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# 5. The center cable of suspension bridge forms parabolic arc The cable is suspended from the tops oftwo support towers,

###### Question:

5. The center cable of suspension bridge forms parabolic arc The cable is suspended from the tops oftwo support towers, 800 feet apart: The tops of the towers are 170 feet above the road and the lowest point ofthe cable is midway between the towers and ten feet above the road. Find theheight ofthe cable above the road at a dis- tance of 100 feet from tower: 6.A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two less bricks than the prior step: How many bricks are required to build the staircase?

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location:  Publications → journals → CJM
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# Structure of modules induced from simple modules with minimal annihilator

Published:2004-04-01
Printed: Apr 2004
• Oleksandr Khomenko
• Volodymyr Mazorchuk
 Format: HTML LaTeX MathJax PDF PostScript

## Abstract

We study the structure of generalized Verma modules over a semi-simple complex finite-dimensional Lie algebra, which are induced from simple modules over a parabolic subalgebra. We consider the case when the annihilator of the starting simple module is a minimal primitive ideal if we restrict this module to the Levi factor of the parabolic subalgebra. We show that these modules correspond to proper standard modules in some parabolic generalization of the Bernstein-Gelfand-Gelfand category $\Oo$ and prove that the blocks of this parabolic category are equivalent to certain blocks of the category of Harish-Chandra bimodules. From this we derive, in particular, an irreducibility criterion for generalized Verma modules. We also compute the composition multiplicities of those simple subquotients, which correspond to the induction from simple modules whose annihilators are minimal primitive ideals.
 Keywords: parabolic induction, generalized Verma module, simple module, Ha\-rish-\-Chand\-ra bimodule, equivalent categories
 MSC Classifications: 17B10 - Representations, algebraic theory (weights) 22E47 - Representations of Lie and real algebraic groups: algebraic methods (Verma modules, etc.) [See also 17B10]

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# Prove that: $\operatorname{cosec}(2A) + \operatorname{cosec}(4A) + \operatorname{cosec}(8A) = \cot(A) - \cot(8A)$ [duplicate]

How can we prove $$\operatorname{cosec}(2A) + \operatorname{cosec}(4A) + \operatorname{cosec}(8A) = \cot(A) - \cot(8A)$$?

• Hi, welcome to MathSE! What have you tried so far? – James O'dare May 19 at 12:37
• Can you prove $\csc 2A=\cot A-\cot2A$? – Lord Shark the Unknown May 19 at 12:44
• – lab bhattacharjee May 19 at 13:20

Put the right side to the left. $$cosec(2A)+cosec(4A)+cosec(8A)-cot(A)+cot(8A)=0$$ Then express the left side by sines and cosines. You get: $$\frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1}{sin(8A)}-\frac{cos(A)}{sin(A)}+\frac{cos(8A)}{sin(8A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{1+cos(8A)}{sin(8A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{2cos^2(4A)}{2sin(4A)cos(4A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1}{sin(4A)}+\frac{cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{1+cos(4A)}{sin(4A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{2cos^2(2A)}{2sin(2A)cos(2A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1}{sin(2A)}+\frac{cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{1+cos(2A)}{sin(2A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{2cos^2(A)}{2sin(A)cos(A)}-\frac{cos(A)}{sin(A)} =$$ $$= \frac{cos(A)}{sin(A)}-\frac{cos(A)}{sin(A)} = 0$$


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# Time Period of a periodic function

Consider the following periodic function: $$f(t) = \sin(ωt) + \cos(2ωt) + \sin(4ωt)$$ What is the time period of the above periodic function?

The following is given in my book:

Period is the least interval of time after which the function repeats. Here, $$\sin(ωt)$$ has a period $$T_o = \frac{2π}{ω}$$, $$\cos(2ωt)$$ has a period $$\frac{T_o}{2}$$ and $$\sin(4ωt)$$ has a period of $$\frac{T_o}{4}$$. The period of the first term is a multiple of the periods of the last two terms. Therefore, the smallest interval of time after which the sum of the three terms repeats is $$T_o$$, and thus, Time period is $$T_o$$.

I don't understand what the above lines mean.

• Have you tried plotting these three functions, each on it's own? This would solve the problem. – Semoi Jan 21 at 18:23

Let's first fix to some number. I'll choose $$\omega=2\pi f$$ with $$f=1Hz$$, but you can choose any other number".
• After the time $$t = T_0 = \frac{2\pi}{\omega} = \frac{1}{f} = 1s$$ the blue function $$\sin{(\omega t)}$$ has done a full oscillation. Thus, it has also done full oscillations after $$t=\{2, 3, 4, \ldots\}$$.
• After the time $$t = T_0/2 = \frac{1}{2}s$$ the red function $$\cos{(2\omega t)}$$ has done a full oscillation. Thus, it has also done full oscillations after $$t=\{1, 1.5, 2, 2.5, \ldots\}$$.
• After the time $$t = T_0/4 = \frac{1}{4}s$$ the green function $$\cos{(2\omega t)}$$ has done a full oscillation. Thus, it has also done full oscillations after $$t=\{0.5, 0.75, 1, 1.25, \ldots\}$$.
Thus, after $$1s$$ the blue curve has done one full oscillation, the red wave has done two and the green curve has done four full oscillations.


