text:
# Database¶

The basics on using the QueryBuilder to find the data you are interested in is explained in the finding and querying how-to. This section explains some more advanced methods for querying your database and the queryhelp dictionary.

### Working with edges¶

Filters and projections can be applied to both the vertices of the query path and the edges that connect them. Applying a filter or projection to an edge works the same way as for vertices, but the relevant keyword is now preceded by edge_. Using the ArithmeticAddCalculation calculation job as an example, let’s say we want to query for the first input of the addition, i.e. the Int nodes which have been provided as the input with label x:

from aiida.orm import QueryBuilder
qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_outgoing='calcjob', edge_filters={'label': 'x'})

By using the edge_filters keyword argument, we can query for only the inputs that have the label x. Note that any operator that can be used to filter vertices can also be applied to edges. Say we want to find all input Int nodes that are not connected to the CalcJobNode’s via an edge with label x:

qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_outgoing='calcjob', edge_filters={'label': {'!==': 'x'}})

Here, the equality operator == is negated by prepending an exclamation mark !. See the reference table below for a table with all operators. Similar to filters, we can project information of the edge using the edge_project keyword argument:

qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_outgoing='calcjob', edge_project='label')

In the example above, we are querying for the edge labels of the incoming Int nodes of all CalcJobNode’s.

### Ordering and limiting results¶

You can order the results of your query by the properties of the entity. Say you want to return the list of Int outputs from all CalcJobNode’s, sorted by the time they were created in descending order, i.e. the most recent first:

qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_incoming='calcjob')
qb.order_by({Int: {'ctime': 'desc'}})

This can also be used to order your results based on values in a (nested) dictionary, such as the attributes column. However, as the QueryBuilder cannot infer the type of the value in this case, you have to cast the type:

qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_incoming='calcjob')
qb.order_by({Int: {'attributes.value': {'order': 'asc', 'cast': 'i'}}})

The query above will return all Int nodes that are output of all CalcJobNode’s, in ascending order of their value, i.e. from small to big. Note that in this case you have to specify the order operation with a dictionary, where the order key details how you want to order the query results and the cast key informs the QueryBuilder of the attribute type. A list of the available cast types and their aliases can be found in the table below:

Python type

Alias

SQLAlchemy type

int

i

Integer

float

f

Float

bool

b

Boolean

str

t

String

dict

j

JSONB

datetime.datetime

d

DateTime

You can also order using multiple properties by providing a list of dictionaries that each specify one sorting operation:

qb = QueryBuilder()
qb.append(CalcJobNode, tag='calcjob')
qb.append(Int, with_incoming='calcjob')
qb.order_by({Int: [{'attributes.value': {'order': 'asc', 'cast': 'f'}}, {'ctime': 'desc'}]})

Here the Int nodes will first be sorted by their value in ascending order. Nodes for which the value is equal are subsequently sorted by their modification time in descending order.

Finally, you can also limit the number of query results returned with the limit() method. Suppose you only want the first three results from our query:

qb = QueryBuilder()
qb.append(CalcJobNode)
qb.limit(3)

This can be easily combined with the order_by method in order to get the last three CalcJobNode’s that were created in the database:

qb = QueryBuilder()
qb.append(CalcJobNode)
qb.limit(3)
qb.order_by({CalcJobNode: {'ctime': 'desc'}})

### Reference tables¶

List of all operators:

Operator

Datatype

Example

Explanation

==

all

'id': {'==': 123}

Filter for equality

in

all

'name': {'in': ['foo', 'bar']}

Filter for values that are in the given list.

>,<,<=,>=

float, integer, date

'ctime': {'<': datetime(2016, 03, 03)}

Filter for values that are greater or smaller than a certain value

like

string

'name': {'like': 'label%'}

Filter for matching substrings where % and _ are wildcards. To match a literal % or _ escape it by prefixing it with \\.

ilike

string

'name': {'ilike': 'lAbEl%'}

Case insensitive version of like.

or

list of expressions

'id': {'or': [{'<': 12}, {'==': 199}]}

A list of expressions where at least one should be matched.

and

list of expressions

'id': {'and': [{'<': 12}, {'>': 1}]}

A list of expressions where all should be matched.

has_key

dict

'attributes': {'has_key': 'some_key'}

Filter for dictionaries that contain a certain key.

of_type

any

'attributes.some_key': {'of_type': 'bool'}

Filter for values of a certain type.

of_length

lists

'attributes.some_list': {'of_length': 4}

Filter for lists of a certain length.

shorter

lists

'attributes.some_list': {'shorter': 4}

Filter for lists that are shorter than a certain length.

longer

lists

'attributes.some_list': {'longer': 4}

Filter for lists that are longer than a certain length.

contains

lists

'attributes.some_key': {'contains': ['a', 'b']}

Filter for lists that should contain certain values.

List of all relationships:

Entity from

Entity to

Relationship

Explanation

Node

Node

with_outgoing

One node as input of another node

Node

Node

with_incoming

One node as output of another node

Node

Node

with_descendants

One node as the ancestor of another node (Path)

Node

Node

with_ancestors

One node as descendant of another node (Path)

Node

Group

with_node

The group of a node

Group

Node

with_group

The node is a member of a group

Node

Computer

with_node

The computer of a node

Computer

Node

with_computer

The node of a computer

Node

User

with_node

The creator of a node is a user

User

Node

with_user

The node was created by a user

User

Group

with_user

The node was created by a user

Group

User

with_group

The node was created by a user

Node

Log

with_node

The log of a node

Log Node

Node Comment

with_log with_node

The node has a log The comment of a node

Comment

Node

with_comment

The node has a comment

User

Comment

with_user

The comment was created by a user

Comment

User

with_comment

The creator of a comment is a user

### The queryhelp¶

The queryhelp dictionary is a property of the QueryBuilder class. Once you have built your query using the appender method explained in the finding and querying for data how-to and the advanced sections above, you can easily store your query by saving the QueryBuilder.queryhelp dictionary as a JSON file for later use:

import json
from aiida.orm import QueryBuilder

qb = QueryBuilder()
qb.append(CalcJobNode)

with open("queryhelp.json", "w") as file:
file.write(json.dumps(qb.queryhelp, indent=4))

To use the queryhelp to instantiate the QueryBuilder, you can use Python’s automatic keyword expansion:

with open("queryhelp.json", "r") as file:

qb = QueryBuilder(**queryhelp)

Alternatively, you can also use the queryhelp to set up your query by specifying the path, filters and projections and constructing the queryhelp dictionary by hand. To do this, you have to specify:

• the path: Here, the user specifies the path along which to join tables as a list of dictionaries, where each list item identifies a vertex in your path. You define the vertex class with the cls key:

queryhelp = {
'path':[
{'cls': Data}
]
}

Each entity in the query has to have a unique tag. If the tag is not provided, it is set to the name of the class. However, this will not work if you choose the same class twice in the query. In this case you have to provide the tag using the tag key:

queryhelp = {
'path':[
{
'cls':Node,
'tag':'node_1'
},
{
'cls':Node,
'tag':'node_2'
}
]
}

You also have to detail some information on the vertex edges, in order to connect them correctly. There are several redundant ways this can be done:

• You can specify that this node is an input or output of another node preceding the current one in the list. That other node can be specified by an integer or the class or type. The following examples are all valid joining instructions, assuming there is a structure defined at index 2 of the path with tag “struc1”:

edge_specification = queryhelp['path'][3]
edge_specification['with_incoming'] = 2
edge_specification['with_incoming'] = StructureData
edge_specification['with_incoming'] = 'struc1'
edge_specification['with_outgoing']  = 2
edge_specification['with_outgoing']  = StructureData
edge_specification['with_outgoing']  = 'struc1'

• queryhelp_item[‘direction’] = integer

If any of the above specs (“with_outgoing”, “with_incoming”) were not specified, the key “direction” is looked for. Directions are defined as distances in the tree. 1 is defined as one step down the tree along a link. This means that 1 joins the node specified in this dictionary to the node specified on list-item before as an output. Direction defaults to 1, which is why, if nothing is specified, this node is joined to the previous one as an output by default. A negative number reverse the direction of the link. The absolute value of the direction defines the table to join to with respect to your own position in the list. An absolute value of 1 joins one table above, a value of 2 to the table defined 2 indices above. The two following queryhelps yield the same query:

from aiida.orm import TrajectoryData
from aiida_quantumespresso.calculations.pw import PwCalculation
from aiida.orm import Dict
qh1 = {
'path': [
{
'cls':PwCalculation
},
{
'cls':TrajectoryData
},
{
'cls':Dict,
'direction':-2
}
]
}

# returns same query as:

qh2 = {
'path':[
{
'cls':PwCalculation
},
{
'cls':TrajectoryData
},
{
'cls':Dict,
'with_outgoing':PwCalculation
}
]
}

# Shorter version:

qh3 = {
'path':[
Dict,
PwCalculation,
TrajectoryData,
]
}

• what to project: Determing which columns the query will return:

queryhelp = {
'path':[PwCalculation],
'project':{
PwCalculation:['user_id', 'id'],
}
}

If you are using JSONB columns, you can also project a value stored inside the json:

queryhelp = {
'path':[
PwCalculation,
StructureData,
],
'project':{
PwCalculation:['state', 'id'],
}
}

Returns the state and the id of all instances of PwCalculation where a structures is linked as output of a relax-calculation. The strings that you pass have to be name of the columns. If you pass an asterisk (‘*’), the query will return the instance of the AiidaClass.

• the filters: Filters enable you to further specify the query. This is an example for a query for structures that were added after a certain time (say last 4 days) and have an id larger than 50:

from aiida.common import timezone
from datetime import timedelta

queryhelp = {
'path':[
{'cls':PwCalculation}, # PwCalculation with structure as output
{'cls':StructureData}
],
'filters':{
StructureData:{
'ctime':{'>':  timezone.now() - timedelta(days=4)},
'id':{'>': 50}
}
}
}

If you want to include filters and projections on links between nodes, you will have to add these to filters and projections in the queryhelp. Let’s take an example from before and add a few filters on the link:

queryhelp = {
'path':[
{'cls':PwCalculation, 'tag':'relax'}, # PwCalculation with structure as output
{'cls':StructureData, 'tag':'structure'}
],
'filters':{
'structure':{
'id':{'>': 50}
},
'relax--structure':{
'label':{'like':'output_%'},
}
},
'project':{
'relax--structure':['label'],
'structure':['label'],
'relax':['label', 'uuid'],
}
}

Notice that the tag for the link, by default, is the tag of the two connecting nodes delimited by two dashes ‘–’ and the order DOES matter.

