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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# Problem Solving with Linear Graphs

## Solve story problems using graphs

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Problem Solving with Linear Graphs

Suppose that you have a part-time job delivering newspapers and that you can deliver 9 newspapers every 15 minutes. If you deliver at a constant rate, how many newspapers could you deliver in 2 hours? Could you construct a graph to represent this situation? How would the graph help you to solve the problem?

### Problem Solving with Linear Graphs

Graphing is a very useful tool when analyzing a situation. This Concept will focus on using graphs to help solve linear problems that occur in real life.

Remember the 4-Step Problem-Solving Plan:

Step 1: Understand the problem and underline or highlight key information.

Step 2: Translate the problem and devise a method to solve the problem.

Step 3: Carry out the plan and solve the problem.

#### Let's solve the following problems using the steps above:

1. A cell phone company is offering its customers the following deal. You can buy a new cell phone for $60 and pay a monthly flat rate of$40 per month for unlimited calls. How much money will this deal cost you after 9 months?

Begin by translating the sentence into an algebraic equation.

Let the number of months and total cost. The equation becomes:

You could use guess and check or solve this equation. However, this Concept focuses on using a graph to problem-solve. This equation is in slope-intercept form. By graphing the line of this equation, you will find all the ordered pairs that are solutions to the cell phone problem.

Finding the cost at month 9, you can see the cost is approximately $425.00. To check if this is approximately correct, substitute 9 in for the variable . Our answer,$425.00, is approximately equal to the exact solution, $420.00. 1. Christine took one hour to read 22 pages of “Harry Potter and the Order of the Phoenix.” She has 100 pages left to read in order to finish the book. Assuming that she reads at a constant rate of pages per hour, how much time should she expect to spend reading in order to finish the book? We do not have enough information to write an equation. We do not know the slope or the intercept. However, we have two points we can graph. We know that if Christine had never picked up the book, she would have read zero pages. So it takes Christine 0 hours to read 0 pages. We also know it took Christine one hour to read 22 pages. The two coordinates we can graph are (0, 0) and (1, 22). Using the graph and finding 100 pages, you can determine it will take Christine about 4.5 hours to read 100 pages. You can also think of this as a direct variation situation and solve it by writing a proportion. By using the Cross Products Theorem, you can find out . It will take Christine about 4.55 hours to read 100 pages, which is very close to your original estimate of 4.5 hours. 1. A cell phone monthly messaging plan costs$10 for the first 100 texts and then $0.25 for each additional text. Make a graph of this relation. Use the graph to determine the cost of sending 123 text messages in a month. The plan costs$10 no matter how many texts you send under 100. Even if you send zero texts, the cost is $10. So,$10 is the -intercept. The slope, or rate, is $0.25 per text, or , but only for texts after the first 100. Also, the independent variable represents the number of texts sent over 100. Since we want to know the cost of sending 123 text messages in a month, we really need to consider when . First, we will graph the equation by using the -intercept and slope: The coordinate pair for when is between 15 and 16. In situations like this, it is helpful to write the equation to solve for the exact value, which in this case is 15.75. It would cost$15.75 to send 123 texts on this plan.

### Examples

#### Example 1

Earlier, you were asked to suppose that you have a part-time job delivering newspapers and that you can deliver 9 newspapers every 15 minutes. If you deliver at a constant rate, how many newspapers could you deliver in 2 hours? Could you construct a graph to represent his situation? How would the graph help you to solve the problem?

You do not have to write an equation to solve this problem. You know that in 0 minutes you can deliver 0 newspapers so you have the -intercept. If you can deliver 9 newspapers every 15 minutes, then the slope of the line will be . The graph would be as shown below:

To determine how many newspapers you could deliver in 2 hours, note that 2 hours is equivalent to 120 minutes. Look at 120 on the -axis and find the corresponding -value.   If you deliver the newspapers at 9 newspapers every 15 minutes, you will deliver 72 newspapers in 2 hours.

#### Example 2

Jerome drove 20 miles an hour for 15 minutes on city streets, and then 40 miles an hour for 30 minutes on the highway, 60 miles an hour for an hour on the freeway, and finally 20 miles an hour for 15 minutes on city streets. Use a graph to determine how far Jerome drove.

Jerome's different speeds are his different rates, in miles per hour. Since his times are given in minutes, convert the rates to miles per minute:

If the change in time is 15 minutes, then the change in distance is . So the rate can be written as

If the change in time is 30 minutes, then the change in distance is . So the rate can be written as

If the change in time is 60 minutes, then the change in distance is . So the rate can be written as

The last interval has the same rate as the first:

Use these rates to draw a graph of distance in miles as a function of time in minutes.

From the graph you can see that Jerome's trip takes 120 minutes, or two hours, and that he drives 90 miles.

### Review

1. Using the following graph, determine these values:
1. The amount of earnings after 40 hours
2. How many hours it takes to earn $250.00 3. The slope of the line and what it represents 4. The intercept of the line and what it represents 2. A stretched spring has a length of 12 inches when a weight of 2 lbs is attached to it. The same spring has a length of 18 inches when a weight of 5 lbs is attached to it. It is known from physics that within certain weight limits, the function that describes how much a spring stretches with different weights is a linear function. What is the length of the spring when no weights are attached? 1. A gym is offering a deal to new members. Customers can sign up by paying a registration fee of$200 and a monthly fee of $39. How much will this membership cost a member by the end of one year? 2. A candle is burning at a linear rate. The candle measures five inches two minutes after it was lit. It measures three inches eight minutes after it was lit. What was the original length of the candle? 3. Tali is trying to find the thickness of a page of his telephone book. To do this, he takes a measurement and finds out that 550 pages measure 1.25 inches. What is the thickness of one page of the phone book? 4. Bobby and Petra are running a lemonade stand and they charge 45 cents for each glass of lemonade. To break even, they must make$25. How many glasses of lemonade must they sell to break even?
5. The tip for a $78.00 restaurant bill is$9.20. What is the tip for a \$21.50 meal?
6. Karen left her house and walked at a rate of 4 miles/hour for 30 minutes. She realized she was late for school and began to jog at a rate of 5.5 miles/hour for 25 minutes. Using a graph, determine how far she is from her house at the end of 45 minutes.

Mixed Review

1. Simplify .
2. Give an example of a direct variation equation and label its constant of variation.
3. Identify the slope and intercept: .
4. Suppose is located in quadrant I. Write a rule that would move into quadrant III.
5. Find the intercepts of .
6. Evaluate when .

To see the Review answers, open this PDF file and look for section 4.15.

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English Spanish

graph of a function

The graph of a function is a visual aid showing the relationship between an independent variable and its dependent variable.


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1. ## Further Maths

1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2

1b) Find the cordinates of the turning point on the curve y = ( 2 - x - x^2 ) / x2

2) sketch y = root of (x-2)

3) Find x such that:
a) x^4 - 5x^2 6 > 0
b) | 3 - 2x < or equal to | x + 4 |

4a) show y = x - 6 is an asymptotoe of the curve y = (x-1)(x-3)/(x+2)

b) Sketch the curve showing cleanly the other asumptote, the points of intersection of the curve with the co-ordinate axes.

1a) Find the set of values of x for which 0 < ( 2 - x - x^2 ) / x^2 < 2

1b) Find the cordinates of the turning point on the curve y = ( 2 - x - x^2 ) / x2

Mr F says: Solve ${\color{red}\frac{dy}{dx} = 0}$. You can get dy/dx using either the quotient rule or seeing that ${\color{red}y = \frac{2}{x^2} - \frac{x}{x^2} - \frac{x^2}{x^2} = 2 x^{-2} - x^{-1} - 1}$.

2) sketch y = root of (x-2) Mr F says: Upper half of sideways parabola. Vertex at (2, 0).

3) Find x such that:
a) x^4 - 5x^2 6 > 0

Mr F says: Do you mean x^4 - 5x^2 + 6 > 0 or x^4 - 5x^2 - 6 > 0? Either way, I suggest drawing a graph and looking for the values of x for which the graph is above the x-axis.

b) | 3 - 2x < or equal to | x + 4 | Mr F says: You might start by drawing graphs of y = |3 - 2x| = |2x - 3| and y = |x + 4|. Then find the x-coordinates of the intersection points ....

4a) show y = x - 6 is an asymptotoe of the curve y = (x-1)(x-3)/(x+2)

Mr F says: ${\color{red} y = \frac{x^2 - 4x + 3}{x+2} = x - 6 + \frac{15}{x+2}}$ using polynomial long division .....

b) Sketch the curve showing cleanly the other asumptote, the points of intersection of the curve with the co-ordinate axes. Mr F says: Vertical asymptote x = -2. You should know how to find x- and y-intercepts (especially y-intercept!!).

format: plain


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Existence of orientation preserving, finite order self homeomorphism on a genus 2 surface without fixed point

Let $M$ be a compact 2-manifold of genus 2. Does there exist an orientation preserving homeomorphism $f:M\to M$, so that $f^n=id$ for some integer $n$, and $f$ doesn't have fixed points?

Using Lefschetz formula and the cohomology ring structure, I can prove that the order of such $f$ must be a multiple of 4 or 6. I vaguely remember that I have seen such an $f$ before, but I'm not sure. Is there anyone familiar with this kind of examples?

• You can view the classification of Seifert fibred spaces as a dictionary for enumerating conjugacy classes of finite-order automorphisms of surfaces, with or without fixed points, it's up to you. With that tool it becomes a simple enumeration problem. You just have to determine precisely which Seifert fiber spaces fiber over the circle -- this means they have a horizonal incompressible surface. – Ryan Budney Sep 4 '14 at 17:22

The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction.

Let $T$ be the sphere $S^2 \subset \Bbb R^3$ with three disks removed. Take these three disks to be centered at the vertices of an equilateral triangle inscribed in the maximal circle $\{z = 0\} \cap S^2$ and with the same radius, so that they are cyclically permuted by the $\frac{2\pi}{3}$-rotation $R$ about the $z$-axis. So, the surface in question is $M = \partial (T \times [-1, 1])$. Let $f : M \to M$ be defined by $f(x, y, z, t) = (R(x,y,-z), -t)$. Hence, $f$ satisfies the conditions in the question, with $n = 6$.

