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## 1. How do I insert a picture in Beamer?

In beamer, figures are inserted essentially in the same way as in any other LaTeX document, with very small differences. In LaTeX, we use the graphicx package to insert images (although in beamer we will not have to import it explicitly, since beamer does it for us), which provides the command:

\includegraphics{file}

where file is the file name (or the file path, if the image is not included in the
current working directory) that we want to insert.

The above command accepts a series of optional arguments, the most important ones being:

• height and width to scale the size of the figure.

When only one of these options is set, the other dimension is scaled so that the aspect ratio of the original image is preserved.

Other less common options are also provided:

• angle can be passed a rotation angle in degrees, and the image will be rotated this amount counter-clockwise,
• keepaspectratio is a boolean that forces the image to preserve its aspect ratio.

But if you want to do very fancy things, like both rescaling and rotating an image, be aware that the order in which the options are given matters. The package graphicx interprets the keys from left to right; this means that rotating and then scaling is not the same as scaling and then rotating.

Illustrative example:

In the following example, you can see the difference:

% Rotate an image in Beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

\begin{document}

\begin{frame}{First images in beamer}
\centering
\includegraphics[width=4cm,angle=45]{Ques}
\includegraphics[angle=45,width=4cm]{Ques}
\end{frame}

\end{document}

which produces:

Observe how the two images are slightly different.

## 2. Add figure caption in Beamer

Although with this we have a totally functional way of inserting images, in practice we don’t insert them this way, since the image has neither a caption nor a way to reference it. It is more convenient to wrap the \includegraphics command inside the figure environment. This is a floating environment that lets us set a caption and a label, and also use position specifiers to control where the image will be placed. However, and this is the main point where beamer differs from other LaTeX documents, the position specifiers have no effect in beamer presentations. They are ignored, and the image is simply placed in the same position as in the source code.

Illustrative example:

In the following example, we show how to use the figure environment to insert images:

% Add numbered caption in Beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% set captions with numbers
\setbeamertemplate{caption}[numbered]

\begin{document}

\begin{frame}{Images in beamer}

Observe Figure \ref{fig:question}: it is the
first figure I have inserted in beamer.

\begin{figure}
\centering
\includegraphics[width=0.25\textwidth]{Ques}
\caption{My first figure in beamer.}
\label{fig:question}
\end{figure}

\end{frame}

\end{document}

Compiling this code yields:

Observe that I have added the declaration:

\setbeamertemplate{caption}[numbered]

in the preamble. The reason for this is that, by default, beamer does not number
the pictures
(this is another key difference with the usual LaTeX documents), but since I wanted the figure numbered to reference it, I had to change how the caption looks in the beamer template.

Usually, you will not need the number, since every frame will contain one, at most two, images to be explained, and it is not convenient to use numbers as references throughout a presentation (as it is in a document, where you can easily go back and forth).

## 3. Customize figure caption

We have already seen how to slightly change the caption appearance, by adding a number to it or not. But we can further customize it by changing its font size and color. The following example shows the beamer theme options that we have to modify in order to customize our caption:

% Customize figure caption in Beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% customize the caption
\setbeamerfont{caption}{size=\large}
\setbeamercolor{caption}{fg=blue}
\setbeamercolor{caption name}{fg=red}

\begin{document}

\begin{frame}{Customized caption}

\begin{figure}
\centering
\includegraphics[width=0.25\textwidth]{Ques}
\caption{My first figure in beamer.}
\label{fig:question}
\end{figure}

\end{frame}

\end{document}

which yields the following result:

## 4. Figure and text side by side LaTeX beamer

It is common to write a frame with a figure next to a certain explanation. For this purpose, we can use beamer’s columns environment, as it is done in the following example:

% Text and figure side by side
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

\begin{document}

\begin{frame}{Figure with other content}

\begin{columns}
% Column 1
\begin{column}{0.5\textwidth}
Here I can explain in detail what the figure represents.
\end{column}

% Column 2
\begin{column}{0.5\textwidth}
\begin{figure}
\centering
\includegraphics[width=0.7\textwidth]{Ques.png}
\caption{A figure that is next to a certain explanation.}
\end{figure}
\end{column}

\end{columns}

\end{frame}

\end{document}

The frame produced by this code is shown below:

Observe that the \textwidth command in the line:

\begin{column}{0.5\textwidth}

represents the whole text width of the frame, whereas the \textwidth in the \includegraphics declaration only represents the text width of the column, which is half of the total text width.

This means that, although we pass to the image the option width=0.7\textwidth, it doesn’t take 70% of the frame, but instead, it takes 70% of the column.

## 5. How to align an image inside a beamer frame?

Almost all of the time we have been using the figure environment to wrap the \includegraphics command so that beamer treats the images as floating objects. However, we can also use the raw \includegraphics command, and align it using pure TeX filling commands. The reason behind this is that the \includegraphics command just creates a TeX’s box with the image inside it; that is, for the TeX system it is just as any other letter.

To easily centre the image, as we have been doing throughout all the tutorial, we can use the \centering command just before \includegraphics. However, if we want to left-align the image, we will have to use the command \hfill just after including the image, so that all the horizontal space after the image is filled by TeX.

Let me show you a complete minimal example of how this would work:

% Right aligned figure
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

\begin{document}

\begin{frame}{Right-aligned figure}
\hfill
\includegraphics[width=0.2\textwidth]{Ques.png}
\end{frame}

\end{document}

Compiling this code yields:

As you may have already guessed, if instead of filling the left space we fill the right space (that is, we put the \hfill command after the \includegraphics command) what we get is a left aligned figure.

## 6. Position an image arbitrarily inside a frame

This part is considered in this step-by-step tutorial “Precise positioning of an image in Beamer“. It uses TikZ to position images at any position relative to the frame.

## 7. Changing the opacity of an image in Beamer

Although we are not going to dive into all of TikZ possibilities, here we are going to explore another functionality that the tikzpicture environment offers: changing the opacity of a figure.

This option is especially interesting when combined with beamer overlay specifications because you can put an image with half its opacity and totally uncover it once you are going to actually talk about it. Even further, you could even decrease the image’s transparency as you get closer to talking about it; this would look very cool.

The following example shows a small implementation of this idea:

% Opacity of images in Beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% Required package
\usepackage{tikz}
\begin{document}

\begin{frame}{Opacity of an image}

\begin{tikzpicture}[remember picture, overlay]
\onslide<1>
\node
[
left=1cm,
opacity=0.3] at (current page.east)
{
\includegraphics[width=0.3\textwidth]{Ques.png}
};

\onslide<2>
\node
[
left=1cm,
opacity=1] at (current page.east)
{
\includegraphics[width=0.3\textwidth]{Ques.png}
};
\end{tikzpicture}
\end{frame}

\end{document}

which yields the following slides:

In the above code, we placed the image 1 cm far from the left side of the frame using absolute positioning provided by TikZ. The first version of the image has 30% opacity and the second one has 100% opacity.

## 8. Overlay specifications and figures

However, if we don’t want such a fancy implementation of opacity, and just want the image to be shown on a given slide, beamer offers us the possibility to pass an overlay specification to the \includegraphics command. For example, the declaration:

\includegraphics<2-4>[width=\textwidth]{image.png}

will make the image.png file appear only on slides 2 to 4. More details are presented in this lesson: “Creating Overlays in Beamer“.

## 9. Wrapping text around an image

In beamer, we can wrap text around a figure, but not with beamer built-in commands. We have to use the external wrapfig package. In the following example, we use the environment wrapfigure that this package provides to wrap text around a figure:

% Wrapping text around image in beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% Required package
\usepackage{wrapfig}

% to generate some dummy text
\usepackage{blindtext}

\begin{document}

\begin{frame}{How to wrap text around a graphic.}

\begin{wrapfigure}{r}{0.5\textwidth}
\centering
\includegraphics[width=0.2\textwidth]{Ques.png}
\caption{Wrapping text around figures}
\end{wrapfigure}

\small \blindtext
\end{frame}

\end{document}

which produces the following output:

Observe that the wrapfigure environment works as the figure environment, in the sense that it makes the image floating, and you can add a \caption and a \label to the figure. However, this command accepts two mandatory arguments:

• the first one is to select where we want the image, it can be r or l, that is, right or left;
• the second one is the width to be reserved for the image.

## 10. Writing text over an image

This can also be done inside a beamer frame, but for that purpose, we have to load the versatile TikZ package. In the following example, we illustrate an easy way to do it:

% Text over an image in beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% Required package
\usepackage{tikz}

\begin{document}

\begin{frame}{Text over an image}

\centering
\begin{tikzpicture}
\node (image)
{
\includegraphics[width=0.7\textwidth]{nature1.png}
};

\node
[
fill=teal,
align=center,
text=white,
font={\huge\bfseries}
] at (image.center) {A beautiful photo!};

\end{tikzpicture}
\end{frame}

\end{document}

Compiling this code yields:

Let’s dissection the commands used in this example:

• First, we insert an image inside the tikzpicture environment as a node called (image).
• Then, we create a second node where its content is aligned at centre, with options to use a white, huge and bold font. Moreover, we added the option fill=teal to fill the node with a teal color.
• The key is that we position this node at the centre of the previous one; this position is easily identified with (image.center).
• Finally, we create the contents of the node itself, which are just the text string
A beautiful photo!.

## 11. Image as a frame background

It is easy to use an image as a frame background in beamer, both globally and locally. In the following example, we put both into practice:

% Background image (local and global options)
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% global background must be put in the preamble
\setbeamertemplate{background canvas}
{%
\includegraphics[width=\paperwidth, height=\paperheight]{nature1.png}%
}

\begin{document}

\begin{frame} {Default background}

\end{frame}

% to use a specific background for a frame or a series
% of frames, we have to enclose the command and the frames
% inside the same group
{
\setbeamertemplate{background canvas}
{%
\includegraphics[width=\paperwidth,height=\paperheight]{nature2.png}
}

\begin{frame} {Special background}

\end{frame}
}

\end{document}

which yields:

You can see that the procedure is very intuitive: we just change the background canvas option of the beamer theme to the image that we want to use as background. However, when importing it we should make sure that the size is adjusted to the frame size; otherwise, we will get undesired results.

## 12. How to produce a title page with an image

To add an image to the title page and get something similar to the following illustration, I recommend the following step-by-step tutorial: “How to add an image to the title page in beamer?“.