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Question

# Write the negation of the following statements: (i) p : For every positive real number x, the number (x − 1) is also positive. (ii) q : For every real number x, either x > 1 or x < 1. (iii) r : There exists a number x such that 0 < x < 1.

Solution

## (i) p: For every positive real number x, the number (x − 1) is also positive.  ~p: At least for one positive real number x, the number $\left(x-1\right)$ is not positive. (ii) q: For every real number x, either x > 1 or x < 1.  ~q: At least for one real number x, neither nor . (iii) r: There exists a number x such that 0 < x < 1.    ~r: For every real number x, either or .MathematicsRD Sharma XI (2015)Standard XI

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Perceptron Training Rule is an algorithm for finding the weights of a single unit (“neuron”).

# What is a Unit (Neuron)?

In machine learning, a “unit” (“single neuron”) is a thing that takes some inputs, applies a weight to each input, and then based on the linear combination of the inputs and an activation function, spits out either on or off. To make our math simple, we say that if it outputs +1, it is on, if it outputs -1, it is off.

Inputs  Weights  Activation Fcn           Output
======  =======  ==============           ======

w1
x1 +------->  +--------------+
|              |
w2    |  activation  |          +1 ("on")
x2 +------->  |   function   |  +---->
|              |          -1 ("off")
w3    |              |
x3 +------->  +--------------+

The inputs are often called features. Vector form of the features is $\vec{x}$. Vector form of the weights is $\vec{w}$.

The activation function is simply a threshold function. A threshold function outputs +1 if the input is greater than some threshold (often 0 is chosen), else it outputs -1.

# Perceptron Training Rule Algorithm

Look at that beauuuutiful, marvelous, ASCII diagram above. We just need to find the weights (i.e. $\vec{w}$) that linearly separates our training data (i.e. the weights that cause our unit to output +1 for our positive training examples, and -1 for our negative training examples).

Here is ze algooooreeeiithm:

• arbitrarily initialize $\vec{w}$
• keep iterating through our training examples
• if with our current weights, we classified this training example correctly, don’t change the weights in response to this example
• if we classified the training example as negative when it should have been positive, 1) increase the weights associated with the positive features and 2) decrease the weights associated with the negative features. In other words, make $\vec{w} \cdot \vec{x}$ produce a greater value, so that when it goes in our activation function, it is more likely to produce +1.
• if we classified the training example as positive when it should have been negative, then we need to make $\vec{w} \cdot \vec{x}$ produce a smaller value, thus we need to 1) decrease the weights associated with positive features and 2) increase the weights associated with negative features

Those last 3 bullets, all those words, can be represented by the following mathematical expression:

$n$ is known as the “learning rate”, and you can just make this a small positive value (like 0.1). This will determine the size of the chunks by which you update $\vec{w}$ after considering each training example.

The $(t-o)\vec{x}$ portion of the expression is really where all of the words from the last 3 bullets went into. $t$ is the actual output for the training example, $o$ is what our unit outputs for the training example using our current weights, and $\vec{x}$ is the features vector.

Let’s consider how our first wordy bullet matches with the expression. If we classified the training example correctly, then $t = o$, thus $t - o = 0$, thus we don’t update $\vec{w}$ at all.

Let’s consider how our second wordy bullet matches with the expression. If we classified the training example as negative, but it should be positive, then $t - o$ is $+1 - -1 = 2$. Thus positive 2 multiplied by $\vec{x}$ will increase the weights for all positive features, and decrease the weights for all negative features. This has the overall effect of increasing $\vec{w} \cdot \vec{x}$, thus increasing the chance that the activation function will subsequently output +1 for this example (which is what it should output for this example).

I will leave the thinking for how the third wordy bullet matches with the expression for you.

# Summary

Perceptron Training Rule is a machine learning algorithm for learning the weights of a single neuron (a single unit). It does so by constantly iterating through your training examples and updating the weights in response to each training example.

The update expression it uses is $\vec{w} = \vec{w} + n(t-o)\vec{x}$.


