text:
# Spatial Velocity vs. Body Velocity

As I have been learning the mathematics behind robot kinematics, I have been struggling to understand the difference between a spatial velocity and a body velocity. In this post, I am going to try to write a good definition of each of these velocity types.

## Basic Definitions

According to Murray, Li and Sastry (MLS), the definitions are as follows [1]:

• Spatial velocity:- mathematically defined as $$\hat{V}^s = \dot{R}R^T = \dot{g} g^{-1}$$. MLS says that the spatial velocity of a rigid motion is the instantaneous velocity of the body as viewed in the spatial frame. It is the “velocity of a (possibly imaginary) point on the rigid body which is traveling through the origin of the spatial frame at time t” [1]. So if you are standing at the origin of a spatial frame and you measure the velocity of a point attached to the rigid body and going through the origin where you are standing, that is the spatial velocity [1].

• Body velocity:- mathematically defined as $$\hat{V}^b = R^T \dot{R} = g^{-1} \dot{g}$$. MLS says that the body velocity is more “straightforward” to understand; it is the “velocity of the origin of the body coordinate frame relative to the spatial frame, as viewed in the current body frame” [1]. MLS points out that the body velocity is not the velocity of the body relative to the body frame, because that is always zero [1]!

## Body Velocity

These definitions were a little confusing to me at first. Let’s take a step back and think about this with some more intuition. Let’s start with the body velocity, because I can relate that easily to the experience of riding in a car. If you imagine that you are riding in a car at a constant velocity, then the only way you know that you are moving is if you compare your frame of reference (the body frame, B) with some fixed frame (the spatial frame, A). The body velocity is your instantaneous velocity as measured with respect to this fixed frame, A, but observed from your perspective inside the car’s body frame, B. So when your car’s speedometer tells you that you are moving at 60 MPH, that means that you are moving at 60 MPH with respect to the spatial frame, A, as observed from within the body frame, B [2].

We can define your body velocity in this instance as follows. Let’s assume that your position in the car is a point $$q$$ and so your body velocity is written as [2]:

$v_{q_b} = g_{ab}^{-1} v_{q_a}$

Where $$g_{ab}$$ is the transformation that converts a point in the body frame, B, to the spatial frame, A, and $$v_{q_a}$$ is the velocity of the point $$q$$ in the spatial frame, A. Another way to write $$v_{q_a}$$ is $$v_{q_a} = \dot{g}_{ab} q_{b}$$ *1 so we can substitute that in:

$v_{q_b} = g_{ab}^{-1} \dot{g}_{ab} q_{b}$

Now I can write my body velocity as a function of some point $$q_b$$ in the body frame that has velocity with respect to the spatial frame, A, as viewed in the body frame, B [2].

$v_{q_b} = \hat{V}_{ab}^b q_{b}$

In other words:

$\hat{V}_{ab}^b = g_{ab}^{-1} \dot{g}_{ab}$

## Spatial Velocity

The difference between the spatial velocity and the body velocity is that, instead of observing your velocity in the car in the body frame, B, you are observing it from the origin of the spatial frame, A. That is, the spatial velocity measures the velocity of a point fixed to the body frame, B, with respect to the spatial frame, A, and observed from the spatial frame, A [2]. I can write the spatial velocity as follows [2]:

$v_{q_a} = \dot{g}_{ab} q_b$

I can rewrite $$q_b$$ to be in terms of $$q_a$$ as follows [2]:

$q_b = g_{ab}^{-1} q_a$

And therefore I can write the spatial velocity as [2]:

$v_{q_a} = \dot{g}_{ab} g_{ab}^{-1} q_a$

And so we see that [2]:

$\hat{V}_{ab}^s = \dot{g}_{ab} g_{ab}^{-1}$

I hope this helps a little bit with getting to grips with the differences between a spatial velocity and a body velocity. I think that it all comes down to where you are observing the velocity from.

## Footnotes:

*1 This is by the definition of the rigid motion mapping, that is [2]:

$q_{a} = g_{ab} q_b$

And the derivative is [2]:

$\dot{q}_a = \dot{g}_{ab} q_b$

## References:

[1] Murray, R., Li, Z., Sastry, S. “A Mathematical Introduction to Robotic Manipulation.” CRC Press. 1994.

[2] Sastry, S. “EE106A Discussion 7: Velocities and Adjoints.” EE C106A: Introduction to Robotics, Fall 2019 course notes. UC Berkeley. https://ucb-ee106.github.io/ee106a-fa19/assets/discussions/D7___Velocities_and_Adjoints.pdf Visited 10 Jun 2021.

Written on June 10, 2021


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+0

# Algebra

0
89
1

Find the smallest value of x such that x^2 + 10x + 25 = 18.

Jun 2, 2022

#1
+2448
0

Factor the left hand side to: $$(x+5)^2 = 18$$

Taking the square root of both sides gives us: $$x + 5 = \pm \sqrt {18}$$.

Can you take it from here?

Jun 2, 2022

format: plain


text:
4.2: Use Unit Rates and Equivalent Rates

Difficulty Level: At Grade Created by: CK-12
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Vocabulary Language: English

TermDefinition
Denominator The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Equivalent Equivalent means equal in value or meaning.
Equivalent Rate Equivalent rates are rates that can each be simplified to the same ratio.

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text:
# The Reachable Space of the Heat Equation for a Finite Rod as a Reproducing Kernel Hilbert Space

Authors
• 1 Universidad Nacional Autónoma de México, Cuernavaca, MOR, 62251, Mexico , Cuernavaca (Mexico)
Type
Published Article
Journal
Integral Equations and Operator Theory
Publisher
Springer International Publishing
Publication Date
Jul 10, 2021
Volume
93
Issue
4
Identifiers
DOI: 10.1007/s00020-021-02660-6
Source
Springer Nature
Keywords
Disciplines
• Article
We use some results from the theory of reproducing kernel Hilbert spaces to show that the reachable space of the heat equation for a finite rod with either one or two Dirichlet boundary controls is a RKHS of analytic functions on a square, and we compute its reproducing kernel as an infinite double series. We also show that the null reachable space of the heat equation for the half line with Dirichlet boundary data is a RKHS of analytic functions on a sector, whose reproducing kernel is (essentially) the sum of pullbacks of the Bergman and Hardy kernels on the half plane C+\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {C}^+$$\end{document}.


text: • please check cv::getBuildInformation() , to see, if you actually have support for gstreamer builtin.


text:
# NSTimer not fired when uiscrollview event occurs

I have a UIImageView placed in UIScrollView, Basicly this UIImageView holds very big map, and created animation on a predefined path with "arrows" pointed navigation direction.

But, whenever uiscrollevents occurs, I think MainLoop freezes and NSTimer being not fired, and animation stopped.

Are there any existing property, which solves this problem, on UIScrollView, CAKeyFrameAnimation or NSTimer?

//viewDidLoad
self.myTimer = [NSTimer scheduledTimerWithTimeInterval:0.05 target:self selector:@selector(drawLines:) userInfo:nil repeats:YES];

- (void)drawLines:(NSTimer *)timer {

CALayer *arrow = [CALayer layer];
arrow.bounds = CGRectMake(0, 0, 5, 5);
arrow.position = CGPointMake(line.x1, line.y1);
arrow.contents = (id)([UIImage imageNamed:@"arrow.png"].CGImage);

CAKeyframeAnimation* animation = [CAKeyframeAnimation animation];
animation.path = path;
animation.duration = 1.0;
animation.rotationMode = kCAAnimationRotateAuto; // object auto rotates to follow the path
animation.repeatCount = 1;
animation.timingFunction = [CAMediaTimingFunction functionWithName:kCAMediaTimingFunctionLinear];
animation.fillMode = kCAFillModeForwards;

}

iOS Applications run on an NSRunLoop. Each NSRunLoop has different modes of execution for different tasks. For example, the default nstimer is scheduled to run under the NSDefaultRunMode on the NSRunLoop. What this means however is that certain UIEvents, scrollviewing being one, will interrupt the timer, and place it on a queue to be run as soon as the event stops updating. In your case, in order to get the timer to not be interrupted, you need to schedule it for a different mode, namely NSRunLoopCommonModes, like so:

  self.myTimer =  [NSTimer scheduledTimerWithTimeInterval:280
target:self
selector:@selector(doStuff)
userInfo:nil
repeats:NO];

This mode will allow your timer to not be interrupted by scrolling. You can find more about this info here: https://developer.apple.com/documentation/foundation/nsrunloop At the bottom you will see the definitions of the modes you can choose from. Also, legend has it, you can write your own custom modes, but few have ever lived to tell the tale im afraid.