## 13. Inserting subfigures in beamer

With the subcaption package, we can build floating figure environments that contain more than one image, each with its corresponding caption and label. The following example shows how to do so:

% Subfigures in Beamer
\documentclass{beamer}

% Theme choice
\usetheme{AnnArbor}

% Required package
\usepackage{subcaption}

% so that the captions are numerated
\setbeamertemplate{caption}[numbered]

\begin{document}

\begin{frame}{Subfigures in Beamer}

You can see in Figure \ref{fig:images} that I have inserted two images, Figures \ref{fig:nature1} and \ref{fig:nature2}, that I can reference independently.

\begin{figure}
\centering
\begin{subfigure}[t]{0.4\textwidth}
\includegraphics[width=0.5\textwidth]{nature2.png}
\caption{Image of the nature V1.}
\label{fig:nature1}
\end{subfigure}
\begin{subfigure}[b]{0.4\textwidth}
\includegraphics[width=\textwidth]{nature1.png}
\caption{Image of the nature V2.}
\label{fig:nature2}
\end{subfigure}
\caption{Two images I want to insert.}
\label{fig:images}
\end{figure}

\end{frame}

\end{document}

Here is the obtained result:

• As you can see, first we create a usual figure environment with its corresponding \caption and \label.
• Then, inside of it, for every subfigure we create a subfigure environment, which works essentially as a figure environment, with its corresponding \includegraphics to insert the image, its \caption and its \label.
• However, this environment takes a mandatory argument, which is the space to be allocated for the corresponding image (in the previous example, 0.4\textwidth) and also an optional argument, which determines the positioning of the image inside its allocated space. This argument can be either c, t or b, standing for centre, top and bottom.
• By default the image is centred, but in the previous example I have used t and b, so that you can see the difference between the two options.
• As you can see, different labels enable us to reference either each one of the subfigures or the figure as a whole.

## Conclusion

• In this lesson, we have addressed a lot of possible scenarios regarding the use of figures that you may face when writing your presentation. However, the options offered by beamer are inscrutable, and you can always combine what you have learned here to produce new and surprising results.


text:
B Discrete definitions of Physics Standard Units

Tags:
1. Sep 8, 2016

Stephanus

Dear PF Forum,
Perhaps it's not a question, just a light discussion.
Time
From this, we get the standard time
https://en.wikipedia.org/wiki/Caesium_standard, it's 9,192,631,770 ticks per second.
I think this number should be discreet. No need to tune it to some figures after decimal point.
Is this right? Although it wouldn't match 1/86400 day length. But, then again, for a very big body like earth with all its dynamic changes (weather, ocean, tectonic movement) it can't be too precise.

Length
And from time, we get the standard length:
https://en.wikipedia.org/wiki/Speed_of_light
Which is 9,192,631,770 / 299,792,458. And I think 299,792,458 is not discreet. Is this true? This is the law of nature, right?Or..., as in https://en.wikipedia.org/wiki/Rømer's_determination_of_the_speed_of_light
We arbitrarily set the speed of light: 299,792,458 meter per second, (whatever "meters" means) and we define that length that time travels when Ca 133 ticks 299,792,458 (discreet?) times and call it 299,792,458 meters (discreet?)

Mass
In junior high, we were taught that 1 kg is the mass of 1 litre (length already defined) water in 3.980C. https://en.wikipedia.org/wiki/Properties_of_water#Density_of_water_and_ice. But there is 1 catch. It's 3.980C in 1 bar atmospheric pressure. But we define pressure in N/m2. And 1 Newton is the force to accelerate 1 kg for 1 meter per second per second. And we meet mass (KG) again .
So since Avogadro had set that 1 mole is 6.022140857(74)×1023 . So can we safely say that 1 gram is the mass of 6.022140857(74)×1023 hydrogen atoms? (Without deuterium/tritium isotop impurity?)
And I think the rest, energy, calorie, volt, etc are derived from this standard unit.

Perhaps my questions are not physical, but rather historical.
Are these numbers:
9,192,631,770, time
299,792,458, length
6.022140857(74)×1023 , mass
discreet?
Just curious to know .
Thank you very much

2. Sep 8, 2016

ZapperZ

Staff Emeritus
It can't be too precise? For whom?

Your GPS had to take into account even the minute time dilation effects to keep its precision. How big of an error do you think you'll end up with if the time clock on your GPS is rounded off to the nearest second? Would you like to fly on an airplane with that kind of an error, especially when it is estimating where the ground is during landing?

Zz.

3. Sep 8, 2016

Stephanus

I mean Caesium clock is very, very precise, but it's the earth rotation that can't be too precise. It's a very big body. But I do believe that GPS is accurate. Or..., the earth rotation is very precise?

4. Sep 8, 2016

Orodruin

Staff Emeritus
No. The 9192631770 oscillations per second is a definition. Just like the definition of the meter in terms of a standard stick is obsolete, so is the definition of the second as a fraction of a day.

What do you mean by discrete? It is a single value and it is a definition.

This is not the definition of the kilogram. It is a property of water.

5. Sep 8, 2016

Stephanus

First of all, English is not my first language. Perhaps I can't express myself clearly.
I'll try the other way.
So we arbitrarily define 1 second is 9192631770 oscillations of Ca 133, which is almost close to 1/86400 the length of day?
I mean Discrete is Integer.
Take hydrogen frequency (as I saw in Contact movie) for example.
Hydrogen frequency is: 1,420,405,751.78 Hz
https://en.wikipedia.org/wiki/Hydrogen_line
After we define the length of 1 second is 9192631770 Hz (integer) of Ca 133, then it's very rarely in nature that Hydrogen frequency coincidently has an integer value for 1 second. So I mean, perhaps we arbitrarily define 1 second is 9192631770 (integer value) Hz of Ca 133 frequency and the other elements frequency follows this standard.

Yes, it is. But I think long ago, people had to define kilogram, and they used this property of water.
http://www.dailymail.co.uk/sciencet...curate-definition-following-breakthrough.html
https://en.wikipedia.org/wiki/Kilogram#Definition

I know that PF Forum Guidelines strongly against personal theories. But if I may indulge myself...
Supposed we have already had the standard for time definiton 9,192,631,770 Hz of Ca 133.
How do we define length? I think we can define it by the distance the light takes when Ca 133 has ticked 9,192,631,770 times.
Because we haven't had the accurate definition of length, right?
And we have metres (1/40000 the circular of latitude lines of earth accross Paris). And perhaps now we arbitrarily (?) define 1 metres is 1/299,792,458 (integer number?) the distance the light takes when Ca 133 has ticked 9,192,631,770 (integer number) times. Which is almost close to the definition of 1/40000 the latitude of earth?

6. Sep 8, 2016

Orodruin

Staff Emeritus
No, it is not. The kilogram is currently defined by the IPK artefact, nothing else. It is likely this definition will be changed to one involving only natural constants in the future but this is the current definition.

Then say integer. Discrete is something else.

This is the definition of the length unit. One meter is the distance that light travels in 1/299 792 458 seconds by definition.
again, this is not how the meter is defined today.

format: plain


text:
# Define Or Explain the Following Concept :Elasticity of Demand . - Economics

#### Question

Define or explain the following concept :

Elasticity of demand .

#### Solution

Elasticity of Demand is the degree of responsiveness of quantity demand for a commodity towards change in its price. It is calculated as  E_d  = (% text { change in demand for × good })/(text { % change in price of × good })

Is there an error in this question or solution?

#### APPEARS IN

2017-2018 (March) (with solutions)
Question 2.1.3 | 2.00 marks

#### Video TutorialsVIEW ALL [1]

Define Or Explain the Following Concept :Elasticity of Demand . Concept: Measurement of Elasticity of Demand.
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text:
Limit of sequence as $n$ goes to infinity with $n$ in every term

I am studying for an exam in introductory analysis and came across this exercise in a practice exam:

Determine the limit of $(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$.

I know that every single fraction in the sequence goes to $0$ so by the addition rule I thought maybe the entire sequence adds to $0$? My intuition tells me I'm wrong since there are infinite terms. I'm not sure where else to start.

Any hints would be appreciated. Thanks!

• Every term goes to $0$ but the number of terms increases with $n$ – Ross Millikan Jun 16 '18 at 14:58
• If every term goes to $0$ but the number of terms goes to $\infty$ then the limit may be some positive number. A simple example is this: $$\underbrace{\frac 1 n + \cdots + \frac 1 n}_{3n \text{ terms}}.$$ The terms approach $0$ but the sum is always $3,$ so the limit as $n$ grows is $3.$ – Michael Hardy Jun 16 '18 at 14:59
• You should not say "infinite terms" if you mean "infinitely many terms". An infinite term is a term that is infinite. If there were some of those, then there would be infinite terms, but there might not be infinitely many terms, for example if there were only six of them. – Michael Hardy Jun 16 '18 at 15:00
• Take a look at this $(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2}) < (\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})$ as $n$ goes to $+\infty$. and the one on the left is simply $\frac{(n)(n+1)}{2n^2}$ which you can probably show goes to $1/2$ as $n \rightarrow \infty$ – Uday Khanna Jun 16 '18 at 15:01

$(\frac{1}{n^2}+\frac{2}{n^2}+\frac{3}{n^2}+ \cdots + \frac{n}{n^2})\le(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})\le(\frac{1}{n^2-n}+\frac{2}{n^2-n}+\frac{3}{n^2-n}+ \cdots + \frac{n}{n^2-n})\ \forall \ n \in N$
$\Rightarrow a_n= \frac{(n)(n+1)}{2(n^2)} \le (\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1}) \le \frac{(n)(n+1)}{2(n^2-n)} =b_n$
By L-Hopital's rule on the upper and lower bouding sequences we have $\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}b_n=\frac{1}{2}$ and therefore by Squeeze theorem we get
$\lim_{n\rightarrow\infty}(\frac{1}{n^2}+\frac{2}{n^2-1}+\frac{3}{n^2-2}+ \cdots + \frac{n}{n^2-n+1})=\frac{1}{2}$


text:
zbMATH — the first resource for mathematics

On partial regularity for the 3D nonstationary Hall magnetohydrodynamics equations on the plane. (English) Zbl 1336.35287
The purpose of this paper is to study the three-dimensional non-stationary Hall magnetohydrodynamics equation, with an extra quadratically nonlinear Hall term, with homogeneous Dirichlet boundary conditions. The main theorem characterizes the existence of a globally weak solution for an approximate system with given initial conditions. The proof consists of several vector analysis calculations and uses Sobolev embedding, Caccioppoli, Cauchy-Schwartz, Poincaré, Hölder’s and Youngs inequalities, and the Vitali convergence theorem.

