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# At the bakery Lew spent a total of 6$for one kind of  Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Manager Joined: 29 Nov 2011 Posts: 81 Followers: 1 Kudos [?]: 197 [0], given: 37 At the bakery Lew spent a total of 6$ for one kind of [#permalink]  15 Apr 2012, 21:29
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Difficulty:

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Question Stats:

49% (02:59) correct 51% (01:37) wrong based on 77 sessions
At the bakery Lew spent a total of 6$for one kind of cupcake and one kind of doughnut. How many donuts did he buy? (1) Price of 2 doughty was$.10 less than 3 cupcakes
(2) Average price of 1 doughnut and 1 cupcake was $.035 [Reveal] Spoiler: OA Senior Manager Joined: 12 Mar 2012 Posts: 369 Concentration: Operations, Strategy Followers: 2 Kudos [?]: 139 [1] , given: 31 Re: Word problem: doughnut [#permalink] 15 Apr 2012, 21:56 1 This post received KUDOS Smita04 wrote: At the bakery lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?
1) price of 2 doughty was $.10 less than 3 cupcakes 2) average price of 1 doughnut and 1 cupcake was$.035

let
cupcakes purchased= x
doughnut purchased= y

price of one cupcake and doughnut be c and d respectively,
then

cx+dy = 6

we need to find y.

statement 1) 2d= 3c - 0.1
INsufficient

statement 2) (c+d)/2 = 0.035
c+d = 0.035*2

Insufficient

1) and 2)

we can have c and d but we have no info about x and y.

Insufficient.

hence E

Hope this helps..!!
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Re: At the bakery Lew spent a total of 6$for one kind of [#permalink] 05 Feb 2014, 21:54 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED Smita04 wrote: At the bakery Lew spent a total of 6$ for one kind of cupcake and one kind of doughnut. How many donuts did he buy?

(1) Price of 2 doughty was $.10 less than 3 cupcakes (2) Average price of 1 doughnut and 1 cupcake was$.035

There is an error in this question.
"Average price of 1 doughnut and 1 cupcake was $0.35" It doesn't make sense if the average price is 3.5 cents. The numbers don't work. It must be 35 cents. Convert everything to cents. Using both statements together, you get 3C - 2D = 10 (C is the price of cupcakes and D is the price of doughnuts) C + D = 35*2 Solving them simultaneously, you get C = 30, D = 40. If Nc is number of cupcakes and Nd is number of doughnuts, 30*Nc + 40*Nd = 600 3*Nc + 4*Nd = 60 The first solution I get is Nd = 3, Nc = 16 We will get many more solutions such as Nd = 6, Nc = 12. Also, Nd = 9, Nc = 8 etc Hence both statements together are not sufficient. Answer (E) Note that in this equation: 3*Nc + 4*Nd = 60, if instead of 60, the total price were say 18, there would have been only one solution. 3*Nc + 4*Nd = 18 Nc = 2, Nd = 3 Other possible solutions will be Nc = 6, Nd = 0; Or Nc = -2, Nd = 6 etc Since we cannot buy negative or 0 number of items (she spends money on doughnuts AND cupcakes so she must have bought at least one of each), no other solution works. Refer to this post for further details on this method: http://www.veritasprep.com/blog/2011/06 ... -of-thumb/ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Word problem: doughnut [#permalink]  05 Feb 2014, 13:41
Smita04 wrote:
At the bakery lew spent a total of 6$for one kind of cupcake and one kind of doughnut. How many donuts did he buy? 1) price of 2 doughty was$.10 less than 3 cupcakes
2) average price of 1 doughnut and 1 cupcake was $.035 let cupcakes purchased= x doughnut purchased= y price of one cupcake and doughnut be c and d respectively, then cx+dy = 6 we need to find y. statement 1) 2d= 3c - 0.1 INsufficient statement 2) (c+d)/2 = 0.035 c+d = 0.035*2 Insufficient 1) and 2) we can have c and d but we have no info about x and y. Insufficient. hence E Hope this helps..!! But I guess still one has to prove whether 'x' and 'y' could be found using the equation Dx + Cy, remember that x,y standing for the number of Donuts and Cupcakes have to be integers and must satisfy Dx + Cy = 6. Therefore, the next question is, how can we do this in the most efficient way? Experts, any thoughts? GMAT Club Legend Joined: 09 Sep 2013 Posts: 8204 Followers: 417 Kudos [?]: 111 [0], given: 0 Re: At the bakery Lew spent a total of 6$ for one kind of [#permalink]  03 Mar 2015, 05:19
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).

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Re: At the bakery Lew spent a total of 6$for one kind of [#permalink] 03 Mar 2015, 05:19 Similar topics Replies Last post Similar Topics: 6 Sam spent a total of$6.00 on pens and pencils. How many pen 3 01 May 2013, 00:59
2 L spends total $6.00 for one kind of D and one kind of C. How many D d 4 18 Nov 2011, 13:06 4 Yesterday Diana spent a total of 240 minutes attending a 4 07 Sep 2010, 14:37 At the bakery , Lew spent a total of$6.00 for one kind of 4 16 May 2010, 05:12


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# The equation of a line is y=mx+1. How do you find the value of the gradient m given that P(3,7) lies on the line?

Jan 9, 2017

#### Answer:

$m = 2$

#### Explanation:

The problem tells you that the equation of a given line in slope-intercept form is

$y = m \cdot x + 1$

The first thing to notice here is that you can find a second point that lies on this line by making $x = 0$, i.e. by looking at the value of the $y$-intercept.

As you know, the value of $y$ that you get for $x = 0$ corresponds to the $y$-intercept. In this case, the $y$-intercept is equal to $1$, since

$y = m \cdot 0 + 1$

$y = 1$

This means that the point $\left(0 , 1\right)$ lies on the given line. Now, the slope of the line, $m$, can be calculated by looking at the ratio between the change in $y$, $\Delta y$, and the change in $x$, $\Delta x$

$m = \frac{\Delta y}{\Delta x}$

Using $\left(0 , 1\right)$ and $\left(3 , 7\right)$ as the two points, you get that $x$ goes from $0$ to $3$ and $y$ goes from $1$ to $7$, which means that you have

$\left\{\begin{matrix}\Delta y = 7 - 1 = 6 \\ \Delta x = 3 - 0 = 3\end{matrix}\right.$

This means that the slope of the line is equal to

$m = \frac{6}{3} = 2$

The equation of the line in slope-intercept form will be

$y = 2 \cdot x + 1$

graph{2x + 1 [-1.073, 4.402, -0.985, 1.753]}

format: plain


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# Consider the following grammarS → m | mn | mnoChoose correct statement from the following:

1. The grammar is LL (2)
2. The grammar is LL (4)
3. The grammar is LL (1)
4. The grammar is LL (3)

Option 4 : The grammar is LL (3)

## LL(1) MCQ Question 1 Detailed Solution

Concept:

In LL(K) grammar, first L means scanning input from left to right and second L stands for left most derivation where k means k symbols are used at each step.

If first(production 1) ꓵ first(production 2) = empty, then it is LL(1). , it must be true for any two prductions.

For LL(2) we will check intersection of second(production) means second element of that production. Moving forward one step with symbol in bracket.

Explanation:

Given grammar:

S → m | mn | mno

Case 1: For LL(1)

First(m) ꓵ first (mn) = m (which is not empty).

So, grammar is not LL(1).

Case 2: For LL(2)

Second (mn) ꓵ second(mno) = n (which is not empty)

So, grammar is not LL(2).

Case 3: For LL(3)

Third(mn) ꓵ third (mno) = empty

So, grammar is LL(3).

## LL(1) MCQ Question 2

### Comprehension:

For the grammar below, a partial LL(1) parsing table is also presented along with the grammar. Entries that need to be filled are indicated as E1, E2, and E3. ε is the empty string, $indicates end of input, and, | separates alternate right hand sides of productions. S → a A b B | b A a B | ε A → S B → S  a b$ S E1 E2 S → ε A A → S A → S Error B B → S B → S E3

# The appropriate entries for E1, E2, and E3 are

1. E1: S → aAbB, A → S

E2: S → bAaB, B → S

E3: B → S

2. E1: S → aAbB, S → ε

E2: S → bAaB, S → ε

E3: S → ε

3. E1: S → aAbB, S → ε

E2: S → bAaB, S → ε

E3: B → S

4. E1: A → S, S → ε

E2: B → S, S → ε

E3: B → S

Option 3 :

E1: S → aAbB, S → ε

E2: S → bAaB, S → ε

E3: B → S

## LL(1) MCQ Question 2 Detailed Solution

The correct answer is option 3

EXPLANATION:

FIRST and FOLLOW:

 FIRST FOLLOW S {a, b, ϵ } {a,b, $} A {a,b, ϵ } {b,a} B {a,b,ϵ } {a,b,$}

Parsing Table:

 a b $S S→ aAbB S→ ϵ S → bAaB S→ ϵ S→ ϵ A A → S A → S Error B B → S B → S B→S Hence the answer is E1: S → aAbB, S → ε E2: S → bAaB, S → ε E3: B → S ## LL(1) MCQ Question 3 ### Comprehension: For the grammar below, a partial LL(1) parsing table is also presented along with the grammar. Entries that need to be filled are indicated as E1, E2, and E3. ε is the empty string,$ indicates end of input, and, | separates alternate right hand sides of productions.

S → a A b B | b A a B | ε

A → S

B → S

 a b $S E1 E2 S → ε A A → S A → S Error B B → S B → S E3 # The FIRST and FOLLOW sets for the non-terminals A and B are 1. FIRST(A) = {a, b, ε} = FIRST(B) FOLLOW(A) = {a, b} FOLLOW(B) = {a, b,$}

2. FIRST(A) = {a, b, $} FIRST(B) = {a, b, ε} FOLLOW(A) = {a, b} FOLLOW(B) = {$}

3. FIRST(A) = {a, b, ε} = FIRST(B)

FOLLOW(A) = {a, b}

FOLLOW(B) = Φ

4. FIRST(A) = {a, b} = FIRST(B)

FOLLOW(A) = {a, b}

FOLLOW(B) = {a, b}

Option 1 :

FIRST(A) = {a, b, ε} = FIRST(B)

FOLLOW(A) = {a, b}

FOLLOW(B) = {a, b, $} ## LL(1) MCQ Question 3 Detailed Solution The correct answer is option 1 EXPLANATION: FIRST(A) = FIRST(s) ={a, b, ε} FIRST(B) = FIRST(s) ={a, b, ε} So,FIRST(A) = {a, b, ε} = FIRST(B) FOLLOW(A) = {First of { bB}, FIRST of {aB}} FOLLOW(A) = {a, b} FOLLOW(B) = FOLLOW(S) ={FOLLOW(A),$}

FOLLOW(B) = FOLLOW(S) ={FOLLOW(A), $} ={a, b,$}

FIRST(A) = {a, b, ε} = FIRST(B)

FOLLOW(A) = {a, b}

# Which one of the following is True?

1. Every regular grammar is LL (1) and every regular has an LR (1) grammar
2. Every regular grammar is not LL (1) and every regular set has an LR (1) grammar
3. Every regular grammar is LL (1) and every regular set does not have-an LR (1) Grammar
4. Every regular grammar is not LL (1) and every regular set does not have-an LR (1) grammar

Option 2 : Every regular grammar is not LL (1) and every regular set has an LR (1) grammar

## LL(1) MCQ Question 5 Detailed Solution

Explanation

LL (1) grammar is those that can be parsed by an LL parser. LL parsers are top-down and work well for a subset of context-free languages.

