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Twice a month, we revisit some of our readers’ favorite posts from through out the history of Nettuts+. This tutorial was first published in October 2010.

Today, we'll dive into jQuery Mobile, which, at the time of this writing, is in a RC1 state. We'll build a simple Tuts+ RSS reader, using PHP and jQuery Mobile. When we're finished, you'll have the ability to add this simple project to your iPhone or Android phone with the click of a button, as well as the skills to build your own custom mobile web apps!

## Step 1: Outline the Project

It's always helpful to first outline what you want your project to do/achieve.

• Display a list of every Tuts+ site, along with its square logo
• Display the feed for each site, when clicked on
• Create a basic *article* stylesheet for rendering each posting
• Create an Apple-touch icon for the users who add the "app" to their phone
• Use YQL to retrieve the desired RSS feed
• Implement a basic form of "text file" caching every three hours

## Step 2: Begin

The next step is to begin creating our project. Go ahead and make a new folder -- name it how you wish -- and add a new header.php file. *Note that this project uses PHP. If you're not familiar with this language, feel free to skip the PHP parts! Within this file, we'll reference jQuery mobile, its stylesheet, and any other assets that we require. If only to stay organized, I've placed my header.php file within an includes/ folder.

There are a handful of things worth noting here.

1. An HTML5 doctype is required. But you should be using that anyways!
2. The X-UA-Compatible tag forces IE to use it most current rendering engine
3. We need to reference jQuery Mobile's stylesheet. You can use their CDN, and save on bandwidth!
4. If you want to designate an icon for when users add your webpage to their iPhone (or Android) home screen, add a link tag, with a rel attribute of apple-touch-icon.
6. Finally, we're loading the jQuery mobile script file (currently in Alpha 1)

### The Basic Structure

The jQuery Mobile framework can be activated by applying unique data-* attributes to your code. The basic structure for most sites will look similar to:

Add the code above to a new index.php file, within the root of your project.

We have to tell jQuery about our project. For example, try not to think of each file as a page. Technically, you can create multiple pages at a time, by adding additional wrapping data-role="page" attributes. These are referred to as inner pages.

Further, the framework has specific settings and stylings in place for the header, main content area, and footer. To inform jQuery Mobile about the locations of these elements, we add the following attributes.

• data-role="content"
• data-role="footer"

No data-role attributes have been applied.

Data-role attributes applied.

## Step 3: Listing the Tutorial Sites

Now that the structure of our index.php page is complete, we can populate each section with our Tuts+ specific mark-up.

• Header: In this section, we're simply inserting the Tuts+ graphic, and providing alternate text if images are turned off.
• Content: In the content area, we need to list all of the tutorial sites, and apply a unique icon next to each heading. We also link to a new page, site.php that will handle the process of retrieving the RSS feed. For convenience, when we link to site.php, we also pass through the name of the selected site, via the querystring: siteName=nettuts.
• Footer: At the bottom, for now, we'll simply add a link to Tuts+.

jQuery Mobile offers a plethora of helpful CSS classes, including ui-li-icon. When applied to an image, it'll float it to the left, and apply 10px worth of margin-right.

At this point, our site should look like the above image.

### Page Transitions

As jQuery will load local pages asynchronously with AJAX, we can specify any number of cool page transitions. The default is the basic slide-left or slide-right effect that most touch-phone users are aware of. To override the default, use the data-transition attribute on the anchor tag.

• slide
• slideup
• slidedown
• pop
• flip

## Step 4: ListViews

Ehh - the image, shown above, still looks like a website. We need to make things a bit more phone-like. The answer is to use the data-role="listview" attribute. Watch what happens when we do nothing more than apply this attribute to the wrapping unordered list.

Wow - what an improvement! Even better, we have access to theme-roller, which allows us, with the change of a single letter, to switch color themes.

### List Dividers

Now, what if we wanted to divide this list of tutorial sites? In these situations, we can take advantage of data-role="list-divider", which can be applied to the
 element.

These, too, can receive lettered theme roller stylings. They can be set within the parent
.

Note that we won't be using dividers for this particular application.

## Step 5: CSS

jQuery Mobile takes care of a great deal of the formatting, however, we still, of course, need our own stylesheet for tweaking. For example, looking at the images above, we can see that the tutorial icons need to be pushed up a bit. Additionally, I'd like to use the Tuts+ red for the background color of the heading and footer, rather than the default black.

Create a new folder, CSS, and add a new stylesheet -- I'll call mine: mobile.css. Within this file, we'll first fix the icon positioning:

Next, we'll create a handful of classes, named after their respective tutorial sites. These classes will contain any specific formatting/colors for the site. For example, Nettuts+ has a darker green color, while MobileTuts+ is yellow.

That should be fine for now. The last step for index.php is to apply the .tuts class to the header and footer elements. That way, the header and footer will render the correct background color.

## Step 6: YQL, PHP, and Caching

Now, it's time to step away from the layout, and work on the functionality. Each of the links we created directed to site.php?siteName="siteName". Let's go ahead and create that file now.

Even though this is a relatively tiny app, we should still strive to follow best practices. In this case, it means that we should keep as little PHP in our document as possible. Instead, we'll use site.php as a controller of sorts. This file will handle the initial logic, and will then, at the bottom, load in our HTML template.

In order to retrieve the desired RSS feed, we first need to capture the name of the site that the user clicked on initially. If you'll refer to a previous step, when we linked to site.php, we also passed the name of the site through the querystring. With PHP, this can easily be retrieved, with $_GET['siteName']. However, what if, for some odd reason, this value doesn't exist? Maybe site.php was accessed directly?? We should set a default site to compensate for these situations. If$_GET['siteName'] is empty, we'll set "nettuts" to the variable, $siteName. Otherwise, it'll be equal to the name of the respective site. #### Security Even though this is a small project, let's also try to set some security in place. To prevent the user from automatically assigning a potentially dangerous value to the siteName key, let's ensure that the value is in fact the name of one of our tutorial sites. The in_array() function allows us to determine if a value -- in our case, the value of$siteName -- is equal to one of the items in the $siteList array. #### Caching Ultimately, we'll be using the excellent YQL to perform our queries. Think of YQL as an API for APIs. Rather than having to learn twenty different APIs, YQL's SQL-like syntax allows you to only learn one. However, though YQL does perform a bit of caching on its own, let's also save the RSS feeds to a text file on our server. That way, we can improve performance a fair bit. We begin by creating a new variable,$cache, and making it equal to the location of where the cached file will be stored.

The code above points to the current directory of the file, and then into a cache folder, and, finally, the name of the selected site.

I've decided that this cached file should be updated every three hours. As such, we can run a quick if statement, and determine the last time that the file was updated. If the file does not exist, or the update was longer than three hours ago, we query YQL.

YQL is ridiculously easy to work with. In our case, we'll use it for a very simple purpose: grab the RSS feed, in JSON form, of the site that was passed through the querystring, via siteName. You can experiment with the various commands by using the YQL console.

To query an RSS feed, we using the command: SELECT * FROM feed WHERE url="path/to/rss/feed".

• Nettuts+ Feed: http://feeds.feedburner.com/nettuts
• Psdtuts+ Feed: http://feeds.feedburner.com/psdtuts
• Vectortuts+ Feed: http://feeds.feedburner.com/vectortuts
• etc.

#### Building the Path

For the sake of readability, we'll build up our YQL query in sections.

The key is the second part above; when the page loaded, we grabbed the name of the site from the querystring. Now, we only need to insert it into the SELECT query. Luckily, all of the tutorial sites use Feedburner! Make sure that you urlencode the query to replace any special characters.

Okay, the path is ready; let's use file_get_contents() to grab the feed!

Assuming that $feed is now equal to the returned JSON, we can store the results in a text file. However, let's first ensure that data was returned. As long as something is returned from the query,$feed->query->count will be equal to a value greater than zero. If it is, we'll open the cached file, write the data to the file, and finally close it.

It seems confusing, but it's really not. The function fopen() accepts two parameters:

• The file to open: We stored this path in the $cache variable at the top of the page. Note that, if this file doesn't exist, it will create the file for you. • Access privileges: Here, we can specify which privileges are available. w stands for "write." Next, we open that file, and write the contents of$feed (the returned RSS JSON data) to the file, and close it.

#### Using the Cached File

Above, we first checked whether the cached file was greater than three hours old.

But what if it wasn't? In that case, we run an else statement, and grab the contents of the text file, rather than using YQL.

Lastly, we can't do much with the JSON RSS feed until we decode it with PHP.

And that should do it for our controller! With the logic out of the way, let's include our HTML template.

Here's our final site.php. Click on the expand icon to view it.

## Step 7: The Site Template

At the end of the previous step, we loaded in our template (or view). Go ahead and create that views folder, and site.tmpl.php file. Feel free to name it how you wish. Next, we'll insert our HTML.

#### Points of Interest Above

• Notice how we follow the same basic layout: header, content area, footer.
• As this template will be used for every Tuts+ tutorial site, we need to set the title dynamically. Luckily, if you remember, the site name was passed through the querystring, and stored in the $siteName variable (like, "nettuts"). To capitalize the first letter, and apply the signature + after the name, we'll run the variable through ucwords() (uppercases the first letter of each word in the string), and append a "+":
<?php echo ucwords($siteName).'+'; ?>

• We'll soon be displaying the number of comments for each posting next to the title. We can, again, use ThemeRoller to style it, via the data-counttheme="e" attribute.

#### Filtering Through the Feed

At this point, we have access to the $feed object that contains our RSS feed. To dissect this object, you can either print_r($feed), or use the YQL console for a prettier view. We'll use the latter in this case. Check it out.

To grab the data for each posting, we need to filter through: $feed->query->results->item. PHP makes this a cinch with foreach(). Within the foreach() statement, we can now access the desired values with$item->title, or \$item->comments, which will display the title, and the comment number, respectively. Add the following within the
 tags.

In the code above, we build up a list item, containing the title of the posting, the number of comments, and a link to article.php that also contains the site name and the permanent link (to the original article on the Tuts+ site) in the query-string.

When we view the updated page in the browser, tada!

Notice how the comment count is in a yellow bubble, and is floated to the right? That's because we applied the data-counttheme="e" attribute to the wrapping unordered list. How convenient.

Hmm...I think the text is too large for these long titles. A quick visit to Firebug shows that I can target the h2 tags with a class of .ui-li-heading. Let's return to our stylesheet (mobile.css), and add a new rule:

That's better.

## Step 8: Displaying the Full Posting

The final step is to build article.php, which will display the entire posting. As with site.php, article.php will serve as our controller, and will query the selected article with YQL, and load the appropriate view.