All finite group actions on a genus 2 surface have been classified; see Classifying finite group actions on surfaces of low genus by A. Broughton, Journal of Pure and Applied Algebra Volume 69, Issue 3, 7 January 1991, Pages 233–270. You can get the article for free from the journal web page. The result is given on page 252 in a table that has fixed point data, from which you can answer your question.

Here is a concrete example: let $C$ be the smooth projective model of the complex plane curve $y^2 = x^6-1$. Then $C$ is genus 2 and has a fixed point free automorphism given by $(x,y)\mapsto (\exp (2\pi i/3)x,-y)$.

Geometrically: let $C$ be the double branched cover of the Riemann sphere, ramified over the 6th roots of unity. The $Z/6Z$ action generated by rotating the Riemann sphere by $2\pi/3$ and simultaneously swapping the sheets of the cover is fixed point free (but the points over 0 and $\infty$ are orbits of size two).

If you are even more topologically minded, you can picture this as follows. Consider the graph with 2 vertices and three edges, each going from one vertex to the other. This graph has an order six automorphism generated by cyclically permuting the three edges and swapping the vertices. Now thicken this graph in three space to be a handle body whose boundary is a genus 2 surface. The previously mentioned automorphism acts on this surface in a fixed point free fashion.

• The projectivization of this curve seems to be singular at $[0,1, 0]$. – Daniele Zuddas Sep 4 '14 at 4:57
• Are you sure of this example? The Euler number of a genus $2$ surface is $-2$, it seems to me that this forbids free cyclic actions on it. Am I missing something? – Francesco Polizzi Sep 4 '14 at 8:38
• The curve is smooth but, as Francesco observes, the automorphism has 2 fixed points at infinity. One should take e.g. $(x,y)\mapsto (\omega x,-y)$ for the same curve, with $\omega$ a nontrivial third root of unity. – abx Sep 4 '14 at 8:52
• @Francesco Polizzi: the Euler number excludes a free action of a finite group, but your automorphism may have no fixed points while some power has fixed points. – abx Sep 4 '14 at 8:54
• @abx: ok, you are right. – Francesco Polizzi Sep 4 '14 at 8:57

The answer is actually yes, and here is an example with $n=6$ (inspired by Broughton's classification quoted by Danny and by Jim's answer).

Take the group $G = \mathbb{Z}/ 6 \mathbb{Z}$ presented as follows:$$G = \langle x, \, y \; | \; x^2=y^3=1, \, [x, y]=1 \rangle.$$ By Riemann Existence Theorem, any Galois cover of $\mathbb{P}^1:=\mathbb{P}^1(\mathbb{C})$ with group $G$ is given by a finite collection of generators $g_1, \ldots, g_k \in G$ whose product is the identity. Now choose $$g_1=x, \quad g_2=x, \quad g_3=y, \quad g_4=y^2.$$

The corresponding Galois cover $\pi \colon M \to \mathbb{P}^1$ has four branch points. The points over two of them have stabilizer of order $2$ (generated by $x$) and the points over the remaining two have stabilizer of order $3$ (generated by $y$).

By Hurwitz formula we have $$2g(M)-2 = |G| \bigg(2g(\mathbb{P}^1)-2+ 2\big(1-\frac{1}{2} \big) + 2 \big(1-\frac{1}{3} \big) \bigg) = 2,$$ that is $M$ is a compact manifold of genus $2$. The automorphism of $M$ corresponding to the generator $g:=xy \in G$ has order $6$ and is fixed point-free (whereas both $g^2=y^2$ and $g^3=x$ have fixed points).

see the paper: A note on the free degrees of homeomorphisms on genus 2 orientable compact surfaces


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# Prove that the Taylor series is the solution to this minimization problem?

The answer to this question says that a MacLaurin polynomial of a function ff is the unique NthN^\text{th} order polynomial that minimizes the following functional:

L[g]=limx→0(f(x)−g(x)xN)2,
L[g] = \lim_{x\rightarrow 0}\left(\frac{f(x)-g(x)}{x^N}\right)^2,

and that the minimum value attained is 00,

\min_{g} L[g] = 0

\min_{g} L[g] = 0

I’m wondering if anyone has a reference for a proof of this fact in a forward sense, i.e., starting from the above, show that gg must be equal to the formula for the MacLaurin series of ff, without directly invoking some form of Taylor’s theorem to expand ff. Basically, assuming we didn’t know anything about Taylor/MacLaurin series or polynomials, can we derive them from the above minimization problem?

Possible path

We know gg is an N^\text{th}N^\text{th} order polynomial:

g(x) = \sum_{n=0}^N a_n x^n = \vec{X}\cdot\vec{a},

g(x) = \sum_{n=0}^N a_n x^n = \vec{X}\cdot\vec{a},

where

\vec{X} = \left[1\;\; x\;\; x^2\;\;\ldots\;\;x^N\right]^\text{T},

\vec{X} = \left[1\;\; x\;\; x^2\;\;\ldots\;\;x^N\right]^\text{T},

and

\vec{a} = \left[a_0\;\; a_1\;\; a_2\;\;\ldots\;\;a_N\right]^\text{T}.

\vec{a} = \left[a_0\;\; a_1\;\; a_2\;\;\ldots\;\;a_N\right]^\text{T}.

Now I know I’ll need to know the sensitivities of that functional to the parameters:

\nabla_\vec{a} L[g] =
\left[
\frac{\partial L[g]}{\partial a_0}\;\;
\frac{\partial L[g]}{\partial a_1}\;\;
\frac{\partial L[g]}{\partial a_2}\;\;\ldots\;\;
\frac{\partial L[g]}{\partial a_N}
\right]^\text{T}

\nabla_\vec{a} L[g] =
\left[
\frac{\partial L[g]}{\partial a_0}\;\;
\frac{\partial L[g]}{\partial a_1}\;\;
\frac{\partial L[g]}{\partial a_2}\;\;\ldots\;\;
\frac{\partial L[g]}{\partial a_N}
\right]^\text{T}

And that through application of the chain rule, I’ll also need that

\nabla_\vec{a} g = \vec{X}.

\nabla_\vec{a} g = \vec{X}.

Setting the gradient of the functional w.r.t. \vec{a}\vec{a} to zero for each component should give me a way to solve for gg, I just don’t see a clear path forward that results in the formula for a Taylor polynomial. I hope someone can illuminate the way.

=================

=================

=================


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## Local PT symmetry violates the no-signaling principle

Yi-Chan Lee, Min-Hsiu Hsieh, Steven T. Flammia, Ray-Kuang Lee

Bender et al. have developed PT-symmetric quantum theory as an extension of quantum theory to non-Hermitian Hamiltonians. We show that when this model has a local PT symmetry acting on composite systems it violates the non-signaling principle of relativity. Since the case of global PT symmetry is known to reduce to standard quantum mechanics, this shows that the PT-symmetric theory is either a trivial extension or likely false as a fundamental theory.

http://arxiv.org/abs/1312.3395
Quantum Physics (quant-ph); High Energy Physics – Theory (hep-th); Mathematical Physics (math-ph)


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## Saturday, December 10, 2011

### HowTo: Root HTC Droid Incredible || Android Phone

Follow the steps to get root access of a HTC droid incredible phone and install CyanogenMode Android OS using windows PC.

1. Install HTC USB Drivers on your PC. Driver can be found here.
3. Connect your phone with the PC via usb cable with USB Debugging enabled in the phone. (do not select file transfer mode for the sdcard).
5. Unzip the revolutionary-x.x.zip inside the folder where adb.exe is available.
6. Run revolutionary.exe from command line.
7. Then follow the steps from here to install CyanogenMode.

## Wednesday, June 29, 2011

### Creating EPS figure in Linux

Use the following steps to create eps figure in linux.
1. Use openoffice to draw the figure
2. Export the file as figure.pdf
3. Run $pdftops -eps figure.pdf. This will create a file figure.eps 4. Run$cat figure.eps | ps2eps > MyFigure.eps. This will fix the bounding box of the eps file.

## Thursday, June 23, 2011

### HowTo: Build GNU RADIO with UHD

Platform: Ubuntu Lucid 10.04/10.10

** All the following commands (in italic) are to be run at the shell.
• CMAKE: Check if CMAKE is installed (just try to execute command cmake). If not installed then do the following.
2. Unzip
3. Goto the folder by $cd ~/cmake-X.X.X 4. Run commands$./bootstrap; make; sudo make install
• GIT: Check if GIT is installed (just try to execute command git). If not then install it from Applications->Ubuntu Software Center
• LibUSB: If LibUSB is not installed...
2. Unzip
3. Goto the folder by $cd ~/libusb-X.X.X 4. Run commands$./configure; make; sudo make install
$sudo apt-get -y install libfontconfig1-dev libxrender-dev libpulse-dev swig g++ automake autoconf libtool python-dev libfftw3-dev \ libcppunit-dev libboost-all-dev libusb-dev fort77 sdcc sdcc-libraries \  libsdl1.2-dev python-wxgtk2.8 git-core guile-1.8-dev \ libqt4-dev python-numpy ccache python-opengl libgsl0-dev \ python-cheetah python-lxml doxygen qt4-dev-tools \ libqwt5-qt4-dev libqwtplot3d-qt4-dev pyqt4-dev-tools python-qwt5-qt4 • Download and Install UHD $git clone git://ettus.sourcerepo.com/ettus/uhd.git
$cd ~/uhd/host/ $mkdir build
$cd build $cmake ../
$make $make test
$sudo make install $sudo ldconfig
$git clone git://gnuradio.org/gnuradio.git $cd ~/gnuradio
$git branch --track next origin/next $git checkout next
$./bootstrap $PKG_CONFIG_PATH=~/uhd/host/build/
$export PKG_CONFIG_PATH $./configure --enable-gr-uhd
$make $make check
$sudo make install $sudo ldconfig
• Configure USRP Support
$sudo addgroup usrp$sudo usermod -G usrp -a <YOUR_USERNAME>
$echo 'ACTION=="add", BUS=="usb", SYSFS{idVendor}=="fffe", SYSFS{idProduct}=="0002", GROUP:="usrp", MODE:="0660"' > tmpfile$sudo chown root.root tmpfile
$sudo mv tmpfile /etc/udev/rules.d/10-usrp.rules$sudo udevadm control --reload-rules

## Thursday, June 2, 2011

### USPR2 and Boost Problem

USRP2 requires boost package for installation. As of today (2nd June, 2011) USPR2 is supposed to work with Boost 1.36 or greater but it doesn't for 1.45. After trying several versions, I got it working with Boost 1.41.

