frac{3}{4}
2\frac{3}{5}
Graph each polynomial function.
f\left(x\right)={\left(x-1\right)}^{4}
f\left(x\right)=\left(4x+3\right){\left(x+2\right)}^{2} |
NewBitGenerator - Maple Help
Home : Support : Online Help : Programming : Random Objects : RandomTools package : BlumBlumShub Subpackage : NewBitGenerator
NewBitGenerator( seed, opt1, opt2, ... )
an integer, the seed for the generator
(optional) argument of the form option=value where option is one of primes, numbits, or output
The NewBitGenerator command outputs a Maple procedure, a pseudo-random bit generator, which when called outputs one pseudo-random bit, a 0 or 1. The generator is a Blum-Blum-Shub (BBS) generator. A BBS generator uses the following quadratic recurrence to generate a sequence of integers
{x}_{1},{x}_{2},\dots
{z}_{1},{z}_{2},\dots
{x}_{\left(k+1\right)}=\mathrm{mod}\left({{x}_{k}}^{2},n\right)
{z}_{\left(k+1\right)}=\mathrm{mod}\left({x}_{\left(k+1\right)},{2}^{m}\right).
n
m=⌊{\mathrm{log}}_{2}\left({\mathrm{log}}_{2}\left(n\right)\right)⌋
{x}_{0}
{z}_{k+1}
m
{x}_{k+1}
m
cryptographically secure bits. The output of the generator depends on the options described below. The default is to output one bit.
\mathrm{gcd}\left(x,n\right)=1
x={y}^{2}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{mod}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}n
n=pq
\mathrm{gcd}\left(z,n\right)=1
p=2s+1
q=2t+1
where s and t are prime and 2 is a primitive element in both Z mod s and Z mod t. This choice guarantees that p and q are both congruent to 3 mod 4 (a requirement for the security of the generator) and also that the period of the generator for
{x}_{0}
s-1
t-1
\mathrm{lcm}\left(s-1,t-1\right)
n=pq
, satisfying these requirements of lengths 512 bits, 768 bits, and 1024 bits have been precomputed and the prime factorization discarded (Maple does not have the factorizations). Thus there are three choices for
n
The argument seed=s determines the seed for the generator. The value used for
{x}_{0}
used must be a "random" quadratic residue which is not equal to 1 (to avoid a short period). The value used for
{x}_{0}
is computed from the seed s. For cryptographic applications the seed should be chosen by the user to be a random integer from a large set, e.g., from 0..10^100. It will be automatically reduced modulo n.
primes =
512
768
1024
The integer l, which must be one of
512
768
1024
specifies the length of the primes p and q in bits. The default is
512
numbits=integer
This specifies how many bits are computed and output by each call to the generator. The default is 1.
output= bits or integer
This specifies how the bits are output. If integer is specified, the output is a Maple integer in the range 0 to 2^b-1. If bits is specified, the output will be a Maple sequence of b bits. The default is bits.
The RandomTools[BlumBlumShub] module also exports the NewGenerator function. This function's interface is compatible with the NewGenerator functions of other RandomTools pseudo-random number generator subpackages.
\mathrm{with}\left(\mathrm{RandomTools}[\mathrm{BlumBlumShub}]\right)
[\textcolor[rgb]{0,0,1}{\mathrm{NewBitGenerator}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{NewGenerator}}]
\mathrm{seed}≔1050109365100103751001961108357097849013652340237134723870:
B≔\mathrm{NewBitGenerator}\left(\mathrm{seed}\right)
\textcolor[rgb]{0,0,1}{B}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathbf{proc}}\left(\textcolor[rgb]{0,0,1}{}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{−}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{]}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{while}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathrm{nops}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{1}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{do}}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{\mathrm{irem}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{^}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{n}\right)\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{z}\textcolor[rgb]{0,0,1}{≔}\left[\textcolor[rgb]{0,0,1}{\mathrm{op}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{z}\right)\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{T}\textcolor[rgb]{0,0,1}{[}\textcolor[rgb]{0,0,1}{\mathrm{irem}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1024}\right)\textcolor[rgb]{0,0,1}{]}\right]\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{end do}}\textcolor[rgb]{0,0,1}{;}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathrm{op}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{..}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{z}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\textcolor[rgb]{0,0,1}{\mathbf{end proc}}
B\left(\right)
\textcolor[rgb]{0,0,1}{1}
B\left(\right)
\textcolor[rgb]{0,0,1}{0}
\mathrm{seq}\left(B\left(\right),i=1..10\right)
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
B≔\mathrm{NewBitGenerator}\left(\mathrm{seed},\mathrm{numbits}=10,\mathrm{primes}=1024\right):
B\left(\right)
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}
B\left(\right)
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}
B≔\mathrm{NewBitGenerator}\left(\mathrm{seed},\mathrm{numbits}=32,\mathrm{output}=\mathrm{integer}\right):
B\left(\right)
\textcolor[rgb]{0,0,1}{3026219245}
\mathrm{seq}\left(B\left(\right),i=1..5\right)
\textcolor[rgb]{0,0,1}{780542672}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3215656114}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4009826535}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4022932036}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3025225627}
RandomTools[BlumBlumShub][NewGenerator] |
isdifferentiable - Maple Help
Home : Support : Online Help : Programming : Logic : Boolean : isdifferentiable
test for piecewise functions
isdifferentiable(expr, var, class)
isdifferentiable(expr, var, class, varparam)
number n telling if the expr is in class
{C}^{n}
isdifferentiable determines if the expression expr containing piecewise or piecewise functions such as abs, signum, max, ..., is of class
C^
class. It returns either true or false.
The optional argument varparam can be passed to the function and in case expr is not of class
{C}^{n}
, varparam contains information of which
{C}^{n}
class the function is, and a list of points where we have discontinuities in the n+1-th derivative.
\mathrm{piecewise}\left(x<-1,-x,x<1,xx,x\right)
{\begin{array}{cc}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{-1}\\ {\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}& \textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{<}\textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{x}& \textcolor[rgb]{0,0,1}{\mathrm{otherwise}}\end{array}
\mathrm{isdifferentiable}\left(,x,1\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{isdifferentiable}\left(,x,2,'\mathrm{la}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{la}
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{-1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}
\mathrm{isdifferentiable}\left(\mathrm{abs}\left(x\right),x,2\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{isdifferentiable}\left(\mathrm{abs}\left(x\right),x,2,'\mathrm{la}'\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{la}
\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}{\textcolor[rgb]{0,0,1}{0}} |
Classical Mechanics Problem: Dark Matter and Orbital Velocity - Matt DeCross | Brilliant
Dark Matter and Orbital Velocity
Which of the following correctly describes the behavior of the orbital velocity of stars in spiral galaxies as a function of radius, including the effects of dark matter?
Grows linearly in the galactic bulge and falls off like
r^{-1/2}
outside the bulge Grows linearly in the galactic bulge and falls off exponentially outside the bulge Grows linearly in the galactic bulge and is constant outside the bulge Falls off exponentially everywhere Is constant in the galactic bulge and grows linearly outside the bulge |
\frac{a}{b}=\left(\frac{a}{b}×100\right)%
\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
P{\left(1+\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}
P{\left(1-\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
A's salary is 40% of B's salary which is 25% of C's salary. What percentage of C's salary is A's salary ?
A's Salary = 40% of B = 40% of (25% of C) = 40/100[(52/100)*100] % of C = 10% of C.
If 75% of a number is added to 75, then the result is the number itself. The number is :
Let the number be x, Then
75% of x + 75 = x
=> x - 75x/100 =75
=> x = 300 .
25% of 25% Is Equal To
A) 0.0625 B) 0.0210
C) 0.03145 D) 0.3210
Answer & Explanation Answer: A) 0.0625
25% of 25% means
25/100 of 25/100
i.e, 1/4 of 1/4
or in other words 1/4 x 1/4=1/16
Hence, the answer is 1/16 = 0.0625.
A salesman gets commission on total sales at 9%. If the sale is exceeded Rs.10,000 he gets an additional commission as bonus of 3% on the excess of sales over Rs.10,000. If he gets total commission of Rs.1380, then the bonus he received is:
C) Rs.480 D) data insufficient
Commission up to 10000 = = 900
Again after 10000,
Commission : Bonus
3x : x
Therefore, Bonus =
56% of 225 + 20% of 150 = ? - 109
Find the value of '?' in the above equation
C) 241 D) None
Given 56% of 225 + 20% of 150 = ? - 109
we know that a% of b = b% of a
225% of 56 + 20% of 150 = ? - 109
200% of 56 + 25% of 56 + 20% of 150 = ? - 109
112 + 14 + 30 = ? - 109
? = 112 + 14 + 30 + 109
45% of 750 - 25% of 480 = ?
C) 376.21 D) 120
Answer & Explanation Answer: B) 217.50
Given expression = (45 x 750/100) - (25 x 480/100) = (337.50 - 120) = 217.50
Gaurav spends 30% of his monthly income on food articles, 40% of the remaining on conveyance and clothes and saves 50% of the remaining. If his monthly salary is Rs. 18,400, how much money does he save every month ?
Saving = 50% of (100 - 40)% of (100 - 30)% of Rs. 18,400 = Rs. (50/100 * 60/100 * 70/100 * 18400) = Rs. 3864.
A housewife saved Rs. 2.50 in buying an item on sale. If she spent Rs. 25 for the item, approximately how much percent she saved in the transaction ?
= % |
DGbiform - Maple Help
Home : Support : Online Help : Mathematics : DifferentialGeometry : Tools : DGbiform
DifferentialGeometry:-Tools[DGbiform, DGform, DGtensor, DGvector]
DGbiform(x, M)
DGform(x, M)
DGtensor(x, indexType, M)
DGvector(y, M)
a positive integer, a list of positive integers, a coordinate variable, or a list of coordinate variables
(optional) the name of defined frame
specifying the index type of the tensor
a positive integer or a coordinate variable
The command DGform will create a single term differential form. Let Theta = [theta_1, theta_2, theta_3, ...] denote the coframe for the current frame or, if the optional argument M is given, the frame M. The list Theta can be obtained from the command DGinfo with the keyword "frameBaseForms" or "frameJetForms". Let V = [x_1, x_2, x_3, ...] denote the local coordinates for the current frame or, if the optional argument M is given, the frame M. The list V can be obtained from the command DGinfo with the keyword "frameIndependentVariables" or "frameJetVariables". If the integer i or coordinate x_i is given, the command returns the corresponding 1-form theta_i. If a list of p integers [i, j, k, ...] or coordinates [x_i, x_j, x_k, ...] is given, the command returns the p-form theta_i &w theta_j &w theta_k...
The commands DGbiform, DGtensor, and DGvector work in a similar fashion.
The command DGform is part of the DifferentialGeometry:-Tools package and so can be used in the form DGform(...) only after executing the commands with(DifferentialGeometry) and with(Tools) in that order. It can always be used in the long form DifferentialGeometry:-Tools:-DGform. DGbiform, DGtensor, and DGvector work in the same way.
\mathrm{with}\left(\mathrm{DifferentialGeometry}\right):
\mathrm{with}\left(\mathrm{Tools}\right):
Define a manifold M with coordinates [x, y, z, w].
\mathrm{DGsetup}\left([x,y,z,w],M\right):
\mathrm{DGvector}\left(x\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"vector"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{}\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGvector}\left(3\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"vector"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{}\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{3}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGform}\left(y\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"form"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGform}\left(4\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"form"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{4}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGform}\left([x,y]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"form"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGform}\left([1,2,3,4]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"form"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGtensor}\left(x,"con_bas"\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"tensor"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{"con_bas"}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{}\right]\right]\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGtensor}\left(2,"cov_bas"\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"tensor"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{"cov_bas"}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{}\right]\right]\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{DGtensor}\left([1,1,1],[["cov_bas","con_bas","cov_bas"],[]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"tensor"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{"cov_bas"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"con_bas"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"cov_bas"}\right]\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{}\right]\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
Define a rank 3 vector bundle E with coordinates [x, y, u, v, w] over a two dimensional base with coordinates [x, y].
\mathrm{DGsetup}\left([x,y,u,v,w],E\right)
\textcolor[rgb]{0,0,1}{\mathrm{frame name: E}}
\mathrm{DGtensor}\left([x,v,v],[["con_bas","cov_bas","con_bas"],[]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"tensor"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{E}\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{"con_bas"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"cov_bas"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"con_bas"}\right]\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{}\right]\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{4}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
Define the jet space J^2(R^2, R^2) for two functions u and v of 2 independent variables x and y.
\mathrm{DGsetup}\left([x,y],[u,v],J,2\right):
\mathrm{\omega }≔\mathrm{DGbiform}\left(u[1,1]\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"biform"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{J}\textcolor[rgb]{0,0,1}{,}\left[\textcolor[rgb]{0,0,1}{0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{9}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right)
\mathrm{convert}\left(\mathrm{\omega },\mathrm{DGform}\right)
\textcolor[rgb]{0,0,1}{\mathrm{_DG}}\textcolor[rgb]{0,0,1}{}\left(\left[\left[\textcolor[rgb]{0,0,1}{"form"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{J}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\left[\textcolor[rgb]{0,0,1}{1}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{2}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{u}}_{\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{2}}\right]\textcolor[rgb]{0,0,1}{,}\left[\left[\textcolor[rgb]{0,0,1}{9}\right]\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\right]\right]\right]\right) |
write the polynomial function f of least degree that had
snajat3l 2022-02-11 Answered
write the polynomial function f of least degree that had rational coefficients of 1, and the given zeros. write the function in standard form. -4, -2,5
Cenedesiidv
y=a(x-x_1)(x-x_2)(x-x_3)
y=1(x+4)(x+2)(x-5)
y=x^2+2x+4x+8(x-5)
y=x^3+x^2-22x-40
Answer is: y=x^3+x^2-22x-40
3{v}^{2}+5v-6
{x}^{2}-6x+9
{x}^{2}-25
3{x}^{2}-21x+30
p\left(t\right)={t}^{5}-6{t}^{4}+6{t}^{3}-6{t}^{2}+5t
{P}_{a}\left(t\right)=p\left(t\right)
Multiply the following polynomials:
\left(4x+4\right)\left(4x+4\right)
Find the difference of the polynomials.
\left(3{x}^{2}+2x-1\right)-\left(-5{x}^{2}+8x+4\right) |
Semisimple module - formulasearchengine
Template:Mergefrom In mathematics, especially in the area of abstract algebra known as module theory, a semisimple module or completely reducible module is a type of module that can be understood easily from its parts. A ring that is a semisimple module over itself is known as an Artinian semisimple ring. Some important rings, such as group rings of finite groups over fields of characteristic zero, are semisimple rings. An Artinian ring is initially understood via its largest semisimple quotient. The structure of Artinian semisimple rings is well understood by the Artin–Wedderburn theorem, which exhibits these rings as finite direct products of matrix rings.
3 Endomorphism rings
4 Semisimple rings
4.2 Simple rings
4.3 Jacobson semisimple
A module over a (not necessarily commutative) ring with unity is said to be semisimple (or completely reducible) if it is the direct sum of simple (irreducible) submodules.
For a module M, the following are equivalent:
M is a direct sum of irreducible modules.
M is the sum of its irreducible submodules.
Every submodule of M is a direct summand: for every submodule N of M, there is a complement P such that M = N ⊕ P.
{\displaystyle 3\Rightarrow 2}
, the starting idea is to find an irreducible submodule by picking any
{\displaystyle x\in M}
{\displaystyle P}
be a maximal submodule such that
{\displaystyle x\notin P}
. It can be shown that the complement of
{\displaystyle P}
is irreducible.[1]
The most basic example of a semisimple module is a module over a field; i.e., a vector space. On the other hand, the ring Z of integers is not a semisimple module over itself (because, for example, it is not an artinian ring.)
Semisimple is stronger than completely decomposable, which is a direct sum of indecomposable submodules.
Let A be an algebra over a field k. Then a left module M over A is said to be absolutely semisimple if, for any field extension F of k,
{\displaystyle F\otimes _{k}M}
is a semisimple module over
{\displaystyle F\otimes _{k}A}
If M is semisimple and N is a submodule, then N and M/N are also semisimple.
If each
{\displaystyle M_{i}}
is a semisimple module, then so is
{\displaystyle \bigoplus _{i}M_{i}}
A module M is finitely generated and semisimple if and only if it is Artinian and its radical is zero.
Endomorphism rings
A semisimple module M over a ring R can also be thought of as a ring homomorphism from R into the ring of abelian group endomorphisms of M. The image of this homomorphism is a semiprimitive ring, and every semiprimitive ring is isomorphic to such an image.
The endomorphism ring of a semisimple module is not only semiprimitive, but also von Neumann regular, Template:Harv.
A ring is said to be (left)-semisimple if it is semisimple as a left module over itself. Surprisingly, a left-semisimple ring is also right-semisimple and vice versa. The left/right distinction is therefore unnecessary, and one can speak of semisimple rings without ambiguity.
A semisimple ring may be characterized in terms of homological algebra: namely, a ring R is semisimple if and only if any short exact sequence of left (or right) R-modules splits. In particular, any module over a semisimple ring is injective and projective. Since "projective" implies "flat", a semisimple ring is a von Neumann regular ring.
Semisimple rings are of particular interest to algebraists. For example, if the base ring R is semisimple, then all R-modules would automatically be semisimple. Furthermore, every simple (left) R-module is isomorphic to a minimal left ideal of R, that is, R is a left Kasch ring.
Semisimple rings are both Artinian and Noetherian. From the above properties, a ring is semisimple if and only if it is Artinian and its Jacobson radical is zero.
If an Artinian semisimple ring contains a field, it is called a semisimple algebra.
A commutative semisimple ring is a finite direct product of fields. A commutative ring is semisimple if and only if it is artinian and reduced.[2]
If k is a field and G is a finite group of order n, then the group ring
{\displaystyle k[G]}
is semisimple if and only if the characteristic of k does not divide n. This is Maschke's theorem, an important result in group representation theory.
By the Artin–Wedderburn theorem, a unital Artinian ring R is semisimple if and only if it is (isomorphic to)
{\displaystyle M_{n}(D_{1})\times M_{n}(D_{2})\times \dots \times M_{n}(D_{r})}
{\displaystyle D_{i}}
is a division ring and
{\displaystyle M_{n}(D)}
is the ring of n-by-n matrices with entries in D.
An example of a semisimple non-unital ring is
{\displaystyle M_{\infty }(K)}
, the row-finite, column-finite, infinite matrices over a field K.
One should beware that despite the terminology, not all simple rings are semisimple. The problem is that the ring may be "too big", that is, not (left/right) Artinian. In fact, if R is a simple ring with a minimal left/right ideal, then R is semisimple.
Classic examples of simple, but not semisimple, rings are the Weyl algebras, such as Q<x,y>/(xy-yx-1), which is a simple noncommutative domain. These and many other nice examples are discussed in more detail in several noncommutative ring theory texts, including chapter 3 of Lam's text, in which they are described as nonartinian simple rings. The module theory for the Weyl algebras is well studied and differs significantly from that of semisimple rings.
{{#invoke:main|main}} A ring is called Jacobson semisimple (or J-semisimple or semiprimitive) if the intersection of the maximal left ideals is zero, that is, if the Jacobson radical is zero. Every ring that is semisimple as a module over itself has zero Jacobson radical, but not every ring with zero Jacobson radical is semisimple as a module over itself. A J-semisimple ring is semisimple if and only if it is an artinian ring, so semisimple rings are often called artinian semisimple rings to avoid confusion.
For example the ring of integers, Z, is J-semisimple, but not artinian semisimple.
semisimple algebra
↑ Nathan Jacobson, Basic Algebra II (Second Edition), p.120
R.S. Pierce. Associative Algebras. Graduate Texts in Mathematics vol 88.
Retrieved from "https://en.formulasearchengine.com/index.php?title=Semisimple_module&oldid=235672" |
herr strathmann | coffee climbing jazz math
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\log p(x)=\sum_{i=1}^n \alpha_i k(z_i, x)
k(x,y)=\exp\left(-\Vert x-y\Vert ^2 / \sigma \right)
{\mathrm{zi}}_{}
z_i
\sigma
{\mathrm{\alpha i}}^{}
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\alpha_i |
\frac{1}{3}\mathrm{rd}
\frac{1}{5}\mathrm{th}
Singh, Jain, Sharma and Gupta were partners in a firm sharing profits in the ratio of 4 : 3 : 2 : 1. On 1.4.2016 their Balance Sheet was as follows: (6)
Balance Sheet of Singh, Jain, Sharma and Gupta
as at 1.4.2016
From the above date the partners decided to share the future profits equally. For this purpose the goodwill of the firm was valued at Rs 60,000. Partners also agreed that:
(i) Claims against Workmen Compensation Reserve was estimated at Rs 40,000 and depreciation of Rs 15,000 will be charged on fixed assets.
(ii) Capitals of the partners will be adjusted according to the new profit sharing ratio for which current accounts will be opened.
On 1.4.2015, Neena Ltd. issued 800, 9% debentures of Rs 100 each at a discount of 5%, redeemable at a premium of 8% after five years. The company closes its books on 31st March every year. Interest on 9% debentures is payable on 30th September and 31st March. Rate of tax deducted at source is 10%.
Pass necessary journal entries for the issue of 9% debentures and payment of interest on 9% debentures for the year ended 31st March, 2016. (6) VIEW SOLUTION
Pass necessary journal entries on the dissolution of a firm in the following cases: (6)
(i) Satish, a partner, agreed to do the dissolution work for which he was allowed a commission of Rs 18,000. He also agreed to bear the dissolution expenses. Actual dissolution expenses paid by Satish were Rs 9,000.
(ii) Suleman, a partner, paid the dissolution expenses Rs 750.
(iii) Dissolution expenses were Rs 500.
(iv) Sandhya was appointed to look after the dissolution work on a remuneration of Rs 3,000. She agreed to bear the dissolution expenses. Actual dissolution expenses Rs 2,750 were paid by Sunil, another partner on behalf of Sandhya.
(v) Seema, a partner, agreed to do the dissolution work for a commission of Rs 4,500. She also agreed to bear the dissolution expenses. Seema took away stock of the same amount as her commission. The stock had already been transferred to realisation account.
(vi) Santosh, a partner, agreed to bear the dissolution expenses for a commission of Rs 6,000. Actual dissolution expenses Rs 4,500 were paid from the firm's bank account. VIEW SOLUTION
\frac{1}{4}\mathrm{th}
Why is separate disclosure of cash flows from 'investing activities' necessary? State. VIEW SOLUTION
What is meant by a non-cash transaction? Give one example of a non-cash transaction. VIEW SOLUTION
What is meant by analysis of financial statements? State any two limitations of such analysis. VIEW SOLUTION |
Cantor Set | Brilliant Math & Science Wiki
The Cantor set is set of points lying on a line segment. It is created by taking some interval, for instance
[0,1],
and removing the middle third
\left(\frac{1}{3},\frac{2}{3}\right)
, then removing the middle third of each of the two remaining sections
\left(\frac{1}{9},\frac{2}{9}\right)
\left(\frac{7}{9},\frac{8}{9}\right)
, then removing the middle third of the remaining four sections, and so on ad infinitum.
It is a closed set consisting entirely of boundary points, and is an important counterexample in set theory and general topology. When learning about cardinality, one is first shown subintervals of the real numbers,
\mathbb{R}
, as examples of uncountably infinite sets. This might suggest that any uncountably infinite subset of
\mathbb{R}
must contain an interval; such an assertion is false. A counterexample to this claim is the Cantor set
\mathcal{C} \subset [0,1]
, which is uncountable despite not containing any intervals. In addition, Cantor sets are uncountable, may have 0 or positive Lebesgue measures, and are nowhere dense. Cantor sets are the only disconnected, perfect, compact metric space up to a homeomorphism.
The Cantor set is constructed by removing increasingly small subintervals from
[0,1]
. In the first step, remove
\left(\frac13, \frac23\right)
[0,1]
. In the second step, remove
\left(\frac19, \frac29\right) \cup \left(\frac79, \frac89\right)
from what remains after the first step. In general, on the
n^\text{th}
step, remove
\left(\frac{1}{3^n}, \frac{2}{3^n}\right) \cup \left(\frac{4}{3^n}, \frac{5}{3^n}\right) \cup \cdots \cup \left(\frac{3^n - 2}{3^n}, \frac{3^n - 1}{3^n}\right)
from what remains after the
(n-1)^\text{th}
step. After all
\mathbb{N}
steps have been taken, what remains is the Cantor set
\mathcal{C}
This construction can be formalized as follows. Let
\mathcal{C}_0 = [0,1]
. Then, after the first step, what remains is
\mathcal{C}_1 = \left[0,\frac13\right] \cup \left[\frac23,1\right]
. After the second step, what remains is
\mathcal{C}_2 = \left[0,\frac{1}{9}\right] \cup \left[\frac{2}{9}, \frac{1}{3}\right] \cup \left[\frac{2}{3},\frac{7}{9}\right] \cup \left[\frac{8}{9}, 1\right].
Continuing in this manner, one obtains an infinite collection of sets
\{\mathcal{C}_i\}_{i=0}^{\infty}
\mathcal{C}_i \subset \mathcal{C}_{i-1}
i\ge 1
. Then, one defines the Cantor set to be
\mathcal{C} = \bigcap_{i=0}^{\infty} \mathcal{C}_i.
Intuitively, this makes sense; since
\bigcap_{i=0}^{n} \mathcal{C}_i = \mathcal{C}_n
n\ge 1
, taking an infinite intersection would provide the "limiting set" in this situation.
After seven iterations of the Cantor set's construction. Going from top to bottom, the image depicts the sets
\mathcal{C}_0
\mathcal{C}_6
There is an alternate characterization that will be useful to prove some properties of the Cantor set:
\mathcal{C}
consists precisely of the real numbers in
[0,1]
whose base-3 expansions only contain the digits 0 and 2.
Base-3 expansions, also called ternary expansions, represent decimal numbers on using the digits
0,1,2
. For instance, the number 3 in decimal is 10 in base-3. Fractions in base-3 are represented in terms of negative powers of 3, where a fraction
N
in decimal can be represented as
N = c_1b^{-1}+c_2b^{-2}+c_3b^{-3}+\cdots+c_n{b^{-n}}+\cdots,
0≤ c_i <3
, and the fraction is written in base-3 as
N = 0.c_1c_2c_3...
\frac14 = 0.25 = 0\times 3^{-1} + 2\times 3^{-2} + 0\times 3^{-3} + 2\times 3^{-4}+\cdots = 0.\overline{02}_3
Similarly, decimal fractions can be converted by multiplying by
3
, taking the result mod 3, then multiplying the remainder by
3
, and then taking mod 3 of that... etc. until the remainder is 0:
\begin{aligned} 0.25 \times 3 = \mathbf{0} + 0.75,\ \ c_1 &= 0\\ 0.75 \times 3 = \mathbf{2} + 0.25,\ \ c_2 &= 2\\ 0.25 \times 3 = \mathbf{0}+ 0.75,\ \ c_3 &= 0\\ 0.75 \times 3 = \mathbf{2} + 0.25,\ \ c_4 &= 2\\ &\vdots\\ &\Rightarrow 0.0202..._3. \end{aligned}
In base-3, some points have more than one notation:
\frac13 \times 3 = \mathbf{1} + 0,\ c_1 = 1
\frac13 = 0.1_3.
But we also have the following:
\begin{aligned} \frac13 \times 3= 0.3333... \times 3 = \mathbf{0} + 0.9999...,\ c_1 &= 0\\ 0.9999... \times 3 = \mathbf{2} + 0.9999...,\ c_2 &= 2\\ 0.9999... \times 3 = \mathbf{2} + 0.9999...,\ c_3 &= 2\\ &\vdots \end{aligned}
\frac13 = 0.\overline{3} = 0.1_3 = 0.0\overline{2}_3.
This sometimes means there is ambiguity in
\mathcal{C}
, for if
\mathcal{C}
expansions only contain the digits 0 and 2, then
\mathcal{C}
0.0\overline{2}_3
0.1_3
even though they are the same. In this case, it's said to contain
0.0\overline{2}_3 = \frac13,
which implies it contains
0.1_3
because of the multiple notations.
x\in [0,1]
contains only the digits 0 and 2 in its base-3 expansion. Let
x_n
be the truncation of
x
n
places after the decimal point. For example, if
x = 0.020202\cdots_{3}
x_1 = 0
x_2 = 0.02_{3}
x_4 = 0.0202_{3}
, etc. Certainly, the sequence
x_n
x
n\to\infty
. In particular, for every
n\ge 1
x_n \le x \le x_n + \frac{1}{3^n}.
Note that the numbers in
[0,1]
whose base-3 expansions go on for exactly
n
digits after the decimal point and
which use only the digits 0 and 2
are precisely the left endpoints of the intervals whose union is
\mathcal{C}_n
. Thus, the interval
\left[x_n, x_n + \frac1{3^n}\right]
\mathcal{C}_n
x\in \mathcal{C}_n
n\ge 0
x\in \mathcal{C}
_\square
Conversely, suppose
x\in \mathcal{C}
x\in \mathcal{C}_n
n\ge 0
. Note that the numbers in
\mathcal{C}_n
are precisely those whose
n^\text{th}
truncation (i.e., the number obtained by taking only the first
n
digits after the decimal point) uses only 0 and 2 as digits in base 3. It follows that every truncation of
x
uses only 0 and 2 as digits. This implies
x
uses only 0 and 2 in its base-3 expansion.
_\square
\frac{107}{364}
\frac{111}{364}
\frac{113}{364}
\frac{115}{364}
Of the given answer choices, which number is in the Cantor set?
From this theorem, the proofs of the following two properties follow:
\mathcal{C}
does not contain any subintervals of
[0,1]
[a,b] \subset [0,1]
be an arbitrary interval. Write
a = 0.a_1 a_2 a_3 \cdots_{3}
b = 0. b_1 b_2 b_3 \cdots_{3}.
If some
a_i
equals 1, then by the previous theorem, we know
a\not \in \mathcal{C}
. Similarly, if some
b_i
equals 1, we know
b\not\in \mathcal{C}
. Otherwise, suppose
k
is the smallest index such that
a_k \neq b_k
. Our casework forces
a_k = 0
b_k = 2
0.a_1 a_2 a_3 \cdots a_{k-1} 1_{3} \in [a,b]
[a,b] \not\subset \mathcal{C}
_\square
\mathcal{C}
f: \mathcal{C} \to [0,1]
x = 0. x_1 x_2 x_3 \cdots_{3}
\mathcal{C}
f(x) = 0 . (x_1 / 2) (x_2 / 2) (x_3 / 2) \cdots _{2}.
In other words, take the ternary expansion of
x
, replace every digit
2
1
, and consider the result as a binary number.
This function is surjective, since any element
y = 0.y_1 y_2 y_3 \cdots_{2}
[0,1]
f\big(0. (2y_1) (2y_2) (2y_3) \cdots_{3}\big) = y
\mathcal{C}
is countable. Write
\mathcal{C} = \{c_1, c_2, c_3, \cdots\}
y\in [0,1]
g(y)
equal the index
j
c_j
f(c_j) = y
(remark: in defining this function
g
, we are implicitly using the axiom of choice). Then, we may construct a sequence
\{d_i\}_{i\in \mathbb{N}}
d_{g(y)} = y
y \in [0,1]
. This constitutes a bijection between
[0,1]
\mathbb{N}
[0,1]
is countable. Contradiction!
Hence, we conclude
\mathcal{C}
_\square
\mathcal{C} + \mathcal{C}
strictly contains
[0,2]
\mathcal{C} + \mathcal{C}
[0,2]
[0,2]
\mathcal{C} + \mathcal{C}
\mathcal{C} + \mathcal{C}
denote the sumset of the Cantor set
\mathcal{C}
\mathcal{C} + \mathcal{C} = \{x+y \, : \, x, y \in \mathcal{C}\}.
Which of the following statements is true about
\mathcal{C} + \mathcal{C}
127 "rect", . Cantor ternary set, in seven iterations. Retrieved September 5th, 2016, from https://commons.wikimedia.org/wiki/File:Cantor_set_in_seven_iterations.svg
Cite as: Cantor Set. Brilliant.org. Retrieved from https://brilliant.org/wiki/cantor-set/ |
A DataSeries is a one-dimensional sequence of data with a label for each data point. For example, you can keep track of nutritional energy values (in
⟦\mathrm{kJ}⟧
⟦g⟧
) of certain types of berries, as follows:
\textcolor[rgb]{0,0,1}{\mathrm{energy}}\textcolor[rgb]{0,0,1}{≔}\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{220}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{288}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{136}\end{array}\right]
\textcolor[rgb]{0,0,1}{288}
\textcolor[rgb]{0,0,1}{136}
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{\mathrm{true}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{\mathrm{true}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{\mathrm{false}}\end{array}\right]
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{220}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{288}\end{array}\right]
\textcolor[rgb]{0,0,1}{\mathrm{berry_data}}\textcolor[rgb]{0,0,1}{≔}\left[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{}& \textcolor[rgb]{0,0,1}{\mathrm{Genus}}& \textcolor[rgb]{0,0,1}{\mathrm{Energy}}& \textcolor[rgb]{0,0,1}{\mathrm{Carbohydrates}}& \textcolor[rgb]{0,0,1}{\mathrm{Total tons}}& \textcolor[rgb]{0,0,1}{\mathrm{Top producer}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{"Rubus"}& \textcolor[rgb]{0,0,1}{220}& \textcolor[rgb]{0,0,1}{11.94}& \textcolor[rgb]{0,0,1}{543421}& \textcolor[rgb]{0,0,1}{\mathrm{Russia}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{"Vitis"}& \textcolor[rgb]{0,0,1}{288}& \textcolor[rgb]{0,0,1}{18.1}& \textcolor[rgb]{0,0,1}{58500118}& \textcolor[rgb]{0,0,1}{\mathrm{China}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{"Fragaria"}& \textcolor[rgb]{0,0,1}{136}& \textcolor[rgb]{0,0,1}{7.68}& \textcolor[rgb]{0,0,1}{4594539}& \textcolor[rgb]{0,0,1}{\mathrm{USA}}\end{array}\right]
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{543421}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{58500118}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{4594539}\end{array}\right]
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{11.94}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{18.1}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{7.68}\end{array}\right]
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{\mathrm{true}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{\mathrm{true}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{\mathrm{false}}\end{array}\right]
\left[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{}& \textcolor[rgb]{0,0,1}{\mathrm{Genus}}& \textcolor[rgb]{0,0,1}{\mathrm{Energy}}& \textcolor[rgb]{0,0,1}{\mathrm{Carbohydrates}}& \textcolor[rgb]{0,0,1}{\mathrm{Total tons}}& \textcolor[rgb]{0,0,1}{\mathrm{Top producer}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{"Rubus"}& \textcolor[rgb]{0,0,1}{220}& \textcolor[rgb]{0,0,1}{11.94}& \textcolor[rgb]{0,0,1}{543421}& \textcolor[rgb]{0,0,1}{\mathrm{Russia}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{"Vitis"}& \textcolor[rgb]{0,0,1}{288}& \textcolor[rgb]{0,0,1}{18.1}& \textcolor[rgb]{0,0,1}{58500118}& \textcolor[rgb]{0,0,1}{\mathrm{China}}\end{array}\right]
\left[\textcolor[rgb]{0,0,1}{\mathrm{Genus}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Energy}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Carbohydrates}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Total tons}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{Top producer}}\right]
\left[\begin{array}{cc}\textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{11.94}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{18.1}\\ \textcolor[rgb]{0,0,1}{\mathrm{Strawberry}}& \textcolor[rgb]{0,0,1}{7.68}\end{array}\right]
\left[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{}& \textcolor[rgb]{0,0,1}{\mathrm{Genus}}& \textcolor[rgb]{0,0,1}{\mathrm{Energy}}& \textcolor[rgb]{0,0,1}{\mathrm{Carbohydrates}}& \textcolor[rgb]{0,0,1}{\mathrm{Total tons}}& \textcolor[rgb]{0,0,1}{\mathrm{Top producer}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Raspberry}}& \textcolor[rgb]{0,0,1}{"Rubus"}& \textcolor[rgb]{0,0,1}{220}& \textcolor[rgb]{0,0,1}{11.94}& \textcolor[rgb]{0,0,1}{543421}& \textcolor[rgb]{0,0,1}{\mathrm{Russia}}\\ \textcolor[rgb]{0,0,1}{\mathrm{Grape}}& \textcolor[rgb]{0,0,1}{"Vitis"}& \textcolor[rgb]{0,0,1}{288}& \textcolor[rgb]{0,0,1}{18.1}& \textcolor[rgb]{0,0,1}{58500118}& \textcolor[rgb]{0,0,1}{\mathrm{China}}\end{array}\right] |
Calculate load reflection coefficient of two-port network - MATLAB gammaml - MathWorks France
gammaml
Load Reflection Coefficient Calculation
Calculate Load Reflection Coefficient of S-Parameters Object
Calculate load reflection coefficient of two-port network
coefficient = gammaml(s_params)
coefficient = gammaml(hs)
coefficient = gammaml(s_params) calculates the load reflection coefficient of a two-port network required for simultaneous conjugate match.
coefficient = gammaml(hs) calculates the load reflection coefficient of the two-port network represented by the S-parameter object hs.
Calculate the load reflection coefficient using network data from a file
coefficient = gammaml(s_params);
Define S-parameters object specified from a file.
Calculate the load reflection coefficient using the gammaml function.
coefficient — Load reflection coefficient
Load reflection coefficient, returned as a M element complex vector.
{\Gamma }_{ML}=\frac{{B}_{2}±\sqrt{{B}_{2}{}^{2}-4|{C}_{2}{}^{2}|}}{2{C}_{2}}
\begin{array}{c}{B}_{2}=1-|{S}_{11}{}^{2}|+|{S}_{22}{}^{2}|-|{\Delta }^{2}|\\ {C}_{2}={S}_{22}-\Delta \cdot {S}_{11}^{*}\\ \Delta ={S}_{11}{S}_{22}-{S}_{12}{S}_{21}\end{array} |
Development of Autonomous VTOL UAV for Wide Area Surveillance
Development of Autonomous VTOL UAV for Wide Area Surveillance ()
Daeil Jo, Yongjin Kwon*
The drone was developed with the use of unmanned aircraft systems in the initial military sector based on the combination of aerospace technology and information and communication technologies in a variety of usability, including the civilian sectors. Developed for the field of reconnaissance, it is used in both civilian and police sectors as traffic monitoring and high altitude reconnaissance missions. It is used in broadcasting and surveillance, while continuously expanding into the areas of courier delivery and rescue missions. Based on the convergence of aviation technology such as various SW, sensor and flight control to utilize unmanned system and information communication technology, commercialization of related technology is being developed as a very diverse route. In this paper, we propose and manufacture of a VTOL UAV. Design process referred to the VTOL development process that has been devised by us, and actual building of a UAV also applied the same VTOL development concept. In order to understand the aerodynamic characteristics of the aircraft, we have applied the aerodynamic design theory and used the CAE method that can replace the actual wind tunnel test. We tested the selection method and criteria for the internal modules that make up the UAV, and we were able to assemble the product. FW coding of flight control computer was conducted for VTOL control. In addition, we developed a LTE communication module for the long distance flight, and carried out flight experiments with GCS to observe and respond to the flight situation from the ground. Flight test results showed that stable transition flight was possible with broadband. We could see that the actual performance results were met, compared to our development target values.
UAV, VTOL, Fixed Wing, Drone, Multi-Copter, Rotary Wing, Aircraft Design, Transition Flight
Jo, D. and Kwon, Y. (2019) Development of Autonomous VTOL UAV for Wide Area Surveillance. World Journal of Engineering and Technology, 7, 227-239. doi: 10.4236/wjet.2019.71015.
With the recent surge of interest in UAVs (unmanned aerial vehicles), there has been an increasing trend in utilization for many industrial sectors, including agricultural control, forest and coastal surveillance, search for victims, broadcasting and photography. The most widely used type of commercial UAV is a multi-copter type UAV, which can take off and land vertically. However, it can’t generate a wing-lift due to the characteristics of propellant types, which shortens the flight time. On the other hand, the fixed-wing type UAV has an advantage of longer flight time and higher speed, but it is difficult to secure a wide space for safe landing. This has been a great limitation in utilization. This study aims to develop a VTOL (vertical takeoff and landing) UAV, utilizing the advantages of both multi-copter type and fixed-wing type UAV [1] [2] [3] .
2. The VTOL Development Process
The overall development process of the vertical takeoff and landing VTOL UAV performed in this study is shown in Figure 1. This is based on our genuine effort to establish a systematic and sound UAV development, which has been mainly based on each developer.
2.1. Body Frame Design
The airframe design is an important part of the UAV development designed to reduce the air resistance. Compared to the quad-X type design (a typical multi-copter design), it is aesthetically and aerodynamically superior. Shown in Figure 2, its design ensures a large mounting space at the center of the fuselage to load various mission equipment and cargo [4] .
Figure 1. VTOL development process, established by the authors’ effort.
Figure 2. Body frame design layout.
2.2. Wing & Fuselage Design
Wing is an important part of determining the performance of UAV. The wing can be designed in various ways, such as low-wing, mid-wing and high-wing, which is mainly dependent on the desired flight characteristics. In the case of the lower wing, it has the merit that it can increase the maneuverability. In the case of the central wing, the balance of frame is said to be the best. Using the CAE (computer-aided engineering) software, the design of the wing shape was analyzed. The flight dynamic characteristics such as air resistance, lift, curvature analysis, airfoil, and air flow, are shown in Figure 3. The experimental conditions are boundary conditions when flying at 0˚C and 15 m/s. In the previous study, we designed the airfoil for the VTOL design as shown in Figure 4, and developed the airfoil which can achieve the efficiency at a low speed, considering that the maximum speed is 80 km/h and the cruising speed is 60 km/h [5] [6] [7] .
2.3. Selection of Motor Thrust
The output is determined by the rotational speed of the motor and the propeller.
Figure 3. CAE simulation Analysis 1.
The thrust determination equation is shown in Equation (1). Also, the equation to determine the propeller for generating the thrust after determining the required thrust is shown in Equation (2) [8] .
Power = Prop Const × rpmPower factor (1)
T=\frac{\pi }{4}{D}^{2}\rho v\Delta v
T: thrust [N].
D: propeller diameter [m].
v: velocity of air the propeller [m/s].
∆v: velocity of air the accelerated by propeller [m/s].
ρ: density of air [1.225 kg/m3].
2.4. Selection of Propellers
The greater the number XX and the larger the number YY, the higher lift can be achieved. The propeller should be large with a high pitch. But, as the weight of the propeller increases, the battery consumption increases accordingly. This leads to the decrease of the flight time. Therefore, it is necessary to select the proper propeller, according to the size of the airframe and the specification of the motor. Based on the motor selected in this way, the propeller is selected using Equation (3) from the source of known reference [9] (Figure 5).