If you've been following along, the code above should look a bit more familiar to you. When we loaded this page, from site.php, we passed through two items, via the query string:

• Site Name: Contains the name of the currently selected tutorial site
• Orig Link: A link to the original posting on the tutorial site

The difference with the YQL query, this time, is that we match the guid (orig link) with the posting that the user clicked on (or pressed). This way, exactly one posting will be returned. Check out this sample YQL query to gain a better idea of what I mean.

#### Article Template

At the bottom of the code above, we loaded the template file for the article page: views/article.tmpl.php. We'll create that file now.

Ah - so familiar. We've already gone over this template. The only difference is that, this time, because there's only one posting from the YQL query to display, we don't need to bother with a foreach() statement.

Unstyled article page

At this point, on your own, the next step would be to begin applying your desired styling to the article. I don't see a need to go over it in this tutorial, as it all comes down to personal taste. Here's my super-minimal version.

Applying a font-size, line-height, padding, and image formatting.

#### Locked Footers

One last thing: in the footer section of the article, we link to the original posting on Nettuts+. In its current state, the reader will only see that when they reach the bottom of the article. Let's lock the footer to the bottom of the current view-point at all times. We can use the data-position attribute to achieve this.

That's better!

## We're Done!

And, with relatively little work, we've successfully built a mobile RSS reader for the Tuts+ sites. It can certainly be expanded to provide additional features, error checking, and performance improvements, but this will hopefully get you started! If you'd like to fork the project and make it better, by all means...do! Thanks for reading, and be sure to refer to the jQuery Mobile documentation for more details. I've no doubt that you'll come across more jQuery mobile tutorials on our sister site, Mobiletuts+.


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Discussion

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Elyse G.

Cornell University

LB

Other Schools

Video Transcript

and this problem. We have a camera with focal length of 15 millimetre that has a CCD sensor. The week of the CCD sensor is 8.7 millimeter and the height of the sea city is 14 millimeter and we have an object of height 1.8 meter equal to 1800 millimeter. So what we're trying to do is we're trying to find the object of stairs perfect object distance so that we can perfectly fit the image of the object at the CCD sensor. So we have to do a couple of things to be able to find that object. This is let's start with the definition of the magnification. So the the definition of the magnification, that magnitude of the magnification is H I over a So which is also D I over t o. And from here, actually, you can write down d I equals each eye over h o times. D'oh! So let's call this equation one. Now we're gonna use the lens equation and using the lens equation we can write down in his distance. D I calls t o times f over d o minus f. And let's call this equation too. All right, so from Equation One and Equation, too. Let's go back here to page one. You can see that D I equals H over H over time CEO and from prison to Diego's duo times f over to you myself. So if you solve this solving equation one into, we're going to get d'oh, that's going to be called to F inside the bracket h over h I plus one. So let's call this equation three. So now, to find the object distance, what we're going to do is we're going to plug in the values of focal length Asia and H i N Equation three. Okay, so we actually have two different cases. One is the portrait more and one is the landscape more so in portrayed, moored in portrait moored. Yeah, we have a chai because h i y. So from here using equation, too. It's our equation. Three dio is going to be cool to h o over h i Y plus one and this is going to be cool to 50 inside the bracket 1800 which was a a Y over 14 plus one, So 1800 was actually a so, um so from here, We're going to get 64 79 millimeter, which is approximately called to 6.48 meter. Okay, so next more is the landscape more okay for the landscape? More? What is gonna happen is a chai is going to be called Thio each i X So from here we'll get d o just similar way as portrait. More Dio is going to be culture 10 3 95 millimeter, which is equal to 10.4 meter.

University of Wisconsin - Milwaukee
Elyse G.

Cornell University

LB

format: plain


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Homotopic maps in the general linear group

Prove that the maps $\mathrm{GL}(n,\mathbb R)\times \mathrm{GL}(n,\mathbb R)\longrightarrow \mathrm{GL}(2n,\mathbb R)$ that send $(A,B)$ to the block matrices $\begin{pmatrix} A & 0 \\ 0 & B \end{pmatrix}$ and $\begin{pmatrix} AB & 0\\ 0 & 1\end{pmatrix}$ are homotopic.

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One thought on “Homotopic maps in the general linear group”

1. The solution we came up with Iván:

The homotopy is given by $(A,B,t)\mapsto \begin{pmatrix} A & 0\\ 0 & B\end{pmatrix} \begin{pmatrix} 1 & tB\\ 0 & 1\end{pmatrix} \begin{pmatrix} 1 & 0\\ -tB^{-1} & 1\end{pmatrix}\begin{pmatrix} 1 & tB\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & -t\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & 0\\ t & 1\end{pmatrix}\begin{pmatrix} 1 & -t\\ 0 & 1\end{pmatrix}$, and it is in fact smooth. This actually proves that the subspace of matrices of the form $\begin{pmatrix} B & 0\\ 0 & B^{-1}\end{pmatrix}$ is contractible in $\mathrm{GL}(2n, \Bbb{R})$.

Some other minor observation: this is some kind of homotopical version of Whitehead’s lemma in K-theory.

Liked by 2 people


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#### Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

#### Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

#### Instant and Unlimited Help

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##### Examples
###### Lessons
1. Introduction to Function Notation
If $f(x) = 5x^2-x+6$ find the following
1. ${f(\heartsuit)}$
2. ${f(\theta)}$
3. ${f(3)}$
4. ${f(-1)}$
5. ${f(3x)}$
6. ${f(-x)}$
7. ${f(3x-4)}$
8. ${3f(x)}$
9. ${f(x)-3}$
2. Express a Function as $f($
$)$

If ${f(x) = \sqrt{x},}$ write the following in terms of the function ${f.}$
1. ${\sqrt{x}+5}$
2. ${\sqrt{x+5}}$
3. ${\sqrt{2x-3}}$
4. ${-8\sqrt{x}}$
5. ${-8\sqrt{2x-3}}$
6. $4\sqrt{x^{5}+9}-1$
3. Find the Value of a Function from Its Graph
Find the value of the following from the given graph

1. ${f(3)}$
2. ${f(0)}$
3. ${f(-5)}$
4. ${f(x)=5,x=?}$
5. ${f(x)=-1,x=?}$
6. ${f(x)=0,x=?}$


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Behind the scenes, ecoevoapps uses the wonderful deSolve package to solve the differential equations in R. If you want to explore how this package works so that you can simulate the dynamics of your own favorite model or tweak existing models in ecoevoapps, read on!

To use deSolve, we first need to write a function that describes the differential equations, which we can then solve using the function deSolve::ode().

## Example 1 - Lotka-Volterra competition

If we want to run the Lotka-Volterra competition model in R, we first write a function using the syntax as follows:

# Hand-coding the Lotka-Volterra competition model
lotka_volterra_competition <- function(time, init, params) {
with (as.list(c(time, init, params)), {
# description of parameters
# r1 = per capita growth rate of species 1
# N1 = population size of species 1
# K1 = carying capacity of species 1
# a = relative per capita effect of species 2 on species 1
# r2 = per capita growth rate of species 2
# N2 = population size of species 2
# K2 = carrying capacity of species 2
# b = relative per capita effect of species 1 on species 2

# Differential equations
dN1 <- r1*N1*(1 - (N1 + a*N2)/K1)
dN2 <- r2*N2*(1 - (N2 + b*N1)/K2)

# Return dN1 and dN2
return(list(c(dN1, dN2)))
})
}

With this function defined, we can now define the parameter values, initial values, and time over which we want to run a simulation:

# define vectors of initial population sizes, time, and parameters
# define initial population sizes for the two species
init <- c(N1 = 10, N2 = 30)
time <- seq(0,500, by = 0.1)
# define the parameter vector -- note that the names should match
# the parameter names we used in the lotka_volterra_competition function above
params <- c(r1 = .2, r2 = .1, K1 = 500, K2 = 600, a = .9, b = 1.1)

We are now ready to simulate the dynamics as follows:

lv_out <- deSolve::ode(func = lotka_volterra_competition,
y = init, times = time, parms = params)
# Plot model output
plot(lv_out[,"N1"]~lv_out[,"time"], type = "l", ylim = c(0, 700))
lines(lv_out[,"N2"]~lv_out[,"time"], type = "l", col = 2)

## Example 2 - predator-prey dynamics

This example shows how the same approach can be used to simulate the Lotka-Volterra predator-prey model with logistic growth in the prey species:

# Define the function
lv_predprey_logPrey <- function(time,init,pars) {
with (as.list(c(time,init,pars)), {
# description of parameters:
# r = per capita growth rate (prey)
# a = attack rate
# e = conversion efficiency
# d = predator death rate
# K = carrying capacity of the prey

dH = r*H*(1 - H/K) - (a*H*P)
dP = e*(a*H*P) - d*P
return(list(c(dH = dH, dP = dP)))

})
}

# Define parameters and initial conditions
time <- seq(0, 250, 0.1)
init <- c(H = 10, P = 10)
params <- c(r = 1, K = 250, a = 0.1, e = 0.2, d = 0.25)

# Now, we are ready to run the mode:

predprey_out <- deSolve::ode(func = lv_predprey_logPrey,
y = init, times = time, parms = params)
# Plot model output
plot(predprey_out[,"H"]~predprey_out[,"time"], type = "l", ylim = c(0,16))
lines(predprey_out[,"P"]~predprey_out[,"time"], type = "l", col = 2)

## More resources

We also strongly recommend that you explore some deSolve tutorials to understand the workflow. (ex. 1; ex. 2; there’s also an example specifically on the Lotka-Volterra predator-prey models!). The package also includes the function deSolve::dede(), which we use in ecoevoapps to simulate the delayed-differential model for logistic growth (see run_logistic_model()).


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Home > INT2 > Chapter 2 > Lesson 2.3.4 > Problem2-117

2-117.

Solve for the missing lengths in the sets of similar figures below.

1. $ABCD \sim JKLM$

If each pair of shapes is similar, then

$\frac{6}{12} = \frac{10}{x} \text{ and }\frac{39}{w} = \frac{3}{7}.$

Start to solve for each variable by multiplying both sides of the equation by the lowest common denominator.

1. $∆NOP \sim ∆XYZ$
$w = 91$


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# Ball trajectory with spin

1. Jan 19, 2016

### RLst

Hi everyone,

Firstly, sorry if this seems a bit rushed, I don’t have much time in between shifts at work. You may have to refer to my links. But if needed, I will add more information later on.

I've been racking my head over this for the last month or so.

I’m kinda doing a private personal project where I need to create a spreadsheet that plots the trajectory of a ball i.e. tennis ball on a graph in the spreadsheet, where the x and y axes of the graph represents the x and y positions of the ball.

The ball needs to be subjected to 3 forces: gravity, air resistance and magnus force (ball spin).

