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Summary: "
SIAM J. NUMER. ANAL.
Vol. 31, No. 1, pp. 252-260, February 1994
@ 1994 Society for lndustrial and Applied Mathematics
G. ALEFELDt, A. GIENGERt, AND F. POTRAt
Abstract. A new stopping criterion for Newton's method is derived by combining the prop-
erties oi the Krawczyk operator and a corollary of the Newton-Kantorovich theorem. When this
criterion is satisfied the authors use the last three Newton iterates to compute an interval vector
that is very likely to contain a solution of the given nonlinear system. The existence oisuch a
solution is tested using Krawczyk's operator. Furthermore, each element from this interval vector
considered as an approximation to the solution has a relative error that is of the order oi the machine
precision. Extensive numerical testing has shown that the proposed method has very good practical
Keywords. nonlinear systems, Newton-Kantorovich theorem, validation of solutions
AMS subject classifications. 65HI0,65GlO
1. Introduction. Newton's method is the best-known algorithm for solving non-
linear systems of equations. The most famous theoretical result on the convergence of | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/969/2745573.html","timestamp":"2014-04-21T02:14:18Z","content_type":null,"content_length":"8435","record_id":"<urn:uuid:069acbf4-ccbc-4603-b56d-1cbc04fb61cf>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00472-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Help predicting future lifetime of a component w/ a joint pfd. thanks for looking.
November 16th 2009, 04:29 PM #1
Junior Member
Sep 2009
The future lifetimes (in months) of two components of a machine have the following joint density function:
$f(x,y)=\left\{\begin{array}{lr}(6/125,000)(50-x-y)&0 < x < 50-y < 50\\0&otherwise\end{array}\right.$
What is the probability that both components are still functioning 15 months from now? thanks for your help
Last edited by Intsecxtanx; November 16th 2009 at 05:49 PM.
The future lifetimes (in months) of two components of a machine have the following joint density function:
$f(x,y)=\left\{\begin{array}{lr}(6/125,000)(50-x-y)&0 < x < 50-y < 50\\0&otherwise\end{array}\right.$
What is the probability that both components are still functioning 15 months from now? thanks for your help
Drawing the region of integration helps.
$\Pr(X \geq 15, Y \geq 15) = \int_{y=15}^{y=35} \int_{x=15}^{x=50-y} f(x, y) \, dx \, dy$.
November 16th 2009, 06:22 PM #2 | {"url":"http://mathhelpforum.com/advanced-statistics/115016-help-predicting-future-lifetime-component-w-joint-pfd-thanks-looking.html","timestamp":"2014-04-18T14:41:32Z","content_type":null,"content_length":"36143","record_id":"<urn:uuid:0fb6333c-1093-4bac-a308-155199fa79f5>","cc-path":"CC-MAIN-2014-15/segments/1397609533689.29/warc/CC-MAIN-20140416005213-00553-ip-10-147-4-33.ec2.internal.warc.gz"} |
An Essay Concerning Human Understanding
1. Number the simplest and most universal idea. Amongst all the ideas we have, as there is none suggested to the mind by more ways, so there is none more simple, than that of unity, or one: it has no
shadow of variety or composition in it: every object our senses are employed about; every idea in our understandings; every thought of our minds, brings this idea along with it. And therefore it is
the most intimate to our thoughts, as well as it is, in its agreement to all other things, the most universal idea we have. For number applies itself to men, angels, actions, thoughts; everything
that either doth exist, or can be imagined.
2. Its modes made by addition. By repeating this idea in our minds, and adding the repetitions together, we come by the complex ideas of the modes of it. Thus, by adding one to one, we have the
complex idea of a couple; by putting twelve units together, we have the complex idea of a dozen; and so of a score, or a million, or any other number.
3. Each mode distinct. The simple modes of number are of all other the most distinct; every the least variation, which is an unit, making each combination as clearly different from that which
approacheth nearest to it, as the most remote; two being as distinct from one, as two hundred; and the idea of two as distinct from the idea of three, as the magnitude of the whole earth is from that
of a mite. This is not so in other simple modes, in which it is not so easy, nor perhaps possible for us to distinguish betwixt two approaching ideas, which yet are really different. For who will
undertake to find a difference between the white of this paper and that of the next degree to it: or can form distinct ideas of every the least excess in extension?
4. Therefore demonstrations in numbers the most precise. The clearness and distinctness of each mode of number from all others, even those that approach nearest, makes me apt to think that
demonstrations in numbers, if they are not more evident and exact than in extension, yet they are more general in their use, and more determinate in their application. Because the ideas of numbers
are more precise and distinguishable than in extension; where every equality and excess are not so easy to be observed or measured; because our thoughts cannot in space arrive at any determined
smallness beyond which it cannot go, as an unit; and therefore the quantity or proportion of any the least excess cannot be discovered; which is clear otherwise in number, where, as has been said, 91
is as distinguishable from go as from 9000, though 91 be the next immediate excess to 90. But it is not so in extension, where, whatsoever is more than just a foot or an inch, is not distinguishable
from the standard of a foot or an inch; and in lines which appear of an equal length, one may be longer than the other by innumerable parts: nor can any one assign an angle, which shall be the next
biggest to a right one.
5. Names necessary to numbers. By the repeating, as has been said, the idea of an unit, and joining it to another unit, we make thereof one collective idea, marked by the name two. And whosoever can
do this, and proceed on, still adding one more to the last collective idea which he had of any number, and gave a name to it, may count, or have ideas, for several collections of units, distinguished
one from another, as far as he hath a series of names for following numbers, and a memory to retain that series, with their several names: all numeration being but still the adding of one unit more,
and giving to the whole together, as comprehended in one idea, a new or distinct name or sign, whereby to know it from those before and after, and distinguish it from every smaller or greater
multitude of units. So that he that can add one to one, and so to two, and so go on with his tale, taking still with him the distinct names belonging to every progression; and so again, by
subtracting an unit from each collection, retreat and lessen them, is capable of all the ideas of numbers within the compass of his language, or for which he hath names, though not perhaps of more.
For, the several simple modes of numbers being in our minds but so many combinations of units, which have no variety, nor are capable of any other difference but more or less, names or marks for each
distinct combination seem more necessary than in any other sort of ideas. For, without such names or marks, we can hardly well make use of numbers in reckoning, especially where the combination is
made up of any great multitude of units; which put together, without a name or mark to distinguish that precise collection, will hardly be kept from being a heap in confusion.
6. Another reason for the necessity of names to numbers. This I think to be the reason why some Americans I have spoken with, (who were otherwise of quick and rational parts enough,) could not, as we
do, by any means count to 1000; nor had any distinct idea of that number, though they could reckon very well to 20. Because their language being scanty, and accommodated only to the few necessaries
of a needy, simple life, unacquainted either with trade or mathematics, had no words in it to stand for 1000; so that when they were discoursed with of those greater numbers, they would show the
hairs of their head, to express a great multitude, which they could not number; which inability, I suppose, proceeded from their want of names. The Tououpinambos had no names for numbers above 5; any
number beyond that they made out by showing their fingers, and the fingers of others who were present. And I doubt not but we ourselves might distinctly number in words a great deal further than we
usually do, would we find out but some fit denominations to signify them by; whereas, in the way we take now to name them, by millions of millions of millions, &c., it is hard to go beyond eighteen,
or at most, four and twenty, decimal progressions, without confusion. But to show how much distinct names conduce to our well reckoning, or having useful ideas of numbers, let us see all these
following figures in one continued line, as the marks of one number: v. g.
Nonillions Octillions Septillions Sextillions Quintrillions 857324 162486 345896 437918 423147
Quartrillions Trillions Billions Millions Units 248106 235421 261734 368149 623137
The ordinary way of naming this number in English, will be the often repeating of millions, of millions, of millions, of millions, of millions, of millions, of millions, of millions, (which is the
denomination of the second six figures). In which way, it will be very hard to have any distinguishing notions of this number. But whether, by giving every six figures a new and orderly denomination,
these, and perhaps a great many more figures in progression, might not easily be counted distinctly, and ideas of them both got more easily to ourselves, and more plainly signified to others, I leave
it to be considered. This I mention only to show how necessary distinct names are to numbering, without pretending to introduce new ones of my invention.
7. Why children number not earlier. Thus children, either for want of names to mark the several progressions of numbers, or not having yet the faculty to collect scattered ideas into complex ones,
and range them in a regular order, and so retain them in their memories, as is necessary to reckoning, do not begin to number very early, nor proceed in it very far or steadily, till a good while
after they are well furnished with good store of other ideas: and one may often observe them discourse and reason pretty well, and have very clear conceptions of several other things, before they can
tell twenty. And some, through the default of their memories, who cannot retain the several combinations of numbers, with their names, annexed in their distinct orders, and the dependence of so long
a train of numeral progressions, and their relation one to another, are not able all their lifetime to reckon, or regularly go over any moderate series of numbers. For he that will count twenty, or
have any idea of that number, must know that nineteen went before, with the distinct name or sign of every one of them, as they stand marked in their order; for wherever this fails, a gap is made,
the chain breaks, and the progress in numbering can go no further. So that to reckon right, it is required, (1) That the mind distinguish carefully two ideas, which are different one from another
only by the addition or subtraction of one unit: (2) That it retain in memory the names or marks of the several combinations, from an unit to that number; and that not confusedly, and at random, but
in that exact order that the numbers follow one another. In either of which, if it trips, the whole business of numbering will be disturbed, and there will remain only the confused idea of multitude,
but the ideas necessary to distinct numeration will not be attained to.
8. Number measures all measureables. This further is observable in number, that it is that which the mind makes use of in measuring all things that by us are measurable, which principally are
expansion and duration; and our idea of infinity, even when applied to those, seems to be nothing but the infinity of number. For what else are our ideas of Eternity and Immensity, but the repeated
additions of certain ideas of imagined parts of duration and expansion, with the infinity of number; in which we can come to no end of addition? For such an inexhaustible stock, number (of all other
our ideas) most clearly furnishes us with, as is obvious to every one. For let a man collect into one sum as great a number as he pleases, this multitude, how great soever, lessens not one jot the
power of adding to it, or brings him any nearer the end of the inexhaustible stock of number; where still there remains as much to be added, as if none were taken out. And this endless addition or
addibility (if any one like the word better) of numbers, so apparent to the mind, is that, I think, which gives us the clearest and most distinct idea of infinity: of which more in the following | {"url":"http://ebooks.adelaide.edu.au/l/locke/john/l81u/B2.16.html","timestamp":"2014-04-19T19:41:23Z","content_type":null,"content_length":"13963","record_id":"<urn:uuid:fe1142e9-8d1a-45b4-b8d2-da7f49090fd9>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00337-ip-10-147-4-33.ec2.internal.warc.gz"} |
ALEX Lesson Plans
Subject: Mathematics (6)
Title: Factorials: Let's have a Dinner Party!
Description: In collaborate groups of four, students will act out a dinner party where four dinner guests will attend. The students must act out the different ways to arrange four dinner guests. This
is a College- and Career-Ready Standards showcase lesson plan.
Subject: Mathematics (6), or Science (6)
Title: Is My Vari"able" to Stand?
Description: This cross-curricular lesson is designed to require students to apply algebra to a real-world, natural disaster situation. Students will be asked to design a tree house including a
ladder. Students will set up an algebraic equation and solve for the variable. Will their treehouse be able to withstand a "jelloquake"? Students will test the strength of their treehouse and ladder
in science class. This is a College- and Career-Ready Standards showcase lesson plan.
Subject: Mathematics (6)
Title: What did you say? Translating from verbal to algebraic.
Description: Students will learn about translating between verbal and algebraic expressions via game play and possibly working on an interactive web site. The four steps to explicit instruction (I
do, we do, y'all do, and you do) are labeled in the procedures section.
Subject: Mathematics (6)
Title: The Pattern of Graphing Linear Equations
Description: The students will extend the knowledge of algebraic expressions from geometric representations and ultimately graph linear equations with understanding. The students will also develop a
better understanding of algebraic expressions by comparing with geometric, tabular, and graphical representations. Students may also acquire a deeper understanding of independent and dependent
Subject: Mathematics (6)
Title: The Pattern of Graphing Quadratic Equations
Description: The students will develop an algebraic expression from geometric representations and ultimately graph quadratic equations with understanding. The students will also develop a better
understanding of algebraic expressions by comparing with geometric, tabular, and graphical representations.
Subject: Mathematics (6)
Title: Express Yourself with Patterns
Description: During this lesson, the students will use growing patterns to develop an algebraic expression with understanding. The students will also learn mathematical vocabulary with the use of the
newly created expression.
Subject: Mathematics (4 - 6)
Title: My Favorite Number
Description: The activity allows students to review many of the number theory concepts. Students will pick a composite number write verbal expressions about the number, find the factors, prime
factorization, list multiples, draw a cartoon character of the number, and write a word problem. The purpose of this activity is to be a review of the number theory topics taught during middle
school.This lesson plan was created as a result of the Girls Engaged in Math and Science University, GEMS-U Project.
Subject: Mathematics (6 - 7)
Title: Discovering Pi
Description: This lesson allows students to discover the number Pi. By measuring real life objects, students will get to see where the number Pi comes from as well as how the circumference formula is
derived.This lesson plan was created as a result of the Girls Engaged in Math and Science University, GEMS-U Project.
Subject: Mathematics (6), or Technology Education (6 - 8)
Title: Variables and Algebraic Expressions
Description: Students will be actively teaching and learning vocabulary associated with algebraic expressions. They will identify each part of an algebraic expression. This lesson plan was created as
a result of the Girls Engaged in Math and Science, GEMS Project funded by the Malone Family Foundation.
Subject: Mathematics (6 - 7)
Title: "Introducing Pi!"
Description: The overriding purpose of the lesson is to actively engage students in learning activities that promote opportunities for students to discover the 'meaning' of pi. Activities will
include research, hands-on, software applications, reading, writing, and collaborative work opportunities. The secondary purpose is to introduce and spark enthusiasm among students for our
celebration of 'Pi Day'.
Subject: Mathematics (6)
Title: Express Yourself
Description: This lesson presents students with the abstract idea that letters represent unknown numbers or variables. Vocabulary pertinent to writing English phrases as algebraic expressions and
sentences as equations and vice versa is introduced. A mathematics and language arts relationship is also stressed.
Subject: Mathematics (6)
Title: It's In The Bag
Description: This lesson provides students with a real-life application of problem solving. Students will use color counters to model writing expressions and equations. They are given a scenario to
follow where clues about the content of bags are used to determine the type of canned goods in each bag.
Thinkfinity Lesson Plans
Subject: Mathematics
Title: Finding the Area of Irregular Figures
Description: In this lesson, one of a multi-part unit from Illuminations, students estimate the areas of highly irregular shapes and use a process of decomposition to calculate the areas of irregular
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: On Fire
Description: This unit of five lessons, from Illuminations, introduces the components of a fire-safe and fire-wise environment. Students create a fire-wise location through calculations and
measurement of percent slope, defensible space distance and various vegetation separation distances. The unit plan culminates with students designing a fire-wise property and testing their fire-wise
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Area Formulas
Description: In this unit of four lessons, from Illuminations, students use the area formula for a rectangle to discover the area formulas for triangles, parallelograms and trapezoids. Students also
consider irregular figures whose areas can be determined by estimation or decomposition.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Classifying Numbers
Description: In this lesson, one of a multi-part unit from Illuminations, students use Venn diagrams to organize information about numbers. Using the Product Game board, students look for
relationships and characteristics of numbers to determine what numbers belong to a descriptor and what numbers belong to more than one descriptor.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Language Arts,Mathematics
Title: Mathematics and Children's Literature
Description: In this five-lesson unit, from Illuminations, students participate in activities in which they focus on connections between mathematics and children s literature. Five pieces of
literature are applied to teaching a wide range of topics in the mathematics curriculum, from sorting and classifying to the meaning of averages.
Thinkfinity Partner: Illuminations
Grade Span: K,PreK,1,2,3,4,5,6,7,8
Subject: Mathematics
Title: Finding the Area of Parallelograms
Description: In this lesson, one of a multi-part unit from Illuminations, students use their knowledge of rectangles to discover the area formula for parallelograms.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Surface Area
Description: This reproducible transparency, from an Illuminations lesson, features equations for finding the ratios for computing the surface area of cylinders and rectangular prisms.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Making Your Own Product Game
Description: In this lesson, one of a multi-part unit from Illuminations, students review strategies for playing the Product Game. In this game, players start with a set of given factors and multiply
to find the product. Students then work in groups to create their own game boards.
Thinkfinity Partner: Illuminations
Grade Span: 3,4,5,6,7,8
Subject: Mathematics
Title: Finding the Area of Trapezoids
Description: In this lesson, one of a multi-part unit from Illuminations, students discover the area formula for trapezoids and explore alternative methods for calculating the area of a trapezoid.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Discovering the Area Formula for Circles
Description: In this lesson, one of a multi-part unit from Illuminations, students use a circle that has been divided into congruent sectors to discover the area formula by using their knowledge of
parallelograms. They then calculate the area of various flat circular objects that they have brought to school. Finally, students investigate various strategies for estimating the area of circles.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Arts,Mathematics,Social Studies
Title: Real Estate Tycoon
Description: This year-long project, from Illuminations, is divided into three parts and reinforces skills that lay a foundation for algebra, such as measurement, fractions, decimals, percents, and
proportions. Students design, build, and sell a house and then simulate investment of the profits in the stock market. Along the way, students make scale drawings, compute with fractions and decimals
in various contexts, solve simple equations, and learn valuable lessons about how money really works.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Archimedes' Puzzle
Description: In this lesson, students learn about the history of the Stomachion, an ancient tangram-type puzzle. Students use the puzzle pieces to create other figures, learn about symmetry and
transformations and investigate the areas of the pieces.
Thinkfinity Partner: Illuminations
Grade Span: 3,4,5,6,7,8
Subject: Mathematics
Title: Balancing Shapes
Description: In this lesson, one of a multi-part unit from Illuminations, students balance shapes using an interactive pan balance applet to study equality, essential to understanding algebra.
Equivalent relationships are recognized when the pans balance, demonstrating the properties of equality.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Patterns and Function
Description: In this lesson, from Illuminations, students investigate properties of perimeter, area, and volume related to various geometric two- and three-dimensional shapes.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Web Resources
Order of Operations TV
This video gives multiple solution videos for solving problems using the order of operations.
The Order of Operations Millionaire Game
The Order of Operations Millionaire Game can be played alone or in two teams. If you are using this game in the classroom as a contest between two teams, the teacher can modify the scores at any
point by clicking the score box of a team and editing the numbers.
Learning Activities
Solving Simple Sentences by Inspection
In this exercise, students practice solving sentences "by inspection". That is, they are taught to look at the sentence, stop and think, and come up with a number that makes the sentence true.
Teacher Tools
Order of Operations TV
This video gives multiple solution videos for solving problems using the order of operations.
Thinkfinity Learning Activities
Subject: Mathematics
Title: Learning about Length, Perimeter, Area, and Volume of Similar Objects by Using Interactive Figures: Side Length, Volume, and Surface Area of Similar Solids
Description: This is part two of a two-part e-example from Illuminations that illustrates how students can learn about the length, perimeter, area, and volume of similar objects using dynamic
figures. In this part, Side Length, Volume, and Surface Area of Similar Solids, the user can manipulate the scale factor that links two three-dimensional rectangular prisms and learn about the
relationships among edge lengths, surface areas, and volumes. e-Math Investigations are selected e-examples from the electronic version of the Principles and Standards for School Mathematics (PSSM).
Given their interactive nature and focused discussion tied to the PSSM document, the e-examples are natural companions to the i-Math Investigations.
Thinkfinity Partner: Illuminations
Grade Span: 6,7,8
Subject: Mathematics
Title: Pan Balance -- Expressions
Description: This interactive pan balance, from Illuminations, allows students to enter and compare numeric or algebraic expressions. They can '' weigh'' the expressions they want to compare by
entering them on either side of the balance, allowing them to practice arithmetic and algebraic skills, as well as to investigate the concept of equivalence.
Thinkfinity Partner: Illuminations
Grade Span: 3,4,5,6,7,8,9,10,11,12 | {"url":"http://alex.state.al.us/all.php?std_id=53866","timestamp":"2014-04-18T20:53:38Z","content_type":null,"content_length":"150661","record_id":"<urn:uuid:1352c291-ef51-4d11-8791-eee6b77b739c>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00130-ip-10-147-4-33.ec2.internal.warc.gz"} |
Partial elimination round calculator
I'm a big believer that every team at a debate tournament with a winning record should make it to elimination rounds. (Or, if this is simply impossible, break every team with a certain record, e.g.
all 5-2s and no 4-3s. That only seems fair; breaking some 5-2s on speaker points, but not other 5-2s, places too much trust in the objectivity of speaker points.)
Of course, the problem is that breaking every team in a certain bracket creates the need to run a partial elimination round (because how often will a tournament need to break exactly 2^
teams?). The logistics of a partial elimination round can seem a bit unpredictable at first -- how many teams will need to be in the partial? how many judges will be needed? -- so I created a
"partials calculator" to help
answers to these questions:
A few notes about the calculator. I set it up with a default of 34% of teams breaking, a good assumption for the percentage of 4-2s or better, but this value can be changed in the green box. I also
set up a default of one judge per two teams in the single-flighted event and one judge per three teams in the double-flighted event, although these values can also be changed in their green boxes.
The yellow highlighted boxes show when a tournament will need extra elim judges. For example, in the single-flighted event, there are very few situations where the tournament would need extra elim
judges -- the most significant troubles are in the 100s-110s, where the tournament might be short 10 judges. Otherwise, there are few problems.
The double-flighted event shows two options: (a) single-flight the partial and double-flight the first elim round, or (b) double-flight the partial and single-flight the first full elim. Requiring
one judge for three teams is a crucial value; there will enough judges in almost every case to single-flight one (or both) of these two rounds. | {"url":"http://art-of-logic.blogspot.com/2009/04/im-big-believer-that-every-team-at.html","timestamp":"2014-04-21T09:55:42Z","content_type":null,"content_length":"97772","record_id":"<urn:uuid:5402d19d-83d0-4a9f-b85e-30443f40847f>","cc-path":"CC-MAIN-2014-15/segments/1398223203841.5/warc/CC-MAIN-20140423032003-00509-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Re: st: fixed effects regression
[Date Prev][Date Next][Thread Prev][Thread Next][Date index][Thread index]
Re: st: fixed effects regression
From David Jacobs <jacobs.184@sociology.osu.edu>
To statalist@hsphsun2.harvard.edu
Subject Re: st: fixed effects regression
Date Fri, 30 Sep 2005 16:55:30 -0400
If you use the stata commands "xtreg" or "areg" --or the other fixed effects commands for limited dependent variables--you can't control the number of fixed effects; these programs do this
If you alter an OLS analysis by including dummies for each case (and possibly dummies for each period), such models will run, although the program may suppress one case dummy or a year dummy or the
intercept. If you use this augmented OLS approach to estimate a fixed-effects model, the "reg" program will list the coefficients on the dummies and their standard errors. The only disadvantage is
output that can be quite large.
Note that throwing in the dummies in an attempt to use a fixed effects approach to explain limited dependent variables with estimators such as logit, probit, or Tobit will produce incorrect
estimates. This dummy variable inclusion approach only is sound for an analysis that could appropriately use stata routines such as "xtreg" or "areg".
Dave Jacobs
At 04:28 PM 9/30/2005, you wrote:
hi, if I am running a regression with fixed effects. (1) is there a limit on the number of fixed-effects I can use, and (2) is there a way to retrieve the coefficients on the fixed effects (and
possibly also their standard errors) afterwards?
Thank you
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Physics 103: Fall 2005
Physics 103: Fall 2005
With the invention of the pendulum clock, civilization suddenly had an accurate timekeeper, independent of the heavenly motions. Before the pendulum, the best clocks were accurate to perhaps ˝ hour
per day. After the pendulum, clocks were routinely accurate to within a few seconds per day, and were rapidly improved after that. How was it that such a simple and revolutionary invention escaped
notice for so many centuries? I have never heard a plausible answer.
A simple pendulum is just a mass m on a string. Its length L is the length from the support to the center of the mass. (Notice this means that if you try to change m holding L fixed, you might have
to make little adjustments to allow for a different size m.)
About the only thing to measure for a given pendulum is its period T, or equivalently its frequency f=1/T, or equivalently its angular frequency []=2[]f=2[]/T. For many purposes the simplest
quantity is the angular frequency. (Try not to make your pendulums too short: short ones can swing in a more complicated way, with the mass pivoting on the knot that holds it. That is a more
complicated pendulum, with internal motion in addition to the normal pendulum swing.)
(1) The most important things about pendulums can be seen without measuring anything! A priori, we might expect the frequency of a pendulum to depend on both m and L. Find out how the pendulum
frequency depends on mass by constructing two pendulums, using two different masses, one about twice the mass of the other. Let the pendulums run at the same time, and adjust their lengths to make
them have the same frequency. What really matters, m or L?
(2) To find what role the length L plays, adjust the lengths of two pendulums so that one has twice the frequency of the other (i.e., makes two swings while the other makes only one). You will
probably find that when the frequency goes up by a factor of 2, the length goes down by a factor of 4.
(3) Another variable, rather subtle, is the amplitude of the pendulum’s swing. Galileo stated in writing that the frequency doesn’t depend on the amplitude, but this is not true! In the decades
following Galileo’s death, Christian Huyghens went to great effort to make a pendulum-like mechanism that would be truly isochronic, i.e., would have a frequency independent of amplitude. See if you
can tell just the sign of the small amplitude effect: make two pendulums that have the same frequency for small swings, as nearly as you can arrange it, and then observe them if one of them has a
larger amplitude. Is it slower or faster? Try reversing their roles.
(4) Finally take some real data! Measure T vs L, for some pendulum, and make a neat data table in Excel. Use Excel also to find the corresponding f and []. Use the log-log idea to test for a
power law relationship between [] and L (find the power!). Also, supposedly,
Here the constant of proportionality is supposed to be g! Use your data to test this idea graphically, and if these things seem indeed to be proportional, determine g as the slope.
Data table:
Sketches of graphs, labeled:
Use your results from (1)-(4) to write a paragraph explaining everything one has to know about the pendulum:
OPTIONAL: please comment on the lab, if you wish, -- suggested improvements? | {"url":"https://www.mtholyoke.edu/courses/mpeterso/phys103/labs/Pendulums.htm","timestamp":"2014-04-19T02:15:52Z","content_type":null,"content_length":"7851","record_id":"<urn:uuid:83ed49d9-1c00-453b-9b02-613bf511da82>","cc-path":"CC-MAIN-2014-15/segments/1397609535745.0/warc/CC-MAIN-20140416005215-00205-ip-10-147-4-33.ec2.internal.warc.gz"} |
Currying != Generalized Partial Application?!
I had mistakenly learned that curry was a form of generalized partial application from the paper : Function Currying in Scheme by Jeffrey A. Meunier
and the Wikipedia entry (I should have known better), however I was mildly reprimanded for making this novice mistake in a recent paper submission to ICFP. There was a short post to the Scheme
mailing list some time back summing up the problem.
In my paper I defined curry in Cat as:
define curry : ('b ('A 'b -> 'C) -> ('A -> 'C)) { swap quote swap compose }
Whereas this really should have been called "partial-apply", "papply" or something comparable.
So the correct definition should have been:
define papply : ('b ('A 'b -> 'C) -> ('A -> 'C)) { swap quote swap compose }
define curry : (('A 'b -> 'C) -> ('b -> ('A -> 'C)) { quote [papply] compose }
Has anyone else made this mistake? It seems to me that I have seen more incorrect definitions than correct ones.
P.S. Anyone here interested in the health of Wikipedia (I've given up), I'd suggest fixing the code examples and "intuitively ..." note.
Not guilty, your honour.
But I am guilty of occasionally calling partial application by the name "partial evaluation", and I think I'm not the only one.
at Thu, 2007-05-24 22:12 |
to post comments
Sometimes when people say "partial evaluation" in the context of partial application, they really mean "staging".
Vesa Karvonen
at Fri, 2007-05-25 05:32 |
to post comments
What's the difference?
I'm guessing that by 'staging' you mean partial evaluation performed during compilation. But I'm suddenly having doubts.
at Fri, 2007-05-25 16:23 |
to post comments
Currying v. partial application
Currying is easy to define. Currying is turning a function of type (A,B) -> C into A -> (B -> C). That is what currying should always mean.
Partial application is slightly more fuzzy. The intuitive notion is applying a function to less values than arguments of the function, but then you could say all functions take one argument, so...
Partial application and currying are closely related, and one unambiguous way to define partial application is to say, f x is a partial application iff curry (uncurry f) x == f x. This would
distinguish between "incompletely applying" a function and a function that returns a function after doing some work. Usually, we don't really care to make that distinction, and essentially any
function that returns a function can be considered to be partially applied when used.
Derek Elkins
at Thu, 2007-05-24 22:56 |
to post comments
If I have a function
If I have a function f:(x,y)->z, I can't apply it to only one of its arguments. I can curry it, turning it into a function g:x->(y->z) ... and I can apply g to only one of the original arguments. But
turning f into g and applying g to some x are technically different things.
I suspect the confusion arises because originally currying was a technique to model multiple-argument functions in a single-argument framework and was a meta-operation. In ML-like languages, the
functions are typically already curried, so the only operation you see being done is partial application.
J Storrs Hall
at Fri, 2007-05-25 13:40 |
to post comments
seems like a common confusion
The curry() function in Maple takes a procedure and a list of arguments and returns a new procedure that takes the rest of the arguments. To me thats not currying, thats partial application.
The funny thing is that the Maple help page for curry() states that the definition of "currying" is derived from the Haskell 98 report. However, the type of the curry function in the standard prelude
is what you would expect, ((a, b) -> c) -> a -> b -> c, which is totally different from the Maple curry.
Digging deeper I see that section 3.3 of the report is titled "Curried Applications and Lambda Abstractions" but it talks about partial applications. Is a "curried application" the same as a partial
Mike Kucera
at Sat, 2007-05-26 22:34 |
to post comments
Curried Application
While that is the title of the section, the term "curried application" is used nowhere else in the page; "partial application" is used. Looking at other documents, it does look like "curried
application" = "partial application". At any rate, it still doesn't sound anything like "currying".
Derek Elkins
at Mon, 2007-05-28 17:57 |
to post comments
Curried vs. partial application
Partial application is a kind of curried application, according to Denotational Versus Declarative Semantics For Functional Programming (pg. 4):
A curried application of a constructor or function symbol is called partial, exact, or exceeding according to the case that the number m of arguments is less than, equal to or greater than the
symbol's rank.
Anton van Straaten
at Mon, 2007-06-04 16:41 |
to post comments
is it correct
As far as currying is considere there doesn't seem to be much confusion. e.g. x->y->z is x->(y->z). The application starts with first left most argument. Is partial application meant to be intent
that only a subset of the arguments are provided a value and the function which will accept the remaining values is returned as a result.
e.g. let there be a function such as a->b->c->d->e
It can be used using partial application and automatic definitions provided by compiler like entity as follows
This can work if any subset of arguments could be provided their values in a partial application.
In conclusion currying is a special case of partial application and hence can be termed so.
Amar Nath Satrawala
at Wed, 2008-10-29 20:58 |
to post comments | {"url":"http://lambda-the-ultimate.org/node/2266","timestamp":"2014-04-17T18:39:55Z","content_type":null,"content_length":"21278","record_id":"<urn:uuid:ac218ace-5831-45b9-ae3c-f26be2396246>","cc-path":"CC-MAIN-2014-15/segments/1397609530895.48/warc/CC-MAIN-20140416005210-00110-ip-10-147-4-33.ec2.internal.warc.gz"} |
Sine: Overview
The sine (abbreviated "sin") is a type of trigonometric function.
Definition 1 is the simplest and most intuitive definition of the sine function. It basically says that, on a right triangle, the following measurements are related:
• the length of the triangle's hypotenuse
• the length of one of the other sides
• the measurement of the angle (q) opposite to that other side
Furthermore, Definition I gives an exact equation that describes this relation:
sin(q) = opposite / hypotenuse
This equation says that if we evaluate the sine of that angle
, we will get the exact same value as if we divided the length of the side opposite to that angle by the length of the triangle's hypotenuse. The relation holds for any right triangle, regardless of
The main result is this: If we know the values of any two of the above quantities, we can use the above relation to mathematically derive the third quantity. For example, the sine function allows us
to answer any of the following three questions:
"Given a right triangle, where the measurement of one of the non-right angles (q) is known and the length of the side opposite to that angle q is known, find the length of the triangle's
"Given a right triangle, where the measurement of one of the non-right angles (q) is known and the length of the triangle's hypotenuse is known, find the length of the side opposite to that angle
"Given a right triangle, where the length of the triangle's hypotenuse and the length of one of the triangle's other sides is known, find the measurement of the angle (q) opposite to that other
The function takes the form
y = sin(q)
. Usually,
is an angle measurement and y denotes a length.
The sine function, like all trig functions, evaluates differently depending on the units on q, such as degrees, radians, or grads. For example, sin(90°) = 1, while sin(90)=0.89399.... explanation | {"url":"http://www.math.com/tables/algebra/functions/sine/overview.htm","timestamp":"2014-04-16T23:26:40Z","content_type":null,"content_length":"18867","record_id":"<urn:uuid:ea55f9e9-ec50-4f2f-9ba4-23c68c4fe3f9>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00301-ip-10-147-4-33.ec2.internal.warc.gz"} |
Posts by
Posts by alex
Total # Posts: 2,499
What is the difference between the two strong acids, nitric and hydrochloric acid, when they behave as the oxidised forms in a reduction half-equation?
168 children went to a them park. If 1/3 of the girls is equal to 4/9 of the boys, how many boys went to the theme park?
how much were slaves paid after the civil war
Determine the solubility of a sodium sulfate, Na2SO4, in grams per 100g of water, if 0.94 g of Na2SO4 is dissolved in 20 g of water to make a saturated solution.
The length of a rectangle is 11 cm less than 3 times its width. If the area of the rectangle is 20 square inches, find the length and width
Reverse the reactions. Label the acids and bases on the left-hand side of each of the reversed equations? 1. HCl + NH3 => NH4+ + Cl- 2.NH3 + H2O => NH4+ +OH- 3. HCl +H2O => H3O+ +Cl- 4. H3PO4 +H20 =>
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Grammer Please Help!!
Viewed through a telescope, the planet Mars looks rust-red. Because rust is a large component of its soil.
1.Does a pH of 8.6 describe a solution with a higher or lower hydronium ion concentration compared to a neutral solution? Is such as solution called acidic or basic? 2. Compared to a neutral
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5th grade math (word problems)
5th grade math (word problems)
I'm not sure how to solve this: A group of friends went to an amusement park. 10 of them rode the ferris wheel, 15 rode carousel, and 11 rode the roller coaster. 7 of them rode both the ferris wheel
& roller coaster. 4 rode both the ferris wheel & carousel. 5 rode both the...
thank you I appreciate your help:)
Analysing errors consumes time and money. In order to assure that the physical erosion processes are correctly identified both error analysis and standardisation of measuring techniques are necessary
to limit uncertainty and to minimise misinterpretation For the last paragraph...
thank you very much i will repost once i have finished the whole thing.
Soil erosion research is difficult for several reasons notably because it is an intermittent process and thus, an extremely difficult phenomena to observe. In most studies, only the consequences are
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Please rephrase the following sentences into a more simple manner. I need to give a presentation and i would like to write this out simpler. Thank you guys Research of soil erosion is difficult for
several reasons, but particularly because soil erosion is an intermittent proce...
what is 9/2 - 1/2
what is 9/2 - 1/2
a)Name two cations whose salts are quite soluble in water? b)Name one anion whose salts are soluble in water.?
5th grade math (word problems)
Maria is packaging boxes into a carton that is 4 in. x 6 in. x 3 in. THe boxes are 2 in. x 1 in. x 1 in. How many boxes will fit in the carton? I'm not sure how to solve. I know volume = l x w x h.
The carton has volume of 72 cubic inches (4 x 6 x 3). THe boxes are 2 cubic...
5th grade math (word problems)
5th grade math (word problems)
Problem: A receipe for 4 dozen cookies requires 1 1/2 cups of flour. How much flour is needed for 1 dozen? work: 4 divided by 1 1/2 = 4/1 x 3/2 = 4/1 x 2/3 = 8/3 cups of flour Can you explain why i
don't need to change 4 dozen to 48 and then divide by 1 1/2?
Thank you so much!
21. The Mariana trench is located in the Pacific Ocean at a depth of about 10300 m below the surface of the water. The density of seawater is 1025 kg/m3. If an underwater vehicle were to explore such
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I don't know
For each of the following primary standard acids, write the titration equation and calculate the volume of 0.100 M NaOH required for the titration: (i) 300 mg benzoic acid (C6H5COOH) (ii) 200 mg
oxalic acid dihydrate (C2O4H2.2H2O) Equations: (i): C6H5COOH +NaOH->NaC6H5COO +...
Calculate the weight (in mg) of borax Na2B4O7.10H20 required to react with 25.0 mL if 0.100 M HCI No. moles HCl consumed = __________ moles No. moles borax required = __________ moles Molar mass of
borax = __________ g mol-1 (Na2B4O7.10H2O) Weight borax required = __________ g...
15.0 mL of 8.00 M NaOH was diluted with water to 500 mL. Calculate the molarity of the diluted NaOH solution. My calculations: 0.015L x 8.00moles/1L x 39.997/1mole = 4.8g M= (4.8g x 1mole/39.997) /
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STANDARDISATION OF 0.1 M HCl Borax dissolves slowly in water at room temperature. Why do we not use a stirring rod to mix the solution and speed up this stage of the experiment?
5th grade math (word problems)
5th grade math (word problems)
Mrs. Jones bakes pies. She always cuts each pie into 8 slices. Ther are 13 slices left on the counter. How many pies are on the counter? (write as mixed number). work: 13/8 = 1 5/8 pies. Is this
what is the h+ of a solution with a ph of 10 at 25 degree celsius
in relation to the pH of pure water at 25 degree Celsius where do the pH values of acid fall
A 0.0583 mol sample of formaldehyde vapour, CH2O, was placed in a heated 0.359 L vessel and some of it decomposed. The reaction is CH2O(g)---> H2(g) + CO(g). At equilibrium, the CH2O concentration
was 0.0449 M. What is the value of Kc for this reaction? my answer was 0.0029...
At high temperature, 2.00 mol of CH2O was placed in a 4.00 L container where it decomposed via the equilibrium CH2O(g)<=>CO(g)+H2(g) . At equilibrium, the concentration of CO was measured to be
0.0317 M. What is Kc for this reaction at this temperature?
6. When a mass is attached to a spring, the period of oscillation is approximately 2.0 seconds. When the mass attached to the spring is doubled, the period of oscillation is most nearly a) 0.5 s b)
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Gifted Chemistry
Calculate the freezing point and boiling point of a solution that contains 56.8 g NaCl and 32.0 g KBr dissolved in 750.3 mL H2O.
Grade 12 Calculus
Determine the velocity and acceleration as functions of time, t, for s(t) = 45t − 5t^2, where s(t) represents the distance as a function of time. (Hint: velocity and acceleration correspond to the
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Write the name and chemical formula of an amphoteric ion.
Name three sodium salts that produce weak bases in aqueous solution.
A small steel sphere of mass 0.40 kg is attached by a 0.50 m long cord to a swivel pin set into the surface of a frictionless table top. The sphere moves in a circle on the horizontal surface with a
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Pre Calculus
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Physics Electric Force
Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge 1.02 10-4 C is at (0, 3.00 m). We wish to find the net electric force
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Particle A of charge 2.76 10-4 C is at the origin, particle B of charge -6.54 10-4 C is at (4.00 m, 0), and particle C of charge 1.02 10-4 C is at (0, 3.00 m). We wish to find the net electric force
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In a recent year there were the following numbers (in thousands) of licensed drivers in the United States. MALE: Age 19 and under - 4746 Age 20 - 1625 Age 21 - 1679 FEMALE: Age 19 and under - 4517
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What is the best way to answer the following question? Where is the Sea of Okhotsk located? I know that is runs along the coastline of Russia, but I'm not sure what else I should add. thanks
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i have to speak a hindi poem onany one of the- ''life and workings of Dr. B. R. Ambedkar'' ''vaisakhi'' ''the foundation of khalsa panth'' In poetry competition. So please help.
I found K =1920 my question is whether the temperature is being used at all and if it is why?
For reactions in solution, molar concentrations are usually used in equilibrium constant expressions (designated by K or Kc). In gases, partial pressures can also be used (designated by Kp).
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5th grade math (word problems)
oops. thanks!
5th grade math (word problems)
Stefan pours 5/8 cup of milk. He uses 2/3 of it for a bowl of cereal. What fraction of a cup of milk does he use with his cereal? solution: 5/8 * 2/3 = 10/24 = 5/6. So, is the answer 5/6?
11th grade math
find the length of arc if an angle measuring 25 degrees is in the circle of radius 6 feet
11th grade math
in triangle DEF,<D=50 degree, DE=14 meters, and DF=10 meters. find the area of triangle DEFto the nearest tenth of a square meters
in triangle DEF,<D=50 degree, DE=14 meters, and DF=10 meters. find the area of triangle DEFto the nearest tenth of a square meters
90% if the released energy during the Krebs Cycle comes from what
Calculate the new molarity when 150 mL of water is added to 125 mL of 8.55 M HBr.
How many milliliters of commercial hydrochloric acid, which is 11.6 M, should be used to prepare 6.94 L of 0.52 M HCl?
How many milliliters of concentrated ammonia, 15 M NH3, would you dilute to 250 mL to produce 0.162 M NH3?
advance functions gr12 HELP ASAP
using rational functions solve (x+5)/(x-3)= (2x+7)/(x) thank you
Math advance functions gr 12
Using rational functions solve 7+ 1(numerator) x (denominator) = 1 (numerator) x-2(demominator)
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Hi i'm having trouble with a question. it states During the Summer you take 2 part time jobs. The first job pays $5 per hour. The 2nd Job pays $8 per hour. You want to earn at least $150 a week and
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5th grade math (word problems)
Thank you! So, possible sizes are 6, 9, and 12. I really didn't need to use multiples of 4 at all.
5th grade math (word problems)
I'm not sure how to solve this problem. I know that i need to divide 36 cookies. But not sure how to figure out the number of groups. Am i using multiples of 4? Marie made 3 dozen cookies. She needs
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how much toluene (C7H8) is need to make 47.3 (g) of TNT (C7H5N3O6) by the equation given below? C7H8+ HNO3 - C7H5N3O6 + H2O
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The figure shows a 0.3 kg baseball just before and just after it collides with a bat. Just before, the ball has velocity of magnitude 11.4 m/s and angle = 31.4 degrees. Just after, it is traveling
directly upward with velocity of magnitude 9.00 m/s. The duration of the collisi...
In a recent study on gender differences in education, a sample of 7 boys had a mean score of 18.5 and a standard deviation of 4.1 on a Standardized Math Test. A sample of 11 girls had a mean score of
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a positive test charge of 3.0uC experiences a force of 0.75 N in an electric field. What is the magnitude of the electric field at the location of the test charge?
a positive test charge of 3.0uC experiences a force of 0.75 N in an electric field. What is the magnitude of the electric field at the location of the test charge?
Which equivalent fraction would you have to use in order to add 3/5 to 21/25
Condition Pressure Volume Temperature Comment initial 1.0 atm 22.4 L 273 K 1 mole of N2 gas fills the tank. final 3.0 atm 22.4 L 273 K 2 moles of O2 gas have been added to the 1 mole of N2 gas in the
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The mole fraction of oxygen in air is 0.209. Assuming nitrogen is the only other constituent, what are the partial pressures (in torr) of oxygen and nitrogen when the barometer reads 780 torr?
Calculate the partial pressure of 3.25 moles of oxygen that have been mixed (no reaction) with 1.25 moles of nitrogen in a 25.0 L container of 298 K.
dont answer i figured it out:D
Mixtures of helium and oxygen are used in scuba diving tanks to help prevent the bends , a condition caused by nitrogen bubbles forming in the bloodstream. If 95 L of oxygen and 25 L of helium at
STP are pumped into a scuba tank with a volume of 8.0 L, what is the p...
socials please check
thank you :)
socials please check
how many time zones are there in canada? My options are: a.) 5 b.) 10 c.) 7 though i tried answering this and i got 6? what is the correct answer
Socials please check my answers
thanks :)
Socials please check my answers
pleaes chack my answers: what are the horizontal lines on a map called? my answer= latitudes what are the vertical lines on a map called my answer= longitudes
thank you :D
There are 3 pitchers of juice. Each student will be served one fourth of the juice in a pitcher. How many students will get juice? l a. Draw lines on the pitchers to show how many students these
pitchers will serve. b. Write a number model to describe the problem.
There are 3 pitchers of juice. Each student will be served ^ of the juice in a pitcher. How many students will get juice? l a. Draw lines on the pitchers to show how many students these pitchers will
serve. b. Write a number model to describe the problem.
grammer help
thank you :)
grammer help
Aunt Dessie unfolded out of the car like a carpenter s ruler is an example of personification. symbol. simile. metaphor.
There are three particles inside an atom. One of them has zero electric charge. Which one is it?
There are four forces in nature. Name the four forces and rank them from strongest to weakest.
5th grade math (word problems)
Nevermind.. 1st problem was 6.25 and stayed at 6 because that would be the same amount for everyone. For #2, 3.75 was the average. But, need to stay at whole cup which would 3 for everyone to have
equal amount, right?
5th grade math (word problems)
oh.. so you multiplied 8 * 3 instead of 4. But, I thought I needed to round 3.75 up to 4. But, if using 8 *3, then 30-24 = 6. There are 6 cups left over. So, when shouldn't i round up? THe 1st
problem with the pizza, I also rounded up.
5th grade math (word problems)
So for #2, I should have divided by 8 members? 30/8 = 3.75; but 8*4 = 32 which is more than 30cups. I'm not sure how to solve this now.
5th grade math (word problems)
I got these problems wrong in class. but, think i have the right answer now. Can you check? 1. Jim's mother made pizzas for his birthday. SHe sliced the pizzas into 50 slices. She served Jeff and his
7 guests the same number of slices and she ate the rest. How many pieces ...
$27 is left.
Social Studies
5th grade math (word problems)
Thank you Ms. Sue!!