{\text{rpm}}_{\text{ideal}}={\left(\frac{2}{\pi }\right)}^{\frac{1}{2\omega }}{\left(\frac{{g}^{\frac{3}{2}}{m}^{\frac{3}{2}}}{\alpha D\sqrt{\rho }}\right)}^{1/\omega }
ω = Power Factor from Aircraft.
α = Power Coefficient from Aircraft.
D = Diameter [m].
ρ = Air Density [1.225 kg/m3].
m = Mass [kg].
g = Gravity [9.8 m/s3].
2.5. Selection of ESC & Power Pack
The ESC (electronic speed control) is a device that can control the direction and speed of the drone motors. It can supply the proper voltage, depending on the motor or battery and is designed to handle the maximum current used by the motor. Figure 6 shows the selected ESC of 40 A.
The battery is directly related to the flight time. Equation (4) represents the flight time, according to the capacity and the consumption current of the battery.
Flight time = Battery capacity/amp (4)
\text{Flighttime}=2000\text{mAh}\ast \frac{60\mathrm{min}}{\left(\text{hour}\right)}\ast \frac{1}{\left(8\text{A}\right)}\ast \frac{A}{\left(10000\text{mA}\right)}=15\text{min}
The battery in this study was selected as 6 Cell, 24 V, 10,000 mAh type, considering flight time and thrust. Here, the “mAh” indicates the capacity of the battery. For example, an indication of 100,000 mAh means that you can use 1 hour when you draw 10,000 mA. The “V” indicates the voltage. One must use a battery with the proper voltage to stabilize the current to the drive motors.
The “Wh” shows how much power the battery has. This value can be calculated
Figure 5. Propeller selection XX & YY.
Figure 6. Selection of 40A ESC.
by multiplying the capacity mAh with the voltage V. The “C” indicates the discharge rate. In this study, we used the Motocalc SW, as shown in Figure 7. This software predicts the flight performance according to the parameters of the motor and propeller. The propeller length was 14 inches and the battery was HBZ-B 5200 22.2 V. The motor was selected as Quantum 400 kv. As a result, it was estimated that the thrust was about 12 kg or more. It is designed to be more than 3 times the weight of the airframe and enough to lift the airframe vertically with a 3 Kg load. Table 1 shows the VTOL specifications produced by the study [10] [11] [12] .
3. VTOL Assembly
3.1. Airframe Manufacturing
The designed UAV parts are produced by using a 3D printer and the UAV is assembled as shown in Figure 8. The 3D printed parts were mainly used to test and evaluate our initial design.
The final frame was produced after identified problems were solved, based on the data obtained from the prototype production. Figure 9 and Figure 10 show the process of fabricating and assembling the final airframe [13] [14] .
3.2. Flight Control Computer FW Coding for VTOL Control
The software setting is the core of FCC (flight control computer) coding. The FCC has built in components such as accelerometer, gyro, geomagnetic, barometer, GPS (global positioning system) and CPU (central processing unit), to maintain the stabilized flight by sensing the equilibrium state of VTOL. The geomagnetic system measures the movement direction of the airframe, while the barometer measures the altitude of the VTOL. The GPS tracks the flight routes of the UAV and performs the mission profile. In this study, FW (firmware) is coded using the Pixhawk controller, which is the most used FCC. The Pixhawk is an open source, and FW coding was developed as shown in Figure 11, using the C-language [15] .
3.3. Development of LTE Communication Module
In a line-of-sight flight, it is easy to manipulate the UAV because it can be adjusted
Table 1. VTOL specifications.
Figure 7. Thrust calculation using Motocalc S/W.
Figure 8. Fuselage sections of the 3D printed parts.
Figure 9. Parts assembly and various wiring work.
Figure 10. Completed VTOL UAV.
Figure 11. C-based navigation system coding.
by looking at the airframe. The UAV can be controlled in real-time, using 5.8 Ghz of for analog video transfer and 2.4 Ghz of frequency for the manipulator. When flying a long distance, it can be controlled through the LTE (long-term evolution) communication by using the automatic flight function. By developing a library to connect FCC and LTE modem, the UAV pilot can remotely check the flight images and UAV status, while controlling the UAV beyond the line-of-sight. As shown in Figure 12, the LTE modem developed in this study can be easily added in the future.
3.4. Ground Control Station Design and Production
For the remote control of UAV, the GCS (ground control station) is developed. The GUI-based (graphical user interface-based) environment and a user-friendly user interface have been developed, so that the flight status can be observed and confirmed on a map screen. The UAV flight control is equipped with both automatic flight and manual flight. As shown in Figure 13, the touch-screen displays the preprogrammed flight path. The control button layout is configured to increase the usability [16] [17] [18] [19] .
4. VTOL Flight Experiments
In this section, we analyzed the flight characteristics of the developed VTOL. Through numerous trials and errors, we have optimally set up the flight parameters
Figure 12. Self-developed LTE modem.
Figure 13. GCS for both automatic and manual flight control.
that can effectively control the VTOL. Figures 14-17 show a part of the test flight process.
Table 2 compares the target performance parameters and the actual performance data of the VTOL developed in this study. All parameters either meet or exceed the target values.
Figure 14. Preparatory work for flight experiments.
Figure 15. Automatic route flight.
Table 2. Comparison of the development goals with the actual performance data.
Figure 16. Profile of the VTOL flight experiments.
Figure 17. Log data extracted from the flight experiments.
In this paper, we proposed the VTOL UAV and studded the design and manufacturing process of the UAV. We developed the VTOL development process and actually followed the process to design and build the UAV. The process is systematic and not ad-hoc based. The developed process has been successfully applied to the new VTOL design and manufacture. In order to understand the aerodynamic characteristics of the VTOL, we applied both the aerodynamic design theory and the CAE that can replace the actual wind tunnel test. We tested the selection method and criteria for the internal modules that make up the UAV, and we were able to assemble the product. Firmware coding of the flight control computer was conducted as well. In addition, we developed a LTE communication module for long distance flight, and carried out flight experiments with the GCS to observe and monitor the flight status from the ground. Flight tests showed that a stable transition flight was possible with our method and we could see that the actual performance results were met compared to the development target.
[1] Jo, D. and Kwon, Y. (2017) Analysis of VTOL UAV Propellant Technology. Journal of Computer and Communications, 5, 76-82.
[2] Jo, D. and Kwon, Y. (2017) Development of Rescue Material Transport UAV. World Journal of Engineering and Technology, 5, 720-729.
[3] Kim, K.B. (2013) Design and Verification of Multi-Rotor Based Unmanned Aerial Vehicle System. Ph.D. Thesis, Department of Mechanical Engineering, Konkuk University, South Korea.
[4] Lee, C.J. (2017) Aircraft Structural Design Practice. Good Land-Company, McMinnville.
[5] kevadiya, M. (2013) CFD Analysis of Pressure Coefficient for NACA 4412. International Journal of Engineering Trends and Technology (IJETT), 4, 2041-2043.
[6] Yun, S.J. (2010) Aerodynamics. Saintian-Company, Pajoo.
[7] SOLIDWORKS 2017 (2017) Structural Analysis & Flow Simulation.
[8] Green, W.E. and Oh, P.Y. (2006) Autonomous Hovering of a Fixed-Wing Micro Air Vehicle. IEEE International Conference of Automation (ICRA), Orlando, 15-19 May 2006, 2164-2169.
[9] https://quadcopterproject.wordpress.com
[10] De la Torre, G.G., Ramallo, M.A. and Cervantes, E.G. (2016) Workload Perception in Drone Flight Training Simulators. Computers in Human Behavior, 64, 449-454.
[11] Liu, Z. and Li, H. (2016) Research on Visual Objective Test Method of High-Level Flight Simulator. Journal of System Simulation, 7, 1609-1614.
[12] Luan, L.-N. (2013) Augmenting Low-Fidelity Flight Simulation Training Devices via Amplified Head Rotations. Loughborough University, Loughborough.
[13] Lee, S. and Lim, K. (2008) PCB Design Guide Book. Sehwa Publishing Co., South Korea.
[14] Kang, M. and Shin, K. (2011) Electronic Circuit. Hanbit Media, Seoul.
[15] Yoon, H. (2015) Mitigation Settings and Their Execution Scenario in Drone Firmware. International Research Journal of Computer Science (IRJCS), 2, 7-11.
[16] Lee, E., Kim, S. and Kwon, Y. (2016) Analysis of Interface and Screen for Ground Control System. Journal of Computer and Communications, 4, 61-66.
[17] Kwon, Y., Heo, J., Jeong, S., Yu, S. and Kim, S. (2016) Analysis of Design Directions for Ground Control Station (GCS). Journal of Computer and Communications, 4, 1-7.
[18] Yu, S., Heo, J., Jeong, S. and Kwon, Y. (2016) Technical Analysis of VTOL UAV. Journal of Computer and Communications, 4, 92-97.
[19] Hong, J., Baek, S., Jung, H., Kim, S. and Kwon, Y. (2015) Usability Analysis of Touch Screen for Ground Operators. Journal of Computer and Communications, 3, 133-139. |
Solve for Y(s), the Laplace transform of the solution y(t)
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+5y=2t^2-9 , y(0)=0 , y'(0)=-3
Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
y"+5y=2{t}^{2}-9,y\left(0\right)=0,{y}^{\prime }\left(0\right)=-3
Aubree Mcintyre
y"+5y=2{t}^{2}-9
y\left(0\right)=0,{y}^{\prime }\left(0\right)=-3
L\left\{y"\right\}+sL\left\{{y}^{\prime }\right\}=L\left\{2{t}^{2}-9\right\}
{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)+5Y\left(s\right)=L\left\{2{t}^{2}\right\}-L\left\{9\right\}
=\frac{4}{{s}^{3}}-\frac{9}{s}
⇒{s}^{2}Y\left(s\right)-sy\left(0\right)+3+5Y\left(s\right)=\frac{4}{{s}^{3}}-\frac{9}{s}
⇒Y\left(s\right)\left[{s}^{2}+5\right]=\frac{4}{{s}^{3}}-\frac{9}{s}-3
⇒Y\left(s\right)=\frac{\frac{4}{{s}^{3}}-\frac{9}{s}-3}{{s}^{2}+5}=\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}
⇒y\left(t\right)={L}^{-1}\left\{\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}\right\}
=\frac{2{t}^{2}}{5}-\frac{3\sqrt{5}\mathrm{sin}\left(\sqrt{5}t\right)}{5}+\frac{49\mathrm{cos}\left(\sqrt{5}t\right)}{25}-\frac{49}{25}
\left[\because {L}^{-1}\left\{\frac{4-9{s}^{2}-3{s}^{3}}{{s}^{3}\left({s}^{2}+5\right)}\right\}=\frac{4}{5{s}^{3}}-\frac{15}{s\left({s}^{2}+25\right)}+\frac{49s}{25\left(25+{5}^{2}\right)}-\frac{49}{25{s}^{2}}\right]
\therefore \frac{2{t}^{2}}{5}-\frac{49}{25}-\frac{3\sqrt{5}}{25}\mathrm{sin}\left(\sqrt{5}t\right)+\frac{49}{25}\mathrm{cos}\left(\sqrt{5}t\right)
g\left(t\right)=\left\{\begin{array}{ll}5\mathrm{sin}\left(3\left[t-\frac{\pi }{4}\right]\right)& t>\frac{\pi }{4}\\ 0& t<\frac{\pi }{4}\end{array}
A steel ball of mass 4-kg is dropped from rest from the top of a building. If the air resistance is 0.012v and the ball hits the ground after 2.1 seconds, how tall is the building? Answer in four decimal places.
Use Laplace transform to solve the initial-value problem
\frac{{d}^{2}y}{{dt}^{2}}+25y=10\mathrm{cos}5t,y\left(0\right)=1,{y}^{\prime }\left(0\right)=0
y=\mathrm{tan}\sqrt{7t}
y=\frac{3-{v}^{2}}{3+{v}^{2}}
Verify that each given function is a solution of the differential equation.
y2{y}^{\prime }-3y=0;\text{ }{y}_{1}\left(t\right)={e}^{-3t},\text{ }{y}_{2}\left(t\right)={e}^{t}
y\left({x}^{2}+xy-2{y}^{2}\right)dx+x\left(3{y}^{2}-xy-{x}^{2}\right)dy=0;\text{ }when\text{ }x=1,y=1
Obtain the Laplace Transform of
L\left\{{e}^{-2x}+4{e}^{-3x}\right\} |
Mole Concept And Stoichiometry, Popular Questions: ICSE Class 10 CHEMISTRY, Concise Chemistry 10 - Meritnation
Name a liquid which is formed by chemical combination of two gases
Give the balanced chemical equation for the conversion from A to D:
Iron Pyrite --(A)--> Colourless Acidic Gas --(B)--> SO3 --(C)--> oleum --(D)--> Sulphuric acid
Please answer as soon as possible...
MnO2 + 4HCl
\to
MnCl2+ 2H2O + Cl2
1.746g of pure MnO2 is heated strongly with conc. HCl. Calculate:
a) Moles of MnO2 used.
b) Mass of salt formed.
c) Moles of chlorine gas formed.
d) Volume of chlorine gas formed at STP.
e) Mass of acid required.
If both concentrated and dilute sulphuric acid on Thermal decomposition produces nascent oxygen, then why does not both act as oxidising agent?
Shraddha Singh asked a question
15. Indium (atomic weight = 114.82) has two naturally occurring isotopes, the predominant one form has isotopic weight 114.9041 and abundance of 95.72%. Which of the following isotopic weights is the most likely for the other isotope?
200 ml of ethylene is burnt in just sufficient air (containing 20% oxygen) to form carbon
dioxide gas and steam. If all measurements are made at constant pressure and 100 0C,find
the composition of the resulting mixture. The equation for the reaction is:
Abhinav Sahu asked a question
Q. The number of
i) molecules [S = 32]
ii) atoms in 192 g. of sulphur. [
{S}_{8}
Calculate the weight of potassium nitrite formed by the thermal decomposition of 15.15g of potassium nitrate. [K = 39 ;N = 14 ;O = 16]
Vany Jain asked a question
24. Solid ammonium dichromate decomposes as under:
{\left(N{H}_{4}\right)}_{2}C{r}_{2}{O}_{7}\to {N}_{2}+C{r}_{2}{O}_{3}+4{H}_{2}O
If 63 g of ammonium dichromate decomposes. Calculate:
\left(a\right) the quantity in moles of {\left(N{H}_{4}\right)}_{2}C{r}_{2}{O}_{7}\phantom{\rule{0ex}{0ex}}\left(b\right) the quantity in moles of nitrogen formed\phantom{\rule{0ex}{0ex}}\left(c\right) the volume of {N}_{2} evolved at STP.\phantom{\rule{0ex}{0ex}}\left(d\right) what will be the loss of mass?\phantom{\rule{0ex}{0ex}}\left(e\right) calculate the mass of chromium \left(III\right) oxide formed at the same time.
Saroj Ku Soren asked a question
Pls tell how to do "problem 3 and 6".
Abhay Bhardwaj asked a question
Calcium hydroxide reacts with ammonium chloride to give ammonia, according to the following equation: Ca(OH)2 + 2NH4Cl --> CaCl2 + 2NH3 + 2H2O . If 5.35g of ammonium chloride are used, calculate: a. The mass of Calcium chloride formed and b. The volume, at S.T.P of NH3 liberated.
Kindly help me by suggesting some practice questions on numerical questions of percentage composition
Harsharan asked a question
calculate the volume of oxygen required for the complete combustion of 8.8 g of propane
Does lead react with hydrochloric acid (dilute) to produce lead chloride?
The gas law which relates the volume of a gas to moles of the gas is:-
1) Avagadro's Law.
2) Gay Lussac's Law.
3) Boyle's Law.
4) Charle's Law.
Himani asked a question
Calculate the % of calcium in a 55% pure sample of calcium carbonate ( answer : 6.6%. Pls tell the solution )
Renu Singh asked a question
When carbon dioxide is passed over Red Hot carbon ,carbon monoxide is produced according to the equation:CO2 +C->CO what volume of carbon monoxide at STP can be obtained from 3 gram of carbon
1g of NaCl and NaNO3 is dissolved in water and treated with excess of silver nitrate solution. 1.43g of precipitate is produced. Calculate the percentage of NaCl in the mixture. Given Ag = 108, Na = 23, N = 14, O =16, Cl = 35.5.
Gayatri Chatterjee asked a question
A sample of zinc oxide contains 80.25% zinc. Calculate the equivalent mass of zinc.
Is there any other way of remembering the use of organic compounds other than practicing them?
Laasya.g asked a question
A gaseous organic compound contains 3.6g of carbon and 0.8 g of hydrogen. The vapour density of this compound is 22. (1) calculate the emperical formula. (2) calculate the molecular formula of the compound. (3) if 4.4g of the above compound are completely burnt in oxygen, calculate the volume of carbon dioxide formed at stp.
My answer is not matching with the options..
Subhransu Dash asked a question
the following experiment was performed in order to determine the formula of a hydrocarbon the hydrocarbon Axis purified by fractional distillation 0.145 grams of his birth certificate from copper oxide into 24 cm cube of carbon dioxide is collected at STP
Can! You help. Me in Q. 8,9,10,11,12,13
Aalu Bukhara asked a question
25 ml N/10 NaOH solution will exactly neutralize which of the following solution :-
Pls tell how to do "problem 3 and problem 6".
Kaustubh asked a question
0.29g of a hydrocarbon are burnt completely in oxygen.on cooling, 448ml of carbon dioxide is collected at STP . calculate weight of carbon dioxide formed. Find weight hydrogen, and empirical formula of the hydrocarbon
Calculate the % of pure aluminium in 10 kg of aluminium oxide of 90% purity
Qb) i) Calculate the mass of residue obtained by heating 2.7 g. Ag2CO1 ? [ Ag = l08 C= 12 O = 16]
ii) How many moles of water are in 720 g of it ?
iii) Calculate the number of atoms in 8 g of helium gas at STP.
iv) KCIO3 on heating decomposes to give KCl and oxygen. What is the volume of O2 at stp liberated by 0.1 mole of KClO3.
M.r.n.varun asked a question
As aqueous solution has molality 11.11. Hence Mike fraction of the solute is
Question 26 please.
Q. (i) If 150 cc of gas A contains X molecules, how many molecules of gas B
will be in 75 cc of B? The gases A and B are under the same conditions of temperature and
(ii) Name the law on which the problem is based.
Q. Aluminium carbide reacts with Water according to the following equation:
A{l}_{4}{C}_{3}+12{H}_{2}O\to 4Al {\left(OH\right)}_{3}+3C{H}_{4}
(i) What mass Of aluminium hydroxide is formed from 12g Of aluminium
carbide?
(ii) What volume of methane at s.t.p. is obtained from 12g of aluminium carbide?
[Relative molecular weight of
A{l}_{4}{C}_{3}=144;Al {\left(OH\right)}_{3}=78
Pls solve Q3 .A
Arshdeep asked a question
A sample of coal gas contains 45% H2 30% ch4 20% CO and 5% C2 H2 by volume 100 cm cube of this gaseous mixture was mixed with 190 cm cube of Oxygen and exploded calculate the volume and composition of the mixture when cool to room temperature and pressure
Are nitrogen and oxygen exceptions of both ionization potential values and electron affinity?
A current of 3 amperes was passed through silver nitrate solution for 125 seconds. The amount of silver deposited at cathode was 0.42 g. Calculate the equivalent mass of silver
Shouldn't this be deliquisence?
A 3g of sample of Na2CO3.10H2O is heated to a constant mass. How much anhydrous salt remains?
All the questions to be solved.
Devesh Sahjwani asked a question
Q. Calculate the percentage of pure aluminium in 10 kg of aluminium oxide
\left[A{l}_{2}{O}_{3}\right]
of 90% purity.
Q. Calculate the percentage of carbon in a 55% pure sample of carbon carbonate.
Calculate the % of calcium in 55% pure sample of calcium carbonate
10 liter of a mixture of propane 60%and butane 40% is burnt.calculate the total volume of carbon dioxide is formed
please solve this first question of gay-lussac based problem
Please solve b)
Pls solve Q2.
Mansha Suresh asked a question
. The reaction : 4NH3 + 3O2 2N2 + 6H2O + heat takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of nitrogen oxide required to give 280cm3 of steam.
Aditya Panigrahi asked a question
112 cm cube of H2S is mixed with 120 cm cube of cl2 at STP to produce HCL and sulphur. Write a balanced equation for this reaction calculate the volume of gaseous products formed and composition of resulting mixture.
Will this this be " under standard conditions of temperature and pressure"?
Hydrated calcium sulphate contains 21% of water crystallisation calculate the number of molecules of water in the hydrated compound
How to solve number 7?
Wara Fatima & 1 other asked a question
Q 24 part (d).
{\left({\mathrm{NH}}_{4}\right)}_{2}{\mathrm{Cr}}_{2}{\mathrm{O}}_{7}\to {\mathrm{N}}_{2}+{\mathrm{Cr}}_{2}{\mathrm{O}}_{3}+4{\mathrm{H}}_{2}\mathrm{O}
If 63 g of ammonium dichromate decomposes, calculate:
(a) The quantity in moles of (NH4)2 Cr2O7
(b) The quantity in moles of nitrogen formed
(c) The volume of N2 evolved at STP.
(d) The loss of mass
State your observations when sodium carbonate is added to acetic acid
Amyra asked a question
2 g of a sample containing NaCl,NaBr nd some inert impurities dissolve in water nd treated with 0.5M sliver nitrate solution to give 2.6 g of mixed precepitate AgCl nd AgBr.the precepitate was treated with concentrate solution sodium bromine giving 3g AgBr precepitate determine volume silver nitrate use nd mass percentage of NaCl nd NaBr.
ZnS is roasted in air, calculate:
a] the number of moles of SO2 liberated by 776g of ZnS and,
b] the weight of ZnS required to produce 22.4 lits of SO2 at s.t.p. (S=32, Zn=65,O=16)
Diptayan Mondal asked a question
1 kg of water contains 10?? drops of water which has dissolved CO2 in it. Find the no of water molecules in 10 to the power 6 drops of water.
Kiranjit Kaur asked a question
24 cc Marsh gas (CH4) was mixed with 106 cc oxygen and then exploded. On cooling, the volume of the mixture became 82 cc, of which 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculations.
Solve Q1. Pls
Adnan asked a question
0.290g of an organic compound containing C, H and O gave on combustion 0.270g of water and 0.66g of CO2. What is emperical formula of the compound.
Pls solve 2.
Yash Pratap asked a question
All empirical formula question.
Monika Dugar asked a question
Please solve Q. 5
Q.(i) A gas of mass 32 gms has a volume of 20 litres at S.T.P. Calculate the gram molecular weight of the gas.
(ii) How much Calcium oxide is formed when 82 g of calcium nitrate is heated? Also find the volume of nitrogendioxide evolved:
2Ca\left(N{O}_{3}{\right)}_{2}\to 2CaO+4N{O}_{2}+{O}_{2}
(Ca = 40, N = 14, O = 16)
Please solve Q)4)a Got the answers as.a)i)0.5mole a)ii)4.48 litres a)iii)22g of CaO. a)iv)65.6g. a)v)656g
The percentage composition of a gas is nitrogen:82.35 and hydrogen 17.64 Find the emperical formula of the gas
1 kg of water has 10 to the power 11 drops of water which has dissolved CO2 in it. Find the number of water molecules in 10 to the power 6 drops of water.
Sanjana Thakur asked a question
If 12.5 g of solid zinc carbonate is heated to constant mass ,what is the mass of the substance formed?
(Zn=65 ,C=12 ,O=16)
radhasuresh31179... asked a question
hydrated CaSO4.xH2o contains 21% of water of crystallization. calculate the no. of molecules of crystallization i.e. x in the hydrated compound.
Bisakha Giri asked a question
Find Vapour density
Please help me in solving these numerical
Karthik Veeranki asked a question
Solve all plz
Chador Wangmo asked a question
Calculate the numbers of molecule present in one litre of water assuming that the density of water is 1g /cm3?
Ayush Chowdhury asked a question
b) A chemical salt gave the following percentage composition: Na = 29.11%, S = 40.51% and O = 30.38%.
Calculate the empirical formu;a of the salt. [Na = 23, S = 32, O = 16].
c) 4.92g of hydrated magnesium sulphate
\left(MgS{O}_{4}.x{H}_{2}O\right)
on strong heating leaves 2.40g of anhydrous magnesium sulphate. Calculate the value of x in
MgS{O}_{4}.x{H}_{2}O . \left[Mg=24, S=32, O=16, H=1\right]
Please anyone solve it and explain..
You are provided with samples of O2, N2, CO2, CO, under similar conditions of pressure and temperature such that they contain the same number of molecules represented by the letter M. The molecules of oxygen occupy a volume of N litres and have a mass of 8g. Under the similar condiyions of temperature and pressure find the mass of CO2 in grams.
If ammonia is not combustible, then why is it saying that Ammonia burns in an atmosphere of oxygen?
Rishaav Raj asked a question
Molecular weight of organic compound
In H2S2O7, what are the oxidation numbers of oxygen?
Pls tell how to do Qn 1
Jayant asked a question
Calculate the mass of a silver atom?
Solve these questions fast
Bhuvi Tiwari asked a question
When carbon monoxide is passed over red hot coke, carbon monoxide is produced according to the equation: CO2+C->2CO. What volume of carbon monoxide at STP can be obtained from 3g of carbon??
Plz solve question b) ii part
\mathrm{c}\right) \mathrm{A} \mathrm{compound} \mathrm{has} \mathrm{the} \mathrm{following} \mathrm{percentage} \mathrm{composition} :\phantom{\rule{0ex}{0ex}}\mathrm{Na}=18.60%, \mathrm{S}=25.80%, \mathrm{H}=4.03%, \mathrm{O}=51.58%.\phantom{\rule{0ex}{0ex}}\mathrm{The} \mathrm{molecular} \mathrm{weight} \mathrm{of} \mathrm{the} \mathrm{compound} \mathrm{is} 248. \mathrm{Calculate} \mathrm{the} \mathrm{molecular} \mathrm{formula} \mathrm{of} \mathrm{the} \mathrm{crystalline} \mathrm{salt}, \mathrm{assuming} \mathrm{all} \mathrm{hydrogen} \mathrm{to} \mathrm{be} \mathrm{present} \mathrm{as} \mathrm{water} \mathrm{of} \mathrm{crysallisatio}\phantom{\rule{0ex}{0ex}}\left[\mathrm{Na}=23, \mathrm{S}=32, \mathrm{O}=16, \mathrm{H}=1\right]
Ciara asked a question
Pls solve for me |
Learning deep kernels for exponential family densities | herr strathmann
coffee climbing jazz math
June 29, 2019 April 19, 2020 ~ karlnapf
An ICML 2019 paper, together with Kevin Li, Dougal Sutherland, and Arthur Gretton. Code by Kevin.
More score matching for estimating gradients using the infinite dimensional kernel exponential family (e.g. for gradient-free HMC)! This paper tackles one of the most limiting practical characteristics of using the kernel infinite dimensional exponential family model in practice: the smoothness assumptions that come with the standard “swiss-army-knife” Gaussian kernel. These smoothness assumptions are blessing and curse at the same time: they allow for strong statistical guarantees, yet they can quite drastically restrict the expressiveness of the model.
To see this, consider a log-density model of the form (the “lite” estimator from our HMC paper) $$\log p(x) = \sum_{i=1}^n\alpha_ik(x,z_i)$$for “inducing points” $z_i$ (the points that “span” the model, could be e.g. the input data) and the Gaussian kernel $$k(x,y)=\exp\left(-\Vert x-y\Vert^2/\sigma\right)$$Intuitively, this means that the log-density is simply a set of layered Gaussian bumps — (infinitely) smooth, with equal degrees of variation everywhere. As the paper puts it
These kernels are typically spatially invariant, corresponding to a uniform smoothness assumption across the domain. Although such kernels are sufficient for consistency in the infinite-sample limit, the
induced models can fail in practice on finite datasets, especially if the data takes differently-scaled shapes in different parts of the space. Figure 1 (left) illustrates this problem when fitting a simple mixture of Gaussians. Here there are two “correct” bandwidths, one for the broad mode and one for the narrow mode. A translation-invariant kernel must pick a single one, e.g. an average between the two, and any choice will yield a poor fit on at least part of the density
How can we learn a kernel that locally adapts to the density? Deep neural networks! We construct a non-stationary (i.e. location dependent) kernel using a deep network $\phi(x)$ on top a Gaussian kernel, i.e. $$k(x,y) = \exp\left(-\Vert \phi(x)-\phi(y)\Vert^2 / \sigma\right)$$ The network $\phi(x)$ is fully connected with softplus nonlinearity, i.e. $$\log(1+e^x)$$Softplus gives us some nice properties such as well-defined loss functions and a normalizable density model (See Proposition 2 in the paper for details).
However, we need to learn the parameters of the network. While kernel methods typically have nice closed-form solution with guarantees (and so does the original kernel exponential family model, see my post). Optimizing the parameters of ϕ(x)
\varphi \left(x\right)
<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></math>
\phi(x)
obviously makes things way more complicated: Whereas we could use a simple grid-search or black-box optimizer for a single kernel parameter, this approach here fails due to the number of parameters in 12∫p0(x)∥∇xlogp(x)−∇xlogp0(x)∥2dx
\varphi \left(x\right)
<math xmlns="http://www.w3.org/1998/Math/MathML"><mi>ϕ</mi><mo stretchy="false">(</mo><mi>x</mi><mo stretchy="false">)</mo></math>
\phi(x)
Could we use naive gradient descent? Doing so on our score-matching objectivewith
\log p(x)=\sum_{i=1}^n \alpha_i k(z_i, x)
k(x,y)=\exp\left(-\Vert x-y\Vert ^2 / \sigma \right)
will always overfit to the training data as the score (gradient error loss) can be made arbitrarily good by moving
{\mathrm{zi}}_{}
z_i
towards a data point and making
\sigma
go to zero. Stochastic gradient descent (the swiss-army-knife of deep learning) on the score matching objective might help, but would indeed produce very unstable updates.
Instead, we employed a two-stage training procedure that is conceptually motivated by cross-validation: we first do a closed-form update for the kernel model coefficients αi
{\mathrm{\alpha i}}^{}
<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>αi</mi></msub></math>
\alpha_i
on one half of the dataset, then we perform a gradient step on the parameters of the deep kernel on the other half. We make use of auto-diff — extremely helpful here as we need to propagate gradients through a quadratic-form-style score matching loss, the closed-form kernel solution, and the network. This seems to work quite well in practice (the usual deep trickery to make it work applies). Take away: By using a two-stage procedure, where each gradient step involves a closed form (linear solve) solutions for the kernel coefficients
{\mathrm{\alpha i}}^{}
<math xmlns="http://www.w3.org/1998/Math/MathML"><msub><mi>αi</mi></msub></math>
\alpha_i
we can fit this model reliably. See Algorithm 1 in the paper for more nitty-gritty details.
A cool extension of the paper would be to get rid of the sub-sampling/partitioning of the data and instead auto-diff through the leave-one-out error, which is closed form for these type of kernel models, see e.g. Wittawat’s blog post.
We experimentally compared the deep kernel exponential family to a number of other approaches based on likelihoods, normalizing flows, etc, and the results are quite positive, see the paper!
Naturally, as I have worked on using these gradients in HMC where the exact gradients are not available, I am very interested to see whether and how useful such a more expressive density model is. The case that comes to mind (and in fact motivated one of the experiments in the this paper) is the Funnel distribution (e.g. Neal 2003, picture by Stan), e.g.$$p(y,x) = \mathsf{normal}(y|0,3) * \prod_{n=1}^9 \mathsf{normal}(x_n|0,\exp(y/2)).$$
The mighty Funnel
This density historically was used as a benchmark for MCMC algorithms. Among others, HMC with second order gradients (Girolami 2011) perform much better due to their ability to adapt their step-sizes to the local scaling of the density — something that our new deep kernel exponential family is able to model. So I wonder: are there cases where such funnel-like densities arise in the context of say ABC or otherwise intractable gradient models? For those cases, an adaptive HMC sampler with the deep network kernel could improve things quite a bit.
kernelspublicationrepresentation learning
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Calculate the area of triangle formed by joining the points (4,0),(0,0) and (0,4) - Maths - Coordinate Geometry - 10544635 | Meritnation.com
Calculate the area of triangle formed by joining the points (4,0),(0,0) and (0,4)
Let A=\left({x}_{1},{y}_{1}\right)=\left(4,0\right),B=\left({x}_{2},{y}_{2}\right)=\left(0,0\right),C=\left({x}_{3},{y}_{3}\right)=\left(0,4\right)\phantom{\rule{0ex}{0ex}}be the given points. Then,\phantom{\rule{0ex}{0ex}}Area of △ABC=\frac{1}{2}\left|{x}_{1}\left({y}_{2}-{y}_{3}\right)+{x}_{2}\left({y}_{3}-{y}_{1}\right)+{x}_{3}\left({y}_{1}-{y}_{2}\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|4\left(0-4\right)+0\left(4-0\right)+0\left(0-0\right)\right|\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left|-16\right|\phantom{\rule{0ex}{0ex}}=8 sq. units
Ramnarayan answered this
8 cm^2.you can find it by drawing x and y axis with a origin and plot the points in the 1 quadrant |
I need to solve this limit: \lim_{x \to 0} \cos x^{\frac{1}{\sin
I need to solve this limit:
\underset{x\to 0}{lim}{\mathrm{cos}x}^{\frac{1}{\mathrm{sin}x}}
1.Call this function f(x)
2. Evaluate the limit
L=\underset{x\to 0}{lim}\mathrm{log}f\left(x\right)=\underset{x\to 0}{lim}\mathrm{log}\left({\left(\mathrm{cos}x\right)}^{\frac{1}{\mathrm{sin}x}}\right)=\underset{x\to 0}{lim}\frac{\left(\mathrm{log}\mathrm{cos}x\right)}{\mathrm{sin}x}
- this is a 0/0 indeterminate form but you can apply L'Hôpital to get
\underset{x\to 0}{lim}\frac{\left(\frac{1}{\mathrm{cos}x}\right)\mathrm{sin}x}{\mathrm{cos}x}=\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{{\mathrm{cos}}^{2}x}=0
{e}^{L}={e}^{0}=1
\underset{x\to 0}{lim}{\left(\mathrm{cos}x\right)}^{\frac{1}{\mathrm{sin}x}}
=\underset{x\to 0}{lim}{\left(1-{\mathrm{sin}}^{2}x\right)}^{\frac{12}{\mathrm{sin}x}}
\begin{array}{}=\left(\underset{x\to 0}{lim}\left(1-{\mathrm{sin}}^{2}x{\right)}^{-\frac{1}{{\mathrm{sin}}^{2}x}}{\right)}^{-\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{2}}\end{array}
\left(\mathrm{cos}x{\right)}^{\frac{1}{\mathrm{sin}x}}
\begin{array}{}\underset{x\to 0}{lim}\mathrm{cos}{x}^{\frac{1}{\mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\mathrm{ln}\mathrm{cos}x}{\mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{x\cdot \mathrm{ln}\mathrm{cos}x}{x\cdot \mathrm{sin}x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\mathrm{ln}x\mathrm{cos}x}{x}}\\ ={e}^{\underset{x\to 0}{lim}\frac{\left(\mathrm{cos}x-1\right)\cdot \mathrm{ln}\left(\mathrm{cos}x+1-1\right)}{\left(\mathrm{cos}x-1\right)x}}\\ v={e}^{\underset{x\to 0}{lim}\frac{\mathrm{cos}x-1}{x}}={e}^{0}=1\end{array}
\mathrm{sin}x+\mathrm{sin}y=a\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}x+\mathrm{cos}y=b
\mathrm{tan}\left(x-\frac{y}{2}\right)
What does x equal when
\mathrm{cos}\left(x\right)=0
\mathrm{tan}×{\mathrm{csc}}^{2}x–\mathrm{tan}x=\mathrm{cot}
\frac{1+{\mathrm{tan}}^{2}A}{1+{\mathrm{cot}}^{2}A}={\left(\frac{1-\mathrm{tan}A}{1-\mathrm{cot}A}\right)}^{2}={\mathrm{tan}}^{2}A
How do you verify the identity
\frac{\mathrm{sin}2x}{\mathrm{sin}x}=\frac{2}{\mathrm{sec}x}?
\frac{1}{{\left(\mathrm{tan}\left(\frac{x}{2}\right)+1\right)}^{2}{\mathrm{cos}}^{2}\left(\frac{x}{2}\right)}=\frac{1}{1+\mathrm{sin}x} |
The Parallel versus Branching Recurrences in Computability Logic
2013 The Parallel versus Branching Recurrences in Computability Logic
Wenyan Xu, Sanyang Liu
This paper shows that the basic logic induced by the parallel recurrence
\Yup
of computability logic (i.e., the one in the signature $\{ ¬, ∧,∨,⅄,\small{𝖸} \}$) is a proper superset of the basic logic induced by the branching recurrence $⫰$ (i.e., the one in the signature $\{ ¬,∧,∨, ⫰, ⫯ \}$). The latter is known to be precisely captured by the cirquent calculus system CL15, conjectured by Japaridze to remain sound—but not complete—with $⅄$ instead of $⫰$. The present result is obtained by positively verifying that conjecture. A secondary result of the paper is showing that $⅄$ is strictly weaker than $⫰$ in the sense that, while $⫰F$ logically implies $⅄F$, the reverse does not hold.
Wenyan Xu. Sanyang Liu. "The Parallel versus Branching Recurrences in Computability Logic." Notre Dame J. Formal Logic 54 (1) 61 - 78, 2013. https://doi.org/10.1215/00294527-1731389
Secondary: 03B70 , 68Q10 , 68T15 , 68T27
Keywords: cirquent calculus , Computability logic , Game semantics , Interactive computation , resource semantics
Wenyan Xu, Sanyang Liu "The Parallel versus Branching Recurrences in Computability Logic," Notre Dame Journal of Formal Logic, Notre Dame J. Formal Logic 54(1), 61-78, (2013) |
3D printed structured porous treatments for flow control with applications for noise and vibration control | JVE Journals
Pranjal Bathla1 , John Kennedy2
Copyright © 2020 Pranjal Bathla, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
The use of porous coatings is one of the passive flow control methods used to reduce turbulence and noise and vibrations generated due to fluid flow. A number of real-world applications rely on single or tandem cylinder systems like aircraft landing gears, chimneys, bridge pillars, etc. Using porous coatings for flow stabilization acts as a light-weight, cost-effective and customizable solution. The design and performance of a porous coating depend on multiple control parameters like lattice size, strut thickness, lattice structure, etc. This study aimed to investigate the suitability of MSLA 3D printers (Anycubic Photon and Prusa SL1) to manufacture porous coatings and to design and optimize porous lattices for flow control behind a cylinder. The Reynolds number used was 60000 and the flow measurements were taken using a hotwire probe. Different sets of experiments were conducted for single and tandem cylinder with varying control parameters to achieve the optimal lattice designs. The results show that the performance primarily depended on porosity, coating thickness, and coating shape. A high porosity thick coating gave a 70-75 % reduction in turbulence rate and also reduced the mean wake velocity. It was also seen that coating thickness is not linearly proportional to turbulence reduction and an optimal thickness exists for best performance.
Investigates manufacturability and control parameters of porous coatings
Higher porosity coating gives lower turbulence rate and mean wake velocity
Turbulence and mean velocity levels affected by lattice type due to difference in airflow resistivity
Turbulence rate affected by surface irregularity and amount of open area on the outermost coating layer
MSLA 3D printers suitable for self-supporting porous coatings
Keywords: porous coatings, lattices, single cylinder, tandem cylinder, 3D printing, turbulence, mean wake velocity, noise, design parameters, flow stabilization, vibrations.
Noise is primarily generated due to turbulence and unstable vortices. Acoustic power of sound is proportional to the 6th-8th power of velocity which leads to an intense increase in noise with a relatively small increase in flow velocity [1]. The vortices produced due to an airflow not only generate noise but can be catastrophic (especially in the aviation industry) and can cause major structural damage to the components in the wake. The strength of vortices depends on the flow momentum. Reducing mean velocity leads to a reduction in wake momentum, wake distance and helps in easy dissipation of the vortices. Therefore, reducing turbulence and mean wake velocity not only help in noise reduction but also help in reducing the damage to components in the wake region and prevent loss in mechanical integrity.
A porous coating over a cylinder deprives the wake region of momentum and leads to suppression of vortices. It also assists in a reduction in drag and pressure fluctuations which can help improve the overall efficiency of a system/component. A porous coating alters the flow field and widens the wake (Fig. 1). Energy dissipation takes place as the flow moves around and through the coating. This includes thermal loss due to skin friction between fluid particles and the surface, and viscous loss due to fluid viscosity and its flow within the coating [2].
The Reynolds number for this study i.e. 60000 falls in the subcritical range with a laminar boundary layer and turbulent vortex street. 3D characteristics of vortices can be neglected and the flow can be accurately represented in two dimensions [3].
Results from [1] suggest that the hardness of the porous material does not make a significant impact on noise reduction. Therefore, this study using 3D printed polymer resin porous coatings will be valid for applications where a metal powder bed fusion process will be used to create the porous material.
Fig. 1. Instantaneous wake vorticity of a) a rigid and b) porous coated cylinder [1]
A variety of software and hardware were used in this study. Porous coatings were designed and controlled using SolidWorks and nTopology Element software. For probe movement and data acquisition, a combination of Matlab, Arduino with Adafruit Motor Shield, and Dantec StreamWare with StreamLine Pro and a P13 hotwire probe was used.
StreamLine Pro is a constant temperature anemometer system for turbulence investigations. It consists of a Wheatstone bridge and works on the principle of cooling effect provided by fluid flow over a heated body. As the air flows over the sensor, the tungsten wire in the sensor tends to heat up and lead to an unbalance in the Wheatstone bridge. Using an advanced feedback loop control, the bridge enables the temperature in the sensor to remain constant. The bridge voltage represents the heat transfer and is in direct relation with the airflow velocity. The Matlab code in conjunction with StreamWare allowed to take mean measurements for turbulence rate and mean wake velocity at each data point. This was done by having number of scans at each point = 200,000 and scans/second or rate = 40,000 i.e. an acquisition time of 5 seconds at each point.