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# number of connected sets [closed]

How many connected sets are there in the following array? $$\begin{array}{cccc} 1& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 1& 1& 0\\ 1& 0& 0& 1 \end{array}$$

"Connected set" can be defined as group of cell(s) which has 1 mentioned on it and has at least one other cell in that set with which they share the neighbor relationship. A cell with 1 in it and no surrounding neighbor having 1 in it can be considered as a set with one cell in it. Neighbors can be defined as all the cells adjacent to the given cell in 8 possible directions ( i.e N , W , E , S , NE , NW , SE , SW direction ). A cell is not a neighbor of itself.

Actually i am stuck on this and not able to understand what the definition of connected set in this question means

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# Synchronization of chaos

Synchronization of chaos is a phenomenon that may occur when two, or more, dissipative chaotic systems are coupled.

Because of the exponential divergence of the nearby trajectories of chaotic systems, having two chaotic systems evolving in synchrony might appear surprising. However, synchronization of coupled or driven chaotic oscillators is a phenomenon well established experimentally and reasonably well-understood theoretically.

The stability of synchronization for coupled systems can be analyzed using master stability. Synchronization of chaos is a rich phenomenon and a multi-disciplinary subject with a broad range of applications.[1][2][3]

Synchronization may present a variety of forms depending on the nature of the interacting systems and the type of coupling, and the proximity between the systems.

## Identical synchronization

This type of synchronization is also known as complete synchronization. It can be observed for identical chaotic systems. The systems are said to be completely synchronized when there is a set of initial conditions so that the systems eventually evolve identically in time. In the simplest case of two diffusively coupled dynamics is described by

${\displaystyle x^{\prime }=F(x)+\alpha (y-x)}$
${\displaystyle y^{\prime }=F(y)+\alpha (x-y)}$

where ${\displaystyle F}$  is the vector field modeling the isolated chaotic dynamics and ${\displaystyle \alpha }$  is the coupling parameter. The regime ${\displaystyle x(t)=y(t)}$  defines an invariant subspace of the coupled system, if this subspace ${\displaystyle x(t)=y(t)}$  is locally attractive then the coupled system exhibit identical synchronization.

If the coupling vanishes the oscillators are decoupled, and the chaotic behavior leads to a divergence of nearby trajectories. Complete synchronization occurs due to the interaction, if the coupling parameter is large enough so that the divergence of trajectories of interacting systems due to chaos is suppressed by the diffusive coupling. To find the critical coupling strength we study the behavior of the difference ${\displaystyle v=x-y}$ . Assuming that ${\displaystyle v}$  is small we can expand the vector field in series and obtain a linear differential equation - by neglecting the Taylor remainder - governing the behavior of the difference

${\displaystyle v^{\prime }=DF(x(t))v-2\alpha v}$

where ${\displaystyle DF(x(t))}$  denotes the Jacobian of the vector field along the solution. If ${\displaystyle \alpha =0}$  then we obtain

${\displaystyle u^{\prime }=DF(x(t))u,}$

and since the dynamics of chaotic we have ${\displaystyle \|u(t)\|\leq \|u(0)\|e^{\lambda t}}$ , where ${\displaystyle \lambda }$  denotes the maximum Lyapunov exponent of the isolated system. Now using the ansatz ${\displaystyle v=ue^{-2\alpha t}}$  we pass from the equation for ${\displaystyle v}$  to the equation for ${\displaystyle u}$ . Therefore, we obtain

${\displaystyle \|v(t)\|\leq \|u(0)\|e^{(-2\alpha +\lambda )t}}$

yield a critical coupling strength ${\displaystyle \alpha _{c}=\lambda /2}$ , for all ${\displaystyle \alpha >\alpha _{c}}$  the system exhibit complete synchronization. The existence of a critical coupling strength is related to the chaotic nature of the isolated dynamics.

In general, this reasoning leads to the correct critical coupling value for synchronization. However, in some cases one might observe loss of synchronization for coupling strengths larger than the critical value. This occurs because the nonlinear terms neglected in the derivation of the critical coupling value can play an important role and destroy the exponential bound for the behavior of the difference.[4] It is however, possible to give a rigorous treatment to this problem and obtain a critical value so that the nonlinearities will not affect the stability.[5]

## Generalized synchronization

This type of synchronization occurs mainly when the coupled chaotic oscillators are different, although it has also been reported between identical oscillators. Given the dynamical variables ${\displaystyle (x_{1},x_{2},...x_{n})}$  and ${\displaystyle (y_{1},y_{2},...y_{m})}$  that determine the state of the oscillators, generalized synchronization occurs when there is a functional, ${\displaystyle \phi }$ , such that, after a transitory evolution from appropriate initial conditions, it is ${\displaystyle [y_{1}(t),y_{2}(t),...,y_{m}(t)=\phi [x_{1}(t),x_{2}(t),...,x_{n}(t)]}$ . This means that the dynamical state of one of the oscillators is completely determined by the state of the other. When the oscillators are mutually coupled this functional has to be invertible, if there is a drive-response configuration the drive determines the evolution of the response, and Φ does not need to be invertible. Identical synchronization is the particular case of generalized synchronization when ${\displaystyle \phi }$  is the identity.