## Monday, May 23, 2011

### NS3: Generate Statistical Result

By default, NS3 generates exactly same result when you run a simulation. If you want to collect statistical behavior by running the simulation multiple times, you need to avoid this default setting of NS3. One way to do this is using different seed for every simulation. Just add the following line before calling the run() method.

SeedManager::SetSeed(time(0));
Here the current system time is used as the seed which ensures that seed will be different for every simulation and hence the simulation result.

## Saturday, May 14, 2011

### NS3: Apply a Patch to Source

Sometimes it is necessary to apply patch to update NS3 source. Use the following steps to apply patch.

• Open shell
• Goto ns3 root directory (cd /...path to .../ns-3***)
• run the command 'patch -p1 < /...path to ../patchFile'

## Friday, May 13, 2011

http://forums.debian.net/viewtopic.php?t=30598

## Wednesday, May 11, 2011

### Bangla Font Problem in Firefox

Some versions of firefox web browser has problem in showing Bangla font properly. It generally occurs after an update of firefox to the next version. Please follow the following steps if you have this problem.

• In the firefox menu, go to option->option ->content
• Select the tab 'Font & Color'
• Select the Bangla font and save.

## Monday, May 9, 2011

For my android app, I needed a UDP receiver which can receive broadcast packets. So I tried to create a UDP socket using the following.

But this socket could only receive unicast packets not the broadcast ones. I had to change it as follows to make it work for broadcast.

mySocket = new DatagramSocket(PORT);

### NS3: Wireless Jamming Model

The basic implementation of NS3 does not include anything for wireless jamming but wireless jamming is interesting research topic for wireless network security. Good news is Network Security Lab, University of Washington is working on a jamming model for NS3. Use the following link for more information.

http://www.nsnam.org/wiki/index.php/NS-3_wireless_jamming_model

## Thursday, May 5, 2011

### Latex: Italic Underline Problem

Some Latex compiler underlines the the italic text, more precisely the text inside \emph{your text}. If you don't want the text underlined, just add "\normalem" after the line "\begin{document}" !

## Wednesday, May 4, 2011

### NS3: Constant Speed Mobility

NS3 does not support constant speed mobility but it has a very good class RandomWalk2dMobilityModel which supports random speed and direction based on defined time or distance. If you need to simulation constant speed mobility, use the following steps:

1. Open file '..../ns-3-dev/src/mobility/random-direction-2d-mobility-model.cc'
2. In function DoStartPrivate()
• comment the line 'double speed = m_speed.GetValue ();'
• add a line 'double speed = your_constant_speed_value'
3. Recompile ns-3 and you are done.


text:
# Projection operators and Weyl spinors

1. Nov 11, 2015

### CAF123

I am working through some course notes where the aim is to derive the equations of motion satisfied by the left handed and right handed components of the Dirac spinor $\psi$. From the Dirac lagrangian, we have $$\mathcal L = \bar \psi (i \not \partial P_L - m P_L)\psi_L + \bar \psi (i \not \partial P_R - m P_R)\psi_R$$ The next line is $$\mathcal L = \bar \psi_L i \not \partial \psi_L - \bar \psi_R m \psi_L + \bar \psi_R i \not \partial \psi_R - \bar \psi_L m \psi_R$$ I understand where the terms involving m come from but I am not sure about the other two. Can anyone help?

Another question is to do with the acting of the projection operators onto Dirac spinors. Since $$P_L \psi = P_L \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \frac{1}{2}(1-\gamma^5)\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix} = \frac{1}{2} \left( \begin{pmatrix} 1 & 0 \\0 & 1 \end{pmatrix} - \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \right) \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$$ which does not project out the left handed component. If I use the Weyl representation of $\gamma^5$, it works but I am trying to understand why not in the Dirac representation of the $\gamma^5$.

Thanks!

2. Nov 11, 2015

### Orodruin

Staff Emeritus
What is your problem with them? They follow directly from the form of the derivative terms.

Because in Dirac representation the field is not represented as
$$\begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix}$$
with $\psi_L$ and $\psi_R$ being the left- and right-handed components of the Dirac spinor.

3. Nov 11, 2015

### CAF123

Hi Orodruin,
If I write out the leftmost term explicitly then I get, $$i\psi^{\dagger} \gamma^0 \gamma^{\mu} \partial_{\mu} P_L \psi_L$$ I don't see the manipulations that made the adjoint spinor go to a left handed adjoint spinor.

I see, how do we write the field in Dirac representation? Just like $\psi = \psi_L + \psi_R$?

Thanks!

4. Nov 11, 2015

### Orodruin

Staff Emeritus
What is the commutation relation of $\gamma^5$ with the other gamma matrices?

That is true in any representation. In Dirac representation, the components just do not have the same interpretation in terms of left and right handed fields.

5. Nov 12, 2015

### CAF123

I think I see: $$\psi^{\dagger} \gamma^0 \gamma^{\mu} \partial_{\mu} (P_L \psi_L)= \psi^{\dagger} \gamma^0 \gamma^{\mu} P_L \partial_{\mu} \psi_L = \frac{1}{2} \psi^{\dagger} \gamma^0 \gamma^{\mu} (1-\gamma^5) \partial_{\mu} \psi_L = \frac{1}{2}\psi^{\dagger} (\gamma^0 \gamma^{\mu} + \gamma^0 \gamma^5 \gamma^{\mu}) \partial_{\mu} \psi_L = \frac{1}{2} \psi^{\dagger} \gamma^0 (1+\gamma^{5}) \gamma^{\mu} \partial_{\mu} \psi_L = \psi^{\dagger} \gamma^0 P_R \gamma^{\mu} \partial_{\mu} \psi_L = \psi^{\dagger} P_L \gamma^0 \gamma^{\mu} \partial_{\mu} \psi_L \Rightarrow \text{result}$$ Is that ok?

Ok, so in Dirac representation, the spinors are those labelled by $u(p,s)$ and $v(p,s)$, where e.g $$u(p,s) = N\begin{pmatrix} \phi^s \\ \frac{\sigma \cdot p}{E+m} \phi^s \end{pmatrix}$$ The components are not written in terms of LH and RH fields. Is that what you meant? So when I (incorrectly) applied the Dirac representation of $\gamma^5$ onto the Weyl spinor $\langle \psi_L, \psi_R \rangle$ I was going to get, as a result, a state with the left and right handed fields mixed and thus no notion of left and right handed states in the Dirac representation?

Thanks!

6. Nov 12, 2015

### Orodruin

Staff Emeritus
Yes.

Yes, this is no stranger than using a rotated representation of SO(2) or any other group. The representations are unitary equivalent, but look different in terms of components.

7. Nov 14, 2015

### CAF123

Ok so with the manipulations in the OP understood I get, in the zero mass limit, a lagrangian like $$\mathcal L = i \bar \psi_L \not \partial \psi_L + i \bar \psi_R \not \partial \psi_R.$$ The first term here is supposed to describe a massless fermion with negative helicity. (or equivalently a massless antifermion with positive helicity). I want to see why this is the case.

The first term gives, for the equation of motion for the LH component of the field, $\not \partial \psi_L = 0$. The generic solution to the Dirac equation is $\psi = \omega(p,s) e^{-ip\cdot x}$, where $\omega$ is four component spinor describing particle (u) or antiparticle (v) state, both of positive energy. A negative helicity means that the momenta is always in direction opposite to that of the spin? which by definition is what left handed means. Are these statements correct? Thanks :)


text:
## Tuesday, November 20, 2012

### The Internal Energy of Air

You know how sometimes, as you're going about your daily life, a completely random thought leads to you suddenly being intensely curious about something and unable to rest until your curiosity has been sated? This weekend I was thinking about nothing in particular while setting up for the morning at work, when I got the burning desire to know how much internal energy a cubic meter of air contained.

The nice part of being a physicist is that I can satisfy these urges, and the nice part of having a blog is that I can share it with other people! So without further ado, let's attempt to calculate the internal energy of 1 cubic meter of air at standard atmospheric pressure and room temperature (around 80 °F, or specifically for ease of calcuation, 300 kelvin).

This is actually fairly simple in theory. There exists a simple equation in thermodynamics for the internal energy of an ideal gas: $$U=\frac{N}{2}nRT\tag{1}$$ In this equation, U is the internal energy locked up in each of the N degrees of freedom of the gas, n is the number of moles of gas, the constant R is the ideal gas constant with value $$8.3144621\ \text{J}/(\text{mol}\cdot\text{K})$$, and T is the absolute temperature in kelvins.

Now, at this point I should probably elaborate on what internal energy is and what it has to do with degrees of freedom. As I'm sure you know, the temperature of a system is merely a measure of the average energy of its constituent particles. This energy is called the internal energy of the system and can be stored in several different ways, each of which is known as a degree of freedom: in the motions of particles (atoms or molecules), in the rotation or vibration of molecules, or in the excitation and relaxation of electrons in the atoms (not all of these actually apply to all systems, as we shall see).