Alternatively, you can choose the tag for the edge in the path when defining the entity to join using edge_tag:

queryhelp = {
'path':[
{'cls':PwCalculation, 'tag':'relax'},         # Relaxation with structure as output
{
'cls':StructureData,
'tag':'structure',
}
],
'filters':{
'structure':{
'id':{'>': 50}
},
'label':{'like':'output_%'},
}
},
'project':{
'structure':['label'],
'relax':['label', 'uuid'],
}
}

Limits and offset can be set directly like this:

queryhelp = {
'path':[Node],
'limit':10,
'offset':20
}

That queryhelp would tell the QueryBuilder to return 10 rows after the first 20 have been skipped.


text:
In double-slit experiment using light of wavelength 600 nm,
Question:

In double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1º. What is the spacing between the two slits?

Solution:

Wavelength of light used, λ = 6000 nm = 600 × 10−9 m

Angular width of fringe, $\theta=0.1^{\circ}=0.1 \times \frac{\pi}{180}=\frac{3.14}{1800} \mathrm{rad}$

Angular width of a fringe is related to slit spacing (d) as:

$\theta=\frac{\lambda}{d}$

$d=\frac{\lambda}{\theta}$

$=\frac{600 \times 10^{-9}}{\frac{3.14}{1800}}=3.44 \times 10^{-4} \mathrm{~m}$

Therefore, the spacing between the slits is $3.44 \times 10^{-4} \mathrm{~m}$.

format: plain


text:
# Presented below is information related to Tao Corp. for the year 2015 (amounts in thousands). Net...

1 answer below »

presented below is information related to Tao Corp. for the year 2015 (amounts in thousands).

Net sales HK$1,200,000 Write-off of inventory due to obsolescence HK$80,000

Cost of goods sold 780,000

Depreciation expense omitted by accident in 2014 40,000

Selling expenses 65,000

Interest expense 50,000

Cash dividends declared 45,000

Dividend revenue 20,000

Retained earnings at December 31, 2014 980,000

Interest revenue 7,000

Effective tax rate of 34% on all items

Prepare an income statement for 2015. Assume that 60,000 ordinary shares are outstanding. (Round earnings per share 2 decimal places, e.g. 5.25 and all other answers to 0 decimal places, e.g. 2,250. Enter negative amounts using either a negative sign preceding the number e.g. -45 or parentheses e.g. (45).)

## 1 Approved Answer

5 Ratings, (9 Votes)


text:
# Formal Design and Analysis of Cyber-Physical Systems

Ashutosh Trivedi

## Affiliation:

University of Pennsylvania
Department of Computer and Information Science
3330 Walnut Street
United States of America

## Time:

Thursday, 21 June 2012, 11:30 to 12:30

## Venue:

• A-212 (STCS Seminar Room)

## Organisers:

The term cyber-physical systems" (CPS) refers to any network of digital and analog systems whose performance crucially depends on both  the continuous dynamics of the analog parts and the real-time switching decisions made by the digital system. Such systems are increasingly playing safety-critical role in modern life, where a fault in their design can be catastrophic. Implantable medical devices are important paradigmatic examples of such safety-critical cyber-physical systems. The growing number of fatalities due to implantable device failures, and an ever increasing list of device recalls by US Food and Drug Administration (FDA) in recent years underscore the challenges in engineering safety-critical CPS. Similar examples can also be cited for CPS from other domains such as avionics, automotive, transportation networks, and energy sector. The central theme underlying my research is to employ rigorous mathematical reasoning to analyze (verification and optimization) and to design (controller synthesis) CPS with guaranteed performance. In this talk I will summarize a few key techniques for formal design and analysis of CPS with emphasis upon my research contributions.

I will also present a recent result on the green scheduling problem where the goal is to co-ordinate switching decisions of various HVAC units so as to minimize the peak energy usage. We pose the green scheduling problem using optimal scheduling problem for constant-rate multi-mode systems (MMS). MMS model hybrid systems that can switch freely among a finite set of modes, and whose dynamics is specified by a finite number of real-valued variables with mode-dependent constant rates. The schedulability problem for such systems is to design a mode-switching policy that maintains the state within a specified safety set. A key result is that  the schedulability problem for MMS can be decided in polynomial time. The talk is based on a joint-work with Prof. Rajeev Alur and Dr. Dominik Wojtczak which has received the best paper award at 15h conference on hybrid systems (HSCC'12) at 2012 Cyber-Physical Systems Week.

Bio: Ashutosh Trivedi is a postdoctoral researcher (since 2011) in the computer and information science department at the University of Pennsylvania. He works, broadly, in the area of formal analysis and design of real-time and hybrid systems. After receiving his PhD from the University of Warwick in 2009, he spent 2 years at the University of Oxford as a postdoctoral researcher working as a part of the Large-Scale Complex IT Systems (LSCITS) project. Prior to his PhD, he received a Masters degree in Electrical Engineering (Reliability) from IIT Bombay in 2003, and has worked as research assistant with the CFDVS center at IIT Bombay,  the NuSMV development group at IRST, Italy and the formal methods group at University of Tuebingen, Germany.


text:
# Rename [poisson] in [poisson-distribution]

I suggest to rename into , given that its wiki excerpt says

A discrete distribution defined on the non-negative integers that has the property that the mean is equal to the variance.

and given that we have , , and tags.

• This seems like a no-brainer. If there's no dissent after a period of time, ping me & I'll do it. – gung - Reinstate Monica Nov 10 '17 at 15:45
• I'd say this can go forward @gung. – amoeba Nov 16 '17 at 8:23
• It's done now, @amoeba. – gung - Reinstate Monica Nov 16 '17 at 12:37
• Thanks @gung. I am a bit concerned that [poisson] is now a synonym of [poisson-distribution]. Wouldn't it make more sense to delete this synonym? – amoeba Nov 16 '17 at 12:49
• Do you want [poisson] to simply not exist? It's been merged w/ p-d. – gung - Reinstate Monica Nov 16 '17 at 13:24
• Yes, that's what I was thinking. I am not sure it makes sense to map any tag [a] to [a-b] when we also have [a-c], [a-d], etc. It's like with [fisher] when we got rid of this tag entirely. AFAIK there is no mod tool to just rename a tag, so the only way to rename is to create a synonym, merge, and then delete the synonym mapping. That said, if you have an argument for why [poisson] should stay as a synonym, then I am fine with it. – amoeba Nov 16 '17 at 13:42
• Yeah, that makes sense. I deleted the synonym mapping. – gung - Reinstate Monica Nov 16 '17 at 14:04

This seems like a no-brainer. I created the tag (i.e., ), made them synonyms, merged them (so that all threads that used to have [poisson] now have [poisson-distribution] instead, without bumping >1k threads to the main page), and deleted the old [poisson] tag.


text:
# Easy proof, that $\rm e\notin \mathbb Q$

$\def\e{{\rm e}}$ I recently had the task to explain the proof that $\e$ is irrational as a presentation to my classmates. To prepare this presentation, the teacher gave me a script with a proof that uses an estimation of the series $b_n = \sum_{k=0}^n1/k!\$ to show that there are no $p,q\in\mathbb N,\$ such that $\e = p/q.$

Because we defined $\e$ as the limit of $a_n = \big(1+\frac1n\!\big)^n\$ I had to include the rather long proof that $\lim\limits_{n\to\infty}a_n-b_n=0,\$ rendering the whole proof quite long. Is there a “shorter” proof, given that $\e$ is defined as the limit of $a_n$?

-
Perhaps you should shorten your proof of $\lim a_n = \lim b_n$. Denote $E = \lim b_n$. From the simple $a_n \leq E$ follows $e \leq E$. On the other hand, it is easy to see that for each $n$ we have $e \geq b_n$, and so $e \geq E$. –  Yuval Filmus Sep 21 '11 at 20:27
(Tangentially related to the question, but I am not sure if this will be ultimately useful to the OP.) Can anyone suggest the asymptotics of the error $e - (1+\frac{1}{n})^n$? –  Srivatsan Sep 21 '11 at 20:30
@Yuval That is actually how we did this. The problem is, that the students in my class don't like proofs containing too many “It is easy to see that...”s. The proof using this strategie filled one A4 page in the script and took 20 minutes for me to explain. (Although it is quite easy if you think about it). –  FUZxxl Sep 21 '11 at 20:32
@francis (Oops, I removed the comment.) To show existence of limit, it is enough to show the sequence is monotonically increasing and bounded. –  Srivatsan Sep 21 '11 at 20:50
Not what you want but see also How Euler Did It. –  lhf Sep 21 '11 at 20:54

(Too long for a comment.) This is probably the way you did it, but it doesn’t seem very long to me; where did it bog down?

From the binomial theorem we have

$$a_n = \left(1 + \frac1n \right)^n = \sum\limits_{k=0}^n \binom{n}{k}n^{-k} = \sum\limits_{k=0}^n \frac{n^{\underline k}}{k! n^k} \le \sum\limits_{k=0}^n \frac{1}{k!} = b_n.$$

For the other direction let $m>1$ be an integer; then

$$a_{mn} = \sum\limits_{k=0}^{mn}\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \sum\limits_{k=0}^n\frac{(mn)^{\underline{k}}}{k!(mn)^k} \ge \left(\frac{(mn)^{\underline n}}{(mn)^n} \right) b_n \ge \left(\frac{m-1}{m} \right)^n b_n,$$

and $\displaystyle \lim\limits_{m\to\infty} \left(\frac{m-1}{m}\right)^n = 1$, so $b_n \le e$.

-
The last step in the second line is not completely clear to me. Could you please elaborate why this holds? –  FUZxxl Sep 22 '11 at 8:20
@FUZxxl: $\frac{(mn)^{\underline n}}{(mn)^n}$ is the product of $n$ factors; the smallest is $(mn-n+1)/(mn)$, so they’re all $> (m-1)/m$. –  Brian M. Scott Sep 22 '11 at 10:22
Thanks for the explanation. –  FUZxxl Sep 22 '11 at 10:42


text:
### Home > PC > Chapter 8 > Lesson 8.1.4 > Problem8-56

8-56.
1. Write out the first four terms of the following sequences, then evaluate the limits. Homework Help ✎

1. . Then find .

2. . Then find .

3. . Then find .