## Phase synchronization

Phase synchronization occurs when the coupled chaotic oscillators keep their phase difference bounded while their amplitudes remain uncorrelated. This phenomenon occurs even if the oscillators are not identical. Observation of phase synchronization requires a previous definition of the phase of a chaotic oscillator. In many practical cases, it is possible to find a plane in phase space in which the projection of the trajectories of the oscillator follows a rotation around a well-defined center. If this is the case, the phase is defined by the angle, φ(t), described by the segment joining the center of rotation and the projection of the trajectory point onto the plane. In other cases it is still possible to define a phase by means of techniques provided by the theory of signal processing, such as the Hilbert transform. In any case, if φ1(t) and φ2(t) denote the phases of the two coupled oscillators, synchronization of the phase is given by the relation nφ1(t)=mφ2(t) with m and n whole numbers.

## Anticipated and lag synchronization

In these cases, the synchronized state is characterized by a time interval τ such that the dynamical variables of the oscillators, ${\displaystyle (x_{1},x_{2},...x_{n})}$  and ${\displaystyle (x'_{1},x'_{2},...x'_{n})}$ , are related by ${\displaystyle x'_{i}(t)=x_{i}(t+\tau )}$ ; this means that the dynamics of one of the oscillators follows, or anticipates, the dynamics of the other. Anticipated synchronization may occur between chaotic oscillators whose dynamics is described by delay differential equations, coupled in a drive-response configuration. In this case, the response anticipates the dynamics of the drive. Lag synchronization may occur when the strength of the coupling between phase-synchronized oscillators is increased.

## Amplitude envelope synchronization

This is a mild form of synchronization that may appear between two weakly coupled chaotic oscillators. In this case, there is no correlation between phases nor amplitudes; instead, the oscillations of the two systems develop a periodic envelope that has the same frequency in the two systems.

This has the same order of magnitude than the difference between the average frequencies of oscillation of the two chaotic oscillator. Often, amplitude envelope synchronization precedes phase synchronization in the sense that when the strength of the coupling between two amplitude envelope synchronized oscillators is increased, phase synchronization develops.

All these forms of synchronization share the property of asymptotic stability. This means that once the synchronized state has been reached, the effect of a small perturbation that destroys synchronization is rapidly damped, and synchronization is recovered again. Mathematically, asymptotic stability is characterized by a positive Lyapunov exponent of the system composed of the two oscillators, which becomes negative when chaotic synchronization is achieved.

Some chaotic systems allow even stronger control of chaos, and both synchronization of chaos and control of chaos constitute parts of what's known as "cybernetical physics".

## Notes

1. ^ Arenas, Alex; Díaz-Guilera, Albert; Kurths, Jurgen; Moreno, Yamir; Zhou, Changsong (2008-12-01). "Synchronization in complex networks". Physics Reports. 469 (3): 93–153. arXiv:0805.2976. Bibcode:2008PhR...469...93A. doi:10.1016/j.physrep.2008.09.002.
2. ^ Wu, Chai Wah (2007). Synchronization in Complex Networks of Nonlinear Dynamical Systems. doi:10.1142/6570. ISBN 978-981-270-973-8.
3. ^ Eroglu, Deniz; Lamb, Jeroen S. W.; Pereira, Tiago (2017). "Synchronisation of chaos and its applications". Contemporary Physics. 58 (3): 207–243. doi:10.1080/00107514.2017.1345844. hdl:10044/1/53479. ISSN 0010-7514.
4. ^ Ashwin, Peter (2006-08-09). "Bubbling transition". Scholarpedia. 1 (8): 1725. Bibcode:2006SchpJ...1.1725A. doi:10.4249/scholarpedia.1725. ISSN 1941-6016.
5. ^ Tiago Pereira, Stability of Synchronized Motion in Complex Networks, arXiv:1112.2297v1, 2011.

## References

• Pikovsky, A.; Rosemblum, M.; Kurths, J. (2001). Synchronization: A Universal Concept in Nonlinear Sciences. Cambridge University Press. ISBN 978-0-521-53352-2.
• González-Miranda, J. M. (2004). Synchronization and Control of Chaos. An introduction for scientists and engineers. Imperial College Press. ISBN 978-1-86094-488-8.
• Fradkov A.L. Cybernetical physics: from control of chaos to quantum control. Springer-Verlag, 2007, (Preliminary Russian version: St.Petersburg, Nauka, 2003).


text:
# In neural networks, what does the term depth generally mean?