For example, S -> a | aa is a regular grammar that cannot be parsed by an LL (1) parser without backtracking. Because LL (1) parser looks at only the first symbol and here the parser cannot always ensure the correct choice of rule. Hence not every regular grammar is LL (1).

LR Grammar is one that can be parsed by LR parsers. These parsers are bottom-up and work efficiently for all deterministic context-free languages in guaranteed linear time.

Since all Regular languages are DCFLs, they can all be parsed using LR (1) grammar.

# Consider the following grammarS → m | mn | mnoChoose correct statement from the following:

1. The grammar is LL (2)
2. The grammar is LL (4)
3. The grammar is LL (1)
4. The grammar is LL (3)

Option 4 : The grammar is LL (3)

## LL(1) MCQ Question 6 Detailed Solution

Concept:

In LL(K) grammar, first L means scanning input from left to right and second L stands for left most derivation where k means k symbols are used at each step.

If first(production 1) ꓵ first(production 2) = empty, then it is LL(1). , it must be true for any two prductions.

For LL(2) we will check intersection of second(production) means second element of that production. Moving forward one step with symbol in bracket.

Explanation:

Given grammar:

S → m | mn | mno

Case 1: For LL(1)

First(m) ꓵ first (mn) = m (which is not empty).

So, grammar is not LL(1).

Case 2: For LL(2)

Second (mn) ꓵ second(mno) = n (which is not empty)

So, grammar is not LL(2).

Case 3: For LL(3)

Third(mn) ꓵ third (mno) = empty

So, grammar is LL(3).

# Consider the following grammar:S → aS'S' → aS' | ϵ Which of the following is/are true for the above grammar?

1. not left recursive
2. ambiguous
3. LL(1)
4. FIRST(S) = {a, ϵ}

Option :

## LL(1) MCQ Question 7 Detailed Solution

Option 1: TRUE

S → aS'  \\ right recursive

S' → aS' | ϵ  \\ right recursive

Therefore given grammar is not left recursive

Option 2: FALSE

Two parse trees cannot be generated for any string in the given grammar and hence it is not ambiguous.

Option 3: TRUE and Option 4: FALSE

A grammar G is LL(1) iff. for all non-terminals A, each distinct pair of productions A → β and A → γ satisfy the condition FIRST(β) $$\cap$$ FIRST(γ) = φ.

A grammar G is LL(1) iff. for each set of productions A→α1 |α2 |···|αn:

1. FIRST(α1), FIRST(α2), ..., FIRST(αn) are all pairwise disjoint

2. If αi ⇒ *ε then FIRST(αj$$\cap$$ FOLLOW(A) = φ, ∀1 ≤ j ≤ n, i $$\neq$$ j

Grammar is LL(1):

 NON- TERMINAL FIRST FOLLOW S a $S’ a, ϵ$

Although  FIRST(S') has ϵ, FIRST(S') and FOLLOW(S') doesn't contain any common element. Therefore a cell in parsing table will contain at most one entry in it and hence it is LL(1)

# Consider the following Grammar G:S → AeA → CbDC → BC | εB → cdD | acdDD → c | ε In LL(1) parse table of above grammar G, How many cells are having multiple entries?

## LL(1) MCQ Question 8 Detailed Solution

EXPLANATION:

FIRST and FOLLOW:

 FIRST FOLLOW S {a, b, c } { $} A {a,b, c } { e } B {c, a } {a,b,c} C {c, a, ε} { b } D {c, ε } { a,b,c,e} Parsing Table:  a b c d e$ S S → Ae S → Ae A A → CbD A → CbD A → CbD B B → acdD B → cdD C C → BC C → ε C → BC D D → ε D → ε D → c D → ε D → ε

In this LL(1) parse table for given grammar G, only one cell is having multiple entries.

# What is/are sufficient to prove that grammar is not LL(1)?

1. Grammar is ambiguous
2. Grammar is left recursive
3. Grammar is right recursive
4. Grammar is not context-free

Option :

## LL(1) MCQ Question 9 Detailed Solution

LL(1) could be right recursive and not context-free

Ambiguous or left recursive grammar cannot be LL(1). Hence it is sufficient to prove that grammar is ambiguous or left recursive in order to prove that it is not LL(1)

Important Points:

A non-ambiguous grammar can also be not LL(1) grammar.

# Consider the following grammar G:A → aA'A' → aA' | ϵ Which of the following is/are TRUE about G?

1. G is right recursive
2. G is ambiguous
3. G is LL(1)


text:
# Unsupervised Learning with PCA

### Principal Component Analysis Using R.

#### 2019-04-23

These note are primarily taken from the Unsupervised Learning in R DataCamp course.

# Background: Unsupervised vs Supervised Machine Learning

Unsupervised machine learning searches for structure in unlabeled data (data without a response variable). The goal of unsupervised learning is to find homogenous subgroups (clusters), and to find patterns (usually through dimensionality reduction). Examples of unsupervised machine learning include k-means clustering, hierarchical cluster analysis (HCA), and principal component analysis (PCA). Unsupervised machine learning often has no single goal beyond gaining insight.

Supervised machine learning makes predictions with labeled data. Supervised learning uses regression for quantitative outcomes and classification for qualitative outcomes.1 Reinforcement machine learning is a third type of learning where the machine learns by operating in an environment. Examples of supervised machine learning include decision trees, random forests, and lasso regression.

# PCA

Dimensionality reduction is a group of methods of finding structure in clusters and aiding visualization. One dimensionality reduction method is principal component analysis (PCA). PCA finds a linear combination of the features to create principal components. PCA maintains as much variance from the original data as possible. The principle features are unrelated to each other (orthogonal).

The prcomp(df, scale = FALSE, center = TRUE) function performs a PCA on the dataframe. As with other unsupervised machine learning algorithms, you must remove or impute data for missing values. If the data are on different scales, scale them to mean zero and unit variance with scale = TRUE. The center parameter shifts the variables to zero-centered and is typically left at TRUE.

## Example

The pokemon dataset contains observations of 800 pokemons2 More information on the dataset at https://www.kaggle.com/abcsds/pokemon on 6 dimensions. The data is unlabeled, meaning there is no response variable, just features. The features here are six pokeon ability measures.

library(readr)

#pokemon$Name <- NULL pokemon$Number <- NULL
pokemon$Type1 <- NULL pokemon$Type2 <- NULL
pokemon$Total <- NULL pokemon$Generation <- NULL
pokemon$Legendary <- NULL head(pokemon) ## # A tibble: 6 x 7 ## Name HitPoints Attack Defense SpecialAttack SpecialDefense Speed ## <chr> <int> <int> <int> <int> <int> <int> ## 1 Bulbasaur 45 49 49 65 65 45 ## 2 Ivysaur 60 62 63 80 80 60 ## 3 Venusaur 80 82 83 100 100 80 ## 4 VenusaurMeg~ 80 100 123 122 120 80 ## 5 Charmander 39 52 43 60 50 65 ## 6 Charmeleon 58 64 58 80 65 80 Before conducting PCA, check whether any preprocessing is required: Are there any NAs? If so, drop these observations, or impute values. Are all of the features comparable? If not, standardize the variables in the model with scale = TRUE. Are the features multi-nomial? If so, create binary variables. In this case, the means and standard deviations are similar, but I am scaling anyway for the exercise. pr.out <- prcomp(x = pokemon[-1], scale = TRUE, center = TRUE) summary(pr.out) ## Importance of components: ## PC1 PC2 PC3 PC4 PC5 PC6 ## Standard deviation 1.6466 1.0457 0.8825 0.8489 0.65463 0.51681 ## Proportion of Variance 0.4519 0.1822 0.1298 0.1201 0.07142 0.04451 ## Cumulative Proportion 0.4519 0.6342 0.7640 0.8841 0.95549 1.00000 PCA models in R produce additional diagnostic and output components: center: the column means used to center to the data, or FALSE if the data weren’t centered scale: the column standard deviations used to scale the data, or FALSE if the data weren’t scaled rotation: the directions of the principal component vectors in terms of the original features/variables. This information allows you to define new data in terms of the original principal components x: the value of each observation in the original dataset projected to the principal components pr.out$center
##      HitPoints         Attack        Defense  SpecialAttack SpecialDefense
##       69.25875       79.00125       73.84250       72.82000       71.90250
##          Speed
##       68.27750
pr.out$scale ## HitPoints Attack Defense SpecialAttack SpecialDefense ## 25.53467 32.45737 31.18350 32.72229 27.82892 ## Speed ## 29.06047 pr.out$rotation
##                      PC1         PC2         PC3        PC4         PC5
## HitPoints      0.3898858  0.08483455 -0.47192614  0.7176913 -0.21999056
## Attack         0.4392537 -0.01182493 -0.59415339 -0.4058359  0.19025457
## Defense        0.3637473  0.62878867  0.06933913 -0.4192373 -0.05903197
## SpecialAttack  0.4571623 -0.30541446  0.30561186  0.1475166  0.73534497
## SpecialDefense 0.4485704  0.23909670  0.56559403  0.1854448 -0.30019970
## Speed          0.3354405 -0.66846305  0.07851327 -0.2971625 -0.53016082
##                       PC6
## HitPoints       0.2336690
## Attack         -0.5029896
## Defense         0.5368986
## SpecialAttack   0.2045304
## SpecialDefense -0.5451707
## Speed           0.2551400

The biplot() function plots both the principal components loadings and the mapping of the observations to their first two principal component values. In the plot below, Attack and HitPoints have approximately the same loadings in the first two principal components.

biplot(pr.out)

The second common plot type for understanding PCA models is a scree plot. A scree plot shows the variance explained as the number of principal components increases. Sometimes the cumulative variance explained is plotted as well.