The design of a porous coating can be controlled via several parameters. These are: lattice type (cubic, diamond cubic, etc.), lattice size/dimensions, lattice/strut thickness, coating thickness (
h
D
ratio, where
h
is the coating thickness and
D
is the diameter of the core rigid cylinder), number of cells in the coating, and coating shape (e.g. helical, spaced discs, full, etc.). As porosity controls the airflow resistivity of the coating, it plays a critical role in regularizing flow around a cylinder, it was necessary to control the parameters which affect the porosity and to accurately calculate porosity for each designed coating.
Since there was no existing method to calculate porosity for a coating, a new method using the existing tools had to be designed. For a selected lattice type/structure, its volume occupied i.e. packing fraction or porosity varies according to lattice size and lattice thickness. However, on changing the lattice thickness in nTopology Element, the dimensions of the thickened cell exceed the original cell dimensions. The solution to this problem was to convert the thickened cell into a CAD model, use reference planes in SolidWorks to create planes that outline the original cell dimensions and use these planes to trim the lattice to its original dimensions (Fig. 2). Using the volume occupied by the trimmed cell, porosity was calculated as:
\mathrm{P}\mathrm{o}\mathrm{r}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{y} \varnothing \left(%\right)=100-\frac{Volum{e}_{occupied inside cell}}{Volum{e}_{total cell}}×100.
For the cubic cell in Fig. 2 of lattice size 1.5 mm×1.5 mm×1.5 mm and thickness = 0.5 mm, the porosity was found to be:
{\varnothing }_{cubic}=100-\frac{0.71}{1.5×1.5×1.5}×100=
Fig. 2. a) Untrimmed thickened cell and b) trimmed thickened cell
To successfully 3D print cylinders with porous coating, the slicing parameters were optimized according to the complexity of the porous lattice.
XY
resolution for printers used in this study was 47 μm [4, 5], and the layer height was varied between 22-29 μm. Sample 3D printed cylinders are shown in Fig. 3.
Fig. 3. a) Rigid cylinder and b) cylinder with porous coating
Coating design parameters were varied in 10 experiment sets of single and tandem cylinder to find their optimal values for best performance.
3.1. Set-1: lattice structure check
Cylinder-1 was a rigid cylinder with no porous coating. Cylinder 2,3 and 4 had a core rigid cylinder of 15 mm diameter and a porous coating of 3.75 mm. For each porous coating, the lattice size was 1.5 mm×1.5 mm×1.5 mm and the thickness was 0.5 mm. The only variable was the porosity of the lattice that occurred due to different structure types. The porosities of cubic, diamond cubic, and kelvin coatings were calculated as 78.96 %, 60.89 %, and 53.78 % respectively.
As per Fig. 4, Just behind the cylinder at
x/D=
1, cubic coating provided nearly a 30 % reduction in turbulence as compared to the turbulence of a rigid cylinder. This reduction is not as compelling at
x/D=
2 but is still better than a rigid cylinder. However, in the case of diamond cubic and kelvin cell coatings, there was no notable turbulence reduction at any position in the wake region. The reason for this was the porosity of different lattice structures. Since cubic lattice had higher porosity, it provided lower airflow resistivity i.e. lower hindrance and enabled airflow to lose momentum and energy when traveling through the channels of the porous medium. In the case of diamond cubic and kelvin cell coatings which had lower porosities, they acted as a rigid coating due to high airflow resistivity. This can be confirmed by their turbulence curves which were similar to the curve of a rigid cylinder. All cylinders with porous coating behaved similarly further away from the cylinder as the wake dispersed and the flow velocities reached closer to the free stream velocity. Based on these results, cubic lattice was chosen as the optimal lattice for further experiments.
Table 1. Cylinders for experiment set-1
Lattice structure check
Rigid cylinder (15mm diameter)
Diamond cubic coating
Kelvin cell coating
h
D
coating thickness, lattice size, lattice thickness
Fig. 4. Turbulence graph for experiment set-1
3.2. Set-2: porosity check
Based on results obtained from set-1, all cylinders (except cylinder-1) used for set-2 had a porous coating of the cubic lattice. All cylinders (except cylinder-1) had a constant lattice size of 2 mm×2 mm×2 mm and a coating thickness of 4 mm. By keeping the lattice size constant, it was possible to vary the porosity by varying lattice thickness or strut diameter. Strut diameter used for cylinder-2, 3, 4 and 5 was 0.3 mm, 0.45 mm, 0.55 mm, and 0.65 mm respectively.
Fig. 5 clearly shows a reduction in turbulence with an increase in porosity. Just behind the cylinder at
x/D=
1, the turbulence rate for cylinder-5 (80 % porosity) was at 15 %, whereas, for cylinder-2 (95 % porosity), it was at a considerably low value of 6 %-7 %. Similar observations can be made at
x/D=
2 where the turbulence rate for cylinder-2 is about 55 % lower than that of cylinder-5. Although the wake at
x/D=
2 is stronger than
x/D=
1, high porosity still delivers a reduction in turbulence rate in comparison to low porosity coatings. Interestingly, at
x/D=
3, the only cylinder which has noticeable reduction is cylinder-2 with 95 % porosity. Even with different porosities, the other 3 cylinders behave similarly at this position with comparable turbulence peaks. Once the wake disperses and the vortices lose their momentum i.e. at
x/D=
4, all 4 cylinders have similar turbulence rates despite their difference in porosities. Based on these results, cubic lattice with 95 % porosity was chosen as the optimal lattice for further experiments.
95 % porosity
Constant lattice size,
h/D
Variable porosity, lattice thickness
Fig. 6 compares the rigid cylinder with a porous-coated cylinder of 95 % porosity and a coating thickness of 9.75 mm. The contour shows the extent to which a porous coating can reduce turbulence in the wake of a cylinder. Near the cylinder, thick porous coating gave a 63-74 % reduction in turbulence. Unlike thin coatings that do not have a huge impact at far away in the wake, a thick coating gave about a 38 % reduction in peak turbulence rate. Due to a thick porous coating, air traveling through the channels of coating has more time and space to lose its kinetic energy. At the same time, the wake region widens and dissipates quickly due to lower momentum.
Fig. 6. Turbulence contour for a rigid cylinder and a cylinder with thick porous coating
This paper investigated the suitability of MSLA 3D printers to manufacture porous coatings and to design and optimize porous lattices for flow control behind a cylinder. 3D printing proved to be successful for porous coatings as the printers had sufficient resolution to print thin and complex lattices. However, depending on the complexity of porous coating, the slicing parameters needed optimization to ensure a good quality print. The performance of these printers degraded when the structure was not self-supporting. Therefore, a thick coating of cubic lattice was more prone to failure than a kelvin cell lattice coating. A high ratio of lattice size to lattice thickness led to print failures as well. Thus, lower porosity coatings were easier to print as they generally had thicker lattices. The quality of prints also depended on the maintenance of the 3D printers. A clean and well-calibrated system resulted in lower print failures.
It was found that different lattice structures affect turbulence rate based on their porosity or packing fraction. A cubic lattice coating had the best performance as it had the simplest structure with no part of the structure inside the cell. It also had the most open area for air to enter the porous medium through the outermost layer of coating. However, for a lattice with a low porosity structure, it acted as a rigid cylinder due to high airflow resistivity and did not provide turbulence reduction.
In the case of porous coatings with varying porosity, it was found that porosity and turbulence reduction have a proportional relationship. Therefore, a high porosity coating gave a maximum reduction in turbulence and mean wake velocity. A thick porous coating gave up to 74 % turbulence reduction close to the cylinder. The distance in wake till which a porous coating had its effect depending on its thickness. Wake velocity for a thin coating at
x/D=
4 reached free stream velocity whereas a thick coating still gave turbulence reduction at this position.
Takeshi Sueki, Takehisa Takaishi, Mitsuru Ikeda, Norio Arai Application of porous material to reduce aerodynamic sound from bluff bodies. Fluid Dynamics Research, Vol. 42, Issue 1, 2010, p. 015004. [Search CrossRef]
Cao Leitao, Fu Qiuxia, Si Yang, Ding Bin, Yu Jianyong Porous materials for sound absorption. Composites Communications, Vol. 10, 2018, p. 25-35. [Publisher]
Lienhard John H. Synopsis of Lift, Drag, and Vortex Frequency Data for Rigid Circular Cylinders. Technical Extension Service, Washington State University, Pullman, 1966. [Search CrossRef]
Anycubic Photon 3D Printer. Anycubic, https://www.anycubic.com/products/anycubic-photon-3d-printer. [Search CrossRef]
Original Prusa SL1 kit. Prusa3D, https://shop.prusa3d.com/en/3d-printers/719-original-prusa-sl1-kit.html. [Search CrossRef] |
You randomly survey students about participating in the science fair.
You randomly survey students about participating in the science fair. The two-way table shows the results. How many male students participate
You randomly survey students about participating in the science fair. The two-way table shows the results. How many male students participate in the science fair?
Looking at the row "Male" and column "Yes", we get the number of male-students participating in science fair as 32.
In a Two-way table, which type of distribution looks at percentages based on one variable only?
Complete the two-way table.
What do you notice about the value that always falls in the cell to the lower right of a two-way table when marginal relative frequencies have been written in? What does this value represent?
\begin{array}{|ccc|}\hline & Like\text{ }Aerobic& Exercise\\ Like\text{ }Weight\text{ }Lifting& Yes& No& Total\\ Yes& 7& 14& 21\\ No& 12& 7& 19\\ Total& 29& 21& 40\\ \hline\end{array}
Find the conditional relative frequency that a student likes to lift weights, given that the student likes aerobics.
How can you use two way-frequency tables to analyze data?
Students and teachers at a school were polled to see if they were in favor of extending the parking lot into part of the athletic fields. The results of the poll are shown in the two-way table. Which of the following statements is false? A. Thirty-nine students were polled in all. B. Fourteen teachers were polled in all. C. Twenty-three students are not in favor of extending the parking lot. D. Nine teachers are in favor of extending the parking lot. In FavorNot in Favor Students1623 Teachers914 |
Evaluate the integral. \int \sec^{2}x\tan x dx
\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx
I=\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx
for evaluating given integral, we substitute
\mathrm{tan}x=t
now, differentiating equation (1) with respect to x
\frac{d}{dx}\left(\mathrm{tan}x\right)=\frac{d}{dx}\left(t\right)\text{ }\text{ }\text{ }\left(\because \frac{d}{dx}\left(\mathrm{tan}x\right)={\mathrm{sec}}^{2}x\right)
{\mathrm{sec}}^{2}x=\frac{dt}{dx}
{\mathrm{sec}}^{2}xdx=dt
now, replacing
{\mathrm{sec}}^{2}xdx
with dt, tanx with t in given integral
\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx=\int tdt\text{ }\text{ }\text{ }\left(\because \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c\right)
=\left(\frac{{t}^{2}}{2}\right)+c
now, replacing t with
\mathrm{tan}x
\int {\mathrm{sec}}^{2}x\mathrm{tan}xdx=\frac{{\mathrm{tan}}^{2}x}{2}+c
hence, given integral is equal to
\frac{{\mathrm{tan}}^{2}x}{2}+c
Jack Maxson
\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx
u={\mathrm{sec}}^{2}\left(x\right)⇒\frac{du}{dx}=2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)⇒dx=\frac{1}{2{\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)}du:
=\frac{1}{2}\int 1du
\int 1du
Integral of a constant:
We substitute the already calculated integrals:
\frac{1}{2}\int 1du
=\frac{u}{2}
Reverse replacement
u={\mathrm{sec}}^{2}\left(x\right):
=\frac{{\mathrm{sec}}^{2}\left(x\right)}{2}
\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx
=\frac{{\mathrm{sec}}^{2}\left(x\right)}{2}+C
\int {\mathrm{sec}}^{2}\left(x\right)\mathrm{tan}\left(x\right)dx
Apply u-substitution:
u=\mathrm{tan}\left(x\right)
=\int udu
=\frac{{u}^{2}}{2}
Substitute back
u=\mathrm{tan}\left(x\right)
=\frac{{\mathrm{tan}}^{2}\left(x\right)}{2}
Add a constant to the solution
=\frac{{\mathrm{tan}}^{2}\left(x\right)}{2}+C
Find indefinite integral.
\int -3{e}^{2x}dx
What is the arc length of f(x)=2x-1 on
x\in \left[0,3\right]
Trigonometric integrals. Evaluate the following integrals.
\int {\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}xdx
\frac{dy}{dx}=\frac{6\mathrm{sin}x}{y}
What is a particular solution to the differential equation
\frac{dy}{dx}=\frac{x}{y}
with y(1)=2?
\int {e}^{x+2}dx
Use partial fractions to evaluate the definite integral.
{\int }_{0}^{1}\frac{{x}^{2}-x}{{x}^{2}+x+1}dx |
Monty Hall Problem | Brilliant Math & Science Wiki
Christopher Williams, Adam Strandberg, Ansh Bhatt, and
Pranav Rs
Aled Burt
daniel kanevsky
The Monty Hall problem is a famous, seemingly paradoxical problem in conditional probability and reasoning using Bayes' theorem. Information affects your decision that at first glance seems as though it shouldn't.
In the problem, you are on a game show, being asked to choose between three doors. Behind each door, there is either a car or a goat. You choose a door. The host, Monty Hall, picks one of the other doors, which he knows has a goat behind it, and opens it, showing you the goat. (You know, by the rules of the game, that Monty will always reveal a goat.) Monty then asks whether you would like to switch your choice of door to the other remaining door. Assuming you prefer having a car more than having a goat, do you choose to switch or not to switch?
The solution is that switching will let you win twice as often as sticking with the original choice, a result that seems counterintuitive to many. The Monty Hall problem famously embarrassed a large number of mathematicians with doctorate degrees when they attempted to "correct" Marilyn vos Savant's solution in a column in Parade Magazine. [1]
Using Bayes' Theorem
Variant: The Random Monty Hall Problem
One way to see the solution is to explicitly list out all the possible outcomes, and count how often you get the car if you stay versus switch. Without loss of generality, suppose your selection was door
1
. Then the possible outcomes can be seen in this table:
In two out of three cases, you win the car by changing your selection after one of the doors is revealed. This is because there is a greater probability that you choose a door with a goat behind it in the first go, and then Monty is guaranteed to reveal that one of the other doors has a goat behind it. Hence, by changing your option, you double your probability of winning.
Another way of seeing the same set of options is by drawing it out as a decision tree, as in the following image:
There are two aspects of the Monty Hall problem that many struggle to agree with. First, why aren’t the odds 50-50 after the host opens the door? Why is it that switching doors has a 2 in 3 chance of winning when sticking with the first pick only has a 1 in 3 chance? Secondly, why is it the case that if Monty opened a door truly randomly and happened to show a goat, then the odds of staying vs. switching doors are now 50-50? Bayes' theorem can answer these questions.
Bayes’ theorem is a formula that describes how to update the probability that a hypothesis is correct, given evidence. In this case, it’s the probability that our initially picked door is the one with the car behind it (that staying is right) given that Monty has opened a door showing a goat behind it: that Monty has shown us that one of the choices we didn’t pick was the wrong choice. Let
H
be the hypothesis "door 1 has a car behind it," and
E
be the evidence that Monty has revealed a door with a goat behind it. Then the problem can be restated as calculating
P(H \mid E)
, the conditional probability of
H
E
Since every door either has a car or a goat behind it, the hypothesis "
\text{not} H
" is the same as "door 1 has a goat behind it."
In this case, Bayes' theorem states that
P(H \mid E) = \frac{P(E \mid H)} {P(E)} P(H) = \frac{ P(E\mid H) \times P(H)} {P(E\mid H) \times P(H) + P(E\mid \text{not} H) \times P(\text{not} H)}.
The problem as stated says that Monty Hall deliberately shows you a door that has a goat behind it.
Breaking down each of the components of this equation, we have the following:
P(H)
is the prior probability that door 1 has a car behind it, without knowing about the door that Monty reveals. This is
\frac{1}{3}
P(\text{not} H)
is the probability that we did not pick the door with the car behind it. Since the door either has the car behind it or not,
P(\text{not} H) = 1 - P(H) = \frac{2}{3}
P(E \mid H)
is the probability that Monty shows a door with a goat behind it, given that there is a car behind door 1. Since Monty always shows a door with a goat, this is equal to
1
P(E\mid \text{not} H)
is the probability that Monty shows the goat, given that there is a goat behind door 1. Again, since Monty always shows a door with a goat, this is equal to
1
Combining all of this information gives
P(H \mid E) = \frac {1 \times \frac13} {1 \times \frac13 + 1\times \frac23} = \frac {\hspace{1mm} \frac13\hspace{1mm} }{1} = \frac{1}{3}.
The probability that the car is behind door 1 is completely unchanged by the evidence. However, since the car can only either be behind door 1 or behind the door Monty didn't reveal, the probability it is behind the door that is not revealed is
\frac{2}{3}
. Therefore, switching is twice as likely to get you the car as staying.
This result depends crucially on the fact that Monty was always guaranteed to open a door with a goat behind it, regardless of what door you picked initially. That is,
P(E \mid H) = P(E \mid \text{not}H)
. Now consider what would happen if Monty randomly opened a door we did not pick and it contained a goat. What is the probability that our first pick is correct, regardless of which specific door we picked?
P(E \mid H),
the probability that Monty will show us a door with a goat behind it given that our first pick was correct and does have a car behind it, is still
1
. If you picked the door with the car, the two other doors are guaranteed to have a goat.
P(E \mid \text{not} H)
changes. This represents the probability that Monty picks a door with a goat behind it given that your original choice of door had a goat behind it. In this case, Monty chooses randomly between one door that has a goat behind it and one door that doesn't have a goat behind it. Therefore, the probability that he picks a door with a goat is
P(E \mid \text{not} H) = \frac{1}{2}
Laying this out, we get
P(H \mid E) = \frac{ P(E\mid H) \times P(H)} {P(E \mid H) \times P(H) + P(E \mid \text{not} H) \times P(\text{not} H)} = \frac{1\times \frac13} {1\times \frac13 + \frac12 \times \frac23} = \frac{\hspace{1mm} \frac13\hspace{1mm} } {\hspace{1mm}\frac23\hspace{1mm}} = \frac12.
\frac12,
or 50%, chance of our first pick being correct when Monty randomly opens a door and it happens to have a goat behind it. The naive reasoning that fooled so many mathematicians works in this case.
Stick Switch Doesn't matter
Many people have probably heard of the Monty Hall problem. Here is a different version of it.
You are playing on a game show and have made it to the last round. You have a chance to win a car! The game show host shows you three doors. He claims that behind one of the doors is the car, and behind the other two are goats. The game show host asks you to pick a door and explains what will happen after you pick your door. Before your door is opened, the host will open one of the other two doors at random. If that door reveals the car, you automatically win it! If it turns out to reveal a goat, you will have the option to swap your chosen door and open the door you didn't choose in the beginning. If the door you open reveals the car, you win the car.
So here is what happens: You pick the door on the right. The host chooses to open the door in the middle. As it turns out, that door was holding a goat. Now you have the option to switch. Should you switch, stick, or does it not matter what you choose to do?
Which action would give you the highest chance of winning the car?
Calvin should switch, the probability of winning is
50.1\%
99.9\%
Calvin should not switch, the probability of winning is a little over
50\%
Calvin should not switch, the probability of winning is
75.0\%
The probability of winning is
50\%
Monty Hall decides to perform his final episode of the Let's Make a Deal series with a little twist, and Calvin is elated when he is the first contestant to be called to the stage. Everyone eagerly watches as Calvin approaches the stage, waiting to see the surprise Mr. Hall will present.
Mr. Hall then hollers into his microphone, "AND THE SURPRISE IS...... CALVIN WILL BE PLAYING WITH 1000 DOORS INSTEAD OF JUST 3." The audience gasps, but Mr. Hall keeps talking, explaining the rules:
999 of the 1000 doors covers a pig, while the last covers the most amazing super special magnificent supercar of your dreams.
Calvin will pick 1 of the 1000 doors, after which I will open 998 doors which cover a pig, leaving two doors for Calvin to choose from.
Calvin will then have to choose whether to open the door he originally chose or the door I left closed.
Should Calvin open the door which he originally chose, or switch to the door which Mr. Hall did not open?
Switch doors Stay at your door Both staying and switching are the same Indeterminate None of the above
In a game show, there are 3 doors. Two doors have nothing behind them, but one door has a brand new shiny red car. The game show host knows which door has the car. You pick a door, and before the host opens it, he opens a door that you did not pick, which has nothing behind it. Now you have the choice of switching to the closed door that hasn't been touched yet, or staying at the door you originally picked.
Which choice is better, if you want the brand new shiny red car?
This is not an original problem.
A host places 100 boxes in front of you, one of which contains the holy grail.
You pick box 3. The host removes all the boxes except 3 and 42, stating that the holy grail is in one of these boxes.
What is the probability that the holy grail is in box 42?
Suppose you're a contestant on a game show and the host shows you 5 curtains. The host informs you that behind 3 of the curtains lie lumps of coal, but behind the other 2 curtains lie separate halves of the same $10,000 bill. He then asks you to choose 2 curtains; if these 2 curtains are the ones with the 2 bill halves behind them then you win the $10,000, otherwise you go home with nothing.
After you choose your 2 curtains, the host opens one of the remaining 3 curtains that he knows has just a lump of coal behind it. He then gives you the following options:
(0) stick with the curtains you initially chose,
(1) swap either one of your curtains for one of the remaining curtains, or
(2) swap both of your curtains for the remaining 2 curtains.
p(k), k = 0, 1, 2
, be the respective probabilities of winning the money in scenarios
(k)
as outlined above. Then
p(2) - p(0) - p(1) = \dfrac{a}{b},
and
b
a + b
You find yourself on a game show and there are 10 doors. Behind nine of them are assorted, random farm animals including a variety of pigs, goats, and geese, but behind one is a brand-new, shiny, gold car made of real gold!
The game proceeds as follows:
The game show host then opens a door you didn't choose that he knows has only farm animals behind it.
You are then given the option to switch to a different, unopened door, after which he opens another door with farm animals behind it.
This process continues until there are only two doors left, one being your current choice. (Every time the host opens a door, you are given another option to switch or stick with the one you currently have selected.)
If you play optimally, your chances of winning the car are of the form
\frac ab
and
b
a+b
You know ahead of time that this process will continue until you are down to only two doors.
Assume that you prefer the car over any of the farm animals!
Photo credit: http://iloboyou.com/
More probability questions:
Check out my other problems.
You find yourself on a game show, and the host presents you with four doors. Behind three doors are an assortment of gummy bears, and the remaining door has a pure gold car behind it!
You pick a door, and then the host reveals a door behind which he knows is only a couple of red gummy bears. Then you are given the choice to stick with your first choice or switch to one of the other two unopened doors.
If the probability that you will win the car if you switch is
\frac ab
and
b
a+b?
Photo credit: http://www.huffingtonpost.com/
On a game show, there is a popular game that is always carried out in the same way. A contestant is given the choice of three doors: behind one door is a car and behind the other two doors are goats. The contestant picks a door, say #1, and the host, who knows what's behind each door, always opens another door, say #3, to reveal a goat. Then, the contestant is given the option to switch to the other unopened door, door #2 in this case.
A long-term fan of the game show has noticed a hint in the staging of the game by the game show host. Thus, this fan can correctly guess the door with the car behind 50% of the time before any door selection is made.
Now, this fan has been selected as a contestant for the game. Using the best possible strategy, what is the probability (as a percentage) that this fan will end up with the car prize?
Question: Can you devise your own variations of the Monty Hall problem?
[1] Crockett, Zachary. The Time Everyone 'Corrected' the World's Smartest Woman. Priceonomics. 19 Feb 2015. Retrieved from http://priceonomics.com/the-time-everyone-corrected-the-worlds-smartest/ on 29 Feb 2016.
Cite as: Monty Hall Problem. Brilliant.org. Retrieved from https://brilliant.org/wiki/monty-hall-problem/ |
Law of Mass Action and Equilibrium Constant Kc and Kp - Course Hero
General Chemistry/Equilibrium Concepts/Law of Mass Action and Equilibrium Constant Kc and Kp
The law of mass action states that Kc is defined as the concentrations of products divided by the concentrations of reactants raised to correct powers. Kp is defined as the partial pressures of products divided by the partial pressures of reactants, raised to respective powers. The chemical coefficients become exponents in the expressions for Kc and Kp.
In the late 19th century, Norwegian chemists Cato Guldberg and Peter Waage proposed the law of mass action, a chemical law that states that when a reaction reaches equilibrium, the product concentrations multiplied together divided by the reactant concentrations multiplied together, with each term raised to the power of its coefficient in the balanced equation, is a constant.
{a{\rm{A}}+b{\rm{B}}}\rightleftarrows{c{\rm{C}}+d{\rm{D}}}
in which a, b, c, and d are the coefficients for a balanced chemical equation. For a chemical system at equilibrium and at a certain temperature, the equilibrium constant (Kc) is the product concentrations multiplied together divided by the reactant concentrations multiplied together, with each term raised to the power of its coefficient in the balanced equation.
{K_{\rm{c}}} = \frac{{{{\left[ {\rm{C}} \right]}^c}{{\left[ {\rm{D}} \right]}^d}}}{{{{\left[ {\rm{A}} \right]}^a}{{\left[ {\rm{B}} \right]}^b}}}
In this formula, the square brackets represent molar concentrations of the compounds. The equilibrium constant, Kc, can be dimensionless or can have a unit, depending on whether the solution (mixture of aqueous or gaseous reactants and products) is highly concentrated, and the unit used for Kc depends on the degree of concentration. For calculations in practice, Kc is often treated as dimensionless. Note that the equilibrium constant, Kc, is calculated using concentrations. Solids and pure liquids do not have a concentration and are not used in calculating the equilibrium constant. Kc describes the forward reaction in which A and B are the reactants. In the reverse reaction, C and D are the reactants and A and B are the products. The equilibrium constant for the reverse reaction, Kc', is the reciprocal of Kc.
{K'_{\rm{c}}} = \frac{{{{\left[ {\rm{A}} \right]}^a}{{\left[ {\rm{B}} \right]}^b}}}{{{{\left[ {\rm{C}} \right]}^c}{{\left[ {\rm{D}} \right]}^d}}}=\frac{1}{K_{\rm{c}}}
When the chemical system is not in equilibrium, a useful expression is the reaction quotient (Q), a ratio that can be used to predict how a system not in equilibrium will change to reach equilibrium. The reaction quotient is calculated in the same way as the equilibrium constant, as the product concentrations multiplied together divided by the reactant concentrations multiplied together, with each term raised to the power of its coefficient in the balanced equation. The reaction quotient, however, uses concentrations for the system not at equilibrium. If Q is less than Kc, the reaction will continue in the forward direction. If Q is greater than Kc, the reaction will proceed in the reverse direction. Consider the following example. This equilibrium reaction forms the basis of the Haber-Bosch process, used for the synthesis of ammonia.
{{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)
The equilibrium constant for the reaction is calculated as
{K_{\rm{c}}} = \frac{{{{\left[ {{\rm{N}}{{\rm{H}}_3}} \right]}^2}}}{{\left[ {{{\rm{N}}_2}} \right]{{\left[ {{{\rm{H}}_2}} \right]}^3}}}
When both the reactants and the products of a reversible reaction are gases, there is another way of calculating an equilibrium constant. This formula uses partial pressures instead of concentrations. If a mixture of gases is present in a closed container, the partial pressure of a gas is the individual pressure exerted by it. If A and B are the only two gases in a system, the total pressure in the system is given by
P_{\rm{total}}=P_{\rm{A}}+P_{\rm{B}}
Similarly, consider a chemical reaction in which all of the reactants and products are gases.
a{\rm{A}} + b{\rm{B}} \rightleftarrows c{\rm{C}} + d{\rm{D}}
The total pressure of the system is
{P_{{\rm{total}}}} = {P_{\rm{A}}} + {P_{\rm{B}}} + {P_{\rm{C}}} + {P_{\rm{D}}}
At constant pressure, another equilibrium constant, Kp, can be defined based on the partial pressure of each gas.
K_{\rm{p}}=\frac{({P_{\rm{C}}}^c)({P_{\rm{D}}}^d)}{({P_{\rm{A}}}^a)({P_{\rm{B}}}^b)}
Note that square brackets are not present in this formula. This helps differentiate Kp from Kc. The temperature must be a constant, or the value of Kp will change. Also, this relation holds true only at equilibrium. Just as
K^{\prime}_{\rm{c}}=K^{-1}_{\rm{c}}
for the reverse reaction, the equilibrium constant based on pressure for the reverse reaction is the reciprocal of the equilibrium constant based on pressure for the forward reaction.
K^\prime_{\rm{p}}=\frac{{({{P_{\rm{A}}}^a})({{P_{\rm{B}}}^b} )}}{{({{P_{\rm{C}}}^c})({{P_{\rm{D}}}^d})}}=\frac{1}{K_{\rm{p}}}
The partial pressure of a gas in a container can be calculated by
\begin{aligned}P_{\rm{A}}&=(\text{mole fraction of }{\rm{A})\text{(total pressure)}}\\&=(\chi_{\rm{A}})(P_{\rm{total}})\end{aligned}
Using this equation, the partial pressure terms in the equilibrium constant can be replaced.
{K_{\rm{p}}} = \frac{{{{{\rm{(}}{\chi _{\rm{C}}}{P_{{\rm{total}}}}{\rm{)}}}^c}{{{\rm{(}}{\chi _{\rm{D}}}{P_{{\rm{total}}}}{\rm{)}}}^d}}}{{{{{\rm{(}}{\chi _{\rm{A}}}{P_{{\rm{total}}}}{\rm{)}}}^a}{{{\rm{(}}{\chi _{\rm{B}}}{P_{{\rm{total}}}}{\rm{)}}}^b}}}
In this version of the equation, the P terms may or may not cancel, depending on the coefficients of the balanced chemical equation. Once again consider the Haber-Bosch process for the synthesis of ammonia.
{{\rm{N}}_2}(g) + 3{{\rm{H}}_2}(g) \rightleftharpoons 2{\rm{N}}{{\rm{H}}_3}(g)
For this reaction, the equilibrium constant in terms of pressure is
{K_{\rm{p}}}=\frac{{{P_{\rm{NH_3}}}^2}}{{P_{\rm{N_2}}}\times{{P_{\rm{H_2}}}^3}}
<Reversible Reactions and Equilibrium>Properties of Chemical Systems and Equilibrium |
18\div\frac{3}{4}
18
to a fraction greater than one, then divide.
24
\frac{2}{3}(9-6)+40
Follow the Order of Operations. First simplify any terms in parentheses, then any exponents, then terms being multiplied or divided, and lastly, combine any terms that are being added or subtracted.
15\div5+\frac{1}{2}·\left(-\frac{4}{7}\right)
2\frac{5}{7}
-\frac{7}{10}+\left(-\frac{5}{12}\right)-\left(\frac{1}{4}\right)
-\frac{164}{120}
-1\frac{11}{30}
4.25-7.06
-2.81
\frac{7}{8}-\left(-\frac{1}{10}\right)+\left(-\frac{3}{5}\right) |
EUDML | Extended affine root systems of type . EuDML | Extended affine root systems of type .
Extended affine root systems of type
BC
Azam, Saeid; Khalili, Valiollah; Yousofzadeh, Malihe
Azam, Saeid, Khalili, Valiollah, and Yousofzadeh, Malihe. "Extended affine root systems of type .." Journal of Lie Theory 15.1 (2005): 145-181. <http://eudml.org/doc/228300>.
@article{Azam2005,
author = {Azam, Saeid, Khalili, Valiollah, Yousofzadeh, Malihe},
keywords = {extended affine root systems},
title = {Extended affine root systems of type .},
AU - Azam, Saeid
AU - Khalili, Valiollah
AU - Yousofzadeh, Malihe
TI - Extended affine root systems of type .
KW - extended affine root systems
extended affine root systems
Articles by Azam
Articles by Khalili
Articles by Yousofzadeh |
EUDML | Extreme points and convolution properties of some classes of multivalent functions. EuDML | Extreme points and convolution properties of some classes of multivalent functions.
Extreme points and convolution properties of some classes of multivalent functions.
Ahuja, O.P.
Ahuja, O.P.. "Extreme points and convolution properties of some classes of multivalent functions.." International Journal of Mathematics and Mathematical Sciences 19.2 (1996): 381-388. <http://eudml.org/doc/47602>.
@article{Ahuja1996,
author = {Ahuja, O.P.},
keywords = {-valent starlike functions of order ; Hadamard product; extreme points; closed convex hulls; -valent starlike functions of order },
title = {Extreme points and convolution properties of some classes of multivalent functions.},
AU - Ahuja, O.P.
TI - Extreme points and convolution properties of some classes of multivalent functions.
KW - -valent starlike functions of order ; Hadamard product; extreme points; closed convex hulls; -valent starlike functions of order
p
-valent starlike functions of order
\alpha
, Hadamard product, extreme points, closed convex hulls,
p
\alpha
Articles by Ahuja |
Home : Support : Online Help : Programming : Logic : Boolean : depends
depends(f, x)
expression, or list or set of expressions
name, or list or set of names
The function depends returns true if any of the expressions contained in f are mathematically dependent on any of the names contained in x.
The expression f is mathematically dependent on the name x if it contains at least one instance of x which is not simply a dummy variable. Examples of dummy variables are index variables in sums, products, integrals, and limits. Dummy variables also appear in integral transforms.
If f is a complicated expression, then depends may not be able to determine that it is independent of x. In such cases, f should simplified before depends is called.
There is a facility for the user to add additional functions to depends. If the function depends/myfunc is defined, then depends will use this function to check for dependence whenever it encounters myfunc. The function depends/myfunc should accept as its parameters the arguments that myfunc expects in the same order, followed by a final parameter which will be a set containing all the names that dependence is being checked against. Note that myfunc may be either a user-defined function or a standard library function.
Unrelated to the above, there is also a procedure parameter modifier, depends, that can be used in declaration to indicate that a parameter's type depends on the value of another parameter.
\mathrm{depends}\left(\mathrm{sin}\left(x\right)+\mathrm{cos}\left(z\right),{x,y}\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{sin}\left(x\right)+\mathrm{cos}\left(z\right),[x,z]\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{sin}\left(x\right)+\mathrm{cos}\left(z\right),[y,z]\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{int}\left(f\left(x\right),x=a..b\right),x\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{depends}\left(\mathrm{int}\left(f\left(x\right),x=a..b\right),a\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{limit}\left(\frac{1}{x},x=0\right),x\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{depends}\left(\mathrm{sum}\left(\frac{1}{x},x=1..n\right),x\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
\mathrm{depends}\left(\mathrm{sum}\left(\frac{1}{x},x=1..n\right),n\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{laplace}\left(f\left(t\right),t,s\right),t\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
In some cases f must be simplified before depends is called to get the correct answer.
\mathrm{depends}\left({\mathrm{sin}\left(x\right)}^{2}+{\mathrm{cos}\left(x\right)}^{2},x\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{depends}\left(\mathrm{simplify}\left({\mathrm{sin}\left(x\right)}^{2}+{\mathrm{cos}\left(x\right)}^{2},\mathrm{trig}\right),x\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}} |
In addition to quadratic and exponential models, another common type
In addition to quadratic and exponential models, another common type of model is called a power model. Power models are models in the form \hat{y}=a \cdot x^{p}.
In addition to quadratic and exponential models, another common type of model is called a power model. Power models are models in the form
\stackrel{^}{y}=a\cdot {x}^{p}.
Here are data on the eight planets of our solar system. Distance from the sun is measured in astronomical units (AU), the average distance Earth is from the sun.
grbavit
We will carry out calculations to build a schedule
The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.
(a) Find a function that models the population t years after 1990.
(b) Find the time required for the population to double.
(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare.
The exponential models describe the population of the indicated country, A, in millions, t years after 2010.Which country has the greatest growth rate? By what percentage is the population of that country increasing each year?
India,
A=1173.1{e}^{0.008t}
Iraq,
A=31.5{e}^{0.019t}
Japan,
A=127.3{e}^{0.006t}
Russia,
A=141.9{e}^{0.005t}
The original number of bacteria found present in a body was 160. Now, after a period of 9 hours, the body has a count of 920 bacteria. Write the exponential equation that models the growth of the bacteria.
By 2009, the models indicate that Canada’s population will exceed Uganda’s by approximately 2.8 million.The exponential growth models describe the population of the indicated country, A, in millions, t years after 2006. Determine whether the statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement.
A=33.1{e}^{0.009t}
Uganda:
A=28.2{e}^{0.034t}
I need help with exponential functions. I know that the derivative of
{e}^{x}
{e}^{x}
, but wolfram alpha shows a different answer to my function below. If you, for example, take the derivative of
{e}^{-2x}
do you get
-2{e}^{-2x}
{e}^{-2x}
Use the factors to identify the zeros of
f\left(x\right)=3{x}^{3}+12{x}^{2}-36x
. Then sketch the graph of the polynomial.
Show that an exponential model fits the data. Then write a recursive rule that models the data.
\begin{array}{|ccccccc|}\hline n& 0& 1& 2& 3& 4& 5\\ f\left(n\right)& 162& 54& 18& 6& 2& \frac{2}{3}\\ \hline\end{array} |
Simplify the expression 12(x^{−6})y^{10⋅3}(x^{7})y A) 15xy^{11} B) 36xy^{10} C) 36x−42y^{10} D) 36xy^{11}
12\left({x}^{-6}\right){y}^{10\cdot 3}\left({x}^{7}\right)y
A\right)15x{y}^{11}
B\right)36x{y}^{10}
C\right)36x-42{y}^{10}
D\right)36x{y}^{11}
12\left({x}^{-6}\right){y}^{10×3}\left({x}^{7}\right)y=\left(12×3\right)\left({x}^{-6}×{x}^{7}\right)\left({y}^{10}×y\right)=36\left({x}^{-6}+7\right){y}^{10+1}=36x{y}^{11}
The correct answer is then
36x{y}^{11}
\left(3\frac{z}{8}\right)-\left(\frac{z}{10}\right)=6
\left(792\right)\left(5.19×{10}^{-5}\right)
A number divided by -8 is 7.
For each systems of equations below choose the best method for solving and solve
loop y=-6
Celine, Devon, and another friend want to purchase some snacks that cost a total of $7.50. They will share the cost of the snacks. Which of these statements is true?
A. An equation that can be used to find x, the amount of money each person will pay is
x+3=7.5
The solution to the equation is 4.5, so each person will pay $4.50.
B. An equation that can be used to find x, the amount of money each person will pay is
x+3=7.5
. The solution to the equation is 10.5, so each person will pay $10.50.
C. An equation that can be used to find x, the amount of money each person will pay is
x\cdot 3=7.5
. The solution to the equation is 2.5, so each person will pay $2.50.
D. B. An equation that can be used to find x, the amount of money each person will pay is
x\cdot 3=7.5
From scnool, you biked 1.2 miles due south and then 0.5 mile due east to your house. If you had biked home on the street that runs directly diagonal from your school to your house, how many fewer miles would you have biked?
\left\{\begin{array}{ccccc}3x& +& 2y& =& -7\\ 5x& -& 2y& =& -1\end{array} |
EUDML | The critical determinant of the double paraboloid and Diophantine approximation in and . EuDML | The critical determinant of the double paraboloid and Diophantine approximation in and .
The critical determinant of the double paraboloid and Diophantine approximation in
{ℝ}^{3}
{ℝ}^{4}
Nowak, Werner Georg. "The critical determinant of the double paraboloid and Diophantine approximation in and .." Mathematica Pannonica 10.1 (1999): 111-122. <http://eudml.org/doc/225101>.
author = {Nowak, Werner Georg},
keywords = {critical determinant; convex bodies; paraboloid; simultaneous approximation; lattice constants; diophantine approximation constant},
title = {The critical determinant of the double paraboloid and Diophantine approximation in and .},
TI - The critical determinant of the double paraboloid and Diophantine approximation in and .
KW - critical determinant; convex bodies; paraboloid; simultaneous approximation; lattice constants; diophantine approximation constant
Werner Georg Nowak, Lower bounds for simultaneous Diophantine approximation constants
Werner Georg Nowak, On the critical determinants of certain star bodies
Werner Georg Nowak, Estimating the critical determinants of a class of three-dimensional star bodies
critical determinant, convex bodies, paraboloid, simultaneous approximation, lattice constants, diophantine approximation constant
Lattices and convex bodies in
Articles by Nowak |
IsElement - Maple Help
Home : Support : Online Help : Connectivity : Web Features : XMLTools : IsElement
determine if an expression is an XML element
IsElement(expr)
The IsElement(expr) command tests whether a Maple expression expr is an XML tree data structure that represents an XML element. This is an XML tree that is not a CDATA structure, comment or processing instruction. If expr is an XML element, the value true is returned. Otherwise, false is returned.
\mathrm{with}\left(\mathrm{XMLTools}\right):
\mathrm{IsElement}\left(\mathrm{XMLElement}\left("a",["b"="c"],"d","e"\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}} |
Electric current questions, solved and explained
Electric current problems with answers
Recent questions in Electric current
{e}^{-x}
What is the statement of Ohm's Law?
A Binomial Coefficient Sum:
\sum _{m=0}^{n}\left(-1{\right)}^{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\genfrac{}{}{0}{}{m-1}{l}\right)
What would happen with the drift velocity of a cylindrical resistor's diameter increases, with a given voltage between its terminals? According to the expression:
\begin{array}{rl}R& =\rho \frac{L}{A}\\ I& =neA{v}_{d}\\ \mathrm{\Delta }V& =IR\\ {v}_{d}& =\frac{\mathrm{\Delta }V}{\rho Lne}\end{array}
The resistivity does not change, neither does the length of the resistor nor the term ne but the resistance does change as well as the current, so the area is eliminated from the expression. I wonder if the drift velocity would be the same after increasing the diameter or if my derivation is wrong.
Nubydayclellaumvcd 2022-05-14 Answered
Suppose we have a projectile of mass m that impacts a material with velocity
{v}_{0}
. As it travels through the material it is subject to the resistive force
F=\alpha {v}^{2}+\beta
. How can I determine
v\left(t\right)
, the velocity of the projectile at time t where t=0 is the impact time, and
v\left(p\right)
, the velocity of the projectile when it reaches a penetration depth of p?
This question addresses v(t) for an exclusively quadratic resistive force (
\beta =0
), but is there a more general formula for when there is a constant component?
Resistivity in terms of temperature coefficient of resistance
R={R}_{0}\ast \left(1+\alpha \left(T-{T}_{0}\right)\right)
{R}_{0}
{T}_{0}
are the reference resistance and temperature.
Is it then safe to say that:
\rho ={\rho }_{0}\ast \left(1+\alpha \left(T-{T}_{0}\right)\right)
{\rho }_{0}
is the reference resistivity)?
How to prove
{\int }_{0}^{1}\frac{1+{x}^{30}}{1+{x}^{60}}dx=1+\frac{c}{31}
, where 0
What is the minimal polynomial of
\alpha =\frac{{3}^{1/2}}{1+{2}^{1/3}}
\mathbb{Q}
Is the resistivity of all (conducting) alloys more than that of all (conducting) metals?