For now, I will assume the spin on the ball will be constant during it’s entire flight path and I’m also ignoring what happens when it bounces on the ground.

I’d like to find the acceleration of the ball, then derive velocity from that, then the displacement/position. All values will be shown on the spreadsheet.

The inputs data will be:
• Initial velocity of ball (m/s)
• Initial angle of trajectory (degrees)
• Spin applied on the ball (rad/s OR rpm)
• Ball radius -> area (m)
• Ball mass (kg
Here are 2 of the references I've been able to find that are the most simplest to me and have helped me the most:
1. Ball trajectory excerpt - Rod Cross
http://www.physics.usyd.edu.au/~cross/TRAJECTORIES/42. Ball Trajectories.pdf
2. Mangeshrai
http://mccabeme.myweb.port.ac.uk/projects2012/MangeshRai.pdf

I’m fairly certain my solution probably lies in the equations of motion in the first article, but unfortunately the math is just a step or two beyond me. I can feel that answer is so close but my trial and error efforts have so far given me results.

Also, the k constant/variable seems to be slightly different on each article, so makes it more difficult to wrap my head around.

Any help would be greatly, greatly appreciated :)

Thank you,
Tony

2. Jan 20, 2016

### Simon Bridge

You'll want a simpler model then - say: modelling weight W, drag D, and magnus effect M ... drag depends on velocity, and magnus effect depends on spin.
Decide on what models you want for those ... say: if +y is up and motion is restricted to the x-y plane:
$\vec W = -mg\hat\jmath$
$\vec D = -\sigma v \vec v$ (quadratic drag)
$\vec M = k\vec S\times\vec v$ ... this give a lift force perpendicular to velocity $\vec v$ and spin $\vec S$.
$k$ and $\sigma$ are the strength of the effect, which you include empirically or just tweek until it feels right.
Commonly $k = \frac{1}{2}\rho C_D A$... but then $C_D$, the coefficient of drag, becomes the empirical parameter.

After that there is just the initial condition ... the velocity and position at time = 0 - and $\sum \vec F = m \vec a$.
Which you want to work as a difference equation for the computer simulation.

3. Jan 21, 2016

### RLst

Hi Simon, thanks for your input. I've seen those formulas in that kinda format somewhere before when I was doing research, so definitely on the right track.

Unfortunately some of the things and symbols in your reply is a little bit over my head... but I'll do my best to try and understand.

FYI, my level of maths and physics is maybe year 11 high school (Australia) @ B or C+ grade, so not the best. Sorry, you might have to dumb down some concepts for me to understand.

So first of all, yes, +Y is up and motion is restricted to the XY plane. X+ is right and so on etc.

Pretty much everything is solved and completed and in plain sight in the first article "42. Ball trajectories".

I can input the initial angle, initial speed, ball mass, etc and get drag and magnus forces... but how do you determine how much displacement or velocity change when a certain x amount of drag or magnus force is applied on the ball?

Since F = ma ∴ a = $\frac {F}{m}$ right?

Then I think my friend said to get velocity from acceleration, you need to integrate... which for me is my roadblock. I can maybe do basic year 10 integrals with some revision, but integrals of gravity+drag+magnus all combined is just way beyond me at the moment.

Ultimately, I just need vx, vy and then the corresponding displacement x, y equations (with reference to time) that I can input into my spreadsheet.
The equations will be something like this:
vx(n) = blah.blah.v(n-1).t - this.that.v(n-1) cos θ...

Attached is what my spreadsheet looks like so far.

I got it off the web somewhere, but the original only took into account gravity and drag, no magnus force.
This is pretty easy since the equations are just the basic physics formulas you see in textbooks i.e. v = at, s = v + at2, etc.
But as soon as I tried to add magnus force to it, it becomes insane and I can no longer keep up with the math.

In the second article I linked (Mangesrai), there was a few equations on page 50. I tried to input the monstrous equation 6.12 and the result was the dotted curve as seen in my spreadsheet. However, adjusting the spin parameters on the spreadsheet essentially does nothing.

In the article it references tait variables of k1 and k2... I just don't get what this is talking about. Also, this only yields the displacement and not velocity at time t.

Not sure what you mean here, don't quite understand the maths notations. Are you saying I should find all the various forces separately and then just subtract them accordingly? Then convert to acceleration, velocity and finally displacement?

Few other things:
• Cd for a tennis ball is 0.55 according to my first link, so I'll set that as a constant and go with that for the whole project
• For tennis balls there are no chaos drag theory thingy happening where once the ball reaches a high enough velocity, the drag force goes down significantly. A tennis ball would never reach those speeds on court and also I think the fuzz on it prevents such phenomenon from happening.
• Tennis ball is 67mm in diameter and mass of 57gm.

#### Attached Files:

• ###### Screen Shot 2016-01-22 at 03.55.17.png
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4. Jan 21, 2016

### Simon Bridge

... or you may have some learning to do ;)

... OK so you already have a way to find the forces ... force is a vector though, so you need magnitude and direction for this to work.
So lets say $\vec W$ = weight (this is the force of gravity on an object), $\vec D$=drag, and $\vec M$=magnus

... that is correct if F is the total force vector and a is the acceleration vector.
Properly, for your case: $\vec W + \vec D + \vec M = m\vec a$
In my prev notation, I was using i,j,k to indicate unit (magnitude 1) vectors pointing along +x,+y, and +z respectively. The "^" on top indicates a unit vector, where an arrow or boldface indicates a regular vector. If you don't know about vectors and vector components, you should stop here and revise: you cannot proceed without that knowledge.

You only need things in the form of a difference equation ... so if T is a small time interval you can use the kinematic (suvat) equations.
You may just want to try modelling basic ballistic motion to start with, to get used to the methods.

If you decide on a small time interval T, and look at times $t=0,T, 2T,3T,\cdots ,nT, \cdots , NT$ so we have N+1 points of time, which you put in cells in your spreadsheet, Then you can get the next velocity from the previous one by doing $v_{n+1} = v_n + a_n T$
Here $v_{n}$ is shorthand for $v(t_n)$ and $t_n = nT: n=0,1,2,3,\cdots, N$ you follow?
For this to work, you need to know $v_0$ ... the initial velocity.

This is all magnitudes, so you have to do this for x and y components.
After that $x_{n+1} = x_n + v_{xn} T$

This process is a form of numerical integration.

... you don't need to know, and I didn't look. The variables will be to account for the relationships between physical quantities - or just to tune the equations - and you can go right from acceleration-time data to displacement time data if you want: just use algebra to include the v-t step.
You want position-time though.

Concentrate on doing your own simulation - don't wast time trying to understand other people's.

Find all the forces separately, then take the vector sum of them, divide by mass, gets acceleration as a vector. From there find velocity and then position. That's not how I'd do it but it is probably the easiest way for you to think about it.

OK - so you also need the density of the ball. The area A in the drag force equation is $A=\pi r^2$ ... but you seem to have a better drag equation for a tennis ball so use that.
The main thing is to remember that forces are vectors.

5. Feb 1, 2016

### RLst

Thanks Simon, that's what I'm here for ;)

I have a ok understanding of vectors so shouldn't be a problem. What I'm doing though is splitting up the forces into x and y components for each force with sin and cos, and solving each component of the force vector separately.

Yep, this is what I ended up doing. I used your simplified WDM method and it seems to work well. I calculated the forces, one for x components, one for y components... added them up, divided the whole thing by m, then used the suvat equations as you mentioned to get $\vec v$ and $\vec s$. Everything is plotted on a graph showing the trajectory of the ball.
Everything seems to work well. . I may post of the spreadsheet later.
I did had slight difficulty finding good k and σ values, but got there eventually through trial and error as you mentioned.
Ideally would've like to use the formulas and values that have been worked out by other people, so that the simulation is as real world as possible. But my attempts were futile and kept coming up with unusual results.
My main concern is the accuracy. If time interval $T$ is relatively large, the results get a bit inaccurate. If I decrease $T$, then it becomes more accurate at the expense of computational power. My project does require the least amount of CPU usage as possible... but it should suffice for now.

Gotcha, good to know.

Can I ask how you would go about doing it? Would your method do away with the inaccuracies of large T time intervals?

What is the density of the ball in this case, and why would I need it? I have the initial velocity + angle of the ball. There is no bounce or rebound to take into effect (yet).

Yes, I do realise this. In my first post I simply wrote the variables without notations etc. But I do realise that each force, does have a magnitude and direction. I'm breaking them down into x and y components in my case as I think it's easier for my project.

6. Feb 1, 2016

### Simon Bridge

You'll have found that the x and y components depend on each other when air resistance is not neglected.

There will be people around interested in it.
I don't use spreadsheets for simulation since GNU-Octave is free. Matlab and Mathematica get used a lot too.

... calculations by other people will be assuming things specific to their models - to be accurate to life you should measure them. You have access to a tennis ball.

It is common to do initial models in a script language like Matlab, just to get the maths right - then do high detail maths using a programming language like C++. There are also a lot of tricks for reducing the computation power needed. Even with all those it is not uncommon to run a program overnight to get a few minutes simulation.

Whole University papers are taught on the subject.
I'd try to use matrix methods... I'd normally work through the error due to finite time-steps by getting a bigger computer xD but there are tricks like having a variable time-step: shorter for times of high curvature.

If you were to calculate the drag from the drag equation, like you said you wanted, then you'd need the density of the ball because it is in the drag equation.

But it looks like you've done well with what you know now.


text:
Home Theory How do you compare two algorithms?

# How do you compare two algorithms?

0 comment

We compare things all the time to determine which one is better over the other. The best example is when you are buying stuff online, we compare prices across websites, two different mobile phones, or even two different outfits. We know that there are different techniques of solving a problem in computer science:

Naturally, when we try to solve a problem with different methods, one would want to compare them. Once they are put side to side, we can then analyze which one performs better, and obviously we would tend to pick up the better one.

##### “Quality” of an Algorithm

When it comes to different outfits or gadgets we have some criteria upon which we evaluate them. So certainly, we want to define some parameters upon which we can compare two algorithms. The two most important factors are:

1. Time: Ever since the invention of computers, programs are written to reduce the time taken to achieve a job. Hence, an algorithm is considered better if it runs faster than the other one.
2. Space: Memory is cheap now days, but everyone loves an application that takes the least amount of RAM. Don’t we all hate when you have 70 tabs open, and a web browser slows down your entire PC. Also, when writing system level code, every extra byte of memory matters. Thus, an algorithm which takes up less RAM would be considered to be a better one.
##### Execution Time:

To put it in plain words, this is the time taken for an algorithm to complete successfully. How do you compare two algorithms based on the execution time? You need to follow a certain set of rules for that:

##### Experimental:
• A fool proof way to compare 2 different algorithms would be to actually run them and observe the results. The one which gives you the output in less time would said to be the better one.
• But, when running these algorithms, you need to ensure that we are using the same hardware for both. A faster processor can give better results for a poorly written algorithm. Also, you cannot rely on a single run. Your computer might be running other workloads in the background and that can affect the performance. Hence, it is recommended that you several iterations of both the techniques and then get an average value before comparing.
• We also need to make sure that the input workload for the algorithms remain the same. Using random input data cannot give us reliable results. It could be possible that certain input loads work unreasonably fast. Example: Trying to sort an already sorted list.