In thermodynamics there is a theorem known as the equipartition theorem that states that the available internal energy of a system is equally divided among all of its degrees of freedom. If we look at an ideal monatomic gas (such as any of the noble gases), it has only three degrees of freedom, corresponding to the three ways the particles making up the gas can move in three dimensions. Technically, the energy stored by electrons being excited in the atoms could count as another degree of freedom, but at the relatively low temperatures we're considering for this problem there is very little excitation going on and we are free to ignore this effect.

This would be fine if we were considering a monatomic gas, but we are interested in air, which is primarily composed of two diatomic gases: nitrogen (78%) and oxygen (21%). (The monatomic gas argon makes up about 90% of the remaining ~1% of the atmosphere, so we will simply assume that it is all argon for simplicity.) Diatomic molecules bring a new factor into the equation, as they can rotate in two dimensions around their long axis, and since energy can be stored in their rotational motion, this gives them another two degrees of freedom. Diatomic molecules can also store energy in the bond between them, and while this could count as another degree of freedom, in practice it takes temperatures much higher than we are considering here for this to be a significant effect.

So, in summary, monatomic gases have 3 degrees of freedom, representing the kinetic energy associated with their motion through three-dimensional space; diatomic gases – at the temperatures we are interested in – have 5 degrees of freedom since they have the ability to rotate as well. (If I was looking at much higher temperatures I'd have to take into account that vibrational mode I neglect here.)

The upshot of that lengthy diversion is that equation ($$1$$) above looks like $$\frac{3}{2}nRT$$ for monatomic gases and $$\frac{5}{2}nRT$$ for diatomic ones. Since R is a constant and we have a temperature in mind already, all that remains is to find n, the amount of each type of gas.

The n in that equation refers to moles of gas. The mole (abbreviated mol) is a unit used in chemistry and physics to represent a quantity of substance in terms of the number of particles (atoms or molecules) that make it up. A closely related concept is that of Avogadro's Number, $$6.022\times10^{23}$$ (named after the Italian scientist Amadeo Avogadro). One mole of a substance is simply the amount of that substance that contains Avogadro's number of particles in it (it's slightly more complicated than that, but this will suffice for our purposes). Avogadro's number may seem arbitrary, but it is actually measured and defined such that if you have an amount of a substance in grams equal to its mean atomic mass in daltons, then you have one mole of that substance. (The name dalton is given to the unit of mass formerly known as the atomic mass unit, a handy measure for measuring the weight of atoms. It is roughly equivalent to the mass of a nucleon.)

For example: a hydrogen atom has a mean atomic mass of $$1.01$$ daltons. Hydrogen typically combines with itself to form dihydrogen gas, H$$_2$$. Thus dihydrogen gas has a mean molecular mass of $$2.02$$ daltons. If you have $$2.02$$ grams of dihydrogen gas, you then have one mole ($$6.022\times10^{23}$$) of dihydrogen gas molecules. Oxygen (mean atomic mass $$16.00$$ daltons) likewise combines to form dioxygen (O$$_2$$) with a mean atomic mass of $$32.00$$ daltons. If you have $$32.00$$ grams of dioxygen gas, you then have one mole of dioxygen molecules. Combining the two to make water, H$$_2$$O, gives water a mean atomic mass of $$18.02$$ ($$2\times1.01+16.00$$), so if you have $$18.02$$ grams of water, you have one mole of water molecules.

Anyway, this lengthy preface should hopefully enable you to follow what should be a fairly straight-forward calculation, which we are finally ready to begin.

First off, we need to find the number of moles of oxygen, nitrogen, and argon in one cubic meter of our theoretical approximation of air. We can do that by first finding the density of air at our specified conditions (300 K, ~80 °F and 1 standard atmosphere of pressure, 101.325 kPa), then multiplying by the fractions established before to find out how much mass of each gas exists, before converting that mass into moles of each gas to fit the equation.

There is a equation for the density of dry air (which we are assuming it is) given by $\rho=\frac{P}{R_{\text{specific}}T}$ In this equation, $$\rho$$ (the Greek letter rho) stands for density (in kg/m$$^3$$), P stands for pressure, R$$_{\text{specific}}$$ is a version of the ideal gas constant specifically for dry air equal to 287.058 J/(kg$$\cdot$$K), and T is again the temperature in kelvins.

Putting in the numbers and doing the math, we get:
\begin{align}\rho&=\frac{101,325 \frac{\text{N}}{\text{m}^2} }{287.058 \frac{\text{N}\cdot\text{m}}{\text{kg}\cdot\text{K}} \cdot300.00\ \text{K}}\\
&=1.1766\frac{\text{kg}}{\text{m}^3}
\end{align}
Since we are assuming a single cubic meter of air, our mass of air consists of 1.1766 kg (about 2.6 pounds of air). Multiplying by the fractions we assumed for each of the ingredients, we get:
\begin{align}
m_{\text{N}_2}&=0.78\cdot1.1766\ \text{kg}=0.9177\ \text{kg}=917.7\ \text{g}\\
m_{\text{O}_2}&=0.21\cdot1.1766\ \text{kg}=0.2471\ \text{kg}=247.1\ \text{g}\\
m_{\text{Ar}}&=0.01\cdot1.1766\ \text{kg}= 0.0118\ \text{kg}=11.8\ \text{g}
\end{align}
Now that we have the masses involved, we can convert to moles using their mean atomic masses:
\begin{align}
n_{\text{N}_2}&=917.7\ \text{g}/28.013\frac{\text{g}}{\text{mol}}=32.762\ \text{moles}\\
n_{\text{O}_2}&=247.1\ \text{g}/31.9988\frac{\text{g}}{\text{mol}}=7.7222\ \text{moles}\\
n_{\text{Ar}}&= 11.8 \ \text{g}/39.948\frac{\text{g}}{\text{mol}}=0.29538 \ \text{moles}
\end{align}
Having now obtained the number of moles of each gas in our hypothetical approximation to air, we can now use equation ($$1$$) to calculate the amount of internal energy each gas contributes to the whole.
\begin{align}
U _{\text{N}_2}&=\frac{5}{2}\cdot32.762\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=204.3\ \text{kJ}\\
U_{\text{O}_2}&= \frac{5}{2}\cdot7.7222\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=48.12\ \text{kJ} \\
U_{\text{Ar}}&=\frac{3}{2}\cdot0.29538\ \text{mol}\cdot8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}=1.105\ \text{kJ}
\end{align}
This gives us a total of
$U _{\text{N}_2}+ U_{\text{O}_2}+ U_{\text{Ar}}=253.5\ \text{kJ}$
That...actually turns out to be a bit more than I was expecting. That's a quarter of a million joules of energy contained in the motion and rotation of the air molecule in a single cubic meter of air.

To put this number in perspective, let's do some conversions to units you may be more familiar with. That many kilojoules is almost exactly 60 kilocalories (or Calories), the unit the energy in food is measured in. Put another way, the normal energy needs of an adult human are typically pegged at around 2,000 Calories per day. If you could somehow extract the energy from air, you'd need only about 33 cubic meters of air per day to survive, a volume smaller than the amount of air in most average-sized homes. Alternatively, the average amount of solar power over a 1 square meter area at the Earth's surface is about 1 kilojoule per second (1 kilowatt), so the amount of energy we calculated is equivalent to the amount hitting an area of 253,500 square meters (a quarter of a square kilometer) every second during full daylight. There's a lot of energy locked up in the air around you.

In a sense, though, I suppose I really shouldn't be too surprised. Gas molecules in the air whiz about at great speed, and this speed comes from the kinetic energy they have. In fact, we can estimate the root-mean-square speed of a typical nitrogen molecule fairly easily (the M is the molar-mass of the gas, in kg/mol):
\begin{align}v_{\text{rms}}&=\sqrt{\frac{3RT}{M}}\\
&=\sqrt{\frac{3\cdot 8.314\frac{\text{J}}{\text{K}\cdot\text{mol}}\cdot300.00\ \text{K}}{0.028013\frac{\text{kg}}{\text{mol}}}}\\
&=516.8\frac{\text{m}}{\text{s}}
\end{align}In case you're wondering, that a whopping 1,156 miles per hour. Those nitrogen molecules are, on average, moving about that fast (oxygen and argon move a bit slower, since they're more massive). So I guess when you consider billions upon billions of tiny atoms all zooming around at speeds comparable to this, it makes sense that there's a lot of energy tied up in their motion as kinetic energy. Wow. Amazing stuff.


text:
# EmotionNet Nano: An Efficient Deep Convolutional Neural Network Design for Real-time Facial Expression Recognition

 Title EmotionNet Nano: An Efficient Deep Convolutional Neural Network Design for Real-time Facial Expression Recognition Publication Type Journal Article Year of Publication 2020 Authors Lee, J. Ren Hou, L. Wang, and A. Wong Journal arXiv preprint arXiv:2006.15759


text:
Table 1:

SG features of trial t. ai, li, and dmpi denote the DM’s action, lottery result, and DM’s payoff in trial t, respectively. s(ht) is the average score of the suggested hotel in trial t, rt is its revealed review, and s(rt) is the revealed review score. * indicates that the feature is taken from Apel et al. (2020).