# Get proportion of variance for scree plot
pr.var <- pr.out\$sdev^2
pve <- pr.var / sum(pr.var)
plot(pve, xlab = "Principal Component",
ylab = "Proportion of Variance Explained",
ylim = c(0, 1), type = "b")

plot(cumsum(pve), xlab = "Principal Component",
ylab = "Cumulative Proportion of Variance Explained",
ylim = c(0, 1), type = "b")


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# Associations to the word «Comparative»

## Wiktionary

COMPARATIVE, adjective. Of or relating to comparison.
COMPARATIVE, adjective. Using comparison as a method of study, or founded on something using it.
COMPARATIVE, adjective. Approximated by comparison; relative.
COMPARATIVE, adjective. (obsolete) Comparable; bearing comparison.
COMPARATIVE, noun. (grammar) A construction showing a relative quality, in English usually formed by adding more or appending -er. For example, the comparative of green is greener; of evil, more evil.
COMPARATIVE, noun. (grammar) A word in the comparative form.
COMPARATIVE, noun. (obsolete) An equal; a rival; a compeer.
COMPARATIVE, noun. (obsolete) One who makes comparisons; one who affects wit.
COMPARATIVE ADVANTAGE, noun. (economics) the ability to produce a particular good at a lower relative opportunity cost than another producer
COMPARATIVE ANATOMY, noun. The study of similarities and differences in the anatomy of organisms.
COMPARATIVE CASE, noun. (grammar) A case used to mark a likeness to something, or that two things have the same quantity of something. It corresponds roughly to the English adverb "like". Languages that use the comparative case include Mari, Nivkh, and Chechen.
COMPARATIVE CASES, noun. Plural of comparative case
COMPARATIVE DEGREE, noun. (grammar) Adverbial or adjectival forms modified by more or ending in -er, used when comparing two things.
COMPARATIVE DEGREES, noun. Plural of comparative degree
COMPARATIVE ETHICS, noun. (anthropology) (sociology) (ethics) A type of descriptive ethics, focussed on studying the nature and origins of beliefs about morality held by people in diverse social or cultural groups.
COMPARATIVE LINGUIST, noun. Linguist whose field is comparative linguistics.
COMPARATIVE LINGUISTICS, noun. (linguistics) A branch of historical linguistics that is concerned with comparing languages in order to establish their historical relatedness.
COMPARATIVE METHOD, noun. (linguistics) A technique for studying the development of languages by performing a feature-by-feature comparison of two or more languages with a common ancestor.
COMPARATIVE SUPERLATIVE, noun. (grammar) In Spanish: the superlative form of an adjective when it is used in a grammatically comparative situation.

## Dictionary definition

COMPARATIVE, noun. The comparative form of an adjective or adverb; "faster' is the comparative of the adjective fast'"; "less famous' is the comparative degree of the adjective famous'"; "more surely' is the comparative of the adverb surely'".
COMPARATIVE, adjective. Relating to or based on or involving comparison; "comparative linguistics".
COMPARATIVE, adjective. Estimated by comparison; not absolute or complete; "a relative stranger".

## Wise words

To use the same words is not a sufficient guarantee of understanding; one must use the same words for the same genus of inward experience; ultimately one must have one's experiences in common.
Friedrich Nietzsche


text:
# Load and solve a model from a file

Filter Content By
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## Load and solve a model from a file

Examples: batchmode, callback, feasopt, fixanddive, lp, lpmethod, lpmod, mip2, params, sensitivity

One of the most basic tasks you can perform with the Gurobi libraries is to read a model from a file, optimize it, and report the result. The lp (lp_c.c, lp_c++.cpp, lp_cs.cs, Lp.java, lp.py, lp_vb.vb) and mip2 (mip2_c.c, mip2_c++.cpp, mip2_cs.cs, Mip2.java, mip2.m, mip2.py, mip2.R, mip2_vb.vb) examples are simple illustrations of how this is done in the various supported Gurobi languages. While the specifics vary from one language to another, the basic structure remains the same for all languages.

After initializing the Gurobi environment, the examples begin by reading the model from the specified file. In C, you call the GRBreadmodel() function:

  error = GRBreadmodel(env, argv[1], &model);

In C++, this is done by constructing a GRBModel object:
    GRBModel model = GRBModel(env, argv[1]);

In C# and Java, this is also done by constructing a GRBModel object:
      GRBModel model = new GRBModel(env, args[0]);

In Python, this is done via the read global function:
model = gp.read(sys.argv[1])

The next step is to invoke the Gurobi optimizer on the model. In C, you call GRBoptimize() on the model variable:

  error = GRBoptimize(model);

In C++, Java, and Python, this is accomplished by calling the optimize method on the model object:
    model.optimize();

In C#, the first letter of the method name is capitalized:
      model.Optimize();

A successful optimize call populates a set of solution attributes in the model. For example, once the call completes, the X variable attribute contains the solution value for each variable. Similarly, for continuous models, the Pi constraint attribute contains the dual value for each constraint.

The examples then retrieve the value of the model Status attribute to determine the result of the optimization. In the lp example, an optimal solution is written to a solution file (model.sol).

There are many other things you can do once you have read and solved the model. For example, lp checks the solution status -- which is highly recommended. If the model is found to be infeasible, this example computes an Irreducible Inconsistent Subsystem (IIS) to isolate the source of the infeasibility.


text:
Mathematics » Functions II » The Tangent Function

# Functions of the Form y = tan(θ + p)

## Functions of the form $$y=\tan(\theta +p)$$

We now consider tangent functions of the form $$y = \tan(\theta + p)$$ and the effects of parameter $$p$$.

## Optional Investigation: The effects of $$p$$ on a tangent graph

1. On the same system of axes, plot the following graphs for $$-\text{360}\text{°} \leq \theta \leq \text{360}\text{°}$$:

1. $$y_1 = \tan \theta$$
2. $$y_2 = \tan (\theta – \text{60}\text{°})$$
3. $$y_3 = \tan (\theta – \text{90}\text{°})$$
4. $$y_4 = \tan (\theta + \text{60}\text{°})$$
5. $$y_5 = \tan (\theta + \text{180}\text{°})$$
2. Use your sketches of the functions above to complete the following table:

 $$y_1$$ $$y_2$$ $$y_3$$ $$y_4$$ $$y_5$$ period domain range $$y$$-intercept(s) $$x$$-intercept(s) asymptotes effect of $$p$$

### The effect of the parameter on $$y = \tan(\theta + p)$$

The effect of $$p$$ on the tangent function is a horizontal shift (or phase shift); the entire graph slides to the left or to the right.

• For $$p > 0$$, the graph of the tangent function shifts to the left by $$p$$.

• For $$p < 0$$, the graph of the tangent function shifts to the right by $$p$$.

 $$p > 0$$ $$p < 0$$

## Example

### Question

1. Sketch the following functions on the same set of axes for $$-\text{180}\text{°} \leq \theta \leq \text{180}\text{°}$$.
1. $$y_1 = \tan \theta$$
2. $$y_2 = \tan (\theta + \text{30}\text{°})$$
2. For each function determine the following:

• Period
• Domain and range
• $$x$$- and $$y$$-intercepts
• Asymptotes

### Examine the equations of the form $$y = \tan (\theta + p)$$

Notice that for $$y_1 = \tan \theta$$ we have $$p = \text{0}\text{°}$$ (no phase shift) and for $$y_2 = \tan (\theta + \text{30}\text{°})$$ we have $$p = \text{30}\text{°} > 0$$ and therefore the graph shifts to the left by $$\text{30}$$$$\text{°}$$.

### Complete a table of values

 θ $$-\text{180}$$$$\text{°}$$ $$-\text{135}$$$$\text{°}$$ $$-\text{90}$$$$\text{°}$$ $$-\text{45}$$$$\text{°}$$ $$\text{0}$$$$\text{°}$$ $$\text{45}$$$$\text{°}$$ $$\text{90}$$$$\text{°}$$ $$\text{135}$$$$\text{°}$$ $$\text{180}$$$$\text{°}$$ $$\tan \theta$$ $$\text{0}$$ $$\text{1}$$ undef $$-\text{1}$$ $$\text{0}$$ $$\text{1}$$ undef $$-\text{1}$$ $$\text{0}$$ $$\tan (\theta + \text{30}\text{°})$$ $$\text{0.58}$$ $$\text{3.73}$$ $$-\text{1.73}$$ $$-\text{0.27}$$ $$\text{0.58}$$ $$\text{3.73}$$ $$-\text{1.73}$$ $$-\text{0.27}$$ $$\text{0.58}$$

### Complete the table

 $$y_1 = \tan \theta$$ $$y_2 = \tan (\theta + \text{30}\text{°})$$ period $$\text{180}\text{°}$$ $$\text{180}\text{°}$$ domain $$\{ \theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{90}\text{°}; \text{90}\text{°} \}$$ $$\{ \theta: -\text{180}\text{°} \leq \theta \leq \text{180}\text{°}, \theta \ne -\text{120}\text{°}; \text{60}\text{°} \}$$ range $$(-\infty;\infty)$$ $$(-\infty;\infty)$$ $$y$$-intercept(s) $$(\text{0}\text{°};0)$$ $$(\text{0}\text{°};\text{0.58})$$ $$x$$-intercept(s) $$(-\text{180}\text{°};0)$$, $$(\text{0}\text{°};0)$$ and $$(\text{180}\text{°};0)$$ $$(-\text{30}\text{°};0) \text{ and } (\text{150}\text{°};0)$$ asymptotes $$\theta = -\text{90}\text{°} \text{ and } \theta = \text{90}\text{°}$$ $$\theta = -\text{120}\text{°} \text{ and } \theta = \text{60}\text{°}$$

### Discovering the characteristics

For functions of the general form: $$f(\theta) = y =\tan (\theta + p)$$:

Domain and range

The domain of one branch is $$\{ \theta: \theta \in (-\text{90}\text{°} – p; \text{90}\text{°} – p) \}$$ because the function is undefined for $$\theta = -\text{90}\text{°} – p$$ and $$\theta = \text{90}\text{°} – p$$.

The range is $$\{ f(\theta): f(\theta) \in \mathbb{R} \}$$.

Intercepts

The $$x$$-intercepts are determined by letting $$f(\theta) = 0$$ and solving for $$\theta$$.

The $$y$$-intercept is calculated by letting $$\theta = \text{0}\text{°}$$ and solving for $$f(\theta)$$. \begin{align*} y &= \tan (\theta + p) \\ &= \tan (\text{0}\text{°} + p) \\ &= \tan p \end{align*} This gives the point $$(\text{0}\text{°};\tan p)$$.


text:
## intro tensors book

whats a good intro book to tensors and manifolds?

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Homework Help Science Advisor There are several threads here in which people have recommended sets of notes on relativity in which tensors are taught. In one of those sets of notes, that I particularly liked, the author Sean Carroll recommended the book by Frank Warner, on Differentiable manifolds and lie groups, as a "standard". I kind of like Michael Spivak's little book Calculus on manifolds, and his much longer series on Differential Geometry, say the first volume for starters. This is a mathematician talking, so I recommend getting some opinions from the physics experts too. Of course Carroll is presumably a physicist. Warnes book is nice because it also has an introduction to "Hodge theory" as I recall. Others here have recommended Tensor analysis on manifolds by Bishop and Goldberg, because it is not only a good classic text, but it is available in paper for a song. Look on the threads "What is a tensor", and Differential geometry lecture notes, and Math "Newb" Wants to know what a Tensor is, and others , for some free sites with downloadable material on tensors. I would warn you of one thing. I myself am primarily educated in the mathematics of manifolds and tensor bundles as in Spivak's calculus on manifolds. As you can see from numerous exchanges I have had with physicists on this forum I have great difficulty understanding what they are talking about. Thus i would suggest that it is not enough to understand only the mathematical concepts of manifolds and tensors, but one should go further and see these concepts in use either in differential geometry, or in physics.
 Their is a book I love and I think is very well suited as an intro : Geometry, Particles, and Fields (Graduate Texts in Contemporary Physics) by Bjorn Felsager http://www.amazon.com/exec/obidos/tg...06466?v=glance It is written by a high-school teacher, and oriented to physics application. However, it is quite rigourous enough (to me). It goes from the very beginning to advanced stuff in physics.

