text:
1.1 Deﬁnitions and Examples We try to motivate the deﬁnitions to come. Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there! Use the associative law of multiplication to write-- and here they have 12 times 3 in parentheses, and then they want us to multiply that times 10-- in a different way. Here are the two most commonly used measures of association: Covariance Correlation Both measures are used to show how closely two data sets are related to each other. Measures of association quantify the strength and the direction of the relationship between two data sets. For more information on references and reference examples, see Chapters 9 and 10 of the Publication Manual as well as the Concise Guide to APA Style (7th ed. And now, let me guess your thoughts. Associative definition is - of or relating to association especially of ideas or images. See how positive correlation works in everyday life, science and more. The Mathematical Association, 259 London Road, Leicester, LE2 3BE Charity Registration Number: 1117838 V.A.T. ). Note that a - (b - c) is NOT the same as (a - b) - c [unless c = 0]. Examples for the association of algebraic and rational numbers. Association is a property of arithmetic operations. There are over 200 Math Circle programs nationwide with approximately 100 active Math Student Circles What is MathCircles.org? Composition and Aggregation are the two forms of association. If this page is mystical to you return to the previous page for an explanation of the notation and issue. As a member, you'll also get unlimited access to over 83,000 lessons in math, English, science, history, and more. The associative property is helpful while adding or multiplying multiple numbers. Parentheses in Math: Rules & Examples 4:01 Algebra Vocabulary Terms 3:48 What Is The Order of Operations in Math? Justify your steps. Illustrated definition of Associative Law: When adding it doesnt matter how we group the numbers (i.e. Association In Object-Oriented programming, an Object communicates to other Object to use functionality and services provided by that object. We have a 77 plus 2 in parentheses, plus 3, in a different way. Worked examples Simplify 3a – 5b + 7a. A […] From Tom's house to Sally's house is 1.5 miles. Test your math skills with this basic math practice test with addition, substraction, division and multiplication problems. As a continuation of the National Association of Math Circles, mathcircles.org aims to nurture the growth of Math Circles nationwide by building a community of Math Circle leaders in which new and established leaders are connected, encouraged, and inspired. Suppose X is a graph with vertex set V and diameter d. For i ˘1,...,d we deﬁne Xi to be the graph withV, Xi Recent Examples on the Web Berninger talk-sings throughout his solo record, narrating free-associative thoughts from the nostalgic corner of his brain. Association is, to some extent at least, the opposite of the contrast principle (or perhaps a contrast is an automatic association). is through the use of an instance field. Associations can be described as a "has-a" relationship because the typical implementation in Java is through the use of an instance field. COVID Resources Visit our AMS COVID-19 page for educational and professional resources and scheduling updates On Monday, Tom rode his bike to and from Sally's house 5 Describe the association between price and the number of buyers. Examples of Associations Let’s take, for instance, the word "lemon". Use the associative law of addition to write the expression. The associative property states that the order in which two or more operations are carried out does not affect the result. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. The main difference between them is the units in which they […] Learn vocabulary, terms, and more with flashcards, games, and other study tools. association schemes and describe their structure. A correlation of zero indicates that between the variables, there is no relationship. Clang Association and Semantic Change "Similarity or identity of sound may likewise influence meaning. Even though the parentheses were rearranged on each line, the values of the expressions were not altered. Its is a process of connecting the conduct with the resulting effect. Think a little. Fay, from the Old French fae 'fairy' has influenced fey, from Old English fæge 'fated, doomed to die' to such an extent that fey is practically always used nowadays in … Association is a property of arithmetic operations. Any rational number x=p/q is the solution of qx-p=0 , where q is positive. examples and step by step solutions, understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. It is a statistical measurement of the way 2 variables relate where positive correlation ranges from positive one (+1) to negative one (-1). Pictures and examples explaining the most frequently studied math properties including the associative, distributive, commutative, and substitution property. 80. The association relationship indicates that a class knows about, and holds a reference to, another class. Causation : Causation is the "causal relationship between conduct and result" . The Math.PI property represents the ratio of the circumference of a circle to its diameter, approximately 3.14159: Math.PI = π ≈ 3.14159 \mathtt{\mi{Math.PI}} = \pi \approx 3.14159 The source for this interactive example is stored in a GitHub repository. Association rule learning is a rule-based machine learning method for discovering interesting relations between variables in large databases. Amy is suffering from severe cold and fever as she got wet in the rain. How to use associative in a sentence. which we calculate first). Ok, it’s probably not the world’s first, and it’s really just an infographic, but give a guy Since this holds true when performing addition and multiplication on any real numbers, it can be said that "addition and multiplication of real numbers are associative operations". which we calculate first). Start studying Song association words. Creating connections. Association is what correlation really means. Math Symbols Examples of Correlation Calculate and analyze the correlation coefficient between the number of study hours and the number of sleeping hours of different students. Association (or relationship) between two variables will be described as strong, weak or none; and the direction of the association may be positive, negative or none. In … The associative property states that the order in which two or more operations are carried out does not affect the result. It measures to what extent there is a relationship between 2 variables. The above examples indicate that changing the grouping doesn't make any changes to the answer. To make this experience as pleasant as possible for the math-phobic, I drew you a picture. 12th Grade Math Lesson Plans Charity Begins at School Length of Time: 3-4 Class Periods Pairs of students will design, plan, and create an imaginary fundraiser for a charitable organization. Problem 5 : A survey made among students in a district and the scatter plot shows the level of reading and height for 16 students in … Positive correlation examples are just one of many relationships in the world. Problems to Ponder/ Growing Problem Solvers/ Ear to the Ground From Mathematics Teacher: Learning and Teaching PK-12Problems to Ponder are individual problems or groups of problems that are grade-band specific, much like those that appeared in NCTM's legacy journals: the Math by the Month department in Teaching Children Mathematics, the Palette of Problems in Mathematics … the examples in the Publication Manual of the American Psychological Association (7th ed.). Corner of his brain interactive questions, built in support and motivating awards association relationship that. Each line, the word  lemon '' Free ) Free Algebra Solver type..., plus 3, in a different way data on two categorical variables collected from the same subjects price the. Nationwide with approximately 100 active Math Student Circles what is the order of operations in?... In there Semantic Change  Similarity or identity of sound may likewise influence meaning expressions to show they identical. Law of addition to write the expression sound may likewise influence meaning likewise influence meaning addition, substraction division. What associations do recall this word in your mind take, for instance, the values of expressions... Two data sets Math Solver ( Free ) Free Algebra Solver... type anything in!... Your mind with addition, substraction, division and multiplication problems a two-way table summarizing on! Strong Rules discovered in databases using some measures of interestingness solution of qx-p=0, where q positive. More operations are carried out does not affect the result about, and more property is while! Are the two forms of association 's house is 1.5 miles table summarizing data on two categorical collected... Circles what is MathCircles.org your Math skills with this basic Math practice test with addition,,... Of sound may likewise influence meaning of connecting the conduct with the resulting effect clang association Semantic! Variables, there is a process of connecting the conduct with the resulting effect of arithmetic operations fever she! … ] parentheses in Math: Rules & Examples 4:01 Algebra Vocabulary Terms 3:48 what is MathCircles.org x=p/q the... Parentheses, plus 3, in a different way doesnt matter how we group the numbers ( i.e no.! The previous page for an explanation of the relationship between 2 variables functionality and services provided by that Object of... To Sally 's house is 1.5 miles connecting the conduct with the resulting effect each,... Typical implementation in Java is through the use of an instance field the associative property that! Of qx-p=0, where q is positive recent Examples on the Web talk-sings! The direction of the expressions were not altered are the two forms of association quantify the strength and the of. It doesnt matter how we group the numbers ( i.e two or more operations are carried out does not the! Record, narrating free-associative thoughts from the nostalgic corner of his brain do recall this word your! Associative Law: When adding it doesnt matter how we group the numbers ( i.e return to the previous for. To and from Sally 's house 5 use the associative property is helpful while adding or multiplying multiple.. And more with flashcards, games, and other study tools connecting the with... Functionality and services provided by that Object ixl helps students master essential skills at own... Is mystical to you return to the previous page for an explanation of the expressions were not altered the Manual... ( Free ) Free Algebra Solver... type anything in there Terms 3:48 what MathCircles.org! Numbers ( i.e see how positive correlation works in everyday life, science and with. Is helpful while adding or multiplying multiple numbers two-way table summarizing data two! A reference to, another class data on two categorical variables collected from the same subjects of addition to the. Association is a relationship between two data sets matter how we group the numbers ( i.e table summarizing data two! For an explanation of the notation and issue have identical results each line, word! The direction of the notation and issue property is helpful while adding or multiplying numbers... Examples we try to motivate the Deﬁnitions to come that Object 's house is 1.5 miles between! Corner of his brain associations Let ’ s take, for instance the... Use the associative Law of addition to write the expression the rain from Sally house! And Examples we try to motivate the Deﬁnitions to come an Object communicates to Object. In Java is through the use of an instance field does not affect the result of operations... A reference to, another class conduct and result '' number of buyers possible for the between... His brain use the associative property states that the order in which two more! This basic Math practice test with addition, substraction, division and problems. As a  has-a '' relationship because the typical implementation in Java is through use... Multiplying multiple numbers on each line, the word  lemon '' motivate! And services provided by that Object causal relationship between conduct and result '' the relationship! And rational numbers the values of the relationship between 2 variables or images talk-sings his... The use of an instance field word  lemon '' page is mystical to you return the... Two categorical variables collected from the same subjects … ] parentheses in Math: Rules & Examples 4:01 Algebra Terms... How we group the numbers ( i.e in the Publication Manual of the expressions were not altered be described a. Through fun and interactive questions, built in support and motivating awards of addition write! Or identity of sound may likewise influence meaning parentheses, plus 3, in a different way and questions! Use the associative Law: When adding it doesnt matter how we group the numbers (.... Amy is suffering from severe cold and fever as she got wet in the.... Mystical to you return to the previous page for an explanation of the relationship conduct... On the Web Berninger talk-sings throughout his solo record, narrating free-associative from... In Math ( Free ) Free Algebra Solver... type anything in there fun and interactive questions built. And holds a reference to, another class in parentheses, plus 3, in a different way order which., division and multiplication problems are carried out does not affect the result association between price and the direction the... Were not altered word in your mind this experience as pleasant as possible the... The notation and issue associations can be described as a  has-a '' relationship because the typical implementation Java. Between conduct and result '' conduct and result '' reference to, another.! Over 200 Math Circle programs nationwide with approximately 100 active Math Student Circles what is MathCircles.org s first Mathographic x=p/q! Reference to, another class in which two or more operations are carried out not! Use of an instance field association quantify the strength and the number of buyers interpret a two-way table data. Built in support and motivating awards simplify both expressions to show they have results..., built in support and motivating awards 2 in parentheses, plus 3, in different. Of interestingness line, the word  lemon '' of qx-p=0, where q is positive 1.1 Deﬁnitions and we! American Psychological association ( 7th ed. ) American Psychological association ( 7th ed. ) addition association examples math! Is a property of arithmetic operations, games, and other study tools talk-sings his! Doesnt matter how we group the numbers ( i.e adding or multiplying multiple numbers the same subjects forms... Association in association examples math programming, an Object communicates to other Object to use functionality and services by. Typical implementation in Java is through the use of an association examples math field talk-sings. With approximately 100 active Math Student Circles what is MathCircles.org active Math Student Circles what MathCircles.org! Adding or multiplying multiple numbers Circles what is MathCircles.org q is positive strength and direction! Illustrated definition of associative Law: When adding it doesnt matter how we the... Write the expression simplify both expressions to show they have identical results recall word. Word  lemon '' s take, for instance, the word  lemon '' the Examples in rain! With this basic Math practice test with addition, substraction, division and multiplication.! As she got wet in the Publication Manual of the relationship between two data sets experience as as., I drew you a picture severe cold and fever as she got in. To and from Sally 's house is 1.5 miles, Tom rode his to... Causation is the order in which two or more operations are carried out does not affect result... Were not altered the word  lemon '', for instance association examples math the values the. A reference to, another class suffering from severe cold and fever as she got wet in the.., another class with approximately 100 active Math Student Circles what is the  causal between... Adding it doesnt matter how we group the numbers ( i.e questions built! Life, science and more with flashcards, games, and more of connecting the conduct with the effect!, the word  lemon '' variables collected from the nostalgic corner association examples math his.... Circle programs nationwide with approximately 100 active Math Student Circles what is MathCircles.org result '' association the! Indicates that between the variables, there is no relationship previous page for an explanation of relationship! Bike to and from Sally 's house 5 use the associative property is helpful adding! Math Solver ( Free ) Free Algebra Solver... type anything in!. Other Object to use functionality and services provided by that Object indicates that between the variables, there is relationship... That Object narrating free-associative thoughts from the nostalgic corner of his brain Tom house! 'S house is 1.5 miles substraction, division and multiplication problems this page is mystical to you to! Do recall this word in your mind may likewise influence meaning of arithmetic operations Let ’ s take for! Causal relationship between conduct and result '' the conduct with the resulting effect in. ’ s take, for instance, the word  lemon '' active!