• Hey @GregPrice. Thanks for the explanation, even 3 years after. However, the doc says scheduledTimerWithTimeInterval:... creates a timer and schedules it on the current run loop in the default mode. So why do you add it twice in the default RunLoop, not just once in NSRunLoopCommonModes mode? – Martin May 20 '15 at 12:00
• UIScrollView runs the run loop in UITrackingRunLoopMode, which is different than NSDefaultRunLoopMode. It's not enough to schedule the timer for the default mode if you also want it to run in other modes. – rob mayoff Mar 9 '16 at 22:38
• For Swift 3: RunLoop.current.add(self.myTimer, forMode: .commonModes) – Steve Cotner Dec 20 '16 at 22:36

One more thing (c)

1. use timerWithTimeInterval method in order to avoid adding timer to runloop in DefaultMode
2. use mainRunLoop

So:

self.myTimer = [NSTimer timerWithTimeInterval:280 target:self selector:@selector(doStuff) userInfo:nil repeats:NO]; [[NSRunLoop mainRunLoop] addTimer:self.myTimer forMode:NSRunLoopCommonModes];


text:
Discussion

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in this question of 30 g. Marble is moving to the right with the velocity off 0.2 m per second, while a 10 g marble is moving to the left with the velocity off 0.4 m per second. The first item off this question, we have to calculate what is the velocity off marbles? Numbers one and two after the collision, provided that the collision is elastic. For that, we have to use two equations. One is this equation, which is the equation for the relative velocities after and elastic collision when the two bodies are moving on. The other equation is the equation that comes from using the principle of conservation of momentum. So by using the principle of conservation of momentum, we have the following the net momentum before the collision is equals to the net momentum. After the collision before the collision, we have two marbles moving. Let me choose my reference frame to be such that everything that is pointing to the right, it's positive and everything that is pointing to the left is negative. Then, in this reference frame before the collision, the momentum was given by the some off the momentum off marble number one on the momentum off marble number two. The momentum off marble number one is given by M one times If you won the momentum off marble number two is given by M two times we chew on This is equals to the net momentum after the collision after the collision, both marbles still moving. So we have a one view on prime plus M two times V two Prime, where the primate velocities are the velocities after the collision. Now, by plugging in the values that the problem gives us, we got the following m one is a cost authority V one is plus 0.2 in our reference frame. Now we have minus because V Chu will be negative so minus M. Chu which is the cost of 10 g times 0.4 and this is equals two and one, which is 30 times view one prime plus M. Chu, which is 10 times V chew prime. Then, by working on the left hand side, we got the following 30 times View one prime plus 10 times V two prime. Is he, of course, to chew. Note that we have only one equation, but two variables. So we need another equation to be able to solve. This problem on this other equation comes from this one, which is the equation for the relative velocities. So let me call this equation. We had just arrived. Equation number one and let's go in the right equation number two. So equation number two thousands. That V two prime minus V one prime is equal to minus V Chew, which is minus 0.4 minus view one, which is plus 0.2. This leads us The following conclusion V two prime minus. View one prime is the cost to minus minus 0.6. Therefore V two prime minus. View one prime is equals to 0.6. Let me call this equation number two. So now we have two equations and two variables. Let me organize my board so I can continue the solution. Okay. Now, using equation number two, we can say the following According to equation number two V two, prime is equals to 0.6 plus view on prime. Now, let us substitute this result in equation number one. By doing that, we get the following 30 times view one prime plus 10 times v Chew Prime which is 0.6 plus view one prime is a cost to chew Then Now we have Onley one variable so we can solve this equation for a view on prime Let's do it So here we have 30 times view on prime plus six plus 10 times view one prime being equals To choose to reform, we have 40 times view one prime being equals to choose minus six So 40 times with you on prime is minus four. The reform view one prime is minus four divided by 40 which results in minus 0.1 m per second. The reform after the collision the big marble is moving to the left with a velocity off 0.1 m per second. Then to calculate the velocity off marble number two after the collision, all you have to do is substitute. This result entered this equation. By doing that, we get the following so it's substituting that equation back into equation. Number two leads us to V two. Prime equals to 0.6 minus 0.1. Then V two prime is equals to 0.5 m per second. So after the collision, marble number two is moving to the right with a velocity off 0.5 m per second. And this is the answer to the first item off this question. Now, before moving into the second item, let me organize my board in the second item off this question, we have to calculate the variations in the momentum's off marbles number one and marbles number two on. Then compare. So let us begin by marble. Number one The variation in the momentum off marble number one is given by the momentum off that marble after the collision, which is m one times V one prime minus the momentum off the marble before the collision and one V one. So, as you can see, we can factor the mass. So we have m one times V one prime minus V one. By plugging in the values that we got, we have the following. Everyone is 30 grands view one prime is minus 0.1 and then we have minus V one, which is 0.2. These results in Delta P one equals to 30 times, minus 0.3 on these results in a variation off minus 9 g times meters per second. This is a variation of the momentum off marble number one. So let me write it here minus 9 g meters per second. Now let's do the same for marble number two for marble number two, we have the following variation in the momentum. Delta Bi Chu is because of the momentum after m two V two prime minus the momentum before m two v chew. Then we can factor m two so that we got em two times V two prime minus V two. Now plug in the values that the problem gives us. We got 10 times V two prime, which is 0.5 minus the velocity number two before the collision which was minus 0.4. This results in 10 times plus 0.9 because here we've got a plus sign on this results in a variation the momentum that is 9 g times meters per second. So as you can see the variation, the momentum off marble number one is because to minus the variation the momentum off marble number two. This is expected because momentum is conserved. So any variation of the momentum off marble number one must be compensated by some variation in the momentum off marble number two Until this precisely. What's happening here now, Before going to the final item, let me erase these calculations that I have done for item B. Okay for items, See, we have to evaluate the changes in the kinetic energies off marbles number one and number two. The variation in the kinetic energy off marble number one is given by the kinetic energy after the collision and one times view one prime squared, divided by two minus the kinetic energy before the collision. And one times we want squared, divided by two plugging in the results that we have, we get the following Delta K one is equal to the mass off marble number one, which is 30 g. Here. I want to convert 2 kg so that we get our answer in jewels. So let's plug in a tanto ministry to convert from grams to kilograms. So this is divided by true and animal to apply by view one prime which is minus 0.1 and then square it for the kinetic energy. After the collision, we have 30 times stand to ministry divided by two times view one squared view one waas zero point shoes. So we have 0.2 squared. This gives us a variation. The kinetic energy off minus 0.0 45 which we can write as minus 4.5 times 10 to minus four Jun's This is a variation the kinetic energy off marble number one. Now let us do the same for marble number two. The variation in the kinetic energy off marble number two is given by the kinetic energy off marble number two After the collision m two V two prime squared divided by two minus the kinetic energy off marble chew before the collision Plugging in the values that we have we got the following m two is 10 times stand to ministry again. I'm converting from grams to kilograms. That's why I'm plugging. Attend to ministry here times V two prime squared So 0.5 squared and this is divided by two. Now we subtract m Chu, which is 10 times stand toe ministry times v chew squared before the collision, virtue was minus 0.4 squared. Now we divide by two These results in a variation, the kinetic energy off marble number two that is given by 0.0 45 Jews, which we can write as 4.5 times, stand to minus four jewels. So the variation in the kinetic energy off marble number two is 4.5 times 10 to minus four Jews. As you can see, the variation in the kinetic energy off marble number two is you cause to minus the variation in the kinetic energy off marble number one Again, this is to be expected because we're talking about an elastic collision on in that kind of collision, the kinetic energy is conserved. So any change in the kinetic energy off marble number one must be compensated by a change in the kinetic energy off marble number two.

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Cornell University

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University of Washington

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Simon Fraser University

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University of Sheffield

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text:
## Self-Assessment questionnaire (online tool)

• ##### Once all "Steps" are checked, you can click on “Submit” at the bottom of the last page. You will then automatically be directed to a page with your results and the SUMP category awarded.

Click here to start

##### And remember, there is very informative background material available about the SUMP Self-Assessment, instructions about how to navigate the tool, a related webinar, a discussion forum and, maybe most importantly, a PDF version of the SUMP questionnaire that will allow you to discuss all questions with colleagues before you fill in and submit your answers.
Background material


text:
# Ruby html href parser working slow

I wrote this simple HTML parser in Ruby.

Basically I have a folder that has files. Every file has crawled web pages. The only addition within these files is before the HTML content there is a page metadata, #URL is one of them.

What I am aiming is to have two files:

1)Will contain: currentURL, destinationURL, description

2)Will contain: currentURL, destinationURL, word, count

So if I did that for this current page and assuming that href that I am looking at is:

Stack Overflow

1st file would have

codereview,stackexchange.com, http://stackoverflow.com, Stack Overflow

2nd file would have

codereview,stackexchange.com, http://stackoverflow.com, Stack, 1

codereview,stackexchange.com, http://stackoverflow.com, Overflow, 2

However my code works really slow. It has been longer than an hour and still parsing a file with size of 31MB.

My code is:

beginning_time = Time.now

folder = "allParts/"
extension = "*.dat"
urlMet = "#URL:"
newLine = "\n"
files = Dir[folder+extension]
hrefTag = ""
hrefEnd = ""
firstOutput = "firstOut.txt"
secondOutput = "secondOut.txt"

File.open(firstOutput, 'w') {}
File.open(secondOutput, 'w') {}

countFile = 1

for file in files do
puts file + "#: "+countFile.to_s
countFile++
File.open(file, "r") do |f|
source = ""
f.each_line do |line|
dest = ""
desc = ""
if line.include? urlMet
source = line[/#{urlMet}(.*?)#{newLine}/m, 1]
end
if line.include? hrefTag
dest = line[/#{hrefTag}(.*?)#{qtMark}/m, 1]
descStIn = line.rindex(descStart)
descEndIn = line.rindex(hrefEnd)
desc = line[descStIn+2..descEndIn-1]
end
if (source != "" && dest != "")
occur = Hash.new(0)
mainEntry = "original-url=\"" + source +
"\", dest-url=\"" + dest + "\""
descEntry = ""
if (desc != nil && desc != "")
descEntry = ", desc=\"" + desc + "\""
words = desc.split(' ')
words.each { |word| occur[word] += 1 }
end
firstEntry = mainEntry+descEntry+"\n\n"
File.open(firstOutput, 'a') { |file|
file.write(firstEntry)
}
occur.each { |word, occurrance|
wordEntry = ", word=\"" + word +
"\", count=" + occurrance.to_s
secondEntry = mainEntry+wordEntry+"\n\n"
File.open(secondOutput, 'a') { |file|
file.write(secondEntry)
}
}
end
end
end
end

end_time = Time.now
puts "Time elapsed #{(end_time - beginning_time)} seconds"

-

You are trying to parse HTML with regexes, which is widely considered to be a bad idea. For example, all these tags would have equivalent link text and target URL:

A line-based parser will fail here, and a regex-based parser will be very difficult to write correctly. So please use an existing HTML parser. The result will be fast and elegant, but as I don't use Ruby I can't give examples.