## intro tensors book

There are notes I highly recommended, because they are free on internet and starting at elementary level:
1) An Introduction To Tensors for Students of Physics and Engineering, by Joseph C. Kolecki
2) Quick Introduction to Tensor Analysis, by R.A. Sharipov
3) An Introduction to Tensor Analysis and Continuum Mechanics, by J.H. Heinbockel

And I would like to thank the authors of these notes.
Thank You!

Will.

 Recognitions: Homework Help Science Advisor THE book on tensor calculus is by Synge and Schild (Tensor Calculus) but I don't know whether it's still in print.

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 Quote by Tide THE book on tensor calculus is by Synge and Schild (Tensor Calculus) but I don't know whether it's still in print.
http://store.yahoo.com/doverpublications/index.html
(enter "tensor" in the search box).

Although Synge and Schild is great, I would have to say that
Schouten's "Ricci Calculus" is THE book on tensor calculus...

 Schaum's Outline of Tensor Calculus is excellent, though it uses the classical, rather than modern notation. If you want to learn to calculate functions on curved spaces this is a good place to go. Another great resource is MTW's Gravitation, though a book on relativity, has some insightful sections on tensors.Of course, all the other books mentioned in this thread are good choices.
 Recognitions: Homework Help Science Advisor A modest proposal: I would suggest that one reason it is hard for physicists and mathematicians to communicate is that some physicists seem to continue to educate themselves via extremely old fashioned mathematics books, teaching a version of tensor calculus that is about 100 years old. I know Einstein used it, but the mathematics Einstein used is not sufficient to understand even today's physics. Modern physicists like (the real, departed) Feynman, and now Witten, have been not only creating new exciting physics, but also new mathematics, and are also inspiring mathematicians to try to catch up with the innovations they are bringing to both subjects. I have been a lecturer at the International Center for Theoretical Physics in Trieste (on Riemann surfaces, and theta functions on abelian varieties) and so I know a little about what at least some of today's physicists want to learn. In particular it is important to learn how geometry is linked to, and illuminated by, modern topology and analysis, such as deRham cohomology, and modern theory of partial differential operators. So I recommend, even if one learns the old versions of Ricci calculus, to also at least look at modern books like Spivak's differential geometry volumes 1 and 2. If that is too mathematical, I suggest trying Misner, Thorne and Wheeler, as recommended above. At some point one might want to look also at books such as the volume of proceedings "Lectures on Riemnann Surfaces" given at ICTP in 1987, pub. by World Scientific in 1989 (for background on string theory), and some works on algebraic geometry. Miles Reid's book is the easiest, and Shafarevich is very nice for a next course. To me many of the references recommended here, although excellent for what they are, nonetheless fall largely into the category: "primarily of historical interest". Feyman's work, in particular the theory of "Feynman integrals" has apparently led recently to the exciting mathematical topic of quantum cohomology, growing out of "quantum gravity" and other topics in which modern physicists lead the way. It seems to me at least, that these leaders are not using 100 year old mathematics, they are using mathematics that has not even been perfected yet. best regards, roy
 Modern tensor notation is in my opinion superior, the problem is that many texts that teach it use abstraction as an excuse to avoid doing any actual calculations - in some cases modern methods step around tedious calculations in a profound and brilliant manner, but in the wrong hands modern tensor notation is obscure gibberish. A case in point is Darling's book - an excellent book overall, and one I recommend, but at times it is degenerates into storm of pretentious machinery. At one point he defines two objects, then proves that they are actually the same object to begin with. What is the point of that? At another point he gives that wrong formula for the wedge product of two forms, and error he would have found easily if he had actually tried to calculate anything with it. If there was a book using modern notation that was not divorced from the nuts and bolts of doing actual calculations I would recommend that, but unfortunately there are none. P.S. DeRham's book on cohomology is excellent - it is strangely obscure. Another excellent book on algebraic geometry is "An Invitation to Algebraic Geometry" by Karen Smith et. al.
 Recognitions: Homework Help Science Advisor You make an excellent point. It is certainly true that many authors of modern math books dwell excessively on the theory and omit useful calculations, in every subject. Now that you have raised this as a key virtue lacked by most modern treatments, maybe someone will recommend a book unknown to us which does have it. In the meantime perhaps your argument implies that physicists need to read both types of books, modern and classical. I have not read Darling, but can suggest a reason for giving two different definitions of the same object in some cases. Each definition may have different advantages, perhaps one is intuitively more appealing while the other admits easier calculation. Or one was used historically, while the other is of more modern acceptance. Just a guess. As I recall, Spivak for example discusses the curvature tensor in progressive degrees of abstraction, starting from Riemann's original version, continuing up through modern incarnations.
 Recognitions: Homework Help Science Advisor What I think this comes down to is the dichotomy between calculating a quantity and understanding the meaning of that quantity. I claim that understanding allows calculation, but not vice versa. For instance, on page 14, of his nice notes on GR, Sean Carroll gives the transformation law, (1.51) in his numbering, for tensors and then says: "Indeed a number of books like to define tensors as collections of numbers transforming according to (1.51). While this is operationally useful, it tends to obscure the deeper meaning of tensors as geometric entities with a life independent of any chosen coordinate system." On page 15 he describes the scalar or dot product as a familiar example of a tensor of type (0,2). I am going to go out on a limb here and try to make a trivial calculation, beginning from a conceptual definition of a tensor of type (0,2) as a bilinear map from pairs of tangent vectors to numbers. I.e. I will try to derive the transformation law from the conceptual meaning. A simple example of such a tensor is a scalar product, i.e. a symmetric, bilinear mapping from pairs of tangent vectors to scalars. Such a thing is often denoted by brackets (or a dot) taking the pair of tangent vectors v,w to the number . Now if f:M-->N is a differentiable mapping from one manifold M to another manifold N, such as a coordinate change, then one can pull back a scalar product from N to M using the derivative of f. I.e. if u,z are two tangent vectors at a point p of M, then applying the derivative of f to them takes them to 2 tangent vectors at the image point f(p) in N, where we can apply <,> to them. I.e. if <,> is the scalar product on N, then the pulled back scalar product f*(<,>) acts on u,z by the obvious, only possible law: f*() = , where f' is the derivative of f, given as a matrix of partials of f with respect to local coordinates in M and N. For example we could denote this matrix as f' = [dyi/dxj]. Now suppose we express the scalar product in N as a matrix, i.e. in local coordinates as A = [akl], sorry about the lack of subscripts. Imagine k and l are subscripts on a. Then if we want to express the pulled back scalar product as a matrix, we just see what it does to the vectors u,z as follows: f*() = = [f'(u)]* [A] [f'(z)] = [u]* [f']* [A] [f'] [z], where now everything is thought of as a matrix, and star means transpose of the matrix. Well since the matrix of partials f' is just [dyi/dxj], and A is [akl], we just multiply out the matrices to get the matrix of the pulled back scalar product as [f*(<,>)] = [f']* [A] [f'] = the matrix whose i,j entry is akl (dyk/dxi)(dyl/dxj), summed over k,l. Now this is exactly the transformation law Carroll calls (1.51) on page 14 of his notes and everyone else also calls the transformation law for a tensor of type or rank (0,2) in the various web sources given here and above. Notice too, if you can imagine my subscripts, that this satisfies the summation convention for subscripts. But I am not dependent on that because I know what it means, so i don't care whether I can see the subscripts or not, whereas someone dependent on seeing where the indices are may not be able to follow this. Anyone who knows conceptually what a tensor is would immediately realize that a homogeneous polynomial of degree d in the entries of a tangent vector, is a (symmetric) tensor of type (0,d), and that the components of the tensor are merely the coefficients of the polynomial (written as a non commutative polynomial, i.e. with a separate coefficient for xy and for yx). It follows of course that they transform via a d dimensional matrix of size n, where n is the dimension of the manifold, i.e. by a collection of n^d numbers. Subscript enthusiasts write this as a symbol like T, with d subscripts. That is an extremely cumbersome way to discuss tensors in my opinion, and leaves me at the mercy of the type setter, whereas knowing what they mean always bails me out eventually. I actually wrote a graduate algebra book, including linear and multilinear alkgebra once, and I discovered to my amusement that I could actually write down tensor products as matrices, and so on, just from the definitions, although I had never needed to do so before in my professional life. peace and love, roy
 Recognitions: Homework Help Science Advisor In 1996, at the end of a chapter on tensors in my graduate algebra notes, after writing out a calculation of the tensor product of two matrices, I wrote the following extremely naive remarks: "The complexity of this sort of calculation may be responsible for the fearsome reputation which "tensor analysis" once enjoyed. In ancient times, books on the topic were filled with lengthy formulas laden with indices. Learning the subject meant memorizing rules for manipulating those indices. Nowadays, confronted with the statement that such and such quantity is "a tensor", I hope we will understand this to mean simply the quantity has certain linearity properties with respect to each of its components. Of course skill in their use will still require an ability to calculate. In this regard, note that we are usually able to recover explicit calculations from our abstract approach, provided we always know exactly what the maps are that yield our isomorphisms. When we know the maps, a choice of bases gives us a calculation. Thus we must resist the tendency to remember only that certain modules are isomorphic, without knowing what the isomorphisms are. Fortunately the maps are virtually always the simplest ones we can think of."
 Recognitions: Homework Help Science Advisor To recommend Spivak again, I read this book (volume II) in one day almost 30 years ago, and have never consulted it again (except for the bet above) until yesterday for about 30 minutes. So it is not going to eat up a lot a lot of your time to give it the once over. It is so well written you can learn something from it very quickly. Although obviously in such a short time I did not come anywhere near mastering anything, still I feel I did learn something. a used copy is available for $20 from http://www.ericweisstein.com/encyclo...lGeometry.html and a new copy for about$40 from "publish or perish". Well after perusing the website of the publisher, I see the first and second editions are no longer available, and I am slightly disappointed to note that apparently the cover art has changed, and there are no longer strange animals on the front of voilume 5 waving flags and and marching in the name of "The generalized Gauss Bonnet theorem and what it means for mankind". You can never have too much nonsense in amthematics. The chapter on p.d.e. called "and now a word from our sponsor" remains however.


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# What is the opposite and reciprocal of -7/2?

The opposite and reciprocal of $- \frac{7}{2}$ is $\frac{2}{7}$.
With a number x, the reciprocal is known as $\frac{1}{x}$.
So we have our x = $- \frac{7}{2}$, our reciprocal $\frac{1}{x}$ $\implies$ $\frac{1}{- \frac{7}{2}}$ $\implies$ $- \frac{2}{7}$.
$- \left(- \frac{2}{7}\right)$ $\implies$ $\frac{2}{7}$.
Since it had asked for the opposite and reciprocal of $- \frac{7}{2}$, we are done, $\frac{2}{7}$.


text:
# How to calculate $z^4 + \frac1{z^4}$ if $z^2 + z + 1 = 0$?

Given that $z^2 + z + 1 = 0$ where $z$ is a complex number, how do I proceed in calculating $z^4 + \dfrac1{z^4}$?

Calculating the complex roots and then the result could be an answer I suppose, but it’s not quite elegant. What alternatives are there?