Even with all these conditions in mind, you cannot certainly reach an exact result, as it is scientifically impossible to run 2 algorithms in the exact same conditions. You wouldn’t be able to compare two algorithms which work nearly the same. Thus, you need to do a little deep dive in the actual flow of the algorithm.

##### Analytical:

Your algorithm may contain a lot of conditional blocks, and control blocks that determine the runtime of your algorithm. Suppose you are given with 2 code blocks. Forget about the implementation details, and assume that both of them give the correct answer.

Algorithm 1:

.wp-block-code {
border: 0;
}

.wp-block-code > div {
overflow: auto;
}

.shcb-language {
border: 0;
clip: rect(1px, 1px, 1px, 1px);
-webkit-clip-path: inset(50%);
clip-path: inset(50%);
height: 1px;
margin: -1px;
overflow: hidden;
position: absolute;
width: 1px;
word-wrap: normal;
word-break: normal;
}

.hljs {
box-sizing: border-box;
}

.hljs.shcb-code-table {
display: table;
width: 100%;
}

.hljs.shcb-code-table > .shcb-loc {
color: inherit;
display: table-row;
width: 100%;
}

.hljs.shcb-code-table .shcb-loc > span {
display: table-cell;
}

.wp-block-code code.hljs:not(.shcb-wrap-lines) {
white-space: pre;
}

.wp-block-code code.hljs.shcb-wrap-lines {
white-space: pre-wrap;
}

.hljs.shcb-line-numbers {
border-spacing: 0;
counter-reset: line;
}

.hljs.shcb-line-numbers > .shcb-loc {
counter-increment: line;
}

.hljs.shcb-line-numbers .shcb-loc > span {
}

.hljs.shcb-line-numbers .shcb-loc::before {
border-right: 1px solid #ddd;
content: counter(line);
display: table-cell;
text-align: right;
-webkit-user-select: none;
-moz-user-select: none;
-ms-user-select: none;
user-select: none;
white-space: nowrap;
width: 1%;
}
for(i = 0; i < 10; i++) {
for(j = 0; j< 5; j++) {
// do something...
}
}
Code language: Java (java)

Algorithm 2:

for(i = 0; i < 10; i++) {
// do something...
}

for(j = 0; j < 5; j++) {
// do something...
}
Code language: Java (java)

Let us analyze both the algorithms:

Algorithm 1:

• We are starting a loop i that runs for 10 times.
• Inside the loop we are starting one more loop j that runs for 5 times.
• This means that for each iteration of i, we run j 5 times.
• Hence, the total number of times we do something is 10 * 5 = 50 times.

Algorithm 2:

• We start a loop i that runs for 10 times.
• Now we do something for 10 times.
• Exit the loop.
• We start another loop j that runs for 5 times.
• Now we do something for 5 times.
• Hence, the total number of times we do something was 10 + 5 = 15 times
##### What did we infer from this?

At the first glance it may seem that both the codes have 2 loops. One runs for 10 times and the other runs for 5 times. So, it is natural to feel that both will take the same time to run. But when we compare both of them analytically, we see that algorithm 1 is doing something for 50 times, while algorithm 2 is doing something for just 15 times. This means that algorithm 1 would be a better choice.

This phenomenon is also known as Runtime Complexity.

##### Space Consumption:

We just discussed in our quality criteria that ‘space’ is an important factory to determine which algorithm performs better. What do you mean by that? Let us understand this with a very basic example. To set up some context, we would consider the block size of a single node in a Linked List as 10kb.

Question: You are given with 2 linked lists, that are already sorted. Merge the two lists and return a single sorted list.

According to the question, we are given with 2 lists. Each node in the linked list would have a block size of 10kb, and hence the setup would look something like this:

The size of first list would be 5 * 10 = 50kb, and the size of the second list would be 3 * 10 = 30kb. Hence, currently you are using 80kb of memory. Let us skip the implementation details and look at 2 algorithms to solve this problem.

##### Algorithm 1:
• Create a new list of 5 + 3 = 8 nodes
• Sort the two lists
• Fill up the new list
• Return the result.

This algorithm would look something like:

In the above algorithm, we create a new list that takes up extra 80kb of space. Hence, the total space required to run this algorithm would be:

50 (list 1) + 30 (list 2) + 80 (new list) = 160kb

Thus, we need a total of 160kb memory for algorithm 1 to run successfully. Let us look at another approach to solve the same problem.

##### Algorithm 2:
• Scan the shorter list one by one.
• Take up an element.
• Scan the longer list and insert it at the right place.
• Repeat the process for all the elements of the smaller list.

This algorithm when completed would give us a result something like:

You see, that in the above algorithm, we do not consume any extra space. We were clever enough and we saved space. This algorithm would just need 80kb of space to run. That is just the space required by the two already present lists.

##### What did we infer from this?

When you have to compare these 2 algorithms, we can easily say that algorithm 2 is the clear winner when it comes to saving space. 80kb of extra space may seem very small, but consider writing codes for gigantic lists. You should always compare your methods to write the code keeping space consumed in mind.

This is also known as the Space Complexity.

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text:
# Help please on complex polynomials

I wanted to know if there's any good approaches to these questions

a)By considering $z^9-1$ as a difference of two cubes, write $1+z+z^2+z^3+z^4+z^5+z^6+z^7+z^8$ as a product of two real factors one of which is a quadratic.

b) Solve $z^9-1=0$ and hence write down the 6 solutions of $z^6+z^3+1=0$

c)By letting $y=z+\frac{1}{z}$ and dividing $z^6+z^3+1=0$ by $z^3$, deduce that: $cos\frac{2\pi}{9}+cos\frac{4\pi}{9}+cos\frac{8\pi}{9}=0$

Any help would be greatly appreciated.

-
It seems like the questions themselves suggest good approaches to answering them – muzzlator Apr 4 '13 at 7:35

## 2 Answers

Hint: $a^3-b^3 = (a-b) (a^2+a b+b^2)$. Think about this with $a=z^3$ and $b=1$.

-

(a) \begin{align*} (z - 1)(z^8 + z^7 + \ldots + 1) = z^9 - 1 & = (z^3 - 1)(z^6 + z^3 + 1) \\ & = (z - 1)(z^2 + z + 1)(z^6 + z^3 + 1) \\ \therefore \qquad z^8 + z^7 + \ldots + 1 & = (z^2 + z + 1)(z^6 + z^3 + 1). \end{align*}

(b) The roots of $z^9 - 1$ are $r_n = e^{\frac{2\pi}9ni}$, for $n = 0, \ldots, 8$. Since $r_0 = 1$ is the root of $z - 1$, the other 8 roots are precisely the roots of $z^8 + z^7 + \ldots + 1 = (z^2 + z + 1)(z^6 + z^3 + 1)$. The quadratic $z^2 + z + 1$ has roots $\frac{-1 \pm \sqrt{3}i}{2}$, which are $r_3$ and $r_6$. Therefore, the 6 roots of $z^6 + z^3 + 1$ are $r_1, r_2, r_4, r_5, r_7, r_8$.

(c) \begin{align*} y^3 & = z^3 + 3z + 3\frac 1z + \frac 1{z^3} \\ y^3 - 3y + 1 & = z^3 + 1 + \frac 1{z^3} = \frac{1}{z^3}(z^6 + z^3 + 1) \\ & = \frac{1}{z^3}(z - r_1)(z - r_2)(z - r_4)(z - r_5)(z - r_7)(z - r_8). \end{align*} The roots of $y^3 - 3y + 1$ are $r_i + \frac{1}{r_i}$, but some $r_i$ give the same root because $r_i = \frac{1}{r_{9-i}}$. More precisely, the three roots are $y_1 = r_1 + \frac{1}{r_1} = r_8 + \frac{1}{r_8}$, $y_2 = r_2 + \frac{1}{r_2} = r_7 + \frac{1}{r_7}$, and $y_3 = r_4 + \frac{1}{r_4} = r_5 + \frac{1}{r_5}$.

We also know that the sum of the roots of $y^3 - 3y + 1$ is $0$ because the coefficient in front of $y^2$ is $0$. This means \begin{align*} y_1 + y_2 + y_3 = r_1 + \frac 1{r_1} + r_2 + \frac 1{r_2} + r_4 + \frac 1{r_4} & = 0 \\ 2\left(\cos\frac{2\pi}9 + \cos\frac{4\pi}9 + \cos\frac{8\pi}9\right) & = 0 \end{align*}

-


text:
# If the dryness fraction of a sample by throttling calorimeter is 0.8 and that by separating calorimeter is also 0.8, then the actual dryness fraction of sample will be taken as

This question was previously asked in
UPPSC AE Mechanical 2013 Official Paper II
View all UPPSC AE Papers >
1. 0.8
2. $$\sqrt {0.8}$$
3. 0.64
4. 0.5

Option 3 : 0.64
Free
CT 1: Growth and Development - 1
46784
10 Questions 10 Marks 10 Mins

## Detailed Solution

Concept:

Throttling Calorimeter: A throttling calorimeter is based on the principle of throttling the wet steam so that it becomes superheated.

Separating Calorimeter: It is a vessel used initially to separate some of the moisture from the steam, to ensure superheat conditions after throttling.

Combined Separating and Throttling Calorimeter

A combined separating and throttling calorimeter is therefore found most suitable for accurate measurement of dryness of steam reducing the drawback of Separating Calorimeter (i.e. water particles from wet steam are not fully separated) and the drawback of Throttling Calorimeter (i.e. not suitable for very wet steam).

x= dryness fraction of steam considering separating calorimeter

x= dryness fraction of steam considering the throttling calorimeter

x = actual dryness fraction of steam in the sample

Calculation:

Actual dryness fraction of steam in the sample is:

X = x1 × x2

Given:

x 1 = 0.8, x2 = 0.8

Calculation:

x = 0.8 × 0.8

∴ X = 0.64


text:
# The Floor Is

### 🔥 | Latest

The Floor Is: oh heck the floor is movin (via: @prkrjb)

oh heck the floor is movin (via: @prkrjb)

The Floor Is: The floor is supporting you even when no one is

The floor is supporting you even when no one is

The Floor Is: The floor is supporting you even when no one is

The floor is supporting you even when no one is

The Floor Is: The floor is supporting you even when no one is

The floor is supporting you even when no one is

The Floor Is: Yes…the floor is made of floor.