Let's move on to coding style. You are using multiple naming styles: beginning_time and newLine. Whatever you do, be consistent, although I'd recommend snake_case for variable names.

Some of your variable names contain abbreviations where none are necessary: quote_character is probably better than qtMark, descStart would probably be better described as description_start.

You used many variables to make your regexes appear more modular, but they are not. Your regexes will become more readable if you write them down directly rather than assembling them. You could also specify them outside of your loops:

href_regex = /href="([^"]+)"/

Points to note here:

• [^"]+ is generally preferable over .*?
• The /m option was highly unnecessary, because you process your input line by line. So there is no way that your . could ever match a newline in your original regex.

In your regexes, you used #{…} interpolation a lot. Strangely you didn't do any of that when assembling other strings, instead using less readable concatenation. E.g. we could rewrite

mainEntry = "original-url=\"" + source +
"\", dest-url=\"" + dest + "\""

as

main_entry = "original-url=\"#{source}\", dest-url=\"#{dest}\""

This is still pretty unreadable because of those pesky escapes – using the alternate %Q() literal syntax we could simplify this to:

main_entry = %Q(original-url="#{source}", dest-url="#{dest}")

… and ditto in the similar cases.

A snippet like

words = desc.split(' ')
words.each { |word| occur[word] += 1 }

could be improved in two ways:

1. get rid of the unnecessary variables. The code is already sufficiently self-documenting with the word parameter to the block
2. rename occur to something sensible like word_count

You declare your source variable outside of the each_line loop. It would be better to declare your variables as close to their point of usage as possible. Furthermore, you initialize them with the empty string. I (as a Ruby newbie) would probably initialize them to nil instead. This has the further advantage that we can remove those line.include? "some string" tests which do not even guarantee that the regex will match. The String#[Regex, capture] method returns either the string matched by that capture group, or nil – which is exactly what we want.

Don't re-open a file all the time inside a tight loop, this is quite expensive. Open the file once, write to it multiple times. This will probably remove your main performance bottleneck.

If we stuff all this advice into your code, we might end up with the following refactoring:

start_time = Time.now

file_count = 1

File.open("firstOut.txt", 'w') do |out1|
File.open("secondOut.txt", 'w') do |out2|
for file_name in Dir["allParts/*.dat"] do
puts "#{file_name}#:#{file_count}"
file_count++

File.open(file_name, "r") do |file|
file.each_line do |line|
source      = line[/#URL:(.*?)\n/, 1]
destination = line[/')   + 2
description_end   = line.rindex('') - 1
description = line[description_start .. description_end]
if (description != nil && description != "")
out1.write(%Q(#{main_entry}, desc="#{desc}"\n\n))
description.split(' ').each do |word|
word_count[word] += 1
end
else
out1.write(%Q(#{main_entry}\n\n))
end

word_count.each do |word, count|
out2.write(%Q(#{main_entry}, word="#{word}", count=#{count}\n\n))
end
end
end
end
end
end
end

puts "Time elapsed #{Time.now - start_time} seconds"

Note that this can be improved further – the high level of indentation suggests we should extract some parts as separate functions. I further want to reiterate that I'm not a proficient Rubyist, so there might be some egregious oversights in the above code.

-
I am new to Ruby. Thanks for pointing out many things. Most of my code(as you might have guessed from having different naming conventions) is copied and pasted from another source. Thank you very much for writing so detailed answer. I have a question, I tried to google HTML parser but all I can find is 3rd party libraries. Are you telling me to use in-built library that comes with Ruby? If so what is the library name/ – Sarp Kaya Mar 3 '14 at 13:12
By the way about this part "You declare your source variable outside of the each_line loop. It would be better to declare your variables as close to their point of usage as possible." The reason why source was declared outside of the scope is because at the beginning of the file you get the source, declaring it right next to where it was used would only return null. – Sarp Kaya Mar 3 '14 at 14:52


text:
Length and Distance

# How many meters equal 23 feet?

###### Wiki User

23 feet is 7.01 meters.

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## Related Questions

### How many meters are there in 23 feet?

There are 0.3048 metres in one foot. Therefore, 23 feet is equal to 23 x 0.3048 = 7.0104 metres.23' is 7.0104 metres.

### How many Cubic feet is 23 Cubic Meters?

23 cubic meters = 812.24 cubic feet (rounded)

### What is 23m equal too?

23 meters is 75 feet and 5.51 inches.

### How many meters is 23 ft?

1 foot is 0.3048 meters. So, 23 feet = 23 x 0.3048 = 7.0104 meters.

### How many meters of square feet would it be of 23 acres?

Meters of square feet makes no sense, but 23 acres = 1 001 880 square feet 23 acres = 93 077.6977 square meters

### How many yards are equal to 23 meters?

23 meters is 25.1531 yards. 1 meter = 1.09361 yards.

### 23 feet equal how many yards?

3 feet = 1 yard so 23 feet = 7.66... yards

## Still have questions?

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text:
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# International Conference on Technology and Instrumentation in Particle Physics 2017(TIPP2017)

21-26 May 2017
Beijing International Convention Center
Asia/Shanghai timezone
Home > Timetable > Session details > Contribution details

# Contribution oral

Beijing International Convention Center - Room 305A
Calorimeters

# High granularity digital Si-W electromagnetic calorimeter for forward direct photon measurements at LHC

• Hongkai WANG

## Content

It is widely expected that the non-linear growth of parton densities at low x predicted from linear QCD evolution will lead to gluon saturation. As a decisive probe of gluon saturation, the measurement of forward (3.5 < y < 5) direct photons in a new region of low x ($10^{−5}$ ∼ $10^{−6}$ ) in proton-nucleus collisions at the LHC is proposed. An extremely high-granularity electromagnetic calorimeter is proposed as a detector upgrade to the ALICE experiment. This Forward Calorimeter (FoCal), is required to discriminate direct photons from decay photons with very small opening angle from neutral pions.

To facilitate the design of the upgrade and to perform generic R&D necessary for such a novel calorimeter, a compact digital Si/W sampling electromagnetic calorimeter prototype using Monolithic Active Pixel Sensors(MAPS) with a granularity of 30 × 30 μm and 28 $X_{0}$ has been built and tested with beams. The test beam results have shown the good energy linearity and very small Moliere radius (∼ 11 mm). We will discuss new results of the R&D with electromagnetic showers, in particular a position resolution of better than 30 μm. This precise position determination and the detailed knowledge of the electromagnetic shower shape obtained will provide the crucial capability for two photon separation down to a few mm. The results also show the successful proof of principle of particle counting calorimetry technology for future calorimeter development.


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# how to calculate the pooled standard error Hyner, Pennsylvania

Die Bewertungsfunktion ist nach Ausleihen des Videos verfügbar. Please help improve this section by adding citations to reliable sources. Note that the quantities s i 2 {\displaystyle s_{i}^{2}} in the right hand sides of both equations are the unbiased estimates. Wird geladen... Über YouTube Presse Urheberrecht YouTuber Werbung Entwickler +YouTube Nutzungsbedingungen Datenschutz Richtlinien und Sicherheit Feedback senden Probier mal was Neues aus!

If, in order to achieve a small variance in y, numerous repeated tests are required at each value of x, the expense of testing may become prohibitive. State a "real world" conclusion.Based on your decision in Step 4, write a conclusion in terms of the original research question. 9.4.1 - Video: Height by Biological Sex (Pooled Method) Example nk are the sizes of the data subsets at each level of the variable x, and S12, S22, . . ., Sk2 are their respective variances. In order to make up for this variation, a special type of standard error is used which accounts for one group of participants contributing more weight to the standard deviation than

Calculate Pooled Variance in Excel 2013 - Dauer: 6:11 Eugene O'Loughlin 4.843 Aufrufe 6:11 Understanding Standard Error - Dauer: 5:01 Andrew Jahn 13.114 Aufrufe 5:01 Pooled-Variance t Tests and Confidence Intervals: Why bash translation file doesn't contain all error texts? Check any necessary assumptions and write null and alternative hypotheses.There are two assumptions for the following test of comparing two independent means: (1) the two samples are independent and (2) each There's a slight difference between standard deviation and pooled sample standard error: When we are talking about a population, we talk about standard deviations.

Sample2 :n=25, s = 6. T Score vs. Wenn du bei YouTube angemeldet bist, kannst du dieses Video zu einer Playlist hinzufügen. Step 2 Calculate the pooled standard deviation, represented by Sp.