Feature NameFeature DescriptionFeature Formulation
Behavioral Features
HotelAcceptance Avg #trials where the hotel was accepted $∑i=1t−11{ai=1}t−1$
HotelAcceptance Earn Avg #trials where the hotel was accepted and the DM achieved a negative payoff.* $∑i=1t−11{ai=1∩dmpi>0}t−1$
HotelAcceptance Lose Avg #trials where the hotel was accepted and the DM achieved a positive payoff.* $∑i=1t−11{ai=1∩dmpi<0}t−1$
¬HotelAcceptance Earn Avg #trials where the hotel was not accepted but the payoff would have been positive if the DM had accepted it.* $∑i=1t−11{ai=0∩dmpi>0}t−1$
¬HotelAcceptance Lose Avg #trials where the hotel was not accepted but the payoff would have been negative if the DM had accepted it.* $∑i=1t−11{ai=0∩dmpi<0}t−1$
BadHotel Acceptance Avg #trials where a hotel with average score lower than 7.5 was accepted. $∑i=1t−11{ai=1∩s(hi)<7.5}t−1$
¬ExcellentHotel Acceptance Avg #trials where a hotel with average score higher than 9.5 was accepted. $∑i=1t−11{ai=0∩s(hi)>9.5}t−1$
DMPayoff Avg DM’s payoff pet trial $∑i=1t−1dmpit−1$

General Features
LotteryLow Avg #trials where the lottery result was lower than 3.* $∑i=1t−11{li<3}t−1$
LotteryMed Avg #trials where the lottery result was between 3 to 5.* $∑i=1t−11{li≥3∩li<5}t−1$
LotteryHigh Avg #trials where the lottery result was higher than 8.* $∑i=1t−11{li≥8}t−1$
CompletedTrials The proportion of trials that have already been played. $t−110$
GoodHotel Avg score of the current hotel is higher than 8.5. $1{s(ht)≥8.5}$
MedHotel Avg score of the current hotel is between 7.8 to 8.5. $1{s(ht)<8.5∩s(ht)≥7.5}$
BadHotel Avg score of the hotel is lower than 7.5. $1{s(ht)≤7.5}$
HighScore The attached score of the presented review is higher than 8.5. $1{s(rt)≥8.5}$
MedScore The attached score of the presented review is between 7.5 to 8.5. $1{s(rt)<8.5∩s(rt)≥7.5}$
LowScore The attached score of the presented review is lower than 7.5. $1{s(rt)<7.5}$
TopReview The attached score of the presented review is in the top 3 scoring reviews. $1{s(rt)∈top 3 scores}$
BottomReview The attached score of the presented review is not in the top 3 scoring reviews. $1{s(rt)∉top 3 scores}$
Feature NameFeature DescriptionFeature Formulation
Behavioral Features
HotelAcceptance Avg #trials where the hotel was accepted $∑i=1t−11{ai=1}t−1$
HotelAcceptance Earn Avg #trials where the hotel was accepted and the DM achieved a negative payoff.* $∑i=1t−11{ai=1∩dmpi>0}t−1$
HotelAcceptance Lose Avg #trials where the hotel was accepted and the DM achieved a positive payoff.* $∑i=1t−11{ai=1∩dmpi<0}t−1$
¬HotelAcceptance Earn Avg #trials where the hotel was not accepted but the payoff would have been positive if the DM had accepted it.* $∑i=1t−11{ai=0∩dmpi>0}t−1$
¬HotelAcceptance Lose Avg #trials where the hotel was not accepted but the payoff would have been negative if the DM had accepted it.* $∑i=1t−11{ai=0∩dmpi<0}t−1$
BadHotel Acceptance Avg #trials where a hotel with average score lower than 7.5 was accepted. $∑i=1t−11{ai=1∩s(hi)<7.5}t−1$
¬ExcellentHotel Acceptance Avg #trials where a hotel with average score higher than 9.5 was accepted. $∑i=1t−11{ai=0∩s(hi)>9.5}t−1$
DMPayoff Avg DM’s payoff pet trial $∑i=1t−1dmpit−1$

General Features
LotteryLow Avg #trials where the lottery result was lower than 3.* $∑i=1t−11{li<3}t−1$
LotteryMed Avg #trials where the lottery result was between 3 to 5.* $∑i=1t−11{li≥3∩li<5}t−1$
LotteryHigh Avg #trials where the lottery result was higher than 8.* $∑i=1t−11{li≥8}t−1$
CompletedTrials The proportion of trials that have already been played. $t−110$
GoodHotel Avg score of the current hotel is higher than 8.5. $1{s(ht)≥8.5}$
MedHotel Avg score of the current hotel is between 7.8 to 8.5. $1{s(ht)<8.5∩s(ht)≥7.5}$
BadHotel Avg score of the hotel is lower than 7.5. $1{s(ht)≤7.5}$
HighScore The attached score of the presented review is higher than 8.5. $1{s(rt)≥8.5}$
MedScore The attached score of the presented review is between 7.5 to 8.5. $1{s(rt)<8.5∩s(rt)≥7.5}$
LowScore The attached score of the presented review is lower than 7.5. $1{s(rt)<7.5}$
TopReview The attached score of the presented review is in the top 3 scoring reviews. $1{s(rt)∈top 3 scores}$
BottomReview The attached score of the presented review is not in the top 3 scoring reviews. $1{s(rt)∉top 3 scores}$
Close Modal


text:
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Please use this identifier to cite or link to this item: http://arks.princeton.edu/ark:/88435/dsp011z40ks95z
 Title: Essays on Incentives and Effort Authors: Lau, Yan Advisors: Mas, Alexandre Contributors: Economics Department Keywords: EffortIncentivesLabor SupplyPersonnel EconomicsSchooling CohortTournament Theory Subjects: EconomicsEconomics, Labor Issue Date: 2013 Publisher: Princeton, NJ : Princeton University Abstract: Incentives motivate economic agents to exert effort in producing outcomes, and the way governments, firms, and educators structure incentive schemes can influence the amount of effort exerted by these agents. This collection of essays examines various factors affecting such effort decisions. Chapter 1 investigates the role of academic competition in student effort decisions. I develop a tournament model to describe how an increase in the size of a particular schooling cohort in Hong Kong can motivate better students to "step up" effort exertion, but discourage less-able students. I use test score data from Hong Kong to empirically test these predictions. While Chapter 1 involves a natural experiment, Chapter 2 examines how changes in tournament structure (the winning cutoff and prize amount) in a laboratory experiment can influence effort decisions. Generating data in a controlled experimental setting where participants are asked to compete with one another in memorization tasks, I find that effort is positively correlated with prize amount, and that only low-performing participants are effort-responsive to movements in the winning cutoff. Moving from tournament incentive schemes to piece-rate incentive schemes, Chapter 3 re-examines an experiment conducted by Fehr and Goette (2007) in which the wage rate of Zurich bicycle messengers is increased for a short period of time. Fehr and Goette use a reference dependent preferences model to explain patterns in effort response. To explain their findings without appealing to behavioral mechanisms, I build a two-step non-time-separable neoclassical model with uncertainty resulting from limited work availability. I conduct additional analyses with their data to argue my case for the neoclassical model. The findings of these three essays suggest that the way incentive schemes are structured does matter in motivating agents to exert effort. URI: http://arks.princeton.edu/ark:/88435/dsp011z40ks95z Alternate format: The Mudd Manuscript Library retains one bound copy of each dissertation. Search for these copies in the library's main catalog Type of Material: Academic dissertations (Ph.D.) Language: en Appears in Collections: Economics

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text:
## CryptoDB

### Paper: Efficient Implementation of the Pairing on Mobilephones using BREW

Authors: Motoi Yoshitomi Tsuyoshi Takagi Shinsaku Kiyomoto Toshiaki Tanaka URL: http://eprint.iacr.org/2007/340 Search ePrint Search Google Pairing based cryptosystems can accomplish novel security applications such as ID-based cryptosystems, which have not been constructed efficiently without the pairing. The processing speed of the pairing based cryptosystems is relatively slow compared with the other conventional public key cryptosystems. However, several efficient algorithms for computing the pairing have been proposed, namely Duursma-Lee algorithm and its variant $\eta_T$ pairing. In this paper, we present an efficient implementation of the pairing over some mobilephones. The processing speed of our implementation in ARM9 processors on BREW achieves under 100 milliseconds using the supersingular curve over $\mathbb F_{3^{97}}$. It has become efficient enough to implement security applications, such as ID-based cryptosystems and broadcast encryption, using the pairing on BREW mobilephones.
##### BibTeX
@misc{eprint-2007-13620,
title={Efficient Implementation of the Pairing on Mobilephones using BREW},
booktitle={IACR Eprint archive},
keywords={implementation / pairing based cryptosystem, mobilephone, BREW},
url={http://eprint.iacr.org/2007/340},
note={This is a full version of WISA 2007 paper. takagi@fun.ac.jp 13769 received 28 Aug 2007, last revised 12 Sep 2007},
author={Motoi Yoshitomi and Tsuyoshi Takagi and Shinsaku Kiyomoto and Toshiaki Tanaka},
year=2007
}


text:
## Elementary Algebra

The standard form of a quadratic equation is $ax^{2}+bx+c=0$. Since the quadratic formulas can be used to solve any type of quadratic equation, this means that quadratic formula can also be used to solve those equations which are easily solved by factoring.


text:
Symbol a0 or rBohr Niels Bohr 5.29×10−11 m 3.27×1024 ℓP

The Bohr radius (a0) is a physical constant, approximately equal to the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It is named after Niels Bohr, due to its role in the Bohr model of an atom. Its value is 5.29177210903(80)×10−11 m.[1][2]

## Definition and value

The Bohr radius is defined as[3] $\displaystyle{ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{e^2 m_{\text{e}}} = \frac{\varepsilon_0 h^2}{\pi e^2 m_{\text{e}}} = \frac{\hbar}{m_{\text{e}} c \alpha} , }$

where

The CODATA value of the Bohr radius (in SI units) is 5.29177210903(80)×10−11 m.[1]

## History

In the Bohr model for atomic structure, put forward by Niels Bohr in 1913, electrons orbit a central nucleus under electrostatic attraction. The original derivation posited that electrons have orbital angular momentum in integer multiples of the reduced Planck constant, which successfully matched the observation of discrete energy levels in emission spectra, along with predicting a fixed radius for each of these levels. In the simplest atom, hydrogen, a single electron orbits the nucleus, and its smallest possible orbit, with the lowest energy, has an orbital radius almost equal to the Bohr radius. (It is not exactly the Bohr radius due to the reduced mass effect. They differ by about 0.05%.)

The Bohr model of the atom was superseded by an electron probability cloud obeying the Schrödinger equation as published in 1926. This is further complicated by spin and quantum vacuum effects to produce fine structure and hyperfine structure. Nevertheless, the Bohr radius formula remains central in atomic physics calculations, due to its simple relationship with fundamental constants (this is why it is defined using the true electron mass rather than the reduced mass, as mentioned above). As such, it became the unit of length in atomic units.