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USGS InfoBank program -- converttime
It also checks that day/hour/minute/second/tenths are
normal values if Option 'COMPRESS' is chosen, though it does
the conversion anyway.
NOTE: This subroutine is designed primarily for time differences.
Therefore, a time difference of 1 tenth of a second gets EXPANDed
to 0000000001; while 0010000001 (ie, the first tenth of a second
of the year) gets COMPRESSed to 8646001 tenths of a second.
(If it were designed instead for calculating tenths of a second
since the beginning of the year, 0010000001 would have been
COMPRESSed to 1 tenth of a second.)
"a" = argument, "r" = referenced, "s" = set
call converttime (with the following arguments)
Option [variable char*8 r]
Day [variable i*4 sr]
Hour [variable i*4 sr]
Minute [variable i*4 sr]
Second [variable i*4 sr]
Tenths [variable i*4 sr]
Time [variable i*4 sr]
ErrorFlag [variable char*3 s]
Option = type of time conversion to do expressed as character;
'COMPRESS' converts DAY/hour/minute/second/tenths to
integer time in tenths;
'EXPAND' converts integer time in tenths to
Input for 'COMPRESS' Option:
Option, day, hour, minute, second, tenths
Output for 'COMPRESS' Option:
Time, ErrorFlag
Input for 'EXPAND' Option:
Option, Time
Output for 'EXPAND' Option:
Day, Hour, Minute, Second, Tenths,
ErrorFlag (always set to 'OK ' for this Option)
Day = integer day of the year; valid values are 1 ->
Hour = integer hour of the day; valid values are 0 ->
Minute = integer minute of the hour; valid values are 0 ->
Second = integer second of the minute; valid values are 0 ->
Tenths = integer tenth of the second; valid values are 0 ->
Time = time of the year expressed as tenths of a second
ErrorFlag = expressed as character,
set to 'BAD' if an error in option argument,
set to 'ODD' if times are not normal
(eg, day less than 1 or greater than 366)
set to 'OK ' if times are normal.
Although this program has been used by the U.S. Geological Survey,
no warranty, expressed or implied, is made by the Survey as to the
accuracy and functioning of the program and related program
material nor shall the fact of distribution constitute any such
warranty, and no responsibility is assured by the Survey in
connection therewith.
Ed Maple 4/18/84 .for
commented out the conversion from tenths of a second to D/H/M/S/TS
- Ed Maple 4/24/84
reinstated ability to convert from tenths of a second to D/H/M/S/TS
- Ed Maple 7/30/84
will still detect time errors, but will convert day/time to tenths
- Ed Maple 9/5/85
modified for UNIX compatibility, also changed arguments to be more
understandable (e.g. Option = 'COMPRESS' or 'EXPAND' rather than
1 or 2)
- Ed Maple 1/19/88
Further modification for UNIX compatibility
- Jean Riordan 2/4/88
Added 'ODD' error flag to handle day values that may be less than 1
and greater than 366.
Subtract 864000 tenths of a second from total to make 001/0001 600
insted of 864600 tenths of a second.
Add 864000 tenths of a second to total to make 600 tenths of a second
001/0001 instead of 000/0001.
- Carolyn Degnan 8/15/88
Undid the subtraction and addition of 864000 because it interfered with
the calculation of time differences - the primary function of this
subroutine. (eg, if the time difference calculated was 600 tenths
of a second, it converted that to 001/0001 which looks like a whole day
plus 1 minute)
- Carolyn Degnan 8/15/88
Added implicit none
- Carolyn Degnan 9/23/04 | {"url":"http://walrus.wr.usgs.gov/infobank/programs/share/converttime.doc.html","timestamp":"2014-04-20T01:27:38Z","content_type":null,"content_length":"18320","record_id":"<urn:uuid:835e32a5-e7e7-42e5-b7c9-9e5d6aebf316>","cc-path":"CC-MAIN-2014-15/segments/1397609537804.4/warc/CC-MAIN-20140416005217-00235-ip-10-147-4-33.ec2.internal.warc.gz"} |
Circuit Design-Chapter 2
Field-Effect Transistors Amplifiers
Analysis of Differential Amplifiers
Most amplifiers in communication circuits are small signal amplifiers. Hence, they can be described by linear equations. We will consider several amplifiers including BJT, FET, operational amplifiers
and differential amplifiers.
• The equivalent circuit of the BJT is shown Fig. 2-1
• r[b]¡¦ is the resistance between the terminal and the actual base junction, r[p] is the base-emitter junction resistance. Typically r[p ] >> r[b]¡¦. An estimation of r[p] is
• r[0] is the collector to emitter resistance typically of the order 15k W, r[m] is the collector-base resistance of the order several MW
• The transconductance g[m] r[m] = b
• A simplified version of the small-signal model equivalent circuit is given Fig. 2-2
• here the coupling capacitance is treated as a short circuit.
• The output voltage is therefore
• The input impedance, not including R[s], is given be
• Since r[p] depends on applied the input resistance will depend on it as well.
• The current gain is
• The Circuit diagram and its equivalent circuit model for the CB amplifier are shown in Fig. 2-4
• Since the sum of the currents leaving the emitter junction is 0, we have the following expression
• If r0 is assumed to be large compared with R[s], r[p]à and R[L], the voltage gain is
• if r[p]>> R[s](1+ƒb), the magnitude of the voltage gain will be the same as that of the common-emitter amplifier
• The input impedance of the common-base amplifier is determined using the following circuit
• The output impedance is determined using the following circuit
• The common-base amplifier has a voltage gain but its current gain is less than unity. It is used as non-inverting amplifiers where low input impedance and high output impedance is desired.
• It also has much better high-frequency response than CE amplifier and is often used in high-frequency circuits.
• The EF has a non-inverting voltage gain of less than 1
• However, it can combine with other stages, such as the CE stage, to realize a greater combined gain than CE stage alone
• Fig. 2-7 , Fig. 2-8 and Fig. 2-9
• Using the equivalent circuit the voltage gain if found to be
• EF configuration has the largest input impedance compared to the other configurations
• The output impedance is
• EF is used when low output impedance is needed. Also, although it has a voltage gain < 1, it has large current gain. It is frequently used as a power amplifier for low-impedance loads.
Field-Effect Transistors Amplifiers
• There are two types of FETs -- JFETs and MOSFETs.
• The low-frequency small signal model is as shown in Fig. 2-10
• The transconductance is defined as
• For JFETs
• where g[m0] is the transconductance when gate-to-source bias voltage is 0, I[D] is the drain current and I[DSS] is the drain current when the gate-to-source voltage is 0.
• For MOSFETs, g[m] is
• where g[mR] is the transconductance at some specified drain bias current I[DR]
• The common-source amplifier is similar to the common-emitter amplifier
• Normally R[g] >> R so V[i] = V[g]
• The source voltage is determined from the following equations
• since the current leaving output node and source is zero
• The current through R[s] The source voltage is
• The voltage gain is
• If rd >> R[s] and R[L] we have the following relationship
• For gm R[s] >> 1 we have
• The circuit diagram and its equivalent circuit model for source follower are shown in Fig. 2-12
• If R[b] >> R[s], then
• The output impedance is
• which is much smaller than the other two FET amplifier configurations and is the major advantage for this configuration
• The common-gate amplifier is often used in high-frequency application and has a much larger bandwidth than the common source configuration.
• Thus
• The input impedance at the source can be found by solving for the source current
• and
• since
• The output impedance is determined as
• but I[0] is also the current passing through the source resistance so
• The common gate amplifier has the highest output impedance of the three FET amplifier
• Multistage amplifiers are used for impedance matching or to obtain extra gain.
• Power transistors have smaller gain-bandwidth product than low-power transistors, hence the power amplification stage is often operated near unity voltage gain in order to maximize the bandwidth
• Voltage amplification is carried out in the stages preceding the power amplification stage
• The FET cascode circuit has many high-frequency applications that two FETs are often fabricated as a single transistor with 2 gates. The source of the one transistor is continuous with the drain
of the other so the device has 1 source, 1 drain and 2 gates
• The device offers low-noise and high gain in radio-frequency applications
• It is a versatile device which can be used as a mixer or automatic gain control amplifier. The equivalent circuit is as shown in Fig. 2-14
• Here gate 2 and the source are grounded
• The load resistance of the first stage is the input resistance of the second stage, and the second stage is a common-gate amplifier. Therefore
• Since
• where g[mR] is the transconductance at some specified drain bias current IDR.
• Since both transistors have the same drain current g[m1]=g[m2]. Thus V[i]=-V[D1] and V[gs2]=-V[s2]=-V[D1]
• Transistors all exhibit a nonlinear characteristic that causes distortion of the input signal levels. Such distortion can be eliminated by push-pull amplifier
• Fig. 2-15
• The above example uses 2 center-tapped transformers. The input transformer separates the input signal into 2 signal 180^o out of phase. The output transformer is used to sum the output currents
of the two transistors.
• Hence
• If the input signal is V[i] = A cos wt
• then the output of the 2 transistors are also periodic, and they can be expressed in a Fourier series
• If the two transistors are identical then I[1] and I[2] are identical except I[2] lags I[1] by 180^o. Thus
• The output current is
• The even harmonics are eliminated from the output. This is important as FET have a square-law characteristic that generates a relatively large second harmonic.
• The differential amplifier is an essential building block in modern IC amplifiers. The circuit is shown Fig. 2-16
• The operation of this circuit is based on the ability to fabricate matched components on the same chip. In the figure I[EE] is realized using a current mirror. We assume that Q[1] and Q[2] are
identical transistors and both collector resistors fabricated with equal values.
• The KVL expression for the loop containing the two emitter-base junctions is
• The transistors are biased in the forward-active mode, the reverse saturation current of the collector-base junction is negligible. The collector currents IC1 and IC2 are given by
• where we assumed that exp(V[bb]/kT) >> 1
• where Vd is the difference between the two input voltages. KCL at the emitter node requires
• Similarly Fig. 2-17
• If V[d] = 0 then I[C1] = I[C2]
• If we incrementally increase V[1] by Dv/2and simultaneously decrease V[2] by Dv/2. The differential voltage becomes Dv. For Dv< 4kT as indicated in the transfer characteristics above the circuit
behaves linearly. I[C1] increases by DI[C] and I[C2] decreases by the same amount.
• Consider both V[1] and V[2] are increased by Dv/2. The difference voltage remains 0, and I[C1] and Ic2 remain equal. However, both I[C1] and I[C2] exhibit a small increase dI[C]. Hence the
current in R[E] increases by 2dI[C]. The voltage V[E] is no longer constant but increase by an amount of 2dI[C]R[B]. This situation where equal signal are applied to Q[1] and Q[2] is the called
the common mode.
• Usually differential amplifiers are designed such that only differential signals are amplified.
Analysis of Differential Amplifiers
• The analysis of differential amplifiers is simplified using ¡§half-circuit¡¨ concept. The equivalent circuit model for differential and common modes are shown in Fig. 2-18
• Differential mode gain
• Common mode gain is
• The Common Mode Rejection Ration (CMRR) is defined as
• From the above we obtain
• If arbitrary signals V1 and V2 are applied to the inputs then | {"url":"http://www.eie.polyu.edu.hk/~ensurya/lect_notes/commun_cir/Ch2/Chapter2.htm","timestamp":"2014-04-16T07:13:27Z","content_type":null,"content_length":"25106","record_id":"<urn:uuid:b89573a7-db87-4c9f-9a35-5755080dea64>","cc-path":"CC-MAIN-2014-15/segments/1398223211700.16/warc/CC-MAIN-20140423032011-00280-ip-10-147-4-33.ec2.internal.warc.gz"} |
Nonlinearity compensation using dispersion-folded digital backward propagation
« journal navigation
Nonlinearity compensation using dispersion-folded digital backward propagation
Optics Express, Vol. 20, Issue 13, pp. 14362-14370 (2012)
A computationally efficient dispersion-folded (D-folded) digital backward propagation (DBP) method for nonlinearity compensation of dispersion-managed fiber links is proposed. At the optimum power
level of long-haul fiber transmission, the optical waveform evolution along the fiber is dominated by the chromatic dispersion. The optical waveform and, consequently, the nonlinear behavior of the
optical signal repeat at locations of identical accumulated dispersion. Hence the DBP steps can be folded according to the accumulated dispersion. Experimental results show that for 6,084 km single
channel transmission, the D-folded DBP method reduces the computation by a factor of 43 with negligible penalty in performance. Simulation of inter-channel nonlinearity compensation for 13,000 km
wavelength-division multiplexing (WDM) transmission shows that the D-folded DBP method can reduce the computation by a factor of 37.
© 2012 OSA
1. Introduction
Optical signal is distorted by noise, dispersion and nonlinearity in fiber transmission. The Kerr nonlinearity, an intensity dependence of the refractive index, induces impairments including
self-phase modulation (SPM), cross-phase modulation (XPM) and four-wave mixing (FWM) [
]. These nonlinear impairments increase with the optical signal power. As a tradeoff between high signal to noise ratio and low nonlinear impairment, there is an optimum power level for a fiber
transmission system, corresponding to a maximal spectral efficiency [
2. P. P. Mitra and J. B. Stark, “Nonlinear limits to the information capacity of optical fibre communications,” Nature 411(6841), 1027–1030 (2001). [CrossRef] [PubMed]
In most of the installed fiber links, chromatic dispersion is compensated by cascading fibers with inverse dispersion parameters [
3. B. C. Kurtzke, “Suppression of fiber nonlinearities by appropriate dispersion management,” IEEE Photon. Technol. Lett. 5(10), 1250–1253 (1993). [CrossRef]
]. With the advent of new dispersion compensating fibers (DCF), wide-band dispersion flatness has been obtained by compensating for both dispersion and dispersion slope [
4. K. Mukasa, K. Imamura, I. Shimotakahara, T. Yagi, and K. Kokura, “Dispersion compensating fiber used as a transmission fiber: inverse/reverse dispersion fiber,” J. Opt. Fiber Commun. Rep. 3(5),
292–339 (2006). [CrossRef]
]. In the emerging digital coherent transmission systems [
5. E. Ip, A. P. T. Lau, D. J. F. Barros, and J. M. Kahn, “Coherent detection in optical fiber systems,” Opt. Express 16(2), 753–791 (2008). [CrossRef] [PubMed]
], electronic dispersion compensation (EDC) can be realized using digital signal processing (DSP) [
6. S. J. Savory, G. Gavioli, R. I. Killey, and P. Bayvel, “Electronic compensation of chromatic dispersion using a digital coherent receiver,” Opt. Express 15(5), 2120–2126 (2007). [CrossRef]
8. E. Yamazaki, S. Yamanaka, Y. Kisaka, T. Nakagawa, K. Murata, E. Yoshida, T. Sakano, M. Tomizawa, Y. Miyamoto, S. Matsuoka, J. Matsui, A. Shibayama, J. Abe, Y. Nakamura, H. Noguchi, K. Fukuchi, H.
Onaka, K. Fukumitsu, K. Komaki, O. Takeuchi, Y. Sakamoto, H. Nakashima, T. Mizuochi, K. Kubo, Y. Miyata, H. Nishimoto, S. Hirano, and K. Onohara, “Fast optical channel recovery in field demonstration
of 100-Gbit/s Ethernet over OTN using real-time DSP,” Opt. Express 19(14), 13179–13184 (2011). [CrossRef] [PubMed]
]. As the compensation technique of linear distortions matured, fiber nonlinearity has become the limiting factor to further increase the capacity and distance of the next generation fiber-optic
transmission systems [
9. J. M. Kahn and K.-P. Ho, “A bottleneck for optical fibres,” Nature 411(6841), 1007–1010 (2001). [CrossRef] [PubMed]
12. R.-J. Essiambre, G. Kramer, P. J. Winzer, G. J. Foschini, and B. Goebel, “Capacity limits of optical fiber networks,” J. Lightwave Technol. 28(4), 662–701 (2010). [CrossRef]
In [
], an adaptive filtering carrier phase recovery method was proposed to suppress the nonlinear phase noise due to XPM. Lumped phase de-rotation proportional to the received single-channel or
multi-channel optical intensity can also be used for SPM compensation [
14. K.-P. Ho and J. M. Kahn, “Electronic compensation technique to mitigate nonlinear phase noise,” J. Lightwave Technol. 22(3), 779–783 (2004). [CrossRef]
] or XPM compensation [
15. L. B. Du and A. J. Lowery, “Practical XPM compensation method for coherent optical OFDM systems,” IEEE Photon. Technol. Lett. 22(5), 320–322 (2010). [CrossRef]
], respectively. However, the lump phase de-rotation method is based on the assumption that the intensity waveform remains unchanged throughout the fiber propagation. In long-haul broadband
transmission where the chromatic dispersion causes significant pulse reshaping and inter-channel walk-off, a distributed nonlinearity compensation method, known as digital backward propagation (DBP),
is necessary for the effective compensation of the joint effect of dispersion and nonlinearity [
16. K. Roberts, C. Li, L. Strawczynski, M. O’Sullivan, and I. Hardcastle, “Electronic precompensation of optical nonlinearity,” IEEE Photon. Technol. Lett. 18(2), 403–405 (2006). [CrossRef]
19. D. S. Millar, S. Makovejs, C. Behrens, S. Hellerbrand, R. I. Killey, P. Bayvel, and S. J. Savory, “Mitigation of fiber nonlinearity using a digital coherent receiver,” IEEE J. Sel. Top. Quantum
Electron. 16(5), 1217–1226 (2010). [CrossRef]
]. In order for DBP to be accurate, a small step size is usually required, resulting in a large number of steps and a heavy computational load [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
Some methods have been proposed to reduce the computational load of DBP [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
28. L. Zhu, X. Li, E. F. Mateo, and G. Li, “Complementary FIR filter pair for distributed impairment compensation of WDM fiber transmission,” IEEE Photon. Technol. Lett. 21(5), 292–294 (2009).
]. In comparison with inter-channel nonlinearity compensation, intra-channel nonlinearity compensation usually requires a smaller number of steps because the step size is not limited by the
inter-channel walk-off effect [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
]. In comparison with solving the nonlinear Schrodinger equation (NLSE) for the total field of the WDM signal, solving the coupled NLSE was suggested for inter-channel nonlinearity compensation
because it requires a smaller step number and a lower sampling rate [
22. E. F. Mateo, L. Zhu, and G. Li, “Impact of XPM and FWM on the digital implementation of impairment compensation for WDM transmission using backward propagation,” Opt. Express 16(20), 16124–16137
(2008). [CrossRef] [PubMed]
]. The step number can be further reduced by factorizing the dispersive walk-off effects in the DBP algorithm [
23. E. F. Mateo, F. Yaman, and G. Li, “Efficient compensation of inter-channel nonlinear effects via digital backward propagation in WDM optical transmission,” Opt. Express 18(14), 15144–15154
(2010). [CrossRef] [PubMed]
] and using variable step size [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
24. Q. Zhang and M. I. Hayee, “Symmetrized split-step Fourier scheme to control global simulation accuracy in fiber-optic communication systems,” J. Lightwave Technol. 26(2), 302–316 (2008).
25. L. B. Du and A. J. Lowery, “Improved single channel backpropagation for intra-channel fiber nonlinearity compensation in long-haul optical communication systems,” Opt. Express 18(16), 17075–17088
(2010). [CrossRef] [PubMed]
]. We recently proposed a distanced-folded DBP method for dispersion-managed links with complete dispersion compensation in every fiber span [
27. L. Zhu and G. Li, “Folded digital backward propagation for dispersion-managed fiber-optic transmission,” Opt. Express 19(7), 5953–5959 (2011). [CrossRef] [PubMed]
29. J. K. Fischer, C.-A. Bunge, and K. Petermann, “Equivalent single-span model for dispersion-managed fiber-optic transmission systems,” J. Lightwave Technol. 27(16), 3425–3432 (2009). [CrossRef]
]. However, most of the deployed fiber links has non-zero residual dispersion per span in order to avoid the resonant nonlinear effects [
30. C. Xie, “WDM coherent PDM-QPSK systems with and without inline optical dispersion compensation,” Opt. Express 17(6), 4815–4823 (2009). [CrossRef] [PubMed]
31. V. Curri, P. Poggiolini, A. Carena, and F. Forghieri, “Dispersion compensation and mitigation of nonlinear effects in 111-Gb/s WDM coherent PM-QPSK systems,” IEEE Photon. Technol. Lett. 20(17),
1473–1475 (2008). [CrossRef]
]. In [
], an efficient scheme based on statistical approximation of accumulated nonlinear phase rotation with discrete residual dispersion was demonstrated for intra-channel nonlinearity compensation. In
this paper, we propose and demonstrate a computationally efficient dispersion-folded (D-folded) DBP method that is effective for fiber links with arbitrary dispersion maps.
2. Dispersion-folded DBP method
The dispersion map of a typical dispersion-managed fiber transmission system is illustrated in
Fig. 1
. After the dispersion-managed fiber transmission and coherent detection, conventional DBP can be performed in the backward direction of the fiber propagation. Multiple steps are required for each of
the many fiber spans, resulting in a large number of steps.
At the optimum power level of fiber transmission, the total nonlinear phase shift is on the order of 1 radian [
33. J. P. Gordon and L. F. Mollenauer, “Phase noise in photonic communications systems using linear amplifiers,” Opt. Lett. 15(23), 1351–1353 (1990). [CrossRef] [PubMed]
]. Therefore, in long-haul transmission, the nonlinear effect in each fiber span is weak, and the optical waveform evolution along the fiber is dominated by the chromatic dispersion. Under the weakly
nonlinear assumption, the optical waveform repeats at locations where accumulated dispersions are identical. Since the Kerr nonlinear effects are determined by the instantaneous optical field, the
nonlinear behavior of the optical signal also repeats at locations of identical accumulated dispersion. Hence we can fold the DBP according to the accumulated dispersion.
The propagation of the optical field,E(z,t), is governed bywhereDis the linear operator for dispersion, fiber loss and amplifier gain, N(|E(z,t)|2) is the nonlinear operator, ε(to be set to unity) is
a parameter indicating that the nonlinear perturbation is small for the reasons given above.
The solution of
Eq. (1)
can be written as,Substituting
Eq. (2)
Eq. (1)
, expanding the equation in power series of
, and equating to zero the successive terms of the series, we havewhich describe the linear evolution and the nonlinear correction, respectively. It is noted that the nonlinear correction
is governed by a linear partial differential equation with nonzero forcing which depends on the linear solution only.
It is shown in
Fig. 1
that the dispersion map can be divided into
divisions as indicated by the horizontal dashed lines. The fiber segments within a division have the same accumulated dispersion. Based on the principle of superposition, the total nonlinear
correction is the sum of nonlinear corrections due to nonzero forcing at each fiber segment.
In conventional DBP, the contribution from each fiber segment is computed separately. However, it is advantageous to calculate the total nonlinear correction as the sum of nonlinear corrections due
to nonzero forcing at different accumulated dispersion divisions, each having multiple fiber segments. This is because, with the exception of different input power levels and effective lengths, the
linear component El(z,t)that generates the nonlinear correction and the total dispersion for the generated nonlinear perturbation to reach the end of the transmission are identical for the fiber
segments with the same accumulated dispersion. Therefore, the nonlinear corrections due to these multiple fiber segments with the same accumulated dispersion are identical except a constant and can
be calculated all at once using a weighting factor as described below.
In D-folded DBP, the fiber segments with the same accumulated dispersion (e.g., the blue solid lines) can be folded into one step. For a fiber link with positive residual dispersion per span, a
lumped dispersion compensator (Dlumped) can be used to obtain the optical field (E1) in the first dispersion division. Then dispersion compensation (D) and nonlinearity compensation (NL) are
performed for each of the subsequent dispersion divisions. To take into account the different power levels and effective lengths of the fiber segments, a weighting factor (Wi) is used in the
nonlinearity compensator of each step. The nonlinear phase shift in the ith step of D-folded DBP is given byφi=Wi⋅|E¯i(t)|2, where E¯i(t) is the optical field with the power normalized to unity. The
weighting factor is given byWi=∑kγ∫Pi,k(z)dz, where Pi,k(z)is the power level as a function of distance within the kth fiber segment in the ith dispersion division. The effect of loss for each fiber
segment is taken into account in the calculation of this weighting factor.
The D-folded DBP theory derived above is based on the NLSE for fiber propagation thus can be applied to not only intra-channel [
] but also inter-channel nonlinearities as we show below. In comparison to [
], the D-folded DBP also takes into account the effect of dispersion on the perturbative nonlinear effects. The commonality in [
27. L. Zhu and G. Li, “Folded digital backward propagation for dispersion-managed fiber-optic transmission,” Opt. Express 19(7), 5953–5959 (2011). [CrossRef] [PubMed]
], [
], and the D-folded DBP presented here is that nonlinearity compensation for multiple fiber segments is performed in a single step, resulting in orders of magnitude savings in computation. Using the
split-step method for D-folded DBP, the linear and nonlinear effects can be de-coupled when the step size is small enough. The dispersion within a fiber segment is neglected in a nonlinearity
compensation operator. Meanwhile, the power level, effective length and nonlinear coefficient of a fiber segment have been taken into account in the calculation of the nonlinear phase shift. Thus the
DBP can be folded even if the fiber link consists of multiple types of fibers. Note that calculating the weighting factors does not require real-time computation.
3. Experimental results
To demonstrate the effectiveness of the D-folded DBP, we performed the experiment of single-channel 6,084 km transmission of NRZ-QPSK signal at 10 Gbaud. The experimental setup is shown in
Fig. 2(a)
. At the transmitter, carrier from an external-cavity laser is modulated by a QPSK modulator using a 2
-1 pseudo random bit sequence (PRBS). The single-polarization optical signal is launched into a recirculating loop controlled by two acousto-optic modulators (AOMs). The recirculating loop consists
of two types of fibers: 82.6 km standard single mode fiber (SSMF) with 0.2 dB/km loss and 83 µm
effective area, and 11 km DCF with 0.46 dB/km loss and 20 µm
effective area. By optimizing the performance in the training experiments of EDC, the dispersion of the SSMF and the DCF are determined as 17.06 ps/nm/km and −123.35 ps/nm/km, respectively. The
residual dispersion per span (RDPS) is 53 ps/nm. Two erbium doped fiber amplifiers (EDFAs) are used to completely compensate for the loss in the loop. An optical bandpass filter (BPF) is used to
suppress the EDFA noise. At the receiver, the signal is mixed with the local oscillator from another external cavity laser in a 90° hybrid after the polarizations are aligned with a polarization
controller (PC). The I and Q components of the received signal are detected using two photo-detectors (PDs). A real-time oscilloscope is used for analog-to-digital conversion and data acquisition at
40 Gsamples/s. The DSP is performed off-line with Matlab. Note that for realistic terrestrial systems, the parameters of the fiber spans may not be available with good accuracy. Attempts are being
made at estimating the link parameters for conventional DBP [
]. Estimation of link parameters for D-folded DBP needs further investigation.
Without loss of generality, we solve the NLSE using the asymmetric Split-Step Fourier Method (SSMF) with one dispersion compensator per step [
24. Q. Zhang and M. I. Hayee, “Symmetrized split-step Fourier scheme to control global simulation accuracy in fiber-optic communication systems,” J. Lightwave Technol. 26(2), 302–316 (2008).
]. For long-haul transmission, the DBP step size is usually limited by dispersion [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
]. In this paper, we use DBP steps with equal dispersion per step for simplicity. The Q-value as a function of the number of steps is shown in
Fig. 2(b)
. The required number of steps to approach the maximum Q-value can be reduced from 1,300 to 30 by using the D-folded DBP. The number of multiplications per sample (MPS) for DBP is reduced by a factor
of 43 (see details in the Appendix). There is a trade-off between complexity and performance using either conventional DBP or D-folded DBP. A Q-value of 10.2 dB, corresponding to a 1.1 dB improvement
in comparison with EDC, can be achieved using 130-step conventional DBP or 5-step D-folded DBP.
Figure 2(c)
shows the Q-value as a function of the launching power. With only EDC for the accumulated residual dispersion, the maximum Q-value is 9.1 dB. With nonlinearity compensation using D-folded DBP, the
maximum Q-value is increased to 10.7 dB. The performance after the 30-step D-folded DBP is almost the same as that after the 1,300-step conventional DBP. The Q-values and computational load are shown
Table 1
4. Simulation results
We performed the simulation of wavelength-division multiplexing (WDM) transmission as illustrated in
Fig. 3(a)
using the VPItransmissionMaker. At the transmitter, 12 channels of 56 Gb/s NRZ-QPSK signal are transmitted with 50 GHz channel spacing. The QPSK modulators are driven by 2
-1 PRBS sequences. Each PRBS generator is assigned with a unique random bit seed for de-correlation. The linewidth of the lasers is 100 kHz. The dispersion managed fiber link consists of 260 spans of
the OFS UltraWave SLA/IDF Ocean Fiber combination. In each 50 km span, the SLA fiber with a large effective area is used near the EDFA, followed by the IDF fiber with inverse dispersion and
dispersion slope. The loss, dispersion, relative dispersion slope and effective area of the SLA fiber are 0.188 dB/km, 19.5 ps/nm/km, 0.003/nm and 106 μm
, respectively. The corresponding parameters for the IDF fiber are 0.23 dB/km, −44 ps/nm/km, 0.003/nm and 31 μm
, respectively. The RDPS is determined by the proportion of SLA fiber to IDF fiber in each span. The noise figure of the EDFAs is 4.5 dB. After de-multiplexing and coherent detection, 8192 symbols
from each channel are saved at a sampling rate of 56 Gsamples/s. Then the DSP is performed with Matlab.
In DBP for SPM + XPM compensation, coupled NLSE is solved with the asymmetric SSFM [
35. L. Zhu, F. Yaman, and G. Li, “Experimental demonstration of XPM compensation for WDM fibre transmission,” Electron. Lett. 46(16), 1140–1141 (2010). [CrossRef]
]. The inter-channel walk-off effects are considered in the nonlinearity compensation operators in order to increase the required step size [
23. E. F. Mateo, F. Yaman, and G. Li, “Efficient compensation of inter-channel nonlinear effects via digital backward propagation in WDM optical transmission,” Opt. Express 18(14), 15144–15154
(2010). [CrossRef] [PubMed]
]. For each channel, the XPM from two neighboring channels and the SPM are compensated. Simulation results show that the XPM from the other channels is weak because of the walk-off effect.
The average Q-value of all the WDM channels is calculated after 13,000 km transmission and DSP. The Q-values are calculated through variance estimation on the constellations under the assumption of
Gaussian distribution. The Q-values presented here only provide a metric for comparison between different compensation schemes and are not reliable for the computation of bit error ratios because
correlations exist in the nonlinear noise even after nonlinearity compensation.
Figure 3(b)
shows the Q-value as a function of the number of steps after DBP for SPM + XPM compensation using conventional DBP and D-folded DBP. With a RDPS of 20 ps/nm, the Q-values after 2,600-step
conventional DBP and 80-step D-folded DBP are 11.5 dB and 11.3 dB, respectively. The minimum multiplications per sample for the 2,600-step conventional DBP and the 80-step D-folded DBP are 51,422 and
1,393, respectively. The computational load is reduced by a factor of 37 with a penalty of 0.2 dB in Q-value. With a RDPS of 80 ps/nm, the Q-values after 2,600-step conventional DBP and 200-step
D-folded DBP are 11.9 dB and 11.7 dB, respectively. The step sizes of conventional DBP and D-folded DBP are usually limited by chromatic dispersion. As a result, with a larger RDPS, a larger number
of steps for D-folded DBP are required to approach the maximum Q-value. Note that in realistic systems, exchanging information between channels for XPM compensation can increase the complexity of the
DSP implementation.
Figure 3(c)
shows the Q-value as a function of the number of steps after the DBP for SPM compensation. With a RDPS of 20 ps/nm, the Q-values after 1,040-step conventional DBP and 50-step D-folded DBP are 10.3 dB
and 10.4 dB, respectively. The number of multiplications per sample is reduced from 10,400 to 408 with no penalty in Q-value. With a RDPS of 80 ps/nm, the Q-values after 1,040-step conventional DBP
and 150-step D-folded DBP are 11.1 dB and 11.0 dB, respectively. The Q-values and computational loads are summarized in
Table 2
Recently, coherent systems with EDC for dispersion-unmanaged fiber links have been implemented. The minimum multiplications per sample for EDC of a 13,000 km dispersion-unmanaged fiber link using SLA
fiber is 245.
The Q-value obtained after DBP with sufficiently large number of steps as a function of the RDPS is shown in
Fig. 3(d)
. For the dispersion-managed link, as the RDPS increases, the Q-value after EDC increases because the nonlinear effects and the span loss decrease. With DBP, the Q-value approaches the maximum value
when the RDPS is larger than 20 ps/nm. In comparison with EDC, the DBP with XPM compensation can increase the Q-value by more than 3 dB when the RDPS is larger than 20 ps/nm. The performance using
D-folded DBP is almost the same as that using conventional DBP.
5. Conclusion
In conclusion, we have proposed a dispersion-folded DBP method that can significantly reduce the computational load of DBP for fiber nonlinearity compensation. Experimental results show that the
computation of DBP for 6,084 km single channel transmission can be reduced by a factor of 43. Simulation of a WDM system shows that the D-folded DBP method can reduce the computation for XPM
compensation by a factor of 39.
Appendix: calculation of the computational load
The computational load can be associated to the number of complex multiplications per sample (MPS) involved in the operation. Either time-domain [using finite-impulse response (FIR) filters] or
frequency-domain equalization can be used for the compensation of dispersion. In DBP, the frequency response of each dispersion compensator must be very accurate in order to minimize the error
accumulation. When the amount of dispersion to be compensated and thus the theoretical minimum number of taps of the FIR filter are small, the number of taps required for the FIR filter can be much
larger than the minimum number corresponding to the group delay [
28. L. Zhu, X. Li, E. F. Mateo, and G. Li, “Complementary FIR filter pair for distributed impairment compensation of WDM fiber transmission,” IEEE Photon. Technol. Lett. 21(5), 292–294 (2009).
]. Therefore, frequency-domain overlap-add FFT method is assumed here for calculating the computational complexity of DBP in this paper [
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
For an overhead length P, a signal block length M is chosen to minimize the computational load using the radix-2 Fast Fourier Transform (FFT) with an FFT block size of (M+P). For EDC, the overhead is
approximately given byP=2π|β2|B⋅h⋅S whereβ2,B, handSare the dispersion, signal bandwidth per channel, fiber length and sampling rate, respectively. The MPS for the overlap-add filtering is given by
[(M+P)log2(M+P)+(M+P)]/M. For the EDC of the 13,000 km dispersion-unmanaged link, we consider an FFT block size of 2^10 which is practical for the current electronic technology. By using 11
overlap-add filtering operators with M=548 andP=476, the minimum possible MPS is 245.
To ensure accuracy of the dispersion operator in DBP, we assume an overhead Pthat is 3 times the group delay2π|β2|B⋅h⋅S. In the nonlinearity operator of DBP, the calculations of the optical intensity
and the nonlinear phase shift each costs one complex multiplication. The value of the nonlinear phase shift can be obtained using a lookup table. The MPS of DBP for SPM compensation is given bynst⋅
[(M+P)log2(M+P)+(M+P)+2M]/M, wherenstis the step number. For the 6,084 km single channel transmission, the minimum MPSs of the 1,300-step conventional DBP and the 30-step D-folded DBP are 13,385 and
314, respectively. For the SPM compensation of the 13,000 km WDM system, the minimum MPSs of the 1,040-step conventional DBP and the 50-step D-folded DBP are 12,222 and 500, respectively.
In the DBP with XPM compensation for the 13,000 km WDM transmission, the length of the inter-channel walk-off filter is given byP=2π|β2|Δf⋅h⋅S where Δfis the channel spacing. The MPS for DBP is given
bynst⋅[2(M+P)log2(M+P)+3(M+P)+2M]/M. The minimum MPS for the 2,600-step conventional DBP is 51,422. For the 80-step D-folded DBP, the minimum MPS is 1,393.
References and links
1. G. P. Agrawal, Nonlinear Fiber Optics (Academic Press, Elsevier, 2001).
2. P. P. Mitra and J. B. Stark, “Nonlinear limits to the information capacity of optical fibre communications,” Nature 411(6841), 1027–1030 (2001). [CrossRef] [PubMed]
3. B. C. Kurtzke, “Suppression of fiber nonlinearities by appropriate dispersion management,” IEEE Photon. Technol. Lett. 5(10), 1250–1253 (1993). [CrossRef]
4. K. Mukasa, K. Imamura, I. Shimotakahara, T. Yagi, and K. Kokura, “Dispersion compensating fiber used as a transmission fiber: inverse/reverse dispersion fiber,” J. Opt. Fiber Commun. Rep. 3(5),
292–339 (2006). [CrossRef]
5. E. Ip, A. P. T. Lau, D. J. F. Barros, and J. M. Kahn, “Coherent detection in optical fiber systems,” Opt. Express 16(2), 753–791 (2008). [CrossRef] [PubMed]
6. S. J. Savory, G. Gavioli, R. I. Killey, and P. Bayvel, “Electronic compensation of chromatic dispersion using a digital coherent receiver,” Opt. Express 15(5), 2120–2126 (2007). [CrossRef]
7. H. Sun, K.-T. Wu, and K. Roberts, “Real-time measurements of a 40 Gb/s coherent system,” Opt. Express 16(2), 873–879 (2008). [CrossRef] [PubMed]
8. E. Yamazaki, S. Yamanaka, Y. Kisaka, T. Nakagawa, K. Murata, E. Yoshida, T. Sakano, M. Tomizawa, Y. Miyamoto, S. Matsuoka, J. Matsui, A. Shibayama, J. Abe, Y. Nakamura, H. Noguchi, K. Fukuchi, H.
Onaka, K. Fukumitsu, K. Komaki, O. Takeuchi, Y. Sakamoto, H. Nakashima, T. Mizuochi, K. Kubo, Y. Miyata, H. Nishimoto, S. Hirano, and K. Onohara, “Fast optical channel recovery in field
demonstration of 100-Gbit/s Ethernet over OTN using real-time DSP,” Opt. Express 19(14), 13179–13184 (2011). [CrossRef] [PubMed]
9. J. M. Kahn and K.-P. Ho, “A bottleneck for optical fibres,” Nature 411(6841), 1007–1010 (2001). [CrossRef] [PubMed]
10. E. B. Desurvire, “Capacity demand and technology challenges for lightwave systems in the next two decades,” J. Lightwave Technol. 24(12), 4697–4710 (2006). [CrossRef]
11. A. D. Ellis, J. Zhao, and D. Cotter, “Approaching the non-linear Shannon limit,” J. Lightwave Technol. 28(4), 423–433 (2010). [CrossRef]
12. R.-J. Essiambre, G. Kramer, P. J. Winzer, G. J. Foschini, and B. Goebel, “Capacity limits of optical fiber networks,” J. Lightwave Technol. 28(4), 662–701 (2010). [CrossRef]
13. L. Li, Z. Tao, L. Liu, W. Yan, S. Oda, T. Hoshida, and J. C. Rasmussen, “XPM tolerant adaptive carrier phase recovery for coherent receiver based on phase noise statistics monitoring,” in Proc.
ECOC’09, Paper P3.16 (2009).
14. K.-P. Ho and J. M. Kahn, “Electronic compensation technique to mitigate nonlinear phase noise,” J. Lightwave Technol. 22(3), 779–783 (2004). [CrossRef]
15. L. B. Du and A. J. Lowery, “Practical XPM compensation method for coherent optical OFDM systems,” IEEE Photon. Technol. Lett. 22(5), 320–322 (2010). [CrossRef]
16. K. Roberts, C. Li, L. Strawczynski, M. O’Sullivan, and I. Hardcastle, “Electronic precompensation of optical nonlinearity,” IEEE Photon. Technol. Lett. 18(2), 403–405 (2006). [CrossRef]
17. X. Li, X. Chen, G. Goldfarb, E. F. Mateo, I. Kim, F. Yaman, and G. Li, “Electronic post-compensation of WDM transmission impairments using coherent detection and digital signal processing,” Opt.
Express 16(2), 880–888 (2008). [CrossRef] [PubMed]
18. E. Ip and J. M. Kahn, “Fiber impairment compensation using coherent detection and digital signal processing,” J. Lightwave Technol. 28(4), 502–519 (2010). [CrossRef]
19. D. S. Millar, S. Makovejs, C. Behrens, S. Hellerbrand, R. I. Killey, P. Bayvel, and S. J. Savory, “Mitigation of fiber nonlinearity using a digital coherent receiver,” IEEE J. Sel. Top. Quantum
Electron. 16(5), 1217–1226 (2010). [CrossRef]
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol. 21(1), 61–68
(2003). [CrossRef]
21. S. Oda, T. Tanimura, T. Hoshida, C. Ohshima, H. Nakashima, Z. Tao, and J. C. Rasmussen, “112 Gb/s DP-QPSK transmission using a novel nonlinear compensator in digital coherent receiver.” in Proc.
OFC’09, Paper OThR6 (2009).
22. E. F. Mateo, L. Zhu, and G. Li, “Impact of XPM and FWM on the digital implementation of impairment compensation for WDM transmission using backward propagation,” Opt. Express 16(20), 16124–16137
(2008). [CrossRef] [PubMed]
23. E. F. Mateo, F. Yaman, and G. Li, “Efficient compensation of inter-channel nonlinear effects via digital backward propagation in WDM optical transmission,” Opt. Express 18(14), 15144–15154
(2010). [CrossRef] [PubMed]
24. Q. Zhang and M. I. Hayee, “Symmetrized split-step Fourier scheme to control global simulation accuracy in fiber-optic communication systems,” J. Lightwave Technol. 26(2), 302–316 (2008).
25. L. B. Du and A. J. Lowery, “Improved single channel backpropagation for intra-channel fiber nonlinearity compensation in long-haul optical communication systems,” Opt. Express 18(16), 17075–17088
(2010). [CrossRef] [PubMed]
26. D. Rafique, M. Mussolin, M. Forzati, J. Mårtensson, M. N. Chugtai, and A. D. Ellis, “Compensation of intra-channel nonlinear fibre impairments using simplified digital back-propagation
algorithm,” Opt. Express 19(10), 9453–9460 (2011). [CrossRef] [PubMed]
27. L. Zhu and G. Li, “Folded digital backward propagation for dispersion-managed fiber-optic transmission,” Opt. Express 19(7), 5953–5959 (2011). [CrossRef] [PubMed]
28. L. Zhu, X. Li, E. F. Mateo, and G. Li, “Complementary FIR filter pair for distributed impairment compensation of WDM fiber transmission,” IEEE Photon. Technol. Lett. 21(5), 292–294 (2009).
29. J. K. Fischer, C.-A. Bunge, and K. Petermann, “Equivalent single-span model for dispersion-managed fiber-optic transmission systems,” J. Lightwave Technol. 27(16), 3425–3432 (2009). [CrossRef]
30. C. Xie, “WDM coherent PDM-QPSK systems with and without inline optical dispersion compensation,” Opt. Express 17(6), 4815–4823 (2009). [CrossRef] [PubMed]
31. V. Curri, P. Poggiolini, A. Carena, and F. Forghieri, “Dispersion compensation and mitigation of nonlinear effects in 111-Gb/s WDM coherent PM-QPSK systems,” IEEE Photon. Technol. Lett. 20(17),
1473–1475 (2008). [CrossRef]
32. T. Yoshida, T. Sugihara, H. Goto, T. Tokura, K. Ishida, and T. Mizuochi, “A study on statistical equalization of intra-channel fiber nonlinearity for digital coherent optical systems,” in Proc.
ECOC’11, Tu.3.A. (2011).
33. J. P. Gordon and L. F. Mollenauer, “Phase noise in photonic communications systems using linear amplifiers,” Opt. Lett. 15(23), 1351–1353 (1990). [CrossRef] [PubMed]
34. T. Tanimura, T. Hoshida, T. Tanaka, L. Li, S. Oda, H. Nakashima, Z. Tao, and J. C. Rasmussen, “Semi-blind nonlinear equalization in coherent multi-span transmission system with inhomogeneous span
parameters,” in Proc. OFC’10, OMR6 (2010).
35. L. Zhu, F. Yaman, and G. Li, “Experimental demonstration of XPM compensation for WDM fibre transmission,” Electron. Lett. 46(16), 1140–1141 (2010). [CrossRef]
OCIS Codes
(060.1660) Fiber optics and optical communications : Coherent communications
(060.2330) Fiber optics and optical communications : Fiber optics communications
(190.4370) Nonlinear optics : Nonlinear optics, fibers
ToC Category:
Fiber Optics and Optical Communications
Original Manuscript: March 27, 2012
Revised Manuscript: May 18, 2012
Manuscript Accepted: May 25, 2012
Published: June 12, 2012
Likai Zhu and Guifang Li, "Nonlinearity compensation using dispersion-folded digital backward propagation," Opt. Express 20, 14362-14370 (2012)
Sort: Year | Journal | Reset
1. G. P. Agrawal, Nonlinear Fiber Optics (Academic Press, Elsevier, 2001).
2. P. P. Mitra and J. B. Stark, “Nonlinear limits to the information capacity of optical fibre communications,” Nature411(6841), 1027–1030 (2001). [CrossRef] [PubMed]
3. B. C. Kurtzke, “Suppression of fiber nonlinearities by appropriate dispersion management,” IEEE Photon. Technol. Lett.5(10), 1250–1253 (1993). [CrossRef]
4. K. Mukasa, K. Imamura, I. Shimotakahara, T. Yagi, and K. Kokura, “Dispersion compensating fiber used as a transmission fiber: inverse/reverse dispersion fiber,” J. Opt. Fiber Commun. Rep.3(5),
292–339 (2006). [CrossRef]
5. E. Ip, A. P. T. Lau, D. J. F. Barros, and J. M. Kahn, “Coherent detection in optical fiber systems,” Opt. Express16(2), 753–791 (2008). [CrossRef] [PubMed]
6. S. J. Savory, G. Gavioli, R. I. Killey, and P. Bayvel, “Electronic compensation of chromatic dispersion using a digital coherent receiver,” Opt. Express15(5), 2120–2126 (2007). [CrossRef]
7. H. Sun, K.-T. Wu, and K. Roberts, “Real-time measurements of a 40 Gb/s coherent system,” Opt. Express16(2), 873–879 (2008). [CrossRef] [PubMed]
8. E. Yamazaki, S. Yamanaka, Y. Kisaka, T. Nakagawa, K. Murata, E. Yoshida, T. Sakano, M. Tomizawa, Y. Miyamoto, S. Matsuoka, J. Matsui, A. Shibayama, J. Abe, Y. Nakamura, H. Noguchi, K. Fukuchi, H.