Your cache administrator is webmaster. Anzeige Autoplay Wenn Autoplay aktiviert ist, wird die Wiedergabe automatisch mit einem der aktuellen Videovorschläge fortgesetzt. Add ellipse with arrow around data points in pgfplots How to get all combinations of length 3 Is the measure of the sum equal to the sum of the measures? A pooled standard error accounts for two sample variances and assumes that both of the variances from the two samples are equal.

Resources by Course Topic Review Sessions Central! With modern technology, is it possible to permanently stay in sunlight, without going into space? Modify the correlation coefficient until you get close to the published result. Population-based statistics The populations of sets, which may overlap, can be calculated simply as follows: N X ∪ Y = N X + N Y − N X ∩ Y X

By using this site, you agree to the Terms of Use and Privacy Policy. Skip to Content Eberly College of Science STAT 200 Elementary Statistics Home » Lesson 9: Comparing Two Groups 9.4 - Comparing Two Independent Means Printer-friendly versionTwo independent means are compared using Transkript Das interaktive Transkript konnte nicht geladen werden. The formula for the pooled SD is: s=sqrt((n1-1)*s1^2+(n2-1)*s2^2)/(n1+n2-2)) (Sorry I can't post pictures and did not find a link that would directly go to the formula) Where 2 is the number

Show that a nonabelian group must have at least five distinct elements IQ Puzzle with no pattern Putting pin(s) back into chain more hot questions question feed about us tour help Pearson's Correlation Coefficient Privacy policy. Correlation Coefficient Formula 6. Wiedergabeliste Warteschlange __count__/__total__ How to calculate pooled standard error Stephanie Glen AbonnierenAbonniertAbo beenden6.0396 Tsd.

Hot Network Questions Moved to acquire Should zero be followed by units? Why is Pablo Escobar not speaking proper Spanish? A matter of style I think. –42- Jun 7 '13 at 3:56 1 Thanks a lot. How to Calculate a Pooled Standard Error by Sky Smith, Demand Media The Classroom » Higher Education Prep » How to Calculate a Pooled Standard Error The field of statistics compares

If so, how can that be corrected? 2) The loop is totally wrong but it is what I could come up with after several hours of no success. The standard error of a sample is another name for the standard deviation of a sample (this is also one of the AP Statistics formulas). Should zero be followed by units? When does bug correction become overkill, if ever?

Welcome to STAT 200! The problem resides in the fact that it is a factorial experiment with two within-subjects factors ($2 \times 3$ levels). The latter one can give a more efficient s p 2 {\displaystyle s_{p}^{2}} to estimate σ 2 {\displaystyle \sigma ^{2}} biasedly. Contents 1 Motivation 2 Definition 2.1 Variants 3 Example 4 Pooled standard deviation 4.1 Population-based statistics 4.2 Sample-based statistics 5 See also 6 References 7 External links Motivation In statistics, many

For calculations, you don't have to worry about that difference: Both are calculated using the same formulas. The square-root of a pooled variance estimator is known as a pooled standard deviation (also known as combined, composite, or overall standard deviation).


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# What are the generalized Gaussian probability laws that are infinitely divisible?

We consider the probability density, often called a generalized Gaussian density, $$p_{\alpha}(t) \propto \exp (- |t|^\alpha),$$ with parameter $0<\alpha<\infty$. For $p = 2$, we recognize a Gaussian distribution and for $p = 1$ a Laplace one. These distributions are known to be infinitely divisible.

For which $\alpha$ is the probability distribution $p_{\alpha}$ infinitely divisible?

Partial answer: The only infinitely divisible distributions that decay faster than $\exp(−O(\lvert t \rvert \log \lvert t \rvert))$ are the Gaussians (see the Theorem 7 of this paper). Hence, for $\alpha>1$ and $\alpha\neq 2$, $p_\alpha$ is not infinitely divisible. I am interested to know what happens in the case $\alpha <1$?

## 1 Answer

It is known that the kernel $\phi(x,y) = \exp(-|x-y|^p)$ for $x, y \in R$ and $0< p < \infty$ is positive definite if and only if $p \le 2$. Thus, the infinite divisibility of your generalized Gaussian should hold for the $0 < p \le 2$ case.

• Could you explain a bit about how positive definiteness of the kernel relates to infinite divisibility? May 15 '14 at 17:03
• The following paper explorers this connection: arxiv.org/pdf/1403.7304 May 15 '14 at 17:53
• If I understand correctly the paper you suggest, the authors show that if a continuous and symmetric density is infinitely divisible, then the kernel you defined is positive-definite. In particular it shows that if $p>2$, the probability density that I defined is not infinitely-divisible. But, if I am not mistaken, I cannot say anything for $p<2$. Is there any kind of converse result to the one of the paper you gave? May 15 '14 at 20:49
• Since the density function actually generates an infinitely divisible kernel---$\phi(x,y)^\alpha = \exp(-\alpha|x-y|^p)$ is ID---I believe we should be able to obtain a converse to conclude ID of the generalized Gaussian distribution; I hope you manage to work out the details (I gotta run). May 16 '14 at 0:23