MSC:
 35Q35 PDEs in connection with fluid mechanics 35Q85 PDEs in connection with astronomy and astrophysics 76W05 Magnetohydrodynamics and electrohydrodynamics 85A30 Hydrodynamic and hydromagnetic problems in astronomy and astrophysics 35D30 Weak solutions to PDEs
Full Text:
References:
 [1] M. Acheritogaray, P. Degond, A. Frouvelle, and J.-G. Liu, Kinetic formulation and global existence for the Hall-magnetohydrodynamic system, Kinet. Relat. Models, 4 (2011), pp. 901–918. · Zbl 1251.35076 [2] L. Caffarelli, R. Kohn, and L. Nirenberg, Partial regularity of suitable weak solutions of the Navier-Stokes equations, Comm. Pure Appl. Math., 35 (1982), pp. 771–831. · Zbl 0509.35067 [3] D. Chae, P. Degond, and J.-G. Liu, Well-posedness for Hall-magnetohydrodynamics, Ann. Inst. H. Poincaré Anal. Non Linéaire, 31 (2014), pp. 555–565. [4] D. Chae and J. Lee, On the blow-up criterion and small data global existence for the Hall-magnetohydrodynamics, J. Differential Equations, 256 (2014), pp. 3835–3858. · Zbl 1295.35122 [5] D. Chae and M. Schonbek, On the temporal decay for the Hall-magnetohydrodynamic equations, J. Differential Equations, 255 (2013), pp. 3971–3982. · Zbl 1291.35212 [6] D. Chae and J. Wolf, On partial regularity for the steady Hall-magnetohydrodynamics system, Comm. Math. Phys., 339 (2015), pp. 1147–1166. · Zbl 1328.35165 [7] E. Dumas and F. Sueur, On the weak solutions to the Maxwell-Landau-Lifshitz equations and to the Hall-magnetohydrodynamic equations, Comm. Math. Phys., 330 (2014), pp. 1179–1225. · Zbl 1294.35094 [8] J. Fan, S. Huang, and G. Nakamura, Well-posedness for the axisymmetric incompressible viscous Hall-magnetohydrodynamic equations, Appl. Math. Lett., 26 (2013), pp. 963–967. · Zbl 1315.35164 [9] T. Forbes, Magnetic reconnection in solar flares, Geophys. Astrophys. Fluid Dyn., 62 (1991), pp. 15–36. [10] M. Giaquinta, Multiple Integrals in the Calculus of Variations and Nonlinear Elliptic Systems, Ann. of Math. Stud. 105, Princeton University Press, Princeton, NJ, 1983. · Zbl 0516.49003 [11] H. Homann and R. Grauer, Bifurcation analysis of magnetic reconnection in Hall-MHD systems, Phys. D, 208 (2005), pp. 59–72. · Zbl 1154.76392 [12] O. A. Ladyzehnskaya and G. A. Seregin, On partial regularity of suitable weak solutions to the three-dimensional Navier-Stokes equations, J. Math. Fluid Mech., 1 (1999), pp. 356–387. · Zbl 0954.35129 [13] M. J. Lighthill, Studies on magneto-hydrodynamic waves and other anisotropic wave motions, Philos. Trans. Roy. Soc. London Ser. A, 252 (1960), pp. 397–430. · Zbl 0097.20806 [14] F. H. Lin, A new proof of the Caffarelli-Kohn-Nirenberg theorem, Comm. Pure Appl. Math., 51 (1998), pp. 241–257. · Zbl 0958.35102 [15] F. H. Lin and C. Liu, Partial regularity of the dynamic system modeling the flow of liquid crystals, Discrete Contin. Dyn. Syst., 2 (1996), pp. 1–23. [16] H. Miura and D. Hori, Hall effects on local structure in decaying MHD turbulence, J. Plasma Fusion Res., 8 (2009), pp. 73–76. [17] J. M. Polygiannakis and X. Moussas, A review of magneto-vorticity induction in Hall-MHD plasmas, Plasma Phys. Control Fusion, 43 (2001), pp. 195–221. [18] G. D. Prato, Spazi $$\mathcal{L}^{p,θ}(Ω,δ)$$ e loro proprietà, Ann. Mat. Pura Appl., 69 (1965), pp. 383–392. · Zbl 0145.16207 [19] V. Scheffer, Partial regularity of solutions to the Navier-Stokes equations, Pacific J. Math., 66 (1976), pp. 535–552. · Zbl 0325.35064 [20] D. Shalybkov and V. Urpin, The Hall effect and the decay of magnetic fields, Astronom. Astrophys., 321 (1997), pp. 685–690. [21] A. N. Simakov and L. Chacón, Quantitative, comprehensive, analytical model for magnetic reconnection in Hall-magnetohydrodynamics, Phys. Rev. Lett., 101 (2008), 105003. [22] M. Wardle, Star formation and the Hall effect, Astrophys. Space Sci., 292 (2004), pp. 317–323. [23] J. Wolf, Regularität schwacher Lösungen elliptischer und parabolischer Systeme partieller Differentialgleichungen mit Entartung. Der fall \(1
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
[This article was first published on Revolutions, and kindly contributed to R-bloggers]. (You can report issue about the content on this page here)
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

A violin plot is a combination box plot and a kernel density plot: it starts with a box plot, and adds a rotated kernel density plot to each side of the box plot. You can create violin plots with the vioplot function (from the vioplot package) package in R

When looking at the volatility of financial instruments, the financial blogger “Milk Trader” suggests Volatility Violins can be helpful. Using the getSymbols function to download VIX and SPX data from Yahoo, he provides code to compare the volatility of the two indices. As suggested by a commenter, I adapted the code to scale the SPX data to match the annualized VIX:

SPX <- getSymbols("^GSPC", auto.assign=FALSE)[,4]
SPX <- Delt(SPX, k=30)
SPX <- na.locf(SPX, na.rm=TRUE)
SPX <- abs(SPX)*sqrt(365/30)

and the corresponding volatility violins are below:

Follow the link below for the code to create charts like these for other symbols.

Milk Trader: Volatility Violins


text:
04-68 Kurt Broderix, Hajo Leschke, Peter Mueller
Continuous integral kernels for unbounded Schroedinger semigroups and their spectral projections (325K, Postscript) Mar 8, 04
Abstract , Paper (src), View paper (auto. generated ps), Index of related papers

Abstract. By suitably extending a Feynman-Kac formula of Simon [Canadian Math. Soc. Conf. Proc. \textbf{28} (2000), 317--321], we study one-parameter semigroups generated by (the negative of) rather general Schr{\"o}dinger operators, which may be unbounded from below and include a magnetic vector potential. In particular, a common domain of essential self-adjointness for such a semigroup is specified. Moreover, each member of the semigroup is proven to be a maximal Carleman operator with a continuous integral kernel given by a Brownian-bridge expectation. The results are used to show that the spectral projections of the generating Schr\"odinger operator also act as Carleman operators with continuous integral kernels. Applications to Schr{\"o}dinger operators with rather general random scalar potentials include a rigorous justification of an integral-kernel representation of their integrated density of states -- a relation frequently used in the physics literature on disordered solids.

Files: 04-68.src( 04-68.comments , 04-68.keywords , jfa.ps )


text:
# SAE AMS7731

## Gold Wire and Ribbon 99.97Au Annealed

active, Most Current
 Organization: SAE Publication Date: 1 March 2018 Status: active Page Count: 6
##### scope:

Form

This specification covers pure gold in the form of wire 0.005 to 0.040 inch (0.012 to 1.000 mm) inclusive, in nominal diameter, and ribbon 0.002 to 0.051 inch (0.05 to 1.28 mm) inclusive, in nominal thickness (see 8.5).

Application

This product has been used typically for electronic components requiring essentially nonmagnetic properties, but usage is not limited to such applications.

### Document History

SAE AMS7731
March 1, 2018
Gold Wire and Ribbon 99.97Au Annealed
Form This specification covers pure gold in the form of wire 0.005 to 0.040 inch (0.012 to 1.000 mm) inclusive, in nominal diameter, and ribbon 0.002 to 0.051 inch (0.05 to 1.28 mm) inclusive, in...
May 1, 2007
Gold Wire and Ribbon 99.97Au Annealed
Form This specification covers pure gold in the form of wire and ribbon. Application This product has been used typically for electronic components requiring essentially nonmagnetic properties,...
May 1, 2007
Gold Wire and Ribbon 99.97Au Annealed
Form This specification covers pure gold in the form of wire and ribbon. Application This product has been used typically for electronic components requiring essentially nonmagnetic properties,...
October 1, 1986
Gold Wire and Ribbon 99.97Au Annealed
1. SCOPE: 1.1 Form: This specification covers pure gold in the form of wire and ribbon. 1.2 Application: Primarily for electronic components requiring essentially nonmagnetic properties.


text:
# Help with calculating probabilities from collected data

I am teaching about frequency tables and reading probabilities off of them. In the first two screens, students answer questions and the data is collected and shows up in the next set of screens. In those following screens, students are asked to find specific probabilities based on the information in the collected data. However, I cannot seem to get those to work. Can anyone help?

I found two errors which may correct it.

On screen 3, I changed the CL to:

table = class2
#you needed to increase the second numbers by 1 because of title column in the other table
a1 = table.cellNumericValue(1,2)
b1 = table.cellNumericValue(1,3)
c1 = table.cellNumericValue(1,4)

a2 = table.cellNumericValue(2,2)
b2 = table.cellNumericValue(2,3)
c2 = table.cellNumericValue(2,4)

a3 = table.cellNumericValue(3,2)
b3 = table.cellNumericValue(3,3)
c3 = table.cellNumericValue(3,4)

#for division, you need to use this \frac{}{} notation
cellContent(1,3): when numericValue("\frac{${a3}}{${c3}}")= class3.cellNumericValue(1,2) "✅"  otherwise "❌"
cellContent(2,3): when numericValue("\frac{${b3}}{${c3}}")= class3.cellNumericValue(2,2) "✅"  otherwise "❌"
cellContent(3,3): when numericValue("\frac{${c1}}{${c3}}")= class3.cellNumericValue(3,2) "✅"  otherwise "❌"
cellContent(4,3): when numericValue("\frac{${c2}}{${c3}}")= class3.cellNumericValue(4,2) "✅"  otherwise "❌"

correct:

class3.cellContent(1,3)="✅" and class3.cellContent(2,3)="✅" and class3.cellContent(3,3)="✅" and class3.cellContent(4,3)="✅"`
1 Like


text:
# Logistic map

The logistic map is a polynomial mapping (equivalently, recurrence relation) of degree 2, often cited as an archetypal example of how complex, chaotic behaviour can arise from very simple non-linear dynamical equations.[1] The map was popularized in a seminal 1976 paper by the biologist Robert May,[2] in part as a discrete-time demographic model analogous to the logistic equation first created by Pierre François Verhulst.[3] Mathematically, the logistic map is written

where is a number between zero and one that represents the ratio of existing population to the maximum possible population. The values of interest for the parameter are those in the interval . This nonlinear difference equation is intended to capture two effects:

• reproduction where the population will increase at a rate proportional to the current population when the population size is small.
• starvation (density-dependent mortality) where the growth rate will decrease at a rate proportional to the value obtained by taking the theoretical "carrying capacity" of the environment less the current population.