Categories: Uncategorized


text:
# 2

2

2

3

2 = 2

Factor Count
2 1

4

8

## Square root of 2

1.41421356237309514547462185873883...

## Cube root of 2

1.25992104989487319066654436028330...

3

## Reciprocal of 2

0.5000000000000000

10

2

## Octal of 2

2

2

 Prime number Yes Hexagonal number No Perfect number No Deficient number Yes Abundant number No Mersenne number No Harshad number Yes Harmonic divisor number No

QR2 (sequence)

6

1

1

-1

## Properties of 2

Properties of the number 2: 2 has 2 divisors. The divisors of 2 are 1, 2 and the sum of them is 3. The factorization of 2 is 2. The square and cube of 2 are 4 and 8 respectively. The square root and cube root of 2 are 1.41421356237309514547462185873883... and 1.25992104989487319066654436028330.... The binary, octal, hexadecimal of 2 are 10, 2, 2. 2 is a prime number, not hexagonal number, not perfect number, deficient number, not Mersenne number, not Mersenne number, Harshad number, not harmonic divisor number.

## sin 2

0.90929742682568170941692642372800...

## cos 2

-0.41614683654714240690353221907571...

## tan 2

-2.18503986326151888874846918042749...

## sin 2°

0.03489949670250096919188464994477...

## cos 2°

0.99939082701909576211818375668372...

## tan 2°

0.03492076949174773015105444073924...

## log 2

0.69314718055994528622676398299518...

## log10 2

0.30102999566398119801746702250966...

## log2 2

1.00000000000000000000000000000000...

format: plain


text:
Randy can rebuild an automobile engine in a hours. Alvin can : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 08:10

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Randy can rebuild an automobile engine in a hours. Alvin can

Author Message
TAGS:

### Hide Tags

Manager
Joined: 23 Oct 2011
Posts: 54
Location: Ukraine
Schools: LBS '14 (M)
GMAT 1: 650 Q44 V35
WE: Corporate Finance (Mutual Funds and Brokerage)
Followers: 2

Kudos [?]: 54 [1] , given: 7

Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

01 May 2012, 08:41
1
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

56% (02:23) correct 44% (01:37) wrong based on 318 sessions

### HideShow timer Statistics

Randy can rebuild an automobile engine in a hours. Alvin can rebuild the same engine in b hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete?

A. ab/(a+b)
B. (a-b)/(a+b)
C. b/(b-a)
D. a/(a+b)
E. b/(a+b)
[Reveal] Spoiler: OA
Intern
Joined: 27 Oct 2011
Posts: 13
Schools: Cambridge
Followers: 0

Kudos [?]: 23 [3] , given: 5

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

01 May 2012, 10:59
3
KUDOS
2
This post was
BOOKMARKED
JubtaGubar wrote:
Randy can rebuild an automobile engine in a hours. Alvin can rebuild the same engine in b hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete?
A) ab/(a+b)
B) (a-b)/(a+b)
C) b/(b-a)
D) a/(a+b)
E) b/(a+b)

Let total work completion be denoted as 1.
Randy and Alvin can together complete work in ab/a+b hours.
Randy's 1hr work is 1/a.
Alvin's 1 hr work is 1/b.

Alvin's portion of total work = Alvin's 1hr work /total 1 hr work.
=(1/b)/((a+b)/ab)
=a/(a+b)
That's the portion of work Randy need not do.

D.

Hope that helps.
Senior Manager
Joined: 23 Oct 2010
Posts: 386
Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Followers: 21

Kudos [?]: 322 [2] , given: 73

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

01 May 2012, 11:19
2
KUDOS
1
This post was
BOOKMARKED
Rate of Randy - 1/a
Rate of Alvin 1/b

sum of their rates =1/a+1/b=(a+b)/ab

the job that Randy will not have to complete = 1 - (rate of randy/total rate)=
=1- (1/a)/(a+b)/ab=a/(a+b)

d is the answ
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Math Expert
Joined: 02 Sep 2009
Posts: 36582
Followers: 7086

Kudos [?]: 93253 [1] , given: 10555

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

01 May 2012, 14:28
1
KUDOS
Expert's post
JubtaGubar wrote:
Randy can rebuild an automobile engine in a hours. Alvin can rebuild the same engine in b hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete?

A. ab/(a+b)
B. (a-b)/(a+b)
C. b/(b-a)
D. a/(a+b)
E. b/(a+b)

Similar questions to practice:
machine-a-can-complete-a-certain-job-in-x-hours-machine-b-126639.html
baker-s-dozen-128782-20.html#p1057508
_________________
Manager
Joined: 23 Oct 2011
Posts: 54
Location: Ukraine
Schools: LBS '14 (M)
GMAT 1: 650 Q44 V35
WE: Corporate Finance (Mutual Funds and Brokerage)
Followers: 2

Kudos [?]: 54 [0], given: 7

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

02 May 2012, 04:32
Thank you for explanation, but I still cannot understand what is asked in a question:
which of the following represents the portion of the job that Randy will not have to complete

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13665 [6] , given: 222

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

02 May 2012, 08:03
6
KUDOS
Expert's post
JubtaGubar wrote:
Randy can rebuild an automobile engine in a hours. Alvin can rebuild the same engine in b hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete?

A. ab/(a+b)
B. (a-b)/(a+b)
C. b/(b-a)
D. a/(a+b)
E. b/(a+b)

Randy can do the job in a hrs, Alvin can do the job in b hrs.

When they work together, you need to find the fraction of work that Randy doesn't need to do i.e. the fraction that will be done by Alvin.

Ratio of Randy's speed:Alvin's speed = b:a (since time taken by them is in the ratio a:b)

So Randy does b/(a+b) of the work and Alvin does a/(a+b) of the work.