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The deleted comb space is not path connected since there is no path from (0,1) to (0,0): 4. ( Let us prove our claim in 2. The deleted comb space is an important variation on the comb space. equipped with the subspace topology is known as the comb space. R Previous question Next question Get more help from Chegg. Let X be a topological space and x a point of X. The comb space and the deleted comb space have some interesting topological properties mostly related to the notion of connectedness. This was on a laptop which is normally not connected to its time machine backup. c) Let C be the comb space. The deleted comb space, D, is defined by: 1. When you try to shrink a volume with disk management, you may get the following error: "There is not enough space available on the disk(s) to complete this operation." A weaker property that a topological space can satisfy at a point is known as ‘weakly locally connected’: Definition. with its standard topology and let K be the set Therefore, Æ(U) is connected. 0 Proof. The comb space and the deleted comb space have some interesting topological properties mostly related to the notion of connectedness. R De ne S= f(x;y) 2R2 jy= sin(1=x)g[(f0g [ 1;1]) R2; so Sis the union of the graph of y= sin(1=x) over x>0, along with the interval [ 1;1] in the y-axis. On the Disk Management window, you will see a list of all connected hard drives to the PC. 1. a)* Prove that the comb space is compact without using the Heine Borel theorem. Make it a rule of thumb to enclose any and all file paths that you enter in Command Prompt in double quotes. Running, walking, cycling, swimming, skiing, triathlons – no matter how you move, you can record your active lifestyle on Garmin Connect. Famous quotes containing the words deleted, comb and/or space: “ There is never finality in the display terminal’s screen, but an irresponsible whimsicality, as words, sentences, and paragraphs are negated at the touch of a key. Configure Run Profiles. × The comb space is path connected (this is trivial) but locally path connected at no point in the set A = {0} × (0,1]. The set C defined by: considered as a subspace of While connector space objects that have not been reported by the data source are deleted during a full import, this is feature was implemented to ensure data consistency - not to track deletions. Therefore, f −1{p} is both open and closed in [0, 1]. Justify your answer. The significance of the past, as expressed in the manuscript by a deleted word or an inserted correction, is annulled in idle gusts of electronic massacre. 2 b. The set Cdefined by: 1.   {\displaystyle \{1/n~|~n\in \mathbb {N} \}} The following command will not run. Prove that both the supremum of A and infimum of A belong to the closure of A and hence to A.). n In PowerShell 2.0, the PSSession is deleted from the remote computer when it's disconnected from the originating session or the session in which it was created ends. The deleted comb space, D, is defined by: This is the comb space with the line segment {\displaystyle \mathbb {R} ^{2}} If it did, there’s obviously something wrong. 0 Consider R 2 It’s the only online community created specifically for … 4. { The path has a space in it and at that space, the command breaks and Command Prompt thinks you’ve entered a new command or parameter. 7.Press Enter to run the command. { The same thing was happening to me -- I deleted 100GB of stuff, Finder was reporting it was gone but Disk Utility showed I hadn't freed up any space. If you do not know how to check wires, do not attempt to plug/unplug any connected cables on the drive. Suppose there is a path from p = (0, 1) to a point q in D, q â  p. Let Æ:[0, 1] â D be this path. Since this ânew setâ is connected, and the deleted comb space, D, is a superset of this ânew setâ and a subset of the closure of this new set, the deleted comb space is also connected. R } 1 Weakly Locally Connected . Sysadmins face some issues when they try to recover disk space by deleting high sized files in a mount point and then they found disk utilization stays the same even after deleting huge files. n ∈ This option is used when you do not want to connect to a forest anymore. By noting that the comb space is path connected and hence connected, and that A must be compact (since C is homeomorphic to A and C is compact by exercise 1.a)), show that A has to be a closed interval. It should say “assuming that Xis path-connected, locally path-connected, and semilocally simply-connected". 1 , d) Show that the comb space cannot be imbedded in R (Hint: Suppose it could be imbedded in R and let A be the subset of R that the comb space, C, is homeomorphic to. The deleted comb space, D, is defined by: is just the comb space with the line segment e) Can the deleted comb space be imbedded in R? Creative Commons Attribution-ShareAlike License. Therefore, A is locally connected by exercise 2.c). with its standard topology and let K be the set 2 Press Win + X and choose the Disk Management selection. SPACES THAT ARE CONNECTED BUT NOT PATH-CONNECTED 3 Theorem 3.1. (Hint: Use part b) and note that a subspace of a Haudorff space is Haudorff, and that a subspace of a space having a countable basis for its topology also has a countable basis for its topology).   b) HENCE show that the set K = {1/n | n is a natural number} U {0} is compact (Hint: Prove that if X X Y is a product space, and Y is compact, then the projection onto the first co-ordinate is a closed map (i.e, maps closed sets in X X Y onto closed sets in X). The set C defined by: considered as a subspace of ( { 0 } × [ 0 , 1 ] ) ∪ ( K × [ 0 , 1 ] ) ∪ ( [ 0 , 1 ] × { 0 } ) {\displaystyle (\{0\}\times [0,1])\cup (K\times [0,1])\cup ([0,1]\times \{0\})} considered as a subspace of R 2 {\displaystyle \mathbb {R} ^{2}} equipped with the subspace topology is known as the comb space. If you are reviewing this article in conjunction with the Deleting the Connector Space document, then you may have already backed up the databases already. The option Delete Connector and connector space removes the data and the configuration. The comb space is an example of a path connected space which is not locally path connected. The trick is the double-quotes. {\displaystyle \mathbb {R} ^{2}} { {\displaystyle \{0\}\times (0,1)} The point (1;0) is a limit point of … that resembles a comb. 2. {\displaystyle \mathbb {R} ^{2}} If you have not, then please think of disaster recovery, we want to be able to get back to the previous setup without too much trouble should the need arise. The interval [0,1] on the x-axis is a deformation retract of the closed infinite broom, but it is not a strong deformation retract. n To prove that Æ −1{p} is open, we proceed as follows: Choose a neighbourhood V (open in We shall note that the comb space is clearly path connected and hence connected. We shall prove that Æ −1{p} is both open and closed in [0, 1] contradicting the connectedness of this set. The topologist's sine curve has similar properties to the comb space. A countably infinite set endowed with the cofinite topology is locally connected (indeed, hyperconnected) but not locally path connected. Change “cover space" to “covering space" §1.3, middle of page 69. Interestingly simply connecting to the drive and letting Time Machine do a backup didn't clear the space, I had to follow your procedure of shutting off time … ( { 0 } × { 0 , 1 } ) ∪ ( K × [ 0 , 1 ] ) ∪ ( [ 0 , 1 … c) Show that every closed interval in R is locally connected. Comb space; Integer broom topology; List of topologies; References {\displaystyle \mathbb {R} ^{2}} If not, that might point toward a deleted file being used by a process. Assume that I = [0,1] is compact and use a theorem from the section on compactness), c) Show that the deleted comb space is not compact. INITIALIZE DISK. b) Let X be locally homeomorphic to Y; that is there is a map f from X to Y that satisfies the following property: For each point x of X, there is a neighbourhood V of x that is homeomorphic to an open subset of Y under the map f (i.e, the map f restricted to V is the homeomorphism), Prove that if Y is locally connected, so is X (Hint: Use part a)). 1. The comb space is an example of a path connected space which is not locally path connected; see the page on locally connected space (next chapter). The topologist's sine curve is connected: All nonzero points are in the same connected component, so the only way it could be disconnected is if the origin and the rest of the space were the two connected components. See the answer. {\displaystyle \mathbb {R} ^{2}} Of course, the main concern here is whether or not the results of these commands come in under the size of the drive. Then there is a basis element U containing Æ −1{p} such that Æ(U) is a subset of V. We know that U is connected since it is a basis element for the order topology on [a, b]. The comb space satisfies some rather interesting properties and provides interesting counterexamples. 2. The deleted comb space, D, is connected: 3. 3. a) Prove that an open subspace of a locally connected space is locally connected. Prove that C is not a manifold (a manifold is a Hasudorff topological space X that has a countable base for its topology and is locally homeomorphic to R^n for some integer n). Example 410 The comb space is not lpc Remark 42 1 Path connected does not imply from MATH MISC at Western Governors University In mathematics, particularly topology, a comb space is a particular subspace of The comb space has properties that serve as a number of counterexamples. Show that the comb space is path connected but not locally connected. ) about p that doesnât intersect the xâaxis. Since both “parts” of the topologist’s sine curve are themselves connected, neither can be partitioned into two open sets.And any open set which contains points of the line segment X 1 must contain points of X 2.So X is not the disjoint union of two nonempty open sets, and is therefore connected. 3. . 1 The deleted in nite broom is connected. ∈ Both options sync all objects and update the metaverse objects. 1 that looks rather like a comb. In general, note that any path connected space must be connected but there exist connected spaces that are not path connected. ) Also, if we deleted the set (0 X [0,1]) out of the comb space, we obtain a new set whose closure is the comb space. 2*. Despite the closed infinite broom being arc connected, the standard infinite broom is not path connected. Not Enough Space Available on The Disk to Shrink Volume. Properties. A comb space is a subspace of It is however locally path connected at every other point. This should paste the path to the MSI file that you copied in Step 2 above. Clearly we have Æ −1{p} is closed in [0, 1] by the continuity of Æ. } See also. This problem has been solved! a) Let A be a connected subset of R. Show that if x is in A, y is in A with x < y, then the whole interval [x,y] is a subset of A. b) Show that a compact subset of R necessarily contains both its supremum and infimum (Hint: If A is a compact subset of R, A is closed. Expert Answer . 0 Right-click in the Command Prompt window, then choose Paste. ( The deleted comb space furnishes such an example, … We assert that Æ(U) = {p} so that Æ −1{p} is open. The comb space is path connected but not locally path connected. | 2 } Let’s consider the plane $$\mathbb{R}^2$$ and the two subspaces: / The option Delete connector space only removes all data, but keep the configuration. { Consider R 2 {\displaystyle \mathbb {R} ^{2}} with its standard topology and let K be the set { 1 / n | n ∈ N } {\displaystyle \{1/n|n\in \mathbb {N} \}} . 2 | The session state changes from Running to Disconnected. Related: Running Bash Commands in the Background the Right Way [Linux] Possible Causes Part 2. https://en.wikipedia.org/w/index.php?title=Comb_space&oldid=994584277, Creative Commons Attribution-ShareAlike License, This page was last edited on 16 December 2020, at 13:55. ATTEMPT QUESTIONS 2.c), 2.d) AND 3 IMMEDIATELY AFTER STUDYING THE NEXT SECTION. This is a contradiction. ) equipped with the subspace topology is known as the comb space. Question: Show That The Comb Space Is Path Connected But Not Locally Connected. Every contractible space is path connected and thus also connected. {\displaystyle \{0\}\times (0,1)} {\displaystyle \{1/n|n\in \mathbb {N} \}} 6. 0 Free disk space not updating after permanently deleting 200 gigs off my drive in one time Hello, The other day i noticed my C partition became almost full for some reason and i looked at all the files in the directory and it said there's only 175 gigs of files in it. Suppose Æ(U) contains a point (1/n, z) other than p. Then (1/n, z) must belong to D. Choose r such that 1/(n + 1) < r < 1/n. R deleted. In the Command Prompt window, type msiexec /i (you need to enter a single space after "/i"). Further examples are given later on in the article. Since Æ(U) doesnât intersect the x-axis, the sets: will form a separation on f(U); contradicting the connectedness of f(U). However, the deleted comb space is not path connected since there is no path from (0,1) to (0,0). This action is a long running operation. Entering paths with spaces. Or, disk management only shows a little space that allows you to shrink when there is actually a lot of free space. Rather, have an expert look at your computer. We want to present the classic example of a space which is connected but not path-connected. A better method to track deletions is to add a delta column to the source file and to populate this attribute with a value that indicates a deletion to ILM. My C partition has 488 gigs, so that's obviously not right. N The topologist's sine curve satisfies similar properties to the comb space. × } Then if C is the comb space, C is a closed subset of I X I (I = [0,1]) given the product topology. , We may not want these folders or files to be completely deleted, but we prefer them to be moved to a different location or copied. R . n Neither are locally connected. Each point on L n can be linked to (0;0) by a path along L n. By concatenating such paths, points onS L m and L n can be linked by a path via (0;0) if m6= n, so the union n 1 L nis path-connected and therefore is connected (Theorem2.1). 2. One of the common issues Linux Unix system users face is disk space is not being released even after files are deleted. But X is connected. Props to Zubie for posting their solution. 1. When you disconnect a PSSession, the PSSession remains active and is maintained on the remote computer. §1.3, page 65, line 12. §1.3, bottom of page 69 (or top of … {\displaystyle \mathbb {R} ^{2}} / In this article, I will describe a subset of the plane that is a connected space while not locally connected nor path connected. The topologist's sine curve is not path-connected: There is no path connecting the origin to any other point on the space. deleted. From Wikibooks, open books for an open world, https://en.wikibooks.org/w/index.php?title=Topology/Comb_Space&oldid=2677169. 2 a. The comb space and the deleted comb space satisfy some interesting topological properties mostly related to the notion of local connectedness (see next chapter). Consider R N connected" has two n’s, not three. {\displaystyle \mathbb {R} ^{2}} This page was last edited on 28 June 2014, at 21:44. 2 The comb space is homotopic to a point but does not admit a deformation retract onto a point for every choice of basepoint. The deleted comb space is a variation on the comb space. Connect to a forest anymore closed infinite broom is not path connected there. Open subspace of a path connected but not locally connected a ) * Prove an... There ’ s the only online community created specifically for … the space...: there is actually a lot of free space this page was last edited 28! Xis path-connected, locally path-connected, and semilocally simply-connected '' but there exist connected spaces that are path! ): 4 IMMEDIATELY after STUDYING the Next SECTION broom being arc connected, standard... ’ s obviously something wrong interesting counterexamples there is no path from 0,1! Not know how to check wires, do not know how to check wires, do not to... Broom is not path connected since there is no path from ( ). My C partition has 488 gigs, so that Æ −1 { p } is both open and closed [... In under the size of the drive only online community created specifically …! Is whether or not the results of these commands come in under the size the... Variation on the space and X a point but does not admit a deformation retract onto a point every! S obviously something wrong 28 June 2014, at 21:44 ) Show that the comb space some! Defined by: 1 's obviously not right point ( 1 ; )... However, the PSSession remains active and is maintained on the drive is not path connected there! F −1 { p } is both open and closed in [ 0, ]... Topology is locally connected ’: Definition size of the drive that allows you to Shrink there. X a point but does not admit a deformation retract onto a point of but!, do not want to connect to a forest anymore but there exist connected spaces that are path. Every closed interval in R retract onto a point of … but X is connected shows a space! Variation on the comb space and X a point of … but X is connected Disk Management selection a! Clearly we have Æ −1 { p } so that Æ −1 { p } is both open and in... Created specifically for … the comb space have some interesting topological properties mostly related to the notion connectedness! Should say “ assuming that Xis path-connected, locally path-connected, locally,! ) = { p } is closed in [ 0, 1 ] hence to a... Does not admit a deformation retract onto a point is known as weakly. Connected: 3 that any path connected and thus also connected choose the Disk Shrink! The origin to any other point on the comb space is path connected since there is no path from 0,1! Borel theorem the results of these commands come in under the size of the drive “ covering ''! Right-Click in the Command Prompt window, you will see a list of all hard. Delete Connector and Connector space removes the data and the deleted comb space is important! The Next SECTION you need to enter a single space after  /i '' ) a! E ) can the deleted comb space is path connected since there is actually a lot of free space,. And infimum of a belong to the comb space, D, is defined by: 1 open,! A PSSession, the PSSession remains active and is maintained on the comb space is a variation on the Management... That 's obviously not right community created specifically for … the comb space has properties serve... Space deleted comb space not path connected is normally not connected to its time machine backup f −1 p... To a forest anymore: there is actually a lot of free.!, note that the comb space has properties that serve as a number of counterexamples and update the metaverse.! Standard infinite broom being arc connected, the deleted comb space on in the Command Prompt double... Shows a little space that allows you to Shrink when there is no path from ( 0,1 ) (. The standard infinite broom is not locally connected attempt to plug/unplug any connected on. The deleted comb space be imbedded in R a variation on the space is however locally path connected hence! A lot of free space you need to enter a single space after  /i )! '' ) connected to its time machine backup rather, have an expert at. Is known as ‘ weakly locally connected: 1 not the results of these commands in... Provides interesting counterexamples to any other point on the drive a variation on the space standard infinite is. Options sync all objects and update the metaverse objects single space after  /i '' ) here is or. By a process exercise 2.c ), 2.d ) and 3 IMMEDIATELY after STUDYING Next... All objects and update the metaverse objects X be a topological space can at... Space have some interesting topological properties mostly related to the notion of connectedness the Next SECTION path connected:.... A laptop which is normally not connected to its time machine backup not path connected but not connected. In double quotes of these commands come in under the size of the drive sync... Heine Borel theorem in [ 0, 1 deleted comb space not path connected by the continuity of Æ that an open of... The comb space and the deleted comb space on 28 June 2014, at 21:44 thumb... In Command Prompt window, then choose Paste = { p } so that 's obviously not right theorem... Not connected to its time machine backup weakly locally connected make it a rule of thumb enclose! The data and the configuration is actually a lot of free space properties to the PC any path connected there! E ) can the deleted comb space, D, is connected ’: Definition if not, might. Is path connected space must be connected but not locally path connected space is an example of a hence... X is connected: 3 satisfy at a point of X the continuity of Æ, 2.d ) and IMMEDIATELY... Open subspace of a belong to the PC X a point is as... Further examples are given later on in the Command Prompt window, then choose Paste in double quotes Chegg... The configuration 0, 1 ] any connected cables on the comb space, f {... '' §1.3, middle of page 69 your computer objects and update the metaverse objects in under the size the... Satisfy at a point but does not admit a deformation retract onto a point but not... That serve as a number of counterexamples expert look at your computer indeed, hyperconnected ) but not locally connected... Every choice of basepoint X be a topological space and the configuration that every closed interval in R results... Infimum of a path connected at every other point on the space Paste the path to the space... 28 June 2014, at 21:44 have Æ −1 { p } is both and... Curve satisfies similar properties to the PC space can satisfy at a point is as., 1 ] from Chegg STUDYING the Next SECTION '' §1.3, middle of page 69 that serve as number! Command Prompt window, you will see a list of all connected hard to. * Prove that an open subspace of a belong to the comb space is a variation the... A topological space and the deleted comb space is not path connected and thus also connected which is path-connected. A topological space and the deleted comb space, D, is connected:.! No path connecting the origin to any other point on the comb.! File being used by a process 28 June 2014, at 21:44 further examples given., f −1 { p } is both open and closed in [,... The deleted comb space is homotopic to a forest anymore connecting the origin to any other point the... Be imbedded in R indeed, hyperconnected ) but not locally connected press Win + and... Must be connected but there exist connected spaces that are deleted comb space not path connected path connected and thus also connected not results. A rule of thumb to enclose any and all file paths that you copied in Step 2 above any! Thumb to enclose any and all deleted comb space not path connected paths that you copied in Step 2 above assuming that Xis path-connected and. ) to ( 0,0 ) not, that might point toward a deleted file being used a! It ’ s the only online community created specifically for … the comb space, and semilocally simply-connected '' on... In [ 0, 1 ] limit point of … but X is connected: 3 then Paste. Point of X despite the closed infinite deleted comb space not path connected being arc connected, the remains! Metaverse objects is used when you disconnect a PSSession, the PSSession active... Expert look at your computer come in under the size of the drive a point is as! An open world, https: //en.wikibooks.org/w/index.php? title=Topology/Comb_Space & oldid=2677169 world, https:?... Of all connected hard drives to the closure of a and hence.. Normally not connected to its time machine backup hence connected should Paste the path to the notion of connectedness more! Properties mostly related to the closure of a path connected space which is path! Broom being arc connected, the PSSession remains active and is maintained on the comb space X. Rather, have an expert look at your computer a forest anymore topology is locally connected ’ Definition... That serve as a number of counterexamples hence connected we shall note that the comb space and X point. The standard infinite broom being arc connected, the PSSession remains active and is maintained on the comb and. Hyperconnected ) but not locally connected path-connected: there is no path from ( 0,1 ) to ( 0,0:!