In Schrödinger's quantum-mechanical theory of the hydrogen atom, the Bohr radius represents the most probable value of the radial coordinate of the electron position, and therefore the most probable distance of the electron from the nucleus.[4]

## Related constants

The Bohr radius is one of a trio of related units of length, the other two being the reduced Compton wavelength of the electron ($\displaystyle{ \lambda_{\mathrm{e}} / 2\pi }$) and the classical electron radius ($\displaystyle{ r_{\mathrm{e}} }$). Any one of these constants can be written in terms of any of the others using the fine-structure constant $\displaystyle{ \alpha }$:

$\displaystyle{ r_{\mathrm{e}} = \alpha \frac{\lambda_{\mathrm{e}}}{2\pi} = \alpha^2 a_0. }$

## Hydrogen atom and similar systems

The Bohr radius including the effect of reduced mass in the hydrogen atom is given by

$\displaystyle{ \ a_0^* \ = \frac{m_\text{e}}{\mu}a_0 , }$

where $\displaystyle{ \mu = m_\text{e} m_\text{p} / (m_\text{e} + m_\text{p}) }$ is the reduced mass of the electron–proton system (with $\displaystyle{ m_\text{p} }$ being the mass of proton). The use of reduced mass is a generalization of the classical two-body problem when we are outside the approximation that the mass of the orbiting body is negligible compared to the mass of the body being orbited. Since the reduced mass of the electron–proton system is a little bit smaller than the electron mass, the "reduced" Bohr radius is slightly larger than the Bohr radius ($\displaystyle{ a^*_0 \approx 1.00054\, a_0 \approx 5.2946541 \times 10^{-11} }$ meters).

This result can be generalized to other systems, such as positronium (an electron orbiting a positron) and muonium (an electron orbiting an anti-muon) by using the reduced mass of the system and considering the possible change in charge. Typically, Bohr model relations (radius, energy, etc.) can be easily modified for these exotic systems (up to lowest order) by simply replacing the electron mass with the reduced mass for the system (as well as adjusting the charge when appropriate). For example, the radius of positronium is approximately $\displaystyle{ 2\,a_0 }$, since the reduced mass of the positronium system is half the electron mass ($\displaystyle{ \mu_{\text{e}^-,\text{e}^+} = m_\text{e}/2 }$).

A hydrogen-like atom will have a Bohr radius which primarily scales as $\displaystyle{ r_{Z}=a_0/Z }$, with $\displaystyle{ Z }$ the number of protons in the nucleus. Meanwhile, the reduced mass ($\displaystyle{ \mu }$) only becomes better approximated by $\displaystyle{ m_\text{e} }$ in the limit of increasing nuclear mass. These results are summarized in the equation

$\displaystyle{ r_{Z,\mu} \ = \frac{m_\text{e}}{\mu}\frac{a_0}{Z} . }$

A table of approximate relationships is given below.

Hydrogen $\displaystyle{ 1.00054\, a_0 }$
Positronium $\displaystyle{ 2 a_0 }$
Muonium $\displaystyle{ 1.0048\, a_0 }$
He+ $\displaystyle{ a_0/2 }$
Li2+ $\displaystyle{ a_0/3 }$


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Marilyn vos Savant

Marilyn vos Savant (/ˌvɒs səˈvɑːnt/; born Marilyn Mach; August 11, 1946) is an American magazine columnist[3] who has the highest recorded intelligence quotient (IQ) in the Guinness Book of Records, a competitive category the publication has since retired. Since 1986, she has written "Ask Marilyn", a Parade magazine Sunday column wherein she solves puzzles and answers questions on various subjects, and which popularised the Monty Hall problem in 1990.

Marilyn vos Savant
BornMarilyn Mach
August 11, 1946 (age 75)[1]
St. Louis, Missouri, U.S.
Occupation
Spouse
(m. 1987)
Children2[2]

Biography

Marilyn vos Savant was born Marilyn Mach[4] on August 11, 1946,[1] in St. Louis, Missouri, to parents Joseph Mach and Marina vos Savant.[citation needed] Savant says one should keep premarital surnames, with sons taking their fathers' and daughters their mothers'.[5][6] The word savant, meaning someone of learning, appears twice in her family: her grandmother's name was Savant; her grandfather's, vos Savant. She is of Italian, Czechoslovak,[7] German,[8] and Austrian ancestry, being descended from the physicist and philosopher Ernst Mach.[9]

As a teenager, Savant worked in her father's general store and wrote for local newspapers using pseudonyms. She married at 16 and divorced ten years later. Her second marriage ended when she was 35.

She went to Meramec Community College and studied philosophy at Washington University in St. Louis but quit two years later to help with a family investment business. Savant moved to New York City in the 1980s to pursue a career in writing. Before starting "Ask Marilyn", she wrote the Omni I.Q. Quiz Contest for Omni, which included intelligence quotient (IQ) quizzes and expositions on intelligence and its testing.

Savant married Robert Jarvik (one of the co-developers of the Jarvik-7 artificial heart) on August 23, 1987,[10][11] and was made Chief Financial Officer of Jarvik Heart, Inc. She has served on the board of directors of the National Council on Economic Education, on the advisory boards of the National Association for Gifted Children and the National Women's History Museum,[12] and as a fellow of the Committee for Skeptical Inquiry.[13] Toastmasters International named her one of "Five Outstanding Speakers of 1999", and in 2003 she was awarded an honorary Doctor of Letters degree from The College of New Jersey.

Rise to fame and IQ score

Savant was listed in the Guinness Book of World Records under "Highest IQ" from 1985 to 1989[4] and entered the Guinness Book of World Records Hall of Fame in 1988.[4][14] Guinness retired the "Highest IQ" category in 1990 after concluding IQ tests were too unreliable to designate a single record holder.[4] The listing drew nationwide attention.[15]

Guinness cited vos Savant's performance on two intelligence tests, the Stanford-Binet and the Mega Test. She took the 1937 Stanford-Binet, Second Revision test at age ten.[8] She says her first test was in September 1956 and measured her mental age at 22 years and 10 months, yielding a 228 score.[8] This figure was listed in the Guinness Book of World Records; it is also listed in her books' biographical sections and was given by her in interviews.

The second test reported by Guinness was Hoeflin's Mega Test, taken in the mid-1980s. The Mega Test yields IQ standard scores obtained by multiplying the subject's normalized z-score, or the rarity of the raw test score, by a constant standard deviation and adding the product to 100, with Savant's raw score reported by Hoeflin to be 46 out of a possible 48, with a 5.4 z-score, and a standard deviation of 16, arriving at a 186 IQ. The Mega Test has been criticized by professional psychologists as improperly designed and scored, "nothing short of number pulverization".[16]

Savant sees IQ tests as measurements of a variety of mental abilities and thinks intelligence entails so many factors that "attempts to measure it are useless".[17] She has held memberships with the high-IQ societies Mensa International and the Mega Society.[18]

Following her listing in the 1986 Guinness Book of World Records, Parade ran a profile of her along with a selection of questions from Parade readers and her answers. Parade continued to get questions, so "Ask Marilyn" was made.

She uses her column to answer questions on many chiefly academic subjects; solve logical, mathematical or vocabulary puzzles posed by readers; answer requests for advice with logic; and give self-devised quizzes and puzzles. Aside from the weekly printed column, "Ask Marilyn" is a daily online column that adds to the printed version by resolving controversial answers, correcting mistakes, expanding answers, reposting previous answers, and solving additional questions.

Three of her books (Ask Marilyn, More Marilyn, and Of Course, I'm for Monogamy) are compilations of questions and answers from "Ask Marilyn". The Power of Logical Thinking includes many questions and answers from the column.

Famous columns

The Monty Hall problem

Savant was asked the following question in her September 9, 1990, column:[19]

Suppose you're on a game show, and you're given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?

This question is called the Monty Hall problem due to its resembling scenarios on the game show Let's Make a Deal; its answer existed before it was used in "Ask Marilyn". She said the selection should be switched to door #2 because it has a 23 probability of success, while door #1 has just 13. To summarize, 23 of the time the opened door #3 will indicate the location of the door with the car (the door you had not picked and the one not opened by the host). Only 13 of the time will the opened door #3 mislead you into changing from the winning door to a losing door. These probabilities assume you change your choice each time door #3 is opened, and that the host always opens a door with a goat. This response provoked letters from thousands of readers, nearly all arguing doors #1 and #2 each have an equal chance of success. A follow-up column reaffirming her position served only to intensify the debate and soon became a feature article on the front page of The New York Times. Parade received around 10,000 letters from readers who thought that her workings were incorrect.[20]

Under the "standard" version of the problem, the host always opens a losing door and offers a switch. In the standard version, Savant's answer is correct. However, the statement of the problem as posed in her column is ambiguous.[21] The answer depends on what strategy the host is following. If the host operates under a strategy of offering a switch only if the initial guess is correct, it would clearly be disadvantageous to accept the offer. If the host merely selects a door at random, the question is likewise very different from the standard version. Savant addressed these issues by writing the following in Parade magazine, "the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. Anything else is a different question."[22]

She expounded on her reasoning in a second follow-up and called on school teachers to show the problem to classes. In her final column on the problem, she gave the results of more than 1,000 school experiments. Most respondents now agree with her original solution, with half of the published letters declaring their authors had changed their minds.[23]

"Two boys" problem

Like the Monty Hall problem, the "two boys" or "second-sibling" problem predates Ask Marilyn, but generated controversy in the column,[24] first appearing there in 1991–1992 in the context of baby beagles:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

When Savant replied "one out of three", readers[25] wrote the odds were 50–50. In a follow-up, she defended her answer, saying, "If we could shake a pair of puppies out of a cup the way we do dice, there are four ways they could land", in three of which at least one is male, but in only one of which none are male.