The wires in (Figure 1) all have the same resistance; the length and radius of each wire are noted.Rank in order, from largest to smallest, the resistivities
{\rho }_{1}
{\rho }_{5}
of these wires.
Logan Lamb 2022-05-13 Answered
I know what resistivity of materials are like, and is given by the formula:
R=\frac{\rho L}{A}
However what is the volume resistivity of materials like copper and aluminium for instance? How do we calculate the volume resistivity of materials, because this was confusing when I read the site:
Method of calculating eddy currents of a conductor, is this correct?
Where there is the answer given by Floris states the equation:
F=\frac{v\cdot {B}^{2}\cdot A}{\rho /t}
\rho
is given to be the volume resisitivity.
how do we measure the volume resistivity of a material for example aluminium?
what happens if the thickness of the material changes?
What is the electric current produced when a voltage of 9V is applied to a circuit with a resistance of
99\mathrm{\Omega }
Under electrostatic conditions, there is no electric field inside a solid conductor. True or false?
What is the electric current produced when a voltage of 24V is applied to a circuit with a resistance of
6\mathrm{\Omega }
llunallenaipg5r 2022-05-10 Answered
In a circuit, the supplied voltage is 1.5 volts, and the resistor value is 3 ohms. What is the current in the circuit?
If a current of 8A passing through a circuit generates 14W of power, what is the resistance of the circuit?
Coaxial cable has radius a of copper core and radius b of copper shield. Between there is an insulator with specific resistivity
\zeta
. What is the resistance of this cable with length L between the core and the shield?
First, I tried to solve this like this:
dR=\zeta \frac{l}{S}
In our case the length is dr, and therefore I suppose that the area of this ring is 2πrdr:
dR=\zeta \frac{dr}{2\pi rdr}
The solution sheet says:
dR=\zeta \frac{dr}{2\pi rL}
I know that something is wrong with my equation, because dr goes away and then I cannot integrate from a to b. But why is in the solution L instead of dr? As I understand problem instruction L= b-a. And therefore L=dr which doesn't make sense to me.
E=\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}\frac{N}{C}
What is the direction of the electric field at the position indicated by the dot in the figure? Specify the direction as an angle above the horizontal line.
\theta ={\overline{\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}}}^{\circ }
When a 120 V potential difference is applied across a
30\mathrm{\Omega }
resistance of electric device. What is the current through the electric device? What is the power of the electric device? |
Formal Charges and Resonance - Course Hero
General Chemistry/Chemical Bonding and Molecular Geometry/Formal Charges and Resonance
Molecules can have more than one valid Lewis structure. Formal charge can be used to determine which form is found in nature. Resonance occurs when two or more Lewis structures have equivalent formal charges.
A molecular structure is the three-dimensional shape of a molecule that takes into account bonding and nonbonding electron pairs and molecular rotations to minimize their interactions. Molecular structure plays an important role in determining the properties of a substance. Lewis structures can be used to predict the molecular structure of a compound. It is possible to draw two different Lewis structures for carbon dioxide (CO2). The first structure has two double bonds. The second structure has a single bond and a triple bond. Both of these structures satisfy the octet rule. One of these structures is a better representation of the molecule, as found in nature.
Two Possible Structures for Carbon Dioxide
Carbon dioxide has two possible resonance structures, each satisfying the octet rule that atoms tend to bond in a way that provides each a valence shell with eight electrons.
A formal charge is a hypothetical charge assigned to an atom in a molecule with the assumption that bonding electrons are shared equally. Formal charges can be used to determine which of two or more possible Lewis representations is better—that is, more likely to exist most of the time. To calculate formal charge, follow these steps:
1. Distribute all shared electrons to the atoms equally between bonded atoms.
2. All nonbonding electrons stay with the atom they are attached to.
3. Count how many electrons are on each atom.
4. Compare with the normal number of valence electrons on the neutral atom.
Formal Charge Calculation for CO2 Structure with a Triple Bond (
{\rm{C{\equiv}O}}
First Oxygen Atom
Second Oxygen Atom
Count the electrons. 7 4 5
Write the number of electrons in an uncharged atom. 6 4 6
Subtract to get the formal charge. 1– 0 1+
If the electrons are arranged in a carbon dioxide molecule so that there is one triple bond, the total charge is zero, but one oxygen atom has a 1– charge and another has a 1+ charge.
Formal Charge Calculation for CO2 Structure with Only Double Bonds (
{\rm{C{=}O}}
Subtract to get the formal charge. 0 0 0
If the electrons are arranged in a carbon dioxide molecule so that there are only double bonds, the total charge is zero, and the formal charge on each atom is zero.
Generally, the structure with the lowest formal charge is a better model. In the two calculations of carbon dioxide, the structure with a triple bond has 1–, 0, and 1+ formal charges, and the total charge is zero. The sum of the formal charges must equal the overall charge of the molecule or ion. In the carbon dioxide structure with only double bonds, each oxygen atom and the carbon atom all have a formal charge of zero, and the total is also zero. The formal charges of all atoms in the double-bond representation is zero. Formal charges predict this is the form of carbon dioxide molecule found in nature. Experimentally, this is proven correct. Carbon dioxide molecules have two double bonds, not a single bond and a triple bond.
In some molecules different Lewis representations have the same formal charge. Consider the ozone (O3) molecule. According to the octet rule, ozone has a double bond and a single bond. Ozone has two possible Lewis structures, one with the double bond on the right and the other with the double bond on the left.
Possible Lewis Structures for Ozone
The Lewis structure of an ozone (O3) molecule can be drawn with a double bond on the right or a double bond on the left.
Formal Charge Calculation for Ozone with Double Bond on the Right (
\rm{O{-}O{=}O}
Left Oxygen Atom
Center Oxygen Atom
Right Oxygen Atom
Subtract to get the formal charge. 1– 1+ 0
If the electrons are arranged in an ozone molecule so that the double bond is on the right, the formal charge is 1– for the left oxygen atom and 1+ for the center oxygen atom. The total charge is zero.
Formal Charge Calculation for Ozone with Double Bond on the Left (
\rm{O{=}O{-}O}
If the electrons are arranged in an ozone molecule so that the double bond is on the left, the formal charge is 1– for the right oxygen atom and 1+ for the center oxygen atom. The total charge is zero.
The formal charges for the two possible structures of ozone are 1–, 1+, 0 and 0, 1+, 1–. There is no difference between these, and both forms are equally valid. One of two or more Lewis structures with multiple equivalent representations is called a resonance structure. When resonance structures exist, they are used at the same time to represent the molecule. They are related by rotation and thus are the same. A double-sided arrow is used between the structures to show they are resonance structures.
Ozone has two resonance structures.
Resonance does not mean the molecule switches back and forth between two forms. In fact, ozone has two equal bonds with the same length. The electrons are shared evenly between the bonds. It does not have a single bond and a double bond, as the models might suggest. Resonance structures are a refinement of two energetically identical Lewis structures. However, they are still not completely accurate representations.
<Lewis Structures>Ion-Ion, Dipole-Dipole, and Ion-Dipole Interactions and Hydrogen Bonding |
EUDML | An Addition Theorem in a Finite Abelian Group. EuDML | An Addition Theorem in a Finite Abelian Group.
An Addition Theorem in a Finite Abelian Group.
Jorgen Cherly
Cherly, Jorgen. "An Addition Theorem in a Finite Abelian Group.." Mathematica Scandinavica 74.1 (1994): 5-8. <http://eudml.org/doc/167275>.
@article{Cherly1994,
author = {Cherly, Jorgen},
keywords = {finite Abelian group; additive bases; basis of order },
title = {An Addition Theorem in a Finite Abelian Group.},
AU - Cherly, Jorgen
TI - An Addition Theorem in a Finite Abelian Group.
KW - finite Abelian group; additive bases; basis of order
finite Abelian group, additive bases, basis of order
h
Additive bases, including sumsets
Articles by Jorgen Cherly |
The description that follows was adopted from Lizal et al. (2012). The model basically comes from two distinct realistic geometries that were carefully connected in the trachea. These are described separately in the following sections. The geometry segments are provided in stl format . (*Broken*) The numbering of the stl segments is the one shown in figure 4. However, the 10 larger outlet segments along with their attached outlet patches are separated from the airway branches (segments 13 – 22) and are designated as segments 23 – 32 (segment 13 is attached to outlet segment 23, segment 14 is attached to outlet segment 24 and so on).
{\displaystyle {20\mu m}}
{\displaystyle {25\mu m}}
{\displaystyle {7.6\mu m}}
{\displaystyle {\,^{85}Kr}}
{\displaystyle {\rho =1.2kg/m^{3}}}
{\displaystyle {\nu =1.7\times 10^{-5}m^{2}/s}} |
Introducing the Boundary-Aware Loss for Deep Image Segmentation - LRDE
Introducing the Boundary-Aware Loss for Deep Image Segmentation
Minh Ôn Vũ Ngoc, Yizi Chen, Nicolas C Boutry, Joseph Chazalon, Edwin Carlinet, Jonathan Fabrizio, Clément Mallet, Thierry Géraud
Proceedings of the 32nd British Machine Vision Conference (BMVC)
Most contemporary supervised image segmentation methods do not preserve the initial topology of the given input (like the closeness of the contours). One can generally remark that edge points have been inserted or removed when the binary prediction and the ground truth are compared. This can be critical when accurate localization of multiple interconnected objects is required. In this paper, we present a new loss function, calledBoundary-Aware loss (BALoss), based on the Minimum Barrier Distance (MBD) cut algorithm. It is able to locate what we call the leakage pixels and to encode the boundary information coming from the given ground truth. Thanks to this adapted loss, we are able to significantly refine the quality of the predicted boundaries during the learning procedure. Furthermore, our loss function is differentiable and can be applied to any kind of neural network used in image processing. We apply this loss function on the standard U-Net and DC U-Net on Electron Microscopy datasets. They are well-known to be challenging due to their high noise level and to the close or even connected objects covering the image space. Our segmentation performance, in terms of Variation of Information (VOI) and Adapted Rank Index (ARI), are very promising and lead to
{\displaystyle \approx {}15\%}
better scores of VOI and
{\displaystyle \approx {}5\%}
better scores of ARI than the state-of-the-art. The code of boundary-awareness loss is freely available at https://github.com/onvungocminh/MBD_BAL
@InProceedings{ movn.21.bmvc,
author = {Minh \^On V\~{u} Ng\d{o}c and Yizi Chen and Nicolas C.
Boutry and Joseph Chazalon and Edwin Carlinet and Jonathan
Fabrizio and Cl\'ement Mallet and Thierry G\'eraud},
title = {Introducing the Boundary-Aware Loss for Deep Image
booktitle = {Proceedings of the 32nd British Machine Vision Conference
(BMVC)},
abstract = {Most contemporary supervised image segmentation methods do
not preserve the initial topology of the given input (like
the closeness of the contours). One can generally remark
that edge points have been inserted or removed when the
binary prediction and the ground truth are compared. This
can be critical when accurate localization of multiple
interconnected objects is required. In this paper, we
present a new loss function, called, Boundary-Aware loss
(BALoss), based on the Minimum Barrier Distance (MBD) cut
algorithm. It is able to locate what we call the {\it
leakage pixels} and to encode the boundary information
coming from the given ground truth. Thanks to this adapted
loss, we are able to significantly refine the quality of
the predicted boundaries during the learning procedure.
Furthermore, our loss function is differentiable and can be
applied to any kind of neural network used in image
processing. We apply this loss function on the standard
U-Net and DC U-Net on Electron Microscopy datasets. They
are well-known to be challenging due to their high noise
level and to the close or even connected objects covering
the image space. Our segmentation performance, in terms of
Variation of Information (VOI) and Adapted Rank Index
(ARI), are very promising and lead to $\approx{}15\%$
better scores of VOI and $\approx{}5\%$ better scores of
ARI than the state-of-the-art. The code of
boundary-awareness loss is freely available at
\url{https://github.com/onvungocminh/MBD_BAL}},
note = {https://www.bmvc2021-virtualconference.com/assets/papers/1546.pdf}
Retrieved from "https://www.lrde.epita.fr/index.php?title=Publications/movn.21.bmvc&oldid=127091" |
Inverse system - formulasearchengine
In mathematics, an inverse system in a category C is a functor from a small cofiltered category I to C. An inverse system is sometimes called a pro-object in C. The dual concept is a direct system.
1 The category of inverse systems
2 Direct systems/Ind-objects
The category of inverse systems
Pro-objects in C form a category pro-C. The general definition was given by Alexander Grothendieck in 1959, in TDTE.[1]
Two inverse systems
F:I
{\displaystyle \to }
G:J
{\displaystyle \to }
C determine a functor
Iop x J
{\displaystyle \to }
Sets,
namely the functor
{\displaystyle \mathrm {Hom} _{C}(F(i),G(j))}
The set of homomorphisms between F and G in pro-C is defined to be the colimit of this functor in the first variable, followed by the limit in the second variable.
If C has all inverse limits, then the limit defines a functor pro-C
{\displaystyle \to }
C. In practice, e.g. if C is a category of algebraic or topological objects, this functor is not an equivalence of categories.
Direct systems/Ind-objects
An ind-object in C is a pro-object in Cop. The category of ind-objects is written ind-C.
If C is the category of finite groups, then pro-C is equivalent to the category of profinite groups and continuous homomorphisms between them.
If C is the category of finitely generated groups, then ind-C is equivalent to the category of all groups.
Retrieved from "https://en.formulasearchengine.com/index.php?title=Inverse_system&oldid=234435" |
EUDML | Two special convolution products of -th derivatives of Dirac delta in hypercone. EuDML | Two special convolution products of -th derivatives of Dirac delta in hypercone.
Two special convolution products of
\left(n/2-k-1\right)
-th derivatives of Dirac delta in hypercone.
Aguirre Téllez, Manuel A.. "Two special convolution products of -th derivatives of Dirac delta in hypercone.." Applied Mathematics E-Notes [electronic only] 1 (2001): 34-39. <http://eudml.org/doc/49348>.
@article{AguirreTéllez2001,
author = {Aguirre Téllez, Manuel A.},
keywords = {delta-function; -function; convolution products; -function},
title = {Two special convolution products of -th derivatives of Dirac delta in hypercone.},
AU - Aguirre Téllez, Manuel A.
TI - Two special convolution products of -th derivatives of Dirac delta in hypercone.
KW - delta-function; -function; convolution products; -function
delta-function,
\text{Γ}
-function, convolution products,
{\Gamma }
Integral transforms in distribution spaces
Articles by Aguirre Téllez |
User:Mattes - Wikimedia Commons
User:Mattes
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Welcome to my user page at Wikimedia Commons!
I'm a 50 y/o German guy.
Originally, I'm from de:Wikipedia, where I'm active since 2004 as Benutzer:Matt1971 (changed account to Mattes and later on to Mateus2019). Riding motorcycles is the third oldest hobby I have, the second oldest one is travelling and the oldest one is visual arts. Other hobbies include classical music, singing, dancing, hiking, sailing, simple photography, documentation and analysing "things". Historic hobbies include skiing, playing recorders, piano and electric organ. I also enjoy to discover new places, museums and being in the nature.
I have performed roundabout 350,000 edits.
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Mattes (talk • contribs • blocks • protections • deletions • moves • rights • rights changes)
Mateus2019 (talk • contribs • blocks • protections • deletions • moves • rights • rights changes).
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{\displaystyle \mathbb {ABOUT} }
{\displaystyle \mathbb {IMAGES} }
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Perimeter | Brilliant Math & Science Wiki
Mei Li, Ariel Gershon, Ashish Menon, and
Jonathan Xian Kuan Chee
The perimeter of a two-dimensional figure is the length of the boundary of the figure. If the figure is a polygon such as a triangle, square, rectangle, pentagon, etc., then the perimeter is the sum of the edge lengths of the polygon. For example, a square with side length 3 has perimeter
3+3+3+3 = 12.
Any closed figure has a perimeter, such as the irregular polygon to the right with perimeter
1+1+2+ 3= 7.
Some shapes do not have a finite number of sides, so calculating the perimeter can be less straighforward. For example, the perimeter (or circumference) of a circle is
2\pi r,
r
is the radius of the circle. Fascinatingly, some shapes (such as certain fractals) have infinite perimeter, despite having a finite area.
Perimeter of Other Shapes (Circles, Fractals, etc.)
Perimeter using Polar Coordinates
Since a square has four sides of equal length and the perimeter of a square is the sum of its side lengths, the perimeter of the square is
6 + 6 + 6 + 6 = 24. \ _\square
6 \times 4
on the count that it is a square.
What is the perimeter of a regular pentagon whose side lengths are all 6?
Since a regular pentagon has five sides and all its side lengths are
6
, the perimeter of the pentagon is
6 + 6 + 6 + 6 + 6= 5 \cdot 6 = 30 . \ _\square
Find the minimum perimeter of a triangle whose area is
\frac{\sqrt{3}}{4}
4
7?
4 + 4 + 7 + 7 = 2(4) + 2(7) = 8 + 14 = 22 . \ _\square
Find the perimeter of a triangle
ABC
AB= 3\text{ cm}, BC=4\text{ cm},
\angle B= 90^{\circ}.
First, use the Pythagorean theorem to find the length of
AC:
AC^{2}=AB^{2}+BC^{2}=3^{2}+4^{2} \implies AC=5.
Then add
AB,BC,CA
to get the perimeter, which is
3+4+5 =12\text{ (cm)}.\ _\square
Suppose an ant walks along the boundary of a triangle with side lengths
4, 6,
7
. If the ant starts at one vertex of the triangle and makes
3
complete trips around the triangle boundary to return to its starting point, what is the total distance traveled by the ant?
Since the perimeter is the sum of edge lengths, the perimeter of the triangle is
4 + 6 + 7 = 17
. Since the ant makes
3
complete trips around the triangle boundary, the total distance traveled by the ant is
17 \times 3 = 51. \ _\square
\begin{aligned} \text{Perimeter} & = \text{Sum of side lengths}\\ & = 9 + 8 + 7 + 3 + 4 + 5\\ & = 36.\ _\square \end{aligned}
Let there be a triangle drawn on a Cartesian plane with vertices at coordinates
(-3,0), (1,-3), (4,1).
Find the perimeter of this triangle to the nearest integer.
The perimeter of a circle with radius
r
2\pi r
. The perimeter of an arc of a circle subtending angle
\theta
(in degrees) at the center is
2\pi r \times \frac {\theta}{360}
What is the perimeter (circumference) of a circle of diameter
14\text{ cm}?
\begin{aligned} \text{Radius} & = \dfrac{\text{Diameter}}{2}\\ & = \dfrac{14}{2}\\ & = 7\text{ (cm)}\\ \\ \text{Circumference} & = 2× \pi × r\\ & = 2 × \dfrac{22}{7} × 7\\ & = 44\text{ (cm)}.\ _\square \end{aligned}
Find the perimeter of a semicircle of radius
21
\begin{aligned} \text{Perimeter of a semicircle} & = \left( \dfrac{1}{2} × 2 × \pi × r \right) + 2r\\ & = \left( \dfrac{1}{2} × 2× \dfrac{22}{7} × 21\right) + 2×21\\ & = 66 + 42\\ & = 106\text{ (cm)}.\ _\square \end{aligned}
In the figure to the right, a circle is circumscribed around a regular hexagon, and the same circle is inscribed within another regular hexagon.
P_1
be the perimeter of the larger hexagon, let
P_2
be the perimeter of the smaller hexagon, and let
C
be the circumference of the circle.
The circumference of the circle can be approximated by finding the mean of the two perimeters:
C\approx \frac{P_1+P_2}{2}.
\pi
is approximated using the approximation for circumference above, then
\pi\approx\frac{a}{b}+\sqrt{c}
a, b, c
and
b
are coprime, and
c
a+b+c
We can calculate the perimeter of an enclosed figure if its graph on a Cartesian plane is a function in polar coordinates. Specifically, suppose the figure has the equation
r = f(\theta)
f
is a continuous, nonnegative function for
\theta \in [0, 2\pi]
f(0) = f(2\pi)
. Then its perimeter is given by the following formula:
P = \int_{0}^{2\pi} \sqrt{r^2 + \left(\dfrac{dr}{d\theta}\right)^2} d\theta.
Suppose we want to calculate the circumference of a circle with radius
r
r
is a constant, and its derivative is
0
. So the formula gives us
P = \int_{0}^{2\pi} \sqrt{r^2 + 0^2} d\theta = \int_{0}^{2\pi} r d\theta = r(2\pi-0) = 2\pi r.
Therefore the circumference is
2\pi r.
Unfortunately, for other shapes, this formula usually leads to complicated integrals which often do not have a simple closed form.
A red rectangle is reshaped into a blue square, as shown below, such that both quadrilaterals have the same perimeter. The area of the square is
4\text{ cm}^2
larger than that of the rectangle.
If all the sides of both quadrilaterals have prime number lengths
(
\text{cm}),
then what is the perimeter of the square?
3
A+B \sqrt{2}+C \sqrt{3},
A+B+C
Find the perimeter of this figure in
\si{\cm}.
In this section, we will learn about applications of the concept of perimeter in real life. They are mostly of fencing-type problems. Here is an example:
Mr. Antony has a rectangular garden of area
630\text{ m}^2
42\text{ m}.
To save his garden from domestic animals grazing them, he wants to fence his whole garden. Find much Mr. Antony has to spend for fencing his field at the rate of $3 per meter?
Given, the area of the rectangle
A = 630\text{ m}^2
and its length
l = 42\text{ m},
A = l \times b \implies b = \dfrac{A}{l} = \dfrac{630}{42} = 15\ (\text{m}).
Now, let's find perimeter:
P = 2(l + b) = 2(42 + 15) = 114\ (\text{m}).
Since the cost of fencing per meter is $3, the total cost of fencing the whole feild is
3 \times P = 3 \times 114 = 342
_\square
Cite as: Perimeter. Brilliant.org. Retrieved from https://brilliant.org/wiki/perimeter/ |
Permanence and Global Attractivity of the Discrete Predator-Prey System with Hassell-Varley-Holling III Type Functional Response
Runxin Wu, Lin Li, "Permanence and Global Attractivity of the Discrete Predator-Prey System with Hassell-Varley-Holling III Type Functional Response", Discrete Dynamics in Nature and Society, vol. 2013, Article ID 393729, 9 pages, 2013. https://doi.org/10.1155/2013/393729
Runxin Wu1 and Lin Li1
1Department of Mathematics and Physics, Fujian University of Technology, Fuzhou, Fujian 350108, China
By constructing a suitable Lyapunov function and using the comparison theorem of difference equation, sufficient conditions which ensure the permanence and global attractivity of the discrete predator-prey system with Hassell-Varley-Holling III type functional response are obtained. An example together with its numerical simulation shows that the main results are verifiable.
Recently, there were many works on predator-prey system done by scholars [1–6]. In particular, since Hassell-Varley [7] proposed a general predator-prey model with Hassell-Varley type functional response in 1969, many excellent works have been conducted for the Hassell-Varley type system [1, 7–13].
Liu and Huang [8] studied the following discrete predator-prey system with Hassell-Varley-Holling III type functional response: where , denote the density of prey and predator species at the th generation, respectively. , , , , , are all periodic positive sequences with common period . Here represents the intrinsic growth rate of prey species at the th generation, and measures the intraspecific effects of the th generation of prey species on their own population; is the death rate of the predator; is the capturing rate; is the maximal growth rate of the predator. Liu and Huang obtained the necessary and sufficient conditions for the existences of positive periodic solutions by applying a new estimation technique of solutions and the invariance property of homotopy. As we know, the persistent property is one of the most important topics in the study of population dynamics. For more papers on permanence and extinction of population dynamics, one could refer to [2–5, 14–17] and the references cited therein. The purpose of this paper is to investigate permanence and global attractivity of this system.
We argue that a general nonautonomous nonperiodic system is more appropriate, and thus, we assume that the coefficients of system (1) satisfy the following:
(A) , , , , , are nonnegative sequences bounded above and below by positive constants.
By the biological meaning, we consider (1) together with the following initial conditions as
For the rest of the paper, we use the following notations: for any bounded sequence , set and .
Now, let us state several lemmas which will be useful to prove our main conclusion.
Definition 1 (see [5]). System (1) said to be permanent if there exist positive constants and , which are independent of the solution of system (1), such that for any positive solution of system (1) satisfies
Lemma 2 (see [14]). Assume that satisfies and for , where and are all nonnegative sequences bounded above and below by positive constants. Then,
Lemma 3 (see [14]). Assume that satisfies , and , where and are all nonnegative sequences bounded above and below by positive constants and . Then,
Theorem 4. Assume that hold, then system (1) is permanent, that is, for any positive solution of system (1), one has where
Proof. We divided the proof into four steps.
Step 1. We show From the first equation of (1), we have By Lemma 2, we have Previous inequality shows that for any , there exists a , such that
Step 2. We prove by distinguishing two cases.
Case 1. There exists a , such that .
By the second equation of system (1), we have which implies The previous inequality combined with (13) leads to . Thus, from the second equation of system (1), again we have We claim that By a way of contradiction, assume that there exists a such that . Then . Let be the smallest integer such that . Then . The previous argument produces that , a contradiction. This proves the claim. Therefore, . Setting in it leads to .
Case 2. Suppose for all . Since is nonincreasing and has a lower bound , we know that exists, denoted by , we claim that By a way of contradiction, assume that .
Taking limit in the second equation in system (1) gives however, which is a contradiction. It implies that . By the fact , we obtain that Therefore, we have Then,
Step 3. We verify Conditions imply that for enough small positive constant , we have For the previous , it follows from Steps 1 and 2 that there exists a such that for all Then, for , it follows from (26) and the first equation of system (1) that According to Lemma 3, one has where
Setting in (28) leads to By the fact that , we see that .
This ends the proof of Step 3.
Step 4. We present two cases to prove that For any small positive constant , from Step 1 to Step 3, it follows that there exists a such that for all
Case 1. There exists a such that , then Hence, and so, Set We claim that By a way of contradiction, assume that there exists a , such that . Then . Let be the smallest integer such that . Then , which implies that , a contradiction, this proves the claim. Therefore, , setting in it leads to .
Case 2. Assume that for all , then, exists, denoted by , then . We claim that By a way of contradiction, assume that . Taking limit in the second equation in system (1) gives which is a contradiction since This proves the claim, then we have So, Obviously, . This completes the proof of the theorem.
Definition 5 (see [18]). System (1) is said to be globally attractive if any two positive solutions and of system (1) satisfy
Theorem 6. Assume that and hold. Assume further that there exist positive constants , , and such that where
Then, system (1), with initial condition (2), is globally attractive, that is, for any two positive solutions and of system (1), we have
Proof. From conditions and , there exists an enough small positive constant such that where
Since and hold, for any positive solutions and of system (1), it follows from Theorem 4 that For the previous and (48), there exists a such that for all , Let Then from the first equation of system (1), we have Using the mean value theorem, we get where lies between and , lies between and .
It follows from (51) and (52) that And so, for , Let Then, from the second equation of system (1), we have Using the mean value theorem, we get where lies between and , respectively. Then, it follows from (56) and (57) that for ,
Now, we define a Lyapunov function as follows: Calculating the difference of along the solution of system (1), for , it follows from (54) and (58) that Summating both sides of the previous inequalities from to , we have which implies It follows that Using the fundamental theorem of positive series, there exists small enough positive constant such that which implies that that is, This completes the proof of Theorem 6.
4. Extinction of the Predator Species
This section is devoted to study the extinction of the predator species .
Theorem 7. Assume that Then, the species will be driven to extinction, and the species is permanent, that is, for any positive solution of system (1), where
Proof. For condition , there exists small enough positive , such that for all , from (69) and the second equation of the system (1), one can easily obtain that Therefore, which yields From the proof of Theorem 4, we have For enough small positive constant , For the previous , from (72) and (73) there exists a such that for all , From the first equation of (1), we have By Lemma 3, we have Setting in (72) leads to The proof of Theorem 7 is completed.
The following example shows the feasibility of the main results.
One could easily see that Clearly, conditions and are satisfied. It follows from Theorem 4 that the system is permanent. Numerical simulation from Figure 1 shows that solutions do converge and system is permanent and globally attractive.
Dynamics behaviors of system (1) with initial conditions , respectively.
In this paper, a discrete predator-prey model with Hassell-Varley-Holling III type functional response is discussed. The main topics are focused on permanence, global attractivity, and extinction of predator species. The numerical simulation shows that the main results are verifiable.
The investigation in this paper suggests the following biological implications. Theorem 4 shows that the coefficients, such as the death rate of the predator, the capturing rate, and the intraspecific effects of prey species, influence permanence. Conditions and imply that the higher the intraspecific effects of prey species are, the more favourable permanence is. Those results have further application on predator-prey population dynamics. However, the conditions for global attractivity in Theorem 4 is so complicated that its application is very difficult. A further study is required to simplify the application.
This work is supported by the Foundation of Fujian Education Bureau (JA11193).
R. Wu and L. Li, “Permanence and global attractivity of discrete predator-prey system with hassell-varley type functional response,” Discrete Dynamics in Nature and Society, vol. 2009, Article ID 323065, 17 pages, 2009. View at: Publisher Site | Google Scholar
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Copyright © 2013 Runxin Wu and Lin Li. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. |
Create extended Kalman filter object for online state estimation - MATLAB extendedKalmanFilter - MathWorks Switzerland
Create Extended Kalman Filter Object for Online State Estimation
Specify Process and Measurement Noise Covariances in Extended Kalman Filter Object
Specify Jacobians for State and Measurement Functions
Specify Nonadditive Measurement Noise in Extended Kalman Filter Object
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState)
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState,Name,Value)
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn)
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,Name,Value)
obj = extendedKalmanFilter(Name,Value)
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState) creates an extended Kalman filter object for online state estimation of a discrete-time nonlinear system. StateTransitionFcn is a function that calculates the state of the system at time k, given the state vector at time k-1. MeasurementFcn is a function that calculates the output measurement of the system at time k, given the state at time k. InitialState specifies the initial value of the state estimates.
After creating the object, use the correct and predict commands to update state estimates and state estimation error covariance values using a first-order discrete-time extended Kalman filter algorithm and real-time data.
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState,Name,Value) specifies additional attributes of the extended Kalman filter object using one or more Name,Value pair arguments.
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn) creates an extended Kalman filter object using the specified state transition and measurement functions. Before using the predict and correct commands, specify the initial state values using dot notation. For example, for a two-state system with initial state values [1;0], specify obj.State = [1;0].
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,Name,Value) specifies additional attributes of the extended Kalman filter object using one or more Name,Value pair arguments. Before using the predict and correct commands, specify the initial state values using Name,Value pair arguments or dot notation.
obj = extendedKalmanFilter(Name,Value) creates an extended Kalman filter object with properties specified using one or more Name,Value pair arguments. Before using the predict and correct commands, specify the state transition function, measurement function, and initial state values using Name,Value pair arguments or dot notation.
extendedKalmanFilter creates an object for online state estimation of a discrete-time nonlinear system using the first-order discrete-time extended Kalman filter algorithm.
\stackrel{^}{x}
\begin{array}{l}x\left[k\right]=f\left(x\left[k-1\right],{u}_{s}\left[k-1\right]\right)+w\left[k-1\right]\\ y\left[k\right]=h\left(x\left[k\right],{u}_{m}\left[k\right]\right)+v\left[k\right]\end{array}
\begin{array}{l}x\left[k\right]=f\left(x\left[k-1\right],w\left[k-1\right],{u}_{s}\left[k-1\right]\right)\\ y\left[k\right]=h\left(x\left[k\right],v\left[k\right],{u}_{m}\left[k\right]\right)\end{array}
When you perform online state estimation, you first create the nonlinear state transition function f and measurement function h. You then construct the extendedKalmanFilter object using these nonlinear functions, and specify whether the noise terms are additive or nonadditive. You can also specify the Jacobians of the state transition and measurement functions. If you do not specify them, the software numerically computes the Jacobians.
After you create the object, you use the predict command to predict state estimate at the next time step, and correct to correct state estimates using the algorithm and real-time data. For information about the algorithm, see Extended and Unscented Kalman Filter Algorithms for Online State Estimation.
You can use the following commands with extendedKalmanFilter objects:
For extendedKalmanFilter object properties, see Properties.
To define an extended Kalman filter object for estimating the states of your system, you first write and save the state transition function and measurement function for the system.
In this example, use the previously written and saved state transition and measurement functions, vdpStateFcn.m and vdpMeasurementFcn.m. These functions describe a discrete-approximation to a van der Pol oscillator with nonlinearity parameter, mu, equal to 1. The oscillator has two states.
Specify an initial guess for the two states. You specify the guess as an M-element row or column vector, where M is the number of states.
Create the extended Kalman filter object. Use function handles to provide the state transition and measurement functions to the object.
obj = extendedKalmanFilter(@vdpStateFcn,@vdpMeasurementFcn,initialStateGuess);
Create an extended Kalman filter object for a van der Pol oscillator with two states and one output. Use the previously written and saved state transition and measurement functions, vdpStateFcn.m and vdpMeasurementFcn.m. These functions are written for additive process and measurement noise terms. Specify the initial state values for the two states as [2;0].
obj = extendedKalmanFilter(@vdpStateFcn,@vdpMeasurementFcn,[2;0],...
Create an extended Kalman filter object for a van der Pol oscillator with two states and one output. Use the previously written and saved state transition and measurement functions, vdpStateFcn.m and vdpMeasurementFcn.m. Specify the initial state values for the two states as [2;0].
The extended Kalman filter algorithm uses Jacobians of the state transition and measurement functions for state estimation. You write and save the Jacobian functions and provide them as function handles to the object. In this example, use the previously written and saved functions vdpStateJacobianFcn.m and vdpMeasurementJacobianFcn.m.
obj.StateTransitionJacobianFcn = @vdpStateJacobianFcn;
obj.MeasurementJacobianFcn = @vdpMeasurementJacobianFcn;
Note that if you do not specify the Jacobians of the functions, the software numerically computes the Jacobians. This numerical computation may result in increased processing time and numerical inaccuracy of the state estimation.
Create an extended Kalman filter object for a van der Pol oscillator with two states and one output. Assume that the process noise terms in the state transition function are additive. That is, there is a linear relation between the state and process noise. Also assume that the measurement noise terms are nonadditive. That is, there is a nonlinear relation between the measurement and measurement noise.
obj = extendedKalmanFilter('HasAdditiveMeasurementNoise',false);
x\left[k\right]=\sqrt{x\left[k-1\right]+u\left[k-1\right]}+w\left[k-1\right]
y\left[k\right]=x\left[k\right]+2*u\left[k\right]+v\left[k{\right]}^{2}
InitialState — Initial state estimate value
The specified value is stored in the State property of the object. If you specify InitialState as a column vector, then State is also a column vector, and the predict and correct commands return state estimates as a column vector. Otherwise, a row vector is returned.
If you want a filter with single-precision floating-point variables, specify InitialState as a single-precision vector variable. For example, for a two-state system with state transition and measurement functions vdpStateFcn.m and vdpMeasurementFcn.m, create the extended Kalman filter object with initial state estimates [1;2] as follows:
obj = extendedKalmanFilter(@vdpStateFcn,@vdpMeasurementFcn,single([1;2]))
Use Name,Value arguments to specify properties of extendedKalmanFilter object during object creation. For example, to create an extended Kalman filter object and specify the process noise covariance as 0.01:
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState,'ProcessNoise',0.01);
extendedKalmanFilter object properties are of three types:
Tunable properties that you can specify multiple times, either during object construction using Name,Value arguments, or any time afterward during state estimation. After object creation, use dot notation to modify the tunable properties.
obj = extendedKalmanFilter(StateTransitionFcn,MeasurementFcn,InitialState);
The tunable properties are State, StateCovariance, ProcessNoise, and MeasurementNoise.
Nontunable properties that you can specify once, either during object construction or afterward using dot notion. Specify these properties before state estimation using correct and predict. The StateTransitionFcn, MeasurementFcn, StateTransitionJacobianFcn, and MeasurementJacobianFcn properties belong to this category.
[] — The Jacobian is numerically computed at every call to the correct command. This may increase processing time and numerical inaccuracy of the state estimation.
function handle — You write and save the Jacobian function and specify the handle to the function. For example, if vdpMeasurementJacobianFcn.m is the Jacobian function, specify MeasurementJacobianFcn as @vdpMeasurementJacobianFcn.
The function calculates the partial derivatives of the measurement function with respect to the states and measurement noise. The number of inputs to the Jacobian function must equal the number of inputs to the measurement function and must be specified in the same order in both functions. The number of outputs of the Jacobian function depends on the HasAdditiveMeasurementNoise property:
HasAdditiveMeasurementNoise is true — The function calculates the partial derivatives of the measurement function with respect to the states (
\partial h/\partial x
HasAdditiveMeasurementNoise is false — The function also returns a second output that is the partial derivative of the measurement function with respect to the measurement noise terms (
\partial h/\partial v
MeasurementJacobianFcn is a nontunable property. You can specify it once before using the correct command either during object construction or using dot notation after object construction. You cannot change it after using the correct command.
[] — The Jacobian is numerically computed at every call to the predict command. This may increase processing time and numerical inaccuracy of the state estimation.
function handle — You write and save the Jacobian function and specify the handle to the function. For example, if vdpStateJacobianFcn.m is the Jacobian function, specify StateTransitionJacobianFcn as @vdpStateJacobianFcn.
The function calculates the partial derivatives of the state transition function with respect to the states and process noise. The number of inputs to the Jacobian function must equal the number of inputs of the state transition function and must be specified in the same order in both functions. The number of outputs of the function depends on the HasAdditiveProcessNoise property:
HasAdditiveProcessNoise is true — The function calculates the partial derivative of the state transition function with respect to the states (
\partial f/\partial x
HasAdditiveProcessNoise is false — The function must also return a second output that is the partial derivative of the state transition function with respect to the process noise terms (
\partial f/\partial w
). The second output is returned as an Ns-by-W Jacobian matrix, where W is the number of process noise terms.
The extended Kalman filter algorithm uses the Jacobian to compute the state estimation error covariance.
StateTransitionJacobianFcn is a nontunable property. You can specify it once before using the predict command either during object construction or using dot notation after object construction. You cannot change it after using the predict command.
obj — Extended Kalman filter object for online state estimation
extendedKalmanFilter object
Extended Kalman filter object for online state estimation, returned as an extendedKalmanFilter object. This object is created using the specified properties. Use the correct and predict commands to estimate the state and state estimation error covariance using the extended Kalman filter algorithm.
Generated code uses an algorithm that is different from the algorithm that the extendedKalmanFilter function uses. You might see some numerical differences in the results obtained using the two methods.
Starting in R2020b, numerical improvements in the extendedKalmanFilter algorithm might produce results that are different from the results you obtained in previous versions.
predict | correct | residual | clone | unscentedKalmanFilter | kalman | kalmd |
NonFreeWiki/Demo/Non-free content - Meta
NonFreeWiki/Demo/Non-free content
This example page is a modified version of the English Wikipedia's non-free content policy. Many of the links are to enwiki but when this wiki is activated these pages would be created specifically for it.
This page in a nutshell: Non-free content can be used in Wikipedia articles only if:
Its usage would be considered fair use under United States copyright law and also complies with the non-free content criteria below;
It has a valid rationale indicating why its usage would be considered fair use within U.S. law and the policy of each Wikipedia that it would be used on.
Wikipedia's goal is to be a free content encyclopedia, with free content defined as content that does not bear copyright restrictions on the right to redistribute, study, modify and improve, or otherwise use works for any purpose in any medium, even commercially. Any content not satisfying these criteria is said to be non-free. This includes all content (including images) that is fully copyrighted, or which is made available subject to restrictions such as "non-commercial use only" or "for use on Wikipedia only". Many images that are generally available free of charge may thus still be "non-free" for Wikipedia purposes. A full definition can be found at http://freedomdefined.org/Definition.
The licensing policy of the Wikimedia Foundation requires all content hosted on Wikipedia to be free content; however, there are exceptions. The policy allows projects to adopt an Exemption Doctrine Policy allowing the use of non-free content within strictly defined limitations. Non-free content can be used on Wikipedia in certain cases (for example, in some situations where acquiring a freely licensed image for a particular subject is not possible), but only within the United States legal doctrine of fair use, and in accordance with each Wikipedia's own non-free content criteria. The use of non-free content on Wikipedia is therefore subject to purposely stricter standards than those laid down in U.S. copyright law.
As per the March 23, 2007 Wikimedia Foundation Licensing policy resolution this document serves as the Exemption Doctrine Policy for the NonFreeWiki and is designed to complement individual EDP's.
Media-specific policy. Non-free content meets Wikipedia's media-specific policy. For example, images must meet w:WP:Image use policy.
Identification of the source of the original copyrighted material, supplemented, where possible, with information about the artist, publisher and copyright holder, and year of copyright; this is to help determine the material's potential market value. See: w:WP:Citing sources#Multimedia.
A copyright tag that indicates which Wikipedia policy provision is claimed to permit the use. For a list of image copyright tags, see w:WP:Image copyright tags/Non-free content.
The name of each article (a link to each article is also recommended) in which fair use is claimed for the item, and a separate, specific non-free use rationale for each use of the item, as explained at w:WP:Non-free use rationale guideline.[1] The rationale is presented in clear, plain language and is relevant to each use.
A file in use in an article and uploaded after NonFreeWiki was activated that does not comply with this policy 48 hours after notification to the uploading editor will be deleted. To avoid deletion, the uploading editor or another Wikipedian will need to provide a convincing non-free-use defense that satisfies all 10 criteria. For a file in use in an article that was uploaded before NonFreeWiki was activated, the 48-hour period is extended to seven days.
You can find a list of these copyright license templates at w:WP:File copyright tags/Non-free.
The rationale to use the non-free content is necessary to show that the non-free content criteria have been met. The rationale should clearly address and satisfy all ten non-free criteria. Template versions to generate such rationales do exist, and include:
w:Template:Non-free use rationale - A generic template applicable for any non-free media.
w:Template:Non-free use rationale 2 - An alternative template to the above.
w:Template:Non-free use rationale logo - A rationale template for logos, assuming they are being used as a header image (standalone or infobox) for the entity the logo represents.
Several other boilerplate rationale templates can be found at w:Category:Non-free use rationale templates, but editors are cautioned that these are generally tenuous in terms of supporting non-free criteria #8, and are encouraged to improve upon rationales if they can do so. You are not required to use the template forms, but whatever form you chose needs to clearly address all 10 non-free criteria.
