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A random sample of n = 15 items are selected
Bruce Partridge 2022-03-01 Answered
A random sample of n = 15 items are selected for measurement. Nothing is known about the distribution of measurements. Are the requirements for constricting a confidence interval for the population mean satisfied? Explain in 1 - 2 complete sentences.
faraidz3i
The sample size considered here is, n = 15.
It is of interest to construct a confidence interval for the population mean measurement.
Now, an unbiased point estimate for the population mean is the sample mean. As a result, the confidence interval is constructed using the sample mean, its standard error, and a confidence level. The construction of the confidence interval for the population mean requires the basic assumption of at least an approximate normal distribution of the variable of interest.
The sampling distribution of the sample mean,
\stackrel{―}{X}
, based on a sample of size n, taken from a population with expectation
\mu
\sigma
, has expectation
{\mu }_{\stackrel{―}{X}}=\mu
{\sigma }_{\stackrel{―}{X}}=\frac{\sigma }{\sqrt{n}}
If the sample size is large
\left(n\ge 30\right)
, or the population distribution is normal, then by the central limit theorem, the sampling distribution of the sample mean is normal, with parameters
{\mu }_{\stackrel{―}{X}}
{\sigma }_{\stackrel{―}{X}}
In this case, no information is available regarding the distribution of the measurements. As a result, it cannot be said that the measurements are normally distributed.
Further, the sample size of n = 15 is not large. Thus, it is not possible to assume even an approximate normal distribution.
Thus, the requirements for constructing a confidence interval for the population mean are not satisfied.
It helped a lot)
1. A data set consists of the ages at death for each of the 41 past presidents of United States.
2. Is this set of measurement a population or a sample?
a. What is the variable being measured?
b. What measurement scale is appropriate for the data?
At work, there is a statistical process going on, which I feel is probably mathematically incorrect but can't quite put my finger on what is wrong:
They are totalling up the number of hours people work per week (in minimum units of 15 minutes), and then producing averages for the whole department per week. Obviously, the results come out to be non-integer numbers of hours with a long number of decimals.
Then, they are judging the result of certain productivity-boosting techniques and displaying the findings in "minutes gained/lost"...in some cases producing productivity gains of as little as a minute or two minutes per week.
So to summarise, they are calculating units of quarters of an hour, but then presenting the average productivity gains in minutes...is this presuming an accuracy which is not present in the initial measurement? I think it is, but don't know how to argue it to my boss.
The following appear on a physician's intake form. Identify the level of measurement of the data.
(a) Disabilities
(b) Change in health (scale of -5 to 5)
(c) Year of birth
(a) What is the level of measurement for "Disabilities"?
(b) What is the level of measurement for "Change in health (scale of - 5 to 5)"?
(c) What is the level of measurement for "Year of birth"?
(d) What is the level of measurement for "Height"?
Suppose the current measurements made on a conductor wire track follow a normal distribution with mean 10 milliamperes and variance 4 (milliamperes)^2
a) what is the probability that the value of a measurement is less than 9 milliamperes?
b) what is the probability that the value of a measurement is greater than 13 milliamps?
c) what is the probability that the value of a current measurement is between 9 and 11 milliamperes?
An ore loader moves 1200 tons/h from a mine to the surface. Convert this rate to lb/s, using 1 ton 2000 lb.
Suppose your weight is 53.81 kilograms. A scale at a health clinic gives your weight as 54.4 kilograms. A digital scale at the gym that gives readings to the nearest 0.01 kilogram gives your weight as 54.19 kilograms. Which measurement is more precise? Which is more accurate?
Which measurement is more precise? Choose the correct answer below.
A. The digital scale at the gym is more precise.
B. The health clinic scale is more precise.
Which measurement is more accurate? Choose the correct answer below.
A. The digital scale at the gym is more accurate.
B. The health clinic scale is more accurate.
X,Y
, I need to show that if
X,Y
are independent, then
E\left(Y\mid X\right)=E\left(Y\right)
. To do so, it suffices to prove that for any
B\subseteq \mathbb{R}
measurable (Borel), we have that
{\int }_{X\in B}EYdP={\int }_{X\in B}YdP
{\int }_{X\in B}EYdP=EYP\left(X\in B\right)
- but how do I show that
{\int }_{X\in B}YdP=EYP\left(X\in B\right) |
International Standard Serial Number - Wikipedia, the free encyclopedia
1Code format
2Code assignment
3Comparison with other identifiers
Toggle Comparison with other identifiers subsection
3.1Media vs Content
5Use in URNs
Toggle Use in URNs subsection
6ISSN variants
Toggle ISSN variants subsection
6.1Print ISSN
6.2Electronic ISSN
6.3Linking ISSN
Example of an ISSN encoded in an EAN-13 barcode, with explanation
An International Standard Serial Number (ISSN) is a unique eight-digit number used to identify a periodical publication at a specific media type.[1] It is internationally accepted as a fundamental identifier for distinguishing between identical serial titles and facilitating checking and ordering procedures, collection management, legal deposit, interlibrary loans etc.[2]
When a periodical is published, with the same content, in two or more different media, a different ISSN is assigned to each media type – in particular the print and electronic media types, named print ISSN (p-ISSN) and electronic ISSN (e-ISSN or eISSN). If a periodical/journal has been assigned with ISSN it does not mean it is an International publication. ISSN is a number that identifies periodicals worldwide, whether in printed form or other media (including online). It does not denote the level of coverage, connectivity and circulation of the journal. A list of major international journals around the world is available on E-journals.org.[3] E-Journals.org is the most honored and trusted source which dates back to Tim Berners-Lee (inventor of the World Wide Web).
The ISSN system was first drafted as an ISO international standard in 1971 and published as ISO 3297 in 1975.[4] The ISO subcommittee TC 46/SC 9 is responsible for the standard. To assign a unique identifier to the serial as content, linking among the different media, "linking ISSN (ISSN-L)" must be used, as defined by ISO 3297:2007.
3.1 Media vs Content
6 ISSN variants
The format of the ISSN is an eight digit code, divided by a hyphen into two four-digit numbers.[1] As an integer number, it can be represented by the first seven digits.[5] The last code digit, which may be 0-9 or an X, is a check digit. The general form of the ISSN code can be expressed by a PCRE regular expression:
\d{4}\-\d{3}[\dX]
The ISSN of the journal Hearing Research, for example, is 0378-5955, where the final 5 is the check digit. To calculate the check digit, the following algorithm may be used:
{\displaystyle 0\cdot 8+3\cdot 7+7\cdot 6+8\cdot 5+5\cdot 4+9\cdot 3+5\cdot 2}
{\displaystyle =0+21+42+40+20+27+10}
{\displaystyle =160}
{\displaystyle {\frac {160}{11}}=14{\mbox{ remainder }}6=14+{\frac {6}{11}}}
{\displaystyle 11-6=5}
An upper case X in the check digit position indicates a check digit of 10 (like a Roman ten).
There is an online ISSN checker that can validate an ISSN, based on the above algorithm.[6][7]
Code assignment[edit]
ISSN codes are assigned by a network of ISSN National Centres, usually located at national libraries and coordinated by the ISSN International Centre based in Paris. The International Centre is an intergovernmental organization created in 1974 through an agreement between UNESCO and the French government. The International Centre maintains a database of all ISSNs assigned worldwide, the ISDS Register (International Serials Data System) otherwise known as the ISSN Register. At the end of 2013[update], the ISSN Register contained records for 1,749,971 items.[8]
Media vs Content[edit]
Separate ISSN are needed for serials in different media (except reproduction microforms). Thus, the print and electronic media versions of a serial need separate ISSN.[9] Also, a CD-ROM version and a web version of a serial require different ISSN since two different media are involved. However, the same ISSN can be used for different file formats (e.g. PDF and HTML) of the same online serial.
This "media-oriented identification" of serials made sense in the 1970s. In the 1990s and onward, with PCs, good screens, and the Web, what makes sense is to consider only content, independent of media. This "content-oriented identification" of serials' was a repressed demand during a decade, but no ISSN's update or initiative occurred. A natural extension for ISSN, the unique-identification of the articles in the serials, was the main demand application. An alternative serials' contents model arrived with the indecs Content Model and its application, the Digital Object Identifier (DOI), as ISSN-independent initiative, consolidated in the 2000s.
The ISSN Register is not freely available for interrogation on the web, but is available by subscription. There are several routes to the identification and verification of ISSN codes for the public:
The print version of a periodical typically will include the ISSN code as part of the publication information.
Most periodical websites contain ISSN code information.
Derivative lists of publications will often contain ISSN codes; these can be found through on-line searches with the ISSN code itself or periodical title.
WorldCat permits searching its catalog by ISSN, by entering "issn:"+ISSN code in the query field. One can also go directly to an ISSN's record by appending it to "https://www.worldcat.org/ISSN/", e.g. https://www.worldcat.org/ISSN/1021-9749. This does not query the ISSN Register itself, but rather shows whether any Worldcat library holds an item with the given ISSN.
An ISSN can be encoded as a Uniform Resource Name (URN) by prefixing it with "urn:ISSN:".[11] For example Rail could be referred to as "urn:ISSN:1534-0481". URN namespaces are case-sensitive, and the ISSN namespace is all caps.[12] If the checksum digit is "X" then it is always encoded in uppercase in a URN.
The util URNs are content-oriented, but ISSN is media-oriented:
ISSN is not unique when the concept is "a journal is a set of contents, generally copyrighted content": the same journal (same contents and same copyrights) have two or more ISSN codes. A URN needs to point to "unique content" (a "unique journal" as a "set of contents" reference).
Examples: Nature has an ISSN for print, 0028-0836, and another for the same content on the Web, 1476-4687; only the oldest (0028-0836) is used as a unique identifier. As the ISSN is not unique, the U.S. National Library of Medicine needed to create, prior to 2007, the NLM Unique ID (JID).[13]
ISSN does not offer resolution mechanisms like a Digital Object Identifier (DOI) or an URN does, so the DOI is used as a URN for articles, with (for historical reasons) no need for an ISSN's existence.
Example: the DOI name "10.1038/nature13777" can represented as an HTTP string by https://dx.doi.org/10.1038/nature13777, and is redirected (resolved) to the current article's page; but there is no ISSN's online service, like http://dx.issn.org/, to resolve the ISSN of the journal (in this sample 1476-4687), that is, a kind of https://dx.issn.org/1476-4687 redirecting to the journal's home.
A unique URN for serials simplifies the search, recovery and delivery of data for various services including, in particular, search systems and knowledge databases.[10] ISSN-L was created to fill this gap.
ISSN variants[edit]
p-ISSN, the "default" ISSN, is the ISSN for the print media (paper) version of a periodical.
e-ISSN (or eISSN) is the ISSN for the electronic media (online) version of a periodical.
ISSN-L is a unique identifier for all versions of the periodical containing the same content across different media. As defined by ISO 3297:2007, the "linking ISSN (ISSN-L)" provides a mechanism for collocation or linking among the different media versions of the same continuing resource.
The ISSN-L is one ISSN number among the existing ISSNs, so, does not change the use or assignment of "ordinary" ISSNs;[14] it is based on the ISSN of the first published medium version of the publication. If the print and online versions of the publication are published at the same time, the ISSN of the print version is chosen as the basis of the ISSN-L.
With ISSN-L is possible to designate one single ISSN for all those media versions of the title. The use of ISSN-L facilitates search, retrieval and delivery across all media versions for services like OpenURL, library catalogues, search engines or knowledge bases.[15]
^ a b "What is an ISSN?". ISSN International Centre. Retrieved 13 July 2014.
^ "Collection Metadata Standards – ISSN UK Centre". British Library. Retrieved 14 July 2014.
^ Scott D. Russell, University of Oklahoma. "E-Journals.Org". e-journals.org.
^ "ISSN, a standardised code". ISSN International Centre. Retrieved 13 July 2014.
^ "Total number of records in the ISSN Register" (PDF). ISSN International Centre. January 2014. Retrieved 13 July 2014.
^ Rozenfeld, Slawek (January 2001). "Using The ISSN (International Serial Standard Number) as URN (Uniform Resource Names) within an ISSN-URN Namespace – RFC 3044". IETF Tools. Retrieved 15 July 2014.
^ "La nueva Norma ISSN facilita la vida de la comunidad de las publicaciones en serie", A. Roucolle. http://www.latindex.unam.mx/biblioteca/nunoiso.html
"Cataloging Part", ISSN Manual (PDF), ISSN International Centre .
Retrieved from "https://en.wikipedia.beta.wmflabs.org/w/index.php?title=International_Standard_Serial_Number&oldid=214708"
This page was last edited 22:12, 31 March 2015 by Wikipedia user Jdforrester (WMF). Based on work by Wikipedia anonymous users imported>Seanoneal and imported>Trackteur. |
LaTeX - Maple Help • Boîte à outils et add-ons Maple • MapleNet • Maple Player gratuit
Home : Support : Online Help : What's New and Release Notes : LaTeX
\mathrm{latex}\left(\mathrm{expression}\right)
followed by pressing Enter produces the display on the screen of the LaTeX form of expression, that you can copy and paste into a .tex file, to be used with any LaTeX application (not provided with the Maple system).
Several settings of the translation to LaTeX can now be adjusted according to preference, using the
\mathrm{latex}:-\mathrm{Settings}
command or its synonym
\mathrm{LaTeX}:-\mathrm{Settings}
. These settings and the corresponding default values can be seen entering the command with no arguments or using the keyword query
\mathrm{latex}:-\mathrm{Settings}\left(\right)
[\textcolor[rgb]{0,0,1}{\mathrm{cacheresults}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{commabetweentensorindices}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{invisibletimes}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{""}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{leavespaceafterfunctionname}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{linelength}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{66}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{powersoftrigonometricfunctions}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{mixed}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{spaceaftersqrt}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usecolor}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usedisplaystyleinput}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{useimaginaryunit}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{I}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{useinputlineprompt}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{userestrictedtypesetting}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usespecialfunctionrules}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usetypesettingcurrentsettings}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}]
\mathrm{cacheresults}=\mathrm{false}
\mathrm{latex}:-\mathrm{Forget}\left(\right)
\mathrm{ab}
\mathrm{invisibletimes}="\\,"
leavespaceafterfunctionname : default value is false; if set to true no \! LaTeX negative spacing command will be placed between a function name, say
F
and the large parenthesis \left( \right) surrounding the function's arguments.
"\\"
{\mathrm{sin}\left(x+y\right)}^{2}
{\mathrm{sin}}^{2}\left(x+y\right)
\mathrm{arcsin}\left(x+y\right)
\mathrm{interface}\left(\mathrm{imaginaryunit}\right)
\mathrm{interface}\left(\mathrm{format}\right)="worksheet"
\mathrm{interface}\left(\mathrm{format}\right)="document"
In the resulting filename.txt file, adjust the new automatic line-breaking of math formulas by placing
\mathrm{\\}
wherever you want an additional line-break, or enclosing a formula's subexpression between
{}
to avoid its automatic line-breaking.
\mathrm{latex}:-\mathrm{Settings}\left(\right)
\mathrm{ee}≔\frac{a}{x+\frac{y}{3}}
\textcolor[rgb]{0,0,1}{\mathrm{ee}}\textcolor[rgb]{0,0,1}{≔}\frac{\textcolor[rgb]{0,0,1}{a}}{\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{+}\frac{\textcolor[rgb]{0,0,1}{y}}{\textcolor[rgb]{0,0,1}{3}}}
\mathrm{latex}\left(\mathrm{ee}\right)
\mathrm{LaTeX}\left(\mathrm{ee}\right)
\mathrm{latex}\left(\mathrm{ee},\mathrm{output}=\mathrm{string}\right)
\textcolor[rgb]{0,0,1}{"\frac\left\{a\right\}\left\{x +\frac\left\{y\right\}\left\{3\right\}\right\}"}
\mathrm{ee}≔\left(\mathrm{Int}=\mathrm{int}\right)\left(\frac{1}{{x}^{2}+1},x\right)
\textcolor[rgb]{0,0,1}{\mathrm{ee}}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{\int }\frac{\textcolor[rgb]{0,0,1}{1}}{{\textcolor[rgb]{0,0,1}{x}}^{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\textcolor[rgb]{0.564705882352941,0.564705882352941,0.564705882352941}{ⅆ}\textcolor[rgb]{0,0,1}{x}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{arctan}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{x}\right)
\mathrm{latex}\left(\mathrm{ee}\right)
\mathrm{query}
\mathrm{latex}:-\mathrm{Settings}\left(\right)
[\textcolor[rgb]{0,0,1}{\mathrm{cacheresults}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{commabetweentensorindices}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{invisibletimes}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{""}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{leavespaceafterfunctionname}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{linelength}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{66}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{powersoftrigonometricfunctions}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{mixed}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{spaceaftersqrt}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usecolor}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usedisplaystyleinput}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{useimaginaryunit}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{I}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{useinputlineprompt}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{userestrictedtypesetting}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usespecialfunctionrules}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{true}}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{\mathrm{usetypesettingcurrentsettings}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}]
\mathrm{latex}:-\mathrm{Settings}\left(\mathrm{usecolor}=\mathrm{false}\right)
[\textcolor[rgb]{0,0,1}{\mathrm{usecolor}}\textcolor[rgb]{0,0,1}{=}\textcolor[rgb]{0,0,1}{\mathrm{false}}]
\mathrm{latex}\left(\mathrm{ee}\right) |
Home : Support : Online Help : Programming : ImageTools Package : FormatFromName
guess image file format from name
FormatFromName( file )
string; name of an image file
The FormatFromName command examines the passed filename and determines the image file format based on the extension. It returns either a member of ImageTools[Formats] or NULL if it was unable to guess.
The parameter file is the name of the image file.
The extensions .jpg, .jpeg, .jpe, and .jeg are recognized as JPEG files.
The extensions .tif, .tiff, and .tff are recognized as TIFF files.
The extensions .bmp and .dib are recognized as BMP files.
The extension .png is recognized as a PNG file.
\mathrm{with}\left(\mathrm{ImageTools}\right):
\mathrm{FormatFromName}\left("photo.jpeg"\right)
\textcolor[rgb]{0,0,1}{\mathrm{JPEG}}
\mathrm{FormatFromName}\left("/tmp/snail.tif"\right)
\textcolor[rgb]{0,0,1}{\mathrm{TIFF}}
\mathrm{FormatFromName}\left("music.mp3"\right) |
Properties of Trapezoids (US) / Trapeziums (UK) | Brilliant Math & Science Wiki
Hemang Agarwal, Aditya Virani, Pranshu Gaba, and
A trapezoid, or trapezium, is a quadrilateral which has a pair of parallel sides. The two parallel sides are called the trapezoid's bases, and the two non-parallel sides are referred to as the legs.
For any trapezoid, the following two correspond:
(1) Like any other quadrilateral, the degree measures of the four angles add up to 360 degrees. Thus the trapezoid
ABCD
shown in the figure above satisfies the following:
\angle A + \angle B + \angle C + \angle D = 360^\circ.
(2) Two angles on the same side of a leg are always supplementary, that is, they add up to 180 degrees. Thus for the trapezoid
ABCD
shown above, we have
\begin{aligned} \angle A + \angle B &= 180^\circ \\ \angle C + \angle D &= 180^\circ. \end{aligned}
The area of a trapezoid is found using the formula
\frac{ 1 } {2} h \times (a + b )
and
b
are length of the parallel sides and
h
is the perpendicular height of the trapezoid.
An isosceles trapezoid is a trapezoid that has congruent legs.
Thus, for the isosceles trapezoid
ABCD
in the figure above, the following correspond:
(1) The bases are parallel.
(2) The legs are equal in length. Thus,
\lvert\overline{BA}\rvert=\lvert\overline{CD}\rvert.
(3) The angles the two legs make with a base are equal. Thus,
\angle BAD=\angle CDA
\angle ABC=\angle DCB.
\angle A + \angle B + \angle C = 290^\circ
in the trapezoid above, what is
\angle D?
Since a trapezoid is a quadrilateral, the sum of its four internal angles must equal
360^\circ .
\begin{aligned} \angle D &= 360^\circ - (\angle A + \angle B + \angle C) \\ &= 360^\circ - 290^\circ \\ &= 70^\circ. \ _\square \end{aligned}
The figure above depicts an isosceles trapezoid.
\overline{AB}
is 5, what is the length of
\overline{CD} ?
An isosceles trapezoid is a trapezoid that has congruent legs. Thus, the lengths of
\overline{CD}
\overline{AB}
are equal, i.e.
\lvert \overline{CD} \rvert = \lvert \overline{AB} \rvert = 5. \ _\square
The above figure depicts a trapezoid where
\overline{AB} \parallel \overline{CD} .
\angle A=120^\circ,
\angle D?
Since two angles on the same side of a trapezoid's leg add up to
180^\circ,
\begin{aligned} \angle A + \angle D = 180^\circ, \\ \angle B + \angle C = 180^\circ. \end{aligned}
\angle D = 180^\circ - \angle A = 180^\circ - 120^\circ = 60^\circ. \ _\square
The above figure depicts an isosceles trapezoid. If
\angle B=65^\circ ,
\angle D ?
In an isosceles trapezoid, the angles on either side of the bases are the same. Thus, it follows that
\angle B = \angle C = 65^\circ.
Since two angles on the same side of a trapezoid's leg are supplementary, we know that
\begin{aligned} \angle C + \angle D &= 180^\circ \\ \Rightarrow \angle D &= 180^\circ - \angle C \\ &= 180^\circ - 65^\circ \\ &= 115^\circ. \end{aligned}
\angle D=115^\circ. \ _\square
The above figure shows an isosceles trapezoid where
\overline{AD} \parallel \overline{BC}.
\lvert \overline{AB} \rvert = \lvert \overline{AD} \rvert
\lvert \overline{BC} \rvert = 2 \lvert \overline{AD} \rvert ,
\angle ABD ?
Let points
A^{'}
D^{'}
be the perpendicular foots on
\overline{BC}
A
D,
\lvert\overline{AD}\rvert=\lvert\overline{A'D'}\rvert.
Since the problem states that
ABCD
is an isosceles trapezoid, we know that
\lvert\overline{BA'}\rvert=\lvert\overline{CD'}\rvert.
Given that the length of
\overline{BC}
is twice the length of
\overline{AD},
\begin{aligned} \lvert \overline{BA^{'}} \rvert &=\frac{1}{2}\cdot \lvert \overline{A^{'}D^{'}} \rvert \end{aligned}
and since the length ratio of the hypotenuse and the base of
\triangle ABA^{'}
2:1 ,
\angle ABA^{'}
60^\circ . \qquad (1)
\triangle ABD
is an isosceles triangle, so
\angle ABD=\angle ADB,
\overline{AD} \parallel\overline{BC},
\angle DBC=\angle ADB.
\angle ABD = \angle ADB = \angle DBC . \qquad (2)
From (1) and (2), we know that
\angle ABD=\angle DBA^{'}
\angle ABA^{'} = 60^\circ.
\begin{aligned} \angle ABD + \angle DBC = 60^\circ \\ 2\cdot \angle ABD = 60^\circ \\ \angle ABD = 30^\circ. \end{aligned}
\angle ABD
30^\circ.\ _ \square
Cite as: Properties of Trapezoids (US) / Trapeziums (UK). Brilliant.org. Retrieved from https://brilliant.org/wiki/properties-of-trapezoids-us-trapeziums-uk/ |
Find parametric equation of the line passing through point (4,\
Weltideepq 2021-12-04 Answered
Find parametric equation of the line passing through point
\left(4,\text{ }2\right)
parallel to vector
\stackrel{\to }{v}=-\stackrel{^}{i}+5\stackrel{^}{j}
A line passes through the point
\left(4,\text{ }2\right)
The line is parallel to the direction vector
\stackrel{\to }{v}=-\stackrel{^}{i}+5\stackrel{^}{j}
We want to find the parametric equations of the line passing through
\left(4,\text{ }2\right)
\stackrel{\to }{v}=-\stackrel{^}{i}+5\stackrel{^}{j}
The parametric equations for a line though the point
\left({x}_{0},\text{ }{y}_{0}\right)
and parallel to direction vector
⟨a,\text{ }b⟩
x={x}_{0}+at
y={y}_{0}+bt
In the given problem, we have
\left({x}_{0},\text{ }{y}_{0}\right)=\left(4,\text{ }2\right)
⟨a,\text{ }b⟩=⟨-1,\text{ }5⟩
x={x}_{0}+at\text{ }ask\text{ }is\text{ }a\text{ }parametric\text{ }equation\text{ },we\text{ }can\text{ }replace\text{ }values\text{ }of\text{ }{x}_{0}
and a in the parametric equation.
x={x}_{0}+at
⇒x=4+\left(-1\right)t
⇒x=4-t
y={y}_{0}+b\text{ }task\text{ }is\text{ }a\text{ }parametric\text{ }equation,\text{ }we\text{ }can\text{ }replace\text{ }values\text{ }of\text{ }{y}_{0}
and b in the parametric equation.
y={y}_{0}+bt
⇒y=2+\left(5\right)t
⇒y=2+5t
x=4-t
y=2+5t
are parametric equations for the line passing through the point
\left(4,\text{ }2\right)
\stackrel{\to }{v}=-\stackrel{^}{i}+5\stackrel{^}{j}
\left(x,\text{ }y\right)=
The three masses shown in the figure are connected by massless, rigid rods.
Find the moment of inertia about an axis that passes through mass A and is perpendicular to the page. Express your answer to two significant figures and include the appropriate units.
Find the moment of inertia about an axis that passes through masses B and C. Express your answer to two significant figures and include the appropriate units.
How do you find the polar coordinates of
\left(-4,\text{ }0\right)
Find the parametrization of the function
f\left(x\right)=2{x}^{3}-7
Represent the plane curve by a vector-valued function.
4x+5y-20=0
Solve the system of equations with elimination method and what is the type of solution?
\left\{\begin{array}{l}x–2y=-3\\ 4x-3y=10\end{array}
Find the gradient at the point (3,-3,2) of the scaler field given by
f=xy-4ze+35 |
Laplacian of scalar function - MATLAB laplacian - MathWorks France
Compute Laplacian of Symbolic Expression
Compute Laplacian of Symbolic Function
Laplacian of Scalar Function
laplacian(f,x)
laplacian(f,x) computes the Laplacian of the scalar function or functional expression f with respect to the vector x in Cartesian coordinates.
laplacian(f) computes the Laplacian of the scalar function or functional expression f with respect to a vector constructed from all symbolic variables found in f. The order of variables in this vector is defined by symvar.
Compute the Laplacian of this symbolic expression. By default, laplacian computes the Laplacian of an expression with respect to a vector of all variables found in that expression. The order of variables is defined by symvar.
laplacian(1/x^3 + y^2 - log(t))
1/t^2 + 12/x^5 + 2
Create this symbolic function:
f(x, y, z) = 1/x + y^2 + z^3;
Compute the Laplacian of this function with respect to the vector [x, y, z]:
L = laplacian(f, [x y z])
L(x, y, z) =
6*z + 2/x^3 + 2
Input, specified as a symbolic expression or function.
Input, specified as a vector of symbolic variables. The Laplacian is computed with respect to these symbolic variables.
The Laplacian of the scalar function or functional expression f with respect to the vector X = (X1,...,Xn) is the sum of the second derivatives of f with respect to X1,...,Xn:
\Delta f=\sum _{i=1}^{n}\frac{{\partial }^{2}f}{\partial {x}_{i}^{2}}
If x is a scalar, laplacian(f, x) = diff(f, 2, x).
The Laplacian of a scalar function or functional expression is the divergence of the gradient of that function or expression:
\Delta f=\nabla \cdot \left(\nabla f\right)
Therefore, you can compute the Laplacian using the divergence and gradient functions:
syms f(x, y)
divergence(gradient(f(x, y)), [x y])
curl | diff | divergence | gradient | hessian | jacobian | potential | vectorPotential |
Pi | Brilliant Math & Science Wiki
Alexander Katz, Andy Hayes, Sharky Kesa, and
\pi
is the ratio between a circle's circumference and diameter. That is,
\dfrac{\text{circumference}}{\text{diameter}}=\pi.
\pi
is a fundamental constant in mathematics, especially in geometry, trigonometry, and calculus. The first 10 digits of
\pi
(sometimes written as "pi" and pronounced as "pie") are
3.141592653...
, but any finite list of digits is an approximation of
\pi
\pi
is an irrational number
\big(
meaning it is not perfectly equivalent to any ratio of whole numbers,
\frac{a}{b}\big),
as well as a transcendental number.
Applications of
\pi
\pi
using Polygons
\pi
by Statistical Techniques
\pi
in Complex Numbers, Trigonometry, and Euler's Formula
Exactly Defining
\pi
as an Infinite Series
\pi
as an Integral
\pi
Common geometric formulae involving
\pi:
\pi = \frac{C}{d},
C
is the circumference of a circle and
is the diameter.
A = \pi r^2,
A
is the area of a circle and
r
V = \frac{4}{3}\pi r^3,
V
is the volume of a sphere and
r
SA = 4\pi r^2,
SA
is the surface area of a sphere and
r
A = \pi a b,
A
is the area of an ellipse and
and
b
are the semi-minor and semi-major axes.
\frac{\pi}{6}
\frac{\pi}{4}
\frac{\pi}{4r^2}
\pi^2
\pi
By measuring the perimeter of these polygons, we can approximate the perimeter of the circle.
The perimeters of these
n
-gons can be obtained from regular polygons trigonometry and we can then use small-angle approximation to find the limit,
2\pi.
An approximation for
\pi
can be derived from the perimeters of a circumscribed square and an inscribed square.
P_1
be the perimeter of the larger square,
P_2
the perimeter of the smaller square, and
C
the circumference of the circle.
The circumference of the circle can then be approximated by the average of the two perimeters:
C\approx \frac{P_1+P_2}{2}.
d
is the diameter of the circle, then the length of a side of the larger square is also
d
. The length of a side of the smaller square can be found using a special right triangle. This length is
\frac{\sqrt{2}}{2}d
\begin{aligned} P_1&=4d\\ P_2&=2\sqrt{2}d. \end{aligned}
So, our approximation for the circumference is
C\approx\big(2+\sqrt{2}\big)d
. Dividing this equation by
d
yields us an approximation for
\pi:
\pi\approx 2+\sqrt{2}.
This isn't a particularly good approximation! However, better approximations can be obtained using a similar method with regular polygons with more sides.
\pi
There are many simulations and statistical techniques that can be used to approximate
\pi
. Long before computers were invented, the French mathematicians Buffon
(1707-1788)
and Laplace
(1749-1827)
proposed using a stochastic simulation to estimate the value of
π
"Buffon's needle" famously approximates
\pi
using the fact that a straight needle is equally likely to land at any angle when it is tossed onto a plane.
needles of length 1 are dropped onto a floor with strips of wood 1 unit apart. The expected value for the number of needles which cross two strips of wood is
\frac{2}{\pi}\cdot n.
The crux of the proof is the idea of breaking up each needle into small segments. Loosely speaking, linearity of expectation tells us that the expected number of times that the needle will cross between two strips of wood is proportional to the length of the needle.
l,
cl
c.
\pi
c\pi = 2,
c = \frac{2}{\pi}.
If one is uncomfortable with the idea of a circle, consider approximating the circle by combining a bunch of very, very small linear segments.
_\square
This results in an easy simulation technique for approximating
\pi.
Simply throw
needles on the floor, count the number of needles
x
which cross two strips of wood, and then
\pi \approx \frac{2n}{x}.
This type of simulation technique is known as a Monte Carlo simulation.
Another approximation is even simpler. Think about inscribing a circle in a square with sides of length
2
, so that the radius
r
of the circle is of length
1
The true area of the circle is
\pi r^{2}=\pi
. Buffon suggested that he could estimate the area of a circle by a dropping a large number of needles (which he argued would follow a random path as they fell) in the vicinity of the square. The ratio of the number of needles with tips lying within the square to the number of needles with tips lying within the circle could then be used to estimate the area of the circle:
\frac { { A }_{ c } }{ { A }_{ s } } =\frac { \pi { r }^{ 2 } }{ 4{ r }^{ 2 } } \implies \pi =4\frac { { A }_{ c } }{ { A }_{ s } }.
\pi
\pi
is important in trigonometry, as it provides a more natural interpretation of angles than degrees do. Specifically, radians are defined so that
2\pi
radians are equivalent to a full circle (in other words,
\pi
, understood as
\pi
radians, is commonly equal to 180 degrees when used in trigonometry); in this way, an angle of
\theta
corresponds to an arc length of
\theta \cdot r
r
is the radius of the circle. Equivalently, radians are defined so that one radian corresponds to an arc length equal to the radius of the circle.
This also allows for easy evaluation of trigonometric functions:
\theta
\sin\theta \hspace{10mm}
\cos\theta \hspace{10mm}
\tan\theta \hspace{10mm}
\frac{\pi}{6}
\frac{1}{2}
\frac{\sqrt{3}}{2}
\frac{\sqrt{3}}{3}
\frac{\pi}{4}
\frac{\sqrt{2}}{2}
\frac{\sqrt{2}}{2}
\frac{\pi}{3}
\frac{\sqrt{3}}{2}
\frac{1}{2}
\sqrt{3}
\frac{\pi}{2}
1 0 undefined
\pi
This also allows complex numbers to be presented in polar coordinates, meaning that any complex number can be written in the form
re^{i\theta}
r
\theta
\pi
plays a key role in their analysis, as this complex number is equivalent to
re^{i\theta}=r(\cos\theta + i\sin\theta)
as both sides represent the same point in the complex plane. Setting
\theta = \pi
gives the famous identity
e^{i\pi} + 1 = 0.
It is conjectured that this is the only nontrivial relation between the numbers
e, i,
\pi
, making this result (known as Euler's formula) even more remarkable.
\pi
\pi
cannot be represented as any finite series of rational numbers (as a consequence of being irrational), there are many ways to express
\pi
as an infinite series. The "first," historically speaking, of these sequences was an infinite product:
\frac{2}{\pi} = \frac{\sqrt{2}}{2} \times \frac{\sqrt{2+\sqrt{2}}}{2} \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2} \times \cdots,
which, since the
n^\text{th}
term of the infinite product is
\cos\frac{\pi}{2^n}
by repeated application of the half-angle formula, is equivalent to the statement
\lim_{n \rightarrow \infty} \frac{1}{2^n\sin\frac{\pi}{2^{n+1}}} = \frac{2}{\pi},
which is true, as
\sin\frac{\pi}{2^{n+1}}\approx \frac{\pi}{2^{n+1}}
n
More useful series involves infinite sums rather than products, because calculating the first few terms of an infinite expression of
\pi
gives good approximations of its value. The simplest is the Gregory-Leibniz series, which uses the evaluation of the Taylor series of
\arctan x
at 1:
\frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots.
However, this converges slowly, meaning that a large number of terms are necessary to get a good approximation of
\pi
. Better series include Machin-like formulae:
\begin{aligned} \frac{\pi}{4} &= 4\arctan\frac{1}{5}-\arctan\frac{1}{239}\\ \frac{\pi}{4} &= 22\arctan\frac{24478}{873121}+17\arctan\frac{685601}{69049993}. \end{aligned}
In the modern computer era, even better series are known:
\frac{1}{\pi} = \frac{2\sqrt{2}}{9801}\sum_{k=0}^{\infty}\frac{(4k)!(1103+26390k)}{k!^4(396^{4k})},
which can be used to calculate millions of digits of
\pi
\pi
\pi
can also be defined in terms of integrals. The simplest are those that represent the area or perimeter of a circle:
\int_{-1}^1\sqrt{1-x^2}dx = \frac{\pi}{2},
\left(x,\sqrt{1-x^2}\right)
represents the top half of a circle over
[-1,1]
More complicated integrals come from statistics, such as the area under the normal distribution:
\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}
and the Cauchy distribution:
\int_{-\infty}^{\infty}\frac{1}{x^2+1}dx=\pi.
\pi
appears in various expressions involving the gamma function, which is an extension of the factorial function defined as
\Gamma(t) = \int_0^{\infty}x^{t-1}e^{-x}dx,
\pi
when evaluated on half-integers. For example,
\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}, ~~\Gamma\left(\frac{5}{2}\right)=\frac{3}{4}\sqrt{\pi}.
The first 500 digits of
\pi
\small 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679 \\ \small 8214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196 \\ \small 4428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273 \\ \small 7245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094 \\ \small 3305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912...
For more digits, follow the link by clicking here.
Cite as: Pi. Brilliant.org. Retrieved from https://brilliant.org/wiki/pi/ |
Hubble volume - Wikipedia
Not to be confused with Hubble bubble (astronomy).
Visualization of the whole observable universe. The inner blue ring indicates the approximate size of the Hubble volume.
In cosmology, a Hubble volume (named for the astronomer Edwin Hubble) or Hubble sphere, subluminal sphere, causal sphere and sphere of causality is a spherical region of the observable universe surrounding an observer beyond which objects recede from that observer at a rate greater than the speed of light due to the expansion of the Universe.[1] The Hubble volume is approximately equal to 1031 cubic light years.
The proper radius of a Hubble sphere (known as the Hubble radius or the Hubble length) is
{\displaystyle c/H_{0}}
{\displaystyle c}
{\displaystyle H_{0}}
is the Hubble constant. The surface of a Hubble sphere is called the microphysical horizon,[2] the Hubble surface, or the Hubble limit.
More generally, the term "Hubble volume" can be applied to any region of space with a volume of order
{\displaystyle (c/H_{0})^{3}}
. However, the term is also frequently (but mistakenly) used as a synonym for the observable universe; the latter is larger than the Hubble volume.[3][4]
The center of the Hubble volume and Observable universe is arbitrary in relation to the overall universe, instead it is centered around the observer.
1 Relationship to age of the universe
2 Hubble limit as an event horizon
Relationship to age of the universe[edit]
The Hubble length
{\displaystyle c/H_{0}}
is 14.4 billion light years in the standard cosmological model, somewhat larger than
{\displaystyle c}
times the age of the universe, 13.8 billion years.
Hubble limit as an event horizon[edit]
For objects at the Hubble limit, the space between us and the object of interest has an average expansion speed of c. So, in a universe with constant Hubble parameter, light emitted at the present time by objects outside the Hubble limit would never be seen by an observer on Earth. That is, the Hubble limit would coincide with a cosmological event horizon (a boundary separating events visible at some time and those that are never visible[5]). See Hubble horizon for more details.
However, the Hubble parameter is not constant in various cosmological models[3] so that the Hubble limit does not, in general, coincide with a cosmological event horizon. For example, in a decelerating Friedmann universe the Hubble sphere expands with time, and its boundary overtakes light emitted by more distant galaxies so that light emitted at earlier times by objects outside the Hubble volume still may eventually arrive inside the sphere and be seen by us.[3] Similarly, in an accelerating universe with a decreasing Hubble constant, the Hubble volume expands with time and can overtake light from sources previously receding relative to us.[3] In both of these circumstances, the cosmological event horizon lies beyond the Hubble Horizon. In a universe with an increasing Hubble constant, the Hubble horizon will contract, and its boundary overtakes light emitted by nearer galaxies so that light emitted at earlier times by objects inside the Hubble sphere will eventually recede outside the sphere and will never be seen by us.[1] If the shrinkage of the Hubble volume does not stop due to some yet unknown phenomenon (one suggestion is the "early phase transition"), the Hubble volume will become nearly a point (due to the uncertainty principle pure singularities are impossible; also a proportion of their self-interactions are energetic enough to produce escaping particles via quantum tunneling), meeting the criteria of big bang. The justification of this view is that no subluminal Hubble volume will exist and pointwise superluminal expansion (the generalization of the Big Bang theory) will prevail everywhere or at least in a vast region of the universe. In this cyclic cosmology (there are many other cyclic versions) the universe always expands and doesn't revert to a smaller default size (non-conformal or expandatory conformal, non-Penrosean expandatory cyclic cosmology).
Observations indicate that the expansion of the universe is accelerating,[6] and the Hubble constant is thought to be decreasing.[7] Thus, sources of light outside the Hubble horizon but inside the cosmological event horizon can eventually reach us. A fairly counter-intuitive result is that photons we observe from the first ~5 billion years of the universe come from regions that are, and always have been, receding from us at superluminal speeds. [3]
Hubble horizon
^ a b Edward Robert Harrison (2003). Masks of the Universe. Cambridge University Press. p. 206. ISBN 978-0-521-77351-5.
^ N. Carlevaro & G. Montani (2009). "Study of the Quasi-isotropic Solution near the Cosmological Singularity in Presence of Bulk-Viscosity". International Journal of Modern Physics D. 17 (6): 881–896. arXiv:0711.1952. Bibcode:2008IJMPD..17..881C. doi:10.1142/S0218271808012553. S2CID 9943577.
^ a b c d e For a discussion of why objects that are outside the Earth's Hubble sphere can be seen from Earth, see TM Davis & CH Linewater (2003). "Expanding Confusion: common misconceptions of cosmological horizons and the superluminal expansion of the universe". Publications of the Astronomical Society of Australia. 21: 97–109. arXiv:astro-ph/0310808. Bibcode:2004PASA...21...97D. doi:10.1071/AS03040. S2CID 13068122.
^ For an example of mistaken usage, see Max Tegmark (2004). "Parallel Universes". In Barrow, J. D.; Davies, J. D.; Harper, C. L. (eds.). Science and Ultimate Reality: From Quantum to Cosmos. Cambridge University Press. pp. 459ff. ISBN 978-0-521-83113-0.
^ Edward Robert Harrison (2000). Masks of the Universe. Cambridge University Press. p. 439. ISBN 978-0-521-66148-5.
^ John L Tonry; et al. (2003). "Cosmological Results from High-z Supernovae". Astrophys J. 594 (1): 1–24. arXiv:astro-ph/0305008. Bibcode:2003ApJ...594....1T. doi:10.1086/376865. S2CID 119080950.
^ "Is the universe expanding faster than the speed of light?". Ask an Astronomer at Cornell University. Archived from the original on 23 November 2003. Retrieved 5 June 2015.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Hubble_volume&oldid=1085185624" |
Choose one technology and explain why it is useful for
Choose one technology and explain why it is useful for modeling complex systems.
A complex system has many variables which interact with each other. Engineering software is one technology which can assist how different technology can perform under different conditions (e.g. weather or resisting corrosion).
An example is how an airplane will perform under different weather environments. Since airplanes are very complex, they can focus on one area such as the wings and model how they can perform against conditions (e.g. high winds, torrential rain, or snow). By focusing on one area to test, they can make better predictions for this model.
A city water department is proposing the construction of a new water pipe, as shown. The new pipe will be perpendicular to the old pipe. Write an equation that represents the new pipe.
y=3.5x+2.8
represents the cost y (in dollars) of a taxi ride of x miles.
a. Identify the independent and dependent variables.
b. You have enough money to travel at most 20 miles in the taxi. Find the domain and range of the function.
Colleen and Martin are modeling the data shown. Colleen thinks the model should be
f\left(x\right)={5.754}^{x}3+2.912{x}^{2}-7.516x0.349
. Martin thinks it should be
f\left(x\right)=3.697{x}^{2}+11.734x-2.476
. Is either of them correct? Explain your reasoning.
f\left(x\right)-2-190.5-2-1511.500.4243
{\left({0.5}^{-2}\right)}^{2}
by hand using properties of exponents.
Write and solve a compound inequality that represents the possible temperatures (in degrees Fahrenheit) of the interior of the iceberg.
Given: The temperatures are given -20 and -15 in degree centigrade.
F=Gm1m\frac{2}{{r}^{2}}
for m2
A single expression for the logarithmic expression
\mathrm{log}10{x}^{5}-\mathrm{log}5x |
Finding maximum and minimum values of \sin(2x) - x \text{
Octavio Miller 2022-01-27 Answered
Finding maximum and minimum values of
\mathrm{sin}\left(2x\right)-x\text{ }\text{ for }\text{ }x\in \left[-\frac{\pi }{2},\frac{\pi }{2}\right]
search633504
First differentiate
y=\mathrm{sin}\left(2x\right)-x
{y}^{\prime }=2\mathrm{cos}\left(2x\right)-1
.At an extrema this value will be zero. So equate the above equation to zero and solve for x. Remember the value of x has to lie in the given interval. It can be seen easily that the
x=\frac{\pi }{6}
Then find the second derivative of y.
Substitute this value of x obtained.
If the resulting value is positive then the function has a minima at that x.
If it is negative then the function has a maxima.
\mathrm{sin}t=,\mathrm{cos}t=,\text{ }\text{and}\text{ }\mathrm{tan}t=
\mathrm{ln}\left(\mathrm{tan}\left(x\right)+\mathrm{sec}\left(x\right)\right)=\frac{\mathrm{ln}\left(1+\mathrm{sin}x\right)-\mathrm{ln}\left(1-\mathrm{sin}x\right)}{2}
is true or false
How can I calculate
\mathrm{sin}\left({10}^{{10}^{100}}-10\right)
2\mathrm{sin}x\le x-\frac{\pi }{3}+\sqrt{3}
|\sum _{r=0}^{3n-1}{\beta }^{{2}^{r}}|=4\sqrt{2}
\beta =\mathrm{exp}\left(\frac{i2\pi }{7}\right)
Proving Trig/Calc Identity
\frac{1}{1-\mathrm{sin}\left(x\right)}=\frac{1+\mathrm{sin}x}{{\mathrm{cos}\left(x\right)}^{2}} |
Tachyonic antitelephone - Wikipedia
Hypothetical device in theoretical physics that could be used to send signals into one's own past
A tachyonic antitelephone is a hypothetical device in theoretical physics that could be used to send signals into one's own past. Albert Einstein in 1907[1][2] presented a thought experiment of how faster-than-light signals can lead to a paradox of causality, which was described by Einstein and Arnold Sommerfeld in 1910 as a means "to telegraph into the past".[3] The same thought experiment was described by Richard Chace Tolman in 1917;[4] thus, it is also known as Tolman's paradox.
A device capable of "telegraphing into the past" was later also called a "tachyonic antitelephone" by Gregory Benford et al. According to the current understanding of physics, no such faster-than-light transfer of information is actually possible. For instance, the hypothetical tachyon particles which give the device its name do not exist even theoretically in the standard model of particle physics, due to tachyon condensation, and there is no experimental evidence that suggests that they might exist. The problem of detecting tachyons via causal contradictions was treated but without scientific verification.[clarification needed][5]
1 One-way example
2 Two-way example
3 Numerical example with two-way communication
One-way example[edit]
This was illustrated in 1911 by Paul Ehrenfest using a Minkowski diagram. Signals are sent in frame B1 into the opposite directions OP and ON with a velocity approaching infinity. Here, event O happens before N. However, in another frame B2, event N happens before O.[6]
Tolman used the following variation of Einstein's thought experiment:[1][4] Imagine a distance with endpoints
{\displaystyle A}
{\displaystyle B}
. Let a signal be sent from A propagating with velocity
{\displaystyle a}
towards B. All of this is measured in an inertial frame where the endpoints are at rest. The arrival at B is given by:
{\displaystyle \Delta t=t_{1}-t_{0}={\frac {B-A}{a}}.}
Here, the event at A is the cause of the event at B. However, in the inertial frame moving with relative velocity v, the time of arrival at B is given according to the Lorentz transformation (c is the speed of light):
{\displaystyle {\begin{aligned}\Delta t'&=t'_{1}-t'_{0}={\frac {t_{1}-vB/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}-{\frac {t_{0}-vA/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\\&={\frac {1-av/c^{2}}{\sqrt {1-v^{2}/c^{2}}}}\Delta t.\end{aligned}}}
It can be easily shown that if a > c, then certain values of v can make Δt' negative. In other words, the effect arises before the cause in this frame. Einstein (and similarly Tolman) concluded that this result contains in their view no logical contradiction; he said, however, it contradicts the totality of our experience so that the impossibility of a > c seems to be sufficiently proven.[1]
Two-way example[edit]
A more common variation of this thought experiment is to send back the signal to the sender (a similar one was given by David Bohm[7]). Suppose Alice (A) is on a spacecraft moving away from the Earth in the positive x-direction with a speed
{\displaystyle v}
, and she wants to communicate with Bob (B) back home. Assume both of them have a device that is capable of transmitting and receiving faster-than-light signals at a speed of
{\displaystyle a}
{\displaystyle c}
{\displaystyle a>1}
. Alice uses this device to send a message to Bob, who sends a reply. Let us choose the origin of the coordinates of Bob's reference frame,
{\displaystyle S}
, to coincide with the reception of Alice's message to him. If Bob immediately sends a message back to Alice, then in his rest frame the coordinates of the reply signal (in natural units so that c=1) are given by:
{\displaystyle (t,x)=(t,at)}
To find out when the reply is received by Alice, we perform a Lorentz transformation to Alice's frame
{\displaystyle S'}
moving in the positive x-direction with velocity
{\displaystyle v}
with respect to the Earth. In this frame Alice is at rest at position
{\displaystyle x'=L}
{\displaystyle L}
is the distance that the signal Alice sent to Earth traversed in her rest frame. The coordinates of the reply signal are given by:
{\displaystyle t'=\gamma \left(1-av\right)t}
{\displaystyle x'=\gamma \left(a-v\right)t}
The reply is received by Alice when
{\displaystyle x'=L}
{\displaystyle t={\tfrac {L}{\gamma (a-v)}}}
{\displaystyle t'={\frac {1-av}{a-v}}L}
Since the message Alice sent to Bob took a time of
{\displaystyle {\tfrac {L}{a}}}
to reach him, the message she receives back from him will reach her at time:
{\displaystyle T={\frac {L}{a}}+t'=\left({\frac {1}{a}}+{\frac {1-av}{a-v}}\right)L}
later than she sent her message. However, if
{\displaystyle v>{\tfrac {2a}{1+a^{2}}}}
{\displaystyle T<0}
and Alice will receive the message back from Bob before she sends her message to him in the first place.
Numerical example with two-way communication[edit]
As an example, imagine that Alice and Bob are aboard spaceships moving inertially with a relative speed of 0.8c. At some point they pass right next to each other, and Alice defines the position and time of their passing to be at position x = 0, time t = 0 in her frame, while Bob defines it to be at position x′ = 0 and time t′ = 0 in his frame (note that this is different from the convention used in the previous section, where the origin of the coordinates was the event of Bob receiving a tachyon signal from Alice). In Alice's frame she remains at rest at position x = 0, while Bob is moving in the positive x direction at 0.8c; in Bob's frame he remains at rest at position x′ = 0, and Alice is moving in the negative x′ direction at 0.8c. Each one also has a tachyon transmitter aboard their ship, which sends out signals that move at 2.4c in the ship's own frame.