Yes…the floor is made of floor.

The Floor Is: Luckey Haskins, the floor is lava champion.

Luckey Haskins, the floor is lava champion.

The Floor Is: Luckey Haskins, the floor is lava champion.

Luckey Haskins, the floor is lava champion.

The Floor Is: andletsbringpearl: The floor is Jasper’s redemption Me irl

andletsbringpearl: The floor is Jasper’s redemption Me irl

The Floor Is: moved-to-in-love-and-liberty: The floor is believing that your student group can overthrow the monarchy This post is canceled

moved-to-in-love-and-liberty: The floor is believing that your student group can overthrow the monarchy This post is canceled


text:
# Operation on Vector fields

I am analyzing a program which transforms vector fields by action of diffeomorphisms and feedbacks.

Here are the operation that I don't understand (it's Mathematica code)

dyn=u F + v G;
tmp=Phi.{x,y,z};
newdyn = Inverse[T1].(dyn/.{x->tmp[[1]],y->tmp[[2]],z->tmp[[3]]});

$F,G$ are two (smooth) vector fields on $\mathbb{R}^3\ni (x,y,z)=q$ and we have a control system of the form $$\dot q = u F(q) + v G(q)$$ where $u,v$ are scalar control functions. The components of $F,G$ are polynomials in $q=(x,y,z)^\intercal$ (they are obtained by Taylor expansions).

Consider:
$$\varphi(Q) = \overbrace{(F,G,-[F,G])_{\mid x=y=z=0}^\intercal}^{A}\, \, Q= q.$$

So $\varphi$ is a linear transformation and $\varphi^{-1}(q) = A^{-1} q$ and $\frac{\partial \varphi}{\partial Q}(Q) = A$.

Then they perform the pushforward of $F,G$ by $\varphi$ so that $$\dot Q = A^{-1} \dot q = A^{-1}\left(u\, F(AQ)+v\, G(AQ)\right).$$

Also, what could be the use of such transformation $\varphi$?

On many examples, I observe that $\varphi$ removes the constant terms in the components $F_2$ and $G_1$.

Edit: Source of the Program : Aloui thesis

• This would make more sense to me if you defined $A=[F,G,-[F,G]]$ (without transposition). In this case, $A^{-1}F\bigg|_{Q=0}=\begin{pmatrix}1\\0\\0\end{pmatrix}$ and $A^{-1}G\bigg|_{Q=0}=\begin{pmatrix}0\\1\\0\end{pmatrix}$ so, in the neighbourhood of $0$, you system would turn to $$\begin{cases}\dot{x}=u\\\dot{y}=v\\\dot{z}=0\end{cases}.$$ – Dmitry Nov 17 '17 at 11:08
• What is the source of the problem? – Dmitry Nov 17 '17 at 11:11
• This paper : sci-hub.cc/https://doi.org/10.1007/BF02269424 and the source file in the question. – Smilia Nov 17 '17 at 12:46
• I think you are right, there is no transposition and your answer makes a lot of sense – Smilia Nov 17 '17 at 12:55


text:
# Malliavin calculus¶

My list of publications which are related to Malliavin calculus per se

## The latest publications¶

### Malliavin calculus for independent random variables¶

Abstract: On any denumerable product of probability spaces, we construct a Malliavin gradient and then a divergence and a number operator. This yields a Dirichlet structure which can be shown to approach the usual structures for Poisson and Brownian processes. We obtain versions of almost all the classical functional inequalities in discrete settings which show that the Efron-Stein inequality can be interpreted as a Poincaré inequality or that Hoeffding decomposition of U-statistics can be interpreted as a chaos decomposition. We obtain a version of the Lyapounov central limit theorem for independent random variables without resorting to ad-hoc couplings, thus increasing the scope of the Stein method.

To appear in Stochastic processes and their applications

@article{decreusefond:hal-01565240,
TITLE = {{Malliavin and Dirichlet structures for independent random variables}},
AUTHOR = {Decreusefond, Laurent and Halconruy, H{'e}l{e}ne},
URL = {https://hal.archives-ouvertes.fr/hal-01565240},
JOURNAL = {{Stochastic Processes and their Applications}},
PUBLISHER = {{Elsevier}},
YEAR = {2018},
KEYWORDS = {Talagrand inequality ; Talagrand   inequality ; Stein method ; Malliavin calculus ; Lyapounov CLT ; log-Sobolev inequality ; Dirichlet structure ; Ewens distribution},
`


text:
26 views

Directions for Questions: Answer the following questions based on the statements given below: Following questions are about $6$ project reports of different heights from $6$ different students and of $6$ different colors displayed on a cupboard.

1. There are three Project Reports on each side of the aisle.
2. These six Project Reports are labeled as Piyush, Qadar, Richa, Sandesh, Tanvi and Urvashi.
3. The Project Reports are of different colours, namely, Red, Blue, Green, Orange, Yellow and White.
4. The Project Reports are of different heights. (v) Tanvi, the tallest Project Report, is exactly opposite to the Red coloured Project Report.
5. The shortest Project Report is exactly opposite to the Green coloured Project Report.
6. Urvsahi, the Orange coloured Project Report, is located between Piyush and Sandesh.
7. Richa, the Yellow coloured Project Report, is exactly opposite to Piyush.
8. Qadar, the Green coloured Project Report, is exactly opposite to Urvashi.
9. Piyush, the White coloured Project Report, is taller than Richa, but shorter than Sandesh and Qadar.

What is the color of the report named Richa?

1. Red
2. Blue
3. Green
4. Yellow


text:
## Xiaojun Liu

Xiaojun Liu is a Professor of physics at the State Key Laboratory of Magnetic Resonance and Atomic and Molecular Physics, Wuhan Institute of Physics and Mathematics, Chinese Acedemy of Sciences, China. He received his Ph.D. degree from Anhui Institute of Optics and Fine Mechanics, Chinese Academy of Sciences, China, in July 2000. His research interests are strong-field atomic physics: ionization dynamics of atoms, molecules subject to ultrashort intense laser pulses; coherent control of strong field, atom interaction; and highly excited Rydberg states in external fields.

Biography Updated on 18 November 2009

#### Articles in Scholarly Journals [Incomplete List]

1. First Report of Powdery Mildew Caused by Oidium neolycopersici on Tomato in China
Plant Disease, vol. 92, no. 9, pp. 1370–1370, 2008
2. Search for the decays J/ψ→γρϕ and J/ψ→γρω
Physical Review D, vol. 77, no. 1, 2008
3. Search for the Invisible Decay of J/ψ in ψ(2S)→π^{+}π^{-}J/ψ
Physical Review Letters, vol. 100, no. 19, 2008
4. Anomalous Line Shape of the Cross Section for e^{+}e^{-}→ Hadrons in the Center-of-Mass Energy Region between 3.650 and 3.872 GeV
Physical Review Letters, vol. 101, no. 10, 2008
5. Classical Trajectory Diagnosis of a Fingerlike Pattern in the Correlated Electron Momentum Distribution in Strong Field Double Ionization of Helium
Physical Review Letters, vol. 101, no. 23, 2008
6. Scattering effect of atoms through a Bose-Einstein condensate in an optical Lattice with a single defect
The European Physical Journal D, vol. 49, no. 2, pp. 161–165, 2008
7. Experimental studies of the π+π-π+π-π0, K+K-π+π-π0 and pp̄π+π-π0 final states produced in e+e- annihilation at $\sqrt{s}= 3.773$ and 3.650 GeV
The European Physical Journal C, vol. 52, no. 4, pp. 805–811, 2007
8. Study of J/ψ decaying into ωpp̄
The European Physical Journal C, vol. 53, no. 1, pp. 15–20, 2007
9. Measurement of ψ(2S) Radiative Decays
Physical Review Letters, vol. 99, no. 1, 2007
10. Complex Dynamics of Correlated Electrons in Molecular Double Ionization by an Ultrashort Intense Laser Pulse
Physical Review Letters, vol. 99, no. 1, 2007
11. Alignment effect in nonsequential double ionization of diatomic molecules in strong laser fields
Physical Review A, vol. 76, no. 2, 2007
12. Measurements of J/ψ and ψ(2S) decays into ΛΛ[over ¯]π^{0} and ΛΛ[over ¯]η
Physical Review D, vol. 76, no. 9, 2007
13. Search for the C-parity violating process J/ψ→γγ via ψ(2S)→π^{+}π^{-}J/ψ
Physical Review D, vol. 76, no. 11, 2007
14. Direct measurements of the non-DD[over ¯] cross section σ_{ψ(3770)→non-DD[over ¯]} at E_{cm}=3.773  GeV and the branching fraction for ψ(3770)→non-DD[over ¯]
Physical Review D, vol. 76, no. 12, 2007
15. Time-frequency analysis of molecular high-harmonic generation spectrum by means of wavelet transform and Wigner distribution techniques
Journal of Physics B: Atomic, Molecular and Optical Physics, vol. 39, no. 22, pp. 4747–4758, 2006
16. Orthogonal transmit beamforming scheme based on semidefinite optimisation
European Transactions on Telecommunications, vol. 17, no. 2, pp. 257–268, 2006
17. A Novel Blind Video Watermarking Scheme Based on Independent Dynamic Component
Multidimensional Systems and Signal Processing, vol. 17, no. 1, pp. 59–74, 2006
18. SVM classification of human intergenic and gene sequences
Mathematical Biosciences, vol. 195, no. 2, pp. 168–178, 2005
19. Polyaniline thin films in situ polymerized under very high pressure
Synthetic Metals, vol. 150, no. 2, pp. 175–179, 2005
20. A kinetic model for simulation of blood coagulation and inhibition in the intrinsic path
Journal of Medical Engineering & Technology, vol. 29, no. 2, pp. 70–74, 2005
21. Finding Keywords for Intergenic and Gene Regions for Human Genome
Nucleosides, Nucleotides and Nucleic Acids, vol. 24, no. 3, pp. 191–198, 2005
22. Rose (Rosa rugosa)-flower extract increases the activities of antioxidant enzymes and their gene expression and reduces lipid peroxidation
Biochemistry and Cell Biology, vol. 83, no. 1, pp. 78–85, 2005
23. Nonlinear tunneling and chaos between two Bose–Einstein condensates trapped in time-dependent potential
Physics Letters A, vol. 291, no. 1, pp. 17–21, 2001
24. Fabrication of silicon nanowires
Applied Physics A: Materials Science & Processing, vol. 66, no. 5, pp. 539–541, 1998
25. Properties of Schottky contact of Al on SiGe alloys
Applied Physics Letters, vol. 68, no. 8, p. 1123, 1996
26. A method for fabricating silicon quantum wires based on SiGe/Si heterostructure
Applied Physics Letters, vol. 68, no. 3, p. 352, 1996
27. Realization of silicon quantum wires based on Si/SiGe/Si heterostructure
Zeitschrift f�r Physik B Condensed Matter, vol. 100, no. 4, pp. 489–491, 1996
28. Embedding in fullerene C60
Chemical Physics Letters, vol. 232, no. 1-2, pp. 27–30, 1995
29. Transport property of Si/Si1−xGex/Si p-type modulation doped double heterostructure
Journal of Applied Physics, vol. 76, no. 4, p. 2544, 1994
30. Antimutagenic effect of eight natural foods on moldy foods in a high liver cancer incidence area
Mutation Research/Genetic Toxicology, vol. 279, no. 1, pp. 35–40, 1992


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Next: Logics of algorithmic knowledge Up: Reasoning about algorithmic knowledge Previous: Reasoning about algorithmic knowledge   Contents

## Axioms for algorithmic knowledge

Let us assume that an agent knows within units of time, i.e., he needs time units to infer . Then naturally he is able to do it when even more time is available. So we can take as axiom any formula where . Note that this axiom does not say that knowledge is persistent in the sense that once established it will be available henceforth. The formula does not entails that will know at time point . It does not even imply that will eventually be known at all. In this aspect the present approach makes a more realistic assumption than the persistence axiom in chapter 4.