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## appleduardo 3 years ago whats the integral of x arctan x dx??

1. ksaimouli

|dw:1363212783612:dw|

2. ksaimouli

u need to remember this $\frac{ 1 }{ \sqrt{x^2+1} }$+c

3. anonymous

How are you doing on this one? Do you still want help?

4. anonymous

$\int x\arctan x\;dx$ Integrate by parts, letting $\begin{matrix}u=\arctan x&dv=x\;dx\\ du=\frac{1}{x^2+1}dx&v=\frac{1}{2}x^2\end{matrix}$ $\frac{1}{2}x^2\arctan x-\frac{1}{2}\int\frac{x^2}{x^2+1}dx$ The new integral requires some trig substitution.

5. anonymous

$\large \frac{x^2}{x^2+1}=\frac{(x^2+1)-1}{x^1+1}=1-\frac{1}{x^2+1}$

6. appleduardo

what trig substitution do I have to apply?

7. anonymous

|dw:1363282174743:dw|Imagine a theta such that tan(theta) = x. Then you can draw this triangle. THen you have these facts: tan(theta) = x, theta = arctan(x) dx = sec(x)^2 d(theta).

8. anonymous

So then you can substitute for your old integral: $\int\limits_{}^{}\theta*\tan(\theta)*\sec(\theta)^2d \theta$ Let me know if you need more help that that.


text:
# How to calculate the mean and variance of a cumulative probability graph?

I'm given a cumulative probability graph, however i can't post any images. Thus, the X and Y coordinates of the graph are:

X:

2.5 7.5 12.5 17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5 57.5

Y:

0 0 0 0.07 0.17 0.27 0.37 0.47 0.57 0.67 0.77 0.87

How do i calculate the mean and variance of this cumulative probability graph?

Thanks alot!!!!

EDIT:

Now that i have enough rep points, i can post the picture of the graph.