When Alice's clock shows that 300 days have elapsed since she passed next to Bob (t = 300 days in her frame), she uses the tachyon transmitter to send a message to Bob, saying "Ugh, I just ate some bad shrimp". At t = 450 days in Alice's frame, she calculates that since the tachyon signal has been traveling away from her at 2.4c for 150 days, it should now be at position x = 2.4×150 = 360 light-days in her frame, and since Bob has been traveling away from her at 0.8c for 450 days, he should now be at position x = 0.8×450 = 360 light-days in her frame as well, meaning that this is the moment the signal catches up with Bob. So, in her frame Bob receives Alice's message at x = 360, t = 450. Due to the effects of time dilation, in her frame Bob is aging more slowly than she is by a factor of
{\displaystyle {\frac {1}{\gamma }}={\sqrt {1-{(v/c)^{2}}}}}
, in this case 0.6, so Bob's clock only shows that 0.6×450 = 270 days have elapsed when he receives the message, meaning that in his frame he receives it at x′ = 0, t′ = 270.
When Bob receives Alice's message, he immediately uses his own tachyon transmitter to send a message back to Alice saying "Don't eat the shrimp!". 135 days later in his frame, at t′ = 270 + 135 = 405, he calculates that since the tachyon signal has been traveling away from him at 2.4c in the −x′ direction for 135 days, it should now be at position x′ = −2.4×135 = −324 light-days in his frame, and since Alice has been traveling at 0.8c in the −x direction for 405 days, she should now be at position x′ = −0.8×405 = −324 light-days as well. So, in his frame Alice receives his reply at x′ = −324, t′ = 405. Time dilation for inertial observers is symmetrical, so in Bob's frame Alice is aging more slowly than he is, by the same factor of 0.6, so Alice's clock should only show that 0.6×405 = 243 days have elapsed when she receives his reply. This means that she receives a message from Bob saying "Don't eat the shrimp!" only 243 days after she passed Bob, while she wasn't supposed to send the message saying "Ugh, I just ate some bad shrimp" until 300 days elapsed since she passed Bob, so Bob's reply constitutes a warning about her own future.
These numbers can be double-checked using the Lorentz transformation. The Lorentz transformation says that if we know the coordinates x, t of some event in Alice's frame, the same event must have the following x′, t′ coordinates in Bob's frame:
{\displaystyle {\begin{aligned}t'&=\gamma \left(t-{\frac {vx}{c^{2}}}\right)\\x'&=\gamma \left(x-vt\right)\\\end{aligned}}}
Where v is Bob's speed along the x-axis in Alice's frame, c is the speed of light (we are using units of days for time and light-days for distance, so in these units c = 1), and
{\displaystyle \gamma ={\frac {1}{\sqrt {1-{(v/c)^{2}}}}}}
is the Lorentz factor. In this case v=0.8c, and
{\displaystyle \gamma ={\frac {1}{0.6}}}
. In Alice's frame, the event of Alice sending the message happens at x = 0, t = 300, and the event of Bob receiving Alice's message happens at x = 360, t = 450. Using the Lorentz transformation, we find that in Bob's frame the event of Alice sending the message happens at position x′ = (1/0.6)×(0 − 0.8×300) = −400 light-days, and time t′ = (1/0.6)×(300 − 0.8×0) = 500 days. Likewise, in Bob's frame the event of Bob receiving Alice's message happens at position x′ = (1/0.6)×(360 − 0.8×450) = 0 light-days, and time t′ = (1/0.6)×(450 − 0.8×360) = 270 days, which are the same coordinates for Bob's frame that were found in the earlier paragraph.
Comparing the coordinates in each frame, we see that in Alice's frame her tachyon signal moves forwards in time (she sent it at an earlier time than Bob received it), and between being sent and received we have (difference in position)/(difference in time) = 360/150 = 2.4c. In Bob's frame, Alice's signal moves back in time (he received it at t′ = 270, but it was sent at t′ = 500), and it has a (difference in position)/(difference in time) of 400/230, about 1.739c. The fact that the two frames disagree about the order of the events of the signal being sent and received is an example of the relativity of simultaneity, a feature of relativity which has no analogue in classical physics, and which is key to understanding why in relativity FTL communication must necessarily lead to causality violation.
Bob is assumed to have sent his reply almost instantaneously after receiving Alice's message, so the coordinates of his sending the reply can be assumed to be the same: x = 360, t = 450 in Alice's frame, and x′ = 0, t′ = 270 in Bob's frame. If the event of Alice receiving Bob's reply happens at x′ = 0, t′ = 243 in her frame (as in the earlier paragraph), then according to the Lorentz transformation, in Bob's frame Alice receives his reply at position x′' = (1/0.6)×(0 − 0.8×243) = −324 light-days, and at time t′ = (1/0.6)×(243 − 0.8×0) = 405 days. So evidently Bob's reply does move forward in time in his own frame, since the time it was sent was t′ = 270 and the time it was received was t′ = 405. And in his frame (difference in position)/(difference in time) for his signal is 324/135 = 2.4c, exactly the same as the speed of Alice's original signal in her own frame. Likewise, in Alice's frame Bob's signal moves backwards in time (she received it before he sent it), and it has a (difference in position)/(difference in time) of 360/207, about 1.739c.
Thus the times of sending and receiving in each frame, as calculated using the Lorentz transformation, match up with the times given in earlier paragraphs, before we made explicit use of the Lorentz transformation. And by using the Lorentz transformation we can see that the two tachyon signals behave symmetrically in each observer's frame: the observer who sends a given signal measures it to move forward in time at 2.4c, the observer who receives it measures it to move back in time at 1.739c. This sort of possibility for symmetric tachyon signals is necessary if tachyons are to respect the first of the two postulates of special relativity, which says that all laws of physics must work exactly the same in all inertial frames. This implies that if it's possible to send a signal at 2.4c in one frame, it must be possible in any other frame as well, and likewise if one frame can observe a signal that moves backwards in time, any other frame must be able to observe such a phenomenon as well. This is another key idea in understanding why FTL communication leads to causality violation in relativity; if tachyons were allowed to have a "preferred frame" in violation of the first postulate of relativity, in that case it could theoretically be possible to avoid causality violations.[8]
Benford et al.[5] wrote about such paradoxes in general, offering a scenario in which two parties are able to send a message two hours into the past:
The paradoxes of backward-in-time communication are well known. Suppose A and B enter into the following agreement: A will send a message at three o'clock if and only if he does not receive one at one o'clock. B sends a message to reach A at one o'clock immediately on receiving one from A at three o'clock. Then the exchange of messages will take place if and only if it does not take place. This is a genuine paradox, a causal contradiction.
They concluded that superluminal particles such as tachyons are therefore not allowed to convey signals.
^ a b c Einstein, Albert (1907). "Über das Relativitätsprinzip und die aus demselben gezogenen Folgerungen" [On the relativity principle and the conclusions drawn from it] (PDF). Jahrbuch der Radioaktivität und Elektronik. 4: 411–462. Retrieved 2 August 2015.
^ Einstein, Albert (1990). "On the relativity principle and the conclusions drawn from it". In Stachel, John; Cassidy, David C; Renn, Jürgen; et al. (eds.). The Collected Papers of Albert Einstein, Volume 2: The Swiss Years: Writings, 1900-1909. Princeton: Princeton University Press. p. 252. ISBN 9780691085265. Retrieved 2 August 2015.
^ Miller, A.I. (1981), Albert Einstein's special theory of relativity. Emergence (1905) and early interpretation (1905–1911), Reading: Addison–Wesley, ISBN 0-201-04679-2
^ a b R. C. Tolman (1917). "Velocities greater than that of light". The theory of the Relativity of Motion. University of California Press. p. 54. OCLC 13129939.
^ a b Gregory Benford; D. L. Book; W. A. Newcomb (1970). "The Tachyonic Antitelephone" (PDF). Physical Review D. 2 (2): 263–265. Bibcode:1970PhRvD...2..263B. doi:10.1103/PhysRevD.2.263. S2CID 121124132. Archived from the original (PDF) on 2020-02-07.
^ Ehrenfest, P. (1911). "Zu Herrn v. Ignatowskys Behandlung der Bornschen Starrheitsdefinition II" [On v. Ignatowsky's Treatment of Born's Definition of Rigidity II]. Physikalische Zeitschrift. 12: 412–413.
^ David Bohm, The Special Theory of Relativity, New York: W.A. Benjamin., 1965
Retrieved from "https://en.wikipedia.org/w/index.php?title=Tachyonic_antitelephone&oldid=1066397170" |
\frac{\left(a+b\right)}{\left(u+v\right)}
\frac{\left(a+b\right)}{\left(u-v\right)}
\left(\sqrt{b}:\sqrt{a}\right)
A train moves past a telegraph post and a bridge 264 m long in 8 seconds and 20 seconds respectively. What is the speed of the train ?
A) 69.5 km/hr B) 70 km/hr
Answer & Explanation Answer: D) 79.2 km/hr
Then, x/y = 8 => x = 8y
Now, (x+264)/20 = y
8y + 264 = 20y
Speed = 22 m/sec = (22*18/5) km/hr = 79.2 km/hr.
Two trains of equal length , running in opposite directions , pass a pole in 18 and 12 seconds. The trains will cross each other in
A) 14.4 sec B) 15.5 sec
C) 18.8 sec D) 20.2 sec
A) 14.4 sec
B) 15.5 sec
C) 18.8 sec
D) 20.2 sec
Answer & Explanation Answer: A) 14.4 sec
Let the length of the train be L metres
Speed of first train = m/hour
Speed of secxond train = m/hour
When running in opposite directions, relative speed = 200 L + 300 L m/hour
Distance to be covered = L + L = 2L metre
Time taken = sec
=14.4 sec
A train covers a distance between station A and station B in 45 min.If the speed of the train is reduced by 5 km/hr,then the same distance is covered in 48 min.what is the distance between the stations A and B ?
Answer & Explanation Answer: B) 60 kms
Let 'd' be the distance and 's' be the speed and 't' be the time
d=sxt
45 mins = 3/4 hr and 48 mins = 4/5 hr
As distance is same in both cases;
s(3/4) = (s-5)(4/5)
3s/4 = (4s-20)/5
15s = 16s-80
=> d = 80 x 3/4 = 60 kms.
Two trains are moving in the same direction at 72 kmph and 36 kmph. The faster train crosses a girl sitting at window seat in the slower train in 32 seconds. Find the length of the faster train ?
Relative speed = (72 - 36) x 5/18 = 2 x 5 = 10 mps.
Distance covered in 32 sec = 32 x 10 = 320 m.
The length of the faster train = 320 m.
Two trains having equal lengths, take 10 seconds and 15 seconds respectively to cross a post. If the length of each train is 120 meters, in what time (in seconds) will they cross each other when traveling in opposite direction?
Speed of train 1 = = 12 m/sec
Speed of train 2 = = 8 m/sec
if they travel in opposite direction, relative speed = 12 + 8 = 20 m/sec
Answer & Explanation Answer: A) 355 mts
Given speed = 63 km/hr = m/s
Let the length of the bridge = x mts
Given time taken to cover the distance of (170 + x)mts is 30 sec.
We know speed = m/s
--> x = 355 mts.
Two trains, one from Hyderabd to Bangalore and the other from Bangalore to Hyderabad, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is ?
Answer & Explanation Answer: B) 4 : 3
Let us name the trains as A and B.
Then, (A's speed) : (B's speed)
= √b : √a = √16 : √9 = 4:3
A train running at the speed of 40 km/hr crosses a signal pole in 9 seconds. Find the length of the train ?
d= 100 mts. |
A proton is moving parallel to a uniform electric field. The electric field acce
A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to 5.0 x 10^-23 kg x m/s from 1.5 x 10^-23 m/s in a time of 6.3 x 10^-6 s. What is the magnitude of the electric field?
A proton is moving parallel to a uniform electric field. The electric field accelerates the proton and increases its linear momentum to
5.0×{10}^{-23}kg\text{ }x\text{ }m/s
1.5×{10}^{-23}m/s
in a time of 6.3 x 10^-6 s. What is the magnitude of the electric field?
F\mathrm{△}t=\mathrm{△}p
qE=\frac{\mathrm{△}p}{\mathrm{△}t}
E=\frac{\mathrm{△}p}{q\mathrm{△}t}
Given that the initial momentum is
{p}_{1}=1.50\ast {10}^{-23}kg×m/s
the final momentum is
{p}_{2}=5.00\ast {10}^{-23}kg×m/s
the time taken is
t=6.3×{10}^{-6}
The charge of the proton is
q=1.602×{10}^{-19}
The force related with momentum is
F=\frac{\left({P}_{2}-{P}_{1}\right)}{t}
{E}_{q}=\frac{\left({P}_{2}-{P}_{1}\right)}{t}
F={E}_{q}
SO electric field ,
E=\frac{\left({P}_{2}-{P}_{1}\right)}{t×q}
=\frac{\left(5-1.5\right)×{10}^{-23}}{\left[1.602×{10}^{-19}×6.3×{10}^{-6}\right]}
0.3468×{10}^{2}
= 34.68 N/C
An adventurous archaeologist crosses between two rock cliffs by slowly going hand-over-hand along a rope stretched between the cliffs. He stops to rest at the middle of the rope (Fig. 1) The rope will break if the tension in it exceeds
4.50\cdot {10}^{4}
N and our hero's mass is 86.5 kg.
Figure 1 a) If the angle
\theta
{13.0}^{\circ }
, find the tension iin the rope.
b) What is the smallest value the angle
\theta
can have if the rope is not to break?
Here is problem, please help
In an industrial cooling process, water is circulated through a system. If the water is pumped with a speed of 0.45 m/s under a pressure of 400 torr from the first floor through a 6.0-cm diameter pipe, what will be the pressure on the next floor 4.0 m above in a pipe with a diameter of 2.0 cm?
a) Calculate the number of cells in a hummingbird assuming the mass of an average cell is ten times the mass of a bacterium. The mass of a hummingbird is of the order of
{10}^{-2}kg
and that of a bacterium is
{10}^{-15}kg
b) aking the same assumption, how many cells are there in a human if his mass is of the order of
{10}^{2}kg?
An indestructible bullet 2.00 cm long is fired straight through a board that is 10.0 cm thick. The bullet strikes the boardwith a speed of 470 m/s and emerges with a speed of 250 m/s. (To simplify,assume that the bullet accelerates only while the front tip is incontact with the wood.)
(a) What is the average acceleration of the bullet through the board?
(b) What is the total time that the bullet is in contact with the board?
(c) What thickness of board (calculated to 0.1 cm) would it take to stop the bullet, assuming that the acceleration through all boards the same?
Find inverse laplace
find y(t) for the following
Y\left(s\right)=\frac{8}{s}+\frac{1}{s-3}\cdot {e}^{-2s} |
Sum-of-the-Parts Valuation – SOTP Definition
Sum-of-the-Parts Valuation – SOTP
What Is the Sum-of-the-Parts Valuation – SOTP?
The sum-of-the-parts valuation (SOTP) is a process of valuing a company by determining what its aggregate divisions would be worth if they were spun off or acquired by another company.
The valuation provides a range of values for a company's equity by aggregating the standalone value of each of its business units and arriving at a single total enterprise value (TEV). The equity value is then derived by adjusting the company's net debt and other non-operating assets and expenses.
The Formula for Sum-of-the-Parts Valuation – SOTP Is
\begin{aligned} &\text{SOTP} = N_1 + N_2 + \dotso + ND - NL + NA \\ &\textbf{where:}\\ &N_1=\text{Value of first segment}\\ &N_2=\text{Value of second segment}\\ &ND=\text{net debt}\\ &NL=\text{nonoperating liabilities}\\ &NA = \text{nonoperating assets} \end{aligned}
SOTP=N1+N2+…+ND−NL+NAwhere:N1=Value of first segmentN2=Value of second segmentND=net debtNL=nonoperating liabilities
SOTP is the process of determining what the individual divisions of a company would be worth if they were spun off or bought by a different company.
SOTP enables a company to establish a useful measure of its value which can be highly relevant in the case of a hostile takeover or a restructuring.
SOTP is often put to use when a company is a conglomerate and has business units in different industries.
How to Calculate Sum-of-the-Parts Valuation – SOTP
The value of each business unit or segment is derived separately and can be determined by any number of analysis methods. For example, discounted cash flow (DCF) valuations, asset-based valuations and multiples valuations using revenue, operating profit or profit margins are methods utilized to value a business segment.
What Does the SOTP Tell You?
Sum-of-the-parts valuation, also known as breakup value analysis, helps a company understand its true value. For example, you might hear that a young technology company is "worth more than the sum of its parts," meaning the value of the company's divisions could be worth more if they were sold to other companies.
In situations such as this one, larger companies have the ability to take advantage of synergies and economies of scale unavailable to smaller companies, enabling them to maximize a division's profitability and unlock unrealized value.
The SOTP valuation is most commonly used to value a company comprised of business units in different industries since valuation methods differ across industries depending on the nature of revenue. It is possible to use this valuation to defend against a hostile takeover by proving the company is worth more as a sum of its parts. It is also possible to use this valuation in situations where a company is being revalued after a restructuring.
The Sum-of-the-parts valuation is also known as the breakup value as it assesses what individual segments would be worth if the company was broken up.
Example of How to Use the Sum-of-the-Parts Valuation – SOTP
Consider United Technologies (NYSE: UTX), which said it will break the company into three units in late 2018—an aerospace, elevator and building systems company. Using the 10-year median enterprise value-to-EBIT (EV/EBIT) multiple for peers and 2019 operating profit projections, the aerospace business is valued at $107 billion, the elevator business at $36 billion and building systems business $52 billion. Thus, the total value is $194 billion. Lessing out net debt and other items of $39 billion, the sum-of-the-parts valuation is $155 billion.
The Difference Between the SOTP and Discounted Cash Flow – DCF
While both are valuation tools, the SOTP valuation can incorporate a discounted cash flow (DCF) valuation. That is, valuing a segment of a company may be done with a DCF analysis. Meanwhile, the DCF uses discounted future cash flows to value a business, project or segment. The present value of expected future cash flows is discounted using a discount rate.
Limitations of Using Sum-of-the-Parts Valuation – SOTP
The sum-of-the-parts (SOTP) valuation involves valuing various business segments, and more valuations come with more inputs. As well, SOTP valuations do not take into account tax implications, notably the implications involved in a spinoff.
Learn More About the SOTP Valuation
For help on choosing the right valuation tool check out this guide to picking the right valuation method.
What Is a Company's Breakup Value?
The breakup value of a corporation is the worth of each of its main business segments if they were spun off from the parent company. |
Solve integrals using residue theorem? \int_0^{\pi}\frac{d\theta}{2+\cos\theta} \int_0^\infty\frac{x}{(1+x)^6}dx
Solve integrals using residue theorem?
{\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }
{\int }_{0}^{\mathrm{\infty }}\frac{x}{{\left(1+x\right)}^{6}}dx
kalfswors0m
Hint for the First one First compute
I={\int }_{0}^{2\pi }\frac{d\theta }{2+\mathrm{cos}\theta }={\int }_{0}^{2\pi }R\left(\mathrm{cos}\left(\theta \right),\text{ }\mathrm{sin}\left(\theta \right)\right)d\theta
Where R is the rational function given by
R\left(x,y\right)=\frac{1}{2+x}
How to do this using the residue theorem? Put
z={e}^{it}=\mathrm{cos}\left(t\right)+i\mathrm{sin}\left(t\right)
\mathrm{cos}\left(t\right)=Re\left(z\right)=\frac{z+{z}^{-1}}{2}
\mathrm{sin}\left(t\right)=Im\left(z\right)=\frac{z-{z}^{-1}}{2i}
dz=i{e}^{it}dt=izdt⇒dt=\frac{1}{iz}dz
Then I can be seen as a contour integral, solve it by using residues
I={\int }_{0}^{2\pi }R\left(\mathrm{cos}\left(t\right),\mathrm{sin}\left(t\right)\right)dt=
{\int }_{|z|=1}R\left(\frac{z+\frac{1}{z}}{2},\frac{z-\frac{1}{z}}{2i}\right)\frac{1}{iz}dz
Hence in your case the integral you will compute is
I={\int }_{|z|=1}\left(\frac{1}{2+\frac{z+\frac{1}{z}}{2}}\right)\frac{1}{iz}dz=\frac{2}{i}
{\int }_{|z|=1}\frac{dz}{{z}^{2}+4z+1}
which can be easily obtain by the residue theorem!
Finally: Note that
{\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }=\frac{I}{2}
and hence your result follows by computing the next integral
{\int }_{0}^{\pi }\frac{d\theta }{2+\mathrm{cos}\theta }=\frac{1}{i}{\int }_{|z|=1}\frac{dz}{{z}^{2}+4z+1}
Note first that this integral is easily done by recognizing that
x=\left(1+x\right)-1
, so the integral is really
{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{\left(1+x\right)}^{5}}-{\int }_{0}^{\mathrm{\infty }}\frac{dx}{{\left(1+x\right)}^{6}}=\frac{1}{4}-\frac{1}{5}=\frac{1}{20}
One may also use the residue theorem. However, one must choose an appropriate contour and integrand. In this case, a useful contour integral to consider is
{\oint }_{C}dz\frac{z\mathrm{log}z}{{\left(1+z\right)}^{6}}
where C is a keyhole contour of outer radius R about the positive real axis. The contour integral is then equal to
{\int }_{ϵ}^{R}dx\frac{x\mathrm{log}x}{{\left(1+x\right)}^{6}}+iR{\int }_{0}^{2\pi }d\theta {e}^{i\theta }\frac{R{e}^{i\theta }\mathrm{log}\left(R{e}^{i\theta }\right\}\left\{{\left(1+R{e}^{i\theta }\right)}^{6}\right\}}{}
+{\int }_{R}^{ϵ}dx\frac{x\left(\mathrm{log}x+i2\pi \right)}{{\left(1+x\right)}^{6}}+iϵ{\int }_{2\pi }^{0}d\varphi
{e}^{i\varphi }\frac{ϵ{e}^{i\varphi }\mathrm{log}\left(ϵ{e}^{i\varphi }\right)}{{\left(1+ϵ{e}^{i\varphi }\right)}^{6}}
R\to \mathrm{\infty }
, the second integral vanishes as
\frac{\mathrm{log}R}{{R}^{4}}
ϵ\to 0
, the fourth integral vanishes as
{ϵ}^{2}\mathrm{log}ϵ
. Thus, the contour integral is, in this limit
-i2\pi {\int }_{0}^{\mathrm{\infty }}dx\frac{x}{{\left(1+x\right)}^{6}}
By the residue theorem, the contour integral is also equal to
i2\pi
times the residue at the pole
x={e}^{i\pi }
. (Note how important it is to get the argument correct.) The residue at this pole is
\frac{1}{5!}{\left[\frac{{d}^{5}}{{dz}^{5}}\left(z\mathrm{log}z\right)\right]}_{z={e}^{i\pi }}=-\frac{3!}{5!}
Putting this altogether, we get that
{\int }_{0}^{\mathrm{\infty }}dx\frac{x}{{\left(1+x\right)}^{6}}=\frac{1}{20}
which agrees with the above.
Hint for the first one:
f\left(z\right)=\frac{2}{{z}^{2}+4z+1}
and find its poles. Then use the known formula for residues:
Under the assumption that f has a pole of order m at
{z}_{0}
Res\left({z}_{0},f\right)=\frac{1}{\left(m-1\right)!}\underset{z\to {z}_{0}}{lim}\frac{{d}^{m-1}}{d{z}^{m-1}}\left(\left(z-{z}_{0}{\right)}^{m}f\left(z\right)\right).
And finally, apply the Residue Theorem.
Integral of a deriative.
I've been learning the fundamental theorem of calculus. So, I can intuitively grasp that the derivative of the integral of a given function brings you back to that function. Is this also the case with the integral of the derivative? And if so, can you please give a intuition for why this is true? Tha
\int \frac{5}{7}{x}^{2}\mathrm{cos}\left(x\right)dx
y=x{e}^{-x}
\int {3}^{-2x}dx
Solving integral:
\int {3}^{x}{e}^{x}dx
a) Write the sigma notation formula for the right Riemann sum
{R}_{n}
f\left(x\right)=4-{x}^{2}
\left[0,\text{ }2\right]
using n subintervals of equal length, and calculate the definite integral
{\int }_{0}^{2}f\left(x\right)dx
as the limit of
{R}_{n}
n\to \mathrm{\infty }
\sum _{k=1}^{n}k=n\frac{n+1}{2},\text{ }\sum _{k=1}^{n}{k}^{2}=n\left(n+1\right)\frac{2n+1}{6}\right)
b) Use the Fundamental Theorem of Calculus to calculate the derivative of
F\left(x\right)={\int }_{{e}^{-x}}^{x}\mathrm{ln}\left({t}^{2}+1\right)dt
Double Integrals involving infinity
{\int }_{0}^{\mathrm{\infty }}{\int }_{0}^{\mathrm{\infty }}xy{e}^{-\left({x}^{2}+{y}^{2}\right)}dxdy |
Turaev hyperbolicity of classical and virtual knots
Colin Adams, Or Eisenberg, Jonah Greenberg, Kabir Kapoor, Zhen Liang, Kate O’Connor, Natalia Pachecho-Tallaj and Yi Wang
By work of W Thurston, knots and links in the
3
–sphere are known to either be torus links; or to contain an essential sphere or torus in their complement; or to be hyperbolic, in which case a unique hyperbolic volume can be calculated for their complement. We employ a construction of Turaev to associate a family of hyperbolic
3
–manifolds of finite volume to any classical or virtual link, even if nonhyperbolic. These are in turn used to define the Turaev volume of a link, which is the minimal volume among all the hyperbolic
3
–manifolds associated via this Turaev construction. In the case of a classical link, we can also define the classical Turaev volume, which is the minimal volume among all the hyperbolic
3
–manifolds associated via this Turaev construction for the classical projections only. We then investigate these new invariants.
Turaev surface, Turaev volume, knot, hyperbolic knot, virtual knot
Natalia Pachecho-Tallaj |
Express the limits as definite integral. lim_(norm(p rarr 0))sum_(k=1)^n(1/c_k
Express the limits as definite integral. lim_(norm(p rarr 0))sum_(k=1)^n(1/c_k)Deltax_k, where P is a partition of[1,4]
Express the limits as definite integral.
\underset{‖p\to 0‖}{lim}\sum _{k=1}^{n}\left(\frac{1}{{c}_{k}}\right)\mathrm{\Delta }{x}_{k},where\text{ }P\text{ }is\text{ }a\text{ }partition\text{ }of\left[1,4\right]
\underset{‖p\to 0‖}{lim}\sum _{k=1}^{n}\left(\frac{1}{{c}_{k}}\right)\mathrm{\Delta }{x}_{k}
To express limits as a definite integrals.
The definition of definite integral is,
{\int }_{a}^{b}f\left(x\right)dx=\underset{n\to \mathrm{\infty }}{lim}\sum _{i=1}^{n}f\left({x}_{i}\right)\mathrm{\Delta }x
f\left({x}_{k}\right)=\frac{1}{{c}_{k}}
And p is a partition of
\left[1,4\right]
\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}\left(\frac{1}{{c}_{k}}\right)\mathrm{\Delta }{x}_{k}={\int }_{1}^{4}f\left(x\right)dx
Solve the integral by substitution.
\int {\left({z}^{2}+2z+1\right)}^{\frac{4}{3}}dz
f\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{\mathrm{sin}\left(\left(2k+1\right)x\right)}
with k a positive integer. It seems that f is strictly increasing in
\left[0,\frac{\pi }{2\left(2k+1\right)}\right]
.Is there some easy proof of this monotonicity property that does not invole differentiation (through suitable inequalities)?
{\int }_{0}^{4}\left(3x-\frac{{x}^{2}}{4}\right)dx
\int {\mathrm{sin}}^{n}\left(x\right){\mathrm{cos}}^{3}\left(x\right)dx
\frac{{\mathrm{sin}}^{8}\left(x\right)}{8}-\frac{{\mathrm{sin}}^{6}\left(x\right)}{6}+\frac{1}{24}=\frac{{\mathrm{cos}}^{8}\left(x\right)}{8}-\frac{{\mathrm{cos}}^{6}\left(x\right)}{3}+\frac{{\mathrm{cos}}^{4}\left(x\right)}{4}
I have integrated the suggested integral successfully, that being:
\frac{{\mathrm{sin}}^{n+1}\left(x\right)}{n+1}-\frac{{\mathrm{sin}}^{n+3}\left(x\right)}{n+3}+c
I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand
{\mathrm{sin}}^{2}\left(x\right)=\left(1-{\mathrm{cos}}^{2}\left(x\right)\right)
but this seems far too tedious.
Evaluate the indefinite integral:
\int {\mathrm{sec}}^{2}x{\mathrm{tan}}^{4}xdx
\int x\mathrm{ln}xdx
Use a table of integrals to find the indefinite integral
\int \frac{{e}^{x}}{1-{\mathrm{tan}e}^{x}}dx |
Nanometric displacement measurement based on interference fringes deflection | JVE Journals
Hely González-Rivera1 , G. E. Sandoval-Romero2
1, 2Centro de Ciencias Aplicadas y Desarrollo Tecnológico, Universidad Nacional Autónoma de México, Apartado Postal 70-186, Coyoacán, Ciudad de México 04510, México
A hybrid method to obtain nanometric displacements based on interference fringes movement by optical beam deflection is proposed. A proof-of-principle experiment is described as a new method for real-time detection and measurement of small vibrations, experimental results shows an increase in the resolution compared with a simply optical beam deflection read out. Also, there is minimum detectable displacement in an assumed noise environment.
Keywords: nanometric displacement, beam deflection sensing, interference, small vibrations.
The quantification of displacement is one of the most important physical variables to determine for science and engineering; in which surface profiles, velocity and vibrations are obtained by means of the change in distance measured. An attribute of relative distance measurement is that it is needed for very long and very short distances. For both cases, optical measurement methods offer plenty advantages compared to other measurement methods; being sensitivity, stability and non-contact measurement its most remarkable characteristics [1]. Since the invention of the laser in the 1960s and the emerging progress into optoelectronics, different optical measurement methods have been developed, where laser interferometry has predominated as the most accurate technique for displacements on nanometer scales. But, on the other hand, through optical beam deflection (OBD) it is possible achieve accurate nanometric displacement measurements, with a less complex setup than the required for an interferometer. The OBD technique introduced by Meyer and Amer [2] is a very reliable and simple detection method developed in the 1980s as a novel optical approach to atomic force microscopy (AFM). In an OBD readout system, as shows Fig. 1, a light beam from a laser is focused at the end of the cantilever and reflected to a position sensitive detector (PSD). The bending of the cantilever results in a large change in the direction of the reflected beam, which can be detected by the PSD as an electric current difference [3, 4].
Fig. 1. Optical beam deflection readout system
For the detection of the cantilever displacements in AFM, as in any optical detection system, the fundamental limits to the resolving power of the OBD are set by diffraction and shot noise due to the particle nature of light [5]. Otherwise, the resolution of the OBD increases with the radius of the waist of the beam, the power carried by the beam, and the reciprocal of the wavelength. Nevertheless, with these parameters in the practice it is not easy to increase the resolution of the OBD, i.e., to achieve a high spatial resolution a small waist beam in needed, but on the other hand a small waist may restrict the optical power. Also, in case of biological samples, with a high optical power there is a high probability of damaging the sample [6]. Then, the aim of this work is to show that it is possible achieve high resolution with a novel OBD setup and readout system notwithstanding the fundamental limits that exist in a conventional OBD system.
Accordingly, a hybrid method based on interference fringes movement by optical beam deflection is proposed to develop a nanometric displacement sensor. Implementing a simply OBD setup, as shows Fig. 2., a diode laser (DL) Coherent 0221-698-01 REV B, which generated 1 mW circular beam with 1/e2 diameter of 1 mm, is focused by a convex lens (L1) in a glass microscope slide (GS) placed at 45° respect to the optical path. The interaction of the light beam with the glass slide, as in all matter, can take four forms; transmission, reflection, scattering and absorption [7]. Only a small portion of the spectrum is scattered and absorbed by the glass slide, being practically non-issue. Otherwise, due to the changes in refractive index that exists between media bounding air-glass and glass-air, the laser beam experiments multiple reflections inside the glass slide. This produces an optical path difference and consequently creates a multiple beam fringes by transmission and reflection [8]. The interference pattern formed by reflection strikes a mirror (M1) attached to an optical post assembly, and it reflection strikes a mirror (M2) attached to an aluminum cantilever with a lumped mass at it free end. The interference pattern is reflected back to the mirror M1, and strikes again the glass slide to be transmitted towards a 37 mm camera lens (L2). This lens expands the interference fringe pattern in order to match it with a grating with opaque and transparent parallel lines uniformly spaced, known as a Ronchi grating (RG). As shows Fig. 3, the opaque lines block partially the constructive interference fringes, then, when the cantilever undergoes deflection, the interference fringes moves horizontally, leading to the constructive interference fringes to gradually leave the blocking lines and be transmitted, and focused by a convex lens (L3), to a silicon detector (PD). As a consequence, there is an incident intensity increment in the photodetector as a function of the deflection.
Fig. 3. Partial blocking of interference fringes as intensity modulation
Most of the analysis instruments are based on voltage measurements, and since the photodetector is a current source, the photocurrent generated is converted to voltage as impedance through a transimpedance amplifier (TIA) with a T network in the feedback loop, in order to get a low dc resistance path in the feedback loop. By means of a DAQ NI-USB 6216 the proportional analog voltage is converted to a digital data to be analyzed in a LabVIEW® virtual instrument. It is important to mention that the experimental setup is placed in a vibration isolated optical table with the purpose of reduce the uncertainty of measurements.
Accordingly, to the main objective, the first measurements were the comparison between the proposed OBD and the conventional OBD. For this experiment the setup shown in Fig. 2 is modified, i.e., the aluminum cantilever is replaced by a Lead Zirconate Titanate piezoelectric (PZT), in order to ensure a controlled displacement and it is placed vertically, this means that the mirror M1 is removed. The conventional OBD setup consist only of three elements; the light source, the PZT cantilever, and a photodetector. As an equivalent of the PSD detection, a knife is placed in front of a half part of the photodetector. A function generator Agilent® 33521A is used to feed the PZT with an amplitude of 20 Vpp and a frequency of 1 Hz in both setups. The Fig. 4 shows the comparison between the conventional and hybrid OBD.
Both signals are not in phase due that were recorded at different times. The amplitude of the hybrid OBD is, approximately, 1.24 times bigger than the conventional OBD; there is only an amplitude difference of 24 %, not big enough to mention the hybrid OBD as a better technique. But on the other hand, it is important to mention that in the conventional OBD the optical power carried by the beam is bigger than the carried in the hybrid OBD, i.e., the optical power measured that strikes the photodetector in the conventional OBD is almost 400 µW, meanwhile in the hybrid OBD the optical power measured at the photodetector is 40 µW. This due to that the interference pattern used in the hybrid OBD is obtained from the reflection of the glass slide, thus there is a low optical power that strikes the mirror attached to the PZT. Then, another conventional OBD setup was develop using the optical power reflected by the glass slide. A convex lens is placed before the photodetector in order to focused the optical power in the knife edge. As shows the Fig. 5, there is an increase in the amplitude of 3 times when both setups were with the same conditions of optical power.
Fig. 4. Hybrid OBD against conventional OBD
Fig. 5. Hybrid OBD against conventional OBD with the same conditions of optical power
3.1. Characterization of the aluminum cantilever
When a cantilever system is subjected to free vibration and the system is considered as a discrete system in which the cantilever is considered as mass-less and the whole mass is concentrated at the free end of the it. The governing equation of motion for such system will be:
m\stackrel{¨}{y}\left(t\right)+ky\left(t\right)=0,
m
is the concentrated mass at the free end of the cantilever, and
k
is the stiffness of the system. The transverse stiffness of a cantilever beam is given as [9]:
k= \frac{3EI}{{l}^{3}},
E
is the Young’s modulus of the cantilever material,
l
is the cantilever length and
I
is the moment of inertia given by [10]:
I=\frac{1}{12} b{d}^{3},
b
and
are the breadth and depth of the cantilever beam cross-section respectively. The fundamental undamped natural frequency of the lumped mass system is given as [9]:
{f}_{n}=\frac{1}{2\pi }\sqrt{\frac{k}{m+0.23 {m}_{c} }},
{m}_{c}
is the cantilever mass. Dimensions and parameters of the cantilever used are given in Table 1.
Table 1. Parameters and dimensions of the cantilever with lumped mass at it free end.
\rho
kg∙m-3
E
N∙m-2
l
Cantilever breadth
b
Cantilever depth
d
Cantilever mass
{m}_{c}
m
Resolving Eq. (2) and Eq. (3) with the values displayed in Table 1, and substituting in Eq. (4), the fundamental natural frequency of the cantilever used in the experiment is
{f}_{n} \approx
73.56 Hz. In the time domain signal, as Fig. 6 shows that there is not an appreciable amplitude from the readout system. But, through Fourier analysis from the signal in time domain, a predominant frequency in 73.53 Hz is appreciated. Other two components are appreciated in 62.10 Hz and 82.69 Hz, when the first one is related to the fundamental frequency of electric power. The predominant frequency match with the natural frequency of the cantilever obtained in Eq. (4).
Fig. 6. Time domain signal with the cantilever in steady state
Fig. 7. Frequency domain analysis from the cantilever in steady state
In the presence of a frictional force, the natural frequency oscillation is determined by the ratio between the damping factor
\delta
and the cantilever natural frequency
{\omega }_{n}
in radians per second. Oscillations with periodic and exponentially decreasing amplitude means
{\omega }_{n}>\delta
. Then, a frictional force is introduced in the fixed end of the cantilever, as shows Fig. 8, and there is an appreciable under damped response. Furthermore, in Fig. 9, the predominant frequency changes to 193.20 Hz, but the natural frequency is still appreciating in a less magnitude.
Fig. 8. Time domain signal with the cantilever perturbed
Fig. 9. Frequency domain analysis from the cantilever perturbed
With the same conditions of optical power, the hybrid OBD have a sensitivity of 3 times bigger than the conventional OBD. The minimum displacement detected, without a lock-in amplifier system, is almost 16 nm in the PZT cantilever. Furthermore, in the last experiments developed, the aluminum cantilever is useful to determinate liquid characteristics when the lumped mass is in contact with the sample, i.e., the damping factor of the cantilever is related with some properties as density. A resonator instead a glass slide to get an interference fringes with less losses in the optical power would be useful as an improvement, but on the other hand, with a low optical power is possible handling biological samples and get information about its behavior.
This work was supported by DGAPA-UNAM through financing of Project PAPIIT IT101515. Besides we thank to Consejo Nacional de Ciencia y Tecnología (CONACyT) for its financial support in form of a scholarship for the Master and Ph.D. Engineering Program at Universidad Nacional Autónoma de México.
Donges A., Roll N. Laser Measurement Technologies Fundamentals and Applications. 1st Edition, Springer Verlag, Berlin, 2015. [Search CrossRef]
Meyer G., Amer N. Novel optical approach to atomic force microscopy. Applied Physics Letters, Vol. 53, Issue 24, 1988, p. 2400-2402. [Search CrossRef]
Sarid D. Scanning Force Microscopy. 1st Edition, Oxford University Press, New York, 1994. [Search CrossRef]
Bashir R., Werely S., Bimolecular Sensing. Processing and Analysis. 1st Edition, Springer, 2006. [Search CrossRef]
Putman C., De Grooth B., Van Hulst N., Greve J. A detailed analysis of the optical beam deflection technique for use in atomic force microscopy. Journal of Applied Physics, Vol. 72, Issue 1, 1992, p. 6-12. [Search CrossRef]
García-Valenzuela A., Díaz-Uribe R. Approach to improve the angle sensitivity and resolution of the optical beam deflection method using a passive interferometer and a Ronchi grating. Optical Engineering, Vol. 36, Issue 6, 1997, p. 1770-1778. [Search CrossRef]
Callister W.D. Jr. Fundamentals of Material Science and Engineering 5th Edition, John Wiley and Sons, New York, 2001. [Search CrossRef]
Hariharan P. Basics of Interferometry. 2nd Edition, Academic Press, San Diego, 2007. [Search CrossRef]
Harris C. M. Shock and Vibration Handbook. 4th Edition, McGraw Hill, New York, 1996. [Search CrossRef]
Gere J.M., Goodno B. J. Mechanics of Materials. 8ht Edition, Cengage Learning, Toronto, 2012. [Search CrossRef] |
Minimizing the vibration amplitude of rotating machinery running through the resonance area by application of magnetorheological squeeze film dampers | JVE Journals
Jaroslav Zapoměl1 , Petr Ferfecki2 , Jan Kozánek3
1, 3Institute of Thermomechanics, Prague, Czech Republic
1, 2VSB-Technical University of Ostrava, Ostrava, Czech Republic
The unbalance of rotating machines excites their lateral vibration that can become excessive in the vicinity of the critical speeds. Placing magnetorheological dampers to the rotor supports is a possible technological solution for reducing its amplitude. Both the analytical studies and practical experience show that behaviour of these damping elements is strongly nonlinear. A novel mathematical model of a magnetorheological squeeze film damper has been developed on the assumptions of the classical theory of lubrication, except that for the lubricant. The magnetorheological oil is represented by a bilinear material, the yielding shear stress of which depends on magnetic induction. The simulation results show that a right control of the damping force arrives at significant reduction of the vibration amplitude. The development and application of a new model of a magnetorheological squeeze film damper and confirmation of its efficiency and computational stability are the principal contributions of this article.
Keywords: magnetorheological dampers, bilinear material, strongly nonlinear damping forces, controllable attenuation of the rotor vibrations.
The unbalance is a source of excitation of lateral vibration of rotors that becomes excessive if they operate near the critical speeds. An often used technological solution for its reducing consists in placing damping devices in the rotor supports. As discussed in [1], to achieve their optimum performance, their damping effect must be adaptable to the current operating conditions. For this purpose, several different types of controllable damping devices based on mechanical, hydraulic, and electromagnetic principles have been developed [2-6].
The enhanced mathematical models of a short and long squeeze film magnetorheological damper were presented by Zapomel et al. [7, 8]. These models utilize Bingham material to represent the magnetorheological oil.
In this paper, the presented model of a magnetorheological squeeze film damper is based on utilization of bilinear material. The model was implemented in the computational procedures for investigation of lateral vibration of flexible rotors. The aim of the analysis was to study efficiency of magnetorheological damping devices used for attenuation of lateral oscillations of rotors operating in the regime when they accelerate and run over the critical speeds.
The development of a novel model of a magnetorheological squeeze film damper, its implementation in the procedures for investigation of the steady state and transient vibrations of flexible rotors, and analysis of efficiency of the magnetorheological damping devices applied in the field of rotor dynamics are the principal contributions of this article.
2. Determination of the controllable hydraulic damping forces
The main parts of a squeeze film magnetorheological damper (Fig. 1) are two concentric rings between which there is a clearance filled with magnetorheological oil. The inner ring is coupled with the rotor journal by a rolling-element bearing and with the damper housing by a squirrel spring. The lateral vibration of the rotor squeezes the oil film, which produces the damping effect. The damper is equipped with electric coils generating magnetic flux that passes through the lubricating layer. As resistance against the flow of magnetorheological oils depends on magnetic induction, the change of the electric current changes the damping force.
The magnetorheological oils belong to the class of fluids with a yielding shear stress. It means the flow occurs only in those areas where the shear stress between two neighbouring layers exceeds a limit value. In the region, called a core, where the shear stress remains lower, the oil behaviour is similar to solid matters.
The developed model of a magnetorheological squeeze film damper is based on assumptions of the classical theory of lubrication except that for the lubricant. The magnetorheological oil is represented by a bilinear material, the yielding shear stress of which depends on magnetic induction. In addition, it is assumed that the geometry and the design of the damper make it possible to consider it as short.
On these conditions, the pressure distribution in the full oil film is described by the Reynolds equation adapted to bilinear material [9]:
\frac{\partial }{\partial Z}\left(\frac{1}{{\eta }_{C}}{h}^{3}p\text{'}\right)=12\stackrel{˙}{h},
\frac{\partial }{\partial Z}\left[\frac{1}{\eta }\left({h}^{3}p\text{'}+3{h}^{2}{\tau }_{y}+8\frac{{\tau }_{C}^{3}}{{p\text{'}}^{2}}-12\frac{{\tau }_{y}{\tau }_{C}^{2}}{{p\text{'}}^{2}}\right)-\frac{8}{{\eta }_{C}}\frac{{\tau }_{C}^{3}}{{p\text{'}}^{2}}\right]=12\stackrel{˙}{h}, \stackrel{˙}{h}<0,
{Z}_{C}=-\frac{{\tau }_{C}{h}^{2}}{6{\eta }_{C}\stackrel{˙}{h}},
{p}_{C}^{\text{'}}=-\frac{2{\tau }_{C}}{h},
h=c-{e}_{H}\mathrm{c}\mathrm{o}\mathrm{s}\left(\phi -\gamma \right),
p
stands for the pressure,
{p}^{\text{'}}
is the pressure gradient in the axial direction,
Z
is the axial coordinate defining position in the oil film (Fig. 2),
h
c
is the width of the damper gap,
{e}_{H}
is the rotor journal eccentricity,
\phi
is the damper circumferential coordinate (Fig. 2),
\gamma
is the position angle of the line of centres (Fig. 2),
{\tau }_{y}
is the yielding shear stress,
{\tau }_{C}
is the shear stress at the core border,
{\eta }_{C}
\eta
are the dynamic viscosities of the oil inside and outside the core area respectively, and (.) denotes the first derivative with respect to time.
{Z}_{C}
is the axial coordinate specifying the location at which the core touches the rings surfaces.
{p}_{C}^{\text{'}}
is the pressure gradient in the axial direction at location specified by axial coordinate
{Z}_{C}
Fig. 1. Scheme of a magnetorheological squeeze film damper
Fig. 2. The damper coordinate systems
The Reynolds Eqs. (1-2) are referred to the extents of the axial coordinates
0\le Z\le {Z}_{C}
{Z}_{C}<Z\le L/2
L
denotes the damper axial length. More details on derivation and solving the Reynolds Eq. (1-2) can be found in [9].
In that part of the damper gap where the thickness of the oil film rises with time a cavitation takes place. In cavitated areas the Reynolds equation does not hold. In accordance with the observations, in the developed mathematical model it is assumed that the pressure of the medium in cavitated regions remains constant and equal to the pressure in the ambient space.
y
z
components of the hydraulic force
{F}_{mry}
{F}_{mrz}
by which the oil film acts on the rotor journal are obtained by integration of the pressure distribution around the circumference and along the length of the damper:
{F}_{mry}=-2R\mathrm{ }\underset{0}{\overset{\frac{L}{2}}{\int }}\underset{0}{\overset{2\pi }{\int }}{p}_{d}\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\phi d\phi dZ,
{F}_{mrz}=-2R\mathrm{ }\underset{0}{\overset{\frac{L}{2}}{\int }}\underset{0}{\overset{2\pi }{\int }}{p}_{d}\mathrm{ }\mathrm{s}\mathrm{i}\mathrm{n}\phi d\phi dZ,
R
is the gap mean radius and
{p}_{d}
is the pressure distribution in the damper clearance.
3. Modelling of the magnetic circuit
At the simplest distinguishing level, the damper housing can be considered as a body composed of a number of parallel branches mutually turned in the circumferential direction (Fig. 3). Each branch represents a divided core of an electromagnet with the gap filled with magnetorheological oil. Then magnetic induction
B
in the damper gap at location specified by the angular coordinate can be expressed by means of a semi-analytical formula:
B={k}_{D}{\mu }_{0}{\mu }_{r}\frac{I}{h},
{k}_{D}
is the design parameter,
{\mu }_{0}
{\mu }_{r}
are the vacuum and relative permeabilities of the magnetorheological fluid, respectively, and
I
is the current.
Fig. 3. One branch representing the damper housing
From the physical point of view, design parameter
{k}_{D}
is a product of the number of the coil turns and magnetic efficiency.
Based on the carried-out experiments at different working places, the dependence of the yielding shear stress on magnetic induction can be approximated by a power function:
{\tau }_{y}={k}_{y}{B}^{{n}_{y}},
{k}_{y}
{n}_{y}
are the proportional and exponential material constants of the magnetorheological oil.
4. The investigated rotor system
The rotor system (Fig. 4) that is a subject of investigations consists of a flexible shaft and of one rigid disc. At both its ends, the shaft is coupled with the frame by magnetorheological squeeze film dampers. The rotor turns at a variable speed and is loaded by its weight and by the disc unbalance. The squirrel springs of both dampers are prestressed to eliminate their deflection caused by the rotor weight. The whole system can be considered as symmetric relative to the disc middle plane.
Fig. 4. Scheme of the studied rotor system
In the computational model, the rotor is represented by a flexibly supported Jeffcott one and both dampers by springs and nonlinear force couplings.
The lateral vibration of the rotor is described by the motion equation:
\mathbf{M}\stackrel{¨}{\mathbf{x}}+\left({\mathbf{B}}_{P}+{\mathbf{B}}_{M}\right)\stackrel{˙}{\mathbf{x}}+\left(\mathbf{K}+\omega {\mathbf{K}}_{C}\right)\mathbf{x}={\mathbf{f}}_{A}+{\mathbf{f}}_{H},
\mathbf{M}
\mathbf{K}
{\mathbf{K}}_{C}
are the mass, stiffness and circulation matrices,
{\mathbf{B}}_{P}
{\mathbf{B}}_{M}
are the matrices of external (environmental) and material (shaft material) damping,
\omega
is the angular velocity of the rotor rotation,
\mathbf{x}
is the vector of displacements of the disc and the journal centres in the
y
z
(vertical) directions,
{\mathbf{f}}_{A}
{\mathbf{f}}_{H}
are the vectors of applied and hydraulic damping forces, and (..) denotes the second derivative with respect to time.
The Adams-Moulton direct integration (transient solutions) and the trigonometric collocation methods (steady state solution) were applied to solve the equation of motion Eq. (10).
5. Results of the simulations
The technological parameters of the studied system are: the disc mass of 250 kg, the shaft stiffness of 20.0 MN/m, the stiffness of each squirrel spring of 5.0 MN/m, the coefficient of the disc external damping of 10 Ns/m, the coefficient of the shaft material viscous damping of 0.00003 s, the damper gap middle radius of 800 μm, the oil dynamical viscosity (not effected by the magnetic field) of 0.3 Pas, the magnetorheological oil proportional and exponential material constants of 10000, 1.1, respectively.