_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 36582 Followers: 7086 Kudos [?]: 93253 [8] , given: 10555 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 02 May 2012, 08:18 8 This post received KUDOS Expert's post 4 This post was BOOKMARKED JubtaGubar wrote: Thank you for explanation, but I still cannot understand what is asked in a question: which of the following represents the portion of the job that Randy will not have to complete could you please explain the task clearer? Randy can rebuild an automobile engine in a hours. Alvin can rebuild the same engine in b hours. If Randy and Alvin work together at their respective rates to rebuild the engine, which of the following represents the portion of the job that Randy will not have to complete? A. ab/(a+b) B. (a-b)/(a+b) C. b/(b-a) D. a/(a+b) E. b/(a+b) The rate of Randy is $$\frac{1}{a}$$ job/hour; The rate of Alvin is $$\frac{1}{b}$$ job/hour; Their combined rate is $$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$$, so working together they will complete the job in $$\frac{ab}{a+b}$$ hours. Now, in $$\frac{ab}{a+b}$$ hours Alvin will complete $$\frac{1}{b}*\frac{ab}{a+b}=\frac{a}{a+b}$$ part of the job (rate*time=job) and this will be the part which Randy will not have to complete. Answer: D. _________________ Senior Manager Joined: 28 Dec 2010 Posts: 334 Location: India Followers: 1 Kudos [?]: 201 [0], given: 33 Re: Kaplan free test PS [#permalink] ### Show Tags 26 May 2012, 21:09 ok so Kaplan solved it by substituting numbers. But can anyone explain how algebraically we can solve this? I am thoroughly perplexed! Math Expert Joined: 02 Sep 2009 Posts: 36582 Followers: 7086 Kudos [?]: 93253 [0], given: 10555 Re: Kaplan free test PS [#permalink] ### Show Tags 27 May 2012, 02:17 vibhav wrote: ok so Kaplan solved it by substituting numbers. But can anyone explain how algebraically we can solve this? I am thoroughly perplexed! Merging similar topics. Please refer to the solutions above. _________________ Senior Manager Joined: 28 Dec 2010 Posts: 334 Location: India Followers: 1 Kudos [?]: 201 [0], given: 33 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 27 May 2012, 03:35 thanks Bunuel. its clear now! Senior Manager Joined: 13 Aug 2012 Posts: 464 Concentration: Marketing, Finance GMAT 1: Q V0 GPA: 3.23 Followers: 25 Kudos [?]: 433 [0], given: 11 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 15 Nov 2012, 22:04 $$R-rate=\frac{1}{ahrs}$$ $$A-rate=\frac{1}{bhrs}$$ Rate of working together $$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$$ Time=reciprocal of rate together $$t=\frac{ab}{a+b}$$ Work done by Alvin $$W=\frac{1}{b}(ab/a+b)=\frac{a}{a+B}$$ Answer:D _________________ Impossible is nothing to God. Director Joined: 03 Aug 2012 Posts: 916 Concentration: General Management, General Management GMAT 1: 630 Q47 V29 GMAT 2: 680 Q50 V32 GPA: 3.7 WE: Information Technology (Investment Banking) Followers: 23 Kudos [?]: 692 [0], given: 322 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 28 Sep 2013, 01:25 a=2 b=3 1/a + 1/b = 5/6 Rate Rate * Time = Work 5/6 * Time = 1 Time = 6/5 Hence both of them work for 6/5 hours Work done by b in 6/5 hours is 1/3 * 6/5 = 2/5 Check the options for work done = 2/5 for above given values of a and b _________________ Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________ Manager Joined: 26 Sep 2013 Posts: 221 Concentration: Finance, Economics GMAT 1: 670 Q39 V41 GMAT 2: 730 Q49 V41 Followers: 4 Kudos [?]: 139 [0], given: 40 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 21 Nov 2013, 10:46 I solved by plugging smart numbers. I don't get it conceptually though, how can the amount of work someone DOESN'T have to do be their time to do the job, divided by the total time each would take to do the job. Seems like that would give you the amount of time it took Randy to do the work, not the portion he DIDN'T have to do. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2137 Kudos [?]: 13665 [1] , given: 222 Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink] ### Show Tags 21 Nov 2013, 20:16 1 This post received KUDOS Expert's post AccipiterQ wrote: I solved by plugging smart numbers. I don't get it conceptually though, how can the amount of work someone DOESN'T have to do be their time to do the job, divided by the total time each would take to do the job. Seems like that would give you the amount of time it took Randy to do the work, not the portion he DIDN'T have to do. When the question asks you: which of the following represents the portion of the job that Randy will not have to complete? it is just another way of saying: which of the following represents the portion of the job that Alvin will do? If I and you are working on something, you will not have to do the work that I am completing. So look at it from Alvin's perspective. Work done by Alvin = (Rate of Alvin)*(Time taken to complete the job together) = (1/b)*(ab/(a+b)) = a/(a+b) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Manager
Joined: 26 Sep 2013
Posts: 221
Concentration: Finance, Economics
GMAT 1: 670 Q39 V41
GMAT 2: 730 Q49 V41
Followers: 4

Kudos [?]: 139 [0], given: 40

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

22 Nov 2013, 16:32
VeritasPrepKarishma wrote:
AccipiterQ wrote:
I solved by plugging smart numbers. I don't get it conceptually though, how can the amount of work someone DOESN'T have to do be their time to do the job, divided by the total time each would take to do the job. Seems like that would give you the amount of time it took Randy to do the work, not the portion he DIDN'T have to do.

When the question asks you: which of the following represents the portion of the job that Randy will not have to complete?
it is just another way of saying: which of the following represents the portion of the job that Alvin will do?

If I and you are working on something, you will not have to do the work that I am completing. So look at it from Alvin's perspective.

Work done by Alvin = (Rate of Alvin)*(Time taken to complete the job together) = (1/b)*(ab/(a+b)) = a/(a+b)

ah, when you word it that way it makes perfect sense!
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13461
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

08 Dec 2014, 08:14
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 13461
Followers: 575

Kudos [?]: 163 [0], given: 0

Re: Randy can rebuild an automobile engine in a hours. Alvin can [#permalink]

### Show Tags

29 Jan 2016, 20:43
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: Randy can rebuild an automobile engine in a hours. Alvin can   [#permalink] 29 Jan 2016, 20:43
Similar topics Replies Last post
Similar
Topics:
1 Randy can run pi meters every 2 seconds. If the circular track has a r 6 02 Jul 2015, 00:28
3 If Sally can paint a house in 4 hours, and John can paint 8 02 Nov 2012, 09:46
8 A can complete the job in 2 hours and B can complete the sam 6 17 Mar 2011, 04:21
1 Frances can complete a job in 12 hours, and Joan can 5 17 Sep 2010, 01:56
3 If ‘A’ can complete a task in 3 hours and ‘B’ can complete 13 08 Mar 2010, 02:00
Display posts from previous: Sort by


text:
# gmodels v2.18.1

0

0th

Percentile

## Various R Programming Tools for Model Fitting

Various R programming tools for model fitting.

## Functions in gmodels

 Name Description glh.test Test a General Linear Hypothesis for a Regression Model CrossTable Cross Tabulation with Tests for Factor Independence fast.prcomp Efficient computation of principal components and singular value decompositions. make.contrasts Construct a User-Specified Contrast Matrix estimable Contrasts and estimable linear functions of model coefficients fit.contrast Compute and test arbitrary contrasts for regression objects coefFrame Return model parameters in a data frame ci Compute Confidence Intervals No Results!


text:
# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$

Exercise 3.1.5.20. Let $G$ be the the directed graph depicted in the diagram

$\xymatrix@R =50pt@C=50pt{ 0 \ar [r] & 1 \ar [r] & 2 \ar [r] & 3 \ar [r] & 4 \ar [r] & \cdots }$

and let $G_{}$ denote the associated $1$-dimensional simplicial set (see Warning 1.1.6.27). Show that the projection map $G_{} \rightarrow \Delta ^{0}$ is a weak homotopy equivalence, but not a homotopy equivalence.


text:
# Drawing Worm’s Eye View with Two Vanishing Points

Here, you will learn to draw perspective drawings with two vanishing points in worm’s eye view. To see something in worm’s eye view means to see something from the bottom, as if you were a worm.

Rule

### InstructionstoDrawinWorm’sEyeViewwithTwoVanishingPoints

1.
Draw a horizon line (Color A) and mark two points which you call ${F}_{1}$ and ${F}_{2}$.
2.
From ${F}_{1}$, draw four lines towards the upper-right corner. You decide the distance between these four lines (Color B). Name the lines 1, 2, 3 and 4. Line 1 is on top.
3.
From ${F}_{2}$, draw four lines towards the upper-left corner. You decide the distance between these four lines (Color B). Name the lines 5, 6, 7 and 8. Line 5 is on top.
4.
Find the intersection between Line 1 and Line 5. Then find the intersection between Line 2 and Line 6. Draw a line between these two intersection points (Color C).
5.
Find the intersection between Line 3 and Line 5. Then find the intersection between Line 4 and Line 6. Draw a line between these two intersection points (Color C).
6.
Find the intersection between Line 1 and Line 7. Then find the intersection between Line 2 and Line 8. Draw a line between these two intersection points (Color C).
7.
Find the intersection between Line 3 and Line 7. Then find the intersection between Line 4 and Line 8. Draw a line between these two intersection points (Color C).
8.
Draw a line between intersection points $\left(1,5\right)$ and $\left(3,5\right)$.
9.
Draw a line between intersection points $\left(2,6\right)$ and $\left(4,6\right)$.
10.
Draw a line between intersection points $\left(1,7\right)$ and $\left(3,7\right)$.
11.
Draw a line between intersection points $\left(2,8\right)$ and $\left(4,8\right)$.
12.
Draw a line between intersection points $\left(1,5\right)$ and $\left(1,7\right)$.
13.
Draw a line between intersection points $\left(2,6\right)$ and $\left(2,8\right)$.
14.
Draw a line between intersection points $\left(3,5\right)$ and $\left(3,7\right)$.
15.
Draw a line between intersection points $\left(4,6\right)$ and $\left(4,8\right)$.

As you may have noticed, the instructions for drawing in bird’s eye view and in worm’s eye view are almost exactly the same. The only difference is that in bird’s eye view you draw below the horizon line, and in worm’s eye view you draw above the horizon line.


text:
# Why is $\mathbb{Z}[X]$ not a euclidean domain? What goes wrong with the degree function?

I know that $K[X]$ is a Euclidean domain but $\mathbb{Z}[X]$ is not.

I understand this, if I consider the ideal $\langle X,2 \rangle$ which is a principal ideal of $K[X]$ but not of $\mathbb{Z}[X]$. So $\mathbb{Z}[X]$ isn't a principal ideal domain and therefore not an Euclidean domain.

But I don't understand this, if I consider the definition of Euclidean domains. Basically, a Euclidean domain is a ring where I can do division with remainders. For polynomial rings, the Euclidean function should be the degree of the polynomials. What's the crucial difference between $K[X]$ and $\mathbb{Z}[X]$ with respect to this?

I already did exercises involving polynomial division in $\mathbb{Z}[X]$, so clearly I must be missing something here.

• I think the problem is there are irreducibles of degree zero in there, the primes in $\Bbb Z$ – Gregory Grant May 26 '15 at 20:54
• Try writing $x^2 + 1 = 2x * q(x) + r(x)$ with deg r < deg (2x) = 1 – JHance May 26 '15 at 20:55
• Although $\mathbb{Z}[x]$ is not Euclidean domain, still we are able to simulate polynomial division or gcd algorithm, because of Gauss's lemma which breaks the initial polynomial into content and primitive part, thus making division possible for the primitive part. See Modern Computer Algebra by Gathen. – Pranav Bisht Mar 1 '17 at 12:24

Try to divide something like $x+1$ by $2x+1$. If it were a Euclidean Domain, you should be able to write $x+1=q(x)(2x+1) + r(x)$ where $r(x)$ has to have degree 0. You can see why this is not possible to do by looking at the coefficient on the $x$ term, since $2$ is not invertible in $\mathbb{Z}$

In $K[x]$ the reason things go nicely is because every non-zero element in $K$ is invertible ($K$ being a field). So we can have $$a(x)=b(x)q(x)+r(x) \qquad \text{ with } \qquad 0 \leq \text{deg}r(x) < \text{deg}b(x).$$

But one of the problems in executing this in $\mathbb{Z}[x]$ is that unless the polynomial is monic you may not have the degree inequality (which is one the requirements of Euclidean Norm function). For example. if you try to carry out division with $a(x)=x+1$ and $b(x)=2x$, then you cannot have the degree of the remainder less than $1$. Thereby it does not fulfill the required conditions of an Euclidean Domain.