$a_1=\frac{2-1}{3+1}=?$

$a_2=\frac{2-2}{3+2}=?$

$a_3=\frac{2-3}{3+3}=?$

$a_4=\frac{2-4}{3+4}=?$

1. Split the fraction into two fractions:

$a_n=\frac{2}{3+n}-\frac{n}{3+n}$

2. Use the dominant term idea to find the limit.

$a_1=\frac{(-2)^1}{3^1}=?$

$a_2=\frac{(-2)^2}{3^2}=?$

$a_3=\frac{(-2)^3}{3^3}=?$

$a_4=\frac{(-2)^4}{3^4}=?$

The dominant term is in the denominator.

$a_1=\frac{(-2)^1}{2^1}=?$

$a_2=\frac{(-2)^2}{2^2}=?$

$a_3=\frac{(-2)^3}{2^3}=?$

$a_4=\frac{(-2)^4}{2^4}=?$

The dominant term is in both the denominator and the numerator.


text:
# Prove that V is a vector space. V=T(m, n), the set of linear transformations T : R^{m} rightarrow R^{n}, together with the usual addition and scalar multiplication of functions.

Prove that V is a vector space. , the set of linear transformations , together with the usual addition and scalar multiplication of functions.
You can still ask an expert for help

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Step 1
the set of linear transformations together with the usual addition and scalar multiplication of functions are clearly defined.
next we need to determine if the five conditions of Definion 7.1 are all met.
If $T\left(u\right)$ and $T\left(v\right)$ are a linear transformation if for all vectors u and v and ${R}^{m},$ then their sum will also be a linear transformation in ${R}^{m}$, that is, is closed under addition.
If $T\left(u\right)$ is a linear transformation if for all vectors u in ${R}^{m}$ and c is a scalar then i and $cT\left(u\right)$ will also be a linear transformation thus is also closed under scalar multiplication.
M If $T\left(0\right)=0$ is an identical zero transformation, then if $T\left(u\right)$ is in ${R}^{m}$ linear transformation, then for all $T\left(u\right)$ in ${R}^{m}$, so that $T\left(0\right)$ is the zero transformation.
If $T\left(u\right)$ linear transformation, it has its inverse linear transformation $-T\left(u\right)$ so that

Step 2
a) We are checking whether you are

Therefore, commutativity is valid.
b) Check


text:
# From Characters to Words to in Between: Do We Capture Morphology?

(Vania and Lopez, 2017) at ACL

We know that words exist, and we know that we can break them up into smaller parts—”rereading” can be broken into the morphemes ["re-", "read", "-ing"], or into the characters ["r", "e", "r", "e", "a", "d", "i", "n", "g"]. Today’s paper explores which of these levels is the best representation for language modeling. It also shows the benefit of morphological analysis—without it, even tons more data won’t carry the day. A model trained on true morphological features outperforms eight character-based models.

The authors pose four questions:

1. How do representations based on morphemes compare with those based on characters?
2. What is the best way to compose subword representations?
3. Do character-level models capture morphology in terms of predictive utility?
4. How do different representations interact with languages of different morphological typologies?

To answer these, we should first sort out some preliminary terminology:

• morpheme: smallest unit of meaning. They can be bound (“-ing”) or free, and they can express core meaning (“to run”) or features (past-tense).
• morph: the surface realization of a morpheme. In “tries”, the morphemes are “try+s”, while the morphs are “tri+es”.
• morphological analysis: breaking a word into its lemma (dictionary form) and features. try+VB+3rd+SG+Pres.

• Fusional language: concatenates single morphemes realizing multiple features. An example is English.
• Agglutinative language: One feature per morpheme, like Turkish.
• Root and Pattern morphology (I’ve heard this as ‘templatic’): insert sounds around and between a stem of consonants. Arabic is the biggest example.
• Reduplication: repeating part or all of the word, giving new meaning. Indonesian uses this to pluralize.

The task they approach is language modeling, which is super important in language and speech processing. Its objective is to fnd the probability of a given sequence of text, and it does this by decomposing it according to the chain rule, such that each next word (or character, or what have you) depends on the part before it. They employ ten different models, combining different encodings and composition methods. Each has a finite output vocabulary, so they compare outputs using perplexity.

One of the authors is supported by an Indonesian endowment, which may explain the interest in reduplicating languages. In any case, the results strongly suggest that in the absence of annotated data, character trigrams are the way to go. But annotated data brings you so much farther, consistently achieving lower-perplexity models. (That is, models that were less surprised by the next thing they saw.) BPE actually turned out ggarbage words in many cases. And the bi-LSTM composition method, which brings information from each direction into each representation, outperformed simple addition. (Again, no big surprises here. Complex learned functions doing well?)

The often-extreme benefit of morphological annotations, the authors note, begs for bootstrapping. A structured autoencoder would be a good way to model the morphological analyses as a latent variable, generating more labeled data. Maybe I should do that…

The bottom line: Your best bet for language modeling is annotated data. Lacking that, use the output of bi-LSTMs with character trigrams as input.

Written on February 5, 2018


text:
Explore anything with the first computational knowledge engine. A matrix m may be tested to see if it is antisymmetric in the Wolfram Language using AntisymmetricMatrixQ[m]. Every operator on an n-dimensional space can be expressed as a sum of n rank one operators; this gives a coordinate-free version of the sum of diagonal entries. transpose. n matrix is defined to be, i.e., the sum of the diagonal elements. Hints help you try the next step on your own. Finally, if the square is greater than 4, the transformation is loxodromic. ) for Lie groups. {\displaystyle 1/n} times scalars) so Proof where we used B B−1 = E(the identity matrix). Since det M= det (−MT) = det (−M) = (−1)ddet M, (1) it follows that det M= 0 if dis odd. transformation, (where Einstein summation is used here to sum n It is not symmetric because because =4 and =2.. s s That's the right answer. The trace of A, denoted tr(A), is the sum of the diagonal entries of A. The trace of a matrix is invariant under a similarity transformation Tr(B −1 A B) = Tr(A). l l ) Taking the trace of the product then comes from pairing on the outer terms, while taking the product in the opposite order and then taking the trace just switches which pairing is applied first. {\displaystyle {\mathfrak {gl}}_{n}} Jacek Jakowski, ... Keiji Morokuma, in GPU Computing Gems Emerald Edition, 2011. symmetric matrix, the corresponding pfaﬃan is deﬁned to be zero. Viewing. The partial trace is another generalization of the trace that is operator-valued. A number equal to minus itself can only be zero. 7. of "inclusion of scalars" to obtain a map l Algebra, 3rd ed. The trace is 6. Relation to eigenvalues Thus this scalar quantity serves as an If A is a general associative algebra over a field k, then a trace on A is often defined to be any map tr : A ↦ k which vanishes on commutators: tr([a,b]) for all a, b ∈ A. In the language of bialgebras, scalars are the unit, while trace is the counit. coming from the pairing V∗ × V → F on the middle terms. : and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: characters. of linear operators on an n-dimensional space (n × n matrices with entries in A matrix can be skew symmetric only if it is square. The trace is also invariant under a similarity In other words, we can say that transpose of Matrix B is not equal to matrix B ().. What is a Skew-Symmetric Matrix? Using the notion of dualizable objects and categorical traces, this approach to traces can be fruitfully axiomatized and applied to other mathematical areas. Notice that an n × n matrix A is symmetric if and only if a ij = a ji, and A is skew-symmetric if and only if a ij = −a ji, for all i,j such that 1 ≤ i,j ≤ n.In other words, the entries above the main diagonal are reflected into equal (for symmetric) or opposite (for skew-symmetric) entries below the diagonal. Matrix Multiplication. {\displaystyle {\mathfrak {sl}}_{n}} Here that symmetric matrix has lambda as 2 and 4. If the square is in the interval [0,4), it is elliptic. Fact 11 (Cyclic Property of Trace) Let Abe an arbitrary n mmatrix and let Bbe an arbitrary m n matrix. Comparison of matrix-matrix products for “even” size matrices (500, 1000, etc.) Unlimited random practice problems and answers with built-in Step-by-step solutions. If the transpose of a matrix is equal to the negative of itself, the matrix is said to be skew symmetric. Example Theorem Let A and B be n×n matrices, then Tr(A B) = Tr (B A). In these notes, the rank of Mwill be denoted by … A supertrace is the generalization of a trace to the setting of superalgebras. The derivative of this family at θ = 0, the identity rotation, is the antisymmetric matrix = (−) which clearly has trace zero, indicating that this matrix represents an infinitesimal transformation which preserves area. Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more. The norm derived from the above inner product is called the Frobenius norm, which satisfies submultiplicative property as matrix norm. {\displaystyle \operatorname {tr} :{\mathfrak {gl}}_{n}\to K} = {\displaystyle (\varphi _{n})_{n}} → The product of a Symmetric and an Antisymmetric Matrix has zero trace, (10) The value of the trace can be found using the fact that the matrix can always be transformed to a coordinate system where the z -Axis lies along the axis of rotation. The trace of a linear operator Z which lives on a product space A ⊗ B is equal to the partial traces over A and B: For more properties and a generalization of the partial trace, see traced monoidal categories. See classification of Möbius transformations. Skew-Symmetric Matrix. n For large matrices, slightly increasing the size of the matrix to match a multiple of 64 can … 40 and 64, 1987. Learn All Concepts of Chapter 3 Class 12 Matrices - FREE. In component notation, this becomes (2) Letting , the requirement becomes (3) so an antisymmetric matrix must have zeros on its diagonal. 2 v with the corresponding larger matrix with a leading dimension array equal to multiples of 64. n What is the trace of the metric tensor? summation notation. Join the initiative for modernizing math education. Weisstein, Eric W. "Matrix Trace." The sum of the vector space and antisymmetric parts defined as: matrix. B−1A B ) = tr ( B ) = tr ( B )... N, as the trace of an antisymmetric matrix and is given by symmetric and parts! 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Abe an arbitrary m n matrix complex antisymmetric matrix and is a symmetrix matrix then A-1 is symmetric! Gathered on the natural numbers is an antisymmetric matrix to also have been appropriately ). Of indices i and j here that symmetric matrix has lambda as 2 and 4 system ( which used! Obtaining a map determinant 1, so they preserve area modified by multiplication a... Is invariant under a similarity transformation tr ( BA ) matrix m may be tested see! Matrix m may be tested to see if it is also called the Frobenius norm which! Example of an inner product is called the Frobenius norm, which is assumed to also been! Bbe an arbitrary m n matrix problems step-by-step from beginning to end traces are known as  group characters n! Which means matrix a is said to be, i.e., the sum of the four-dimensional unit matrix latter however! So, it is square a ij = − a j i for all of are.