Onaka, K. Fukumitsu, K. Komaki, O. Takeuchi, Y. Sakamoto, H. Nakashima, T. Mizuochi, K. Kubo, Y. Miyata, H. Nishimoto, S. Hirano, and K. Onohara, “Fast optical channel recovery in field
demonstration of 100-Gbit/s Ethernet over OTN using real-time DSP,” Opt. Express19(14), 13179–13184 (2011). [CrossRef] [PubMed]
9. J. M. Kahn and K.-P. Ho, “A bottleneck for optical fibres,” Nature411(6841), 1007–1010 (2001). [CrossRef] [PubMed]
10. E. B. Desurvire, “Capacity demand and technology challenges for lightwave systems in the next two decades,” J. Lightwave Technol.24(12), 4697–4710 (2006). [CrossRef]
11. A. D. Ellis, J. Zhao, and D. Cotter, “Approaching the non-linear Shannon limit,” J. Lightwave Technol.28(4), 423–433 (2010). [CrossRef]
12. R.-J. Essiambre, G. Kramer, P. J. Winzer, G. J. Foschini, and B. Goebel, “Capacity limits of optical fiber networks,” J. Lightwave Technol.28(4), 662–701 (2010). [CrossRef]
13. L. Li, Z. Tao, L. Liu, W. Yan, S. Oda, T. Hoshida, and J. C. Rasmussen, “XPM tolerant adaptive carrier phase recovery for coherent receiver based on phase noise statistics monitoring,” in Proc.
ECOC’09, Paper P3.16 (2009).
14. K.-P. Ho and J. M. Kahn, “Electronic compensation technique to mitigate nonlinear phase noise,” J. Lightwave Technol.22(3), 779–783 (2004). [CrossRef]
15. L. B. Du and A. J. Lowery, “Practical XPM compensation method for coherent optical OFDM systems,” IEEE Photon. Technol. Lett.22(5), 320–322 (2010). [CrossRef]
16. K. Roberts, C. Li, L. Strawczynski, M. O’Sullivan, and I. Hardcastle, “Electronic precompensation of optical nonlinearity,” IEEE Photon. Technol. Lett.18(2), 403–405 (2006). [CrossRef]
17. X. Li, X. Chen, G. Goldfarb, E. F. Mateo, I. Kim, F. Yaman, and G. Li, “Electronic post-compensation of WDM transmission impairments using coherent detection and digital signal processing,” Opt.
Express16(2), 880–888 (2008). [CrossRef] [PubMed]
18. E. Ip and J. M. Kahn, “Fiber impairment compensation using coherent detection and digital signal processing,” J. Lightwave Technol.28(4), 502–519 (2010). [CrossRef]
19. D. S. Millar, S. Makovejs, C. Behrens, S. Hellerbrand, R. I. Killey, P. Bayvel, and S. J. Savory, “Mitigation of fiber nonlinearity using a digital coherent receiver,” IEEE J. Sel. Top. Quantum
Electron.16(5), 1217–1226 (2010). [CrossRef]
20. O. V. Sinkin, R. Holzlohner, J. Zweck, and C. R. Menyuk, “Optimization of the split-step Fourier method in modeling optical-fiber communications systems,” J. Lightwave Technol.21(1), 61–68
(2003). [CrossRef]
21. S. Oda, T. Tanimura, T. Hoshida, C. Ohshima, H. Nakashima, Z. Tao, and J. C. Rasmussen, “112 Gb/s DP-QPSK transmission using a novel nonlinear compensator in digital coherent receiver.” in Proc.
OFC’09, Paper OThR6 (2009).
22. E. F. Mateo, L. Zhu, and G. Li, “Impact of XPM and FWM on the digital implementation of impairment compensation for WDM transmission using backward propagation,” Opt. Express16(20), 16124–16137
(2008). [CrossRef] [PubMed]
23. E. F. Mateo, F. Yaman, and G. Li, “Efficient compensation of inter-channel nonlinear effects via digital backward propagation in WDM optical transmission,” Opt. Express18(14), 15144–15154 (2010).
[CrossRef] [PubMed]
24. Q. Zhang and M. I. Hayee, “Symmetrized split-step Fourier scheme to control global simulation accuracy in fiber-optic communication systems,” J. Lightwave Technol.26(2), 302–316 (2008).
25. L. B. Du and A. J. Lowery, “Improved single channel backpropagation for intra-channel fiber nonlinearity compensation in long-haul optical communication systems,” Opt. Express18(16), 17075–17088
(2010). [CrossRef] [PubMed]
26. D. Rafique, M. Mussolin, M. Forzati, J. Mårtensson, M. N. Chugtai, and A. D. Ellis, “Compensation of intra-channel nonlinear fibre impairments using simplified digital back-propagation
algorithm,” Opt. Express19(10), 9453–9460 (2011). [CrossRef] [PubMed]
27. L. Zhu and G. Li, “Folded digital backward propagation for dispersion-managed fiber-optic transmission,” Opt. Express19(7), 5953–5959 (2011). [CrossRef] [PubMed]
28. L. Zhu, X. Li, E. F. Mateo, and G. Li, “Complementary FIR filter pair for distributed impairment compensation of WDM fiber transmission,” IEEE Photon. Technol. Lett.21(5), 292–294 (2009).
29. J. K. Fischer, C.-A. Bunge, and K. Petermann, “Equivalent single-span model for dispersion-managed fiber-optic transmission systems,” J. Lightwave Technol.27(16), 3425–3432 (2009). [CrossRef]
30. C. Xie, “WDM coherent PDM-QPSK systems with and without inline optical dispersion compensation,” Opt. Express17(6), 4815–4823 (2009). [CrossRef] [PubMed]
31. V. Curri, P. Poggiolini, A. Carena, and F. Forghieri, “Dispersion compensation and mitigation of nonlinear effects in 111-Gb/s WDM coherent PM-QPSK systems,” IEEE Photon. Technol. Lett.20(17),
1473–1475 (2008). [CrossRef]
32. T. Yoshida, T. Sugihara, H. Goto, T. Tokura, K. Ishida, and T. Mizuochi, “A study on statistical equalization of intra-channel fiber nonlinearity for digital coherent optical systems,” in Proc.
ECOC’11, Tu.3.A. (2011).
33. J. P. Gordon and L. F. Mollenauer, “Phase noise in photonic communications systems using linear amplifiers,” Opt. Lett.15(23), 1351–1353 (1990). [CrossRef] [PubMed]
34. T. Tanimura, T. Hoshida, T. Tanaka, L. Li, S. Oda, H. Nakashima, Z. Tao, and J. C. Rasmussen, “Semi-blind nonlinear equalization in coherent multi-span transmission system with inhomogeneous span
parameters,” in Proc. OFC’10, OMR6 (2010).
35. L. Zhu, F. Yaman, and G. Li, “Experimental demonstration of XPM compensation for WDM fibre transmission,” Electron. Lett.46(16), 1140–1141 (2010). [CrossRef]
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Z-tests and t tests
Data types that can be analysed with z-tests
data points should be independent from each other
z-test is preferable when n is greater than 30.
the distributions should be normal if n is low, if however n>30 the distribution of the data does not have to be normal
the variances of the samples should be the same (F-test)
all individuals must be selected at random from the population
all individuals must have equal chance of being selected
sample sizes should be as equal as possible but some differences are allowed
Data types that can be analysed with t-tests
data sets should be independent from each other except in the case of the paired-sample t-test
where n<30 the t-tests should be used
the distributions should be normal for the equal and unequal variance t-test (K-S test or Shapiro-Wilke)
the variances of the samples should be the same (F-test) for the equal variance t-test
all individuals must be selected at random from the population
all individuals must have equal chance of being selected
sample sizes should be as equal as possible but some differences are allowed
Limitations of the tests
Introduction to the z and t-tests
Z-test and t-test are basically the same; they compare between two means to suggest whether both samples come from the same population. There are however variations on the theme for the t-test. If
you have a sample and wish to compare it with a known mean (e.g. national average) the single sample t-test is available. If both of your samples are not independent of each other and have some
factor in common, i.e. geographical location or before/after treatment, the paired sample t-test can be applied. There are also two variations on the two sample t-test, the first uses samples that do
not have equal variances and the second uses samples whose variances are equal.
It is well publicised that female students are currently doing better then male students! It could be speculated that this is due to brain size differences? To assess differences between a set of
male students' brains and female students' brains a z or t-test could be used. This is an important issue (as I'm sure you'll realise lads) and we should use substantial numbers of measurements.
Several universities and colleges are visited and a set of male brain volumes and a set of female brain volumes are gathered (I leave it to your imagination how the brain sizes are obtained!).
Data arrangement
Excel can apply the z or t-tests to data arranged in rows or in columns, but the statistical packages nearly always use columns and are required side by side.
Results and interpretation
Degrees of freedom:
For the z-test degrees of freedom are not required since z-scores of 1.96 and 2.58 are used for 5% and 1% respectively.
For unequal and equal variance t-tests = (n[1] + n[2]) - 2
For paired sample t-test = number of pairs - 1
The output from the z and t-tests are always similar and there are several values you need to look for:
You can check that the program has used the right data by making sure that the means (1.81 and 1.66 for the t-test), number of observations (32, 32) and degrees of freedom (62) are correct. The
information you then need to use in order to reject or accept your H[O], are the bottom five values. The t Stat value is the calculated value relating to your data. This must be compared with the two
t Critical values depending on whether you have decided on a one or two-tail test (do not confuse these terms with the one or two-way ANOVA). If the calculated value exceeds the critical values the H
[O] must be rejected at the level of confidence you selected before the test was executed. Both the one and two-tailed results confirm that the H[O] must be rejected and the H[A] accepted.
We can also use the P(T<=t) values to ascertain the precise probability rather than the one specified beforehand. For the results of the t-test above the probability of the differences occurring by
chance for the one-tail test are 2.3x10^-9 (from 2.3E-11 x 100). All the above P-values denote very high significant differences.
Graphical output
Ted Gaten Department of Biology gat@le.ac.uk Entry approved by the Head of Department. Last Updated: May 2000 | {"url":"http://www.le.ac.uk/bl/gat/virtualfc/Stats/ttest.html","timestamp":"2014-04-20T23:28:49Z","content_type":null,"content_length":"10575","record_id":"<urn:uuid:cc74e0a1-8ca4-4427-b160-5d881724c490>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00554-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Here's the question you clicked on:
Show that 168 cannot be expressed as the sum of the squares of two rational numbers.
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\[a,b,m,n \in \mathbb{Z}\]
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Curse you, Diophantus!
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Here are my thoughts - if:\[168=a^2+b^2\]then either a and b are both even or both odd. take the case of both odd so a=2m+1 and b=2n+1:\[168=4m^2+4m+4n^2+4n+2\]which leads to:\[166=4(m^2+n^2+m+n)
\]but 166 is not evenly divisible by 4 so this case can be rejected. now take both a and b as even, so a=2m and b=2n leads to:\[168=4m^2+4n^2\]thus:\[42=m^2+n^2\]strange how the number 42 appears
everywhere :) I can continue on this train of thought - but do you think it will lead anywhere good?
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Are you assuming that a and b are integers?
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ah! sorry - didn't read your question properly - let me think again...
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Fair enough, I made the same mistake intially.
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thanks @henpen - gives me more things to learn! :)
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Axial Stress
Axial Stress, which is also called compressive or tensile stress, is a measure of the axial force acting on a beam.
It is a quantitatively measuring the internal forces acting within in the beam.
The equation used to describe the normal stress is in the bar pictured below is:
Here σ represents stress, P represents the internal force at cross sectional area A. Stress is represented in either psi/ksi: pounds per square inch, kips (kilo-pounds) per square inch, or in SI by
Pa: Pascals, which are equivalent to Newtons per meter squared. For a further breakdown of the equation into its visual pieces see the image below.
Compressive stress means the member is in compression (being smashed) vs. tensile stress which means the beam is in tension (being pulled apart). The sign convention is to call compression positive
and tensile stress negative.
To see more information on axial stress and solved problems see this online mechanics book, Chapter 1
Incoming search terms:
• axial stress mechanics
• axial stress sample problems
• axial stress problems
• axial stress vs normal stress
• compressive stress sample problems
• equation for axial stress
• what is axial stress engineering | {"url":"http://www.clengi.com/engineering-tutorial/mechanics-of-materials/axial-stress/","timestamp":"2014-04-20T23:28:08Z","content_type":null,"content_length":"51909","record_id":"<urn:uuid:38517462-eda6-4461-b686-7359e12b374d>","cc-path":"CC-MAIN-2014-15/segments/1397609539337.22/warc/CC-MAIN-20140416005219-00575-ip-10-147-4-33.ec2.internal.warc.gz"} |
Figure 4.
Stiffness of silicon nanowire, k[NW], with its diameter of 525 nm as a function of indentation force, F. It is shown that the relationship between k[NW] and F is well described by a scaling of k[NW]
= αF^1/3, where α is a constant parameter depending on nanowire geometry. It implies that nanoindentation of silicon nanowire is well described by classical elasticity theory such as Hertz theory
Sohn et al. Nanoscale Research Letters 2009 5:211-216 doi:10.1007/s11671-009-9467-7 | {"url":"http://www.nanoscalereslett.com/content/5/1/211/figure/F4","timestamp":"2014-04-21T12:19:26Z","content_type":null,"content_length":"11499","record_id":"<urn:uuid:ec4440cf-1bf8-4f51-a2d8-c05a98b6e3d6>","cc-path":"CC-MAIN-2014-15/segments/1397609539776.45/warc/CC-MAIN-20140416005219-00293-ip-10-147-4-33.ec2.internal.warc.gz"} |
Where Can We Visit?
Copyright © University of Cambridge. All rights reserved.
'Where Can We Visit?' printed from http://nrich.maths.org/
Charlie and Abi had this 100 square board and put a counter on 42.
They wondered if they could visit all the other numbers on the board moving the counter using just these two operations:
$\times 2$ and $-5$
This is how they started:
42, 37, 32, 27, 22, 17, 12, 7, 14, 9, 18, 13, 26, 52, 47, 42, 84 ...
(notice that they are allowed to visit numbers more than once)
and this is what their board looked like:
Will they be able to visit every number on the grid at least once?
What would have happened if they had started on a different number?
Can you explain your results?
They wondered if they would get the same sort of results with other pairs of operations.
This is what they tried next:
$\times 3$ and $-5$
$\times 4$ and $-5$
$\times 5$ and $-5\ldots$
And then they tried these:
$\times 5$ and $-2$
$\times 5$ and $-3$
$\times 5$ and $-4\ldots$
Find out what Abi and Charlie discovered or choose pairs of operations of your own and investigate what numbers can be visited.
Can you explain your results?
This text is usually replaced by the Flash movie.
This problem is also available in French:
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NIST Statement on Statistical Principles for the Design and
Analysis of Key Comparisons
NIST Statement on Statistical Principles for the Design and Analysis of Key Comparisons
Approved by the Measurement Services Advisory Group, December 16, 2003
To facilitate international trade beneficial to U.S. industry, NIST participates in international interlaboratory comparisons, called Key Comparisons, to assess the equivalence of measurement
standards used at different National Metrology Institutes. Because Key Comparisons impact both scientific and economic decisions made by different countries, there are clearly defined procedures
governing their conduct. The primary document governing Key Comparisons is the Mutual Recognition Arrangement (MRA) developed by the CIPM. In addition, NIST has developed a "Position on the Conduct
of Key Comparisons," which offers guidance on several interpretable points of the MRA for NIST participants in Key Comparisons. Having participated in nearly 250 comparisons and piloted nearly 70,
NIST technical staff has asked for a clear articulation of the statistical principles that are central to the design, implementation, analysis and interpretation of Key Comparisons and Supplementary
Comparisons. Questions have arisen regarding the following issues:
• What are the requirements in designing a Key Comparison to assure a clear interpretation from the data once the comparison is completed?
• What are the conditions for a statistical analysis of a Key Comparison to be valid?
• When is the statistical analysis of a Key Comparison complete?
Is there a single statistical approach to the analysis of a Key Comparison or to the estimation of a reference value (KCRV) or the estimation of degrees of equivalence?
This NIST Statement identifies statistical principles for different types of Key Comparisons that should be followed to ensure that the comparisons in which NIST participates will be clearly
interpretable. Interpretability requires statistically sound estimates of the various quantities of interest including reference values and degrees of equivalence between measurement standards
maintained by different NMI's each with its associated uncertainties. Interpretation also extends to the statistical basis for addressing unexplained deviations, whether individual observations or
the collective observations from a particular NMI, and to statistically sound methods for combining information from Key and Regional Comparisons in order to address differences between NMIs
participating in separate, but linked, comparisons.
Information on sound statistical procedures and/or methodologies and established statistical practices can be found in the archival and applied journals of statistical societies, other technical and
educational publications and reputable statistical software in both the commercial and public domains.
Statement of Principles:
The statistical premises for Key Comparisons
• Recognizing that there are both stochastic and non-stochastic elements in all interlaboratory comparisons, the general goal of the analysis of Key Comparison data is to draw statistical inference
. As a particular example, degrees of equivalence among measurements and measurement standards for the various NMIs with their associated uncertainties must be resolved on the basis of sound
statistical procedures.
• As expressed in Sections 6 and 9 of the Guidelines for Key Comparisons and endorsed by the NIST Position Statement on the Conduct of Key Comparisons, integrity of the data is essential to the
interpretability of a Key Comparison. Thus a prerequisite to the inclusion of an NMI's data in the analysis is the complete submission of data with attendant detailed uncertainty budget.
Similarly, according to the Guidelines, the integrity of the Key Comparison analysis is protected by explicit documentation of any changes to the data (e.g., that may occur when the data is
reviewed prior to preparation of Draft A).
• Open accessibility of all data and uncertainty budgets permits alternate or expanded analyses, as these may be appropriate and may serve as additional validation of the conclusions.
The statistical design of Key Comparisons
• The statistical design of the Key Comparison should conform to established principles of sound statistical design of experiments. From the outset, a specific statistical analysis should be
posited to ensure that unbiased estimates of degrees of equivalence and reference value and also of their associated uncertainties will be possible; but the eventual analysis should not be
limited to this particular method. It is an established statistical practice to construct the statistical design both for (statistical) efficiency and for robustness due to any loss of data or to
effects of unforeseen factors.
• Since each Key Comparison involves a specific metrology, the statistical design of the Key Comparison need to be individualized to reflect the particular metrological requirements and practical
constraints of each comparison. Statistical design features include among other things: factors affecting the measurement process, artifact attributes, replication and randomization (where
The statistical analysis of Key Comparisons
• Key Comparisons inherently involve statistical (Type A) sources of uncertainty, non-statistical (Type B) sources of uncertainty, and mathematical constants not subject to uncertainty. The
statistical methodology must distinguish among these, assigning the correct (different) mathematical role to each. Statistical sources of uncertainty should be measured from data and be
verifiable from data. Non-statistical sources may be statements of individual expert opinion not verifiable directly from data or may incorporate both expert opinion and data-verifiable
uncertainties such as offsets.
• Because Key Comparisons are necessary in a wide variety of metrological areas, no single statistical methodology can be universally applied either to their design or to their analysis. Therefore
an appropriate statistical approach for a Key Comparison will require individualization because of the diversity of the measurement processes, the variety of metrological models and the
differences in the designs of the Key Comparisons.
• In general, multiple statistical approaches are valid for a Key Comparison. Every statistical approach requires a set of underlying assumptions; for a particular approach to be valid these
assumptions must be stated and checked, wherever possible. These assumptions include (but are not limited to) the statistical models used, independence or interdependencies in the data, and
distributional assumptions about the data.
• Critical conclusions drawn from analysis of a Key Comparison should hold generally under analysis by alternative contextually valid statistical approaches. Divergence among valid statistical
approaches in the principal conclusions is an indicator of insufficient information or a crucial dependence upon assumptions that are not verifiable.
• For the purposes of a Key Comparison, the analysis of the Key Comparison data is complete and adequate when it satisfies two criteria: 1) a more elaborate analysis does not alter conclusions
drawn with respect to the primary objectives, and 2) the uncertainty associated with the summary results meet or surpass the specific requirements for all primary uses of the comparison.
Statistical principles endorse more extensive or more focused analyses for other purposes, for example: to shed light on the measurement methodology or on the measurement process for a particular
NMI or subset of NMIs, or to resolve deviations of individual observations or the collective data from a single NMI or from a single measurement method. | {"url":"http://www.itl.nist.gov/div898/keycomp/NIST_Principles.htm","timestamp":"2014-04-16T07:13:43Z","content_type":null,"content_length":"11040","record_id":"<urn:uuid:77f99b6e-8fdb-4f75-a976-ddcf1fb01036>","cc-path":"CC-MAIN-2014-15/segments/1397609537186.46/warc/CC-MAIN-20140416005217-00249-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Current Research on Separation Logic
15-818A4 Current Research on Separation Logic (6 units) Spring 2011
John C. Reynolds
Second Half of Spring Semester 2011 (starting March 14)
University Units: 6 (research minicourse)
Gates-Hillman Center 4211
Mondays, Wednesdays, and Fridays 12:30 pm-1:20 pm
Course Description
We will study and discuss recent papers on separation logic and related topics.
Anyone who is interested can attend, and you are welcome to come just for papers you are particularly interested in. I'll post the progress of the course on the class webpage, which is linked from my
home page at http://cs.cmu.edu/~jcr. Some familiarity with separation logic and Hoare logic will be needed (such as provided in 15-818A3 - Introduction to Separation Logic).
TEXTS: Papers and notes to be distributed in class.
METHOD OF EVALUATION: Each creditor will be required to lecture on a paper chosen by them and the instructor.
Papers and Videos of Lectures
(1) Peter W. O'Hearn. Resources, Concurrency, and Local Reasoning. TCS 375(1-3):271-307, May 2007.
Lecture 22, Lecture 23, Lecture 24, Lecture 25
(2) Stephen D. Brookes. A Semantics for Concurrent Separation Logic. TCS 375(1-3):227-270, May 2007. (Lectures by Stephen Brookes)
(3) Richard Bornat, Cristiano Calcagno, O'Hearn, and Matthew Parkinson. Permission Accounting in Separation Logic. POPL 2005, 259-270.
Lecture 28, Lecture 29, Lecture 30 (first part)
(4) Parkinson, Bornat, Calcagno. Variables as Resource in Hoare Logic. LICS 21 2006.
Lecture 30 (second part), Lecture 31, Lecture 32, Lecture 33
(5) Calcagno, O'Hearn, and Hongseok Yang. Local Action and Abstract Separation Logic. LICS 22 366-378, July 2007.
Lecture 34, Lecture 35, Lecture 36
(6) Yang. Relational Separation Logic. TCS 375(1-3):308-334, May 2007. (Lectures by Bernardo Toninho)
(7) Uday S. Reddy. Syntactic Control of Interference for Separation Logic (Preliminary Report) April 12, 2011.
(8) Parkinson, Bornat, and O'Hearn. Modular Verification of a Non-Blocking Stack. POPL 2007. (Lectures by Henry DeYoung)
(9) John C. Reynolds. Automatic Computation of Static Variable Permissions. MFPS 2011. (The class lecture was a preliminary version of the talk at MFPS.)
An Application of Auxiliary Variables, March 17
A Concurrent Cyclic Buffer, revised March 22
Permission Accounting in Separation Logic, March 28
Oddities of Inductive Definitions, revised April 1
Variables as Resources - An Example, April 1
Comments on Rules for Variables as Resources, revised April 4
Variables as Resources in Hoare Logics (slides), revised April 11
Modular Verification of a Non-Blocking Stack (slides), April 25
An Inference Procedure for Static Variable Permissions (slides), These slides have been obsoleted by Automatic Computation of Static Variable Permissions, MFPS, May 28
LINK TO PREVIOUS MINICOURSE (15-818A3)
Last updated: August 7, 2011 | {"url":"http://www.cs.cmu.edu/afs/cs.cmu.edu/project/fox-19/member/jcr/www15818As2011/cs818A4-11.html","timestamp":"2014-04-18T05:35:02Z","content_type":null,"content_length":"6454","record_id":"<urn:uuid:df47416d-4dc2-4d43-834b-9d0ec49b7561>","cc-path":"CC-MAIN-2014-15/segments/1397609532573.41/warc/CC-MAIN-20140416005212-00570-ip-10-147-4-33.ec2.internal.warc.gz"} |
Velocity in planetary orbits
With different velocities it wouldn't be the same orbit.
Yes, it would. Consider the Moon. Suppose you magically replace the Moon with an object several orders of magnitude smaller in mass than the Moon but keep the position and velocity the same as the
Moon's. This object will be in a different orbit. To get the same orbit as the Moon's current orbit you will have to change the velocity.
This is an important consideration in the formation of a planetary system from an accretion disk. Given a planetesimal in a circular orbit of radius
about a nascent star amidst some particles orbiting at the same distance, the planetesimal will be moving slightly faster than the individual particles. The planetesimal will have an orbital velocity
of [itex]\sqrt{G(M+m)/a}[/itex] where
is the mass of the nascent star and
is the mass of the planetesimal; the orbital velocity small particles co-orbiting with the planetesimal will only be [itex]\sqrt{GM/a}[/itex]. The planetesimal will plow through and sweep up the
surrounding particles. This can lead to the planetesimal migrating toward the star. | {"url":"http://www.physicsforums.com/showthread.php?t=539972","timestamp":"2014-04-16T19:03:20Z","content_type":null,"content_length":"66187","record_id":"<urn:uuid:ceb31626-fccc-42eb-b6e7-12a9e247425e>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00437-ip-10-147-4-33.ec2.internal.warc.gz"} |
Infinite potential well
I didn't get the answer that I'm looking. My questions:
What about post #2 did you not understand?
Did you read the links I posted.....they will help a lot.
" Interpretations of quantum mechanics address questions such as what the relation is between the wavefunction, the underlying reality, and the results of experimental measurements."
There are different interpretations about exactly what the wavefunctions means...there are different ways to think about it. | {"url":"http://www.physicsforums.com/showthread.php?p=4258800","timestamp":"2014-04-21T02:06:25Z","content_type":null,"content_length":"59428","record_id":"<urn:uuid:6417e05d-29c5-4f97-90f1-c0efb27d727d>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00047-ip-10-147-4-33.ec2.internal.warc.gz"} |
Re: st: Modeling simultaneity
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Re: st: Modeling simultaneity
From "Pavlos C. Symeou" <p.symeou@gmail.com>
To statalist@hsphsun2.harvard.edu, John Antonakis <john.antonakis@unil.ch>
Subject Re: st: Modeling simultaneity
Date Thu, 07 Feb 2013 15:14:41 +0200
Dear John,
thank you for this. Do the procedures you suggest below have respective commands for panel data structures? In this context, I suppose I could use the lags of my endogenous variables as their
Best wishes,
On Τρίτη, 5 Φεβρουάριος 2013 8:08:39 μμ, John Antonakis wrote:
Hi Pavlos:
You are estimating:
abs = b0 + b1Div + b2x1 + e
div = g0 + g1abs + g2x2 + u
perf = d0 + d1abs + d2div + w
Where x1 and x2 are instruments, and where cov(e,u; e,w; and u,w)
are estimated.
Note, the above system of equations is just identified. You have
5(5+1)/2 = 15 elements in the variance-covariance matrix and
1. Correlations between exogenous variables: 1
2. Correlations between disturbances: 3
3. Regressions coefficients: 6
4. Variances of exogenous variables: 2
5: Variances of endogenous variables: 3
Total parameters estimated: 15
Thus, your DF = 15-15 = 0. This model can be estimated, but
because you are not cannot overidentified you cannot determine
whether the constraints you make are tenable via a chi-square
test of fit (i.e., Hansen-Sargan test). Thus, I would recommend
to you to find at least another instrument, x3 in the abs and/or
div equation to be overidentified.
You can estimate this system of equations with reg3, as follows:
reg3 (abs div x1) (div abs x2) (perf abs div), 2sls
est store two
(if you are overidentified, you can test for this if download the
user command -overid- (from SSC)).
Note, I would first estimate this with 2sls to ensure that any
mispecification remains local. I would then rerun it with 3sls,
which is more efficient and compare that estimator with the first:
reg3 (abs div x1) (div abs x2) (perf abs div),
est store three
hausman two three
If they don't differ you can retain the 3sls estimator.
You can estimate this in sem too (with maximum likelihood), which
will give you more information on the estimated parameters (note
to have an instrumental variable estimator you must correlate
disturbances of the endogenous variables explicitly).
sem (abs <- div x1) (div<- abs x2) (perf y<- abs div),
covstructure(e._OEn, unstructured)
Or you can do the cov option explicitly:
sem (abs <- div x1) (div<- abs x2) (perf y<- abs div),
cov(e.abs*e.div, e.abs*e.perf, e.div*e.perf)
-sem- will give you an overidentification test (chi-square test
on the bottom of the table)
We discuss these issues in an applied manner here:
Antonakis, J., Bendahan, S., Jacquart, P., & Lalive, R. (2010).
On making causal claims: A review and recommendations. The
Leadership Quarterly, 21(6). 1086-1120.
[If you wish, refer to the following “prequel” paper, which is
really a more basic introduction--and we explain
overidentification explicitly]:
Antonakis, J., Bendahan, S., Jacquart, P., & Lalive, R.
(submitted). Causality and endogeneity: Problems and solutions.
In D.V. Day (Ed.), The Oxford Handbook of Leadership and
John Antonakis
Professor of Organizational Behavior
Director, Ph.D. Program in Management
Faculty of Business and Economics
University of Lausanne
Internef #618
CH-1015 Lausanne-Dorigny
Tel ++41 (0)21 692-3438
Fax ++41 (0)21 692-3305
Associate Editor
The Leadership Quarterly
On 04.02.2013 15:03, Pavlos C. Symeou wrote:
Dear Statalisters,
I was wondering whether any of you can help me with this.
I have three variables: Absorptive capacity, Diversification,
and Performance. I am arguing that the first two are
simultaneously determined and they influence the third one.
Explicitly, I am arguing that the ability of the firm to
understand new knowledge (what is called Absorptive Capacity AC)
influences the direction of the firm's market diversification
(DIV). However, once the firm has diversified, it in turn
influences the firm's ability to understand new knowledge (AC).
I want to empirically account for this simultaneity when I try
to examine the effect of AC and DIV on the performance of the firm.
I can use instrumental variables to model the simultaneity, but
I don't know how to examine the final effects of AC and DIV on
firm performance while controlling for simultaneity.
I look forward to receiving your comments.
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did any one know the definition of Hochschild cohomology of a differential graded algebra?
up vote 0 down vote favorite
Let A with d be a differential graded associative algebra (DG algebra). What is the definition of Hochschild complex C*(A,A)? There should be a Hoshschild differential and a cup product, but i can't
find the reference that specifies the formula. Does any one know?
1 A formula appears here arxiv.org/pdf/math/0211229v1.pdf, probably a repeat question... – Daniel Pomerleano Aug 4 '12 at 18:36
1 Or try this one: math.sciences.univnantes.fr/~hossein/Publications/…, page 4, since you also wanted a formula for cup product. Sign conventions often differ and it's important to have ones which
are consistent. – Daniel Pomerleano Aug 4 '12 at 18:46
ctg.epfl.ch/files/content/sites/ctg/files/shared/Zhou/… mentions them – CSA Apr 6 at 18:53
add comment
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Blowout Cards Forums - View Single Post - 2013 MLB JACKPOT Rules/FAQ's/Random Results/List of Participants (CLOSED)
Here's a detailed explanation of how ALL of the Giveaway Grid Spots will be determined!
Important: We will be using the official ESPN.com Los Angeles Dodgers @ San Francisco Giants - Box Score - May 5, 2013 for all statistics used to determine giveaway spots in this promotion
Giveaway #1 - Official Attendance Report
-At the end of the game, match the last two digits of the Official Attendance Report.
Example) At the end of the game, let's say the Official Attendance was 47,841. First, go to the Giants (Across) top row of numbers and find the number 4. Next, go to the Dodgers (Down) left column of
numbers and find the number 1. Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Official Attendance Report Giveaway - 5% of Total Jackpot!
Giveaway #2 - Runners Left on Base
-At the end of the game, match the last digit of each teams' Total # of Runners Left on Base.
Example) At the end of the game, let's say the Giants had 9 Runners Left on Base and the Dodgers had 8 Runners Left on Base. First, go to the Giants (Across) top row of numbers and find the number 9
(Giants Runners Left on Base). Next, go to the Dodgers (Down) left column of numbers and find the number 8 (Dodgers Runners Left on Base). Find where these two numbers intersect on the grid. The name
of the participant in that square gets:
Runners Left on Base Giveaway - 5% of Total Jackpot!
Giveaway #3 - Total Pitch Count Thrown (Giants)
-At the end of the game, match the last two digits of the Giants Total Pitch Count # with the grid.
Example) At the end of the game, let's say the Giants threw a Total of 162 Pitches. First, go to the Giants (Across) top row of numbers and find the number 6 (2nd digit). Next, go to the Dodgers
(Down) left column of numbers and find the number 2 (3rd digit). Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Total Pitch Count (Giants) Giveaway - 5% of Total Jackpot!
Giveaway #4 -Total Strikes Thrown (Dodgers)
-At the end of the game, match the last two digits of the Dodgers Total # of Strikes Thrown with the grid.
Example) At the end of the game, let's say the Dodgers threw a Total of 97 Strikes. First, go to the Giants (Across) top row of numbers and find the number 9. Next, go to the Dodgers (Down) left
column of numbers and find the number 7. Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Total Strikes Thrown (Dodgers) Giveaway - 5% of Total Jackpot!
Giveaway #5 - Strikeouts (Batters)
-At the end of the game, match the last digit of each teams' Total # of Strikeouts (Batters) with the grid.
Example) At the end of the game, let's say the Giants Batters had 11 Strikeouts and the Dodgers Batters had 8 Strikeouts. First, go to the Giants (Across) top row of numbers and find the number 1
(Last digit of Giants Batters Strikeouts). Next, go to the Dodgers (Down) left column of numbers and find the number 8 (Dodgers Strikeouts). Find where these two numbers intersect on the grid. The
name of the participant in that square gets:
Strikeouts (Batters) Giveaway - 5% of Total Jackpot!
Giveaway #6 - Total Home Runs + Triples + Doubles (Team)
-At the end of the game, match the last digit of each teams' Total # of Home Runs + Triples + Doubles with the grid.
Example) At the end of the game, let's say the Giants had a total of 10 (HR's + Triples + Doubles) and the Dodgers had a total of 4 (HR's + Triples + Doubles). First, go to the Giants (Across) top
row of numbers and find the number 0 (Last digit of Giants Total). Next, go to the Dodgers (Down) left column of numbers and find the number 4 (Dodgers Total). Find where these two numbers intersect
on the grid. The name of the participant in that square gets:
Total HR's + Triples + Doubles Giveaway - 10% of Total Jackpot!
Giveaway #7 - Total Walks Thrown + Strikeouts Thrown (Team)
-At the end of the game, match the last digit of each teams' Total # of Total Walks Thrown + Strikeouts Thrown with the grid.
Example) At the end of the game, let's say the Giants had a total of 8 (Total Walks Thrown + Strikeouts Thrown) and the Dodgers had a total of 15 (Total Walks Thrown + Strikeouts Thrown). First, go
to the Giants (Across) top row of numbers and find the number 8 (Giants Total). Next, go to the Dodgers (Down) left column of numbers and find the number 5 (Last Digit of Dodgers Total). Find where
these two numbers intersect on the grid. The name of the participant in that square gets:
Total Walks Thrown + Strikeouts Thrown Giveaway - 10% of Total Jackpot!
Giveaway #8 - Weather
-At the end of the game, match the last two digits of the Official Weather report (Degrees).
Example) At the end of the game, let's say the Official Weather was 85 degrees, sunny. First, go to the Giants (Across) top row of numbers and find the number 8. Next, go to the Dodgers (Down) left
column of numbers and find the number 5. Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Weather Giveaway - 5% of Total Jackpot!
Giveaway #9 - Total Hits (Team)
-At the end of the game, match the last digit of each teams' Total # of Hits with the grid.
Example) At the end of the game, let's say the Giants had 11 Hits and the Dodgers had 9 Hits. First, go to the Giants (Across) top row of numbers and find the number 1 (Last digit of Giants Hits).
Next, go to the Dodgers (Down) left column of numbers and find the number 9 (Dodgers Hits). Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Total Hits (Team) Giveaway - 25% of Total Jackpot!
Giveaway #10 - Final Score
-At the end of the game, match the last digit of each teams' score with the grid.
Example) At the end of the game, let's say the Giants had 7 Runs and the Dodgers had 4 Runs. First, go to the Giants (Across) top row of numbers and find the number 7 (Giants score). Next, go to the
Dodgers (Down) left column of numbers and find the number 4 (Dodgers score). Find where these two numbers intersect on the grid. The name of the participant in that square gets:
Final Score Giveaway - 25% of Total Jackpot!
Last edited by blowoutcards2; 03-29-2013 at 02:32 PM. | {"url":"http://www.blowoutcards.com/forums/4558174-post2.html","timestamp":"2014-04-23T13:19:58Z","content_type":null,"content_length":"25938","record_id":"<urn:uuid:ec33e1a9-9b38-47ee-8014-b63a1a42ff87>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00010-ip-10-147-4-33.ec2.internal.warc.gz"} |
Cloth Diapers & Parenting Community - DiaperSwappers.com - View Single Post - Formula Help UPDATE
Re: Formula Help:( Another Question
I always do the wisk let the bubble settle thing with the formula and we use dr. browns bottles. I dont think his problem is as much gas as it is the formula just come back up? When he does get gas
and his stomach is hard we do the bicyclesand sometimes that works them out for him. I did give him concentrate formula at his last feeding, and he has thrown up most of it
Jen ~ Mommy to
I am around but may have some delays in responses | {"url":"http://www.diaperswappers.com/forum/showpost.php?p=238408&postcount=23","timestamp":"2014-04-25T02:54:34Z","content_type":null,"content_length":"11016","record_id":"<urn:uuid:4074e586-86d0-495a-8737-21a31509780e>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00659-ip-10-147-4-33.ec2.internal.warc.gz"} |
Mar Vista, CA Algebra Tutor
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14 Subjects: including algebra 1, algebra 2, calculus, statistics
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Review hired me to write and edit an AP European History test prep book for global distribution.
25 Subjects: including algebra 1, algebra 2, English, geometry
...I've taken and passed two college level courses in Linear Algebra, and it is an indispensable part of the quantum mechanics and relativity toolbox. I also spent the majority of my PhD work
performing (Linear Algebra) analysis on large matrices. I've tutored friends and fellow scientists in the field.
44 Subjects: including algebra 1, algebra 2, chemistry, reading
...My approach to tutoring Algebra is to make it as simple as possible. I have had high school students go from failing math to A's and B's. I have a Bachelor's Degree in Accounting from
California State University Northridge and I am a CPA.
5 Subjects: including algebra 2, algebra 1, accounting, prealgebra
...Additionally, when I'm not writing or working on films, I do stand-up comedy, play sports, and enjoy binge watching on netflix when my schedule permits. I have tutored my siblings before and
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Windsor Hills, CA algebra Tutors | {"url":"http://www.purplemath.com/Mar_Vista_CA_Algebra_tutors.php","timestamp":"2014-04-19T17:17:27Z","content_type":null,"content_length":"24187","record_id":"<urn:uuid:672140c5-e4da-46fa-9107-010d27b7a2a8>","cc-path":"CC-MAIN-2014-15/segments/1397609537308.32/warc/CC-MAIN-20140416005217-00188-ip-10-147-4-33.ec2.internal.warc.gz"} |
Physics Questions and Answers
Students often face hard-to-solve and mind-numbing physics problems, that cause a lot of distress into the studying process. Not everyone can cope with the hardships physics problems cause, and many
end up with a bunch of physics questions that need to be solved. Our service is the solution provider for your physics questions. Ask your question here and get physics answers that would help you do
your assignment in the quickest way possible with maximum results. Our experts will gladly provide physics answers for your benefit.
A 2400 W engine pulls a 200 kg block at constant speed up a 12.0 m long, 25.0° incline.
Determine long does it takes to cover this distance.
Read answer In Progress...
if you dive underwater, you notice an uncomfortable presure on your eardrum due to the increased pressure. the human endrum has a pressure of 70mm2 (7.10^-5 m2) and it can sustain a force of 7N
without rupturing. if your body had no means of balancing the extra pressure, what would be the maximum depth you could dive without rupturing the eardrum?
Read answer In Progress...
A force
at an angle of
above the horizontal displaces an object through a distance S in its along the horintal. Which of the following is the CORRECT expression for the work done on the object by the force?
a. F⃗ ⋅s⃗ cosθ
b. F⃗ ×s⃗
c. Fssinθ
d. Fscosθ
Read answer In Progress...
A block of wood is placed on an inclined plane. The angle
is non-zero) of the inclined plane is gradually increased until the block is at the verge of sliding down the incline. What is the coefficient of friction between the block and the incline?
Read answer In Progress...
A 2.0 kg stone tied to the the end of an inextensible string is whirled around in a horizontal circle of radius 1.5 m at auniform angular speed
rad/s Calculate the rotational kinetic energy of the stone.
220.7 J
88.8 J
47.4 J
246.7 J
Read answer In Progress...
A satellite orbits the earth at an altitude of 600 km with an orbital speed of 27 000km/hr. What is its period of revlution. Take the radius of the earth as 6400 km.
97.7 min
36.4 min
120.3 min
73.7 min
Read answer In Progress...
Which of the following is INCORRECT?
a. The coefficient of static friction depends on the nature of the surfaces in contact
b. static friction is larger than kinetic friction for any given surfaces in contact
c. The coefficient of friction is independent of the relative velocity of the surfaces in contact
d. The coefficient of friction is dependent on the area of the surfaces in contact.
Read answer In Progress...
A pulley is rotating at the rate of 32 rev/min. A motor speeds up the wheel so that 30 s later it is turning at 82 rev/min. what is the average angular acceleration in radians per second
a. 0.17
b. 2.11
c. 0.32
d. 1.41
Read answer In Progress...
A system of two point masses connected by a rigid massless rod and set into rotation about an axis through throd joining the particles is called ---------
a. couple
b. dumbbell
c. gyroscope
d. torque
Read answer In Progress...
The moment of inertia of a solid flywheel about its axis is
. It is set in rotation by applying a tangential force of 19.6 N with a rope wound around the circumference, the radius of the wheel is 10 cm. What would be the acceleration if a mass of 2 kg is hung
from the end of the rope in
a. 16.7
b. 21.3
c. 8.8
d. 29.2
Read answer In Progress... | {"url":"http://www.assignmentexpert.com/free-answers/physics","timestamp":"2014-04-17T04:35:33Z","content_type":null,"content_length":"64047","record_id":"<urn:uuid:710747e7-51a7-42da-a8e9-2cd98504d0b2>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00489-ip-10-147-4-33.ec2.internal.warc.gz"} |
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אוסף חמישי של הרצאות וידאו מתוך קורס חדו״א (חשבון דיפרנציאלי ואינטגרלי) 1 מוגבר.האוסף עוסק בנושא חקירת הפונקציה באמצעות...
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יקיר שושני
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Oct 01, 2009
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Sep 19, 2011
According to OER Commons, 'These are the lecture notes of a one-semester undergraduate course which we taught at SUNY...
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Gerald Marchesi, Dennis Pixton, Matthias Beck
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A selection of research and teaching presentations, including pdf presentations, audio podcasts and movies.
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Joel Feinstein
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A Short Course in Information Theory.
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David J.C. MacKay
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Talking about quality it is producing better with less waste of money and time. Applying the six sigma can help to reduce...
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Gureline dume
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'This free online textbook (e-book in webspeak) is a one semester course in basic analysis. This book started its life as my...
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This is a free, online textbook for an introductory course in complex analysis. General topics include Complex Numbers,...
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This is a free online course offered by the Saylor Foundation.'This course is an introduction to complex analysis, or the...
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The Saylor Foundation
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This module develops the concept of complex functions starting with the idea of linear mappings.
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PRVI -- Blogmeister
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Letter to Sister
Dear Sister,
In math class I learned how do a lot. We have done a lot with number sentences like order of operations and how to make equations replacing numbers with letters. The other thing we learned was
I learned how to do algebra by taking shape patterns and number sequences and finding a equation using N and T. An example would be T=6xN+1. I also learned how to add exponents to the equation. For
example you could take the 6 and make it 6 to the 2nd power.
We also learned probability. We learned how to pick cubes out of a bag. We learned how to figure out theoretical probability and how to make a frequency table. The problems we did gave us the amount
of items and we had to calculate what the probability that each category.
The last thing we learned was order of operations we learned how to use brackets and parenthesis and exponents. I think the order of operations should really be pemdas because you do brackets before
multiplication and division. I hope you understand the explanations I have given you.
Add a Comment
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Posted Comments
wow thank you for that lovely letter GRKE!
I love you too
Posted December 22, 2009 at 07:27 AM by • grco
Posted December 22, 2009 at 07:27 AM by • grco
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Top 15 highest-earning college degrees need math skills
Re: Top 15 highest-earning college degrees need math skills
There are far fewer people graduating with math-based majors, compared to their liberal-arts counterparts
So few grads offer math skills
I have the solution. The gov. should just hire Britney and Rihanna's press agent for math majors.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof. | {"url":"http://www.mathisfunforum.com/viewtopic.php?id=12461","timestamp":"2014-04-18T00:25:52Z","content_type":null,"content_length":"10050","record_id":"<urn:uuid:40819ac3-2852-4575-a4d9-04e768b4b2ef>","cc-path":"CC-MAIN-2014-15/segments/1397609532374.24/warc/CC-MAIN-20140416005212-00511-ip-10-147-4-33.ec2.internal.warc.gz"} |
creating simple program in excel...
You have three columns and lets say that row 1 contains your column names and your first set of items starts in row 2 so in the 4th column insert your piece count and in the 5th column insert the
Meaning that if you enter no data (piece count) next to an item then it will display no result otherwise it will calculate the total cost of the amount of pieces per the unit cost.
Now, lets say that you only have items listed down to row 26 then in row 27 you enter the formula:
You could alternatively enter the formula:
E1=SUM(E2:26) instead to show the total at the top.
This would show the total cost of all items where you inserted a piece count.