However, as a demographic model the logistic map has the pathological problem that some initial conditions and parameter values lead to negative population sizes. This problem does not appear in the older Ricker model, which also exhibits chaotic dynamics.

The case of the logistic map is a nonlinear transformation of both the bit-shift map and the case of the tent map.

## Behavior dependent on

The image below shows the amplitude and frequency content of some logistic map iterates for parameter values ranging from 2 to 4.

By varying the parameter , the following behavior is observed:

• With between 0 and 1, the population will eventually die, independent of the initial population.
• With between 1 and 2, the population will quickly approach the value , independent of the initial population.
• With between 2 and 3, the population will also eventually approach the same value , but first will fluctuate around that value for some time. The rate of convergence is linear, except for , when it is dramatically slow, less than linear (see Bifurcation memory).
• With between 3 and , from almost all initial conditions the population will approach permanent oscillations between two values. These two values are dependent on .
• With between 3.44949 and 3.54409 (approximately), from almost all initial conditions the population will approach permanent oscillations among four values. The latter number is a root of a 12th degree polynomial (sequence A086181 in the OEIS).
• With increasing beyond 3.54409, from almost all initial conditions the population will approach oscillations among 8 values, then 16, 32, etc. The lengths of the parameter intervals that yield oscillations of a given length decrease rapidly; the ratio between the lengths of two successive such bifurcation intervals approaches the Feigenbaum constant . This behavior is an example of a period-doubling cascade.
• At (sequence A098587 in the OEIS) is the onset of chaos, at the end of the period-doubling cascade. From almost all initial conditions, we no longer see oscillations of finite period. Slight variations in the initial population yield dramatically different results over time, a prime characteristic of chaos.
• Most values of beyond 3.56995 exhibit chaotic behaviour, but there are still certain isolated ranges of that show non-chaotic behavior; these are sometimes called islands of stability. For instance, beginning at [4] (approximately 3.82843) there is a range of parameters that show oscillation among three values, and for slightly higher values of oscillation among 6 values, then 12 etc.
• The development of the chaotic behavior of the logistic sequence as the parameter varies from approximately 3.56995 to approximately 3.82843 is sometimes called the Pomeau–Manneville scenario, characterized by a periodic (laminar) phase interrupted by bursts of aperiodic behavior. Such a scenario has an application in semiconductor devices.[5] There are other ranges that yield oscillation among 5 values etc.; all oscillation periods occur for some values of . A period-doubling window with parameter is a range of -values consisting of a succession of sub-ranges. The th sub-range contains the values of for which there is a stable cycle (a cycle that attracts a set of initial points of unit measure) of period . This sequence of sub-ranges is called a cascade of harmonics.[6] In a sub-range with a stable cycle of period , there are unstable cycles of period for all . The value at the end of the infinite sequence of sub-ranges is called the point of accumulation of the cascade of harmonics. As rises there is a succession of new windows with different values. The first one is for ; all subsequent windows involving odd occur in decreasing order of starting with arbitrarily large .[6][7]
• Beyond , almost all initial values eventually leave the interval and diverge.

For any value of there is at most one stable cycle. If a stable cycle exists, it is globally stable, attracting almost all points.[8]:13 Some values of with a stable cycle of some period have infinitely many unstable cycles of various periods.

The bifurcation diagram at right summarizes this. The horizontal axis shows the possible values of the parameter while the vertical axis shows the set of values of visited asymptotically from almost all initial conditions by the iterates of the logistic equation with that value.

Bifurcation diagram for the logistic map. The attractor for any value of the parameter is shown on the vertical line at that .

The bifurcation diagram is a self-similar: if you zoom in on the above-mentioned value and focus on one arm of the three, the situation nearby looks like a shrunk and slightly distorted version of the whole diagram. The same is true for all other non-chaotic points. This is an example of the deep and ubiquitous connection between chaos and fractals.

## Chaos and the logistic map

Two- and three-dimensional Poincaré plots show the stretching-and-folding structure of the logistic map
A cobweb diagram of the logistic map, showing chaotic behaviour for most values of
Logistic function (blue) and its iterated versions (orange), , and for . For example, for any initial value on the horizontal axis, gives the value of the iterate four iterations later.

The relative simplicity of the logistic map makes it a widely used point of entry into a consideration of the concept of chaos.[1] A rough description of chaos is that chaotic systems exhibit a great sensitivity to initial conditions—a property of the logistic map for most values of between about 3.57 and 4 (as noted above).[9] A common source of such sensitivity to initial conditions is that the map represents a repeated folding and stretching of the space on which it is defined. In the case of the logistic map, the quadratic difference equation (1) describing it may be thought of as a stretching-and-folding operation on the interval .[10]

The following figure illustrates the stretching and folding over a sequence of iterates of the map. Figure (a), left, shows a two-dimensional Poincaré plot of the logistic map's state space for , and clearly shows the quadratic curve of the difference equation (1). However, we can embed the same sequence in a three-dimensional state space, in order to investigate the deeper structure of the map. Figure (b), right, demonstrates this, showing how initially nearby points begin to diverge, particularly in those regions of corresponding to the steeper sections of the plot.

This stretching-and-folding does not just produce a gradual divergence of the sequences of iterates, but an exponential divergence (see Lyapunov exponents), evidenced also by the complexity and unpredictability of the chaotic logistic map. In fact, exponential divergence of sequences of iterates explains the connection between chaos and unpredictability: a small error in the supposed initial state of the system will tend to correspond to a large error later in its evolution. Hence, predictions about future states become progressively (indeed, exponentially) worse when there are even very small errors in our knowledge of the initial state. This quality of unpredictability and apparent randomness led the logistic map equation to be used as a pseudo-random number generator in early computers.[10]

Since the map is confined to an interval on the real number line, its dimension is less than or equal to unity. Numerical estimates yield a correlation dimension of 0.500 ± 0.005 (Grassberger, 1983), a Hausdorff dimension of about 0.538 (Grassberger 1981), and an information dimension of 0.5170976... (Grassberger 1983) for (onset of chaos). Note: It can be shown that the correlation dimension is certainly between 0.4926 and 0.5024.

It is often possible, however, to make precise and accurate statements about the likelihood of a future state in a chaotic system. If a (possibly chaotic) dynamical system has an attractor, then there exists a probability measure that gives the long-run proportion of time spent by the system in the various regions of the attractor. In the case of the logistic map with parameter and an initial state in , the attractor is also the interval and the probability measure corresponds to the beta distribution with parameters and . Specifically,[11] the invariant measure is . Unpredictability is not randomness, but in some circumstances looks very much like it. Hence, and fortunately, even if we know very little about the initial state of the logistic map (or some other chaotic system), we can still say something about the distribution of states arbitrarily far into the future, and use this knowledge to inform decisions based on the state of the system.

## Solution in some cases

The special case of can in fact be solved exactly, as can the case with [12] however the general case can only be predicted statistically.[13] The solution when is,[12][14]

where the initial condition parameter is given by . For rational , after a finite number of iterations maps into a periodic sequence. But almost all are irrational, and, for irrational , never repeats itself it is non-periodic. This solution equation clearly demonstrates the two key features of chaos stretching and folding: the factor shows the exponential growth of stretching, which results in sensitive dependence on initial conditions, while the squared sine function keeps folded within the range .

For an equivalent solution in terms of complex numbers instead of trigonometric functions is[15]

where is either of the complex numbers

with modulus equal to 1. Just as the squared sine function in the trigonometric solution leads to neither shrinkage nor expansion of the set of points visited, in the latter solution this effect is accomplished by the unit modulus of .

By contrast, the solution when is[15]

for . Since for any value of other than the unstable fixed point 0, the term goes to 0 as goes to infinity, so goes to the stable fixed point

## Finding cycles of any length when

For the case, from almost all initial conditions the iterate sequence is chaotic. Nevertheless, there exist an infinite number of initial conditions that lead to cycles, and indeed there exist cycles of length for all integers . We can exploit the relationship of the logistic map to the dyadic transformation (also known as the bit-shift map) to find cycles of any length. If follows the logistic map and follows the dyadic transformation

then the two are related by

.

The reason that the dyadic transformation is also called the bit-shift map is that when is written in binary notation, the map moves the binary point one place to the right (and if the bit to the left of the binary point has become a "1", this "1" is changed to a "0"). A cycle of length 3, for example, occurs if an iterate has a 3-bit repeating sequence in its binary expansion (which is not also a one-bit repeating sequence): 001, 010, 100, 110, 101, or 011. The iterate 001001001... maps into 010010010..., which maps into 100100100..., which in turn maps into the original 001001001...; so this is a 3-cycle of the bit shift map. And the other three binary-expansion repeating sequences give the 3-cycle 110110110... → 101101101... → 011011011... → 110110110.... Either of these 3-cycles can be converted to fraction form: for example, the first-given 3-cycle can be written as 1/7 → 2/7 → 4/7 → 1/7. Using the above translation from the bit-shift map to the logistic map gives the corresponding logistic cycle .611260467... → .950484434... → .188255099... → .611260467... . We could similarly translate the other bit-shift 3-cycle into its corresponding logistic cycle. Likewise, cycles of any length can be found in the bit-shift map and then translated into the corresponding logistic cycles.

However, since almost all numbers in are irrational, almost all initial conditions of the bit-shift map lead to the non-periodicity of chaos. This is one way to see that the logistic map is chaotic for almost all initial conditions.

The number of cycles of (minimal) length for the logistic map with (tent map with ) is a known integer sequence (sequence A001037 in the OEIS): 2, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335, 630, 1161 ... This tells us that the logistic map with has 2 fixed points, 1 cycle of length 2, 2 cycles of length 3 and so on. This sequence takes a particularly simple form for prime : . For example: is the number of cycles of length 13. Since this case of the logistic map is chaotic for almost all initial conditions, all of these finite-length cycles are unstable.