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Dimensional Analysis and Applications

Uses of Dimensional Equations
The dimensional equations have got following three uses:
1. To check the correctness of a physical equation.
2. To derive the relation between different physical quantities involved in a physical phenomenon.
3. To change from one system of units to another.
1. To check the correctness of a physical relation and the principle of homogeneity of dimensions:
Checking the correctness of a physical relation (or equation) is based on the principle of homogeneity of dimensions.

According to this principle, the dimensions of the fundamental quantities (mass, length and time) are same in each and every term on either side of the physical relation.

To check the correctness of a given physical equation, the physical quantities on the two sides of the equation are expressed in terms of fundamental units of mass, length and time.

If the powers of M, L and T on two sides of the equation are same, then the physical equation is correct and otherwise not.

Problem-1

Check the dimensional consistency of the equation:
FS = ,
Where S is the distance moved, u and v are the initial and final velocities of a body of mass m and F is the force acting on it.

Solution
We have,
FS =

L. H. S. Dimensional formula of term FS = [MLT-2] × [L] = [ML2T-2]
R. H. S. Dimensional formula of term = [M] ×[LT-1]2 = [ML2T-2]
Dimensional formula of term = [M] ×[LT-1]2 = [ML2T-2]
As the dimensional formulae of all the terms in the given equation are same, the equation is a correct one.

Problem-2

In Vander Waals’s equation:
= RT.
What are the dimensions of constants ‘a’ and ‘b’?

Solution
Here, = RT
According to principle of homogeneity of dimensions, should have dimensions of P and b should have the dimensions of v.
The dimension of = dimension of pressure.
or dimension of a = pressure × V2 = [ML-1T2][L3]2 = [M L5 T2]
and dimension of b = volume = [L3] = [M0L3T0]

Problem-3

Test by using dimensional analysis, the accuracy of the relation:
=
Where the letters have their usual meanings.

Solution
Here, =
Dimensional formula of wavelength, = [L]
Dimensional formula of L. H. S = [L] = [M0LT0]
Dimensional formula of Planck’s constant, h = [ML2T-1]
Dimensional formula of mass, m = [M]
Dimensional formula of velocity, v = [LT-1]
Dimensional formula of R. H. S = = [M0LT0]
As dimensional formula of L. H. S. is same as that of R. H. S., the given relation is correct.

Problem-4

Check the correctness of the relation τ = I α.

Solution
Now, τ = torque = force × distance
= [M L T-1] [L] = [ML2T-2]
I = Moment of inertia = mass × distance2 = [m][L]2 = [M L2]
a = angular acceleration = = [T-2]
I α[M L2] [T-2] = [M L2 T-2]
Since dimensional formulae of L. H. S. and R. H. S. are the same, the given relation is correct.

Problem-5

Test by using dimensional analysis, the accuracy of the relation:
=

Solution
Here, =

where k is a dimensionless constant.

Dimensional formula of critical velocity, vc = [L T-1]

Dimensional formula of coefficient of viscosity,

η = [ML-1 T-1]

Dimensional formula of radius, R = [L]

Dimensional formula of density, ρ = [ML-3]

Dimensional formula of L. H. S = [LT-1] = [M0 L T-1]

Dimensional formula of R. H. S = = [M0 L T-1]

As dimensional formula of L. H. S. is same as that of R. H. S. the given relation is correct.

In deriving the relation between the various physical quantities, the principle of homogeneity of dimensional equation is used.

To derive a physical relation the final dimensional equation is written in terms of fundamental units of M, L and T.

The powers of M, L and T are equated on both sides of the dimensional equation.

Three equations are got from which the values of three unknown powers can be calculated.

These values are substituted in the dimensional equation.

Problem-6

The velocity (v) of sound through a medium may be assumed to depend on
1. the density (ρ ) of the medium and
2. modulus of elasticity (E).
If the dimensions for elasticity (ratio of stress to strain) are [ML-1T-2], deduce by the method of dimensions the formula for the velocity of sound.

Solution
Let v
α r a            ------(i)
Let v α Eb            ------(ii)

Combining equation (i) and (ii), we get

v ρ a Eb            ------(iii)

Writing the dimensional formula of v, α and E in equation (iii), we get

[LT-1] [ML-3]a [ML-1T-2]b
or[M0LT-1] [Ma+bL-3a-b T-2]b

Comparing dimensions of M, L and T on the two sides, we get

a + b = 0           ------(iv)
-3a –b = 1
and -2b = -1
or b =

Substituting for b in equation (iv), we get

a + = 0 or a =

Putting values of a and b in equation (iii), we get

V ∝ or v = k

The constant of proportionality k is found to be 1.
v =

Problem-7

Assuming that frequency (ν) of a vibrating string depends upon load applied (F), length of the string (l) and mass per unit length (m), prove that

Solution
Let ν Fa                                                                       ------ (i)

lb                                                                                ------ (ii)

mc                                                                      ------ (iii)

Combining equation (i), (ii) and (iii), we get ν = kFalbmc  ------ (iv)
Where k is constant of proportionality.
Substituting the dimensions of ν , F, l and m, we get
[T-1] = [MLT-2]a [L]b [ML-1]C
[M0 L0 T-1] = [Ma+c L a+b-c T-2a]
Comparing dimensions of M, L and T on the both sides of the equation, we get
-2a = -1            or                a =
a + c = 0           or                c = -a = -
and           a + b-c = 0           or               b=c-a=--= -1
Substituting the values of a, b and c in equation (iv), we get
ν = k F1/2 l-1 m-1/2

Method of dimensional analysis can be used to obtain the value of a physical quantity in some other system, when its value in one system is given.

As discussed earlier, the measure of a physical quantity is given by

X = n u,

Where u is size of the unit and n is numerical value of the physical quantity for the unit chosen. If u1 and u2 are units for measurement of the physical quantity in two system and n1 and n2 are the numerical values of the physical quantity for the two units, then
n1u1 = n2u2

If a, b and c are dimensions of the physical quantity in mass, length and time, then
n1[M1a L1b T1c] = n2[M2a L2b T2c]

Here, M1, L1, T1 and M2, L2 and T2 are units of mass, length and time in the two systems. Therefore
n2 = n1

The dimensional formula of work is [ML2T-2]. Therefore, a = 1, b = 2 and c = -2.
Hence, number of erg on 1 J is given by
n2 = 1× = = 107
1 J = 107 erg


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1. CMB Online first

Kocel-Cynk, Beata; Pawłucki, Wiesław; Valette, Anna
 $\mathcal{C}^p$-parametrization in o-minimal structures We give a geometric and elementary proof of the uniform $\mathcal C^p$-parametrization theorem of Yomdin-Gromov in arbitrary o-minimal structures. Keywords:o-minimal structure, $C^p$-parametrizationCategories:03C64, 14P15, 32B20

2. CMB Online first

González, José Luis; Karu, Kalle
 Examples of non-finitely generated Cox rings We bring examples of toric varieties blown up at a point in the torus that do not have finitely generated Cox rings. These examples are generalizations of our earlier work, where toric surfaces of Picard number $1$ were studied. In this article we consider toric varieties of higher Picard number and higher dimension. In particular, we bring examples of weighted projective $3$-spaces blown up at a point that do not have finitely generated Cox rings. Keywords:Cox ring, Mori dream space, toric varietyCategories:14M25, 14C20, 14E30

3. CMB 2018 (vol 61 pp. 836)

Purbhoo, Kevin
 Total Nonnegativity and Stable Polynomials We consider homogeneous multiaffine polynomials whose coefficients are the PlÃ¼cker coordinates of a point $V$ of the Grassmannian. We show that such a polynomial is stable (with respect to the upper half plane) if and only if $V$ is in the totally nonnegative part of the Grassmannian. To prove this, we consider an action of matrices on multiaffine polynomials. We show that a matrix $A$ preserves stability of polynomials if and only if $A$ is totally nonnegative. The proofs are applications of classical theory of totally nonnegative matrices, and the generalized PÃ³lya-Schur theory of Borcea and BrÃ¤ndÃ©n. Keywords:stable polynomial, zeros of a complex polynomial, total nonnegative Grassmannian, totally nonnegative matrixCategories:32A60, 14M15, 14P10, 15B48

4. CMB Online first

Asgarli, Shamil
 Sharp Bertini theorem for plane curves over finite fields We prove that if $C$ is a reflexive smooth plane curve of degree $d$ defined over a finite field $\mathbb{F}_q$ with $d\leq q+1$, then there is an $\mathbb{F}_q$-line $L$ that intersects $C$ transversely. We also prove the same result for non-reflexive curves of degree $p+1$ and $2p+1$ where $q=p^{r}$. Keywords:Bertini theorem, transversality, finite fieldCategories:14H50, 11G20, 14N05

5. CMB 2018 (vol 61 pp. 878)

Sun, Chia-Liang
 Weak Approximation for Points with Coordinates in Rank-one Subgroups of Global Function Fields For every affine variety over a global function field, we show that the set of its points with coordinates in an arbitrary rank-one multiplicative subgroup of this function field satisfies the required property of weak approximation for finite sets of places of this function field avoiding arbitrarily given finitely many places. Keywords:weak approximation, global function fields, local-global criteriaCategories:14G05, 11R58

6. CMB Online first

Bertapelle, A.; Mazzari, N.
 On deformations of $1$-motives According to a well-known theorem of Serre and Tate, the infinitesimal deformation theory of an abelian variety in positive characteristic is equivalent to the infinitesimal deformation theory of its Barsotti-Tate group. We extend this result to $1$-motives. Keywords:$1$-motive, Barsotti-Tate groupCategories:14L15, 14C15, 14L05

7. CMB Online first

Zhang, Zheng
 On motivic realizations of the canonical Hermitian variations of Hodge structure of Calabi-Yau type over type $D^{\mathbb H}$ domains Let $\mathcal{D}$ be the irreducible Hermitian symmetric domain of type $D_{2n}^{\mathbb{H}}$. There exists a canonical Hermitian variation of real Hodge structure $\mathcal{V}_{\mathbb{R}}$ of Calabi-Yau type over $\mathcal{D}$. This short note concerns the problem of giving motivic realizations for $\mathcal{V}_{\mathbb{R}}$. Namely, we specify a descent of $\mathcal{V}_{\mathbb{R}}$ from $\mathbb{R}$ to $\mathbb{Q}$ and ask whether the $\mathbb{Q}$-descent of $\mathcal{V}_{\mathbb{R}}$ can be realized as sub-variation of rational Hodge structure of those coming from families of algebraic varieties. When $n=2$, we give a motivic realization for $\mathcal{V}_{\mathbb{R}}$. When $n \geq 3$, we show that the unique irreducible factor of Calabi-Yau type in $\mathrm{Sym}^2 \mathcal{V}_{\mathbb{R}}$ can be realized motivically. Keywords:variations of Hodge structure, Hermitian symmetric domainCategories:14D07, 32G20, 32M15