Meeting the no free equivalent criterionEdit
Meeting the previous publication criterionEdit
Meeting the contextual significance criterionEdit
Meeting the minimal usage criterionEdit
Number of itemsEdit
Articles are structured and worded, where reasonable to do so, in order to minimize the total number of times items of non-free content are included within the encyclopedia.
For example, an excerpt of a significant artistic work, notable for both its production and performance, is usually included only in an article about the work, which is then referenced in those about its performer and its producer (where these are notable enough for their own articles).
An item of non-free content that conveys multiple points of significant understanding within a topic, is used in preference to multiple non-free items where each conveys fewer such points. This is independent of whether the topic is covered by a single article, or is split across several.
For example, an article about an ensemble may warrant inclusion of a non-free image identifying the ensemble: this is preferable to including non-free images for each member of the ensemble, even if the article has been split with sub-articles for each member's part in the overall topic. Note however, that an article covering an ensemble member's notable activities outside those of the ensemble, is usually treated as being a separate topic.
{\displaystyle {\text{new width}}=\left\lfloor {\sqrt {\tfrac {{\text{target pixel count}}\times {\text{original width}}}{\text{original height}}}}\right\rfloor }
If you believe an image is oversized, either re-upload a new version at the same file location, or tag the image file page with a non-free reduce template, which will place it in a maintenance category to be reduced by volunteers or a bot like w:User:Theo's Little Bot.
Guideline examplesEdit
Non-free content that meets all of the policy criteria above but does not fall under one of the designated categories below may or may not be allowable, depending on what the material is and how it is used. These examples are not meant to be exhaustive, and depending on the situation there are exceptions. When in doubt as to whether non-free content may be included, please make a judgement based on the spirit of the policy, not necessarily the exact wording. If you want help in assessing whether a use is acceptable, please ask at w:WP:Media copyright questions. w:Wikipedia talk:Copyrights, w:Wikipedia talk:Copyright problems, and w:Wikipedia talk:Non-free content may also be useful. These are places where those who understand copyright law and Wikipedia policy are likely to be watching.
See also: w:WP:Wikipedia Signpost/2008-09-22/Dispatches, a guide to evaluating the acceptability of non-free images.
Acceptable useEdit
Brief quotations of copyrighted text may be used to illustrate a point, establish context, or attribute a point of view or idea. In all cases, a citation is required. Copyrighted text that is used verbatim must be attributed with quotation marks or other standard notation, such as block quotes. Any alterations must be clearly marked, i.e., [brackets] for added text, an ellipsis (...) for removed text, and emphasis noted after the quotation as "(emphasis added)" or "(emphasis in the original)". Extensive quotation of copyrighted text is prohibited. Please see both w:WP:QUOTE for use and formatting issues in using quotations, and w:WP:MOSQUOTE for style guidelines related to quoting.
Audio clipsEdit
All non-free audio files must meet each of the non-free content criteria; failure to meet those overrides any acceptable allowance here. Advice for preparing non-free audio files for Wikipedia can be found at w:Wikipedia:Music samples. The following list is non-inclusive but contains the most common cases where non-free audio samples may be used.
Music clips may be used to identify a musical style, group, or iconic piece of music when accompanied by appropriate sourced commentary and attributed to the copyright holder. Samples should generally not be longer than 30 seconds or 10% of the length of the original song, whichever is shorter (see w:Wikipedia:Music samples).
Some non-free images may be used on Wikipedia, providing they meet both the legal criteria for fair use, and Wikipedia's own guidelines for non-free content. Non-free images that reasonably could be replaced by free content images are not suitable for Wikipedia. All non-free images must meet each non-free content criterion; failure to meet those overrides any acceptable allowance here. The following list is not exhaustive but contains the most common cases where non-free images may be used.
Team and corporate logos: For identification. See w:Wikipedia:Logos.
Screenshots from software products: For critical commentary. See w:Wikipedia:Software screenshots.
Iconic and historical images which are not subject of commentary themselves but significantly aid in illustrating historical events may be used if they meet all aspects of the non-free content criteria, particularly no free alternatives (non-free criterion 1), respect for commercial opportunity (criterion 2), and contextual significance (criterion 8).
Unacceptable useEdit
In considering the ability to take a free photograph, it is expected that the photographer respects all local property and privacy laws and restrictions. For example, we would not expect a free photograph of a structure on inaccessible private property that is not visible from public locations.
Non-free image use in list articlesEdit
Non-free image use in galleries or tablesEdit
Montages consisting of some or all non-free images created by a user should be avoided for similar reasons. Within the scope of NFCC#3a, such montages are considered as multiple non-free images based on each non-free image that contributes towards the montage. If a montage is determined to be appropriate, each contributing non-free item should have its source described (such as w:File:Versions_of_the_Doctor.jpg). A montage created by the copyright holder of the images used to create the montage is considered a single non-free item and not separate items.
Exemptions from non-free content policy are made for the use of non-free content on certain administrative, non-article space pages as necessary to creating or managing the encyclopedia, specifically for those that are used to manage questionable non-free content. Those pages that are exempt are listed in w:Category:Wikipedia non-free content criteria exemptions.[4]
Explanation of policy and guidelinesEdit
Legal positionEdit
Certain works have no copyright at all. Most material published in the United States before 1923, works published before 1978 without a copyright notice, with an expired copyright term, or produced by the U.S. federal government, among others, is public domain, i.e., has no copyright. Some do not have a sufficient degree of creativity apart from their functional aspects to have a copyright e.g. photos and scans of 2-dimensional objects and other "slavish reproductions", short text phrases, typographic logos, and product designs.
If material does have a copyright, it may only be copied or distributed under a license (permission) from the copyright holder, or under the doctrine of fair use. If there is a valid license, the user must stay within the scope of the license (which may include limitations on amount of use, geographic or business territory, time period, nature of use, etc.). Fair use, by contrast, is a limited right to use copyrighted works without permission, highly dependent on the specific circumstances of the work and the use in question. It is a doctrine incorporated as a clause in United States copyright code, arising out of a concern that strict application of copyright law would limit criticism, commentary, scholarship, and other important free speech rights. A comparable concept of w:fair dealing:fair dealing exists in some other countries, where standards may vary.
Anything published 1923 or later in other countries and still copyrighted there, is typically also copyrighted in the United States. See Wikipedia:Non-U.S. copyrights.[8]
Applied to WikipediaEdit
Reporting inappropriate use of non-free contentEdit
This policy is specific to the English language Wikipedia. Other Wikimedia projects, including Wikipedias in other languages, may have different policies on non-free content. A list of some of the projects and their policies on fair use can be read here. Specific information about different language versions of Wikipedia and their rules on fair use can be found here.
↑ NFCI#1 relates to the use of cover art within articles whose main subject is the work associated with the cover. Within such articles, the cover art implicitly satisfies the "contextual significance" NFCC criterion (NFCC#8) by virtue of the marketing, branding, and identification information that the cover conveys. The same rationale does not usually apply when the work is described in other articles, such as articles about the author or musician; in such articles, the NFCC criteria typically require that the cover art itself be significantly discussed within the article.
↑ The Wikimedia Foundation's associate counsel advised in March 2011 that while the courts have not firmly established precedence on the matter, polls are likely to be protectable as well because the parameters of the survey are chosen by those who conduct the polls and the selection of respondents indicates "at least some creativity." She recommended using polls in accordance with fair use principles, reminding that "Merely republishing them without any commentary or transformation is not fair use." The counsel also recommends that the use of even uncopyrightable lists be considered with regards to licensing agreements that may "bind the user/reader from republishing the list/survey results without permission", noting that "Absent a license agreement, you may still run afoul of state unfair competition and/or misappropriation laws if you take a substantial portion of the list or survey results."
↑ Due to software limitations, TimedText pages for non-free video files will automatically include the video file, and as such, pages in the TimedText namespace are presumed exception from NFCC#9.
↑ To find out how to search for copyright registrations and renewals see How to Investigate the Copyright Status of a Work, Stanford's Copyright Renewal Database, Project Gutenberg and Iinformation about The Catalog of Copyright Entries.
↑ Non-US copyrights apply in the US under the w:URAA.
Coverage of U.S. fair use law by w:Cornell University
Coverage of U.S. fair use law by w:Stanford University
Code of Best Practices in Fair Use for Online Video(PDF) - w:American University Center for Social Media. Jan 14, 2012 (updated link).
Retrieved from "https://meta.wikimedia.org/w/index.php?title=NonFreeWiki/Demo/Non-free_content&oldid=15474715" |
Find an answer to any fluid mechanics questions
Get help with fluid mechanics questions
Recent questions in Fluid Mechanics
\alpha \text{ }{m}^{2}
\beta
\beta
\mu \text{ }N
{r}^{4}
r
{r}^{2}
{r}^{2}
{r}^{2}
The continuity equation states that flow rate should be conserved in different areas of a pipe:
Q={v}_{1}{A}_{1}={v}_{2}{A}_{2}=v\pi {r}^{2}
We can see from this equation that velocity and pipe radius are inversely proportional. If radius is doubled, velocity of flow is quartered.
Another way I was taught to describe flow rate is through Poiseuilles Law:
Q=\frac{\pi {r}^{4}\mathrm{\Delta }P}{8\eta L}
So if I were to plug in the continuity equations definition of flow rate into Poiseuilles Law:
vA=v\pi {r}^{2}=\frac{\pi {r}^{4}\mathrm{\Delta }P}{8\eta L}
v=\frac{{r}^{2}\mathrm{\Delta }P}{8\eta L}
Now in this case, the velocity is proportional to the radius of the pipe. If the radius is doubled, then velocity is qaudrupled.
What am I misunderstanding here? I would prefer a conceptual explanation because I feel that these equations are probably used with different assumptions/in different contexts.
Do you need to increase pressure to pump a fluid from a large pipe through a small pipe?
The application of this question is towards clogged arteries and blood flow. I am wondering why people with clogged arteries have high blood pressure if a smaller artery (assuming it is smooth and cylindrical) means less pressure exerted by the blood according to Bernoulli's Equation. Is this because the heart has to pump the blood harder to be able to travel through a smaller arterial volume?
Matthew Hubbard 2022-05-10 Answered
Power generation with a long, horizontal pipe, using atmospheric pressure differences
If I float a pipe starting from a point in the ocean that has the atmospheric pressure of 101 kPa (Call it Point A) all the way to another geographic position on the ocean with an atmospheric pressure of 96 kPa (point B). Lets say the distance between the two points is 1000 km. The air temperature at point A is 28 °C while the air temp at point is 18 °C.
If you opened both ends of the pipe,would the atmospheric pressure cause air to flow from Point A to Point B? (factoring in reasonable friction co-eff for a pipe.)
Can you please provide factors that would hinder the flow of air!
Could such a pipe, practically be used to generate a flow of air to generate electrical power?
Pressure changes in Continuity equations and Poiseuille's Law?
Continuity says Q=AV, and we know that velocity and pressure are inversely related. So if we are in a closed system, like vasculature for example, Q is constant and any decrease in vessel radius would be expected to raise velocity, which would result in lower pressure.
If we look at Poiseuille's Law, on the other hand, we see the opposite! If Q is constant, then a decrease in radius/cross sectional area we should expect pressure to be raised!
Physical meaning of
\frac{\pi }{8}
in Poiseuille's equation
Recently I read about the Poiseuille's equation which relates the flow rate of a viscous fluid to coefficient of viscosity (
\nu
), pressure per unit length(
\frac{P}{l}
) and radius of the tube (
r
) in which the fluid is flowing. The equation is
\frac{V}{t}=\frac{\pi P{r}^{4}}{8\nu l},
V
denotes volume of the fluid.
One thing which I don't understand in this equation is that why do we have the constant term as
\frac{\pi }{8}
. Is there some physical significance of this specific number here or is it just a mathematical convention?
Blood as Newtonian fluid
In some of the literature I read that blood can be considered as Newtonian fluid when a larger vesses with high shear stress is considered... How is the shear stress calculated for aorta and how do they claim that shear stress is more for larger vessel when compared to the smaller ones . What is the relationship between the shear stress and the size of the vessel
\begin{array}{}\text{(1)}& {z}_{1}+\frac{{v}_{1}^{2}}{2g}+\frac{{p}_{1}}{\rho g}={z}_{2}+\frac{{v}_{2}^{2}}{2g}+\frac{{p}_{2}}{\rho g}+{h}_{L}\end{array}
{h}_{L}
is the head loss (due to viscosity) calculated using Hagen Poiseuille law.
\begin{array}{}\text{(2)}& {h}_{L}=\rho g\frac{8\eta L\overline{v}}{{R}^{2}}\end{array}
If the flow is not laminar, I cannot use
\left(2\right)
, but can I still write
\left(1\right)
in that way?
\left(1\right)
valid only along the singular streamline or between different ones (assuming the fluid irrotational)?
Deriving the scaling law for the Reynolds number
I'm trying to derive a scaling law for the Reynolds number, to get a better understanding of how it changes for microsystem applications, but I'm getting stuck. From textbook tables I should end up with
{l}^{2}
, but can't get there. The goal is to find a simplistic relation of Reynolds number and the system dimension.
This is my reasoning so far:
Considering a tube
Re=\frac{\rho v2R}{\mu }
v=\frac{Q}{A}=\frac{Q}{\pi {R}^{2}}
From Hagen-Poiseuille:
Q=\frac{\delta P\pi {R}^{4}}{8\mu L},
v=\frac{\frac{\delta P\pi {R}^{4}}{8\mu L}}{\pi {R}^{2}}=\frac{\delta P\pi {R}^{4}}{8\mu L\pi {R}^{2}}=\frac{\delta P{R}^{2}}{8\mu L}
v
Re
Re=\frac{\frac{\rho \delta P{R}^{2}2R}{8\mu L}}{\mu }=\frac{\rho \delta P{R}^{3}2}{8{\mu }^{2}L}
From this, if I consider the
\delta P
as constant with the size scaling,
\rho
\mu
are material properties that are constant at different scales, I get:
Re\sim \frac{{R}^{3}}{L}
I can only think that I could consider it as
\frac{{l}^{3}}{l}={l}^{2}
but doesn't seem right.
Intuitive explanation of Poiseuille's law -- why
{r}^{4}
According to Poiseuille's law, the effective resistance of a tube is inversely proportional to the fourth power of its radius (as given by the following equation).
R=\frac{8\eta \mathrm{\Delta }x}{\pi {r}^{4}}
I can go through the derivation of the law but is there an intuitive explanation for why it is the fourth power of radius, or a thought experiment that might make it more obvious?
So I have a steady isothermal flow of an ideal gas through a smooth duct (no frictional losses) and need to compute the mass flow rate (per unit area) as a function of pressures at any two different arbitrary points, say 1 and 2. I have the following momentum equation in differential form:
\rho vdv+dP=0
v
is the gas the flow velocity and
P
is static pressure. The mass flow rate per unit cross section
G
, can be calculated by integrating this equation between points 1 and 2. This is where it gets confusing. I do the integration by two ways:
1) Use the ideal gas equation
P=\rho RT
right away and restructure the momentum equation:
vdv+RT\frac{dP}{P}=0
, integrate it between points 1 and 2 and arrive at:
{v}_{1}^{2}-{v}_{2}^{2}+2RTln\frac{{P}_{1}}{{P}_{2}}=0
Since the flow is steady, I can write
G={\rho }_{1}{v}_{1}={\rho }_{2}{v}_{2}
, again use the ideal gas law to write density in terms of gas pressure and finally arrive at the mass flow rate expression:
{G}^{2}=\frac{2ln\frac{{P}_{2}}{{P}_{1}}}{RT\left(\frac{1}{{P}_{1}^{2}}-\frac{1}{{P}_{2}^{2}}\right)}
2) In another way of integrating (which is mathematically correct), I start by multiplying the original momentum equation by
\rho
{\rho }^{2}vdv+\frac{1}{RT}PdP=0
{\rho }^{2}v={G}^{2}/v
, integrate between points 1 and 2 to arrive at
{G}^{2}ln\frac{{v}_{2}}{{v}_{1}}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RT}
Using the ideal gas law the velocity ratio can be written as the pressure ratio to finally arrive at the mass flow rate equation
{G}^{2}=\frac{{P}_{1}^{2}-{P}_{2}^{2}}{2RTln\frac{{P}_{1}}{{P}_{2}}}
Both the expressions are dimensionaly sound and I know that the second expression is the correct one. My question is, whats wrong with first expression.
What determines the dominant pressure-flow relationship for a gas across a flow restriction?
If one measures the pressure drop across any gas flow restriction you can generally fit the relationship to
\mathrm{\Delta }P={K}_{2}{Q}^{2}+{K}_{1}Q
\mathrm{\Delta }P
is the pressure drop and
Q
is the volumetric flow
and what I've observed is that if the restriction is orifice-like,
{K}_{2}>>{K}_{1}
and if the restriction is somewhat more of a complex, tortuous path,
{K}_{1}>>{K}_{2}
{K}_{2}
tends towards zero.
I get that the Bernoulli equation will dominate when velocities are large and so the square relationship component. But what's determining the
{K}_{1}
component behavior? Is this due to viscosity effrects becoming dominant? Does the Pouiselle relationship become dominant?
Yaritza Oneill 2022-05-08 Answered
Pneumatic flow formula
I'm looking for a pneumatic formula in order to have the flow. I found many formula, but only for hydraulic :
\mathrm{\Delta }P={R}_{p}Q
{R}_{p}
is the resistance,
Q
the flow, and
\mathrm{\Delta }P
the pressure potential:
{R}_{p}=\frac{8\eta L}{\pi {R}^{4}}
\eta =
dynamic viscosity of the liquid ;
L=
Length of the pipe ;
R=
radius of the pipe
I don't know if I can use them, since I'm in pneumatic ! After doing research on the internet, I found some other variables like : sonic conductance, critical pressure coefficient, but no formula...
I think that I have all the information in order the calculate the flow : Pipe length = 1m ; Pipe diameter = 10mm ;
\mathrm{\Delta }P=
2 bars
Hydraulics, flow rate, and tubing size questions
Here are a couple assumptions / things I believe to be true:
1.Pumps create a certain pressure in a piping circuit.
2.Pressure is like voltage and constant across the piping circuit, with differences in piping size increasing flow rate by reducing friction loss ie resistance
3.Given constant pressure, flow rate is variable as a fuction of pipe radius per the Poiseuille law.
Imagine the case that a 2" pump outlet, runs for 1 foot of 2" piping and then connects in with a 4" pipe for reduced friction.
So if the pressure across both pipes (the system) are the same, the gpm for the larger pipe is something like 16x bigger for the larger pipe than the smaller pipe. The larger pipe is fed by the smaller pipe, so where does this water come from?
Understand what I am getting at?
One of my assumptions seems erroneous.
The figure shows four charges at the corners of a square of side L. What magnitude and sign of charge Q will make the force on charge q zero?
For a fluid with viscosity to flow through a pipe that has the same cross-sectional area at both ends, at a constant velocity, there has to be a pressure difference according to Poiseuille's Law. Why exactly is there a change in pressure required to keep the velocity constant?
Is it because according to Bernoulli's principle that Pressure or Pressure-Energy gets converted to Kinetic Energy to speed up the fluid so the mass flow rate at both ends of the pipe stays the same?
So if that's the case in light of Bernoulli's principle, that means the change in pressure in Poiseullie's Law is there so that pressure energy gets converted to Kinetic Energy to fight off the viscosity of the fluid to keep the velocity of the fluid constant?
What exactly is Pressure? I know it is the Force divided by the area. I understand that concept, but I've seen the terms Pressure and Pressure Energy used interchangeably when talking about fluids, which creates for some amount of confusion. Aren't Pressure and Pressure Energy different? But when we talk about fluids in light of Bernoulli's principles, it seems as if Pressure and Pressure Energy are the same, which is pretty confusing. |
Some steps in solving an equation are more efficient than others. Complete parts (a) through (d) to determine the most efficient first step to solve the equation
34=5x-21
If both sides of the equation were divided by
5
, then the equation would be
\frac{34}{5}=x-\frac{21}{5}
. Does this make the problem simpler? Why or why not?
Fractional constants are more complex and can be harder to deal with than whole number constants.
34
from both sides, the equation becomes
0=5x-55
. Does this make the equation simpler to solve? Why or why not?
All terms are on one side of the equation.
If you add
21
to both sides, the equation becomes
55=5x
. Does this suggestion make this a problem you can solve more easily? Why or why not?
x
has been isolated on one side of the equation. What would you do next to solve the equation?
Is this one less step than you would take for the other first steps from parts (a) and (b)?
All three suggestions are legal moves, but which method will lead to the most efficient solution? Why?
Part (c) is the most efficient method. Can you see why?
Try using each method to see which is the quickest with the fewest steps to solve the equation. |
Draw the Hasse diagram representing the partial ordering {(a, b) | a divides b}
Draw the Hasse diagram representing the partial ordering {(a, b) | a divides b} on {1, 2, 3, 4, 6, 8, 12}.
Want to know more about Discrete math?
gwibdaithq
R=(1.1),(1.2),(1.3),(1.4),(1.6),(1.8),(1.12),(2.2),(2.4),(2.6),(2.8),(2.12),(3.3),(3.6),(3.12),(4.4),(4.8),(4.12),(6.6),(6.12),(8.8),(12.12).
Discrete Mathematics Basics
1) Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where
\left(a,b\right)\in R
I) everyone who has visited Web page a has also visited Web page b.
II) there are no common links found on both Web page a and Web page b.
III) there is at least one common link on Web page a and Web page b.
Suppose that A is the set of sophomores at your school and B is the set of students in discrete mathematics at your school. Express each of these sets in terms of A and B.
a) the set of sophomores taking discrete mathematics in your school
b) the set of sophomores at your school who are not taking discrete mathematics
c) the set of students at your school who either are sophomores or are taking discrete mathematics
Use these symbols:
\cap \cup
Let A, B, and C be sets. Show that
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
How many elements are in the set { 0, { { 0 } }?
Let R be the relation from X={1,2,3,5} to Y={0,3,4,9} defined by xRy if and only if
{x}^{2}=y
n\ge 8
. Give a
big-\Theta
estimate for the number of circuits of length 8 in Kn.
\Theta \left(n\right)\Theta \left(n8\right)\Theta \left(8\right)\Theta \left(8n\right)\Theta \left(8n\right)
Express the following in set-builder notation:
a) The set A of natural numbers divisible by 3.
b) The set B of pairs (a,b) of real numbers such that
a+b
c) The open interval
C=\left(-2,2\right)
d) The set D of 20 element subsets of N. |
EWheelEn < Velo < TWiki
There are things you don't need but are fun to use, build, and learn.
Yet another LED blinker (YALB). There are so many LED blinking projects out there, why another boring blinker? This is a different one. At least, it has more LEDs (32 RGB-LEDs ) than an average blinker and the LEDs are controlled in real time (reaction time less than 270 μs to switch 96 LEDs). But there are more technical concepts to explore. I2C communication between main MCU and peripherals (pressure meter, accelerometer). Asynchronous serial communication between main MCU and the bluetooth subsystem. USB serial communication (terminal console) for testing purposes. Applied low power techniques (control and charging LiPol battery, real time clock, wake up through motion detection). Definitely a good project to learn how embedded systems work.
The first project name (working title) was eWheel. The e stands for Euler :
\displaystyle \begin{matrix} \mathrm{e}^{\mathrm{i}\,\pi} = \end{matrix}-1\;
See also on Hackster .
Cool video on youtube Building a replica of the Velo Bling-Bling (Bicycle POV DIY project) from Dmitry Pakhomenko.
Built in cyclocomputer, w/ pressure altimeter
Configuration via Bluetooth Low Energy (BLE, Bluetooth 4.0) by Smartphone (Android 4.4+, iPhone)
Integrated LiPol battery charger (Micro-USB connector)
Push button for start and stop trip
50 custom pictures (upload through USB or BLE)
Software and hardware is Open Source.
Bling-Bling history
Betatest & installation party
Mechanical design (frame)
Nonstop 1230 km endurance test at Paris-Brest-Paris
jpg kmh-veloblingbling-thumb.jpg r1 manage 44.0 K 2016-03-30 - 20:56 PeterSchmid
png kmh-veloblingbling-thumb.png r1 manage 36.7 K 2016-03-30 - 21:00 PeterSchmid |
Vehicle Grid Integration (VGI) | RangL
Vehicle Grid Integration (VGI)
Vehicle and grid integration (VGI) is the the latest data centred challenge for Rangl, the new exciting AI challenge platform dedicated to control problems.
Electric vehicles (EVs) are by now part of our daily lives, with the likes of Tesla and Nissan helping to promote green and sustainable transportation. EVs are powered by efficient lithium-ion batteries that can be charged by the electricity coming from the grid, which in many countries is cheaper and has less carbon footprint than conventional fuels. Everything sounds great right?
However, as EVs become mainstream, our local distribution systems (DSs) will be faced with the ambitious challenge of supplying these powerful storages. This is some tricky business. As fast charging seems almost an unavoidable reality, DSs will be increasingly loaded with the power required for transportation and distribution system operators (DSOs) will have to choose between upgrading the current grid infrastructure or hindering the EV revolution. But what if there was another way?
Rangl presents VGI, which aims at providing solutions for optimising the operation of DSs while supplying and promoting e-mobility. This is a challenge for the current generation of data scientists, engineers, applied mathematicians to help promoting green transportation and safeguard local networks.
The issue of uncontrolled charging
Let us consider a generic DS as shown below.
Where each household
h
is connected to a certain phased
p
of a connection bus
b
, and will have a certain electricity demand, given by the household appliances for each time step
t
. The householder may ov and EV, identified with the index
e
, which will be connected to the household's low voltage network. A three-phase transformer will supply this DS. In this challenge
p\in[1,2,3]
b\in[1,2,3,4]
h\in[1,\dots,96]
e\in[1,\dots,75]
Each household will consume electricity each day,
P^H_{h,t}
and EVs need to be charged for their journeys. In fact, EVs can depart for trips and come back afterwards. Therefore, the electricity used to charge the EVs,
P^{EV}_{e,t}
will be provided by the DS along with the electricity consumed by the household appliances, i.e.
P^{TOT}_{b,p,h,t}=P^H_{h,t}+P^{EV}_{e,t}
is the total electricity demand of the house and the EV. In this challenge, each day is represented by
t\in[1,\dots,48]
half hourly time steps.
In power systems, especially in DSs, electricity consumption leads to voltage drop. Hence, the higher the electricity consumption, the lower the voltage. There are statutory voltage limits that have been set to ensure a reliable operation of DSs; in the UK these limits are -6% from the nominal voltage of 230 V, to +10% of the nominal voltage. Therefore, for a safe, reliable and efficient utilisation of DSs, the voltage must always be kept within these limits.
As EVs are additional loads to the DS, charging them will inevitably cause voltage drop. The key is to minimise the voltage drop in order to comply with the statutory limits. This is what VGI is all about: smart charging that satisfy EV drivers and protects the DS.
There are two key objectives that need to be fulfilled in this challenge:
minimising the voltage drop in a DS;
sufficiently charging all EVs prior to their departure.
The architecture of the challenge includes two essential parties, namely the environment and the agent, which are typical elements of reinforcement learning (RL) algorithms. RL is a key part of Rangl, where an agent interacts with an environment to improve its performance. The environment provides feedback to every action of the agent in the form of rewards, and observations of the current state of the system.
To this end, the environment fulfils the following duties:
implement any action performed by an agent and calcuate an associated scalar reward;
provide an observation of the current state of the system.
The agent must attain the following tasks:
intake the reward and observations provided by the environment;
formulate an action based on historical observations and rewards.
Environment - a typical distribution system
In VGI, the environment is represented by a DS with 96 houses, distributed accross the three phases of a four-buses low voltage network. An arbitrary (known) number of houses include an EV. The environment updates its state every time step of half an hour; hence, there are 48 time steps in the simulation of one full day. The key parameter in identifying the state of the environment is the voltage. Other parameters are the prediction of the electricity power consumed by each household, arrival (at home) and departure (from home) times of EVs and the energy stored in the EVs. All the observable elements are provided in .csv files for each simulated time step. The observation package at each time step consists of the following elements:
Arrival times of EVs,
\{t^{a}_e\}_h
Departure times of EVs,
\{t^{d}_e\}_h
Required energy for next trip,
\{E^{d}_e\}_h
Energy stored in the EV,
\{E^{EV}_{e,t}\}_h
Voltage of the DS at the current time step,
V_t
Prediction of household electricity demand
\{P^H_{h,\tau}\}_h
\tau=[t,\dots,t+48]
is the forward looking time period.
Each reading is an array of 96 elements, as same as the number of houses,
h
; the houses that are not equipped with EVs will display suitable information that will help to discern this unavailability. One exception is the prediction of the household electricity demands, provided for the next 48 time steps. Naturally, this predicted infromation has uncertainties.
Arrival and departure (at and from home, respectively) times are randomly generated from normal distributions:
Arrival times have a mean
\mu=34.55
\sigma=0.97
Departure times have a mean
\mu=16.43
\sigma=0.89
At the beginning of each simulation day, the initial energies stored in the EVs are generated randomly. In VGI, the maximum energy capacity of any EV is 30 kWh.
Once the total electricity demand of each household has been determined, a power flow simulation is run to determine the system voltage. In VGI, OpenDSS is used to calculate power flows: OpenDSS is an open-source software for DS analysis, that allow power flow calculation, transient analysis and fault analysis (more information is available here).
The agent must use the available information, i.e. observations and reward to formulate an action policy. In VGI, the action policy is a set of
h=96
variables, representing the charging strategy for the EVs in each household, i.e.
\{P^{EV}_t\}_h
. If a household, with index,
h
is not equipped with an EV than any variable at index
h
of the array will not be considered towards the formulation of the total demand.
At each simulated time step, the following steps are undertaken by the environment:
Read action policy,
\{P^{EV}_t\}_h
, submitted by the agent.
Calculate total power demand of each household as
P^{TOT}_{b,p,h,t}=P^H_{h,t}+P^{EV}_{e,t}
Calculate total electricity demand at each phase
p
of bus
b
at current time step
t
P_{b,p,t}=\sum_{h}P^{TOT}_{b,p,h,t}, \forall p,b
Run OpenDSS to calculate voltage
V_t
Check if EVs are sufficiently charged;
E^{EV}_{e,t}\geq E^{d}_e
t=t^{d}_e,\:\forall e
? Calculate mismatch penalty.
Update prediction of all household electrcity demands
\{P^H_{h,\tau}\}_h
The reward is calculated based on the two key objectives of VGI, namely system voltages and energy condition of the EVs prior to departure. The following expression is employed:
\Psi_t=\frac{\Delta_t+\Xi_t}{2}
\Delta_t
is the reward related to the voltage deviation in the DS, formulated as follows:
\Delta_t=\begin{cases} \max\left(0,\frac{1}{0.04}\left(\frac{V_t}{V^{nom}}-0.96\right)\right),\: \text{ if }\:V_t< V^{nom}\\ \max\left(0,\frac{1}{0.1}\left(1.1-\frac{V_t}{V^{nom}}\right)\right) ,\: \text{ if } \:V_t\geq V^{nom} \end{cases}
V^{nom}
is the nominal voltage of 1 per unit. The above expression enforces the statutory voltage limits: if the voltage exceeds the limits
[0.96,1.1]
then the associated reward is zero. On the other hand, if the voltage is within the said limits, then the reward is porportional to the associated deviation from
V^{nom}
. It can be seen that the voltage reward can assume utmost unitary value.
\Xi_t
is the reward associated to the satisfaction of the charging requirement fro travelling, as expressed hereby:
\Xi_t = - \sum_{i \in\mathcal{D}} \max\left(0,\frac{E^{d}_i-E^{EV}_{i,t}}{E^{d}_i}\right),
\mathcal{D}=\left\{e|t^{d}_e\leq t\right\}
is the set of indeces identifying the EVs that have not yet departed at the current time step. The reward formulated above provides a penalty if the energy stored in a departing EV is lower than the energy required for the trip. As also the energy related reward can assume utmost unitary value, the total reward can be utmost unitary.
Docker (get it here)
git clone https://gitlab.com/rangl/challenges/vgi.git
Run tests to ensure the correct operation and parameters
cd vgi/envs
make clean && make test
Install the custom gym environment
./example.py
Training and evaluation of models
Currently, the most recent branch is linearised_voltage, to which I am refering in these notes. Specifically, we are using the code in two files: src/example.py and src/plots.py.
The code that can be used for getting started with training is located in the file src/example.py, which needs to be executed from within docker as explained above. Because of some permission issues on the linux/docker line I couldn't put various RL and heuristic agents into septate files so depending on what you want to do you must uncomment an appropriate section in src/example.py.
Currently, it has some introductory code (section 0.) and clearly marked 4 sections: 1. training, 2. evaluation, 3. do-nothing agent, 4. full-charging agent. If you want to play with it, execute one secton at a time.
We are using RL algorithms implemented in the library stable_baselines such as ACKTR. We also used TRPO in the previous challenge (OPF).
While debugging I set N_STEPS = 12 to speed up learning, taking into account that we are mostly interested in what happens before the vehicles departs. Later on we may consider one whole day by setting N_STEPS = 48. Also, currently the network consists of 1 house only.
We crete an environment, reset it and record its initial state as observation by executing
env = gym.make("vgi-v0")
We specify the number of epochs (days) that we want to use for training e.g. STEPS_MULT. It is beneficial to check how the model is performing every now and then, so after the learning is completed we simulate N_test days, save the rewards in a data frame and pickle the data frame. The whole process is repeated, for instance N_train times. If we take
STEPS_MULT = 100
the procedure takes about 1h to complete. In summary, with these values, we train the model on 4000 days, evaluating the model every 100 days using a sample of 10 days.
We select a model from stable_baselines e.g.
model = ACKTR(MlpPolicy, env, verbose=1)
The training is performed using the following code. Note that the trained models are saved periodically.
rewards_df = pd.DataFrame(columns = ['iteration_'+ str(x) for x in range(N_test)])
model.learn(total_timesteps=TOTAL_TIMESTEPS)
model.save('TRPO_'+str(i)) # can't choose a different directory because of the issues with docker
rewards_list = []
for j in range(N_test):
rewards_list.append(run())
rewards_df.loc[i] = rewards_list
print(np.mean(rewards_list))
rewards_df.to_pickle('rewards_df')
Plots of the learning curve
We can now plot the learning curve, which gives average reward versus number of iterations and enables us to see whether the agent was improving during the learning process or not. This can be plotted using the code in src/plots.py (which I normally execute outside the docker container in my Python editor's console).
rewards_df = pd.read_pickle(current_directory + '/vgi/src/rewards_df')
plt.plot(rewards_df.mean(axis=1))
plt.xlabel('number of iterations x100')
plt.ylabel('mean reward')
Agent sanity check
The next step is to see what actions is the trained agent performing and what effects on the environment they have. In the file src/example.py, use the following to read one of the models saved during training
model = ACKTR.load('TRPO_39.zip')
The function observations_data_frame_model() simulates one day and outputs 3 data frames: observations, actions, rewards. Additionally the two components of the reward can be directly read from the environment:
rewards['voltage_rewards'] = env.voltage_rewards
rewards['EV_rewards'] = env.EV_rewards
The three data frames (observations, actions, rewards) are then saved. Once more, outside the container (code in the file src/plots.py) the data frames are loaded and the plot() can be executed producing 6 graphs, which show the most important elements of the environment:
state of the EV's battery
E_{e,t}^{EV}
E_e^d
and predicted departure time
t_e^d
\Phi_t
and the EV part of the reward
\Xi_t
V_t
hausehold demand in the next step
P_{h,t+1}^H
actions taken by the agent.
observations = pd.read_pickle(current_directory + '/vgi/src/observations_random_agent')
actions = pd.read_pickle(current_directory + '/vgi/src/actions_random_agent')
rewards = pd.read_pickle(current_directory + '/vgi/src/rewards_random_agent')
plot(observations,actions,rewards,0,True)
Comparing with the do-nothing agent
It is instructive (especially while debugging the environment) to compare the reward and overall behaviour of the environment with an agent which does nothing all the time i.e. submits action equal to 0 at all times. This can me done in a similar manner as evaluating the trained RL agent. In the file src/example.py the function observations_data_frame_zero() needs to be executed and the observations, actions, rewards data frames need to be saved. Then as before, the plot() function in src/plot.py produces 6 graphs of various components of the environment and calculates the cumulative reward (to be compared with the RL agent).
Comparing with the max-charging agent
Another simple agent, which can be treated as a benchmark is the one which submits maximal action (equal to 3) at all times, regardless of whether the EV requires charging or not. Similarly as before, in the file src/example.py the function observations_data_frame_full() needs to be executed and the observations, actions, rewards data frames need to be saved. Then the plot() function in src/plot.py should be executed and the results compared with the RL and do-nothing agents). |
Wikizero - Perimeter
In geometry, perimeter is the distance around a flat object. For example, all four sides of a square rhombus have the same length, so a rhombus with side length 2 inches would have a perimeter of 8 inches (2+2+2+2=8).
For a polygon, the perimeter is simply the sum of the length of all of its sides.[1] For a rectangle, the perimeter is twice the sum of its length and width (
{\displaystyle P=2\ell +2w}
).[2] Perimeter can also be calculated for other planar figures, such as circle, sector and ellipse.[3]
Real-life objects have perimeters as well. A football field, including the end zones, is 360 feet long and 160 feet wide. This means that the perimeter of the field is 360+160+360+160=1040 feet.
The perimeter of a circle is usually called the circumference.[3] It may be calculated by multiplying the diameter times "Pi". Pi is a constant which is approximately equal to 3.14159; however, the places to the right of the decimal are endless. The number of places used depends on the accuracy required for the result.
↑ "Perimeter and Area". www.montereyinstitute.org. Retrieved 2020-09-25.
↑ 3.0 3.1 "Perimeter". www.mathsisfun.com. Retrieved 2020-09-25.
Retrieved from "https://simple.wikipedia.org/w/index.php?title=Perimeter&oldid=7163710" |
On the ECI and CEI of Boron-Nitrogen Fullerenes
On the ECI and CEI of Boron-Nitrogen Fullerenes ()
School of Mathematics and Statistics, Qinghai Nationalities University, Xining, China.
The eccentric connectivity index and connective eccentricity index are important topological indices for chemistry. In this paper, we investigate the eccentric connectivity index and connective eccentricity index of boron-nitrogen fullerenes, respectively. And we give computing formulas of eccentric connectivity index and connective eccentricity index of all boron-nitrogen fullerenes with regular structure.
Eccentric Connectivity Index, Connective Eccentricity Index, Boron-Nitrogen Fullerenes
Wu, T. and Wu, Y. (2018) On the ECI and CEI of Boron-Nitrogen Fullerenes. Applied Mathematics, 9, 891-896. doi: 10.4236/am.2018.98061.
All graphs considered in this paper are simple connected graphs. Let G be a graph with vertex set
V\left(G\right)
and edge set
E\left(G\right)
v\in V\left(G\right)
d\left(v\right)
denotes the degree of v. For vertices
u,v\in V\left(G\right)
d\left(u,v\right)
is defined as the length of a shortest path between u and v in G. The eccentricity
\epsilon \left(v\right)
of a vertex v is the maximum distance from v to any other vertex.
The chemical information derived through the topological index has been found useful in chemical documentation, isomer discrimination, structure-property correlations, etc. For quite some time there has been rising interest in the field of computational chemistry in topological indices. The interest in topological indices is mainly related to their use in nonempirical quantitative structure-property relationships and quantitative structure-activity relationships. Among various indices, the eccentric connectivity index and connective eccentricity index involving eccentricity have attracted much attention.
Sharma et al. [1] introduced a distance-based molecular structure descriptor, the eccentric connectivity index (ECI for short) defined as
{\xi }^{c}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }d\left(v\right)\epsilon \left(v\right).
A novel graph invariant for predicting biological and physical properties―connective eccentricity index (CEI briefly) was introduced by Gupta et al. [8] , which was defined as:
{\xi }^{ce}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }\frac{d\left(v\right)}{\epsilon \left(v\right)}.
Some recent results on the CEI of graphs can be found in [9] - [14] .
Boron-nitrogen fullerene is a member of the fullerene family, it has been extensively studied [15] [16] [17] . A boron-nitrogen fullerene is also called (4,6)-fullerene graph which is a plane cubic graph whose faces have sizes 4 and 6. Let G be a (4,6)-fullerene graph with n vertices. By Euler’s formula, G has exactly
six faces of size 4 and
\frac{n}{2}-4
faces of size 6.
A (4,6)-fullerene graph is said to be of dispersive structure if has neither three squares arranged in a line nor a square-cap. According to the quadrilateral positional relationship, Wei and Zhang [18] given a classification of all (4,6)-fullerenes as follows.
Lemma 1 [18] Let G be a (4,6)-fullerene graph. Then G can be of one of the following five types:
1) a cube,
2) a hexagonal prism,
3) a tubular graph with at least one hexagon-layer,
4) a (4,6)-fullerene graph of lantern structure,
5) a (4,6)-fullerene graph of dispersive structure,
where the resulting graphs see Figure 1.
In this paper, we aim to investigate the ECI and CEI of a boron-nitrogen fullerene. In next section, we give computing formulas of ECI and CEI of a boron-nitrogen fullerene.
2. Computing Formula of ECI and CEI of a (4,6)-Fullerene
In this section, according the classification of all boron-nitrogen fullerenes in Lemma 1, we will give computing formula of CEI of a (4,6)-fullerene.
Theorem 2 Let G be a (4,6)-fullerene graph. Then
{\xi }^{ce}\left(G\right)=\left(\begin{array}{ll}8\hfill & \text{if}G\text{isacube},\hfill \\ 9\hfill & \text{if}G\text{ishexagonalprism}.\hfill \end{array}
{\xi }^{c}\left(G\right)=\left(\begin{array}{ll}72\hfill & \text{if}G\text{isacube},\hfill \\ 144\hfill & \text{if}G\text{ishexagonalprism}.\hfill \end{array}
Figure 1. The (4,6)-fullerenes in Lemma 1.
Proof. Let G be a cube. Checking the structure of G, we obtain that the eccentricity of every vertex in G is 3. By (1) and (2), we have
{\xi }^{ce}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }\frac{d\left(v\right)}{\epsilon \left(v\right)}=\underset{v\in V\left(G\right)}{\sum }\frac{3}{3}=8,
{\xi }^{c}\left(G\right)={\displaystyle \underset{v\in V\left(G\right)}{\sum }}d\left(v\right)\epsilon \left(v\right)={\displaystyle \underset{v\in V\left(G\right)}{\sum }}3\times 3=72.