The confusion arises here because the bather is not asked if the puppy he is holding is a male, but rather if either is a male. If the puppies are labeled (A and B), each has a 50% chance of being male independently. This independence is restricted when at least A or B is male. Now, if A is not male, B must be male, and if B is not male, A must be male. This restriction is introduced by the way the question is structured and is easily overlooked – misleading people to the erroneous answer of 50%. See Boy or Girl paradox for solution details.

The problem re-emerged in 1996–97 with two cases juxtaposed:

Say that a woman and a man (who are unrelated) each have two children. We know that at least one of the woman's children is a boy and that the man's oldest child is a boy. Can you explain why the chances that the woman has two boys do not equal the chances that the man has two boys? My algebra teacher insists that the probability is greater that the man has two boys, but I think the chances may be the same. What do you think?

Savant agreed with the teacher, saying the chances were only 1 out of 3 that the woman had two boys, but 1 out of 2 the man had two boys. Readers argued for 1 out of 2 in both cases, prompting follow-ups. Finally she began a survey, asking female readers with exactly two children, at least one of them male, to give the sex of both children. Of the 17,946 women who responded, 35.9%, about 1 in 3, had two boys.[26]

Woman has young boy, older girl young girl, older boy 2 boys 1/3 1/3 1/3 0

Man has young boy, older girl young girl, older boy 2 boys 0 1/2 1/2 0

Errors in the column

On January 22, 2012, Savant admitted a mistake in her column. In the original column, published on December 25, 2011, a reader asked:

I manage a drug-testing program for an organization with 400 employees. Every three months, a random-number generator selects 100 names for testing. Afterward, these names go back into the selection pool. Obviously, the probability of an employee being chosen in one quarter is 25 percent. But what is the likelihood of being chosen over the course of a year?

Her response was:

The probability remains 25 percent, despite the repeated testing. One might think that as the number of tests grows, the likelihood of being chosen increases, but as long as the size of the pool remains the same, so does the probability. Goes against your intuition, doesn't it?

The correctness of the answer depends on how the question is asked. The probability of being chosen each time is 25% but probability of being chosen at least once across the 4 events is higher. In this case, the correct answer is around 68%, calculated as the complement of the probability of not being chosen in any of the four quarters: 1 – (0.754).[27]

On June 22, 2014, Savant made an error in a word problem. The question was: "If two people could complete a project in six hours, how long would it take each of them to do identical projects on their own, given that one took four hours longer than the other?" Her answer was 10 hours and 14 hours, reasoning that if together it took them 6 hours to complete a project, then the total effort was 12 "man hours". If they then each do a separate full project, the total effort needed would be 24 hours, so the answer (10+14) needed to add up to 24 with a difference of 4.[28] Savant later issued a correction, as the answer ignored the fact that the two people get different amounts of work done per hour: if they are working jointly on a project, they can maximize their combined productivity, but if they split the work in half, one person will finish sooner and cannot fully contribute. This subtlety causes the problem to require solving a quadratic equation and does not have a rational solution. Instead, the answer is ${\displaystyle 4+{\sqrt {40}}}$  (approximately 10.32) and ${\displaystyle 8+{\sqrt {40}}}$  (approximately 14.32) hours.[29]

Fermat's Last Theorem

A few months after Andrew Wiles said he had proved Fermat's Last Theorem, Savant published the book The World's Most Famous Math Problem (October 1993),[30] which surveys the history of Fermat's Last Theorem as well as other mathematical problems. Reviewers questioned her criticism of Wiles' proof; asking whether it was based on a correct understanding of mathematical induction, proof by contradiction, and imaginary numbers.[31]

Especially contested was Savant's statement that Wiles' proof should be rejected for its use of non-Euclidean geometry. Savant stated that because "the chain of proof is based in hyperbolic (Lobachevskian) geometry", and because squaring the circle is seen as a "famous impossibility" despite being possible in hyperbolic geometry, then "if we reject a hyperbolic method of squaring the circle, we should also reject a hyperbolic proof of Fermat's last theorem."

Specialists[who?] flagged discrepancies between the two cases, distinguishing the use of hyperbolic geometry as a tool for proving Fermat's Last Theorem from its use as a setting for squaring the circle: squaring the circle in hyperbolic geometry is a different problem from that of squaring it in Euclidean geometry, whereas Fermat's Last Theorem is not inherently geometry-specific. Savant was criticized for rejecting hyperbolic geometry as a satisfactory basis for Wiles' proof, with critics pointing out that axiomatic set theory (rather than Euclidean geometry) is now the accepted foundation of mathematical proofs and that set theory is sufficiently robust to encompass both Euclidean and non-Euclidean geometry as well as geometry and adding numbers.[citation needed]

Savant retracted the argument in a July 1995 addendum, saying she saw the theorem as "an intellectual challenge – 'to find another proof using only tools available to Fermat in the 17th century.'"

The book came with a glowing introduction by Martin Gardner.[citation needed]

Publications

• 1985 – Omni I.Q. Quiz Contest
• 1990 – Brain Building: Exercising Yourself Smarter (co-written with Leonore Fleischer)
• 1993 – The World's Most Famous Math Problem: The Proof of Fermat's Last Theorem and Other Mathematical Mysteries
• 1994 – More Marilyn: Some Like It Bright!
• 1994 – "I've Forgotten Everything I Learned in School!": A Refresher Course to Help You Reclaim Your Education
• 1996 – Of Course I'm for Monogamy: I'm Also for Everlasting Peace and an End to Taxes
• 1996 – The Power of Logical Thinking: Easy Lessons in the Art of Reasoning...and Hard Facts about Its Absence in Our Lives
• 2000 – The Art of Spelling: The Madness and the Method
• 2002 – Growing Up: A Classic American Childhood

References

1. ^ a b "MILESTONES: August 11 birthdays for Viola Davis, Tomi Lahren, Joe Rogan". Brooklyn Eagle. August 11, 2020. Retrieved October 3, 2020.
2. ^ LLC, New York Media (February 6, 1989). "New York Magazine". New York Media, LLC. Retrieved July 24, 2022.
3. ^ The Time Everyone "Corrected" the World's Smartest Woman". Priceonomics, February 19, 2015
4. ^ a b c d Knight, Sam (April 10, 2009). "Is a high IQ a burden as much as a blessing?". Financial Times. Financial Times Ltd. Retrieved October 7, 2013.
5. ^ vos Savant, Marilyn (November 25, 2007). "Ask Marilyn". Parade. Archived from the original on April 23, 2008.{{cite news}}: CS1 maint: bot: original URL status unknown (link)
6. ^ vos Savant, Marilyn (January 23, 2008). "Keeping It in the Family". Parade.
7. ^ vos Savant, Marilyn (May 4, 2013). "Ask Marilyn: The 'First Sandwich Generation': True Trend or Marketing Invention?". Parade. Retrieved August 15, 2013.
8. ^ a b c Baumgold, Julie (February 6, 1989). "In the Kingdom of the Brain". New York Magazine. New York Media, LLC.
9. ^ Vitez, Michael (October 12, 1988). "Two of a Kind". The Chicago Tribune.
10. ^ "Love Stories We Love". Parade. February 6, 2015. Retrieved February 6, 2022. Vos Savant and Jarvik were married Aug. 23, 1987, a year to the day after they first met, at New York's Plaza Hotel. They wore rings the groom made of gold and pyrolitic carbon, a substance used in artificial heart valves. Isaac Asimov walked the bride down the aisle; the best man was Tom Gaidosh, the seventh recipient of the Jarvik 7 artificial heart.
11. ^ McCleary, Kathleen (September 11, 2018). "Marilyn's Rules for a Happy Marriage". Parade. Retrieved February 6, 2022.
12. ^ "About – National Women's History Museum – NWHM". Retrieved February 19, 2016.
13. ^ "CSI Fellows and Staff". Center for Inquiry. Retrieved June 20, 2012.
15. ^ Knight, Sam (April 10, 2009). "Is a high IQ a burden as much as a blessing?". Financial Times. Financial Times Ltd. Castles, Elaine E. (June 6, 2012). Inventing Intelligence. ABC-CLIO. p. 3. ISBN 978-1-4408-0338-3. Retrieved August 31, 2013. And what is that makes Marilyn vos Savant so uniquely qualified to answer such questions? There is only one reason: she is listed in the Guinness Book of World Records as having the highest IQ ever recorded. Never mind that this record is based on a non-standardized test put out by an obscure group known as Mega, supposedly the world's most selective organization of geniuses. Ignore the fact that test scores at the extreme ends of any distribution are notoriously unreliable. . . . None of this is meant to downplay her very real accomplishments; by all accounts, vos Savant is a sensible and grounded woman, and she has won several awards for her work in the fields of education and communications. But her fame came, in the words of journalist Julie Baumgold, 'only because of the glory of that number.' (citing New York magazine 22 (1989):36–42)
16. ^ Carlson, Roger D. (1991). Keyser, Daniel J.; Sweetland, Richard C. (eds.). Test Critiques. Test Critique: The Mega Test (Volume VIII ed.). PRO-ED. pp. 431–435. ISBN 0-89079-254-2. Although the approach that Hoeflin takes is interesting, it violates good psychometric principles by overinterpreting the weak data of a self-selected sample.
17. ^ vos Savant, Marilyn (July 17, 2005). "Ask Marilyn: Are Men Smarter Than Women?". Parade. Archived from the original on October 11, 2007. Retrieved February 25, 2008.
18. ^ Thompson, D. (July 5, 1986). "Marilyn's Most Vital Statistic". The Courier-Mail.
19. ^ vos Savant, Marilyn. "Game Show Problem". marilynvossavant.com. Archived from the original on March 10, 2010. Retrieved August 7, 2010.
20. ^ Tierney, John (July 21, 1991). "Behind Monty Hall's Doors: Puzzle, Debate and Answer?". The New York Times. Retrieved August 7, 2008.
21. ^ Krauss, Stefan and Wang, X. T. (2003). "The Psychology of the Monty Hall Problem: Discovering Psychological Mechanisms for Solving a Tenacious Brain Teaser", Journal of Experimental Psychology: General 132(1). Retrieved from "Archived copy" (PDF). Archived from the original (PDF) on May 30, 2009. Retrieved May 30, 2009.{{cite web}}: CS1 maint: archived copy as title (link)
22. ^ "Game Show Problem". marilynvossavant.com. Archived from the original on March 10, 2010. Retrieved June 2, 2008.


text:
## Tag Archives: absolute continuous

### The volume of uniform hyperbolic sets

This is a note of some well known results. The argument here may be new, and may be complete.