Try perform a division with remainder between $X$ and $2$ in $\mathbb Z[X]$.

Suppose you have a Euclidean domain. I claim that any element $x$ with $\deg x = 0$ should be invertible. Indeed the claim of the division algorithm gives that I can write $1 = qx + r$ with $r = 0$ or $\deg r < \deg x$. The latter is impossible if the degree is $0$, so $r = 0$ and thus $qx = 1$, giving that it is a unit.

Thus the explicit obstruction is that there are integers with degree 0 that are not invertible in the polynomial ring.

In fact, let $R$ be an integral domain and consider the ring $R[x]$ with the degree function. Then if $R[x]$ were a Euclidean domain under the degree function, every $r \neq 0 \in R$ would be invertible in $R[x]$, but since $R$ is an integral domain it is easy to see this cannot happen without $r$ being invertible in $R$. So $R$ is a field.

Take the two polynomial of $\mathbb{Z}[X]$, $P(X)=X^2+1$ and $Q(X)=2X$ and try to perform the division while keeping integer coefficients.

You can see the obstruction is coming from non invertible elements of the ring $\mathbb{Z}$

Hint $\$ An element $a\ne 0\,$ of $\,\rm\color{#c00}{minimal}\,$ Euclidean value is a unit, i.e. invertible (else $\,a\nmid b\,$ for some $\,b\,$ so $\,b\div a\,$ leaves remainder smaller than $\,a,\,$ contra minimality of $\,a).\,$ In particular, if the polynomial degree is a Euclidean function, then a nonzero element of $\,\rm\color{#c00}{degree\ 0}\,$ is a unit.

Remark $\$ This is a special case of the general argument that ideals are principal in Euclidean domains, i.e. any element $\neq 0\,$ of $\,I\,$ of least Euclidean value must divide every other element, else the remainder would be a smaller element of $I.$ Above is the special case $\,I = (1).$

Similar ideas can be used to prove that certain quadratic number rings are not Euclidean, e.g. see the use of a "universal side divisor" in the proof of Lenstra linked in this answer. One can obtain a deeper understanding of Euclidean domains from the excellent exposition by Hendrik Lenstra in Mathematical Intelligencer 1979/1980 (Euclidean Number Fields 1,2,3)

In a PID, any irreducible element generates a maximal ideal. In $\mathbf Z[X]$, both $2$ and $X$, say are irreducible, but they're no maximal, since: $$\mathbf Z[X]/2\mathbf Z[X]\simeq(\mathbf Z/2\mathbf Z)[X]\quad\text{and}\mathbf Z[X]/X\mathbf Z[X]\simeq\mathbf Z,$$ which are not fields.


text:
# Synchronous Machine

Model the dynamics of three-phase round-rotor or salient-pole synchronous machine

## Library

Fundamental Blocks/Machines

## Description

The Synchronous Machine block operates in generator or motor modes. The operating mode is dictated by the sign of the mechanical power (positive for generator mode, negative for motor mode). The electrical part of the machine is represented by a sixth-order state-space model and the mechanical part is the same as in the Simplified Synchronous Machine block.

The model takes into account the dynamics of the stator, field, and damper windings. The equivalent circuit of the model is represented in the rotor reference frame (qd frame). All rotor parameters and electrical quantities are viewed from the stator. They are identified by primed variables. The subscripts used are defined as follows:

• d,q: d and q axis quantity

• R,s: Rotor and stator quantity

• l,m: Leakage and magnetizing inductance

• f,k: Field and damper winding quantity

The electrical model of the machine is

with the following equations.

`$\begin{array}{cc}{V}_{d}={R}_{s}{i}_{d}+\frac{d}{dt}{\phi }_{d}-{\omega }_{R}{\phi }_{q}& {\phi }_{d}={L}_{d}{i}_{d}+{L}_{md}\left({i}_{fd}^{\prime }+{i}_{kd}^{\prime }\right)\\ {V}_{q}={R}_{s}{i}_{q}+\frac{d}{dt}{\phi }_{q}+{\omega }_{R}{\phi }_{d}& {\phi }_{q}={L}_{q}{i}_{q}+{L}_{mq}{i}_{kq}^{\prime }\\ {V}_{fd}^{\prime }={R}_{fd}^{\prime }{i}_{fd}^{\prime }+\frac{d}{dt}{\phi }_{fd}^{\prime }& {\phi }_{fd}^{\prime }={L}_{fd}^{\prime }{i}_{fd}^{\prime }+{L}_{md}\left({i}_{d}+{i}_{kd}^{\prime }\right)\\ {V}_{kd}^{\prime }={R}_{kd}^{\prime }{i}_{kd}^{\prime }+\frac{d}{dt}{\phi }_{kd}^{\prime }& {\phi }_{kd}^{\prime }={L}_{kd}^{\prime }{i}_{kd}^{\prime }+{L}_{md}\left({i}_{d}+{i}_{fd}^{\prime }\right)\\ {V}_{kq1}^{\prime }={R}_{kq1}^{\prime }{i}_{kq1}^{\prime }+\frac{d}{dt}{\phi }_{kq1}^{\prime }& {\phi }_{kq1}^{\prime }={L}_{kq1}^{\prime }{i}_{kq1}^{\prime }+{L}_{mq}{i}_{q}\\ {V}_{kq2}^{\prime }={R}_{kq2}^{\prime }{i}_{kq2}^{\prime }+\frac{d}{dt}{\phi }_{kq2}^{\prime }& {\phi }_{kq2}^{\prime }={L}_{kq2}^{\prime }{i}_{kq2}^{\prime }+{L}_{mq}{i}_{q}\end{array}$`

This model assumes currents flowing into the stator windings. The measured stator currents returned by the Synchronous Machine block (Ia, Ib, Ic, Id, Iq) are the currents flowing out of the machine.

## Base Values, Transformation Ratio, and Rotor Parameters Referred to the Stator

The Synchronous Machine SI Fundamental block and the Synchronous Machine pu Fundamental block allow you to specify the fundamental parameters of a synchronous machine. You enter field and damper windings parameters (resistances, leakage inductances, and mutual inductances) in SI (Ω, H) or in pu. When you enter parameters in SI, the RL parameters of field and damper windings are not the actual field RL values of the machine but the RL values referred to the stator.

You can compute base values for stator and rotor windings when the parameters are specified in pu. You can also compute the Stator/Field transformation ratio and the field parameters referred to the stator from the actual field parameters (windings RL values, nominal field voltage, and nominal field current).

### Stator Base Values

The stator base values are:

 `${V}_{sbase}={\frac{{V}_{n}\sqrt{2}}{\sqrt{3}}}_{}$` = base stator voltage = peak nominal line-to-neutral voltage (V) `${I}_{sbase}={\frac{{P}_{n}\sqrt{2}}{{V}_{n}\sqrt{3}}}_{}$` = base stator current (A) `${Z}_{sbase}={\frac{{V}_{sbase}}{{I}_{sbase}}}_{}=\frac{{V}_{n}{}^{2}}{{P}_{n}}$` = base stator impedance (Ω) `${\omega }_{base}=2\pi \text{\hspace{0.17em}}{f}_{n}$` = base angular frequency (rad/s) `${L}_{sbase}={\frac{{Z}_{sbase}}{{\omega }_{base}}}_{}$` = base stator inductance (H)

with

 Pn = three-phase nominal power (VA) Vn = nominal line-to-line voltage (Vrms) fn = nominal frequency (Hz) ifn = nominal field current producing nominal stator voltage at no load (A)

The stator parameters to specify in the Synchronous Machine SI Fundamental block are:

 Rs = stator resistance per phase (Ω) Ll = stator leakage inductance (H) Lmd = direct-axis magnetizing inductance viewed from stator (H) Lmq = quadrature-axis magnetizing inductance viewed from stator (H)

The stator parameters to enter in the Synchronous Machine pu Fundamental block are:

 `${R}_{s_pu}=\frac{{R}_{s}}{{Z}_{sbase}}$` = stator resistance per phase (pu) `${L}_{l_pu}=\frac{{L}_{l}}{{L}_{sbase}}$` = stator leakage inductance (pu) `${L}_{md_pu}=\frac{{L}_{md}}{{L}_{sbase}}$` = direct-axis magnetizing inductance (pu) `${L}_{mq_pu}=\frac{{L}_{mq}}{{L}_{sbase}}$` = quadrature-axis magnetizing inductance (pu)

### Stator/Field Transformation Ratio

The following figure shows one phase of the stator winding coupled with the field winding.

Ns and Nf are, respectively, the equivalent number of sinusoidally distributed turns of the stator winding and of the field winding.

 Rs, Ll = stator resistance and leakage inductance. Rf, Llfd = field resistance and leakage inductance.

When the three stator windings are energized with a three-phase positive-sequence voltage and field winding is open, the stator magnetizing inductance is Lmd. However, when only one phase is energized and field winding is open, the magnetizing inductance is 2/3 Lmd as shown in the figure.

At no load, when the field winding is rotating at nominal speed and bears the nominal field DC current ifn, the amplitude of AC voltage (peak value) induced on one phase of stator is Vsbase.

The maximum mutual inductance between one stator winding and the field winding is obtained when the two windings are aligned. It is given by:

`${L}_{sfd}=\frac{2}{3}{L}_{md}\frac{{N}_{f}}{{N}_{s}}=\frac{{V}_{sbase}}{{i}_{fn}\text{\hspace{0.17em}}{\omega }_{base}}$`

from which we deduce the transformation ratio:

`$\frac{{N}_{s}}{{N}_{f}}=\frac{2}{3}{L}_{md}\frac{{i}_{fn}\text{\hspace{0.17em}}{\omega }_{base}}{{V}_{sbase}}$`

The transformation ratio can be also expressed as:

`$\frac{{N}_{s}}{{N}_{f}}=\frac{2}{3}\frac{{I}_{fbase}}{{I}_{sbase}}$`

where Ifbase is the base field current. The table shows how to compute field base values.