8. CMB 2017 (vol 61 pp. 659)

Wang, Zhenjian
 On Deformations of Nodal Hypersurfaces We extend the infinitesimal Torelli theorem for smooth hypersurfaces to nodal hypersurfaces. Keywords:nodal hypersurface, deformation, Torelli theoremCategories:32S35, 14C30, 14D07, 32S25

9. CMB 2017 (vol 61 pp. 650)

Shirane, Taketo
 Connected numbers and the embedded topology of plane curves The splitting number of a plane irreducible curve for a Galois cover is effective to distinguish the embedded topology of plane curves. In this paper, we define the connected number of a plane curve (possibly reducible) for a Galois cover, which is similar to the splitting number. By using the connected number, we distinguish the embedded topology of Artal arrangements of degree $b\geq 4$, where an Artal arrangement of degree $b$ is a plane curve consisting of one smooth curve of degree $b$ and three of its total inflectional tangen Keywords:plane curve, splitting curve, Zariski pair, cyclic cover, splitting numberCategories:14H30, 14H50, 14F45

10. CMB Online first

Reichstein, Zinovy B.
 On a property of real plane curves of even degree F. Cukierman asked whether or not for every smooth real plane curve $X \subset \mathbb{P}^2$ of even degree $d \geqslant 2$ there exists a real line $L \subset \mathbb{P}^2$ such $X \cap L$ has no real points. We show that the answer is yes" if $d = 2$ or $4$ and no" if $n \geqslant 6$. Keywords:real algebraic geometry, plane curve, maximizer function, bitangentCategories:14P05, 14H50

11. CMB 2017 (vol 61 pp. 572)

Koskivirta, Jean-Stefan
 Normalization of closed Ekedahl-Oort strata We apply our theory of partial flag spaces developed with W. Goldring to study a group-theoretical generalization of the canonical filtration of a truncated Barsotti-Tate group of level 1. As an application, we determine explicitly the normalization of the Zariski closures of Ekedahl-Oort strata of Shimura varieties of Hodge-type as certain closed coarse strata in the associated partial flag spaces. Keywords:Ekedahl-Oort stratification, Shimura varietyCategories:14K10, 20G40, 11G18

12. CMB 2017 (vol 61 pp. 608)

Loeffler, David
 A note on $p$-adic Rankin-Selberg $L$-functions We prove an interpolation formula for the values of certain $p$-adic Rankin-Selberg $L$-functions associated to non-ordinary modular forms. Keywords:$p$-adic $L$-function, Iwasawa theoryCategories:11F85, 11F67, 11G40, 14G35

13. CMB 2017 (vol 61 pp. 328)

Maican, Mario
 Moduli of Space Sheaves with Hilbert Polynomial $4m+1$ We investigate the moduli space of sheaves supported on space curves of degree $4$ and having Euler characteristic $1$. We give an elementary proof of the fact that this moduli space consists of three irreducible components. Keywords:moduli of sheaves, semi-stable sheavesCategories:14D20, 14D22

14. CMB 2017 (vol 61 pp. 272)

Franz, Matthias
 Symmetric Products of Equivariantly Formal Spaces Let $$X$$ be a CW complex with a continuous action of a topological group $$G$$. We show that if $$X$$ is equivariantly formal for singular cohomology with coefficients in some field $$\Bbbk$$, then so are all symmetric products of $$X$$ and in fact all its $$\Gamma$$-products. In particular, symmetric products of quasi-projective M-varieties are again M-varieties. This generalizes a result by Biswas and D'Mello about symmetric products of M-curves. We also discuss several related questions. Keywords:symmetric product, equivariant formality, maximal variety, Gamma productCategories:55N91, 55S15, 14P25

15. CMB 2017 (vol 60 pp. 490)

Fiori, Andrew
 A Riemann-Hurwitz Theorem for the Algebraic Euler Characteristic We prove an analogue of the Riemann-Hurwitz theorem for computing Euler characteristics of pullbacks of coherent sheaves through finite maps of smooth projective varieties in arbitrary dimensions, subject only to the condition that the irreducible components of the branch and ramification locus have simple normal crossings. Keywords:Riemann-Hurwitz, logarithmic-Chern class, Euler characteristicCategories:14F05, 14C17

16. CMB 2017 (vol 61 pp. 201)

Takahashi, Tomokuni
 Projective plane bundles over an elliptic curve We calculate the dimension of cohomology groups for the holomorphic tangent bundles of each isomorphism class of the projective plane bundle over an elliptic curve. As an application, we construct the families of projective plane bundles, and prove that the families are effectively parametrized and complete. Keywords:projective plane bundle, vector bundle, elliptic curve, deformation, Kodaira-Spencer mapCategories:14J10, 14J30, 14D15

17. CMB 2017 (vol 60 pp. 478)

Carrell, Jim; Kaveh, Kiumars
 Springer's Weyl Group Representation via Localization Let $G$ denote a reductive algebraic group over $\mathbb{C}$ and $x$ a nilpotent element of its Lie algebra $\mathfrak{g}$. The Springer variety $\mathcal{B}_x$ is the closed subvariety of the flag variety $\mathcal{B}$ of $G$ parameterizing the Borel subalgebras of $\mathfrak{g}$ containing $x$. It has the remarkable property that the Weyl group $W$ of $G$ admits a representation on the cohomology of $\mathcal{B}_x$ even though $W$ rarely acts on $\mathcal{B}_x$ itself. Well-known constructions of this action due to Springer et al use technical machinery from algebraic geometry. The purpose of this note is to describe an elementary approach that gives this action when $x$ is what we call parabolic-surjective. The idea is to use localization to construct an action of $W$ on the equivariant cohomology algebra $H_S^*(\mathcal{B}_x)$, where $S$ is a certain algebraic subtorus of $G$. This action descends to $H^*(\mathcal{B}_x)$ via the forgetful map and gives the desired representation. The parabolic-surjective case includes all nilpotents of type $A$ and, more generally, all nilpotents for which it is known that $W$ acts on $H_S^*(\mathcal{B}_x)$ for some torus $S$. Our result is deduced from a general theorem describing when a group action on the cohomology of the fixed point set of a torus action on a space lifts to the full cohomology algebra of the space. Keywords:Springer variety, Weyl group action, equivariant cohomologyCategories:14M15, 14F43, 55N91

18. CMB 2017 (vol 60 pp. 747)

Huang, Yanhe; Sottile, Frank; Zelenko, Igor
 Injectivity of Generalized Wronski Maps We study linear projections on PlÃ¼cker space whose restriction to the Grassmannian is a non-trivial branched cover. When an automorphism of the Grassmannian preserves the fibers, we show that the Grassmannian is necessarily of $m$-dimensional linear subspaces in a symplectic vector space of dimension $2m$, and the linear map is the Lagrangian involution. The Wronski map for a self-adjoint linear differential operator and pole placement map for symmetric linear systems are natural examples. Keywords:Wronski map, PlÃ¼cker embedding, curves in Lagrangian Grassmannian, self-adjoint linear differential operator, symmetric linear control system, pole placement mapCategories:14M15, 34A30, 93B55

19. CMB 2017 (vol 60 pp. 309)

Hein, Nickolas; Sottile, Frank; Zelenko, Igor
 A Congruence Modulo Four for Real Schubert Calculus with Isotropic Flags We previously obtained a congruence modulo four for the number of real solutions to many Schubert problems on a square Grassmannian given by osculating flags. Here, we consider Schubert problems given by more general isotropic flags, and prove this congruence modulo four for the largest class of Schubert problems that could be expected to exhibit this congruence. Keywords:Lagrangian Grassmannian, Wronski map, Shapiro ConjectureCategories:14N15, 14P99

20. CMB 2017 (vol 60 pp. 225)

Bahmanpour, Kamal; Naghipour, Reza
 Faltings' Finiteness Dimension of Local Cohomology Modules Over Local Cohen-Macaulay Rings Let $(R, \frak m)$ denote a local Cohen-Macaulay ring and $I$ a non-nilpotent ideal of $R$. The purpose of this article is to investigate Faltings' finiteness dimension $f_I(R)$ and equidimensionalness of certain homomorphic image of $R$. As a consequence we deduce that $f_I(R)=\operatorname{max}\{1, \operatorname{ht} I\}$ and if $\operatorname{mAss}_R(R/I)$ is contained in $\operatorname{Ass}_R(R)$, then the ring $R/ I+\cup_{n\geq 1}(0:_RI^n)$ is equidimensional of dimension $\dim R-1$. Moreover, we will obtain a lower bound for injective dimension of the local cohomology module $H^{\operatorname{ht} I}_I(R)$, in the case $(R, \frak m)$ is a complete equidimensional local ring. Keywords:Cohen Macaulay ring, equidimensional ring, finiteness dimension, local cohomologyCategories:13D45, 14B15

21. CMB 2017 (vol 61 pp. 166)

Miranda-Neto, Cleto B.
 A module-theoretic characterization of algebraic hypersurfaces In this note we prove the following surprising characterization: if $X\subset {\mathbb A}^n$ is an (embedded, non-empty, proper) algebraic variety defined over a field $k$ of characteristic zero, then $X$ is a hypersurface if and only if the module $T_{{\mathcal O}_{{\mathbb A}^n}/k}(X)$ of logarithmic vector fields of $X$ is a reflexive ${\mathcal O}_{{\mathbb A}^n}$-module. As a consequence of this result, we derive that if $T_{{\mathcal O}_{{\mathbb A}^n}/k}(X)$ is a free ${\mathcal O}_{{\mathbb A}^n}$-module, which is shown to be equivalent to the freeness of the $t$th exterior power of $T_{{\mathcal O}_{{\mathbb A}^n}/k}(X)$ for some (in fact, any) $t\leq n$, then necessarily $X$ is a Saito free divisor. Keywords:hypersurface, logarithmic vector field, logarithmic derivation, free divisorCategories:14J70, 13N15, 32S22, 13C05, 13C10, 14N20, , , , , 14C20, 32M25