The rotor turns at constant angular velocity of 150 rad/s. At the specified point of time, it starts to accelerate so that its speed increases uniformly to 250 rad/s during the span of time of 1.0 s. The task was to propose dependence of the applied current on time to minimize amplitude of the rotor vibrations during its passing over the critical speed.
A simple analysis gives the natural frequency of the rotor system of 163 rad/s. It implies the rotor must pass through the resonance area during its acceleration.
Fig. 5 shows the frequency response of the disc center (vibration in the horizontal direction) for three magnitudes of the applied current (0.0, 1.0, 2.0 A). If the rotor turns at angular speed lower than approximately 230 rad/s, the maximum reduction of the oscillation amplitude is achieved if the current is maximum. To achieve the same effect for higher operating velocities, the current must be minimum (0.0 A).
Fig. 5. The disc center frequency response (direction
y
), current: 0.0 A, 1.0 A, 2.0 A
In Fig. 6, the time histories of the disc center displacement in the horizontal direction are depicted during running the rotor over the critical angular velocity for the case when the damping elements in the rotor supports work in the passive regime for two magnitudes of the electric current of 0.0 A, 2.0 A. The increase of the oscillation amplitude is significant especially if no current is applied. The observed behavior is in full correspondence with the frequency response characteristic (Fig. 5).
Fig. 6. Time history of the disc center displacement (
y
direction), current 0.0 A, 2.0 A
y
direction), controlled current
The optimum variant of reducing the rotor vibration during its acceleration consists in a gradual decrease of the electric current feeding the damper coils from 2.0 A to zero in the time interval from 600 to 900 ms after the point of time when the rotor starts to accelerate. As evident from Fig. 7, this operation minimizes the oscillation amplitude both during the transient period when the rotor passes through the resonance area and after finishing the rotor acceleration when its vibration becomes steady state.
This article presents a novel mathematical model of a magnetorheological squeeze film damper for reducing lateral oscillations of flexible rotors and its implementation in the computational procedures. The principle of work of this damping device is based on combination of several physical phenomena (mechanical, hydraulic, electrical, magnetic). Because of the hydraulic damping forces, the governing equations of the rotor have a strongly nonlinear character. The performed simulations showed that the developed procedures, which are based on modelling the magnetorheological oil by bilinear material, are computationally stable, confirmed that the right control of the applied current enables to minimize amplitude of the rotor vibrations in the vicinity of the critical speeds, and enabled the influence of magnetorheological damping devices on behaviour of flexible rotors to be learnt more on.
The research work reported in this article was made possible by the Projects 15-06621S (Czech Science Foundation) and LQ1602 (IT4Innovations Excellence in Science). The support is highly acknowledged.
Zapoměl J., Ferfecki P., Kozánek J. Determination of the transient vibrations of a rigid rotor attenuated by a semiactive magnetorheological damping device by means of computational modelling. Applied and Computational Mechanics, Vol. 7, Issue 2, 2013, p. 223-234. [Search CrossRef]
Mu C., Darling J., Burrows C. R. An appraisal of a proposed active squeeze film damper. Journal of Tribology, Vol. 113, Issue 4, 1991, p. 750-754. [Search CrossRef]
El-Shafei A., El-Hakim M. Experimental investigation of adaptive control applied to HSFD supported rotors. Journal of Engineering for Gas Turbines and Power, Vol. 122, Issue 4, 2000, p. 685-692. [Search CrossRef]
Takayama Y., Sueoka A., Kondou T. Modeling of moving-conductor type eddy current damper. Journal of System Design and Dynamics, Vol. 2, Issue 5, 2008, p. 1148-1159. [Search CrossRef]
Morishita S., Mitsui J. Controllable squeeze film damper (an application of electro-rheological fluid). Journal of Vibration and Acoustics, Vol. 114, Issue 3, 1992, p. 354-357. [Search CrossRef]
Wang J., Meng G. Magnetorheological fluid devices: principles, characteristics and applications in mechanical engineering. Proceedings of the Institution of Mechanical Engineers, Part L: Journal of Materials: Design and Applications, Vol. 215, Issue 3, 2001, p. 165-174. [Search CrossRef]
Zapoměl J., Ferfecki P., Forte P. A computational investigation of the transient response of an unbalanced rigid rotor flexibly supported and damped by short magnetorheological squeeze film dampers. Smart Materials and Structures, Vol. 21, Issue 10, 2012, p. 1-12. [Search CrossRef]
Zapoměl J., Ferfecki P. Mathematical modelling of a long squeeze film magnetorheological damper for rotor systems. Modelling and Optimization of Physical Systems, Vol. 9, 2010, p. 97-102. [Search CrossRef]
Zapoměl J., Ferfecki P. 2015. A 2D mathematical model of a short magnetorheological squeeze film damper based on representing the lubricating oil by bilinear theoretical material. Proceedings of the IFToMM World Congress, Taipei, Taiwan, 2015, p. 1-6. [Search CrossRef] |
AbsoluteDeviation - Maple Help
Home : Support : Online Help : Statistics and Data Analysis : Statistics Package : Quantities : AbsoluteDeviation
compute the average absolute deviation from a given point
AbsoluteDeviation(A, b, ds_options)
AbsoluteDeviation(M, bs, ds_options)
AbsoluteDeviation(X, p, rv_options)
real number; base point
real number or list of real numbers; base points
algebraic expression; base point
(optional) equation(s) of the form option=value where option is one of ignore, or weights; specify options for computing the absolute deviation of a data set
(optional) equation of the form numeric=value; specifies options for computing the absolute deviation of a random variable
The AbsoluteDeviation function computes the average absolute deviation of the specified random variable or data set from the specified base point.
The first parameter can be a data set (e.g. a Vector), a Matrix data set, a distribution (see Statistics[Distribution]), a random variable, or an algebraic expression involving random variables (see Statistics[RandomVariable]).
The parameter b must be a real number in the first calling sequence. In the second calling sequence, M has to be a Matrix; then bs can be a real number or a list of real numbers. A list gives the base points for respective columns of the Matrix data set. If bs is a single real number, then the base point is the same for all columns. In the third calling sequence, p can be any expression of type/algebraic.
If M is a DataFrame object, then b has to be a single real number as a base point.
ignore=truefalse -- This option controls how missing data is handled by the AbsoluteDeviation command. Missing items are represented by undefined or Float(undefined). So, if ignore=false and A contains missing data, the AbsoluteDeviation command will return undefined. If ignore=true all missing items in A will be ignored. The default value is false.
1
numeric=truefalse -- By default, the absolute deviation is computed symbolically. To compute the absolute deviation numerically, specify the numeric or numeric = true option.
\mathrm{with}\left(\mathrm{Statistics}\right):
Compute the average absolute deviation of the beta distribution with parameters 3 and 5 from point
\frac{1}{2}
\mathrm{AbsoluteDeviation}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2}\right)
\frac{\textcolor[rgb]{0,0,1}{175}}{\textcolor[rgb]{0,0,1}{1024}}
\mathrm{AbsoluteDeviation}\left('\mathrm{Β}'\left(3,5\right),\frac{1}{2},\mathrm{numeric}\right)
\textcolor[rgb]{0,0,1}{0.1708984375}
Generate a random sample of size 100000 drawn from the above distribution and compute the sample absolute deviation from
\frac{1}{2}
A≔\mathrm{Sample}\left('\mathrm{Β}'\left(3,5\right),{10}^{5}\right):
\mathrm{AbsoluteDeviation}\left(A,\frac{1}{2}\right)
\textcolor[rgb]{0,0,1}{0.171592502667439}
Compute the standard error of the sample absolute deviation from
\frac{1}{2}
for the normal distribution with parameters 5 and 2.
X≔\mathrm{RandomVariable}\left(\mathrm{Normal}\left(5,2\right)\right):
B≔\mathrm{Sample}\left(X,{10}^{6}\right):
[\mathrm{AbsoluteDeviation}\left(X,\frac{1}{2}\right),\mathrm{StandardError}[{10}^{6}]\left(\mathrm{AbsoluteDeviation},X,\frac{1}{2}\right)]
[\frac{\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{81}}{\textcolor[rgb]{0,0,1}{32}}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{erf}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{8}}\right)\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}}{\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}}\textcolor[rgb]{0,0,1}{,}\frac{\sqrt{\textcolor[rgb]{0,0,1}{97}\textcolor[rgb]{0,0,1}{-}\frac{{\left(\textcolor[rgb]{0,0,1}{4}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}\textcolor[rgb]{0,0,1}{}{\textcolor[rgb]{0,0,1}{ⅇ}}^{\textcolor[rgb]{0,0,1}{-}\frac{\textcolor[rgb]{0,0,1}{81}}{\textcolor[rgb]{0,0,1}{32}}}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{erf}}\textcolor[rgb]{0,0,1}{}\left(\frac{\textcolor[rgb]{0,0,1}{9}\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{8}}\right)\textcolor[rgb]{0,0,1}{}\sqrt{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}\right)}^{\textcolor[rgb]{0,0,1}{2}}}{\textcolor[rgb]{0,0,1}{\mathrm{\pi }}}}}{\textcolor[rgb]{0,0,1}{2000}}]
[\mathrm{AbsoluteDeviation}\left(X,\frac{1}{2},\mathrm{numeric}\right),\mathrm{StandardError}[{10}^{6}]\left(\mathrm{AbsoluteDeviation},X,\frac{1}{2},\mathrm{numeric}\right)]
[\textcolor[rgb]{0,0,1}{4.516938351}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.001961445368}]
\mathrm{AbsoluteDeviation}\left(B,\frac{1}{2}\right)
\textcolor[rgb]{0,0,1}{4.51491322247257}
Y
and compute the average absolute deviation of
\frac{1}{Y+2}
\frac{1}{2}
Y≔\mathrm{RandomVariable}\left('\mathrm{Β}'\left(5,2\right)\right):
\mathrm{AbsoluteDeviation}\left(\frac{1}{Y+2},\frac{1}{2}\right)
\textcolor[rgb]{0,0,1}{584}\textcolor[rgb]{0,0,1}{+}\textcolor[rgb]{0,0,1}{1440}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{2}\right)\textcolor[rgb]{0,0,1}{-}\textcolor[rgb]{0,0,1}{1440}\textcolor[rgb]{0,0,1}{}\textcolor[rgb]{0,0,1}{\mathrm{ln}}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{3}\right)
\mathrm{AbsoluteDeviation}\left(\frac{1}{Y+2},\frac{1}{2},\mathrm{numeric}\right)
\textcolor[rgb]{0,0,1}{0.1302443242}
C≔\mathrm{Sample}\left(\frac{1}{Y+2},{10}^{5}\right):
\mathrm{AbsoluteDeviation}\left(C,\frac{1}{2}\right)
\textcolor[rgb]{0,0,1}{0.130266795070712}
Compute the average absolute deviation of a weighted data set.
V≔〈\mathrm{seq}\left(i,i=57..77\right),\mathrm{undefined}〉:
W≔〈2,4,14,41,83,169,394,669,990,1223,1329,1230,1063,646,392,202,79,32,16,5,2,5〉:
\mathrm{AbsoluteDeviation}\left(V,60,\mathrm{weights}=W\right)
\textcolor[rgb]{0,0,1}{Float}\textcolor[rgb]{0,0,1}{}\left(\textcolor[rgb]{0,0,1}{\mathrm{undefined}}\right)
\mathrm{AbsoluteDeviation}\left(V,60,\mathrm{weights}=W,\mathrm{ignore}=\mathrm{true}\right)
\textcolor[rgb]{0,0,1}{7.02737332556785}
M≔\mathrm{Matrix}\left([[3,1130,114694],[4,1527,127368],[3,907,88464],[2,878,96484],[4,995,128007]]\right)
\textcolor[rgb]{0,0,1}{M}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{1130}& \textcolor[rgb]{0,0,1}{114694}\\ \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{1527}& \textcolor[rgb]{0,0,1}{127368}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{907}& \textcolor[rgb]{0,0,1}{88464}\\ \textcolor[rgb]{0,0,1}{2}& \textcolor[rgb]{0,0,1}{878}& \textcolor[rgb]{0,0,1}{96484}\\ \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{995}& \textcolor[rgb]{0,0,1}{128007}\end{array}]
We compute the average absolute deviation from a fixed number.
\mathrm{AbsoluteDeviation}\left(M,10000\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{9996.80000000000}& \textcolor[rgb]{0,0,1}{8912.60000000000}& \textcolor[rgb]{0,0,1}{101003.400000000}\end{array}]
It might be more useful to take the average absolute deviation from three different numbers.
\mathrm{AbsoluteDeviation}\left(M,[3,1000,100000]\right)
[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{0.600000000000000}& \textcolor[rgb]{0,0,1}{175.400000000000}& \textcolor[rgb]{0,0,1}{17024.2000000000}\end{array}]
The M and bs parameters were introduced in Maple 16. |
Predict tracks to a time stamp - MATLAB predictTracksToTime - MathWorks Nordic
Predict Track State in radarTracker
Predict tracks to a time stamp
predictedtracks = predictTracksToTime(tracker,trackID,time)
predictedtracks = predictTracksToTime(tracker,category,time)
predictedtracks = predictTracksToTime(tracker,category,time,'WithCovariance',tf)
predictedtracks = predictTracksToTime(tracker,trackID,time) returns the predicted tracks, predictedtracks, of the tracker, at the specified time, time. The tracker or fuser must be updated at least once before calling this object function. Use isLocked(tracker) to test whether the tracker or fuser has been updated.
This function only outputs the predicted tracks and does not update the internal track states of the tracker.
predictedtracks = predictTracksToTime(tracker,category,time) returns all predicted tracks for a specified category, category, of tracked objects.
predictedtracks = predictTracksToTime(tracker,category,time,'WithCovariance',tf) also allows you to specify whether to predict the state covariance of each track or not by setting the tf flag to true or false. Predicting the covariance slows down the prediction process and increases the computation cost, but it provides the predicted track state covariance in addition to the predicted state. The default is false.
Create a track from a detection at time
\mathit{t}=0
tracker = radarTracker;
Predict the track to
\mathit{t}=1
predictedtracks = predictTracksToTime(tracker,'all',1)
Track identifier, specified as a positive integer. Only the track specified by the trackID is predicted in the tracker.
List of tracks or branches, returned as:
An array of objectTrack objects in the MATLAB interpreted mode.
An array of structures in the code generation mode. The field names of the structures are the same as the names of properties in objectTrack. |
Leverage Model - Goldfinch Docs | Goldfinch
The Leverage Model determines how much capital the Senior Pool allocates toward each Borrower Pool, based on how much it trusts each Borrower Pool. Currently, the leverage model as described here is not yet live, but the community put forward a proposal, linked here.
Trust Through Consensus
In order to determine how to allocate capital from the Senior Pool, the protocol uses a principle of "trust through consensus." This means that while the protocol doesn't trust any individual Backer or Auditor, it does trust the collective actions of many of them. At a high level: when more Backers supply to a given Borrower Pool, the Senior Pool increases the ratio with which it adds leverage.
Because this approach relies on counting individual Backers, the protocol must ensure they are in fact represented by different people. Therefore, all Backers, Borrowers, and Auditors require a unique entity check to participate (see the Unique Entity Check section).
Leverage Model Formula
The leverage amount,
A
, that the Senior Pool allocates is determined by the formula,
A = S * D * L
S
is the total capital supplied by Backers.
D
is the distribution adjustment on a scale of
0
1
, which accounts for how evenly distributed the Backers are.
D
0
when the distribution is skewed and closer to
1
when the Backers are more equally distributed. This ensures no single Backer has an outsized influence. The formula for
D
uses the percent supplied by each Backer,
s_{n}
, and is based on the Herfindahl-Hirschman Index:
D = 1-\sum_{i=1}^n s_n^2
L is the leverage ratio on a scale of
0
to the maximum potential leverage ratio. Based on the number of Backers,
b
, the leverage ratio increases linearly from
B_{min}
, the minimum number of Backers necessary for leverage, to
B_{max}
, the maximum number of Backers necessary to achieve the maximum potential leverage,
L_{max}
L=L_{max}*\frac{max(0, b-B_{min})}{B_{max}-B_{min}} |
Examine the integrals below. Consider the multiple tools available for integrating and use the best strategy for each part. Evaluate each integral and briefly describe your method.
\int ( \operatorname { ln } ( 3 ) ) 3 ^ { x } d x
Remember that the derivative of
3^x = (\ln 3) · 3^x
2 \pi \int _ { 0 } ^ { 2 } x ( 2 x + 5 ) d x
Distribute the '
x
' then try evaluating the integral.
\int 4 e ^ { \operatorname { ln } ( x ) } d x
How can you rewrite
e^{\ln x}
e^{\ln x} = x
\int ( 4 x ) d x = 2 x ^ { 2 } + C
\int 2 ^ { m + 2 } d m
2^{m + 2} = 2^2 · 2^m |
A girl drops a ball off a cliff into the
A girl drops a ball off a cliff into the ocean. The polynomial
-16{t}^{2}+250
gives the height, in feet, of the ball t seconds after it is dropped from a 250ft tall cliff. Find the height, in feet, after t=2 seconds.
The height as a function of time
h\left(t\right)=250-16{t}^{2}
Height after 2 sec
h\left(2\right)=250-16×{2}^{2}
-16{t}^{2}+250
To simplify the expression, by substitute the value of variable by given term.
So simplify as follow:
Substitute the value of variable t=2.
-16{t}^{2}+250=-16\cdot {\left(2\right)}^{2}+250
[Substitute the variable.]
=-16*4+250 [Write the value of square.]
=-64+250 [Multiply them.]
=186 [Simplify the term.]
Therefore, the simplified term is 186.
g\left(x\right)=3-\frac{{x}^{2}}{4}
Factor each polynomial completely. If the polynomial cannot be factored, say it is prime.
4{x}^{2}-8x+32
81-25{z}^{2}
To explain: The reason behind to not use the SAS congruence criteria even though it looks like angle BAC and DAC are equal.
What is the standard form of y=(x+13)(x-4)(x-1)?
\left[\begin{array}{ccc}-2& 1& 2\\ 4& 1& -2\\ -6& -3& 4\end{array}\right]
Complete Factorization Factor the polynomial completely, and find all its zeros.State the multiplicity of each zero.
P\left(x\right)={x}^{4}+2{x}^{2}+1 |
Find \frac{\partical f}{\partical x}, \frac{\partical f}{\partical y} and \frac{
Find \frac{\partical f}{\partical x}, \frac{\partical f}{\partical y} and \frac{\partical f}{\partical z} for the function f(x,y,z)=x^{2}y+y^{2}+xz^{2}
\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\text{ }\text{and}\text{ }\frac{\partial f}{\partial z}
f\left(x,y,z\right)={x}^{2}y+{y}^{2}+x{z}^{2}
f\left(x,y,z\right)={x}^{2}y+{y}^{2}+x{z}^{2}
We need to find some partial derivatives.
f\left(x,y,z\right)={x}^{2}y+{y}^{2}+x{z}^{2}
{f}_{x}=2xy+{z}^{2}
{f}_{y}={x}^{2}+2y
{f}_{z}=2xz
Thus, we obtained all our partial derivatives.
Find the following derivatives.
\mathrm{ln}\left(\frac{x+3}{{x}^{2}}\right)
Find all first and the second partial derivatives.
f\left(x,y\right)=2{x}^{5}{y}^{2}+{x}^{2}y
f\left(x\right)=\frac{4{x}^{3}+\sqrt[3]{x}-6x}{{x}^{\frac{1}{2}}}
Use derivatives to find the critical points and inflection points.
f\left(x\right)={x}^{3}-9{x}^{2}+24x+5
Second derivatives Find
{d}^{2}\frac{y}{{dx}^{2}}
for the following functions.
y=2{x}^{2}+2 |
Disposal of Fixed Assets - Course Hero
Principles of Accounting/Long-Term Assets/Disposal of Fixed Assets
Over time a fixed asset may decline in usefulness or discontinue providing a benefit to the business entity. Once this occurs business entities will either dispose of or sell the asset. To dispose of a fixed asset, the business entity must remove the asset from the accounting records. The entry to remove the asset will vary depending on the remaining net book value of the asset, whether it is scrapped or sold, and if sold, how much it is sold for.
Big Truck Company has an old piece of equipment it no longer wishes to use. The equipment was originally purchased for $28,000, has an estimated useful life of 10 years, and has no residual value. Big Truck Company uses straight-line depreciation. At this time 11 years have passed, and Big Truck Company has depreciated the equipment down to its residual value of $0. The machine has no value in the market, so the company is not able to sell it. The equipment is scrapped. Big Truck Company will make an entry to remove the asset when it is scrapped. Disposing of a fully depreciated fixed asset will not have an effect on net assets as the book value is already $0.
Fully Depreciated, Scrapped, Asset Disposal Journal Entry
April 18 Accumulated Depreciation Machinery $28,000
To write off discarded machinery
In the event that an asset is not fully depreciated, the depreciation for the current period should be recorded prior to removing the asset from the accounting records. For example, Big Truck Company’s machinery with a cost of $28,000 with no residual value and estimated useful life of 10 years is set to be discarded after 8 years of service on April 1, 2018. In this case straight-line depreciation was used.
\$28\rm{,}000/10 \text{\;Years}=\$2\rm{,}800
of depreciation had been recorded each year for the first 8 years of service. Accumulated depreciation thus has a balance of $22,400 at the beginning of this year. Big Truck Company must record depreciation for January through April 1 of this year before making the entry to dispose of the machine. Depreciation for January, February, and March totals $700. Depreciation for these 3 months is $2,800 times 3 out of 12 months.
Update Depreciation Journal Entry
April 25 Depreciation Expense Machinery $700
Accumulated Depreciation - Machinery $700
To record depreciation for the period
Accumulated depreciation has a balance of $22,400 at the beginning of the year, and an additional $700 is recorded for the current year, bringing the balance to $23,100. Then Big Truck Company will make an entry to discard the machine.
Asset Disposal Journal Entry
April 25 Accumulated Depreciation $23,100
Loss on Disposal of Machinery $4,900
To write off machinery discarded
The loss of $4,900 is determined by subtracting the accumulated depreciation from the machinery account balance. The loss should be reported on the income statement in the period in which the asset is disposed of.
Business entities may also sell fixed assets they no longer use. The assets are removed from the accounting records at book value. If the cash received is equal to book value, no gain or loss is recorded. If the cash received is greater than the book value, the asset is sold at a gain. Conversely, if the asset is sold for a price less than the book value, the asset is sold at a loss.
Big Truck Company is selling its machine on April 1 this year with a cost of $28,000 with no residual value and estimated useful life of 10 years. Straight-line depreciation is used on this asset and has been recorded up to date as of April 1, bringing the accumulated depreciation for the asset to $23,100. Thus, current book value is $4,900 ($28,000 initial cost less $23,100 accumulated depreciation). The machine is being sold for $4,900. Big Truck Company will make an entry to sell the machine at book value.
Sale of Machinery at Book Value Journal Entry
April 25 Cash $4,900
Accumulated Depreciation Machinery $23,100
Big Truck Company will record a loss if it sold the machine below book value at $4,000.
Loss on Sale of Machinery Journal Entry
Loss on the Sale of Machinery $900
Big Truck Company will record a gain if it sold the machine above book value at $6,000.
Gain on Sale of Machinery Journal Entry
<Accounting for Repairs and Improvements>Natural Resources and Intangible Assets |
Demodulate FSK-modulated data - Simulink - MathWorks Switzerland
M-FSK Demodulator Baseband
Demodulate FSK-modulated data
FM, in Digital Baseband sublibrary of Modulation
The M-FSK Demodulator Baseband block demodulates a signal that was modulated using the M-ary frequency shift keying method. The input is a baseband representation of the modulated signal. The input and output for this block are discrete-time signals. This block accepts a scalar value or column vector input signal of type single or double. For information about the data types each block port supports, see Supported Data Types.
The M-ary number parameter, M, is the number of frequencies in the modulated signal. The Frequency separation parameter is the distance, in Hz, between successive frequencies of the modulated signal.
The M-FSK Demodulator Baseband block implements a non-coherent energy detector. To obtain the same BER performance as that of coherent FSK demodulation, use the CPFSK Demodulator Baseband block.
The Symbol set ordering parameter indicates how the block maps a symbol to a group of K output bits. When you set the parameter to Binary, the block maps the integer, I, to [u(1) u(2) ... u(K)] bits, where the individual u(i) are given by
I\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\sum _{i=1}^{K}u\left(i\right){2}^{K-i}
u(1) is the most significant bit.
For example, if M = 8, you set Symbol set ordering to Binary, and the demodulated integer symbol value is 6, then the binary output word is [1 1 0].
When you set Symbol set ordering to Gray, the block assigns bit outputs from points of a predefined Gray-coded signal constellation. The predefined M-ary Gray-coded signal constellation assigns the bit representation in the pth row of bit matrix b to the pth integer, where the left-most bit is the most significant bit (MSB).
M = 8; P = [0:M-1]';
b = int2bit(bitxor(P,floor(P/2)),nBits);
b = reshape(b,[],8)';
The typical Binary to Gray mapping for M = 8 is shown in the following tables.
Binary to Gray Mapping for Integers
Whether the output is an integer or a binary representation of an integer, the block maps the highest frequency to the integer 0 and maps the lowest frequency to the integer M-1. In baseband simulation, the lowest frequency is the negative frequency with the largest absolute value.
To run the M-FSK Demodulator block in multirate mode, clear the Treat each discrete rate as a separate task checkbox (in Simulation > Configuration Parameters > Solver).
The number of frequencies in the modulated signal.
Frequency separation (Hz)
The distance between successive frequencies in the modulated signal.
The number of input samples that represent each modulated symbol.
Enforce single-rate processing — When you select this option, the input and output signals have the same port sample times. The block implements the rate change by making a size change at the output when compared to the input. The output width is the number of symbols (which is given by dividing the input length by the Samples per symbol parameter value when the Output type parameter is set to Integer).
The output type of the block can be specified here as boolean, int8, uint8, int16, uint16, int32, uint32, or double. By default, the block sets this to double.
M-FSK Modulator Baseband
[1] Sklar, Bernard. Digital Communications: Fundamentals and Applications. Upper Saddle River, NJ: Prentice-Hall, 2001.
M-FSK Modulator Baseband | CPFSK Demodulator Baseband |
Count-distinct problem - Wikipedia
In computer science, the count-distinct problem[1] (also known in applied mathematics as the cardinality estimation problem) is the problem of finding the number of distinct elements in a data stream with repeated elements. This is a well-known problem with numerous applications. The elements might represent IP addresses of packets passing through a router, unique visitors to a web site, elements in a large database, motifs in a DNA sequence, or elements of RFID/sensor networks.
3 HyperLogLog algorithm
4.1 Min/max sketches
4.2 Bottom-m sketches
5 Weighted count-distinct problem
6 Solving the weighted count-distinct problem
Instance: A stream of elements
{\displaystyle x_{1},x_{2},\ldots ,x_{s}}
with repetitions, and an integer
{\displaystyle m}
{\displaystyle n}
be the number of distinct elements, namely
{\displaystyle n=|\left\{{x_{1},x_{2},\ldots ,x_{s}}\right\}|}
, and let these elements be
{\displaystyle \left\{{e_{1},e_{2},\ldots ,e_{n}}\right\}}
Objective: Find an estimate
{\displaystyle {\widehat {n}}}
{\displaystyle n}
{\displaystyle m}
storage units, where
{\displaystyle m\ll n}
An example of an instance for the cardinality estimation problem is the stream:
{\displaystyle a,b,a,c,d,b,d}
. For this instance,
{\displaystyle n=|\left\{{a,b,c,d}\right\}|=4}
The naive solution to the problem is as follows:
Initialize a counter, c, to zero,
{\displaystyle c\leftarrow 0}
Initialize an efficient dictionary data structure, D, such as hash table or search tree in which insertion and membership can be performed quickly.
{\displaystyle x_{i}}
, a membership query is issued.
{\displaystyle x_{i}}
is not a member of D (
{\displaystyle x_{i}\notin D}
{\displaystyle x_{i}}
Increase c by one,
{\displaystyle c\leftarrow c+1}
{\displaystyle x_{i}\in D}
) do nothing.
{\displaystyle n=c}
As long as the number of distinct elements is not too big, D fits in main memory and an exact answer can be retrieved. However, this approach does not scale for bounded storage, or if the computation performed for each element
{\displaystyle x_{i}}
should be minimized. In such a case, several streaming algorithms have been proposed that use a fixed number of storage units.
HyperLogLog algorithm[edit]
Main article: HyperLogLog
To handle the bounded storage constraint, streaming algorithms use a randomization to produce a non-exact estimation of the distinct number of elements,
{\displaystyle n}
. State-of-the-art estimators hash every element
{\displaystyle e_{j}}
into a low-dimensional data sketch using a hash function,
{\displaystyle h(e_{j})}
. The different techniques can be classified according to the data sketches they store.
Min/max sketches[edit]
Min/max sketches[2][3] store only the minimum/maximum hashed values. Examples of known min/max sketch estimators: Chassaing et al. [4] presents max sketch which is the minimum-variance unbiased estimator for the problem. The continuous max sketches estimator [5] is the maximum likelihood estimator. The estimator of choice in practice is the HyperLogLog algorithm.[6]
The intuition behind such estimators is that each sketch carries information about the desired quantity. For example, when every element
{\displaystyle e_{j}}
is associated with a uniform RV,
{\displaystyle h(e_{j})\sim U(0,1)}
, the expected minimum value of
{\displaystyle h(e_{1}),h(e_{2}),\ldots ,h(e_{n})}
{\displaystyle 1/(n+1)}
. The hash function guarantees that
{\displaystyle h(e_{j})}
is identical for all the appearances of
{\displaystyle e_{j}}
. Thus, the existence of duplicates does not affect the value of the extreme order statistics.
There are other estimation techniques other than min/max sketches. The first paper on count-distinct estimation by Flajolet et al. [7] describes a bit pattern sketch. In this case, the elements are hashed into a bit vector and the sketch holds the logical OR of all hashed values. The first asymptotically space- and time-optimal algorithm for this problem was given by Daniel M. Kane, Jelani Nelson, and David P. Woodruff.[8]
Bottom-m sketches[edit]
Bottom-m sketches [9] are a generalization of min sketches, which maintain the
{\displaystyle m}inimal values, where
{\displaystyle m\geq 1}
. See Cosma et al.[2] for a theoretical overview of count-distinct estimation algorithms, and Metwally [10] for a practical overview with comparative simulation results.
Weighted count-distinct problem[edit]
In its weighted version, each element is associated with a weight and the goal is to estimate the total sum of weights. Formally,
Instance: A stream of weighted elements
{\displaystyle x_{1},x_{2},\ldots ,x_{s}}
{\displaystyle m}
{\displaystyle n}
{\displaystyle n=|\left\{{x_{1},x_{2},\ldots ,x_{s}}\right\}|}
{\displaystyle \left\{{e_{1},e_{2},\ldots ,e_{n}}\right\}}
. Finally, let
{\displaystyle w_{j}}
be the weight of
{\displaystyle e_{j}}
{\displaystyle {\widehat {w}}}
{\displaystyle w=\sum _{j=1}^{n}w_{j}}
{\displaystyle m}
{\displaystyle m\ll n}
An example of an instance for the weighted problem is:
{\displaystyle a(3),b(4),a(3),c(2),d(3),b(4),d(3)}
{\displaystyle e_{1}=a,e_{2}=b,e_{3}=c,e_{4}=d}
, the weights are
{\displaystyle w_{1}=3,w_{2}=4,w_{3}=2,w_{4}=3}
{\displaystyle \sum {w_{j}}=12}
As an application example,
{\displaystyle x_{1},x_{2},\ldots ,x_{s}}
could be IP packets received by a server. Each packet belongs to one of
{\displaystyle n}
IP flows
{\displaystyle e_{1},e_{2},\ldots ,e_{n}}
. The weight
{\displaystyle w_{j}}
can be the load imposed by flow
{\displaystyle e_{j}}
on the server. Thus,
{\displaystyle \sum _{j=1}^{n}{w_{j}}}
represents the total load imposed on the server by all the flows to which packets
{\displaystyle x_{1},x_{2},\ldots ,x_{s}}
belong.
Solving the weighted count-distinct problem[edit]
Any extreme order statistics estimator (min/max sketches) for the unweighted problem can be generalized to an estimator for the weighted problem .[11] For example, the weighted estimator proposed by Cohen et al.[5] can be obtained when the continuous max sketches estimator is extended to solve the weighted problem. In particular, the HyperLogLog algorithm [6] can be extended to solve the weighted problem. The extended HyperLogLog algorithm offers the best performance, in terms of statistical accuracy and memory usage, among all the other known algorithms for the weighted problem.
^ Ullman, Jeff; Rajaraman, Anand; Leskovec, Jure. "Mining data streams" (PDF). {{cite journal}}: Cite journal requires |journal= (help)
^ a b Cosma, Ioana A.; Clifford, Peter (2011). "A statistical analysis of probabilistic counting algorithms". Scandinavian Journal of Statistics. arXiv:0801.3552.
^ Giroire, Frederic; Fusy, Eric (2007). 2007 Proceedings of the Fourth Workshop on Analytic Algorithmics and Combinatorics (ANALCO). pp. 223–231. CiteSeerX 10.1.1.214.270. doi:10.1137/1.9781611972979.9. ISBN 978-1-61197-297-9.
^ Chassaing, Philippe; Gerin, Lucas (2006). "Efficient estimation of the cardinality of large data sets". Proceedings of the 4th Colloquium on Mathematics and Computer Science. arXiv:math/0701347. Bibcode:2007math......1347C.
^ a b Cohen, Edith (1997). "Size-estimation framework with applications to transitive closure and reachability". J. Comput. Syst. Sci. 55 (3): 441–453. doi:10.1006/jcss.1997.1534.
^ a b Flajolet, Philippe; Fusy, Eric; Gandouet, Olivier; Meunier, Frederic (2007). "HyperLoglog: the analysis of a near-optimal cardinality estimation algorithm" (PDF). Analysis of Algorithms.
^ Flajolet, Philippe; Martin, G. Nigel (1985). "Probabilistic counting algorithms for data base applications" (PDF). J. Comput. Syst. Sci. 31 (2): 182–209. doi:10.1016/0022-0000(85)90041-8.
^ Kane, Daniel M.; Nelson, Jelani; Woodruff, David P. (2010). "An Optimal Algorithm for the Distinct Elements Problem". Proceedings of the 29th Annual ACM Symposium on Principles of Database Systems (PODS).
^ Cohen, Edith; Kaplan, Haim (2008). "Tighter estimation using bottom k sketches" (PDF). PVLDB.
^ Metwally, Ahmed; Agrawal, Divyakant; Abbadi, Amr El (2008), Why go logarithmic if we can go linear?: Towards effective distinct counting of search traffic, Proceedings of the 11th international conference on Extending Database Technology: Advances in Database Technology, CiteSeerX 10.1.1.377.4771
^ Cohen, Reuven; Katzir, Liran; Yehezkel, Aviv (2014). "A Unified Scheme for Generalizing Cardinality Estimators to Sum Aggregation". Information Processing Letters. 115 (2): 336–342. doi:10.1016/j.ipl.2014.10.009.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Count-distinct_problem&oldid=1032348949" |
Maximizing Conditional Probability Practice Problems Online | Brilliant
At a party, 10 guests including Winnie and Looney are requested to draw lots for a prize. Looney thinks that if someone draws the prize, then there will be no chance for the rest to win, so whoever goes first has an advantage. Winnie, on the other hand, thinks that if someone draws a blank, then there will be greater probability for the rest to win, so whoever goes last has an advantage. Whose reasoning is correct?
Neither Winnie Looney
There are two identical boxes in front of you. In box
X,
there are 3 white balls and 7 black balls. In box
Y,
there are 7 white balls and 3 black balls. You randomly reach one of the boxes and draw three balls from it one by one. If every time you draw a ball you find it is white, at which of the following stages is the probability the highest that you chose box
X?
(A) You drew one ball, and it was white.
(B) You drew two balls, and they were both white.
(C) You drew three balls, and they were all white.
Suppose that you are playing 5 card poker. You have 4 cards in your hand and are to receive one more card. If you want to make a full house with the
5^\text{th}
card, which of the following is the most preferable situation?
(A) You have a three of a kind in your hand.
(B) You have a two pair in your hand.
(C) You have a one pair in your hand.
Do not take other players' cards into consideration.
Suppose that you are playing 5 card poker. You already have 4 cards in your hand and are waiting for the last card. In which of the following situations do you have a higher probability of success?
A:
You have a 7 of spade, a 7 of diamond, a 7 of heart, and an ace of club. You want a full house.
B:
You have a 6, a 7, a 9, and a 10 of spade. You want a straight.
Do not take the other players' cards into consideration.
Equal probability B A
Jack Potter has studied how to win a jackpot in Lotto for a long time and recently found out that the numbers 11, 21, 31, 41 (call them "Highs") showed very high winning rates while the numbers 13, 23, 33, 43 (call them "Lows") showed very low winning rates.
Now he is going to buy a Lotto ticket including the Highs in his choice of 6 numbers, because he thinks that their excellent performance in the past must be a good indicator of a higher winning chance in the future.
His brother Harry, however, after listening to Jack's plan, thinks otherwise: Highs showed higher winning rates in the past, so they will show lower winning rates in the future because if all the numbers are equally likely, their winning rates should be averaged out.
Whose reasoning is correct?
Lotto is a form of lottery in which six numbers are drawn from a larger pool (say, 54 or 56). You have to match all six numbers drawn to win the top prize.
Neither Jack Harry |
{\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}
I={\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}
\int \frac{dx}{\sqrt{1-{x}^{2}}}={\mathrm{sin}}^{-1}\left(x\right)+C
\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C
I={\int }_{0}^{1}\frac{dx}{\sqrt{1-{x}^{2}}}
={\left[{\mathrm{sin}}^{-1}\left(x\right)\right]}_{0}^{1}
={\mathrm{sin}}^{-1}\left(1\right)-{\mathrm{sin}}^{-1}\left(0\right)
={\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(\frac{\pi }{2}\right)\right)-{\mathrm{sin}}^{-1}\left(\mathrm{sin}\left(0\right)\right)
=\frac{\pi }{2}-0
=\frac{\pi }{2}
{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{\sqrt{x+yi+2}}dy
where i is the imaginary number. How to compute this integral?
{\int }_{6}^{7}\frac{x}{x-6}dx
-Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
-Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
-The integral is a proper integral.
{\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{x}^{3}}dx
Since the integral has an infinite interval of integration, it is a Type 1 improper integral.
Since the integral has an infinite discontinuity, it is a Type 2 improper integral.
The integral is a proper integral.
{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{x}^{2}{e}^{-{x}^{2}}dx
{\int }_{0}^{\frac{\pi }{4}}\mathrm{cot}xdx
{\int }_{2}^{9}9vdv
\mathrm{tan}u+2\mathrm{cos}u\mathrm{tan}u
{\int }_{0}^{x}\left(2x-y\right)dy
\int \left(2-\frac{3}{{x}^{10}}\right)dx
\int x{2}^{{x}^{2}}dx |
Polynomial End Behavior | Brilliant Math & Science Wiki
Polynomial end behavior is the direction the graph of a polynomial function goes as the input value goes "to infinity" on the left and right sides of the graph. There are four possibilities, as shown below.
With end behavior, the only term that matters with the polynomial is the one that has an exponent of largest degree. For example, if you have the polynomial
5x^4 + 12x^2 - 3x ,
5x^4
matters in terms of end behavior.
This term will be of the form
ax^n .
a
n
is an even number, the left and right sides of the graph both go to +infinity.
a
n
is an even number, the left and right sides of the graph both go to -infinity.
C) When
a
n
is an odd number, the left side goes to -infinity and the right side goes to +infinity.
D) When
a
n
is an odd number, the left side goes to +infinity and the right side goes to -infinity.
What is the end behavior of
f(x) = -55x^4 - 3x^3 + 2x - 1 ?
The largest exponent is 4, so the relevant term is
-55x^4 .
-55
4
is an even number so the end behavior matches that of B above.
Falls left, rises right Rises left, falls right Falls left, falls right Rises left, rises right
What is the end behavior of the graph of the following polynomial function?
f(x)=-4x^5-8x^3-3x^2+7x
Cite as: Polynomial End Behavior. Brilliant.org. Retrieved from https://brilliant.org/wiki/polynomial-end-behavior/ |
Free solutions to force, motion and energy problems
Force, motion and energy problems and answers
Recent questions in Force, Motion and Energy
1.Does the equivalence principle imply that there is some fundamental difference between acceleration due to gravity and acceleration by other means (because there is no way to 'feel' free fall acceleration for a uniform gravitational field)?
2.Does General Relativity allow you to describe the acceleration due to gravity without Newton's second law (because every other source of 'push or pull' outside the nucleus involves the electromagnetic field)?
3.Is the acceleration due to gravity a result of changes in time dilation/length contraction as opposed to an actual push or pull?\
An electron and a positron are separated by distance r. Findthe ratio of the gravitational force to the electric force betweenthem. From the result, what can you conclude concerning theforces acting between particles detected in a bubble chamber?(Should gravitational interactions be considered?)
How is the gravitational force between two point masses affected when they are dipped in water keeping the seperation between them the same?
(a) Calculate the gravitational force exerted on a 5.00 kg baby by a 90 kg father 0.200 m away at birth (assisting so he is close).
(b) Calculate the force on the baby due to Jupiter if it is at its closest to the earth, some
6.29×{10}^{11}m
away, showing it to be comparable to that of the father. The mass of Jupiter is about
1.90×{10}^{27}\text{ }kg
. Other objects in the room and the hospital building also exert similar gravitational forces.
The only known force a planet exerts on Earth is gravitational. (a) Calculate the magnitude of the gravitational force exerted on a 4.20 kg baby by a 100 kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)
Two balls have their centers 2.0 m apart. One ball has a mass of 8.0 kg. The other has a mass of 6.0 kg. What is the gravitational force between them?
What is the electric potential at the point indicated with the dot in the figure? Express your answer to two significant figures and include the appropriate units.
mars6svhym 2022-05-13 Answered
What is the jerk due to gravity with ascent?
Acceleration is defined as the rate of change of velocity with time. Jerk is defined as the rate of change of acceleration with time. What is the jerk due to gravity with ascent?
When an object, say a ball, is attracted by the black hole it gets acceleration due to gravity. Suppose light is moving towards the black hole vertical to it... then does it gain acceleration due to gravity? If yes then won't be the speed of light increase and get violated?
Why do we use gravitational force in earth by relating just the mass of an object with the acceleration produced by the gravitational field:
{F}_{g}=m\cdot \stackrel{\to }{g}
And when we're dealing with planets, we use a relation defined by the masses of two planets, distance squared and gravitational constant:
{F}_{g}=G\cdot \frac{{M}_{1}\cdot {M}_{2}}{{d}^{2}}
I really don't get why we use just the first relation here on earth, because we're dealing with a interction between two objects... It's because our mass is irrelevant??
Is acceleration due to gravity constant?
I was taught in school that acceleration due to gravity is constant. But recently, when I checked Physics textbook, I noted that
F=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}.
So, as the body falls down,
r
must be changing, so should acceleration due to gravity.
fetsBedscurce4why1 2022-05-10 Answered
Why do we integrate along the whole length while finding the gravitational force between a object of mass (m) and a rod of length L and mass M? Can't we simply use
F=\frac{GmM}{{r}^{2}}\phantom{\rule{thinmathspace}{0ex}}?
Two objects are attracted to each other with 36 N of gravitational force. What would the force between them be if the distance between them were doubled?
Question regarding calculating acceleration due to gravity on planet Mercury
I was asked to calculate the acceleration due to gravity on planet Mercury, if the mas of Mercury is
2,99×{10}^{22}kg
2,42×{10}^{3}\text{ }km
. The mass of the object is
10kg
and the mass of Earth is
6×{10}^{24}kg
and the Radius of the Earth is
3,82×{10}^{3}km
This question rather puzzled me because I was not sure if my answer is correct or not but let me proceed :
\stackrel{\to }{F}=m\stackrel{\to }{a}={F}_{g}=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}
{m}_{1}g=G\frac{{m}_{1}{m}_{2}}{{r}^{2}}
{m}_{2}
= mass of mercury)
g=\frac{\left(6.673×{10}^{-11}\frac{{m}^{2}}{k{g}^{2}}\right)\left(2,99×{10}^{22}kg\right)}{\left(2,42×{10}^{6}\text{ }m{\right)}^{2}}
I compute my answer to be
0.34\frac{m}{{s}^{2}}
What really is confusing me is that when I look at my textbook, it shows me the gravitational acceleration due to gravity on mercury to be
3.59\frac{m}{{s}^{2}}
Can someone please explain to me what the answer that I am getting is giving me? My computation was marked correct in a test but I do not understand what this value is giving me.
Two objects, with masses
{m}_{1}
{m}_{2}
, attract each other with force
F
. If the mass of
{m}_{1}
increased by a factor of two then the new gravitational force would be:
I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.
According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.
BUT Newton's second law states that
a=F/m
a
the acceleration of a particle,
m
its mass and
F
the sum of forces applied to it.
I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".
Can someone explain this to me? I feel pretty dumb right now :)
Does Acceleration due to Gravity take into consideration the centrifugal acceleration due to Earth's spin?
Is the gravitational acceleration we consider only the attraction due to the Earth's gravity or is it that of gravity plus the attraction due to Earth spinning?
We know that earth produces an acceleration towards the centre on any body near it due to gravitational attraction.
We denote this acceleration as
g
But we also know that any body on Earth is also undergoing rotational motion due to earth's Spin. So every body should experience an outward centrifugal acceleration.
Does acceleration due to gravity
g
take the centrifugal acceleration into consideration?
A book is lifted upward a vertical distance of 0.800 m. Duringthis displacement, does the gravitational force acting on the book do positivework or negative work? Does the gravitational potential energy of thebook increase or decrease?
Compare the gravitational attraction between objects on earth and interaction of celestial bodies in space. Which gravitational force is almost negligible? Why?
1. Why is gravitational force always an attractive force?
2. And is the Newtonian formula of gravitational force applicable for very small particles like electrons and protons etc.? From Formula of Gravitational Force, I'm referring to:
{F}_{G}=\frac{GMm}{{R}^{2}},
where M and m are the masses of objects.
A logical explanation will be much appreciated. |
Directed Angles | Brilliant Math & Science Wiki
Hua Zhi Vee and Jimin Khim contributed
Directed angles are angles that are directed. There are 2 ways of counting the directed angle
\measuredangle ABC:
the angle is positive when the points
A, B, C
are in clockwise order, and negative otherwise, or
the angle is positive when the lines
AB
BC
are in counterclockwise order, and negative otherwise.
You should see that both ways are identical to each other. As an example, see the diagram below:
\measuredangle ABC= 50^\circ, \measuredangle CBA= -50^\circ
We take every angle modulo
180^\circ
\cdots=-110^\circ=70^\circ=250^\circ=\cdots.
This means that in the diagram above,
\measuredangle ABC = 50^\circ = 230 ^\circ = -130^\circ = \cdots
Directed angles might seem really annoying and useless at first, but soon you should find it natural and sometimes better than the normal angles.
Directed angles are useful in combining multiple cases in a statement into one. An example is shown below.
The theorem below has configuration issues; the two cases seem so different and hence we have to differentiate them. By using directed angles, we can significantly reduce the number of words used.
Theorem (Cyclic Quadrilaterals)
A, B, C, D
be any four points, no three collinear.
A
B
lie on the same side of
CD,
then the four points are concyclic if and only if
∠CAD = ∠CBD.
A
B
lie on different sides of
CD,
∠CAD + ∠CBD = 180^\circ.
Theorem (Directed Cyclic Quadrilaterals)
A, B, C, D
be four points, no three collinear. Then they are concyclic if and only if
\measuredangle CAD = \measuredangle CBD.
Task: Make sure it works.
X
be any point. Points
A, B, C
are collinear if and only if
\measuredangle XBC = \measuredangle XBA.
Task: Prove this yourself.
(
Hint: Show that the assertion is equivalent to
\measuredangle ABC = 0.)
Always remember that this doesn't work for every single case. You should always check your answer after writing it out. An example where directed angles will destroy the solution is shown below:
ABCD
I_1
I_2
denote the incenters of
\triangle{ABC}
{DBC}
, respectively. Prove that
I_1I_2BC
Do not use directed angles; the problem is false if
A, C, B, D
lie in that order.
How to Use Directed Angles 1 Introduction - Evan Chen
Cite as: Directed Angles. Brilliant.org. Retrieved from https://brilliant.org/wiki/directed-angles/ |
EUDML | Kazhdan constants and matrix coefficients of . EuDML | Kazhdan constants and matrix coefficients of .
Kazhdan constants and matrix coefficients of
\text{Sp}\left(n,ℝ\right)
Neuhauser, Markus. "Kazhdan constants and matrix coefficients of .." Journal of Lie Theory 13.1 (2003): 133-154. <http://eudml.org/doc/122850>.
@article{Neuhauser2003,
author = {Neuhauser, Markus},
keywords = {Kazhdan constant; locally compact group; Kazhdan's property ; Kazhdan pair; Kazhdan’s property },
title = {Kazhdan constants and matrix coefficients of .},
AU - Neuhauser, Markus
TI - Kazhdan constants and matrix coefficients of .
KW - Kazhdan constant; locally compact group; Kazhdan's property ; Kazhdan pair; Kazhdan’s property
Kazhdan constant, locally compact group, Kazhdan's property
T
, Kazhdan pair, Kazhdan’s property
T
Articles by Neuhauser |
EUDML | Dual partial quadrangles embedded in . EuDML | Dual partial quadrangles embedded in .
Dual partial quadrangles embedded in
PG\left(3,q\right)
De Clerck, F.; Durante, N.; Thas, J. A.
Volume: 2003, page S224-S231
De Clerck, F., Durante, N., and Thas, J. A.. "Dual partial quadrangles embedded in .." Advances in Geometry 2003 (2003): S224-S231. <http://eudml.org/doc/123741>.
@article{DeClerck2003,
author = {De Clerck, F., Durante, N., Thas, J. A.},
keywords = {partial geometries; quadrangles},
title = {Dual partial quadrangles embedded in .},
AU - De Clerck, F.
AU - Durante, N.
AU - Thas, J. A.
TI - Dual partial quadrangles embedded in .
EP - S231
KW - partial geometries; quadrangles
partial geometries, quadrangles
Incidence structures imbeddable into projective geometries
Generalized quadrangles, generalized polygons
Articles by De Clerck
Articles by Durante
Articles by Thas |
EUDML | Notable curves in geometrized framework. EuDML | Notable curves in geometrized framework.
Notable curves in geometrized
{J}^{1}\left(T,M\right)
framework.