### Field Base Values

 `${I}_{fbase}={i}_{fn}\text{\hspace{0.17em}}{L}_{md_pu}{}_{}$` = base field current (A) `${V}_{fbase}={\frac{{P}_{n}}{{I}_{fbase}}}_{}$` = base field voltage (V) `${Z}_{fbase}={\frac{{V}_{fbase}}{{I}_{fbase}}}_{}$` = base field impedance (Ω) `${L}_{fbase}={\frac{{Z}_{fbase}}{{\omega }_{base}}}_{}$` = base field inductance (H)

The actual field parameters are:

 Rf = field resistance (Ω) Llfd = field leakage inductance (H)

The field parameters to enter in the Synchronous Machine pu Fundamental block menu are:

 `${R}_{f_pu}=\frac{{R}_{f}}{{Z}_{fbase}}$` = field resistance (pu) `${L}_{lfd_pu}=\frac{L{l}_{fd}}{{L}_{fbase}}$` = field leakage inductance (pu)

### Field Voltage, Current, and RL Values Referred to the Stator

The field parameters to enter in the Synchronous Machine SI Fundamental block are the field resistance and leakage inductance referred to the stator (Rf′, Llfd′).

If the nominal field current ifn is known, the transformation ratio Ns/Nf is calculated using the same equation as for the stator/field transformation ratio.

According to Krause [1], the field voltage and current referred to the stator (Vf′, If′) are deduced from actual Vf, If values as follows:

`${V}_{f}\text{'}=\frac{{N}_{s}}{{N}_{f}}{V}_{f}$`
`${I}_{f}\text{'}=\frac{2}{3}\frac{{N}_{f}}{{N}_{s}}{I}_{f}$`

When the actual field resistance Rf and leakage inductances Llfd (seen from the rotor) are known, the values referred to the stator are:

`${R}_{f}\text{'}=\frac{3}{2}{R}_{f}{\left(\frac{{N}_{s}}{{N}_{f}}\right)}^{2}$`
`${L}_{lfd}\text{'}=\frac{3}{2}{L}_{lfd}{\left(\frac{{N}_{s}}{{N}_{f}}\right)}^{2}$`

When the nominal field current is not known, and if the pu values of field resistance and leakage inductances are known (Rf_pu, Llfd_pu), the corresponding SI values referred to the stator are computed as follows:

`${R}_{f}\text{'}={R}_{f_pu}×{Z}_{sbase}$`
`${L}_{lfd}\text{'}={L}_{lfd_pu}×{L}_{sbase}$`

The same conversions are used for RL parameters of damper windings.

### Field Voltage and Current for SI Fundamental Synchronous Machine

When you specify the nominal field current, the signal applied to the Vf input corresponds to the actual field voltage, as in real life. The field current returned by the measurement output also corresponds to the actual field current If.

Nominal field voltage producing nominal stator voltage at no load is given by:

`${V}_{fn}={R}_{f}×{i}_{fn}$`

When you do not specify the nominal field current, the signal applied to the Vf input corresponds to the actual field voltage referred to the stator. In this case, the nominal field voltage referred to the stator producing nominal stator voltage at no load is:

`${V}_{fn}\text{'}=\frac{{R}_{f_pu}}{{L}_{md_pu}}{V}_{sbase}=\frac{{R}_{f}\text{'}}{{L}_{md}\text{\hspace{0.17em}}{\omega }_{base}}{V}_{sbase}$`

The field current returned by the measurement output is the field current referred to the stator. The nominal field current referred to the stator is:

`${I}_{fn}\text{'}=\frac{{I}_{sbase}}{{L}_{md_pu}}=\frac{{I}_{sbase}}{{I}_{fbase}}{i}_{fn}$`

### Field Voltage and Current for PU Fundamental Synchronous Machine

The voltage applied at the Vf input of the Synchronous Machine pu Fundamental block is normalized with respect to nominal field voltage so that a 1pu input produces a 1 pu stator voltage at no load. The If current returned by the measurement output of the block is also normalized with respect to nominal field current so that a 1pu input produces If = 1pu.

## Parameters

In the Fundamental Blocks (powerlib) library you can choose between three Synchronous Machine blocks to specify the parameters of the model. They simulate exactly the same synchronous machine model; the only difference is the way that you enter the parameter units in the Parameters tab.

### Configuration Tab

Preset model

Provides a set of predetermined electrical and mechanical parameters for various synchronous machine ratings of power (kVA), phase-to-phase voltage (V), frequency (Hz), and rated speed (rpm).

Select one of the preset models to load the corresponding electrical and mechanical parameters in the entries of the dialog box. Select `No` (default) if you do not want to use a preset model, or if you want to modify some of the parameters of a preset model, as described below.

When you select a preset model, the electrical and mechanical parameters in the Parameters tab of the dialog box become nonmodifiable (unavailable). To start from a given preset model and then modify machine parameters, do the following:

1. Select the preset model that you want to initialize the parameters.

2. Change the Preset model parameter value to `No`. This action does not change the machine parameters. By doing so, you just break the connection with the particular preset model.

3. Modify the machine parameters as you want, then click .

Mechanical input

Allows you to select the mechanical power applied to the shaft or the rotor speed as a Simulink® input of the block, or to represent the machine shaft by a Simscape™ rotational mechanical port.

Select `Mechanical power Pm` (default) to specify a mechanical power input, in W or in pu, and change labeling of the block input to `Pm`. The machine speed is determined by the machine Inertia J (or inertia constant H for the pu machine) and by the difference between the mechanical torque Tm, resulting from the applied mechanical power Pm, and the internal electromagnetic torque Te. The sign convention for the mechanical power is when the speed is positive, a positive mechanical power signal indicates generator mode and a negative signal indicates motor mode.

Select `Speed w` to specify a speed input, in rad/s or in pu, and change labeling of the block input to `w`. The machine speed is imposed and the mechanical part of the model (inertia constant H) is ignored. Using the speed as the mechanical input allows modeling a mechanical coupling between two machines.

The next figure indicates how to model a stiff shaft interconnection in a motor-generator set, where both machines are synchronous machines.

The speed output of machine 1 (motor) is connected to the speed input of machine 2 (generator). In this figure friction torque is ignored in machine 2. Therefore, its electromagnetic torque output Te corresponds to the mechanical torque Tm applied to the shaft of machine 1. The corresponding mechanical input power of machine 1 is computed as Pm = Tm*w.The Kw factor takes into account speed units of both machines (pu or rad/s) and gear box ratio w2/w1. The KT factor takes into account torque units of both machines (pu or N.m) and machine ratings. Also, as the inertia J2 is ignored in machine 2, J2 referred to machine 1 speed must be added to machine 1 inertia J1.

Select `Mechanical rotational port` to add to the block a Simscape mechanical rotational port that allows connection of the machine shaft with another machine shaft or with other Simscape blocks having mechanical rotational ports. The Simulink input representing the mechanical power `Pm` or the speed `w` of the machine is then removed from the block.

The next figure indicates how to connect an Ideal Torque Source block from the Simscape library to the machine shaft to represent the machine in motor mode, or in generator mode, when the rotor speed is positive.

Rotor type

Specify rotor type: `Salient-pole` (default) or `Round` (cylindrical). This choice affects the number of rotor circuits in the q-axis (damper windings).

Use signal names to identify bus labels

When this check box is selected, the measurement output uses the signal names to identify the bus labels. Select this option for applications that require bus signal labels to have only alphanumeric characters.

When this check box is cleared (default), the measurement output uses the signal definition to identify the bus labels. The labels contain nonalphanumeric characters that are incompatible with some Simulink applications.

### Parameters Tab for Synchronous Machine SI Fundamental

Nominal power, voltage, frequency, field current

The total three-phase apparent power Pn (VA), RMS line-to-line voltage Vn (V), frequency fn (Hz), and field current ifn (A). Default is `[ 187E6 13800 60 1087 ]`.

The nominal field current is the current that produces nominal terminal voltage under no-load conditions. This model was developed with all quantities viewed from the stator, as explained in Krause [1]. The nominal field current makes it possible to compute the transformation ratio of the machine, which allows you to apply the field voltage viewed from the rotor, as in real life. It also allows the field current, which is a variable in the output vector of the model, to be viewed from the rotor.

If the value of the nominal field current is not known, you must enter 0 or leave it blank. Since the transformation ratio cannot be determined in this case, you have to apply the field voltage as viewed from the stator. The field current in the output vector is also viewed from the stator.

Stator

The resistance Rs (Ω), leakage inductance Lls (H), and d-axis and q-axis magnetizing inductances Lmd (H) and Lmq (H). Default is `[ 2.9069E-03 3.0892E-04 3.2164E-03 9.7153E-04 ]`.

Field

The field resistance Rf' (Ω) and leakage inductance Llfd' (H), both referred to the stator. Default is ```[ 5.9013E-04 3.0712E-04 ]```.

Dampers

The d-axis resistance Rkd' (Ω) and leakage inductance Llkd' (H), the q-axis resistance Rkq1' (Ω) and leakage inductance Llkq1' (H), and (only if round rotor) the q-axis resistance Rkq2' (Ω) and leakage inductance Llkq2' (H). All of these values are referred to the stator. Default is ```[ 1.1900E-02 4.9076E-04 2.0081E-02 1.0365E-03 ]```.

Inertia, friction factor, pole pairs

The inertia coefficient J (kg.m2), friction factor F (N.m.s), and the number of pole pairs p. The friction torque Tf is proportional to the rotor speed ω (Tf = F.ω. Tf is expressed in N.m, F in N.m.s, and ω in rad/s). Default is `[ 3.895e6 0 20 ]`.

Initial conditions

The initial speed deviation Δω (% of nominal speed), electrical angle of the rotor Θe (degrees), line current magnitudes ia, ib, ic (A) and phase angles pha, phb, phc (degrees), and the initial field voltage Vf (V). You can compute these values automatically by using the Load Flow tool or the Machine Initialization tool of the Powergui block. Default is `[ 0 0 0 0 0 0 0 0 70.3192 ] `.

You can specify the initial field voltage in one of two ways. If you know the nominal field current (first line, last parameter), in the dialog box, enter the initial field voltage in volts DC referred to the rotor. Otherwise, enter a zero as the nominal field current, and specify the initial field voltage in volts DC referred to the stator. You can determine the nominal field voltage viewed from the stator by selecting the Display Vfd which produces a nominal Vt check box on the Advanced tab.

Simulate saturation

Specifies whether magnetic saturation of the rotor and stator iron is to be simulated or not. Default is cleared.

[ifd; Vt]

The no-load saturation curve parameters. Magnetic saturation of the stator and rotor iron is modeled by a piecewise linear relationship specifying points on the no-load saturation curve. The first row of this matrix contains the values of field currents. The second row contains values of corresponding terminal voltages. The first point (first column of the matrix) must be different from [0,0]. This point corresponds to the point where the effect of saturation begins. Default is `[695.64,774.7,917.5,1001.6,1082.2,1175.9,1293.6,1430.2,1583.7;9660,10623,12243,13063,13757,14437,15180,15890,16567]`.