#### Solutions Collecting From Web of "How to calculate $z^4 + \frac1{z^4}$ if $z^2 + z + 1 = 0$?"

From $z^2 + z + 1 = 0$, we have
$$z +\frac{1}{z}=-1.$$
Taking square, we get
$$z^2 +\frac{1}{z^2}+2=1,$$
which implies that
$$z^2 +\frac{1}{z^2}=-1.$$
I think you will know what to do next.

Essentially the same calculation also follows from the observation that $x^2+x+1=\phi_3(x)$ is the third cyclotomic polynomial. So $z^3-1=(z-1)(z^2+z+1)=0$, and hence $z^3=1$ for any solution $z$. Therefore
$$z^4+\frac{1}{z^4}=z\cdot z^3+\frac{(z^3)^2}{z^4}=z+z^2=-1.$$

\begin{align*} z^4+\frac1{z^4}&=(-z-1)^2+\frac1{(-z-1)^2}\\ &=z^2+2z+1+\frac1{z^2+2z+1}\\ &=(-z-1)+2z+1+\frac1{(-z-1)+2z+1}\\ &=z+\frac1{z}=\frac{z^2+1}{z}=\frac{-z-1+1}{z}=-1 \end{align*}

Here is an alternative approach: let’s consider $z^{8}+1$ , and then divide by $z^{4}$.

By using geometric series, notice that $$z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=\left(z^{2}+z+1\right)\left(z^{6}+z^{3}+1\right)=0.$$ Now, as $z^{2}+z+1=0$, we know that both $z^{7}+z^{6}+z^{5}=0$ and $z^{3}+z^{2}+z=0$, and hence $$z^{8}+z^{4}+1=0$$ so that $z^{8}+1=-z^{4}$. Thus, we conclude that $$z^{4}+\frac{1}{z^{4}}=-1.$$

Different people see different things in the relation $z^2+z+1=0$. Just as Jyrki did, I see the third cyclotomic polynomial. Its roots are $-1/2 \pm \sqrt{-3}/2$, the two primitive cube roots of unity. Call one of these $\omega$ and see that $\omega^3=1$, so that $\omega^4=\omega$ and $\omega^{-4}=\omega^2$. Their sum is $-1$, from the defining relation.

Hint $\$ Exploit innate symmetry: for $\rm\ y = z^{-1}$ you know $\rm\ yz\ (=\: 1)\:$ and $\rm\ y+z\ (=\: z^{-1}\!+z\: =\: -1)\$

Thus you know $\rm\ \ \ y^2 + z^2\ =\ (y\ +\ z)^2-\ 2\:(y\:z)$

hence you know $\rm\ y^4 + z^4\ =\ (y^2\! + z^2)^2 – 2\:(y\:z)^2$

For more on symmetric polynomials see the Wikipedia article on Newton’s identities.

$$z^2+z+1=0$$ Clearly $z \not= 0$ and therefore $z+\dfrac{1}{z}=-1.$
Note that $$z^{n+1}+\frac{1}{z^{n+1}}=\left(z+\frac{1}{z}\right)\left(z^n+\frac{1}{z^n}\right)-\left(z^{n-1}+\frac{1}{z^{n-1}}\right).$$
Therefore if we define the function $U:\Bbb{N}\cup\{0\} \to \Bbb{C}$ as $$U_n=z^n+\frac{1}{z^n}$$ then we have $U(0)=2,$ $U(1)=-1$ and $$U(n+1)+U(n)+U(n-1)=0\,\,\,\,\,\,\,\,\,\forall n \in \Bbb{N}.$$
Using this recurrence you can calculate $z^n+\dfrac{1}{z^n}$ for any $n \in \Bbb{N}.$

Alos you can follow this approach.


text:
Volume 34, Issue 2
Spanning Pre-Disks in a Compression Body

Commun. Math. Res., 34 (2018), pp. 177-183.

Published online: 2019-12

Cited by

Export citation
• Abstract

A properly embedded essential planar surface $P$ (not a disk) in a compression body $V$ is called a spanning pre-disk with respect to $J$, if one boundary component of $P$ is lying in $\partial_+V$ and all other boundary components of $P$ are lying in $\partial_-V$ and coplanar with $J$. In this paper, we show that the number of boundary components of spanning pre-disks in a compression body is unbounded. But the number of a maximal collection of spanning pre-disks is bounded.

• Keywords

spanning pre-disk, curve complex, compression body, maximal collection

57N10, 57M50

liang_liang@aliyun.com (Liang Liang)

• BibTex
• RIS
• TXT
@Article{CMR-34-177, author = {Liang and Liang and liang_liang@aliyun.com and 6012 and School of Mathematics, Liaoning Normal University, Dalian, Liaoning, 116029 and Liang Liang and Youfa and Han and and 6013 and School of Mathematics, Liaoning Normal University, Dalian, Liaoning, 116029 and Youfa Han and Fengling and Li and and 5416 and School of Mathematical Sciences, Dalian University of Technology, Dalian, 116024 and Fengling Li and Luying and Zhao and and 6014 and School of Mathematics, Liaoning Normal University, Dalian, Liaoning, 116029 and Luying Zhao}, title = {Spanning Pre-Disks in a Compression Body}, journal = {Communications in Mathematical Research }, year = {2019}, volume = {34}, number = {2}, pages = {177--183}, abstract = {

A properly embedded essential planar surface $P$ (not a disk) in a compression body $V$ is called a spanning pre-disk with respect to $J$, if one boundary component of $P$ is lying in $\partial_+V$ and all other boundary components of $P$ are lying in $\partial_-V$ and coplanar with $J$. In this paper, we show that the number of boundary components of spanning pre-disks in a compression body is unbounded. But the number of a maximal collection of spanning pre-disks is bounded.

}, issn = {2707-8523}, doi = {https://doi.org/10.13447/j.1674-5647.2018.02.10}, url = {http://global-sci.org/intro/article_detail/cmr/13522.html} }
TY - JOUR T1 - Spanning Pre-Disks in a Compression Body AU - Liang , Liang AU - Han , Youfa AU - Li , Fengling AU - Zhao , Luying JO - Communications in Mathematical Research VL - 2 SP - 177 EP - 183 PY - 2019 DA - 2019/12 SN - 34 DO - http://doi.org/10.13447/j.1674-5647.2018.02.10 UR - https://global-sci.org/intro/article_detail/cmr/13522.html KW - spanning pre-disk, curve complex, compression body, maximal collection AB -

A properly embedded essential planar surface $P$ (not a disk) in a compression body $V$ is called a spanning pre-disk with respect to $J$, if one boundary component of $P$ is lying in $\partial_+V$ and all other boundary components of $P$ are lying in $\partial_-V$ and coplanar with $J$. In this paper, we show that the number of boundary components of spanning pre-disks in a compression body is unbounded. But the number of a maximal collection of spanning pre-disks is bounded.

Liang Liang, Youfa Han, Fengling Li & Luying Zhao. (2019). Spanning Pre-Disks in a Compression Body. Communications in Mathematical Research . 34 (2). 177-183. doi:10.13447/j.1674-5647.2018.02.10
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text:
# Problem: What volume of 12.0 M HCl is required to make 75.0 ml of 3.50 M HCl?

🤓 Based on our data, we think this question is relevant for Professor Davis' class at UCF.

###### FREE Expert Solution

We are being asked to determine the volume of 12.0 M HCl is needed to make a 75 mL of 3.50 M solution.

When dealing with dilution we will use the following equation:

$\overline{){{\mathbf{M}}}_{{\mathbf{1}}}{{\mathbf{V}}}_{{\mathbf{1}}}{\mathbf{=}}{{\mathbf{M}}}_{{\mathbf{2}}}{{\mathbf{V}}}_{{\mathbf{2}}}}$

M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume

Calculate the initial volume of the HCl solution.

###### Problem Details

What volume of 12.0 M HCl is required to make 75.0 ml of 3.50 M HCl?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Dilution concept. If you need more Dilution practice, you can also practice Dilution practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Davis' class at UCF.


text:
# Motivation for normed space definition?

So if $X$ is a vector space, and you define a norm, $x \mapsto \| x \|$, on it, then the bounded subset, $V = \{ x \in X: \|x\| < \infty \}$ is automatically a subspace. This follows from the definition of a norm, so for all $x,y \in V, c \in \mathbb{C}$, $\|x + c y\| \le \|x\| + \|cy\| = \|x\|+ |c| \|y\| < \infty$, so $x+cy \in V$.

Is this the reasoning behind the definition of a norm?

thanks

-
In a normed vector space every vector has finite norm. –  Rasmus Jan 10 '11 at 7:47
Note that for your definition $V=X$. –  mpiktas Jan 10 '11 at 8:14


text:
Why can't I access a public function that has a protected overload?

Given the code:

class Foo
{
public:
Foo() {}

int foo() const { return 6; }
protected:
int foo() { return 5; }
};

int main(int argc, char* argv[])
{
Foo foo;
foo.foo();
return 0;
}

I get a compile error:

'Foo::foo' : cannot access protected member declared in class 'Foo'

In actual usage, I'm trying to provide a function that returns a pointer to an object that the class is storing - modifiable for the members, and const for everyone else. I figured that if I say

const Item *i = foo.foo();

in a non-member function, the correct, const variant will be called. Somehow though, the compiler insists on accessing the non-const variant in the protected section. Any idea why? Thanks.

-
IIRC correctly the compiler pretends to be "dumb" when looking for the function to call and ignores visibility until the very end(where it then throws an error). –  Lalaland Feb 28 '12 at 3:25
possible duplicate of Const Overloading: Public-Private Lookup in C++ Class –  Rob Kennedy Feb 28 '12 at 3:26
Sorry, didn't notice, voted to close. –  neuviemeporte Feb 28 '12 at 7:00

Your variable is non-const, so the non-const overload of the function is selected. After overload resolution occurs, then accessibility is checked.

When const and non-const functions both exist, the const version is only called on const objects. Your foo variable is not const. You can make it const with a const_cast if you want:

const_cast<Foo const&>(foo).foo();

That's cumbersome to use, though. A better solution would be to just give the internal function a different name so it doesn't clash with the external API.

-

For a non-const object, non-const function overload is chosen over the const function, if both are existing.

foo is of type Foo and not const Foo, that's why compiler chooses the protected version which is non-const.

In 3 ways the code will compile:

1. Either, Make Foo::foo() in public
2. Or, use const Foo, so that compiler chooses the const-correct version
3. Remove the non-const version
-


text:
## What is Curvature? (II)

[ Background required: rudimentary vector calculus. ]

The aforementioned definition of curvature is practical but a little aesthetically displeasing. Specifically, one seeks a definition which is independent of the parametrization of the curve. The advantage of such a definition is not merely conceptual clarity, but also the ease at which we can generalise to higher dimensions.

Suppose the curve is given by $\mathbf x(t) = (f(t), g(t))$, where all vectorial values are given in boldface. One imagines this as a particle travelling along the curve such that at time t, its position is at x(t). Recall that the velocity of the particle is then given by $\mathbf{x}'(t) = (f'(t), g'(t))$ which is parallel to the gradient of the curve at x(t).