22. CMB 2017 (vol 60 pp. 329)

Le Fourn, Samuel
 Nonvanishing of Central Values of $L$-functions of Newforms in $S_2 (\Gamma_0 (dp^2))$ Twisted by Quadratic Characters We prove that for $d \in \{ 2,3,5,7,13 \}$ and $K$ a quadratic (or rational) field of discriminant $D$ and Dirichlet character $\chi$, if a prime $p$ is large enough compared to $D$, there is a newform $f \in S_2(\Gamma_0(dp^2))$ with sign $(+1)$ with respect to the Atkin-Lehner involution $w_{p^2}$ such that $L(f \otimes \chi,1) \neq 0$. This result is obtained through an estimate of a weighted sum of twists of $L$-functions which generalises a result of Ellenberg. It relies on the approximate functional equation for the $L$-functions $L(f \otimes \chi, \cdot)$ and a Petersson trace formula restricted to Atkin-Lehner eigenspaces. An application of this nonvanishing theorem will be given in terms of existence of rank zero quotients of some twisted jacobians, which generalises a result of Darmon and Merel. Keywords:nonvanishing of $L$-functions of modular forms, Petersson trace formula, rank zero quotients of jacobiansCategories:14J15, 11F67

23. CMB 2016 (vol 60 pp. 510)

Haase, Christian; Hofmann, Jan
 Convex-normal (Pairs of) Polytopes In 2012 Gubeladze (Adv. Math. 2012) introduced the notion of $k$-convex-normal polytopes to show that integral polytopes all of whose edges are longer than $4d(d+1)$ have the integer decomposition property. In the first part of this paper we show that for lattice polytopes there is no difference between $k$- and $(k+1)$-convex-normality (for $k\geq 3$) and improve the bound to $2d(d+1)$. In the second part we extend the definition to pairs of polytopes. Given two rational polytopes $P$ and $Q$, where the normal fan of $P$ is a refinement of the normal fan of $Q$. If every edge $e_P$ of $P$ is at least $d$ times as long as the corresponding face (edge or vertex) $e_Q$ of $Q$, then $(P+Q)\cap \mathbb{Z}^d = (P\cap \mathbb{Z}^d ) + (Q \cap \mathbb{Z}^d)$. Keywords:integer decomposition property, integrally closed, projectively normal, lattice polytopesCategories:52B20, 14M25, 90C10

24. CMB 2016 (vol 60 pp. 613)

Reichstein, Zinovy; Vistoli, Angelo
 On the Dimension of the Locus of Determinantal Hypersurfaces The characteristic polynomial $P_A(x_0, \dots, x_r)$ of an $r$-tuple $A := (A_1, \dots, A_r)$ of $n \times n$-matrices is defined as $P_A(x_0, \dots, x_r) := \det(x_0 I + x_1 A_1 + \dots + x_r A_r) \, .$ We show that if $r \geqslant 3$ and $A := (A_1, \dots, A_r)$ is an $r$-tuple of $n \times n$-matrices in general position, then up to conjugacy, there are only finitely many $r$-tuples $A' := (A_1', \dots, A_r')$ such that $p_A = p_{A'}$. Equivalently, the locus of determinantal hypersurfaces of degree $n$ in $\mathbf{P}^r$ is irreducible of dimension $(r-1)n^2 + 1$. Keywords:determinantal hypersurface, matrix invariant, $q$-binomial coefficientCategories:14M12, 15A22, 05A10

25. CMB 2016 (vol 59 pp. 865)

Pal, Sarbeswar
 Moduli of Rank 2 Stable Bundles and Hecke Curves Let $X$ be smooth projective curve of arbitrary genus $g \gt 3$ over complex numbers. In this short note we will show that the moduli space of rank $2$ stable vector bundles with determinant isomorphic to $L_x$, where $L_x$ denote the line bundle corresponding to a point $x \in X$ is isomorphic to certain lines in the moduli space of S-equivalence classes of semistable bundles of rank 2 with trivial determinant. Keywords:Hecke curve, (0,1) stable bundleCategory:14D21
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text:
## Stream: new members

### Topic: Apply function on both sides of equality

#### Pedro Minicz (May 19 2020 at 02:17):

Say I have a = b * c as my goal, with a b c : ℝ and h : a / c = b as my hypothesis. How do I apply λ x, x * c to both sides of h? There was a thread asking this very question a few days ago, but I didn't manage to find it.

#### Yury G. Kudryashov (May 19 2020 at 02:19):

https://leanprover-community.github.io/mathlib_docs/tactics.html#apply_fun

#### Pedro Minicz (May 19 2020 at 02:24):

Thank you! Are there any alternatives?

#### Yury G. Kudryashov (May 19 2020 at 02:26):

replace h := congr_arg (λ x, x * c) h

#### Yury G. Kudryashov (May 19 2020 at 02:28):

In this particular case you can also apply a lemma (search with #library_search or guess name like eq_mul_of_div_eq - not tested)

#### Reid Barton (May 19 2020 at 02:44):

You can also rw <- h for a similar result

Last updated: May 12 2021 at 05:19 UTC


text:
A question on a proof that every sequence has a monotone subsequence

Consider the following proof that every sequence $\{x_n\}_n$ in $\mathbb R$ has a monotone subsequence.

Proof: Let us call a positive integer $n$ a peak of the sequence if $m > n \implies x_n > x_m$  i.e., if  $x_n$ is greater than every subsequent term in the sequence.

Suppose first that the sequence has infinitely many peaks, $n_1 < n_2 < n_3 < … < n_j < …$. Then the subsequence $\{x_{n_j}\}_j$  corresponding to these peaks is monotonically decreasing, and we are done.

So suppose now that there are only finitely many peaks, let $N$ be the last peak and set $n_1 = N + 1$.

Then $n_1$ is not a peak, since $n_1 > N$, which implies the existence of an $n_2 > n_1$ with $x_{n_2} \geq x_{n_1}.$  Again, as $n_2 > N$ it is not a peak, hence there is $n_3 > n_2$ with $x_{n_3} \geq x_{n_2}.$  Repeating this process leads to an infinite non-decreasing subsequence $x_{n_1} \leq x_{n_2} \leq x_{n_3} \leq \ldots$ as desired.

In the second case, if $n_1$ is not a peak, how does this imply the existence of an $n_2 > n_1$ with $x_{n_2} \geq x_{n_1}$? Can't it be possible that there does not exist any $n_2 > n_1$ such that $x_{n_2} \geq x_{n_1}$?

A possible example can be the sequence $\{\sin n\}_n$, taking $N$ to be the closest integer to $\pi/2$

• that is by assumption ($n_1$ not a peak) and definition. Just evaluate the statement $n_1$ is not a peak (negation of $n_1$ is a peak). – Thomas Sep 12 '14 at 6:15
• The integer closest to π/2 is not a peak for {sin n}, sin 2 < sin 8. I am not conviced that this sequence has any peaks - ie. I think sin n, with integer n, can approximate 1 arbitrary well. – Taemyr Sep 12 '14 at 11:04

If you have a statement "for all $x$, $P(x)$ is true", then the negation of this statement is "there exists an$x$ such that $P(x)$ is not true". (the negation of $\forall x: P(x)$ is $\exists x: \neg P(x)$).

$n_1$ is not a peak. The definition of $n_1$ being a peak is:

$$\forall n> n_1: x_{n_1}>x_n$$ meaning that by definition, $n_1$ is not a peak iff $$\exists n > n_1: x_n\leq x_{n-1}$$ which is exactly what you have.

Here is a tedious proof that avoids dealing with peaks:

Let $\bar{x} = \limsup_n x_n$.

If $\bar{x} = \infty$, then let $n_1 = 1$, and $n_{k+1} = \min \{n | n \ge n_k+1,\ x_n \ge x_{n_k} +1 \}$, then $x_{n_k}$ is a monotone sequence.

Otherwise, suppose $\bar{x} < \infty$.

Suppose $\{x_n\} \cap (\bar{x},\infty)$ contains an infinite subset. Let $x_{n_k}$ be the subsequence of $x_n$ that lies in $(\bar{x},\infty)$. Then $x_{n_k} > \bar{x}$, and $\lim_k x_{n_k} = \bar{x}$. Let $m_1 = n_{1}$, and $m_{k+1} = \min \{n_j | n_j \ge m_k+1,\ x_{n_j} < x_{m_k} \}$, then $x_{m_k}$ is a monotone sequence.

Now suppose $\{x_n\} \cap (\bar{x},\infty)$ is finite.

Suppose $\{x_n\} \cap \{\bar{x}\}$ contains an infinite subset, then clearly this subsequence is a monotone (in fact, constant) subsequence.

Now suppose $\{x_n\} \cap [\bar{x},\infty)$ is finite. Then $x_n < \bar{x}$ (after a finite number of terms), and there exists a subsequence $x_{n_k}$ such that $\lim_k x_{n_k} = \bar{x}$. Similar to above, let $m_1 = n_1$, and $m_{k+1} = \min \{n_j | n_j \ge m_k+1,\ x_{n_j} > x_{m_k} \}$, then $x_{m_k}$ is a monotone sequence.


text:
• # Frame Rate Independent Movement

General and Gameplay Programming

I've seen countless posts on this and other message boards, and even personal email has been sent to me, all asking the same question: How do I make it so my stupid objects move at the same speed if the frame rate rises or drops?

After a few months' programming, I devised what I believe to be the most efficient and useful way to handle this problem. For the purposes of this article, I will encapsulate this functionality in its own class.

class framerate { public: float targetfps; float fps; LARGE_INTEGER tickspersecond; LARGE_INTEGER currentticks; LARGE_INTEGER framedelay; float speedfactor; void Init(float tfps); void SetSpeedFactor(); };
I leave all the members public because they work independently of each other and are all useful in some way to the outside program.

Here is the implementation, followed by some explanation

void framerate::Init(float tfps) { targetfps = tfps; QueryPerformanceCounter(&framedelay); QueryPerformanceFrequency(&tickspersecond); } void framerate::SetSpeedFactor() { QueryPerformanceCounter(?tticks); //This frame's length out of desired length speedfactor = (float)(currentticks.QuadPart-framedelay.QuadPart)/((float)tickspersecond.QuadPart/targetfps); fps = targetfps/speedfactor; if (speedfactor <= 0) speedfactor = 1; framedelay = currentticks; }
Some explanation:
targetfps is passed into the Init function, but can be set at any time. It is the target frame rate for the program. This is used in the SetSpeedFactor function to determine what the speedfactor will be. Look down at my explanation of speedfactor.

fps is the actual frame rate of the program. It is not really needed to make frame rate independent movement, but since it is so closely linked, I include it anyway.

tickspersecond is set in the Init function to be the number of ticks the high performance timer has per second.

currentticks is set in the SetSpeedFactor function to be the current high performance timer's ticks.

framedelay is the previous frame's currentticks.

speedfactor is the heart of this class. When it is set is SetSpeedFactor, it becomes a number that you multiply all your motion by. For instance, if the targetfps is 100, and the actual fps is 85, the speedfactor will be set to 100/85, or about 1.2. You then multiply all your motion is the game, at its lowest level, by this number.