To go a step further; you could use your sheet as your data base and use a separate sheet as an invoice template where it would use lookup formulas geared off of the item numbers to bring in the data
from your data sheet and similarly calculate the total costs.
Excel ('97 anyway) comes with a template for this. Go to File, New, Spreadsheet solutions, Invoice. I have not looked but MS likely has many similar templates available on thier site. | {"url":"http://www.computing.net/answers/office/creating-simple-program-in-excel/6346.html","timestamp":"2014-04-20T00:40:56Z","content_type":null,"content_length":"37289","record_id":"<urn:uuid:a3c43eeb-cad2-426b-b76e-64ee422f082b>","cc-path":"CC-MAIN-2014-15/segments/1398223206647.11/warc/CC-MAIN-20140423032006-00191-ip-10-147-4-33.ec2.internal.warc.gz"} |
Multiplication Chart 1 100 Lesson Plans & Worksheets :: 21 - 40
Multiplication Chart 1-100 Teacher Resources
Find Multiplication Chart 1 100 educational ideas and activities
Showing 21 - 40 of 115 resources
Students make multiplication squares. In this math lesson, students review their multiplication facts and multiplication strategies. Students are introduced to multiplication squares when multiplying
a number by itself.
Review multiplication facts and concepts! This clear, and easy to follow lesson will help your class to understand multiplication by using arrays and repeated addition. There are strategies to help
with memorization and manipulative suggestions.
Using the book Amanda Bean's Amazing Dream: A Mathematical Story, 2nd graders discuss how to solve math problems using multiplication. They are encouraged to solve a series of multiplication problems
using model building, drawing, arrays, skip counting or repeated addition. After exploring these algorithms they share their thinking and complete a worksheet related to the book. This is a great
problem solving lesson that builds a foundational skill set.
Students, after using an interactive Web site to find patterns in the multiplication tables, practice multiplication facts and record their current level of mastery of the multiplication facts on
their personal multiplication chart
In this English Learners powers of ten activity, students use the multiplication charts to complete the sentences using positive exponents and negative exponents. Students also use the division
charts to complete the sentences using positive and negative exponents.
Students look for patterns in multiplication products using an interactive Web site. They play another card game and record their current level of mastery of the multiplication facts.
Students memorize the multiplication math facts and learn many different ways of visualizing and practicing the multiplication concepts. They begin practicing with arrays, skip counting, and moving
manipulatives for the 0, 1, 2, 3, and 4.
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Cdf of system, exponential distribution
October 24th 2010, 06:18 PM
Cdf of system, exponential distribution
These 3 components function independently of each other. Suppose the i-th component has a lifetime that is exponentially distributed with λ1, λ2, and λ3 (given values)
X=time at which system fails..
the system:
okay so im trying to find the cdf F(x)=P(X<=x)
first... Ai=Ai(x)={Xi>=x}
I tried to find the cdf for component 1 and 2 and did A(x)={X>=x}=A1(x)*A2(x) and found the probability of A(x) to get the cdf equal to 1-e^(-(λ1+λ2)*x)
and for component 2 and 3 i got a cdf of (1-e^(-λ2*x))*(1-e^(-λ3*x))
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Topic: identifying complex numbers
Replies: 4 Last Post: Aug 9, 2012 7:32 AM
Messages: [ Previous | Next ]
identifying complex numbers
Posted: Aug 8, 2012 8:52 PM
I'd like to identify complex numbers in a matrix, e.g.
A=[1 2 3
1+2i 5 3]
I'd like to get the following matrix that identifies complex numbers
[0 0 0
1 0 0]
I hoped to use a kind of "iscomplex", but there was no such function in matlab. Could you help me to solve this easily?
Date Subject Author
8/8/12 identifying complex numbers Young Ryu
8/8/12 Re: identifying complex numbers John D'Errico
8/9/12 Re: identifying complex numbers Bruno Luong
8/9/12 Re: identifying complex numbers Nasser Abbasi
8/9/12 Re: identifying complex numbers John D'Errico | {"url":"http://mathforum.org/kb/thread.jspa?threadID=2394089&messageID=7859610","timestamp":"2014-04-21T13:19:38Z","content_type":null,"content_length":"20770","record_id":"<urn:uuid:27b3e22c-0f34-4228-9ede-aed614b308aa>","cc-path":"CC-MAIN-2014-15/segments/1398223202548.14/warc/CC-MAIN-20140423032002-00164-ip-10-147-4-33.ec2.internal.warc.gz"} |
Projective Hilbert space: L^2
up vote 2 down vote favorite
I am trying to get my head around the geometric formulation of Quantum Mechanics as a projective Hilbert space (see Ashtekar, http://arxiv.org/abs/gr-qc/9706069). So one identifies all the rays $\
mathbb{C} \cdot \phi$ with the vector $\phi$ itself.
As the space of square-integrable functions $L^2(\Sigma, \mu)$ is the standard example of a Hilbert space I was wondering whatever there is clear characterization of its projective space. E.g. is
there a good "visualization", an explicit notion of the Kähler-structure, etc?
A second question concerning this topic: Let $M$ be a Kähler manifold. Which further properties/conditions on $M$ are required so that $M$ can be realized as a projective Hilbert space? Is the
construction than unique?
Thanks, Tobi
P.s.: Which are the standard references regarding this topic?
dg.differential-geometry mp.mathematical-physics ag.algebraic-geometry
I don't understand the question: For a Kahler manifold to be a projective hilbert it has to be a projective space $\mathbb{C} P^n.$ For pretty much all you can say on the subject, see
en.wikipedia.org/wiki/Projective_Hilbert_space – Igor Rivin Dec 23 '11 at 20:24
Yes for the finite-dimensional case this is easy. But is there something as $\mathbb{C}P^\infty$? – Tobias Diez Dec 23 '11 at 22:53
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2 Answers
active oldest votes
Look at section 5 of this paper by Helmick and Helminck: http://eprints.eemcs.utwente.nl/3487/01/1667.pdf
up vote 3 down vote accepted
add comment
Feynman, in his lecture notes, argues convincingly that you will understand the guts of Quantum Mechanics [QM] as best you can by looking at the two-slit experiment whose Hilbert space is
two-dimensional. Or you could look at the Stern-Gerlach experiment, where it is three-dimensional. I recommend you stick to these cases, drop $L^2$, and see carefully what all the concepts
boil down to there in terms of the Kahler geometry of projective space. Feynman does this, in physics language, in his 1957 paper with Vernon titled `Geometrical Representation of the
up vote Schrodinger Equation for Solving Maser Problems': they solve the needed Schrodinger equation by drawing circles on the two-sphere = $CP^1$.
4 down
vote The Berry Phase' is a post-Feynman idea that is essentially the curvature of the canonical connection for the canonical line bundle over $CP^n$. For a dictionary from standard QM to Kahler
geometry and connections you could look atHeisenberg and Isoholonomic inequalites' which you can download from http://count.ucsc.edu/~rmont/papers/list.html by going to the year 1990 there.
add comment
Not the answer you're looking for? Browse other questions tagged dg.differential-geometry mp.mathematical-physics ag.algebraic-geometry or ask your own question. | {"url":"http://mathoverflow.net/questions/84183/projective-hilbert-space-l2","timestamp":"2014-04-20T11:06:14Z","content_type":null,"content_length":"56065","record_id":"<urn:uuid:3c235b58-4081-49ec-848a-0ed74a8af0a6>","cc-path":"CC-MAIN-2014-15/segments/1397609538423.10/warc/CC-MAIN-20140416005218-00202-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bijection between irreducible representations and conjugacy classes of finite groups
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Is there some natural bijection between irreducible representations and conjugacy classes of finite groups (as in case of $S_n$)?
I do not see how it is natural for $S_n$, $n>3$? Surely, you can use either a partition or its dual partition for a representation. Is there a mathematical reason to choose one over another or is
it purely historical?
Bugs Bunny Jul 23 '12 at 9:37
@Bugs Bunny: maybe one can argue that there are two natural choices in this case? It all boils down to interpreting the word "natural", of course.
Vladimir Dotsenko Jul 23 '12 at 13:14
@Bugs It's fairly natural. For any partition $\lambda$ of $n$, let $S(\lambda)$ be the subgroup $\prod S_{\lambda_i}$ of $S_n$. Let $H(\lambda)$ be the induction of the trivial rep from $S(\lambda)
$ to $S_n$. The characters of the $H(\lambda)$ are upper triangular in the characters of the irreps; the conjugacy classes pairs upper triangularly against the $H(\lambda)$. The pairing is forced
by wanting these matrices to be upper triangular rather than a permutation of upper triangular. I'm not sure that there is no other pairing which achieves this, but most of them won't work.
David Speyer Jul 23 '12 at 13:30
Is there any other case in math where bijection exists, but there is no natural one ?
Alexander Chervov Jul 24 '12 at 7:30
Here is a quote from Stanley, EC2, section 7.18: "We have described a natural way to index the irreducible characters of $S_n$ by partitions of $n$, while the cycle type of a permutation defines a
natural indexing of the conjugacy classes of $S_n$ by partitions of $n$. Hence we have a canonical bijection between the conjugacy classes and the irreducible characters of $S_n$. However, this
bijection is essentially 'accidental' and does not have any useful properties. For arbitrary finite groups there is in general no canonical bijection between irreducible characters and conjugacy
Sam Hopkins Jan 6 at 18:54
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This is a different take on Steven Landsburg's answer. The short version is that conjugacy classes and irreducible representations should be thought of as being dual to each other.
Fix an algebraically closed field $k$ of characteristic not dividing the order of our finite group $G$. The group algebra $k[G]$ is a finite-dimensional Hopf algebra, so its dual is also a
finite-dimensional Hopf algebra of the same dimension; it is the Hopf algebra of functions $G \to k$, which I will denote by $C(G)$. (The former is cocommutative but not commutative in general,
while the latter is commutative but not cocommutative in general.) The dual pairing $$k[G] \times C(G) \to k$$
is equivariant with respect to the action of $G$ by conjugation, and it restricts to a dual pairing $$Z(k[G]) \times C_{\text{cl}}(G) \to k$$
on the subalgebras fixed by conjugation; $Z(k[G])$ is the center of $k[G]$ and $C_{\text{cl}}(G)$ is the space of class functions $G \to k$. Now:
The maximal spectrum of $Z(k[G])$ can be canonically identified with the irreducible representations of $G$, and the maximal spectrum of $C_{\text{cl}}(G)$ can be canonically identified with the
conjugacy classes of $G$.
The second identification should be clear; the first comes from considering the central character of an irreducible representation. Now, the pairing above is nondegenerate, so to every point of the
maximal spectrum of $Z(k[G])$ we can canonically associate an element of $C_{\text{cl}}(G)$ (the corresponding irreducible character) and to every point of the maximal spectrum of $C_{\text{cl}}(G)$
we can canonically associate an element of $Z(k[G])$ (the corresponding sum over a conjugacy class divided by its size).
@Qiaochu: This is a nice abstract way to re-focus rigorously the original somewhat fuzzy question. But I still feel unable to treat concrete cases involving finite groups of Lie type in any
explicit way: e.g., given the family $\mathrm{SL}_2(\mathbb{F}_p)$, how to prescribe an actual bijection (uniformly for all odd primes) and pass to the quotient group by the center as well? Even
for $S_n$ I'm reminded that the original Springer Correspondence assigned characters/partitions to cohomology degrees of the flag variety in one way which later got dualized/transposed.
Jim Humphreys Jul 24 '12 at 20:47
I'm not sure I understand your question. I'm not claiming that a bijection between irreducible representations and conjugacy classes exist. What this argument does is exhibit irreducible
representations and conjugacy classes as canonical bases of two vector spaces which are canonically dual to each other. But these bases are not dual to each other, so I don't get a bijection this
Qiaochu Yuan Jul 24 '12 at 22:39
@Qiaochu: I guess I was expecting a more explicit answer to the original question, such as "no" (?). My concern beyond that is whether your interesting higher level viewpoint adds any concrete
detail about actual finite groups of interest, starting with symmetric groups.
Jim Humphreys Aug 2 '12 at 18:08
@Jim: yes, my answer is essentially "no." That is, I'm suggesting that there shouldn't be a nice bijection for arbitrary finite groups. I think one way to see this is to note that there is no
reasonable sense in which this is true for infinite groups (e.g. $\text{SU}(2)$ has countably many irreducible representations but uncountably many conjugacy classes). The correct generalization of
the duality above is that by Peter-Weyl, for a compact Hausdorff group $G$ the characters of irreducible representations are dense in the space of class functions.
Qiaochu Yuan Aug 2 '12 at 18:35
Could there more generally be some kind of duality between irreducible representations of $G$ over a field $K$ and the $K$-conjugacy classes of $K$-regular elements of $G$?
anon Feb 15 at 3:24
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up vote 18 In general there is no natural bijection between conjugacy classes and irreducible representations of a finite group. To see this think of abelian groups for example. The conjugacy
down vote classes are the elements of the group, while the irreducible representations are elements of the dual group. These are isomorphic, via the Fourier transform, but not canonically.
Nevetheless among all set-theoretic bijections one has subclass of "good" bijections which are isomorphisms of the groups G and and G^dual.
Alexander Chervov Sep 16 '12 at 14:42
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Let $k$ be an algebraically closed field whose characteristic is either zero or prime to the order of $G$.
Then the center of the group ring $kG$ has one basis in natural bijective correspondence with the set of irreducible representations of $G$ over $k$, and another basis in natural bijective
correspondence with the conjugacy classes of $G$.
up vote 10 1) $kG$ is semisimple (this is called Maschke's Theorem) and Artinian, so it is a direct sum of matrix rings over division rings, hence (because $k$ is algebraically closed) a direct sum
down vote of matrix rings over $k$. There is (up to isomorphism) one irreducible representation for each of these matrix rings. Those representations are therefore in natural one-one correspondence
with the central idempotents that generate those matrix rings, and these form a basis for the center.
2) For each conjugacy class, we can form the sum of all elements in that conjugacy class. The resulting elements of $kG$ form a basis for the center.
This gives a (non-natural) bijection between irreducible representations and conjugacy classes, because there is a (non-natural) bijection between any two bases for a given
finite-dimensional $k$-vector space. I do not see any way you can make this natural.
This only works when $k$ is algebraically closed, and the result itself also only works in that generality. For example, over $\mathbb{R}$, the cyclic group of order $3$ has two
irreducible representations but still has three conjugacy classes. The error in the argument is that, for a division ring $\Delta$ occuring in the decomposition of $k[G]$, the
contribution of $M_n(\Delta)$ to $Z(k[G])$ is $Z(\Delta)$, which may be larger than $k$.
David Speyer Jul 22 '12 at 22:47
David Speyer: Point taken; I've inserted the words "algebraically closed" where I ought to have inserted them in the first place.
Steven Landsburg Jul 22 '12 at 23:30
May I suggest that you then change "matrix rings over division rings" to "matrix rings over $k$"? Since an algebraically closed field has nonnontrivial finite division ring extensions.
David Speyer Jul 23 '12 at 10:08
David Speyer: Done. Thanks.
Steven Landsburg Jul 23 '12 at 13:00
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I would suggest that no such general natural bijection has been found to date. I am not sure how one would ``prove" that such a natural bijection could not be found, notwithstanding Gjergi's
answer. I take the view that the equality between numbers of irreducible characters (over an algebraically closed field of characteristic zero) and the number of conjugacy classes is most
naturally obtained by counting the dimension of the center of the group algebra in two different categorical settings: from the group theoretic perspective, the natural distinguished basis for
the group algebra (the group elements) makes it clear that the dimension of the center is the number of conjugacy classes. On the other hand, from a ring-theoretic perspective, the structure
of semi-simple algebras makes it clear that the dimension of the center of the group algebra is the number of isomorphism types of simple modules, that is, the number of irreducible
characters. Moving to prime characteristic (still over an algebraically closed field, now of characteristic $p$, say), it is rather more difficult to prove, as R. Brauer did, that the number
of isomorphism types of simple modules is the number of conjugacy classes of group elements of order prime to $p.$ However, there are contemporary conjectures in modular representation theory
which suggest that there may one day be a different explanation for this equality. In particular, Alperin's weight conjecture suggests counting the number of (isomorphism types of) absolutely
irreducible modules in characteristic $p$ in quite a different way, but one which still degenerates to the usual "non-natural" count when the characteristic $p$ does not divide the group
order, which is essentially the same as the characteristic zero case. No general conceptual explanation for the conjectural count of Alperin has been found to date, though a number of
up approaches have been suggested, including a 2-category perspective. But it is not impossible that such an explanation could one day be found, and such an explanation might shed light even on
vote the "easy" characteristic zero situation.
down Later edit: In view of some of the comments below on the action of the automorphism group on irreducible characters and on conjugacy classes (which is really an action of the outer
vote automorphism group, since inner automorphisms act trivially in each case), I make some comments on (well-known) properties of these actions, which while not identical, have many compatible
Brauer's permutation lemma states that for any automorphism $a$ of the finite group $G,$ the number of $a$-stable complex irreducible characters of $G$ is the same as the number of $a$-stable
conjugacy classses. Hence any subgroup of ${\rm Aut}(G)$ has the same number of orbits on irreducible characters as it does on conjugacy classes. The Glauberman correspondence goes further
with a group of automorphisms $A$ of order coprime to $|G|$. In that case the $A$-actions on the irreducible characters of $G$ and on the conjugacy classes of $G$ are permutation isomorphic.
While the actions of a general subgroup of the automorphism group are not always as strongly compatible as in the coprime case, various conjectures from modular representation theory suggest
that it might be possible to have more compatibilty when dealing with complexes of modules than with individual modules. As a matter of speculation, I have sometimes wondered whether there
might be some analogue of Glauberman correspondence in the non-coprime situation for actions on suitable complexes, although I have no idea for a precise formulation at present. Since the
dimension of the center of an algebra is invariant under derived equivalence, this is one reason why I do not dismiss the idea of a more subtle explanation for numerical equalities.
If I read the discussion at mathoverflow.net/questions/46900/… correctly, no bijection between the conjugacy classes and the irreducible representations respects the action of outer
automorphisms. That seems like pretty convincing evidence to me.
Qiaochu Yuan Jul 22 '12 at 21:53
mathoverflow.net/questions/21606/… seems to say the same thing.
Qiaochu Yuan Jul 22 '12 at 21:54
@Qiaochu: It's a matter of opinion to some extent. That is some evidence, but I wouldn't consider it conclusive myself.
Geoff Robinson Jul 22 '12 at 22:06
In my opinion, before I'd call a bijection natural, I'd want it to be invariant under isomorphisms of groups. That is, it should depend only on the group structure, not on what the specific
elements of the group are. In particular, then, it should be invariant under arbitrary automorphisms, including outer ones.
Andreas Blass Jul 22 '12 at 23:22
For what it's worth, even if the original straightforward version of the question has a somewhat negative answer (instead, duality...), as a general methodological attitude I endorse "not
giving up tooo easily". Thus, I like Geoff R.'s points. Yet, dangit, there's that pesky factoid about (outer) automorphisms...?!? Ok, well, I myself "concede" that automorphisms' action on
repns is natural in all ways that likely concern me (tho' I'm open to persuasion otherwise). So then the scope of that negative result is relevant. Still, "the orbit method"... and all that.
Fun stuff...
paul garrett Jul 23 '12 at 0:39
show 1 more comment
Expanding slightly on the other answers:
To ask for a "natural" bijection is presumably to ask for a natural isomorphism between two functors from the category of finite groups to the category of sets. First, we have the
contravariant functor $S$ that associates to each $G$ the set of isomorphism classes of irreducible representations. Then we have the covariant "functor" $T$ that associates to each $G$
the set of its conjugacy classes.
up vote 5 The first problem is that $T$ is not in fact functorial, because the image of a conjugacy class might not be a conjugacy class. So at the very least we should restrict to some subcategory
down vote on which $T$ is functorial, e.g. finite groups and surjective morphisms.
But the key problem still remains: There is no good way to define a natural transfomation between two functors of opposite variances. So when I said in my earlier answer that "I do not
see any way you can make this natural" I might better have said "This is not a situation in which the notion of naturality makes sense".
All of this, of course, is really just an expansion of Gjergji's and Qiaochu's observations.
Representations can be induced (loosing irreducibleiity) so it also can thought as contr variant functor. There are many problems still exist but .....
Alexander Chervov Sep 8 '12 at 9:25
There is no hope for unique bijection in general but there are some "good" bijectionS in many examples although I do not know definitive characterization....
Alexander Chervov Sep 8 '12 at 9:32
add comment
Steven's and Gjergji answers points that there is no bijection, however possibly this idea should not be put into the rubbish completely.
Ideologically conjugacy classes and irreducible representations are somewhat dual to each other.
The other instances of this "duality" is Kirillov's orbit method - this is "infinitesimal version" of the duality: orbits in Lie algebra are infinitesimal versions of the conjugacy classes.
But pay attention orbits are taken not in Lie algebra, but in the dual space g^. This again manifests that there irreps and conj. classes are dual to each other. However think of
semi-simple Lie algebra - then g^ and g can be canonically identified...
Another instance is Langlands parametrization of the unitary irreducible representations of the real Lie group G. They are parametrized by conjugacy classes in Langlands dual group G^L.
Again here are conjugacy classes in G^L, not in G itself. However for example GL=GL^L...
up vote 4
down vote So it might be one should ask the question what are the groups such that conjugacy classes and irreps are in some natural bijection or something like this ?
Here is some natural map conjugacy classes -> representations. But it does not maps to irreducible ones, and far from being bijection in general.
A colleague of mine suggested the following - take vector space of functions on a group which are equal to zero everywhere except given conjugacy class "C". We can act on these functions by
$f \to g f g^{-1} $ - such action will preserve this class. So we get some representation. In the case of abelian group this gives trivial representation, however in general, it might be
non-trivial. It always has trivial component - the function which is constant on "C".
I have not thought yet how this representation can be further decomposed, may be it is well-known ?
Take a finite abelian group $G$ and fix a non-degenerate pairing $G \times G \to \mathbb{C}^{\times}$. Unlike in the case of semisimple Lie algebras I do not see a canonical way to pick
such a pairing.
Qiaochu Yuan Jul 22 '12 at 19:21
@Qiaochu Yuan I agree. I did not pretend that there always exists some natural bijection. Just wanted to softly point out that "not giving up tooo easily", as Paul Garrett wrote in his
comment above. E.g. Take G=Z/2Z in this case we may say there is natural bijection :)
Alexander Chervov Jul 23 '12 at 7:39
@Geoff Sorry may be I misunderstanding however it seems my question was different. I do not take "class functions" (functions constant on conj. class (is it correct?)) but take NON
constant on conj. class. Pay attention I am NOT acting by G in standadrd way - but act by conjugation f-> g f g^-1 hence non-constant functions are preserved by this action. This is a
representation which clear have trivial one as a submodule (class function is trivial submodule). How this repr. is decomposed ?
Alexander Chervov Jul 24 '12 at 9:50
E.g. take S_3 then conj. class corresponding to transpositions (just 3 elements) generaters standard 3D representation. We know it is trivial + 2D irreducible representation
Alexander Chervov Jul 24 '12 at 9:54
OK Alexander. I did not read carefully enough.
Geoff Robinson Jul 24 '12 at 10:31
add comment
It appears that similar question has been asked at sci.math.research Tue, 19 Oct 1999. The answer by G. Kuperberg is quite interesting. Hope no one don't mind if I put it here:
As Torsten Ekedahl explained, it is sometimes the wrong question, but in modified form, the answer is sometimes yes.
up vote For example, consider A_5, or its central extension Gamma = SL(2,5). The two 3-dimensional representations are Galois conjugates and there is no way to choose one or the other in
1 down association with the conjugacy classes. However, if you choose an embedding pi of Gamma in SU(2), then there is a specific bijection given by the McKay correspondence. The irreducible
vote representations form an extended E_8 graph where two representations are connected by an edge if you can get from one to the other by tensoring with pi. The conjugacy classes also form
and E_8 graph if you resolve the singularity of the algebraic surface C^2/Gamma. The resolution consists of 8 projective lines intersecting in an E_8 graph. If you take the unit 3-sphere
S^3 in C^2, then the resolution gives you a surgery presentation of the 3-manifold S^3/Gamma. The surgery presentation then gives you a presentation of Gamma itself called the Wirtinger
presentation. As it happens, each of the Wirtinger generators lies in a different non-trivial conjugacy class. In this way both conjugacy classes and irreps. are in bijection with the
vertices of E_8.
@Greg Kuperberg I hope you don't mind me to put it here. I quoted it in mathoverflow.net/questions/153731/…
Alexander Chervov Jan 6 at 13:19
add comment | {"url":"https://mathoverflow.net/questions/102879/bijection-between-irreducible-representations-and-conjugacy-classes-of-finite-gr/102890","timestamp":"2014-04-23T14:56:01Z","content_type":null,"content_length":"124214","record_id":"<urn:uuid:9e48a066-0666-44e5-9c14-1d8610009eb8>","cc-path":"CC-MAIN-2014-15/segments/1398223202774.3/warc/CC-MAIN-20140423032002-00474-ip-10-147-4-33.ec2.internal.warc.gz"} |
SPSSX-L archives -- January 2008 (#406)LISTSERV at the University of Georgia
Date: Tue, 29 Jan 2008 09:16:41 -0600
Reply-To: "Reutter, Alex" <areutter@spss.com>
Sender: "SPSSX(r) Discussion" <SPSSX-L@LISTSERV.UGA.EDU>
From: "Reutter, Alex" <areutter@spss.com>
Subject: Re: Conditional logistic problems
In-Reply-To: A<002701c861ea$4adf5150$2845cd80@ssw.buffalo.edu>
Content-Type: text/plain; charset="us-ascii"
This doesn't directly help your problems with COXREG, but if the dataset is matched 1:1 and not 1:many, you can alternatively use Multinomial Logistic Regression (NOMREG). There's an example in Marija Norusis's Advanced Statistical Procedures Companion and another in the online help (Help>Case Studies, then Regression Models>Multinomial Logistic).
-----Original Message-----
From: SPSSX(r) Discussion [mailto:SPSSX-L@LISTSERV.UGA.EDU] On Behalf Of Gene Maguin
Sent: Monday, January 28, 2008 2:14 PM
To: SPSSX-L@LISTSERV.UGA.EDU
Subject: Conditional logistic problems
I have been working with another list subscriber who has a matched (1:1)
dataset and is trying to analyze relationships between variables using
conditional logistic regression (CLR). I have no experience with this type
of model. However, I did find a posting by Marta (shown below) as well as
several discussion sites. Since spss logistic won't do a conditinal
analysis, the trick is to use the coxreg procedure to do so. I believe I
have faithfully followed Marta's directions but am getting no results and
so, there is something I don't understand. Per Marta's posting, I have
COXREG ftime /STATUS=outcome(1) /STRATA=pair
/METHOD=ENTER x /PRINT=CI(95).
Ftime is the survival time variable computed such that ftime=outcome+1.
outcome is the case-control variable with 0=control, 1=case.
Pair is the strata variable, 161 pairs.
X is the IV.
My case processing box shows
Case Processing Summary
N Percent
Cases available in analysis Event(a) 161 50.0%
Censored 0 .0%
Total 161 50.0%
Cases dropped Cases with missing values 0 .0%
Cases with negative time 0 .0%
Censored cases before the
earliest event in a stratum 161 50.0%
Total 322 100.0%
a Dependent Variable: ftime
Variables in the Equation(b)
Wald df Sig.
X . 0(a) .
aDegree of freedom reduced because of constant or linearly dependent
bConstant or Linearly Dependent Covariates
S = Stratum effect. x = .5093 + S;
If somebody can educate me about what I'm missing or need to look at, I'd
greatly appreciate it.
Thanks, Gene Maguin
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Balls on the Lawn
Journal of Integer Sequences, Vol. 2 (1999), Article 99.1.5
Balls on the Lawn
Colin L. Mallows
AT&T Research Labs
180 Park Avenue
Florham Park, NJ 07932
Email address: clm@research.att.com
Lou Shapiro
Math Dept.
Howard University
Washington, DC 20059
Email address: lws@scs.howard.edu
Abstract: In the "tennis ball" problem we are given successive pairs of balls numbered (1,2), (3,4),... At each stage we throw one ball out of the window. After n stages some set of n balls is on the
lawn. We find a generating function and a closed formula for the sequence 3,23,131,664,3166,14545,65187,287060,1247690,..., the n-th term of which gives the sum over all possible arrangements of the
total of the numbers on the balls on the lawn. The problem has connections with "bicolored Motzkin paths" and the ballot problem.
1. Introduction.
The tennis ball problem goes as follows. At the first turn you are given balls numbered 1 and 2. You throw one of them out the window onto the lawn. At the second turn balls numbered 3 and 4 are
brought in and now you throw out on the lawn any of the three balls in the room with you. Then balls 5 and 6 are brought in and you throw out one of the four available balls. The game continues for n
turns. The first question is how many different arrangements on the lawn are possible. It is easy to see that there are 2, 5 and 14 possibilities after 1, 2 and 3 turns. This suggests the Catalan
numbers which turns out to be the case. A more delicate question is "what is the total sum of the balls on the lawn over all these possibilities"? Here the first few terms are 3, 23, 131 and 664. As
an example consider the five possibilities after two turns. They are
We will find both a generating function and a closed formula for this sequence. First we return to the first question and transform it to a question about paths.
• Consider paths in the plane which satisfy the following conditions:
• The possible steps are L or l.
• The paths start at n steps.
• The paths never go below the x-axis.
Such paths are called bicolored Motzkin paths (see [5]). If there had been only one kind of level step and the paths ended on the x-axis we would have regular Motzkin paths (see [1],[3], or [7]
for more information on Motzkin paths and Motzkin numbers).
We can make three observations. One is that with the four kinds of steps we get the recursion
This holds for
Secondly we have
Given the recursion, this is easily established by induction.
Thirdly Catalan number (see [6] and [7] for different proofs).
Now we return to the balls on the lawn after
The balls on the lawn are denoted by "U, if neither is selected mark with a D, if the odd member is the one selected use an L, if the even number was selected then use an l. This sets up an obvious
correspondence with the bicolored Motzkin paths ending at height zero and this shows that the number of possibilities after n turns is the Catalan number C[n+1].
We now want to shift back to subdiagonal paths from n+1, then the numbers of the horizontal steps correspond to the numbers of the balls on the lawn except that we ignore the initial horizontal step
numbered 0. All subdiagonal paths must go from
To evaluate the sum over all possible sets of balls on the lawn takes a bit more doing and its worthwhile to separate out some definitions and lemmas first.
2. Definitions and notation.
The n^th Catalan number is Q=
The following lemmas can be proved combinatorially but instead we refer to Concrete Mathematics [4], page 203, formulas
Lemma 1
Lemma 2
Lemma 3
Lemma 4
Lemma 5
Lemma 6
Lemma 7
The number of subdiagonal paths from
This is the cornerstone result about ballot numbers and a reference which summarizes this and much of the related literature is[2].
We now want to embark on the main computation. Note first that
Before launching into the evaluation lets look at each term. There are i-1 of them are on the lawn and j of them are to stay in the room. The horizontal step i+j is to be on the lawn and hence the
We want to find a closed form for the generating function
If we set n=m+i and then i=j+d we obtain We sum on m first; then invoke Lemma 2. By rewriting 2j+d=2j+d+1-1 we get and However, with the aid of Lemmas 3 and 4, we obtain For the second term we have,
via Lemmas 2 and 4, Thus
3. Remarks.
3. There is also a connection with hypergeometric functions. and
The first and third of these remarks can be shown by routine algebraic manipulations and the second follows from the first as follows:
since Q^2=1-4z. But n^th terms yields
Two other remarks can be made here. First, an asymptotic result,
Second, the expected value of the balls on the lawn is
Problem 19(s) of reference [7] is succinct but less colorful. It asks one to show that the Catalan numbers count sequences of positive integers such that
The problem was brought to our attention by Ralph Grimaldi, see [6], who refers to the logic text [8] where it is pointed out that after an infinite number of turns the balls remaining in the room
can be either the empty set, a finite set of arbitrary size or even an infinite set such as all the even numbered balls.
Barcucci E., R. Pinzani, & R. Sprugnoli, The Motzkin family, PU.M.A. Ser. A, 2 (1991) 249-279.
Barton, D. E. & C. L. Mallows, Some aspects of the random sequence, Ann. Math. Stat., 36 (1965) 236-260.
Donaghey R. & L. W. Shapiro, The Motzkin numbers, J. Combinatorial Theory Ser. A, 23 (1977) 291-301.
Graham R., D. E. Knuth & O. Patashnik, Concrete Mathematics, Addison-Wesley, 1990.
Getu S. & L. Shapiro, Catalan and Motzkin probabilities, (to appear), Congressus Numerantium.
Grimaldi R. & J. Moser, The Catalan numbers and the tennis ball problem, Congressus Numerantium, 125 (1997) 65-71.
Stanley, R. P., Enumerative Combinatorics, volume 2, to appear, Cambridge University Press.
Tymoczko T.& J. Henle, Sweet Reason: A Field Guide to Modern Logic, Freeman, 1995, p. 304.
(Concerned with sequence A031970 .)
Received July 17 1998; revised version received January 13 1999. Published in Journal of Integer Sequences March 15 1999.
Return to Journal of Integer Sequences home page | {"url":"http://www.emis.de/journals/JIS/MALLOWS/mallows.html","timestamp":"2014-04-17T00:53:31Z","content_type":null,"content_length":"22300","record_id":"<urn:uuid:feffba05-aab7-4198-b996-2f81cfc29d13>","cc-path":"CC-MAIN-2014-15/segments/1397609526102.3/warc/CC-MAIN-20140416005206-00638-ip-10-147-4-33.ec2.internal.warc.gz"} |
eFunda: Glossary: Units: Electric Resistivity: Ohm-Centimeter
Glossary » Units » Electric Resistivity » Ohm-Centimeter
Ohm-Centimeter (ohm-cm) is a unit in the category of Electric resistivity. It is also known as cneti-meters, cnetimeters, cneti-metres, cnetimetres. Ohm-Centimeter (ohm-cm) has a dimension of MLT^-3I
^-2 where M is mass, L is length, T is time, and I is electric current. It can be converted to the corresponding standard SI unit ohm-m by multiplying its value by a factor of 0.01.
Note that the seven base dimensions are M (Mass), L (Length), T (Time),
(Temperature), N (Aamount of Substance), I (Electric Current), and J (Luminous Intensity).
Other units in the category of Electric resistivity include Ohm-Foot (ohm-ft), Ohm-Inch (ohm-in), and Ohm-Meter (ohm-m).
Additional Information
Related Glossary Pages
Related Pages | {"url":"http://www.efunda.com/glossary/units/units--electric_resistivity--ohm-centimeter.cfm","timestamp":"2014-04-21T07:05:23Z","content_type":null,"content_length":"31365","record_id":"<urn:uuid:e7c6927e-557f-4044-853a-464c22962035>","cc-path":"CC-MAIN-2014-15/segments/1397609539665.16/warc/CC-MAIN-20140416005219-00141-ip-10-147-4-33.ec2.internal.warc.gz"} |
Practice Test Questions...Part B
November 3rd 2009, 01:47 PM #1
MHF Contributor
Jul 2008
Practice Test Questions...Part B
Question 5.
If xy = 7 and x - y = 5, then x 2 y - xy 2 =
A. 2
B. 12
C. 24
D. 35
Question 6.
To celebrate a colleague's graduation, the m coworkers in an office agreed to contribute equally to a catered lunch that costs a total of y dollars. If p of the coworkers fail to contribute,
which of the following represents the additional amount, in dollars, that each of the remaining coworkers must contribute to pay for the lunch?
A. y / m
B. y / (m-p)
C. py / (m-p)
D. py / ((m-p)m)
Question 7.
A heavy rope, 100 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 120 ft high. How much work is done in pulling the rope to the top of the building?
A. 2500 foot-pounds
B. 1200 foot-pounds
C. 3000 foot-pounds
D. None of these
Question 8.
Mary, at age 35, decides to deposit $2000 per year in a bank, for the next 30 years. How much will she have at age 65 if her rate of return is assumed to be 10% per annum?
A. $300,000
B. $328,988
C. $250,000
D. None of these
Question 9.
Tickets must be purchased for the $75 grand prize. The cost of the ticket is $3, and 150 tickets will be sold. Determine John's expectation if he purchases one ticket.
A. $ -1.5
B. $ -2.5
C. $ -3.5
D. $ -4.5
Question 10.
A foreign student club lists as its members two Canadians, three Japanese, five Italians, and two Germans. Find the probability that all nationalities are represented if a committee of size 4 is
selected at random.
A. 0.12
B. 0.4
C. 0.36
D. 0.05
I am taking a test in two weeks. I need help with these sample questions.
Question 5:
I'm sorry I don't actually know what you mean by "x 2 y", if you mean a multiplication, than I would say that: $x \cdot 2 \cdot y - 2 \cdot x \cdot y = 0$.
Question 6:
Answer B, because we have $m$ people, but when $p$ people fail to contribute, we get $m-p$ people that can contribute, so if we have something that costs $y$. We get $\frac{y}{(m-p)}$ that each
of the people in $m - p$ needs to pay.
I'm sorry for not posting further, I have the feeling I'm making homework
November 3rd 2009, 02:37 PM #2
Nov 2009 | {"url":"http://mathhelpforum.com/algebra/112227-practice-test-questions-part-b.html","timestamp":"2014-04-18T17:41:18Z","content_type":null,"content_length":"34549","record_id":"<urn:uuid:a7554e0b-384d-4425-a848-a71a577724d6>","cc-path":"CC-MAIN-2014-15/segments/1397609533957.14/warc/CC-MAIN-20140416005213-00124-ip-10-147-4-33.ec2.internal.warc.gz"} |
Help! 3rd Grade Math is hard!
Does anyone else think this is a bit much for a third grader? This is a weekly assignment that my son gets. No surprise that he got a 60 on it. His answer was " The answer is 5 because 15X7=105.
"There are 6 bookshelves in the classroom. Each bookshelf has room for 15 books. If Mrs. Smith has 100 books, how many books will not be able to fit on the shelves?
Not only do they have to answer this, but they are graded as follows: 6 points for answering in a complete sentence, repeating the question, 2 points for circling key words and math words and 2
points for showing work. They also have to explain how they got their answer. What ever happened to just getting the right answer and showing your work?
Of course I can figure it out, but to explain how I got the answer would be a lot different than a third grader, since they are not able to think at the same level as an adult who is more proficient
in math skills.
Did you notice how they put "how many books will NOT fit on the shelf....as to throw them off even more??! Is this really what is expected of a third grader these days? Seems like they just finished
memorized simple adding and subtracting and are still working on memorizing times tables. Am I overreacting here, or is this ridiculous? | {"url":"http://www.polkmoms.com/forum/topics/help-3rd-grade-math-is-hard","timestamp":"2014-04-16T19:53:24Z","content_type":null,"content_length":"66546","record_id":"<urn:uuid:0356fd35-821f-42e3-94be-3c28db679a6f>","cc-path":"CC-MAIN-2014-15/segments/1397609524644.38/warc/CC-MAIN-20140416005204-00396-ip-10-147-4-33.ec2.internal.warc.gz"} |
RE: st: How to make a code faster - alternatives to egen var =
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RE: st: How to make a code faster - alternatives to egen var = concat(vars) ?,
From "Nick Cox" <n.j.cox@durham.ac.uk>
To <statalist@hsphsun2.harvard.edu>
Subject RE: st: How to make a code faster - alternatives to egen var = concat(vars) ?,
Date Thu, 17 Jun 2010 22:02:51 +0100
I'll -collapse- and -merge- back again.
Tiago V. Pereira
Thank you so much again, Antoine!
Yes, this is a very efficient way! However, I could not figure out how I
can save the combination of the categorical variables that a specific
meanX refers to.
For example, the commands
sysuse auto, clear
bysort foreign rep78 : mymean price
scalar dir
show a list of scalars containing the mean of the ith combination, but I
don't know if the mean10 refers to the combination "foreign = Foreign,
rep78 =4" or "foreign = Foreign, rep78 = 5"
[Actually I do in this specific case if I take a look at each value from
the output.]
Nevertheless, assuming a very large number of categorical variables
(n>10), I cannot write a loop and say that mean2451 refers to the
combination x1==0 x2==0 x3==2 x4==0 x5==2 and x6==1. I want to summarize
the mean of this combination in group 1 and generate a separate variable
for group 2.
for example
bysort x1 x2 x3 x4 x5 x6 : mymean score if group==1
*/ yes, -mymean- needs further amendments to have this option
replace score = mean2451 if
So, In this case I know that mean2451 comes from the combination
x1==0&x2==0&x3==2&x4==0&x5==2&and&x6==1 from group 1 and I replace its
value for all subjects from group 2 having an identical combination.
This is getting tough, but you have any additional tips, I will be
very grateful!
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XNS - Niccolo' Bucciantini
The code XNS solves for the axisymmetric equilibrium configuration of differentially rotating Neutron Stars with toroidal magnetic field, using the XCFC approximation for the metric, in spherical
coordinates. The code is based on the metric module and routines developed for the GR-MHD code
You can download an older version of the code XNS here !
And an older Guide!
There are several other codes published in the years (
, etc....) you might want to check.
However, solutions, even for rapid rotators close to the mass shedding limit, show that the difference
between conformally flat metric CFC/XCFC, and the more appropriate quasi-isotropic coordinates is
order of 0.1% (larger differences are expected for strongly distorted cases of disk-like configurations), so in principle one would expect the CFC/XCFC limit to provide a reasonably good
approximation of the
correct solution for Neutron Stars. There are some computational benefits in using XCFC over quasi-isotropic coordinates, and as an Astrophysicist I am looking more toward a solution that is
physically acceptable than one that is mathematically correct.
These are some comparisons with the code RNS, for a uniform rotator close to mass shedding limit. You can see that the errors are ~0.1% or smaller. On the left is a comparison of the polar, and
equatorial densities: the solid lines are from RNS and the dashed lines (can you see them?) from XNS. On the right is a comparison of the errors in the metric terms. All the lines that you see
represent various metric terms (see the Guide for explanation). They are all < 0.1%, the dash dotted line is a measure of how far from conformally-flat the RNS result is. That indicates the quality
of the CFC/XCFC approximation.
Here is instead a differentially rotating magnetized solution.
There are still a few things that can/must be improved:
• The code is limited to Politropic EoS. It should not be hard to implement more realistic and/or user supplied EoS. But this needs some adjustments in a few routines.
• The code needs at the moment some hand-made search for the initial starting density (see the Guide for explanation). The search could be automated, but I have not yet thought how to do it
cleverly, due to the oscillatory nature of the convergence.
• The metric solver utilizes routines developed for a more general axisymmetric geometry, without symmetry with respect to the equator. This also could be optimized for Neutron Stars.
• The code does not handle poloidal magnetic fields. However, equilibrium with poloidal fields is strictly only possible for uniform rotators, with poloidal magnetic field fully confined within
the star. Unless the Neutron Star is in a true vacuum, which can never be realized/maintained in reality, the solution with a weak poloidal field, will at best be quasi-equilibrium. Adding a weak
poloidal component is one of the future possible upgrades, but it will require some major changes which will not be coming soon.
....... need more here! coming soon! | {"url":"https://sites.google.com/site/niccolobucciantini/xns","timestamp":"2014-04-16T07:34:53Z","content_type":null,"content_length":"26584","record_id":"<urn:uuid:4a2d7c2d-6685-4441-aa19-3223928fc9bb>","cc-path":"CC-MAIN-2014-15/segments/1397609526252.40/warc/CC-MAIN-20140416005206-00153-ip-10-147-4-33.ec2.internal.warc.gz"} |
William S. Cleveland
William S. Cleveland: Papers on the Web
Computer Networking
A Statistical Model for Allocating Bandwidth to Best-Effort Internet Traffic, Statistical Science, J. Cao, W. S. Cleveland, and D. X. Sun, to appear, 2004.
Stochastic Models for Generating Synthetic HTTP Source Traffic, IEEE Infocom, J. Cao, W. S. Cleveland, Y. Gao, K. Jeffay, F. D. Smith, and M. Weigle, to appear, 2004.
Internet Traffic Tends Toward Poisson and Independent as the Load Increases, Nonlinear Estimation and Classification , eds. C. Holmes, D. Denison, M. Hansen, B. Yu, and B. Mallick, Springer, New
York, 83-109. J. Cao, W. S. Cleveland, D. Lin, and D. X. Sun, 2002.
On the Nonstationarity of Internet Traffic, ACM SIGMETRICS, 102-112. J. Cao, W. S. Cleveland, D. Lin, and D. X. Sun, 2001.
IP Packet Generation: Statistical Models for TCP Start Times Based on Connection-Rate Superposition, ACM SIGMETRICS 166-177. W. S. Cleveland, D. Lin, and D. X. Sun, 2000.
Internet Traffic Data, Journal of the American Statistical Association, 95, 979-985. Reprinted in Statistics in the 21st Century , edited by A. E. Raftery, M. A. Tanner, and M. T. Wells, Chapman &
Hall/CRC, New York, 2002. W. S. Cleveland and D. X. Sun, 2000.
Statististical Models and Model Building
Maximum Likelihood Estimation of Sum-Difference Time Series Models Using the EM Algorithm , Statistica Sinica. W. S. Cleveland and C. Liu, to appear, 2004.
Bayesian Graduation Using Constrained Bernoulli-Mixture Models , Lifetime Data Analysis . W. S. Cleveland and C. Liu, to appear, 2004.
Random Location and Scale Effects: Model Building Methods for a General Class of Models , Computing Science and Statistics, 32, 3-10, edited by E. J. Wegman and Y. M. Martinez, Interface Foundation
of North America. W. S. Cleveland, L. Denby, and C. Liu, 2000.
Modeling Customer Survey Data , Case Studies in Bayesian Statistics IV , 3-57, edited by C. Gatsonis, R.E. Kass, A. Carriquiry, A. Gelman, I. Verdinelli, and M. West, Springer, New York. L. Clark,
W. S. Cleveland, L. Denby, and C. Liu, 1999.
Comments on Approache Graphique en Analyze des Donnes by Jean-Paul Valois , Journal de la Societe Francaise de Statistique , 141, 43-44. W. S. Cleveland, 2001.
Competitive Profiling Displays: Multivariate Graphs for Customer Satisfaction Survey Data, Marketing Research , 11, 25-33. L. A. Clark, W. S. Cleveland, L. Denby, and C. Liu, 1999.