Thus the speed of the particle is the magnitude of its velocity, $||\mathbf{x}'(t)|| = \sqrt{f'(t)^2 + g'(t)^2}$, from which it’s clear that the distance travelled by the particle during a time interval [a,b] is given by:

$s = \int_a^b || \mathbf{x}'(t) || dt = \int_a^b \sqrt{f'(t)^2 + g'(t)^2} dt.$

Example: consider the parabola $y = x^2$ parametrized via $(x, x^2)$. The arc length of the curve from (0,0) to (1,1) is given by $s = \int_0^1 \sqrt{1 + 4x^2} dx$ which is $\frac 1 4 (2\sqrt 5 + \sinh^{-1}(2))$, i.e. around 1.479, which is only slightly longer than the straight line joining the two points.

Exercise: find the length of the curve $y = x^{3/2}$ from (0,0) to (1,1).

Exercise: for the curve in polar coordinates $r = f(\theta)$, find the formula of the arc-length.

Now we’re ready to state our definition of curvature “intrinsically” (i.e. without reference to external coordinates or parametrization). At each point of time t, take the unit tangent vector at the curve, i.e. let T(t) be the unit vector along $\mathbf{x}'(t)$. Since regardless of parametrization, $\mathbf{x}'(t)$ is always parallel to the tangent at the curve at x(t), we see that T(t) depends only on the point x(t) and not on the parametrization t. [ E.g. if we let t = 2u, then the new parametrization x(u) would be “twice as fast”: from u=0 to u=1, the particle would have traversed the path from t=0 to t=2. ]

Next, we fix a standard parametrization of the curve: namely via the arc length s, i.e. we assume that the particle is travelling along the curve at unit speed.

Example. Consider the semi-circle in the upper-half plane $y = \sqrt{25 - x^2}$. Let’s calculate its parametrization by arc-length s, via the hard way. [ The easy way is via polar coordinates. ] Starting from the point (-5, 0), the arc-length to the point $(a, f(a))$ is given by:

$\int_{-5}^a \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = \int_{-5}^a \frac {5} {\sqrt{25-x^2}} dx = 5(\frac\pi 2 + \sin^{-1} (\frac a 5)).$

This gives the parametrization $x(s) = 5\sin(\frac s 5 - \frac \pi 2) = -5 \cos \frac s 5$, $y(s) = 5\sin \frac s 5$, where $0 \le s \le 5\pi$. This is consistent with our intuition that the arc-length is a linear function of the angle θ.

Exercise. Take the curve $y= \frac 2 3 x^{3/2}$. Parametrize the curve by arc-length, starting from the origin (0, 0).

Exercise. Take the curve defined by the polar equation $r = e^\theta$ and parametrize it by arc-length, starting from $\theta = 0$.

Exercise. (*) Prove that if $\mathbf{x}(s) = (f(s), g(s))$ is parametrized by arc-length, then $||\frac{d\mathbf{x}}{ds}|| = 1$. [ Note: if your concept is right, this should be a 2-line proof. We will need this result later. ]

Now we’re ready to unveil the new definition of curvature.

Intrinsic Definition of Curvature. Let $\mathbf{T}(s)$ be the unit tangent vector of the curve $\mathbf{x}(s)$, parametrized by arc-length. The curvature $\kappa$ of a curve is defined by $||\frac{d\mathbf{T}}{ds}||$.

While we’re at it, we also define the curvature vector to be $\frac{d\mathbf{T}}{ds}$.

Our most pressing need is to check that this definition of curvature is consistent with our last. To be specific, for a function $y = f(x)$ we need to show that after parametrizing by arc-length that:

$||\frac{d\mathbf{T}}{ds}|| = \frac{f''(x)}{(1+f'(x)^2)^{3/2}}.$

To do that, first rotate and translate the figure so that the graph is tangent to the x-axis at the origin (i.e. $f(0) = f'(0) = 0$). The RHS remains unchanged thanks to the way we derived it. The LHS is intrinsically defined, so it has nothing to do with orientation. Now we have:

$\frac{d\mathbf{T}}{dx} = \frac{d\mathbf{T}}{ds} \cdot \frac{ds}{dx}$ and $\frac{ds}{dx} = \sqrt{1 + f'(x)^2}$.

At x=0, we thus have $\frac{ds}{dx} = 1$ and it suffices to compute $\frac{d\mathbf{T}}{dx}$. To do that, note that the vector $(1, f(x))$ is parallel to the tangent of the curve at the point $(x, f(x))$, so we can express the unit tangent vector in terms of x :

$\mathbf{T}(x) = \frac 1 {\sqrt{f'(x)^2+1}} (1, f'(x))$.

Now just differentiate both sides wrt x and put x=0: we get $\left.\frac{d\mathbf{T}}{dx}\right|_{x=0} = (0, f''(0))$. So our definition is consistent with the earlier one. Notice that the curvature vector $(0, f''(0))$ is perpendicular to tangent vector (x-axis). This always holds.

Generalisation

Armed with this, we can generalise the definition of curvature to a curve in arbitrary dimensions. Let $\mathbf{x} : \mathbb R \to \mathbb R^n$ be a curve in n-space, with components $\mathbf{x}(t) = (x_1(t), x_2(t), \dots, x_n(t))$. Everything follows as before:

• The velocity vector is $\mathbf{v}(t) = \mathbf{x}'(t) = (x_1'(t), x_2'(t), \dots, x_n'(t))$.
• The unit tangent vector of the curve at a point is $\mathbf{T} = \frac{\mathbf{x}'}{||\mathbf{x}'||}$.
• The arc-length along $a \le t \le b$ is $s = \int_a^b ||\mathbf{v}(t)|| dt$, or $s = \int_a^b \sqrt{\sum_{i=1}^n x_i'(t)^2} dt$.

So naturally, we also define the curvature vector to be $\frac{d\mathbf{T}}{ds}$ and the curvature to be its length. This definition is consistent with all our earlier definitions of curvature (for n=2 case).

Example. Consider the helix given by $\mathbf{x}(t) = (\cos t, \sin t, 2t)$.

Then the velocity vector is $\mathbf{v}(t) = \mathbf{x}'(t) = (-\sin t, \cos t, 2)$, so the tangent vector is $\mathbf{T}(t) = \frac 1 {\sqrt 5} (-\sin t, \cos t, 2)$. To compute the curvature vector, we apply:

$\frac{d\mathbb{T}}{ds} = \frac{d\mathbb{T}}{dt} \left(\frac {ds}{dt}\right)^{-1}$.

And $\frac {ds}{dt} = ||\mathbf{v}(t)|| = \sqrt 5$. Hence, the curvature vector is given by $\frac 1 5 (-\cos t, -\sin t, 0)$. and the curvature is 1/5. In other words, the helix has constant curvature, which is consistent with our geometric figure.

Exercise. Take the curve $\mathbf x(t) = (2e^t \cos t, 2e^t \sin t, e^t)$. Compute the curvature and curvature vector at a general point on the curve.

Application. Take a cylinder with base radius R and height H. A strake is a curved strip which bends around the cylinder, following the shape of its contour:

(Image from britannica.com)

What should the radius r be for the strip to wrap around the cylinder effectively? The correct answer is that the curvature must match that of the strake. The graph of the helix is given by: $\mathbf{x}(\theta) = (R \cos\theta, R\sin\theta, h\theta)$, where $0 \le \theta \le 2\pi$ and $h = \frac{H}{2\pi}$. Hence:

• $\mathbf{x}'(\theta) = (-R\sin\theta, R\cos\theta, h)$;
• so $\frac{ds}{d\theta} = ||\mathbf{x}'(\theta)|| = \sqrt{R^2 + h^2}$;
• $\mathbf{T}(\theta) = \frac{\mathbf{x}'}{||\mathbf{x}'||} = \frac 1 {\sqrt{R^2 + h^2}} (-R\sin\theta, R\cos\theta, h)$;
• $\frac{d\mathbf{T}}{d\theta} = \frac 1 {\sqrt{R^2+h^2}} (-R\cos\theta, -R\sin\theta, 0)$;
• so curvature vector is $\frac{d\mathbf{T}}{ds} = \frac{d\mathbf{T}}{d\theta}\left(\frac{ds}{d\theta}\right)^{-1} = \frac 1 {R^2+h^2} (-R\cos\theta, -R\sin\theta, 0)$ and the scalar curvature is $\frac R {R^2 + h^2}$.

So the correct radius for r is $\frac {R^2 + h^2} R$ where $h = \frac{H}{2\pi}$.

This entry was posted in Extra and tagged , , . Bookmark the permalink.

### 1 Response to What is Curvature? (II)

1. Rayna says:

Reblogged this on Rayna's Blog and commented:
A follow-up of an earlier article, this post looks at “curvature” using concepts of path and velocity. The last question on helix and strakes is very interesting.


text:
# Platinum as a Metallic Material

## Processing Platinum and Alloys (forming, welding)

Fig.1
Rolling

Rolling a platinum sheet

Physically or chemically pure platinum is very soft and ductile. The cast ingots can therefore be readily formed by cold rolling and other conventional forming processes. To achieve the optimum surface quality of the semi-finished products, it is advantageous to skim machine the surface of the cast ingot, e.g. by planing.

### Hot working

As the alloying elements rhodium and iridium increase the mechanical strength of platinum, the forming of platinum-rhodium and platinum-iridium alloys is – depending on the rhodium or iridium content – somewhat more difficult than it is the case for pure platinum. Hot working by forging or rolling can be advantageous both for the microstructural homogeneity and for subsequent cold working to semifinished products. Because the work hardening causes an increase in strength during cold forming the alloys, annealing treatments are necessary which are normally carried out at 900-1,000 $°C$. As the strength and the work hardening increase with increasing alloy content, the frequency and the temperature and duration of the annealing treatments must be adapted to meet the particular alloy composition.

Finished products are manufactured from the semifinished products by means of all conventional forming processes, e.g. spinning, deep drawing, tube and wire drawing.

### Welding

Fig.2
TIG welding

TIG welding a component made of PtRh90/10 alloy for use in the glass industry

Furthermore, platinum materials can be readily welded. The most commonly used method is the tungsten-inert gas technique (TIG). Other methods, such as laser, electron beam and plasma welding can also be used without difficulty.