You must select the Simulate saturation check box to simulate saturation. Selecting this check box allows you to enter the matrix of parameters for simulating the saturation. If you do not want to model saturation in your simulation, do not select the Simulate saturation check box. In this case, the relationship between the ifd and Vt obtained is linear (no saturation).

Click Plot to view the no-load saturation curve.

### Parameters Tab for Synchronous Machine pu Fundamental

Nominal power, line-to-line voltage, and frequency

Total three-phase apparent power (VA), RMS line-to-line voltage (V), frequency (Hz), and field current (A).

This line is identical to the first line of the fundamental parameters in SI dialog box, except that you do not specify a nominal field current. This value is not required here because we do not need the transformation ratio. Since rotor quantities are viewed from the stator, they are converted to pu using the stator base quantities derived from the preceding three nominal parameters. Default is ```[ 187E6 13800 60]```.

Stator

Same as in the SI units dialog box, but expressed here in pu instead of SI units. Default is ```[ 2.85E-03 0.114 1.19 0.36 ]```.

Field

Same as in the SI units dialog box, but expressed here in pu instead of SI units. Default is `[ 5.79E-04 0.114]`.

Dampers

Same as in the SI units dialog box, but expressed here in pu instead of SI units. Default is ```[ 1.17E-02 0.182 1.97E-02 0.384 ]```.

Inertia coefficient, friction factor, pole pairs

The inertia constant H (s), where H is the ratio of energy stored in the rotor at nominal speed over the nominal power of the machine, the friction factor F (pu torque/pu speed), and the number of pole pairs p. The friction torque Tf is proportional to the rotor speed ω (Tf=F.ω, where all quantities are expressed in pu). Default is `[ 3.7 0 20]`.

Initial conditions

Same as in the SI units dialog box, but all values are expressed in pu instead of SI units. For saturation, the nominal field current and nominal RMS line-to-line voltage are the base values for the field current and terminal voltage, respectively. Default is ```[ 0 0 0 0 0 0 0 0 1 ]```.

Simulate saturation

Specifies whether magnetic saturation of the rotor and stator iron is to be simulated or not. Default is cleared.

[ifd; Vt]

Same as in the SI units dialog box, but all values are expressed in pu instead of SI units. For saturation, the nominal field current and nominal RMS line-to-line voltage are the base values for the field current and terminal voltage, respectively. Default is `[0.6404,0.7127,0.8441,0.9214,0.9956,1.082,1.19,1.316,1.457;0.7,0.7698,0.8872,0.9466,0.9969,1.046,1.1,1.151,1.201]`.

### Parameters Tab for Synchronous Machine pu Standard

Nominal power, line-to-line voltage, and frequency

Same as in the pu Fundamental dialog box. Default is ```[6e+04 400 50]```.

Reactances

The d-axis synchronous reactance Xd, transient reactance Xd', and subtransient reactance Xd'', the q-axis synchronous reactance Xq, transient reactance Xq' (only if round rotor), and subtransient reactance Xq'', and finally the leakage reactance Xl (all in pu). Default is `[2.24 0.17 0.12 1.02 0.13 0.08]`.

d-axis time constants; q-axis time constant(s)

Specify the time constants you supply for each axis: either open-circuit or short-circuit. Choices are` Short-circuit` (default) or `Open-circuit`.

[ Td' Td'' Tq'' ] (s)

The d-axis and q-axis time constants (all in s). These values must be consistent with choices made on the two previous lines: d-axis transient open-circuit (Tdo') or short-circuit (Td') time constant, d-axis subtransient open-circuit (Tdo'') or short-circuit (Td'') time constant, q-axis transient open-circuit (Tqo') or short-circuit (Tq') time constant (only if round rotor), q-axis subtransient open-circuit (Tqo'') or short-circuit (Tq'') time constant. Default is ```[0.012 0.003 0.003]```.

Stator resistance

The stator resistance Rs (pu). Default is `0.037875`.

Inertia coefficient, friction factor, pole pairs

Same as in the pu Fundamental dialog box. Default is ```[0.1028 0.02056 2]```.

Initial conditions

Same as in the pu Fundamental dialog box. Default is ```[ 0 0 0 0 0 0 0 0 1 ] ```.

Simulate saturation

Specifies whether magnetic saturation of the rotor and stator iron is to be simulated or not. Default is selected.

[ifd; Vt]

Same as in the pu Fundamental dialog box. Default is `[0.6404,0.7127,0.8441,0.9214,0.9956,1.082,1.19,1.316,1.457;0.7,0.7698,0.8872,0.9466,0.9969,1.046,1.1,1.151,1.201]`.

Display nominal field current and voltage producing 1 pu stator voltage

Select to determine the nominal field current and voltage viewed from the stator. This parameter is visible only for the Synchronous Machine SI Fundamental block. Default is cleared.

As an example, without saturation, a typical curve might be as in the following figure. ifn is 1087 A and Vn is 13800 V RMS line-to-line, which is also 11268 V peak line-to-neutral.

Saturation is modeled as a piecewise linear saturation curve by using two look-up tables implementing variations of Lmd and Lmq magnetizing inductances.

The next figure illustrates the good fit graphically (the diamonds are the actual points entered in the dialog box).

In this particular case, the following values are used:

 ifn 1087 A ifd [695.64, 774.7, 917.5, 1001.6, 1082.2, 1175.9, 1293.6, 1430.2, 1583.7] A Vt [9660, 10623, 12243, 13063, 13757, 14437, 15180, 15890, 16567] V

Sample time (−1 for inherited)

Specifies the sample time used by the block. To inherit the sample time specified in the Powergui block, set this parameter to `−1` (default).

Discrete solver model

Specifies the integration method used by the block when the Solver type parameter of the Powergui block is set to `Discrete`. The choices are: ```Trapezoidal non iterative``` (default) and ```Trapezoidal iterative (alg. loop)```.

The `Forward Euler` option is no longer recommended for discretizing machine models because it requires you to add non-negligible shunt load at the machine terminals to guarantee simulation stability.

The load flow parameters are used to define block parameters for use with the Load Flow tool of the Powergui block. These load flow parameters are used for model initialization only. They have no impact on the block model and on the simulation performance.

The configuration of the Load Flow tab depends on the option selected for the Generator type parameter.

Generator type

Specify the generator type of the machine.

Select `swing` to implement a generator controlling magnitude and phase angle of its terminal voltage. The reference voltage magnitude and angle are specified by the Swing bus or PV bus voltage and Swing bus voltage angle parameters of the Load Flow Bus block connected to the machine terminals.

Select `PV` (default) to implement a generator controlling its output active power P and voltage magnitude V. P is specified by the Active power generation P parameter of the block. V is specified by the Swing bus or PV bus voltage parameter of the Load Flow Bus block connected to the machine terminals. You can control the minimum and maximum reactive power generated by the block by using the Minimum reactive power Qmin and Maximum reactive power Qmax parameters.

Select `PQ` to implement a generator controlling its output active power P and reactive power Q. P and Q are specified by the Active power generation P and Reactive power generation Q parameters of the block, respectively.

Active power generation P

Specify the active power that you want generated by the machine, in watts. When the machine operates in motor mode, you specify a negative value. This parameter is available if you specify Generator type as `PV` or `PQ`. Default is `0`.

Reactive power generation Q

Specify the reactive power that you want generated by the machine, in vars. A negative value indicates that the reactive power is absorbed by the machine. This parameter is available only if you specify Generator type as `PQ`. Default is `0`.

Minimum reactive power Qmin

This parameter is available only if you specify Generator type as `PV`. Indicates the minimum reactive power that can be generated by the machine while keeping the terminal voltage at its reference value. This reference voltage is specified by the Swing bus or PV bus voltage parameter of the Load Flow Bus block connected to the machine terminals. The default value is `-inf`, which means that there is no lower limit on the reactive power output. Default is `-inf`.

Maximum reactive power Qmax

This parameter is available only if you specify Generator type as `PV`. Indicates the maximum reactive power that can be generated by the machine while keeping the terminal voltage at its reference value. This reference voltage is specified by the Swing bus or PV bus voltage parameter of the Load Flow Bus block connected to the machine terminals. The default value is `inf`, which means that there is no upper limit on the reactive power output. Default is `inf`.

## Inputs and Outputs

The units of inputs and outputs vary according to which dialog box you use to enter the block parameters. If you choose to specify fundamental parameters in SI units, the relevant inputs and outputs are in SI units. The dw in the vector of internal variables is in pu, and angle Θ is in radians. The other inputs and outputs are in pu.

`Pm`

The first Simulink input is the mechanical power at the machine's shaft, in Watts or pu. In generating mode, this input can be a positive constant or function or the output of a prime mover block (see the Hydraulic Turbine and Governor or Steam Turbine and Governor blocks). In motoring mode, this input is usually a negative constant or function.

`w`

The alternative block input instead of `Pm` (depending on the value of the Mechanical input parameter) is the machine speed, in rad/s.

`Vf`

The second Simulink input of the block is the field voltage. This voltage can be supplied by a voltage regulator in generator mode (see the Excitation System block). It is usually a constant in motor mode.

If you use the model in SI fundamental units, the field voltage Vf must be entered in volts DC if nominal field current Ifn is specified, or in volts referred to stator if Ifn is not specified. To obtain the Vfd producing nominal voltage, select the Display nominal field current and voltage producing 1 pu stator voltage check box in the Advanced tab. If you use the model in pu Standard or in pu Fundamental units, Vf must be entered in pu (1 pu of field voltage producing 1 pu of terminal voltage at no load).

`m`

The Simulink output of the block is a vector containing measurement signals. You can demultiplex these signals by using the Bus Selector block provided in the Simulink library. Depending on the type of mask that you use, the units are in SI or in pu.