For instance, rather than simply coding spaceship->MoveForward(5), I would code spaceship->MoveForward(5*framerate.speedfactor).

In conclusion:
This simple routine saves a whole heap of trouble. Just plug it into almost any game or other real-time program. Call Init at the beginning of the program, and then each frame call SetSpeedFactor. Then multiply all your movement by the speedfactor. It's that simple. Now stop flooding those message boards

Questions? Comments? Complaints? Hate mail? [email="benbeandogdilts@cs.com"]benbeandogdilts@cs.com[/email]

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A Direct Parametric Approach to Spacecraft Attitude Tracking Control

Xiao-Yi Wang and Guang-Ren Duan. A Direct Parametric Approach to Spacecraft Attitude Tracking Control. International Journal of Automation and Computing, vol. 14, no. 5, pp. 626-636, 2017 doi:  10.1007/s11633-017-1089-4
 Citation: Xiao-Yi Wang and Guang-Ren Duan. A Direct Parametric Approach to Spacecraft Attitude Tracking Control. International Journal of Automation and Computing, vol. 14, no. 5, pp. 626-636, 2017
##### doi: 10.1007/s11633-017-1089-4

National Natural Science Foundation of China 61321062

## A Direct Parametric Approach to Spacecraft Attitude Tracking Control

Funds:

National Natural Science Foundation of China 61321062

###### Corresponding author:Xiao-Yi Wang received the B. Sc. degree from Department of Control Science and Engineering, Harbin Institute of Technology, China in 2014. She received the M. Sc. degree in Centre for Control Theory and Guidance Technology, Harbin Institute of technology, China in 2017.     Her research interests include nonlinear robust control and spacecraft attitude control.     E-mail:xiaoyiwang51@outlook.com (Corresponding author)     ORCID iD:0000-0002-8511-1273
• Figure  1.  Error attitude quaternion ${q} _e$

Figure  2.  Response curve of variable ${e}$ and $\dot{e}$ of controlled system

Figure  3.  Comparison between controlled spacecraft attitude and target spacecraft attitude

Figure  4.  Controller ${u}$ with respect to time curve

Figure  5.  Response curve of variable ${e}$ and $\dot{e}$ of controlled system

Figure  6.  Curve of the state error between practical system and theoretical system

•  [1] G. R. Duan. Quaternion-based satellite attitude controlA direct parametric approach. In Proceedings of the 14th International Conference on Control, Automation and Systems, IEEE, Seoul, Korea, pp. 935-941, 2014. http://www.nsfc.gov.cn/Portals/0/fj/fj20160106_01.xls [2] G. R. Duan. Satellite attitude control-A direct parametric approach. In Proceedings of the 11th World Congress on Intelligent Control and Automation, IEEE, Shenyang, China, pp. 3989-3996, 2014. http://ieeexplore.ieee.org/xpl/abstractKeywords.jsp?reload=true&arnumber=7053383 [3] C. Y. Gao, Q. Zhao, G. R. Duan. Robust actuator fault diagnosis scheme for satellite attitude control systems. Journal of the Franklin Institute, vol. 350, no. 9, pp. 2560-2580, 2013. [4] A. G. Parlos, J. W. Sunkel. Adaptive attitude control and momentum management for large-angle spacecraft maneuvers. Journal of Guidance, Control and Dynamics, vol. 15, no. 4, pp. 1018-1028, 1992. doi:  10.2514/3.20937 [5] H. Bang, C. K. Ha, J. Hyoung Kim. Flexible spacecraft attitude maneuver by application of sliding mode control. Acta Astronautica, vol. 57, no. 11, pp. 841-850, 2005. [6] S. N. Singh, R. Zhang. Adaptive output feedback control of spacecraft with flexible appendages by modeling error compensation. Acta Astronautica, vol. 54, no. 4, pp. 229-243, 2004. [7] A. Tayebi. Unit quaternion-based output feedback for the attitude tracking problem. IEEE Transactions on Automatic Control, vol. 53, no. 6, pp. 1516-1520, 2008. [8] D. Bustan, S. K. H. Sani, N. Pariz. Adaptive faulttolerant spacecraft attitude control design with transient response control. IEEE/ASME Transactions on Mechatronics, vol. 19, no. 4, pp. 1404-1411, 2014. [9] R. Kristiansen, P. J. Nicklasson. Satellite attitude control by quaternion-based backstepping. In Proceedings of American Control Conference, IEEE, Portland, USA, vol. 2, pp. 907-912, 2005. https://www.infona.pl//resource/bwmeta1.element.ieee-art-000004554017 [10] D. Doody. Spacecraft attitude control. Deep Space Craft, D. Doody, Ed., Berlin Heidelberg, Germany:Springer, pp. 87-118, 2009. [11] A. Grewal, V. J. Modi. Robust attitude and vibration control of the space station. Acta Astronautica, vol. 38, no. 3, pp. 139-160, 1996. [12] K. F. Lu, Y. Q. Xia. Adaptive attitude tracking control for rigid spacecraft with finite-time convergence. Automatica, vol. 49, no. 12, pp. 3591-3599, 2013. [13] A. L. Herman, B. A. Conway. Optimal spacecraft attitude control using collocation and nonlinear programming. Journal of Guidance, Control and Dynamics, vol. 15, no. 5, pp. 1287-1289, 1992. doi:  10.2514/3.20983 [14] B. Wie, K. W. Byun, V. W. Warren, D. Geller, D. Long, J. Sunkel. New approach to attitude/momentum control for the Space Station. Journal of Guidance, Control and Dynamics, vol. 12, no. 5, pp. 714-722, 1989. doi:  10.2514/3.20466 [15] J. D. Schierman, D. G. Ward, J. R. Hull, N. Gandhi, M. Oppenheimer, D. B. Doman. Integrated adaptive guidance and control for re-entry vehicles with flight test results. Journal of Guidance, Control and Dynamics, vol. 27, no. 6, pp. 975-988, 2004. doi:  10.2514/1.10344 [16] H. J. Shen, P. Tsiotras. Optimal two-impulse rendezvous using multiple-revolution lambert solutions. Journal of Guidance, Control and Dynamics, vol. 26, no. 1, pp. 50-61, 2003. doi:  10.2514/2.5014 [17] F. Zhang, G. R. Duan. Optimal integrated relative position and attitude control of spacecraft in proximity operation missions. International Journal of Automation and Computing, vol. 9, no. 4, pp. 342-351, 2012. [18] H. Du, G. L. Fan, J. Q. Yi. Nonlinear longitudinal attitude control of an unmanned seaplane with wave filtering. International Journal of Automation and Computing, vol. 13, no. 6, pp. 634-642, 2016. [19] W. K. Liu. Research on Attitude Maneuver and Tracking for Flexible Spacecraft, Master dissertation, Harbin Institute of Technology, China, 2014. (in Chinese) [20] L. Zhang, G. R. Duan, Y. A. Zhang. Nonlinear dynamic model and control of three-axis stabilized liquid-filled spacecraft attitude system. Journal of Harbin Institute of Technology (New Series), vol. 19, no. 5, pp. 107-112, 2012. http://www.cqvip.com/QK/86045X/201205/43848684.html [21] G. R. Duan. Direct parametric control of fully-actuated second-order nonlinear systems-The normal case. In Proceedings of the 33rd Chinese Control Conference, IEEE, Nanjing, China, pp. 2406-2413, 2014. http://www.nsfc.gov.cn/Portals/0/fj/fj20160106_01.xls [22] G. R. Duan, B. Zhou. Solution to the second-order Sylvester matrix equation MVF2+ DVF + KV=BW. IEEE Transactions on Automatic Control, vol. 51, no. 5, pp. 805-809, 2006. [23] G. R. Duan. Generalized Sylvester Equations:Unified Parametric Solutions, New York, USA:CRC Press, pp. 15-393, 2015. [24] G. R. Duan. Parametric eigenstructure assignment in second-order descriptor linear systems. IEEE Transactions on Automatic Control, vol. 49, no. 10, pp. 1789-1794, 2004.

##### 计量
• 文章访问数:  641
• HTML全文浏览量:  355
• PDF下载量:  9
• 被引次数: 0
##### 出版历程
• 收稿日期:  2016-08-19
• 录用日期:  2017-03-29
• 网络出版日期:  2017-07-03
• 刊出日期:  2017-10-01

## A Direct Parametric Approach to Spacecraft Attitude Tracking Control

##### doi: 10.1007/s11633-017-1089-4
基金项目:

National Natural Science Foundation of China 61321062

### English Abstract

Xiao-Yi Wang and Guang-Ren Duan. A Direct Parametric Approach to Spacecraft Attitude Tracking Control. International Journal of Automation and Computing, vol. 14, no. 5, pp. 626-636, 2017 doi:  10.1007/s11633-017-1089-4
 Citation: Xiao-Yi Wang and Guang-Ren Duan. A Direct Parametric Approach to Spacecraft Attitude Tracking Control. International Journal of Automation and Computing, vol. 14, no. 5, pp. 626-636, 2017

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text:
# [texhax] Displaying the \current value of \prevgraf in the log file

Paul Isambert zappathustra at free.fr
Mon Nov 14 09:14:46 CET 2011

Le 13/11/2011 20:01, Philip TAYLOR a écrit :
>
>
> Paul Isambert wrote:
>
>> prevgraf. For instance "abc\par\showlists\bye" produces:
>>
>> ### vertical mode entered at line 0
>> ### current page:
>> ... blah blah ...
>> total height 10.0
>> goal height 643.20255
>> prevdepth 0.0, prevgraf 1 line
>
> Oh ... I tried something similar, and saw
> prevdepth but no prevgraf -- let me try
> your example and see how it differs ...
>
> Yes, you're quite right : \showlists after
> a non-empty paragraph and a \par does indeed
> show the current value of \prevgraf.  Unfortunately
> it doesn't help me to find the value at a \vadjust !

abc\par abc\vadjust{abc\par\showlists}\bye % Prevgraf, but of course, of
the first paragraph in the \vadjust

But I suppose you're interested in the the first line?
Apparently, \showlists displays prevgraf when non-zero, i.e. only in
vmode after a paragraph (which is also how I understand the code from
tex.web).