Page 12 of 20


text:
### \newenvironment

 \newenvironment{nam}[args]{begdef}{enddef}
\renewenvironment{nam}[args]{begdef}{enddef}

These commands define or redefine an environment.

nam
The name of the environment. For \newenvironment there must be no currently defined environment by that name, and the command \nam must be undefined. For \renewenvironment the environment must already be defined.
args
An integer from 1 to 9 denoting the number of arguments of the newly-defined environment. The default is no arguments.
begdef
The text substituted for every occurrence of \begin{nam}; a parameter of the form #n in cmd is replaced by the text of the nth argument when this substitution takes place.
enddef
The text substituted for every occurrence of \end{nam}. It may not contain any argument parameters.


text:
ratinterp - Maple Programming Help

Home : Support : Online Help : System : Libraries and Packages : Deprecated Packages and Commands : Deprecated commands : ratinterp(deprecated)

ratinterp

rational interpolation

 Calling Sequence ratinterp(xpoints, ypoints, z, mn)

Parameters

 xpoints - list of independent values, x0,..xk ypoints - list of dependent values, y0,..yk z - variable to be used in polynomial mn - list of numerator and denominator degrees, [m, n]

Description

 • Important: The ratinterp function has been deprecated.  Use the function CurveFitting[RationalInterpolation] instead.  A call to ratinterp automatically generates a call to CurveFitting[RationalInterpolation].
 • The ratinterp function computes a rational function u/v that satisfies

$\mathrm{yi}v\left(\mathrm{xi}\right)-u\left(\mathrm{xi}\right)=0\mathrm{for}i=0\mathrm{..}m+n.$

 with u and v being polynomials in z having degrees at most m and n. In most (but not all) cases, this is equivalent to finding a rational function in z having certain degree constraints that interpolate the points (x0,y0), (x1,y1), ..., (xk,yk). Here $\mathrm{mn}=\left[m,n\right]$. At least m+n+1 points are required, with all the independent points distinct.
 • Sometimes the linearized equation gives u and v satisfying the linearized equation, but with at least one of the xi a common zero. In this case, the resulting rational function does not interpolate the point xi (called an unattainable'' point), since common factors are removed from the rational function. In such cases ratinterp finds the rational function that interpolates a maximum number of points with the given degree constraints. The case where there are unattainable points is recognizable by the fact that the degrees of both the numerator and the denominator are less than the degree constraints.
 • Two procedures are available in the case of exact input: one that avoids singularities in the corresponding table of interpolants by a look-around'' procedure, and another that jumps singularities by a look-ahead'' procedure. Both algorithms are fraction-free. The look-ahead procedure uses a recusion similar to that used for subresultant computation. The second procedure can be specified by using a fifth argument, subresultant. The procedure produces an error message in the case of inexact data.

Examples

Important: The ratinterp function has been deprecated.  Use the function CurveFitting[RationalInterpolation] instead.  A call to ratinterp automatically generates a call to CurveFitting[RationalInterpolation].

 > $\mathrm{xpoints}≔\left[0,1,2,3\right]:$$\mathrm{ypoints}≔\left[0,3,1,3\right]:$
 > $\mathrm{f1}≔\mathrm{ratinterp}\left(\mathrm{xpoints},\mathrm{ypoints},x,\left[1,2\right]\right)$
 ${\mathrm{f1}}{:=}{-}\frac{{3}{}{x}}{{4}{}{{x}}^{{2}}{-}{17}{}{x}{+}{12}}$ (1)
 > $\mathbf{for}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{to}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}4\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{normal}\left(\genfrac{}{}{0}{}{\mathrm{f1}}{\phantom{x={\mathrm{xpoints}}_{i}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\mathrm{f1}}}{x={\mathrm{xpoints}}_{i}}-{\mathrm{ypoints}}_{i}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end do}$
 ${0}$ (2)
 > $\mathrm{xpoints}≔\left[0,1,2,3,4,-1\right]:$$\mathrm{ypoints}≔\left[0,3,1,3,a,\frac{1}{11}\right]:$
 > $\mathrm{f2}≔\mathrm{ratinterp}\left(\mathrm{xpoints},\mathrm{ypoints},x,\left[2,3\right]\right)$
 ${\mathrm{f2}}{:=}{-}\frac{{3}{}{x}}{{4}{}{{x}}^{{2}}{-}{17}{}{x}{+}{12}}$ (3)
 > $\mathbf{for}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}i\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{to}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}6\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{do}\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathrm{normal}\left(\genfrac{}{}{0}{}{\mathrm{f1}}{\phantom{x={\mathrm{xpoints}}_{i}}}\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}|\phantom{\rule[-0.0ex]{0.1em}{0.0ex}}\genfrac{}{}{0}{}{\phantom{\mathrm{f1}}}{x={\mathrm{xpoints}}_{i}}-{\mathrm{ypoints}}_{i}\right)\phantom{\rule[-0.0ex]{0.5em}{0.0ex}}\mathbf{end do}$
 ${0}$ (4)

Implies that x=4 is an unattainable point.

References

 Beckermann, B., and Labahn, G. Fraction-free Computation of Matrix GCD's and Rational Interpolants. CS Tech Report, University of Waterloo, 1997.
 See also: SIAM Journal on Matrix Analysis and Applications, 2000.
 Beckermann, B., and Labahn, G. Numeric and Symbolic Computation of Problems Defined by Structured Linear Systems. CS Tech Report, University of Waterloo, 1999.


text:
# Total angular momentum Q

1. Dec 21, 2005

### ElDavidas

hey everyone, I've been working through a past paper and I'm stuck on a question:

"(a) Consider a system consisting of n distinct particles P1, P2, : : :, Pn, with masses m1, m2, ...., mn and position vectors r1, r2, : : :, rn, relative to the origin O of an inertial frame, respectively.

For each i = 1, 2, : : :, n, suppose that the only forces acting on Pi are internal forces Fij , which always act along the line joining Pi and Pj , for j = 1, 2, : : :, n and j doesn't equal i.

Assume that Fij = ¡Fji for i = 1, 2, : : :, n, j = 1, 2, : : :, n and i doesn't equal j.

Define what is meant by the total angular momentum of the system about O.

Show, as a consequence of Newton's second law, that the total angular momentum of the system about O is constant."

I can define the total angular momentum of the system, it's the 2nd part of the Q I have trouble with.

Thanks

2. Dec 21, 2005

### Tide

Use the defintion of angular momentum:

$$\vec L = \vec r \times \vec p$$

Take the time derivative and sum over all particles noting that

$$\frac {d \vec p}{dt} = \vec F$$

You will find $d\vec L /dt = 0$.


text:
# General method for under and over determined systems?

Suppose I have a system:

$$Ax = b$$

where $A$ is a $m$ by $n$ matrix which is less than full rank (neither full column nor row rank). In my particular case $m<n$.

I'd like a combination of a minimum norm solution (for elements of $x$ which are not determined and a least squares solution for those that are determined or over determined).

I know I can approximate this with Tikhonov regularization:

$$\mathop{\text{minimize}}\limits_x \|Ax-b\|^2 + \alpha\|x\|^2.$$

But this is not ideal as the regularizer tugs on all elements of $x$ not just those left undetermined. If I make $\alpha$ too large then my $x$ at determined or over determined elements will not be optimal (w.r.t. $\|Ax-b\|^2$). If $\alpha$ is too small, I imagine I run into conditioning issues.

Is there no way to formulate this correctly using some sort null space/QR decomposition?

Update: See accepted answer, but for completeness here's a full recipe: \begin{align} Ax &= b\\ AE &= QR &\text{rank-revealing QR decomposition of A}\\ AE &= Q\left(\begin{array}{c}I\\0\end{array}\right)\left(\begin{array}{cc}R_1&R_2\end{array}\right) & \text{only keep $r$ (rank) many rows of $R$}\\ \left(\begin{array}{c}R_1^T\\R_2^T\end{array}\right)F &= VT_\text{full} &\text{rank-revealing QR decomposition of $\left(\begin{array}{c}R_1^T\\R_2^T\end{array}\right)$} \\ \left(\begin{array}{c}R_1^T\\R_2^T\end{array}\right)F &= V\left(\begin{array}{c}I\\0\end{array}\right)T & \text{only keep $r$ many rows of $T$}\\ AE &= Q\left(\begin{array}{c}I\\0\end{array}\right)FT^T\left(\begin{array}{cc}I&0\end{array}\right)V^T \\ x &= A^+b\\ x &= EV\left(\begin{array}{c}I\\0\end{array}\right)T^{-T}F^T\left(\begin{array}{cc}I&0\end{array}\right)Q^Tb \end{align}

• Preemptive strike: Here's an example of an under and over determined system: $$\left(\begin{array}{ccc}1&0&0\\1&0&0\\0&1&0\end{array}\right)\left(\begin{array}x_1\\x_2\\x_3\end{array}\right) = \left(\begin{array}{c}4\\5\\6\end{array}\right)$$ May 6 '14 at 22:12
• Does SVD not give you what you want? May 7 '14 at 8:54
• My matrices are very large and sparse, so complete SVD is not really an option (I assume). May 7 '14 at 14:32
• The SVD will give the answer, but it's not always the minimal amount of work to get a desired datum. They are implementations for sparse cases. stats.stackexchange.com/questions/41259/… I expect that using one of them is probably better than trying to implement one's own minimalist approach (with QR rather than full SVD). May 8 '14 at 4:06
• I'm happy to implement my own approach with QR if it avoids a dense matrix of singular vectors... May 8 '14 at 12:00

The Moore-Penrose pseudoinverse is probably what you're looking for. The pseudoinverse solution $A^+b$ is the smallest norm $x$ such that $\|Ax-b\|_2$ is minimized. It can be computed using QR decomposition although you have to use rank-revealing QR when $A$ does not have full column rank. A more complete explanation is given here.
• So if I have $x = A^+b$ and $A^+=(A^TA)^{-1}A^T$ and the reduced QR decomposition of $A$ is $A=Q_1R_1$, then I'll want to solve using $R_1^TR_1x = A^Tb$ apply back-,forward- substitution with $R_1$? But if $A$ doesn't have full column rank, won't $R_1$ be rank deficient? Or what do you mean by "reduced QR"? May 6 '14 at 22:55
• I think you do want the pseudoinverse. An alternative way to compute, take random (orthogonal) right multiplicators $T$ until from QR you decompose $AT=QR$ where $R$ is of the special form $\Bigl({U|B\atop Z}\Bigr)$ where $Z$ is numericaly zero (possibly empty) and $U$ is invertible and upper triangular. When $B$ is empty (full column rank), you are done (take $U^{-1}$, suitably zero-extended). Else you can recurse (only once) to compute the pseudo inverse of $(U|B)$ by taking the transpose before and after, and if the same parameters are used for zero-detection, this has full column rank. May 7 '14 at 4:25
• @Conder can you elaborate? Is this the same rank-revealing QR that AlexBecker writes of? How can $B$ be empty and full column rank? What does suitably zero-extended mean? Does $B$ participate in the final solve? May 8 '14 at 2:03


text:
# Sladjan Stankovik's Inequality With Constraint II

### Solution

We need to prove that $3(a^4+b^4+c^4+d^4)+12abcd\ge 24.\,$ By Surányi's Inequality,

$3(a^4+b^4+c^4+d^4)+4abcd\ge (a+b+c+d)(a^3+b^3+c^3+d^3).$

By the Cauchy-Schwarz inequality,

$3(a+b+c+d)(a^3+b^3+c^3+d^3)\ge (a^2+b^2+c^2+d^2)^2.$

Thus, suffice it to show that

$(a^2+b^2+c^2+d^2)^2+8abcd\ge 24.$

Since $a+b+c+d=4,\,$ $a^2+b^2+c^2+d^2=4(1+3t^2),\,$ with $0\le t\le 1.\,$ Hence, we need to prove that

$2(1+3t^2)^2+abcd\ge 3.$

By one of Leonard Giugiuc and Daniel Sitaru's results (Problem 4121, Crux Mathematicorum), for $0\le t\le\displaystyle\frac{1}{3},\,$

$abcd\ge (1+t)^3(1-3t)=-3t^4-8t^3-6t^2+1.$

Hence, in this case, suffice it to show that

$2(1+3t^2)^2-3t^4-8t^3-6t^2+1\ge 3.$

But this is equivalent to $t^2(15t^2-8t+6)\ge 0,\,$ which is true.