Similarly, let G be a hexagonal prism. Checking the structure of G, we obtain that the eccentricity of every vertex in G is 4. By (1) and (2), we obtain
{\xi }^{ce}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }\frac{d\left(v\right)}{\epsilon \left(v\right)}=\underset{v\in V\left(G\right)}{\sum }\frac{3}{4}=9,
{\xi }^{c}\left(G\right)={\displaystyle \underset{v\in V\left(G\right)}{\sum }}d\left(v\right)\epsilon \left(v\right)={\displaystyle \underset{v\in V\left(G\right)}{\sum }}3\times 4=144.
Theorem 3 Let G be a tubular graph, and let the number of hexagon-layer in G be
k\ge 1
{\xi }^{ce}\left(G\right)=18\underset{i=3}{\overset{k+4}{\sum }}\frac{1}{k+i}+\frac{6}{2k+5};
{\xi }^{c}\left(G\right)=27{k}^{2}+129k+156.
Proof. Let v be a vertex of G. By the structure of G (see Figure 1), we know that the eccentricity of v equal to the distance between v and u (or u’). Thus, the eccentricity sequence of G is (
\stackrel{2}{\stackrel{︷}{2k+5,2k+5}}
\stackrel{6}{\stackrel{︷}{2k+4,\cdots ,2k+4}}
\stackrel{6}{\stackrel{︷}{2k+3,\cdots ,2k+3}}
\cdots
\stackrel{6}{\stackrel{︷}{k+4,\cdots ,k+4}}
\stackrel{6}{\stackrel{︷}{k+3,\cdots ,k+3}}
). By (1) and (2), we obtain
{\xi }^{ce}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }\frac{d\left(v\right)}{\epsilon \left(v\right)}
=3\left[\left(\frac{1}{2k+4}+\frac{1}{2k+3}+\cdots +\frac{1}{k+3}\right)\times 6+\frac{1}{2k+5}\times 2\right]
=18\underset{i=3}{\overset{k+4}{\sum }}\frac{1}{k+i}+\frac{6}{2k+5}.
{\xi }^{c}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }d\left(v\right)\epsilon \left(v\right)
=3\left[2\left(2k+5\right)+6\left[\left(k+3\right)+\left(k+4\right)+\cdots +\left(2k+4\right)\right]\right]
=3\left[2\left(2k+5\right)+6\frac{\left[\left(k+3\right)+\left(2k+4\right)\right]\left(k+2\right)}{2}\right]
=27{k}^{2}+129k+156.
Theorem 4 Let G be a (4,6)-fullerene graph having lantern structure, and let the number of hexagon-layer in G be
k\ge 1
, see Figure 1. Then
{\xi }^{ce}\left(G\right)=24\underset{i=3}{\overset{k+3}{\sum }}\frac{1}{k+i};
{\xi }^{c}\left(G\right)=36{k}^{2}+108k+72.
Proof. Checking G, we see that G is symmetry. By the symmetry of G, it is easy to obtain that the eccentricity sequence of G is (
\stackrel{8}{\stackrel{︷}{2k+3,\cdots ,2k+3}}
\stackrel{8}{\stackrel{︷}{2k+2,\cdots ,2k+2}}
\cdots
\stackrel{8}{\stackrel{︷}{k+4,\cdots ,k+4}}
\stackrel{8}{\stackrel{︷}{k+3,\cdots ,k+3}}
). By (1) and (2), we have
{\xi }^{ce}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }\frac{d\left(v\right)}{\epsilon \left(v\right)}
=3\left[\left(\frac{1}{2k+3}+\frac{1}{2k+2}+\cdots +\frac{1}{k+3}\right)\times 8\right]
=24\underset{i=3}{\overset{k+3}{\sum }}\frac{1}{k+i}.
{\xi }^{c}\left(G\right)=\underset{v\in V\left(G\right)}{\sum }d\left(v\right)\epsilon \left(v\right)
=3\left[8\left[\left(k+3\right)+\left(k+4\right)+\cdots +\left(2k+2\right)+\left(2k+3\right)\right]\right]
=3\left[8\frac{\left[\left(k+3\right)+\left(2k+3\right)\right]\left(k+1\right)}{2}\right]
=36{k}^{2}+108k+72.
The proof of the theorem is now complete. □
In this note, we give calculation formulas of ECI and CEI of all (4,6)-fullerenes with regular structures, respectively. In the future, we will discuss the calculation of ECI and CEI of a (4,6)-fullerene graph of dispersive structure, and try to give an algorithm.
The work is supported by the Natural Science Foundation of Qinghai Province (2016-ZJ-947Q), the Ministry of Education Chunhui Project (No. Z2017047) and the High-level Personnel of Scientific Research Project of QHMU (2016XJG07).
[1] Sharma, V., Goswami, R. and Madan, A.K. (1997) Eccentric Connectivity index: A Novel Highly Discriminating Topological Descriptor for Structure-Property and Structure-Activity Studies. Journal of Chemical Information and Computer Sciences, 37, 273-282.
[2] Ashrafi, A.R., Doslic, T. and Saheli, M. (2011) The Eccentric Connectivity Index of TUC4C8(R) Nanotubes. MATCH Communications in Mathematical and in Computer Chemistry, 65, 221-230.
[3] Doslic, T., Saheli, M. and Vukicevic, D. (2010) Eccentric Connectivity Index: Extremal Graphs and Values. Iranian Journal of Mathematical Chemistry, 1, 45-56.
[4] Dureja, H., Gupta, S. and Madan, A.K. (2008) Predicting Anti-HIV-1 Activity of 6-Arylbenzonitriles: Computational Approach Using Superaugmented Eccentric Connectivity Topochemical Indices. Journal of Molecular Graphics, 26, 1020-1029.
[5] Yu, G., Feng, L. and Ilic, A. (2011) On the Eccentric Distance Sum of Trees and unicyclic Graphs. Journal of Mathematical Analysis and Applications, 375, 99-107.
[6] Zhang, J., Zhou, B. and Liu, Z. (2012) On the Minimal Eccentric Connectivity Indices of Graphs. Discrete Mathematics, 312, 819-829.
[7] Zhou, B. and Du, Z. (2010) On Eccentric Connectivity Index. MATCH Communications in Mathematical and in Computer Chemistry, 63, 181-198.
[8] Gupta, S., Singh, M. and Madan, A.K. (2000) Connective Eccentricity Index: A Novel Topological Descriptor for Predicting Biological Activity. Journal of Molecular Graphics and Modelling, 18, 18-25.
[9] Azari, M. and Iranmanesh, A. (2013) Computing the Eccentric-Distance Sum for Graph Operations. Discrete Applied Mathematics, 161, 2827-2840.
[10] Li, S., Zhang, M., Yu, G. and Feng, L. (2012) On the Extremal Values of the eccentric Distance Sum of Trees. Journal of Mathematical Analysis and Applications, 390, 99-112.
[11] Yu, G. and Feng, L. (2013) On the Connective Eccentricity Index of Graphs. MATCH Communications in Mathematical and in Computer Chemistry, 69, 611-628.
[12] Yu, G., Qu, H., Tang, L. and Feng, L. (2014) On the Connective Eccentricity Index of Trees and Unicyclic Graphs with Given Diameter. Journal of Mathematical Analysis and Applications, 420, 1776-1786.
[13] Xu, K., Das, K. and Liu, H. (2016) Some Extremal Results on the Connective Eccentricity Index of Graphs. Journal of Mathematical Analysis and Applications, 433, 803-817.
[14] Xu, K., Liu, M., Das, K., Gutman, I. and Furtula, B. (2014) A Survey on Graphs Extremal with Respect to Distance-Based Topological Indices. MATCH Communications in Mathematical and in Computer Chemistry, 71, 461-508.
[15] Gao, Y. and Zhang, H. (2014) Clar Structure and Fries Set of Fullerenes and (4,6)-Fullerenes. Journal of Applied Mathematics, 2014, Article ID: 196792.
[16] Jiang, X. and Zhang, H. (2011) On Forcing Matching Number of Boron-Nitrogen Fullerene Graphs. Discrete Applied Mathematics, 159, 1581-1593.
[17] Zhang, H. and Liu, S. (2010) 2-Resonance of Plane Bipartite Graphs and Its Applications to Boron-Nitrogen Fullerenes. Discrete Applied Mathematics, 158, 1559-1569.
[18] Wei, Z. and Zhang, H. (2017) 2-Resonance of Plane Bipartite Graphs and Its Applications to Boron-Nitrogen Fullerenes. MATCH Communications in Mathematical and in Computer Chemistry, 77, 707-724. |
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6×{10}^{8}
{5}^{n}
{2}^{n}
975436×{5}^{4}
\frac{9754360000}{{2}^{4}}
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\left({x}^{n}-{a}^{n}\right)
\left({x}^{n}+{a}^{n}\right)
\frac{n}{2}\left[2a+\left(n-1\right)d\right]
\frac{n}{2}\left[a+l\right]
\left(1+2+3+...+n\right)=\frac{1}{2}n\left(n+1\right)
\left({1}^{2}+{2}^{2}+{3}^{2}+...+{n}^{2}\right)=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)
\left({1}^{3}+{2}^{3}+{3}^{3}+...+{n}^{3}\right)=\frac{1}{4}{n}^{2}\left(n+1\right){ }^{2}
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\left\{\begin{array}{ll}\frac{a\left(1-{r}^{n}\right)}{\left(1-r\right)}& where r<1\\ \frac{a\left({r}^{n}-1\right)}{\left(r-1\right)}& where r>1\end{array}\right\
If one-fifth of one-fourth of a number is 35, then what is seven-eighth of that number?
A) 624.5 B) 612.5
C) 723.5 D) 715
Answer & Explanation Answer: B) 612.5
If a nine-digit number 1263487xy is divisible by both 8 and 5, then the greatest possible values of x and y, respectively, are
Answer & Explanation Answer: C) 6 and 0
If the number 687x29 is divisible by 9, then the value of 2x is:
The sum of 17 consecutive numbers is 289. The sum of another 10 consecutive numbers, whose first term is 5 more than the average of the first set of consecutive numbers, is:
If the sum of two numbers is 11 and the sum of their squares is 65, then the sum of their cubes will be:
What number must be added to each of the numbers 8, 13, 26 and 40 so that the numbers obtained in this order are in proportion? |
Friedmann–Lemaître–Robertson–Walker metric - formulasearchengine
{{#invoke:Hatnote|hatnote}}Template:Main other {{#invoke: Sidebar | collapsible }}
The Friedmann–Lemaître–Robertson–Walker (FLRW) metric is an exact solution of Einstein's field equations of general relativity; it describes a homogeneous, isotropic expanding or contracting universe that may be simply connected or multiply connected.[1][2][3] (If multiply connected, then each event in spacetime will be represented by more than one tuple of coordinates.) The general form of the metric follows from the geometric properties of homogeneity and isotropy; Einstein's field equations are only needed to derive the scale factor of the universe as a function of time. Depending on geographical or historical preferences, a subset of the four scientists — Alexander Friedmann, Georges Lemaître, Howard P. Robertson and Arthur Geoffrey Walker — may be named (e.g., Friedmann–Robertson–Walker (FRW) or Robertson–Walker (RW) or Friedmann–Lemaître (FL)). This model is sometimes called the Standard Model of modern cosmology.[4] It was developed independently by the named authors in the 1920s and 1930s.
1 General metric
1.1 Reduced-circumference polar coordinates
1.2 Hyperspherical coordinates
2.3 Newtonian interpretation
4 Einstein's radius of the universe
The FLRW metric starts with the assumption of homogeneity and isotropy of space. It also assumes that the spatial component of the metric can be time-dependent. The generic metric which meets these conditions is
{\displaystyle -c^{2}\mathrm {d} \tau ^{2}=-c^{2}\mathrm {d} t^{2}+{a(t)}^{2}\mathrm {d} \mathbf {\Sigma } ^{2}}
{\displaystyle \mathbf {\Sigma } }
ranges over a 3-dimensional space of uniform curvature, that is, elliptical space, Euclidean space, or hyperbolic space. It is normally written as a function of three spatial coordinates, but there are several conventions for doing so, detailed below.
{\displaystyle \mathrm {d} \mathbf {\Sigma } }
does not depend on t — all of the time dependence is in the function a(t), known as the "scale factor".
Reduced-circumference polar coordinates
In reduced-circumference polar coordinates the spatial metric has the form
{\displaystyle \mathrm {d} \mathbf {\Sigma } ^{2}={\frac {\mathrm {d} r^{2}}{1-kr^{2}}}+r^{2}\mathrm {d} \mathbf {\Omega } ^{2},\quad {\text{where }}\mathrm {d} \mathbf {\Omega } ^{2}=\mathrm {d} \theta ^{2}+\sin ^{2}\theta \,\mathrm {d} \phi ^{2}.}
k is a constant representing the curvature of the space. There are two common unit conventions:
k may be taken to have units of length−2, in which case r has units of length and a(t) is unitless. k is then the Gaussian curvature of the space at the time when a(t) = 1. r is sometimes called the reduced circumference because it is equal to the measured circumference of a circle (at that value of r), centered at the origin, divided by 2π (like the r of Schwarzschild coordinates). Where appropriate, a(t) is often chosen to equal 1 in the present cosmological era, so that
{\displaystyle \mathrm {d} \mathbf {\Sigma } }
measures comoving distance.
Alternatively, k may be taken to belong to the set {−1,0,+1} (for negative, zero, and positive curvature respectively). Then r is unitless and a(t) has units of length. When k = ±1, a(t) is the radius of curvature of the space, and may also be written R(t).
A disadvantage of reduced circumference coordinates is that they cover only half of the 3-sphere in the case of positive curvature—circumferences beyond that point begin to decrease, leading to degeneracy. (This is not a problem if space is elliptical, i.e. a 3-sphere with opposite points identified.)
In hyperspherical or curvature-normalized coordinates the coordinate r is proportional to radial distance; this gives
{\displaystyle \mathrm {d} \mathbf {\Sigma } ^{2}=\mathrm {d} r^{2}+S_{k}(r)^{2}\,\mathrm {d} \mathbf {\Omega } ^{2}}
{\displaystyle \mathrm {d} \mathbf {\Omega } }
is as before and
{\displaystyle S_{k}(r)={\begin{cases}{\sqrt {k}}^{\,-1}\sin(r{\sqrt {k}}),&k>0\\r,&k=0\\{\sqrt {|k|}}^{\,-1}\sinh(r{\sqrt {|k|}}),&k<0.\end{cases}}}
As before, there are two common unit conventions:
k may be taken to have units of length−2, in which case r has units of length and a(t ) is unitless. k is then the Gaussian curvature of the space at the time when a(t ) = 1. Where appropriate, a(t ) is often chosen to equal 1 in the present cosmological era, so that
{\displaystyle \mathrm {d} \mathbf {\Sigma } }
Alternatively, as before, k may be taken to belong to the set {−1,0,+1} (for negative, zero, and positive curvature respectively). Then r is unitless and a(t ) has units of length. When k = ±1, a(t ) is the radius of curvature of the space, and may also be written R(t ). Note that, when k = +1, r is essentially a third angle along with θ and φ. The letter χ may be used instead of r.
Though it is usually defined piecewise as above, S is an analytic function of both k and r. It can also be written as a power series
{\displaystyle S_{k}(r)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}k^{n}r^{2n+1}}{(2n+1)!}}=r-{\frac {kr^{3}}{6}}+{\frac {k^{2}r^{5}}{120}}-\cdots }
{\displaystyle S_{k}(r)=r\;\mathrm {sinc} \,(r{\sqrt {k}})}
where sinc is the unnormalized sinc function and
{\displaystyle {\sqrt {k}}}
is one of the imaginary, zero or real square roots of k. These definitions are valid for all k.
When k = 0 one may write simply
{\displaystyle \mathrm {d} \mathbf {\Sigma } ^{2}=\mathrm {d} x^{2}+\mathrm {d} y^{2}+\mathrm {d} z^{2}.}
This can be extended to k ≠ 0 by defining
{\displaystyle x=r\cos \theta \,}
{\displaystyle y=r\sin \theta \cos \phi \,}
{\displaystyle z=r\sin \theta \sin \phi \,}
where r is one of the radial coordinates defined above, but this is rare.
Template:General relativity sidebar {{#invoke:main|main}}
Einstein's field equations are not used in deriving the general form for the metric: it follows from the geometric properties of homogeneity and isotropy. However, determining the time evolution of
{\displaystyle a(t)}
does require Einstein's field equations together with a way of calculating the density,
{\displaystyle \rho (t),}
such as a cosmological equation of state.
This metric has an analytic solution to Einstein's field equations
{\displaystyle G_{\mu \nu }+\Lambda g_{\mu \nu }={\frac {8\pi G}{c^{4}}}T_{\mu \nu }}
giving the Friedmann equations when the energy-momentum tensor is similarly assumed to be isotropic and homogeneous. The resulting equations are:[5]
{\displaystyle \left({\frac {\dot {a}}{a}}\right)^{2}+{\frac {kc^{2}}{a^{2}}}-{\frac {\Lambda c^{2}}{3}}={\frac {8\pi G}{3}}\rho }
{\displaystyle 2{\frac {\ddot {a}}{a}}+\left({\frac {\dot {a}}{a}}\right)^{2}+{\frac {kc^{2}}{a^{2}}}-\Lambda c^{2}=-{\frac {8\pi G}{c^{2}}}p.}
These equations are the basis of the standard big bang cosmological model including the current ΛCDM model. Because the FLRW model assumes homogeneity, some popular accounts mistakenly assert that the big bang model cannot account for the observed lumpiness of the universe. In a strictly FLRW model, there are no clusters of galaxies, stars or people, since these are objects much denser than a typical part of the universe. Nonetheless, the FLRW model is used as a first approximation for the evolution of the real, lumpy universe because it is simple to calculate, and models which calculate the lumpiness in the universe are added onto the FLRW models as extensions. Most cosmologists agree that the observable universe is well approximated by an almost FLRW model, i.e., a model which follows the FLRW metric apart from primordial density fluctuations. Template:As of, the theoretical implications of the various extensions to the FLRW model appear to be well understood, and the goal is to make these consistent with observations from COBE and WMAP.
The pair of equations given above is equivalent to the following pair of equations
{\displaystyle {\dot {\rho }}=-3{\frac {\dot {a}}{a}}\left(\rho +{\frac {p}{c^{2}}}\right)}
{\displaystyle {\frac {\ddot {a}}{a}}=-{\frac {4\pi G}{3}}\left(\rho +{\frac {3p}{c^{2}}}\right)+{\frac {\Lambda c^{2}}{3}}}
{\displaystyle k}
, the spatial curvature index, serving as a constant of integration for the first equation.
The first equation can be derived also from thermodynamical considerations and is equivalent to the first law of thermodynamics, assuming the expansion of the universe is an adiabatic process (which is implicitly assumed in the derivation of the Friedmann–Lemaître–Robertson–Walker metric).
The second equation states that both the energy density and the pressure cause the expansion rate of the universe
{\displaystyle {\dot {a}}}
to decrease, i.e., both cause a deceleration in the expansion of the universe. This is a consequence of gravitation, with pressure playing a similar role to that of energy (or mass) density, according to the principles of general relativity. The cosmological constant, on the other hand, causes an acceleration in the expansion of the universe.
The cosmological constant term can be omitted if we make the following replacements
{\displaystyle \rho \rightarrow \rho +{\frac {\Lambda c^{2}}{8\pi G}}}
{\displaystyle p\rightarrow p-{\frac {\Lambda c^{4}}{8\pi G}}.}
Therefore the cosmological constant can be interpreted as arising from a form of energy which has negative pressure, equal in magnitude to its (positive) energy density:
{\displaystyle p=-\rho c^{2}.\,}
Such form of energy—a generalization of the notion of a cosmological constant—is known as dark energy.
In fact, in order to get a term which causes an acceleration of the universe expansion, it is enough to have a scalar field which satisfies
{\displaystyle p<-{\frac {\rho c^{2}}{3}}.\,}
Such a field is sometimes called quintessence.
Newtonian interpretation
The Friedmann equations are equivalent to this pair of equations:
{\displaystyle -a^{3}{\dot {\rho }}=3a^{2}{\dot {a}}\rho +{\frac {3a^{2}p{\dot {a}}}{c^{2}}}\,}
{\displaystyle {\frac {{\dot {a}}^{2}}{2}}-{\frac {G{\frac {4\pi a^{3}}{3}}\rho }{a}}=-{\frac {kc^{2}}{2}}\,.}
The first equation says that the decrease in the mass contained in a fixed cube (whose side is momentarily a) is the amount which leaves through the sides due to the expansion of the universe plus the mass equivalent of the work done by pressure against the material being expelled. This is the conservation of mass-energy (first law of thermodynamics) contained within a part of the universe.
The second equation says that the kinetic energy (seen from the origin) of a particle of unit mass moving with the expansion plus its (negative) gravitational potential energy (relative to the mass contained in the sphere of matter closer to the origin) is equal to a constant related to the curvature of the universe. In other words, the energy (relative to the origin) of a co-moving particle in free-fall is conserved. General relativity merely adds a connection between the spatial curvature of the universe and the energy of such a particle: positive total energy implies negative curvature and negative total energy implies positive curvature.
The cosmological constant term is assumed to be treated as dark energy and thus merged into the density and pressure terms.
During the Planck epoch, one cannot neglect quantum effects. So they may cause a deviation from the Friedmann equations.
The main results of the FLRW model were first derived by the Soviet mathematician Alexander Friedmann in 1922 and 1924. Although his work was published in the prestigious physics journal Zeitschrift für Physik, it remained relatively unnoticed by his contemporaries. Friedmann was in direct communication with Albert Einstein, who, on behalf of Zeitschrift für Physik, acted as the scientific referee of Friedmann's work. Eventually Einstein acknowledged the correctness of Friedmann's calculations, but failed to appreciate the physical significance of Friedmann's predictions.
Friedmann died in 1925. In 1927, Georges Lemaître, a Belgian priest, astronomer and periodic professor of physics at the Catholic University of Leuven, arrived independently at similar results as Friedmann and published them in Annals of the Scientific Society of Brussels. In the face of the observational evidence for the expansion of the universe obtained by Edwin Hubble in the late 1920s, Lemaître's results were noticed in particular by Arthur Eddington, and in 1930–31 his paper was translated into English and published in the Monthly Notices of the Royal Astronomical Society.
Howard P. Robertson from the US and Arthur Geoffrey Walker from the UK explored the problem further during the 1930s. In 1935 Robertson and Walker rigorously proved that the FLRW metric is the only one on a spacetime that is spatially homogeneous and isotropic (as noted above, this is a geometric result and is not tied specifically to the equations of general relativity, which were always assumed by Friedmann and Lemaître).
Because the dynamics of the FLRW model were derived by Friedmann and Lemaître, the latter two names are often omitted by scientists outside the US. Conversely, US physicists often refer to it as simply "Robertson–Walker". The full four-name title is the most democratic and it is frequently used.{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }} Often the "Robertson–Walker" metric, so-called since they proved its generic properties, is distinguished from the dynamical "Friedmann-Lemaître" models, specific solutions for a(t) which assume that the only contributions to stress-energy are cold matter ("dust"), radiation, and a cosmological constant.
Einstein's radius of the universe
Einstein's radius of the universe is the radius of curvature of space of Einstein's universe, a long-abandoned static model that was supposed to represent our universe in idealized form. Putting
{\displaystyle {\dot {a}}={\ddot {a}}=0}
in the Friedmann equation, the radius of curvature of space of this universe (Einstein's radius) is{{ safesubst:#invoke:Unsubst||date=__DATE__ |$B= {{#invoke:Category handler|main}}{{#invoke:Category handler|main}}[citation needed] }}
{\displaystyle R_{E}=c/{\sqrt {4\pi G\rho }}}
{\displaystyle c}
{\displaystyle G}
is the Newtonian gravitational constant, and
{\displaystyle \rho }
is the density of space of this universe. The numerical value of Einstein's radius is of the order of 1010 light years.
By combining the observation data from some experiments such as WMAP and Planck with theoretical results of Ehlers–Geren–Sachs theorem and its generalization,[6] astrophysicists now agree that the universe is almost homogeneous and isotropic (when averaged over a very large scale) and thus nearly a FLRW spacetime.
↑ For an early reference, see Robertson (1935); Robertson assumes multiple connectedness in the positive curvature case and says that "we are still free to restore" simple connectedness.
↑ See pp. 351ff. in {{#invoke:citation/CS1|citation |CitationClass=citation }}. The original work is Ehlers, J., Geren, P., Sachs, R.K.: Isotropic solutions of Einstein-Liouville equations. J. Math. Phys. 9, 1344 (1968). For the generalization, see {{#invoke:citation/CS1|citation |CitationClass=citation }}.
|CitationClass=citation }} English trans. in 'General Relativity and Gravitation' 1999 vol.31, 31–
|CitationClass=citation }}. (See Chapter 23 for a particularly clear and concise introduction to the FLRW models.)
|CitationClass=citation }} translated from {{#invoke:citation/CS1|citation |CitationClass=citation }}
Template:Relativity
Retrieved from "https://en.formulasearchengine.com/index.php?title=Friedmann–Lemaître–Robertson–Walker_metric&oldid=228787"
Coordinate charts in general relativity |
Decompose signal into high-frequency and low-frequency subbands - Simulink - MathWorks España
Two-Channel Analysis Subband Filter
Creating Multilevel Dyadic Analysis Filter Banks
Decompose signal into high-frequency and low-frequency subbands
The Two-Channel Analysis Subband Filter block decomposes the input into high-frequency and low-frequency subbands, each with half the bandwidth and half the sample rate of the input.
The block filters the input with a pair of highpass and lowpass FIR filters, and then downsamples the results by 2, as illustrated in the following figure.
The block implements the FIR filtering and downsampling steps together using a polyphase filter structure, which is more efficient than the straightforward filter-then-decimate algorithm shown in the preceding figure. Each subband is the first phase of the respective polyphase filter. You can implement a multilevel dyadic analysis filter bank by connecting multiple copies of this block or by using the Dyadic Analysis Filter Bank block. See Creating Multilevel Dyadic Analysis Filter Banks for more information.
You must provide a vector of filter coefficients for the lowpass and highpass FIR filters. Each filter should be a half-band filter that passes the frequency band that the other filter stops.
H\left(z\right)=B\left(z\right)={b}_{1}+{b}_{2}{z}^{-1}+\dots +{b}_{m}{z}^{-\left(m-1\right)}
Each filter should be a half-band filter that passes the frequency band that the other filter stops. You can use the Two-Channel Synthesis Subband Filter block to reconstruct the input to this block. To do so, you must design perfect reconstruction filters to use in the synthesis subband filter.
The best way to design perfect reconstruction filters is to use the Wavelet Toolbox™ wfilters function in to design both the filters both in this block and in the Two-Channel Synthesis Subband Filter block. You can also use other DSP System Toolbox™ and Signal Processing Toolbox™ functions.
The Two-Channel Analysis Subband Filter block initializes all filter states to zero.
When you set the Input processing parameter to Columns as channels (frame based), the block accepts an M-by-N matrix. The block treats each column of the input as the high- or low-frequency subbands of the corresponding output channel. You can use the Rate options parameter to specify how the block resamples the input:
When you set the Rate options parameter to Enforce single-rate processing, the input to the block can be an M-by-N matrix, where M is a multiple of two. The block treats each column of the input as an independent channel and decomposes each channel over time. The block outputs two matrices, where each column of the output is the high- or low-frequency subband of the corresponding input column. To maintain the input sample rate, the block decreases the output frame size by a factor of two.
When you set the Rate options parameter to Allow multirate processing, the block treats an Mi-by-N matrix input as N independent channels and decomposes each channel over time. The block outputs two M-by-N matrices, where each column of the output is the high- or low-frequency subband of the corresponding input column. The input and output frame sizes are the same, but the frame rate of the output is half that of the input. Thus, the overall sample rate of the output is half that of the input.
When you set the Input processing parameter to Elements as channels (sample based), the block treats an M-by-N matrix input as M · N independent channels. The block decomposes each channel over time and outputs two M-by-N matrices whose sample rates are half the input sample rate. Each element in the output matrix is the high- or low-frequency subband output of the corresponding element of the input matrix.
When you set the Input processing parameter to Columns as channels (frame based) and the Rate options parameter to Enforce single-rate processing, the Two-Channel Analysis Subband Filter block always has zero-tasking latency. Zero-tasking latency means that the block propagates the first input sample (received at time t=0) as the first output sample.
When you set the Rate options parameter to Allow multirate processing, the Two-Channel Analysis Subband Filter block may exhibit latency. The amount of latency depends on the setting of the Input processing parameter of this block, and the setting of the Simulink Treat each discrete rate as a separate task configuration parameter. The following table summarizes the conditions that produce latency when the block is performing multirate processing.
The Two-Channel Analysis Subband Filter block is the basic unit of a dyadic analysis filter bank. You can connect several of these blocks to implement an n-level filter bank, as illustrated in the following figure. For a review of dyadic analysis filter banks, see the Dyadic Analysis Filter Bank block reference page.
The frame size of the signal you are decomposing is a multiple of 2n.
You are decomposing the signal into n+1 or 2n subbands.
In all other cases, use Two-Channel Analysis Subband Filter blocks to implement your filter banks.
The Dyadic Analysis Filter Bank block allows you to specify the filter bank filters by providing vectors of filter coefficients, just as this block does. The Dyadic Analysis Filter Bank block provides an additional option of using wavelet-based filters that the block designs by using a wavelet you specify.
The Two-Channel Analysis Subband Filter Bank block is composed of two FIR Decimation blocks as shown in the following diagram.
For fixed-point signals, you can set the coefficient, product output, accumulator, and output data types of the FIR Decimation blocks as discussed in Parameters. For a diagram showing the usage of these data types, see the FIR Decimation block reference page.
Specify a vector of lowpass FIR filter coefficients, in descending powers of z. The lowpass filter should be a half-band filter that passes the frequency band stopped by the filter specified in the Highpass FIR filter coefficients parameter. The default values of this parameter specify a filter based on a third-order Daubechies wavelet. When you use the Two-Channel Synthesis Subband Filter block to reconstruct the input to this block, you need to design perfect reconstruction filters to use in the synthesis subband filter. For more information, see Specifying the FIR Filters.
Specify a vector of highpass FIR filter coefficients, in descending powers of z. The highpass filter should be a half-band filter that passes the frequency band stopped by the filter specified in the Lowpass FIR filter coefficients parameter. The default values of this parameter specify a filter based on a third-order Daubechies wavelet. When you use the Two-Channel Synthesis Subband Filter block to reconstruct the input to this block, you need to design perfect reconstruction filters to use in the synthesis subband filter. For more information, see Specifying the FIR Filters.
Enforce single-rate processing — When you select this option, the block treats each column of the input as an independent channel and decomposes each channel over time. The output has the same sample rate as the input, but the output frame size is half that of the input frame size. To select this option, you must set the Input processing parameter to Columns as channels (frame based).
Allow multirate processing — When you select this option, the input and output of the block are the same size, but the sample rate of the output is half that of the input.
Select the rounding mode for fixed-point operations. The filter coefficients do not obey this parameter; they are always rounded to Nearest.
DWT | Dyadic Analysis Filter Bank | FIR Decimation | IDWT | Two-Channel Synthesis Subband Filter |
Extract endmember signatures using N-FINDR - MATLAB nfindr - MathWorks India
nfindr
Extract Endmembers Using N-FINDR Method
Set N-FINDR Parameter Values for Finding Endmembers
Extract endmember signatures using N-FINDR
endmembers = nfindr(inputData,numEndmembers)
endmembers = nfindr(inputData,numEndmembers,Name,Value)
endmembers = nfindr(inputData,numEndmembers) extracts endmember signatures from hyperspectral data cube by using the N-finder (N-FINDR) algorithm. numEndmembers is the number of endmember signatures to be extracted using N-FINDR algorithm. For more information about the N-FINDR method, see Algorithms.
endmembers = nfindr(inputData,numEndmembers,Name,Value) specifies options using one or more name-value pair arguments in addition to the input arguments in the previous syntax. Use this syntax to set the options for number of iterations and dimensionality reduction.
Compute the endmembers using the N-FINDR method. By default, the nfindr function uses maximum noise fraction (MNF) transform for preprocessing. The default value for number of iterations is 3 times the number of estimated endmembers.
endmembers = nfindr(hcube.DataCube,numEndmembers);
Compute the endmembers using the N-FINDR method. Specify the value for number of iterations as 1000. Select principal component analysis (PCA) as the dimensionality reduction method for preprocessing.
endmembers = nfindr(hcube.DataCube,numEndmembers,'NumIterations',1000,'ReductionMethod','PCA');
Input hyperspectral data, specified as an 3-D numeric array or a hypercube object. If the input is a hypercube object, then the function reads the hyperspectral data from its DataCube property.
The hyperspectral data is an numeric array of size M-by-N-by-C. M and N are the number of rows and columns in the hyperspectral data respectively. C is the number of spectral bands in the hyperspectral data.
Number of endmembers to extract, specified as a positive scalar integer. The value must be in the range [1 C]. C is the number of spectral bands in the input hyperspectral data.
Example: nfindr(cube,7,'NumIterations',100,'Method','None')
3P (default) | positive scalar integer
Number of iterations, specified as a positive scalar integer. The default value is 3P. P is the number of endmember signatures to be extracted. The computation time of the algorithm increases with the increase in the number of iterations.
'MNF' (default) | 'PCA'
Dimensionality reduction method, specified as one of these values:
'MNF' — To perform dimensionality reduction by using the maximum noise fraction (MNF) method. This is the default.
'PCA' — To perform dimensionality reduction by using the principal component analysis (PCA) method.
If you specify this argument, the function first reduces the spectral dimension of the input data by using the specified method. Then, it computes the endmember signatures from the reduced data.
Endmember signatures, returned as a matrix of size C-by-P and datatype same as the datatype of the input hyperspectral data.
N-FINDR is an iterative approach for finding the endmembers of a hyperspectral data. The method assumes that the volume of a simplex formed by the endmembers (purest pixels) is larger than any other volume defined by any other combination of pixels [1]. The steps involved are as follows:
Compute principal component bands and reduce the spectral dimensionality of the input data by using MNF or PCA. The number of principal component bands to be extracted is set equal to the number of endmembers to be extracted. The endmembers are extracted from the principal component bands.
Randomly select n number of pixel spectra from the reduced data as initial set of endmembers.
For iteration 1, denote the initial set of endmembers as
\text{\hspace{0.17em}}\left\{{\text{e}}_{1}^{\left(1\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{2}^{\left(1\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{p}^{\left(1\right)}\right\}
Consider the endmembers as vertices of a simplex and compute the volume by using
\text{\hspace{0.17em}}V\left({\text{E}}^{\left(1\right)}\right)\text{\hspace{0.17em}}=\text{\hspace{0.17em}}|\mathrm{det}\left({\text{E}}^{\left(1\right)}\right)|
{\text{E}}^{\left(1\right)}\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ {\text{e}}_{1}^{\left(1\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{2}^{\left(1\right)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{p}^{\left(1\right)}\end{array}\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}
For iteration 2, Select a new pixel spectra r, such that
\text{\hspace{0.17em}}\text{r}\text{\hspace{0.17em}}\notin \left\{{\text{e}}_{1}^{\left(1\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{2}^{\left(1\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{p}^{\left(1\right)}\right\}
Replace each endmember in the set with r and compute the volume of the simplex V(E(2)).
Replace the ith endmember in the set with r, if the computed volume V(E(2)) is greater than V(E(1)). This results in an updated set of endmembers. For example, if i = 2, the new set of endmembers derived at the end of the second iteration is
\text{\hspace{0.17em}}\left\{{\text{e}}_{1}^{\left(2\right)},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{2}^{\left(2\right)}=\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{r},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{e}}_{p}^{\left(2\right)}\right\}
For each iteration, select a new pixel spectra r and repeat steps 5 and 6. Each iteration results in an update set of endmembers. The iteration ends when the total number of iterations reaches the specified value NumIterations.
[1] Winter, Michael E. “N-FINDR: An Algorithm for Fast Autonomous Spectral End-Member Determination in Hyperspectral Data.” Proc. SPIE Imaging Spectrometry V 3753, (October 1999): 266–75. https://doi.org/10.1117/12.366289.
hypercube | ppi | countEndmembersHFC | fippi |
Find the area of the largest isosceles triangle that can
Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 6.
a) Solve by writing the area as a function of h.
b) Solve by writing the area as a function of a.
c) Identify the type of triangle of maximum area.
temnimam2
See the figure above
Using Pythagoras theorem, we can write
b=\sqrt{{R}^{2}-{\left(h-R\right)}^{2}}
b=\sqrt{{R}^{2}-\left(h-{R}^{2}-2Rh\right)}
b=\sqrt{2Rh-{h}^{2}}
Length of the Base
=2b=2\sqrt{2Rh-{h}^{2}}
In the given problem, radius of the circle is 6. Therefore
=2\sqrt{12Rh-{h}^{2}}
Area of a triangle
=\frac{1}{2}\text{ }Base\text{ }×\text{ }Height
=\frac{1}{2}\cdot 2\sqrt{12h-{h}^{2}}×h
A\left(h\right)=h\sqrt{12Rh-{h}^{2}}
{A}^{\prime }\left(h\right)=\sqrt{12Rh-{h}^{2}}+\frac{h\left(12-2h\right)}{2\sqrt{12h-{h}^{2}}}
{A}^{\prime }\left(h\right)=\sqrt{12Rh-{h}^{2}}+\frac{6h-{h}^{2}}{2\sqrt{12h-{h}^{2}}}
Equate A’(h) to 0
\sqrt{12Rh-{h}^{2}}+\frac{6h-{h}^{2}}{2\sqrt{12h-{h}^{2}}}=0
Multiply throughout by
\sqrt{12Rh-{h}^{2}}
\left({\left(12h-h\right)}^{2}+\left(6h={h}^{2}\right)=0
18h—2{h}^{2}=0
\mid {n}^{2}=9
h=3
Area cannot be minimum when h = 3, because minimum area is 0.
deginasiba
For solving (b), we can find the total height
\left({h}_{T}\right)
of the isosceles triangle by adding the value for h found in (a) to the other height 6:
{h}_{T}=6+3=9
So, we can say that the total height of the triangle is 9 units. Then we can find the base:
b=\sqrt{36-{3}^{2}}=\sqrt{36-9}=\sqrt{27}=3\sqrt{3}
We can then divide the base by 2 to get the length of the base for half the triangle (by the AAS.
Similar Triangles Theorem) to get the base to be
\frac{3\sqrt{3}}{2}
, We can now focus on the right half of the
triangle divided by the dotted line going through the vertex point of the circle, We can tell that the
dotted line through the vertex makes a right angle. Thus, we ean use Trigonometry to solve for a.
a=\mathrm{tan}\frac{opposite}{adjacent}=\mathrm{tan}\frac{\frac{3\sqrt{3}}{2}}{9}=\mathrm{tan}\frac{27\sqrt{3}}{2}
Since we have found what a is, we can find the function of it with respect to a:
\phantom{\rule{0ex}{0ex}}tan\left(\begin{array}{c}\frac{\left(6+h\right)\left(\sqrt{36+{h}^{2}\right)}}{2}\end{array}\right)
or as an incorporated function:
\frac{1}{2}bh=\frac{1}{2}\left(6\mathrm{sin}a\right)\left(12\mathrm{cos}a\right)=36\mathrm{sin}a\mathrm{cos}a
And we are done with (b)
For (c), By the Isoperimetric Theorem, it states
Theorem: Among all triangles inscribed in a given circle, the equilateral one has the largest area.
Therefore, the equilateral triangle has the maximum area.
A marble moves along the x-axis. The potential-energy functionis shown in Fig. 1
a) At which of the labeled x-coordinates is the force on the marble zero?
b) Which of the labeled x-coordinates is a position of stable equilibrium?
c) Which of the labeled x-coordinates is a position of unstable equilibrium?
A rifle with a weight of 30 N fires a 5.0 g bullet with a speed of 300m/s. (a) Find the recoil speed of the rifle. (b) If a700 N man holds the rilfe firmly against his shoulder, find there coil speed of the man and his rifle.
Prove that if a line is drawn parallel to the base of a triangle, it cuts off a triangle similar to the given triangle.
\mathrm{△}ABC,\text{ }DE\parallel AB
\mathrm{△}DEC\sim \mathrm{△}ABC
\begin{array}{|cc|}\hline \text{Statements}& \text{Reasons}\\ \text{1. (see above)}& \text{1. Given}\\ 2.& \text{2. If two parellel lines are cut by a transversal, then the corresponding angles are equal.}\\ 3.\mathrm{△}DEC\sim \mathrm{△}ABC& 3.\\ \hline\end{array}
{f}^{\prime }\left(t\right)=\frac{20}{1+{t}^{2}},\text{ }f\left(1\right)=0
Identify each substance as an acid or a base and write a chemical equation showing how it is an acid or a base in aqueous solution according to the Arrhenius definition.
a. NaOH(aq)
{H}_{2}S{O}_{4}\left(aq\right)
c. HBr(aq)
Sr{\left(OH\right)}_{2}\left(aq\right) |
College physics problems, solved and explained
Get help with college physics
A new linear temperature scale, degrees Zunzol
{\left(}^{\circ }Z\right)
, is based on the freezing point and boiling point of a newly discovered compound zunzol. The freezing point of zunzol
-{117.3}^{\circ }C
{0}^{\circ }Z
, and the boing point of zunzol,
{78.4}^{\circ }C
{100}^{\circ }Z
. What is the freezing point of water in degrees Zunzol?
{e}^{-x}
The temperature of
100
grams of water is to be raised from
{24}^{\circ }
C to
{90}^{\circ }
C by adding steam to it. Find the mass of steam required.
\sum _{m=0}^{n}\left(-1{\right)}^{n-m}\left(\genfrac{}{}{0}{}{n}{m}\right)\left(\genfrac{}{}{0}{}{m-1}{l}\right)
Suppose a syringe (placed horizontally) contains a liquid with the density of water, composed of a barrel and a needle component. The barrel of the syringe has a cross-sectional area of
\alpha \text{ }{m}^{2}
, and the pressure everywhere is
\beta
atm, when no force is applied.
The needle has a pressure which remains equal to
\beta
atm (regardless of force applied). If we push on the needle, applying a force of magnitude
\mu \text{ }N
, is it possible to determine the medicine's flow speed through the needle?
Poiseuille's law states that the rate of flow of water is proportionate to
{r}^{4}
r
is the radius of the pipe. I don't see why.