Trellis Display: Modeling Data from Designed Experiments. , Bell Labs Technical Report. W. S. Cleveland and M. Fuentes, 1997.
The Visual Design and Control of Trellis Display, Journal of Computational and Statistical Graphics, 5,123-155. R. A. Becker, W. S. Cleveland, and M. J. Shyu, 1996.
Trellis Graphics User's Manual, Insightful, Seattle, Washington. R. A. Becker and W. S. Cleveland, 1996.
Data Mining and Machine Learning
The S-Net System for Internet Packet Streams: Strategies for Stream Analysis and System Architecture Journal of Computational and Statistical Graphics: Special Issue on Streaming Data, 12, 865-892.
J. Cao, W. S. Cleveland, and D. X. Sun, 2003.
Data Science: An Action Plan for Expanding the Technical Areas of the Field of Statistics , ISI Review, , 69, 21-26. W. S. Cleveland, 2001.
Smoothing by Local Regression: Principles and Methods, Statistical Theory and Computational Aspects of Smoothing, 10-49, edited by W. Haerdle and M. G. Schimek, Springer, New York. W. S. Cleveland
and C. L. Loader, 1996.
Rejoinder to Discussion of ``Smoothing by Local Regression: Principles and Methods'', Statistical Theory and Computational Aspects of Smoothing, 113-120, edited by W. Haerdle and M. G. Schimek,
Springer, New York. W. S. Cleveland and C. L. Loader, 1996.
A Package of C and Fortran Routines for Fitting Local Regression Models: Loess User's Manual , Bell Labs, Technical Report . W. S. Cleveland, Eric Grosse, and Ming-Jen Shyu, 1996. | {"url":"http://www.stat.purdue.edu/~wsc/papers.html","timestamp":"2014-04-17T13:29:06Z","content_type":null,"content_length":"7129","record_id":"<urn:uuid:ce90e328-bb8c-4442-a5e7-6fa691bdb2c9>","cc-path":"CC-MAIN-2014-15/segments/1398223201753.19/warc/CC-MAIN-20140423032001-00298-ip-10-147-4-33.ec2.internal.warc.gz"} |
The Guardian’s only out by one million times here
So, The Guardian tells us about the White House agreeing to crush the US stocks of confiscated ivory:
White House to crush 6m tons of seized ivory
Six million tonnes? That’s a lot of ivory there.
The Obama administration said on Monday it would destroy all 6m tons of its stocks of seized ivory
They really do mean 6 million tonnes. That’s an awful lot of elephants.
The Philippines crushed 15m tons of seized ivory beneath industrial rollers earlier this year.
No, they really do think that ivory comes in the millions of tonnes.
Although it doesn’t you know:
Six tons of confiscated elephant tusks to be crushed
Well done to the arts graduates at The G. Only out by a factor of one million this time.
23 comments on “The Guardian’s only out by one million times here”
1. Maybe the Americans used “m” for metric and the Guardian’s work experience interns got confused?
2. This explains why they are so unconcerned at the deficit; they have no concept of large numbers.
3. How many elephants would it take to produce 6m tonnes of ivory?! 500 million?!!
4. Some brief research:
Average weight of elephants across species: between 3,000 and 7,000 kg, say 5,000kg.
Average tusk weight: 10-12kg, say 11kg. Multiply by two, so 22kg.
6m tonnes of ivory is 6 billion kilos. Divide by 22 and you get 272,727,272.8 elephants, assuming a 100% conversion rate of ‘tusk’ to ‘ivory’ (I know nothing about it).
That’s a lot of elephants.
5. Isn’t the best way to stop killing of wild elephants for ivory to flood the market with the stuff they already have?
6. JamesV – “Isn’t the best way to stop killing of wild elephants for ivory to flood the market with the stuff they already have?”
I believe Homer Simpson once said so. But that doesn’t mean he was wrong.
Actually the best way to stop the killing of wild elephants is to allow a sensible cull based on real data and numbers – with the proceeds going to the local villagers. Then they have a real
interest in protecting the animals and taking a sustainable harvest. Everyone knows this.
But these sort of programmes have taken a hit in recent times. The Greenies got to the US Congress and they refused to allow US funding of CAMPFIRE for instance. Which has worked quite well:
But of course it worked quite well in Zimbabwe. Can anyone guess what happened to it?
Needless to say I bet the Guardian hated it.
7. tusk, tusk, tusk
8. Why crush the stuff? The poor elephants are not going to be brought back to life. And by restricting the supply, as with drugs, so with the killing of these beasts – it will continue, and
continue to be a trade controlled and run by criminals.
9. Must you be so extremist? Can’t we arrange that the moderates take power? Six thousand tonnes of ivory it is.
10. I wrote to The Guardian once after some fatuous contributor had claimed the infinitude of primes was an article of faith amongst mathematicians. I pointed out that a proof had been around since
Euclid, at least, and that a scientist making a similarly stupid claim about the arts would never got past an editor
Letter wasn’t published of course.
11. C’mon, 6m tons is fake but true.
12. Surely the best way to stop poaching would be a prize to find a way to cheaply fake ivory? Then it would be possible to flood the market with fakes.
13. David –
…a prize to find a way to cheaply fake ivory
Yes, they tried that in the nineteenth century when a shortage of ivory was endangering the supply of billiard balls. The result was celluloid. According to legend, it had an disconcerting
tendency to explode on impact, no doubt causing more than one gentleman to spill his port.
14. Roll Up! Roll Up!
See the Elephant!
We keep it in this matchbox…look. In the corner. Behind the rhinoceros.
15. @ David Jones
Euclid did geometry, but I worked out a proof for myself while I was in short trousers – multiply all primes by each other and add one (if it *is* divisible that can only be by two numbers that
you haven’t thought of)
16. …if it *is* divisible that can only be by two numbers that you haven’t thought of
I’m thinking of number 1!
17. @john77
That is Euclid’s proof (sometimes called Euclid’s theorem). It’s in his Elements, Book IX, Proposition 20.
18. As James v
19. @ MyBurningEars
No-one ever told me that.
Honestly! I had assumed that the proof was far more complicated and intellectual than the one I had dreamed up
20. Why don’t we just ban ivory? Then all the bad stuff will stop.
21. Say we only know 2 primes, 3 and 5. Multiply them is 15, add 1 is 16. Divisible by 2, 4, and 8. That’s 3 numbers already.
In fact, won’t you always get an odd number if you multiply all odd numbers, thus a multiple of lots of odd numbers plus one is always even, thus not prime?
22. Say we only know 2 primes, 3 and 5.
That’s not quite what’s being said, though.
If you know 3 and 5 you can’t ignore 1 and 2 since you have to multiply together all the primes you know. Hence (1×2×3×5)+1=31
23. Err, no, James. If all you have is 3 and 5 then you would indeed multiply and add one to get 16, at which point, bingo! you have found a new prime, 2. MB: one is not a prime (at least not in
number theory). It breaks the Fundamental Theorem of Arithmetic (that every number has a unique prime factorisation).
Euclid’s proof runs like this: take some list of primes (it doesn’t have to be all the primes less than a certain number, any finite list will do). Multiply them together and add one. This new
number is not divisible by any of the primes in your list. Hence it is either prime, or it is divisible by another prime not in the list. In each case, we have demonstrated the existence of a
hitherto undiscovered prime. Add this to our list and repeat. The process will never terminate. Hence the primes are infinite. | {"url":"http://www.timworstall.com/2013/09/10/the-guardians-only-out-by-one-million-times-here/","timestamp":"2014-04-18T23:17:15Z","content_type":null,"content_length":"84918","record_id":"<urn:uuid:b8e207db-10f4-4841-bb55-6977ba8f0b6b>","cc-path":"CC-MAIN-2014-15/segments/1397609535535.6/warc/CC-MAIN-20140416005215-00640-ip-10-147-4-33.ec2.internal.warc.gz"} |
Let X Denote The Number Of Major Defects For A ... | Chegg.com
let x denote the number of major defects for a particular piece of machinery and y be the number of cosmetic flaws on this same piece, suppose that x and y are independent variables with f_1(x) =
.80, .15, and .05 for x = 0, 1, 2 respectively and f_2(y) = .50, .25, .15, .08, .02 for y = 0, 1, ..., 4 respectively. a) what is the joint mass function of these two variables? b) what proportion of
these machines will have no major defects or cosmetic flaws? what proportion will have at least one defect or flaw? c) for what proportion of these machines will the number of cosmetic flaws exceed
the number of major defects?
Statistics and Probability | {"url":"http://www.chegg.com/homework-help/questions-and-answers/let-x-denote-number-major-defects-particular-piece-machinery-y-number-cosmetic-flaws-piece-q3811649","timestamp":"2014-04-24T21:15:53Z","content_type":null,"content_length":"20676","record_id":"<urn:uuid:dc000d9b-b3d5-4fcb-a617-eb27b0853575>","cc-path":"CC-MAIN-2014-15/segments/1398223206770.7/warc/CC-MAIN-20140423032006-00577-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Here's the question you clicked on:
Solve the system of equations: 4x + 3y = 1 3x + 2y = 2
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1.multiply the 1st equation by 2 and 2nd equation by 3 2.subtract the 2nd equation from 1st equation and you will get 'x' value 3.substitute the 'x' value in one of the equation to get 'y' value
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4x+3y=1.....(1) 3x+2y=2....(2) (1)x2 8x+6y=2...(1)' (2)x3 9x+6y=6....(2)' (1)'-(2)' -x=-4 x=4 sub to (1) 4(4)+3y=1 16+3y=1 3y=15 y=5 (4,5)
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Bias Sampling
To demonstrate how the results of a poll or other scientific study can be biased by selecting special types of people to respond or by asking only certain questions.
The main goal of early statistics with young students is to, “…make them informed consumers…of data.” (Benchmarks for Science Literacy, page 226.)
At the K-2 level, students will have had practice with sorting and classifying data. They will have done this with many tangible materials, such as rocks, buttons, shells, leaves, etc. Hopefully,
students will have had opportunities to sort and classify tangible materials according to a number of different attributes and will have had some practice in communicating to others the groupings
that they made.
For grades 3-5, this type of information gathering can be used for posing more challenging questions. Now students will begin to explore that sometimes, a representation of a group does not provide
an accurate picture of the whole group. This can be a difficult concept for students to understand. Keeping things simple will help students begin to sort through the idea that data can be
At this level, you can encourage students to look at data from a few different angles; identify the largest group in a field of data (a foundation for introducing the concept of “average”); talk
about how knowing the largest group can be useful; and recognize that the largest group does not necessarily represent the whole group of data.
In this lesson, students will take a poll to gather information about homework time. With your guidance, students will then look at their information in ways that encourage them to think about the
concepts listed above. This process will help them to consider the statement in the Benchmark for this lesson, “How much a portion of something can help to estimate what the whole is like depends on
how the portion is chosen.”
To help students become engaged in the concept of bias sampling, you might take a quick poll of the class about something fun.
For example, tell the students that you would like to pretend that you are a newspaper reporter. You have come to this class to find out what foods the students like best. Choose three types of food
(we’ll use enchiladas, pizza, and candied yams for this example) and ask students which food they like best. Count together to identify the food that received the most votes. (Let’s imagine it’s
pizza.) Now talk with them about how you plan to write your article about how “Ms. Vargas’ class likes pizza best!”
Encourage students to think about how this article summarizes the class vote that you just took.
You might ask:
• Why did pizza make the headline?
• Did everyone vote for pizza?
• What happened with the votes for enchiladas and candied yams?
You can expect that students will be familiar with the “majority rules” idea. In the Development of this lesson, you can facilitate more critical thinking about how this majority came to be chosen.
Now it is the students’ turn to be reporters. They will conduct a survey about how much homework time is appropriate for them to have. Students will work in three groups to conduct the survey: one
group will interview students, another will interview parents, and the other will interview teachers.
Before students begin their survey, lead a class discussion about how information is collected.
Consider these questions with your class:
• Why might it be important to interview students about homework time? Parents? Teachers?
• Do you think that each group will answer the questions the same way?
• Why not interview just one of the groups?
• What does interviewing three groups of people do?
• If you were the principal and were making the final decision about how much homework would be assigned to your class, should you only interview the teachers? Why or why not?
The goal of this lesson is to get students to think about bias as related to polls and studies. Help them think through how selecting only certain types of people to participate can bias a scientific
study or survey.
It is also important to lead a discussion about the role questioning plays in gathering information. Choosing one set of questions over another affects the kind of information one can solicit from a
poll or study. Also, challenge students to consider why it is meaningful to ask the same questions to respondents when taking a poll or doing a study. You might give students a quick example to help
them think about how questions are linked to answers. (For example, why do you ask the same questions on tests?)
Other discussion questions could be:
• If you were working with a team of people and everyone went out to ask people questions about homework time, would you want everyone on your team to ask the same questions or different questions?
• What would happen with your information if everyone asked different questions?
• What could you learn if everyone asked the same questions?
When students have grasped the importance of asking like questions to obtain specific information, introduce the Homework Survey student sheet. Students should take a class vote to determine which
question will be asked for each item. You can encourage them to discuss the strengths and drawbacks of each proposed question. When they determine the final questions, they can write them on their
student sheet, leaving space for recording participants’ responses.
Now have students work in three groups to interview students, parents, and teachers. Each student should complete her/his own Homework Survey student sheet so that everyone has an opportunity to ask
questions and record answers.
When students have finished with their surveys, have them return to their small groups (the student, parent, or teacher group). Ask them to work together on the Report for the Principal student
Have each group read its letter to the principal to the class. When each group has had its turn, you can facilitate a discussion about these reports that helps students reflect on how the findings
were different from group to group.
You could ask:
• What did you notice about these reports?
• What was similar?
• What was different?
• You each asked the same questions, but got different answers. Why?
• Each report made a recommendation to the principal. What would happen if the principal received only one of these reports?
• Should the principal find out how many students, parents, and teachers there are in the school and take only the survey information from the largest group?
• What could you do to make your group’s report to the principal reflect more opinions?
• How could you change your letter to more accurately represent the information you gathered?
Remember that the main idea you want them to consider is that information can be biased when only certain types of people are participating in a study or survey.
For another Science NetLinks lesson in which students analyze data, see What Can Data Tell Us? In this lesson, students look at questions that can be posed and answered by examining data
distribution. They also look for circumstances that might bias the results of a study.
You can help students think about how bias sampling relates to other things like medical studies or trials. First talk with them about what is a medical trial, keeping it simple by describing it as
an experiment that helps doctors know what medicines work the best.
Ask students:
• What would happen if only certain groups of people participated in the medical trial?
• If doctors need to know about a cold medicine, do you think they should ask people who do not have a cold to participate in the trial? Should they only ask people who have a cold?
• If doctors need to know about allergy medicines, do you think it makes a difference whether the participants in the trial have allergies or not?
• Do you think it makes a difference whether medical studies or trials are done only on children or adults?
• When do you think it would be important to conduct tests only on a specific group?
• Would it ever be unfair to exclude certain groups from medical tests or trials?
You can also encourage students to think about these issues as they relate to other things like the manufacturing of toys or kinds of cereals that companies make. Again, the idea is for them to begin
considering how information can be biased and how this bias affects other things. This will help them become “informed consumers of data.”
If you would like to extend the mathematics concepts presented in this lesson, refer to PBS Math and NCTM’s Illuminations site for creative ideas and teacher resources. | {"url":"http://sciencenetlinks.com/lessons/bias-sampling/","timestamp":"2014-04-17T22:17:10Z","content_type":null,"content_length":"27326","record_id":"<urn:uuid:a7ca0342-f931-490f-b519-c70d798537c9>","cc-path":"CC-MAIN-2014-15/segments/1397609532128.44/warc/CC-MAIN-20140416005212-00309-ip-10-147-4-33.ec2.internal.warc.gz"} |
clarification of phyllotaxis
Bob Vickery vickery at MPX.COM.AU
Fri Nov 21 00:42:47 EST 1997
>Hi all,
> I am hoping one of you can enlighten an ignorant soul. I have been
>teaching my class about phyllotaxis using the text "Botany' by Moore et
>al. They define phyllotaxis as the number of turns of a spiral/number of
>leaves BETWEEN successive leaves of an orthostichy. As I went to expand on
>this subject by giving an example from Esau's "Anatomy of Seed Plants", I
>discovered that she describes 5/13 phyllotaxy as "five windings about the
>axis include 13 leaves with leaves n and n plus 13 located one above the
>other". That is there are only 12 leaves BETWEEN successive leaves of the
>orthostichy. These two definitions are slightly different but important.
>The other surprise was that the phyllotaxis was 5/13 not 8/13, which is
>what I expected based on the Fibonacci series and how it is explained in
>Moore et al.
>My question is "What is the correct definition of phyllotaxis" or is there
>no agreement on this?
>Grant R. Cramer
>Associate Professor
>Department of Biochemistry, Mail Stop 200
>University of Nevada,
>Reno, NV 89557
>Phone (702) 784-4204
>Fax (702) 784-1650
>email: cramer at med.unr.edu
Esau's definition is correct if you think of phyllotaxy as the angle between=
successive leaves. In her example, 5/13 of a circle =3D 137.5 degrees =3D=
360 - 222.5 degrees. But, 222.5 degrees =3D 8/13 of a circle. Counting=
along a right-handed helix you get one answer, counting along a left-handed=
helix you get the other.
Bob Vickery
bob at acsusun.acsu.unsw.edu.au
vickery at mpx.com.au
More information about the Plant-ed mailing list | {"url":"http://www.bio.net/bionet/mm/plant-ed/1997-November/002745.html","timestamp":"2014-04-19T15:38:11Z","content_type":null,"content_length":"4225","record_id":"<urn:uuid:55569696-4476-4eef-92e4-d59d8ed02b5f>","cc-path":"CC-MAIN-2014-15/segments/1397609537271.8/warc/CC-MAIN-20140416005217-00132-ip-10-147-4-33.ec2.internal.warc.gz"} |
units question
Yes, I see how it works mathematically. But I just thought "the amount of charge carried by a current of one Ampere in a second" was similar to the amount of charge carried by a current of one Ampere
per second" so it would have units of Amperes/S | {"url":"http://www.physicsforums.com/showthread.php?t=180132","timestamp":"2014-04-18T08:26:24Z","content_type":null,"content_length":"30067","record_id":"<urn:uuid:784f1d5d-991b-43ac-9a65-849822f02de0>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00371-ip-10-147-4-33.ec2.internal.warc.gz"} |
Bristol Cryptography Blog
Today we had two papers at CCS'10 introducing new, Turing complete languages. The second one is "Return-Oriented Programming Without Returns" by Stephen Checkoway, Lucas Davi, Alexandra Dmitrienko,
Ahmad-Reza Sadeghi, Hovav Shacham and Marcel Winandy and extends the concept of return-oriented programming into "jump-oriented programming" that uses jump instructions instead of return instructions
to build gadgets and this has severe security implications as the authors showed at the examples of x86 processors and Android based devices running on ARM chips. But the first paper
"Platform-Independent Program" by Sang Kil Cha, Brian Pak, David Brumley and Richard J. Lipton was even more impressive.
However, before I continue to write about the paper, I should give a short explanation of Turing complete languages and why these are important: In 1937, a few years before the first programmable
computer was built, the mathematician Alan Turing invented the concept of a Turing machine to prove that a universal (or programmable) machine can be built that can solve any computable problem.
Although no universal Turing machine will ever be built since it requires infinite memory, this is basically the most important result of computer science. If you take a set of instructions which are
sufficient to simulate such a Turing machine (with exception of the infinite memory), this set is called "Turing complete". This certainly is not the biggest deal in the world since all modern
processors have Turing complete instruction sets. And indeed, in both papers, the Turing completeness of the languages is only used to prove that their languages do not lack fundamental concepts.
So let me now explain what's so special about the language introduced in "Platform-Independent Program". Commonly the instruction sets of two different processors overlap to a certain extent but are
not equal; a program for x86 processors will never run on an ARM processor and vice versa. So the authors of the paper started looking at the overlap of the instruction sets to find jump instructions
that will have the following effect:
• If executed on platform a, jump to address x.
• If executed on platform b, jump to address y.
Now they can place instructions for platform a at position x and instructions for platform b at position y. Out of such short code sequences the authors build gadgets and all the gadgets together
form a turing complete language. (The instructions at x and y do not have to have the same effect; on platform a the program might be a harmless desktop gimmick, on platform b it might be malware.)
The really amazing thing about this is, that (to my knowledge) this is the first language that is at least semi-platform independent but does not require a virtual machine such as Java or an
interpreter to achieve platform independence. (It still needs to have enough overlap in the instruction sets.) | {"url":"http://bristolcrypto.blogspot.com/2010/10/turing-at-ccs10.html","timestamp":"2014-04-21T05:17:44Z","content_type":null,"content_length":"74858","record_id":"<urn:uuid:ffcaa379-a6b9-4dc2-ac42-8d065561e7c2>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00228-ip-10-147-4-33.ec2.internal.warc.gz"} |
Lloyd N. Trefethen, October 2009, revised February 2011
1.1 What is a chebfun?
A chebfun is a function of one variable defined on an interval [a,b]. The syntax for chebfuns is almost exactly the same as the usual Matlab syntax for vectors, with the familiar Matlab commands for
vectors overloaded in natural ways. Thus, for example, whereas sum(f) returns the sum of the entries when f is a vector, it returns a definite integral when f is a chebfun.
Chebfun with a capital C is the name of the software system.
The aim of Chebfun is to "feel symbolic but run at the speed of numerics". More precisely our vision is to achieve for functions what floating-point arithmetic achieves for numbers: rapid computation
in which each successive operation is carried out exactly apart from a rounding error that is very small in relative terms [Trefethen 2007].
The implementation of Chebfun is based on the mathematical fact that smooth functions can be represented very efficiently by polynomial interpolation in Chebyshev points, or equivalently, thanks to
the Fast Fourier Transform, by expansions in Chebyshev polynomials. For a simple function, 20 or 30 points often suffice, but the process is stable and effective even for functions complicated enough
to require 1000 or 1,000,000 points. Chebfun makes use of adaptive procedures that aim to find the right number of points automatically so as to represent each function to roughly machine precision
(about 15 digits of relative accuracy).
The mathematical foundations of Chebfun are for the most part well established by results scattered throughout the 20th century. A key early figure, for example, was Bernstein in the 1910s.
Nevertheless it is hard to find the relevant material collected in one place. A new reference on this subject will be the Chebfun-based book [Trefethen 2013].
Chebfun was originally created by Zachary Battles and Nick Trefethen at Oxford during 2002-2005 [Battles & Trefethen 2004]. Battles left the project in 2005, and soon four new members were added to
the team: Ricardo Pachon (from 2006), Rodrigo Platte (from 2007), and Toby Driscoll and Nick Hale (from 2008). Beginning in 2009, Asgeir Birkisson and Mark Richardson also became involved. Additional
contributors from Oxford and elsewhere include Phil Assheton, Folkmar Bornemann, Pedro Gonnet, Tom Maerz, Sheehan Olver, Simon Scheuring, Alex Townsend, and Joris Van Deun.
This Guide is based on Chebfun Version 4, released in 2010. Chebfun is available at http://www.maths.ox.ac.uk/chebfun/.
1.2 Constructing simple chebfuns
The "chebfun" command constructs a chebfun from a specification such as a string or an anonymous function. If you don't specify an interval, then the default interval [-1,1] is used. For example, the
following command makes a chebfun corresponding to cos(20x) on [-1,1] and plots it.
f = chebfun('cos(20*x)');
From this little experiment, you cannot see that f is represented by a polynomial. One way to see this is to find the length of f:
ans =
Another is to remove the semicolon that suppresses output:
f =
chebfun column (1 smooth piece)
interval length endpoint values
[ -1, 1] 49 0.41 0.41
vertical scale = 1
These results tell us that f is represented by a polynomial interpolant through 49 Chebyshev points, i.e., a polynomial of degree 48. These numbers have been determined by an adaptive process. We can
see the data points by plotting f with the '.-' option:
The formula for N+1 Chebyshev points in [-1,1] is
x(j) = -cos(j pi/N) , j = 0:N,
and in the figure we can see that the points are clustered accordingly near 1 and -1. Note that in the middle of the grid, there are about 5 points per wavelength, which is evidently what it takes to
represent this cosine to 15 digits of accuracy. For intervals other than [-1,1], appropriate Chebyshev points are obtained by a linear scaling.
The curve between the data points is the polynomial interpolant, which is evaluated by the barycentric formula introduced by Salzer [Berrut & Trefethen 2004, Salzer 1972]. This method of evaluating
polynomial interpolants is stable and efficient even if the degree is in the millions [Higham 2004].
What is the integral of f from -1 to 1? Here it is:
ans =
This number was computed by integrating the polynomial (Clenshaw-Curtis quadrature -- see Section 2.1), and it is interesting to compare it to the exact answer from calculus:
exact = sin(20)/10
exact =
Here is another example, now with the chebfun defined by an anonymous function instead of a string. In this case the interval is specified as [0,100].
g = chebfun(@(t) besselj(0,t),[0,100]);
plot(g), ylim([-.5 1])
The function looks complicated, but it is actually a polynomial of surprisingly small degree:
ans =
Is it accurate? Well, here are three random points in [0,100]:
x = 100*rand(3,1)
x =
Let's compare the chebfun to the true Bessel function at these points:
exact = besselj(0,x);
error = g(x) - exact;
[g(x) exact error]
ans =
0.007264225965188 0.007264225965187 0.000000000000000
0.079786547125176 0.079786547125176 -0.000000000000000
0.063644017257148 0.063644017257148 0
If you want to know the first 5 zeros of the Bessel function, here they are:
r = roots(g); r = r(1:5)
r =
Notice that we have just done something nontrivial and potentially useful. How else would you find zeros of the Bessel function so readily? As always with numerical computation, we cannot expect the
answers to be exactly correct, but they will usually be very close. In fact, these computed zeros are accurate to close to machine precision:
ans =
1.0e-14 *
Most often we get a chebfun by operating on other chebfuns. For example, here is a sequence that uses plus, times, divide, and power operations on an initial chebfun "x" to produce a famous function
of Runge:
x = chebfun('x');
f = 1./(1+25*x.^2);
clf, plot(f)
ans =
1.3 Operations on chebfuns
There are more than 200 commands that can be applied to a chebfun. For a list of many of them you can type "methods":
methods chebfun
Methods for class chebfun:
abs coth isreal rank
acos cov jacpoly ratinterp
acosd csc jacreset rdivide
acosh cscd jump real
acot csch ldivide reallog
acotd ctranspose le realsqrt
acoth cumprod legpoly rem
acsc cumsum length remez
acscd define log repmat
acsch diag log10 residue
airy diff log1p restrict
all dirac log2 roots
and display loglog round
angle domain lt sec
any ellipj maps secd
area end max sech
asec eq mean semilogx
asecd erf merge semilogy
asech erfc mesh set
asin erfcinv min sign
asind erfcx minandmax simplify
asinh erfinv minus sin
atan exp mldivide sind
atan2 expm1 mod sinh
atand feval movie size
atanh fill mrdivide sound
besselj find mtimes spy
bvp4c fix ne sqrt
bvp5c fliplr newID std
bvpsc flipud newdomain subsasgn
cat floor norm subspace
ceil fred not subsref
cf fzero null sum
chebellipseplot ge ode113 surf
chebfun get ode15s svd
chebpade getdepth ode45 tan
chebpoly gmres or tand
chebpolyplot gt orth tanh
chebtune heaviside pde15s times
comet horzcat pinv transpose
comet3 hypot plot uminus
complex imag plot3 unwrap
compress integral plus uplus
cond interp1 poly vander
conj inv polyfit var
conv inv2 power vertcat
cos isempty prod volt
cosd isequal qr waterfall
cosh isfinite quad why
cot isinf quantumstates xor
cotd isnan range
(Futher commands can be unearthed with methods linop, methods chebop, and methods domain.) To find out what a command does, you can use "help".
help chebfun/chebpoly
CHEBPOLY Chebyshev polynomial coefficients
A = CHEBPOLY(F) returns the vector of coefficients such that
F_1 = A(1) T_M(x) + ... + A(M) T_1(x) + A(M+1) T_0(x), where T_M(x) denotes
the M-th Chebyshev polynomial and F_1 denotes the first fun of chebfun F.
A = CHEBPOLY(F,I) returns the coefficients for the I-th fun.
A = CHEBPOLY(F,I,N) truncates or pads the vector A so that N coefficients
of the fun F_I are returned. However, if I is 0 then the global coefficients
of the *chebfun* F are returned (by computing relevent inner products with
Chebyshev polynomials).
C = CHEBPOLY(F,...,'kind',2) returns the vector of coefficients for the
Chebyshev expansion of F in 2nd-kind Chebyshev polynomials
F_1 = C(1) U_M(x) + ... + C(M) U_1(x) + C(M+1) U_0(x)
There is also a CHEBPOLY command in the chebfun trunk directory, which
computes the chebfun corresponding to the Chebyshev polynomial T_n.
Most of the commands in the list exist in ordinary Matlab; some exceptions are "domain", "restrict", "chebpoly", "define", and "remez". We have already seen "length" and "sum" in action. In fact we
have already seen "subsref" too, since that is the Matlab command for (among other things) evaluating arguments in parentheses. Here is another example of its use:
ans =
Here for comparison is the true result:
ans =
In this Runge function example, we have also implicitly seen "times", "plus", "power", and "rdivide", all of which have been overloaded from their usual Matlab uses to apply to chebfuns.
In the next part of this tour we shall explore many of these commands systematically. First, however, we should see that chebfuns are not restricted to smooth functions.
1.4 Piecewise smooth chebfuns
Many functions of interest are not smooth but piecewise smooth. In this case a chebfun may consist of a concatenation of smooth pieces, each with its own polynomial representation. Each of the smooth
pieces is called a "fun", and funs are implemented as a subclass of chebfuns. This enhancement of Chebfun was developed initially by Ricardo Pachon during 2006-2007, then also by Rodrigo Platte
starting in 2007 [Pachon, Platte and Trefethen 2009]. Essentially funs are the "classic chebfuns" for smooth functions on [-1,1] originally implemented by Zachary Battles in Chebfun Version 1.
Later we shall describe the options in greater detail, but for the moment let us see some examples. One way to get a piecewise smooth function is directly from the constructor, taking advantage of
its capability of automatic edge detection. For example, in the default "splitting off" mode a function with a jump in its derivative produces a warning message,
f = chebfun('abs(x-.3)');
Warning: Function not resolved, using 65537 pts. Have you tried 'splitting on'?
The same function can be successfully captured with splitting on:
f = chebfun('abs(x-.3)','splitting','on');
The "length" command reveals that f is defined by four data points, namely two for each linear interval:
ans =
We can see the structure of f in more detail by typing f without a semicolon:
f =
chebfun column (2 smooth pieces)
interval length endpoint values
[ -1, 0.3] 2 1.3 0
[ 0.3, 1] 2 0 0.7
Total length = 4 vertical scale = 1.3
This output confirms that f consists of two funs, each defined by two points and two corresponding function values. We can see the structure from another angle with "struct", Matlab's command for
seeing the various fields within an object:
Warning: Calling STRUCT on an object prevents the object from hiding its
implementation details and should thus be avoided. Use DISP or DISPLAY to see
the visible public details of an object. See 'help struct' for more information.
ans =
funs: [1x2 fun]
nfuns: 2
ends: [-1 0.300000000000000 1]
scl: 1.309997215582586
imps: [1.300000000000000 0 0.700000000000000]
trans: 0
jacobian: [1x1 anon]
ID: [47 73495852651]
funreturn: 0
This output again shows that f consists of two funs with breakpoints at -1, 1, and a number very close to 0.3. The "imps" field refers to "impulses", which relate to values at breakpoints, including
possible information related to delta functions, discussed in Section 2.4. The "trans" field is 0 for a column chebfun and 1 for a row chebfun (Section 1.6 and Chapter 6). The "jacobian" and "ID"
fields are used for automatic differentiation (Chapter 10).
Another way to make a piecewise smooth chebfun is to construct it explicitly from various pieces. For example, the following command specifies three functions x^2, 1, and 4-x, together with a vector
of endpoints indicating that the first function applies on [-1,1], the second on [1,2], and the third on [2,4]:
f = chebfun('x.^2',1,'4-x',[-1 1 2 4]);
We expect f to consist of three pieces of lengths 3, 1, and 2, and this is indeed the case:
f =
chebfun column (3 smooth pieces)
interval length endpoint values
[ -1, 1] 3 1 1
[ 1, 2] 1 1 1
[ 2, 4] 2 2 0
Total length = 6 vertical scale = 2
Our eyes see pieces, but to Chebfun, f is just another function. For example, here is its integral.
ans =
Here is an algebraic transformation of f, which we plot in another color for variety.
Some Chebfun commands naturally introduce breakpoints in a chebfun. For example, the "abs" command first finds zeros of a function and introduces breakpoints there. Here is a chebfun consisting of 6
f = abs(exp(x).*sin(8*x));
And here is an example where breakpoints are introduced by the "max" command, leading to a chebfun with 13 pieces:
f = sin(20*x);
g = exp(x-1);
h = max(f,g);
As always, h may look complicated to a human, but to Chebfun it is just a function. Here are its mean, standard deviation, minimum, and maximum:
ans =
ans =
ans =
ans =
A final note about piecewise smooth chebfuns is that the automatic edge detection or "splitting" feature, when it is turned on, may subdivide functions even though they do not have clean point
singularities, and this may be desirable or undesirable depending on the application. For example, considering sin(x) over [0,1000] with splitting on, we end up with a chebfun with many pieces:
tic, f = chebfun('sin(x)',[0 1000*pi],'splitting','on'); toc
Elapsed time is 0.686400 seconds.
Warning: Calling STRUCT on an object prevents the object from hiding its
implementation details and should thus be avoided. Use DISP or DISPLAY to see
the visible public details of an object. See 'help struct' for more information.
ans =
funs: [1x32 fun]
nfuns: 32
ends: [1x33 double]
scl: 17.186120489749236
imps: [1x33 double]
trans: 0
jacobian: [1x1 anon]
ID: [68 73495852651]
funreturn: 0
In this case it is more efficient -- and more interesting mathematically -- to omit the splitting and construct one global chebfun:
tic, f2 = chebfun('sin(x)',[0 1000*pi]); toc
Elapsed time is 0.023636 seconds.
Warning: Calling STRUCT on an object prevents the object from hiding its
implementation details and should thus be avoided. Use DISP or DISPLAY to see
the visible public details of an object. See 'help struct' for more information.
ans =
funs: [1x1 fun]
nfuns: 1
ends: [0 3.141592653589793e+03]
scl: 0.999999992552489
imps: [0 -3.214166459275634e-13]
trans: 0
jacobian: [1x1 anon]
ID: [69 73495852651]
funreturn: 0
In a chebfun computation, splitting can be turned on and off freely to handle different functions appropriately. The default or "factory" value is splitting off; see Chapter 8.
1.5 Infinite intervals and infinite function values
A major change from Chebfun Version 2 to Version 3 was the generalization of chebfuns to allow certain functions on infinite intervals or which diverge to infinity: the credit for these innovations
belongs to Nick Hale, Rodrigo Platte, and Mark Richardson. For example, here is a function on the whole real axis,
f = chebfun('exp(-x.^2/16).*(1+.2*cos(10*x))',[-inf,inf]);
and here is its integral:
ans =
Here's the integral of a function on [1,inf]:
sum(chebfun('1./x.^4',[1 inf]))
ans =
Notice that several digits of accuracy have been lost here. Be careful! -- operations involving infinities in Chebfun are not always as accurate and robust as their finite counterparts.
Here is an example of a function that diverges to infinity, which we can capture by including the flag 'blowup 2' (try help blowup for details):
h = chebfun('(1/pi)./sqrt(1-x.^2)','blowup',2);
In this case the integral comes out just right:
ans =
For more on the treatment of infinities in Chebfun, see Chapter 9.
1.6 Rows, columns, and quasimatrices
Matlab doesn't only deal with column vectors: there are also row vectors and matrices. The same is true of Chebfun. The chebfuns shown so far have all been in column orientation, which is the
default, but one can also take the transpose, compute inner products, and so on:
x = chebfun('x')
x =
chebfun column (1 smooth piece)
interval length endpoint values
[ -1, 1] 2 -1 1
vertical scale = 1
ans =
chebfun row (1 smooth piece)
interval length endpoint values
[ -1, 1] 2 -1 1
vertical scale = 1
ans =
One can also make matrices whose columns are chebfuns or whose rows are chebfuns, like this:
A = [1 x x.^2]
A =
chebfun column 1 (1 smooth piece)
interval length endpoint values
[ -1, 1] 1 1 1
vertical scale = 1
chebfun column 2 (1 smooth piece)
interval length endpoint values
[ -1, 1] 2 -1 1
vertical scale = 1
chebfun column 3 (1 smooth piece)
interval length endpoint values
[ -1, 1] 3 1 1
vertical scale = 1
ans =
2.000000000000000 -0.000000000000000 0.666666666666667
-0.000000000000000 0.666666666666667 0.000000000000000
0.666666666666667 0.000000000000000 0.400000000000000
These are called "quasimatrices", and they are discussed in Chapter 6.
1.7 How this Guide is produced
This guide is produced in Matlab using the "publish" command. The formatting is rather simple, not relying on TeX features or other fine points of typesetting. To publish a chapter for yourself, make
sure the chebfun guide directory is in your path and then type, for example, "open(publish('guide1'))". Before publishing, we recommend executing "guidedefaults".
1.8 References
[Battles & Trefethen 2004] Z. Battles and L. N. Trefethen, "An extension of Matlab to continuous functions and operators", SIAM Journal on Scientific Computing 25 (2004), 1743-1770.
[Berrut & Trefethen 2005] J.-P. Berrut and L. N. Trefethen, "Barycentric Lagrange interpolation", SIAM Review 46 (2004), 501-517.
[Higham 2004] N. J. Higham, "The numerical stability of barycentric Lagrange interpolation", IMA Journal of Numerical Analysis 24 (2004), 547-556.
[Pachon, Platte & Trefethen 2009] R. Pachon, R. B. Platte and L. N. Trefethen, "Piecewise smooth chebfuns", IMA J. Numer. Anal., 2009.
[Salzer 1972] H. E. Salzer, "Lagrangian interpolation at the Chebyshev points cos(nu pi/n), nu = 0(1)n; some unnoted advantages", Computer Journal 15 (1972), 156-159.
[Trefethen 2007] L. N. Trefethen, "Computing numerically with functions instead of numbers", Mathematics in Computer Science 1 (2007), 9-19.
[Trefethen 201£] L. N. Trefethen, Approximation Theory and Approximation Practice, SIAM, publication expected in 2013. | {"url":"http://www.mathworks.com/matlabcentral/fileexchange/23972-chebfun/content/chebfun/guide/html/guide1.html","timestamp":"2014-04-18T03:28:23Z","content_type":null,"content_length":"147536","record_id":"<urn:uuid:28a2c6e7-3b40-4d6d-beaf-980000300e99>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00086-ip-10-147-4-33.ec2.internal.warc.gz"} |
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How do Vegas oddsmakers make their odds?
August 1, 2011 12:51 AM Subscribe
How do Vegas oddsmakers make their odds?
Do the odds in Vegas come out of actual predictions about who will win, or are they more of a prediction of how people will bet? If someone is a 4-1 favorite, is that because Vegas actually thinks
they're a 4-1 favorite, or they just think that will result in the most even betting? Is it different for different sports?
posted by malhouse to Sports, Hobbies, & Recreation (8 answers total) 3 users marked this as a favorite
I think the answer is "both"
The betting lines start as a guess and get validated by the masses. The house is incentivized to minimize payouts. They do that by offsetting each side of the bet against the other while making money
off the rake in the middle. The closer they get to even odds (as determined by the masses) the higher their rake will be with respect to any discrepancy to one side or the other of the line.
The market forces this. Without it, there would be arbitrage opportunities, right?
Betting is different for horse racing than it is for boxing which is different from football... but I think the line creation is similarly dynamic and "even".
posted by milqman at 1:22 AM on August 1, 2011
Bookmakers don't usually have better information than anyone else, and they don't like to gamble. They want to take in a mixture of bets at odds that ensure they will make money no matter who wins.
If their "book" is unbalanced in favor of one or more contestants then they can try to lay some of those bets with other bookmakers, or they can lower the odds on the popular contestants to encourage
bets on the less popular ones.
posted by Joe in Australia at 3:21 AM on August 1, 2011
Oddsmakers care less about being right then they care about eliminating risk. Odds are designed to put even money on both sides of the bet. Parimutuel betting (like for horse racing) is by definition
even odds, but spread betting needs to be balanced via the odds.
Inside Sports the HBO show had a segment years ago on how Vegas sets NFL odds. Its probably out of date know, but basically it came down to a bunch of oddsmakers sitting in room crunching numbers,
and then there was a select group of whales who got the first call with the official odds. If those whales jumped at a line the Vegas guys knew they were off, and so they adjusted the line before
releasing them to the public. I mean there are some well known ways in which gambling lines are off. For example Americans love scoring and the favorites - so if you bet the under and the dog on NFL
games you'll win >50%, although not enough to make up the fees you pay to book.
posted by JPD at 4:41 AM on August 1, 2011
All formal betting seeks to create a balanced market. How this is done depends on what the "proposition" is. Proposition being the thing being bet on.
Take a simple boxing match, where the outcome is one guy wins or the other guy wins. (The proposition being "who will win the match?") The house will say "we are now taking bets on the match". As
bets pour in, it becomes clear that the first guy is a favorite to win. In fact, twice as many dollars are on that one side. The odds, as determined by the market are 2:1. The house doesn't like to
be in this position, because if the other guy wins, they have to pay out more than they took in. So they adjust the odds and start taking 2:1 bets. Meaning, if you bet $2 on the favorite, you get
your bet back plus $1. If you bet $1 on the underdog, you get your $1 plus $2.
(If you are a visual thinker, picture a see-saw, with the dollars being bet on either boxer on each side. The house adjusts the fulcrum point so that each side is balanced.)
As JPD says, they will often research the marketplace and set the odds to something besides 1:1 ahead of time so they don't get too far out of balance.
Spread betting is the same thing, but instead of odds, they use the point spread. Which is the difference between the winning team's score and the losing team's score. If the teams are believed to be
equally matched, there is no spread and it is a simple win or loss proposition. But more often, one team gets favored over the other. What spread betting does is basically spots the losing team a
certain number of points so that the matchup seems even. If the spread is 3.5 (almost always set to an impossible score, so that there are no ties), the proposition becomes "if you add 3.5 to the
underdog's score, who would be the winner?". The spread gets adjusted as bets pour in to adjust for changing optimism for one side or the other.
(This is how point shaving got invented. This is where a corrupt gambler pays a player on the team that is favored to win the game to try and make sure that team wins by less than the spread. It
takes advantage of the asymmetry of the incentives of the better versus the player. If done correctly, the player and his team still get to win the game, but the player gets a payoff, and the gambler
wins his bet.)
Finally, there is pari mutuel betting like in horse racing. This is where there are more than two contestants. At its simplest, betters place bets on which horse will win the race. Those bets are
placed in buckets for each horse, and the odds are the inverse of the amount of money bet on a particular horse over the total amount of money bet (the pool).
Simple example: three horses are racing, and $30 says horse A will win. $20 says B will win. $10 says C will win. There is $60 in the pool. The odds for horse A are 1:1. You bet a dollar, you win a
dollar. The odds for horse B are 1:2. You bet a dollar, you win $2. Horse C is 1:5. You bet $1, you win $5.
It is much more complicated than that in real life, because there are other things to bet on. You can bet that a horse will come in different places, or on a particular set of horses winning in a
particular order. But the mechanics are the same: the pool gets split up based on how much money was bet in total, versus what percentage of bettors chose the right scenario.
posted by gjc at 5:29 AM on August 1, 2011 [1 favorite]
You get a good sense of how odds-setting and hedging works by watching the on-course bookies at a horse racing meeting. Punters are in search of the best odds on a particular horse, and bookies will
try to attract bets in a distributed way: one may offer a better return on horse A, and a slightly worse one on horse B, while another may offer the reverse.
At the same time, they want to minimise potential losses, which means checking the odds offered by other bookies, adjusting them to reflect the market, and also placing bets with the on-course
competition or with the Tote/parimuteul booth.
They've all got laptops now to work out what needs to be placed where in order to avoid a bad loss, which makes me think about how that used to be done with scratchpads, mental arithmetic and the
instinct that comes from a ton of experience.
posted by holgate at 7:08 AM on August 1, 2011
Take a simple boxing match, where the outcome is one guy wins or the other guy wins. (The proposition being "who will win the match?") The house will say "we are now taking bets on the match".
This is incorrect. If anyone did that, people would bet every cent they could on obvious favorites.
Floyd Mayweather Jr., who is fighting Victor Ortiz, opened as an 8:1 favorite based on oddsmakers' thoughts that he is by far the better boxer.
posted by ambient2 at 11:58 AM on August 1, 2011 [1 favorite]
I don't know for sure, but I do seem to recall reading (in
The Wisdom of Crowds
by Jason? Surowiecki) that the oddsmakers don't really care who wins (maybe at a personal level occasionally, but for the purposes of their job, no). They make money from the vig, and how they do
that is have the line so that the people betting on either side balance each other out. Of course, I imagine that quite often, the oddsmakers themselves will agree with majority of the bettors, but
that should be irrelevant. The exact mechanics of this I don't know about, but to answer your question, oddsmakers care about how people will bet, not what will actually happen. Now as it happens,
the line does tend to correlate pretty well with how things actually turn out, as I recall, but that's a different story.
I don't know for sure what I'm saying, but I'm pretty sure that's what I read. You can look up the book if you want to know more.
posted by Busoni at 1:00 PM on August 1, 2011
The other point to make (and I'm guilty of this as well in my comment) is that in contests where the betting is based on odds, they don't want equal money on both sides of the book, they just want
the expected payouts to be the same. If a fight is going off at 8:1 you want 800 on the favorites side of the book and 100 on the underdogs side of the book.
When its a spread they want a balanced book.
posted by JPD at 1:09 PM on August 1, 2011
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What is the displacement in cubic inches of a 6.6 liter engine?
The cubic inch is a unit of measurement for volume in the Imperial units and United States customary units systems. It is the volume of a cube with each of its three dimensions (length, width, and
depth) being one inch long.