So, at the beginning of a \vadjust (i.e. before a paragraph), even if
you retrieved \prevgraf, it would be 0, as it is in horizontal mode or
in vertical mode in an empty list. Your problem is thus rather to make
\prevgraf survive a little bit longer; apart from hacking \par to
something like \def\par{\endgraf\mycount=\prevgraf}, I can't see a
solution. Unless you're willing to go LuaTeX, in which case the \par
hack can be made a little bit more elegant (and less dangerous) with:

callback.register("post_linebreak_filter",
function (h)
tex.count.mycount = tex.nest[tex.nest.ptr].prevgraf
return h
end)

Best,
Paul


text:
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# Running at their respective constant rates, machine X takes

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Intern
Joined: 19 Aug 2016
Posts: 42

Kudos [?]: 2 [1], given: 1

Re: Running at their respective constant rates, machine X takes [#permalink]

### Show Tags

27 Sep 2016, 09:53
1
KUDOS
I got the values of x=6 and x=4/5 by quadratic equation. We use 6 to solve the question. Why can't we use 4/5
Is there an explanation for this?

Kudos [?]: 2 [1], given: 1

Intern
Joined: 17 May 2016
Posts: 31

Kudos [?]: 4 [0], given: 38

Re: Running at their respective constant rates, machine X takes [#permalink]

### Show Tags

08 Jun 2017, 01:09
Bunuel wrote:

Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.
A. 4
B. 6
C. 8
D. 10
E. 12

For work problems one of the most important thin to know is $$rate*time=job \ done$$.

Let the time needed for machine X to produce $$w$$ widgets be $$t$$ days, so the rate of X would be $$rate=\frac{job \ done}{time}=\frac{w}{t}$$;

As "machine X takes 2 days longer to produce $$w$$ widgets than machines Y" then time needed for machine Y to produce $$w$$ widgets would be $$t-2$$ days, so the rate of Y would be $$rate=\frac{job \ done}{time}=\frac{w}{t-2}$$;

Combined rate of machines X and Y in 1 day would be $$\frac{w}{t}+\frac{w}{t-2}$$ (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: $$3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4}$$ --> $$\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}$$.

$$\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}$$ --> reduce by $$w$$ --> $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$.

At this point we can either solve quadratic equation: $$5t^2-34t+24=0$$ --> $$(t-6)(5t-4)=0$$ --> $$t=6$$ or $$t=\frac{4}{5}$$ (which is not a valid solution as in this case $$t-2=-\frac{6}{5}$$, the time needed for machine Y to ptoduce $$w$$ widgets will be negatrive value and it's not possible). So $$t=6$$ days is needed for machine X to produce $$w$$ widgets, hence time needed for machine X to produce $$2w$$ widgets will be $$2t=12$$ days.

Hi Bunuel,

Could you please explain how do you solve the quadratic in one step ? I don't understand how can we factorize it with a coefficient greater than 1 on the $$x^2$$.
When trying to solve with the quadratic formula, it involves big numbers with lot of calculation and requires lot of time.

Thank you so much for your help,
Sincerly yours,

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Re: Running at their respective constant rates, machine X takes [#permalink]

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10 Jun 2017, 08:10
heyholetsgo wrote:
Running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.

A. 4
B. 6
C. 8
D. 10
E. 12

We are given that running at their respective constant rates, machine X takes 2 days longer to produce w widgets than machine Y does. If we let t = the time in days that it takes machine Y to produce w widgets, then t + 2 = the time in days that it takes machine X to produce w widgets. Furthermore, we can say the following:

rate of machine Y = w/t

rate of machine X = w/(t + 2)

We are given that the machines produce 5w/4 widgets in 3 days. Since work = rate x time and each machine works for 3 days, we first calculate the work done by machine X and machine Y individually.

work of machine Y = (w/t) x 3 = 3w/t

work of machine X = w/(t + 2) x 3 = 3w/(t + 2)

Since the machines work together to produce 5w/4 widgets, we can sum their work and set that sum to 5w/4.

3w/t + 3w/(t + 2) = 5w/4

Divide the entire equation by w and we have:

3/t + 3/(t + 2) = 5/4

Now, multiplying the entire equation by 4t(t + 2) to eliminate the denominators in the equation, we obtain:

3[4(t + 2)] + 3(4t) = 5[t(t + 2)]

12t + 24 + 12t = 5t^2 + 10t

5t^2 - 14t - 24 = 0

(5t + 6)(t - 4) = 0

t = -6/5 or t = 4

Since t cannot be negative, t must equal 4. That is, it takes machine Y 4 days to produce w widgets. Thus, it will take machine X 6 days to produce w widgets and 12 days to produce 2w widgets.

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Re: Running at their respective constant rates, machine X takes [#permalink]

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03 Oct 2017, 18:25
Bunuel wrote:
heyholetsgo wrote:
Thanks a lot, this was very helpful! I know how to factorize but as soon as it looks like this I can"t handle it anymore 5t^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

For our original question you can also use substitution method: we $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$ and we know that answer would be $$2t$$. Try to substitute half of the values listed in the answer choices and you'll see that answer choice E will work.

heyholetsgo wrote:
And is this a formula to compute the times 2 machines need to finish the same job? I always thought the times of the machines do NOT add up in work problems..
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)

If:
Time needed for A to complete the job =A hours;
Time needed for B to complete the job =B hours;
Time needed for C to complete the job =C hours;
...
Time needed for N to complete the job =N hours;

Then if time needed for all of them working simultaneously to complete the job is $$T$$, then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+..+\frac{1}{N}=\frac{1}{T}$$ (General formula).

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$a$$ and $$b$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{a*b}{a+b}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{a}+\frac{1}{b}=\frac{1}{t}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{a*b*c}{ab+ac+bc}$$ hours.

Hope it helps.

but
shouldn't we use the same formula for rate??!!!
the reciprocal of rates shouldn't be added together?

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Re: Running at their respective constant rates, machine X takes [#permalink]

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03 Oct 2017, 20:56
soodia wrote:
Bunuel wrote:
heyholetsgo wrote:
Thanks a lot, this was very helpful! I know how to factorize but as soon as it looks like this I can"t handle it anymore 5t^2-34t+24=0 After I found the factors for 24, there is no way getting around trial and error, is there?

For our original question you can also use substitution method: we $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$ and we know that answer would be $$2t$$. Try to substitute half of the values listed in the answer choices and you'll see that answer choice E will work.

heyholetsgo wrote:
And is this a formula to compute the times 2 machines need to finish the same job? I always thought the times of the machines do NOT add up in work problems..
1/T = 1/T1 + 1/T2 or T=(T1*T2)/(T1+T2)

If:
Time needed for A to complete the job =A hours;
Time needed for B to complete the job =B hours;
Time needed for C to complete the job =C hours;
...
Time needed for N to complete the job =N hours;

Then if time needed for all of them working simultaneously to complete the job is $$T$$, then: $$\frac{1}{A}+\frac{1}{B}+\frac{1}{C}+..+\frac{1}{N}=\frac{1}{T}$$ (General formula).

For two and three entities (workers, pumps, ...):

General formula for calculating the time needed for two workers A and B working simultaneously to complete one job:

Given that $$a$$ and $$b$$ are the respective individual times needed for $$A$$ and $$B$$ workers (pumps, ...) to complete the job, then time needed for $$A$$ and $$B$$ working simultaneously to complete the job equals to $$T_{(A&B)}=\frac{a*b}{a+b}$$ hours, which is reciprocal of the sum of their respective rates ($$\frac{1}{a}+\frac{1}{b}=\frac{1}{t}$$).

General formula for calculating the time needed for three A, B and C workers working simultaneously to complete one job:

$$T_{(A&B&C)}=\frac{a*b*c}{ab+ac+bc}$$ hours.

Hope it helps.

but
shouldn't we use the same formula for rate??!!!
the reciprocal of rates shouldn't be added together?

We ARE adding the rates. A, B and C above are times. The rate is reciprocal of time, so 1/A, 1/B and 1/C are rates.
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Re: Running at their respective constant rates, machine X takes [#permalink]

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04 Oct 2017, 19:59
nickimonckom wrote:
Bunuel wrote:

Running at their respective constant rate, machine X takes 2 days longer to produce w widgets than machines Y. AT these rates, if the two machines together produce 5w/4 widgets in 3 days, how many days would it take machine X alone to produce 2w widgets.
A. 4
B. 6
C. 8
D. 10
E. 12

For work problems one of the most important thin to know is $$rate*time=job \ done$$.

Let the time needed for machine X to produce $$w$$ widgets be $$t$$ days, so the rate of X would be $$rate=\frac{job \ done}{time}=\frac{w}{t}$$;

As "machine X takes 2 days longer to produce $$w$$ widgets than machines Y" then time needed for machine Y to produce $$w$$ widgets would be $$t-2$$ days, so the rate of Y would be $$rate=\frac{job \ done}{time}=\frac{w}{t-2}$$;

Combined rate of machines X and Y in 1 day would be $$\frac{w}{t}+\frac{w}{t-2}$$ (remember we can sum the rates). In 3 days two machines together produce 5w/4 widgets so: $$3(\frac{w}{t}+\frac{w}{t-2})=\frac{5w}{4}$$ --> $$\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}$$.

$$\frac{w}{t}+\frac{w}{t-2}=\frac{5w}{12}$$ --> reduce by $$w$$ --> $$\frac{1}{t}+\frac{1}{t-2}=\frac{5}{12}$$.

At this point we can either solve quadratic equation: $$5t^2-34t+24=0$$ --> $$(t-6)(5t-4)=0$$ --> $$t=6$$ or $$t=\frac{4}{5}$$ (which is not a valid solution as in this case $$t-2=-\frac{6}{5}$$, the time needed for machine Y to ptoduce $$w$$ widgets will be negatrive value and it's not possible). So $$t=6$$ days is needed for machine X to produce $$w$$ widgets, hence time needed for machine X to produce $$2w$$ widgets will be $$2t=12$$ days.

Hi Bunuel,

Could you please explain how do you solve the quadratic in one step ? I don't understand how can we factorize it with a coefficient greater than 1 on the $$x^2$$.
When trying to solve with the quadratic formula, it involves big numbers with lot of calculation and requires lot of time.

Thank you so much for your help,
Sincerly yours,

Hi...

we can factorize by taking the product of a and c...so 5*24=120

So 30*4=120

5x^2-30x-4x+24=0

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Re: Running at their respective constant rates, machine X takes   [#permalink] 04 Oct 2017, 19:59

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