Balan, Vladimir. "Notable curves in geometrized framework.." Balkan Journal of Geometry and its Applications (BJGA) 8.2 (2003): 1-10. <http://eudml.org/doc/124632>.
@article{Balan2003,
author = {Balan, Vladimir},
title = {Notable curves in geometrized framework.},
AU - Balan, Vladimir
TI - Notable curves in geometrized framework.
Articles by Balan |
EUDML | On the Uniform (mod 1) of the Farey Fractions and lP Spaces. EuDML | On the Uniform (mod 1) of the Farey Fractions and lP Spaces.
On the Uniform (mod 1) of the Farey Fractions and lP Spaces.
P. Codecà; A. Perelli
Codecà, P., and Perelli, A.. "On the Uniform (mod 1) of the Farey Fractions and lP Spaces.." Mathematische Annalen 279.3 (1987/88): 413-422. <http://eudml.org/doc/164331>.
@article{Codecà1987/88,
author = {Codecà, P., Perelli, A.},
keywords = {Farey fractions; spaces; discrepancy},
title = {On the Uniform (mod 1) of the Farey Fractions and lP Spaces.},
AU - Codecà, P.
AU - Perelli, A.
TI - On the Uniform (mod 1) of the Farey Fractions and lP Spaces.
KW - Farey fractions; spaces; discrepancy
S. Kanemitsu, M. Yoshimoto, Farey series and the Riemann hypothesis
Farey fractions,
{\ell }^{p}
spaces, discrepancy
1
1
Articles by P. Codecà
Articles by A. Perelli |
EUDML | Discrete groups with Kazhdan' s property T and factorization property are residually finite. EuDML | Discrete groups with Kazhdan' s property T and factorization property are residually finite.
Discrete groups with Kazhdan' s property T and factorization property are residually finite.
Kirchberg, Eberhard. "Discrete groups with Kazhdan' s property T and factorization property are residually finite.." Mathematische Annalen 299.3 (1994): 551-564. <http://eudml.org/doc/165222>.
@article{Kirchberg1994,
author = {Kirchberg, Eberhard},
keywords = {property (F); discrete group with property of Kazhdan; factorization property; two-sided regular representation; maximal almost periodicity; residual finiteness; nuclear; Lie group; faithful unitary representation; hyperfinite factor},
title = {Discrete groups with Kazhdan' s property T and factorization property are residually finite.},
AU - Kirchberg, Eberhard
TI - Discrete groups with Kazhdan' s property T and factorization property are residually finite.
KW - property (F); discrete group with property of Kazhdan; factorization property; two-sided regular representation; maximal almost periodicity; residual finiteness; nuclear; Lie group; faithful unitary representation; hyperfinite factor
property (F), discrete group with property
T
of Kazhdan, factorization property, two-sided regular representation, maximal almost periodicity, residual finiteness, nuclear, Lie group, faithful unitary representation, hyperfinite
{\text{II}}_{1}
{C}^{*}
{W}^{*}
{C}^{*}
{W}^{*}
{C}^{*}
Articles by Eberhard Kirchberg |
Find an answer to any elementary geometry problems
Elementary geometry questions and answers
Recent questions in Elementary geometry
What's the ratio between the segments
\frac{AF.BG}{FG}
"The sum of the squares of the diagonals is equal to the sum of the squares of the four sides of a parallelogram."
I find this property very useful while solving different problems on Quadrilaterals & Polygon,so I am very inquisitive about a intuitive proof of this property.
Find the area of a triangle with vertices
\left(0,1,1\right),\left(-1,-1,2\right),\left(2,3,1\right)
Given a lattice
\mathrm{\Gamma }\subset \mathbb{C}
, a Theta function
\vartheta :\mathbb{C}\to \mathbb{C}
is a holomorphic function with the following property:
\vartheta \left(z+\gamma \right)={e}^{2i\pi {a}_{\gamma }z+{b}_{\gamma }}\vartheta \left(z\right)
\gamma \in \mathrm{\Gamma }
{a}_{\gamma },{b}_{\gamma }\in \mathbb{C}
Exercise: A Theta function never vanishes iff
\vartheta \left(z\right)={e}^{p\left(z\right)}
p\left(z\right)
a polynomial of degree at most 2.
Hint: The "only if" part is trivial. The hint is: show that
\mathrm{log}\left(\vartheta \left(z\right)\right)=O\left(1+|z{|}^{2}\right)
. I tried to apply log on both sides, or derive one and two times, or everything I could have thought of. I don't get where the square comes from.
I have the following problem: In a triangle ABC, the measure of the angle formed by the external angle bisectors of B and C is equal to twice the measure of the angle A. Find out the value of angle A.
Can anyone suggest how to approach this example?
Suppose I have 4 unit vectors in 3D and I know all the
{}^{4}{C}_{2}=6
angles between them. These angles provide the complete description of this group of vectors. Now, I want to add anther unit vector to the mix. How many additional angles do I need to uniquely identify this new vector?
A convex quadrilateral ABCD is inscribed and circumscribed. If the diagonals AC and BD are perpendicular, show that one of them divides the quadrilateral into two congruent right triangles.
If K is the midpoint of AH,
P\in AB
Q\in AC
K\in PQ
OK\perp PQ
OP=OQ
Given: ABCD is a parallelogram,
\overline{AM},\overline{BN}
angle bisectors,
DM=4\phantom{\rule{thinmathspace}{0ex}}\text{ft.}
MN=3\phantom{\rule{thinmathspace}{0ex}}\text{ft.}
Find: the perimeter of ABCD
5<n\in \mathbb{N}
there is at least two prime numbers different in the opening interval
\left(n,2n\right)
.Use reinforcement of Bertrand's postulate in order to prove that :
10<n\in \mathbb{N}
Exists that has at least two Prime factors in the factorization of
n!
which appear with a power of 1.
n=11
the primes 7,11 are such prime's meet the conditions.
But , for
n=10
it dosen't meet the conditions because only 7 appears with power 1 in the factorization of 10!.
Hint : Consider two cases,
n
even or
n
we need to use the claim in order to Implement the solution
n
n
is even we can rewrite
n=2t
if we use the claim we can get
\left(2t,4t\right)
Case (2): n odd
n
is odd we can rewrite
n=2t+1
\left(2t+1,2\left(2t+1\right)\right)
The internal and external bisectors of angle
A
ABC
BC
(or the extension of
BC
X
{X}^{\prime }
respectively, show that the circles
ABC
AX{X}^{\prime }
intersect orthogonally. I have shown that the centre of the circle
AX{X}^{\prime }
lies on (line)
BC
, but cannot proceed further. Any advice would be appreciated.
Is there an elegant way to prove that the midpoint is
\left(\frac{{x}_{1}+{y}_{1}}{2},\frac{{x}_{2}+{y}_{2}}{2}\right)
using non-right angled similar triangles?
I came up with this question- how would you show 4 not equal to 6 (or m not equal to m+n ( n not 0)), using only Peano's Postulates?
I can see a number of things go wrong- for instance the Principle of Mathematical Induction seems to fail. Also possibly 0 seems to be in the image of the successor function in that case.
A
B
be two points on opposite sides of a line
l
. Then the line segment
AB
l
1. Using only the 4 postulates of Euclid, is there a way to make precise the meaning of ``opposite sides"?
2. Is the intersection guaranteed to exist using only the 4 postulates? If so, why is it true?
I\left(f\right)={\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx
. The midpoint rule (open Newton-Cotes for
n=0
) is I0(f)=(b−a)f(a+b2)
{I}_{0}\left(f\right)=\left(b-a\right)f\left(\frac{a+b}{2}\right)
Show: For
f\in {C}^{1}\left(\left[a,b\right]\right)
|I\left(f\right)-{I}_{0}\left(f\right)|\le \frac{\left(b-a{\right)}^{2}}{4}‖{f}^{\prime }{‖}_{\mathrm{\infty }}
{d}_{1}
\frac{1}{2}\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}
Suppose you have a pair of lines passing through origin,
a{x}^{2}+2hxy+b{y}^{2}=0
, how would you find the equation of pair of angle bisectors for this pair of lines. I can do this for 2 separate lines, but I am not able to figure it out for a pair of lines. Can someone please help.
If you draw a parallelogram its diagonals will form four triangles of which at least the opposite pairs will be congruent, which is to say, each triangle is a reflection of its opposite one, why is this? What property of parallelograms am I missing here? because if I draw a trapezium its diagonals will obviously not form congruent triangles.
Let K be a nonempty closed, convex subset of
{R}^{d}
x\notin K
y\in K
that is closest to x. |
Weak topology - formulasearchengine
In mathematics, weak topology is an alternative term for initial topology. The term is most commonly used for the initial topology of a topological vector space (such as a normed vector space) with respect to its continuous dual. The remainder of this article will deal with this case, which is one of the concepts of functional analysis.
One may call subsets of a topological vector space weakly closed (respectively, weakly compact, etc.) if they are closed (respectively, compact, etc.) with respect to the weak topology. Likewise, functions are sometimes called weakly continuous (respectively, weakly differentiable, weakly analytic, etc.) if they are continuous (respectively, differentiable, analytic, etc.) with respect to the weak topology.
1 The weak and strong topologies
2 The weak-* topology
2.1 Weak-* convergence
4 Operator topologies
The weak and strong topologies
Let K be a topological field, namely a field with a topology such that addition, multiplication, and division are continuous. In most applications K will be either the field of complex numbers or the field of real numbers with the familiar topologies. Let X be a topological vector space over K. Namely, X is a K vector space equipped with a topology so that vector addition and scalar multiplication are continuous.
We may define a possibly different topology on X using the continuous (or topological) dual space X*. The topological dual space consists of all linear functions from X into the base field K that are continuous with respect to the given topology. The weak topology on X is the initial topology with respect to X*. In other words, it is the coarsest topology (the topology with the fewest open sets) such that each element of X* remains a continuous function. In order to distinguish the weak topology from the original topology on X, the original topology is often called the strong topology.
A subbase for the weak topology is the collection of sets of the form φ−1(U) where φ ∈ X* and U is an open subset of the base field K. In other words, a subset of X is open in the weak topology if and only if it can be written as a union of (possibly infinitely many) sets, each of which is an intersection of finitely many sets of the form φ−1(U).
More generally, if F is a subset of the algebraic dual space, then the initial topology of X with respect to F, denoted by σ(X,F), is the weak topology with respect to F . If one takes F to be the whole continuous dual space of X, then the weak topology with respect to F coincides with the weak topology defined above.
If the field K has an absolute value
{\displaystyle |\cdot |}
, then the weak topology σ(X,F) is induced by the family of seminorms,
{\displaystyle \|x\|_{f}{\overset {\text{def}}{=}}|f(x)|}
for all f∈F and x∈X. In particular, weak topologies are locally convex. From this point of view, the weak topology is the coarsest polar topology; see weak topology (polar topology) for details. Specifically, if F is a vector space of linear functionals on X which separates points of X, then the continuous dual of X with respect to the topology σ(X,F) is precisely equal to F Template:Harv.
The weak topology is characterized by the following condition: a net (xλ) in X converges in the weak topology to the element x of X if and only if φ(xλ) converges to φ(x) in R or C for all φ in X* .
In particular, if xn is a sequence in X, then xn converges weakly to x if
{\displaystyle \phi (x_{n})\to \phi (x)}
as n → ∞ for all φ ∈ X*. In this case, it is customary to write
{\displaystyle x_{n}{\overset {\mathrm {w} }{\longrightarrow }}x}
{\displaystyle x_{n}\rightharpoonup x.}
If X is equipped with the weak topology, then addition and scalar multiplication remain continuous operations, and X is a locally convex topological vector space.
If X is a normed space, then the dual space X* is itself a normed vector space by using the norm ǁφǁ = supǁxǁ≤1|φ(x)|. This norm gives rise to a topology, called the strong topology, on X*. This is the topology of uniform convergence. The uniform and strong topologies are generally different for other spaces of linear maps; see below.
The weak-* topology
A space X can be embedded into the double dual X** by
{\displaystyle x\mapsto T_{x}}
{\displaystyle T_{x}(\phi )=\phi (x).\ }
Thus T : X → X** is an injective linear mapping, though not necessarily surjective (spaces for which this canonical embedding is surjective are called reflexive). The weak-* topology on X* is the weak topology induced by the image of T: T(X) ⊂ X**. In other words, it is the coarsest topology such that the maps Tx, defined by Tx(φ) = φ(x) from X* to the base field R or C remain continuous.
Weak-* convergence
A net φλ in X* is convergent to φ in the weak-* topology if it converges pointwise:
{\displaystyle \phi _{\lambda }(x)\rightarrow \phi (x)}
for all x in X. In particular, a sequence of φn ∈ X* converges to φ provided that
{\displaystyle \phi _{n}(x)\to \phi (x)}
for all x in X. In this case, one writes
{\displaystyle \phi _{n}{\overset {w^{*}}{\rightarrow }}\phi }
Weak-* convergence is sometimes called the topology of simple convergence or the topology of pointwise convergence. Indeed, it coincides with the topology of pointwise convergence of linear functionals.
By definition, the weak* topology is weaker than the weak topology on X*. An important fact about the weak* topology is the Banach–Alaoglu theorem: if X is normed, then the closed unit ball in X* is weak*-compact (more generally, the polar in X* of a neighborhood of 0 in X is weak*-compact). Moreover, the closed unit ball in a normed space X is compact in the weak topology if and only if X is reflexive.
In more generality, let F be locally compact valued field (e.g., the reals, the complex numbers, or any of the p-adic number systems). Let X be a normed topological vector space over F, compatible with the absolute value in F. Then in X*, the topological dual space X of continuous F-valued linear functionals on X, all norm-closed balls are compact in the weak-* topology.
If a normed space X is separable, then the weak-* topology is metrizable on the norm-bounded subsets of X* when the normed space X is separable. If X is a Banach space, the weak-* topology is not metrizable on all of X* unless X is finite-dimensional.[1]
Consider, for example, the difference between strong and weak convergence of functions in the Hilbert space L2(Rn). Strong convergence of a sequence ψk∈L2(Rn) to an element ψ means that
{\displaystyle \int _{\mathbf {R} ^{n}}|\psi _{k}-\psi |^{2}\,{\rm {d}}\mu \,\to 0\,}
as k→∞. Here the notion of convergence corresponds to the norm on L2.
In contrast weak convergence only demands that
{\displaystyle \int _{\mathbf {R} ^{n}}{\bar {\psi }}_{k}f\,\mathrm {d} \mu \to \int _{\mathbf {R} ^{n}}{\bar {\psi }}f\,\mathrm {d} \mu }
for all functions f∈L2 (or, more typically, all f in a dense subset of L2 such as a space of test functions, if the sequence {ψk} is bounded). For given test functions, the relevant notion of convergence only corresponds to the topology used in C.
For example, in the Hilbert space L2(0,π), the sequence of functions
{\displaystyle \psi _{k}(x)={\sqrt {2/\pi }}\sin(kx)}
form an orthonormal basis. In particular, the (strong) limit of ψk as k→∞ does not exist. On the other hand, by the Riemann–Lebesgue lemma, the weak limit exists and is zero.
One normally obtains spaces of distributions by forming the strong dual of a space of test functions (such as the compactly supported smooth functions on Rn). In an alternative construction of such spaces, one can take the weak dual of a space of test functions inside a Hilbert space such as L2. Thus one is led to consider the idea of a rigged Hilbert space.
If X and Y are topological vector spaces, the space L(X,Y) of continuous linear operators f:X → Y may carry a variety of different possible topologies. The naming of such topologies depends on the kind of topology one is using on the target space Y to define operator convergence Template:Harv. There are, in general, a vast array of possible operator topologies on L(X,Y), whose naming is not entirely intuitive.
For example, the strong operator topology on L(X,Y) is the topology of pointwise convergence. For instance, if Y is a normed space, then this topology is defined by the seminorms indexed by x∈X:
{\displaystyle f\mapsto \|f(x)\|_{Y}.}
More generally, if a family of seminorms Q defines the topology on Y, then the seminorms pq,x on L(X,Y) defining the strong topology are given by
{\displaystyle p_{q,x}:f\mapsto q(f(x)),}
indexed by q∈Q and x∈X.
In particular, see the weak operator topology and weak* operator topology.
Eberlein compactum, a compact set in the weak topology
Weak convergence of measures
Vague topology
↑ Proposition 2.6.12, p. 226 in {{#invoke:citation/CS1|citation |CitationClass=citation }}.
Retrieved from "https://en.formulasearchengine.com/index.php?title=Weak_topology&oldid=221378" |
A bin contains 71 light bulbs, of which 9 are
assumingsun 2022-02-11 Answered
There are 71 light bulbs in the bin of which 9 are defective. So 62 of them are good.6 bulbs are selected and we want all of them to be good.
So, the good bulbs must come from the 62 bulbs and this is possible in (62c6) ways. The total possibilities are (71c6) since we select 6 bulbs from the 71 bulbs available for selection. Finally, the probability is obtained as the ratio mentioned.
This Problem is to be solved using Binomial Distribution.
If a ball is taken randomly from the bin the ball being defective is
q=\frac{9}{71}
The ball not being defective is
p=1-\frac{9}{91}=\frac{62}{71}
The desired event is not being defective.
Let x be the number of desired event.
We expect
x=6
{P}_{x=6}=6{C}_{6}\left({p}^{x}\right)\left({q}^{n-x}\right)
{P}_{x=6}=6{C}_{6}{\left(\frac{62}{71}\right)}^{6}{\left(\frac{9}{71}\right)}^{6-6}
{P}_{x=6}=6{C}_{6}{\left(\frac{62}{71}\right)}^{6}{\left(\frac{9}{71}\right)}^{0}
{P}_{x=6}=\left(1\right){\left(\frac{62}{71}\right)}^{6}\left(1\right)
{P}_{x=6}=\left(1\right){\left(\frac{62}{71}\right)}^{6}\left(1\right)
{P}_{x=6}=\left(1\right)\left(0.44\right)\left(1\right)
{P}_{x=6}=0.44
P\left(A\right)=0.7,P\left(B\right)=0.9\text{ }\text{and}\text{ }P\left(A\cap B\right)=0.4
P\left(B\mid A\right)?
P\left(A\mid B\right)
A fair quarter is flipped three times. For each of the following probabilities, use the formula for the binomial distribution and a calculator to compute the requested probability. Next, look up the probability in the binomial probability distribution table. (Enter your answers to three decimal places.) Find the probability of getting exactly three tails. (d) Find the probability of getting exactly three tails. (d) Find the probability of getting exactly three tails.
John charges his cell phone only when it is fully drained. The length of time, in hours, between charges of his cell phone is normally distributed with mean 31 and variance 64.
Suppose that the cell phone lasted 48.84 hours since the most recent charge. z-score for this value
If X is an exponential random variable parameter
\lambda =1
, compute the probability density function of the random variable Y defined by
Y=\mathrm{log}X |
Home : Support : Online Help : Mathematics : Numerical Computations : Matlab : transpose
compute the transpose of a MapleMatrix or MatlabMatrix using MATLAB(R)
The transpose command computes the transpose of a matrix using MATLAB®.
Executing the transpose command returns a Matrix, Vector or constant.
\mathrm{with}\left(\mathrm{Matlab}\right):
\mathrm{maplematrix_a}≔\mathrm{Matrix}\left([[1,3,5],[6,4,2],[7,8,1],[3,7,3]]\right)
\textcolor[rgb]{0,0,1}{\mathrm{maplematrix_a}}\textcolor[rgb]{0,0,1}{≔}[\begin{array}{ccc}\textcolor[rgb]{0,0,1}{1}& \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{5}\\ \textcolor[rgb]{0,0,1}{6}& \textcolor[rgb]{0,0,1}{4}& \textcolor[rgb]{0,0,1}{2}\\ \textcolor[rgb]{0,0,1}{7}& \textcolor[rgb]{0,0,1}{8}& \textcolor[rgb]{0,0,1}{1}\\ \textcolor[rgb]{0,0,1}{3}& \textcolor[rgb]{0,0,1}{7}& \textcolor[rgb]{0,0,1}{3}\end{array}]
The transpose of this MapleMatrix is computed as
\mathrm{Matlab}[\mathrm{transpose}]\left(\mathrm{maplematrix_a}\right)
The matrix is set in MATLAB® and transposed as a MatlabMatrix.
\mathrm{Matlab}[\mathrm{setvar}]\left("matlabmatrix_a",\mathrm{maplematrix_a}\right)
\mathrm{Matlab}[\mathrm{transpose}]\left("matlabmatrix_a"\right) |
\large \frac ab \ , \ ab \ , \ a -b \ , \ a+b
Above shows real numbers that belong to an arithmetic progression in order. Find the next term of this sequence.
Find the number of
6
-term strictly increasing geometric progressions, such that all terms are positive integers less than
1000.
{2^x + 2^{x+1} + 2^{x+2} + \ldots + 2^{x+2015} = 4^x + 4^{x+1} + 4^{x+2} + \ldots + 4^{x+2015}}
x
satisfies the equation above and it can be represented as
\log_D \left(\dfrac{A}{1+B^C} \right)
A
B
C
D
B
is prime, determine the smallest value of
A + B + C +D
The image above shows a broken line (a series of connected line segments) starting at the origin, O. The nth segment in the broken line has length
\frac{1}{n}
, and at the end of each segment, the broken line turns
60^{\circ}
counter-clockwise.
As the number of segments in the broken line approaches infinity, the final endpoint of the broken line approaches a point P. The distance OP can be written as
\frac{a}{b}\pi
, where a and b are positive coprime integers. Find
a+b
\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \cot ^ {-1} 31 + \ldots
Evaluate the sum above in degrees. |
EUDML | Gaussian quadrature rules and -stability of Galerkin schemes for ODE. EuDML | Gaussian quadrature rules and -stability of Galerkin schemes for ODE.
Gaussian quadrature rules and
A
-stability of Galerkin schemes for ODE.
Bensebah, Ali; Dubeau, François; Gélinas, Jacques
Bensebah, Ali, Dubeau, François, and Gélinas, Jacques. "Gaussian quadrature rules and -stability of Galerkin schemes for ODE.." International Journal of Mathematics and Mathematical Sciences 2003.31 (2003): 1947-1959. <http://eudml.org/doc/50405>.
@article{Bensebah2003,
author = {Bensebah, Ali, Dubeau, François, Gélinas, Jacques},
keywords = {initial value problems; Gaussian quadrature rules; Galerkin schemes; A-stability; collocation methods},
title = {Gaussian quadrature rules and -stability of Galerkin schemes for ODE.},
AU - Bensebah, Ali
AU - Gélinas, Jacques
TI - Gaussian quadrature rules and -stability of Galerkin schemes for ODE.
KW - initial value problems; Gaussian quadrature rules; Galerkin schemes; A-stability; collocation methods
initial value problems, Gaussian quadrature rules, Galerkin schemes, A-stability, collocation methods
Finite elements, Rayleigh-Ritz, Galerkin and collocation methods
Approximate quadratures
Articles by Bensebah
Articles by Dubeau
Articles by Gélinas |
Free vibration of circular annular plate with different boundary conditions | JVE Journals
Yash Jaiman1 , Baij Singh2
1Bennett University, Greater Noida, Uttar Pradesh, India
2Indian Institute of Technology (ISM), Dhanbad, India
Copyright © 2019 Yash Jaiman, 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.
This paper deals with the numerical simulation of free vibration analysis of a thin circular annular plate for various boundary conditions at the outer edge and inner edge. Classical plate theory is used to derive the governing differential equation for the transverse deflection of the thin isotropic plate. The finite element method is used to evaluate the first six natural frequencies and mode shapes of the thin uniform circular annular plate with radius ratios
\left({r}_{1}/{r}_{2}\right)
for different boundary conditions. These natural frequencies results are compared with those available in the literature. The results are verified with classical plate theory with our Abaqus results and checked with the previous research literature on the topic.
Keywords: circular annular plate, free vibration, numerical simulation.
Plates are widely used as a structural element and have vast practical applications in many engineering fields such as aerospace, mechanical, civil, nuclear, electronic, automotive, marine and heavy machinery, etc. Various researchers have analyzed the free vibration behavior of circular annular plates of different shapes, sizes, thickness for different boundary conditions. Leissa [1] used the Ritz method to estimate the natural frequencies of the isotropic plate for different boundary conditions. Kim and Dickinson [2] used the Rayleigh-Ritz approximation method for free vibration of a thin plate to extract natural frequencies. Rajalingham et al. [3] used a Rayleigh-Ritz method to analyze the plate characteristics parameter as shape functions and continued his work to formulate a variational reduction expression to analyze frequencies and mode shapes. Liew et al. [4] used the polynomials-Ritz method for the vibration of circular plates by using three-dimensional elasticity solutions. Zhou et al. [5] used the Chebyshev-Ritz method for three-dimensional vibration and mode shapes of the circular plate. Lim et al. [6] used the state-space method to analyze transverse vibration and mode of a thick circular plate. Zhou et al. [7] used the Hamiltonian principle to solve governing equations for free vibration analysis by using the variational principle of mixed energy method. Kumar et al. [8] use a dynamic stiffness method to extract the natural frequency and mode shapes of a thin plate. Piyush et al. [9] used the Rayleigh-Ritz method to compute the natural frequencies of the thin plate.
2. Basic formulation
Consider a homogeneous, isotropic circular annular plate in cylindrical coordinates
\left(r,\theta ,z\right)
with uniform thickness
h
as shown in Fig. 1.
Classical plate theory is used to derive the governing differential equation for transverse vibration in the polar coordinate system is defined as:
D{\nabla }^{4}W\left(r,\theta ,t\right)+\rho hw\left(r,\theta \right)=0, {\nabla }^{4}={\nabla }^{2}{\nabla }^{2},
where Laplacian operator:
{\nabla }^{2}=\frac{{\partial }^{2}}{{\partial r}^{2}}+\frac{1}{r}\frac{\partial }{\partial r}+\frac{1}{{r}^{2}}\frac{{\partial }^{2}}{{\partial \theta }^{2}}
\rho
is the mass density,
D=E{h}^{3}/\left[12\left(1-{\nu }^{2}\right)\right]
is the flexural rigidity and
\nu
is Poisson’s ratio.
Transverse deflection of natural vibrations for thin circular plate is assumed to be:
W\left(r,\theta ,t\right)=w\left(r,\theta \right){e}^{i\omega t},
\omega
is the natural frequency and
w\left(r,\theta \right)
is natural mode.
Substituting Eqs. (2) in (1), we get:
{\nabla }^{4}W-{\gamma }^{4}W=0, \text{or} {\left(\nabla }^{2}+{\gamma }^{2}\right)\left({\nabla }^{2}+{\gamma }^{2}\right)W=0,
{\gamma }^{4}=\frac{\rho h{\omega }^{2}}{D}.
The general finite element equation for the transverse deflection of thin plate is given by:
\left[M\right]\left\{\stackrel{¨}{q}\right\}+\left[K\right]\left\{q\right\}=0,
\left[M\right]
\left[K\right]
is stiffness matrix,
\left\{\stackrel{¨}{q}\right\}
is the nodal acceleration vector, and
\left\{q\right\}
is nodal displacement vector.
The non-dimension natural frequency parameter
\varpi
are calculated as:
\varpi =2\pi \omega \left[\sqrt{\rho h/D}\right]{r}_{1}^{2},
\omega
is the natural frequency in Hz.
Fig. 1. Schematic diagram and coordinate system of annular circular plate
In this section, the first six non-dimensional natural frequencies and mode shapes of the circular annular plate are estimated by using the finite element method. Here, we calculated different eigenvalues for different boundary conditions with different radii ratio
\left({r}_{1}/{r}_{2}\right)
. Different combinations of boundary conditions are applied to compute the natural frequencies and mode shapes of the circular annular plate. 2820 elements and 5922 nodes are used to estimate the natural frequencies and mode shape function of thin circular annular plates after convergence study. The present natural frequencies results are compared with those available in the literature.
Tables 1-4 shows that the first six non-dimensional natural frequencies values for the circular annular plate. These present results are nearly the same as Leissa [1] and Zhou [8] under different boundary conditions.
Fig. 2. Natural modes of a clamped annular plate with a free inner boundary,
{r}_{1}/{r}_{2}
Table 5-7 presents the effect of the radii ratio on the non-dimensional frequency parameter of thin plates. It is observed from these tables that as the radii ratio increases non-dimensional frequency parameter increases.
Table 1. Comparison of the non-dimensional natural frequency parameter with Leissa [1] and Zhou [8] for clamped outer and free inner boundary (
\nu =1/3
{r}_{1}/{r}_{2} =\text{0.4}
Leissa [1]
Zhou [8]
Table 2. Comparison of the non-dimensional natural frequency parameter with Leissa [1] and Zhou [8] for free outer and clamped inner boundary (
\nu =1/3
{r}_{1}/{r}_{2} =0.4
Table 3. Comparison of the non-dimensional natural frequency parameter with Leissa [1] and Zhou [8] for free outer and free inner boundary (
\nu =1/3
{r}_{1}/{r}_{2} =0.4
Table 4. Comparison of the non-dimensional natural frequency parameter with Leissa [1] and Zhou [8] for clamped outer and clamped inner boundary (
\nu =1/3
{r}_{1}/{r}_{2} =0.4
Table 5. Non-dimensional frequency parameter
\omega {r}_{2}^{2}\sqrt{\rho h/D }
for the annular circular plate with clamped outer and free inner edge (
\nu =1/3
n
{r}_{1}/{r}_{2}
\omega {r}_{2}^{2}\sqrt{\rho h/D }
for the annular circular plate with free outer and clamped inner edge (
\nu =1/3
n
{r}_{1}/{r}_{2}
\omega {r}_{2}^{2}\sqrt{\rho h/D }
for the annular circular plate with clamped outer and clamped inner edge (
\nu =1/3
n
{r}_{1}/{r}_{2}
In this paper, numerical analysis for free vibration analysis of a thin annular solid plate is carried out using the finite element method for different boundary conditions at the inner and outer radius. It is found that those natural frequency results are quite close to those reported in previous works of literature. The novelty of this paper is the effect of the radii ratio on natural frequency is discussed and found that with increasing radii ratio, natural frequency increases and another novelty is by using shell element modeling as per Abaqus convention the dimensionless frequency parameter as found in the literature are completely validated.
Leissa A. W. Vibration of Plates. Office of Technology Utilization, Washington, 1969. [Search CrossRef]
Kim C. S., Dickinson S. M. On the free, transverse vibration of annular and circular, thin, sectorial plates subject to certain complicating effects. Journal of Sound and Vibration, Vol. 134, Issue 3, 1989, p. 407-421. [Publisher]
Rajalingham C., Bhat R. B., Xistris G. D. Vibration of rectangular plates using plate characteristic functions as shape functions in the Rayleigh-Ritz method. Journal of Sound and Vibration, Vol. 193, Issue 2, 1996, p. 497-509. [Publisher]
Liew K. M., Yang B. Three-dimensional elasticity solutions for free vibrations of circular plates: a polynomials-Ritz analysis. Computer Methods in Applied Mechanics and Engineering, Vol. 175, Issues 1-2, 1999, p. 189-201. [Publisher]
Zhou D., Au F. T. K., Cheung Y. K., Lo S. H. Three-dimensional vibration analysis of circular and annular plates via the Chebyshev-Ritz method. International Journal of Solids and Structures, Vol. 40, Issue 12, 2003, p. 3089-3105. [Publisher]
Lim C. W., Li Z. R., Xiang Y., Wei G. W., Wang C. M. On the missing modes when using the exact frequency relationship between Kirchhoff and Mindlin plates. Advances in Vibration Engineering, Vol. 4, Issue 3, 2005, p. 221-248. [Search CrossRef]
Zhou Z. H., Wong K. W., Xu X. S., Leung A. Y. T. Natural vibration of circular and annular thin plates by Hamiltonian approach. Journal of Sound and Vibration, Vol. 330, Issue 5, 2011, p. 1005-1017. [Publisher]
Kumar S., Vinayak Ranjan, Jana P. Free vibration analysis of thin functionally graded rectangular plates using the dynamic stiffness method. Composite Structures, Vol. 197, 2018, p. 39-53. [Publisher]
Pratap Singh P., Azam M. S., Vinayak Ranjan Vibration analysis of a thin functionally graded plate having an out of plane material inhomogeneity resting on Winkler-Pasternak foundation under different combinations of boundary conditions. Proceedings of the Institution of Mechanical Engineers, Part C: Journal of Mechanical Engineering Science, 2018. [Search CrossRef]
Study of modes and deflection for ring shaped plate with varying thickness
Manu Gupta, Ankit Kumar, Ajendra Kumar, Robin, Anuj Kumar
Hilal Koç, Ömer Ekim Genel, Mertol Tüfekci, Ekrem Tüfekci |
Pushdown Automata | Brilliant Math & Science Wiki
Pushdown automata are nondeterministic finite state machines augmented with additional memory in the form of a stack, which is why the term “pushdown” is used, as elements are pushed down onto the stack. Pushdown automata are computational models—theoretical computer-like machines—that can do more than a finite state machine, but less than a Turing machine.
Pushdown automata accept context-free languages, which include the set of regular languages. The language that describes strings that have matching parentheses is a context-free language. Say that a programmer has written some code, and in order for the code to be valid, any parentheses must be matched. One way to do this would be to feed the code (as strings) into a pushdown automaton programmed with transition functions that implement the context-free grammar for the language of balanced parentheses. If the code is valid and all parentheses are matched, the pushdown automata will "accept" the code. If there are unbalanced parentheses, the pushdown automaton will be able to return to the programmer that the code is not valid. This is one of the more theoretical ideas behind computer parsers and compilers.
Pushdown automata can be useful when thinking about parser design and any area where context-free grammars are used, such as in computer language design. Since pushdown automata are equal in power to context-free languages, there are two ways of proving that a language is context-free: provide the context-free grammar or provide a pushdown automaton for the language.
\delta
represents transition functions (the program of the pushdown automaton),
A
is the stack symbol,
a
is the tape symbol, and
p
represents the state[1].
Translating between Context-free Grammars and Pushdown Automata
A stack can be thought of as a stack of plates, one stacked on top of the other, and plates can be taken off of the top of the stack. To get to the bottom of the stack of plates, all others must be removed first. Stacks are a last-in-first-out, or LIFO, data structure. In pushdown automata, state transitions include adding a symbol to the string generated, as in FSMs, but state transitions can also include instructions about pushing and popping elements to and from the stack.
One can walk through the pushdown automata diagram to see what kinds of strings can be produced by the transition functions describing the language the pushdown automata generate, or you can feed it an input string and verify that there exists a set of transitions that end in an accepting state that creates the input string.
At each transition, a pushdown automaton can push a symbol to the stack, pop a symbol from the stack, do both, or do no operations to the stack. This transition symbol is
\epsilon
\epsilon
also represents the empty string and can be used as a symbol. If the instructions say that
\epsilon
is the symbol read, this means that the stack/input is empty. If the instructions say to replace the symbol on top of the stack with an
\epsilon,
this means to delete the symbol on top of the stack (this is popping).
The pushdown automaton starts with an empty stack and accepts if it ends in an accepting state at the end. The contents of the stack at the end do not matter unless the problem specifies that the stack must be empty at the end. If no transition from the current state can be made, reject. For example, if the transition from state A to state B requires popping an
x
from the stack, if there is no
x
on the top of the stack to pop, reject.
Pushdown automata can be modeled as a state machine diagram with added instructions about the stack. Where in the finite state machine the arrows between states were labeled with a symbol that represented the input symbol from the string, a pushdown automaton includes information in the form of input symbol followed by the symbol that is currently at the top of the stack, followed by the symbol to replace the top of the stack with. These instructions are sometimes separated by commas, slashes, or arrows:
The exception to the "replace with this symbol" command is during the first step after we write the $ symbol: we do not overwrite (i.e. pop/delete) the $ symbol. We need to keep this so that as we reach the end of the string, we know when we've reached the bottom of our stack. Instead of overwriting this symbol, simply place the next stack symbol on top of the $.
For this example, assume that
s_5
s_6
are the accepting states. This pushdown automaton only shows instructions for the stack; usually, the pushdown automata diagrams will also contain information about which symbols are needed to move from one state to another, but let's use this example to get a feel for how the stack works. Assume the stack starts off empty, with the symbol
\$
, which indicates the bottom of the stack: so the stack is initially set to
[\$]
What does the stack look like after following these transitions:
s_1
s_2
s_3
The push down automaton pushes
and then pushes
b,
so the stack at this point is
[\$, a, b]
Starting with the empty stack, what does the stack look like after the transitions
s_1
s_2
s_3
s_3
s_4
s_4
a,
pushes
b,
b,
b,
and pops
b,
so the stack looks like
[\$,a]
$ $00 $0000 $000011
Given the following pushdown automata and input string 00001111, what does the PDA's stack look like once it has gotten through and read the second 1 of the string (i.e. the substring 000011 has been read)?
What's the point of a stack?
A stack allows pushdown automata a limited amount of memory. A pushdown automaton can read from, push (add) to, or pop (remove) the stack. The transition functions that describe the pushdown automaton (usually represented by labels on the arrows between the state circles) tell the automaton what to do.
Pushdown automata accept context-free languages. This means that a context-free language can be represented by a pushdown automaton or a context-free grammar.
For example, the language containing all strings of 0's followed by an equal number of 1's is a context-free language, and it was proved on the regular languages page that this language is not a regular language, so it is possible to represent this language using a pushdown automaton.
Here is a push down automaton that accepts strings in the language
L = \{0,1 \big| 0^n1^n \text{ for n } \geq 0\}
Note: in the transition from A to B, do not overwrite the $ symbol with an empty string (i.e. don't remove the $), just write the new symbol on top of that.
A pushdown automaton is formally defined as a 7-tuple:
(Q, \Sigma, \Gamma, \delta, q_0, Z, F)
Q
is the finite set of states.
\Sigma
is the finite alphabet of the input alphabet.
\Gamma
is the finite stack alphabet.
\delta
is the set of transition functions.
q_0
is the starting state. This is a member of the set of states,
Q.
Z
is the initial stack symbol which is a member of
\Gamma.
F
is the set of accepting states which is a subset of
Q.
The transition functions in
\delta
can be represented as another tuple:
(p, a, A, q, \alpha).
p
represents the current state
(
which is a member of
Q).
a
(
\Sigma \cup \epsilon)
on which the transition is triggered.
A
is the top stack symbol
(
\Gamma)
which may read
a
, change the state to
q
, pop
A
replacing it by pushing
\alpha
q
(
Q)
that the stack symbol might cause us to switch to.
\alpha
is the symbol
(
\Gamma)
the stack symbol might cause us to push to the top of the stack.
Can you come up with a diagram and formal description of a pushdown automaton that recognizes strings containing only parentheses and accepts on strings that have matched parentheses? The context-free grammar for this language is here.
\Sigma = \{(,)\}
\Gamma = \{\$,X\}\ \ \
X
could be any symbol you want
Q = \{A, B, C, D\}
F = \{D\}
q_0 = A
Z = \$
\delta = \big\{(A, \epsilon, \epsilon, A, \$), (A(,\$,B,X), (B, (, X, B, X), (B, ), X, C, \epsilon), (C, ), X, C, \epsilon), (C, \epsilon, \$, D, \epsilon)\big\}
We just showed that the problem above can be represented by either a context-free grammar or a pushdown automaton. Here is how to convert between the two:
Context-free grammar to pushdown automaton: [3]
Each derivation or sequence of production rules that results in a given string is made up of intermediate strings (which are made at each step of the derivation).
The pushdown automata's non-determinism helps it to guess the sequence of steps in the derivation that will result in the desired string. So at each step in the derivation, one of the production rules for a given variable is selected non-deterministically and substituted in for the variable.
The pushdown automaton begins by pushing a symbol onto the stack and then goes through the series of intermediate strings until it arrives at a string that contains only the terminal symbols (this will happen if the string is actually in the grammar, otherwise it will reject).
Push the start symbol, $, to the stack.
Then the following steps are repeated until the automaton finishes:
If there is a variable
X
on top of the stack, non-deterministically pick one of the production rules for
X
X
with the string on the right-hand side of the production rule.
If there is a terminal variable
a
on the input, read the next symbol from the input and compare it to
a
. If they are the same, repeat, and if they are not, reject on this branch of the non-determinism.
If it is the end of the input and the top of the stack has the start symbol, $, then accept.
\text{}
Pushdown automaton to context-free grammar:[3]
The goal is to make a pushdown automaton that will accept all of the input strings that the context-free grammar accepts.
For every pair of states,
p
q
, there will be a variable
A_{pq}
from the grammar. This variable will generate all of the strings that can take the pushdown automaton from state
p
q
The pushdown automaton needs to have these three features:
A single accepting state
The stack needs to be empty before it accepts.
Each transition must only push or pop—it can't do both or neither. To do this, to mimic a transition that does neither, we would need a transition that pushes and then another that pops, resulting in a net action of nothing done. Each of these transitions must go to a new state, so this pushdown automaton may have more states than it might need if we were converting from a context-free grammar to a pushdown automaton.
The first move on a string
x
must be a push since the stack starts off empty, and the last move must be a pop since the stack must be empty at the end.
There are two cases for
x:
The symbol that is popped at the end is the symbol that was pushed at the beginning. If this is the case, then we know that the stack was only ever empty at the beginning and the end. We make the production rule
A_{pq} \rightarrow aA_{rs}b,
a
is an input symbol,
b
is the symbol read at the last step,
r
is the state after
p
s
is the state preceding
q
The symbol that is popped at the end is not the symbol that was pushed at the beginning. If this is the case, then we know that the stack must have been empty sometime between the beginning and end. We make the production rule
A_{pq} \rightarrow A_{rs}A_{rq},
a
r
is the state where the stack becomes empty, and
s
q
Since pushdown automata recognize context-free languages, and context-free languages include regular languages, pushdown automata accept regular languages. A pushdown automaton is just a finite state machine augmented with a stack. Another way to think about this is that a finite state machine is a pushdown automaton that ignores its stack.
If the pushdown automata had access to two stacks, making it a queue automaton, it would create a machine equal in power to a Turing machine.
, J. Pushdown-overview. Retrieved June 18, 2016, from https://en.wikipedia.org/wiki/File:Pushdown-overview.svg
Abdulla, P., Atig, M., & Stenman, J. PDA Example. Retrieved June 18, 2016, from https://www.semanticscholar.org/paper/Adding-Time-to-Pushdown-Automata-Abdulla-Atig/1c95cfb83c281a5d443e23fc38e378aed053efba
Sipser, M. (1997). Introduction to the Theory of Computation (pp. 107). PWS Publishing Company.
Cite as: Pushdown Automata. Brilliant.org. Retrieved from https://brilliant.org/wiki/pushdown-automata/ |
You are playing a game with your friend. You win
You are playing a game with your friend. You win with 55% probability,
arneguet9k 2021-11-19 Answered
You are playing a game with your friend. You win with 55% probability, your friend 30%. There is no draw. You decide to play a series of games. What is your probability of winning the series if the player with
a) 5 wins?
b) 6 wins?
Want to know more about Probability?
Sarythe
Here let the event of winning me is
{E}_{1}
Let the event of winning my friend is
{E}_{2}
The probability of winning mine
P\left({E}_{1}\right)=\frac{55}{100}=0.55
The probability of winning of my friend
P\left({E}_{2}\right)=\frac{30}{100}=0.3
a) the probability of winning of mine if my friend win 5 times
P\left({E}_{1}\right)-P\left({E}_{2}\right)×5=0.55-0.3×5=-0.95
b) the probability of winning mine if my friends win 6 times
P\left({E}_{1}\right)-P\left({E}_{2}\right)×6=0.55-0.3×6=-1.25
Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P{X>c}=.10.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed n times.
What are the possible values of X?
Based on the Normal model N(100, 16) describing IQ scores, what percent of peoples
What are the mean, median, and mode of the following numbers?: 1, 4, 5, 6, 10, 25?
At a restaurant, each dinner special consists of an appetizer, a salad, and an entrée.
The choices of appetizer, salad, and entrée are shown below. Use the choices to answer parts (a) through (d).
\begin{array}{|ccc|}\hline Appetizer& Salad& Entree\\ Meatballs& House& Ravioli\\ Antipasto& Greens& Fettuccine\\ Soup& Greek\\ \hline\end{array}
a) Determine the number of points in the sample space.
The number of points in the sample space is...?
b) If a customer randomly selects one of the dinner specials, determine the probability the customer selects meatballs for the appetizer.
The probability is...?
c) Find the probability that the customer selects the greens Salad and fettuccine.
d) Find the probability that the customer selects an entrée other than ravioli.
Consider two products, A and B, both of which have the same purchase probability.
Product A has a population size of 1000 and a 50 percent awareness probability. Product B has a population size of 10,000 and a 5 percent awareness probability. Which product do you expect to sell more?
If A and B are two independent random variables, is
P\left(A\mid B\right)<P\left(A\right) |
EUDML | Control functions and total boundedness in the space . EuDML | Control functions and total boundedness in the space .
Control functions and total boundedness in the space
{L}_{0}
Caponetti, Diana; Lewicki, Grzegorz; Trombetta, Giulio
Caponetti, Diana, Lewicki, Grzegorz, and Trombetta, Giulio. "Control functions and total boundedness in the space .." Novi Sad Journal of Mathematics 32.2 (2002): 109-123. <http://eudml.org/doc/123257>.
@article{Caponetti2002,
author = {Caponetti, Diana, Lewicki, Grzegorz, Trombetta, Giulio},
keywords = {function space; Hausdorff measure of noncompactness; equimeasurability; control functions},
title = {Control functions and total boundedness in the space .},
AU - Lewicki, Grzegorz
AU - Trombetta, Giulio
TI - Control functions and total boundedness in the space .
KW - function space; Hausdorff measure of noncompactness; equimeasurability; control functions
function space, Hausdorff measure of noncompactness, equimeasurability, control functions
A
Articles by Caponetti
Articles by Lewicki
Articles by Trombetta |
EUDML | Linear systems with multiple base points in . EuDML | Linear systems with multiple base points in .
Linear systems with multiple base points in
{ℙ}^{2}
Harbourne, Brian; Roé, Joaquim
Harbourne, Brian, and Roé, Joaquim. "Linear systems with multiple base points in .." Advances in Geometry 4.1 (2004): 41-59. <http://eudml.org/doc/125534>.
@article{Harbourne2004,
author = {Harbourne, Brian, Roé, Joaquim},
title = {Linear systems with multiple base points in .},
AU - Harbourne, Brian
TI - Linear systems with multiple base points in .
Articles by Harbourne
Articles by Roé |
Board Paper Solutions for ICSE Class 12-science CHEMISTRY Board Paper 2020 MeritNation Set 1
Question 1 is of 20 marks having four sub parts, all of which are compulsory.
Question numbers 2 to 8 carry 2 marks each, with two questions having internal choice.
Question numbers 9 to 15 carry 3 marks each, with two questions having an internal choice.
Question numbers 16 to 18 carry 5 marks each, with an internal choice.
Balanced equations must be given wherever possible and diagrams where they are helpful.
When solving numerical problems, all essential working must be shown.
In working out problems, use the following data:
Gas constant R = 1.987 cal deg–1 mol–1 = 8.314 JK–1 mol–1 = 0.0821 dm3 atm K–1mol–1
1 l atm = 1 dm3 atm = 101.3 J. 1 Faraday = 96500 coulombs.
Avogadro’s number = 6.023 ×1023.
(a) Fill in the blanks by choosing the appropriate word/words from those given in the brackets:
(iodoform, volume, mass, haloform, gram equivalent, chloroform, carbylamine, sp3d2, high, coke, d2sp3, low, gram mole, carbon monoxide)
(i) Equivalent conductivity is the conducting power of all the ions furnished by one ___________ of an electrolyte present in a definite ________ of the solution.
(ii) Bleaching powder, on treatment with ethanol or acetone gives _______. This is an example of _________ reaction.
(iii) Outer orbital complexes involve ________ hybridization and are _______ spin complexes.
(iv) Zinc oxide is reduced by _________ at 1673K to form zinc and __________.
(b) Select the correct alternative from the choices given:
(i) The packing efficiency of simple cubic structure, body centered cubic structure and face centered cubic structure respectively is:
(1) 52·4%, 74%, 68%
(2) 74%, 68%, 52·4%
(ii) When acetone is treated with Grignard’s reagent, followed by hydrolysis, the product formed is:
(1) Secondary alcohol
(2) Tertiary alcohol
(3) Primary alcohol
(4) Aldehyde
(iii) Which of the following electrolytes is least effective in causing flocculation of positively charged ferric hydroxide sol?
(1) K3[Fe(CN)6]
(2) K2CrO4
(4) KBr
(iv) On heating an aliphatic primary amine with chloroform and alcoholic potassium hydroxide, the organic compound formed is an:
(1) Alkyl isocyanide
(2) Alkanol
(3) Alkanal
(4) Alkyl cyanide
(c) Match the following:
(i) Silicon and phosphorous (a) Acetaldehyde
(ii) Iodoform test (b) Xenon hexafluoride
(iii) Arrhenius equation (c) n-type of semiconductors
(iv) Distorted octahedral structure (d) Frequency factor
(d) Answer the following questions:
(i) What is the common name of the polymer obtained by the polymerization of caprolactam? Is it addition polymer or condensation polymer?
(ii) Why Zn2+ ions are colourless while Ni2+ ions are green and Cu2+ ions are blue in colour?
(iii) The molar conductivity of NaCl, CH3COONa and HCl at infinite dilution is 126·45, 91·0 and 426·16 ohm–1 cm2 mol–1 respectively.
Calculate the molar conductivity
\left({\mathrm{\lambda }}_{\mathrm{m}}^{\infty }\right)
for CH3COOH at infinite dilution.
(iv) Identify the compounds A, B, C and D.
{\mathrm{C}}_{6}{\mathrm{H}}_{5}\mathrm{COOH}\stackrel{{\mathrm{SOCl}}_{2}}{\to } \mathrm{A} \stackrel{{\mathrm{NH}}_{3}}{\to } \mathrm{B} \stackrel{{\mathrm{Br}}_{2}/\mathrm{KOH}}{\to } \mathrm{C} \underset{0-5° \mathrm{C}}{\overset{{\mathrm{NaNO}}_{2}+\mathrm{HCl}}{\to }}\mathrm{D}
(a) An element has atomic weight 93 g mol–1 and density 11·5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(NA = 6·023×1023 mol–1)
(b) Calculate the radius of copper atom. The atomic weight of copper is 63·55 g mol–1. It crystallises in face centered cubic lattice and has density of 8·93 g cm–3 at 298K.