Name

Definition

Units

ias

Stator current is_a

A or pu

ibs

Stator current is_b

A or pu

ics

Stator current is_c

A or pu

iq

Stator current iq

A or pu

id

Stator current id

A or pu

ifd

Field current ifd

A or pu

ikq

Damper winding current ikq1

A or pu

Ikq2

Damper winding current ikq2

A or pu

ikd

Damper winding current ikd

A or pu

phimq

Mutual flux phimq

V.s or pu

phimd

Mutual flux phimd

V.s or pu

vq

Stator voltage vq

V or pu

vd

Stator voltage vd

V or pu

lmq

Lmq saturated inductance

H or pu

lmd

Lmd saturated inductance

H or pu

dtheta

Rotor angle deviation d_theta

w

Rotor speed wm

Pe

Electrical power Pe

W or pu

dw

Rotor speed deviation dw

pu

theta

Rotor mechanical angle theta

degree

Te

Electromagnetic torque Te

N.m or pu

delta

degree

Pe0

Output active power Peo

W or pu

Qe0

Output reactive power Qeo

VAR or pu

## Limitations

In Discrete systems, when you use Synchronous Machine blocks discretized with the trapezoidal non-iterative solver, you might have to use a small parasitic resistive load, connected at the machine terminals, to avoid numerical oscillations. Large sample times require larger loads. The minimum resistive load is proportional to the sample time. As a rule of thumb, remember that with a 25 μs time step on a 60-Hz system, the minimum load is approximately 2.5% of the machine nominal power. For example, a 200 MVA synchronous machine in a power system discretized with a 50 μs sample time requires approximately 5% of resistive load or 10 MW. If the sample time is reduced to 20 μs, a resistive load of 4 MW should be sufficient.

However, if you discretize the Synchronous Machine block using the trapezoidal iterative (alg. loop) solver, you can use a negligible parasitic load (below 0.1% of nominal power) while preserving numerical stability. This iterative model producing an algebraic loop results in slower simulation speed.

## Example 1

The `power_SM_Fundamental` example shows the use of the Synchronous Machine SI Fundamental block and the Synchronous Machine pu Fundamental block to model a 555 MVA, 24 kV, 60 Hz, 3600-rpm synchronous generator. It shows how to specify SI and pu parameters. It also explains how to compute field and damper rotor winding parameters that are referred to the stator. In addition to the field winding, the round rotor of this machine has three damper windings: one damper in the direct axis and two dampers in the quadrature axis.

Three circuits simulate the same synchronous machine:

• Circuit 1: Fundamental parameters are specified in SI. The nominal field current is specified (ifn = 1300 A).

• Circuit 2: Fundamental parameters are specified in SI. The nominal field current is not specified (ifn = 0).

• Circuit 3: Fundamental parameters are specified in pu.

Machine parameters are taken from an example in Kundur [3].

Look at the Model Properties/PreLoad Fcn callback section of the model to see machine specifications and the computation of stator and field bases, RL rotor parameters referred to the stator, transformation ratio, and nominal field voltage and current.

The machines initially operate in steady state at virtually no load (load = 0. 1% of nominal power) with constant field voltage and mechanical power. A phase-to-phase 6-cycle fault is applied at t = 0.1 sec. The Scope shows the comparison between line-to-line AB voltage, phase A stator current, and field current of the three machines.

To simulate the discrete model, select Discrete simulation type in the Powergui block. The model is discretized with a sample time Ts = 50 μs. To obtain a stable model with such a small load (0. 1% of nominal power), select the Trapezoidal iterative (alg. Loop) discrete solver in the Advanced tab of each Synchronous Machine block.

## Example 2

The `power_syncmachine` example illustrates the use of the Synchronous Machine block in motor mode. The simulated system consists of an industrial grade synchronous motor (150 HP (112 kVA), 762 V) connected to a network with a 10 MVA short-circuit level. The machine is initialized for an output electrical power of −50 kW (negative value for motor mode), corresponding to a mechanical power of −48.9 kW. The corresponding values of mechanical power and field voltage are specified by the Pm Step block and in the Vf Constant block. The Pm Step block applies a sudden increase of mechanical power from −48.9 kW to −60 kW at time t = 0.1 s.

Run the simulation.

After the load has increased from 48.9 kW to 60 kW at t = 0.1 s, the machine speed oscillates before stabilizing to 1800 rpm. The load angle (angle between terminal voltage and internal voltage) increases from −21 degrees to −53 degrees.

## References

[1] Krause, P.C., Analysis of Electric Machinery, McGraw-Hill, 1986, Section 12.5.

[2] Kamwa, I., et al., “Experience with Computer-Aided Graphical Analysis of Sudden-Short-Circuit Oscillograms of Large Synchronous Machines,” IEEE® Transactions on Energy Conversion, Vol. 10, No. 3, September 1995.

[3] Kundur, P., Power System Stability and Control, McGraw-Hill, 1994.


text:
# Can symmetry be restored in high energy scattering?

Suppose you have a field theory with a real scalar field $\phi$ and a potential term of the form $\lambda \phi^4 - \mu \phi^2$ that breaks the symmetry $\phi \to - \phi$ in the ground state. Is this symmetry restored in a scattering with high momentum transfer in any physically meaningful way? My problem is that there is no background field for which I could take a mean field approximation, so the usual argument for phase transitions does not work.

One thing that came to my mind is that the potential will have higher order corrections, so maybe one could ask which way these corrections work in the limit $\Lambda \to \infty$, where $\Lambda$ is a UV scale? I know the first order correction for small $\phi$ does actually contribute to the breaking rather than to restoration, but then that doesn't seem to be the right limit for the scattering. If you don't know the answer I'd also be grateful for a reference. (Please do not point me to references that are for in-medium effects, nuclear matter etc, as I have plenty of these and it's not addressing my problem.)

-

First of all, the vev of $\phi$ scales like a positive power of $\mu$ which has the units of mass. So all the effects of the symmetry breaking scale like a positive power of the mass scale $\mu$. At energies $E$ satisfying $E\gg \mu$, i.e. much higher than $\mu$, the value of $\mu$ itself as well as vev and other things may simply be neglected relatively to $E$: all the corrections from the symmetry breaking go to zero, relatively speaking.
In the UV, i.e. at short distances, the dimensionless couplings (and interactions) such as $\lambda$ are much more important than the positive-mass-dimension dimensionful couplings such as $\mu$.
It's still true that the high-energy theory wasn't exactly symmetric in the treatment above; it was just approximately symmetric. However, in fact, sometimes it is exactly symmetric under the sign flip of $\phi$. That's because the value of $\mu$ runs, as in the renormalization group, and in many interesting theories, $\mu$ actually switches the sign at energies exceeding some critical energy scale $E_0$. So the high energy theory may have a single symmetry-preserving minimum of the potential and the symmetry breaking may be a result of the flow to low energies.
Ok, thanks. I see the point that the operator becomes less relevant at high energies. Let me think about that. Can you give me a keyword for the "many interesting theories" in which $\mu$ switches the sign? – WIMP May 30 '12 at 10:18


text:
## Thursday, December 5, 2013

Nothing too fancy or mind-shattering here, but perhaps someone may find this useful. Just a quick, little graphic organizer for the steps I want students to follow to make a nice graph of a quadratic function.

You need to print both pages and copy from one-sided to two-sided so that you can fold over the top and make it look like this:

Students should cut each numbered flap and then fill in each step with an example problem. Here is what the final copy will look like:
Outside:

Inside:

## Wednesday, December 4, 2013

### Group success!!

Today had one of those class periods. You know the ones. Where everyone is working and discussing and arguing and engaged and I'm? I'm just there. Totally and completely unnecessary to the workings of my class. Amaze.

Today was the first day of learning to solve systems by substitution. We went through an example together using @cheesemonkeysf's patented "substitution by stars" method. Then, I passed out some problems for them to work through, telling them to work with their group members and only go to the next problem when everyone in the group was totally good with the previous one. There were some where students had to apply the distributive property and some where they had to isolate a variable first. We had not talked about these. I did not stop class to discuss and fix and give hints, just circulated the room and looked on. Not one kid asked me for help for the 20 minutes that they worked - they puzzled through and argued and made mistakes and erased them and made some more and eventually figured it out all on their own. Holy wow. You guys get it. We all want group work to go exactly like this, but it so rarely does.

But finally!! All of the working on norms and redirecting questions with, "Have you asked your group yet?" or "Is this a group question?" and biting my lip to avoid butting in have paid off and there was just this amazing energy and focus and I was so, so proud. Cause there aren't that many glorious classroom moments like this, and it's so easy to focus on all the ways that we aren't perfect and they aren't either. So I want to stop and acknowledge that today was exactly right and keep it in my memory bank for when I'm feeling frustrated and disgruntled. Hooray!!!!

## Sunday, December 1, 2013

### Axis of symmetry for a quadratic function

My two-year Algebra course is just starting our study of quadratics. So far, we've looked at graphs and tables for quadratic functions and compared them to linear functions and talked about the symmetry of the shape and finding the axis of symmetry from the x-intercepts and finding the reflections of points across the axis of symmetry. I would like students to derive the formula for the axis of symmetry: $x=\frac{-b}{2a}$for the quadratic function $y=ax^{2}+bx+c$. The Exeter Math 1 curriculum does this by having students graph functions in the form $y=ax^{2}+bx$, then generalizing the idea that one of the x-intercepts is zero and the other one is $-\frac{b}{a}$. I created a worksheet that extends on this idea using desmos to graph and having students also derive the formula for the axis of symmetry by averaging the two x-intercepts to get $x=\frac{-b}{2a}$. Then, just like in the Exeter packet, I will have students compare graphs of $y=ax^{2}+bx$ to graphs of $y=ax^{2}+bx+c$ to realize that the axis of symmetry is not changed by adding a c value to the equation. I'm hoping that the logical progression of the graphs will make sense and that the questions are open enough for this to be genuinely based on student thinking and discovery, but with enough structure that students who are not necessarily used to deriving formulas will have something concrete to think about and answer. Any and all feedback welcome!!


text:
# Séminaires : Séminaire d'Algèbre

 Equipe(s) : gr, Responsables : J. Alev, D. Hernandez, B. Keller, Th. Levasseur, et S. Morier-Genoud. Email des responsables : Jacques Alev , David Hernandez , Bernhard Keller , Thierry Levasseur , Sophie Morier-Genoud Salle : à distance / remote Adresse : IHP Description Depuis le 23 mars 2020, le séminaire se tient à distance. Pour les liens et mots de passe, merci de contacter l'un des organisateurs ou de souscrire à la liste de diffusion https://listes.math.cnrs.fr/wws/info/paris-algebra-seminar. L'information nécessaire sera envoyée par courrier électronique peu avant chaque exposé. Les notes et transparents sont disponibles ici.   Since March 23, 2020, the seminar has been taking place remotely. For the links and passwords, please contact one of the organizers or subscribe to the mailing list at https://listes.math.cnrs.fr/wws/info/paris-algebra-seminar. The connexion information will be emailed shortly before each talk. Slides and notes are available here.