Intuitively I would expect rate of flow of fluid to vary with
{r}^{2}
as the volume of a cylinder varies with
{r}^{2}
for a constant length. The volume that flows past a point is equal to rate of flow fluid, and thus it should vary with
{r}^{2}
{41}^{\circ } |
Solving Elastic Net
Implementing Elastic Net Regression
Parameter Sparsity Testing for Elastic Net
Finding the optimal value for \alpha and the L1-ratio
A Summary of Scikit-Learn-Classes
Background image by Pawel Czerwinski (link)
You have probably heard about linear regression. Most likely you have also heard about ridge and lasso. Maybe you have even read some articles about ridge and lasso. Ridge and lasso are the two most popular variations of linear regression which try to make it a bit more robust. Nowadays it is actually very uncommon to use regular linear regression, and not one of its variations like ridge or lasso. In previous articles we have seen how ridge and lasso operate, what their differences, as well as strengths and weaknesses are, and how you can implement them in practice. But what should you use? Ridge or lasso? The good news is that you don’t have to choose! With elastic net, you can use both the ridge penalty as well as the lasso penalty at once. And in this article, you will learn how!
This article is the third article in a series where we take a deep dive into ridge and lasso regression. Let’s start!
This article is a direct follow up to the articles about ridge and lasso, so ideally you should read the articles about ridge and lasso before reading this article. Elastic net is based on ridge and lasso, so it’s important to understand those models first. With that being said, let’s take a look at elastic net regression!
So what is wrong with linear regression? Why do we need more machine learning algorithms that do the same thing? And why are there two of them? We’ve explored this question in the articles about ridge and lasso. Here’s a lightning-quick recap:
We had a dataset of figure prices, where each entry in the dataset contained the age of the figure as well as its price for that age in € (or any other currency). We then wanted to predict the price of a figure given its age using linear regression, to see how much the figures depreciate over time. The dataset looked like this:
We then split our dataset into a train set and a test set, and trained our linear regression (OLS regression) model on the training data. Here’s how that looked like:
We then noticed that this model had a very low training error but a rather high testing error and thus we concluded that our linear regression model is overfit. We then tried to come up with an imaginary, better model that was less overfit and looked more like this:
This imaginary model turned out to be ridge regression. We analyzed what exactly lead to our linear regression model overfitting and we noticed that the main cause of overfitting were large model parameters. After discovering this insight, we developed a new loss function that penalizes large model parameters by adding a penalty term to our mean squared error. It looked like this (where
m
is the number of model parameters):
New\ Loss(y,y_{pred}) = MSE(y,y_{pred}) + \sum_{i=1}^m{\theta_i}
There’s just one problem with this loss function. Since our model parameters can be negative, adding them might decrease our loss instead of increasing it. In order to circumvent this, we can either square our model parameters or take their absolute values:
\begin{aligned} {\color{#26a6ed}RidgeMSE}(y,y_{pred}) &= MSE(y,y_{pred}) + \alpha \sum_{i=1}^m{{\color{#26a6ed}\theta_i^2}} \\ &= MSE(y,y_{pred}) + \alpha {\color{#26a6ed}||\boldsymbol{\theta}||_2^2} \\ \end{aligned} \\ \begin{aligned} {\color{#26a6ed}LassoMSE}(y,y_{pred}) &= MSE(y,y_{pred}) + \alpha \sum_{i=1}^m{{\color{#26a6ed}|\theta_i|}} \\ &= MSE(y,y_{pred}) + \alpha {\color{#26a6ed}||\boldsymbol{\theta}||_1} \\ \end{aligned}
The first function is the loss function of ridge regression, while the second one is the loss function of lasso regression.
What we can do now is combine the two penalties, and we get the loss function of elastic net:
\begin{aligned} {\color{#26a6ed}ElasticNetMSE} &= MSE(y,y_{pred}) + \alpha_1 \sum_{i=1}^m{{\color{#26a6ed}|\theta_i|}} + \alpha_2 \sum_{i=1}^m{{\color{#26a6ed}|\theta_i|}} \\ &= MSE(y,y_{pred}) + \alpha_1 {\color{#26a6ed}||\boldsymbol{\theta}||_1} + \alpha_2 {\color{#26a6ed}||\boldsymbol{\theta}||_2^2} \\ \end{aligned}
And that’s pretty much it! Instead of one regularization parameter
\alpha
we now use two parameters, one for each penalty.
\alpha_1
controls the L1 penalty and
\alpha_2
controls the L2 penalty. We can now use elastic net in the same way that we can use ridge or lasso. If
\alpha_1 = 0
, then we have ridge regression. If
\alpha_2 = 0
, we have lasso. Alternatively, instead of using two
\alpha
-parameters, we can also use just one
\alpha
and one L1-ratio-parameter, which determines the percentage of our L1 penalty with regard to
\alpha
\alpha = 1
and L1-ratio = 0.4, our L1 penalty will be multiplied with 0.4 and our L2 penalty will be multiplied with
1 - L1-ratio = 0.6
. Here’s the equation:
ElasticNetMSE = MSE(y,y_{pred}) + {\color{#26a6ed}\alpha \cdot (1 - L1Ratio)} \sum_{i=1}^m{{\color{#26a6ed}|\theta_i|}} + {\color{#26a6ed}\alpha \cdot L1Ratio} \sum_{i=1}^m{{\color{#26a6ed}|\theta_i|}}
Ok, looks good! In this case we have ridge regression if L1-ratio = 0 and lasso regression if L1-ratio = 1. In most cases, unless you already have some information about the importance of your features, you should use elastic net instead of lasso or ridge. You can then use cross-validation to determine the best ratio between L1 and L2 penalty strength. Now let’s look at how we determine the optimal model parameters
\boldsymbol{\theta}
for our elastic net model.
If L1-ratio = 0, we have ridge regression. This means that we can treat our model as a ridge regression model, and solve it in the same ways we would solve ridge regression. Namely, we can use the normal equation for ridge regression to solve our model directly, or we can use gradient descent to solve it iteratively.
If L1-ratio = 1, we have lasso regression. Then we can solve it with the same ways we would use to solve lasso regression. Since our model contains absolute values, we can’t construct a normal equation, and neither can we use (regular) gradient descent. Instead, we can use an adaptation of gradient descent like subgradient descent or coordinate descent.
If we are using both the L1 and the L2-penalty, then we also have absolute values, so we can use the same techniques as the ones we would use for lasso regression, like subgradient descent or coordinate descent.
If you’re interested in implementing elastic net from scratch, then I recommend that you take a look at the articles about subgradient descent or coordinate descent, where we do exactly that! In this article, we will use scikit-learn to help us out. Scikit-learn provides a ElasticNet-class, which implements coordinate descent under the hood. We can use it like this:
elastic_pipeline = make_pipeline(StandardScaler(),
ElasticNet(alpha=1, l1_ratio=0.1))
elastic_pipeline.fit(X_train, y_train)
print(elastic_pipeline[1].intercept_, elastic_pipeline[1].coef_)
# output: 41.0 [-1.2127174]
Just like with lasso, we can also use scikit-learn’s SGDRegressor-class, which uses truncated gradients instead of regular ones. Here’s the code:
elastic_sgd_pipeline = make_pipeline(StandardScaler(), SGDRegressor(alpha=1, l1_ratio=0.1, penalty = "elasticnet"))
elastic_sgd_pipeline.fit(X_train, y_train)
print(elastic_sgd_pipeline[1].intercept_, elastic_sgd_pipeline[1].coef_)
# output: [40.69570804] [-1.21309447]
Cool! In practice, you should probably stick to ElasticNet instead of SGDRegressor since coordinate descent converges more quickly than the truncated SGD in this scenario.
Coordinate descent for lasso in particular is extremely efficient. The article about coordinate descent goes into more depth as to why this is, but in general coordinate descent is the preferred way to train lasso or elastic net models.
Since we’re using regularized models like lasso or elastic net it is important to first standardize our data before feeding it into our regularized model! If you’re interested in what happens when we don’t standardize our data, check out When You Should Standardize Your Data. There you will learn all about standardization as well as pipelines in scikit-learn, which is what we’ve used in the above code to make our lives a bit easier.
The most important property of lasso is that lasso produces sparse model weights, meaning weights can be set all the way to 0. Whenever you are presented with an implementation of lasso (or any model that incorporates an L1-penalty, like elastic net), you should verify that this property actually holds. The easiest way to do so is to generate a randomized dataset, fit the model on it, and see whether or not all of the parameters are zeroed-out. Here it goes:
elastic_rand_pipeline = make_pipeline(StandardScaler(),
elastic_rand_pipeline.fit(X_rand, y_rand)
print(elastic_rand_pipeline[1].intercept_, elastic_rand_pipeline[1].coef_)
# 0.4881255425051216 [-0. 0. -0. -0. -0. -0. -0. -0. -0. 0. -0. 0. -0. -0. 0. 0. -0. -0.
# -0. 0. -0. 0. -0. -0. 0. -0. 0. 0. 0. -0. -0. 0. -0. -0. 0. 0.
# -0. 0. -0. -0. 0. 0. 0. 0. -0. 0. 0. 0. -0. -0.]
Nice, the weights are all zeroed out! We can perform the same test for SGDRegressor:
elastic_sgd_rand_pipeline = make_pipeline(StandardScaler(), SGDRegressor(alpha=1, l1_ratio=0.1, penalty = "elasticnet"))
elastic_sgd_rand_pipeline.fit(X_rand, y_rand)
print(elastic_sgd_rand_pipeline[1].intercept_, elastic_sgd_rand_pipeline[1].coef_)
# [0.46150165] [0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
# 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0. 0.
# 0. 0.]
Finding the optimal value for
\alpha
and the L1-ratio
Here, we can use the power of cross-validation to compute the most optimal parameters for our model. Scikit-learn even provides a special class for this called ElasticNetCV. It takes in an array of
\alpha
-values to compare and select the best of. If no array of
\alpha
-values is provided, scikit-learn will automatically determine the optimal value of
\alpha
. We can use it like so:
elastic_cv_pipeline = make_pipeline(StandardScaler(),
ElasticNetCV(l1_ratio=0.1))
elastic_cv_pipeline.fit(X_train, y_train)
print(elastic_cv_pipeline[1].alpha_)
Ok that’s nice, but how can you find an optimal value for the L1-ratio? ElasticNetCV only determines the optimal value for
\alpha
, so if we want to determine the optimal value for the L1-ratio as well, we’ll have to do an additional round of cross-validation. For this, we can use techniques such as grid or random search, which you can learn more about by reading the article Grid and Random Search Explained, Step by Step.
We’ve looked at quite a few models so far. To make it easier to remember when you should use which scikit-learn-class, I’ve created this little table. Here you can find the corresponding scikit-learn class for every model and every solver. This should make it a bit more organized.
Model / Solver
Gradient Descent variant
OLS Regression LinearRegression SGDRegressor
Ridge Ridge SGDRegressor with penalty=“l2”
Lasso / Lasso [Coordinate Descent] or
SGDRegressor with penalty=“l1” [Truncated SGD]
Elastic Net / ElasticNet [Coordinate Descent] or
SGDRegressor with penalty=“elasticnet” [Truncated SGD]
We’ve looked at ridge, lasso, and elastic net in the context of regression, but we can also take the corresponding penalties and apply them to other models, like logistic regression or polynomial regression, as well. If you’re interested in these regularized models, I recommend you take a look at the articles Logistic Regression Explained, Step by Step and Polynomial Regression Explained, Step by Step respectively. In those articles you will learn everything about the named models as well as their regularized variants!
Classes like RidgeCV, LassoCV or ElasticNetCV are handy, but not every model has a CV-variant. But if you know how cross-validation works, you can make these models yourself! Cross-validation is an extremely important method to train machine learning models effectively and to optimize hyperparameters. You can read more about this technique in the article Cross-Validation Explained, Step by Step, where you will learn everything you need to know to start using cross-validation in your own projects!
More in Machine Learning Models
When, Why, And How You Should Standardize Your Data
Standardization is one of the most useful transformations you can apply to your dataset. What is even more important is that many models, especially regularized ones, require the data to be standardized in order to function properly. In this article, you will learn everything you need to know about standardization. You will learn why it works, when you should use it, and how you can do so with just a few lines of code.
Boris Giba 06 Jul 2021 • 10 min read |
Answer true or false to each of the following statement
Answer true or false to each of the following statement and explain your answer.
Cordazzoyn 2021-11-19 Answered
In using the method of transformations, we should only transform the predictor variable to straighten a scatterplot.
See my photo below
Give a counterexample to show that the given transformation is not a linear transformation.
T\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}xy\\ x+y\end{array}\right]T
Provide answers to all tasks using the information provided.
a) Find the parent function f.
g\left(x\right)=-2|x-1|-4
b) Find the sequence of transformation from f to g.
f\left(x\right)=\left[x\right]
c) To sketch the graph of g.
g\left(x\right)=-2|x-1|-4
d) To write g in terms of f.
g\left(x\right)=-2|x-1|-4\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}f\left(x\right)=\left[x\right]
What do you mean by the Karl Pearsons’ coefficient of correlation? What are the properties of the coefficient of correlation?
To find: The solution of the given initial value problem.
It is not always possible to fnd a power transformation of the response variable or the predictor variable (or both) that will straighten the scatterplot.
To determine the solution of the initial value problem
y{}^{″}+4y=\mathrm{sin}t+{u}_{\pi }\left(t\right)\mathrm{sin}\left(t-\pi \right):
y\left(0\right)=0,
{y}^{\prime }\left(0\right)=0.
Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..
Explain how to find the value of
\mathrm{cos}\left(-{45}^{\circ }\right)
using even-odd properties. |
Assume that T is a linear transformation. Find the standard
pogonofor9z 2021-12-26 Answered
Assume that T is a linear transformation. Find the standard matrix of T.
{\mathbb{R}}^{2}\to {\mathbb{R}}^{4},\text{ }T\left({e}_{1}\right)=\left(3,1,3,1\right),\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }T\left({e}_{2}\right)=\left(-5,6,0,0\right)
{e}_{1}=\left(1,0\right)
{e}_{2}=\left(0,1\right)
A=
Want to know more about Matrix transformations?
x\in {F}^{n}
\left[\begin{array}{c}{x}_{1}\\ ⋮\\ {x}_{n}\end{array}\right]
T\left(x\right)=T\left(\sum _{j}{e}_{j}{x}_{j}\right)=\sum _{j}T\left({e}_{j}\right){x}_{j}
=\sum _{j}{A}_{j}{x}_{j}=Ax
A=\left[T\left({e}_{1}\right)\dots T\left({e}_{n}\right)\right]
{\mathbb{R}}^{2}\to {\mathbb{R}}^{4},\text{ }T\left({e}_{1}\right)=\left(3,1,3,1\right),\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }T\left({e}_{2}\right)=\left(-5,6,0,0\right)
So, the standart matrix of T is
\left[T\left({e}_{1}\right)T\left({e}_{2}\right)\right]
=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]
- Answer
=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]
T=\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]
T\left({e}_{1}\right)=\left(3,1,3,1\right)
⇒T\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right)
⇒\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right]
⇒\left[\begin{array}{c}a\\ c\\ e\\ g\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 3\\ 1\end{array}\right]⇒a=3,c=1,e=3,g=1
T\left({e}_{2}\right)=\left(-5,6,0,0\right)
⇒T\left(\begin{array}{c}1\\ 0\end{array}\right)=\left(\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right)
⇒\left[\begin{array}{cc}a& b\\ c& d\\ e& f\\ g& h\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right]
⇒\left[\begin{array}{c}b\\ d\\ f\\ h\end{array}\right]==\left[\begin{array}{c}-5\\ 6\\ 0\\ 0\end{array}\right]⇒b=-5,d=6,f=h=0
⇒T=\left[\begin{array}{cc}3& -5\\ 1& 6\\ 3& 0\\ 1& 0\end{array}\right]
For the matrix A below, find a nonzero vector in Nul A and a nonzero vector in Col A.
A=\left[\begin{array}{cccc}2& 3& 5& -9\\ -8& -9& -11& 21\\ 4& -3& -17& 27\end{array}\right]
Find a nonzero vector in Nul A.
A=\left[\begin{array}{c}-3\\ 2\\ 0\\ 1\end{array}\right]
Assume that A is row equivalent to B. Find bases for Nul A and Col A.
A=\left[\begin{array}{ccccc}1& 2& -5& 11& -3\\ 2& 4& -5& 15& 2\\ 1& 2& 0& 4& 5\\ 3& 6& -5& 19& -2\end{array}\right]
B=\left[\begin{array}{ccccc}1& 2& 0& 4& 5\\ 0& 0& 5& -7& 8\\ 0& 0& 0& 0& -9\\ 0& 0& 0& 0& 0\end{array}\right]
Find an explicit description of Nul A by listing vectors that span the null space.
A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]
Соnstruct a nonzerо
3×3
matriх А and a nonzerа vector b such that b is in Nul A.
Use an inverse matrix to solve each system of linear equations.
Use the given inverse of the coefficient matrix to solve the following system
5{x}_{1}+3{x}_{2}=6
-6{x}_{1}-3{x}_{2}=-2
{A}^{-1}=\left[\begin{array}{cc}-1& -1\\ 2& \frac{5}{3}\end{array}\right]
Find k
such that the following matrix M
M=⎡⎣⎢−225+k−5933−4114⎤⎦⎥ |
Hodge_bundle Knowpia
In mathematics, the Hodge bundle, named after W. V. D. Hodge, appears in the study of families of curves, where it provides an invariant in the moduli theory of algebraic curves. Furthermore, it has applications to the theory of modular forms on reductive algebraic groups[1] and string theory.[2]
{\displaystyle {\mathcal {M}}_{g}}
be the moduli space of algebraic curves of genus g curves over some scheme. The Hodge bundle
{\displaystyle \Lambda _{g}}
is a vector bundle[note 1] on
{\displaystyle {\mathcal {M}}_{g}}
whose fiber at a point C in
{\displaystyle {\mathcal {M}}_{g}}
is the space of holomorphic differentials on the curve C. To define the Hodge bundle, let
{\displaystyle \pi \colon {\mathcal {C}}_{g}\rightarrow {\mathcal {M}}_{g}}
be the universal algebraic curve of genus g and let
{\displaystyle \omega _{g}}
be its relative dualizing sheaf. The Hodge bundle is the pushforward of this sheaf, i.e.,[3]
{\displaystyle \Lambda _{g}=\pi _{*}\omega _{g}}
^ Here, "vector bundle" in the sense of quasi-coherent sheaf on an algebraic stack
^ van der Geer, Gerard (2008), "Siegel modular forms and their applications", in Ranestad, Kristian (ed.), The 1-2-3 of modular forms, Universitext, Berlin: Springer-Verlag, pp. 181–245 (at §13), doi:10.1007/978-3-540-74119-0, ISBN 978-3-540-74117-6, MR 2409679
^ Liu, Kefeng (2006), "Localization and conjectures from string duality", in Ge, Mo-Lin; Zhang, Weiping (eds.), Differential geometry and physics, Nankai Tracts in Mathematics, vol. 10, World Scientific, pp. 63–105 (at §5), ISBN 978-981-270-377-4, MR 2322389
^ Harris, Joe; Morrison, Ian (1998), Moduli of curves, Graduate Texts in Mathematics, vol. 187, Springer-Verlag, p. 155, doi:10.1007/b98867, ISBN 978-0-387-98429-2, MR 1631825 |
Free solutions to light and optics problems
Get help with light and optics problems
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What happens to the angle of refraction as the angle of incidence increases?
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{\theta }_{c}
{40.7}^{\circ }
, what is the index of refraction of the solid?
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Home : Support : Online Help : Mathematics : Factorization and Solving Equations : solve : function
solve/functions
for a variable which is used as a function name
solve(eqn, fcnname)
single equation (as for solve)
fcnname
single variable (as for solve)
When the unknown is a name which is only used as a function in the equation, the solver constructs a function or procedure which solves the equation.
\mathrm{solve}\left(f\left(x\right)-x+2,f\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{↦}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
This is more or less equivalent to:
\mathrm{solve}\left(f\left(x\right)-x+2,f\left(x\right)\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
\mathrm{unapply}\left(,x\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{↦}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
This in turn is more or less equivalent to:
\mathrm{solve}\left(y-x+2,y\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
\mathrm{unapply}\left(,x\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{↦}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{2}
More complicated inputs will generally return answers involving RootOfs
\mathrm{solve}\left({f\left(x\right)}^{4}-{f\left(x\right)}^{3}+x+1,f\right)
\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{↦}\textcolor[rgb]{0,0,1}{\mathrm{RootOf}}\textcolor[rgb]{0,0,1}{}\left({\textcolor[rgb]{0,0,1}{\mathrm{_Z}}}^{\textcolor[rgb]{0,0,1}{4}}\textcolor[rgb]{0,0,1}{-}{\textcolor[rgb]{0,0,1}{\mathrm{_Z}}}^{\textcolor[rgb]{0,0,1}{3}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}\right) |
Galois/Counter Mode - Wikipedia
(Redirected from GCM mode)
Authenticated encryption mode for block ciphers
In cryptography, Galois/Counter Mode (GCM) is a mode of operation for symmetric-key cryptographic block ciphers which is widely adopted for its performance. GCM throughput rates for state-of-the-art, high-speed communication channels can be achieved with inexpensive hardware resources.[1] The operation is an authenticated encryption algorithm designed to provide both data authenticity (integrity) and confidentiality. GCM is defined for block ciphers with a block size of 128 bits. Galois Message Authentication Code (GMAC) is an authentication-only variant of the GCM which can form an incremental message authentication code. Both GCM and GMAC can accept initialization vectors of arbitrary length.
Different block cipher modes of operation can have significantly different performance and efficiency characteristics, even when used with the same block cipher. GCM can take full advantage of parallel processing and implementing GCM can make efficient use of an instruction pipeline or a hardware pipeline. By contrast, the cipher block chaining (CBC) mode of operation incurs pipeline stalls that hamper its efficiency and performance.
2 Mathematical basis
Like in normal counter mode, blocks are numbered sequentially, and then this block number is combined with an initialization vector (IV) and encrypted with a block cipher E, usually AES. The result of this encryption is then XORed with the plaintext to produce the ciphertext. Like all counter modes, this is essentially a stream cipher, and so it is essential that a different IV is used for each stream that is encrypted.
Mathematical basis[edit]
GCM combines the well-known counter mode of encryption with the new Galois mode of authentication. The key-feature is the ease of parallel-computation of the Galois field multiplication used for authentication. This feature permits higher throughput than encryption algorithms, like CBC, which use chaining modes. The GF(2128) field used is defined by the polynomial
{\displaystyle x^{128}+x^{7}+x^{2}+x+1}
The authentication tag is constructed by feeding blocks of data into the GHASH function and encrypting the result. This GHASH function is defined by
{\displaystyle {\text{GHASH}}(H,A,C)=X_{m+n+1}}
where H = Ek(0128) is the Hash Key, a string of 128 zero bits encrypted using the block cipher, A is data which is only authenticated (not encrypted), C is the ciphertext, m is the number of 128-bit blocks in A (rounded up), n is the number of 128-bit blocks in C (rounded up), and the variable Xi for i = 0, ..., m + n + 1 is defined below.[2]
First, the authenticated text and the cipher text are separately zero-padded to multiples of 128 bits and combined into a single message Si:
{\displaystyle S_{i}={\begin{cases}A_{i}&{\text{for }}i=1,\ldots ,m-1\\A_{m}^{*}\parallel 0^{128-v}&{\text{for }}i=m\\C_{i-m}&{\text{for }}i=m+1,\ldots ,m+n-1\\C_{n}^{*}\parallel 0^{128-u}&{\text{for }}i=m+n\\\operatorname {len} (A)\parallel \operatorname {len} (C)&{\text{for }}i=m+n+1\end{cases}}}
where len(A) and len(C) are the 64-bit representations of the bit lengths of A and C, respectively, v = len(A) mod 128 is the bit length of the final block of A, u = len(C) mod 128 is the bit length of the final block of C, and
{\displaystyle \parallel }
denotes concatenation of bit strings.
Then Xi is defined as:
{\displaystyle X_{i}=\sum _{j=1}^{i}S_{j}\cdot H^{i-j+1}={\begin{cases}0&{\text{for }}i=0\\\left(X_{i-1}\oplus S_{i}\right)\cdot H&{\text{for }}i=1,\ldots ,m+n+1\end{cases}}}
The second form is an efficient iterative algorithm (each Xi depends on Xi−1) produced by applying Horner's method to the first. Only the final Xm+n+1 remains an output.
If it is necessary to parallelize the hash computation, this can be done by interleaving k times:
{\displaystyle {\begin{aligned}X_{i}^{'}&={\begin{cases}0&{\text{for }}i\leq 0\\\left(X_{i-k}^{'}\oplus S_{i}\right)\cdot H^{k}&{\text{for }}i=1,\ldots ,m+n+1-k\\\end{cases}}\\[6pt]X_{i}&=\sum _{j=1}^{k}\left(X_{i+j-2k}^{'}\oplus S_{i+j-k}\right)\cdot H^{k-j+1}\end{aligned}}}
If the length of the IV is not 96, the GHASH function is used to calculate Counter 0:
{\displaystyle {\begin{aligned}Counter0={\begin{cases}IV\parallel 0^{31}\parallel 1&{\text{for }}len(IV)=96\\{\text{GHASH}}(\ IV\parallel 0^{s}\ \parallel \ 0^{64}\parallel len_{64}(IV)\ ){\text{ with }}s=(128-(len(IV)\mod 128))\mod 128&{\text{otherwise}}\end{cases}}\end{aligned}}}
GCM was designed by John Viega and David A. McGrew to be an improvement to Carter–Wegman counter mode (CWC mode).[citation needed]
In November 2007, NIST announced the release of NIST Special Publication 800-38D Recommendation for Block Cipher Modes of Operation: Galois/Counter Mode (GCM) and GMAC making GCM and GMAC official standards.[3]
GCM mode is used in the IEEE 802.1AE (MACsec) Ethernet security, IEEE 802.11ad (also dubbed WiGig), ANSI (INCITS) Fibre Channel Security Protocols (FC-SP), IEEE P1619.1 tape storage, IETF IPsec standards,[4][5] SSH,[6] TLS 1.2[7][8] and TLS 1.3.[9] AES-GCM is included in the NSA Suite B Cryptography and its latest replacement in 2018 Commercial National Security Algorithm (CNSA) suite.[10] GCM mode is used in the SoftEther VPN server and client,[11] as well as OpenVPN since version 2.4.
GCM requires one block cipher operation and one 128-bit multiplication in the Galois field per each block (128 bit) of encrypted and authenticated data. The block cipher operations are easily pipelined or parallelized; the multiplication operations are easily pipelined and can be parallelized with some modest effort (either by parallelizing the actual operation, by adapting Horner's method per the original NIST submission, or both).
Intel has added the PCLMULQDQ instruction, highlighting its use for GCM.[12] In 2011, SPARC added the XMULX and XMULXHI instructions, which also perform 64 × 64 bit carry-less multiplication. In 2015, SPARC added the XMPMUL instruction, which performs XOR multiplication of much larger values, up to 2048 × 2048 bit input values producing a 4096-bit result. These instructions enable fast multiplication over GF(2n), and can be used with any field representation.
Impressive performance results are published for GCM on a number of platforms. Käsper and Schwabe described a "Faster and Timing-Attack Resistant AES-GCM"[13] that achieves 10.68 cycles per byte AES-GCM authenticated encryption on 64-bit Intel processors. Dai et al. report 3.5 cycles per byte for the same algorithm when using Intel's AES-NI and PCLMULQDQ instructions. Shay Gueron and Vlad Krasnov achieved 2.47 cycles per byte on the 3rd generation Intel processors. Appropriate patches were prepared for the OpenSSL and NSS libraries.[14]
When both authentication and encryption need to be performed on a message, a software implementation can achieve speed gains by overlapping the execution of those operations. Performance is increased by exploiting instruction-level parallelism by interleaving operations. This process is called function stitching,[15] and while in principle it can be applied to any combination of cryptographic algorithms, GCM is especially suitable. Manley and Gregg[16] show the ease of optimizing when using function stitching with GCM. They present a program generator that takes an annotated C version of a cryptographic algorithm and generates code that runs well on the target processor.
GCM has been criticized for example by Silicon Labs in the embedded world because the parallel processing isn't suited for performant use of cryptographic hardware engines and therefore reduces the performance of encryption for some of the most performance-sensitive devices.[17]
According to the authors' statement, GCM is unencumbered by patents.[18]
GCM is proven secure in the concrete security model.[19] It is secure when it is used with a block cipher that is indistinguishable from a random permutation; however, security depends on choosing a unique initialization vector for every encryption performed with the same key (see stream cipher attack). For any given key and initialization vector combination, GCM is limited to encrypting 239−256 bits of plain text (64 GiB). NIST Special Publication 800-38D[3] includes guidelines for initialization vector selection.
The authentication strength depends on the length of the authentication tag, like with all symmetric message authentication codes. The use of shorter authentication tags with GCM is discouraged. The bit-length of the tag, denoted t, is a security parameter. In general, t may be any one of the following five values: 128, 120, 112, 104, or 96. For certain applications, t may be 64 or 32, but the use of these two tag lengths constrains the length of the input data and the lifetime of the key. Appendix C in NIST SP 800-38D provides guidance for these constraints (for example, if t = 32 and the maximal packet size is 210 bytes, the authentication decryption function should be invoked no more than 211 times; if t = 64 and the maximal packet size is 215 bytes, the authentication decryption function should be invoked no more than 232 times).
Like with any message authentication code, if the adversary chooses a t-bit tag at random, it is expected to be correct for given data with probability measure 2−t. With GCM, however, an adversary can increase their likelihood of success by choosing tags with n words – the total length of the ciphertext plus any additional authenticated data (AAD) – with probability measure 2−t by a factor of n. Although, one must bear in mind that these optimal tags are still dominated by the algorithm's survival measure 1 − n⋅2−t for arbitrarily large t. Moreover, GCM is neither well-suited for use with very short tag-lengths nor very long messages.
Ferguson and Saarinen independently described how an attacker can perform optimal attacks against GCM authentication, which meet the lower bound on its security. Ferguson showed that, if n denotes the total number of blocks in the encoding (the input to the GHASH function), then there is a method of constructing a targeted ciphertext forgery that is expected to succeed with a probability of approximately n⋅2−t. If the tag length t is shorter than 128, then each successful forgery in this attack increases the probability that subsequent targeted forgeries will succeed, and leaks information about the hash subkey, H. Eventually, H may be compromised entirely and the authentication assurance is completely lost.[20]
Independent of this attack, an adversary may attempt to systematically guess many different tags for a given input to authenticated decryption and thereby increase the probability that one (or more) of them, eventually, will be considered valid. For this reason, the system or protocol that implements GCM should monitor and, if necessary, limit the number of unsuccessful verification attempts for each key.
Saarinen described GCM weak keys.[21] This work gives some valuable insights into how polynomial hash-based authentication works. More precisely, this work describes a particular way of forging a GCM message, given a valid GCM message, that works with probability of about n⋅2−128 for messages that are n × 128 bits long. However, this work does not show a more effective attack than was previously known; the success probability in observation 1 of this paper matches that of lemma 2 from the INDOCRYPT 2004 analysis (setting w = 128 and l = n × 128). Saarinen also described a GCM variant Sophie Germain Counter Mode (SGCM) based on Sophie Germain primes.
^ Lemsitzer, S.; Wolkerstorfer, J.; Felber, N.; Braendli, M. (2007). "Multi-gigabit GCM-AES Architecture Optimized for FPGAs". In Paillier, P.; Verbauwhede, I. (eds.). Cryptographic Hardware and Embedded Systems — CHES 2007. Lecture Notes in Computer Science. Vol. 4727. Springer. pp. 227–238. doi:10.1007/978-3-540-74735-2_16. ISBN 978-3-540-74734-5.
^ McGrew, David A.; Viega, John (2005). "The Galois/Counter Mode of Operation (GCM)" (PDF). p. 5. Retrieved 20 July 2013. Note that there is a typo in the formulas in the article.
^ a b Dworkin, Morris (2007–2011). Recommendation for Block Cipher Modes of Operation: Galois/Counter Mode (GCM) and GMAC (PDF) (Technical report). NIST. 800-38D. Retrieved 2015-08-18.
^ RFC 4106 The Use of Galois/Counter Mode (GCM) in IPsec Encapsulating Security Payload (ESP)
^ RFC 4543 The Use of Galois Message Authentication Code (GMAC) in IPsec ESP and AH
^ RFC 5647 AES Galois Counter Mode for the Secure Shell Transport Layer Protocol
^ RFC 5288 AES Galois Counter Mode (GCM) Cipher Suites for TLS
^ RFC 6367 Addition of the Camellia Cipher Suites to Transport Layer Security (TLS)
^ RFC 8446 The Transport Layer Security protocol version 1.3
^ "Algorithm Registration - Computer Security Objects Register | CSRC | CSRC". 24 May 2016.
^ "Why SoftEther VPN – SoftEther VPN Project".
^ Gueron, Shay; Kounavis, Michael (April 2014). "Intel Carry-Less Multiplication Instruction and its Usage for Computing the GCM Mode (Revision 2.02)". Retrieved 2018-04-29.
^ Käsper, E.; Schwabe, P. (2009). "Faster and Timing-Attack Resistant AES-GCM". In Clavier, C.; Gaj, K. (eds.). Cryptographic Hardware and Embedded Systems – CHES 2009. Lecture Notes in Computer Science. Vol. 5747. Springer. pp. 1–17. doi:10.1007/978-3-642-04138-9_1. ISBN 978-3-642-04138-9.
^ Gueron, Shay. "AES-GCM for Efficient Authenticated Encryption – Ending the Reign of HMAC-SHA-1?" (PDF). Workshop on Real-World Cryptography. Retrieved 8 February 2013.
^ Gopal, V., Feghali, W., Guilford, J., Ozturk, E., Wolrich, G., Dixon, M., Locktyukhin, M., Perminov, M. "Fast Cryptographic Computation on Intel Architecture via Function Stitching" Intel Corp. (2010)
^ Manley, Raymond; Gregg, David (2010). "A Program Generator for Intel AES-NI Instructions". In Gong, G.; Gupta, K.C. (eds.). Progress in Cryptology – INDOCRYPT 2010. Lecture Notes in Computer Science. Vol. 6498. Springer. pp. 311–327. doi:10.1007/978-3-642-17401-8_22. ISBN 978-3-642-17400-1.
^ "IoT Security Part 6: Galois Counter Mode". 2016-05-06. Retrieved 2018-04-29.
^ McGrew, David A.; Viega, John. "The Galois/Counter Mode of Operation (GCM) Intellectual Property Statement" (PDF). Computer Security Resource Center, NIST.
^ McGrew, David A.; Viega, John (2004). "The Security and Performance of the Galois/counter mode (GCM) of Operation". Proceedings of INDOCRYPT 2004. Lecture Notes in Computer Science. Vol. 3348. Springer. CiteSeerX 10.1.1.1.4591. doi:10.1007/978-3-540-30556-9_27. ISBN 978-3-540-30556-9.
^ Niels Ferguson, Authentication Weaknesses in GCM, 2005-05-20
^ Markku-Juhani O. Saarinen (2011-04-20). "Cycling Attacks on GCM, GHASH and Other Polynomial MACs and Hashes". FSE 2012. {{cite journal}}: Cite journal requires |journal= (help)
NIST Special Publication SP800-38D defining GCM and GMAC
IEEE 802.1AE – Media Access Control (MAC) Security
IEEE Security in Storage Working Group developed the P1619.1 standard
INCITS T11 Technical Committee works on Fibre Channel – Security Protocols project.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Galois/Counter_Mode&oldid=1050311983"
Authenticated-encryption schemes |
Families of analytic discs in $\mathbf {C}^n$ with boundaries on a prescribed $CR$ submanifold
{𝐂}^{n}
CR
author = {Hill, C. Denson and Taiani, Geraldine},
title = {Families of analytic discs in $\mathbf {C}^n$ with boundaries on a prescribed $CR$ submanifold},
AU - Hill, C. Denson
AU - Taiani, Geraldine
TI - Families of analytic discs in $\mathbf {C}^n$ with boundaries on a prescribed $CR$ submanifold
Hill, C. Denson; Taiani, Geraldine. Families of analytic discs in $\mathbf {C}^n$ with boundaries on a prescribed $CR$ submanifold. Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Série 4, Tome 5 (1978) no. 2, pp. 327-380. http://www.numdam.org/item/ASNSP_1978_4_5_2_327_0/
[1] A. Andreotti - C.D. Hill - S. Łojasiewicz - B. Mackichan, Complexes of differential operators. The Mayer-Vietoris sequence, Inventiones Math., 35 (1976), pp. 43-86. | Zbl 0332.58016
[2] R. Bartle, The elements of real analysis, John Wiley and Sons, New York, 1964. | MR 393369
[3] E. Bishop, Differentiable manifolds in complex Euclidean space, Duke Math. J., (1965), pp. 1-22. | MR 200476 | Zbl 0154.08501
[4] R. Courant - D. Hilbert, Methods of mathematical physics, Vol. II, Interscience Publishers, New York, 1962. | MR 65391 | Zbl 0099.29504
[5] D. Ellis - C.D. Hill - C. Seabury, The maximum modulus principle. - I: Necessary conditions, Indiana Univ. Math. J., 25 (1976), pp. 709-717. | MR 590086 | Zbl 0336.32013
[6] S.J. Greenfield, Cauchy-Riemann equations in several variables, Ann. Scuola Norm. Sup. Pisa, 22 (1968), pp. 275-314. | Numdam | MR 237816 | Zbl 0159.37502
[7] C.D. Hill, A Kontinuitätssatz for ∂M and Lewy extendibility, Indiana Univ. Math. J., 22 (1972), pp. 339-347. | Zbl 0247.32009
[8] C.D. Hill - B. Mackichan, Hyperf-unction cohomology classes and their boundary values, Ann. Scuola Norm. Sup. Pisa, 4 (1977), pp. 577-597. | Numdam | MR 467847 | Zbl 0367.46037
[9] K. Hoffman, Banach spaces of analytic functions, Prentice-Hall, Englewood Cliffs, N. J., 1962. | MR 133008 | Zbl 0117.34001
[10] L. Hormander, An introduetion to complex analysis in several variables, Van Nostrand, Princeton, N. J., 1966. | MR 203075 | Zbl 0138.06203
[11] L.R. Hunt - R.O. Wells, Extensions of CR-functions, Amer. J. Math., 98 (1976), pp. 805-820. | MR 432913 | Zbl 0334.32014
[12] H. Lewy, On the local character of the solutions of an atypical linear differential equation in three variables and a related theorem for regular functions of two complex variables, Ann. Math., 64 (1956), pp. 514-522. | MR 81952 | Zbl 0074.06204
[13] H. Lewy, An example of a smooth linear partial differential equation without solution, Ann. Math., 66 (1957), pp. 155-158. | MR 88629 | Zbl 0078.08104
[14] H. Lewy, On hulls of holomorphy, Comm. Pure Appl. Math., 13 (1960), pp. 587-591. | MR 150339 | Zbl 0113.06102
[15] A. Nijenhuis, Strong derivatives and inverse mappings, Amer. Math. Monthly, 81 (1974), pp. 969-980. | MR 360958 | Zbl 0296.58002
[16] R. Nirenberg, On the H. Lewy extension phenomenon, Trans. Amer. Math. Soc., 168 (1972), pp. 337-356. | MR 301234 | Zbl 0241.32006
[17] M. Sato, On a generalization of the concept of functions, Proc. Japan Acad., 34 (1958), pp. 126-130 and 604-608. | MR 96122
[18] M. Sato, Theory of hyperfunctions, J. Fac. Sci., Univ. Tokyo, Sect. I, 8 (1959-60), pp. 139-193 and 387-436. | MR 114124 | Zbl 0087.31402
[19] M. Sato - T. Kawai - M. Kashiwara, Microfunctions and pseudodifferential equations, in Hyperfunctions and pseudodifferential equations, Lecture Notes in Math., 287 (1973), pp. 265-529. | MR 420735 | Zbl 0277.46039
[20] P. Schapira, Théorie des hyperfonctions, Lecture Notes in Math., 126 (1970). | MR 270151 | Zbl 0192.47305
[21] B. Weinstock, On holomorphic extension from real submanifolds of complex Euclidean space, Ph. D. Thesis, M.I.T., Cambridge, Mass., 1966.
[22] R.O. Wells, On the local holomorphic hull of a real submanifold in several complex variables, Comm. Pure Appl. Math., 19 (1966), pp. 145-165. | MR 197785 | Zbl 0142.33901 |
\frac{a}{b}=\left(\frac{a}{b}×100\right)%
\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
P{\left(1+\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}
P{\left(1-\frac{R}{100}\right)}^{n}
\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}
\left[\frac{R}{\left(100+R\right)}×100\right]%
\left[\frac{R}{\left(100-R\right)}×100\right]%
Of the 1000 inhabitants of a town, 60 % are males of whom 120 % are literate. If, of all the inhabitants, 25% are literate, then what percent of the females of the town are literate ?
A) 32.5 % B) 43 %
Number of males = 60% of 1000 = 600. Number of females = (1000 - 600) = 400.
Number of literates = 25% of 1000 = 250.
Number of literate males = 20% of 600 = 120.
Number of literate females = (250 - 120) = 130.
Required pecentage = (130/400 * 100 ) % = 32.5 %.
A candidate got 35% of the votes polled and he lost to his rival by 2250 votes. How many votes were cast ?
Let the candidates be A & B
35%-----------A
65%-----------B
Difference of votes (65-35)% = 30%
=> 30% ------- 2250
Then for 100% -------- ? => 7500
If x is 80% of y, then what percent of 2x is y?
Answer & Explanation Answer: D) 62.5 %
x = 80 % of y
=> x =(80/100 )y
Required percentage = [(y/2x)*100] % = (5/8*100) % =62.5%
The population of a town was 1,60,000 three years ago, If it increased by 3%, 2.5% and 5% respectively in the last three years, then the present population in
Answer & Explanation Answer: D) 177366
Present population = 160000 * (1 + 3/100)(1 + 5/200)(1 + 5/100)= 177366.
Entry fee in an exhibition was Rs. 1. Later, this was reduced by 25% which increased the sale by 20%. The percentage increase in the number of visitors is :
Answer & Explanation Answer: C) 60 %
Let the total original sale be Rs. 100. Then, original number of visitors = 100.
New number of visitors = 120/0.75 = 160.
Increase % = 60 %.
In a City, 35% of the population is composed of migrants, 20% of whom are from rural areas. Of the local population, 48% is female while this figure for rural and urban migrants is 30% and 40% respectively. If the total population of the city is 728400, what is its female population ?
C) 509940 D) None of these
Answer & Explanation Answer: A) 324138
Migrants = 35 % of 728400 = 254940
local population = (728400 - 254940) = 473460.
Rural migrants = 20% of 254940 = 50988
Urban migrants = (254940 - 50988) = 203952
Female population = 48% of 473460 + 30% of 50988 + 40% of 203952 = 324138
What percent decrease in salaries would exactly cancel out the 20 percent increase ?