\left(\mathrm{i}\right) {\mathrm{P}}_{4}+\mathrm{NaOH}+{\mathrm{H}}_{2}\mathrm{O} \underset{\mathrm{Inert} \mathrm{atm}.}{\overset{\mathrm{heat}}{\to }} \mathrm{_________} + \mathrm{_________}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right) \underset{ \mathrm{dil}.}{\mathrm{Cu}+{\mathrm{HNO}}_{3} }\to \mathrm{________} + \mathrm{________} + \mathrm{_________}
(i) Write the chemical equation for the reaction of glucose with bromine water.
(ii) Write the zwitter ion structure of glycine. VIEW SOLUTION
(i) How do antiseptics differ from disinfectants?
(ii) Name a substance that can be used as an antiseptic as well as a disinfectant. VIEW SOLUTION
An alloy of gold (Au) and cadmium (Cd) crystallises with a cubic structure in which gold atoms occupy the corners and cadmium atoms fit into the face centres. What is the formula of this alloy? VIEW SOLUTION
(a) State reasons for the following:
(i) Ethylamine is soluble in water whereas aniline is insoluble in water.
(ii) Aliphatic amines are stronger bases than aromatic amines.
(b) Complete and balance the following equations:
(i) C6H5NH2 + CH3COCl ⎯⎯→ ________+ _________
(ii) C2H5NH2 + HNO2 ⎯⎯→ ________+ __________ + __________
Draw the structure of xenon tetrafluoride molecule. State the hybridisation of the central atom and the geometry of the molecule. VIEW SOLUTION
(a) Calculate the emf and ΔG for the given cell at 25°C:
{\mathrm{Cr}}_{\left(\mathrm{s}\right)}/{\mathrm{Cr}}^{3+}\left(0.1\mathrm{M}\right)//{\mathrm{Fe}}^{2+}\left(0.01\mathrm{M}\right)/{\mathrm{Fe}}_{\left(\mathrm{s}\right)}\phantom{\rule{0ex}{0ex}}\mathrm{Given}: {\mathrm{E}}_{{\mathrm{cr}}^{3+}/\mathrm{Cr}}^{°}=-0.74\mathrm{V}, {\mathrm{E}}_{{\mathrm{Fe}}^{2+}/\mathrm{Fe}}^{°}=-0.44\mathrm{V}\phantom{\rule{0ex}{0ex}}\left(1\mathrm{F}=96500 \mathrm{C}, \mathrm{R}=8.314 {\mathrm{JK}}^{-1} {\mathrm{mol}}^{-1}\right)
(b) Calculate the degree of dissociation (∝) of acetic acid, if its molar conductivity (Ʌm) is 39·05 S cm2 mol–1
\left(\mathrm{Given}: {\mathrm{\lambda }}_{\left({\mathrm{H}}^{+}\right)}^{0}=349.6 \mathrm{S} {\mathrm{cm}}^{2} {\mathrm{mol}}^{-1} \mathrm{and} {\mathrm{\lambda }}_{\left({\mathrm{CH}}_{3}{\mathrm{COO}}^{-}\right)}^{0} =40.95 \mathrm{S} {\mathrm{cm}}^{2} {\mathrm{mol}}^{-1}\right)
Name an important ore of silver. How is silver extracted from its sulphide ore? Give balanced chemical equations involved in the extraction of pure silver. VIEW SOLUTION
How will you convert the following:
(i) Chlorobenzene to biphenyl.
(ii) Propene to 1- bromopropane.
(iii) Chlorobenzene to aniline. VIEW SOLUTION
Explain what is observed when:
(i) A beam of light is passed through a colloidal solution.
(ii) An electric current is passed through a colloidal solution.
(iii) An electrolyte (AlCl3) is added to a colloidal solution of arsenious sulphide (As2S3). VIEW SOLUTION
(a) How will you convert the following: (Give balanced equation)
(i) Benzoyl chloride to benzaldehyde.
(ii) Methyl chloride to acetic acid.
(iii) Acetic acid to methane.
(b) A ketone A (C4H8O) which undergoes Iodoform reaction gives compound B on reduction.
B on heating with conc. H2SO4 at 443 K gives a compound C which forms ozonide D. D on hydrolysis with Zn dust gives only E. Identify the compounds A to E. Write the Iodoform reaction with compound A.
A first order reaction is 50% completed in 30 minutes at 300 K and in 10 minutes at 320K. Calculate the activation energy of the reaction. (R = 8·314 JK–1 mol–1) VIEW SOLUTION
(i) Transition metals and their compounds generally exhibit a paramagnetic behaviour.
(ii) There is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the 3d series of transition elements.
(iii) K2Cr2O7 acts as a powerful oxidising agent in acidic medium. VIEW SOLUTION
(a) (i) The elevation in boiling point when 0·30 g of acetic acid is dissolved in 100 g of benzene is 0·0633°C.
Calculate the molecular weight of acetic acid from this data. What conclusion can you draw about the molecular state of the solute in the solution?
(Given Kb for benzene = 2·53 K kg mol–1, at. wt. of C = 12, H = 1, O = 16)
(ii) Determine the osmotic pressure of a solution prepared by dissolving 0·025 g of K2SO4 in 2 litres of water at 25°C, assuming that K2SO4 is completely dissociated.
(R = 0·0821 Lit–atm K–1 mol–1 , mol. wt. of K2SO4 = 174 g mol–1)
(b) (i) An aqueous solution of a non–volatile solute freezes at 272·4 K, while pure water freezes at 273·0 K. Determine the following:
(Given Kf = 1·86 K kg mol–1 , Kb = 0·512 K kg mol–1 and vapour pressure of water at 298 K = 23·756 mm of Hg)
(1) The molality of solution
(2) Boiling point of solution
(3) The lowering of vapour pressure of water at 298 K
(ii) A solution containing 1·23g of calcium nitrate in 10g of water, boils at 100·975ºC at 760 mm of Hg. Calculate the van’t Hoff factor for the salt at this concentration.
(Kb for water = 0·52 K kg mol–1, mol. wt. of calcium nitrate = 164 g mol–1)
(a) (i) Write the IUPAC names of the following complexes:
(1) [Cu(NH3)4]SO4
(2) [Co(en)2Cl2]
(3) K3[Al(C2O4)3]
(ii) With reference to the coordination complex ion [Fe(H2O)6]2+ answer the following: (at. no. of Fe = 26)
(1) Give the IUPAC name of the complex ion.
(2) What is the oxidation number of the central metal atom?
(3) How many unpaired electrons are there in the complex ion?
(4) State the type of hybridisation of the complex ion.
(b) (i) Name of the type of isomerism exhibited by the following pairs of compounds:
(1) [Co(ONO)(NH3)5]2+ and [Co(NO2)(NH3)5]2+
(2) [Cr(H2O)4Cl2] Cl.2H2O and [Cr(H2O)5Cl]Cl2.H2O
(3) [Co(NH3)6] [Cr(CN)6] and [Cr(NH3)6] [Co(CN)6]
(ii) Using the valence bond approach, predict the shape, hybridisation and magnetic behaviour of [Ni(CO)4]. (at. no. of Ni = 28)
(a) (i) Give balanced chemical equations for the following reactions:
(1) Phenol is treated with ice cold alkaline solution of benzene diazonium chloride.
(2) Diethyl ether is treated with phosphorous pentachloride.
(3) Ethyl alcohol is treated with thionyl chloride.
(ii) Give one chemical test each to distinguish between the following pairs of compounds:
(1) Ethanol and dimethyl ether
(2) Propan-1-ol and propan-2-ol
(b) (i) Write chemical equations to illustrate the following name reactions:
(1) Williamson’s synthesis.
(2) Esterification reaction
(3) Reimer-Tiemann reaction.
(ii) Identify the compounds A and B in the given reactions:
{\mathrm{C}}_{2}{\mathrm{H}}_{5}\mathrm{OH} \underset{∆}{\overset{\mathrm{Cu}}{\to }} \mathrm{A} \underset{}{\overset{\mathrm{dil}.\mathrm{NaOH}}{\to }} \mathrm{B} |
Freefem - Wikibooks, open books for an open world
1 FREEFEM: a PDE solver
1.3 User Example 1
1.3.1 Analytical Description
1.3.2 freefem script
FREEFEM: a PDE solverEdit
The project freefem has two branches: the 2D solver (freefem++) is mature and well documented and used by many while the 3D solver (freefem3D) is still under construction but one is welcome to use it.
The code is open source and can be downloaded from the freefem web site
To solve a Partial Differential Equation (PDE) or set of equations with freefem one must first write an algorithm which uses only linear systems (one or more) of PDE as intermediate steps.
Then each system must be well posed, i.e. provided with admissible boundary conditions and written in variational form. For instance Dirichlet's problem
{\displaystyle -\Delta v=f~{\hbox{ in }}D,~~v|_{\partial D}=g}
defines in an open set D of the plane a scalar function v(x,y) at every point (x,y) of D; v would be the electrostatic potential in the vacuum with charge density f(x,y) around objects of boundary əD at potential g(x,y).
The automatic triangulation, the finite element solver and the graphics are integrated in freefem and driven by a freefem script.
The Dirichlet problem above can be solved with freefem++ by the following script
border C(t=0,2*pi){x=5*cos(t); y=5*sin(t);}
mesh Th = buildmesh (C(50));
fespace Vh(Th,P1); // says that we will use linear triangular elements
Vh u,v; // defines u and v as piecewise-P1 continuous functions
func f= x*y; // definition of a function called f
real cpu=clock(); // optional
solve Poisson(u,v)
= int2d(Th)(dx(u)*dx(v) + dy(u)*dy(v)) // bilinear part
- int2d(Th)( f*v) // right hand side of the variational form
+ on(C,u=0) ; // Dirichlet boundary condition (g=0)
cout << " CPU time = " << clock()-cpu << endl;
User Example 1Edit
Analytical DescriptionEdit
freefem scriptEdit
Retrieved from "https://en.wikibooks.org/w/index.php?title=Freefem&oldid=3450227" |
EUDML | A Pieri-Chevalley formula in the K-theory of a -bundle. EuDML | A Pieri-Chevalley formula in the K-theory of a -bundle.
A Pieri-Chevalley formula in the K-theory of a
G/B
-bundle.
Pittie, Harsh; Ram, Arun
Pittie, Harsh, and Ram, Arun. "A Pieri-Chevalley formula in the K-theory of a -bundle.." Electronic Research Announcements of the American Mathematical Society [electronic only] 5.14 (1999): 102-107. <http://eudml.org/doc/231407>.
@article{Pittie1999,
author = {Pittie, Harsh, Ram, Arun},
keywords = {flag variety; Schubert subvarieties; -theory; Schubert class; -theory},
title = {A Pieri-Chevalley formula in the K-theory of a -bundle.},
AU - Pittie, Harsh
TI - A Pieri-Chevalley formula in the K-theory of a -bundle.
KW - flag variety; Schubert subvarieties; -theory; Schubert class; -theory
flag variety, Schubert subvarieties,
K
-theory, Schubert class,
K
K
-theory in geometry
K
-theory of schemes
K
Articles by Pittie
Articles by Ram |
Note that, since there are only
255
distinct characters (byte values), the Pigeon-Hole Principle implies that no string permutation can exceed
255
characters in length.
8
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{IsPermutation}\left(""\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{IsPermutation}\left("abc"\right)
\textcolor[rgb]{0,0,1}{\mathrm{true}}
\mathrm{IsPermutation}\left("abcba"\right)
\textcolor[rgb]{0,0,1}{\mathrm{false}}
r≔\mathrm{rand}\left(1..10\right):
S≔{\mathrm{seq}}\left(\mathrm{Random}\left(r\left(\right),'\mathrm{lower}'\right),i=1..1000\right):
\mathrm{select}\left(\mathrm{IsPermutation}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathbf{and}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}\mathrm{IsPalindrome},S\right)
{\textcolor[rgb]{0,0,1}{"a"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"b"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"c"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"d"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"e"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"f"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"g"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"h"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"i"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"j"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"k"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"l"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"m"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"n"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"o"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"p"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"q"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"r"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"s"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"t"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"u"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"v"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"w"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"x"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"y"}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{"z"}} |
Study Of Acids Bases And Salts, Popular Questions: ICSE Class 10 CHEMISTRY, Concise Chemistry 10 - Meritnation
Faraz Ahmed asked a question
Viraf j Dalal chemistry class 10 Solutions
24. Which among the following does not undergoes hydrolysis?
(a) Na2CO3 (b) CH3COOH
(c) CH3COONH4 (d) K2SO4
25. Which of the following acts as both Bronsted and Lewis acid?
(a) Cu+2 (b) SO2
(c) Fe+3 (d) NH4+
List the colour of 20 important salts.
Anshumaan Pandey asked a question
what do you understand by PH value? 2 solutions X and Y have pH values of 4 and 10 respectively. which one of these two will give a pink colour with phenolphthalein indicator?
Name 5 strong alkali and weak alkali.
Abhiroop Sujay asked a question
Calculate the pH of a solution which contains 9.9 ml of 1 M HCl and 100 ml of 0.1 M NaOH
Carbonic acid gives an acid salt but hydrochloric acid does not.
In this question I was told to write the answer in terms of basicity( monobasic , dibasic, etc.).....if I write it in terms of partial replacement will the answer be wrong ??
Swayam Ahuja asked a question
Name An alkali which on dissociation produces a high concentration of Hydroxyl ions
Please solve question 1a
Manya asked a question
PbCl2 is an acid or not
Mpv Krishna Kumar asked a question
How to calculate OH- ions in solution having values for H+ ions?
Q.21. The compound that is not Lewis acis is:
BaC{l}_{2}
AlC{l}_{3}
BC{l}_{3}
SnC{l}_{4}
Which type of nitric acid is required to convert phosphorus into phosphoric acid?... Dilute or concentrated ? Pls answer fast.. its urgent...
Solution P has a pH of 13. Solution Q has a pH of 6 and solution R has a pH of 2. Which solution:
(a) will liberate ammonia from ammonium sulphate on heating?
(b) is a strong acid?
(c) contains molecules as well as ions?
dry HCL gas does not affect a try strip of blue litmus paper but it turns red in the presence of a drop of water.
Anshika Taner asked a question
a normal salt formed when basic oxide and acidic oxide react?
Please solve question no.a.
Vanshika Agarwal asked a question
a salt whose solubility increase with temperature
Plz solve this question no 44 and 45
44. The pH value of human blood is 7.3. What would happen if the pH of his blood becomes 7.6?
45. Two solutions X and Y have pH values 3 and 9 respectively Which of these two give a pink colour with phenolphthalein?
Apoorvi Jain asked a question
Which is a more volatile acid HCl or nitric acid ?
So, please state the reactions of chlorides with dil. H2So4 and dil.HCL
Taking sodium carbonate as an example give the meaning of the following terms
b)Effloresence
In both, the methods of reaction do HEAT is required or not? This is the methods of preparation of a base. It's from the chapter Acid, Bases and Salts.
Please answer clearly and according to the question.
How to convert sulphur dioxide to sodium sulphite.give proper chemical equation
1. Metals are extracted from their ores by
2. The unwanted material in an ore is known as
Barium chloride is acid, base or salt
what do you understand by PH value? two solutions X and Y have pH values of 4 and 10 respectively which one of these two will give a pink colour with phenolphthalein indicator
Name the acid which on mixing with lead nitrate solution produces a white precipitate which is soluble in excess ammonium hydroxide.
Kunal Dutta asked a question
by what means other than heating could i dehydrate blue vitriol to obtain anhydrous copper sulphate
Salt is made up of metal only, it does not consist of non-metal. Is my statement right? If not please correct me... ANSWER FAST
A metal which gives out hydrogen gas on reacting with dilute acid and an alkali.name the metal
Raaz asked a question
Why is a solution of hydrogen chloride an acid,but a solution of glucose is not,though both contain hydrogen
Shaavi Gupta asked a question
Can anyone tell that what the table is showing?Plz tell the full form of poH.
Ayushi ❤️❤️ asked a question
Hello guys how are you all....
During oxidation of Ethanol to acetaldehyde, oxygen is used but the number of oxygen atoms in both remains the same. Then where does the extra oxygen go?
How to distinguish between hcl hno3 and h2so4.give chemical tests
What is the product formed when iron(2) & (3) oxide reacts with sulphuric acid dilute each?
Lolly asked a question
Which bases are insoluble in water
Name an amphoteric metallic nitrate which on heating gives a white Residue?
Please say 100% sure answers
Answer Q no. 10 (c,d)
Himanshu Mamgain asked a question
FeSO4+NaOH gives
Please solve question no.c.
Injeev Acharya asked a question
Name the alkali that react with skin
How to identify each and every acid by just looking at it's molecular formula?
Q. Ions of an element are chemically inactive because of the octet configuration of their valence shells .then how do this reaction takes place?
M{n}^{2+}\left(aq\right)+{H}_{2}S\left(g\right) \to MnS+{H}^{+}\left(aq\right)
Yash bhai yha aao please
Please solve this question no 19
19. What happens when copper chloride reacts with an aqueous solution of sodium hydroxide? State the colour of the precipitate formed.
Aditya Chatterjee asked a question
An acid of sulphur reacting with water to give Sulphuric acid?
Are the oxides of all elements of period 4 colourless except copper ?
In the brown ring test, can we use any sulphate salt instead of sulphuric acid?
Yashi Mishra asked a question
Name two salts which hydrolysis in water
This question is from topography map 45d/10 please answer it
Q 1. Is the dam in this part of the map natural or artificial.Give reasons
What is the meaning of plastic
Which is the type of structure below ( structure of phosphoric acid) and how to draw it.
64. Calculate the pH of a solution which contains 9.9 ml of 1 M HCI and 100 ml of 0.1 M NaOH.
65. Calculate the pH of a solution by mixing 0.1 litre of pH - 4 and 0.2 litre of pH = 10.
65. Calculate the pH of a solution obtained by mixing 10 ml of 0.1 M HCI and 40 ml of 0.2 M H2SO4-
What do u mean by hydrolysis of salt? Give me a brief description of the question asked. NO LINKS, ONLY HANDWRITTEN ANSWER.
Solve NAME THE FOLLOWING
(a) A basic solution which does not contain a metallic element.
(b) An alkali which on dissociation produces a high concentration of hydroxyl ions.
(c) A complex salt solution used for testing a basic gas lighter than air.
(d) A base which reacts with hydrochloric acid to give a salt which on hydrolysis gives a slightly acidic solution.
Please answer all the give reasons...
Solve q no 6
Is this a synthesis reaction?
State your observations for: concentrated nitric acid is poured on alcohol
-- asked a question
differentiate between zinc nitrate and copper nitrate with a chemical test
Sneh asked a question
Please solve question 1
Solve it quickly now I have exam
Name the oxidising agent present in aquaregia.
Please explain me the NOTE box
Difference between common acid base indicator and universal indicator
Please answer the last copper one.. .
please help in queaton 67,68,69
Please explain experts.....
Name the acidic oxide of a metal?
Please let me know the answer of question no.11.
Plzz help this
Which of the following species can act as a Bronsted base as well as a Lewis base?
(A) NO3- (B) CN- (C) NH3 (D) BF3
Please solve the 28 number.
Q.28. State what will be the effect of each of the following solutions on blue litmus -
i]
{K}_{2}C{O}_{3}
soln. ii] KCl soln. iii]
N{H}_{4}N{O}_{3}
soln.
Please solve this matching
Barium (Ba) is a divalent element. It is placed between sodium and calcium in the activity series. State the outcome of the following:
(i) The exposure of barium to air.
(ii) The addition of barium to water.
(iii) The treatment of barium with dil. HCl.
What is the solution of year 2004
Uday Krishna asked a question
Correct order of acidity
Diff between natural water and treated water
Solve the year 2004
Old tell Q3
Answer both questions...
Name 2 cations that do not form a precipitate with ammonium hydroxide but form a precipitate with sodium hydroxide.
Please explain the answer?
Please answer 2006 second one
Solve q mo 6 with solution:
25. In the reaction, I2 + I-
\to
I3-, the Lewis base is :-
(A) I2 (B) I-
(C) I3- (D) None. |
What is the probability that a card selected at random from a standard deck of 5
What is the probability that a card selected at random from a standard deck of 52 cards is a number 5 or 6 or 10?
In a standard deck of 52 cards, there are four 5’s, four 6’s, and four 10’s. By Addition Rule, the probability is:
\frac{4}{52}+\frac{4}{52}+\frac{4}{52}=\frac{12}{52}=\frac{3}{13}
\left(A-B\right)-C=\left(A-C\right)-\left(B-C\right)
\left(a,b\right)\in R
Use proof by Contradiction to prove that the sum of an irrational number and a rational number is irrational.
HELP needed on HW Questions ASAP
11. E is the relation defined on Z as follows: for all
m,n,\in Z,mEn⇔4\mid \left(m-n\right)
a) Prove that E is equivalence relation.
b) List five elements in [5].
c) Find a partition of set Z based on relation E.
Discrete Mathematics Division Algorithm proof
I'm not quite sure how to do this problem if anyone can do a step by step to help me understand it I would appreciate it a lot.
Let a and b be positive integers with
b>a
, and suppose that the division algorithm yields
b=a\cdot q+r
0\le r<a
. (note: its a zero)
\mathrm{l}\mathrm{c}\mathrm{m}\left(a,b\right)-\mathrm{l}\mathrm{c}\mathrm{m}\left(a,r\right)=\frac{{a}^{2}\cdot q}{gcd\left(a,b\right)}. |
Fermat_point Knowpia
In geometry, the Fermat point of a triangle, also called the Torricelli point or Fermat–Torricelli point, is a point such that the sum of the three distances from each of the three vertices of the triangle to the point is the smallest possible.[1] It is so named because this problem was first raised by Fermat in a private letter to Evangelista Torricelli, who solved it.
Fig 1. Construction of the first isogonic center, X(13). When no angle of the triangle exceeds 120°, this point is the Fermat point.
The Fermat point of a triangle with largest angle at most 120° is simply its first isogonic center or X(13), which is constructed as follows:
Construct an equilateral triangle on each of two arbitrarily chosen sides of the given triangle.
Draw a line from each new vertex to the opposite vertex of the original triangle.
On each of two arbitrarily chosen sides, construct an isosceles triangle, with base the side in question, 30-degree angles at the base, and the third vertex of each isosceles triangle lying outside the original triangle.
For each isosceles triangle draw a circle, in each case with center on the new vertex of the isosceles triangle and with radius equal to each of the two new sides of that isosceles triangle.
When a triangle has an angle greater than 120°, the Fermat point is sited at the obtuse-angled vertex.
In what follows "Case 1" means the triangle has an angle exceeding 120°. "Case 2" means no angle of the triangle exceeds 120°.
Location of X(13)Edit
Fig. 2 shows the equilateral triangles ARB, AQC and CPB attached to the sides of the arbitrary triangle ABC. Here is a proof using properties of concyclic points to show that the three lines RC, BQ and AP in Fig 2 all intersect at the point F and cut one another at angles of 60°.
The triangles RAC and BAQ are congruent because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF. By the converse of the inscribed angle theorem applied to the segment AF, the points ARBF are concyclic (they lie on a circle). Similarly, the points AFCQ are concyclic.
∠ARB = 60°, so ∠AFB = 120°, using the inscribed angle theorem. Similarly, ∠AFC = 120°.
So ∠BFC = 120°. So, ∠BFC and ∠BPC add up to 180°. Using the inscribed angle theorem, this implies that the points BPCF are concyclic. So, using the inscribed angle theorem applied to the segment BP, ∠BFP = ∠BCP = 60°. Because ∠BFP + ∠BFA = 180°, the point F lies on the line segment AP. So, the lines RC, BQ and AP are concurrent (they intersect at a single point). Q.E.D.
This proof applies only in Case 2 since if ∠BAC > 120°, point A lies inside the circumcircle of BPC which switches the relative positions of A and F. However it is easily modified to cover Case 1. Then ∠AFB = ∠AFC = 60° hence ∠BFC = ∠AFB + ∠AFC = 120° which means BPCF is concyclic so ∠BFP = ∠BCP = 60° = ∠BFA. Therefore, A lies on FP.
The lines joining the centers of the circles in Fig. 2 are perpendicular to the line segments AP, BQ and CR. For example, the line joining the center of the circle containing ARB and the center of the circle containing AQC, is perpendicular to the segment AP. So, the lines joining the centers of the circles also intersect at 60° angles. Therefore, the centers of the circles form an equilateral triangle. This is known as Napoleon's Theorem.
Location of the Fermat pointEdit
Traditional geometryEdit
Given any Euclidean triangle ABC and an arbitrary point P let d(P) = PA+PB+PC, with PA denoting the distance between P and A. The aim of this section is to identify a point P0 such that d(P0) < d(P) for all P ≠ P0. If such a point exists then it will be the Fermat point. In what follows Δ will denote the points inside the triangle and will be taken to include its boundary Ω.
[If AB is the common side extend AC to cut the polygon at X. Then by the triangle inequality the polygon perimeter > AB + AX + XB = AB + AC + CX + XB ≥ AB + AC + BC.]
Without loss of generality suppose that the angle at A is ≥ 120°. Construct the equilateral triangle AFB and for any point P in Δ (except A itself) construct Q so that the triangle AQP is equilateral and has the orientation shown. Then the triangle ABP is a 60° rotation of the triangle AFQ about A so these two triangles are congruent and it follows that d(P) = CP+PQ+QF which is simply the length of the path CPQF. As P is constrained to lie within ABC, by the dogleg rule the length of this path exceeds AC+AF = d(A). Therefore, d(A) < d(P) for all P є Δ, P ≠ A. Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(A) ≤ d (P') it follows that d(A) < d(P) for all P outside Δ. Thus d(A) < d(P) for all P ≠ A which means that A is the Fermat point of Δ. In other words, the Fermat point lies at the obtuse-angled vertex.
Construct the equilateral triangle BCD and let P be any point inside Δ and construct the equilateral triangle CPQ. Then CQD is a 60° rotation of CPB about C so d(P) = PA+PB+PC = AP+PQ+QD which is simply the length of the path APQD. Let P0 be the point where AD and CF intersect. This point is commonly called the first isogonic center. Carry out the same exercise with P0 as you did with P, and find the point Q0. By the angular restriction P0 lies inside Δ moreover BCF is a 60° rotation of BDA about B so Q0 must lie somewhere on AD. Since CDB = 60° it follows that Q0 lies between P0 and D which means AP0Q0D is a straight line so d(P0) = AD. Moreover, if P ≠ P0 then either P or Q won't lie on AD which means d(P0) = AD < d(P). Now allow P to range outside Δ. From above a point P' є Ω exists such that d(P') < d(P) and as d(P0) ≤ d(P') it follows that d(P0) < d(P) for all P outside Δ. That means P0 is the Fermat point of Δ. In other words, the Fermat point is coincident with the first isogonic center.
Vector analysisEdit
Let O, A, B, C, X be any five points in a plane. Denote the vectors
{\displaystyle {\overrightarrow {\mathrm {OA} }},{\overrightarrow {\mathrm {OB} }},{\overrightarrow {\mathrm {OC} }},{\overrightarrow {\mathrm {OX} }}}
by a, b, c, x respectively, and let i, j, k be the unit vectors from O along a, b, c.
Now |a| = a⋅i = (a − x)⋅i + x⋅i ≤ |a − x| + x⋅i and similarly |b| ≤ |b − x| + x⋅j and |c| ≤ |c − x| + x⋅k.
Adding gives |a| + |b| + |c| ≤ |a − x| + |b − x| + |c − x| + x⋅(i + j + k).
If a, b, c meet at O at angles of 120° then i + j + k = 0 so |a| + |b| + |c| ≤ |a − x| + |b − x| + |c − x| for all x.
In other words, OA + OB + OC ≤ XA + XB + XC and hence O is the Fermat point of ABC.
This argument fails when the triangle has an angle ∠C > 120° because there is no point O where a, b, c meet at angles of 120°. Nevertheless, it is easily fixed by redefining k = − (i + j) and placing O at C so that c = 0. Note that |k| ≤ 1 because the angle between the unit vectors i and j is ∠C which exceeds 120°. Since |0| ≤ |0 − x| + x⋅k the third inequality still holds, the other two inequalities are unchanged. The proof now continues as above (adding the three inequalities and using i + j + k = 0) to reach the same conclusion that O (or in this case C) must be the Fermat point of ABC.
Lagrange multipliersEdit
Another approach to finding the point within a triangle, from which the sum of the distances to the vertices of the triangle is minimal, is to use one of the mathematical optimization methods; specifically, the method of Lagrange multipliers and the law of cosines.
We draw lines from the point within the triangle to its vertices and call them X, Y and Z. Also, let the lengths of these lines be x, y, and z, respectively. Let the angle between X and Y be α, Y and Z be β. Then the angle between X and Z is (2π − α − β). Using the method of Lagrange multipliers we have to find the minimum of the Lagrangian L, which is expressed as:
L = x + y + z + λ1 (x2 + y2 − 2xy cos(α) − a2) + λ2 (y2 + z2 − 2yz cos(β) − b2) + λ3 (z2 + x2 − 2zx cos(α + β) − c2)
Equating each of the five partial derivatives δL/δx, δL/δy, δL/δz, δL/δα, δL/δβ to zero and eliminating λ1, λ2, λ3 eventually gives sin(α) = sin(β) and sin(α + β) = − sin(β) so α = β = 120°. However the elimination is a long and tedious business, and the end result covers only Case 2.
The circumcircles of the three constructed equilateral triangles are concurrent at X(13).
The points X(13), X(14), the circumcenter, and the nine-point center lie on a Lester circle.
The line X(13)X(14) meets the Euler line at midpoint of X(2) and X(4).[6]
The Fermat point lies in the open orthocentroidal disk punctured at its own center, and could be any point therein.[7]
The isogonic centers X(13) and X(14) are also known as the first Fermat point and the second Fermat point respectively. Alternatives are the positive Fermat point and the negative Fermat point. However these different names can be confusing and are perhaps best avoided. The problem is that much of the literature blurs the distinction between the Fermat point and the first Fermat point whereas it is only in Case 2 above that they are actually the same.
This question was proposed by Fermat, as a challenge to Evangelista Torricelli. He solved the problem in a similar way to Fermat's, albeit using the intersection of the circumcircles of the three regular triangles instead. His pupil, Viviani, published the solution in 1659.[8]
^ Entry X(13) in the Encyclopedia of Triangle Centers Archived April 19, 2012, at the Wayback Machine
^ Kimberling, Clark. "Encyclopedia of Triangle Centers".
^ Christopher J. Bradley and Geoff C. Smith, "The locations of triangle centers", Forum Geometricorum 6 (2006), 57--70. http://forumgeom.fau.edu/FG2006volume6/FG200607index.html
^ Weisstein, Eric W. "Fermat Points". MathWorld.
"Fermat-Torricelli problem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
A practical example of the Fermat point
iOS Interactive sketch |
Triangular matrix - Wikipedia
In the mathematical discipline of linear algebra, a triangular matrix is a special kind of square matrix. A square matrix is called lower triangular if all the entries above the main diagonal are zero. Similarly, a square matrix is called upper triangular if all the entries below the main diagonal are zero.
2 Forward and back substitution
2.1 Forward substitution
4.1 Unitriangular matrix
4.2 Strictly triangular matrix
4.3 Atomic triangular matrix
5 Triangularisability
5.1 Simultaneous triangularisability
6 Algebras of triangular matrices
6.1 Borel subgroups and Borel subalgebras
{\displaystyle L={\begin{bmatrix}\ell _{1,1}&&&&0\\\ell _{2,1}&\ell _{2,2}&&&\\\ell _{3,1}&\ell _{3,2}&\ddots &&\\\vdots &\vdots &\ddots &\ddots &\\\ell _{n,1}&\ell _{n,2}&\ldots &\ell _{n,n-1}&\ell _{n,n}\end{bmatrix}}}
{\displaystyle U={\begin{bmatrix}u_{1,1}&u_{1,2}&u_{1,3}&\ldots &u_{1,n}\\&u_{2,2}&u_{2,3}&\ldots &u_{2,n}\\&&\ddots &\ddots &\vdots \\&&&\ddots &u_{n-1,n}\\0&&&&u_{n,n}\end{bmatrix}}}
is called an upper triangular matrix or right triangular matrix. A lower or left triangular matrix is commonly denoted with the variable L, and an upper or right triangular matrix is commonly denoted with the variable U or R.
A matrix that is both upper and lower triangular is diagonal. Matrices that are similar to triangular matrices are called triangularisable.
{\displaystyle {\begin{bmatrix}1&4&1\\0&6&4\\0&0&1\\\end{bmatrix}}}
{\displaystyle {\begin{bmatrix}1&0&0\\2&96&0\\4&9&69\\\end{bmatrix}}}
Forward and back substitutionEdit
A matrix equation in the form
{\displaystyle L\mathbf {x} =\mathbf {b} }
{\displaystyle U\mathbf {x} =\mathbf {b} }
is very easy to solve by an iterative process called forward substitution for lower triangular matrices and analogously back substitution for upper triangular matrices. The process is so called because for lower triangular matrices, one first computes
{\displaystyle x_{1}}
, then substitutes that forward into the next equation to solve for
{\displaystyle x_{2}}
, and repeats through to
{\displaystyle x_{n}}
. In an upper triangular matrix, one works backwards, first computing
{\displaystyle x_{n}}
, then substituting that back into the previous equation to solve for
{\displaystyle x_{n-1}}
, and repeating through
{\displaystyle x_{1}}
Forward substitutionEdit
{\displaystyle {\begin{matrix}\ell _{1,1}x_{1}&&&&&&&=&b_{1}\\\ell _{2,1}x_{1}&+&\ell _{2,2}x_{2}&&&&&=&b_{2}\\\vdots &&\vdots &&\ddots &&&&\vdots \\\ell _{m,1}x_{1}&+&\ell _{m,2}x_{2}&+&\dotsb &+&\ell _{m,m}x_{m}&=&b_{m}\\\end{matrix}}}
Observe that the first equation (
{\displaystyle \ell _{1,1}x_{1}=b_{1}}
) only involves
{\displaystyle x_{1}}
, and thus one can solve for
{\displaystyle x_{1}}
directly. The second equation only involves
{\displaystyle x_{1}}
{\displaystyle x_{2}}
, and thus can be solved once one substitutes in the already solved value for
{\displaystyle x_{1}}
. Continuing in this way, the
{\displaystyle k}
-th equation only involves
{\displaystyle x_{1},\dots ,x_{k}}
, and one can solve for
{\displaystyle x_{k}}
using the previously solved values for
{\displaystyle x_{1},\dots ,x_{k-1}}
. Note: errors in formulas below: The resulting formulas are:
{\displaystyle {\begin{aligned}x_{1}&={\frac {b_{1}}{\ell _{1,1}}},\\x_{2}&={\frac {b_{2}-\ell _{2,1}x_{1}}{\ell _{2,2}}},\\&\ \ \vdots \\x_{m}&={\frac {b_{m}-\sum _{i=1}^{m-1}\ell _{m,i}x_{i}}{\ell _{m,m}}}.\end{aligned}}}
A matrix which is both symmetric and triangular is diagonal. In a similar vein, a matrix which is both normal (meaning A*A = AA*, where A* is the conjugate transpose) and triangular is also diagonal. This can be seen by looking at the diagonal entries of A*A and AA*.
The determinant and permanent of a triangular matrix equal the product of the diagonal entries, as can be checked by direct computation.
In fact more is true: the eigenvalues of a triangular matrix are exactly its diagonal entries. Moreover, each eigenvalue occurs exactly k times on the diagonal, where k is its algebraic multiplicity, that is, its multiplicity as a root of the characteristic polynomial
{\displaystyle p_{A}(x)=\det(xI-A)}
of A. In other words, the characteristic polynomial of a triangular n×n matrix A is exactly
{\displaystyle p_{A}(x)=(x-a_{11})(x-a_{22})\cdots (x-a_{nn})}
that is, the unique degree n polynomial whose roots are the diagonal entries of A (with multiplicities). To see this, observe that
{\displaystyle xI-A}
is also triangular and hence its determinant
{\displaystyle \det(xI-A)}
is the product of its diagonal entries
{\displaystyle (x-a_{11})(x-a_{22})\cdots (x-a_{nn})}
Unitriangular matrixEdit
If the entries on the main diagonal of a (upper or lower) triangular matrix are all 1, the matrix is called (upper or lower) unitriangular.
Other names used for these matrices are unit (upper or lower) triangular, or very rarely normed (upper or lower) triangular. However, a unit triangular matrix is not the same as the unit matrix, and a normed triangular matrix has nothing to do with the notion of matrix norm.
All finite unitriangular matrices are unipotent.
Strictly triangular matrixEdit
If all of the entries on the main diagonal of a (upper or lower) triangular matrix are also 0, the matrix is called strictly (upper or lower) triangular.
All finite strictly triangular matrices are nilpotent of index n as a consequence of the Cayley-Hamilton theorem.
Atomic triangular matrixEdit
Main article: Frobenius matrix
An atomic (upper or lower) triangular matrix is a special form of unitriangular matrix, where all of the off-diagonal elements are zero, except for the entries in a single column. Such a matrix is also called a Frobenius matrix, a Gauss matrix, or a Gauss transformation matrix.
TriangularisabilityEdit
A matrix that is similar to a triangular matrix is referred to as triangularizable. Abstractly, this is equivalent to stabilizing a flag: upper triangular matrices are precisely those that preserve the standard flag, which is given by the standard ordered basis
{\displaystyle (e_{1},\ldots ,e_{n})}
and the resulting flag
{\displaystyle 0<\left\langle e_{1}\right\rangle <\left\langle e_{1},e_{2}\right\rangle <\cdots <\left\langle e_{1},\ldots ,e_{n}\right\rangle =K^{n}.}
All flags are conjugate (as the general linear group acts transitively on bases), so any matrix that stabilises a flag is similar to one that stabilizes the standard flag.
Any complex square matrix is triangularizable.[1] In fact, a matrix A over a field containing all of the eigenvalues of A (for example, any matrix over an algebraically closed field) is similar to a triangular matrix. This can be proven by using induction on the fact that A has an eigenvector, by taking the quotient space by the eigenvector and inducting to show that A stabilizes a flag, and is thus triangularizable with respect to a basis for that flag.
Simultaneous triangularisabilityEdit
A set of matrices
{\displaystyle A_{1},\ldots ,A_{k}}
are said to be simultaneously triangularisable if there is a basis under which they are all upper triangular; equivalently, if they are upper triangularizable by a single similarity matrix P. Such a set of matrices is more easily understood by considering the algebra of matrices it generates, namely all polynomials in the
{\displaystyle A_{i},}
{\displaystyle K[A_{1},\ldots ,A_{k}].}
Simultaneous triangularizability means that this algebra is conjugate into the Lie subalgebra of upper triangular matrices, and is equivalent to this algebra being a Lie subalgebra of a Borel subalgebra.
The basic result is that (over an algebraically closed field), the commuting matrices
{\displaystyle A,B}
or more generally
{\displaystyle A_{1},\ldots ,A_{k}}
are simultaneously triangularizable. This can be proven by first showing that commuting matrices have a common eigenvector, and then inducting on dimension as before. This was proven by Frobenius, starting in 1878 for a commuting pair, as discussed at commuting matrices. As for a single matrix, over the complex numbers these can be triangularized by unitary matrices.
The fact that commuting matrices have a common eigenvector can be interpreted as a result of Hilbert's Nullstellensatz: commuting matrices form a commutative algebra
{\displaystyle K[A_{1},\ldots ,A_{k}]}
{\displaystyle K[x_{1},\ldots ,x_{k}]}
which can be interpreted as a variety in k-dimensional affine space, and the existence of a (common) eigenvalue (and hence a common eigenvector) corresponds to this variety having a point (being non-empty), which is the content of the (weak) Nullstellensatz.[citation needed] In algebraic terms, these operators correspond to an algebra representation of the polynomial algebra in k variables.
More generally and precisely, a set of matrices
{\displaystyle A_{1},\ldots ,A_{k}}
is simultaneously triangularisable if and only if the matrix
{\displaystyle p(A_{1},\ldots ,A_{k})[A_{i},A_{j}]}
is nilpotent for all polynomials p in k non-commuting variables, where
{\displaystyle [A_{i},A_{j}]}
is the commutator; for commuting
{\displaystyle A_{i}}
the commutator vanishes so this holds. This was proven in (Drazin, Dungey & Gruenberg 1951); a brief proof is given in (Prasolov 1994, pp. 178–179). One direction is clear: if the matrices are simultaneously triangularisable, then
{\displaystyle [A_{i},A_{j}]}
is strictly upper triangularizable (hence nilpotent), which is preserved by multiplication by any
{\displaystyle A_{k}}
or combination thereof – it will still have 0s on the diagonal in the triangularizing basis.
Algebras of triangular matricesEdit
Binary lower unitriangular Toeplitz matrices, multiplied using F2 operations. They form the Cayley table of Z4 and correspond to powers of the 4-bit Gray code permutation.
Upper triangularity is preserved by many operations:
The inverse of an upper triangular matrix, if it exists, is upper triangular.
The product of an upper triangular matrix and a scalar is upper triangular.
Together these facts mean that the upper triangular matrices form a subalgebra of the associative algebra of square matrices for a given size. Additionally, this also shows that the upper triangular matrices can be viewed as a Lie subalgebra of the Lie algebra of square matrices of a fixed size, where the Lie bracket [a, b] given by the commutator ab − ba. The Lie algebra of all upper triangular matrices is a solvable Lie algebra. It is often referred to as a Borel subalgebra of the Lie algebra of all square matrices.
All these results hold if upper triangular is replaced by lower triangular throughout; in particular the lower triangular matrices also form a Lie algebra. However, operations mixing upper and lower triangular matrices do not in general produce triangular matrices. For instance, the sum of an upper and a lower triangular matrix can be any matrix; the product of a lower triangular with an upper triangular matrix is not necessarily triangular either.
The set of strictly upper (or lower) triangular matrices forms a nilpotent Lie algebra, denoted
{\displaystyle {\mathfrak {n}}.}
This algebra is the derived Lie algebra of
{\displaystyle {\mathfrak {b}}}
, the Lie algebra of all upper triangular matrices; in symbols,
{\displaystyle {\mathfrak {n}}=[{\mathfrak {b}},{\mathfrak {b}}].}
{\displaystyle {\mathfrak {n}}}
is the Lie algebra of the Lie group of unitriangular matrices.
Algebras of upper triangular matrices have a natural generalization in functional analysis which yields nest algebras on Hilbert spaces.
See also: Affine group
Borel subgroups and Borel subalgebrasEdit
The set of invertible triangular matrices of a given kind (upper or lower) forms a group, indeed a Lie group, which is a subgroup of the general linear group of all invertible matrices. A triangular matrix is invertible precisely when its diagonal entries are invertible (non-zero).
Over the real numbers, this group is disconnected, having
{\displaystyle 2^{n}}
components accordingly as each diagonal entry is positive or negative. The identity component is invertible triangular matrices with positive entries on the diagonal, and the group of all invertible triangular matrices is a semidirect product of this group and the group of diagonal matrices with
{\displaystyle \pm 1}
on the diagonal, corresponding to the components.
The Lie algebra of the Lie group of invertible upper triangular matrices is the set of all upper triangular matrices, not necessarily invertible, and is a solvable Lie algebra. These are, respectively, the standard Borel subgroup B of the Lie group GLn and the standard Borel subalgebra
{\displaystyle {\mathfrak {b}}}
of the Lie algebra gln.
The group of 2×2 upper unitriangular matrices is isomorphic to the additive group of the field of scalars; in the case of complex numbers it corresponds to a group formed of parabolic Möbius transformations; the 3×3 upper unitriangular matrices form the Heisenberg group.
^ a b c (Axler 1996, pp. 86–87, 169)
^ (Herstein 1975, pp. 285–290)
Retrieved from "https://en.wikipedia.org/w/index.php?title=Triangular_matrix&oldid=1087495116" |
Write an equation to represent the situation below and answer the question. You may use the 5-D Process to help you.
Andrew just opened a new office-supply store. He has been keeping track of how many customers visit his store. During his second week, he had
18
more customers than he did the first week. The third week, he had four less than twice as many customers as he had during the second week. In his fourth week of business, he had
92
customers. If he had
382
customers in total during his first four weeks of business, how many customers did he have during the second week?
Create an expression to represent this problem using a variable,
x
. Try to write and solve for an equation, or if necessary, use the 5-D Process you learned earlier.
Each of the weeks ultimately relates to the first week, so we can define the first week's customers as
x
. Write expressions for each of the other weeks based on the variable.
x
x+18
2(x+18)-4
92
Create an expression by summing up each of the week's customers to total
382
as given by the problem, then solve for the variable. Remember that the problem asks you to find the number of customers in the second week.
78 |
Basic Mathematics(pre Requisite), Revision Notes: CBSE Class 11-science PHYSICS, Physics Part I - Meritnation
-
-
-
-
\frac{1}{x}
h\text{'}\left(x\right) = -\frac{f\text{'}\left(x\right)}{\left(f\left(x\right){\right)}^{2}}
(v) The quotient rule: If h(x) = f(x)/g(x) and g(x) ≠ 0 for all x, then
h\text{'}\left(x\right) = \frac{g\left(x\right)f\text{'}\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{\left(g\left(x\right){\right)}^{2}}
(vi) The chain rule: If h(x) = (f o g)(x) = f(g(x)) for all x, then h′(x) = f′(g(x))g′(x).
Integration is the process of findin… |
Quaternion point rotation - MATLAB rotatepoint - MathWorks Benelux
Rotate Point Using Quaternion Vector
Rotate Group of Points Using Quaternion
cartesianPoints
rotationResult
Quaternion point rotation
rotationResult = rotatepoint(quat,cartesianPoints)
rotationResult = rotatepoint(quat,cartesianPoints) rotates the Cartesian points using the quaternion, quat. The elements of the quaternion are normalized before use in the rotation.
Define a point in three dimensions. The coordinates of a point are always specified in order x, y, z. For convenient visualization, define the point on the x-y plane.
Create a quaternion vector specifying two separate rotations, one to rotate the point 45 and another to rotate the point -90 degrees about the z-axis. Use rotatepoint to perform the rotation.
quat = quaternion([0,0,pi/4; ...
0,0,-pi/2],'euler','XYZ','point');
rotatedPoint = rotatepoint(quat,[x,y,z])
rotatedPoint = 2×3
Plot the rotated points.
plot(rotatedPoint(1,1),rotatedPoint(1,2),'bo')
plot(rotatedPoint(2,1),rotatedPoint(2,2),'go')
Define two points in three-dimensional space. Define a quaternion to rotate the point by first rotating about the z-axis 30 degrees and then about the new y-axis 45 degrees.
quat = quaternion([30,45,0],'eulerd','ZYX','point');
Use rotatepoint to rotate both points using the quaternion rotation operator. Display the result.
rP = rotatepoint(quat,[a;b])
rP = 2×3
Visualize the original orientation and the rotated orientation of the points. Draw lines from the origin to each of the points for visualization purposes.
plot3(a(1),a(2),a(3),'bo');
plot3(b(1),b(2),b(3),'ro');
plot3(rP(1,1),rP(1,2),rP(1,3),'bd')
plot3(rP(2,1),rP(2,2),rP(2,3),'rd')
plot3([0;rP(1,1)],[0;rP(1,2)],[0;rP(1,3)],'k')
plot3([0;a(1)],[0;a(2)],[0;a(3)],'k')
plot3([0;b(1)],[0;b(2)],[0;b(3)],'k')
quat — Quaternion that defines rotation
Quaternion that defines rotation, specified as a scalar quaternion, row vector of quaternions, or column vector of quaternions.
cartesianPoints — Three-dimensional Cartesian points
1-by-3 vector | N-by-3 matrix
Three-dimensional Cartesian points, specified as a 1-by-3 vector or N-by-3 matrix.
rotationResult — Repositioned Cartesian points
Rotated Cartesian points defined using the quaternion rotation, returned as a vector or matrix the same size as cartesianPoints.
Quaternion point rotation rotates a point specified in R3 according to a specified quaternion:
{L}_{q}\left(u\right)=qu{q}^{*}
where q is the quaternion, * represents conjugation, and u is the point to rotate, specified as a quaternion.
For convenience, the rotatepoint function takes in a point in R3 and returns a point in R3. Given a function call with some arbitrary quaternion, q = a + bi + cj + dk, and arbitrary coordinate, [x,y,z], for example,
rereferencedPoint = rotatepoint(q,[x,y,z])
the rotatepoint function performs the following operations:
Converts point [x,y,z] to a quaternion:
{u}_{q}=0+xi+yj+zk
Normalizes the quaternion, q:
{q}_{n}=\frac{q}{\sqrt{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}}
Applies the rotation:
{v}_{q}=q{u}_{q}{q}^{*}
Converts the quaternion output, vq, back to R3
rotateframe |
Determine the following indefinite integral. \int (x^{8}-3x^{3}+1)dx
\int \left({x}^{8}-3{x}^{3}+1\right)dx
First we separate the integral.
\int \left({x}^{8}-3{x}^{3}+1\right)dx
=\int {x}^{8}dx-3\int {x}^{3}dx+\int dx
Then we integrate each part
\int \left({x}^{8}-3{x}^{3}+1\right)dx
=\int {x}^{8}dx-3\int {x}^{3}dx+\int dx
=\frac{{x}^{9}}{9}-3\frac{{x}^{4}}{4}+x+C
=\frac{{x}^{9}}{9}-\frac{3{x}^{4}}{4}+x+C
\frac{{x}^{9}}{9}-\frac{3{x}^{4}}{4}+x+C
\int \left({x}^{8}-3{x}^{3}+1\right)dx
\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+C
\frac{{x}^{9}}{9}-\frac{3{x}^{4}}{4}+x
\frac{{x}^{9}}{9}-\frac{3{x}^{4}}{4}+x+C
I am trying to evaluate the integral
{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{log}\left({e}^{ix}+{e}^{-ix}\right)}{-}\mathrm{log}\left\{2\right\}\right\}\left\{{e}^{2ix}-1\right\}{e}^{ix}dx
I am wondering if a complex variable substitution the likes of
iu={e}^{ix}
du={e}^{ix}dx
is justifiable? In my mind. when x=0,u=0 and when
x=\frac{\pi }{2},u=1
which would make the integral
-{\int }_{0}^{1}\frac{\mathrm{log}\left(iu-\frac{i}{u}\right)-\mathrm{log}\left(2\right)}{1+{u}^{2}},du
{\int }_{1,1,2}^{3,5,0}yzdx+xzdy+xydz
\int \frac{2}{3t\left(\mathrm{ln}\left(4t\right)-1\right)}dt
Evaluate the integral and check answer by differentiating.