Given D = 0.07

-
Hint: For your problem, where $X$ takes on only positive values, $E[X]$ equals the area between the curve $F(x)$ and the line $y = 1$ for nonnegative $x$ from $0$ to $\infty$. Thus, area of rectangular region with opposing corners at $(0,0)$ and $(2.5,1)$ plus area of rectangular region with opposing corners $(2.5,0)$ and $(7.5, 1)$ plus... Note that what the area is between $x=7.5$ and $x=12.5$ depends on the graph of $F(x)$ which may increase linearly or as some smooth rise from $(7.5,0)$ to $(12.5,0.07)$, or in a sudden jump at $x=12.5$, etc. –  Dilip Sarwate Oct 19 '12 at 20:49
As well as Dilip's points, you need to know about the right tail of the distribution (or be able to estimate it), i.e. when the cumulative probability reaches $1$ –  Henry Oct 19 '12 at 22:05

One possibility is that what you actually have is a uniform distribution on the interval $[14,64]$ in which case the mean would be $\dfrac{14+64}{2}=39$ and variance $\dfrac{(64-14)^2}{12}\approx 208.3333$.
In most cases life will not be as simple as this, and you will need numerical methods. You have incomplete data, so suppose your $X_i$ finish with 62.5 67.5 and your $Y_i$ finish with 0.97 1. Then you can estimate, by putting the probability at the centre of each interval, the first moment with $$\sum_i (Y_i-Y_{i-1})\left(\frac{X_{i-1}+X_i}{2}\right)$$ which in this case would be $39$, and estimate the second moment with $$\sum_i (Y_i-Y_{i-1})\left(\frac{X_{i-1}+X_i}{2}\right)^2$$ which in this case would be $1732.5$, giving an estimate of the variance of $1732.5-39^2=211.5.$


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Multiplying matrix by itself [closed]

Hi

There is any generic mathematical expression (formula) to represent a Matrix squared ?

I search it before, but i didn't find.

thanks

-
This is not research level - try math.stackexchange.com. And the answer is no, there isn't beyond A^2. – Jacques Carette Nov 18 2011 at 20:14
$A$ squared = $A.A$ – Anthony Quas Nov 19 2011 at 2:06


text:
Experience the CMOS Annealing Machine

# Building network robustness

This tutorial explains the processing of the annealing machine in the demonstration application building network robustness

### Overview

Taking the vertex cover problem for graphs as an example, let's consider solving combinatorial optimization problem with an annealing machine.

Consider the situation where the server is connected with the network (green line in the figure) as shown below. Under these circumstances, we will apply security measures to the server. It is best to apply the same measures to all servers in the same way, but it costs more, so we want to reduce the number to as few as possible. Therefore, we decide to select a server so that there is a countermeasure server at least next to even unprotected servers. The part surrounded by blue circles in the figure is an example of such a combination.

In this way, a subset of vertices where at least one vertex is included regardless of which side in the graph is taken is called "vertex cover". There are many combinations of vertex covers, but the one with the smallest number of vertices among them is particularly called the min vertex cover.

The problem this time is finding the minimum vertex cover.

### Create a cost function

In this problem, we correspond "included" or "not included" in the cover for each vertex of the graph to the spin value -1 or +1 of CMOS annealing machine. In order to obtain the minimum vertex cover, the following two conditions must be met.

• (a) The spin to be +1 (the vertex included in the vertex cover) should be as small as possible.
• (b) The vertices on both sides of each side of the graph must not be -1.

The cost function for (a) is as follows.

#### $$\tag*{Equation(1)} \sum_{i \in V}s_i$$

Here $V$ is the set of all the vertices of the input graph. For (b), we call $s_i$ , $s_j$ for the spins corresponding to the vertices on both sides of the edge. At this time, you can think of a function that cost is high only when it is -1, and it is the same cost in all other cases, but since it is hard to think as it is, think variables $x_i$ or $x_j$ which takes binary values 0 or 1 instead of $s$. At this time, if included in the vertex cover, the value of $x_i$, $x_j$ will be 1, otherwise it will be 0. In this case, it is only necessary to increase the cost only when $x_i$ and $x_j$ are both 0.

#### $$\tag*{Equation(2)} (1−x_i)(1-x_j)$$

Considering the cost function of above, the value of the expression is 1 only when both $x_i$ and $x_j$ are 0, otherwise it is 0, so the desired condition is satisfied.
Since it is a spin that takes a value of ±1 only on an annealing machine, you need to convert equation (2). Because $s_i$ , $s_j$ also $x_i$, $x_j$ takes only binary values, they can be converted to each other by the following expression.

#### $$x_i=\frac{s_i+1}2$$

Substituting this into equation (2) and rearranging it will give the following form.

#### $$\tag*{Equation(3)} −s_i−s_j+s_i s_j$$

By the way, in equation (3), the total coefficient of 1/4 was dropped, which is equivalent to changing the energy scale (unit). At this time, the constant term does not affect the combination of the spin values giving the ground state, so we ignore it.
When equation (3) is added to all sides of the input graph, it becomes as follows.

#### $$\tag*{Equation(4)} \sum_{(i,j)\in E}(-s_i-s_j+s_is_j)$$

However, $E$ is a set of all sides of the input graph.
By adding equation (1) and equation (4), the following cost function corresponding to the vertex cover problem can be obtained.

#### $s_i \in \{\pm1\}$: Spin value. $+ 1$ when the vertex i of the input graph is included in the vertex cover.

We introduced the parameter ($\eta$>0) before the first term. This is to adjust the strength of the two conditions.
If $\eta=0$, the cost function is only the second term. In this case, there is no effect of suppressing the number of vertices included in the vertex cover. Conversely, if $\eta$ is made very large, it is practically only the first term, so the cost function is minimized when the number of vertices in the vertex cover is zero.
In the demo of this site, it is $\eta=1$.


text:
Browse Questions

# Both B and T-cells of immune system are produced in

$\begin{array}{1 1}(A)\;\text{Spleen}\\(B)\;\text{Lymphoid nodes}\\(C)\;\text{Bone marrow}\\(D)\;\text{Thymus}\end{array}$

Can you answer this question?

Both B and T-cells of immune system are produced in bone marrow
Hence (C) is the correct answer.
answered Jul 10, 2014


text:
Other Open Access

# Two Methods of the Clock Jitter Measurement Aimed at Embedded TRNG Testing

Oto Petura; Marek Laban; Elie Noumon Allini; Viktor Fischer

### JSON-LD (schema.org) Export

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In modern cryptographic systems, security is based on quality and unpredictability of confidential keys. These keys are generated in random number generators using random physical phenomena appearing inside the cryptographic system on chip. The most frequently used source of randomness in digital devices is the jitter of clock signals generated inside the device in ring oscillators, self-timed rings, RC oscillators, phase-locked loops (PLLs), etc. The quality and unpredictability of generated numbers depends on the quality and the size of the clock jitter. It is therefore a good practice to monitor this jitter continuously using some embedded jitter measurement method. The measured jitter parameters can be then used as input parameters of the stochastic model used to estimate entropy, which characterizes unpredictability of generated numbers. In this paper, we present and compare two methods of embedded jitter assessment based on the measurement of the variance of counter values, obtained by counting the periods of the jittery clock during a time interval defined by a reference clock generated in the same device. Besides comparing obvious design results such as area, speed, and power consumption, we observe and discuss the impact of the two embedded variance measurement methods on the clock jitter itself, and compare the behavior of the two clock generators used as sources of randomness with and without clock variance measurement circuitry, and with and without additional logic such as an AES cipher, which perturbs the variance computation, as it is the case in most cryptographic embedded systems. This comparison is very important for a good estimation of the low entropy bound from the measurement results.
\n
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text:
# Computer simulation of protein-carbohydrate complexes Application to concanavalin A and L-arabinose-binding protein

Rao, VSR and Reddy, BVS and Mukhopadhyay, C and Biswas, M (1990) Computer simulation of protein-carbohydrate complexes Application to concanavalin A and L-arabinose-binding protein. [Book Chapter]

Full text not available from this repository.

## Abstract

The CCEM (Contact Criteria and Energy Minimization) method has been developed and applied to the study of protein-carbohydrate interactions. The method uses available x-ray data on the native protein to generate realistic models of a variety of proteins with various ligands. The examples discussed in this paper are Con A and L-arabinose-binding protein (ABP). Complexes of Con A with $\alpha$-methyl-mannoside ($\alpha$MeMan) and a trimannoside,3,6-di-O-($\alpha$-D-mannopyranosyl)-D-mannoside, which is present as a structural moiety in all complex carbohydrates and glycopeptides, were generated using the 2.4 .ANG.-resoln. x-ray data of native Con A. It is suggested that the higher affinity of Con A for the trimannoside may be due to extended interactions rather than to an extended sugar binding site of Con A. The x-ray crystal structure data reported on the complex of ABP with L-arabinose were used to generate complexes of ABP with other sugars. The results are in good agreement with the exptl. data.

Item Type: Book Chapter Publisher Copyright of this article belongs to American Chemical Society. Computer simulation;protein-carbohydrate complexes; concanavalin Division of Biological Sciences > Molecular Biophysics Unit 28 Dec 2007 08 Feb 2012 07:34 http://eprints.iisc.ernet.in/id/eprint/12580


text:
## annas Group Title Hello dear friends I am going to post my first php tutorial: (Do comment your views). 2 years ago 2 years ago

1. annas Group Title

PHP is a general-purpose server-side scripting language originally designed for Web development to produce dynamic Web pages. It is one of the first developed server-side scripting languages to be embedded into an HTML source document rather than calling an external file to process data. The code is interpreted by a Web server with a PHP processor module which generates the resulting Web page. It also has evolved to include a command-line interface capability and can be used in standalone graphical applications.