A) (16 + 2/3) % B) (15 + 2/3)%
C) (14 +1/3)% D) (13 + 4/5)%
A) (16 + 2/3) %
B) (15 + 2/3)%
C) (14 +1/3)%
D) (13 + 4/5)%
Answer & Explanation Answer: A) (16 + 2/3) %
Let the orginal salary = Rs. 100. New's salary = Rs.120.
Decrease on 120 = 20. Decrease on 100 = [(20/120)*100]% = (16 + 2/3) %.
If the single discount equivalent to successive discounts of 20% and x% is 24%, then the value of x is:
Answer & Explanation Answer: B) 5% |
Determine the equation of the quadrac function with vertex (2,4)
rompentskj 2021-12-03 Answered
Determine the equation of the quadrac function with vertex (2,4) and passing through the point (-1, -14).
Your answer should be in vertex form.
f\left(x\right)=
Let the vertex form of quadratic function
f\left(x\right)=a{\left(x-h\right)}^{4}+k
, whese (h,k) is
We are given: vertex is (2, 4)
\therefore n=2,k=4
\therefore f\left(x\right)=a{\left(x-2\right)}^{2}+4
We are aslo given that function
\left(-1,-14\right)
\therefore Plug\text{ }in\text{ }f\left(x\right)=-14\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }x=-1
-14=a{\left(-1-2\right)}^{2}+4
-14=9a+4
-14-4=9⇒9a=-18
a=-2
\therefore f\left(x\right)=-2{\left(x-2\right)}^{2}+4
A\left(\alpha \right)=\left(\alpha I+\left(1-\alpha \right)B\right)-1
be the inverse of a convex combination of the n-by-n identity matrix I and a positive semidefinite symmetric matrix B, where the combination is given by
\alpha \in \left(0,1\right]
. The matrix B is not invertible. For some n-dimensional vector c, I want to find
\alpha
{c}^{T}A\left(\alpha \right)\left(I-B\right)A\left(\alpha \right)c=0
Is there a closed-form solution for
\alpha
Factor the following quadratics.
2{x}^{2}+7x+6
4{x}^{2}-13x+9
3{x}^{2}-10x+8
2{x}^{2}-3x-2
3{x}^{2}+17x+20
4{x}^{2}-25x+25
Is an equation only a parabola if it is quadratic? Could the graph of
y={x}^{1.65}
be described as parabolic in shape?
Range of p satisfying quadratic inequality
What is the set of values of p for which
p\left({x}^{2}+2\right)<2{x}^{2}+6x+1
, for all real values of x?
A quadratic function has its vertex at the point (1,6). The function passes through the point (-2, -3). Find the quadratic and linear coefficients and the constant team of the function.
-The quadratic coefficients is
-The linear coefficients is
-The constant term is
\sqrt{{x}^{2}+8x+7}+\sqrt{{x}^{2}+3x+2}=\sqrt{6{x}^{2}+19x+13}
\frac{{\left(t+2\right)}^{2}}{3}+3\frac{{\left(t+2\right)}^{1}}{3}-10=0 |
Make fractions out of the following information, reduce, if possible, 1 foot is
Make fractions out of the following information, reduce, if possible, 1 foot is divided into 12 inches. Make a fraction of the distance from 0 to a-d 0 to a. = ___ 0 to b. = ___ 0 to c. = ___ 0 to d. = ___
Make fractions out of the following information, reduce, if possible,
1 foot is divided into 12 inches. Make a fraction of the distance from 0 to a-d
0 to a. = ___
0 to b. = ___
0 to c. = ___
0 to d. = ___
From 0 to a., there are 3 inches so the fraction is:
\frac{3}{12}=\frac{1}{4}ft
\frac{5}{12}ft
\frac{8}{12}=\frac{2}{3}ft
\frac{11}{12}ft
\cap \cup
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
\left(a,b\right)\in R
Prove the following fact by induction:
For every non-negative integer n, 2 evenly divides
{n}^{2}-7n+4
The number of terms in
{\left(x+y+z\right)}^{30}
How many terms would there be if the same terms were combined? How can we find out?
Prove: Any postage of 24 or more can be constructed using 5 and 7 cent stamps.
Use K and
K+1
for the induction hypothesis. |
Linear Regression in Machine Learning-plot-algorithms-explain 02 - pythonslearning
June 1, 2020 by sachin Pagar
Linear Regression in Machine Learning: part02
X and y array
Now let’s splits the data into a training set and a testing set. We will train out models on the training set and then use the test set to evaluate the modes
Predictions from our Model
Regression Evaluation Metric
WELCOME TO SECOND PART OF YOUR LINEAR REGRESSION POST. IN THE FIRST POST WE SEE OPERATION ON DATASETS. IN THIS POST WE SEE LINEAR REGRESSION OPERATION. LET’S START:
Let’s now begin to train out regression models. We will need to first split up our data into an X array that contain the feature to train on, and a y array with the target variables, in this case the Price column. We will toss out the Address columns because it only has text info that the linear regression model can not use.
Let’s evaluate the model by checking out it is coefficient and how we can interpret them
# print the intercept
coeff_df = pd.DataFrame(lm.coef_,X.columns,columns=['Coefficient'])
CoefficientAvg. Area Incomes21.528276Avg. Area House Ages164883.282027Avg. Area Number of Room122368.678027Avg. Area Number of Bedroom2233.801864Area Populations15.150420
- Holding all other features fixed, a 1 unit increase in **Avg. Area Income** is associated with an **increase of $21.52 **.
- Holding all other features fixed, a 1 unit increase in **Avg. Area House Age** is associated with an **increase of $164883.28 **.
- Holding all other features fixed, a 1 unit increase in **Avg. Area Number of Rooms** is associated with an **increase of $122368.67 **.
- Holding all other features fixed, a 1 unit increase in **Avg. Area Number of Bedrooms** is associated with an **increase of $2233.80 **.
- Holding all other features fixed, a 1 unit increase in **Area Population** is associated with an **increase of $15.15 **.
Does this make sense? Probably not because I made up this data. If you want real data to repeat this sort of analysis, check out the [boston dataset](http://scikit-learn.org/stable/modules/generated/sklearn.datasets.load_boston.html):
boston_df = boston.data
Let's grab prediction off our test set and see how well it did.
<matplotlib.collections.PathCollection at 0x142622c88>
sns.distplot((y_test-predictions));
Here are three common evaluation metric for regression problem:
Mean Absolute Error (MAE) is the mean of the absolute value of the error:
\frac{1}{n}\sum _{i=1}^{n}|{y}_{i}-{\stackrel{^}{y}}_{i}|
Mean Squared Error (MSE) is the mean of the squared error:
\frac{1}{n}\sum _{i=1}^{n}\left({y}_{i}-{\stackrel{^}{y}}_{i}{\right)}^{2}
Root Mean Squared Error (RMSE) is the square root of the mean of the squared error:
\sqrt{\frac{1}{n}\sum _{i=1}^{n}\left({y}_{i}-{\stackrel{^}{y}}_{i}{\right)}^{2}}
Comparing these metric:
MAE is the easiest to understand, because it is the average error.
MSE is more popular than MAE, because MSE “punishes” larger error, which tends to be useful in the real world.
RMSE is more popular than MSE, because RMSE is interpretableS in the “y” units.
All of these are loss function, because we want to minimize them.
MAE: 82288.2225191
MSE: 10460958907.2
RMSE: 102278.829223
This was your Machine Learning Project!
Tags: Linear Regression in Machine Learning-plot-algorithms-explain
FOR THE FIRST PART OF PROJECT CLICK HERE.
Linear Regression in Machine Learning -algorithms 03 |
Clustering of ECG segments for patients before sudden cardiac death based on Lagrange descriptors | JVE Journals
Mantas Landauskas1 , Minvydas Ragulskis2
1, 2Kaunas University of Technology, Centre for Nonlinear Systems, Studentu 50-146, Kaunas LT-51368, Lithuania
Copyright © 2019 Mantas Landauskas, et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Novel approach to clustering of ECG segments based on Lagrange descriptors is presented in this paper. The approach starts by extracting 2D features with the help of Lagrange descriptors. Then the features are transformed to latent vectors which are clustered using K-means algorithm. The object of the research is to visualize the dynamics of clusters of 2D features of segments of ECG before sudden cardiac death happens to a patient.
Keywords: ECG, Lagrange descriptors, feature extraction, convolutional autoencoder, clustering.
Sudden cardiac death (SCD) is often described as the result of a change of typical sinus rhythm of a heart to a rhythm which does not support adequate pumping of the blood to the brain [1]. A number of techniques were used for predicting SCD and they achieve very different accuracy [2]. Despite the numerous methods and attempts to predict SCD there still exists challenges. Pacemakers and other devices requires high level of accuracy while interacting with human beings. More on these challenges could be found in [3].
Clustering and classification approaches in ECG data analysis is not a new direction [4-6]. But the novelty of this paper arises from the fact that we incorporate here Lagrangian descriptors (LD) as the first step in feature extraction. LD were introduced in [7]. The methodology is able to visualize the structure of a phase space. The method is based on computing the length of the trajectory a particle (trajectory) travels.
Convolutional autoencoder neural network (CNN) is employed here for extracting latent vectors. More on such techniques could be found in [8, 9].
In this paper we use Sudden Cardiac Death Database available at PhysioNet [10]. The database contains 23 complete Holter recordings. Each recording comprises 2 time series – two lead ECG. For short description of the data used refer to Table 1.
Table 1. ECG data annotations for Sudden Cardiac Death Database
Underlying cardiac rhythm
30m.mat
Sinus, intermittent VP
We have analyzed each data file of the database but provide results here for the first entry i.e. file “30m.mat”.
Consider ECG signal
{X}_{t}
t=1, 2, \dots , N
. Since the aim here is to investigate the dynamics of extracted features in different segments of ECG, the signal is divided into 5 data sets of equal length:
{x}_{ t}^{\left(q\right)}={X}_{ \left(q-1\right)\bullet ⌊N/5⌋+t}^{ }
t=1, 2, \dots , ⌊N/5⌋
q=1, 2, \dots , 5
. Each segment is investigated separately using the same methodology.
{x}_{ t}^{\left(q\right)}
are further divided into a number of non-overlapping vectors
{y}_{ i}^{\left(q,c\right)}\text{,}
c=⌊⌊N/5⌋/256⌋
i=1, 2, \dots , 256
. Now, time embedding is performed for each vector and collection of triples of ECG values are computed:
{\left({u}_{i},{v}_{i},{w}_{i}\right)}_{{\tau }_{1},{\tau }_{2}}:=\left({y}_{ i}^{\left(q,c\right)},{y}_{ i+{\tau }_{1}}^{\left(q,c\right)},{y}_{ i+{\tau }_{1}+{\tau }_{2}}^{\left(q,c\right)}\right)
{\tau }_{1},{\tau }_{2}=1, 2, \dots , 48
i=1, 2, \dots , 166
. This is in fact a standard non-uniform time series embedding approach. Thus, the triples obtained could be treated as points of a trajectory in 3D space. Lengths
L\left({\tau }_{1},{\tau }_{2}\right)
of the vectors
{\left({u}_{i},{v}_{i},{w}_{i}\right)}_{{\tau }_{1},{\tau }_{2}}
are the simplest LD estimates and are computed by Eq. (1).
L\left({\tau }_{1},{\tau }_{2}\right)=\sqrt{{\sum }_{i=1}^{165}{\left({\left({u}_{i+1},{v}_{i+1},{w}_{i+1}\right)}_{{\tau }_{1},{\tau }_{2}}-{\left({u}_{i},{v}_{i},{w}_{i}\right)}_{{\tau }_{1},{\tau }_{2}}\right)}^{2}}.
LD estimates are then used for 2D feature computation. We place a dot of grayscale intensity
L\left({\tau }_{1},{\tau }_{2}\right)
in the two-dimensional plane at coordinates
\left({\tau }_{1},{\tau }_{2}\right)
. After considering
{\tau }_{1},{\tau }_{2}=1, 2, \dots , 48
two-dimensional features are obtained for ECG segments
{x}_{ t}^{\left(q\right)}
. Clustering of the features is a rather straightforward task. So CNN is employed for extracting second level features – latent vectors. These are later fed to dimensionality reduction by PCA (leaving only 2 components, just for the visualization purposes) and clustered by K-means. Fig. 1 shows full diagram of the methodology proposed for clustering of ECG segments.
Fig. 1. Full diagram of the methodology presented in this paper together with the structure of CNN for extracting latent vectors (second level features) from 2D features of ECG
At first 2D feature maps must be computed. Actually, this step of preprocessing implies a very big number of numerical experiments to be carried out. We have used two-dimensional time embedding together with the simplest LD estimate as mentioned before. One can try different dimensions as well as different LD estimates but that would only lead to optimal feature extraction technique and not tell more details about the methodology itself.
First three 2D feature maps for the first data file and the measurements of the first lead are shown in Fig. 2.
Fig. 2. A subset of 2D features of ECG segments before sudden cardiac death for “30m.mat” data file
Visual differences in Fig. 2 implies that clustering of such mages would be a logical step to try to find different clusters for the beginning of the recordings and for the end of the recordings. This is the ultimate goal of the research presented here.
Next, optimal CNN must be trained on the obtained 2D feature maps and latent vectors computed. It should be noted that we linearly transform the values in the maps to the interval
\left[0;1\right]
before applying CNN. For training of CNN we have tested filter sizes 4 and 8, 4 and 8 for number of convolutional filters in the first layer of the network, 2 and 4 for number of filters in the second layer as well as 2 and 4 for parameters in downsampling (MaxPooling) layers. We have used python with keras for training of CNN. Computations took significant amount of time to complete. For that matter cuda hardware acceleration was employed. Training performance was measured by the mean squared error (MSE). Table 2 shows best CNN architectures for the first data file.
Table 2. Best CNN architectures for the data file “30m.mat”. Depending on the network parameters the length
{n}_{latent}
of the latent vectors is different. Training and validation sets were selected randomly with ratio 70/30
Conv. dim.
{n}_{c1}
{n}_{c2}
{n}_{p1}
{n}_{p2}
{n}_{latent}
{MSE}_{train}
{MSE}_{val}
Table 2 shows that bigger filter sizes are preferred in terms of MSE. It could be also noted that MSE in validation set is always lower compared to the MSE in the training set. This ensures that there was no overfitting during the training process. Since there are no big differences between the MSEs for different optimal CNN architectures we fix parameters
{n}_{c1}=\text{8}
{n}_{c2}=\text{4}
{n}_{p1}=\text{2}
{n}_{p2}=\text{2}
and 8 as the dimension of convolutional filters for the experiments presented in this paper.
The change in training performance of CNN is depicted in Fig. 3. Note that the plots of training performance for other data files are of almost identical shape.
Fig. 3. Training performance of CNN
a) The case for the signal of the lead 1
b) The case for lead 2
Optimal CNN network was employed and sets of latent vectors were computed for each segment of ECG. Clustering of the latent vectors for the measurements from the first lead from “30m.mat” data file is depicted in Fig. 4.
Fig. 4. Clusters of latent vectors corresponding to different segments of ECG before sudden cardiac death (1st lead of the “30m.mat” data file). Colors depict different clusters and are not related in any manner between different parts a) – e)
It is known that 2 hours before sudden cardiac death is the time window in which an expert can spot unnatural and potentially risky ECG activity [1]. In Fig. 4 we clearly see that 5 segments during the last 25 minutes have visibly different clusters of latent space vectors. Same applies to the lead 2 (Fig. 5). Unfortunately, the data sets used do not have more measurements which could span more than 2 hours. That would be especially valuable for the validation if the methodology proposed actually signals something unusual before the SCD happens.
Note that shapes of clusters of the vectors of latent space in general have some similar areas between different segments of ECG and even between different data files. This suggests that the proposed methodology might not be completely random. Of course, bigger data sets are required to prove or disprove this.
Fig. 5. Clusters of latent vectors corresponding to different segments of ECG before sudden cardiac death (2nd lead of the “30m.mat” data file). Colors depict different clusters and are not related in any manner between different parts a) – e)
The main finding of this research is the fact that LD based feature extraction is capable of differently labeling different ECG segments although some similarities also exist.
The methodology is an interesting tool which can be tested for more different CNN parameters, different clustering approaches or different LD estimates. And that is a huge testbed for possible research outcomes which could potentially be an alternative to visual analysis of ECG performed by an expert.
Keeping track of the extracted features in the way showed here could also be considered as a universal tool or even an expert system for anomaly detection in other dynamical systems. One just needs to note cluster centers and/or sizes for the expert system to be completely automated.
This research was supported by the Research, Development and Innovation Fund of Kaunas University of Technology (project acronym DDetect).
Lerma C., Glass L. Predicting the risk of sudden cardiac death. The Journal of Physiology, Vol. 594, Issue 9, 2016, p. 2445-2458. [Publisher]
Goldberger A. L. Nonlinear Dynamics, Fractals, Cardiac Physiology and Sudden Death. Temporal Disorder in Human Oscillatory Systems. Springer, Berlin, Heidelberg, 1987, p. 118-125. [Publisher]
Murukesan L., Murugappan M., Iqbal M. Sudden cardiac death prediction using ECG signal derivative (heart rate variability): a review. 9th International Colloquium on Signal Processing and its Applications, 2013, p. 269-274. [Publisher]
Abawajy J. H., Kelarev A. V., Chowdhury M. Multistage approach for clustering and classification of ECG data. Computer Methods and Programs in Biomedicine, Vol. 112, Issue 3, 2013, p. 720-730. [Publisher]
Zhang C., Wang G., Zhao J., Gao P., Lin J., Yang H. Patient-specific ECG classification based on recurrent neural networks and clustering technique. 13th IASTED International Conference on Biomedical Engineering (BioMed), 2017, p. 63-67. [Search CrossRef]
He H., Tan Y. Automatic pattern recognition of ECG signals using entropy-based adaptive dimensionality reduction and clustering. Applied Soft Computing, Vol. 55, 2017, p. 238-252. [Publisher]
Mendoza C., Mancho A. M. Hidden geometry of ocean flows. Physical review letters, Vol. 105, Issue 3, 2010, p. 038501. [Publisher]
Das D., Ghosh R., Bhowmick B. Deep representation learning characterized by inter-class separation for image clustering. Winter Conference on Applications of Computer Vision (WACV), 2019, p. 628-637. [Search CrossRef]
Bhamare D., Suryawanshi P. Review on reliable pattern recognition with machine learning techniques. Fuzzy Information and Engineering, 2019, https://doi.org/10.1080/16168658.2019.1611030. [Publisher]
Goldberger A. L., Amaral L. A. N., Glass L., Hausdorff J. M., Ivanov P. Ch., Mark R. G., Mietus J. E., Moody G. B., Peng C-K, Stanley H. E. PhysioBank, PhysioToolkit, and PhysioNet: components of a new research resource for complex physiologic signals. Circulation, Vol. 101, Issue 23, 2000, p. 215-220. [Publisher] |
Find the covariance s_(xy) using bivariate data.
Find the covariance
{s}_{xy}
using bivariate data.
The given bivariate data is(1,6),(3,2),and(2,4).
First determine
\sum {x}_{i}{y}_{i},\sum {x}_{i}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sum {y}_{i}
{x}_{i}\to
first coordinates of the ordered pairs.
{y}_{i}\to
second coordinates of the ordered pairs.
\sum {x}_{i}{y}_{i}=1\cdot 6+3\cdot 2+2\cdot 4=20
\sum {x}_{i}=1+3+2=6
\sum {y}_{i}=6+2+4=12
For covariance
{s}_{xy}using\text{ }f\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\mu \text{ }{s}_{xy}=\frac{\sum {x}_{i}{y}_{i}-\frac{\sum {x}_{i}-\sum {y}_{i}}{n}}{n-1}
Where n = number of ordered pairs.
{s}_{xy}=\frac{20-\frac{6\cdot 12}{3}}{3-1}
=\frac{20-\frac{72}{3}}{3-1}
=\frac{2-24}{2}
=-\frac{4}{2}=-2
{s}_{xy}=-2
One hundred adults and children were randomly selected and asked whether they spoke more than one language fluently. The data were recorded in a two-way table. Maria and Brennan each used the data to make the tables of joint relative frequencies shown below, but their results are slightly different. The difference is shaded. Can you tell by looking at the tables which of them made an error? Explain. Maria's table YesNo Children0.150.25 Adults0.10.6 Brennan’s table YesNo Children0.150.25 Adults0.10.5
Give an example of each or state that the request is impossible. For any that are impossible, give a compelling argument for why that is the case.
a) A sequence with an infinite number of ones that does not converge to one.
b) A sequence with an infinite number of ones that converges to a limit not equal to one.
c) A divergent sequence such that for every
n\in N
it is possible to find nn consecutive ones somewhere in the sequence.
The average credit card debt for a recent year was $9205. Five years earlier the average credit card
debt was $6618. Assume sample sizes of 35 were used and the population standard deviations of
both samples were $1928. Find the 95% confidence interval of the difference in means.
lim\left\{n\to \mathrm{\infty }\right\}\frac{a\left\{2n\right\}}{{a}_{n}}<\frac{1}{2}
\sum {a}_{n}
lim\left\{n\to \mathrm{\infty }\right\}\frac{a\left\{2n+1\right\}}{{a}_{n}}>\frac{1}{2}
\sum {a}_{n}
$ diverges. Let
{a}_{n}=\frac{{n}^{\mathrm{ln}n}}{{\left(\mathrm{ln}n\right)}^{n}}
\frac{{a}_{2n}}{{a}_{n}}\to 0
n\to \mathrm{\infty }
What is the correlation coefficient and what is its significance? Explain why correlation coefficient is bound between -1 and 1. How correlation coefficient depicts itself in scatterplots? Aside from probabilistic models, explain what is least squares line fitting.
Consider binary strings with n digits (for example, if n=4, some of the possible strings are 0011,1010,1101, etc.)
Let Zn be the number of binary strings of length n that do not contain the substring 000
Find a recurrence relation for Zn
You are not required to find a closed form for this recurrence.
\frac{2}{5}
\frac{3}{5}
\frac{1}{10} |
Erratum: “Corner-Filleted Flexure Hinges” [ASME J. Mech. Des., 123, No. 3, pp. 346–352] | J. Mech. Des. | ASME Digital Collection
Erratum: “Corner-Filleted Flexure Hinges” [ASME J. Mech. Des., 123, No. 3, pp. 346–352]
Nicolae Lobontiu,,
Nicolae Lobontiu,
Jeffrey S. N. Paine,,
Jeffrey S. N. Paine,
Ephrahim Garcia, and,
Ephrahim Garcia, and
J. Mech. Des. Jun 2002, 124(2): 358 (1 pages)
This is a correction to: Corner-Filleted Flexure Hinges
Lobontiu, , N., Paine, , J. S. N., Garcia, and , E., and Goldfarb , M. (May 16, 2002). "Erratum: “Corner-Filleted Flexure Hinges” [ASME J. Mech. Des., 123, No. 3, pp. 346–352]." ASME. J. Mech. Des. June 2002; 124(2): 358. https://doi.org/10.1115/1.1431262
flexible structures, finite element analysis, design engineering, stress effects
Bending (Stress), Corners (Structural elements), Design engineering, Finite element analysis, Flexible structures, Hinges, Stress
The correct equations (15), (16) and (17) are:
C11=12/Ebt3{l−2r+2r[t4r+t6r2+4rt+t2+6r2r+t2t1/24r+t1/2a tan1+4r/t1/2]2r+t−14r+t−3}
C12=−6l/[Ebt32r+t4r+t2]{4r+t[l2r+t4r+t2−4r216r2+13rt+3t2]+12r22r+t2t1/24r+t1/2a tan1+4r/t1/2}
C22=3/Eb{4l−2rl2−lr+r23t3−1+1/2{{t1/24r+t1/2[−80r4+24r3t+83+2πr2t2+41+2πrt3+πt4]+42r+t36r2−4rt−t2a tan1+4r/t1/2}2t5/24r+t5/2−1+[−40r4+8lr22r−t+12r3t+43+2πr2t2+21+2πrt3+πt4/2]t−24r+t−2+8l2r6r2+4rt+t22r+t−1t−24r+t−2−{22r+t[−24l−r2r2−8r3t+14r2t2+8rt3+t4]a tan1+4r/t1/2}t−5/24r+t−5/2}}.
Corner-Filleted Flexure Hinges |
EUDML | A computational investigation of an integro-differential inequality with periodic potential. EuDML | A computational investigation of an integro-differential inequality with periodic potential.
Brown, B.M.; Kirby, V.G.
Brown, B.M., and Kirby, V.G.. "A computational investigation of an integro-differential inequality with periodic potential.." Journal of Inequalities and Applications [electronic only] 5.4 (2000): 321-342. <http://eudml.org/doc/121356>.
author = {Brown, B.M., Kirby, V.G.},
keywords = {integral inequality; Sturm-Liouville operator; Titchmarsh-Weyl -function; periodic potential; Titchmarsh-Weyl -function},
title = {A computational investigation of an integro-differential inequality with periodic potential.},
AU - Kirby, V.G.
TI - A computational investigation of an integro-differential inequality with periodic potential.
KW - integral inequality; Sturm-Liouville operator; Titchmarsh-Weyl -function; periodic potential; Titchmarsh-Weyl -function
integral inequality, Sturm-Liouville operator, Titchmarsh-Weyl
m
-function, periodic potential, Titchmarsh-Weyl
m
Articles by Brown
Articles by Kirby |
Specify Layers of Convolutional Neural Network - MATLAB & Simulink - MathWorks España
Cross Channel Normalization (Local Response Normalization) Layer
Max and Average Pooling Layers
Softmax and Classification Layers
Regression Layer
The first step of creating and training a new convolutional neural network (ConvNet) is to define the network architecture. This topic explains the details of ConvNet layers, and the order they appear in a ConvNet. For a complete list of deep learning layers and how to create them, see List of Deep Learning Layers. To learn about LSTM networks for sequence classification and regression, see Long Short-Term Memory Networks. To learn how to create your own custom layers, see Define Custom Deep Learning Layers.
The network architecture can vary depending on the types and numbers of layers included. The types and number of layers included depends on the particular application or data. For example, classification networks typically have a softmax layer and a classification layer, whereas regression networks must have a regression layer at the end of the network. A smaller network with only one or two convolutional layers might be sufficient to learn on a small number of grayscale image data. On the other hand, for more complex data with millions of colored images, you might need a more complicated network with multiple convolutional and fully connected layers.
To specify the architecture of a deep network with all layers connected sequentially, create an array of layers directly. For example, to create a deep network which classifies 28-by-28 grayscale images into 10 classes, specify the layer array
layers is an array of Layer objects. You can then use layers as an input to the training function trainNetwork.
To specify the architecture of a neural network with all layers connected sequentially, create an array of layers directly. To specify the architecture of a network where layers can have multiple inputs or outputs, use a LayerGraph object.
Create an image input layer using imageInputLayer.
An image input layer inputs images to a network and applies data normalization.
Specify the image size using the inputSize argument. The size of an image corresponds to the height, width, and the number of color channels of that image. For example, for a grayscale image, the number of channels is 1, and for a color image it is 3.
A 2-D convolutional layer applies sliding convolutional filters to 2-D input. Create a 2-D convolutional layer using convolution2dLayer.
Create a batch normalization layer using batchNormalizationLayer.
Create a ReLU layer using reluLayer.
f\left(x\right)=\left\{\begin{array}{cc}x,& x\ge 0\\ 0,& x<0\end{array}.
Create a cross channel normalization layer using crossChannelNormalizationLayer.
x
{x}^{\text{'}}
{x}^{\text{'}}=\frac{x}{{\left(K+\frac{\alpha *ss}{windowChannelSize}\right)}^{\beta }},
A 2-D max pooling layer performs downsampling by dividing the input into rectangular pooling regions, then computing the maximum of each region. Create a max pooling layer using maxPooling2dLayer.
A 2-D average pooling layer performs downsampling by dividing the input into rectangular pooling regions, then computing the average values of each region. Create an average pooling layer using averagePooling2dLayer.
A max pooling layer returns the maximum values of rectangular regions of its input. The size of the rectangular regions is determined by the poolSize argument of maxPoolingLayer. For example, if poolSize equals [2,3], then the layer returns the maximum value in regions of height 2 and width 3.An average pooling layer outputs the average values of rectangular regions of its input. The size of the rectangular regions is determined by the poolSize argument of averagePoolingLayer. For example, if poolSize is [2,3], then the layer returns the average value of regions of height 2 and width 3.
Create a dropout layer using dropoutLayer.
Create a fully connected layer using fullyConnectedLayer.
W{X}_{t}+b
{X}_{t}
A softmax layer applies a softmax function to the input. Create a softmax layer using softmaxLayer.
A classification layer computes the cross-entropy loss for classification and weighted classification tasks with mutually exclusive classes. Create a classification layer using classificationLayer.
{y}_{r}\left(x\right)=\frac{\mathrm{exp}\left({a}_{r}\left(x\right)\right)}{\sum _{j=1}^{k}\mathrm{exp}\left({a}_{j}\left(x\right)\right)},
0\le {y}_{r}\le 1
\sum _{j=1}^{k}{y}_{j}=1
P\left({c}_{r}|x,\theta \right)=\frac{P\left(x,\theta |{c}_{r}\right)P\left({c}_{r}\right)}{\sum _{j=1}^{k}P\left(x,\theta |{c}_{j}\right)P\left({c}_{j}\right)}=\frac{\mathrm{exp}\left({a}_{r}\left(x,\theta \right)\right)}{\sum _{j=1}^{k}\mathrm{exp}\left({a}_{j}\left(x,\theta \right)\right)},
0\le P\left({c}_{r}|x,\theta \right)\le 1
\sum _{j=1}^{k}P\left({c}_{j}|x,\theta \right)=1
{a}_{r}=\mathrm{ln}\left(P\left(x,\theta |{c}_{r}\right)P\left({c}_{r}\right)\right)
P\left(x,\theta |{c}_{r}\right)
P\left({c}_{r}\right)
\text{loss}=-\frac{1}{N}\sum _{n=1}^{N}\sum _{i=1}^{K}\text{}{w}_{i}{\text{t}}_{ni}\mathrm{ln}{y}_{ni},
{w}_{i}
{t}_{ni}
{y}_{ni}
{y}_{ni}
Create a regression layer using regressionLayer.
\text{MSE}=\sum _{i=1}^{R}\frac{{\left({t}_{i}-{y}_{i}\right)}^{2}}{R},
\text{loss}=\frac{1}{2}\sum _{i=1}^{R}{\left({t}_{i}-{y}_{i}\right)}^{2}.
\text{loss}=\frac{1}{2}\sum _{p=1}^{HWC}{\left({t}_{p}-{y}_{p}\right)}^{2},
\text{loss}=\frac{1}{2S}\sum _{i=1}^{S}\sum _{j=1}^{R}{\left({t}_{ij}-{y}_{ij}\right)}^{2},
[3] LeCun, Y., Boser, B., Denker, J.S., Henderson, D., Howard, R.E., Hubbard, W., Jackel, L.D., et al. ''Handwritten Digit Recognition with a Back-propagation Network.'' In Advances of Neural Information Processing Systems, 1990.
[4] LeCun, Y., L. Bottou, Y. Bengio, and P. Haffner. ''Gradient-based Learning Applied to Document Recognition.'' Proceedings of the IEEE. Vol 86, pp. 2278–2324, 1998.
[5] Nair, V. and G. E. Hinton. "Rectified linear units improve restricted boltzmann machines." In Proc. 27th International Conference on Machine Learning, 2010.
convolution2dLayer | batchNormalizationLayer | dropoutLayer | averagePooling2dLayer | maxPooling2dLayer | classificationLayer | regressionLayer | softmaxLayer | crossChannelNormalizationLayer | fullyConnectedLayer | reluLayer | leakyReluLayer | clippedReluLayer | imageInputLayer | trainingOptions | trainNetwork |
Multiphase Quasi-RMS Potential Sensor - MapleSim Help
Home : Support : Online Help : MapleSim : MapleSim Component Library : Electrical : Multiphase Systems : Sensors : Multiphase Quasi-RMS Potential Sensor
Potential Quasi-RMS Sensor
The Potential Quasi-RMS Sensor component measures the continuous quasi-RMS value of a multiphase potential.
If the potential waveform deviates from a sine curve, the output of the sensor will not be exactly the average RMS value.
{V}_{\mathrm{rms}}=\sqrt{\frac{1}{m}\sum _{k=1}^{m}{\mathrm{\phi }}_{k}^{2}}
i={i}_{p}=0
\mathrm{\phi }={v}_{p}
{\mathrm{plug}}_{p}
m
V
Real output; quasi-RMS of potential in
V
m
3 |
Car FAQ Assistant Based on BILSTM-Siamese Network
Car FAQ Assistant Based on BILSTM-Siamese Network ()
Faculty of Science, Yanbian University, Yanji, China.
With the development of artificial intelligence, the automatic question answering technology has been paid more and more attention. Along with the technology’s maturing, the problem is gradually exposed. This technique has several major difficulties. There is semantic recognition. How accurate are the answers to the questions? Because Chinese grammar is relatively complex. There are many different ways of saying the same thing. There is a powerful challenge to semantic recognition. Due to poor semantic recognition, the accuracy of the corresponding questions and answers is low. Aiming at this kind of questions, we build a Q & A library for common use of cars. Firstly, TFIDF method is used to build a basic FAQ retrieval system. Then we use BILSTM-siamese network to construct a semantic similarity model. The accuracy rate on the test set was 99.52%. The final FAQ system: first retrieves 30 most similar questions using the TFIDF model, then uses BILSTM-siamese network matching and returns the answer of the most similar question. The system has a lot to be improved. For example, how to combine this system with speech recognition? How to realize speech recognition will be a great challenge.
TFIDF, BILSTM-Siamese, FAQ
Jin, B. and Jin, Z. (2019) Car FAQ Assistant Based on BILSTM-Siamese Network. Open Access Library Journal, 6, 1-7. doi: 10.4236/oalib.1105817.
With the gradual improvement of people’s living standards, cars have gradually entered our door to door. There will also be a variety of confusion. When we have a problem, we usually open “Baidu”. However, we live in an era of rapid development of the Internet, and there is more and more complicated information, so it is difficult to find really effective and valuable information among them. Therefore, the traditional way of searching information can no longer meet people’s needs [1]. In the past two years, with the development of artificial intelligence, intelligent questions and answers have gradually received attention. We built an intelligent question and answer system, which can give effective answers to questions in mandarin. It greatly facilitates our life. The most difficult problem in intelligent question answering is to identify the meaning of sentences. For example, many different sentences express the same problem. How to identify them and then accurately correspond to the answers. It is one of the criteria for testing the quality of this question and answer system [2].
The automatic question answering system began to be studied by people in the 1960s. In the 1990s, with the development of natural language processing technology and the application of semantic information, the automatic question answering system has been greatly improved, from about 30% in the past to more than 50%. In 1993, MIT released START on the Internet, which greatly improved the number and types of questions answered, and made a staged breakthrough in the automatic question answering system technology [3]. Other countries have also begun to invest in research, including the CLEF cross-language question and answer system and Japan’s NTCIR [4]. There’s also a question and answer system for turby. Its name is FREYA. It is specifically designed to train semantic models. The development of our country’s intelligent question-answering system began in 1970. China’s first human-computer dialogue system was invented in 1980 by the Research Institute of the Chinese Academy of Sciences. Intelligent question-and-answer technology is becoming more and more mature. It gradually replaces human customer service, liberates a large amount of labor, and can solve people’s problems all day long.
2. Construct Corpus
We collect the problems that car owners often encounter during their use. And expand it to build a question-answer corpus. The same Id is the same semantic question, and the first one is the standard question. There are 123 common questions in the corpus. The structure is shown in Table 1.
Table 1. Question and answer table.
3. TFIDF-Based FAQ System
TFIDF Value
TF-IDF Full name is Term Frequency/Inverse Document Frequency. In 1973 Salton [5] came up with TFIDF. It is a word frequency-inverse document frequency algorithm and is a statistical method. The number of times a given word appears in a document, we define it as the word frequency. In how many documents a word appears in, we define it as a reverse word frequency. Then multiplying the two parts of the word frequency is the TFIDF value [6].
T{F}_{ij}=\frac{{n}_{ij}}{{N}_{j}}
ID{F}_{ij}=\mathrm{log}\frac{N}{{N}_{i}}
{\omega }_{ij}=T{F}_{ij}\ast ID{F}_{ij}
We can use “jieba” word segmentation to solve the problem of corpus, and then extract the keywords and part of speech tagging, and then statistical word frequency and reverse word frequency, calculate the TFIDF value. Its value can be approximated to a probability between 0 and 1. Arrange the value of TFIDF from small to large, the larger the value is, the more important it is. We can make the problem the row of the matrix and the keyword the column of the matrix. Form a TFIDF matrix and save it locally. When the user enters a question, the system participles the question and calculates the TFIDF. This value is mapped to the vector space, and then the inner product space cosine is applied to calculate their similarity with the previously saved TFIDF matrix locally. Eventually we return the answer to the most similar question to the user. The structural framework is shown below (Figure 1) [8].
Figure 1. System framework.
4. Semantic Similarity Model Based on BILSTM-Siamese
We convert the problem pair (Q1, Q2) into a list of characters of length 15. According to previous research scholars, character-based models perform better. So here, we use a pre-trained 100-dimensional character vector to convert the Q1, Q2 character list into a character vector matrix. Use the BILSTM-32 network to extract the timing characteristics of the statement, then connect the neural node to the fully connected layer of 64 and output the advanced features. Finally, we use cosine similarity to calculate the similarity of advanced features. [9] In the entire model, the left subnetwork shares the weight parameters with the right subnetwork. Therefore, when Q1 is exactly the same as Q2, the model output is 1. The result is shown in Figure 2.
1) Under the same problem, different questions are combined between two pairs to form a synonymous statement pair (positive sample set);
2) Using 123 standard problem sets, use tfidf to retrieve the k problems that are closest to each problem (removing themselves), and compose non-synonymous statement pairs (negative sample).
4.3. Positive Sample Expansion
1) Commonly used consultation words to expand, how to swear, how; where ® where, where, where, what position; why ® how, oh, for example: How to adjust the seat? Expanded to:
(How to adjust the seat? How to adjust the seat?)
Where is the speedometer? Expanded to:
(Where is the speedometer? Where is the speedometer? Where is the speedometer? Where is the speedometer? Where is the speedometer?)
Figure 2. Model framework.
Why is the warning light on? Expanded to:
(Why is the warning light on?, how does the warning light light up? Is the warning light on?)
2) Randomly exchange the position of two words. E.g.:
How to adjust the seat? Expanded to:
(How to adjust the seat, how to adjust the seat, how to adjust the seat, …)
4.4. Negative Sample Expansion
The negative sample is expanded by the unexpanded positive sample, that is, if q1 and q2 are negative, the positive examples of q1 and q2 can form a negative example.
The sample set built is shown in Table 2.
The constructed sample set is divided into a training set and a test set according to a ratio of 7:3, as shown in Table 3.
The model training batch size is 128, and the epoch (iteration period) is 30. Each layer of the model is processed by dropout, which speeds up the training iteration and prevents over-fitting.
The performance of the model on the test set is shown in Table 4.
Table 2. Expand problem.
Table 3. Sample collection.
Table 4. Performance table.
Accuracy (precision): (19,894 + 8590)/28,622 = 99.52%.
Recall rate (recall): 8590/8590 = 100%.
F1:2 * precision * recall/(precision + recall) = 99.20%.
6. FAQ System Structure
Use the data in Figure 1 to build a QA corpus. As shown in Figure 3, for a user input, we first use the TFIDF to recall k questions as candidate sets, and then use the BILSTM-Siamese semantic similarity model constructed above to calculate the semantic similarity between the input and k candidate sets. Eventually return the answer to the semantically most similar candidate question.
We feed the problem into the system. Run according to the program in our system. Can quickly output the most similar answer. The corresponding accuracy rate is high, as shown in Figure 4.
This paper introduces the automatic question answering system model in the
Figure 4. Actual presentation.
field of mathematical computing, and the specific operation process improves the accuracy of answering questions. In this paper, there are still some problems in expanding the vocabulary of automobile proper nouns and solving various names of the same parts. The characteristics of interaction between the two sentences of BILSTM are not perfect. These are the places that need to be improved next.
[1] Guo, H. (2008) Semantic-Based Online Book Automatic Question Answering System Research. Taiyuan University of Technology, Taiyuan.
[2] Dong, Z.T., Bao, Y.Q. and Ma, X.H. (2010) Method for Calculating Similarity of Question Sentences in Intelligent Question Answering System. Journal of Wuhan University of Technology (Information and Management Engineering Edition), 32, 31-34.
[3] Shi, J.X. (2010) Research on Automatic Question and Answer Technology Based on FAQ. Nankai University, Tianjin.
[4] Sen, Z., Wang, B. and Jones, G.J.F. (2007) ICT-DCU Question Answering Task at NTCIR-6. Proceedings of the 6th Ntcir Workshop on Research in Information Access Technologies, Tokyo, 15-18 May 2007, 154-167.
[5] Salton, G. and Clement, T.Y. (1973) On the Construction of Effective Vocabularies for Information Retrieval. Proceedings of the 1973 Meeting on Programming Languages and Information Retrieval, Gaithersburg, 4-6 November 1973, 48-60.
[6] Liu, Z. (2019) Application of TFIDF Algorithm in Article Recommendation System. Computer Knowledge and Technology, 15, 17-20.
[7] Yang, Y., Yan, D.B., Xu, M., Wan, L., Li, Q. and Qiu, D. (2019) Clas-sification of 95598 Complaint Work Order Based on Improved TFIDF Feature Weighting Algorithm. Electricity and Energy, 40, 205-207+226.
[8] Sun, H.F. and Hou, W. (2014) Application Research of Improved TFIDF Algorithm in Potential Cooperative Relationship Mining. Modern Book Information Technology, No. 10, 84-92.
[9] Lin, S.M., Chen, T.Y. and Liang, W. (2019) DGA Domain Name Detection Method Based on BILSTM Neural Network. Network Security Technology and Application, No. 1, 15-17. |
Kandi has a bag of marbles. She has
5
3
2
green, and
4
orange marbles. Kandi reaches into the bag without looking and pulls out a marble.
What is the probability that she will pull out a green marble?
How many of these are green?
\frac{2}{14}=\frac{1}{7}
If she does get a green marble and does not put it back in the bag, what is the probability that she will now pull the other green marble from the bag?
Since she has removed a marble, the bag now has only
13
marbles. What portion of these are green?
Assume that Kandi does get the second green marble and does not return it to the bag. What is the probability that she will now pull another green marble from the bag?
Are there any green marbles left to pull? |
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