\int x\left(1+{x}^{3}\right)dx
\frac{4{e}^{{\mathrm{tan}}^{-1}\left(t\right)}}{7\left({t}^{2}+1\right)}
\int \frac{x{e}^{2x}}{{\left(2x+1\right)}^{2}}dx
\int \frac{x}{\sqrt{x-4}}dx |
Machine Learning Summer School 2015 Kyoto #MLSSKYOTO | Takuya Kitazawa
Home > Blog > Machine Learning > Machine Learning Summer School 2015 Kyoto #MLSSKYOTO
Hi, I am takuti, a master's student in Japan. Currently, I am working on matrix factorization and approximation. Also, my research interests are in web engineering, mining and their applications such as recommender systems.
I have attended to Machine Learning Summer School 2015 Kyoto (MLSS'15) from August 23 to September 4. This entry briefly reviews each of 14 exciting lectures in the summer school. Note that there might be mistakes in the content. In that case, please let me know via comments or twitter.
Slides for the lectures are available at: http://www.iip.ist.i.kyoto-u.ac.jp/mlss15/doku.php?id=schedule
(with strong emphasis on regularization and LASSO)
MLSS Sensu
Stephen P. Boyd, Stanford
This is one of the most impressive lectures in MLSS'15 because Prof. Boyd provided us really interesting and stimulating lecture; it was like a show. He interactively discussed key ideas and specific applications of convex optimization problems. Also, thanks to Convex Optimization Short Course, we can easily learn and try to solve real convex optimization problems on our laptop after MLSS.
As mentioned in the above tweet, optimization problems are everywhere in machine learning field. How can we find optimal weights? Which cluster assignment is the best? What is the best approximation? All of the answers can be found from optimization problems. Hence, the MLSS'15 lecture schedule seems to be well-organized (that's why this lecture is the first one in this summer school).
Next: Try Convex Optimization Short Course
Gábor Lugosi, Pompeu Fabra
This topic seemed to be super advanced mathematics, and the lecture was really hard to follow. In particular, computer scientists who are familiar with engineering aspects were easily left behind, because we were not able to imagine specific applications from lots of equations/proofs.
While at the same time, I can feel how concentration inequalities, inequalities for bounding random fluctuations $P\{Z \leq t\}$, are important in machine learning field. In fact, modern machine learning techniques strongly depend on probabilistic approaches (random fluctuations), so importance of guarantees for the fluctuations is clear.
Next: Read Concentration Inequalities: A Nonasymptotic Theory of Independence
Emmanuel Candès, Stanford
His lecture started from a discussion about reliability of science. In my opinion, this was good introduction to show how selective inference can be an important field to discuss experimental results on so-called "BigData". Currently, we will first collect huge data before asking questions and planning experiments, so reliability of inference from BigData is a crucial problem.
I am especially interested in Knockoff Machines, a simple algorithm to select statistically favorable hypothesis from many different hypotheses. For the reliability of inference, we need to decrease false discovery rate (FDR), and Knockoff Machines allow us to select hypotheses which give low FDR. This algorithm is also promising in terms of feature selection. When we would like to choose better features from many different features, Knockoff Machines-like algorithm can choose features which give low false positives. In the future, I would like to survey on this application.
Next: Learn more about Knockoff Machines from the lecture slide and related online sources
This lecture was really different from the others. At first, I was afraid of a term "programming"; I imagined this indicates something like "linear programming" stuff, but this was not true. The content focused on writing probabilistic code, especially in ProbLog, extension of Prolog. Since I am already familiar with Prolog, I really enjoyed the lecture.
One of the most important things of ProbLog is that this programming language can infer from predicates with uncertainty. When we tackle AI problems, programming languages need to deal uncertainty in it; that is, dealing uncertainty is important problem to represent human intuition.
Next: Try probabilistic programming in ProbLog
Stefanie Jegelka, MIT
I heard a term submodular many times before, but I did not know what submodular actually is. The answer was in this lecture; this lecture started from definition of set functions and submodularity with good intuitive examples. After such introduction, she gradually moved to mathematical details, optimization and approximation of submodular functions.
One of the most interesting applications I have learnt from this lecture is summarization. This idea assumes that summarization is just choosing subset of words from target article; that is, we can summarize articles by maximizing set function for words in terms of coverage or relevance. This idea is very interesting, and I am also interested in incremental summary updating and application to recommender systems. I would like to learn more about this field.
Next: Read one curious paper found on the web "Streaming Submodular Maximization: Massive Data Summarization on the Fly"
Statistical and Computational Aspects of High-Dimensional Learning
We needed to highly focus on Statistical and Computational Aspects in the title. The content consisted of mathematical discussion in these aspects. Even though the professor first introduced well-known technique, Principal Component Analysis, I was not able to follow the lecture due to difficult mathematical discussion.
I was only able to understand what high-dimensinal setting is; high-dimensional setting is situation such that the number of parameters is much bigger than the number of observations. However, what is more concrete difference between high-dimensional learning and usual machine learning? What is the key idea in high-dimensional learning? Which points are new in the field? To grasp these points, I should learn more before listing this lecture.
Next: First and foremost, I need to review this lecture :(
Lorenzo Rosasco, MIT / Genoa
\mathbf{\Phi}
X \rightarrow F
, mapping data space
X
to representation space
F
. Keeping in mind this point really helps us to understand a wide variety of ideas in the field. This course was really well-organized because the professor taught several different representations one-by-one and summarized them from practical viewpoint.
Representations discussed in the lecture were: Dictionary, Random Projection, Kernel and Deep Representation. All of them were scientifically stimulating, but which representation is appropriate to my problem? This lecture covered such practical points, and important thing is Good representation decreases complexity of samples. Currently, I am curious about online dictionary learning, so this lecture was good introduction.
Next: Review dictionary learning from online sources such as Sparse Coding Course, and its online setting
Alexander J. Smola, CMU
Are there anybody who are studying in engineering department? OK, it's time for you!
Two-thirds of this lecture discussed computer and system architecture in the BigData era. Since I am definitely curious about engineering and applicational aspects in machine learning rather than theoretical, mathematical points, the lecture was quite exciting.
This course has three parts: (1) efficient machine learning on single machine, (2) distributed machine learning, and (3) system architecture for Deep Learning. The last part was pale compared to the others, because it was just overview of well-known recent studies on Deep Learning. During the introduction, the professor said things like You don't have to be a professional hardware researcher, but please care about the efficiency of memory usage. This is very important point in recent large-scale machine learning.
Next: Check his lectures on YouTube channel, and try Deep Learning techniques by my hand on cloud environment
In a weekend, I visited Arashiyama and saw famous bamboo forest.
Csaba Szepesvári, Alberta
Even though I can see many interesting articles and videos related to reinforcement learning on the web, I have never learnt about theoretical aspect of the technique. So, this field was a huge black box for me.
The professor said that reinforcement learning has all of machine learning problems such as regression, classification and density estimation. This is true; I confirmed this statement through the lecture. However, after taking this short course, my understanding of reinforcement learning is still unclear.
Next: Read Algorithms for Rainforcement Learning, free pdf written by the lecturer
Zaid Harchaoui, NYU/INRIA
What kind of lecture are you expect from this title? Your thought is probably wrong.
This lecture focused on some specific theories such as Convolutional Kernel Networks and Stochastic Gradient Descent, so we were not able to obtain practical techniques in computer vision. Of course, these theories are really important to realize better application, but the title "Machine Learning for Computer Vision" was definitely inappropriate.
After this course, I shortly discussed with one Japanese researcher who is working on computer vision in a research institute. He also said the same things, and he actually expected more practical topic like 3D reconstruction. Due to this disappointing situation, I was not able to enjoy this lecture.
Next: Learn and try Convolutional Neural Networks and Convolutional Kernel Networks
Ryota Tomioka, TTI Chicago
As I said at the beginning, I am currently working on machine learning for matrices, so I really looked forward to this lecture. And, this lecture was actually scientifically stimulating for me. Most importantly, the lecture was well-organized and really easy to follow. For example, he used enough time to explain what rank of matrices is, and he repeatedly emphasized this fundamental point is important to understand tensors.
However, tensor decomposition and its related theory/application are wide-ranging, and, in this lecture, the professor just focused on one of them, theoretical analysis of tensor decomposition. When we need to use tensors in a practical problem, you should learn more from different studies beyond this lecture.
Next: Try tensor decomposition on real-world data set and learn more applications
Taiji Suzuki, Tokyo Tech
Since I am interested in online learning, Stochastic Gradient Descent is one of my curious topics. However, this lecture has too much mathematical formulas. I was not able to follow at all.
In fact, most lectures in MLSS strongly emphasized theoretical, mathematical aspects of recent machine learning studies, but this lecture was too much. I would like to learn more about this topic from the other sources.
Next: Understand Stochastic Gradient Descent theoretically
Vincent Vanhoucke, Google
This was very good introductory lecture on Neural Networks and Deep Learning. We were able to obtain brief overview of recent significant achievements and theoretical basics in the field.
Most importantly, he first emphasized Deep Learning is not the only way to get sufficient results in real-world machine learning problems. Actually, on Kaggle, other techniques quite often win. So, try Rogistic Regression, Gradient Boosting and Random Forest first! This is the message from the lecture.
Next: Review basic techniques (rogistic regression, random forest, gradient boosting) and attend data competision with them :)
Statistical Guarantees in Optimization
Martin Wainwright, Berkeley
The last lecture focused on sketching, and first started from Johnson–Lindenstrauss lemma. During MLSS, since some professors introduced J-L lemma in their lecture, I understood how the lemma has an important role to approximate data. So, this notice strongly motives for learning these theoretical fundamentals.
He introduced his latest work called Newton Sketch in the lecture. This paper seems very interesting and exciting, so I would like to read later.
Next: Learn more about statistical guarantees with starting from Johnson–Lindenstrauss lemma, and read Newton Sketch
Overall, MLSS was really, really exciting two weeks. We were able to briefly see all of machine learning field, and these knowledge definitely leads a new research idea!
MLSS also posed a shade of anxiety to us by difficult math, but I guess we do not have to think too much about that. In machine learning, key ideas and essential problems we need to solve are usually quite simple. So, when we tackle own problem, let us start from the simplest approach, and gradually extend it.
Finally, one thing I would like to introduce is one cool podcast channel talking about machine learning called Talking Machines. I found this podcast after MLSS, and listening the excellent talks is intellectually stimulating. If you have not subscribed it yet, go ahead!
MLSS Masu, Japanese traditional sake glass |
The solution of the equations or inequalities and express it
The solution of the equations or inequalities and express it in interval notatio
The solution of the equations or inequalities and express it in interval notation where approptiate.
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ y+1
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ 2y+1
a. Inequality is
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ 2y+1
Solution is given as:
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-3y-6\succ 2y+1
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-3y+2y>6+1
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-y>7
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }y<-7
Solution in interval notation form is
8,-7?\left(2,8\right)
b. Inequality is
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ 2y+1
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-3y-6\succ 2y+1
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-3y+2y>6+1
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-y>7
y>2\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }y<-7
8,-7n\left(2,8\right)
Conclusion: Hence, the solutions for the below equations and inequalities are as follows:
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ 2y+1
Solution is
y<-7\text{ }\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\text{ }y>2
4<2y\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }-3\left(y+2\right)\succ 2y+1
y\succ 7\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }y>2
Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a 1.50 kg ball swings downward and strikes a 4.60kg ballthat is at rest, as the drawing shows. a. using the principle of conservation of mechanicalenergy,find the speed of the 1.50 kg ball just before impact b. assuming that the collision is elastic, find the velocities( magnitude and direction ) of both balls just after thecollision c. how high does each abll swing after the collision, ignoringair resistance?
The boom weighs 2600N. The boom is attached with a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from the pivot is 35% of its length. a) Find the tension in the guy wire and the horizontal and vertical components of the force exerted on the boom at its lower end. b) Does the line of action of the force exerted on the boom a tits lower end lie along the boom?
Find the tension in each cord in Figure if the weight ofthe suspended object is w.
An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers. the plane is traveling horizontally at
30.0m/s
at a height of
200.0m
above the ground.
A)What horizontal distance does the package fall before landing?
B)Find the velocity of the package just before it hits the ground.
\frac{x}{x+1}+\frac{x+1}{x}=\frac{34}{15},x\ne 0,x\ne -1
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by the equation
x\left(t\right)=\alpha {t}^{2}-\beta {t}^{3}
\alpha =1.50\frac{m}{{s}^{2}}\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }\beta =0.0500\frac{m}{{s}^{3}}
. Calculate the average velocity of the car for each time interval: (a)
t=0\text{ }\to \text{ }t=2.00s
t=0\text{ }\to \text{ }t=4.00s
t=2.00s\text{ }\to \text{ }t=4.00s
A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence. With 800 m of wire at your disposal, what is the largest area you can enclose, and what are its dimensions? |
Understanding Basic Traveling Waves | Brilliant Math & Science Wiki
A Former Brilliant Member, Matt DeCross, Jimin Khim, and
After learning the basics of periodic motion, it's time to take up the study of oscillations to the next level. By the next level, I mean the dependence of the motion of the considered particle on not just the time variable, but also on the distance variable. Let's dive right into the topic by first differentiating waves.
What do we mean by travelling waves?
Equation Representing a Travelling Wave
We all have heard about various kinds of of waves traveling around us. Starting from the most familiar waves of light and sound, to the complex matter waves, all of them follow a common feature, i.e. oscillation of energy.
That is what a wave is.
A wave is just the phenomenon of oscillation of energy, using various properties of a medium such as physical, electro-magnetic, etc. For example, sound waves and light waves are both the carriers of energy, but a sound wave propagates through pressure variations, whereas a light wave travels by making use of electro-magnetic phenomena, which we'll discuss shortly.
Differentiating waves on alignment of propagation with respect to oscillation:
Longitudinal waves: The type of waves that move in such a way that the oscillation of energy is along the direction of motion of the wave is defined as a longitudinal wave. Such kind of wave can be viewed by imagining two friends, who are walking forward, one ahead of the other, such that they continue tossing a ball between each other. In this, their motion can be seen as the motion of a wave and the ball acts as the packet of energy. These waves are also known as pressure waves.
Transverse waves: The type of waves that move in such a way that the oscillation of energy is perpendicular to the direction of the motion of the wave is defined as transverse wave. Such kind of wave can be pictured by imagining the same two friends, walking forward, but one beside another, such that they continue tossing a ball between each other. In this, their motion can be seen as the motion of a wave and the ball acts as the packet of energy.
Imagine stretching a string and fixing both its ends on two points. Now grab the mid point of the string and pull it down, before letting it go. Chances are that you'll see the mid point of the string oscillating with an amplitude, with the end points fixed at their respective positions. These kind of waves are what we call the standing waves.
Now, for the next experiment, get into a hall with your friend and call out to him. If you shout loud enough and your friend hears well, chances are that he'll hear your call. Your voice reached him by the motion of sound waves, i.e. travelling waves. Had the sound waves been stationary, your voice would have never reached him.
We all have read about the basic mathematical equation that governs the trajectory or the position of a particle in SHM, or in other words, a particle belonging to a stationary wave. It is given by
y(t)=a\sin \omega t
a
represents the maximum displacement of the particle from the mean position (amplitude), and
\omega
represents the angular frequency for the SHM.
Now, since a travelling wave also moves forward while changing with time, a similar equation in case of a travelling wave must definitely include a function of both the direction of propagation (let it be
z
) and time. So, we get:
y\left( z,t \right) =a\sin [z,t]
, where the symbols have their usual meanings.
To find this function that will explain the oscillation of the particle, we use the basic property of linearity of the function, i.e. the function enclosed inside the
\sin
block must be a linear function of
z
t,
or else the oscillating graph would lose its shape and the wave would compress or stretch out in different locations.
So, let that be given by
\Phi \left( z,t \right)=\alpha z+\beta t
, but since we assume the wave to be travelling towards
z=+\infty
\alpha
\beta
must have opposite signs. Therefore,
\Phi \left( z,t \right)=|\alpha |z-|\beta |t.
Also, we already know that this function must have the dimensions of radians, so
\beta
T^{-1}
\alpha
L^{-1}
. As it turns out,
\alpha
is given by a constant
k
that is known as the wave-number for the wave, and equals
\frac{2\pi}{\lambda}
\lambda
is the wavelength of the wave. On the other hand,
\beta
is our age old friend
\omega
2\pi\nu
\nu
is the frequency.
So, we end up with:
y\left( z,t \right) =a\sin ({k z-\omega t}).
Note: If the wave travels towards
-\infty
, the function would change to
y\left( z,t \right) =a\sin ({k z+\omega t}) .
Cite as: Understanding Basic Traveling Waves. Brilliant.org. Retrieved from https://brilliant.org/wiki/understanding-basic-traveling-waves/ |
Flower Garden - NZOI Training
Hannah's flower garden is a row of N
N
flowers. All of the flowers in her garden are either red, green, or blue in colour.
She would like to replace some of these flowers so that no two neighbouring flowers share the same colour.
A single replacement consists of replacing any flower with a new flower of a different colour (either red, green, or blue).
Please help Hannah find the minimum number of replacements required.
The first line contains a single integer N
N
, the number of flowers in Hannah's garden.
The second line contains a string of length N
N
, representing the colours of the flowers in the garden in order from left to right. Each character of the string is either 'R', 'G', or 'B', corresponding to red, green, and blue respectively.
You should print a single integer – the minimum number of replacements required to change the garden in such a way that no two neighbouring flowers share the same colour.
For all subtasks, 1 \le N \le 100
1 \le N \le 100
Subtask 1 (30%): All flowers are the same colour.
Subtask 2 (30%): No flower shares the same colour as both of its neighbours.
Subtask 3 (40%): No further constraints apply.
In the first sample case, only a single replacement is necessary – the garden can be changed to "RGR" or "RBR" by replacing the second flower with either a green or blue flower.
In the second sample case, at least two replacements are required. One possible solution is to change the third flower to blue, and the fourth flower to green, resulting in "RGBGB".
RGGBB |
Gamma probability density function - MATLAB gampdf - MathWorks Deutschland
Compute Gamma pdf
Gamma pdf
Gamma probability density function
y = gampdf(x,a)
y = gampdf(x,a,b)
y = gampdf(x,a) returns the probability density function (pdf) of the standard gamma distribution with the shape parameter a, evaluated at the values in x.
y = gampdf(x,a,b) returns the pdf of the gamma distribution with the shape parameter a and the scale parameter b, evaluated at the values in x.
Compute the density of the observed value 5 in the standard gamma distribution with shape parameter 2.
y1 = gampdf(5,2)
Compute the density of the observed value 5 in the gamma distributions with shape parameter 2 and scale parameters 1 through 5.
y2 = gampdf(5,2,1:5)
If one or more of the input arguments x, a, and b are arrays, then the array sizes must be the same. In this case, gampdf expands each scalar input into a constant array of the same size as the array inputs. Each element in y is the pdf value of the distribution specified by the corresponding elements in a and b, evaluated at the corresponding element in x.
The gamma distribution is a two-parameter family of curves. The parameters a and b are shape and scale, respectively.
The gamma pdf is
y=f\left(x|a,b\right)=\frac{1}{{b}^{a}\Gamma \left(a\right)}{x}^{a-1}{e}^{\frac{-x}{b}},
The standard gamma distribution occurs when b = 1.
For more information, see Gamma Distribution.
gampdf is a function specific to the gamma distribution. Statistics and Machine Learning Toolbox™ also offers the generic function pdf, which supports various probability distributions. To use pdf, create a GammaDistribution probability distribution object and pass the object as an input argument or specify the probability distribution name and its parameters. Note that the distribution-specific function gampdf is faster than the generic function pdf.
GammaDistribution | pdf | gamcdf | gaminv | gamstat | gamfit | gamlike | gamrnd |
Design parametric equalizer - MATLAB designParamEQ - MathWorks Australia
designParamEQ
Design Two-Band Parametric Equalizer
Filter Audio Using SOS Parametric Equalizer
Filter Audio Using FOS Parametric Equalizer
Design parametric equalizer
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth)
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth,mode)
[B,A] = designParamEQ(___,Name,Value)
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth) designs an Nth-order parametric equalizer with specified gain, center frequency, and bandwidth. B and A are matrices of numerator and denominator coefficients, with columns corresponding to cascaded second-order section (SOS) filters.
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth,mode) specifies whether the parametric equalizer is implemented with second-order sections or fourth-order sections (FOS).
[B,A] = designParamEQ(___,Name,Value) specifies options using one or more Name,Value pair arguments.
Specify the filter order, peak gain in dB, normalized center frequencies, and normalized bandwidth of the bands of your parametric equalizer.
N = [ 2, ...
gain = [ 6, ...
centerFreq = [ 0.25, ...
bandwidth = [ 0.12, ...
Generate the filter coefficients using the specified parameters.
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth,"Orientation","row");
fvtool([B,A]);
Design a second-order sections (SOS) parametric equalizer using designParamEQ and filter an audio stream.
Create audio file reader and audio device writer System objects. Use the sample rate of the reader as the sample rate of the writer.
fileReader = dsp.AudioFileReader("RockGuitar-16-44p1-stereo-72secs.wav","SamplesPerFrame",frameSize);
deviceWriter = audioDeviceWriter("SampleRate",sampleRate);
Design an SOS parametric equalizer suitable for use with dsp.BiquadFilter.
N = [4,4];
gain = [-25,35];
centerFreq = [0.01,0.5];
bandwidth = [0.35,0.5];
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth);
Visualize your filter design. Call designParamEQ with the same design specifications. Specify the output orientation as "row" so that it is suitable for use with fvtool.
[Bvisualize,Avisualize] = designParamEQ(N,gain,centerFreq,bandwidth,"Orientation","row");
fvtool([Bvisualize,Avisualize], ...
"Fs",fileReader.SampleRate, ...
Create a spectrum analyzer to visualize the original audio signal and the audio signal passed through your parametric equalizer.
"SampleRate",sampleRate, ...
"PlotAsTwoSidedSpectrum",false, ...
"FrequencyScale","Log", ...
"FrequencyResolutionMethod","WindowLength", ...
"WindowLength",frameSize, ...
"Title","Original and Equalized Signals", ...
"ChannelNames",{'Original Signal','Equalized Signal'});
Design a fourth-order sections (FOS) parametric equalizer using designParamEQ and filter an audio stream.
Construct audio file reader and audio device writer System objects. Use the sample rate of the reader as the sample rate of the writer.
Design FOS parametric equalizer coefficients.
gain = [5,10];
centerFreq = [0.025,0.65];
bandwidth = [0.025,0.35];
mode = "fos";
[B,A] = designParamEQ(N,gain,centerFreq,bandwidth,mode,Orientation="row");
Construct FOS IIR filters.
myFilter = dsp.FourthOrderSectionFilter(B,A);
Visualize the frequency response of your parametric equalizer.
fvtool(myFilter)
Construct a spectrum analyzer to visualize the original audio signal and the audio signal passed through your parametric equalizer.
Title="Original and Equalized Signals", ...
Play the filtered audio signal and visualize the original and filtered spectra.
y = myFilter(x);
scope([x(:,1),y(:,1)]);
Filter order, specified as a scalar or row vector the same length as centerFreq. Elements of the vector must be even integers.
Peak gain in dB, specified as a scalar or row vector the same length as centerFreq. Elements of the vector must be real-valued.
centerFreq — Normalized center frequency of equalizer bands
Normalized center frequency of equalizer bands, specified as a scalar or row vector of real values in the range 0 to 1, where 1 corresponds to the Nyquist frequency (π rad/sample). If centerFreq is specified as a row vector, separate equalizers are designed for each element of centerFreq.
bandwidth — Normalized bandwidth
Normalized bandwidth, specified as a scalar or row vector the same length as centerFreq. Elements of the vector are specified as real values in the range 0 to 1, where 1 corresponds to the Nyquist frequency (π rad/sample).
Normalized bandwidth is measured at gain/2 dB. If gain is set to -Inf (notch filter), normalized bandwidth is measured at the 3 dB attenuation point:
10×{\mathrm{log}}_{10}\left(0.5\right)
To convert octave bandwidth to normalized bandwidth, calculate the associated Q-factor as
Q=\frac{\sqrt{{2}^{\left(octave\text{\hspace{0.17em}}bandwidth\right)}}}{{2}^{\left(octave\text{\hspace{0.17em}}bandwidth\right)}-1}\text{\hspace{0.17em}}.
Then convert to bandwidth
bandwidth=\frac{centerFreq}{Q}\text{\hspace{0.17em}}.
mode — Design mode
'sos' (default) | 'fos'
Design mode, specified as 'sos' or 'fos'.
'sos' –– Implements your equalizer as cascaded second-order filters.
'fos' –– Implements your equalizer as cascaded fourth-order filters. Because fourth-order sections do not require the computation of roots, they are generally more computationally efficient.
Numerator filter coefficients, returned as a matrix. The size and interpretation of B depends on the Orientation and mode:
If 'Orientation' is set to "column" and mode is set to "sos", then B is returned as an L-by-3 matrix. Each column corresponds to the numerator coefficients of your cascaded second-order sections.
If 'Orientation' is set to "column" and mode is set to "fos", then B is returned as an L-by-5 matrix. Each column corresponds to the numerator coefficients of your cascaded fourth-order sections.
If 'Orientation' is set to "row" and mode is set to "sos", then B is returned as a 3-by-L matrix. Each row corresponds to the numerator coefficients of your cascaded second-order sections.
If 'Orientation' is set to "row" and mode is set to "fos", then B is returned as a 5-by-L matrix. Each row corresponds to the numerator coefficients of your cascaded fourth-order sections.
Denominator filter coefficients, returned as a matrix. The size and interpretation of A depends on the Orientation and mode:
If 'Orientation' is set to "column" and mode is set to "sos", then A is returned as an L-by-2 matrix. Each column corresponds to the denominator coefficients of your cascaded second-order sections. A does not include the leading unity coefficients.
If 'Orientation' is set to "column" and mode is set to "fos", then A is returned as an L-by-4 matrix. Each column corresponds to the denominator coefficients of your cascaded fourth-order sections. A does not include the leading unity coefficients.
If 'Orientation' is set to "row" and mode is set to "sos", then A is returned as a 3-by-L matrix. Each row corresponds to the denominator coefficients of your cascaded second-order sections.
If 'Orientation' is set to "row" and mode is set to "fos", then A is returned as a 5-by-L matrix. Each row corresponds to the denominator coefficients of your cascaded fourth-order sections.
designVarSlopeFilter | designShelvingEQ | multibandParametricEQ | dsp.BiquadFilter |
Hybrid nanofillers consisting of MWCNT (multiwalled carbon nanotubes) functionalized by ZrO
{}_{2}
nanoparticles have been infused in an epoxy matrix to fabricate MWCNT/ZrO
{}_{2}
hybrid epoxy nanocomposites (MNC) using an ultrasonication mixing technique.
The structural morphology of the hybrid nanofillers and their dispersion in the epoxy matrix, were studied using transmission electron microscopy and field emission scanning electron microscopy, respectively. The surface chemical composition of the MWCNT, ZrO
{}_{2}
nanoparticles and hybrid nanofillers were studied using infrared spectroscopy and the elemental composition of the hybrid nanofillers was determined using EDX analysis. |
Reduce order of differential equations to first-order - MATLAB odeToVectorField
\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}+{\mathit{y}}^{2}\mathit{t}=3\mathit{t}.
\left(\begin{array}{c}{Y}_{2}\\ 3 t-t {{Y}_{1}}^{2}\end{array}\right)
The elements of V represent the system of first-order differential equations, where V[i] =
{{Y}_{i}}^{\prime }
{Y}_{1}=y
. Here, the output V represents these equations:
\frac{\mathit{d}{\mathit{Y}}_{1}}{\mathit{dt}}={\mathit{Y}}_{2}
\frac{{\mathit{dY}}_{2}}{\mathit{dt}}=3\mathit{t}-\mathit{t}{\mathit{Y}}_{1}^{2}.
\left(\begin{array}{c}{Y}_{2}\\ {Y}_{1}+{Y}_{3}\\ {Y}_{3}-{Y}_{1}\end{array}\right)
\left(\begin{array}{c}f\\ \mathrm{Df}\\ g\end{array}\right)
{{Y}_{i}}^{\prime }
. The output S shows the substitutions being made, S[1] =
{Y}_{1}=f
, S[2] =
{Y}_{2}
= diff(f), and S[3] =
{Y}_{3}=g
\frac{{\mathit{d}}^{2}\mathit{y}}{\mathit{d}{\mathit{t}}^{2}}=\left(1-{\mathit{y}}^{2}\right)\frac{\mathit{dy}}{\mathit{dt}}-\mathit{y}.
\left(\begin{array}{c}{Y}_{2}\\ -\left({{Y}_{1}}^{2}-1\right) {Y}_{2}-{Y}_{1}\end{array}\right)
Specify the solution interval to be [0 20] and the initial conditions to be
{y}^{\prime }\left(0\right)=2
{y}^{\prime \prime }\left(0\right)=0
. Solve the system of first-order differential equations by using ode45.
Next, plot the solution
y\left(t\right)
t
= [0 20]. Generate the values of t by using linspace. Evaluate the solution for
y\left(t\right)
, which is the first index in ySol, by calling the deval function with an index of 1. Plot the solution using plot.
Convert the second-order differential equation
{y}^{\prime \prime }\left(x\right)=x
y\left(0\right)=a
to a first-order system.
\left(\begin{array}{c}{Y}_{2}\\ x\end{array}\right)
{a}_{n}\left(t\right){y}^{\left(n\right)}+{a}_{n-1}\left(t\right){y}^{\left(n-1\right)}+\dots +{a}_{1}\left(t\right){y}^{\prime }+{a}_{0}\left(t\right)y+r\left(t\right)=0
\begin{array}{l}{Y}_{1}=y\\ {Y}_{2}={y}^{\prime }\\ {Y}_{3}={y}^{″}\\ \dots \\ {Y}_{n-1}={y}^{\left(n-2\right)}\\ {Y}_{n}={y}^{\left(n-1\right)}\end{array}
\begin{array}{l}{Y}_{1}{}^{\prime }={y}^{\prime }={Y}_{2}\\ {Y}_{2}{}^{\prime }={y}^{″}={Y}_{3}\\ \dots \\ {Y}_{n-1}{}^{\prime }={y}^{\left(n-1\right)}={Y}_{n}\\ {Y}_{n}{}^{\prime }=-\frac{{a}_{n-1}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{n}-\frac{{a}_{n-2}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{n-1}-...-\frac{{a}_{1}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{2}-\frac{{a}_{0}\left(t\right)}{{a}_{n}\left(t\right)}{Y}_{1}+\frac{r\left(t\right)}{{a}_{n}\left(t\right)}\end{array} |
Properties of Real Numbers | Brilliant Math & Science Wiki
Ivan Koswara, Mehul Arora, Rajdeep Bharati, and
This page is under construction. You can help us by adding examples of what you think should be on this page.
This page was the subject of a [wiki collaboration party].
A real number is one whose square is positive.
(https://docs.google.com/spreadsheets/d/1t6AckNp5MJ115DJ6Wo7kifqVGUkvzx6NEpSu_Fk1nQE/edit#gid=133220531) at #mathematics, on 20 September 02.30am UTC.
\huge{.999 \ldots = 1}
The above equation is a very common question that has misled many people over years. It even got a lengthy Wikipedia page. Is it true? They are two different numbers, no?
Surprisingly, it is true! You might have known this popular problem and hence memorized the answer. But do you know why this is true? More surprisingly, we can make this equation false! If we work on something other than the real numbers, that is. The above equation hinges on a very important property of the real numbers.
Outline dependencies
Geometric progression formula and
.999 \ldots = 1
x\in R
. Show that there exists an integer
m
m\le x < m+1
l
x<l\le x+1
\alpha,\beta
are two real numbers satisfying
\alpha<\beta
. Show that there exist
n_1,n_2 \in N
\alpha<\alpha+\frac{1}{n_1}<\beta
\alpha<\beta-\frac{1}{n_2}<\beta.
.999 \ldots = 1
0.999 \ldots
, being a number represented in decimal form, has the value
\displaystyle 0 + \sum_{n=1}^\infty 9 \cdot \left( \frac{1}{10} \right)^n
. This is a consequence of any number that is represented in decimal: if the number is
\overline{k.b_1 b_2 b_3 \ldots},
k
b_1, b_2, b_3, \ldots \in \{0,1,2,\ldots,9\},
the value of any such number is
\displaystyle k + \sum_{n=1}^\infty b_n \cdot \left( \frac{1}{10} \right)^n
by definition. So we'd like to compute the value of
\displaystyle S = \sum_{n=1}^\infty 9 \cdot \left( \frac{1}{10} \right)^n
A geometric sequence is a sequence of real numbers
\{a_n\}_{n=1}^\infty
that satisfies the property
a_{n+1} = a_n \cdot r
for some real number
and all positive integers
n
. It can be shown by simple induction that if
\{a_n\}
forms a geometric sequence, there exist real numbers
and
r
a_n = ar^{n-1}
n
. (Here we take
0^0 = 1
.) A geometric series (or geometric progression) is a series whose elements form a geometric sequence, i.e. a series in the form
\displaystyle \sum_{n=1}^\infty ar^{n-1}
S
is a geometric sequence with
a = \frac{9}{10}, r = \frac{1}{10}
To compute the value of this series, we can take a look at its partial sums
\displaystyle \sum_{n=1}^k ar^{n-1}
. This form is called a finite geometric series. Being the sum of a finite number of real numbers, we can apply the commutative and associative properties of addition and the distributive property of multiplication over addition, to manipulate the sum as such:
\displaystyle\begin{aligned} S_k &= \sum_{n=1}^k ar^{n-1} \\ r S_k &= r \cdot \sum_{n=1}^k ar^{n-1} \\ &= \sum_{n=1}^k \left(r \cdot ar^{n-1}\right) \\ &= \sum_{n=1}^k ar^n \\ &= \sum_{n=2}^{k+1} ar^{n-1} \\ S_k - r S_k &= \left( \sum_{n=1}^k ar^{n-1} \right) - \left( \sum_{n=2}^{k+1} ar^{n-1} \right) \\ &= ar^0 + \sum_{n=2}^k \left(ar^{n-1} - ar^{n-1}\right) - ar^k \\ &= a\left(1-r^k\right) \\ S_k &= \frac{a\left(1-r^k\right)}{1-r} \\ &= \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k. \end{aligned}
By definition, the value of the series is simply the limit of its partial sums. Thus we'd like to compute
\displaystyle \lim_{k \to \infty} S_k
\displaystyle \lim_{k \to \infty} \left( \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k \right)
It's an easy exercise to show that if
\{b_n\}
is a sequence of real numbers that converges to a limit
\displaystyle \lim_{n \to \infty} b_n
c
is a constant, then
\displaystyle \lim_{n \to \infty} (c + b_n) = c + \lim_{n \to \infty} b_n
\displaystyle \lim_{n \to \infty} (c \cdot b_n) = c \cdot \lim_{n \to \infty} b_n
\frac{a}{1-r}
-\frac{a}{1-r}
are constants, applying these properties gives:
\displaystyle\begin{aligned} \lim_{k \to \infty} S_k &= \lim_{k \to \infty} \left( \frac{a}{1-r} - \frac{a}{1-r} \cdot r^k \right) \\ &= \frac{a}{1-r} + \lim_{k \to \infty} \left( \left(- \frac{a}{1-r} \right) \cdot r^k \right) \\ &= \frac{a}{1-r} + \left(- \frac{a}{1-r} \right) \cdot \lim_{k \to \infty} r^k. \end{aligned}
Our problem has been reduced to determining
\displaystyle \lim_{k \to \infty} r^k
For the purpose of proving
0.999 \ldots = 1
, it suffices to restrict our attention to
0 < r < 1
. It is simple induction again to show that
0 < r^{k+1} < r^k
k
r^k > r^{k+1}
, we know that the sequence
\{r^k\}
is monotonically decreasing, and thus converges to a limit
L
by monotone convergence theorem. From the fact that
r^k > 0
L
cannot be less than
0
L < 0
, then simply take
\epsilon = -L
, and now there is no element of
\{r^k\}
that lies in
(L-\epsilon, L+\epsilon)
L+\epsilon = 0 < r^k
). Hence there can be no
N
r^k \in (L-\epsilon, L+\epsilon)
k > N
(the condition necessary for a convergent sequence to converge to the limit).
todo: show that
L > 0
is impossible
L < 0
L > 0
lead to a contradiction. The Archimedean property tells us that
L
cannot be infinitesimal either, since there is no non-zero infinitesimal in the real numbers. The remaining option is thus
L = 0
\displaystyle \lim_{k \to \infty} S_k = \frac{a}{1-r} + \left(- \frac{a}{1-r} \right) \cdot 0
\displaystyle \lim_{k \to \infty} S_k = \frac{a}{1-r}
; the geometric series, for
0 < r < 1
, gives the value
\frac{a}{1-r}
S
is a geometric series, with
a = \frac{9}{10}, r = \frac{1}{10}
0 < r < 1
, we can apply the result we obtained above to get
S = \frac{9/10}{1 - 1/10} = 1
0.999 \ldots = 1
Density of Q in R
Cite as: Properties of Real Numbers. Brilliant.org. Retrieved from https://brilliant.org/wiki/properties-of-real-numbers/ |
Detect small changes in mean using cumulative sum - MATLAB cusum - MathWorks Australia
cusum Default Values
Unstable Motion Detection
iupper,ilower
uppersum,lowersum
[iupper,ilower] = cusum(x)
[iupper,ilower] = cusum(x,climit,mshift,tmean,tdev)
[iupper,ilower] = cusum(___,'all')
[iupper,ilower,uppersum,lowersum] = cusum(___)
cusum(___)
[iupper,ilower] = cusum(x) returns the first index of the upper and lower cumulative sums of x that have drifted beyond five standard deviations above and below a target mean. The minimum detectable mean shift is set to one standard deviation. The target mean and standard deviations are estimated from the first 25 samples of x.
[iupper,ilower] = cusum(x,climit,mshift,tmean,tdev) specifies climit, the number of standard deviations that the upper and lower cumulative sums are allowed to drift from the mean. It also specifies the minimum detectable mean shift, the target mean, and the target standard deviation.
[iupper,ilower] = cusum(___,'all') returns all the indices at which the upper and lower cumulative sums exceed the control limit.
[iupper,ilower,uppersum,lowersum] = cusum(___) also returns the upper and lower cumulative sums.
cusum(___) with no output arguments plots the upper and lower cumulative sums normalized to one standard deviation above and below the target mean.
Generate and plot a 100-sample random signal with a linear trend. Reset the random number generator for reproducible results.
rnds = rand(1,100);
trnd = linspace(0,1,100);
fnc = rnds + trnd;
plot(fnc)
Apply cusum to the function using the default values of the input arguments.
cusum(fnc)
Compute the mean and standard deviation of the first 25 samples. Apply cusum using these numbers as the target mean and the target standard deviation. Highlight the point where the cumulative sum drifts more than five standard deviations beyond the target mean. Set the minimum detectable mean shift to one standard deviation.
mfnc = mean(fnc(1:25));
sfnc = std(fnc(1:25));
cusum(fnc,5,1,mfnc,sfnc)
Repeat the calculation using a negative linear trend.
nnc = rnds - trnd;
cusum(nnc)
Generate a signal resembling motion about an axle that becomes unstable due to wear. Add white Gaussian noise of variance 1/9. Reset the random number generator for reproducible results.
dr = airy(2,linspace(-14.9371,1.2,sz));
rd = dr + sin(2*pi*(1:sz)/5) + randn(1,sz)/3;
Plot the growing background drift and the resulting signal.
plot(dr)
plot(rd,'.-')
Find the mean and standard deviation if the drift is not present and there is no noise. Plot the ideal noiseless signal and its stable background.
id = 0.3*sin(2*pi*(1:sz)/20);
st = id + sin(2*pi*(1:sz)/5);
mf = mean(st)
mf = -3.8212e-16
sf = std(st)
sf = 0.7401
plot(id)
plot(st,'.-')
Use the CUSUM control chart to pinpoint the onset of instability. Assume that the system becomes unstable when the signal is three standard deviations beyond its ideal behavior. Specify a minimum detectable shift of one standard deviation.
cusum(rd,3,1,mf,sf)
Make the violation criterion more strict by increasing the minimum detectable shift. Return all instances of unwanted drift.
cusum(rd,3,1.2,mf,sf,'all')
Every hole in golf has an associated "par" that indicates the expected number of strokes needed to sink the ball. Skilled players usually complete each hole with a number of strokes very close to par. It is necessary to play several holes and let scores accumulate before a clear winner emerges in a match.
Ben, Jen, and Ken play a full round, which consists of 18 holes. The course has an assortment of par-3, par-4, and par-5 holes. At the end of the game, the players tabulate their scores.
hole = 1:18;
par = [4 3 5 3 4 5 3 4 4 4 5 3 5 4 4 4 3 4];
nms = {'Ben';'Jen';'Ken'};
Ben = [4 3 4 2 3 5 2 3 3 4 3 2 3 3 3 3 2 3];
Jen = [4 3 4 3 4 4 3 4 4 4 5 3 4 4 5 5 3 3];
Ken = [4 3 4 3 5 5 4 4 4 4 5 3 5 4 5 4 3 5];
T = table(hole',par',Ben',Jen',Ken', ...
'VariableNames',['hole';'par';nms])
hole par Ben Jen Ken
____ ___ ___ ___ ___
The winner of the round is the player whose lower cumulative sum drifts the most below par at the end. Compute the sums for the three players to determine the winner. Make every shift in mean detectable by setting a small threshold.
[~,b,~,Bensum] = cusum(Ben-par,1,1e-4,0);
[~,j,~,Jensum] = cusum(Jen-par,1,1e-4,0);
[~,k,~,Kensum] = cusum(Ken-par,1,1e-4,0);
plot([Bensum;Jensum;Kensum]')
legend(nms,'Location','best')
Ben wins the round. Simulate their next game by adding or subtracting a stroke per hole at random.
Ben = Ben+randi(3,1,18)-2;
Jen = Jen+randi(3,1,18)-2;
Ken = Ken+randi(3,1,18)-2;
Example: reshape(rand(100,1)*[-1 1],1,200)
climit — Control limit
Control limit, specified as a real scalar expressed in standard deviations.
mshift — Minimum mean shift to detect
Minimum mean shift to detect, specified as a real scalar expressed in standard deviations.
tmean — Target mean
mean(x(1:25)) (default) | real scalar
Target mean, specified as a real scalar. If tmean is not specified, then it is estimated as the mean of the first 25 samples of x.
tdev — Target standard deviation
std(x(1:25)) (default) | real scalar
Target standard deviation, specified as a real scalar. If tdev is not specified, then it is estimated as the standard deviation of the first 25 samples of x.
iupper,ilower — Out-of-control point indices
integer scalars | integer vectors
Out-of-control point indices, returned as integer scalars or vectors. If all signal samples are within the specified tolerance, then cusum returns empty iupper and ilower arguments.
uppersum,lowersum — Upper and lower cumulative sums
Upper and lower cumulative sums, returned as vectors.
The CUSUM control chart is designed to detect small incremental changes in the mean of a process.
Given a sequence x1, x2, x3, …, xn with estimated average mx and estimated standard deviation σx, define upper and lower cumulative process sums using:
Upper cumulative sum
{U}_{i}=\left\{\begin{array}{cc}0,& i=1\\ \mathrm{max}\left(0,{U}_{i-1}+{x}_{i}-{m}_{x}-\frac{1}{2}n{\sigma }_{x}\right),& i>1\end{array}
{L}_{i}=\left\{\begin{array}{cc}0,& i=1\\ \mathrm{min}\left(0,{L}_{i-1}+{x}_{i}-{m}_{x}+\frac{1}{2}n{\sigma }_{x}\right),& i>1\end{array}
The variable n, represented in cusum by the mshift argument, is the number of standard deviations from the target mean, tmean, that make a shift detectable.
A process violates the CUSUM criterion at the sample xj if it obeys Uj > cσx or Lj < –cσx. The control limit c is represented in cusum by the climit argument.
By default, the function returns the first violation it detects. If you specify the 'all' flag, the function returns every violation.
If supplied, input argument 'all' must be a compile-time constant.
findchangepts | mean |
\begin{array}{cccccc}Ten-Crores& Crores& Ten-Lacs& Lacs& Ten-Thousands& Thousands\\ 6& 8& 9& 7& 4& 5\end{array}\begin{array}{ccc}Hundreds& Tens& Units\\ 1& 3& 2\end{array}
6×{10}^{8}
{5}^{n}
{2}^{n}
975436×{5}^{4}
\frac{9754360000}{{2}^{4}}
\left({x}^{n}-{a}^{n}\right)
\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}
a,ar,a{r}^{2},a{r}^{3},...
a{r}^{n-1}
\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\
Let the number be v. Then, 3 (2x + 9) = 75.
If the number 59a44b is divisible by 36, then the maximum value of a + b is:
The least number that must be subtracted from 7577 to get a perfect square is
The square root of which of the numbers below will be rational?
Which of the numbers given below is NOT a perfect square?
Half the people on a bus get off at each stop after the first, and no one gets on after the first stop. If only one person gets off at stop number 7, how many people got on at the first stop?
7th stop only one was left and in each preceding stop twice the nos... there were so 6 stops before 7th.
If the 8-digit number 1a765b12 is to be divisible by 72, the least value of (2a + 3b)is:
LCM of 4, 6, 8 and 10 = 120
The least number of four digits which is divisible by 4, 6, 8 and 10 => 1000 + 120 - 40 = 1080. |
Orthodiagonal_quadrilateral Knowpia
Orthodiagonal quadrilateral
In Euclidean geometry, an orthodiagonal quadrilateral is a quadrilateral in which the diagonals cross at right angles. In other words, it is a four-sided figure in which the line segments between non-adjacent vertices are orthogonal (perpendicular) to each other.
An orthodiagonal quadrilateral (yellow). According to the characterization of these quadrilaterals, the two red squares on two opposite sides of the quadrilateral have the same total area as the two blue squares on the other pair of opposite sides.
A kite is an orthodiagonal quadrilateral in which one diagonal is a line of symmetry. The kites are exactly the orthodiagonal quadrilaterals that contain a circle tangent to all four of their sides; that is, the kites are the tangential orthodiagonal quadrilaterals.[1]
A rhombus is an orthodiagonal quadrilateral with two pairs of parallel sides (that is, an orthodiagonal quadrilateral that is also a parallelogram).
A square is a limiting case of both a kite and a rhombus.
Orthodiagonal equidiagonal quadrilaterals in which the diagonals are at least as long as all of the quadrilateral's sides have the maximum area for their diameter among all quadrilaterals, solving the n = 4 case of the biggest little polygon problem. The square is one such quadrilateral, but there are infinitely many others. An orthodiagonal quadrilateral that is also equidiagonal is a midsquare quadrilateral because its Varignon parallelogram is a square. Its area can be expressed purely in terms of its sides.
For any orthodiagonal quadrilateral, the sum of the squares of two opposite sides equals that of the other two opposite sides: for successive sides a, b, c, and d, we have [2][3]
{\displaystyle \displaystyle a^{2}+c^{2}=b^{2}+d^{2}.}
This follows from the Pythagorean theorem, by which either of these two sums of two squares can be expanded to equal the sum of the four squared distances from the quadrilateral's vertices to the point where the diagonals intersect. Conversely, any quadrilateral in which a2 + c2 = b2 + d2 must be orthodiagonal.[4] This can be proved in a number of ways, including using the law of cosines, vectors, an indirect proof, and complex numbers.[5]
The diagonals of a convex quadrilateral are perpendicular if and only if the two bimedians have equal length.[5]
According to another characterization, the diagonals of a convex quadrilateral ABCD are perpendicular if and only if
{\displaystyle \angle PAB+\angle PBA+\angle PCD+\angle PDC=\pi }
where P is the point of intersection of the diagonals. From this equation it follows almost immediately that the diagonals of a convex quadrilateral are perpendicular if and only if the projections of the diagonal intersection onto the sides of the quadrilateral are the vertices of a cyclic quadrilateral.[5]
A convex quadrilateral is orthodiagonal if and only if its Varignon parallelogram (whose vertices are the midpoints of its sides) is a rectangle.[5] A related characterization states that a convex quadrilateral is orthodiagonal if and only if the midpoints of the sides and the feet of the four maltitudes are eight concyclic points; the eight point circle. The center of this circle is the centroid of the quadrilateral. The quadrilateral formed by the feet of the maltitudes is called the principal orthic quadrilateral.[6]
If the normals to the sides of a convex quadrilateral ABCD through the diagonal intersection intersect the opposite sides in R, S, T, U, and K, L, M, N are the feet of these normals, then ABCD is orthodiagonal if and only if the eight points K, L, M, N, R, S, T and U are concyclic; the second eight point circle. A related characterization states that a convex quadrilateral is orthodiagonal if and only if RSTU is a rectangle whose sides are parallel to the diagonals of ABCD.[5]
There are several metric characterizations regarding the four triangles formed by the diagonal intersection P and the vertices of a convex quadrilateral ABCD. Denote by m1, m2, m3, m4 the medians in triangles ABP, BCP, CDP, DAP from P to the sides AB, BC, CD, DA respectively. If R1, R2, R3, R4 and h1, h2, h3, h4 denote the radii of the circumcircles and the altitudes respectively of these triangles, then the quadrilateral ABCD is orthodiagonal if and only if any one of the following equalities holds:[5]
{\displaystyle m_{1}^{2}+m_{3}^{2}=m_{2}^{2}+m_{4}^{2}}
{\displaystyle R_{1}^{2}+R_{3}^{2}=R_{2}^{2}+R_{4}^{2}}
{\displaystyle {\frac {1}{h_{1}^{2}}}+{\frac {1}{h_{3}^{2}}}={\frac {1}{h_{2}^{2}}}+{\frac {1}{h_{4}^{2}}}}
Furthermore, a quadrilateral ABCD with intersection P of the diagonals is orthodiagonal if and only if the circumcenters of the triangles ABP, BCP, CDP and DAP are the midpoints of the sides of the quadrilateral.[5]
Comparison with a tangential quadrilateralEdit
A few metric characterizations of tangential quadrilaterals and orthodiagonal quadrilaterals are very similar in appearance, as can be seen in this table.[5] The notations on the sides a, b, c, d, the circumradii R1, R2, R3, R4, and the altitudes h1, h2, h3, h4 are the same as above in both types of quadrilaterals.