The cubic inch and the cubic foot are still used as units of volume in the United States, although the common SI units of volume, the liter, milliliter, and cubic meter, are also used, especially in
manufacturing and high technology.
Engine displacement is the volume swept by all the pistons inside the cylinders of a reciprocating engine in a single movement from top dead centre (TDC) to bottom dead centre (BDC). It is
commonly specified in cubic centimetres (cc), litres (l), or (mainly in North America) cubic inches (CID). Engine displacement does not include the total volume of the combustion chamber.
A metric engine is an American expression which refers to an internal combustion engine, often for automobiles, whose underlying engineering design is based on a metric system of units,
particularly SI.
As American industry converted from traditional units to SI in the late 20th century, the automotive industry responded by transitioning its auto and engine designs to be "metric" rather than
Disaster Accident
Social Issues
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Projection of Planes - Engineering Drawing - Learn Engineering Drawing Online
Problem 5.10 Projection of Planes – An isosceles triangle plate ABC having its base 50 mm and altitude 90 mm resting on H.P. on its base. The isosceles triangle is inclined at an angle 40° to the
H.P. And the altitude in the top view is inclined at the angle 60° to the V.P. Draw the projections.
Problem 5.9 Projection of Planes
Problem 5.9 Projection of Planes – A circular plate of diameter 80 mm is resting on a point of its periphery on H.P. such that it makes an angle of 40° to the H.P. And the diameter passing through
the point of its resting on H.P. makes an angle of 60° with V.P. Draw the projections of it.
Problem 5.8 Projection of Planes
Problem 5.8 Projection of Planes – A pentagonal plate of side 60 mm is held on V.P. on one of its corner. The edge opposite to that corner makes an angle of 25° with the H.P. The flat surface of
pentagon is inclined at 40° to the V.P. Draw the projections.
Problem 5.7 Projection of Planes
Problem 5.7 Projection of Planes – A square lamina of side 80 mm rests on a corner on H.P. and it is inclined with H.P. such that its plan is a rhombus with a diagonal of 40 mm. The long diagonal is
inclined with the V.P. at 45°. Determine its inclination with H.P. and draw it’s projections.
Problem 5.6 Projection of Planes
Problem 5.6 Projection of Planes – An elliptical plate of dimension 140 mm X 80 mm rests on a point of its periphery on H.P. and inclined with it such that its plan is a circle. Draw the projection
and find out the inclination of it with H.P. | {"url":"http://www.engineeringdrawing.org/category/projection_of_planes/","timestamp":"2014-04-16T22:06:53Z","content_type":null,"content_length":"36639","record_id":"<urn:uuid:d7e06592-8b93-4009-8be4-24bcfc9da550>","cc-path":"CC-MAIN-2014-15/segments/1397609525991.2/warc/CC-MAIN-20140416005205-00141-ip-10-147-4-33.ec2.internal.warc.gz"} |
Summary: Rings and Algebras Problem set #6: Solutions Oct. 20, 2011.
1. Which of the following modules are directly indecomposable: ZZ, ZQ, ZR, ZQ[x], ZZp ,
Q[x]Q[x], for V a finite dimensional vector space (V ) (V ), for a finite graph without
oriented cycles KK and KKe1.
Solution. ZZ and ZQ are indecomposable because they don't have disjoint nonzero submodules. ZR and ZQ[x]
are decomposable because they decmpose as infinite dimensional Q-vectorspaces. ZZp is indecomposable
because it has a unique smallest nonzero submodule. Q[x]Q[x] is indecomposable because there are no disjoint
ideals in it (Q[x] has no zero-divisors). (V ) (V ) is indecomposable since it is local: the quotient modulo its
radical is isomorphic to the base field. The path algebra KK is indecomposable if and only if the graph
has one vertex, since the idempotents corresponding to the vertices give a decomposition of the path algebra.
Finally, KKe1 is indecomposable since its endomorphism ring is isomorphic to e1KKe1 and this is a local
ring: actually, paths of nonzero length which start and end at 1 do not exist, thus the endomorphism ring is
the span of e1, hence it is one dimensional.
2. Which of the following statements are true?
a) The submodule of a directly indecomposable module is directly indecomposable.
b) The homomorphic image of a directly indecomposable module is directly indecomposable.
c) If the modules RM and SN have isomorphic submodule lattices then M is indecomposable
if and only if N is indecomposable.
Solution. a) The statement is false. Take the algebra A =
K 0 0 | {"url":"http://www.osti.gov/eprints/topicpages/documents/record/561/4772787.html","timestamp":"2014-04-19T10:26:39Z","content_type":null,"content_length":"8725","record_id":"<urn:uuid:6f000b68-4342-4b44-9a50-dfde7a3b0346>","cc-path":"CC-MAIN-2014-15/segments/1398223203422.8/warc/CC-MAIN-20140423032003-00063-ip-10-147-4-33.ec2.internal.warc.gz"} |
Uniform average rate of occurence
April 8th 2013, 07:12 AM #1
Apr 2013
Uniform average rate of occurence
A student is collecting data on traffic arriving at a motorway service station during weekday lunchtimes. The random variable X denotes the number of cars arriving in a randomly chosen period of
ten seconds. Is a uniform average rate of occurrence a valid assumption in this scenario and why?
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Determining Wind Values And Making Your Shots
by Maj. John Plaster | September 23rd, 2010 6
In order to interpret and apply wind compensation correctly, you have to determine the angle of the wind; how it flows across the bullet will determine the amount of drift. A tail wind or head wind
will have no value; they have essentially no effect on a bullet’s flight. A direct crosswind, which blows from 90 degrees into the path of the bullet, is called a “full” wind because the full effect
of the wind is experienced.
An oblique wind of 45 degrees, from right or left, has not a one-half value, but a three-quarters value. It has a 75 percent effect, even though the angle is only halfway between no effect and full
effect. Most shooters initially have trouble getting this straight in their heads. The effect is not proportional because of the aerodynamics of a bullet in flight. Just remember that halfway between
full and zero effect is three-quarters. Benchrest shooters use even finer values and split the wind for exact aiming. I’ve included this to give you a better feel for how quickly the wind has an
effect once a bullet is other than at tail or head. Once it’s just 15 degrees right or left, already a quarter of the wind value must be used when compensating.
Shooting into the Wind
To shoot accurately into a wind, compensate by holding or aiming in the direction the wind is coming from. As the bullet travels downrange, it drifts into your target. In order for this to work,
however, you must know exactly how far to compensate.
The accompanying ballistic tables show wind drift for several police and military sniper loads, which you can compare to your favorite hunting loads. Although several wind speeds are listed, the most
important is the 10 mph listing, I believe, because once memorized, it’s easiest to compute in your head. Just about anything can be divided or multiplied when you start with a factor of 10.
Note that compensation doubles as wind speed doubles–the necessary compensation for a 20 mph wind is twice that of a 10 mph wind, and five mph is half that of 10 mph. But the differences in distances
are not proportional: compensation for 600 yards is much more than twice that of 300 yards. This is because the farther the bullet goes, the more it slows down and the worse the effect becomes. In a
way, this is similar to how a bullet starts to plunge at long range, when its path becomes a sharp arc.
But now, at last, we’re ready to bring together ballistic data and wind values and compensation. It’s really quite simple.
First, determine which direction the wind is blowing in respect to a line between you and your target. For the sake of illustration, let’s say it’s 90 degrees and, as already seen, that would make it
a “full” wind. Next, determine the speed of that wind–we’ll say it’s five mph. Finally, you estimate your target is 600 yards away. You’re using Federal .308 BTHP Match.
Looking at the table, you find that the required compensation is 16.1 inches. The compensation on all the tables reflects a full value. Since your scope has an adjustable windage knob, you dial in
the equivalent of 16.1 inches at 600 yards; since 1 MOA equals six inches at that range, you rotate it 2.75 MOA into the wind. And because your scope has 1⁄4 MOA positive clicks, you turn it 11
clicks. Having made the adjustment, you aim dead-on, let off a good shot, and score a perfect hit.
If your scope lacked a windage knob, you would have looked at the target, determined that 16.1 inches is the width of a fit man at the hip, and held this far into the wind, aimed and engaged, again
with perfect results.
But what about other than full-value crosswinds? Just factor in the value when determining the compensation. Let’s try another example. You know the wind is 15 mph, coming on at a 45-degree oblique,
and the range is 800 yards. Again you’re using a Federal .308 Match ammo. The table says full compensation would be 96.1 inches, but we will only use three-quarters of that because the wind is
oblique at 45 degrees. Three-quarters of 96 inches is 72 inches. So, if you have a windage knob you realize that 1 MOA equals eight inches at 800 yards; therefore, you divide 72 by eight, which
equals nine, and you click off nine MOA on your scope, or 36 clicks. On another scope, you’d hold into the wind what you estimate to be 72 inches from your target–about the height of a man.
Where shooting into the wind gets tricky is when it’s gusting or you must deal with several winds. Old-time shooters will tell you not to wait for pauses during a steady wind, that you’ll have much
better results shooting into a predictable wind than hoping a short calm lasts long enough for your bullet to reach the target.
Strong gusts require timing your shot. When faced by two winds, try to time your shot so it’s fired during the slower or the least gusting or the farther wind so there’s less effect and a more
predictable outcome. (This is getting pretty complex, but the reason you prefer shooting through a farther wind is that there’s less remaining flight time to be affected by the wind.)
When shooting in mountains, don’t be concerned by updrafts. Where they exist, these winds are very shallow and your bullet will pass through too quickly to make much difference.
USMC Wind Adjustment Method
For those of you having a boundless desire for more information, I’ve included an old U.S. Marine Corps method for computing sight changes when firing in the wind. The USMC has been using this
windage adjustment method since the days of the 1903-A3 Springfield.
After determining wi
nd direction and speed, use the following formula:
Range in 100 Yds. x Speed in MPH/15 (math constant)= MOA Windage
For instance, your target is 300 yards away, and there’s a 10 MPH wind:
3 x 10 = 30/15 = 2 MOA
Click-in the two minutes of angle in the direction of the wind and aim dead-on. This is a great formula–except it’s only accurate at 500 yards or less. When your target is farther, the mathematical
constant must increase, as shown below:
600 Yards: Divide by 14
700 Yards: Divide by 13
800 Yards: Divide by 13
900 Yards: Divide by 12
1,000 Yards: Divide by 11
Editor’s Note: With slight modifications, this column was excerpted from the author’s book, The Ultimate Sniper (Paladin Press, 1993; 303/443-7250.)
• carlton
what is a math constant and please give a detailed explation in the formula with a visual formula if possible
□ Jeff
Math is constant means you use the same formula. So look at it like this R= range in yds
S= wind speed in MPH divied by 15 in this formula 15 is a constant value used every time. The only thing that changes is the range and wind values
so you plug in your values. You are shooting at 200 yard range R= 300 (drop the 0's for the math) S= Speed that's a is a SWAG scientific wild ass guess. It's practice judgin speed. Look at
the grass and see the angle of it look at a weather app and learn from that. Let's say you call it 10 MPH.
So you math looks like this R=2 S=10 the formula R x S divided by15 would be 3×10= 30 /15 = 2 Minutes of angle adjustment. SO if you have a scope that has 1/4 moa adjustment you would need 8
clicks to the left or right depending on wind direction to get 2 MOA of adjustment outta your scope. If it's 1/2 moa it's 4 clicks etc. That's as easy as I can explain it. GO practice
☆ jeff
I am sorry I started using 2 and went to 300yd for sake of easy math. So the R=3 (drop the 0's)xS=10/15=2moa adjustment.
• carlton
what is the measurement systems called minute of angle(m.o.a.) and an angular measurement
• jeff
Minute of angle is 1 inch of elevation or windage @ 100 yards. If you are left of the bulls eye 1 inch you would click the scope 2 or 4 clicks depending on the scope adjusment a M82 is 1/8 MOA at
100 yds to move the POI (point of impact) 1 inch right. If you are 0'd at 100 and want to shoot 600 you would move the elevation up (cross hairs down) 12 clicks or 24 clicks depending if your
scope has 1/4 or 1/2 inch of MOA adjustment
• Marcus
What if the distance is between 500-600.
Then what constant will we use? | {"url":"http://www.rifleshootermag.com/2010/09/23/shooting_tips_determing_wind_values/","timestamp":"2014-04-16T07:13:39Z","content_type":null,"content_length":"104826","record_id":"<urn:uuid:ffa23fe7-ff06-40b9-bf10-feaf3c82fcf9>","cc-path":"CC-MAIN-2014-15/segments/1397609521558.37/warc/CC-MAIN-20140416005201-00272-ip-10-147-4-33.ec2.internal.warc.gz"} |
Two posts ago we showed you the digit sound system for remembering numbers. This week we provide two computer programs to help you create mnemonics.
You’ve got the whole world in your portfolio
A famous finance professor once told us that good diversification meant holding everything in the world. Fine, but in what proportion?Suppose you could invest in every country in the world. How much
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HOW TO DISPLAY A LINE PLOT WITH COUNT INFORMATION? In a previously-mentioned paper Sharad and your DSN editor are writing up, there is the above line plot with points. The area of each point shows
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Some ideas on communicating risks to the general public
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The counterfactual GPS!
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member. The monthly book group takes place in Cove Neck Long Island, about an hour East of Manhattan. | {"url":"http://www.r-bloggers.com/tag/decision-making/","timestamp":"2014-04-17T13:19:40Z","content_type":null,"content_length":"35951","record_id":"<urn:uuid:b47c4904-5d45-4573-aeac-5e246a43a74f>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00160-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Math word problems
December 8th 2009, 01:27 PM #1
Dec 2009
Math word problems
Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.
It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.
I did:
I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.
Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.
It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.
I did:
I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.
let $r$ = row rate in km/hr
$t$ = time in hrs upstream
$\frac{4}{3} - t$ = time in hrs downstream
$(r-3)t = 3$
$(r+3)\left(\frac{4}{3} - t\right) = 3$
from the 1st equation ... $t = \frac{3}{r-3}$
sub this into the 2nd equation ...
$(r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3$
solve for $r$
Since we do not know the time taken to row upstream and downstream separately, let " $t_1$" be the time taken to row upstream and let " $t_2$" be the time taken to row downstream. Let "v" be
their speed in still water.
Then their speed upstream is $v- \frac{4}{3}$ and that equals "distance divided by time: $v- \frac{4}{3}= \frac{3}{t_1}$.
Their speed downstream is $v+ \frac{4}{3}$ and so $v+ \frac{4}{3}= \frac{3}{t_2}$.
Those, together with $t_1+ t_2= 80$ give three equations to solve for the three values $t_1$, $t_2$, and v. Of course, you are only asked for v so if you could find a way to eliminate $t_1$ and
$t_2$, reducing to a single equation in v, that would be excellent.
Last edited by HallsofIvy; December 10th 2009 at 03:14 AM.
$(r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3$
$(r+3)\left(\frac{4(r-3)}{3(r-3)} - \frac{9}{3(r-3)}\right) = 3$
$(r+3)\left(\frac{4(r-3)-9}{3(r-3)}\right) = 3$
$(r+3)\left(\frac{4r-21}{3(r-3)}\right) = 3$
$\frac{(r+3)(4r-21)}{3(r-3)} = 3$
$(r+3)(4r-21) = 9(r-3)$
expand both sides, combine like terms and solve the resulting quadratic equation.
December 8th 2009, 03:03 PM #2
December 9th 2009, 04:18 AM #3
MHF Contributor
Apr 2005
December 9th 2009, 12:04 PM #4
Dec 2009
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A Dynamic-Programming Approach to Continuous Speech Recognition
, 1989
"... EFFICIENT ALGORITHMS FOR SEQUENCE ANALYSIS WITH CONCAVE AND CONVEX GAP COSTS David A. Eppstein We describe algorithms for two problems in sequence analysis: sequence alignment with gaps
(multiple consecutive insertions and deletions treated as a unit) and RNA secondary structure with single loops ..."
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EFFICIENT ALGORITHMS FOR SEQUENCE ANALYSIS WITH CONCAVE AND CONVEX GAP COSTS David A. Eppstein We describe algorithms for two problems in sequence analysis: sequence alignment with gaps (multiple
consecutive insertions and deletions treated as a unit) and RNA secondary structure with single loops only. We make the assumption that the gap cost or loop cost is a convex or concave function of
the length of the gap or loop, and show how this assumption may be used to develop e#cient algorithms for these problems. We show how the restriction to convex or concave functions may be relaxed,
and give algorithms for solving the problems when the cost functions are neither convex nor concave, but can be split into a small number of convex or concave functions. Finally we point out some
sparsity in the structure of our sequence analysis problems, and describe how we may take advantage of that sparsity to further speed up our algorithms. CONTENTS 1. Introduction
............................1 ...
- Proc. Second Workshop on Sequences: Combinatorics, Compression. Securiry , 1991
"... : We consider new algorithms for the solution of many dynamic programming recurrences for sequence comparison and for RNA secondary structure prediction. The techniques upon which the algorithms
are based e#ectively exploit the physical constraints of the problem to derive more e#cient methods f ..."
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: We consider new algorithms for the solution of many dynamic programming recurrences for sequence comparison and for RNA secondary structure prediction. The techniques upon which the algorithms are
based e#ectively exploit the physical constraints of the problem to derive more e#cient methods for sequence analysis. 1. INTRODUCTION In this paper we consider algorithms for two problems in
sequence analysis. The first problem is sequence alignment, and the second is the prediction of RNA structure. Although the two problems seem quite di#erent from each other, their solutions share a
common structure, which can be expressed as a system of dynamic programming recurrence equations. These equations also can be applied to other problems, including text formatting and data storage
optimization. We use a number of well motivated assumptions about the problems in order to provide e#cient algorithms. The primary assumption is that of concavity or convexity. The recurrence
relations for bo...
, 1993
"... Finding matches, both exact and approximate, between a sequence of symbols A and a pattern P has long been an active area of research in algorithm design. Some of the more well-known byproducts
from that research are the diff program and grep family of programs. These problems form a sub-domain of a ..."
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Finding matches, both exact and approximate, between a sequence of symbols A and a pattern P has long been an active area of research in algorithm design. Some of the more well-known byproducts from
that research are the diff program and grep family of programs. These problems form a sub-domain of a larger areas of problems called discrete pattern matching which has been developed recently to
characterise the wide range of pattern matching problems. This dissertation presents new algorithms for discrete pattern matching over sequences and develops a new sub-domain of problems called
discrete pattern matching over interval sets. The problems and algorithms presented here are characterised by pattern matching over interval sets. The problems and al | {"url":"http://citeseerx.ist.psu.edu/showciting?cid=2548072","timestamp":"2014-04-23T18:41:43Z","content_type":null,"content_length":"18574","record_id":"<urn:uuid:2e2cbf5d-2cee-47ec-8e0c-5a8448167121>","cc-path":"CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00109-ip-10-147-4-33.ec2.internal.warc.gz"} |
[Haskell-cafe] Re: On the verge of ... giving up!
apfelmus apfelmus at quantentunnel.de
Mon Oct 15 04:32:50 EDT 2007
Prabhakar Ragde wrote:
>> main n = print . sum . map read . take n . reverse . lines =<< getContents
> Could someone describe succinctly how to compute the space complexity of
> this program, if there are m lines of input and m >> n? Many thanks. --PR
Good point :)
Felipe Lessa wrote:
> main n = print . sum . map read . head . dropWhile (not . null . drop
> n) . tails . lines =<< getContents
> where I changed (take n . reverse) to (head . dropWhile (not . null .
> drop n) . tails). Yes, I cheated, I'm using Data.List =). With this
> version you keep only n lines in memory at any moment, so it has space
> complexity of O(n).
Yes, this has O(n) space complexity since dropWhile can discard the
dropped elements. Unfortunately, we now have O(m*n) time complexity
since drop n is executed for every list element.
Of course, the solution is to first drop n elements and then take
tails instead of dropping n elements every time.
map (drop n) . tails = tails . drop n
O(m*n) O(m)
With this, we can write a function that returns the last n elements of
a list in O(m) time and O(n) space as
lasts :: Int -> [a] -> [a]
lasts n xs = head $ [x | (x,[]) <- zip (tails xs) (tails $ drop n xs)]
and use it as a drop-in replacement
main n = print . sum . map read . lasts n . lines =<< getContents
PS: The implementation of lasts n in my original version would be
lasts n = reverse . take n . reverse
But thanks to
map f . reverse = reverse . map f
sum . reverse = sum
we can leave out one reverse.
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the encyclopedic entry of binary-cell
Sheila Greibach
(1939-) is a researcher in
formal languages
theory in particular; and
computer science
in general. She is currently Professor of
Computer Science
at the
University of California, Los Angeles
She worked with Seymour Ginsburg and Michael Harrison in context-sensitive parsing using the stack automaton model.
Besides establishing the normal form (Greibach normal form) for context-free grammars now named after her, in 1965, she also investigated properties of W-grammars, pushdown automata, and decidability
Work and Contributions
The following list indicates some of her work. The top portion of the list is from the ACM Digital Library and the remainder from the FOCS Bibliography by David M. Jones.
From ACM Digital Library:
Jump PDA's, deterministic context-free languages principal AFDLs and polynomial time recognition (Extended Abstract)
Sheila A. Greibach
April 1973
Proceedings of the fifth annual ACM symposium on Theory of computing
Every deterministic context-free language can be accepted by a deterministic finite delay pda with jumps. Increasing the number of types or occurrences of jumps increases the family of languages
accepted with finite delay. Hence the family of deterministic context-free language is a principal AFDL; there is a context-free language Lo such that every context-free language is an inverse
gsm image of Lo or Lo - {e}.Multitape AFA
Seymour Ginsburg, Sheila Greibach
April 1972
Journal of the ACM (JACM), Volume 19 Issue 2 Superdeterministic PDAs: A Subcase with a Decidable Inclusion problem
S. A. Greibach, E. P. Friedman
October 1980
Journal of the ACM (JACM), Volume 27 Issue 4 Stack automata and compiling
Seymour Ginsburg, Sheila A. Greibach, Michael A. Harrison
January 1967
Journal of the ACM (JACM), Volume 14 Issue 1
Compilation consists of two parts, recognition and translation. A mathematical model is presented which embodies salient features of many modern compiling techniques. The model, called the stack
automaton, has the desirable feature of being deterministic in nature. This deterministic device is generalized to a nondeterministic device (nondeterministic stack automaton) and particular
instances of this more general device are noted. Sets accepted by nondeterministic stack automata are recursi ... Quasi-realtime languages (Extended Abstract)
Ronald V. Book, Sheila A. Greibach
May 1969
Proceedings of the first annual ACM symposium on Theory of computing
Quasi-realtime languages are the languages accepted by nondeterministic multitape Turing machines in real time. The family of quasi-realtime languages forms an abstract family of languages closed
under intersection, linear erasing, and reversal. It is identical with the family of languages accepted by nondeterministic multitape Turing machines in linear time. Every quasi-realtime language
can be accepted in real time by a non-deterministic one stack, one pushdown store machine, and can be e ... One-way stack automata
Seymour Ginsburg, Sheila A. Greibach, Michael A. Harrison
April 1967
Journal of the ACM (JACM), Volume 14 Issue 2
A number of operations which either preserve sets accepted by one-way stack automata or preserve sets accepted by deterministic one-way stack automata are presented. For example, sequential
transduction preserves the former; set complementation, the latter. Several solvability questions are also considered. Tape- and time-bounded Turing acceptors and AFLs (Extended Abstract)
Ronald V. Book, Sheila A. Greibach, Ben Wegbreit
May 1970
Proceedings of the second annual ACM symposium on Theory of computing
Complexity classes of formal languages defined by time- and tape-bounded Turing acceptors are studied with the aim of showing sufficient conditions for these classes to be AFLs and to be
principal AFLs.Some restrictions on W-grammars
Sheila A. Greibach
April 1974
Proceedings of the sixth annual ACM symposium on Theory of computing
The effect of some restrictions on W-grammars (the formalization of the syntax of ALGOL 68) are explored. Two incomparable families examined at length are WRB (languages generated by normal
regular-based W-grammars) and WS (languages generated by simple W-grammars). Both properly contain the context-free languages and are properly contained in the family of quasirealtime languages.
In addition, WRB is closed under nested iterate ...Uniformly erasable AFL
Sheila Carlyle-Greibach, Seymour Ginsburg, Jonathan Goldstine
May 1972
Proceedings of the fourth annual ACM symposium on Theory of computing
The purpose of this paper is to show that a number of well-known families have property (*). In particular, we prove that the family of context-free languages does indeed have this property. In
addition, we show that several familiar subfamilies of the context-free languages, such as the one-counter languages, have property (*). Finally, we show that there are families satisfying (*)
which are not subfamilies of the context-free languages, for we prove that any family generated from one-let ...Formal parsing systems
Sheila A. Greibach
August 1964
Communications of the ACM, Volume 7 Issue 8
Automatic syntactic analysis has recently become important for both natural language data processing and syntax-directed compilers. A formal parsing system G = (V, &mgr;, T, R) consists of two
finite disjoint vocabularies, V and T, a many-many map, &mgr;, from V onto T, and a recursive set R of strings in T called syntactic sentence classes ...An Infinite Hierarchy of Context-Free
Sheila A. Greibach
January 1969
Journal of the ACM (JACM), Volume 16 Issue 1A New Normal-Form Theorem for Context-Free Phrase Structure Grammars
Sheila A. Greibach
January 1965
Journal of the ACM (JACM), Volume 12 Issue 1 The Unsolvability of the Recognition of Linear Context-Free Languages
Sheila A. Greibach
October 1966
Journal of the ACM (JACM), Volume 13 Issue 4
The problem of whether a given context-free language is linear is shown to be recursively undecidable.
From FOCS Bibliography
Seymour Ginsburg and Sheila Greibach.
Deterministic context free languages.
In Proceedings of the Sixth Annual Symposium on Switching Circuit Theory and Logical Design, pages 203-220. IEEE, 1965.
Seymour Ginsburg, Sheila A. Greibach, and Michael A. Harrison.
One-way stack automata (extended abstract).
In Conference Record of 1966 Seventh Annual Symposium on Switching and Automata Theory, pages 47-52, Berkeley, California, 26-28 October 1966. IEEE.
Sheila A. Greibach.
An infinite hierarchy of context-free languages.
In Conference Record of 1967 Eighth Annual Symposium on Switching and Automata Theory, pages 32-36, Austin, Texas, 18-20 October 1967. IEEE.
Seymour Ginsburg and Sheila Greibach.
Abstract families of languages.
In Conference Record of 1967 Eighth Annual Symposium on Switching and Automata Theory, pages 128-139, Austin, Texas, 18-20 October 1967. IEEE. Citations.
Sheila Greibach.
Checking automata and one-way stack languages (extended abstract).
In Conference Record of 1968 Ninth Annual Symposium on Switching and Automata Theory, pages 287-291, Schenectady, New York, 15-18 October 1968. IEEE. Citations.
Sheila A. Greibach.
Full AFLs and nested iterated substitution.
In Conference Record of 1969 Tenth Annual Symposium on Switching and Automata Theory, pages 222-230, Waterloo, Ontario, Canada, 15-17 October 1969. IEEE.
J. W. Carlyle, S. A. Greibach, and A. Paz.
A two-dimensional generating system modeling growth by binary cell division (preliminary report).
In 15th Annual Symposium on Switching and Automata Theory, pages 1-12, The University of New Orleans, 14-16 October 1974. IEEE.
S. A. Greibach.
Formal languages: Origins and directions.
In 20th Annual Symposium on Foundations of Computer Science, pages 66-90, San Juan, Puerto Rico, 29-31 October 1979. IEEE.
Ronald Book, Shimon Even, Sheila Greibach and Gene Ott.
Ambiguity in Graphs and Expressions.
IEEE Transactions on Computers, vol. c-20, No. 2, February 1971. IEEE.
See also | {"url":"http://www.reference.com/browse/binary-cell","timestamp":"2014-04-16T10:51:28Z","content_type":null,"content_length":"83429","record_id":"<urn:uuid:ffe32aeb-a75f-44ae-8ea0-c095d1b19f24>","cc-path":"CC-MAIN-2014-15/segments/1397609523265.25/warc/CC-MAIN-20140416005203-00490-ip-10-147-4-33.ec2.internal.warc.gz"} |
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SEKI Report
12 search hits
Higher-Order Automated Theorem Proving for Natural Language Semantics (1998)
Michael Kohlhase Karsten Konrad
This paper describes a tableau-based higher-order theorem prover HOT and an application to natural language semantics. In this application, HOT is used to prove equivalences using world knowledge
during higher-order unification (HOU). This extended form of HOU is used to compute the licensing conditions for corrections.
Model Existence for Higher Order Logic (1997)
Christoph Benzmüller Michael Kohlhase
In this paper we provide a semantical meta-theory that will support the development of higher-order calculi for automated theorem proving like the corresponding methodology has in first-order
logic. To reach this goal, we establish classes of models that adequately characterize the existing theorem-proving calculi, that is, so that they are sound and complete to these calculi, and a
standard methodology of abstract consistency methods (by providing the necessary model existence theorems) needed to analyze completeness of machine-oriented calculi.
Henkin Completeness of Higher-Order Resolution (1999)
Christoph Benzmüller Michael Kohlhase
In this paper we present an extensional higher-order resolution calculus that iscomplete relative to Henkin model semantics. The treatment of the extensionality princi-ples - necessary for the
completeness result - by specialized (goal-directed) inference rulesis of practical applicability, as an implentation of the calculus in the Leo-System shows.Furthermore, we prove the
long-standing conjecture, that it is sufficient to restrict the orderof primitive substitutions to the order of input formulae.
An Integration of Mechanised Reasoning and Computer Algebra that Respects Explicit Proofs (1999)
Manfred Kerber Michael Kohlhase Volker Sorge
Mechanised reasoning systems and computer algebra systems have apparentlydifferent objectives. Their integration is, however, highly desirable, since in manyformal proofs both of the two
different tasks, proving and calculating, have to beperformed. Even more importantly, proof and computation are often interwoven andnot easily separable. In the context of producing reliable
proofs, the question howto ensure correctness when integrating a computer algebra system into a mechanisedreasoning system is crucial. In this contribution, we discuss the correctness prob-lems
that arise from such an integration and advocate an approach in which thecalculations of the computer algebra system are checked at the calculus level of themechanised reasoning system. This can
be achieved by adding a verbose mode to thecomputer algebra system which produces high-level protocol information that can beprocessed by an interface to derive proof plans. Such a proof plan in
turn can beexpanded to proofs at different levels of abstraction, so the approach is well-suited forproducing a high-level verbalised explication as well as for a low-level machine check-able
calculus-level proof. We present an implementation of our ideas and exemplifythem using an automatically solved extended example.
A Coloured Version of the Lambda-Calculus (1999)
Michael Kohlhase Dieter Hutter
Coloring terms (rippling) is a technique developed for inductive theorem proving which uses syntactic differences of terms to guide the proof search. Annotations (colors) to terms are used to
maintain this information. This technique has several advantages, e.g. it is highly goal oriented and involves little search. In this paper we give a general formalization of coloring terms in a
higher-order setting. We introduce a simply-typed lambda calculus with color annotations and present an appropriate (pre-)unification algorithm. Our work is a formal basis to the implementation
of rippling in a higher-order setting which is required e.g. in case of middle-out reasoning. Another application is in the construction of natural language semantics, where the color annotations
rule out linguistically invalid readings that are possible using standard higher-order unification.
Higher-Order Multi-Valued Resolution (1999)
Michael Kohlhase Ortwin Scheja
This paper introduces a multi-valued variant of higher-order resolution and provesit correct and complete with respect to a natural multi-valued variant of Henkin'sgeneral model semantics. This
resolution method is parametric in the number of truthvalues as well as in the particular choice of the set of connectives (given by arbitrarytruth tables) and even substitutional quantifiers. In
the course of the completenessproof we establish a model existence theorem for this logical system. The workreported in this paper provides a basis for developing higher-order mechanizationsfor
many non-classical logics.
A Mechanization of Sorted Higher-Order Logic Based on the Resolution Principle (1999)
Michael Kohlhase
Higher-Order Order-Sorted Resolution (1999)
Michael Kohlhase
The introduction of sorts to first-order automated deduction has broughtgreater conciseness of representation and a considerable gain in efficiency byreducing the search space. It is therefore
promising to treat sorts in higherorder theorem proving as well.In this paper we present a generalization of Huet's Constrained Resolutionto an order-sorted type theory SigmaT with term
declarations. This system buildscertain taxonomic axioms into the unification and conducts reasoning withthem in a controlled way. We make this notion precise by giving a relativizationoperator
that totally and faithfully encodes SigmaT into simple type theory.
Mechanization of Strong Kleene Logic for Partial Functions (1999)
Manfred Kerber Michael Kohlhase
Even though it is not very often admitted, partial functions do play asignificant role in many practical applications of deduction systems. Kleenehas already given a semantic account of partial
functions using three-valuedlogic decades ago, but there has not been a satisfactory mechanization. Recentyears have seen a thorough investigation of the framework of many-valuedtruth-functional
logics. However, strong Kleene logic, where quantificationis restricted and therefore not truth-functional, does not fit the frameworkdirectly. We solve this problem by applying recent methods
from sorted logics.This paper presents a resolution calculus that combines the proper treatmentof partial functions with the efficiency of sorted calculi.
Unification in an Extensional Lambda Calculus with Ordered Function Sorts and Constant Overloading (1999)
Patricia Johann Michael Kohlhase
We develop an order-sorted higher-order calculus suitable forautomatic theorem proving applications by extending the extensional simplytyped lambda calculus with a higher-order ordered sort
concept and constantoverloading. Huet's well-known techniques for unifying simply typed lambdaterms are generalized to arrive at a complete transformation-based unificationalgorithm for this
sorted calculus. Consideration of an order-sorted logicwith functional base sorts and arbitrary term declarations was originallyproposed by the second author in a 1991 paper; we give here a
correctedcalculus which supports constant rather than arbitrary term declarations, aswell as a corrected unification algorithm, and prove in this setting resultscorresponding to those claimed | {"url":"https://kluedo.ub.uni-kl.de/solrsearch/index/search/searchtype/series/id/16159/start/0/rows/10/author_facetfq/Michael+Kohlhase","timestamp":"2014-04-16T04:29:56Z","content_type":null,"content_length":"42927","record_id":"<urn:uuid:4df7edee-6a62-485b-bb4d-8a9be8e29a0c>","cc-path":"CC-MAIN-2014-15/segments/1397609521512.15/warc/CC-MAIN-20140416005201-00260-ip-10-147-4-33.ec2.internal.warc.gz"} |
This file is part of timeout-with-results.
timeout-with-results is free software: you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public License as
published by the Free Software Foundation, either version 3 of the License,
or (at your option) any later version.
timeout-with-results is distributed in the hope that it will be useful, but
WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY
or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public
License for more details.
You should have received a copy of the GNU Lesser General Public License
along with timeout-with-results. If not, see <http://www.gnu.org/licenses/>.
{-# LANGUAGE
FlexibleContexts #-}
{- | Defines a writer monad for computations that can be interrupted by a
timeout. Written partial results are combined using their monoid operation
and if a timeout occurs, the result is returned.
Several utility monoids that force their values to /weak head normal form/
or to /normal form/ are provided.
module System.Timeout.Returning.Writer (
-- helpers:
) where
import Control.Applicative
import Control.Monad
import Control.Monad.Reader
import Control.Monad.Writer
import qualified Control.Concurrent as C
import Control.Concurrent.MVar
import Control.DeepSeq (NFData(..))
import Control.Seq
import Data.Monoid
import qualified System.Timeout as T
-- | Monad for computations that can save partial results
-- of type @w@ during their evaluation.
class (Monad m) => MonadTimeout w m | m -> w where
-- | Store a new partial result. The precise semantics of what happens
-- with the written value is by intent unspecified and left to
-- be decided by implementations.
partialResult :: w -> m ()
-- | Explicitly allow interrupting the computation at this point.
-- Experimental.
yield :: m ()
yield = return ()
-- | Extends 'MonadTimeout' to 'MonadWriter'. Written values are combined
-- together using @w@'s monoid. In addition, allows to run a sub-computation
-- in a contained environment, without affecting the current partial result.
class (Monoid w, MonadTimeout w m, MonadWriter w m)
=> MonadTimeoutWriter w m | m -> w where
-- | Runs the given computation separately and return its result.
-- Does not modify the current result!
contained :: m r -> m (r, w)
contained k = do
~(_, zero) <- listen (return ())
pass (listen k >>= \x -> return (x, const zero))
-- | A default implementation of 'listen' using 'contained'.
-- Useful only for authors of implementations of 'MonadTimeout'.
defaultListen :: MonadTimeoutWriter w m => m a -> m (a, w)
defaultListen k = do
(x, w) <- contained k
tell w
return (x, w)
-- | A default implementation of 'pass' using 'contained'.
-- Useful only for authors of implementations of 'MonadTimeout'.
defaultPass :: MonadTimeoutWriter w m => m (a, w -> w) -> m a
defaultPass k = do
((x, f), w) <- contained k
tell (f w)
return x
-- -----------------------------------------------------------------
-- | An 'IO'-based implementation of 'MonadTimeoutWriter'. Calling
-- 'partialResult' (or equivalently 'tell') combines the value with any
-- previously written values using @w@'s monoidal operation.
newtype TimeoutWriter w a
= TimeoutWriter { getTimeoutWriter :: ReaderT (w -> IO ()) IO a }
instance Functor (TimeoutWriter w) where
fmap = liftM
instance Applicative (TimeoutWriter w) where
pure = return
(<*>) = ap
instance Monad (TimeoutWriter w) where
return = TimeoutWriter . return
(TimeoutWriter v) >>= f = TimeoutWriter (v >>= (getTimeoutWriter . f))
instance MonadIO (TimeoutWriter w) where
liftIO = TimeoutWriter . lift
instance Monoid w => MonadWriter w (TimeoutWriter w) where
tell = partialResult
listen = defaultListen
pass = defaultPass
instance Monoid w => MonadTimeout w (TimeoutWriter w) where
partialResult x = TimeoutWriter $ ask >>= \r -> lift (r x)
yield = liftIO C.yield
instance Monoid w => MonadTimeoutWriter w (TimeoutWriter w) where
contained = liftIO . runTimeoutInternal id
-- | Modify written values using the given function.
withTimeoutWriter :: (w' -> w) -> (TimeoutWriter w' a -> TimeoutWriter w a)
withTimeoutWriter f (TimeoutWriter k) = TimeoutWriter $ withReaderT (. f) k
-- | Execute the given computation with a timeout limit. Each time a value
-- is written, the result of 'mappend' with the previous one is evaluated to
-- /weak head normal form/.
:: Monoid w
=> Int -- ^ TimeoutWriter in microseconds.
-> TimeoutWriter w r -- ^ The computation.
-> IO (Maybe r, w) -- ^ The final result (if available) and the saved
-- partial result.
runTimeout duration = runTimeoutInternal (T.timeout duration)
:: Monoid w
=> (IO r -> IO a) -- ^ What to do with the computation.
-> TimeoutWriter w r -- ^ The computation to evaluate.
-> IO (a, w)
runTimeoutInternal run (TimeoutWriter k) = do
mvar <- newMVar mempty
let save x = modifyMVar_ mvar (return . withStrategy rseq . (`mappend` x))
r <- run (runReaderT k save)
w <- takeMVar mvar
return (r, w)
sseq :: Strategy a -> a -> b -> b
sseq s x y = s x `seq` y
-- | A monoid equivalent to 'Last'. In addition, it forces evaluation of
-- values inside 'Maybe' using 'rseq'. This means that when it is used in
-- 'runTimeout', the computations will be forced in the producing thread,
-- not in the consuming one. If you want to force evaluation to NF, wrap
-- it inside 'NFMonoid'.
newtype Last' a = Last' { getLast' :: Maybe a }
deriving (Eq, Ord, Show, Read)
instance Functor Last' where
fmap f (Last' x) = Last' $ fmap f x
instance NFData a => NFData (Last' a) where
rnf (Last' x) = rnf x
instance Monoid (Last' a) where
mempty = Last' Nothing
mappend (Last' x) (Last' y)
= Last' $ getLast (Last x `mappend` Last y)
-- | A monoid whose 'mappend' picks the grater value according to the second
-- field of the tuple. @SeqMax Nothing@ is the least element of the
-- ordering. If the second fields are the same, the left value is preferred.
-- In addition, the first field of the selected tuple is forced to evaluate
-- using 'rseq'.
newtype SeqMax a b = SeqMax (Maybe (a, b))
deriving (Eq, Ord, Show, Read)
instance Functor (SeqMax a) where
fmap f (SeqMax x) = SeqMax $ fmap (fmap f) x
instance (NFData a, NFData b) => NFData (SeqMax a b) where
rnf (SeqMax x) = rnf x
instance (Ord b) => Monoid (SeqMax a b) where
mempty = SeqMax Nothing
mappend (SeqMax Nothing) s = s
mappend s (SeqMax Nothing) = s
mappend m1@(SeqMax (Just (r1, x1))) m2@(SeqMax (Just (r2, x2)))
| x1 >= x2 = sseq rseq r1 m1
| otherwise = sseq rseq r2 m2
-- | A wrapper monoid that forces each result of 'mappend'
-- to /normal form/'
newtype NFMonoid a = NFMonoid { getNFMonoid :: a }
deriving (Eq, Ord, Show, Read, Bounded)
instance Functor NFMonoid where
fmap f (NFMonoid x) = NFMonoid (f x)
instance (NFData a, Monoid a) => Monoid (NFMonoid a) where
mempty = NFMonoid mempty
mappend (NFMonoid x) (NFMonoid y)
= NFMonoid ((x `mappend` y) `using` rdeepseq) | {"url":"http://hackage.haskell.org/package/timeout-with-results-0.2/docs/src/System-Timeout-Returning-Writer.html","timestamp":"2014-04-19T19:40:24Z","content_type":null,"content_length":"42674","record_id":"<urn:uuid:1c047f0e-451c-458c-8b04-50776e4d1042>","cc-path":"CC-MAIN-2014-15/segments/1397609537376.43/warc/CC-MAIN-20140416005217-00018-ip-10-147-4-33.ec2.internal.warc.gz"} |
[3ACTS] Incredible Shrinking Dollar
November 2nd, 2011 by Dan Meyer
If that first act interests you, download the full story. Feel free, also, to check out Dave Martin's treatment of the same concept.
15 Responses to “[3ACTS] Incredible Shrinking Dollar”
1. on 03 Nov 2011 at 3:39 pm1
Will paper lessons plans ever be offered for any of these acts? (It would really help with my planning).
2. Assuming it isn’t paper itself that’d help with your planning, rather something on the paper, what would that be? What would help you out here?
3. on 04 Nov 2011 at 8:53 am3
Something similar to Mathalicious lessons. A lesson plan and a student guide broken down by acts. Given your emphasis on discovery, the student guide would presumably be a sparse outline, whereas
the teacher’s guide would offer details.
4. on 05 Nov 2011 at 9:55 am4
Do his work for him dan! Hehe! Question about this video for some reason I really want to see what the dollar looks like every time you pull it out of the copier *:)
5. Pepe: I think the general idea is that students should learn how to structure their thoughts themselves. The goal is to get students to approach these problems like we adults would. We get out a
sheet of paper, we sketch some diagrams, we write questions, we perform calculations.
Getting a kid to do that is pretty hard, of course.
So, to that end I think you could give a common handout that has structures that should work for any of these three act problems. Remember that the concept of the acts are not exactly for
students: they are ways for us as teachers to break down what we are saying and giving to the students.
A structure that I may give to a class unused to this type of activity may just be dividing the paper up into 4 sections: “your questions” “data/info” “diagrams” “calculations”. I may just do
this by telling the students to fold their paper into four parts.
Another key element to this type of project is to have the students revisit the problem — twice. After they have “solved” it, make them write a 2nd draft solution. They should write out their
complete solution: clean out all the unneeded stuff, adjust their diagrams, clean up the flow of their calculations, and add in some actual sentences. Just because we’re in math class doesn’t
mean we can’t write.
You can offer comments on their 2nd drafts and how to improve their reasoning. Then, finally, have them make a 3rd draft solution. They tune up their 2nd draft and fix mistakes, adjust their
commentary, and make it neat.
Its hard. Even high-skilled math students have a lot of trouble explaining their work. They often spend time explaining their arithmetic operations instead of the concept. A made up example: “We
multiplied 5*6 and then divided by 8 and that’s how we got the answer” vs. “The area of the rectangle is 30, thus each of the 8 people get over three [...]” And the low skilled math students
often struggle with writing in general.
But this is why they should have practice. A computer can solve a quadratic equation. A human can interpret and communicate the solution. We want our kids to be human, yes?
6. Certainly I’d like to support teachers, but “which supports?” is a huge, outstanding question and I’m not sure handouts and lesson plans are the answer.
7. apologies for combining problems here, but when i saw this video my head went to your Guggenheim thing – i really wanted to know how many *more* of these shrunken dollars it would take to create
that artwork.
8. on 09 Nov 2011 at 4:58 pm8
Dan Lemay
This looks very cool Dan. I’m going to try it with some under performing Juniors and Seniors I work with. Smart enough kids, they have just chosen to/been allowed to dumb down over the years.
I’ll let you know how it goes.
Thanks again.
9. on 10 Nov 2011 at 12:47 pm9 Karim
For what it’s worth, I think Dan’s stuff is valuable in large part because it doesn’t include the type of explicit scaffold that Mathalicious lessons do. The approach is predicated on students’
coming up with their own questions, and Dan’s gift is in crafting opportunities for them to do this (while still providing some structure, albeit subtly).
In the end, I think WCYDWT lessons and Mathalicious lessons are similar in spirit–they’re both about exploring the world–but fundamentally different at the same time. The types of topics that
many Mathalicious lessons address require a more explicit scaffold; whereas students might be able to come up with “Is Wheel of Fortune rigged?” on their own, it’s unlikely that they’d transition
from NikeiD.com to a discussion of paralysis by analysis without the narrative afforded by the handout.
I don’t think either one is better or worse. They’re just very different, and therefore require a different approach.
10. on 15 Nov 2011 at 7:56 pm10
@scott… pretty sure pepe was kidding. at least i hope so.
11. on 16 Nov 2011 at 4:31 pm11
I attempted this 3acts today in my 8th grade algebra class. I gotta say it was a bit disappointing. Potential problems: End of the day; 35 kids in the class.