We have remarked previously that the formula contains more information than . While the latter formula only says that agent is able to derive in finite time if he chooses to, the former one also specifies the amount of time needed by to complete that task. Thus, the implication can be assumed as an axiom.

Let be a provable formula of the logic used by agent . We have argued previously that it cannot be assumed that knows automatically (i.e., without any reasoning.) However, he may know it after some course of reasoning. The interesting question is whether or not he is able to compute reliably within finite time. That would be the case if the agent has a general strategy which fulfills the following two requirements. First, it selects for any formula of the language an algorithm to compute that formula. Second, if the formula is provable then the selected algorithm will terminate with the correct answer. I shall argue that under reasonably weak assumptions, such a strategy exists and can be adopted by any intelligent agent, so that can be safely postulated if is provable.

The set of axioms that any agent presupposes is decidable -- in the normal case even finite. Because the number of permissible inference rules is also limited, all proofs can be generated algorithmically. Hence, there is a general-purpose theorem prover that can validate any theorem in finitely many steps. If the agent's selection mechanism always returns that general algorithm for computing knowledge, he is able to validate every theorem . That is, when presented with a theorem he can select an algorithm which runs on and outputs the answer Yes'' after a finite number of steps. Although the described strategy (always use the same algorithm'') satisfies the stated conditions, it may not be the best: specific problems may be solved much more quickly by special algorithms than by a general-purpose theorem prover. Hence, the following would be a more reasonable strategy. First, the agent analyzes the query and tries to select one of his special algorithms to infer it. If no such algorithm can be found, then the general algorithm is selected. In this way, he can always find an algorithm to verify . (If the selection mechanism is not reliable, i.e., if it could return a wrong algorithm for some queries, then several algorithms can be selected and executed concurrently or interleavingly.)

A strategy to successfully prove every provable formula can be acquired by rational agents. An intelligent agent may learn to use some algorithms to compute knowledge. Such algorithms (together with a suitable selection scheme) can be built into artificial agents. Hence, the rule to infer from can be assumed to be valid.

The statement contains some uncertainty. It is not clear how long agent will need to infer . Can a more definite statement be made? That is, can a natural number (which typically depends on the structure of the theorem ) be determined such that can be assumed as a postulate? The discussion of this question will be postponed until section 5.2.3.

Now suppose that is provable and that each of the formulae can be computed reliably by an agent in finite time, i.e., are regarded to be true. Is it reasonable to infer that can compute reliably if he chooses to derive it? I shall argue that the conclusion can be justified.

When presented with a question , an agent naturally attempts to derive from all what he knows5.3. It is reasonable to assume that the consequence relation used by a rational agent has a certain transitivity property: if all the formulae are derivable from some knowledge base and can be inferred from , then can also be inferred from that knowledge base. Thus we can assume that follows from all what knows. Because agents are assumed to process only a limited amount of information, every consequence of their knowledge can be computed algorithmically. With a suitable selection strategy, e.g., one of the strategies outlined previously, an agent can find an algorithm to compute his knowledge successfully. Consequently, agent is able to compute after a finite number of steps. So, if is a theorem then we can assume that is valid.

As a special case we can assume that is valid, which says that agent can use modus ponens in his reasoning: if he can derive both and then he is also able to derive . Because explicit knowledge implies implicit knowledge, the formula is valid, provided that is a theorem. Thus, agents are able to compute all logical consequences of their explicit knowledge.

Recall that in chapter 4, the persistence axiom (everything that has been once deduced will be available henceforth'') plays a prominent role in justifying the postulates stating that agents are able to use logical laws in their reasoning. In the current approach such an assumption is not necessary because we argue at a higher abstraction level. Only the final result is important, not the intermediate ones.

Since the formula is merely a definitional stipulation, it seems uncontroversial. If that axiom is adopted, the formula can be proved for all natural numbers . More interesting are variants of the consistency axiom. A weak consistency criterion is that agents do not believe apparent contradictions, i.e., their explicit knowledge is consistent: . A stronger requirement is that an agent's implicit knowledge be consistent, which is captured by the schema . That is, the agent's explicit knowledge is free of contradictions and his inference procedures are sound, so that consistency is preserved. (The latter formula is indeed a stronger condition than the former one because follows from .)

What about self-introspection? If an agent knows or does not know something explicitly, he only needs to reflect about himself to find it out. The cost of reflection is usually low, so it can be assumed that self-reflection can be performed in constant time. Hence, the formulae and seem plausible.

The situation is quite different when an agent tries to compute the truth value of where . Although he may actually compute and reflect about that, the result of his computation does not say much about his ability to infer . He may succeed to compute within time units, but there is still a chance that this success is only accidental and not reproducible. On the other hand, even if is false (i.e., agent cannot compute reliably within time units), may still happen to successfully infer after less than time units. Consequently, it is not sound to infer that he can or cannot compute reliably within time units. So, generally neither nor is valid. The same argument applies to the formulae and . Those principles can only be assumed for agents who know their abilities well. They can, for example, be postulated for an agent who works deterministically. For such an agent, a small number of tests may suffice to determine if he can perform a task under certain conditions.

Next: Logics of algorithmic knowledge Up: Reasoning about algorithmic knowledge Previous: Reasoning about algorithmic knowledge   Contents
2001-04-05


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$f$ entire, $g(x, y) := |f(x+iy)|$ integrable on $\mathbb{R}^{2}$ $\implies$ $f \equiv 0$. [duplicate]

Let $$f \in H(\mathbb{C})$$ and $$g(x, y) := |f(x+iy)|$$, and assume $$g$$ is integrable on $$\mathbb{R}^{2}$$. Show then that $$f \equiv 0$$.

Things I Know: $$f$$ is bounded on each $$D(0, R)$$ for $$R > 0$$, but the bounds must increase as $$R \rightarrow \infty$$. Intuitively, this should mean $$\int_{\partial D(0, R)} |f(z)| dz \rightarrow \infty$$ as $$R \rightarrow \infty$$ (assuming $$f \not\equiv 0$$), which could say something about $$g$$. But here I'm talking about upper bounds, which don't help. I could consider the infimum $$f$$ takes on $$\partial D(0, r)$$ to bound below, but these inf's don't necessarily grow larger as $$R$$ does.

My intuition tells me Liouville will come into play, but I'm not sure... Any ideas? Thank you!

Edit: Though this is the same as this question essentially, I want to include proofs that use any complex analytic tools, while the linked question wants a solution based only in elementary calculus.

By the mean-value property of harmonic functions, $$f(z)$$ is the average of $$f$$ over any circle centred at $$z$$, and therefore over any disk centred at $$z$$. A disk $$D$$ of radius $$r$$ centred at $$z$$ has area $$\pi r^2$$, so $$\int_D |f(x+iy)|\; dx\; dy \ge |f(z)| \pi r^2$$ and as $$r \to \infty$$ the only way to avoid an infinite limit is $$f(z)=0$$.


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## Basic College Mathematics (10th Edition)

Published by Pearson

# Chapter 8 - Geometry - 8.6 Circles - 8.6 Exercises - Page 573: 2

#### Answer

To use the formula for finding the area of a circle, you need to know the length of the radius. If you are given the diameter of a circle, what should you do to find the radius. Divide by two.

#### Work Step by Step

Area of a Circle: $A$ = $\pi$$r$$^{2}$ Therefore, to find the area the radius is required. Diameter = 2$r$ Therefore: $r$ = Diameter / 2

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# 10.5 Polar form of complex numbers  (Page 2/8)

 Page 2 / 8

Given $\text{\hspace{0.17em}}z=1-7i,\text{\hspace{0.17em}}$ find $\text{\hspace{0.17em}}|z|.$

$|z|=\sqrt{50}=5\sqrt{2}$

## Writing complex numbers in polar form

The polar form of a complex number    expresses a number in terms of an angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and its distance from the origin $\text{\hspace{0.17em}}r.\text{\hspace{0.17em}}$ Given a complex number in rectangular form expressed as $\text{\hspace{0.17em}}z=x+yi,\text{\hspace{0.17em}}$ we use the same conversion formulas as we do to write the number in trigonometric form:

$\begin{array}{l}\text{\hspace{0.17em}}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ \text{\hspace{0.17em}}y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$

We review these relationships in [link] .

We use the term modulus to represent the absolute value of a complex number, or the distance from the origin to the point $\text{\hspace{0.17em}}\left(x,y\right).\text{\hspace{0.17em}}$ The modulus, then, is the same as $\text{\hspace{0.17em}}r,\text{\hspace{0.17em}}$ the radius in polar form. We use $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ to indicate the angle of direction (just as with polar coordinates). Substituting, we have

$\begin{array}{l}z=x+yi\hfill \\ z=r\mathrm{cos}\text{\hspace{0.17em}}\theta +\left(r\mathrm{sin}\text{\hspace{0.17em}}\theta \right)i\hfill \\ z=r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\hfill \end{array}$

## Polar form of a complex number

Writing a complex number in polar form involves the following conversion formulas:

$\begin{array}{l}\hfill \\ x=r\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \\ y=r\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ r=\sqrt{{x}^{2}+{y}^{2}}\hfill \end{array}$

Making a direct substitution, we have

$\begin{array}{l}z=x+yi\hfill \\ z=\left(r\mathrm{cos}\text{\hspace{0.17em}}\theta \right)+i\left(r\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\hfill \\ z=r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\hfill \end{array}$

where $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ is the modulus    and $\theta$ is the argument    . We often use the abbreviation $\text{\hspace{0.17em}}r\text{cis}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ to represent $\text{\hspace{0.17em}}r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right).$

## Expressing a complex number using polar coordinates

Express the complex number $\text{\hspace{0.17em}}4i\text{\hspace{0.17em}}$ using polar coordinates.