{\displaystyle a+c=b+d}
{\displaystyle a^{2}+c^{2}=b^{2}+d^{2}}
{\displaystyle R_{1}+R_{3}=R_{2}+R_{4}}
{\displaystyle R_{1}^{2}+R_{3}^{2}=R_{2}^{2}+R_{4}^{2}}
{\displaystyle {\frac {1}{h_{1}}}+{\frac {1}{h_{3}}}={\frac {1}{h_{2}}}+{\frac {1}{h_{4}}}}
{\displaystyle {\frac {1}{h_{1}^{2}}}+{\frac {1}{h_{3}^{2}}}={\frac {1}{h_{2}^{2}}}+{\frac {1}{h_{4}^{2}}}}
The area K of an orthodiagonal quadrilateral equals one half the product of the lengths of the diagonals p and q:[7]
{\displaystyle K={\frac {p\cdot q}{2}}.}
Conversely, any convex quadrilateral where the area can be calculated with this formula must be orthodiagonal.[5] The orthodiagonal quadrilateral has the biggest area of all convex quadrilaterals with given diagonals.
Orthodiagonal quadrilaterals are the only quadrilaterals for which the sides and the angle formed by the diagonals do not uniquely determine the area.[3] For example, two rhombi both having common side a (and, as for all rhombi, both having a right angle between the diagonals), but one having a smaller acute angle than the other, have different areas (the area of the former approaching zero as the acute angle approaches zero).
If squares are erected outward on the sides of any quadrilateral (convex, concave, or crossed), then their centres (centroids) are the vertices of an orthodiagonal quadrilateral that is also equidiagonal (that is, having diagonals of equal length). This is called Van Aubel's theorem.
Each side of an orthodiagonal quadrilateral has at least one common point with the Pascal points circle.[8]
Properties of orthodiagonal quadrilaterals that are also cyclicEdit
Circumradius and areaEdit
For a cyclic orthodiagonal quadrilateral (one that can be inscribed in a circle), suppose the intersection of the diagonals divides one diagonal into segments of lengths p1 and p2 and divides the other diagonal into segments of lengths q1 and q2. Then[9] (the first equality is Proposition 11 in Archimedes Book of Lemmas)
{\displaystyle D^{2}=p_{1}^{2}+p_{2}^{2}+q_{1}^{2}+q_{2}^{2}=a^{2}+c^{2}=b^{2}+d^{2}}
where D is the diameter of the circumcircle. This holds because the diagonals are perpendicular chords of a circle. These equations yield the circumradius expression
{\displaystyle R={\tfrac {1}{2}}{\sqrt {p_{1}^{2}+p_{2}^{2}+q_{1}^{2}+q_{2}^{2}}}}
or, in terms of the sides of the quadrilateral, as[2]
{\displaystyle R={\tfrac {1}{2}}{\sqrt {a^{2}+c^{2}}}={\tfrac {1}{2}}{\sqrt {b^{2}+d^{2}}}.}
It also follows that[2]
{\displaystyle a^{2}+b^{2}+c^{2}+d^{2}=8R^{2}.}
Thus, according to Euler's quadrilateral theorem, the circumradius can be expressed in terms of the diagonals p and q, and the distance x between the midpoints of the diagonals as
{\displaystyle R={\sqrt {\frac {p^{2}+q^{2}+4x^{2}}{8}}}.}
A formula for the area K of a cyclic orthodiagonal quadrilateral in terms of the four sides is obtained directly when combining Ptolemy's theorem and the formula for the area of an orthodiagonal quadrilateral. The result is[10]: p.222
{\displaystyle K={\tfrac {1}{2}}(ac+bd).}
In a cyclic orthodiagonal quadrilateral, the anticenter coincides with the point where the diagonals intersect.[2]
Brahmagupta's theorem states that for a cyclic orthodiagonal quadrilateral, the perpendicular from any side through the point of intersection of the diagonals bisects the opposite side.[2]
If an orthodiagonal quadrilateral is also cyclic, the distance from the circumcenter (the center of the circumscribed circle) to any side equals half the length of the opposite side.[2]
In a cyclic orthodiagonal quadrilateral, the distance between the midpoints of the diagonals equals the distance between the circumcenter and the point where the diagonals intersect.[2]
Infinite sets of inscribed rectanglesEdit
{\displaystyle ABCD}
is an orthodiagonal quadrilateral,
{\displaystyle P_{1}X_{1}Z_{1}Y_{1}}
{\displaystyle P_{2}X_{2}Z_{2}Y_{2}}
are rectangles whose sides are parallel to the diagonals of the quadrilateral.
{\displaystyle ABCD}
is an orthodiagonal quadrilateral.
{\displaystyle P_{1}}
{\displaystyle Q_{1}}
are Pascal points formed by the circle
{\displaystyle \omega _{1}}
{\displaystyle \sigma _{P_{1}Q_{1}}}
is Pascal-points circle which defines the rectangle
{\displaystyle P_{1}V_{1}Q_{1}W_{1}}
{\displaystyle P_{2}}
{\displaystyle Q_{2}}
{\displaystyle \omega _{2}}
{\displaystyle \sigma _{P_{2}Q_{2}}}
{\displaystyle P_{2}V_{2}Q_{2}W_{2}}
For every orthodiagonal quadrilateral, we can inscribe two infinite sets of rectangles:
(i) a set of rectangles whose sides are parallel to the diagonals of the quadrilateral
(ii) a set of rectangles defined by Pascal-points circles.[11]
^ Josefsson, Martin (2010), "Calculations concerning the tangent lengths and tangency chords of a tangential quadrilateral" (PDF), Forum Geometricorum, 10: 119–130 .
^ a b c d e f g Altshiller-Court, N. (2007), College Geometry, Dover Publications . Republication of second edition, 1952, Barnes & Noble, pp. 136-138.
^ a b Mitchell, Douglas, W. (2009), "The area of a quadrilateral", The Mathematical Gazette, 93 (July): 306–309 .
^ Ismailescu, Dan; Vojdany, Adam (2009), "Class preserving dissections of convex quadrilaterals" (PDF), Forum Geometricorum, 9: 195–211 .
^ a b c d e f g h i Josefsson, Martin (2012), "Characterizations of Orthodiagonal Quadrilaterals" (PDF), Forum Geometricorum, 12: 13–25 .
^ Mammana, Maria Flavia; Micale, Biagio; Pennisi, Mario (2011), "The Droz-Farny Circles of a Convex Quadrilateral" (PDF), Forum Geometricorum, 11: 109–119 .
^ Harries, J. (2002), "Area of a quadrilateral", The Mathematical Gazette, 86 (July): 310–311
^ David, Fraivert (2017), "Properties of a Pascal points circle in a quadrilateral with perpendicular diagonals" (PDF), Forum Geometricorum, 17: 509–526 .
^ Posamentier, Alfred S.; Salkind, Charles T. (1996), Challenging Problems in Geometry (second ed.), Dover Publications, pp. 104–105, #4–23 .
^ Josefsson, Martin (2016), "Properties of Pythagorean quadrilaterals", The Mathematical Gazette, 100 (July): 213–224 .
^ David, Fraivert (2019), "A Set of Rectangles Inscribed in an Orthodiagonal Quadrilateral and Defined by Pascal-Points Circles", Journal for Geometry and Graphics, 23: 5–27 . |
Propositional Logic Using Algebra | Brilliant Math & Science Wiki
Michael Mendrin contributed
As this wiki Propositional Logic explains, propositions are treated as atomic units. Such propositions can be denoted by letters such as
P, Q, R,...
, which have a value of TRUE or FALSE. Expressions can be built up using the three primary logical operations AND, OR and NOT, commonly represented by symbols
\wedge
\vee
\sim
\neg
). and such expressions can be related by or even include conditionals and biconditionals represented by symbols
\rightarrow
\leftrightarrow
. So, for example, we have the expression
\left( P\wedge Q \right) \rightarrow P
which is a logical identity, "decomposing a conjunction", meaning that if
\left( P\wedge Q \right)
P
is true. We can have something more complicated,
\left( P\leftrightarrow Q \right) \leftrightarrow \left( \left( P\rightarrow Q \right)\wedge \left( Q\rightarrow P \right) \right)
which in fact is the "definiton of the biconditional"
\leftrightarrow
being the symbol. Propositional Logic explains more in detail, and, in practice, one is expected to make use of such logical identities to prove any expression to be true or not. This can be a cumbersome exercise, for one not familiar working with this.
It seems much like algebra, so is there a way to work these things out algebraically? Yes, sort of, one can. First of all, all propositions and expressions necessarily have a value of either TRUE or FALSE. We can use numeric values
1
0
to mean the same thing (although other schemes are possible), and for this wiki, lowercase letters
p, q, r...
shall denote propositions and expressions that have numeric values. We can immediately see that AND and NOT are simply
P\wedge Q\Longleftrightarrow pq
\sim P\Longleftrightarrow 1-p
\Longleftrightarrow
means similar, or homologous, operations. Now, but the OR operation is a bit more cumbersome
\left( P\vee Q \right) \Longleftrightarrow p+q-pq
We round this out with the conditional and biconditional
\left( P\rightarrow Q \right) \leftrightarrow \left( \sim P\vee Q \right)
\left( \sim P\vee Q \right) \Longleftrightarrow 1-p+pq
\left( P\leftrightarrow Q \right) \leftrightarrow \left( \sim P\vee Q \right) \wedge \left( P\vee \sim Q \right)
\left( \sim P\vee Q \right) \wedge \left( P\vee \sim Q \right) \Longleftrightarrow 1-p-q+2pq
Now, let's try to put this into practice, say, Modus Ponens. Is it true?
\left( \left( P\rightarrow Q \right) \wedge P \right) \rightarrow Q
Using the algebraic equivalents given above, we can work out the corresponding algebraic expression
\left( \left( P\rightarrow Q \right) \wedge P \right) \rightarrow Q\Longleftrightarrow 1-p+{ p }^{ 2 }+pq-2{ p }^{ 2 }q+{ p }^{ 2 }{ q }^{ 2 }
Now, but here's we depart from usual algebraic convention. Every proposition and expression always have a value of either TRUE or FALSE, either
1
0
. This means that all the exponents in the algebraic expression can be reduced to
1
, and we're left with
1-p+{ p }+pq-2{ p }q+{ p }{ q }=1
which means it's always true, and therefore a logical identity.
While this "trick" could be a handy way of double-checking logical identities or working out complicated combinations of logic gates (see Logic Gates), this approach helps one understand that even a "If ... therefore ...." statement can be represented by a logical gate, as well as biconditionals. This puts all of these things on an equal footing, so that conditionals and biconditionals are not fundamentally any different from logical operations.
\sim \left( P\vee Q \right) \leftrightarrow (\sim P\wedge \sim Q)
Find the equivalent algebraic expression for the logical expressions on both sides of the bicondtional.
1-p-q+pq\quad
Note that, when worked out, the algebraic form yields the same truth table as the logical expressions on either side of the biconditional, if we say that
TRUE \Longleftrightarrow 1
FALSE \Longleftrightarrow 0
Also note that had the problem asked for the equivalent algebraic expression for the entire logical expression, including the biconditional, the result would have simply been
1
, as it is a well-known logical identity.
Helpful note: Given an arbitrary truth table, say, of two variables
P
Q
, the logical expression yielding this truth table can easily be found by performing an
OR
operation on all the instances of
TRUE
, each one being an
AND
expression based on its location, each of which are
\sim P \wedge \sim Q
\sim P \wedge Q
P \wedge \sim Q
P \wedge Q
. So, for instance, for the biconditional, defined by a truth table where if
P
Q
TRUE
FALSE
TRUE
, otherwise it returns
FALSE
, to generate the logical expression we simply write out
\left( \sim P\wedge \sim Q \right) \vee \left( P\wedge Q \right)
This might look different from the expression used above for the same thing, but in fact both are logically equivalent, and have the same simplest algebraic expression, which is
1-p-q+2pq
This method of finding logical expressions for arbitrary truth tables can be generalized for any number of variables.
Here are algebraic equivalents for commonly used logic gates
NOT\Longleftrightarrow 1-p
AND\Longleftrightarrow pq
OR\Longleftrightarrow p+q-pq
NAND\Longleftrightarrow 1-pq
NOR\Longleftrightarrow 1-p-q+pq
XOR\Longleftrightarrow p+q-2pq
XNOR\Longleftrightarrow 1-p-q+2pq
so that it can be said that a biconditional is logically the same as
XNOR
. Meanwhile, a conditional doesn't correspond to any of the classic logic gates, as it is asymmetrical with respect to the two inputs, having the algebraic equivalent of
1-p+pq
Cite as: Propositional Logic Using Algebra. Brilliant.org. Retrieved from https://brilliant.org/wiki/propositional-logic-using-algebra/ |
Faster Data Fitting Solver | nag
Faster Data Fitting Solver
Calibrating the parameters of complex numerical models to fit real-world observations is one of the most common problems found in industries such as finance, physics, simulations, engineering, etc. NAG introduces at Mark 27.1 of the NAG Library a novel nonlinear least squares (NLN-LSQ) trust-region solver handle_solve_bxnl (e04gg) for unconstrained and bound-constrained fitting problems which implements various algorithms and regularization techniques. It is aimed at small to medium sized fitting problems (up to 1000s of parameters) of the form
\begin{array}{llll}\hfill \underset{x\in {ℝ}^{n}}{\text{minimize}}& \sum _{i=1}^{m}{\left(\varphi \left({t}_{i},x\right)-{y}_{i}\right)}^{2}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{subject to}& \phantom{\rule{2.77695pt}{0ex}}\ell \le x\le u,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}
where the idea is to find optimal values for the parameters,
x
, of the model represented by the smooth function
\varphi \left(t,x\right)
(blue line) which fit
m
observed data points
\left({t}_{i},{y}_{i}\right)
(black dots). That is, to minimize the squared error between the model and the data (vertical red bars).
e04gg should present a significant improvement over the current nonlinear least squares solvers in the NAG Library. In addition, this solver fills the gap between unconstrained solvers such as lsq_uncon_quasi_deriv_comp (e04gb) and the fully constrained ones, such as lsq_gencon_deriv (e04us).
e04gg is part of the NAG Optimization Modelling Suite which offers clarity and consistency on the interface of the solvers within the suite.
The new solver stems from a collaboration with the Rutherford Appleton Laboratory [1] and demonstrates NAG’s ongoing effort to expand and improve its offering in mathematical optimization.
Well established trust-region method with a variety of implemented algorithms, ranging from simple Powell’s dogleg or Gauss-Newton methods to sophisticated Tensor–Newton schemes which overcome convergence difficulties present in simpler methods;
Avoids data over-fitting by incorporating different types of regularization for both the problem formulation and the trust-region subproblem;
Ability to recover when it is not possible to evaluate the function
\varphi \left(t,x\right)
or its gradient at a given point;
Account for uncertainty in the observed data by using optional residual weights;
Flexible stopping criteria that can suit the problem and data tolerances.
Modern Replacement for e04gb and e04us
The new solver e04gg offers unprecedented robustness and a significant speed-up over current alternatives in the Library, namely e04gb for unconstrained nonlinear least squares problems and e04us for problems with simple variable bounds. You are highly recommended to upgrade to the new solver.
Our benchmarks comparing e04gg to e04gb on 68 unconstrained nonlinear least squares CUTEst problems are reported in Figure 1 (a)–(c) using performance profiles. The three plots show that the new solver is faster: solving 60% of the problems in less time (a) and is more robust, solving 25% more problems. Requires fewer user callbacks: 55% of problems require fewer function calls (b) and 65% require fewer gradient evaluations (c).
As the e04gb solver was not designed to tackle simple bounds directly, typically e04us would be used instead for such problems. However, e04us being a more general solver does not fully exploit the structure of NLN-LSQ problems as e04gg does in the presence of simple bounds. That’s why a speed-up on 45% of problems can be observed (d) as well as a reduction in user callbacks: 65% of problems require fewer function and gradient calls (e and f) when comparing e04us and e04gg on 112 unconstrained and bound-constrained nonlinear least square CUTEst problems.
(a) Time (b) Function calls (c) Gradient calls
(d) Time (e) Function calls (f) Gradient calls
Figure 1: (a)–(c) Performance profiles comparing solvers e04gg and e04gb over 68 CUTEst unconstrained nonlinear least squares problems, while (d)–(f) report the performance profiles of e04gg and e04us for 112 CUTEst unconstrained and bound constrained nonlinear least squares problems. Performance measures are time in seconds (a and d), number of function calls (b and e), and number of gradient calls (c and f). For the time plots, the higher line indicates the faster solver. For the centre and right column plots, the higher line represents fewer functions and gradients calls respectively.
Real-world example: a particle track data-fitting
The following example illustrates the usage of e04gg to fit PADC etched nuclear track data to a convoluted distribution. A target sheet is scanned and track diameters are recorded (red wedges in left image of Figure 2) into a histogram (blue bars in right plot of Figure 2) and a mixed Normal and log-Normal model is to be fitted to the obtained experimental histogram. e04gg is used to fit the six parameter model
\begin{array}{llll}\hfill \varphi \left(\rightt,x=\left(µ,s,{A}_{g},a,b,{A}_{l}\right)\left)\right& =\text{Normal}\left(µ,s,{A}_{g}\right)+\text{log-Normal}\left(a,b,{A}_{l}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill \text{subject to}& \phantom{\rule{1em}{0ex}}0\le x,\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}
using the histogram heights. Thanks to the use of regularization and residual weights, e04gg provided a robust solution
{x}^{*}
to unfold the parameters for the two distributions (red and blue curves in right plot of Figure 2). Adding these together produce the green curve which is the one used to perform the fitting. Data and complete python source code for this example are available here.
Figure 2: Left: example of a PADC target with
a
particle etched tracks, wedges in red show the track diameter. Right: diameter histogram from experimental data (blue bars), aggregated model used for the fitting (green curve) and unfolded models (blue and red curves). Optimal parameter values are reported in the legend.
For further examples and reading visit our GitHub Local Optimization page.
Local Optimization on NAG GitHub
[1] Gould N I M, Rees T, and Scott J A (2017) A higher order method for solving nonlinear least-squares problems. Technical report, RAL-P-1027-010 RAL Library. STFC Rutherford Appleton Laboratory. Link to report.
[2] Kanzow C, Yamashita N, and Fukushima M (2004) Levenberg-Marquardt methods with strong local convergence properties for solving nonlinear equations with convex constraints. Journal of Computational and Applied Mathematics 174 375–397.
[3] Sajo-Bohus L. (2020) Data provided in private communication. |
octahedron - Maple Help
Home : Support : Online Help : Graphics : Packages : Plot Tools : octahedron
generate 3-D plot object for an octahedron
octahedron([x, y, z], s, options)
location of the octahedron
(optional) scale of the octahedron; default is 1
The octahedron command creates a three-dimensional plot data object, which when displayed is a scaled octahedron located at point [x,y,z]. Note that the scale factor is applied in each dimension. This command is an interface to the plots[polyhedraplot] command.
The plot data object produced by the octahedron command can be used in a PLOT3D data structure, or displayed using the plots[display] command.
\mathrm{with}\left(\mathrm{plottools}\right):
\mathrm{with}\left(\mathrm{plots}\right):
\mathrm{display}\left(\mathrm{octahedron}\left([0,0,0],0.8\right),\mathrm{orientation}=[-75,70]\right)
\mathrm{display}\left(\mathrm{octahedron}\left([0,0,0],0.8\right),\mathrm{octahedron}\left([1,1,1],0.5\right),\mathrm{orientation}=[0,70]\right) |
Plot residuals of generalized linear mixed-effects model - MATLAB - MathWorks Deutschland
Create Plots of Residuals
Plot residuals of generalized linear mixed-effects model
plotResiduals(glme,plottype)
plotResiduals(glme,plottype,Name,Value)
plotResiduals(glme,plottype) plots the raw conditional residuals of the generalized linear mixed-effects model glme in a plot of the type specified by plottype.
plotResiduals(glme,plottype,Name,Value) plots the conditional residuals of glme using additional options specified by one or more Name,Value pair arguments. For example, you can specify to plot the Pearson residuals.
h = plotResiduals(___) returns a handle, h, to the lines or patches in the plot of residuals.
plottype — Type of residual plot
Type of residual plot, specified as one of the following.
'histogram' Histogram of residuals
'caseorder' Residuals versus case order. Case order is the same as the row order used in the input data tbl when fitting the model using fitglme.
'fitted' Residuals versus fitted values
'lagged' Residuals versus lagged residual (r(t) versus r(t – 1))
Example: plotResiduals(glme,'lagged')
'raw' (default) | 'Pearson'
{r}_{ci}={y}_{i}-{g}^{-1}\left({x}_{i}^{T}\stackrel{^}{\beta }+{z}_{i}^{T}\stackrel{^}{b}+{\delta }_{i}\right)
{r}_{ci}^{pearson}=\frac{{r}_{ci}}{\sqrt{\frac{\stackrel{^}{{\sigma }^{2}}}{{w}_{i}}{v}_{i}\left({\mu }_{i}\left(\stackrel{^}{\beta },\stackrel{^}{b}\right)\right)}}
In each of these equations:
yi is the ith element of the n-by-1 response vector, y, where i = 1, ..., n.
g-1 is the inverse link function for the model.
xiT is the ith row of the fixed-effects design matrix X.
ziT is the ith row of the random-effects design matrix Z.
δi is the ith offset value.
σ2 is the dispersion parameter.
wi is the ith observation weight.
vi is the variance term for the ith observation.
μi is the mean of the response for the ith observation.
\stackrel{^}{\beta }
\stackrel{^}{b}
are estimated values of β and b.
Raw residuals from a generalized linear mixed-effects model have nonconstant variance. Pearson residuals are expected to have an approximately constant variance, and are generally used for analysis.
h — Handle to residual plot
Handle to the residual plot, returned as a graphics object. You can use dot notation to change certain property values of the object, including face color for a histogram, and marker style and color for a scatterplot. For more information, see Access Property Values.
The number of defects can be modeled using a Poisson distribution:
{\text{defects}}_{ij}\sim \text{Poisson}\left({\mu }_{ij}\right)
\mathrm{log}\left({\mu }_{ij}\right)={\beta }_{0}+{\beta }_{1}{\text{newprocess}}_{ij}+{\beta }_{2}{\text{time}\text{_}\text{dev}}_{ij}+{\beta }_{3}{\text{temp}\text{_}\text{dev}}_{ij}+{\beta }_{4}{\text{supplier}\text{_}\text{C}}_{ij}+{\beta }_{5}{\text{supplier}\text{_}\text{B}}_{ij}+{b}_{i},
{\text{defects}}_{ij}
i
j
{\mu }_{ij}
i
i=1,2,...,20
j
j=1,2,...,5
{\text{newprocess}}_{ij}
{\text{time}\text{_}\text{dev}}_{ij}
{\text{temp}\text{_}\text{dev}}_{ij}
i
j
{\text{newprocess}}_{ij}
i
j
{\text{supplier}\text{_}\text{C}}_{ij}
{\text{supplier}\text{_}\text{B}}_{ij}
i
j
{b}_{i}\sim N\left(0,{\sigma }_{b}^{2}\right)
i
Create diagnostic plots using Pearson residuals to test the model assumptions.
Plot a histogram to visually confirm that the mean of the Pearson residuals is equal to 0. If the model is correct, we expect the Pearson residuals to be centered at 0.
plotResiduals(glme,'histogram','ResidualType','Pearson')
The histogram shows that the Pearson residuals are centered at 0.
Plot the Pearson residuals versus the fitted values, to check for signs of nonconstant variance among the residuals (heteroscedasticity). We expect the conditional Pearson residuals to have a constant variance. Therefore, a plot of conditional Pearson residuals versus conditional fitted values should not reveal any systematic dependence on the conditional fitted values.
plotResiduals(glme,'fitted','ResidualType','Pearson')
The plot does not show a systematic dependence on the fitted values, so there are no signs of nonconstant variance among the residuals.
Plot the Pearson residuals versus lagged residuals, to check for correlation among the residuals. The conditional independence assumption in GLME implies that the conditional Pearson residuals are approximately uncorrelated.
plotResiduals(glme,'lagged','ResidualType','Pearson')
There is no pattern to the plot, so there are no signs of correlation among the residuals.
GeneralizedLinearMixedModel | fitglme | fitted | plot | residuals |
Irradiance - Wikipedia
(Redirected from W/m2)
In radiometry, irradiance is the radiant flux received by a surface per unit area. The SI unit of irradiance is the watt per square metre (W⋅m−2). The CGS unit erg per square centimetre per second (erg⋅cm−2⋅s−1) is often used in astronomy. Irradiance is often called intensity, but this term is avoided in radiometry where such usage leads to confusion with radiant intensity. In astrophysics, irradiance is called radiant flux.[1]
Spectral irradiance is the irradiance of a surface per unit frequency or wavelength, depending on whether the spectrum is taken as a function of frequency or of wavelength. The two forms have different dimensions: spectral irradiance of a frequency spectrum is measured in watts per square metre per hertz (W⋅m−2⋅Hz−1), while spectral irradiance of a wavelength spectrum is measured in watts per square metre per metre (W⋅m−3), or more commonly watts per square metre per nanometre (W⋅m−2⋅nm−1).
Irradiance[edit]
Irradiance of a surface, denoted Ee ("e" for "energetic", to avoid confusion with photometric quantities), is defined as[2]
{\displaystyle E_{\mathrm {e} }={\frac {\partial \Phi _{\mathrm {e} }}{\partial A}},}
Φe is the radiant flux received;
If we want to talk about the radiant flux emitted by a surface, we speak of radiant exitance.
Spectral irradiance[edit]
Spectral irradiance in frequency of a surface, denoted Ee,ν, is defined as[2]
{\displaystyle E_{\mathrm {e} ,\nu }={\frac {\partial E_{\mathrm {e} }}{\partial \nu }},}
Spectral irradiance in wavelength of a surface, denoted Ee,λ, is defined as[2]
{\displaystyle E_{\mathrm {e} ,\lambda }={\frac {\partial E_{\mathrm {e} }}{\partial \lambda }},}
Irradiance of a surface is also, according to the definition of radiant flux, equal to the time-average of the component of the Poynting vector perpendicular to the surface:
{\displaystyle E_{\mathrm {e} }=\langle |\mathbf {S} |\rangle \cos \alpha ,}
⟨ • ⟩ is the time-average;
S is the Poynting vector;
α is the angle between a unit vector normal to the surface and S.
For a propagating sinusoidal linearly polarized electromagnetic plane wave, the Poynting vector always points to the direction of propagation while oscillating in magnitude. The irradiance of a surface is then given by[3]
{\displaystyle E_{\mathrm {e} }={\frac {n}{2\mu _{0}\mathrm {c} }}E_{\mathrm {m} }^{2}\cos \alpha ={\frac {n\varepsilon _{0}\mathrm {c} }{2}}E_{\mathrm {m} }^{2}\cos \alpha ,}
Em is the amplitude of the wave's electric field;
n is the refractive index of the medium of propagation;
c is the speed of light in vacuum;
ε0 is the vacuum permittivity.
This formula assumes that the magnetic susceptibility is negligible, i.e. that μr ≈ 1 where μr is the magnetic permeability of the propagation medium. This assumption is typically valid in transparent media in the optical frequency range.
Point source[edit]
A point source of light produces spherical wavefronts. The irradiance in this case varies inversely with the square of the distance from the source.
{\displaystyle E={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}.\,}
r is the distance;
P is the radiant power;
A is the surface area of a sphere of radius r.
For quick approximations, this equation indicates that doubling the distance reduces irradiation to one quarter; or similarly, to double irradiation, reduce the distance to 0.7. When it is not a point source, for real light sources, the irradiance profile may be obtained by the image convolution of a picture of the light source.[4]
The global irradiance on a horizontal surface on Earth consists of the direct irradiance Ee,dir and diffuse irradiance Ee,diff. On a tilted plane, there is another irradiance component, Ee,refl, which is the component that is reflected from the ground. The average ground reflection is about 20% of the global irradiance. Hence, the irradiance Ee on a tilted plane consists of three components:[5]
{\displaystyle E_{\mathrm {e} }=E_{\mathrm {e} ,\mathrm {dir} }+E_{\mathrm {e} ,\mathrm {diff} }+E_{\mathrm {e} ,\mathrm {refl} }.}
The integral of solar irradiance over a time period is called "solar exposure" or "insolation".[5][6]
SI radiometry units[edit]
^ Carroll, Bradley W. (2017-09-07). An introduction to modern astrophysics. p. 60. ISBN 978-1-108-42216-1. OCLC 991641816.
^ a b c "Thermal insulation — Heat transfer by radiation — Physical quantities and definitions". ISO 9288:1989. ISO catalogue. 1989. Retrieved 2015-03-15.
^ Griffiths, David J. (1999). Introduction to electrodynamics (3. ed., reprint. with corr. ed.). Upper Saddle River, NJ [u.a.]: Prentice-Hall. ISBN 0-13-805326-X.
^ I. Moreno, P. X. Viveros-Méndez, "Modeling the irradiation pattern of LEDs at short distances," Opt. Express 29, 6845 (2021). https://doi.org/10.1364/OE.419428
^ a b Quaschning, Volker (2003). "Technology fundamentals—The sun as an energy resource". Renewable Energy World. 6 (5): 90–93.
^ Liu, B. Y. H.; Jordan, R. C. (1960). "The interrelationship and characteristic distribution of direct, diffuse and total solar radiation". Solar Energy. 4 (3): 1. Bibcode:1960SoEn....4....1L. doi:10.1016/0038-092X(60)90062-1.
Retrieved from "https://en.wikipedia.org/w/index.php?title=Irradiance&oldid=1063366895" |
The circles x2 + y2 + x + y = 0 and x2 + y2 + x - y = - Maths - Three Dimensional Geometry - 11580826 | Meritnation.com
The circles x2 + y2 + x + y = 0 and x2 + y2 + x - y = 0 intersect at an angle?
If S = x2 + y2 + 2g1x + 2f1y + c1 = 0
and S“ ≡ x2 + y2 + 2g2x + 2f2y + c2 = 0
\mathrm{then} \mathrm{angle} \mathrm{between} \mathrm{them} \mathrm{is} \mathrm{same} \mathrm{as} \mathrm{angle} \mathrm{between} \mathrm{their} \mathrm{tangent} \mathrm{and} \mathrm{formula} \phantom{\rule{0ex}{0ex}}\mathrm{is} \phantom{\rule{0ex}{0ex}}\mathrm{cos\theta }=\frac{2\left({\mathrm{g}}_{1}{\mathrm{g}}_{2}+{\mathrm{f}}_{1}{\mathrm{f}}_{2}\right)-{\mathrm{c}}_{1}-{\mathrm{c}}_{2}}{\sqrt{{{\mathrm{g}}_{1}}^{2}+{{\mathrm{f}}_{1}}^{2}-{\mathrm{c}}_{1}}×\sqrt{{{\mathrm{g}}_{2}}^{2}+{{\mathrm{f}}_{2}}^{2}-{\mathrm{c}}_{2}}}\phantom{\rule{0ex}{0ex}}=\frac{2\left(\frac{1}{2}×\frac{1}{2}+\frac{1}{2}×\left(\frac{-1}{2}\right)\right)-0}{\sqrt{{{\mathrm{g}}_{1}}^{2}+{{\mathrm{f}}_{1}}^{2}-{\mathrm{c}}_{1}}×\sqrt{{{\mathrm{g}}_{2}}^{2}+{{\mathrm{f}}_{2}}^{2}-{\mathrm{c}}_{2}}}\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}\mathrm{so} \mathrm{cos\theta }=0 = \mathrm{cos}90\phantom{\rule{0ex}{0ex}}\mathrm{so} \mathrm{\theta }=90 \mathrm{and} \mathrm{hence} \mathrm{they} \mathrm{are} \mathrm{at} \mathrm{right} \mathrm{angle} |
EUDML | Some Perturbation Results for Analytic Semigroups. EuDML | Some Perturbation Results for Analytic Semigroups.
Some Perturbation Results for Analytic Semigroups.
W. Desch; W. Schappacher
Desch, W., and Schappacher, W.. "Some Perturbation Results for Analytic Semigroups.." Mathematische Annalen 281.1 (1988): 157-162. <http://eudml.org/doc/164407>.
@article{Desch1988,
author = {Desch, W., Schappacher, W.},
keywords = {analytic semigroups; infinitesimal generator; compact linear operator; graph norm; -semigroup},
title = {Some Perturbation Results for Analytic Semigroups.},
AU - Desch, W.
AU - Schappacher, W.
TI - Some Perturbation Results for Analytic Semigroups.
KW - analytic semigroups; infinitesimal generator; compact linear operator; graph norm; -semigroup
analytic semigroups, infinitesimal generator, compact linear operator, graph norm,
{C}_{0}
Articles by W. Desch
Articles by W. Schappacher |
Elvin found the box plot below in the school newspaper.
Number of hours spent watching TV each week
Based on the plot, what percent of students watch more than
10
hours of television each week?
The box plot illustrates the quartiles of the data represented by the box in the middle.
If the right side of the box over the
10\text{th}
hour represents the
3\text{rd}
quartile of the data or the
75\text{th}
percentile, what percent of students watch more than
10
25\%
Based on the plot, what percent of students watch less than
5
If the middle line of the box represents the middle half of all the data, what percent of students watch less than
5
Can Elvin use the plot to find the mean (average) number of hours of television students watch each week? If so, what is it? Explain your reasoning.
Box plots are useful for representing the distribution of data, but they do not express any specific data other than the minimum and maximum.
Since the mean would be the average number of hours that students watched television each week, could we calculate the mean with the information given or are specific data points necessary? Remember, mean and median are not the same.
No, because individual data points are not given.
Can he use the plot to find the median? If so, what is it? Explain your reasoning.
The median is the middle half of all the data. Where on the box plot can this be represented? Does this make sense? |
Portfolio Margin - Delta Exchange - User Guide
Portfolio margining is a risk based approach to margining that allows for effective margin coverage while ensuring efficient use of capital. In this method, the risk of a group of positions and orders in futures and options with the same underlying is analysed together to compute the combined margin requirement for the entire group. Hence the name portfolio margin.
Portfolio margin tends to be more capital efficient than isolated or cross margin, i.e. requires less margin for the same set of positions. This capital efficiency emerges when a portfolio has positions/ orders with offsetting risks. In cross or isolated margin, the margin requirement for a group of positions is simply the sum of the margin requirement for each position individually. So, recognition of offsetting risks is just not possible. Portfolio margin overcomes this limitation by assessing the risk of the entire group together.
Obvious examples of such portfolios are option spreads and futures calendar spreads. The combination of long and short positions in spread trades makes them much less risky than standalone long or short positions in the same contracts. This lowering of risk is taken into account in portfolio margin, unlike in the case of isolated or cross margin.
Key points about portfolio margin
Portfolio margin is available only on USDT settled futures and options contracts. This means futures, including perpetual contracts, and options on BTC, ETH, XRP, SOL, AVAX, MATIC and BNB are eligible for portfolio margining
Portfolio margin is not available on MOVE contracts
Portfolio margin can be enabled only on a single underlying at a time. For e.g. if portfolio margin is enabled on BTC, then: (a) all USDT settled BTC futures and options contracts will be in portfolio margin mode, and (b) contracts for any other underlying cannot be moved to portfolio margin mode
The margin mode of any position or open order cannot be changed. This means that if you wish to enable/ disable portfolio margin for a particular underlying, you must not have any open positions or orders in all the USDT settled futures and options on that contract
Your entire account balance except what is in use for isolated margined orders and positions is used for portfolio margining
The margin requirement for a portfolio with offsetting positions (e.g. futures/ options spreads) is likely to be much lower than the margin requirement for positions individually. Consequently, you might end up in a situation, wherein closing one position might leave the rest of your portfolio insufficiently margined. In such cases, you may need to close all the portfolio margined positions together
Because margin requirements are measured and maintained for the entire portfolio, liquidation prices for individual positions in a portfolio are not available
Portfolio margin liquidation process is quite complex. The Liquidation engine attempts to reduce the risk and hence margin requirement of your portfolio through a combination of scaling down existing positions and acquiring new futures positions on your behalf
We may periodically update the parameters used in the computation of portfolio margin or make changes to the portfolio margin methodology to better reflect market conditions. The exchange will provide sufficient time for traders to manage their portfolio margined positions/ orders in the event of such changes.
How to enable/ disable portfolio margin
You can enable portfolio margin either from the margin mode selector on the top of the order placement panel or from the preferences page. Please note that if you choose to enable portfolio margin from the preferences page, it will be enabled by default on BTC.
Once portfolio margin for an account is enabled, it will stay enabled on one of the eligible underlying for portfolio margin until it is disabled. You can disable portfolio margin on an account from the preferences page.
Portfolio margin calculation methodology
Portfolio margin is computed by stress testing the portfolio in a range of simulated market conditions. The margin requirement for the portfolio is set at a level that portfolio remains sufficiently margined in all the stress test scenarios.
Portfolio margin is comprised risk margin and contingency margins as per the following equation
Portfolio margin = Risk margin + Futures contingency margin + Options contingency margin
Risk margin is the maximum likely loss that the portfolio will incur in a range of simulated price and volatility scenarios.
Underlying price range: The price stress range is defined around the current price of the underlying with the percentage span varying with the underlying.
Price stress - down
Price stress - up
Example: If current BTC price is 35000, then in risk margin computation, BTC price will be varied from 35000 * ( 1 - 10%) = 31500 to 35000 * (1 +10%) = 38500.
Implied volatility (IV) range: The volatility stress range is defined around the current mark IV and is dependent on the time to expiry of an options.
IV max up = 45% * (30/DTE)^0.30
IV max down = 30% * (30/DTE)^0.30
where DTE = days to expiry
Days to expiry (DTE)
IV max up
IV max down
Example: If the current mark IV for a BTC option due to expire in 90 days is 60%, then in the risk margin computation, IV will be varied between 38.42% and 92.37%.
We create 27 scenarios, each with a unique combination of underlying price movement and IV movement. The underlying price is changed in steps of 0%, 33%, 50%, 67% and 100% of the price stress range, in both up and down directions. For IV, three values are considered: unchanged, up ( IV max) and down (IV min). For each scenario, PNL of the portfolio is computed.
Risk margin = max portfolio loss across the 27 scenarios
Underlying price change as % of price stress range
Contingency Margins
Contingency margins aim to consider risk factors that are not adequately captured in change in underlying price or volatility. For example, a futures calendar spread (long BTC June futures/ Short BTC March futures) position will remain unperturbed by underlying price and volatility stress testing and will have no risk margin requirement. Similarly, the risk of a portfolio having short deep OTM options may be underestimated in risk margin calculations.
Contingency margin consists of future and option contingency margin.
Futures contingency margin
Futures contingency margin is equal to sum of 1% of absolute notional position size for all the futures contracts in the portfolio
Futures\ contingency\ margin = 1\% * \sum|Futures\ Notional|
Option contingency margin
Option contingency margin is equal to 1% of absolute Option sum notional position. Here, Option sum is the sum of net short option positions across strikes for a given maturity. In doing this computation, short positions at a far OTM strike is nullified by long options positions at near ATM strikes.
Option\ contingency\ margin = 1\% * \sum_{Expiries} \sum_{Strikes} Option\ sum
To calculate the Option sum for a maturity, options are divided into two categories; one with strikes above the current price, and other with strike below the current price.
For each strike, Net option position is equal to sum of put and call positions discounted by a factor which depends on distance of the strike from the prevailing underlying price.
Net\ Option\ Position = DF* Option\ Sum
where,\ DF (discount\ factor) = min (1, \%OTM/ 0.1)
and,\ \%OTM = |Strike\ Price - Underlying\ Price|/Underlying\ Price
Example of Option Sum calculation
Spot = 50000
Category1 : Option positions with strikes above 50000
Rolled over longs
Option Sum
Category2 : Option positions with strikes below 50000
Option Sum = Option Sum for category 1 + option Sum for Category 2 = 44
Margin Floor
Margin floor is applied to ensure a minimum margin is charged for all portfolios. Margin floor scales with the notional size of the portfolio and is bigger for bigger portfolios.
Sum of notional sizes of short option positions and orders across all options is the Options notional for margin floor
Options\ notional\ for\ margin\ floor = \sum_{Options} Short \ positions + Sell\ orders
Futures notional for margin floor is the higher of the combined long or short positions across all futures and perpetual contracts in the relevant underlying
Futures\ notional\ for\ margin\ floor = max (\sum_{Futures} Long\ positions + Buy\ orders, \sum_{Futures} Short \ positions+ Sell\ orders
Notional for margin floor
Total\ notional = Options\ notional \ for\ margin\ floor+ Futures\ notional\ for\ margin\ floor
Margin\ floor = mm\% * Total\ Notional
where, mm% is the minimum margin% which is dependent on both Total notional and the Underlying
mm\% = 0.2\% + Slope * max (0, Total\ notional - Max\ leverage\ notional)
Slope and Max leverage notional values are as follows:
Max leverage notional
In the discussion thus far, the implicit assumption has been that all the positions in the portfolio are entered into at current prices. We introduce a term, Unrealised cashflows (UCF) to factor in mark to market gains/ losses. For futures, UCF is equal to the unrealised PnL and for options, it is equal to the expected pay-off.
Portfolio margin requirements are computed using the following equations:
Initial margin = min (Risk margin + Futures contingency margin + Option contingency margin, Margin floor) + UCF
Maintenance margin = 80% * (Initial margin - UCF) + UCF
If you do not have sufficient collateral to meet the Initial margin requirement, you cannot open a new position. And, if you do not have sufficient collateral to meet the Maintenance margin requirement, your portfolio goes into liquidation.
Order margins
The margin requirement for an order is equal to the increase in Risk margin requirement of the portfolio after the order is added to it. The order limit prices are taken into consideration when computing the losses for the portfolio in various stress testing scenarios.
An order which will not get executed within the Price or volatility stress testing range does not incur any margin. Furthermore, Contingency margins are not applicable for orders.
Portfolio margined positions go into liquidation if the available collateral is not sufficient to meet the maintenance margin requirement. The key idea in liquidation is to reduce the margin requirement of the portfolio through a combination of reducing the portfolio delta and scaling down open positions.
Steps in portfolio margin liquidation
After each step, portfolio margin requirement is recalculated. The liquidation process stops as soon as a state is achieved in which collateral available for portfolio margining is more than the initial margin requirement.
All open orders in portfolio margined contracts are canceled
Margin requirement for the portfolio is recomputed after assuming that the delta risk of the portfolio has been completely hedged by taking appropriate position in the perpetual contract underlying
The percentage reduction in the sizes of all positions in the hypothetical delta hedged portfolio required to make the portfolio sufficiently margined is computed
Margin requirement reduction through delta hedging
Risk margin is the typically the biggest contributor to the margin requirement of a portfolio. When a portfolio is delta hedged, the portfolio value stays broadly constant as underlying price is simulated through the price stress range. This helps to reduce the margin requirement.
Theoretically, a portfolio could be delta hedged by trading either futures or options contracts. However, since liquidity in futures is typically greater than in options, only futures are traded to make the portfolio delta neutral.
Therefore, if your portfolio goes into liquidation, our Liquidation engine may take a position in the perpetual contract of the underlying on your behalf.
It is important to note that the actual execution of the trading actions of the Liquidation engine only once the final state which would be sufficiently margined is known. This means that if the Liquidation engine estimates that making the portfolio delta neutral would be sufficient, that step is executed. If not, the Liquidation engine executes trades for both delta hedging and position size reduction together.
Immediately after the Liquidation engine is done executing the above-mentioned trades, the portfolio is checked for margin sufficiency, i.e. is the initial margin required for the new portfolio is less than the available collateral. If yes, liquidation process stops. If not, steps 2 and/ or 3 of the liquidation process are repeated. This loop continues until the portfolio is sufficiently margined or completely liquidated.
What to expect in a portfolio liquidation
Please note that you will not have access to your portfolio while it is in liquidation. This means you will not be able to place close existing portfolio margined positions or orders or place new portfolio margined orders. Typically, the liquidation process should not take more than a few seconds.
Once the liquidation process is complete, we will send you an email which will have full details of the action taken by the Liquidation engine. We strongly encourage you to thoroughly review your updated portfolio. If you so wish, you could close the positions the Liquidation engine may have acquired to delta hedge your portfolio. |
EUDML | Invariant subspaces for polynomially compact almost superdiagonal operators on . EuDML | Invariant subspaces for polynomially compact almost superdiagonal operators on .
Invariant subspaces for polynomially compact almost superdiagonal operators on
l\left({p}_{i}\right)
Grainger, Arthur D.
Grainger, Arthur D.. "Invariant subspaces for polynomially compact almost superdiagonal operators on .." International Journal of Mathematics and Mathematical Sciences 2003.31 (2003): 1961-1971. <http://eudml.org/doc/50426>.
@article{Grainger2003,
author = {Grainger, Arthur D.},
keywords = {nonstandard analysis; invariant subspaces},
title = {Invariant subspaces for polynomially compact almost superdiagonal operators on .},
AU - Grainger, Arthur D.
TI - Invariant subspaces for polynomially compact almost superdiagonal operators on .
KW - nonstandard analysis; invariant subspaces
nonstandard analysis, invariant subspaces
Nonstandard functional analysis
Nonstandard models in mathematics
Articles by Grainger |
Error, (in Units:-RemoveSystem) the system `SI` of units is currently set to the default (see ?UseSystem) - Maple Help
Home : Support : Online Help : Error, (in Units:-RemoveSystem) the system `SI` of units is currently set to the default (see ?UseSystem)
\mathrm{with}\left(\mathrm{Units}\right):
\mathrm{AddSystem}\left('\mathrm{FHF}','\mathrm{check}','\mathrm{furlong}','\mathrm{hundredweight}[\mathrm{long}]','\mathrm{fortnight}'\right)
\mathrm{UseSystem}\left('\mathrm{FHF}'\right)
\mathrm{RemoveSystem}\left('\mathrm{FHF}'\right)
\mathrm{UseSystem}\left('\mathrm{SI}'\right)
\mathrm{RemoveSystem}\left('\mathrm{FHF}'\right)
\mathrm{RemoveSystem}\left('\mathrm{SI}'\right)
\mathrm{RemoveSystem}\left('\mathrm{CGS}'\right) |
Why is this solution incorrect? Prove that on the axis of any parabola there is a certain point
brecruicswbp 2022-03-09 Answered
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then
\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}
is the same for all positions of the chord.
If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for
{y}^{2}=4ax
let the point K be (c,0)
Equation of line PQ using parametric coordinates:
x=c+r\mathrm{cos}\theta
y=r\mathrm{sin}\theta
From equation 1 and 2:
{\left(x-c\right)}^{2}+{y}^{2}={r}^{2}
Using Equation 3 and
{y}^{2}=4ax
, we get this quadratic in r:
{r}^{2}-4ax-{\left(x-k\right)}^{2}=0
Roots of this quadratic
\left({r}_{1}\text{ }\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\text{ }{r}_{2}\right)
would be the lengths of PK and QK
{r}_{1}+{r}_{2}=0
{r}_{1}{r}_{2}=-\left(4ax+{\left(x-k\right)}^{2}\right)
\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{1}{{r}_{1}^{2}}+\frac{1}{{r}_{2}^{2}}=\frac{{\left({r}_{1}+{r}_{2}\right)}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{-2}{{r}_{1}{r}_{2}}
As value of
{r}_{1}{r}_{2}
is not constant thus
\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}
does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know why this solution is incorrect?
Lilliana Rich
\left({r}_{1}\text{ }\text{and}\text{ }{r}_{2}\right)
would be the lengths of PK and QK.
This part is incorrect. Roots of this equation correspond to the same value of x. P and Q can have different x.
x=c+r\mathrm{cos}\theta
{y}^{2}=4ax⇒{r}^{2}{\mathrm{sin}}^{2}\theta -4ar\mathrm{cos}\theta -4ac=0
Absolute values of roots of this equation would be the PK and QK.
{r}_{1}+{r}_{2}=\frac{4a\mathrm{cos}\theta }{{\mathrm{sin}}^{2}\theta }
\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{{\left({r}_{1}+{r}_{2}\right)}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{16{a}^{2}{\mathrm{cos}}^{2}\theta +8ac{\mathrm{sin}}^{2}\theta }{16{a}^{2}{c}^{2}}
This value does not depend on
\theta \text{ }\text{at}\text{ }c=2a
. This is the answer.
To convert the given radical expression
\sqrt[3]{{x}^{12}}
to its rational exponent form and then simplify
A latus rectum of a conic section is a chord through a focus parallel to the directrix. Find the area bounded by the parabola
y={x}^{2}\text{/}\left(4c\right)
and its latus rectum.
How will look the general conic sections equation
A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0
in case of rotated and shifted from coordinates origin ellipse, parabola and hyperbola?
I need a formulas for coefficients A, B, C, D, E and F for ellipse, hyperbola and parabola. I did it for not-rotated conic sections at the origin of coordinates but have a difficults with rotated and shifted.
For i.e. standart ellipse equation
\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1
gives me general equation
\frac{1}{{a}^{2}}{x}^{2}+0xy+\frac{1}{{b}^{2}}{y}^{2}+0x+0y-1=0
I need the same in case if ellipse located in position (h;k) and rotated on some angle
\alpha
from positive X axis. And the same for parabola and hyperbola.
Write the following equation in standard form and sketch its graph
9{x}^{2}+72x-64{y}^{2}+128y+80=0
The type of the conic section using the Discriminant Test and plot the curve using a computer algebra system.
{x}^{2}\text{ }-\text{ }2xy\text{ }+\text{ }{y}^{2}\text{ }+\text{ }24x\text{ }-\text{ }8=0
Word problem involving comic sections. A hall 100 feet in length is designed as a whispering gallery (See example 9 for Ellipses). If the focus is 25 feet from the center, how high will the ceiling be at the center?
Where is the vertical directrix of a conic if the denominator of its polar equation has the form
1-e\mathrm{cos}q |
FAQs - Orca
Orca is the easiest place to exchange cryptocurrency on the Solana blockchain. On Orca, you can exchange tokens cheaply, quickly, and confidently (thanks to our Fair Price Indicator). Additionally, you can provide liquidity to one of our liquidity pools, including our concentrated liquidity pools (Whirlpools) to earn trading fees and token emissions.
Of the solutions competing to scale the blockchain ecosystem, Solana's performance stands out: 50,000 transactions per second, 400ms block times, and $0.01 average transaction fees. What’s more, we've been continually impressed by the value, vision, and engineering chops of the Solana team. For these reasons, we believe that Solana is the best choice to support the next generation of DeFi apps.
Do you have a governance token?
The ORCA governance token was launched on Monday, August 9, 2021. Details on the ORCA token can be found on the Tokenomics page.
ORCA token mint address: orcaEKTdK7LKz57vaAYr9QeNsVEPfiu6QeMU1kektZE (Solana Explorer)
For details of our governance process please see the Govenance v0 User Guide.
Was there an airdrop?