## Notes

1. Boeing, G. (2016). "Visual Analysis of Nonlinear Dynamical Systems: Chaos, Fractals, Self-Similarity and the Limits of Prediction". Systems. 4 (4): 37. doi:10.3390/systems4040037. Retrieved 2016-12-02.
2. May, Robert M. 1976. "Simple mathematical models with very complicated dynamics." Nature 261(5560):459-467.
3. Zhang, Cheng, "Period three begins", Mathematics Magazine 83, October 2010, 295-297.
4. Carson Jeffries; Jose Perez (1982). "Observation of a Pomeau–Manneville intermittent route to chaos in a nonlinear oscillator". Physical Review A. 26 (4): 2117–2122. Bibcode:1982PhRvA..26.2117J. doi:10.1103/PhysRevA.26.2117.
5. R.M. May (1976). "Simple mathematical models with very complicated dynamics". Nature. 261 (5560): 459–67. Bibcode:1976Natur.261..459M. doi:10.1038/261459a0. PMID 934280.
6. Baumol, William J; Benhabib, Jess (February 1989). "Chaos: Significance, Mechanism, and Economic Applications". Journal of Economic Perspectives. 3 (1): 77–105. doi:10.1257/jep.3.1.77.
7. Collet, Pierre, and Jean-Pierre Eckmann, Iterated Maps on the Interval as Dynamical Systems, Birkhauser, 1980.
8. May, Robert M. 1976. "Simple mathematical models with very complicated dynamics." Nature 261(5560):459-467.
9. Gleick, James (1987). Chaos: Making a New Science. Penguin Books.
10. Jakobson, M.,"Absolutely continuous invariant measures for one-parameter families of one-dimensional maps," Communications in Mathematical Physics 81, 1981, 39-88.
11. Schröder, Ernst (1870). "Über iterierte Funktionen". Math. Ann. 3 (2): 296322. doi:10.1007/BF01443992.
12. Little, M.; Heesch, D. (2004). "Chaotic root-finding for a small class of polynomials" (PDF). Journal of Difference Equations and Applications. 10 (11): 949–953. doi:10.1080/10236190412331285351.
13. Lorenz, Edward (1964), "The problem of deducing the climate from the governing equations," Tellus 16 (February): 1-11.
14. Schröder, Ernst (1870). "Ueber iterirte Functionen". Math. Ann. 3 (2): 296322. doi:10.1007/BF01443992.

## References

• P. Grassberger and I. Procaccia (1983). "Measuring the strangeness of strange attractors". Physica D. 9 (1–2): 189–208. Bibcode:1983PhyD....9..189G. doi:10.1016/0167-2789(83)90298-1.
• P. Grassberger (1981). "On the Hausdorff dimension of fractal attractors". Journal of Statistical Physics. 26 (1): 173–179. Bibcode:1981JSP....26..173G. doi:10.1007/BF01106792.
• Sprott, Julien Clinton (2003). Chaos and Time-Series Analysis. Oxford University Press. ISBN 0-19-850840-9.
• Strogatz, Steven (2000). Nonlinear Dynamics and Chaos. Perseus Publishing. ISBN 0-7382-0453-6.
• Tufillaro, Nicholas; Tyler Abbott; Jeremiah Reilly (1992). An experimental approach to nonlinear dynamics and chaos. Addison-Wesley New York. ISBN 0-201-55441-0.


text:
### Hiến pháp

You can fool all of the people some of the time, and some of the people all of the time, but you can't fool all of the people all of the time.
by Abraham Lincoln

## Stratified affine spaces

These are quick notes for a joint work with Ch. Wacheux

Here, by an affine manifold, we mean a paracompact manifold with a locally flat affine structure. If the manifold is of dimension n, then near each point we have a chart whose coordinate functions are affine functions, and any local affine function is . . . → Read More: Stratified affine spaces

## Three lectures on singular foliations

For those who might be interested: This PDF file contains the slides of my 3-hour minicourse on singular foliations, given in the Workshop “GAP XII: Geometric Mechanics”, Internatinal Mathematical Forum of Tsing Hua University, Sanya, 10-14/March/2014:

http://zung.zetamu.net/Maths/Talks/FoliationsSanya2014.pdf

The main topic of this minicourse is the correspondence Singular Foliations <–> Nambu structures, and how . . . → Read More: Three lectures on singular foliations

## Notes on random systems 3: Liouville theorem for RDS

Extension of Liouville’s theorem to the case of RDS

The case without first integrals:

X_1, …, X_p are p generators of RDS on a compact manifold M of dimension p, which commute with each other, and whose deterministic parts are linearly independent everwhere. Then M is a  p-dimensional torus T^p with a periodic coordinate . . . → Read More: Notes on random systems 3: Liouville theorem for RDS

## Notes on random systems 2

What is a random fixed point ?

A reference: Ochs – Oseledets: Examples of  RDS on a closed unit ball without random fixed points. (So the topological fixed point theorem is NOT valid for RDS)

Definition. A random fixed point of a RDS \Phi over noise space (\Omega, \theta) (theta is the dynamics in . . . → Read More: Notes on random systems 2

## Notes on random systems 1: RDS vs NDS

Edited: 25/02/2014

These series of notes are for a research project that I’m doing with a student of mine. Some of what I write here will look quite stupid, because I myself know little about random dynamical systems.

I’ll explain why we’re interested in these systems later. But first, here . . . → Read More: Notes on random systems 1: RDS vs NDS

## Monodromy can kill global convexity!

We just got a rather surprising result in our joint research project with Christophe Wacheux (currently post-doc at EPFL) about the intrinstic convexity of singular affine spaces.

The problem is to study the intrinsic local and global convexity of the affine structure of the base space of integrable Hamiltonian systems whose singularities are nondegenerate . . . → Read More: Monodromy can kill global convexity!

## Thuật toán của người mất trí nhớ

Có một bài toán “khá đơn giản” sau về vấn đề thuật toán, tôi biết chắc chắn là giải được (vì có định lý về vấn đề này), có điều tôi thử tự tìm lời giải mà loay hoay mãi chưa ra:

Có một người ở một làng bị mất trí và được cho . . . → Read More: Thuật toán của người mất trí nhớ

## Commuting Foliations

Our paper with Truong Hong Minh entitled “Commuting Foliations” has been accepted for publication in a special isue dedicated to Alain Chenciner in the journal  Regular and Chaotic Dynamics

The most imporant part of this paper is actually the exposition of the relationship between singular foliations and Nambu structures: how to go from a . . . → Read More: Commuting Foliations

## Geometry and Physics XII, Hainan (China), 10-14/03/2014

Next year I’ll come to Sanya, Hainan, China to attend “GAP XII”, an international conference on geometry and physics.

The organizers said they would have money to cover local expenses for a number of participants, so if anyone is interested in visiting Sanya, please contact them (or contact me if you don’t know them)

. . . → Read More: Geometry and Physics XII, Hainan (China), 10-14/03/2014

## Integrable p-vector fields and singular foliations

There seems to be a lot of confusion (among my colleagues, and also of myself) concerning the relationships between singular foliations and integrable p-vector fields (a.k.a. Nambu structures). The aim of this note is to make some clarifications.

1) How to construct a singular foliation from an integrable p-vector field ?

The obvious (but . . . → Read More: Integrable p-vector fields and singular foliations

## Talk at AlanFest 07/2013

This week I’m at the AlanFest in EPFL, Switzerland, on the occasion of Alan Weinstein’s 70th birthday. I gave  a talk on Thursday entitled:

A normalization toolbox, with applications to singular foliations

Here are the slides of my talks (with some typographical errors), for people who might be interested:

NTZ_AlanFest2013

. . . → Read More: Talk at AlanFest 07/2013

## Linearization of smooth integrable systems

I’m writing down here the ideas for proving that smooth nondegenerate integrable dynamical systems are smoothly linearizable. The analytic case can be proved using analytic torus actions (my paper about that will appear in Ergodic Th Dyn Sys). I think the smooth case is also true, but the proof is much more complicated than . . . → Read More: Linearization of smooth integrable systems

## Anti-canonical bundle of singular foliations

Take a singular foliation say of dimension k.

Locally at each point there exists $k$ vector fields X1, …, Xk which are tangent to the foliation and which are linearly independent almost everywhere (we will only consider foliations which satisfy this property)

The wedge product L = X1…Xk is a Nambu structure tangent to . . . → Read More: Anti-canonical bundle of singular foliations


text:
# Infinite Monkey Theorem flaw?

#### chris1235

##### New Member
Right. Of course, the point of the theorem is to show by analogy that DNA could have arisen by chance. That discussion involves Dawkins versus creative design, and involves biology as well as statistical probability. Probably a different forum!

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#### Dason

A maths solution has meaning. Are your infinite monkeys ever going to produce, for instance, a solution to the Goldbach conjecture?
If a proof of the Goldbach conjecture exists then that will be produced eventually.

#### Dason

It doesn't have to be written down yet. But since the conjecture hasn't been proven yet we can't be sure if it's true or not. If the conjecture isn't true then we can't write down a proof of the theorem itself (and have it be correct). And due to the Incompleteness theorem it theoretically might not even be possible to write down a proof even if it is true. A disproof would just need to show a counter-example so there would be no issues writing that down. Basically all I was saying was "assuming it's possible to write down that proof - then yes it will be written almost surely".

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#### chris1235

##### New Member
I still think that because it's possible that any given string of characters might NOT be produced randomly during infinity it can't be statistically right to say that it will be. And I still think that the original questioner had a point about randomness not being able to produce language. That involves the philosophy of meaning. Another different forum. Thanks for taking me seriously. I'll leave you all to your maths!

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# DocTable Overview¶

A DocTable acts as an object-oriented interface to a single database table. It will maintain a connection to the database and sqlalchemy metadata about the table schema. Schemas are provided as classes that act much like dataclasses (see the doctable schema guide for more explanation), and indices and constraints are often specified in a definition of a subclass of DocTable.

You may also want to see the vignettes for more examples, the DocTable docs for more information about the class, or the schema guide for more information about creating schemas. I also recommend looking examples for insert, delete, select, and update methods.

Here I'll just provide some examples to give a sense of how DocTable works.

# Defining a Database Schema¶

DocTable schemas are created using the doctable.schema decorator on a class that uses doctable.Col for defaulted parameters. Check out the schema guide for more detail about schema classes. Our demonstration class will include three columns: id, name, and age, with an additional .is_old property derived from age for example.

We can instantiate a DocTable by passing a target and schema (Record in our example) parameters, and I show the resulting schema using .schema_table(). Note that the type hints were used to describe column types, and id was used as the auto-incremented primary key.

Probably a more common use case will be to subclass DocTable to provide some basic definitions.

# Object Interface¶

The main goal of doctable was to create an object-oriented interface to working with database tables. To that end, I'll show some of the more common use cases in the following examples.

First, note that subscripting the table object allows you to access sqlalchemy Column objects, which, as I will show a bit later, can be used to create where conditionals for select and update queries.

As we'll show later, these column objects also have some operators defined such that they can be used to construct complex queries and functions.

## Inserting Rows¶

We use the .insert() method to insert a row passed as a dictionary of column name -> value entries.

## Select Statements¶

Now we show how to select data from the table. Use the .count() method to check the number of rows. It also accepts some column conditionals to count entries that satisfy a given criteria

Use the .select() method with no arguments to retrieve all rows of the table. You can also choose to select one or more columns to select.

The SUM() and COUNT() SQL functions have been mapped to .sum and .count attributes of columns.

Alternatively, to see the results as a pandas dataframe, we can use .select_df().

Now we can select specific elements of the db using the where argument of the .select() method.