On the complex plane, the number $\text{\hspace{0.17em}}z=4i\text{\hspace{0.17em}}$ is the same as $\text{\hspace{0.17em}}z=0+4i.\text{\hspace{0.17em}}$ Writing it in polar form, we have to calculate $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ first.

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{0}^{2}+{4}^{2}}\hfill \\ r=\sqrt{16}\hfill \\ r=4\hfill \end{array}$

Next, we look at $\text{\hspace{0.17em}}x.\text{\hspace{0.17em}}$ If $\text{\hspace{0.17em}}x=r\mathrm{cos}\text{\hspace{0.17em}}\theta ,\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}x=0,\text{\hspace{0.17em}}$ then $\text{\hspace{0.17em}}\theta =\frac{\pi }{2}.\text{\hspace{0.17em}}$ In polar coordinates, the complex number $\text{\hspace{0.17em}}z=0+4i\text{\hspace{0.17em}}$ can be written as $\text{\hspace{0.17em}}z=4\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}4\text{cis}\left(\text{\hspace{0.17em}}\frac{\pi }{2}\right).\text{\hspace{0.17em}}$ See [link] .

Express $\text{\hspace{0.17em}}z=3i\text{\hspace{0.17em}}$ as $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}\text{cis}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ in polar form.

$z=3\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)$

## Finding the polar form of a complex number

Find the polar form of $\text{\hspace{0.17em}}-4+4i.$

First, find the value of $\text{\hspace{0.17em}}r.$

$\begin{array}{l}r=\sqrt{{x}^{2}+{y}^{2}}\hfill \\ r=\sqrt{{\left(-4\right)}^{2}+\left({4}^{2}\right)}\hfill \\ r=\sqrt{32}\hfill \\ r=4\sqrt{2}\hfill \end{array}$

Find the angle $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ using the formula:

$\begin{array}{l}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\hfill \\ \mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{-4}{4\sqrt{2}}\hfill \\ \mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{1}{\sqrt{2}}\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\theta ={\mathrm{cos}}^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3\pi }{4}\hfill \end{array}$

Thus, the solution is $\text{\hspace{0.17em}}4\sqrt{2}\text{cis}\left(\frac{3\pi }{4}\right).$

Write $\text{\hspace{0.17em}}z=\sqrt{3}+i\text{\hspace{0.17em}}$ in polar form.

$z=2\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)$

## Converting a complex number from polar to rectangular form

Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. In other words, given $\text{\hspace{0.17em}}z=r\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right),\text{\hspace{0.17em}}$ first evaluate the trigonometric functions $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Then, multiply through by $\text{\hspace{0.17em}}r.$

## Converting from polar to rectangular form

Convert the polar form of the given complex number to rectangular form:

$z=12\left(\mathrm{cos}\left(\frac{\pi }{6}\right)+i\mathrm{sin}\left(\frac{\pi }{6}\right)\right)$

We begin by evaluating the trigonometric expressions.

$\mathrm{cos}\left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\pi }{6}\right)=\frac{1}{2}\text{\hspace{0.17em}}$

After substitution, the complex number is

$z=12\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)$

We apply the distributive property:

The rectangular form of the given point in complex form is $\text{\hspace{0.17em}}6\sqrt{3}+6i.$

## Finding the rectangular form of a complex number

Find the rectangular form of the complex number given $\text{\hspace{0.17em}}r=13\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{12}.$

If $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{12},\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{y}{x},\text{\hspace{0.17em}}$ we first determine $\text{\hspace{0.17em}}r=\sqrt{{x}^{2}+{y}^{2}}=\sqrt{{12}^{2}+{5}^{2}}=13\text{.}$ We then find $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =\frac{x}{r}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{y}{r}.$

$\begin{array}{l}z=13\left(\mathrm{cos}\text{\hspace{0.17em}}\theta +i\mathrm{sin}\text{\hspace{0.17em}}\theta \right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=13\left(\frac{12}{13}+\frac{5}{13}i\right)\hfill \\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=12+5i\hfill \end{array}$

The rectangular form of the given number in complex form is $\text{\hspace{0.17em}}12+5i.$

Convert the complex number to rectangular form:

$z=4\left(\mathrm{cos}\frac{11\pi }{6}+i\mathrm{sin}\frac{11\pi }{6}\right)$

$z=2\sqrt{3}-2i$

## Finding products of complex numbers in polar form

Now that we can convert complex numbers to polar form we will learn how to perform operations on complex numbers in polar form. For the rest of this section, we will work with formulas developed by French mathematician Abraham de Moivre (1667-1754). These formulas have made working with products, quotients, powers, and roots of complex numbers much simpler than they appear. The rules are based on multiplying the moduli and adding the arguments.

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Miranda
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I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
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Miranda
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jai
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Propessor
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algebra is a branch of the mathematics to calculate expressions follow.
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Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
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(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
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what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function


text:
Shape Convergence for Aggregate Tiles in Conformal Tilings

13 Mar 2017 Kenyon Richard Stephenson Kenneth

Given a substitution tiling $T$ of the plane with subdivision operator $\tau$, we study the conformal tilings $\mathcal{T}_n$ associated with $\tau^n T$. We prove that aggregate tiles within $\mathcal{T}_n$ converge in shape as $n\rightarrow \infty$ to their associated Euclidean tiles in $T$...

PDF Abstract


text:
Map $F$ open and $F(\mathbb{R})$ is a submanifold

Let $F: \mathbb{R} \rightarrow S^1 \times S^1$ defined by $$F(t) = ((\cos2\pi t, \sin2 \pi t), (\cos2\pi \alpha t, \sin2 \pi \alpha t))$$.

For which values of $\alpha$ the map $F$ is open? For which values of $\alpha$ the set $F(\mathbb{R})$ is a submanifold?

ps.: my initial idea was to verify what is required for the coordinate functions $F_1$ and $F_2$ to be a local diffeomorphismos. Hence he would find conditions on the $\alpha$. And it would use that local diphomorphism product is local diffeomorphism and therefore open. But I did not succeed.

• This map is called Kronecker's flux on $2$-torus. – Armando j18eos May 6 '18 at 23:38

For general values of $\alpha$, each coordinate of $F$ wraps $\Bbb R$ around $S^1$ (if $\alpha=0$, then $F$ is just a circle "parallel" to the first "axis"). So, if we imagine a point going to the right in $\Bbb R$ and its respective image point on the torus $S^1\times S^1$, this means that each coordinate (the projection on each of the "axes" $S^1$) keeps spinning around. So, intuitively, we can hope the image $F(\Bbb R)$ to be something like in the figure below.

In fact, we are being optimistic. In general, $F$ is not a closed (i.e. periodic) curve as in the figure. Maybe $F$ keeps "infinitely" going around $S^1\times S^1$ and in this case $F(\Bbb R)$ is dense in $S^1\times S^1$. The fact is that $F$ is a closed curve if and only if $\alpha \in \Bbb Q$. But let's answer the first question first:

For which values of $\alpha$ the map $F$ is open?

Answer: for none of them. Indeed, by the coordinates of $F$ we see that $F$ is continuous ($C^\infty$, in fact). Suppose that for some value of $\alpha$ the map $F$ is also open. Suppose also that $\alpha\neq 0$ (the case $\alpha=0$ is similar). Put $\epsilon=\frac{1}{2}\min\{1,\,1/|\alpha|\}$>0. Then $G:(-\epsilon,\epsilon)\to F(-\epsilon,\epsilon)$, with $G=F|_{(-\epsilon,\epsilon)}$ is open, bijective and continuous, therefore, a homeomorphism. But $(-\epsilon,\epsilon)$ has dimension 1, while $F(-\epsilon,\epsilon)$ has dimension 2 (because it is open in $S^1\times S^1$). This is impossible (because of this). This proves that $F$ is never open.

Comment: I am doubtful about this question. Shouldn't it be: "For which values of $\alpha$ the map $F$ is open onto its image?" Then this would be very related to the cases below. I think $F$ is open onto its image if and only if $\alpha\in \Bbb Q$...

For which values of $\alpha$ the set $F(\Bbb R)$ is a submanifold?

Answer: for and only for rational values of $\alpha$. Let's prove that.

Suppose $\alpha\in \Bbb Q$: suppose also $\alpha> 0$, $\alpha\neq 1$, the cases $\alpha\leq 0$ or $\alpha=1$ are similar. Denote $\alpha=\frac{p}{q}$, with $p,q\in \Bbb N$ having no common factors other than $\pm 1$. Then \begin{align}F(t+q)&=((\cos 2\pi(t+q),\sin 2\pi(t+q)),(\cos 2\pi\alpha(t+q),\sin 2\pi\alpha(t+q)))\\ &=((\cos (2\pi t+2\pi q),\sin (2\pi t+2\pi q)),(\cos (2\pi\alpha t+2\pi p),\sin (2\pi\alpha t+2\pi p)))\\ &=((\cos 2\pi t,\sin 2\pi t),(\cos 2\pi\alpha t,\sin 2\pi\alpha t))\\ &=F(t), \end{align} so $F$ is $q$-periodic. Therefore, parametrizing $S^1$ by $(\cos 2\pi \theta,\sin 2\pi \theta),\,\, \theta \in [0,1]$, we define \begin{align} \begin{matrix} G:&S^1&\to&S^1\times S^1\\ &(\cos 2\pi \theta,\sin 2\pi\theta)&\mapsto&((\cos 2\pi q \theta,\sin 2\pi q\theta),\,(\cos 2\pi p \theta,\sin 2\pi p\theta)). \end{matrix} \end{align} One verifies that $G$ is an immersion from $S^1$ into $S^1\times S^1$. Also, because $p$ and $q$ have no common factors, $G$ is injective. Since $S^1$ is compact and $G(S^1)\subset S^1\times S^1$ is Hausdorff, we have that $G:S^1\to S^1\times S^1$ is a homeomorphism onto its image $G(S^1)$ (by this result). Therefore, $G$ is an embedding or, in other words, $G(S^1)$ is a submanifold of $S^1\times S^1$. So, we are done if we prove that $G(S^1)=F(\Bbb R)$.