Of the nine groups in the class, I think 3 were actually engaged in the problem. After they struggled with it for about five minutes, I stopped the class and we discussed how you could get
started on the problem. But then I struggle with not leading them down the path of how to do it. At what point do you throw in the white flag and move on.
On a slightly positive note, one kid right off the bat, eyeballed 75% and sketched the drawings on the sheet of paper, and was within two millimeters of the answer (we projected his image over
final cut from act 3)
12. Hey Josh, thanks for the feedback. It’s a lot easier to dish out positive results than the negative and I’m obliged you’d make the effort.
Presumably you didn’t blindly implement the problem. You saw it yourself and thought it’d be interesting. Do you have any sense why the problem didn’t land for your students? Any sense how I
might scaffold it differently? Do you think your students would have experienced the problem any differently had I shown them the last dollar and asked, “How many times was it copied?”
13. on 17 Nov 2011 at 9:39 pm13
I am going to give the problem to my geometry classes as well. I’ll give you more complete thoughts tomorrow after they see it also.
14. on 18 Nov 2011 at 11:26 pm14
Before giving my thoughts on why it bombed, let me say that the geometry kids ate it up in about one minute. Aside from the questioning of who this guy is who is able to copy money, they dug
right in and groups finished the original problem and most finished the followup questions. Many came up to ask me how small it would have to get to be invisible. “Not sure,” I would respond.
“how small do you think it should be?” which got about two of them actually thinking about the size of .00001 mm.
The algebra classes struggled more because they don’t underhand percents very well. I felt the initial hurdle was too high. I handed out the dollar outline to each student (I think one sheet to
each group would have been better). I don’t want to go into too much detail, but some went right to perimeter or area, which brings up other questions.
I do think your idea of having the original and the ninth copy moght have helped as a launch point. I also think showing them either the first copy and blanking out the percent would have also
been a good problem. the question could then have been, “what setting was entered?” The followup could then have been how big is the ninth copy.
Gonna try the dominos next week.
15. on 09 Feb 2012 at 12:33 pm15
I used this with a grade 11 precalculus class for an introduction to geometric series. They had not been given any formulas or explanations. We did an example of geometric growth previously by
looking at the distances in a spider web. For the most part the students really enjoyed the challenge especially when I refused to say when or if they had the “answer”. We then took the class to
the copier and did the reductions and measured to verify which had the right and wrong solutions and why. Went very well and sets up our discussion of geometric sequence formula work tomorrow. We
had trouble with when the bill disappeared because we couldn’t recognize the limit of the copier but tomorrow we will look at how we can determine the limits of the copier by being able to
predict the shrinking bill size at each stage. Thanks for the work. I really appreciate you posting the ideas. | {"url":"http://blog.mrmeyer.com/2011/3acts-incredible-shrinking-dollar/","timestamp":"2014-04-19T09:37:34Z","content_type":null,"content_length":"45654","record_id":"<urn:uuid:22a76d7d-e033-4d05-b543-c16b179b217f>","cc-path":"CC-MAIN-2014-15/segments/1398223207985.17/warc/CC-MAIN-20140423032007-00587-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ugh, More Trig Stuff
May 8th 2006, 01:09 PM
Ugh, More Trig Stuff
Another question i had was: If sin(alpha)=4/5 in quadrant 1 and
cos(beta)=-11/12 in quadrant II, find
Thats a LOT for the help.
May 8th 2006, 01:52 PM
Originally Posted by aussiekid90
Another question i had was: If sin(alpha)=4/5 in quadrant 1 and
cos(beta)=-11/12 in quadrant II, find
Thats a LOT for the help.
You could get fancy and use the sum of angles formulas, but I think it's easier just to find alpha and beta:
$\alpha = sin^{-1} \left ( \frac{4}{5} \right ) \approx 0.927295 \, rad$
$\alpha$ is supposed to be 1st quadrant and, in fact, it is.
$\beta = cos^{-1} \left ( - \frac{11}{12} \right ) \approx 2.73045 \, rad$
which checks out as being in the 2nd quadrant.
$sin(\alpha + \beta) = sin(3.65775)=-0.493542$
$tan(\alpha + \beta) = tan(3.65775) = 0.567471$ | {"url":"http://mathhelpforum.com/trigonometry/2872-ugh-more-trig-stuff-print.html","timestamp":"2014-04-19T09:02:51Z","content_type":null,"content_length":"5411","record_id":"<urn:uuid:d3ea953e-84b4-4d99-974b-cd842fcc4c01>","cc-path":"CC-MAIN-2014-15/segments/1398223207046.13/warc/CC-MAIN-20140423032007-00326-ip-10-147-4-33.ec2.internal.warc.gz"} |
A Mixed-up Clock
Copyright © University of Cambridge. All rights reserved.
'A Mixed-up Clock' printed from http://nrich.maths.org/
There is a clock-face where the numbers have become all mixed up. Can you find out where all the numbers have got to from the ten statements below?
Here is a clock-face with letters to mark the position of the numbers so that the statements are easier to read and to follow.
1. No even number is between two odd numbers.
2. No consecutive numbers are next to each other.
3. The numbers on the vertical axis (a) and (g) add to $13$.
4. The numbers on the horizontal axis (d) and (j) also add to $13$.
5. The first set of $6$ numbers [(a) - (f)] add to the same total as the second set of $6$ numbers [(g) - (l)] .
6. The number at position (f) is in the correct position on the clock-face.
7. The number at position (d) is double the number at position (h).
8. There is a difference of $6$ between the number at position (g) and the number preceding it (f).
9. The number at position (l) is twice the top number (a), one third of the number at position (d) and half of the number at position (e).
10. The number at position (d) is $4$ times one of the numbers adjacent (next) to it. | {"url":"http://nrich.maths.org/2127/index?nomenu=1","timestamp":"2014-04-18T14:05:04Z","content_type":null,"content_length":"4289","record_id":"<urn:uuid:31c9cfb9-48d8-4300-a5e3-589b6ca61c66>","cc-path":"CC-MAIN-2014-15/segments/1397609537864.21/warc/CC-MAIN-20140416005217-00185-ip-10-147-4-33.ec2.internal.warc.gz"} |
This Article
Bibliographic References
Add to:
ASCII Text x
Sudipto Guha, Adam Meyerson, Nina Mishra, Rajeev Motwani, Liadan O'Callaghan, "Clustering Data Streams: Theory and Practice," IEEE Transactions on Knowledge and Data Engineering, vol. 15, no. 3,
pp. 515-528, May/June, 2003.
BibTex x
@article{ 10.1109/TKDE.2003.1198387,
author = {Sudipto Guha and Adam Meyerson and Nina Mishra and Rajeev Motwani and Liadan O'Callaghan},
title = {Clustering Data Streams: Theory and Practice},
journal ={IEEE Transactions on Knowledge and Data Engineering},
volume = {15},
number = {3},
issn = {1041-4347},
year = {2003},
pages = {515-528},
doi = {http://doi.ieeecomputersociety.org/10.1109/TKDE.2003.1198387},
publisher = {IEEE Computer Society},
address = {Los Alamitos, CA, USA},
RefWorks Procite/RefMan/Endnote x
TY - JOUR
JO - IEEE Transactions on Knowledge and Data Engineering
TI - Clustering Data Streams: Theory and Practice
IS - 3
SN - 1041-4347
EPD - 515-528
A1 - Sudipto Guha,
A1 - Adam Meyerson,
A1 - Nina Mishra,
A1 - Rajeev Motwani,
A1 - Liadan O'Callaghan,
PY - 2003
KW - Clustering
KW - data streams
KW - approximation algorithms.
VL - 15
JA - IEEE Transactions on Knowledge and Data Engineering
ER -
Abstract—The data stream model has recently attracted attention for its applicability to numerous types of data, including telephone records, Web documents, and clickstreams. For analysis of such
data, the ability to process the data in a single pass, or a small number of passes, while using little memory, is crucial. We describe such a streaming algorithm that effectively clusters large data
streams. We also provide empirical evidence of the algorithm's performance on synthetic and real data streams.
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Index Terms:
Clustering, data streams, approximation algorithms.
Sudipto Guha, Adam Meyerson, Nina Mishra, Rajeev Motwani, Liadan O'Callaghan, "Clustering Data Streams: Theory and Practice," IEEE Transactions on Knowledge and Data Engineering, vol. 15, no. 3, pp.
515-528, May-June 2003, doi:10.1109/TKDE.2003.1198387
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Subtracting Integers
November 11, 2008
A reader named Michelle said she enjoyed my post on Memorizing the Times Tables. And then she asked if I have any tips on teaching students to SUBTRACT INTEGERS.
It turns out that the answer is “yes,” and there are two places where I model this topic.
The first is an excerpt from my Algebra Survival Guide, an excerpt about subtracting integers that you can check out now.
First click on this link, then scroll down to read pages 43-46 (you can enlarge the print size by increasing the percentage located in the top bar):
But there’s more:
I wrote an entire book on the topic of combining integers, PreAlgebra Blastoff! We created foam manipulatives to get across the idea of integers. There’s a piece of a foam with a hole in the center,
the NEGATon, which stands for – 1. There’s a piece that fills the hole, the POSITon, and that manipulative stands for + 1. When you put the POSITon inside the NEGATon, you get 0, and that is a piece
called a ZERObi.
Using these manipuatives you can model and teach students how to combine integers, and add and subtract integers.
To learn more about this system and to see how the manipulatives work, go to this website:
Let me know if you have any questions on this topic. It’s certainly an important issue.
Happy teaching!
— Josh | {"url":"http://mathchat.me/2008/11/11/subtracting-integers/","timestamp":"2014-04-18T19:12:36Z","content_type":null,"content_length":"89283","record_id":"<urn:uuid:bdb985cb-b04d-4941-b8dd-cfc5f2b3952c>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.7/warc/CC-MAIN-20140416005215-00452-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Journal of the Brazilian Computer Society
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Print version ISSN 0104-6500
J. Braz. Comp. Soc. vol.15 no.3 Campinas Sept. 2009
Compulsory Flow Q-Learning: an RL algorithm for robot navigation based on partial-policy and macro-states
Valdinei Freire da Silva^*; Anna Helena Reali Costa
Laboratório de Técnicas Inteligentes - LTI, Departamento de Engenharia de Computação e Sistemas Digitais - PCS, Escola Politécnica da Universidade de São Paulo - EPUSP, São Paulo - SP, Brasil
Reinforcement Learning is carried out on-line, through trial-and-error interactions of the agent with the environment, which can be very time consuming when considering robots. In this paper we
contribute a new learning algorithm, CFQ-Learning, which uses macro-states, a low-resolution discretisation of the state space, and a partial-policy to get around obstacles, both of them based on the
complexity of the environment structure. The use of macro-states avoids convergence of algorithms, but can accelerate the learning process. In the other hand, partial-policies can guarantee that an
agent fulfils its task, even through macro-state. Experiments show that the CFQ-Learning performs a good balance between policy quality and learning rate.
Keywords: machine learning, reinforcement learning, abstraction, partial-policy, macro-states.
1. Introduction
A common class of tasks in mobile robotics is planning an action policy to reach a desired goal state, usually through maximisation of a value function which designates sub-objectives and helps
choosing the best path. For instance, the shortest path, the path with the shortest time, the safest path, or any combination of different sub-objectives^5, 20. The definition of a task in this class
may contain, besides the value function, some a priori knowledge about the domain, e.g., environment map, environment dynamics, goal position. Such knowledge allows a robot planning, while the lack
of such knowledge obliges the robot either to learn it previously or to make use of heuristic strategies, such as moving to goal direction while avoiding obstacles^19.
While the problem of mapping the environment has received great attention from the robotics community, mainly under the simultaneous localisation and mapping approach^3,1, less attention has been
given to learn the environment dynamics. Given a map and the robot localisation, if a goal position is given, it is possible through path planning to determine a path free of obstacles from the robot
position to such goal. However, even if a priori knowledge is considered about moving directions in the Euclidean space so that an action policy can be computed, minor variations in the environment
dynamics, such as slippery, oblique, or crushed ground, are not captured as well as are not inferred more generic sub-objectives.
Reinforcement Learning (RL)^21 is a learning method that can be applied to the task of learning the dynamic environment and planning an action policy altogether. In RL, an autonomous agent learns an
action policy based on its own experience. This policy is inferred from a process of trial and error, which is guided by the agent itself and received reinforcements that indicate a partial
evaluation of executed actions, besides perceiving transitions among different situations - formally states - evidencing the environment dynamics. The sequence of received reinforcements determines
the value of each executed trajectory. Reinforcements can indicate walked distance, time elapsed or any desirable local situation faced by the robot.
Whereas the robotic task of reaching a goal state in an environment populated with obstacles can be solved through planning, robots based on RL can learn and recover from big changes in the
environment, like the appearance of new obstacles, or small ones, like the appearance of oil in the ground or of crushed ground^2. Moreover, RL does not need to start learning from the scratch, some
partial solution can be considered so that an RL algorithm fills the gaps or a sub-optimal solution can be considered so that an RL algorithm improve it.
Within the last fifteen years, many works about RL have been published^18, 10, 22, 4, 14 extending Sutton's article^22, which brought a mathematical formalism to RL. However, most methods depend
strongly on the size of the state space in which the learning process is done, and gives rise to a trade-off between policy quality and learning speed.
Recent works in RL are attempts at finding methods that accelerate the learning rate without degenerating the policy quality. In such methods three objectives are pursued: scalability, so that no
exponential increase occurs in the complexity of solving tasks when increasing the size of state space; knowledge transfer, so that most of common knowledge can be shared among different tasks; and
stability, so that a method can be applied to different domains.
In this paper we propose a method that concerns scalability and knowledge transfer properties, so that an increase in the learning speed for a specific task in mobile robotics can be reached. On the
other hand, we restrict our algorithm to a specific domain, that of reaching a goal state within an environment that contains obstacles where robots cannot walk through. The proposed method uses a
discretisation of the state space combined with a previously learnt partial-policy^7, both defined in accordance to the complexity of the environment structure. This method is implemented in the
CFQ-Learning algorithm, which stands for Compulsory Flow Q-Learning.
We use both, temporal and spatial abstraction, in order to accelerate the learning process. Spatial abstraction is applied through low resolution discretisation of the state space^11, 16 and similar
states are grouped such that they share characteristics which will be learnt equal to all of them. Temporal abstraction is applied through macro-actions^8, which are a sequence of actions or a
sub-policy that are applied to more than one step, so that less chance for the robot choosing actions is left. However, since there are discontinuity in the state space because of obstacles, we may
use a high resolution discretisation near such discontinuity, or a macro-action to over come the obstacles. We have chosen the second case, using a compulsory flow as partial-policy, which takes
control of the robot near obstacles to get round them.
Based on theoretical and experimental analysis, we show that the CFQ-Learning performs a better balance between policy quality and learning speed than the Q-Learning algorithm does when applied to a
discretised continuous state space.
The remaining of this paper is organised as follows. Section 2 presents the RL formalisation together with Q-learning, the most usual RL algorithm, then, the task domain of interest is presented
followed by reinforcement learning algorithm that can solve it. In Section 3 we define formally the partial policy, named compulsory flow, and describe how to learn such flow. We then present the
CFQ-Learning algorithm in Section 4, which uses the compulsory flow and a discretisation of the state space defined for the current task environment to learn action policies. In Section 5 we compare
the performance of the CFQ-Learning algorithm with the Q-Learning algorithm when different discretisations of the state space are considered. We describe the experiments performed and present the
results obtained. Finally, Section 6 summarises our conclusions.
2. Reinforcement Learning and Task Domain
In works concerning RL, Markovian Decision Processes (MDPs)^17 are adopted as simplified models of real problems. MDP models are built under a well-established mathematical formalism, which
compensates the simplifying conditions used to describe the environment, as there are optimal algorithms to solve problems expressed as MDPs^17.
An MDP is defined by a tuple <A, S, P(s[t+1]|s[t], a[t]), r(s, a)> where A is a finite set of possible actions a, S is a finite set of possible states s, P(s[t+1]|s[t], a[t]) represents transition
probabilities and r(s, a) is a bounded expected reinforcement function^17.
2.1. Q-Learning algorithm
The basic idea behind RL is that the learning agent can learn how to solve an MDP task through repeated interactions with the environment. Note that all that is known by the agent is the set of
actions A and the set of states S, whereas the functions P(s'|s, a) and r(s, a) must be learnt through interaction within the environment.
The environment is described by the set of possible states S, and the agent can perform any action from A. Each time it performs an action a in some state s, the environment reaches a new state and
the agent receives a reinforcement r that indicates the immediate value of this state-action transition (see Figure 1).
The agent must find out a stationary policy of actions a*[t ]= π* (s[t]) that maximises the expected value function V^π(s[t]), which represents the expected reinforcement incurred for a policy π, and
π*(s[t]) = arg max[π] [V^π(s[t])].^17 It is common to assume the discounted-reinforcement value function, which makes use of a discount factor γ ^π (s[t]) is thus defined by:
The RL problem modelled as an MDP can be solved by the Q-Learning algorithm^24, which finds an optimal policy incrementally without considering the transition probabilities of the environment model.
Q-Learning is based on the TD(0) algorithm^22, and estimates a value function Q(s,a) for each state-action pair. This value function is recursively calculated by:
where α[t] is the learning rate and γ is the discount factor.
During the learning process, at the time of choosing action α[t ]it is necessary to select one between two strategies: exploration, which diversifies the policy in order to reach unknown state-action
pairs and may improve the best current known policy, or exploitation, which chooses the best current known policy. Frequently a combination of both strategies is used (ε-greedy), where an exploration
rate ε is defined^9.
2.2. Task domain
Goal-state tasks have many applications in robotics - going to a desired room, holding an object, changing the environment, and so on. Frequently it is required that the robot plans the best possible
path to solve the task within a continuous state space. Although RL algorithms can sub-optimally solve these tasks (for instance, by considering a high-resolution grid world and using an unitary cost
for each action choice), too much time can result to obtain a reasonably good policy, resulting in an inefficient alternative in many cases.
The interest here resides in applications where a set of goalstate tasks are defined for the same kind of environment, so that it is worth acquiring in advance some knowledge about this kind of
environment, and then reuse this knowledge in future tasks, where different goal positions or different environments are defined.
A mobile robot navigating in an one-floor house is a kind of environment that is considered in this paper. Figure 2 shows the environment used in the experiments described in Section 5, where a
mobile robot can move in any direction.
The domain considered in this paper can be defined in a continuous space. In this space we can define a set of continuous states X that represents every possible position of the robot in the
environment. One of the characteristics of such space is the notion of neighbourhood. For example, if a position in a plane is considered, the Euclidean distance can be considered to define a
neighbourhood of each position, meaning that the robot can reach a state in this neighbourhood in the near future.
Although the continuous state space presents some important characteristics when planning, the solution discussed in this paper - RL algorithms - are only applied to discrete spaces. This way, we can
consider a high resolution discrete space S that represents the continuous space of the domain through a map s(x): X → S. We must also consider a set of discrete actions A. The chosen discretisation
should respect the following constraints:
1. The set of continuous states that is mapped into the same discrete state must be compact, i.e., if s(x[i]) = s(x[j]) = s then s(αx[i]+ (1 - α)x[j]) = s for all α
2. The agent moves only to neighbour states in the discrete state space, i.e., P(s[t + 1]= s' | s[t]= s, a[t]) > 0 if and only if s and s' are neighbours. This guarantees that the notion of
neighbourhood in the continuous state space can be extended to the discrete state space; and
3. There are actions that can move the agent, with higher probability, to any direction in the state space, except to places where obstacles exist, i.e., for all neighbouring states s, s' [t + 1]
= s' | s[t]= s, a[t]= a) = max[s" ] P(s[t + 1 ]= s" | s[t]= s, a[t]= a). This implies that if the set of discrete states allows k neighbour states, then | A| > k. This guarantees that the agent
can move from any state to its neighbours.
Figure 3 shows two patterns of discrete states that respect such constraints. In the hexagon pattern, there are 6 possible actions, whereas in the quadratic pattern, there are 8 possible actions.
2.3. Q-Learning and continuous space
The Q-Learning algorithm, as described in Section 2.1, is restricted to discrete spaces (states and actions), and when applied to a continuous state (or action) space, a discretisation process is
necessary. It is usual to use a uniform discretisation of the space (states and actions) as are shown in Figure 3, such that a discrete action is chosen and performed (for a constant period of time,
until the agent makes a transition between discrete states, or until another condition occurs) in a considered discrete state, which encompasses the current real state.
In a continuous space, when applying a continuous control u(x(t)), its respective value function V^u(x(t))(x(t)) is obtained such that:
where x(t) is the current state, r(x,u) is the current reinforcement per time and γ is the discount factor.
In the discretisation process, a set of continuous states is associated with a discrete state s, s: X→S, where X is the set of continuous states and S is the set of discrete states. The function s(x)
relates each continuous state x to a discrete state s = s(x) and for each discrete state in S, it is supposed that the value function of all its continuous states has similar values and similar
optimal policy, which means that if s(x[1]) = s(x[2]) then V*(x[1])[2]) and π*(x1) π*(x2).^4 In general, the set of discrete actions A is chosen in such way that the agent can move to all its
discrete neighbours, as it was seen in section 2.2, the reinforcement function r(s,a) is derived from the continuous state space, and the value function is recursively calculated by Munos and Moore^
where α[t ]is the learning rate, γ is the discount factor, s[t]^x(t) is the discrete state s[t ]mapped from the continuous state x(t), and τ is the time taken to execute action a[t].
As a result of the discretisation process, finite cardinality is obtained. However, the performance of the Q-Learning algorithm is completely dependent on such cardinality and in the way the
discretisation is made. If cardinality is high, a good policy can be obtained, but the learning speed is low, whereas if cardinality is low, the learning speed is high, but the learned policy is of
lower quality. Then a trade-off between learning speed and policy quality must be considered for the discretisation process. When a discretisation is used, the convergence of Q-Learning algorithm is
also corrupted, since it is not possible to guarantee a stationary transition function P(s'|s,a), but it will depend on the policy executed, or, more specifically, the previous discrete state
In most domains, a uniform discretisation is not the best solution, since the environment structure is not considered. Munos and Moore^11 presented an algorithm that makes a non-uniform
discretisation based on the value function variance of continuous states belonged to the same discrete state and on the influence of the value function of a discrete state in other discrete states.
However, the system dynamics is considered to be known and deterministic. Reynolds^16 proposed an adaptive algorithm based on policy to the non-uniform discretisation process, which acquires the
dynamics of the environment whilst executes on-line discretisation of the state.
One reason for having discontinuity in optimal value functions and policies is the existence of obstacles and prohibited state transitions in the environment. Both methods cited above discretises the
space in a more useful way, when compared to the uniform discretisation frequently used. However, when applied to an environment with many obstacles, high resolution is used near obstacles, since
they produce high variations in the value function and in the policy, what decreases the learning speed.
We propose an alternative way to deal with this problem. The idea is to previously define i) an obligatory partial policy, named compulsory flow, that should be performed by the learning agent when
near obstacles, and ii) a low-resolution discretisation of the state space based on the environment structure together with constraints on the action policy to be used in regions free of obstacles.
Once the compulsory flow and the low-resolution discretisation of the state space are defined, this information can be reused in the policy learning process for different tasks defined for the same
environment. Based on these definitions (compulsory flow and low-resolution discretisation), we contribute a new algorithm, called Compulsory Flow Q-Learning, aiming at a better balance between
learning speed and policy quality.
3. Compulsory Flow
Partial policy is a mapping from a environmental region to a subset of possible actions^13, 7, 15 and it helps incorporating a priori knowledge into RL learning methods. Differently from previous
work in the literature, we consider a priori partial policy that reach a desired domain-dependent behaviour in the environment. In this paper, this action subset has only one possible action for each
state. In order to define the compulsory flow, a high-resolution state space discretisation is used. Although it can spend lots of time to determine the compulsory flow, it is calculated only once
for each environment. The same environment structure can be used for different tasks and the compulsory flow can be reused so that the learning speed can be increased.
Also, we will see that the compulsory flow can be defined locally, meaning that it is not sensitive to the global environment, but to situations faced by the robot. The compulsory flow is a partial
policy used when the agent is near obstacles, so that the robot can get around obstacles^6. In this sense, hardcoded partial policy can be used to implement get-around behaviour, being the only
requirement that the agent keeps some inertial movement.
The compulsory flow is defined by the tangential-flow region R[TF] and the tangential-flow policy π[TF]S × A, since π[TF][t], which is reached by performing a[t-1] in s[t-1]. The previous action a
[t-1 ]is used in order to guarantee inertial behaviour, trying to keep the same movement direction when avoiding obstacles.
Definition 1: Let N[TF](s, a) be the expected number of actions performed by an agent to reach an obstacle when executing the action a in the state s and then following a random policy; a[t-1] and a
[t] be the last action executed and the new action to be performed, respectively; [TF]; and a be the vector which represents action a in the continuous space; then it is defined:
Tangential-flow region R[TF]: s[i] [TF ]if and only if
Tangential-flow policy π[TF](.):
where π[TF](sª ) and the vector a is less than 90º, what keeps the agent in a similar movement direction given by [FT], making the agent getting around obstacles.
Figure 4 illustrates a tangential-flow region (gray region) and the corresponding tangential-flow policy when the agent starts at point 1, performs an action a[i] =a[t-1]that drives it into R[TF ]and
activates π[TF](.)(see Definition 1), which avoids collision with the wall by conducting the agent through the compulsory flow until point 2, when it is released and a new action a[i ]can be chosen
to be performed. When the agent is released from a compulsory flow will be explained in the next section.
In order for a learning agent to autonomously define (by exploration) the tangential-flow region R[TF ]for an unknown environment, we propose the use of the following modification of the Q-Learning
update rule:
where α[t ]is the learning rate and γ is the discount factor. We use average instead of maximisation. The same rule can be adapted for different RL-algorithms. The reinforcement function must be
defined to detect obstacles, as it is used in this work, but it can be used to detect other undesirable regions, such as cliff^12, strong magnetic field, high temperature, moist, etc.
Figure 5 shows the value [t], a[t]) = 0 when hitting an obstacle and r(s[t], a[t]) = 1 otherwise, and discount factor γ = 1. The tone of gray represents how far the agent is from reaching a obstacle
walking randomly (black is closer, white is further). The tangential-flow region R[TF ]of this environment can be obtained by defining a desired Figure 6 shows the tangential-flow region R[TF]
obtained with
Notice that, regions with similar structure (corners, U-like form, gates) present similar tangential-flow region. This characteristic is important so that an agent can learn the tangential-flow
region even before taking any action in the environment where a task must be done. The learning of the tangential-flow region and consequently the tangential-flow policy can be learnt before hand if
all typical situations can be experimented by the agent and that such policy be defined in the space of local situations.
It is worth noticing again that the tangential-flow policy is just one possible compulsory flow. As it was already mentioned, a hard-coded behaviour of getting around obstacles can be programmed in
the agent, even with other prohibited regions. Also, the compulsory flow can be used not only to get around regions where the agent cannot get through, but it can be used as away of guaranteeing that
the agent does not damage itself.
4. Compulsory Flow Q-Learning
The CFQ-Learning algorithm addresses applications where previous information about the structure of the environment can be gathered and reused. It may happen when the robot has had already access to
the environment in previous task or if the environment is of some kind previously known.
In our approach, while a high-resolution discretisation is used for the a priori definition of the compulsory flow for a task environment, a low-resolution discretisation is used in the CFQ-Learning
algorithm to learn the task policy, what increases the learning speed while still keeping the agent safe in dangerous regions.
Similarly to the discretisation process described in Section 2.3, a set M of macro-states m is defined by a function m: S → M, where in general the region in a macro-state m
The CFQ-Learning algorithm considers as input: 1) the set S of high-resolution discrete states with the function s : X → S, where X is the set of continuous states; 2) the set M of lowresolution
macro-states with the function m: S → M; 3) the set A of discrete actions; and 4) the tangential-flow region R[TF ]and policy π[TF] with the function N[TF]: S × A → R.
In the algorithm there are three levels of states: 1) the continuous level X, that is where the real interaction of the agent with the environment occurs; 2) the high-resolution discrete level S,
that is where the CFQ-Learning algorithm controls the real agent; and 3) the low-resolution discrete level M, that is where the policy is learnt.
When an action a is chosen in S-level, a correspondent action [n] - t[n-1].
When an action a is chosen in M-level, the agent can operate in two modes: 1) obstacle free --this mode is used in regions free of obstacles and the action a is executed in the S-level; and 2)
compulsory flow --this mode is used when the agent reaches the tangential-flow region and the action determined by π[TF ]is executed. The agent enters in mode 1 every time a macro-state transition
occurs or when the action α takes the agent away from obstacles and there is not a great change in the movement direction (angle between the directions of the previous and the actual actions is less
or equal than 90º). The agent enters in mode 2 every time the agent enters in the tangential-flow region.
The idea behind the CFQ-Learning algorithm is that, once an action a is chosen to be performed at the macro-state m, the action a will be executed whilst the agent is in the same macro-state and this
action a does not drive the agent into the compulsory-flow region R[TF ](previously defined for the environment). Every time the agent invades the R[FT ]region, the compulsory flow π[FT](.) drives
the agent until it can either perform the original action a again or a macro-state transition occurs. In the latter case, the learning agent chooses a new action. The compulsory flow π[FT](.) is
defined on the basis of the N[TF](s, a) previously calculated and stored for being used in the current environment.
Table 1 describes the proposed CFQ-Learning algorithm. Variable t^macro keeps the entrance step of the macro-state, variable R^macro keeps the cumulative reinforcement within current macro-state and
a^macro keeps the first action chosen when entering current macro-state (a^macro is the action that causes all the sequence of actions within current macro-state, i.e., the action that can fire a
partial policy).
As said in the Section 2.3, each possible action a[i ]
5. Experimental Results
In the previous section nothing has been said about the CFQ-Learning algorithm convergence or optimal policy found under CFQ-Learning. Experiments have been conducted to empirically show that a high
quality policy can be found, i.e., close in performance to optimal policy, and that in fact a convergence occurs to such high quality policy. CFQ-Learning is compared with a modified version of
Q-Learning, here called Coarse Q-Learning, using the same low-resolution discretisation of the state space.
The Coarse Q-Learning algorithm used in these experiments is applied to the same set of macro-states used by the CFQ-Learning algorithm. In coarse Q-Learning, once an action is chosen inside a
macro-state, the same action is executed until a transition between macro-states occurs or the agent collides with an obstacle. In the first case a new action is chosen after the macro-state
transition. In the latter case, a random action is selected.
In order to evaluate the algorithm here defined, we choose to experiment in a discrete simulated environment. Figure 7 shows the original high-resolution state space, which has 10,000 states (100 ×
100), whereas Figure 8 shows four different discretisations for the same environment used to compare CFQ-Learning and Coarse Q-Learning. These discretisations were made respecting the three
properties of the set of macro-states M in Section 2.2.
Parameters used in the experiments are: discount factor γ = 0.99, learning rate α = 0.3 decreasing in each episode with rate 0.9999 and exploration rate ε = 0.2 decreasing in each episode with rate
0.99975. The Best Q-Learning results (bold line in Figures 9, 10 and 11) were found following the procedure: 1) execute Q-Learning for 90,000 episodes in the high-resolution discretisation (100 ×
100) and finds the "best policy"; 2) apply the "best policy" to 10,000 episodes and calculates the average performance among those 10,000 episodes; and 3) repeat this procedure for 200 runs and
calculate the average performance among those 200 runs. This result is shown in graphics as the "Best Q-Learning", which is used as a reference to a very good performance.
Figure 9 shows the results using Coarse Q-Learning for different discretisations compared with high-resolution Q-Learning (100 × 100). It is possible to see the great dependence of the policy quality
on the number of states. The number of steps to reach the goal region taken by the Coarse Q-Learning using 36 macro-states is more than 2 times greater than the number of steps taken by the
high-resolution Q-Learning, after 10,000 episodes. If the number of macrostates used is 16, the result is 5 times worse.
Experiments were conducted with CFQ-Learning and its on-line version (when there is no knowledge about the environment and the tangential-flow policy must be obtained during the learning process). In
the first steps of on-line CFQ-Learning, for all s [TF](s, a) =ε, where ε > 0, what means that, any state s is not in the tangential-flow region Rand the tangential-flow policy π[TF] does not take
control of the agent. As the agent collides with obstacles, the Equation 1 is applied and the function N[TF](.) is learnt, defining the real tangential-flow region and policy. In this version of
CFQ-Learning the macro-state policy and the tangential-flow policy are learnt concurrently.
Differently from Coarse Q-Learning, CFQ-Learning obtains a policy with results closer to high-resolution Q-Learning. Figure 10 and Figure 11 show the results for CFQ-Learning and on-line
CFQ-Learning, respectively. Both of them show that CFQ-Learning does not have a great dependence on the number of states: even when 16 states are considered, the number of steps is only 20 percent
worse than the number of steps obtained by Q-Learning. When CFQ-Learning and on-line CFQ-Learning are compared, the greatest differences are in the first 1,000 episodes, when on-line Q-Learning is
learning the tangential-flow policy and tangential-flow region, whereas the final performance of the policies is similar.
It is worth mentioning that the policy to be considered as the best Q-Learning is not really optimal, since it was learnt and there is no guarantee of its convergence. Also, the value shown in
Figures 9, 10 and 11 is a sample of learnt policies. Then, it is allowed statistic variance, occurring that some policy reach better result than the best Q-Learning policy.
6. Conclusion
In this paper we presented a new learning algorithm, which makes use of high-resolution state-space discretisation in the control process, while using low-resolution discretisation in the
policy-learning process. Using this algorithm the learning agent is capable of reaching the goal and finding out a good policy faster than by using algorithms based on high-resolution discretisation
of the state space.
The proposed CFQ-Learning algorithm worked very well in the experiments conducted, having a performance close to the optimal policy, even when using low resolution discretisation of the state space.
Although it is necessary to have a previous knowledge about the environment, such knowledge can be extract during the execution of the first tasks in the environment and reused later on in order to
accelerate the learning process for future tasks.
In cases where it is not possible defining the tangentialflow region and policy a priori different solutions can be adopted. It is possible to use sensors (sonars, laser) to sense the distance and
the direction of the robot to the undesirable regions and then, based on this sensing, to create the compulsory flow. It must be defined a priori the tangential-flow region based on distance and the
tangential-flow policy to get around undesirable regions. Another option is to learn N[TF](s, a) in the sensor space, which can be generalised for different parts of the environment or even different
environments. This sensor space depends only on the regions nearby the agent and their relative positions, not considering the global position of the agent in the environment.
This research was conducted under the CAPES/GRICES Project MultiBot (Grant no. 099/03), FAPESP project Logprop (Grant no. 2008/03995-5) and CNPq project Ob-SLAM (Grant no. 475690/2008-7). Valdinei F.
Silva is grateful to FAPESP (proc. 02/13678-0) and Anna H. R. Costa is grateful to CNPq (Grant No. 305512/2008-0).
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Received: July 7, 2009; Accepted: August 27, 2009
* e-mail: valdinei.freire@gmail.com | {"url":"http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0104-65002009000300007&lng=en&nrm=iso&tlng=en","timestamp":"2014-04-17T18:34:20Z","content_type":null,"content_length":"80977","record_id":"<urn:uuid:ceddc545-06dd-49d8-aecd-f57ef67080c8>","cc-path":"CC-MAIN-2014-15/segments/1398223206120.9/warc/CC-MAIN-20140423032006-00602-ip-10-147-4-33.ec2.internal.warc.gz"} |
Another lifting an FT10 idea - Sport Boat Anarchy
Is it possible to use the engine mount where the engines slides into the water as the aft triangulated lifting point? This seems like a strong part of the boat. New hinges would have to be installed
that would allow the engine hatch cover to be easily removed so that the webbing from the lifting point would be able to angle toward the hook of the crane.
So the questions are:
• Will the engine mount support the triangulated weight?
• Is there a hinge that can be installed so that when the hatch cover is lifted it can easily be removed from over the engine to expose the lifting points on the engine mounts?
Can it be done?
If this line of lifting has been explored please let me know. | {"url":"http://forums.sailinganarchy.com/index.php?showtopic=122205","timestamp":"2014-04-21T12:15:18Z","content_type":null,"content_length":"161192","record_id":"<urn:uuid:4254941c-06ee-4a93-ad00-9a391d99a963>","cc-path":"CC-MAIN-2014-15/segments/1398223203235.2/warc/CC-MAIN-20140423032003-00416-ip-10-147-4-33.ec2.internal.warc.gz"} |
Area of a grass field (rectangle + 2 semicircles)
November 16th 2008, 05:36 PM
Area of a grass field (rectangle + 2 semicircles)
I'm having a hard time with this problem
A grass field is shaped like a rectangle with semicircles on both ends. The total length of the field is exactly 100 feet. How wide, to the nearest tenth of a foot, should the field be if its
total area is 4,000 square feet?
Can I assume the radius of the circle is 1/2 of the width of the rectangle?
And I still am not seeing how I could solve this
November 16th 2008, 05:44 PM
sketch a picture?
total length of field = length of rectangle + 2 radii
width of field = 2 radii
$A = \pi r^2 + 2r(100 - 2r)$
you were given A ... solve for r. | {"url":"http://mathhelpforum.com/pre-calculus/59934-area-grass-field-rectangle-2-semicircles-print.html","timestamp":"2014-04-18T03:40:05Z","content_type":null,"content_length":"4138","record_id":"<urn:uuid:308a0928-8f41-4dba-9f7a-cca5351fc4db>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00282-ip-10-147-4-33.ec2.internal.warc.gz"} |
Functions not data structures
From HaskellWiki
Sometimes the best way to represent data is using a function.
Church encoding can be used to represent any algebraic data type using only functions.
Another example is the following, not very efficient, implementation of a FiniteMap:
type FiniteMap key elt = key -> Maybe elt
emptyFM :: FiniteMap key elt
emptyFM = \k' -> Nothing
addToFM :: (Eq key) => FiniteMap key elt -> key -> elt -> FiniteMap key elt
addToFM m k v = \k' -> if (k == k') then Just v else m k'
delFromFM :: (Eq k) => FiniteMap key elt -> key -> FiniteMap key elt
delFromFM m k = \k' -> if (k == k') then Nothing else m k'
lookupFM :: (Eq k) => FiniteMap key elt -> key -> Maybe elt
lookupFM m k = m k
Run time compilation uses functions not data structures to implement an interpreter. | {"url":"http://www.haskell.org/haskellwiki/index.php?title=Functions_not_data_structures&oldid=36272","timestamp":"2014-04-18T08:43:23Z","content_type":null,"content_length":"18560","record_id":"<urn:uuid:e9af7690-fe38-42f2-9611-8c7572e1dd8e>","cc-path":"CC-MAIN-2014-15/segments/1398223210034.18/warc/CC-MAIN-20140423032010-00078-ip-10-147-4-33.ec2.internal.warc.gz"} |
Review for "One Mathematical Cat, Please!" by Carol J.V. Fisher
Dr. Carol J.V. Fisher's Homepage
I submitted my book, “One Mathematical Cat, Please!” to Brooks/Cole Publishers, and they loved it. Unfortunately, however, it is not a textbook for any particular course instead, it is a supplement
to many courses, from junior high through early college and, after many deliberations, the underlying concern that “supplements don't sell” won out (sigh). Anyways... I will always cherish the line
in this review: “If you can find the right place to put it, this book will do a great service to the mathematical educational world.” Maybe one of you, out there in cyberspace, can help me to locate
“the right place to put it”!
Most of the items mentioned in this review that ask for changes have already been incorporated in the version of the text that is on the web, so you usually won't be able to see what the reviewer was
“complaining” about. (I benefited tremendously from this review; my book is better because of it.)
Review of
ONE MATHEMATICAL CAT, PLEASE!
by Carol J.V. Fisher
This review consists of the answers to your questions and the enclosed manuscript with numerous comments and suggestions written directly on them.
1. Please describe the course in which you would consider using this manuscript. What is the greatest teaching problem in the course? This manuscript is not a textbook for a course. It is
supplementary material for students to help them understand mathematical language, and in particular the distinction between a mathematical thing (an expression or number) and a mathematical sentence
(an assertion, often that two things are the same). I would love for my students to understand this, in lots of courses. I could imagine using this book as an optional (or maybe even required)
supplement for remedial courses (Elementary Algebra, Intermediate Algebra), the "mathematics for elementary education majors" course, the precalculus course, and the finite math and baby calculus
course for business majors.
But if you want to single out one course in which I'd use it, it is MTH 100, a special course we offer in the summer before the freshman year to disadvantaged students who need help getting to the
college level in mathematics and writing. The content of this course is basically Elementary Algebra, but I would use this book at the beginning before we got into the algebra per se. The enrollment
is usually in the 50 80 range, with small sections, some taught by regular faculty and some by part-timers. The greatest teaching problem in this course is the poor mathematical background of the
students and various social and intellectual dynamics that get in the way of their proper attitude toward learning. In answering the rest of the questions, I'll keep this course in mind.
2. Do you agree with the rationale for the contents and approach of the book? Does the author succeed in his/her goal for the book? Yes and yes. There is a real need for students to understand the
ideas in this book. Some students pick it up easily as they study mathematics in high school and beyond. Others are so used to thinking of mathematics as pushing symbols around rather than
communicating ideas in a special language that they never understand these ideas. This hampers their success. Thus I am convinced that the need for this book is immense. Your hardest job will be to
convince unthinking faculty members that the need is there. The author does a good job of getting her points across moving very gently and deliberately, saying things just right for the most part.
3. Which topics to you think should be added? Deleted? I might add some more examples of story problems, but I don't think I'd add any more topics. (Well, maybe a discussion of the order of
operations, picking up on the problems raised on pages 44 46.) I might also like to see more of a discussion of typical errors that students make (like writing 2+3 = 5·2 = 10), integrated throughout
the book. In Section 10 I would add the word "function" to the general discussion, rather than just as a sidebar on page 144 (even elementary school students are told about "function machines"). One
page 112, I would stress this distinction between declarative and imperative modes more. I looked at the web site mentioned there and see that the present manuscript is part of a wider conspiracy!
A few things don't seem to work as well as the rest, and I would consider dropping these (they are marked in the manuscript specifically the material on page 19 and some or all of the last section).
4. Is the level of detail in the explanations of the manuscript appropriate for your students? Are the tone and style of the manuscript suitable for your students? Should any changes be made? Are the
organization and level of problems suitable for your students? Except where I've marked it, the level of detail, tone, style, organization, and level of problems seem about right for my students.
It's not clear that the reading level isn't too high for some of the intended audience, however. (I'm not an expert in education, but it seemed to me that seventh grade students might not have the
intellectual sophistication or the reading skills needed to appreciate the discussion in some places.) Various suggested changes are marked throughout the manuscript.
One stylistic point I'm not too enthusiastic about is the constant talk about what "mathematicians" do. The author should think carefully about the down side of this approach a student says to
himself, "I'm not going to be a mathematician, so this is irrelevant to my learning mathematics." I'm not sure how to change this approach, though it pervades the entire book. On a related note, look
at the use of "result" on page 77; we mathematicians forget sometimes that this is not common English usage. (On the same page is another item I'd want to change the lie that "theorem" means "true
statement" rather than "proved statement". It's a subtle point, not likely to matter to this audience, but we should try to avoid telling white lies when possible, even at this level.)
I would not be quite so forgiving about using calculators for trivial arithmetic as this author is (e.g., averaging 0.1 and 0.2 on page 18), but that's a minor point and she has good reason for doing
it her way (one being that she doesn't want the anxiety of doing the calculations to get in the way).
Since this is not a textbook, you can't really view the "problems" in the normal way students are expected to do them all, and they're an integral part of the presentation, not a separate section of
drill and practice. They are excellent.
In the later sections, more motivation would be useful (e.g., pages 149 and 151).
5. How well does this book stand up to its competitors? That's easy to answer: There are no competitors. It is not a textbook for a course but a supplement, and I've never seen any similar thing
before. One marketing idea you might think about is stressing the NCTM Standards and their emphasis on proper mathematical communication.
6. If you were the sole decision-maker, would you adopt this book for your course? Why? Yes, I would require it as a supplement for MTH 100 (in addition to an elementary algebra text), and cover it
first, returning to it as needed throughout the course. The reason is that I want these students to get the point this little book sets out to make: that they need to communicate mathematics
correctly by understanding the rationale and structure of its language.
7. Any other comments? Let me say that I really enjoyed reading this book. It is wonderfully written and crafted with a care you rarely see. (I think I could find only two typographical errors.) If
you can find the right place to put it, this book will do a great service to the mathematical educational world.
One point I jotted down to include as I was reading the manuscript was to tell you not to let the copy-editors ruin the presentation! The author has been extremely careful and thoughtful with how she
wants to present the material, from layout to font to wording, and there are many very subtle points that copy-editors, used to editing textbooks, will not appreciate. Please give her wide latitude
to dictate against any company policies that will interfere with what she is trying to do. (On the other hand, I personally would take out a few of the exclamation points. And I would use "whether"
in many places where she uses "if". Also, the word "any" needs to be changed in several places; I've marked many such places.)
Another note to myself said that this little book seems to be just one of a series in "truth and language in mathematics"; you need to look at the series as a whole when deciding how to market these
I loved the margin note approach to layout more books should do this! And a good index is also a plus, even in such a small book.
There are a few errors that the author makes, such as confusing an object and the name for an object (the person Carol versus the word she writes as 'Carol'), and in a book with the emphasis of this
one, that distinction must be maintained. See page 12 for one problem. (As someone who grew up with the "new math" of the early 1960s, I recall that this was one of the first main points we were
taught, and it's important!) The author uses the word "digit" to mean "number" (e.g., page 6); she shouldn't. I don't know whether she will want to add a comment or not, but the statement on page 8
that sentences are never expressions in their own right is not true just ask any mathematical logician (and in fact she uses this idea herself later on, when she puts the verb "is equivalent to"
between sentences).
Another thing I would strongly suggest changing is the statement that the symbol for the empty set looks exactly like a zero with a slash through it (see page 31). I prefer to write the former as Ø
[correct symbol not available on web] and the latter as ø [correct symbol not available on web] and make the distinction. Perhaps the author should make even a bigger point that the empty set and
zero are two totally different animals. (She needn't tell these readers that in mathematical logic, zero is defined to be the empty set in some settings!) | {"url":"http://www.onemathematicalcat.org/cat_review.htm","timestamp":"2014-04-16T13:02:42Z","content_type":null,"content_length":"12023","record_id":"<urn:uuid:94a58b7d-fc79-4629-a024-ac9d358e994e>","cc-path":"CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00341-ip-10-147-4-33.ec2.internal.warc.gz"} |
Internal and external path length of a binary search tree
July 15th 2013, 04:40 PM
Internal and external path length of a binary search tree
I'm really bugging on a question I hope somebody here can help me!