Proposition 1. Let $f\in\mathrm{Diff}^2_m(M)$. Then $m(\Lambda)=0$ for every closed, invariant hyperbolic set $\Lambda\neq M$.

See Theorem 15 of Bochi–Viana’s paper. Note that Proposition 1 also applies to Anosov case, in the sense that $m(\Lambda)>0$ implies that $\Lambda=M$ and $f$ is Anosov.

Proof. Suppose $m(\Lambda)>0$ for some hyperbolic set. Then the stable and unstable foliations/laminations are absolutely continuous. Hopf argument shows that $\Lambda$ is (essentially) saturated by stable and unstable manifolds. Being a closed subset, $\Lambda$ is in fact saturated by stable and unstable manifolds, and hence open. So $\Lambda=M$.

Proposition 2. There exists a residual subset $\mathcal{R}\subset \mathrm{Diff}_m^1(M)$, such that for every $f\in\mathcal{R}$, $m(\Lambda)=0$ for every closed, invariant hyperbolic set $\Lambda\neq M$.

Proof. Let $U\subset M$ be an open subset such that $\overline{U}\neq M$, $\Lambda_U(f)=\bigcap_{\mathbb{Z}}f^n\overline{U}$, which is always a closed invariant set (maybe empty). Given $\epsilon>0$, let $\mathcal{D}(U,\epsilon)$ be the set of maps $f\in\mathrm{Diff}_m^1(M)$ that either $\Lambda_U(f)$ is not a uniformly hyperbolic set, or it’s hyperbolic but  $m(\Lambda_U(f))<\epsilon$. It follows from Proposition 1 that $\mathcal{D}(U,\epsilon)$ is dense. We only need to show the openness. Pick an $f\in \mathcal{D}(U,\epsilon)$. Since $m(\Lambda_U(f))<\epsilon$, there exists $N\ge 1$ such that $m(\bigcap_{-N}^N f^n\overline{U})<\epsilon$. So there exists $\mathcal{U}\ni f$ such that $m(\bigcap_{-N}^N g^n\overline{U})<\epsilon$. In particular, $m(\Lambda_U(g))<\epsilon$ for every $g\in \mathcal{U}$. The genericity follows by the countable intersection of the open dense subsets $\mathcal{D}(U_n,1/k)$.

The dissipative version has been obtained in Alves–Pinheiro’s paper

Proposition 3. Let $f\in\mathrm{Diff}^2(M)$. Then $m(\Lambda)=0$ for every closed, transitive hyperbolic set $\Lambda\neq M$. In particular, $m(\Lambda)>0$ implies that $\Lambda=M$ and $f$ is Anosov.

See Theorem 4.11 in R. Bowen’s book when $\Lambda$ is a basic set.

4. Ledrappier-Young’s criterion for measures with Absolutely Continuous Conditional measures on Unstable manifolds. Let $f:M\to M$ be a $C^2$ diffeomorphism and $\mu$ be an invariant probability measure. For $\mu$-a.e. $x$, if there exist negative Lyapunov exponent(s) at $x$, then the set of points with exponentially approximating future of $x$ is a $C^2$-submanifold, $W^u(x)$. Then $\mu$ is said to have ACCU if for each measurable partition $\xi$ with $\xi(x)\subset W^u(x)$ and contains an unstable plaque for $\mu$-a.e. $x$, the conditional measure $\mu_{\xi(x)}\ll m_{W^u(x)}$ for $\mu$-a.e. $x$.
Theorem. $\mu$ has ACCU if and only if $h_\mu(f)=\Lambda^+(\mu)$, where $\Lambda^+(\mu)=\int\sum\lambda^+_i(x)d\mu(x)$.
Moreover, the density $\frac{d\mu_{\xi(x)}}{dm_{W^u(x)}}$ is strictly positive and $C^1$ on $\xi(x)$ for $\mu$-a.e. $x$.


text:
Volume 14, issue 1 (2014)

 Recent Issues
 The Journal About the Journal Editorial Board Subscriptions Editorial Interests Editorial Procedure Submission Guidelines Submission Page Ethics Statement ISSN (electronic): 1472-2739 ISSN (print): 1472-2747 Author Index To Appear Other MSP Journals
The bumping set and the characteristic submanifold

Genevieve S Walsh

Algebraic & Geometric Topology 14 (2014) 283–297
Abstract

We show here that the Nielsen core of the bumping set of the domain of discontinuity of a Kleinian group $\Gamma$ is the boundary of the characteristic submanifold of the associated $3$–manifold with boundary. Some examples of interesting characteristic submanifolds are given. We also give a construction of the characteristic submanifold directly from the Nielsen core of the bumping set. The proofs are from “first principles”, using properties of uniform domains and the fact that quasi-conformal discs are uniform domains.

Keywords
Kleinian group, characteristic submanifold
Mathematical Subject Classification 2010
Primary: 30F40, 57M60


text:
I understand that BatchNorm (Batch Normalization) centers to (mean, std) = (0, 1) and potentially scales (with $\gamma$) and offsets (with $\beta$) the data which is input to the layer. BatchNorm follows this formula:

However, when it comes to 'adaptive BatchNorm', I don't understand what the difference is. What is adaptive BatchNorm doing differently? It is described as follows:

I think the original batch normalization paper proposes to use mean and standard deviation estimated on the train set. The adaptive batch normalization simply re-estimates them on the target domain (could be the test set, or some unlabeled data from the target domain).

Please correct me if I am wrong.

• After a long time having this question, I also think this is the only difference. Thanks a lot for the answer! :) Jun 13 '17 at 9:53


text:
# two pages on one and exact crop marks

the page size of my document is A5 and has about 100 pages.

For printing I need to put two pages on one in A4 and rearrange them in a specific way due to my thumb index at eachside of the page.

That's the code I got so far:

\documentclass[a4paper]{article}
\usepackage{pdfpages}
\usepackage{forloop}

\begin{document}

\pagestyle{plain}
\newcounter{i}
\newcounter{p}
\newcounter{pp}
\newcounter{ppp}
\newcounter{pppp}

\forloop{i}{1}{\value{i} < 34}{%
\setcounter{p}{\thei*4+2}
\setcounter{pp}{\thei*4-1}
\setcounter{ppp}{\thei*4}
\setcounter{pppp}{\thei*4+1}

\includepdfmerge[nup=1x2, landscape]{file.pdf, \thep,\thepp,\theppp,\thepppp}%
}

\end{document}

I'd like to add crop marks like this:

3mm margin for printing area and 3mm crop marks.

Instead of the option a5paper you can use any setting for your paper dimension. The package crop allows several crop marks, see documentation (run texdoc crop),. You can adopt this solution for using pdfpages and two on one.

\documentclass[twoside]{article}
\usepackage{geometry}
\geometry{a5paper}
\usepackage[center,a4,cross]{crop}
\usepackage{blindtext}
\begin{document}
\Blinddocument

\end{document}

• that didn't answer my question. I didn't want to have such big crop marks. Just 3mm like in my solution; besides it doesn't have x-shaped marks – Moldevort May 5 '14 at 7:57

Had finally an idea over night and solved it:

\documentclass[twoside,landscape]{scrreprt}
\usepackage{forloop}
\usepackage{pdfpages}
\usepackage{background}
\usetikzlibrary{calc}
\usepackage[a4paper]{geometry}
\geometry{top=6mm, left=6mm, right=0mm, bottom=0mm}

\backgroundsetup{
scale=1,
opacity=1,
angle=0,
color=black,
contents={%
\begin{tikzpicture}[remember picture,overlay]
\draw[thick] ($(current page.north west)+(0mm,-9mm)$) -- ++(\paperwidth,0mm); %o
\draw[thick] ($(current page.south west)+(0mm,9mm)$) -- ++(\paperwidth,0mm); %u
\draw[thick] ($(current page.north east)+(-9mm,0mm)$) -- ++(0,-\paperheight); %r
\draw[thick] ($(current page.north west)+(9mm,0mm)$) -- ++(0,-\paperheight); %l
\draw[thick] ($(current page.north)+(0mm,0mm)$) -- ++(0,-\paperheight); %m
\end{tikzpicture}%
}}

\newcommand{\ing}[1]{\includegraphics[width=(\textwidth-6mm)/2,height=\textheight-6mm,page=#1]{empty}}

\newcommand{\pdf}[4]{
\hspace*{-11pt}%
\ing{\numexpr#1\relax}%
\ing{\numexpr#2\relax}%
\clearpage%
\hspace*{6pt}%
\ing{\numexpr#3\relax}%
\ing{\numexpr#4\relax}%
\clearpage%
}

\pagestyle{plain}
\newcounter{i}
\begin{document}

%%% Cover
\pdf{1}{3}{3}{3}

%%% Inhalt
\forloop{i}{1}{\value{i} < 34}{%
\pdf{\thei*4+2}{\thei*4-1}{\thei*4}{\thei*4+1}
}

%%% letzte 3 Seiten
\pdf{3}{137}{138}{139}

\null\clearpage

\end{document}


text:
# If f(x) varies directly with x and f(x) = 24 when x = –4, then what is f(x) when x = 12?

$f \left(12\right) = - 72$

#### Explanation:

Given that $f \left(x\right)$ varies directly with $x$ hence let

$f \left(x\right) = k x$

Where $k$ is some arbitrary constant

Now, given that $f \left(- 4\right) = 24$ hence

$k \left(- 4\right) = 24$

$k = - 6$

$\setminus \implies f \left(x\right) = - 6 x$

$\setminus \therefore f \left(12\right) = - 6 \left(12\right)$

$= - 72$