Note that $\theta\in [0,1]$, then putting $t=q\theta$, we have \begin{align} G(S^1)&\ni ((\cos 2\pi q \theta,\sin 2\pi q\theta),\,(\cos 2\pi p \theta,\sin 2\pi p\theta))\\ &=((\cos 2\pi t,\sin 2\pi t),\,(\cos 2\pi \alpha t,\sin 2\pi \alpha t))\in F(\Bbb R). \end{align}

Therefore, $G(S^1)\subset F(\Bbb R)$. Conversely, if $t\in \Bbb R$ is arbitrary, then because $F$ is $q$-periodic, there exists some $t'\in [0,q]$ such that $F(t)=F(t')$. Then, putting $\theta=t'/q\in [0,1]$, by the same equation above with $t'$ replaced for $t$ we get that $F(\Bbb R)\ni F(t)=F(t')=G(\theta)\in G(S^1)$. This shows that $F(\Bbb R)=G(S^1)$ is a submanifold of $S^1\times S^1$.

Suppose now $\alpha\notin \Bbb Q$: I will prove that $F(\Bbb R)$ is not submanifold a bit more intuitively than I has been doing in the item above (it's pretty geometric but somewhat troublesome to write down; maybe I can improve with details later). If $\alpha\notin \Bbb Q$, then $F$ wraps $\Bbb R$ around $S^1\times S^1$ just like the figure above, BUT never closing (i.e. with no period at all). It keeps going on and on, always going spinning around $S^1\times S^1$, in such a way that $F(\Bbb R)$ is dense in $S^1\times S^1$. Therefore, for any point $p\in F(\Bbb R)$, every sufficiently small open neighborhood of $p$ in $F(\Bbb R)$ has infinitely many path-connected components (remember, an open neighborhood of $p$ in $F(\Bbb R)$ is some $U\cap F(\Bbb R)$, for some $U\ni p$ open in $S^1\times S^1$). Therefore, $F(\Bbb R)$ with its induced topology from $S^1\times S^1$ is not locally path-connected. Since every submanifold is itself a manifold, $F(\Bbb R)$ cannot be a submanifold because every manifold is locally path-connected.

It is for this helpful to view the torus $S^1 \times S^1$ as a square $[0,1 ]^2$ with opposite edges identified. In one dimension, for example, $S^1$ can be viewed as $[0, 1]$ with the "edges" 0 and 1 identified, via the map $\theta \in [0,1 ] \mapsto (\cos(2\pi \theta), \sin(2\pi \theta))$.

With this said, viewed on the square with edges identified (call this $A$, even though really it is the same space up to homeomorphism), your map $F : \mathbb{R} \to S^1 \times S^1$ becomes $\tilde{F}(t) = (t, \alpha t)$. This is a much easier map to visualize; consider the question in this setting. For then the image of $\tilde{F}$ is just the graph of the function $f(t) = \alpha t$, taken back to the fundamental domain. Now try some cases.

If $\alpha = 1$, then we just get the diagonal line $(t, t)$ through the unit square. This is clearly a submanifold. What about for $\alpha = 1/2$? 2? Any rational number? If $\alpha$ is irrational, what happens?


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Question: ath1121501 and GO
0
13.0 years ago by
Martin Olivier70 wrote:
Hi all, I need to extract information from the library ath1121501. I would like to obtain a list to have a correspondance between GO terms and the gene names. The list should be such that the names of the list are the gene names (AT4G19650" ,"AT5G47010", etc...) and each member of the list contains the vector of all the terms GO in which the gene is affected. The result should be similar to : $AT4G36260 [1] "GO:0048479" ....$AT5G47010 [1] "GO:000814" ... etc... Thanks for your help, Olivier.
go ath1121501 • 526 views
modified 13.0 years ago by James W. MacDonald51k • written 13.0 years ago by Martin Olivier70
Answer: ath1121501 and GO
0
13.0 years ago by
United States
James W. MacDonald51k wrote:
Martin Olivier wrote: > Hi all, > > I need to extract information from the library ath1121501. > I would like to obtain a list to have a correspondance between GO terms and > the gene names. > > The list should be such that the names of the list are > the gene names (AT4G19650" ,"AT5G47010", etc...) and each member of the > list > contains the vector of all the terms GO in which the gene is affected. > The result should be similar to : > > $AT4G36260 > [1] "GO:0048479" .... > >$AT5G47010 > [1] "GO:000814" ... > > etc... > a <- mget(ls(ath1121501GO)[1:10], ath1121501GO) > b <- sapply(a, names) > names(b) <- sapply(mget(ls(ath1121501GO)[1:10], ath1121501SYMBOL), function(x) x[1]) > b $ATMG00640.1 [1] "GO:0000276" "GO:0005739" "GO:0015986" "GO:0046933"$ATMG00650.1 [1] "GO:0005739" "GO:0003954" "GO:0006118" "GO:0045333" "GO:0045279" $ATMG00660.1 [1] "GO:0005739"$ATMG00670.1 [1] "GO:0012505" $ATMG00680.1 [1] "GO:0005739"$ATMG00690.1 [1] "GO:0005739" $ATMG00710.1 [1] "GO:0008372" "GO:0005554" "GO:0000004"$ATMG00720.1 [1] "GO:0008372" "GO:0005554" "GO:0000004" $ATMG00740.1 [1] "GO:0008372" "GO:0005554" "GO:0000004"$multiple [1] "multiple" HTH, Jim > > Thanks for your help, > Olivier. > > _______________________________________________ > Bioconductor mailing list > Bioconductor at stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > Search the archives: http://news.gmane.org/gmane.science.biology.informatics.conductor -- James W. MacDonald, M.S. Biostatistician Affymetrix and cDNA Microarray Core University of Michigan Cancer Center 1500 E. Medical Center Drive 7410 CCGC Ann Arbor MI 48109 734-647-5623 ********************************************************** Electronic Mail is not secure, may not be read every day, and should not be used for urgent or sensitive issues.


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# Level statistics of one-dimensional Schrödinger operators with random decaying potential

Level statistics of one-dimensional Schrödinger operators with random decaying potential - Descarga este documento en PDF. Documentación en PDF para descargar gratis. Disponible también para leer online.

Descargar gratis o leer online en formato PDF el libro: Level statistics of one-dimensional Schrödinger operators with random decaying potential
We study the level statistics of one-dimensional Schr\-odinger operator with random potential decaying like $x^{-\alpha}$ at infinity. We consider the point process $\xi L$ consisting of the rescaled eigenvalues and show that : iac spectrum case for $\alpha \frac 12$, $\xi L$ converges to a clock process, and the fluctuation of the eigenvalue spacing converges to Gaussian. iicritical case for $\alpha = \frac 12$, $\xi L$ converges to the limit of the circular $\beta$-ensemble.

Autor: Shinichi Kotani; Fumihiko Nakano

Fuente: https://archive.org/


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# Countries of the World - an Analysis of Sporcle Quiz Data

## or: what's a Burkina Faso?

This post comes as a result of a comment a friend made about what he thought was a good idea for a /r/dataisbeautiful post: “Some kind of graph showing the correlation between the population of a country and the number of people that answered it in the countries of the world sporcle quiz.” he said in a WhatsApp message. I think he knew exactly what would happen once I saw that suggestion…

My grandfather used to work as a painter in a car garage. I remember him telling me that the sorts of people he worked with, mechanics and engineery types, had this quirk where if you were stuck with something, you simply had to stand in the middle of the shop looking puzzled. “What’s that you’ve got there? Oh let me have a look. Ah yeah, you see, what you need there…” And solutions to your problem would be offered up by these inquisitive minds.

Apparently, I suffer from the same problem, except with data.

Anyway…

Sporcle gives you access to the global results at the end of the quiz (I got 124 out of 197, managing to forget that Afghanistan and Colombia exist along the way). I’m well aware that this isn’t a representative sample of the global population. Judging by the Alexa results, Sporcle’s users are mainly in English-speaking countries like the UK, USA and Australia, so if anything this is a look at how well the English-speaking world knows the rest.

Next, population data: Wikipedia was my starting point. I headed on over to the List of countries by population (United Nations) article, and then converted the table to a CSV with this handy tool. There’s a couple of different articles with similar titles, but this one also includes labels of UN continental region and UN statistical region which will be useful later on when looking for regional effects on the “memorability” of a country.

After a bit of manual cleaning up of the data - things like removing British Overseas Territories and Crown Dependencies, adding Kosovo, renaming “Côte d’Ivoire” and “Macedonia” to “Ivory Coast” and “Republic of Macedonia” for consistency etc. - it was a simple case of joining the two tables together in R and outputting to a CSV file.

Here’s a sample:

Country Percentage Answered UN continental region UN statistical region Population2017
Afghanistan 79.5% Asia Southern Asia 35530081
Albania 64.5% Europe Southern Europe 2930187
Algeria 71.6% Africa Northern Africa 41318142
Andorra 66.2% Europe Southern Europe 76965
Angola 58.9% Africa Middle Africa 29784193

Full table is available here.

And finally some plots:

These are the same plots with differing labels.

The grey lines are a linear-regression of “Memorability” against population: countries above the line are more memorable than their population would predict, countries below the line, less memorable.

No real surprises here, the upper right region is dominated by the G20 countries with the bizarre outlier that is the Vatican over on the top left, and then the African states that apparently people forget exist.

I posted these to /r/dataisbeautiful, where another user suggested that using GDP in place of population might yield some interesting results. I tried it out. Here’s the same plots as above but with Nominal GDP on the X-axis:

There’s a lot less spread here which suggets that GDP makes for a better predictor than population. Another user suggested that the reason for this is that Nominal GDP is relevant for international trade.

The R scripts for the two above sets of plots are here:
Sporcle-population.R
Sporcle-GDP.R

Finally, I did a bit of simple modelling in Stan to determine what the continent-level effects were, just a fixed-effects model with a Student-T residual:

$$\text{Percentage Answered} \sim \text{Student}(\nu, \beta^\text{Continent} + \beta^\text{Population} \cdot log(\text{Population}), \sigma)$$

And similarly, using Nominal GDP:

$$\text{Percentage Answered} \sim \text{Student}(\nu, \beta^\text{Continent} + \beta^\text{GDP} \cdot log(\text{Nominal GDP}), \sigma)$$

The Stan model file is here for interested parties, it’s the same for both models but with differing input data.

What’s important is the results. First, with population:

Here general geographic effects are evident, negative effects for Africa, positive effects for Europe and North America.

And then with GDP:

Arguably these results are more interesting: once the effect of GDP is removed, there’s not an awful lot of distance between geographic regions.

Finally, R scripts for doing the Stan analysis and plots are here:
Sporcle-stan-population.R
Sporcle-stan-GDP.R

I originally did this all back in June, but only just bothered to write it up.