2. annas Group Title

Now friends I am going to show you how to write your first code in php. <?php //to start a php code you will put '<?php' it'll tell compiler that the code is started ?> //to end the php code you will put '?>' . It'll tell compiler when to stop. between these starting and end delimiters you'll write the php instructions

3. annas Group Title

now the first php code, //this code will display hello world <?php echo "Hello world"; ?> //echo => its an output instruction output: Hello world Now, this code can be written by embedding html in it. <?php echo "Hello World<br><b>This text is bold"; ?> //BR => it's a line break instruction //B => it is used for making the text bold output: Hello World $$\mathbf{This}$$ $$\mathbf{text}$$ $$\mathbf{is}$$ $$\mathbf{bold}$$

4. annas Group Title

we can also use a variable for this purpose: to initialize a variable in php we use $variable_name <?php$name = "Aproov"; echo "My friend name is \$name"; ?> output: My friend name is Aproov

5. annas Group Title

$\color{maroon}{\mathtt{\text{<Comments>}}}$ like C language, php also have comments. Comments are used to increase the readability of a program. There are two types of comments 1. single line 2. multi line For single line comments we type // and text put after these are ignored by the compiler For multi line comments we type /* text after this is ignored until compiler finds */

6. annas Group Title

Thats it for this tutorial. I hope you'll learn from this tutorial. Good luck.

7. lgbasallote Group Title

i see this has a lot of resemblance to C language. i think if a person has background on C language then this is not that hard yes? interesting. one question though. where can i type php? i mean does notepad support it? or dev c++? or a completely new software?

8. annas Group Title

@lgbasallote yes a person with C language background can easily learn this language. for php you'll need a server to execute you code or if you dont like to pay for server then download & install xampp. to write php code you can use a notepad or a notepad++ and save your file with an extension .php. link to download xampp and notepad++ are as under: 1.http://www.apachefriends.org/en/xampp.html 2.http://notepad-plus-plus.org/

9. i-luv-choolates Group Title

It is very well written Annas

10. annas Group Title

@Eyad see the first tutorial and do comment about it :)

11. Eyad Group Title

Good Job Mate! ,and btw u can just use Dreamweaver ...

12. Ron.mystery Group Title

Good job anns

13. annas Group Title

thank you

14. annas Group Title

i'll post my other tutorials very soon :)

15. Muskan Group Title

v well :)) God job,,,carry on..

16. moha_10 Group Title

good job annas carry on

17. annas Group Title

thank you guyz :)

18. ash2326 Group Title

@annas Good work

19. annas Group Title

thanks ash :)


text:
# If the selling price is doubled, the profit triples. Find the profit percent

A. 50%
B. 70%
C. 100%
D. 120%

The correct answer is (C) 100%
Step by step Solution:
We solve the problem by assuming that the original cost price be CP and the selling price be SP. Now, we have the profit (p), we have,
p = SP – CP ⇒ (1)
It is given in the problem that on doubling the selling price, the profit triples, we have,
3p = 2SP – CP ⇒ (2)
Now, we need to find the profit percent given by

\frac{S p-C p}{C p} \times 100

We need to find SP in terms of CP to find the profit percent. We use equations (1) and (2) and solve them to find SP in terms of CP. Firstly, we subtract (1) from (2),
we get ⇒ 3p-p = SP
2p = SP ⇒ (A)
Putting this in (1), we get,
p = 2p – CP
p = CP ⇒ (B)
From (A) and (B), we get that SP = 2CP. Now, we put this in the formula of profit percentage given by

\frac{S p-C p}{C p} \times 100

We get,

\frac{2C p-C p}{C p} \times 100
We will be happy to hear your thoughts


text:
Article Contents
Article Contents

# Asset price volatility and price extrema

• The relationship between price volatility and expected price market extremum is examined using a fundamental economics model of supply and demand. By examining randomness through a microeconomic setting, we obtain the implications of randomness in the supply and demand, rather than assuming that price has randomness on an empirical basis. Within a general setting of changing fundamentals, the volatility is maximum when expected prices are changing most rapidly, with the maximum of volatility reached prior to the maximum of expected price. A key issue is that randomness arises from the supply and demand, and the variance in the stochastic differential equation governing the logarithm of price must reflect this. Analogous results are obtained by further assuming that the supply and demand are dependent on the deviation from fundamental value of the asset.

Mathematics Subject Classification: 91B25, 91B70, 91G80.

 Citation:

•  [1] P. Billingsley, Probability and Measure, Wiley Series in Probability and Statistics. John Wiley & Sons, Inc., Hoboken, NJ, 2012. [2] F. Black and M. Scholes, The pricing of options and corporate liabilities, J. Political Economy, 81 (1973), 637-654.  doi: 10.1086/260062. [3] Z. Bodie, A. Kane and A. Marcus, Investments, Ed. McGraw Hill, New York, 2010. [4] G. Caginalp and D. Balenovich, Asset flow and momentum: Deterministic and stochastic equations, R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci., 357 (1999), 2119-2133.  doi: 10.1098/rsta.1999.0421. [5] C. Caginalp and G. Caginalp, The quotient of normal random variables and application to asset price fat tails, Physica A, 499 (2018), 457-471.  doi: 10.1016/j.physa.2018.02.077. [6] C. Caginalp and G. Caginalp, Stochastic asset price dynamics and volatility using a symmetric supply and demand price equation, Physica A, 523 (2019), 807-824.  doi: 10.1016/j.physa.2019.02.049. [7] G. Caginalp and M. Desantis, Multi-group asset flow equations and stability, Disc. and Cont. Dynam. Systems B, 16 (2011), 109-150.  doi: 10.3934/dcdsb.2011.16.109. [8] G. Caginalp, D. Porter and V. Smith, Initial cash/asset ratio and asset prices: An experimental study, Proc. Nat. Acad. Sciences USA, 95 (1998), 756-761.  doi: 10.1073/pnas.95.2.756. [9] E. Díaz-Francés and F. J. Rubio, On the existence of a normal approximation to the distribution of the ratio of two independent normal random variables, Stat. Papers, 54 (2013), 309-323.  doi: 10.1007/s00362-012-0429-2. [10] D. V. Hinkley, On the ratio of two correlated normal random variables, Biometrika, 56 (1969), 635-639.  doi: 10.1093/biomet/56.3.635. [11] J. Hirshleifer and A. Glazer, Price Theory and its Applications, Prentice Hall, Engelwood Cliffs, NJ, 1997. [12] J. Hull, Risk Management and Financial Institutions, John Wiley & Sons, 2012. [13] H. H. Merdan and M. Alisen, A mathematical model for asset pricing, Applied Mathematics and Computation, 218 (2011), 1449-1456.  doi: 10.1016/j.amc.2011.06.028. [14] H. Merdan and H. Cakmak, Liquidity effect on the asset price forecasting, Journal of Nonlinear Systems and Applications, (2012), 82–87. [15] M. Nerouppos, D. Saunders, C. Xiouros and S. A. Zenios, Risk Management in Emerging Markets: Practical Methodologies and Empirical Tests, 2015. [16] C. R. Plott, Markets as information gathering tools, Southern Economic Journal, 10 (2006), 179-221. [17] C. Plott and T. Salmon, The simultaneous, ascending auction: Dynamics of price adjustment in experiments and in the UK3G spectrum auction, Journal of Economic Behavior and Organization, 53 (2004), 353-383. [18] W. Rudin, Real and Complex Analysis, Third edition, McGraw-Hill Book Co., New York, 1987. [19] Z. Schuss, Theory and Applications of Stochastic Processes, An analytical approach. Applied Mathematical Sciences, 170. Springer, New York, 2010. doi: 10.1007/978-1-4419-1605-1. [20] H. Shefrin, A Behavioral Approach to Asset Pricing, Elsevier, 2008. [21] V. L. Smith, G. L. Suchanek and A. W. Williams, Bubbles, crashes, and endogenous expectations in experimental spot asset markets, Econometrica, (1988), 1119–1151. [22] D. Sornette, P. Cauwels and G. Smilyanov, 2017, Can We Use Volatility to Diagnose Financial Bubbles?, Lessons from 40 Historical Bubbles, Swiss Finance Institute Research Paper No. 17-27, (2017), Available at SSRN: https://ssrn.com/abstract=3006642 or http://dx.doi.org/10.2139/ssrn.3006642. [23] S. Stojanovic, Computational Financial Mathematics Using Mathematica: Optimal Trading in Stocks and Options, SBirkhäuser Boston, Inc., Boston, MA, TELOS. The Electronic Library of Science, Santa Clara, CA, 2003. doi: 10.1007/978-1-4612-0043-7. [24] S. D. Stojanovic, Risk premium and fair option prices under stochastic volatility: The HARA solution, C. R. Math. Acad. Sci. Paris, 340 (2005), 551-556.  doi: 10.1016/j.crma.2004.11.002. [25] Y. L. Tong, The Multivariate Normal Distribution, Springer Series in Statistics, Springer-Verlag, New York, 1990. doi: 10.1007/978-1-4613-9655-0. [26] D. Watson and M. Getz, Price Theory and its Uses, 5th Ed., University Press of America, Lanham, MD, 1981. [27] P. Wilmott,  S. Howison and  J. Dewynne,  The Mathematics of Financial Derivatives. A Student Introduction, Cambridge University Press, Cambridge, 1995.  doi: 10.1017/CBO9780511812545.

• on this site

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