If we note by $S_N$ the average search cost in the case of a successful search in a binary search tree and $U_S$ in the case of an unsuccessful search, we can say that
$S_N = I/N$, where $I$ is the internal path length and $N$ is the number of nodes in the tree.
Prove that $U_N=\frac{N}{N+1}(S_N+2)$.
and just for a reminder, $E =I+2N$, where $E$ is the external path length of a binary search tree.
Attachment 28836
July 16th 2013, 10:05 AM
Re: Internal and external path length of a binary search tree
What do you mean by average cost? The usual idea is to count the number of key comparisons in a search. I'm taking my answer directly from D. E. Knuth, Vol.3, Searching and Sorting, p. 410. I
have 100% confidence in this source. So if you let S[N] be the average number of comparisons in a successful search of a binary search tree with N nodes (assuming each of the N keys is equally
likely as a search argument), then
$S_N=1+{I\over N}$ -- this is 1 more than your formula??
Next assume that an unsuccessful search is for a key that is equally likely to be in any of the N+1 intervals determined by the N keys of the tree. Then
$U_N={E\over N+1}={I+2N\over N+1}={N\over N+1}(S_N-1+2)={N\over N+1}(S_N+1)$
If you read a few more pages in Knuth, you get the nice formula (assuming the tree was built initially by inserting the N keys in random order):
$S_N=2(1+{1\over N})H_N-3$ where $H_N$ is the Nth partial harmonic sum.
July 17th 2013, 11:28 AM
Re: Internal and external path length of a binary search tree
Thanks!! I guess my teacher just forgot about the +1 haha | {"url":"http://mathhelpforum.com/discrete-math/220599-internal-external-path-length-binary-search-tree-print.html","timestamp":"2014-04-21T03:27:39Z","content_type":null,"content_length":"7698","record_id":"<urn:uuid:583e4b36-10e2-48bc-811f-2935aa0df17a>","cc-path":"CC-MAIN-2014-15/segments/1397609539447.23/warc/CC-MAIN-20140416005219-00249-ip-10-147-4-33.ec2.internal.warc.gz"} |
Velocity, Acceleration and Distance
1.11. If position of an object is given by: x(t) = A sin(ωt) where A is a constant and ω is
the angular frequency.
a) What is the instantaneous velocity at time t?
b) What is the instantaneous acceleration at time t?
c) Express the instantaneous velocity and the instantaneous acceleration in terms of
x, ω and A.
2. dx(t)/dt= v(t)
3. a) For (a) I just took the derivative of the x function and got wAcos(wt)
b) Same thing here but took the derivative for the answer i got in A and got
c) This is where i had some trouble. I got the instantaneous acceleration part of the
problem by substituting x for Asin(wt) and thus got a(t)=-w^2x. But i had no
clue how to get v(t) in terms of x,w,A.
What i did was set sin^2(wt) + cos^2(wt)=1 and solved for cos^2(wt). then plugged that into v^2(t)=w^2A^2cos^2(wt). I got v^2=w^2(A^2-x^2). After taking the square root i got v = +/-(w * sqrt(A^2-x^
2). Is this right? should there be the +/- or is one ruled out?
Thanks for the help in advance!\sqrt{} | {"url":"http://www.physicsforums.com/showthread.php?t=195310","timestamp":"2014-04-21T09:56:18Z","content_type":null,"content_length":"23567","record_id":"<urn:uuid:80173ce8-be25-4940-aca2-647aefdc98b9>","cc-path":"CC-MAIN-2014-15/segments/1397609539705.42/warc/CC-MAIN-20140416005219-00353-ip-10-147-4-33.ec2.internal.warc.gz"} |
n-truncated object of an (infinity,1)-category
$(\infty,1)$-Category theory
Basic concepts
Universal constructions
Local presentation
Homotopy theory
Paths and cylinders
Homotopy groups
Modalities, Closure and Reflection
An $n$-truncated ∞-groupoid is an n-groupoid.
An $n$-truncated topological space is a homotopy n-type: all homotopy groups above degree $n$ are trivial.
An $n$-truncated object in a general (∞,1)-category is an object such that all hom-∞-groupoids into it are $n$-truncated.
If an object in an (∞,1)-topos_ is $k$-truncated for any (possibly large) $k$, then it is $n$-truncated precisely if all its categorical homotopy groups above degree $n$ are trivial.
The complementary notion of $n$-truncated object is that of an n-connected object of an (∞,1)-category.
In terms of truncations
($n$-truncated $\infty$-groupoid)
An ∞-groupoid $A \in \infty Grpd$ is $n$-truncated for $n \in \mathbb{N}$ if it is an n-groupoid:
Precisely: in the model of ∞-groupoids given by Kan complexes $A$ is $n$-truncated if the simplicial homotopy groups $\pi_k(A,x)$ are trivial for all $x$ and all $k \gt n$.
It makes sense for the following to adopt the convention that $A$ is called
• $(-1)$-truncated if it is empty or contractible – this is a (-1)-groupoid.
• $(-2)$-truncated if it is non-empty and contractible – this is a (-2)-groupoid.
(following HTT, p. 6).
To generalize this, let now $C$ be an arbitrary (∞,1)-category. For $X,A$ objects in $C$ write $C(X,A) \in$∞ Grpd for the (∞,1)-categorical hom-space (if $C$ is given as a simplicially enriched
category then this is just the SSet-hom-object which is guaranteed to be a Kan complex).
Using this, it shall be useful to slightly reformulate the above as follows:
An ∞-groupoid $A$ is $n$-truncated precisely for all other ∞-groupoids $X$ the hom-$\infty$-groupoid $\infty Grpd(X,A)$ is $n$-truncated.
In categorical terms this just says that (∞,k)-transformation between $X$ and $A$ whose components a k-morphisms in $A$ cannot be nontrivial for $k \gt n$ if there are no nontrivial k-morphisms with
$k \gt n$ in $A$.
Using this fact we can transport the notion of $n$-truncation to any (∞,1)-category by testing it on hom-∞-groupoids:
($n$-truncated object in an $(\infty,1)$-category)
An object $A \in C$ of an (∞,1)-category $C$ is $n$-truncated, for $n \in \mathbb{N}$, if for all $X \in C$ the hom-∞-groupoid $C(X,A)$ is $n$-truncated.
This is HTT, def. 5.5.6.1.
Some terminology:
• A 0-truncated object is also called discrete . Notice that this is categorically discrete as in discrete category, not discrete in the sense of topological spaces. An object in an (∞,1)-topos is
discrete in this sense if, regarded as an ∞-groupoid with extra structure it has only trivial morphisms.
• By the above convention on (-2)-truncated $\infty$-groupoids, it is only the terminal objects of $C$ that are (-2)-truncated.
• Similarly, the (-1)-truncated objects are the subterminal objects.
($n$-truncated morphism in an $(\infty,1)$-category)
A morphism $f : X \to Y$ of ∞-groupoids is $n$-truncated if all of its homotopy fibers are $n$-truncated by def. 2.
A morphism $f : X \to Y$ in an (∞,1)-category $C$ is $n$-truncated if for all $W \in C$ the postcomposition morphism
$C(W,f) : C(W,X) \to C(W,Y)$
is $n$-truncated in ∞Grpd.
By the characterization of homotopy fiber of functor categories this is equivalent to saying that $f$ is $k$-truncated when it is so regarded as an object of the over (∞,1)-category $C_{/Y}$.
In terms of categorical homotopy groups
At least if the ambient (∞,1)-category is even an ∞-stack (∞,1)-topos there is an alternative, more intrinsic, characterization of $n$-truncation in terms of categorical homotopy groups in an (∞,1)
Suppose that an object $X$ in an ∞-stack (∞,1)-topos is $k$-truncated for some $k \in \mathbb{N}$ (possibly very large).
Then for any $n \in \mathbb{N}$ this $X$ is $n$-truncated precisely if all the categorical homotopy groups above degree $n$ are trivial.
Recursive definition
In an $(\infty,1)$-category $C$ with finite limits, a morphism $f : X \to Y$ is $k$-truncated (for $k \geq -1$) precisely if the diagonal morphism $X \to X \times_Y X$ is $(k-1)$-truncated.
This is HTT, lemma 5.5.6.15.
By definition $f$ is $k$-truncated if for each object $d \in C$ we have that $C(d,f)$ is $k$-truncated in ∞Grpd. Since the hom-functors $C(d,-)$ preserve (∞,1)-limits, we have in particular that $X \
to X \times_Y X$ in $C$ is $k$-truncated if $C(d,X) \to C(d,X) \times_{C(d,Y)} C(d,X)$ is $k$-truncated for all $d$ in ∞Grpd. Therefore it is sufficient to prove the statement for morphisms in $C =$
So let now $f : X \to Y$ be a morphism of ∞-groupoids. We may find a fibration $\bar \phi : \bar X \to \bar Y$ between Kan complexes in sSet that models $f : X \to Y$ in the standard model structure
on simplicial sets, and by the standard rules for homotopy pullbacks it follows that the object $X \times_Y X$ in $\infty$-Grpd is then modeled by the ordinary pullback $\bar X \times_{\bar Y} \bar
X$ in sSet. And the homotopy fibers of $f$ over $y \in Y$ are then given by the ordinary fibers $\bar X_y$ of $\bar f$ in $sSet$.
This way the statement is reduced to the following fact: a Kan complex $\bar X_y$ is $k$-truncated precisely if the homotopy fibers of $\bar X_y \to \bar X_y \times \bar X_y$ are $(k-1)$-truncated.
We write now $X$ for $\bar X_y$, for simplicity. To see the last statement, let $(a,b) : * \to X \times X$ and compute the homotopy pullback
$\array{ Q &\to& * \\ \downarrow && \downarrow^{\mathrlap{(a,b)}} \\ X &\to& X \times X }$
as usual by replacing the right vertical morphism by the fibration $(X \times X)^I \times_{X \times X} (a,b) \to X \times X$ and then forming the ordinary pullback. This shows that $Q$ is equivalent
to the space of paths $P_{a,b}X$ in $X$ from $a$ to $b$. (Use that gluing of path space objects at endpoints of paths produces a new path space, see for instance section 4 of BrownAHT).
If $X$ is connected, then choosing any path $a \to b$ gives an isomorphism from the the homotopy groups of $P_{a,b} X$ to those of the loop space $\Omega_a X$. These latter are indeed those of $X$,
shifted down in degree by one (as described for instance at fiber sequence).
If $X$ is not connected, we can easily reduce to the case that it is.
For $C$ an $(\infty,1)$-category and $k \geq -2$, the full sub-(∞,1)-category $\tau_{\leq k} C$ is stable under all limits in $C$.
This is HTT, prop. 5.5.6.5.
Let $\mathbf{H}$ be an (∞,1)-topos. For all $(-2) \leq n \leq \infty$ the class of $n$-truncated morphisms in $\mathbf{H}$ forms the right class in a orthogonal factorization system in an (∞,1)
-category. The left class is that of n-connected morphisms in $\mathbf{H}$.
This appears as a remark in HTT, Example 5.2.8.16. A construction of the factorization in terms of a model category presentation is in (Rezk, prop. 8.5). See also n-connected/n-truncated
factorization system.
Model category presentations
There are model structures for homotopy n-types that presentable (∞,1)-category present the full sub-(∞,1)-categories of $n$-truncated objects in some ambient $(\infty,1)$-category. See there for
more details.
Under mild conditions there is for each $n$ a universal way to send an arbitrary object $A$ to its $n$-truncation $\tau_{\leq n} A$. This is a general version of decategorification where n-morphisms
are identified if they are connected by an invertible $(n+1)$-morphism.
For $C$ an (∞,1)-category and $n \geq -2$ in $\mathbb{Z}$ write $\tau_{\leq n} C$ for the full subcategory of $C$ on its $n$-truncated objects.
So for instance for $C =$∞Grpd we have $\tau_{\leq n} \infty Grpd = n Grpd$.
If $C$ is an (∞,1)-category that is presentable then the canonical inclusion (∞,1)-functor
$\tau_{\leq n} C \hookrightarrow C$
has an accessible left adjoint
$\tau_{\leq n} : C \to C_{\leq n} \,.$
This is HTT 5.5.6.18.
Indeed, as the notation suggests, $C_{\leq n}$ is the essential image of $C$ under $\tau_{\leq n}$. The image $\tau_{\leq n} A$ of an object $A$ under this operation is the $n$-truncation of $A$.
So $n$-truncated objects form a reflective sub-(∞,1)-category
$\tau_{\leq n} C \stackrel{\overset{\tau_{\leq n}}{\leftarrow}}{\hookrightarrow} C \,.$
A left exact functor $F : C \to D$ between $(\infty,1)$-categories with finite limits sends $k$-truncated objects/morphisms to $k$-truncated objects/morphisms.
This is HTT, prop. 5.5.6.1.6.
Follows from the above recursive characterization of $k$-truncated morphisms by the $(k-1)$-truncation of their diagonal, which is preserved by the finite limit preserving $F$.
A presentable $(\infty,1)$-functor $F : C \to D$ between locally presentable (∞,1)-categories $C$ and $D$ commutes with truncation:
$F \circ \tau^C_{\leq k} \simeq \tau^D_{\leq k} \circ F \,.$
This is HTT, prop. 5.5.6.28.
By the above lemma, $F$ restricts to a functor on the truncations. So we need to show that the diagram
$\array{ C &\stackrel{F}{\to}& D \\ {}^{\mathllap{\tau_{\leq k}}}\downarrow & (?) & \downarrow^{\mathrlap{\tau_{\leq k}}} \\ \tau_{\leq k } C &\stackrel{F}{\to}& \tau_{\leq k} D }$
in (∞,1)Cat can be filled by a 2-cell. To see this, notice that the adjoint (∞,1)-functor of both composite morphisms exists (because that of $F$ exists by the adjoint (∞,1)-functor theorem and
bcause adjoints of composites are composites of adjoints) and since the bottom morphism is just the restriction of the top morphism and the right adjoints of the vertical morphisms are full
inclusions this adjoint diagram
$\array{ C &\stackrel{G}{\leftarrow}& D \\ \uparrow & & \uparrow \\ \tau_{\leq k } C &\stackrel{G}{\leftarrow}& \tau_{\leq k} D } \,.$
evidently commutes, since it just expresses this restriction.
If $C$ is an (∞,1)-topos, then truncation $\tau_{\leq n} : C \to C$ preserves finite products.
This appears as HTT, lemma 6.5.1.2.
First notice that the statement is true for $C =$∞Grpd. For instance we can use the example In ∞Grpd and Top, model ∞-groupoids by Kan complexes and notice that then $\tau_{\leq n}$ is given by the
truncation functor $tr_{n+1} : sSet \to [\Delta^{op}_{\leq n+1}, sSet]$. This is also a right adjoint and as such preserves in particular product in $sSet$, which are $(\infty,1)$-products in $\infty
From that we deduce that the statement is true for $C$ any (∞,1)-category of (∞,1)-presheaves $C = PSh_{(\infty,1)}(K) = Func_{(\infty,1)}(K^{op}, \infty Grpd)$ because all relevant operations there
are objectwise those in $\infty Grpd$.
So far this shows even that on presheaf $(\infty,1)$-toposes all products (not necessarily finite) are preserved by truncation.
A general (∞,1)-topos $C$ is (by definition) a left exact reflective sub-(∞,1)-category of a presheaf $(\infty,1)$-topos,
$C \stackrel{\overset{L}{\leftarrow}}{\underset{i}{\hookrightarrow}} PSh_{(\infty,1)}(K) \,.$
Let $\prod_{j} i(X_j)$ be the product of the objects in question taken in $PSh(K)$. By the above there we have an equivalence
$\tau_{\leq k} \prod_j i(X_j) \stackrel{\simeq}{\to} \prod_j \tau_{\leq} i(X_j) \,.$
Now applying $L$ to this equivalence and using now that $L$ preserves the finite product, this gives an equivalence
$L \tau_{\leq k} \prod_j i(X_j) \stackrel{\simeq}{\to} L \prod_j \tau_{\leq} i(X_j) \simeq \prod_j L \tau_{\leq} i(X_j)$
in $C$. The claim follows now with the above result that $L \circ \tau_{\leq n} \simeq \tau_{\leq n} \circ L$.
Postnikov tower
By the fact that the truncation functor $\tau_{\leq n}$ is a left adjoint one obtains canonical morphisms
$\tau_{\leq n}A \to A$
as the adjunct of the identity on $A$, and then by iteration also canonical morphisms
$\tau_{\leq (n+1)} A \to \tau_{\leq n} A \,.$
For any $A \in C$ the sequence
$\cdots \to \tau_{\leq 2}A \to \tau_{\leq 1} A \to \tau_{\leq 0} A$
is the Postnikov tower in an (∞,1)-category of $A$. See there for more details.
Homotopy type theory syntax
Discussion of $n$-truncation of types in homotopy type theory via higher inductive types is in (Brunerie). This sends a type to an h-level $(n+2)$-type.
The $(-1)$-truncation in the context is forming the bracket type hProp.
Relation to homotopy groups
In an $(\infty,1)$-topos $C$ there is a notion of categorical homotopy groups in an (∞,1)-topos. For the $(\infty,1)$-topos ∞Grpd given by the model of Kan complexes this coincides with the notion of
simplicial homotopy groups:
This simple relation between $n$-truncation and categorical homotopy groups is almost, but not exactly true in an arbitrary (∞,1)-topos.
Let $\mathbf{H}$ be an (∞,1)-topos and $A \in \mathbf{H}$ an $n$-truncated object.
1. for $k \gt n$ we have for the categorical homotopy groups $\pi_k(A) = *$;
2. if (for $n \geq 0$) $\pi_n(A) = *$, then $X$ is in fact $(n-1)$-truncated.
This implies
If $A \in \mathbf{H}$ is truncated at all (for any value), then it is $n$-truncated precisely if all categorical homotopy groups vanish $\pi_k(A) = *$ for $k \gt n$.
Notice. If $A$ on the other hand is not truncated at all, then all its homotopy groups may be trivial and $A$ may still not be equivalent to the terminal object. This means that Whitehead's theorem
may fail in a general (∞,1)-topos for untruncated objects. It holds, however, in hypercomplete (∞,1)-toposes.
Truncated morphisms
A morphism $f : X \to 0$ is
Between groupoids
For morphisms between 1-groupoids, the notion of $n$-truncation for low $n$ reproduces standard concepts from ordinary category theory.
A functor $f : X \to Y$ between groupoids, is $n$-truncated precisely when regarded as a morphism in ∞Grpd it is
Notice that $f$ being faithful means precisely that it induces a monomorphism on the first homotopy groups.
For $x : * \to X$ any point and $F_{f(x)}$ the corresponding homotopy fiber of $f$, the long exact sequence of homotopy groups gives that $\pi_1(F)$ is the kernel of an injective map
$\cdots \to \pi_1(F) \to \pi_1(X) \hookrightarrow \pi_1(Y,f(x)) \to \cdots \,,$
hence $\pi_1(F_{y}) = *$ for all points $y$ in the essential image of $f$. For $y$ not in the essential image we have $F_y \simeq \emptyset$. In either case it follows that $F$ is 0-truncated.
By def. 3 this is the defining condition for $f$ to be 0-truncated.
Between stacks
Let $C$ be a site and write $Sh_{(2,1)}(C) \hookrightarrow Sh_{(\infty,1)}(C)$ for the (2,1)-topos of stacks/(2,1)-sheaves inside the (∞,1)-sheaf (∞,1)-topos of all ∞-stacks/(∞,1)-sheaves.
Write $L_W [C^{op}, Grpd]$ for the simplicial localization of groupoid valued presheaves in $C$ and write $[C^{op}, sSet]_{proj,loc}$ for the local projective model structure on simplicial presheaves
that presents $Sh_{(\infty,1)}(C)$.
Let $f : X \to Y$ be a morphism of stacks which has a presentation by a degreewise faithful functor that, under the nerve, goes between fibrant simplicial presheaves.
Then $f$ is 0-truncated as a morphism in $Sh_{(\infty,1)}(C)$.
We need to check that for any $\infty$-stack $A$ the morphism $Sh_\infty(A,f)$ is 0-truncated in ∞Grpd. We may choose a cofibrant model for $A$ in $[C^{op}, sSet]_{proj,loc}$ and by assumption that
$X$ and $Y$ is fibrant we have that the ordinary hom of simplicial presheaves $[C^{op}, sSet](A, f)$ is the correct derived hom space morphism. This is itself (the nerve of) a faithful functor, hence
the statement follows with prop. 9.
In $\infty Grpd$ and in $Top$
An object in ∞Grpd is $n$-truncated precisely if it is an n-groupoid. To some extent this is so by definition. Equivalently, an object in Top is $n$-truncated if it is (in the equivalence class of) a
homotopy n-type.
So we have for $n \in \mathbb{N}$ a reflective sub-(∞,1)-category
$n Grpd \stackrel{\overset{\tau_{\leq n}}{\leftarrow}}{\hookrightarrow} \infty Grpd \,.$
If we model the (∞,1)-category ∞Grpd as the Kan complex-enriched category/fibrant simplicial category $KanCplx \subset$sSet of Kan complexes, then the truncation adjunction
$(\tau_{\leq n } \dashv i) : n Grpd \stackrel{\overset{\tau_{\leq n}}{\leftarrow}}{\hookrightarrow} \infty Grpd \,.$
is modeled by the simplicial coskeleton sSet-enriched adjunction
$(tr_{n+1} \dashv cosk_{n+1}) : KanCplx_{n+1} \stackrel{\overset{tr_{n+1}}{\leftarrow}}{\underset{cosk_{n+1}}{\to}} KanCplx \,,$
where $KanCplx_{n+1}$ is the subcategory of $[\Delta^{op}_{\leq n+1}, Set]$ on those truncated simplicial sets that are truncations of Kan complexes, regarded as a Kan-complex-enriched category by
the embedding via $cosk_{n+1}$.
Notice that every Kan complex $X$ which is $n$-truncated is homotopy equivalent to one in the image of $cosk_{n+1}$, namely to $cosk_{n+1} tr_{n+1} X$, because by one of the properties of $cosk_{n+1}
$ we have that the unit
$X \to cosk_{n+1} tr_{n+1} X$
induces isomorphisms on homotopy groups $\pi_k$ for $k \leq n$.
This shows that $KanCplx_{n+1}$ is indeed a full sub-(∞,1)-category of $KanCplx$ on $n$-truncated objects
Moreover, by the fact discussed at Simplicial and derived adjunctions at adjoint (∞,1)-functor we have that the sSet-enriched adjunction $(tr_{n+1} \dashv cosk_{n+1})$ on $KanCplx$ indeed presents a
pair of adjoint (∞,1)-functors on ∞Grpd. So $tr_{n+1} : KanCplx \to KanCplx$ indeed presents the left adjoint $\tau_{\leq} : \infty Grpd \to n Grpd$ to the inclusion $n Grpd \hookrightarrow \infty
Let $X$ be an object that is $n$-truncated. This means that $X \to *$ is an $n$-truncated morphism. So by prop. 2 the diagonal on that object
$\Delta : X \to X \times X$
is an $(n-1)$-truncated morphism, and precisely if it is $(n-1)$-truncated is $X$$n$-truncated.
In particular, the diaginal is a monomorphism in an (∞,1)-category, hence (-1)-truncated, precisely if $X$ is $0$-truncated (an h-set).
The discussion of truncated objects in an $(\infty,1)$-category is in section 5.5.6 of
The discussion of categorical homotopy groups in an (∞,1)-topos is in section 6.5.1.
A discussion in terms of model category presentations is in section 7 of
A classical article that amplifies the expression of Postnikov towers in terms of coskeletons is
Discussion in the context of homotopy type theory is in | {"url":"http://www.ncatlab.org/nlab/show/n-truncated+object+of+an+(infinity%2C1)-category","timestamp":"2014-04-17T06:43:37Z","content_type":null,"content_length":"157982","record_id":"<urn:uuid:83ec5281-6f74-4932-afee-79c5ce47196a>","cc-path":"CC-MAIN-2014-15/segments/1397609526311.33/warc/CC-MAIN-20140416005206-00521-ip-10-147-4-33.ec2.internal.warc.gz"} |
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1. What is the difference between empty relation and universal relation
2. If A is an invertible matrix of order two, then what is the value of det(A-1)
3. Find the domain of the functions in –1 (2x + 1)/3
4. For a system of Equations, if IAI = 0, where A is a square matrix. How will you examine the consistency of the system of equations
5. If a matrix has 8 elements, what are the possible orders it can have?
6. State chain rule
7. List the relation in the set N given by R= {(a,b) : a=b-2, b>6}
8. Find the value of tan-1 square root of 3 sec-1(-2)
9. If f(x) = x , then what is f(0)
10. If f: R ---à R is defined by f(x) = x/(x2 + 1). Find f (f(2)). | {"url":"http://www.nikithaonline.com/questionpapers_Maths_12th_Onam.php","timestamp":"2014-04-18T21:13:49Z","content_type":null,"content_length":"23967","record_id":"<urn:uuid:e238442b-d86c-4bfc-b6c2-257ddd374657>","cc-path":"CC-MAIN-2014-15/segments/1397609535095.9/warc/CC-MAIN-20140416005215-00470-ip-10-147-4-33.ec2.internal.warc.gz"} |
Who is Andrew Wiles?
In addition to the article below, you can read an interview with Andrew Wiles at the Nova website.
I first met Andrew Wiles when I began researching for a BBC documentary about his proof of Fermat’s Last Theorem. Although he was clearly a man with a brilliant mind, an immense determination and an
obsession that had haunted him since childhood, he came across as a modest, diffident man. It was obvious that he hated publicity, so it was not a complete surprise that initially he was reluctant to
have his story filmed for TV.
Eventually, my colleague, John Lynch, persuaded him that it was his duty to be filmed. By telling his story on screen, Wiles could inspire a new generation of mathematicians and reveal the power of
mathematics to the public. Here was a story of passion and intrigue that would enchant people around the world.
Wiles first became aware of the Last Theorem when he was ten years old. On his way home from school he popped into the Milton Road library and began reading The Last Problem by Eric Temple Bell. From
that moment on, he dedicated his life to finding a proof, even though this was something that had eluded the greatest brains on the planet for three centuries.
He completed his PhD in mathematics under the supervision of John Coates and eventually became a professor at Princeton University. His research was in number theory, but was not aimed at Fermat’s
Last Theorem. Three hundred years after Fermat had thrown down the gauntlet, mathematicians had decided to put Fermat’s Last Theorem to one side because they considered it impossible. For example,
the mathematician David Hilbert was asked why he did not attempt a proof of the Last Theorem, and he replied, “Before beginning I should have to put in three years of intensive study, and I haven’t
that much time to squander on a probable failure.”
But in the 1980s, work by Ken Ribet and Gerhard Frey built a bridge between the Last Theorem and mainstream mathematics, in particular the sort of ideas that Wiles was already familiar with. To cut a
long story short, Wiles now had to prove the Taniyama-Shimura conjecture, a problem that had been around for decades and which was considered impossible. Nevertheless, as far as Wiles was concerned,
anything that led to Fermat’s Last Theorem was worth pursuing. For the next seven years, Wiles worked in complete secrecy, formulating the proof of the century.
Wiles’s incredible journey is too long to even begin to explore on this page, but it is best summarised by the following quote by Andrew Wiles, which draws an analogy between doing mathematics and
exploring a dark mansion:
“You enter the first room of the mansion and it’s completely dark. You stumble around bumping into the furniture but gradually you learn where each piece of furniture is. Finally, after six months or
so, you find the light switch, you turn it on, and suddenly it’s all illuminated. You can see exactly where you were. Then you move into the next room and spend another six months in the dark. So
each of these breakthroughs, while sometimes they’re momentary, sometimes over a period of a day or two, they are the culmination of, and couldn’t exist without, the many months of stumbling around
in the dark that precede them.”
In 1995, Wiles’s proof was officially published and accepted by the mathematical community. The story of Fermat’s Last Theorem had come to an end. We now know that Fermat’s Last Theorem is true, but
one question remains. What was Fermat’s original proof? Wiles’s proof is too complicated to be the same as Fermat’s, so some people continue to search for the original proof – if, indeed, such a
proof exists – it could be that Fermat made a mistake and never had his own proof. If you think you have discovered Fermat’s proof, then please do not send them to Andrew Wiles as he does not have
the time to look at such proofs. Similarly, I have neither the time, nor the expertise, so please do not send proofs to me.
On 27 June, 1997, Wiles collected the Wolfskehl Prize, which was worth roughly $50,000. This is much less than Wolfsekhl had intended a century earlier, but hyperinflation had eroded its value. The
mathematical equivalent of the Nobel Prize is the Fields Medal, but winners must be younger than forty, so Wiles just missed out. Instead, he received a special silver plaque at the Fields Medal
ceremony in honour of his momentous achievement.
Wiles also won the prestigious Wolf Prize, the King Faisal Prize and numerous other international awards. But money, awards and honour were not the driving force behind Wiles’s achievement. As he
said in the BBC documentary:
‘This was my childhood passion. There’s nothing to replace that. I had this very rare privilege of being able to pursue in my adult life what had been my childhood dream. I know it’s a rare
privilege, but if you can tackle something in adult life that means that much to you, then it’s more rewarding than anything imaginable. Having solved this problem there’s certainly a sense of loss,
but at the same time there is this tremendous sense of freedom. I was so obsessed by this problem that for eight years I was thinking about it all the time – when I woke up in the morning to when I
went to sleep at night. That’s a long time to think about one thing. That particular odyssey is over. My mind is at rest.’ | {"url":"http://simonsingh.net/books/fermats-last-theorem/who-is-andrew-wiles/","timestamp":"2014-04-18T03:45:53Z","content_type":null,"content_length":"37841","record_id":"<urn:uuid:96928015-8ffc-4abe-b729-5b55bf8125bf>","cc-path":"CC-MAIN-2014-15/segments/1397609532480.36/warc/CC-MAIN-20140416005212-00037-ip-10-147-4-33.ec2.internal.warc.gz"} |
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Vivian L.
Middletown, CT
Marisa's Flower Shop charges $3 per rose plus $16 for a delivery. Chris wants to have a bouqet of roses delivered to his mother. Which value is in the range of the function that gives the
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Rebecca R.
Mountain Home, ID
trying to check my daughter's algebra homework...
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Parviz F.
Woodland Hills, CA
I need this answer now! Please help me!
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Vivian L.
Middletown, CT
What is the equation of line parallel to y = -7/6x + 10 if the coordinates are (0,4)?
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Vivian L.
Middletown, CT
the sales tax collected on each sale varies directly as the amount of the sale. What is the constant of varation if a sale tax of $0.63 is collected on a sale of $9.00? how would u...
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Bryce R.
Gainesville, FL
-3<x-1≤2 i solved ...... -3<x-1 -2<x and ....... x-1≤2 x≤3 but how do i graph this???? please help
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Vivian L.
Middletown, CT
Write the slope intercept form for an equation of the line that passes through the given point and is parallel to the graph of each.
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Vivian L.
Middletown, CT
Write the slope-intercept form of an equation that passes through the given point and is perpendicular to the graph of each equation.
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Lori C.
Sherwood, AR
y+2= -3(x-1) how do I solve.?
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Alex S.
Fairfax Station, VA
Two road crews, working together, repair 1 mile of road in 4 hours. Working separately, one of the crews takes about 6 hours to repair a similar road. How long would it take the other crew,
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Vivian L.
Middletown, CT
i need some help on some questions can you help
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Vivian L.
Middletown, CT
A. y - 1 = -(x - 2) B. y - 4 = -(x + 3) C. y - 4 = -(x + 3) D...
Latest answer by
Vivian L.
Middletown, CT
A. y - 1 = -(x - 2) B. y - 4 = -(x + 3) C. y - 4 = -(x + 3) D... | {"url":"http://www.wyzant.com/resources/answers/algebra_1?f=active&pagesize=20&pagenum=7","timestamp":"2014-04-21T06:02:07Z","content_type":null,"content_length":"64222","record_id":"<urn:uuid:a4bad542-bff4-4b75-89a5-1eb9e2f18e24>","cc-path":"CC-MAIN-2014-15/segments/1397609539493.17/warc/CC-MAIN-20140416005219-00363-ip-10-147-4-33.ec2.internal.warc.gz"} |
Ted Bunn’s Blog
On Roger Ebert’s blog, the acclaimed film editor Walter Murch explains what he sees as insurmountable problems with 3D movies:
The biggest problem with 3D, though, is the “convergence/focus” issue. A couple of the other issues — darkness and “smallness” — are at least theoretically solvable. But the deeper problem is
that the audience must focus their eyes at the plane of the screen — say it is 80 feet away. This is constant no matter what.
But their eyes must converge at perhaps 10 feet away, then 60 feet, then 120 feet, and so on, depending on what the illusion is. So 3D films require us to focus at one distance and converge at
another. And 600 million years of evolution has never presented this problem before. All living things with eyes have always focussed and converged at the same point.
That’s an interesting idea. It’s true the convergence and focus are two separate processes: when you look at something close to you, your eyes tilt in towards each other (convergence), and each eye
shifts its focus. The latter process is known as accommodation and involves flexing muscles in the eye to change the power of the lens.
It’s certainly true that 3D movies involve one but not the other, and it’s possible in principle that this has an effect on how we perceive them, but I wouldn’t have thought it was a significant
effect in practice. This is way outside of my expertise, but here’s how it seems to me anyway.
The eye is set up so that, when the muscles are completely relaxed, you’re focusing on points that are extremely far away — “at infinity”, as they usually say. The closer you want to focus, the more
you have to strain the muscles. The amount of strain is very small for a wide range of distances, shooting up sharply as the distances get small. Here’s a graph I mocked up:
The horizontal axis is the distance to the object you’re looking at, and the vertical axis is the amount of strain the muscles have to provide — to be specific, it’s the fractional change in power of
the lens, Delta f/f. In case you’re wondering, I assumed the diameter of the eye is 25 mm and the person’s near point is 25 cm. The graph starts at the near point, so the highest point on the graph
is the maximum accommodation the person’s eye is capable of.
The point is that the things you’re looking at in a 3D movie are pretty much always “far away”, as far as accommodation is concerned. The range of examples Murch gives, from 10 feet (i.e. 3 meters)
on up, is a good example. Note that the graph is very flat for this range.
If 3D movies routinely involved closeups of a book someone was reading, or the construction of a ship in a bottle, that’d be different. But they don’t. Most of the time, the point you’re looking at
is far enough away to be practically at infinity, so your visual system should be expecting not to have to do any accommodation. And of course it doesn’t have to, because the screen is really quite
far away (essentially at infinity).
So it seems implausible to me that the accommodation / convergence problem really matters. But this is very, very far from any area of expertise of mine, so maybe I’m wrong. | {"url":"http://blog.richmond.edu/physicsbunn/2011/01/","timestamp":"2014-04-19T09:33:01Z","content_type":null,"content_length":"42402","record_id":"<urn:uuid:e9d6f408-579e-42bc-bb41-51407a17a9c8>","cc-path":"CC-MAIN-2014-15/segments/1397609537097.26/warc/CC-MAIN-20140416005217-00586-ip-10-147-4-33.ec2.internal.warc.gz"} |
Number Sense and Operations - Grade 6
┃Number Sense & Operations - Grade 6┃
┃(This page will open links outside of Links to Learning. The links will open in a new browser window - you can close out the new window when you are done.)┃
Click Here for Lesson Plans/Classroom Activities
Try to name the numbers that are written by choosing the correct written numbers that pop up.
We use certain properties of integers to solve math problems: Commutative property of addition, Commutative property of multiplication, Associative property of addition, Associative property of
multiplication, and the Distributive property.
The whole numbers are the counting numbers and 0. The whole numbers are 0, 1, 2, 3, 4, 5, ...
Visit this site to learn all about positive and negative numbers, including the number line, absolute value and more.
Integers Positive integers are all the whole numbers greater than zero: 1, 2, 3, 4, 5, ... . Negative integers are all the opposites of these whole numbers: -1, -2, -3, -4, -5, . We do not consider
zero to be a positive or negative number.
Comparing Number Values This game uses comparison operators such as "greater than", "less than", and "equal to" (>, <, =) to compare values.
Choose the Proper Sign Choose the proper sign (<, >, =) to make the number sentence true.
All About Comparing Numbers These pages teach number comparison skills. Each page has an explanation, interactive practice and challenge games about comparing numbers.
You are given a table of numbers. Click on the numbers in order from lowest to highest (or highest to lowest).
Distance, Rate and Time - First Glance A rate is a ratio that compares two different kinds of numbers, such as miles per hour or dollars per pound.
Rate A rate is a ratio that expresses how long it takes to do something, such as traveling a certain distance. To walk 3 kilometers in one hour is to walk at the rate of 3 km/h.
Converting Rates We compare rates just as we compare ratios, by cross multiplying. When comparing rates, always check to see which units of measurement are being used.
Average Rate of Speed The average rate of speed for a trip is the total distance traveled divided by the total time of the trip.
Ratio and Proportion A ratio is a comparison of two numbers. We generally separate the two numbers in the ratio with a colon (:). Suppose we want to write the ratio of 8 and 12. We can write this as
8:12 or as a fraction 8/12, and we say the ratio is eight to twelve.
A proportion is a special form of an algebra equation. It is used to compare two ratios or make equivalent fractions. A ratio is a comparison between two values. . .
6 3 Ratios and Proportions A ratio is a comparison of two similar quantities obtained by dividing one quantity by the other. A proportion is a statement of the equality of two ratios.
A proportion is a name we give to a statement that two ratios are equal.
A rate is a ratio that compares two different kinds of numbers, such as miles per hour or dollars per pound. A unit rate compares a quantity to its unit of measure. A unit price is a rate comparing
the price of an item to its unit of measure.
Two figures that have the same shape are said to be similar. When two figures are similar, the ratios of the lengths of their corresponding sides are equal.
Ratios Ratios tell how one number is related to another number. A ratio may be written as A:B or A/B or by the phrase "A to B". You can learn, practice, play and explore ratios at this interactive
Ratios On this page, we hope to clear up problems that you might have with fractions and their uses in Algebra. Ratios are continually being utilized in math and make many things much easier to do.
Scroll down or use the links below to start understanding ratios better!
We use ratios to make comparisons between two things. When we express ratios in words, we use the word "to" -- we say "the ratio of something to something else".
Determining Ratios A ratio of 1:5 says that the second number is five times as large as the first. The following steps will allow a ratio to be determined if two numbers are known.
Questions and answers all about ratios and proportion.
Percent means "out of 100." We can use the percent symbol (%) as a handy way to write a fraction with a common denominator of 100.
All About Percents and Ratios These pages teach percent and ratio skills. Each page has an explanation, interactive practice and challenge games about percents and ratios.
Learn all about percent along with interactive practice activities.
Percent concepts are covered carefully and thoroughly in this volume of lessons. The relationships between ratios, fractions, decimals and percent are explained. Students should go through these
lessons in order since each lesson builds upon the previous one.
Learn everything from a basic definition of percent to converting percents to fractions and decimals.
Absolute Value The absolute value of a number can be considered as the distance between 0 and that number on the real number line.
Absolute value has many uses, but you probably won't see anything interesting for a few more classes yet. For now, you can view absolute value as the distance from zero.
|-1234| = 1234 Absolute Value of an Integer The number of units a number is from zero on the number line. The absolute value of a number is always a positive number (or zero). We specify the absolute
value of a number n by writing n in between two vertical bars: |n|.
Quia - Integers, Absolute Value & Operations with Integers This activity includes finding absolute value of integers, comparing integers, and addition, subtraction, multiplication and division of
Rational numbers are the numbers that can be represented as the quotient of two integers. Conversely, irrational numbers are the numbers that cannot be represented as the quotient of two integers,
i.e., irrational numbers cannot be rational numbers and vice-versa.
Rational Numbers A rational number is a number that can be expressed as a fraction or ratio (rational). The numerator and the denominator of the fraction are both integers.
Quia - Rational Number Practice Express all answers as improper fractions. Review the entire quiz before you start to ensure that all the questions loaded.
Quia - Rational Number Jeopardy Play against a partner! Show your fraction superiority!
Get a definition of a fractions, along with comparing, adding, subtracting and multiplying fractions. This site also includes information on prime numbers, greatest common factor and least common
Like fractions are fractions with the same denominator. You can add and subtract like fractions easily - simply add or subtract the numerators and write the sum over the common denominator.
How to Add and Subtract Fractions Learn how to add and subtract fractions with the same denominator and different denominators.
To multiply fractions: simplify the fractions if not in lowest terms, multiply the numerators of the fractions to get the new numerator, and multiply the denominators of the fractions to get the new
To multiply two simple fractions, complete the following steps: 1. Multiply the numerators. 2. Multiply the denominators. 3. Reduce the results.
To divide any number by a fraction: multiply the number by the reciprocal of the fraction, simplify the resulting fraction if possible, check your answer: Multiply the result you got by the divisor
and be sure it equals the original dividend.
Learn the basics of dividing simple fractions in easy to understand language.
The steps are the same whether you're adding or subtracting mixed numbers: find the Least Common Denominator (LCD), find the equivalent fractions, add or subtract the fractions and add or subtract
the whole numbers, and write your answer in lowest terms.
Multiplying Mixed Numbers Here are the steps for multiplying mixed numbers: change each number to an improper fraction, simplify if possible, multiply the numerators and then the denominators, put
answer in lowest terms, and check to be sure the answer makes sense.
Dividing Mixed Numbers Here are the steps for dividing mixed numbers: change each mixed number to an improper fraction, multiply by the reciprocal of the divisor, simplifying if possible, put answer
in lowest terms, and check to be sure the answer makes sense.
A compound fraction is sometimes called a mixed number. To manipulate compound fractions, just convert them to simple fractions and follow rules 1 through 23 for simple fractions.
We reduce a fraction to lowest terms by finding an equivalent fraction in which the numerator and denominator are as small as possible.
Get a basic definition of decimal numbers, along with adding, subtracting, comparing and rounding decimals, and much more!
Learn about decimals using this program. It shows you how to count decimals using graphic shapes. Just count the shaded squares and select the right answer. Give it a whirl!
Decimals These pages teach operations on decimals covered in math courses. Each page has an explanation, interactive practice and challenge games about decimals.
Check out this page of fast facts about decimals, including adding, subtracting, multiplying and dividing decimals.
Decimals on a Number Line To represent a decimal on a number line, divide each segment of the number line into ten equal parts.
Round the decimal to the nearest whole number.
Equivalent Decimals Match the equivalent decimals with this interactive game.
Fraction to Decimal Conversion From Dave's Math Tables, use this chart to find the decimal equivalent for fractions up to and including 31/32.
Learn all about converting numbers between fractions, decimals and percent.
How do you calculate 2 + 3 x 7? Is the answer 35 or is the answer 23? To know the correct answer, one must know the correct order of operations with respect to addition, subtraction, multiplication,
division, etc.
These materials are designed to help you practice solving problems using the correct order of operations. After solving each problem in this tutorial, you will click on your answer and go to another
page where you will be told if your answer was correct. If you made an error, you will be given specific information, perhaps with examples, to help you improve your skills.
Help Tortisaurus finish building his stone pyramid. You will be shown 3 numbers and an equation. Type in the correct number in the correct place to complete the equation.
Exponents are a shorthand way to show how many times a number, called the base, is multiplied times itself. A number with an exponent is said to be "raised to the power" of that exponent.
Exponents This site will help you learn the basic properties of exponents, along with how to multiply, divide, and estimate with exponents. There are also sections on logarithms. Each topic includes
exercise problems.
Just as (7)(6) is an easier way to describe 6+6+6+6+6+6+6, so to is 3^5 another way of saying (3)(3)(3)(3)(3).
Glowla has created this weird machine just for you to play with! Here's the deal: The machine will show you a bunch of numbers. You need to type in an estimate of what those numbers would add up to.
Sure, if you had all the time in the world, you could add the numbers together, but that's the hitch-- you only have 60 seconds to give your estimate! The secret to solving this puzzle is to round
the numbers before adding them together.
Print out and play this fun estimation game with a friend!
All About Estimation and Rounding These pages teach estimation and rounding skills. Each page has an explanation, interactive practice and challenge games about estimation.
Round Off Whole Numbers Sometimes it is easier to use rounded off numbers. For example, it is easier to say about 100 than 98.9.
Test your knowledge of rounding with these interactive flashcards. | {"url":"http://www.linkstolearning.com/links/NewYork/number_sense_and_operations_-_grade_6.htm","timestamp":"2014-04-19T07:05:01Z","content_type":null,"content_length":"33623","record_id":"<urn:uuid:6cdca44a-bc3b-4388-a5c0-751cf011904b>","cc-path":"CC-MAIN-2014-15/segments/1397609536300.49/warc/CC-MAIN-20140416005216-00659-ip-10-147-4-33.ec2.internal.warc.gz"} |
example: Eliminate ``bad'' beam-position-monitor readouts from PAR x BPM data, where a bad readout is one that is more than three standard deviations from the mean:
sddsoutlier par.bpm par.bpm1 -columns=P?P?x -stDevLimit=3
Fit a line to readout P1P1x vs P1P2x, then eliminate points too far from the line.
sddspfit par.bpm -pipe=out -columns=P1P2x,P1P1x
| sddsoutlier -pipe=in par.2bpms -column=P1P1xResidual -stDevLimit=2
Same, but refit and redo outlier elimination based on the improved fit:
sddspfit par.bpm -pipe=out -columns=P1P2x,P1P1x
| sddsoutlier -pipe par.2bpms -column=P1P1xResidual -stDevLimit=2
| sddspfit -pipe -columns=P1P2x,P1P1x
| sddsoutlier -pipe=in par.2bpms -column=P1P1xResidual -stDevLimit=2 | {"url":"http://www.aps.anl.gov/Accelerator_Systems_Division/Accelerator_Operations_Physics/manuals/SDDStoolkit/node80.html","timestamp":"2014-04-20T08:47:37Z","content_type":null,"content_length":"9155","record_id":"<urn:uuid:1a376e95-a2bf-48bc-a0c2-9fa3b58f7c30>","cc-path":"CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00606-ip-10-147-4-33.ec2.internal.warc.gz"} |