For $t\ge\displaystyle\frac{1}{3},\,$ $2(1+3t^2)^2+abcd\ge 2(1+3t^2)^2\ge\displaystyle\frac{32}{9}\ge 3.\,$ The proof is complete.

### Acknowledgment

The problem and the above solution have been kindly communicated to me by Leo Giugiuc.

[an error occurred while processing this directive]


text:
# [pdftex] Undocumented hyperref commands

Heiko Oberdiek oberdiek at uni-freiburg.de
Wed Jul 21 13:06:31 CEST 2004

On Wed, Jul 21, 2004 at 12:09:48AM +0100, Roboco Sanchez wrote:

> On Tue, 20 Jul 2004 11:48:11 +0200, Heiko Oberdiek
> <oberdiek at uni-freiburg.de> wrote:
> > On Tue, Jul 20, 2004 at 12:29:38AM +0100, Roboco Sanchez wrote:
> >
> > > A number of hyperref commands can't be found in the manual. For example,
> > >
> > > \pdfbookmark
> >
> > It *is* documented:
> >  hyperref/doc/slides.pdf or hyperref/doc/paper.pdf
>
> Thanks. I expected it to be in the manual.pdf

Then rewrite slides.pdf/paper.pdf as chapter for manual.pdf.

> > > \hyperlink{page.5}{go to p.5}
> > >
> > > I'm particularly interested to know what else I can do with the
> > > "page.5" part of \hyperlink. Also I would love to know other
> > > hidden/undocumented commands of hyperref.
> >
> > \hyperlink is documented in hyperref/doc/manual.pdf, section 4
>
> It's mentioned in manual.pdf that it's to be used together with
> \hypertarget. But as I was asking, \hyperlink's used alone. I'd like
> to know what else I can put in there. For example, I happened to try:
>
> \hyperlink{section.5.2}{where are you taking me to?}
>
> and found that it jumped to Section 5.2. But I couldn't find anywhere
> in the manual that I could put "section.5.2" in that bracket so that I


text:
# dual isogeny for abelian varieties over a general field

Let $\phi: A \rightarrow B$ be a separable isogeny between two abelian varieties over a field $k$. One knows that there is a dual isogeny $\hat {\phi} : B \rightarrow A$ such that $\hat{\phi} \circ \phi =$ multiplication by $\mathrm{deg}(\phi)$.

When I studied elliptic curves and abelian varieties, most of the references deal with a base field which is perfect. In this case, the proof for the existence of the dual isogeny is as following:

One makes a base change to work on the algebraic closure $\overline{k}$ of $k$. One considers $\mathrm{ker} (\phi)$, the closed points of the fiber of $\phi$ at the origin of $B$. It's a finite group which acts on $A$ and we have $\mathrm{ ker } (\phi) \subset \mathrm{ker} ( \times \mathrm{deg} (\phi))$. Then using the results about the quotien of a scheme by a finite group, one get the dual isogeny $\hat{\phi}$. Finally, one uses the action of $\mathrm{Gal}(\overline{k}/k)$ to get the result on the original base filed $k$.

Now if $k$ is not perfect, I didn't figure out how to do this. I have a feeling that the reason we work on the algebraic closure $\overline{k}$ is to get a finite group action. Of course, one can think about the action of $\mathrm{ker} (\phi)(k)$, but I think it doesn't work, because it's too small (We need all its closed points).

I would like to know if the dual isogeny exists in a general base field. If yes, what's the idea to see it, (and some reference)

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For elliptic curves, one can treat the inseparable case (with the base field $k$ being perfect or not.) The idea is as following:

(1) For every morphism $\psi : C_1 \rightarrow C_2$ of smooth curves (geometrically integral, complete) over a filed of char($k$) $=p > 0$, it factors as $C_1 \ \xrightarrow{\phi} \ C_1^{(q)} \ \xrightarrow{\lambda} \ C_2$, here $\phi$ is the $q^{\mathrm{th}}$-power Frobenius morphism, $q=$ inseparable degree of $\psi$, and $\lambda$ is separable.

(2) Using (1), one only need to treat with separable case and Frobenius morphism case. Furthermore, one only need to treat with the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ only. In elliptic curves case, one shows that the morphism $\times p$ is not separable, hence using (1), its factroization must contains $\phi$, i.e $\times p = \lambda \circ \phi^r$ and we are done.

Does this method can be used to higher dimensional abelian varieties? I have some difficulities to di this.

(a) The factorization in (1): In elliptic curves case, the degree of the $p^{\mathrm{th}}$-power Frobenius morphism $\phi$ is $p$, (not a power of $p$), which make it possible to construct this factorization. (Silverman, The Arithmetic of Elliptic Curves, p.30, Corollary 2.12). But for an abelian varieties of dimension $g$, the degree is $p^g$. In order to construct this factorization, one should have a result "the inseparable degree of an isogeny is a power of $p^g$", which I don't know if this is correct.

(b) One need to show that $\times p$ is not separable, which is ok.

So for (a), is it possible to do this for higher dimensional abelian varieties?

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I have learned the quotient of a scheme by a finite group scheme action, but still can't give a proof for this for purely inseparable case. Let $K$ be the kernel of $\phi$, and consider the action $m: K \times_k A \rightarrow A$ which is induced by the multiplication on $A$ and let $\pi : K \times_k A \rightarrow A$ be the natural projection. Let $\psi : A \rightarrow A$ be the multiplication by $\mathrm{deg}(\phi)$ map. Then one needs to show that $\psi \circ m = \psi \circ \pi$ to get the desired induced morphism, but I can't get a proof for this. If $k$ is perfect, then by working on the function fields of $A$ and $B$, one only need to deal with $\phi =$ Frobenius morphism. (But still need to show that the multiplication by $p$ factors through by the Frobeinus morphism.) I need help to give a proof for this case ( $k$ not necessary being perfect), i.e the equality $\psi \circ m = \psi \circ \pi$. I think one just need to show that the induced morphism of $\psi$ on $K$ factors throguth the structure morphism $K \rightarrow k$ followed by the constant morphism $e: k \rightarrow K$.

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Thank Qing Liu for the reference. Today I thought I had a way to prove this. Suppose $\phi : A \rightarrow B$ has degree $p^r$. Let $\psi := \times p^r : A \rightarrow A$. In order to show that the existence of $\hat{ \phi}$, one only need to show that $\psi^* (K(A))$ is contained in $\phi^* (K(B))$, here $K(A)$ and $K(B)$ are the function fields of $A$ and $B$. (Then we have a rational map from $B$ to $A$ and extends to a morphism $\hat{ \phi }$.) But notice that for each element $g \in K(A)$, $g^{p^r} \in \phi^* (K(B))$. If one can show that for each $f \in K(A)$, there exists $g \in K(A)$ such that $\psi^* (f) = g^{p^r}$, then we are done.

Since the multiplication by $p$ morphism $[p] : A \rightarrow A$ is the composition of the Frobenius morphism $F: A \rightarrow A^{(p)}$ with the Verschiebung morphism $V : A^{(p)} \rightarrow A$, one sees that for any $f \in K(A)$, $[p]^* (f) = g^p$ for some $g \in K(A)$, if $k$ is perfect. In particular, this holds for an algebraically closed field $k$. So in this case, the statement in the end of the previous paragraph is true.

Finally, for $k$ not necessary being algebraically closed, we take a base change and work on $\overline{k}$ first. We then get $\overline{\psi} = \Phi \circ \overline{\phi}$, here $\Phi : \overline{B} \rightarrow \overline{A}$ and the $\ \overline{ \ }$ means the schemes and the morphisms obtained from the base change. For any $f \in K(A)$, if one can show that $\Phi^* (f) \in \phi^* (K(B))$, then one completes the proof. But we have $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B)) = K(A) \cap \phi^* (K(B)) \otimes_k \overline{k}$. Consider the subfield extension $\phi^* (K(B)) (\Phi^* (f))$ of $K(A)/\phi^* (K(B))$, then $\Phi^* (f) \in K(A) \cap \overline{\phi}^* (\overline{K}(B))$ gives $\phi^* (K(B)) (\Phi^* (f)) = \phi^* (K(B))$, and we are done.

Just a remark about the factorization of the morphism $[p] = V \circ F$. This is somehow different with our $\phi$ and $\hat{ \phi }$, since $\mathrm{deg}(F) = p^g$, not $p$, so it's a stronger result. Also one can show that the kernel of $F$ is of this form (via a $k$-isomorphism) $k[X_1, \cdots, X_n]/(X_1^p, \cdots, X_n^p)$. For a commutative group scheme of this form, I think one can show that the morphism $[p]$ on it is the zero morphism $0$ by dealing with symmetric functions. (I am not sure about this, but I saw somewhere that the Verschiebung morphism $V$ also uses the symmetric functions.) I like this way of approach since it uses the description of $[p]$ by the Frobenius morphism $F$, and the kernel $K$ of $F$ is explicit (if one chooses a coordinate), also the factorization $[p] = V \circ F$ seems can be proved without further knowledge. However, in Mumford's book, there is a proof that $[p] = 0$ for height one commutative group scheme by using the Lie-algebra. It would be appreciated for any comment on this.

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The dual isogeny should be $\vee{\phi} : B^{\vee} \to A^{\vee}$. I think the map $\phi \mapsto \hat{\phi}$ you want to define will not be well-behaved, even in characteristic 0. For example it won't be involutive if the dim is >1 (look at the degrees of the isogenies which are involved). –  François Brunault Jul 4 '11 at 9:58
If the isogeny is separable, then the kernel of $\phi$ is split after passing to the separable closure of $k$. Now you can form the group quotient, etc., over the separable closure. And you can perform Galois descent for the extension of $k$ to the separable closure to get $\phi^\vee$ defined over $k$. Possibly the references are imposing "perfect" as a hypothesis because they want a result that applies to all isogenies, not just separable isogenies. –  Jason Starr Jul 4 '11 at 10:06
If the characteristic of the base field divides $\deg(\phi)$ then you need to work with group schemes even if the field is perfect. In general, one can define the quotient of an abelian variety by any finite subgroup scheme, so the same proof works over any field. See for example, SGA 3, Expose VI_A. (As pointed out by Francois Brunault, the map you want is not what is usually called the dual isogeny (in higher dimensions) but it does exist.) –  ulrich Jul 4 '11 at 10:14
As I have already mentioned, you can get a proof for abelian varieties by working with group schemes. One cannot use just Frobenius in this case; for example, by taking products of elliptic curves it is easy to find inseparable isogenies of degree $p$ between abelian varieties of dimension $g$ for any $g \geq 1$. –  ulrich Jul 5 '11 at 10:14
It's been a while since I looked (and I don't have a copy handy), but doesn't Mumford give a construction of the dual isogeny over arbitrary fields in his book on Abelian Varieties? Probably Lang does, too, in his *Abelian Varieties" book, but that's in the older Weil-style language, so probably harder to read. –  Joe Silverman Jul 26 '11 at 0:40
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If $100\; J$ of work is done in carrying a charge of $2C$ from a place where the potential is $-10V$ to another point where potential is V, find the value of V.

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$V= 40 volt$
$V_2 = 3.6 \times 10^6 V$
$W= 2.88 \times 10^{-13}\;J$
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