Is it

1. number of units in a layer
2. number of layers
3. overall complexity of the network (both 1 and 2)
• Usually it refers to the number of layers that are sequential (i.e. one after the other). However, it isn't used much like a quantity (e.g. this network has a depth of 8), rather a qualitative characteristic of the network (e.g. this network is deeper than the other). Feb 18 '20 at 9:33


text:
# We can prepare for earthquakes, but we can't predict them

Release Date:

In the last six weeks, Hawaii residents have been shaken by two major earthquakes and numerous aftershocks.

Note: HVO normally emails our weekly column, called Volcano Watch, on Thursdays and the Hawaii Tribune-Herald publishes them in their Sunday edition. Because of the continuing aftershocks from the October 15 quakes and the false rumor being passed around regarding earthquake predictions, HVO scientists feel the need to share more information than usual. Last week's Volcano Watch dealt with aftershocks and what to expect and this one discusses earthquake prediction. We will resume our normal weekly schedule this Thursday.

(Public domain.)

Then, on the last Sunday in November, they were subjected to a new danger: Rumors predicting that a magnitude-9 earthquake and tsunami would strike sometime that night rocked the state of Hawaii. With their memories of the two damaging earthquakes on October 15 still fresh, residents were understandably concerned and even fearful.

People stocked up on gasoline, provisions, and emergency supplies, and gathered to compare notes about what they had heard. Some evacuated from areas they perceived as dangerous. Police, television, and radio stations were flooded with telephone calls for confirmation of the rumors. NOAA's Pacific Tsunami Warning Center was also hit hard with incoming calls requesting information.

Accurate and reliable information was absolutely critical. The fast-spreading rumors were potentially harmful, and public officials and broadcast media quickly responded to dispel them. The official messages were clear and concise: It is not scientifically possible, at present, to predict earthquakes. There was no official earthquake or tsunami prediction. And, certainly, no government entities were withholding information about a prediction that didn't exist.

The United States Geological Survey (USGS) is the federal agency responsible by law for routine monitoring of, and providing notifications relating to, earthquakes, volcanoes, and landslides. Where appropriate, the notifications include warnings and forecasts. The Hawaiian Volcano Observatory is the office of the USGS whose duties include earthquake and volcano monitoring in Hawaii.

Much research has been done by the USGS and university scientists to understand earthquakes. Besides improving our understanding of earthquakes, this research might eventually result in the ability to predict them, but so far that goal has not been reached. So-called alternate (non-scientific) methods of earthquake prediction have also been tested and found to be unreliable, despite proponent's claims.

To be testable, a prediction should include a statement of the future earthquake's specific location, magnitude, and time of occurrence. A prediction should also include the methodology or basis for prediction, to allow independent testing and verification. Sunday's rumor-borne prediction was difficult to test but was unfulfilled.

To keep abreast of new approaches and applications for earthquake prediction, the USGS has recently reestablished a National Earthquake Prediction Evaluation Council. The Council is made up of a number of USGS and university scientists and provides objective and expert analysis and opinions on earthquake prediction. The Council reviews predictions and resolves scientific issues relating to earthquake prediction.

Even if we can't predict the time and place of future earthquakes, we can reduce the damage they inflict through strong ground shaking by recognizing the areas where damaging earthquakes occur most frequently. The USGS publishes seismic hazard maps for the entire United States. Based on earthquake statistics, geology, and knowledge of the way earthquake energy spreads, the national seismic hazards maps show areas of likely strong, earthquake-generated ground shaking. As we learn more about earthquakes, the maps are updated.

Seismic hazards are expressed in terms of probabilities that specific levels of strong ground shaking will be exceeded in a 50-year interval. The USGS National Seismic Hazards maps for Hawaii Island show that it is one of the highest hazard areas in the US. These maps are used to develop modern building codes. Thus, even without the ability to predict earthquakes, safer buildings and structures in earthquake-prone regions will hopefully save lives and reduce losses due to earthquakes.

Scientists have also calculated earthquake probabilities for specific regions, using new geological and geophysical data, theories, and modeling techniques. Quite noteworthy is the work done for the San Francisco Bay Area which has determined that there is a 62% chance that a magnitude 6.7 earthquake will strike there before the year 2032.

Large earthquakes are certain to strike Hawaii, and we should not be completely caught off guard by earthquakes like last month's. In fact, work published in 1992, determined that there was a 75% chance that a magnitude 6.5 or greater earthquake would strike the Hawaiian Archipelago before the year 2010. The magnitude-6.7 Kiholo Bay earthquake on October 15, 2006 proved the validity of this approach.