We can update the results in a similar way, using the where argument.

And we can delete elements using the .delete() method.

# Notes on DB Interface¶

DocTable2 allows you to access columns through direct subscripting, then relies on the power of sqlalchemy column objects to do most of the work of constructing queries. Here are a few notes on their use. For more demonstration, see the example in examples/dt2_select.ipynb


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# Why didn’t I think of that? Sonar Math

This blog is for those who are curious about how mathematics shapes our world. Some of the most powerful applications of mathematics come from the need to develop military technologies in which the lives of entire nations might be at stake. Here, necessity is truly the mother of invention. The atomic bomb for example, could not have been discovered without Albert Einstein’s famous equation, $\displaystyle\ E = mc^2$. Similarly, there would be no aircraft radar and naval sonar techniques without the help of math.

I want to tell the story sonar technology unfolded during the 1940’s. It’s a story that began rather improbably with Hollywood and finished with scientists deriving a beautiful “sonar equation” of great help to the navy.

Our story begins with Hedy Lamarr (1914 – 2000), a legend from the golden era of moving pictures. Unusual for an actress, Lemarr took an interest in solving engineering problems. In the 1940s, she was mulling over the problem how to control a torpedo by radio without being jammed by the enemy. Her brilliant answer: by frequency hopping – that is, synchronize both the torpedo transmitter and its receiver as they hop from frequency to frequency, thus confusing the enemy who doesn’t know which frequency to jam. Although Lemarr’s idea didn’t take off (the US Naval folks had another approach to the problem), she was hailed as a pioneer of modern communications technology, which is not far from the truth since all mobile phones today use a more advanced version of her frequency hopping idea. However, instead of hopping, signals are spread over several frequencies simultaneously.

Rather than avoiding jamming, the priority of the US navy was to develop a system to distinguish between reflected signals, so that both the distance to an object and its speed can be measured precisely. But there was a major problem, illustrated in figures 1 and 2 which are simplified frequency-time plots of echo reflections.

In Figure 1, the gray cells on the 6-by-6 grid represents a sequence of echo reflections from a stationary object. The chart on the right shows the sequence shifted in time (along the horizontal axis).

The problem facing the navy engineers is that it is not always possible to distinguish between a reflection shifted in time from that of a closer moving object shifted in frequency (along the vertical axis) as shown on the right-hand side of Figure 2.

For sonar to work precisely, the navy needed to focus on a particular frequency-time pattern that enables them to distinguish between various kinds of echo reflections and to read them with ease. This pattern would have the property that any two distinct shifts (horizontal or vertical or a combination of both) have at most one “ping” in common. Such a pattern is called a Costas Array, named American engineer, John P. Costa (1923 – 2008). The question is how to locate a Costas Array in real time. The solution calls for some mathematical thinking. For starters, we need to interpret a frequency-time pattern in the language of mathematics. Then, we have to derive an equation to pin down the coordinates of a Costas Array.

Note to readers: you can skip the math below and jump to the conclusion at the end. Or, (as I would recommend), follow the steps which are mostly based on concepts familiar from high school math.

Let’s proceed. If your device detects a frequency-time pattern like Figure 3 below, bingo – you have found a Costas Array!  You easily see that the pattern in this figure has the property that any two distinct shifts (vertical, or horizontal or a combination of both) have at most one “ping” in common.

The next task is to find an equation that exactly describes this pattern. We begin by interpreting a frequency-time pattern as a function:

(1) $\displaystyle\ y = f(x) }$

defined over a short initial segment of natural numbers, say,

X = {1,2,….,n}

where for simplicity, n = 6 to mirror the plots. In Figure 3, the graph shifted b units to the right and a units in the vertical direction. The following function can be used to represent these shifts.

(2) $\displaystyle\ y = a + f{(x-b)}$

where at least one of the parameters a, b is different from zero.

For sonar precision, we want the function to have at most one solution. What kind of function satisfy this criterion? The answer, very simply, is a linear equation

(3) $\displaystyle\ Ax + B = 0$

Therefore, the idea is to make equation (2) to be equivalent to a linear equation after some rearrangements. How to do that is the problem. Since we want to solve equations involving the unknown function f, it would be nice if $y = f(x)$ has an inverse function such that:

(4) $\displaystyle\ x = g(f(x))$

We can then apply g to both sides of equation (2) to get

(5) $\displaystyle\ x = g(a + f(x-b))$

It would be nice too, if this equation can be simplified into a linear equation. For this to happen, g must satisfy the identity: $g(x+t) = g(s) \times g(t).$ If we let g(1) = d, a constant, then this identify implies that g is an exponential function of the form:

(6) $\displaystyle\ g(x) = d^{x}$ for all x in X.

At this point, we shall call upon some nifty mathematical tricks known as modular arithmetic. The idea is to reduce every result modulo some fixed integer, m.

(7) $\displaystyle\ g(x) = d^x \mod m$

The basics of modular arithmetic are explained in the end notes. For now, I’ll ask you to trust the results shown below. Since our Costa Array is a 6 x 6 grid, we want n = 6.  Look what happens when we let m = 7 and d = 3.

$3^1 = 3 \mod 7$
$3^2 = 9 = 2 \mod 7$
$3^3 = 27 = 6 \mod 7$
$3^4 = 81 = 4 \mod 7$
$3^5 = 5 \mod 7$
$3^6 = 1 \mod 7$

Behold! We get a running sequence of 6 numbers from 1 to 6, as desired. From high school algebra, recall that the opposite of exponentiation is logarithms (or logs in short). The following log operations to base 3 leads to the sequence 1, 2, …, 6.
$log_{3} 3 =1$
$log_{3} 2 = 2$
$log_{3} 6 = 3$
$log_{3} 4 = 4$
$log_{3} 5 = 5$
$log_{3} 1 = 6$

You can almost see where we’re heading. With the dignified entrance of the log operation, we get the discrete logarithm function $y = log_{3} x \mod 7$ which produces the exact frequency-time pattern in Figure 3!

A final note: What we just did is a construction known as the Welch-Costa array. Our example can be easily generalized to a larger grid. Moreover, the equation works for every prime number p in place of the number 7.

Note: The ABCs of Modular Arithmetic

This note is a primer on modular arithmetic. Modular arithmetic is a special (if strange) type of arithmetic that involves only whole numbers or integers. While it may seem confusing at first, counting using this type of arithmetic helps mathematicians handle a large array of number theory problems more easily.

When we divide two integers, we have an equation that looks like:

A/B = Q + R, where

A is the dividend
B is the divisor
Q is the quotient
R is the remainder

Sometimes, we are only interested in what the remainder is when A is divided by B. For these cases, there is an operator called modulo operator (shorthand: mod).

For example, 13/5 gives a Q of 2 and R of 3 or “2 remainder 3”. Thus, we say that

13 mod 5 = 2

For a given mod, m, the remainder starts with 0, increases by one as the dividend increases by one until it reaches m-1. Then it begins with 0 again and continues the same cycle. Here is an example using mod 3.

0 mod 3 = 0 remainder 0
1 mod 3 = 0 remainder 1
2 mod 3 = 0 remainder 2 (i.e., m-1)
3 mod 3 = 1 remainder 0
4 mod 3 = 1 remainder 1
5 mod 3 = 1 remainder 2
6 mod 3 = 2 remainder 0 (m-1 again)

Conclusion
If we have integers A and B and we increase A by a multiple of B, we end up in the same spot. Thus,

3 mod 10= 3
13 mod 10 = 3
23 mod 10 = 3
33 mod 10 = 3

Problem: Find
(a) 311 mod 3
(b) 3 mod 4


text:
# Simplifying a trigonometry expression

mileena

## Homework Statement

Simplify:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

none

## The Attempt at a Solution

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - tan x (1 -1) sec x ] /
sec2 x

[(sec x)(sec2 x) - 0 ] /
sec2 x

sec x

Thanks! I hope the above is correct, but it probably isn't.

Homework Helper
Gold Member
2022 Award
[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - tan x (1 -1) sec x ] /
sec2 x

That step is wrong. You've dropped a tan x. Try it again.
But I would start by getting rid of the denominator. What is another way of expressing 1/sec(θ)?

Homework Helper

## Homework Statement

Simplify:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x
So
$$\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}$$?

Is there any particular reason for writing "$sec(x)sec^2(x)$" rather than $sec^3(x)$?

none

## The Attempt at a Solution

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - tan x (1 -1) sec x ] /
Where did that "(1- 1)" come from?
sec2 x

[(sec x)(sec2 x) - 0 ] /
sec2 x

sec x

Thanks! I hope the above is correct, but it probably isn't.[/QUOTE]
It's easy to check that it is NOT correct. If x= 30 degrees ($\pi/6$ radians), the original expression is $(2/3)\sqrt{3}+ 2/3$ while sec(x) is $(2/3)\sqrt{3}$.

Homework Helper

## Homework Statement

Simplify:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x
So
$$\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}$$?

Is there any particular reason for writing "$sec(x)sec^2(x)$" rather than $sec^3(x)$?

none

## The Attempt at a Solution

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - tan x (1 -1) sec x ] /
Where did that "(1- 1)" come from?

sec2 x

[(sec x)(sec2 x) - 0 ] /
sec2 x

sec x

Thanks! I hope the above is correct, but it probably isn't.
It's easy to check that it is NOT correct. If x= 30 degrees ($\pi/6$ radians), the original expression is $(2/3)\sqrt{3}+ 2/3$ while sec(x) is $(2/3)\sqrt{3}$.

mileena
That step is wrong. You've dropped a tan x. Try it again.
But I would start by getting rid of the denominator. What is another way of expressing 1/sec(θ)?

Thanks haruspex, and sorry for my late reply. I have a hard time getting online in the evenings.

cos x = 1/sec x

So:

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
sec2 x

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)] /
[(sec x)(sec x)]

(cos x)(cos x) [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]

cos2 x [sec3 x - (sec2 x tan x - sec x tan x)]

[sec3 x/sec2 x] - [[(sec x)(tan2 x - tan x)]/sec2 x]

sec x - [[(tan2 x - tan x)]/sec x]

sec x - [cos x[(tan2 x - tan x)]]

That's it. I don't know how to reduce this more.

Last edited:
janhaa
Thanks haruspex, and sorry for my late reply. I have a hard time getting online in the evenings.
1 - [[(tan2 x - tan x)]/sec2 x]
That's it. I don't know how to reduce this more.
maybe you can simplify it this way:

$1-\left(\frac{\tan^2(x)-\tan(x)}{1+\tan^2(x)}\right)$

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

mileena
So
$$\frac{sec(x)(sec^2(x))- (tan(x)- 1)(sec(x)tan(x))}{sec^2(x)}$$?

Is there any particular reason for writing "$sec(x)sec^2(x)$" rather than $sec^3(x)$?