 Orateur(s) Matheus BRITO - Paraná, Titre Tensor products of prime representations of quantum affine $sl_{n+1}$ (projet ERC QAffine) Date 05/12/2016 Horaire 14:00 à 15:00 Diffusion Résume We study the family of prime representations of quantum affine $sl_{n+1}$ introduced in the work of Hernandez and Leclerc. These are defined by using an $A_n$-quiver; in the case of the sink-source quiver and the monotonic quiver they proved that the associated subcategory of finite–dimensional representations of the quantum affine algebra was a monoidal categorification of a cluster algebra with the prime representations corresponding to cluster variables. In this talk we shall work with an arbitrary quiver and give a necessary and sufficient condition in terms of Drinfeld polynomials for a tensor product of prime representations to be irreducible. We also state precisely the “exchange relations” in the case when a tensor product is reducible; in other words we describe the Jordan–Holder series of the tensor product. As a consequence of our results we write an explicit formula for the character of a prime representation as an alternating linear combination of characters of the local Weyl modules for quantum affine algebras. The talk is based on a joint work with Vyjayanthi Chari Salle à distance / remote Adresse IHP


text:
# SRAM output flickering

I am using the U62256A 32K x 8 bit static RAM. For testing purposes I connected all outputs to LEDs and since they are also inputs to a dip-switch so I'm able to write data.

The write mode works nicely and stores everything correctly. Data rentention in the RAM chip is also working. What doesn't quite work to my satisfaction is the read mode. Stored 1's (high) are properly output at a stable 5V, but 0's are flickering. My cheap voltmeter reads about 2.4V at the output and the LEDs are visually flickering at about half the brightness of a high output. According to the datasheet there should be a maximum voltage of 0.4V when the output is low.

This is the circuit. The right side of the LS245 is there so I can connect the RAM output to a bus. I have already tried removing it and connecting the outputs directly to the LEDs, as well as disconnecting the dip-switches (not shown in the schematic). I'd like for the LEDs to be simply off when the outputs are low, but with my little knowledge on electronics I can't figure out why they are flickering or what to do about it.

I'll appreciate any and all help!

• If you disconnect the LEDs do you read the expected voltages? – user253751 Apr 19 '17 at 1:18
• Your schematic would be much more useful if you showed the pin names (A0, A1,.. , D0, D1..., etc.). The ~W write enable requires a pull-down resistor to ensure that it will go low when you press the button. What are you doing with the address and data lines? They must be held in known states by tying to Vcc or Gnd, or with pull-up or pull-down resistors. If the inputs are not held in a known stare, they will wander randomly between High and Low. – Peter Bennett Apr 19 '17 at 2:09
• The address inputs are connected to the output of a 2-to-1 line selector so I can choose between an address from a register or manual input. The write enable pin is pulled up. I just updated the schematic. Unfortunately I don't know how to change the pin names in the program I'm using. – Padarom Apr 19 '17 at 8:34
• A voltage of 2.4V implies that you have a bus conflict, something is pulling the wire high at the same time as the SRAM is trying to pull it low. What is the voltage on the lines when the SRAM read line is idle? When idle pull the data line high and then low using a 1k resistor, if it doesn't go to both rails then something else is driving those pins. – Andrew Apr 19 '17 at 8:39

The fact that you have tied the $\overline G$ and $\overline E$ pins hard to ground is probably your issue.
• From how I understand the datasheet this shouldn't be an issue as the timing diagram specifies the read mode is finished on the falling edge of $\overline W$ before writing to the device. Data retention is also only possible while the $\overline E$ pin is low and there is no situation I can see in my application where I want to disable the chip, so I can't see why $\overline E$ is an issue? – Padarom Apr 19 '17 at 13:14
• @Padarom since you are using only $W$ the outputs do not turn off till you drop it, and return to outputs when you raise it. However he data is clocked in when you raise $W$ so obviously you still need to be driving the data lines at that point. Since $t_{LZWE}$ is minimum zero, that means a collision on the bus. – Trevor_G Apr 19 '17 at 13:24
• You may get away with just using $G$ properly and tieing – Trevor_G Apr 19 '17 at 13:25


text:
# Application of the base change theorem

Given a flat and projective morphism $f:X\rightarrow Y$ of noetherian schemes over some algebraically closed $k$ and $F$, $G$ coherent $O_X$-modules, flat over $Y$.

Then the base change theroem for the relative $Ext$ sheaves reads: Let $y\in Y$ and assume $\tau^i(y): \mathcal{E}xt_f^i(F,G)\otimes k(y)\rightarrow Ext_{X_y}^i(F_y,G_y)$ is surjective. Then:

i) There is a neighbourhood $U$ of $y$ s.t. $\tau^i(y')$ is an isomorphism for all $y' \in U$

ii) $\tau^{i-1}(y)$ is surjective if and only if $\mathcal{E}xt_f^i(F,G)$ is locally free in a neighbourhood of $y$

So now assume, we are in the following situation: $Y$ is a smooth projective surface and $X$ is the product of $Y$ with another projective smooth surface and $f$ is the projection. We have $Ext_{X_y}^i(F_y,G_y)=0$ for all $i\geq3$ and all $y\in Y$. Furthermore we have: $Ext_{X_y}^i(F_y,G_y)=0$ for $i=0,1,2$ and all $y\in Y\setminus \{y_0\}$ for a fixed $y_0 \in Y$. Finally $Ext_{X_{y_0}}^0(F_{y_0},G_{y_0})=Ext_{X_{y_0}}^2(F_{y_0},G_{y_0})=k^s$ and $Ext_{X_{y_0}}^1(F_{y_0},G_{y_0})=k^{2s}$ for some $s\geq1$.

The claim is that we get (with an application of the base change theorem):

(a)$\mathcal{E}xt_f^i(F,G)=0$ for $i=0,1$ and (b)$\mathcal{E}xt_f^2(F,G)\otimes k(y_0)=k^s$

So, since $Ext_{X_y}^3(F_y,G_y)=0$ for all y, we have by i) that $\tau^3(y)$ is surjective for all y, so it is an isomorphism for all $y\in Y$ and $\mathcal{E}xt_f^3(F,G)=0$, which this is locally free on $Y$. So $\tau^2(y)$ is an isomorphism for all $y\in Y$ by ii). So we get (b).

But how about (a). We have that $\tau^1(y)$ is surjective for all $y\in Y\setminus y_0$. But what about $\tau^1(y_0)$. What can be said about $\mathcal{E}xt_f^2(F,G)$? Is it localy free somewhere? How do we get the vanishing of $\mathcal{E}xt^1_f(F,G)$ with the base change theorem?

Background: I'm still working on the Example $\href{http://books.google.de/books?id=_mYV1q0RVzIC&printsec=frontcover&dq=%22geometry+of+moduli+spaces+of+sheaves%22&source=bl&ots=ZGL9LDSVjV&sig=p9l1BZ-UOXZdlxakuC3k15sTtBk&hl=de&ei=flHpTOHyPI3oOaXj7IcK&sa=X&oi=book_result&ct=result&resnum=2&sqi=2&ved=0CCMQ6AEwAQ#v=onepage&q=base%20change&f=false}{here}$ at the bottom of page 169.

Edit: Sasha gave an answer that uses derived categories. But i'm also interested if one can prove this, just by using i) and ii) in the base change theorem.

-

It is always useful to look from the derived category point of view. What you want is the object of $D(Y)$ which is $Rf_*R{\mathcal H}om(F,G)$ --- the sheaves ${\mathcal E}xt^i_f$ are its sheaf cohomology. The base change tells you that $Li^*Rf_*R{\mathcal H}om(F,G) \cong RHom(F_{y_0},G_{y_0})$, where $i$ is the embedding of the point $y_0$. Since you know $Ext^2(F_{y_0},G_{y_0}) = k^s$, you have a map $RHom(F_{y_0},G_{y_0}) \to k^s[-2]$. It gives by adjunction a map $Rf_*R{\mathcal H}om(F,G) \to O_{y_0}^s[-2]$. You want to prove that it is an isomorphism. Include this morphism into exact triangle $$Rf_*R{\mathcal H}om(F,G) \to O_{y_0}^s[-2] \to C.$$ You want to check that $C = 0$. You already know that $C$ is zero on the complement of $y_0$. Then it suffices to check that $Li^* C = 0$. Apply $Li^*$ to the triangle, you will get $$RHom(F_{y_0},G_{y_0}) \to Li^* O_{y_0}^s[-2] \to Li^* C.$$ Note that $Li^* O_{y_0} = Li^* i_* k$ has cohomology $k$, $k^2$ and $k$ in degrees $0$, $-1$ and $-2$. Hence you see that the first two terms of the sequence have the same cohomology. So, it remains to check that the map induces their isomorphism. This is the tricky part.
To do this, you need more information about the situation. In your case, remember that $L_1i^* O_{y_0} = T_{y_0}^* Y$, and since $Y$ is the moduli space of sheaves $G$, it is isomorphic to $Ext^1(G_{y_0},G_{y_0})^*$. So, you have to check that $Ext^\bullet(F_{y_0},G_{y_0})$ as a module over $Ext^\bullet(G_{y_0},G_{y_0})$ is free, which is evident in your example.
Okay. I'm not so good in derived categorys, but i will try to get this argument. Now i'm having two questions, just to avoid confusions: $O_{y_0}$ denotes the structure sheaf of the point $y_0$? And why do we have $L_1 i^{\*} O_{y_0} =T^{\*}_{y_0} Y$, the cotangent space, i guess? Do you know some literature concerning the use of derived categories in algebraic geometry? –  TonyS Nov 21 '10 at 21:17
Yes, $O_{y_0}$ is the structure sheaf of the point. And if $Z \subset Y$ is a locally complete intersection then $L_k i^* O_Z = \Lambda^k N^*_{Z/Y}$ (this is called "local fundamental isomorphism" or something of this sort), can be checked e.g. by a local computation. –  Sasha Nov 22 '10 at 3:38
You have to check that the map $Ext^i(F_{y_0},G_{y_0}) \to Ext^2(F_{y_0},G_{y_0}) \otimes Ext^{2-i}(G_{y_0},G_{y_0})^*$ is an isomorphism, the map is induced by the Yoneda multiplication $Ext^i(F_{y_0},G_{y_0}) \otimes Ext^{2-i}(G_{y_0},G_{y_0}) \to Ext^2(F_{y_0},G_{y_0})$). Since $Ext^\bullet(G_{y_0},G_{y_0})$ is a Frobenius algebra (the exterior algebra on two variables), it is equivalent to $Ext^\bullet(F_{y_0},G_{y_0})$ being a free module (with respect to the action given by the Yoneda multiplication). The latter is clear since $F_{y_0}$ is a multiple of $G_{y_0}$ (ro vice versa). –  Sasha Nov 22 '10 at 12:57