It is very important for government, business, and the general public to use earthquake probabilities and seismic hazards maps, as well as other information available about earthquakes, to plan for future damaging quakes. It is equally important to improve earthquake monitoring and to continue to pursue basic earthquake research to improve our ability to understand and live more safely with earthquakes. skip past bottom navigational bar.

————————————————————————————————————————————————————————————————

### Volcano Activity Update

An update is not included with this Volcano Watch article.


text:
# How many ways to write 4 as the sum of 5 non-negative integers?

1. Sep 8, 2014

### fleazo

the problem:
In how many ways can we write the number 4 as the sum of 5 non-negative integers?

I realize this is a generalized combinations problem. I can plug it in using a formula, but I want to understand the logic behind why the generalizaed combination formula works. More specifically, my book gives a description of the problem and I don't understand why it works at all.

I have taken a screen cap of the solution that my book provides. Here it is

http://imgur.com/8BhxXPq

So I understand the concept of how they're explaining this - for example - one possible selection is I chose "box 1" 2 times, "box 2" 1 time, "box 3" 0 times, "box 4" 2 times, and "box 5" 0 times, this corresponds to : 2 + 1 + 2 = 5.

The confusion for me is, it seems like this approach would have a lot of repeats. For example - the selection of "box 1" 0 times, "box 2" 0 times, "box 3" 2 times, "box 4" 1 time, and "box 5" 2 times, would also produce : 2 + 1 + 2 = 5.

I realize I'm missing something in the logic, but I'm not sure what.

2. Relevant equations

C(r + n -1, r) (generalized combinations with repetitions allowed formula)

3. The attempt at a solution

see the above section, this thread is about trying to understand a solution attempt in a text book

2. Sep 8, 2014

### Staff: Mentor

You have "overselected" - you can't select 5 boxes, as you have only 4 balls (and each ball is used to select a box).

I am not stating I understand the solution, but your initial approach is already wrong.

3. Sep 8, 2014

### fleazo

Oops, yes it seems in the examples I gave I was using 5 rather than 4. That was just a dumb mistake on my part. But if I redo the examples using 4 balls, I still have the same problem right?

(example - "box 1" 2 times, "box 2" 2 times and all the rest 0 equates to 2 + 2, "box 3" 2 times, "box 4" 2 times and all the rest 0 also equates to 2 + 2.)

The reason I'm asking specifically about this approach is because I googled around and found this exact approach the box uses to solve this problem many times. I just can't wrap my head around it because I'm wondering where all these repeats go :(

4. Sep 8, 2014

### PeroK

You're right about the repeats, but it says in the solution that the order does not matter in the box analogy. So, these repeats get bundled up in the calculation.

If the order did matter, then you'd have $5^4$ choices.

So, order does matter in the original setting of the problem, but does not matter when you use the box mapping.

5. Sep 8, 2014

### fleazo

Hi PeroK, perhaps you could elaborate a little bit, because this is where my disconnect is happening.

I understand that in a "combinations" problem (as opposed to a permutation problem), the order doesn't matter, and so repeats come out in the calculation. My confusion is seeing how their visual analogy works, because I think it winds up giving an additional set of repeats which don't get filtered out. Let me rephrase their approach to a similar problem where it makes sense. This should highlight why I don't think the analogy works on their problem.

I believe the setup their using would work for this problem: "How many ways can you select 4 cans of soda, from a cooler stocked full of Coke, Pepsi, 7UP, Dr. Pepper, and Diet Coke?" I'm going to use their visual approach, only with bit strings instead of boxes, since I can't draw boxes here. So if I were to set up this problem I'd first recognize two things: (1) order doesn't matter (2) repeats are allowed. I could envision the problem this way: I have a bit string of 4 0's and 4 1's, where 0's represent "column partitions", (a column for each type of soda), and the 1's in between these column partitions correspond to how many selections of that column's soda I've made. Here's an example of such a bit string:

10101001

Now suppose my columns are like this:

---------------------------------------------------
Coke | Pepsi | 7UP | Dr. Pepper | Diet Coke
---------------------------------------------------

Then the bit string above translates to: 1 Coke, 1 Pepsi, 1 7up column, then I skip Dr. Pepper column, and there's 1 for Diet Coke. So this corresponds to selecting 1 Coke, 1 Pepsi, 1 7UP, and 1 Diet Coke. I can visualize it like this:

-------------------------------------------
Coke | Pepsi | 7UP | Dr. Pepper | Diet Coke
--------------------------------------------
1----0-1----0-1---0-----------0-1----------

Similarly:

01001110 would correspond to 3 Dr. Peppers and a Pepsi.

--------------------------------------------
Coke | Pepsi | 7UP | Dr. Pepper | Diet Coke
--------------------------------------------
-----0-1----0-----0-111-------0------------

Using this setup, I can now think of the problem as "how do I rearrange 4 1's, in a bit string of length 8, where order doesn't matter? This is C(8,4) = 70 choices. in fact, this is the answer they arrive at in the book for their problem.