That is the way the problem was presented to me on the homework!

Where did that "(1- 1)" come from?

I tried to distribute the tan x out of both factors!

Homework Helper
Gold Member
2022 Award
(cos x)(cos x) [(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]

cos2 x [sec3 x - (sec2 x tan x - sec x tan x)]
There's a mistake in how you multiplied out the (tan x - 1)(sec x tan x) term.
[sec3 x/sec2 x] - [[(sec x)(tan2 x - tan x)]/sec2 x]
Why flip back to dividing by sec x again? You can use sec x * cos x = 1 without having to turn it back into a division.

mileena
Ok, I decided to redo the problem. Part of my problem is it's hard to check the work on here because I can't write fractions they way they are supposed to be written without complex computer operations.

Thank you harupex for the tip: sec x * cos x = 1

I knew they were inverses, but it didn't dawn on me to use that equation!

[(sec x)(sec2 x) - (tan x - 1)(sec x tan x)]
/
sec2 x

cos2 x [(sec3 x) - (tan x - 1) (sec x tan x)]

sec x - cos2 x (sec x tan2 x - sec x tan x)

sec x - (cos x tan2 x + cos x tan x)

sec x - [(cos x tan x)(tan x + 1)]

mileena
maybe you can simplify it this way:

$1-\left(\frac{\tan^2(x)-\tan(x)}{1+\tan^2(x)}\right)$

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

Thanks janhaa! I am not sure what you did though to get rid of the initial "1 -".

Homework Helper
Gold Member
2022 Award
sec x - cos2 x (sec x tan2 x - sec x tan x)

sec x - (cos x tan2 x + cos x tan x)
There's a sign error in that step.
What does cos x tan x simplify to?
Do you know sec2 = 1 + tan2 ? (Can derive this easily from cos2+sin2=1.)

mileena
Ok, I see my error. I become very flustered with trig for some reason. I guess I get anxious when the pressure is on.

Ok, since tan x = sin x/cos x,
cos x tan x reduces to just sin x.

Here it is corrected:

sec x - cos2 x (sec x tan2 x - sec x tan x)

sec x - cos x tan2 x + cos x tan x

sec x - cos x tan2 x + sin x

And yes, I know that sec2 = 1 + tan2 is one of the three Pythagorean identities, but I don't see any sec2 or 1's above.

Homework Helper
Gold Member
2022 Award
sec x - cos x tan2 x + sin x
I don't see any sec2 or 1's above.
No, but there's a tan2.

mileena
Oh my gosh. I think I have it!

sec x - cos x tan2 x + sin x

sec x - cos x (sec2 x -1) + sin x

sec x - cos x sec2 x - cos x + sin x

sec x -sec x - cos x + sin x

sin x - cos x

I hope it's that easy.

For some reason, I am not seeing all the basic trigonometric equivalents.


text:
# Minimum number of observations for logistic regression?

I'm running a binary logistic regressions with 3 numerical variables. I'm suppressing the intercept in my models as the probability should be zero if all input variables are zero.

What's minimal number of observations I should use?

• You would need an intercept of $-\infty$ to do that! An intercept of 0 corresponds to a probability of $\frac{1}{1 + \exp(-0)} = 1/2$, not $0$, when all independent variables are zero. – whuber Jun 8 '11 at 18:57
• There is a related discussion here: sample-size-for-logistic-regression. – gung - Reinstate Monica Dec 13 '12 at 14:18

There is one way to get at a solid starting point. Suppose there were no covariates, so that the only parameter in the model were the intercept. What is the sample size required to allow the estimate of the intercept to be precise enough so that the predicted probability is within 0.1 of the true probability with 95% confidence, when the true intercept is in the neighborhood of zero? The answer is n=96. What if there were one covariate, and it was binary with a prevalence of 0.5? One would need 96 subjects with x=0 and 96 with x=1 to have an upper bound on the margin of error for estimating Prob[Y=1 | X=x] not exceed 0.1. The general formula for the sample size required to achieve a margin of error of $\delta$ in estimating a true probability of $p$ at the 0.95 confidence level is $n = (\frac{1.96}{\delta})^{2} \times p(1-p)$. Set $p = 0.5$ for the worst case.

• I appreciate your help in this forum. I have ~90000 events and ~2000000 non-events. I need to a logistic model with 65 predictors. Now how and how many sample can I take. in fact my question is related to stats.stackexchange.com/questions/268201/… – SIslam Mar 22 '17 at 13:12
• No problem with fitting 65 simultaneously with your effective sample size. – Frank Harrell Mar 22 '17 at 14:00
• but I was suggested that too many samples can cause problem since I was getting psudo r squared as low. – SIslam Mar 22 '17 at 14:31
• Are you kidding? If the $R^2$ is low using a large sample, that is the most accurate estimate of the true $R^2$ and dropping observations will not improve the performance of the model; it will only make it worse. Supplement the $R^2$ with other easier to understand metrics such as the $c$-index (concordance probability; ROC area). And above all, ignore any advice to "balance" the outcome category frequencies. – Frank Harrell Mar 22 '17 at 15:29
• Do I need use glmnet for this to find most useful predictor at this stage? – SIslam Mar 23 '17 at 11:39

There isn't really a minimum number of observations. Essentially the more observations you have the more the parameters of your model are constrained by the data, and the more confident the model becomes. How many observations you need depends on the nature of the problem and how confident you need to be in your model. I don't think it is a good idea to rely too much on "rules of thumb" about this sort of thing, but use the all the data you can get and inspect the confidence/credible intervals on your model parameters and on predictions.

• no minimum number! I have ~90000 event and ~2000000 non-events. I need to a logistic model with 65 regressor. I am told that this is too many samples, since I am taking this whole ~90000 events and ~90000 non events randomly selected from ~2000000 , try to lessen sample while samples are representative. at this stage how many sample can I take and how. In fact I am referring stats.stackexchange.com/questions/268201/… – SIslam Mar 22 '17 at 13:31
• No don't do that – Frank Harrell Mar 22 '17 at 15:30
• I agree with @FrankHarrell (possibly for a different reason?). The "class imbalance" problem tends to go away the more data you collect, and if you artificially balance the training data you are telling the model that the operational class frequencies are 50-50, which probably isn't true, and you will over classify the minority class in operational use. If you do do this, then post-process the output probabilities to adjust for the difference in training and operational class frequencies (at which point you will probably get essentially the same result as training with all the data). – Dikran Marsupial Mar 23 '17 at 8:25

Update: I didn't see the above comment, by @David Harris, which is pretty much like mine. Sorry for that. You guys can delete my answer if it is too similar.

I'd second Dikran Marsupail post and add my two cents.

Take in consideration your prior knowledge about the effects that you expect from your independent variables. If you expect small effects, than you will need a huge sample. If the effects are expected to be big, than a small sample can do the job.

As you might know, standard errors are a function of sample size, so the bigger the sample size, the smaller the standard errors. Thus, if effects are small, i.e., are near zero, only a small standard error will be able to detect this effect, i.e, to show that it is significantly different from zero. On the other hand, if the effect is big (far from zero), than even a large standard error will produce significant results.

If you need some reference, take a look at Andrew Gelmans' Blog.

• Gelman's blog is getting pretty big :-). Do you have a particular post in mind? – whuber Jun 9 '11 at 14:06
• @Whuber, you are right, I should have pointed to something more specific. He has some recent talk presentations about small effects and multiple comparisons, but I think the following link is enough: stat.columbia.edu/~gelman/research/published/power4r.pdf – Manoel Galdino Jun 9 '11 at 21:45
• that link in the comments is dead and there's no reference to a particular post of the mentioned blog – baxx Apr 17 '19 at 23:44

It seems that to get an acceptable estimation we have to apply the rules that have been examined by other researchers. I agree with the two rules of thumb above (10 obs for each var. and the formula by Harrell). Here, there is another question that the data are revealed or stated preference. Hosmer and Lemeshow in their book have provided a rule for revealed and Louviere and Hensher in their book (The methods of Stated preference) provided a rule for stated preference data

• This would benefit from a fuller explanation and also complete and precise references. – Nick Cox Jul 20 '13 at 22:45


text:
# Understanding the difference between physical and chemical changes

Which of the following is a chemical change?

A. Hydrolysis of a dipeptide.

B. Stretching of a balloon.

C. Freezing of water.

D. Breaking a stick.

I'm pretty sure that the answer is (A), but I'm not 100% positive.

I know that, for (C), you can freeze water, and then you can melt it back to liquid water.

For (D), you can break a stick, and the stick is still now two sticks.

For (B), however, I think that it is a physical change, but I'm not entirely certain.

Can someone please explain this for me?

• It seems you're getting it more or less right. – Mithoron Oct 15 '16 at 22:57
• Hopefully! I'm getting graded on this! – 关一骏 Oct 15 '16 at 23:00


text:
## Friday, May 1, 2020

### Next talk

#### May 28, 2020 17:00 Bergen time (GMT+2)

Daniel MarxMax-Planck-Institut für Informatik

#### Abstract: Given a graph G and an integer k, the H-free Edge Editing problem is to find whether there exists at most k pairs of vertices in G such that changing the adjacency of the pairs in G results in a graph without any induced copy of H. The existence of polynomial kernels for H-free Edge Editing received significant attention in the parameterised complexity literature. Nontrivial polynomial kernels are known to exist for some graphs H with at most 4 vertices, but starting from 5 vertices, polynomial kernels are known only if H is either complete or empty. This suggests the conjecture that there is no H with at least 5 vertices where H-free Edge Editing admits a polynomial kernel. Towards this goal, we obtain a set of nine 5-vertex graphs such that if H-free Edge Editing is incompressible for every H in this set, then H-free Edge Editing is incompressible for every graph H with at least five vertices that is neither complete nor empty (under the usual complexity assumptions). That is, proving incompressibility for these nine graphs would give a complete classification of the kernelization complexity of H-free Edge Editing for every H with at least 5 vertices.

We obtain similar result also for H-free Edge Deletion. Here the picture is more complicated due to the existence of another infinite family of graphs H where the problem is trivial (graphs with exactly one edge). We obtain a larger set of 19 graphs whose incompressibility would give a complete classification of the kernelization complexity of H-free Edge Deletion for every graph H with at least 5 vertices. Analogous results follow also for the H-free Edge Completion problem by simple complementation.

(Joint Work with R. B. Sandeep)

Link to the zoom meeting: https://uib.zoom.us/j/4231169675
Meeting ID: 423 116 9675
Password: name of the W[1]-complete problem, 6 letters, all capital. Also known as a set of pairwise adjacent vertices.

## Sunday, April 12, 2020

### Talks schedule

By default, the talks start at 17:00 Bergen time (GMT+2)

Link to the zoom meeting: https://uib.zoom.us/j/4231169675
Meeting ID: 423 116 9675
Password: name of the W[1]-complete problem, 6 letters, all capital. Also known as a set of pairwise adjacent vertices.