The ORCA token was Fair Launched (retroactively airdropped) to liquidity providers and traders on Monday, August 9, 2021. Read more about Orca's tokenomics and airdrop details on the Tokenomics page.
Several of our collectibles have already been distributed. If you are lucky enough to have earned one, you can view it here.
Was there a public sale?
No, there has not been any public sales of the ORCA token.
There was a private fundraising round that took place in September 2021, during which 9.6% of the ORCA token was allocated. All investors are subject to a three-year vest and a linear unlock after the one-year lockup.
Public information about the fundraise can be found here:
Which wallets can I use with Orca?
To connect to Orca you will need a compatible wallet, we currently support: Phantom (available on desktop and mobile); Solflare (a multiplatform wallet available on desktop and mobile); Math Wallet (a multichain wallet available on desktop and mobile); Coin98 extension (a multichain wallet); Slope (available on desktop and mobile); or Sollet.io (the Solana wallet created by Project Serum). We plan to integrate other wallets that support Solana program execution, as they are released.
Can I use Orca on my phone?
You can use Orca on the phone with Phantom, Solflare, Math Wallet or Slope.
Is the Orca smart contract open-source?
Yes! The code for our Constant Product pools are here, our Stable curve pools are here, and Whirlpools are here. Note that the Orca smart contract uses the token-swap program developed by the Solana team. The Constant Product pool and Whirlpool smart contracts have undergone formal audits.
Has Orca been audited?
We use the Solana team's token-swap program. Orca's Constant Product pools use a version which was audited by crypto security firm Kudelski. This deployment uses the commit hash: 3613cea3c.
Orca's Stable curve pools use a more recent version which has not been through a formal audit. However, it has been reviewed by Orca and Solana's engineers. The deployment uses the commit hash: 813aa3.
Orca's Concentrated Liquidity product Whirlpools has been double audited by Kudelski and Neodyme.
What is the Orca Treasury wallet address?
Orca strives for transparency and in that spirit we disclose relevant wallet addresses as follows:
Orca Treasury and Orca Impact Fund https://solscan.io/account/2YM8LrJGRtsDcWeqsjX2EQwJfhArxyDdtDzgt7vrwwbV
ORCA token initial source wallet (see our most recent Treasury Report for more details): https://solscan.io/account/GjKxauMroc9MT27xXYyjLYNfy9nSS5EuHTmzhdkjCAR1
What are the risks of using Orca?
Orca is a new application on a relatively new blockchain. There are a number of risks to using Orca:
1. Smart contract vulnerabilities: The Solana mainnet is still in beta, and there is always the possibility of an exploit in the smart contract. To mitigate these risks, Orca's smart contract uses the Solana Foundation's token-swap program, which has been audited by crypto security firm Kudelski. It is also very similar in structure to well-understood AMMs on Ethereum, such as Uniswap. Nonetheless, we strongly encourage taking the time to understand the risks before trading.
2. Divergence loss (impermanent loss): As token prices diverge from their prices at deposit your liquidity becomes lower in value when compared to its value if tokens were held outside the pool. Depending on volatility, your liquidity stake may become lower than your deposit when you withdraw. Though this has historically been rare on Uniswap, large price swings could cause liquidity providers to lose money. For more information, this blog post from the Uniswap team is a great primer on impermanent loss. The risk of divergence loss is amplified in Whirlpools, please see here.
3. Wallet providers: Orca is compatible with a wide range of wallets, including Phantom, Solflare, MathWallet, Coin98, Slope and an open-source Solana wallet called Sollet.io. A wallet exploit could affect the user.
What does "Not Enough SOL" mean?
SOL is required to pay network fees. The actual fees are likely to be lower, but for simplicity, a small minimum balance of SOL is required to transact on Orca.
(For more details, see the below FAQ: What fees do I pay when I exchange tokens?)
What fees do I pay when I exchange tokens?
Liquidity provider fee: When swapping, you pay a percentage of the trade value as a fee. These fees vary depending on the route the AMM uses, see below for details of the fees. (NOTE: a "double-hop swap" such as SOL -> USDT -> ETH would pay 0.3% to swap SOL -> USDT and a further 0.3% to swap USDT -> ETH.) Regular Pools have a swap fee of 0.3%, composed of 0.25% paid as earnings for liquidity providers and the remaining 0.05% split 80/20 into the Treasury and Impact Fund respectively. Stable Pools (such as USDC/USDT or SOL/mSOL) have a lower swap fee, 0.07%, than other Constant Product pool, composed of 0.06% paid to liquidity providers and the remaining 0.01% split 80/20 into the Treasury and Impact Fund respectively. Whirlpools have a 0.25% swap fee, of which 0.24% goes to liquidity providers and the remaining 0.01% to the Impact Fund.
Stable Whirlpools have a swap fee of 0.01% which goes to liquidity providers.
Network fee: when trading a nominal amount of SOL is also paid in the form of Solana network fees. The exact amount varies depending on the parameters of the trade; when trading a token for the first time, more SOL is required to add that token to your wallet. In the past, we've found that most trades cost between 0.0001 — 0.001 SOL.
Orca does not charge any additional fees.
How does the Fair Price Indicator work?
When you enter a trade, there are two factors that determine whether we show a Fair price label:
Is the price per token within 1% of the rate quoted by CoinGecko?
Is the price impact caused by this trade less than 1%?
If either of these two conditions is not met, you'll see either the "Rate Warning!" or "Great Price!" alert in the UI. When a Rate Warning label is displayed you can still trade after acknowledging the warning.
The price you get on Orca depends on the size of the order. As the amount of tokens you buy from the pool increases, the price of the token increases as well. This increase in price is called price impact.
Which curves do your trading pools use?
Our Pools use the Constant Product curve (x * y = k) popularized by Uniswap, as well as the stable curve popularized by curve.fi for Stablecoin pools (currently USDC/USDT).
Our Whirlpools are concentrated liquidity pools, similar to Uniswap v3. Each liquidity provider chooses the range within which they want to provide liquidity and swaps are executed using the combined liquidity of all individual curves currently in range.
If you have ideas for other curves, please don't hesitate to reach out!
Why did my exchange fail?
The exchange will fail if the price of the underlying pools moves past your Slippage Tolerance setting. Increasing the tolerance in your local settings will raise the chances of your exchange succeeding, but also increase the probability of another party front-running your trade.
Due to current limitations of the smart contract, exchanges that route through multiple pools have a higher likelihood of failing due to slippage. The transaction may succeed if sent again. We recognize that this could be frustrating, so we have plans in the works to improve the logic for trades that use multiple pools.
During periods of Solana network congestion there is an increased likelihood of transaction failure.
How do you decide which tokens to list on Orca?
Orca's team will list tokens based on information provided by the project and demonstrated community demand (e.g., volume or interest). Orca protocol is a DEX and and will aim to support every asset that is tradable on Solana.
Are tokens on Orca wrapped?
Some tokens on Orca are wrapped, including BTC, ETH, AVAX, LUNA, LidoDAO, Celo, FTM and UST. There are several source protocols for wrapped tokens, you can find more details in our Tokenpedia.
I have ERC-20 tokens in MetaMask. How can I trade them on Orca?
To trade ERC-20s on Orca, you’ll need to convert them to SPL tokens by transferring them from MetaMask to your native SPL wallet. For a step-by-step guide, see How to Exchange Tokens.
Why should I provide liquidity on Orca?
Users typically provide liquidity to Orca’s pools to earn trading fees. Example: you provide liquidity to the SOL/USDC Pool and receive liquidity tokens (LP tokens) in return, these act like a receipt for your deposited tokens.
Each time a swap routes through the SOL/USDC pool, 0.3% of the trade value is added to the pool. Over time these fees accumulate increasing the Pool's value relative to the value of its original deposits.
When you redeem your liquidity tokens, you will receive the share of the SOL/USDC Pool that you LP tokens represent, this will include your share of any accumulated fees. (Note: due to divergence loss, you are not guaranteed a positive return).
NOTE: in Whirlpools fees are not added to the pool and can be harvested at any time.
How do I earn fees on Orca?
Trading fees are automatically added to Pools after each swap. These fees increase the amount of tokens that you get back when you withdraw liquidity. While you cannot harvest trading fees, without withdrawing at least some liquidity, you can harvest ORCA rewards from Aquafarms and other tokens from eligible Double-Dip pools at any time. All earned yields are harvestable from Whirlpools at any time.
How do I track trading fees earned from Pools?
For now, we recommend manual tracking. To do this, record how many of each token you've deposited, along with how many tokens you're able to currently withdraw, then use the following formula:
growth = \dfrac{\sqrt{currentTokenA * currentTokenB}}{\sqrt{initialTokenA*initialTokenB}}
This number represents the percentage growth of your deposit from trading fees. NOTE: you will need to subtract divergence loss to determine net earnings.
What are liquidity tokens?
Liquidity tokens (also called LP Tokens) represent a share of a liquidity pool. For instance, if you contribute to the SOL/USDC pool, you will receive SOL/USDC liquidity tokens. If you have deposited liquidity, you will be able to see liquidity tokens in your wallet.
Given a pool with: pool token supply, token A, and token B: we use the following formula to calculate the ratio: (token A * token B)^(1/2) / (pool token supply). This ratio increases when trading occurs in the pool, regardless of how the pool size changes. We sample this value every 10 minutes and the data is used to calculate the ratio growth over time. This allows us to project an APY value for any sampled time period.
Can I withdraw my liquidity at any time?
Yes, Orca allows you to redeem liquidity tokens for your share of tokens at any time.
Why aren't stable pool pairs balanced like other Pools?
The deposits and withdraws for stable pools are often imbalanced, but this is intended behavior.
Stable pools are designed to maintain a relatively stable price despite large imbalances in the pool. For example, the mSOL/SOL pool may contain far more mSOL than SOL, but the exchange rate will remain close to even (e.g. 1 mSOL for 1 SOL). However, this means that liquidity providers will deposit and withdraw more mSOL than SOL, since the amount they provide must be proportional to the amount of tokens currently in the pool.
This is in contrast to other Constant Product pools where changes in the balance of the pool directly effect the price of the tokens. Since the balance of the pool is proportional to the price of the tokens, liquidity providers can generally expect the value of each token to be the same when depositing or withdrawing.
Can I yield farm / stake on Orca?
Yes, Aquafarms are Orca’s yield farming program. View the How to Provide Liquidity on Orca page for a step-by-step guide on how to participate as an LP.
Aquafarm LPs earn both trading fees and ORCA emissions. You will see ORCA rewards accrue continuously and can harvest any time with no lock-ups. Trading fees accrue directly to your LP tokens and are available when you withdraw. Double-Dip Pools are a special kind of Aquafarm that allow LP holders to earn additional tokens, please see How to Use Double-Dip Pools.
Update: For a limited time between 28th September and 12th October 2021, ORCA holders were able to stake their ORCA in single-sided pools and earn a share of 65k ORCA in total rewards. Over 2m ORCA were staked by ORCA holders, earning roughly 72% APR.
To learn more about how to stake on Orca.so, check out our video tutorial.
Collectible tokens, featuring creatures from our marine ecosystem have been given as a token of appreciation to users at various times. They have no intrinsic value, but can be transferred like any other SPL token. On desktop when connecting to Orca using a wallet that contains collectibles you will enjoy an enhanced background with more abundant sealife.
I previously staked ORCA or a Collectible. How do I unstake?
Visit https://www.orca.so/staking and toggle to "Ended." You'll be able to unstake from that screen!
Can I exchange programmatically on Orca?
Of course! You can interact with our pools using the token-swap Javascript client from the Solana engineering team. You can also view the deployed program on the Solana Explorer.
Can I integrate with Orca?
Yes! You can read more in Integrations, but also check out #integrate-with-orca.
Please tell us about it in #bug-reports.
I have feedback or an idea for a feature, how can I share it with Orca?
We value your feedback! Share it directly with us in #feedback.
You can find us on Discord, Telegram, or Twitter (@orca_so). We look forward to chatting with you! |
Swappers - THORChain Docs
THORChain's value proposition for Swappers.
On THORChain, users can swap their digital assets for other digital assets. The network aims to give users access to:
A large variety of assets through cross-chain compatibility and simple asset listing
Superior user experience through open finance protocols and permissionless access
1-transaction access to fast chains (Binance Chain), smart chains (Ethereum), censorship-resistant chains (Bitcoin) and private chains (Monero).
See an overview of Swapping in THORChain
Users can swap any assets which are on connected chains and which have been added to the network. Users can swap from any connected asset to any other connected asset. They can also swap from any connected asset to RUNE.
Learn more about how chains and assets get added to the network in the Governance section.
THORChain manages the swaps in accordance with the rules of the state machine - which is completely autonomous. Every swap that it observes is finalised, ordered and processed. Invalid swaps are refunded, valid swaps ordered in a transparent way that is resistant to front-running. Validators can not influence the order of trades, and are punished if they fail to observe a valid swap.
Swaps are completed as fast as they can be confirmed, which is around 5-10 seconds.
Swaps on THORChain are made possible by liquidity pools. These are pools of assets deposited by Liquidity providers, where each pool consists of 1 connected asset, for example Bitcoin, and THORChain's own asset, RUNE. They're called Continuous Liquidity Pools because RUNE, being in each pool, links all pools together in a single, continuous liquidity network.
When a user swaps 2 connected assets on THORChain, they swap between two pools:
Swap to the desired asset in the second pool with the RUNE from (2)
The THORChain state machine handles this swap in one go, so the user never handles RUNE.
See this example for further detail and the page below for broader detail on Continuous Liquidity Pools.
Calculating Swap Output
The output of a swap can be worked out using the formula
y = \frac{ xYX} {(x+X)^2 }
x is input asset amount
X is input asset balance
y is output asset amount
Y is output asset balance
The BTC.RUNE pool has 100 BTC and 2.5 million RUNE. A user swaps 1 BTC into the pool. Calculate how much RUNE is output:
\frac {1 * 2500000 * 100 } {(1 + 100)^2} = 24,507.40
This user swaps 1 BTC for 24,507.40 RUNE.
The cost of a swap is made up of two parts:
All swaps are charged a network fee. The network fee is dynamic – it's calculated by averaging a set of recent gas prices. Learn more about Network Fees.
Note that users who force their swaps through quickly cause large slips and pay larger fees to liquidity providers. |
Distance (graph theory) - Wikipedia
"Geodesic distance" redirects here. For distances on the surface of a sphere, see Great-circle distance. For distances on the surface of the Earth, see Geodesics on an ellipsoid. For geodesics in differential geometry, see Geodesic.
In the mathematical field of graph theory, the distance between two vertices in a graph is the number of edges in a shortest path (also called a graph geodesic) connecting them. This is also known as the geodesic distance or shortest-path distance.[1] Notice that there may be more than one shortest path between two vertices.[2] If there is no path connecting the two vertices, i.e., if they belong to different connected components, then conventionally the distance is defined as infinite.
In the case of a directed graph the distance
{\displaystyle d(u,v)}
{\displaystyle u}
{\displaystyle v}
is defined as the length of a shortest directed path from
{\displaystyle u}
{\displaystyle v}
consisting of arcs, provided at least one such path exists.[3] Notice that, in contrast with the case of undirected graphs,
{\displaystyle d(u,v)}
does not necessarily coincide with
{\displaystyle d(v,u)}
—so it is just a quasi-metric, and it might be the case that one is defined while the other is not.
2 Algorithm for finding pseudo-peripheral vertices
The eccentricity
{\displaystyle \epsilon (v)}
of a vertex
{\displaystyle v}
is the greatest distance between
{\displaystyle v}
and any other vertex; in symbols that is
{\displaystyle \epsilon (v)=\max _{u\in V}d(v,u)}
. It can be thought of as how far a node is from the node most distant from it in the graph.
{\displaystyle r}
of a graph is the minimum eccentricity of any vertex or, in symbols,
{\displaystyle r=\min _{v\in V}\epsilon (v)=\min _{v\in V}\max _{u\in V}d(v,u)}
The diameter
{\displaystyle d}
of a graph is the maximum eccentricity of any vertex in the graph. That is,
{\displaystyle d}
is the greatest distance between any pair of vertices or, alternatively,
{\displaystyle d=\max _{v\in V}\epsilon (v)=\max _{v\in V}\max _{u\in V}d(v,u)}
. To find the diameter of a graph, first find the shortest path between each pair of vertices. The greatest length of any of these paths is the diameter of the graph.
A central vertex in a graph of radius
{\displaystyle r}
is one whose eccentricity is
{\displaystyle r}
—that is, a vertex that achieves the radius or, equivalently, a vertex
{\displaystyle v}
{\displaystyle \epsilon (v)=r}
A peripheral vertex in a graph of diameter
{\displaystyle d}
is one that is distance
{\displaystyle d}
from some other vertex—that is, a vertex that achieves the diameter. Formally,
{\displaystyle v}
is peripheral if
{\displaystyle \epsilon (v)=d}
A pseudo-peripheral vertex
{\displaystyle v}
has the property that for any vertex
{\displaystyle u}
{\displaystyle v}
is as far away from
{\displaystyle u}
as possible, then
{\displaystyle u}
{\displaystyle v}
as possible. Formally, a vertex u is pseudo-peripheral, if for each vertex v with
{\displaystyle d(u,v)=\epsilon (u)}
{\displaystyle \epsilon (u)=\epsilon (v)}
The partition of a graph's vertices into subsets by their distances from a given starting vertex is called a level structure of the graph.
The weighted shortest-path distance generalises the geodesic distance to weighted graphs. In this case it is assumed that the weight of an edge represents its length or, for complex networks the cost of the interaction, and the weighted shortest-path distance
{\displaystyle d^{W}(u,v)}
is the minimum sum of weights across all the paths connecting
{\displaystyle u}
{\displaystyle v}
. See the shortest path problem for more details and algorithms.
Algorithm for finding pseudo-peripheral vertices[edit]
Choose a vertex
{\displaystyle u}
Among all the vertices that are as far from
{\displaystyle u}
as possible, let
{\displaystyle v}
be one with minimal degree.
{\displaystyle \epsilon (v)>\epsilon (u)}
{\displaystyle u=v}
and repeat with step 2, else
{\displaystyle u}
is a pseudo-peripheral vertex.
^ Bouttier, Jérémie; Di Francesco,P.; Guitter, E. (July 2003). "Geodesic distance in planar graphs". Nuclear Physics B. 663 (3): 535–567. arXiv:cond-mat/0303272. doi:10.1016/S0550-3213(03)00355-9. By distance we mean here geodesic distance along the graph, namely the length of any shortest path between say two given faces
^ Weisstein, Eric W. "Graph Geodesic". MathWorld--A Wolfram Web Resource. Wolfram Research. Retrieved 2008-04-23. The length of the graph geodesic between these points d(u,v) is called the graph distance between u and v
^ F. Harary, Graph Theory, Addison-Wesley, 1969, p.199.
^ Øystein Ore, Theory of graphs [3rd ed., 1967], Colloquium Publications, American Mathematical Society, p. 104
Retrieved from "https://en.wikipedia.org/w/index.php?title=Distance_(graph_theory)&oldid=1084540210" |
\textcolor[rgb]{0.501960784313725,0.501960784313725,0.501960784313725}{\mathbf{R}}″
\mathbf{R}″\left(p\right)=\mathrm{ρ}\prime \mathbf{T}+\mathrm{κ} {\mathrm{ρ}}^{2}\mathbf{N}
\mathbf{R}″\left(p\right)=\mathrm{\rho }\prime \mathbf{T}+\mathrm{\kappa } {\mathrm{\rho }}^{2}\mathbf{N}
follows from the Frenet formulas
\mathbf{T}\left(p\right)=\mathbf{R}\prime /\mathrm{ρ}
\mathbf{N}=\frac{1}{\mathrm{κ}} \frac{d\mathbf{T}}{\mathrm{ds}}=\frac{1}{\mathrm{κ} \mathrm{ρ}} \mathbf{T}\prime \left(p\right)
if they are written as
\mathbf{R}\prime =\mathrm{ρ} \mathbf{T}
\frac{d\mathbf{T}}{\mathrm{ds}}=k \mathbf{N}
Differentiating, with respect to
p
, the first of these, gives
\mathbf{R}″\left(p\right)=\left(\mathrm{ρ} \mathbf{T}\right)\prime =\mathrm{ρ}\prime \left(p\right)\mathbf{T}+\mathrm{ρ} \mathbf{T}\prime \left(p\right)
By the chain rule, the second term on the right becomes
\mathbf{T}\prime \left(p\right)=\frac{d\mathbf{T}}{\mathrm{dp}}=\frac{d\mathbf{T}}{\mathrm{ds}}\frac{\mathrm{ds}}{\mathrm{dp}}=\frac{d\mathbf{T}}{\mathrm{ds}} \mathrm{ρ}
\mathbf{R}″\left(p\right)
\mathbf{R}″\left(p\right)=\mathrm{ρ}\prime \left(p\right)\mathbf{T}+\mathrm{ρ} \frac{d\mathbf{T}}{\mathrm{ds}} \mathrm{ρ}
which, upon replacing
\frac{d\mathbf{T}}{\mathrm{ds}}
\mathrm{κ} \mathbf{N}
, and dropping the explicit display of the independent variable
p
on the right, becomes the desired result
\mathbf{R}″\left(p\right)=\mathrm{\rho }\prime \mathbf{T}+\mathrm{\kappa } {\mathrm{\rho }}^{2}\mathbf{N} |
Asymptotic behaviour of the porous media equation in domains with holes | EMS Press
Asymptotic behaviour of the porous media equation in domains with holes
The paper deals with the asymptotic behaviour of solutions to the porous media equation,
u_t=\Delta u^m
m>1
, in an exterior domain,
\Omega
, which excludes one or several holes, and with zero Dirichlet data on
\partial\Omega
. When the space dimension is three or more this behaviour is given by a Barenblatt function away from the fixed boundary
\partial\Omega
and near the free boundary. The asymptotic behaviour of the free boundary is given by the same Barenblatt function. On the other hand, if the solution is scaled according to its decay factor, away from the free boundary and close to the holes it behaves like a function whose
m
-th power is harmonic and vanishes on
\partial\Omega
. The height of such a function is determined by matching with the Barenblatt solution representing the outer behaviour. The inner and the outer behaviour can be presented in a unified way through a suitable global approximation.
Juan Luis Vázquez, Cristina Brändle, Fernando Quirós, Asymptotic behaviour of the porous media equation in domains with holes. Interfaces Free Bound. 9 (2007), no. 2, pp. 211–232 |
Home : Support : Online Help : Programming : Names and Strings : StringTools Package : Statistics : Entropy
compute the Entropy of a string
Entropy( s )
The Entropy(s) command returns the Shannon entropy of the string s. A floating-point number, the entropy of the string, is returned.
Shannon's entropy is defined as -add( P( ch ) * log[ 2 ]( P( ch ) ), ch = Support( s ) ), where
P\left(\mathrm{ch}\right)=\frac{\mathrm{CountCharacterOccurrences}\left(s,\mathrm{ch}\right)}{\mathrm{length}\left(s\right)}
. It is a measure of the information content of the string, and can be interpreted as the number of bits required to encode each character of the string given perfect compression. The entropy is maximal when each character is equally likely. For arbitrary non-null characters, this maximal value is
{\mathrm{log}}_{2}\left(255\right)=7.99435
(The null byte, with code point
0
, cannot appear in a Maple string. If all 256 single byte code points could appear, then the maximal entropy would be
{\mathrm{log}}_{2}\left(256\right)=8
, which is the number of bits per byte).
Note that the entropy is computed as a floating-point number, at hardware (double) precision.
8
\mathbf{use}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{StringTools}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{in}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{Entropy}\left("Mathematics"\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end use}
\textcolor[rgb]{0,0,1}{3.09579525500093355}
\mathrm{with}\left(\mathrm{StringTools}\right):
\mathrm{Entropy}\left("aaaaaaaaaaaaaaaaaaaaaaaaaa"\right)
\textcolor[rgb]{0,0,1}{-0.}
\mathrm{Entropy}\left("aaaaaaaaaaaaaaaaaaaaaaaaaaB"\right)
\textcolor[rgb]{0,0,1}{0.228538143953528006}
Entropy( Iota( 1, 255 ) );
\textcolor[rgb]{0,0,1}{7.99435343685886934}
\mathrm{Entropy}\left(\mathrm{Random}\left(1000000\right)\right)
\textcolor[rgb]{0,0,1}{7.99417106407216149}
\mathrm{evalf}\left(\mathrm{log}[2]\left(255\right)\right)
\textcolor[rgb]{0,0,1}{7.994353436}
\mathrm{Entropy}\left(\mathrm{Random}\left(1000000,'\mathrm{lower}'\right)\right)
\textcolor[rgb]{0,0,1}{4.70042263084046397}
\mathrm{evalf}\left(\mathrm{log}[2]\left(26\right)\right)
\textcolor[rgb]{0,0,1}{4.700439718}
\mathrm{Entropy}\left(\mathrm{Repeat}\left("ab",100\right)\right)
\textcolor[rgb]{0,0,1}{1.}
\mathrm{Entropy}\left(\mathrm{Repeat}\left("abc",100\right)\right)
\textcolor[rgb]{0,0,1}{1.58496250072115585}
\mathrm{Entropy}\left(\mathrm{Repeat}\left("abcde",100\right)\right)
\textcolor[rgb]{0,0,1}{2.32192809488736218}
\mathrm{Entropy}\left(\mathrm{Repeat}\left(\mathrm{Random}\left(10\right),10000\right)\right)
\textcolor[rgb]{0,0,1}{3.32192809488736218}
The following steps illustrate the definition of Entropy.
s≔\mathrm{Random}\left(30,'\mathrm{lower}'\right)
\textcolor[rgb]{0,0,1}{s}\textcolor[rgb]{0,0,1}{≔}\textcolor[rgb]{0,0,1}{"rbygsggdjijjtiqelzxehfnojeorwr"}
\mathrm{occ}≔[\mathrm{seq}]\left(\mathrm{CountCharacterOccurrences}\left(s,\mathrm{ch}\right),\mathrm{ch}=\mathrm{Support}\left(s\right)\right)
\textcolor[rgb]{0,0,1}{\mathrm{occ}}\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}{3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\textcolor[rgb]{0,0,1}{,}\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}{4}\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}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{3}\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}\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}]
L≔\mathrm{map}\left(\mathrm{`/`},\mathrm{occ},\mathrm{length}\left(s\right)\right)
\textcolor[rgb]{0,0,1}{L}\textcolor[rgb]{0,0,1}{≔}[\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{2}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{15}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{10}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}\textcolor[rgb]{0,0,1}{,}\frac{\textcolor[rgb]{0,0,1}{1}}{\textcolor[rgb]{0,0,1}{30}}]
U≔\mathrm{map}\left(p↦-\mathrm{evalf}\left(p\cdot \mathrm{log}[2]\left(p\right)\right),L\right)
\textcolor[rgb]{0,0,1}{U}\textcolor[rgb]{0,0,1}{≔}[\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.3321928095}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.3321928095}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2604593730}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.3875854127}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2604593730}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.3321928095}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.1635630199}]
\mathrm{convert}\left(U,'\mathrm{`+`}'\right)
\textcolor[rgb]{0,0,1}{4.031401848}
\mathrm{Entropy}\left(s\right)
\textcolor[rgb]{0,0,1}{4.03140184539217117}
StringTools[CountCharacterOccurrences]
StringTools[Repeat] |
Dividend Yield Calculator | Dash Calculator
A quick and simple dividend yield calculator that helps you calculate the dividend yield on any stock. Just enter the stock price and dividend per share to see what the dividend yield would be.
How often is the dividend paid? Monthly Quarterly Semi-annually Annually
The dividend yield is
\textrm{Dividend yield} = \frac{\textrm{Dividend} \times \textrm{Dividend frequency}}{\textrm{Share price}}
\textrm{Dividend yield} = \frac{\$3.25 \times 4}{\$125}
\textrm{Dividend yield} = 10.40\%
The dividend yield is a financial ratio used by investors to understand how much a company is paying out in dividends relative to its share price.
The dividend yield is calculated as:
It is generally expressed as a percentage, and for stocks that pay a dividend, is generally in the range of 1% to 15%.
Calculating the dividend yield
To calculate the dividend yield, we generally look at the most recent year’s dividends, or the total dividends for the most recent four quarters. We can also use the expected dividends for the next four quarters or for the coming year. This is call the prospective dividend yield.
We take the the annual dividend per share and divide it by the current stock price for the company to arrive at the dividend yield.
Why the dividend yield matters more than the dividend
Why is the dividend yield important? Why don’t we just look at the dividends paid by the company?
A large dividend payout may look attractive, but this can be misleading without taking into account the share price. The more important metric is the dividend relative to the price paid for share, or the dividend yield.
Let’s look at an example to understand why the dividend yield is more important than the dividend.
We have two companies: Company A and Company B. The annual dividend per share is $12 for Company A and $25 for Company B. The current share price is $125 for Company A and $500 for Company B.
If we only look at the annual dividend, then Company B seems more attractive. But is this really true?
Let’s look at the dividend yield. The dividend yield for Company A is $12 / $125, which is 10%. The dividend yield for Company B is $25 / $500, which is 5%.
The dividend yield for Company A is higher, which means that Company A is actually more attractive than Company B.
To understand this more concretely, let’s say you have $5,000 to invest in either Company A or Company B. If you invest all of the $5,000 in Company A, you would have 40 shares of Company A. If you invest all of the $5,000 in Company B, you would have 10 shares of Company B.
What are the dividends that you would receive? Since the dividend per share for Company A is $12, you would receive 40 x $12, which is $480. For Company B, you would receive $25 x 10, which is $250. As suggested by the dividend yield, you would make more money investing in Company A solely based on the dividend yield.
The following table summarizes this example.
Annual dividend per share $12 $25
Dividend yield 10% 5%
My investment $5,000 $5,000
Shares purchased 40 10
Dividends received per year $480 $250
Why do investors care about dividends? They care about it is one of two ways to make money in the stock market. The way you earn money on a stock is either through:
Stock appreciation: You earn money by buying a stock at a low price and selling it at a high price. The difference between the two prices is your capital gains.
Dividend payout: You can also earn money by receiving dividend payouts from the company while you hold their shares.
Dividends are cash returns that you actually receive on hand. Stock prices fluctuate — if you buy or sell at the wrong time, you could make no money or lose money — but you can lock in some of your gains through dividend income.
Put another way, capital gains are theoretical, but dividend income is very much real.
What’s a good dividend yield?
When investing, what dividend yields are considered to be good? First, dividend yields are open to interpretation.
A high dividend yield may signal a good investment opportunity, but it may also reflect a low share price, driven by expectations that the company may not be able to sustain future dividend payouts.
A low dividend yield could indicate a poor investment opportunity, or it could indicate that the company is planning to reinvest more of its earnings into the company rather than paying out dividends.
Generally, the dividend yield of a company should be compared within the sector it is in and against the market as a whole. The S&P 500 has an overall dividend yield of around 2%. This can give you a sense if the dividend yield is high or low.
Investments by Zvi Bodie, Alex Kane, and Alan Marcus
"Dividend yield and expected returns: The zero-dividend puzzle" by William G. Christie
"The effects of dividend yield and dividend policy on common stock prices and returns" by Fischer Black and Myron Scholes |
Noise Spectrum Plots - MATLAB & Simulink - MathWorks France
What Does a Noise Spectrum Plot Show?
When you estimate the noise model of your linear system, you can plot the spectrum of the estimated noise model. Noise-spectrum plots are available for all linear parametric models and spectral analysis (nonparametric) models.
For nonlinear models and correlation analysis models, noise-spectrum plots are not available. For time-series models, you can only generate noise-spectrum plots for parametric and spectral-analysis models.
y\left(t\right)=G\left(z\right)u\left(t\right)+v\left(t\right)
In this equation, G is an operator that takes the input to the output and captures the system dynamics, and v is the additive noise term. The toolbox treats the noise term as filtered white noise, as follows:
v\left(t\right)=H\left(z\right)e\left(t\right)
where e(t) is a white-noise source with variance λ.
The toolbox computes both H and
\lambda
during the estimation of the noise model and stores these quantities as model properties. The H(z) operator represents the noise model.
Whereas the frequency-response plot shows the response of G, the noise-spectrum plot shows the frequency-response of the noise model H.
For input-output models, the noise spectrum is given by the following equation:
{\Phi }_{v}\left(\omega \right)=\lambda {|H\left({e}^{i\omega }\right)|}^{2}
For time-series models (no input), the vertical axis of the noise-spectrum plot is the same as the dynamic model spectrum. These axes are the same because there is no input for time series and
y=He
You can avoid estimating the noise model by selecting the Output-Error model structure or by setting the DisturbanceModel property value to 'None' for a state space model. If you choose to not estimate a noise model for your system, then H and the noise spectrum amplitude are equal to 1 at all frequencies.
In addition to the noise-spectrum curve, you can display a confidence interval on the plot. To learn how to show or hide confidence interval, see the description of the plot settings in Plot the Noise Spectrum Using the System Identification App.
The confidence interval corresponds to the range of power-spectrum values with a specific probability of being the actual noise spectrum 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 where there is a 95% chance that the true response belongs.. 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. |
2\left(3p+1\right)>-4
2\left(3p+1\right)>-4
Test values on both sides of the boundary point
\left(p=-1\right)
to determine the solution region.
\begin{array}{r} 2 \left(3\left(-2\right) + 1\right) > - 4 \\ 2 \left(-5\right) > -4 \\ -10 > -4 \\ \text{False} \qquad \;\;\; \end{array}
\begin{array}{r} 2 \left(3\left(0\right) + 1\right) > - 4 \\ 2 \left(1\right) > -4 \\ 2 > -4 \\ \text{True} \qquad \;\;\; \end{array}
9k-2<3k+10
5-h\ge4 |
Delay-Dependent Robust L 2 - L ∞ Filtering for a Class of Fuzzy Stochastic Systems
2014 Delay-Dependent Robust
{L}_{2}-{L}_{\infty }
Filtering for a Class of Fuzzy Stochastic Systems
Ze Li, Xinhao Yang
This paper is concerned with the
{L}_{2}-{L}_{\infty }
filtering problem for a kind of Takagi-Sugeno (T-S) fuzzy stochastic system with time-varying delay and parameter uncertainties. Parameter uncertainties in the system are assumed to satisfy global Lipschitz conditions. And the attention of this paper is focused on the stochastically mean-square stability of the filtering error system, and the
{L}_{2}-{L}_{\infty }
performance level of the output error with the disturbance input. The method designed for the delay-dependent filter is developed based on linear matrix inequalities. Finally, the effectiveness of the proposed method is substantiated with an illustrative example.
Ze Li. Xinhao Yang. "Delay-Dependent Robust
{L}_{2}-{L}_{\infty }
Filtering for a Class of Fuzzy Stochastic Systems." Abstr. Appl. Anal. 2014 (SI63) 1 - 10, 2014. https://doi.org/10.1155/2014/673956
Ze Li, Xinhao Yang "Delay-Dependent Robust
{L}_{2}-{L}_{\infty }
Filtering for a Class of Fuzzy Stochastic Systems," Abstract and Applied Analysis, Abstr. Appl. Anal. 2014(SI63), 1-10, (2014) |
Banach spaces which are $M$-ideals in their bidual have property $(u)$
Banach spaces which are
M
-ideals in their bidual have property
\left(u\right)
Godefroy, Gilles ; Li, D.
Nous montrons que tout espace de Banach qui est
M
-idéal de son bidual a la propriété
\left(u\right)
de A. Pelczynski, et mentionnons quelques conséquences.
We show that every Banach space which is an
M
-ideal in its bidual has the property
\left(u\right)
of Pelczynski. Several consequences are mentioned.
author = {Godefroy, Gilles and Li, D.},
title = {Banach spaces which are $M$-ideals in their bidual have property $(u)$},
AU - Godefroy, Gilles
AU - Li, D.
TI - Banach spaces which are $M$-ideals in their bidual have property $(u)$
Godefroy, Gilles; Li, D. Banach spaces which are $M$-ideals in their bidual have property $(u)$. Annales de l'Institut Fourier, Tome 39 (1989) no. 2, pp. 361-371. doi : 10.5802/aif.1170. http://www.numdam.org/articles/10.5802/aif.1170/
[1] E. M. Alfsen, E. G. Effros, Structure in real Banach spaces I, Ann. of Math., 96 (1972), 98-128. | MR 50 #5432 | Zbl 0248.46019
[2] E. Behrends, M-structure and the Banach-Stone theorem, Lecture Notes in Mathematics 736, Springer-Verlag (1977). | MR 81b:46002 | Zbl 0436.46013
[3] E. Behrends, P. Harmand, Banach spaces which are proper M-ideals, Studia Mathematica, 81 (1985), 159-169. | MR 87f:46031 | Zbl 0529.46015
[4] G. A. Edgar, An ordering of Banach spaces, Pacific J. of Maths, 108, 1 (1983), 83-98. | MR 84k:46012 | Zbl 0533.46007
[5] G. Godefroy, On Riesz subsets of abelian discrete groups, Israel J. of Maths, 61, 3 (1988), 301-331. | MR 89m:43011 | Zbl 0661.43003
[6] G. Godefroy, P. Saab, Weakly unconditionally convergent series in M-ideals, Math. Scand., to appear. | Zbl 0676.46006
[7] G. Godefroy, M. Talagrand, Nouvelles classes d'espaces de Banach à predual unique, Séminaire d'Ana. Fonct. de l'École Polytechnique, Exposé n° 6 (1980/1981). | Numdam | Zbl 0475.46013
[8] G. Godefroy, Existence and uniqueness of isometric preduals : a survey, in Banach space Theory, Proceedings of a Research workshop held July 5-25, 1987, Contemporary Mathematics vol. 85 (1989), 131-194. | Zbl 0674.46010
[9] G. Godefroy, D. Li, Some natural families of M-ideals, to appear. | Zbl 0687.46010
[10] P. Harmand, A. Lima, On spaces which are M-ideals in their biduals, Trans. Amer. Math. Soc., 283-1 (1984), 253-264. | MR 86b:46016 | Zbl 0545.46009
[11] A. Lima, M-ideals of compact operators in classical Banach spaces, Math. Scand., 44 (1979), 207-217. | EuDML 166642 | MR 81c:47047 | Zbl 0407.46019
[12] J. Lindenstrauss, L. Tzafriri, Classical Banach spaces, Vol. II, Springer-Verlag (1979). | MR 81c:46001 | Zbl 0403.46022
[13] F. Lust, Produits tensoriels projectifs d'espaces de Banach faiblement sequentiellement complets, Coll. Math., 36-2 (1976), 255-267. | EuDML 263584 | MR 55 #11072 | Zbl 0356.46058
[14] A. Pelczynski, Banach spaces on which every unconditionally convergent operator is weakly compact, Bull. Acad. Pol. Sciences, 10 (1962), 641-648. | MR 26 #6785 | Zbl 0107.32504
[15] R. R. Smith, J. D. Ward, Applications of convexity and M-ideal theory to quotient Banach algebras, Quart. J. of Maths. Oxford, 2-30 (1978), 365-384. | Zbl 0412.46042 |
Represent the solutions to the inequalities below on a number line.
3x-2<10
\begin{array}{r} 3x - 2 = 10 \\ 3x = 12 \\ x = 4 \end{array}
\begin{array}{r} 3\left(0\right) - 2 < 10 \\ 0 - 2 <10 \\ -2 < 10 \\ \text{True} \qquad \end{array} \qquad \begin{array}{r} 3\left(5\right) - 2 < 10 \\ 15 - 2 <10 \\ 13 < 10 \\ \text{False} \qquad \end{array}
5x-1-3x\ge4x+5
2\left(x+2\right)>10-x
x
4\left(x-3\right)+5\ge-7 |
Futures Guide - Delta Exchange - User Guide
Futures: Motivation & Use Cases
A derivative is a financial security which derives its value from an underlying asset or group of assets. The derivative itself is a contract between two parties and its price is driven by fluctuations in the underlying asset.
Futures is a type of derivative that is essentially an agreement between two parties to buy/ sell an asset (e.g. Ether) at a predetermined future date and price. There is a deterministic but complex relation between the price of the underlying asset and its Futures contract. However, simplistically, we can assume that both spot price and Futures price tend to move in the same direction.
Trading Futures vs. the underlying
Futures offer several advantages over trading the underlying asset directly. Futures ..
enable you to benefit from both price increases (long position) as well as declines (short position)
provide financial leverage
can be used to hedge price risk
tend to incur smaller transaction fees
View on the price of an underlying can be expressed via a Futures contract for it. Going long (buying) futures will generally lead to a profit if the price of the underlying rises. Conversely, shorting (selling) Futures will generally result in a profit if the price of the underlying declines.
A Futures contract can be used to hedge the price risk of the underlying asset. Let’s say you own
1 BTC
(Bitcoin). If you want to lock the current BTC price of
, then you can short sell
8000 BTC/ USD
Futures (assuming each contract size is
1 USD
). Once this hedge is in place, regardless of bitcoin price movement, the effective value of your position will remain constant at
Math for Futures
The math behind Futures is better illustrated with an example. Let’s say, we would like to trade Ether-Bitcoin (ETH-BTC) Futures. We need to define a few parameters:
Underlying: This is the asset over which a Futures contract is defined. In our example the underlying is Ether.
Quote currency: This is the currency in which the price of the underlying is quoted. In our example, price of Ether is quoted in Bitcoin terms. Hence, quote currency is BTC.
Base currency: This is the currency in which the PnL of a Futures position is calculated. In our example, the base currency is same as the quote currency, i.e. Bitcoin. However, this need not always be true.
Multiplier: This refers to the quantity of the underlying that the two parties trading a single Futures contract agree to buy/ sell in future. If our ETH-BTC Futures contract entails agreement to buy/ sell m Ethers at a future date, then m is the multiplier.
Given above, if you buy (or even sell)
n
ETH-BTC Futures contracts at
Fut\_EntryPrice
, then your overall position size (nominal exposure) is:
n ∗ m ∗ F u t \_ E n t r y P r i c e ( i n B T C )
To understand why this the position size consider this: the buyer of the Futures contract has agreed to buy
n*m
Ethers at contract expiry, with each Ether priced at
Fut\_EntryPrice
BTC. Thus, at the time of settlement, he will need to have
n ∗ m ∗ F u t \_ E n t r y P r i c e
BTC to honor the agreement.
However, to be able to enter into a Futures trade, you are not required to have resources equalling the position size. This is where financial leverage kicks in.
Margin%: This is the fraction of your position size that you are required to post as collateral when entering into a Futures trade. So, in our example, to buy/ sell \(n\) contracts, you need to have only:
n*m* Margin\% * Fut\_EntryPrice \ (in \ BTC)
This also means that maximum position size that you can afford is
1/Margin\%
times the collateral (in BTC) that you have available. Thus,
^1/(Margin\%) = Financial\ leverage
PnL Computation
Unrealised PnL For a long position in a Futures contract:
PnL = n*m*(Fut\_CurrentPrice - Fut\_EntryPrice) \ (in \ BTC)
For a short position in a Futures contract:
PnL = - n*m*(1/ Fut\_EntryPrice - 1/ Fut\_CurrentPrice) \ (in \ BTC)
It is worth noting that your (realised or unrealised) profits and losses are adjusted from the margin you post. Profit on a Futures position add to the Margin (collateral). Conversely, loss erodes margin and you might need to top it up to continue holding your position.
Realised PnL In case of Futures, PnL can be realised either by exiting the position in market or via settlement process at the maturity of the contract
Exit long position in market
PnL = n*m*(Fut\_ExitPrice - Fut\_EntryPrice) \ (in \ BTC)
PnL = n*m*(Fut\_ExitPrice)
Exit long position via settlement
Exit short position in market
PnL = -n*m*(Fut\_ExitPrice - Fut\_EntryPrice)\ (in \ BTC)
Exit short position via settlement
PnL = -n*m*(Fut\_SettlementPrice - Fut\_EntryPrice) \ (in \ BTC)
Settlement price is determined at the maturity of the contract through a pre-defined method described in the contract specifications . All open positions at the time of contract maturity are closed at the settlement price.
The Futures contract discussed above are also known as Vanilla Futures. Inverse Futures are similar to Vanilla Futures, but only with one key distinction: the relationship between Futures price and position PnL is sort of inverted.
For a long position in an inverse Futures contract
PnL = n*m*(1/ Fut\_EntryPrice - 1/ Fut\_CurrentPrice) \ (in \ BTC)
For a short position in an inverse Futures contract
PnL
Leverage has a multiplier effect on your trading returns. This is best illustrated with an example. Let’s say you have
, and you think BTC price is likely to go up. Currently,
1 BTC
No Leverage (trade the underlying)
0.01 BTC
\$100
. If after a week, BTC rises by
10\%
\$11,000
, your position in BTC is now worth
\$110
and your PnL is
\$10
. And, the return on your equity (RoE),
\$100
in this case is:
RoE_{NoLeverage} = PnL / Equity = \$10/\$100 = 10\%
With Leverage (trade the Futures contract)
Long USD - BTC Futures contract using as the margin money. Let’s assume that the contract size is
0.01 BTC
and Margin% is
10\%
\$100
as margin is sufficient for
10
Let’s further assume that the Futures contract price broadly follows BTC price movement. Now, we can approximate the PnL as
So, in this case, we have generated a PnL of
with an equity of
RoE_{Leverage} = \$100/ \$100 = 100\%
It is also easy to see that,
RoE_{Leverage}= RoE_{NoLeverage}/ Margin\% = RoE_{NoLeverage}*Financial Leverage
Do note that leverage is a double edged sword. Just as it will amplify your profits, it will also have the same multiplier effect on your losses. |
DividedDifferenceTable - Maple Help
Home : Support : Online Help : Education : Student Packages : Numerical Analysis : Computation : DividedDifferenceTable
compute the divided difference table
DividedDifferenceTable(p)
DividedDifferenceTable(p, pt)
(optional) numeric; a point to evaluate the divided difference table
The DividedDifferenceTable command takes an interpolation structure and computes the associated divided difference table.
This command can only be used on interpolation structures that were computed with Hermite or Newton methods.
\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):
\mathrm{xy}≔[[1.0,0.7651977],[1.3,0.6200860],[1.6,0.4554022],[1.9,0.2818186]]
\textcolor[rgb]{0,0,1}{\mathrm{xy}}\textcolor[rgb]{0,0,1}{≔}[[\textcolor[rgb]{0,0,1}{1.0}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.7651977}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1.3}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.6200860}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1.6}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.4554022}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1.9}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2818186}]]
\mathrm{p1}≔\mathrm{PolynomialInterpolation}\left(\mathrm{xy},\mathrm{independentvar}='x',\mathrm{method}=\mathrm{newton}\right):
\mathrm{DividedDifferenceTable}\left(\mathrm{p1}\right)
[\begin{array}{cccc}\textcolor[rgb]{0,0,1}{0.7651977}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.6200860}& \textcolor[rgb]{0,0,1}{-0.4837056667}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.4554022}& \textcolor[rgb]{0,0,1}{-0.5489460000}& \textcolor[rgb]{0,0,1}{-0.1087338888}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.2818186}& \textcolor[rgb]{0,0,1}{-0.5786120000}& \textcolor[rgb]{0,0,1}{-0.04944333333}& \textcolor[rgb]{0,0,1}{0.06587839497}\end{array}]
\mathrm{p1a}≔\mathrm{AddPoint}\left(\mathrm{p1},[1.8,0.3920223]\right):
\mathrm{DividedDifferenceTable}\left(\mathrm{p1a}\right)
[\begin{array}{ccccc}\textcolor[rgb]{0,0,1}{0.7651977}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.6200860}& \textcolor[rgb]{0,0,1}{-0.4837056667}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.4554022}& \textcolor[rgb]{0,0,1}{-0.5489460000}& \textcolor[rgb]{0,0,1}{-0.1087338888}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.2818186}& \textcolor[rgb]{0,0,1}{-0.5786120000}& \textcolor[rgb]{0,0,1}{-0.04944333333}& \textcolor[rgb]{0,0,1}{0.06587839497}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{0.3920223}& \textcolor[rgb]{0,0,1}{-1.102037000}& \textcolor[rgb]{0,0,1}{-2.617125000}& \textcolor[rgb]{0,0,1}{-5.135363334}& \textcolor[rgb]{0,0,1}{-6.501552161}\end{array}]
\mathrm{xyyp}≔[[1,1.105170918,0.2210341836],[1.5,1.252322716,0.3756968148],[2,1.491824698,0.5967298792]]
\textcolor[rgb]{0,0,1}{\mathrm{xyyp}}\textcolor[rgb]{0,0,1}{≔}[[\textcolor[rgb]{0,0,1}{1}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1.105170918}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.2210341836}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{1.5}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1.252322716}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.3756968148}]\textcolor[rgb]{0,0,1}{,}[\textcolor[rgb]{0,0,1}{2}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{1.491824698}\textcolor[rgb]{0,0,1}{,}\textcolor[rgb]{0,0,1}{0.5967298792}]]
\mathrm{p2}≔\mathrm{PolynomialInterpolation}\left(\mathrm{xyyp},\mathrm{method}=\mathrm{hermite},\mathrm{function}=\mathrm{exp}\left(0.1{x}^{2}\right),\mathrm{independentvar}='x',\mathrm{errorboundvar}='\mathrm{\xi }',\mathrm{digits}=5\right):
\mathrm{DividedDifferenceTable}\left(\mathrm{p2}\right)
[\begin{array}{cccccc}\textcolor[rgb]{0,0,1}{1.1052}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1.1052}& \textcolor[rgb]{0,0,1}{0.22103}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1.2523}& \textcolor[rgb]{0,0,1}{0.29420}& \textcolor[rgb]{0,0,1}{0.14634}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1.2523}& \textcolor[rgb]{0,0,1}{0.37570}& \textcolor[rgb]{0,0,1}{0.16300}& \textcolor[rgb]{0,0,1}{0.033320}& \textcolor[rgb]{0,0,1}{0}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1.4918}& \textcolor[rgb]{0,0,1}{0.47900}& \textcolor[rgb]{0,0,1}{0.20660}& \textcolor[rgb]{0,0,1}{0.043600}& \textcolor[rgb]{0,0,1}{0.010280}& \textcolor[rgb]{0,0,1}{0}\\ \textcolor[rgb]{0,0,1}{1.4918}& \textcolor[rgb]{0,0,1}{0.59673}& \textcolor[rgb]{0,0,1}{0.23546}& \textcolor[rgb]{0,0,1}{0.057720}& \textcolor[rgb]{0,0,1}{0.014120}& \textcolor[rgb]{0,0,1}{0.0038400}\end{array}]
Student[NumericalAnalysis][NevilleTable] |
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