Now here's why I don't see the connection between this problem and the one they've given.

In my problem, suppose I have bit strings:

10001110 and 1110100

This first bitstring is 1 Coke and 3 Dr. Peppers, and the second one is 3 Cokes and a Pepsi. These are distinct answers.

Now in THEIR problem, 10001110 and 1110100 both correspond to 1 + 3, which are NOT distinct answers.

So see my confusion is, it seems like in the "soda approach", each bit string arrangement is a *unique* answer, whereas in this "integer problem", there are bit string arrangements which are repeats of each other. Now, the fact that the "soda problem" reduces to a combinations problem (arranging bit strings), is what allows repeats to be filtered (i.e., 1's are indistinguishable, so this is why it's not a permutation problem). But what I'm getting at is they don't seem to be the same repeats I'm concerned about. When this approach is used on the "integer problem" there seem to be an additional set of repeats going on, and I don't see how they're filtered out.

Last edited: Sep 8, 2014
6. Sep 8, 2014

### PeroK

Their analogy works like this:

If you have 5 integers that add to 4, where the order does matter. These are all strings of 0, 1, 2, 3 and 4. E.g. 0-2-1-0-1 or 0-0-0-4-0.

Each of these corresponds to a combination (where the order does not matter) of placing 4 balls in 5 boxes.

0-2-1-0-1 corresponds to 2 balls in the 2nd box, 1 in the 3rd box and 1 in the 5th box

0-0-0-4-0 corresponds to 4 balls in the 4th box.

You need to convince yourself that there is a 1-1 correspondence here.

The second problem, the one with the boxes, has an already known solution in terms of binomials.

I'd forget the soda analogy. I don't think that helps.

7. Sep 8, 2014

### fleazo

This is where I'm struggling. Based on my understanding, 0-2-1-1-0 is equivalent to the "answer" 2 + 1 + 1. But how would the answers 0-2-1-1-0 and 2-1-0-1-0 be different from each other? This is where I'm failing to see the 1 - 1 correspondence, as both are valid combinations that would come out using the "box method", but here, 0-2-1-1-0 is equivalent to 2 + 1 + 1 and so is 2-1-0-1-0.

8. Sep 8, 2014

### PeroK

You've misunderstood the original problem. You're looking for 5 numbers that add to 4 where:

0-0-0-0-4 and 0-0-0-4-0 are different possibilities.

This is what is meant by "the order matters".

If the order (in the original problem) did not matter, you would only have 5 possibilities:

4
1+3
2+2
1+1+2
1+1+1+1

To get 70 possibilities, you need to add in the 0's and allow any order.

9. Sep 8, 2014

### fleazo

So you're saying the problem is, assume x_1, x_2, x_3, x_4, and x_5 are 5 numbers such that when you add them together, they equal 4, and the problem in the book is asking: how many different ways can we select sets of 4 of these variables, where repeats are allowed? (so in their analogy, each "box" is equivalent to one of the variables, where the number of balls in that box is saying how many times we're picking that specific variable).

Yes, I thought the original question was literally asking "how solutions are there to the problem x_1 + x_2 + x_3 + x_4 + x_5 = 4, where x_1, x_2, x_3, x_4, and x_5 are all non-negative integers.

Last edited: Sep 8, 2014
10. Sep 8, 2014

### PeroK

No, the problem is how many ways to add x_1 ... x_5 to equal 4. That's the problem. The equivalent problem is how many ways can 4 balls be put in 5 boxes, where the number of balls in box b_1 = the value of the variable x_1.

It is!

11. Sep 8, 2014

### fleazo

I have to be honest, I feel like a bit of a dunce here because I don't seem to understand what the problem is even asking :( I'm sorry to be beating a dead horse.

If this is what it is asking ("how many solutions are there to the problem x_1 + x_2 + x_3 + x_4 + x_5 = 4, where x_1, x_2, x_3, x_4, and x_5 are all non-negative integers?) then how is it possible that order doesn't matter, when answering this question? It seems like there are only 5 possibilities: 4, 1 + 1 + 1 + 1, 1 + 2 + 1, 1 + 3, 2 + 2. I get that you're saying these 5 answers are only when order matters. But I just can't wrap my head around order NOT mattering - how can 2 + 2 (coming from the "box solution" of 0-0-2-2-0) be distinct from 2 + 2 (coming from the "box solution" of 2-0-2-0-0)?

12. Sep 8, 2014

### PeroK

Well, suppose each number was picked by a different person? Let's call them x_1 to x_5. Then x_1 picking 4 and the rest picking 0 is different from x_2 picking 4 and the rest picking 0.