April 23, 2020
• Jesper Nederlof, Utrecht University
• Title: Bipartite TSP in O(1.9999^n) Time, Assuming Quadratic Time Matrix Multiplication
• Abstract:
The symmetric traveling salesman problem (TSP) is the problem of finding the shortest Hamiltonian cycle in an edge-weighted undirected graph. In 1962 Bellman, and independently Held and Karp, showed that TSP instances with n cities can be solved in O(n^2*2^n) time. Since then it has been a notorious problem to improve the runtime to O((2-eps)^n) for some constant eps >0. In this work we establish the following progress: If (s x s)-matrices can be multiplied in s^{2+o(1)} time, than all instances of TSP in bipartite graphs can be solved in O(1.9999^n) time by a randomized algorithm with constant error probability.  We also indicate how our methods may be useful to solve TSP in non-bipartite graphs.

On a high level, our approach is via a new problem called the MINHAMPAIR problem: Given two families of weighted perfect matchings, find a combination of minimum weight that forms a Hamiltonian cycle. As our main technical contribution, we give a fast algorithm for MINHAMPAIR based on a new sparse cut-based factorization of the `matchings connectivity matrix', introduced by Cygan et al. [JACM'18].

The recording of this talk is now available at https://www.youtube.com/watch?v=BHvegY-xYBE. The slides of the talk are available here.

April 30, 2020
• Daniel Lokshtanov, UCSB
• Title: A Parameterized Approximation Scheme for Min k-Cut
• Abstract:

#### In the Min k-Cut problem, input is an edge weighted graph G and an integer k, and the task is to partition the vertex set into k non-empty sets, such that the total weight of the edges with endpoints in different parts is minimized.  When k is part of the input, the problem is NP-complete and hard to approximate within any factor less than 2. Recently, the problem has received significant attention from the perspective of parameterized approximation.  Gupta {\em et al.} [SODA 2018] initiated the study of FPT-approximation for  the Min k-Cut problem and  gave an 1.9997-approximation algorithm running in time 2^{O(k^6}n^{O(1)}. Later, the same set of authors~[FOCS 2018]  designed an (1 +\epsilon)-approximation algorithm that runs in time (k/\epsilon)^{O(k)}n^{k+O(1)}, and a 1.81-approximation algorithm  running in time 2^{O(k^2)}n^{O(1)}. More, recently, Kawarabayashi and Lin~[SODA 2020] gave a (5/3 + \epsilon)-approximation for  Min k-Cut running in time  2^{O(k^2 \log k)}n^{O(1)}.  In this paper we give a parameterized approximation algorithm with best possible approximation guarantee, and best possible running time dependence on said guarantee (up to Exponential Time Hypothesis (\ETH) and constants in the exponent). In particular, for every \epsilon > 0, the algorithm obtains a  (1 +\epsilon)-approximate solution in time (k/\epsilon)^{O(k)}n^{O(1)}. The main ingredients of our algorithm are: a simple sparsification procedure, a new polynomial time algorithm for decomposing a graph into highly connected parts, and a new exact algorithm with running time s^{O(k)}n^{O(1)} on unweighted (multi-) graphs. Here, s denotes the number of edges in a minimum k-cut. The later two are of independent interest.  The recording of this talk is now available at https://www.youtube.com/watch?v=6EvV8Ljn8VI. May 7, 2020

• Jason Li, CMU
• Title: Detecting Feedback Vertex Sets of Size k in O*(2.7^k) Time
• Abstract:
In the Feedback Vertex Set problem, one is given an undirected graph $G$ and an integer $k$, and one needs to determine whether there exists a set of $k$ vertices that intersects all cycles of $G$ (a so-called feedback vertex set). Feedback Vertex Set is one of the most central problems in parameterized complexity: It served as an excellent test bed for many important algorithmic techniques in the field such as Iterative Compression~[Guo et al. (JCSS'06)], Randomized Branching~[Becker et al. (J. Artif. Intell. Res'00)] and Cut\&Count~[Cygan et al. (FOCS'11)].
In particular, there has been a long race for the smallest dependence $f(k)$ in run times of the type $O^\star(f(k))$, where the $O^\star$ notation omits factors polynomial in $n$. This race seemed to be run in 2011, when a randomized $O^\star(3^k)$ time algorithm based on Cut\&Count was introduced.
In this work, we show the contrary and give a $O^\star(2.7^k)$ time randomized algorithm. Our algorithm combines all mentioned techniques with substantial new ideas: First, we show that, given a feedback vertex set of size $k$ of bounded average degree, a tree decomposition of width $(1-\Omega(1))k$ can be found in polynomial time. Second, we give a randomized branching strategy inspired by the one from~[Becker et al. (J. Artif. Intell. Res'00)] to reduce to the aforementioned bounded average degree setting. Third, we obtain significant run time improvements by employing fast matrix multiplication.

The recording of this talk is now available at https://www.youtube.com/watch?v=mhyS7cqYoYw.

#### May 14, 2020

• Title: On the Parameterized Complexity of Various Clustering Problems
• Abstract:
Clustering problems are fundamental computational problems. On the one hand they are at the heart of various data analysis, bioinformatics or machine learning approaches, on the other hand, they can be seen as variants of 'set cover' problems (involving e.g. metrics, graphs with a special structure, etc.) and so are very basic problems.
For many applications, finding efficient fixed-parameter algorithms is a very natural approach. Popular choices of parameters are either parameters of the problem itself (e.g. the number of clusters) or on the underlying structure of the input (e.g. the dimension in the metric case). We will focus on classic metric clustering objectives such as k-median and k-means and review recent results on the fixed parameter complexity of these problems, and the most important open problems.

The recording of this talk is now available at https://www.youtube.com/watch?v=Fgh-20qHzfg. The slides of the talk are available here.

#### May 21, 2020

The last decade has witnessed impressive progress on the understanding of structure in sparse graphs. Much of this progress can be attributed to the systematic exploration of abstract definitions of structural sparsity --- concepts of bounded expansion and nowhere denseness. From the parameterized perspective, these two notions have brought new techniques to the toolbox and helped to explain the parameterized complexity of problems definable in First Order logic. However, the contemporary graph theory offers multiple quantitative notions (aka parameters) of "well-structuredness" that reach beyond sparse graphs. To name just a few, we may bound the rankwidth of a graph, exclude a vertex-minor, or stipulate VC parameters like VC dimension or VC density. In the talk we will survey the recent advances in lifting the developments of sparsity to classes of well-structured dense graphs. We will particularly focus on connections with concepts from model theory: stable and NIP classes, and the notion of FO transductions.

The slides of the talk are available here.

#### May 28, 2020

• Daniel MarxMax-Planck-Institut für Informatik
• Title: Incompressibility of H-free edge modification problems: Towards  dichotomy
• Abstract:
Given a graph G and an integer k, the H-free Edge Editing problem is to find whether there exists at most k pairs of vertices in G such that changing the adjacency of the pairs in G results in a graph without any induced copy of H. The existence of polynomial kernels for H-free Edge Editing received significant attention in the parameterised complexity literature. Nontrivial polynomial kernels are known to exist for some graphs H with at most 4 vertices, but starting from 5 vertices, polynomial kernels are known only if H is either complete or empty. This suggests the conjecture that there is no H with at least 5 vertices where H-free Edge Editing admits a polynomial kernel. Towards this goal, we obtain a set of nine 5-vertex graphs such that if H-free Edge Editing is incompressible for every H in this set, then H-free Edge Editing is incompressible for every graph H with at least five vertices that is neither complete nor empty (under the usual complexity assumptions). That is, proving incompressibility for these nine graphs would give a complete classification of the kernelization complexity of H-free Edge Editing for every H with at least 5 vertices.

We obtain similar result also for H-free Edge Deletion. Here the picture is more complicated due to the existence of another infinite family of graphs H where the problem is trivial (graphs with exactly one edge). We obtain a larger set of 19 graphs whose incompressibility would give a complete classification of the kernelization complexity of H-free Edge Deletion for every graph H with at least 5 vertices. Analogous results follow also for the H-free Edge Completion problem by simple complementation.

(Joint Work with R. B. Sandeep)

#### June 04, 2020

• Hans Bodlaender, Utrecht University
• Title: Typical Sequences Revisited - Computing Width Parameters of Graphs
• Abstract:
In this talk, we revisit a key element of the currently asymptotic fasters parameterized algorithms for treewidth and pathwidth: typical sequences.These were introduced in 1991, by Lagergren and Arnborg, and independently by Bodlaender and Kloks. We give a structural result on thejoin of typical sequences, which speeds up one step in these algorithms for treewidth and pathwidth, but does not yet give better time bounds. The result is used to obtain polynomial time algorithms for some problems on directed series parallel graphs. The talk also discusses the current (open) status of the parameterized complexity of pathwidth and treewidth. (This is  a joint work with Lars Jaffke and Jan Arne Telle.)

#### June 11, 2020

• Edith Elkind, University of Oxford
• Title: tba
• Abstract:

#### June 18, 2020

• Martin Grohe, RWTH Aachen University
• Title: Polylogarithmic Parameterized Algorithms for the Graph Isomorphism Problem: From Bounded Degree to Excluded Minors
• Abstract:
Building on Babai's quasipolynomial graph isomorphism test, we designed a number of parameterized isomorphism tests that exhibit an unusual running time that is quasipolynomial in the parameter (that is, n^{polylog(k)}). The first parameter for which we could find such an algorithm is the maximum degree, the latest the Hadwiger number, that is, the maximum h such that K_h is a minor.

In this talk, I'll give a high level introduction into the techniques used in these algorithms. (Joint work with Daniel Neuen, Pascal Schweitzer, and Daniel Wiebking.)

## Saturday, April 11, 2020

This blog contains the list of talks on Frontiers of Parameterized Complexity. The initial plan to have talks every week, on Thursdays, at 17.00 Bergen time (GMT+2). The recordings of the talks have be found at our YouTube channel Frontiers of Parameterized Complexity https://www.youtube.com/channel/UCdfML-PShQNSCeqbz9Ol_oA/


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# Category: Mathematics Q&A

## L’Hopital’s Limit From A Phone Game

One of my recently graduated students was playing a phone riddle game that required an answer to this level: Turns out the solution to the game was just simply the word “answer” as the instruction was to simply enter that in. However, the integral and the limit shown there...

## No, You Don’t Need Quadratic Formula

On the MANSW Facebook page, a fellow teacher from another school threw this question to the brain’s trust. All the answers provided to the problem by others used the Quadratic Formula, but the Algebra from those solutions was not exactly clean. Personally, I don’t think getting the answer in...