13. Sep 8, 2014

### fleazo

ok, this does make sense when you put it this way! Thank you very much :)

(1) If you were to see the original problem statement (how many ways to write 4 as the sum of 5 non-negative integers?), which version would you jump to, the one where order matters (where you get 5 as the answer), or the one where order doesn't matter? (as in the book, where you get 70 as an answer). The reason I ask is because, the way that problem is asked, I would immediately jump to wanting to get 5 as the answer, and so I'm worried there's something off in the way I'm interpreting the question. I seem to misinterpreting a lot of word problems when it comes to combinations/permutations.

(2) Just out of curiosity, is there a permutation/combination formula I could use to answer the problem I thought it was asking - the one where order DOES matter and you get 5 as the answer? Because that to me seems like the more "practical" problem. As in, I could see myself really wanting to know the answer to such a question, but I'm not sure how I'd compute it other than brute force.

14. Sep 8, 2014

### PeroK

(1) I thought initially as you did, but soon realised what was intended. The question, as stated, is ambiguous.

(2) I don't know. You could try working it out for a few examples and see if you can spot a pattern that could be proved inductively.

15. Sep 8, 2014

### fleazo

Thanks PeroK, as long as I know the problem statement is a little ambiguous, I feel better. As for (2), I'll play around with it, I just wasn't sure if there was just a super common formula that I just didn't know which is why I had asked.

Thanks for all the help, especially since it took me so many replies to understand what you were getting at!


text:
# The Bounded L2 Curvature Conjecture

@article{Klainerman2012TheBL,
title={The Bounded L2 Curvature Conjecture},
author={S. Klainerman and I. Rodnianski and J. Szeftel},
journal={arXiv: Analysis of PDEs},
year={2012}
}
• Published 2012
• Mathematics, Physics
• arXiv: Analysis of PDEs
This is the main paper in a sequence in which we give a complete proof of the bounded $L^2$ curvature conjecture. More precisely we show that the time of existence of a classical solution to the Einstein-vacuum equations depends only on the $L^2$-norm of the curvature and a lower bound on the volume radius of the corresponding initial data set. We note that though the result is not optimal with respect to the standard scaling of the Einstein equations, it is nevertheless critical with respect… Expand
86 Citations


text:
# Constructing a context-free grammar

I want to design a context-free grammar that generates words that either both start and end with $$c$$, or contain the same amount of $$a$$-s and $$b$$-s. Here is what I have. The nonterminals are $$S,X,Y$$, the terminals are $$a,b,c$$, the start symbol is $$S$$, and the productions are

\begin{align} &S \to aYS \mid cXc \mid bXS \mid cYc \mid \varepsilon \\ &X \to a \mid bXX \\ &Y \to b \mid aYY \\ \end{align}

Your grammar doesn't seem to generate $$caac$$. Indeed, in order to generate this work, a simple case analysis shows that the derivation must start $$S \to cYc \to caYYc,$$ at which point it is clear that we cannot generate $$caac$$.
I assume you know how to generate the language $$c\Sigma^*c$$, as well as the language of all words with an equal number of $$a$$-s and $$b$$-s. Instead of combining them in a complicated way, write grammars for both, and then join them together using a rule of the sort $$S \to S_1 \cup S_2$$.


text:
## Duke Mathematical Journal

### On the classification of Heegaard splittings

#### Abstract

The long-standing classification problem in the theory of Heegaard splittings of $3$-manifolds is to exhibit for each closed $3$-manifold a complete list, without duplication, of all its irreducible Heegaard surfaces, up to isotopy. We solve this problem for non-Haken hyperbolic $3$-manifolds.

#### Article information

Source
Duke Math. J., Volume 167, Number 15 (2018), 2833-2856.

Dates
Revised: 10 January 2018
First available in Project Euclid: 3 October 2018

https://projecteuclid.org/euclid.dmj/1538532048

Digital Object Identifier
doi:10.1215/00127094-2018-0023

Mathematical Reviews number (MathSciNet)
MR3865653

Zentralblatt MATH identifier
06982208

Subjects
Primary: 57M50: Geometric structures on low-dimensional manifolds

#### Citation

Colding, Tobias Holck; Gabai, David; Ketover, Daniel. On the classification of Heegaard splittings. Duke Math. J. 167 (2018), no. 15, 2833--2856. doi:10.1215/00127094-2018-0023. https://projecteuclid.org/euclid.dmj/1538532048

#### References

• [1] J. W. Alexander, “Some problems in topology” in Verhandlungen des Internationalen Mathematiker-Kongresses, Zurich 1932, Vol. 1, Kraus Reprint, Nendeln, 1967, 249–257.
• [2] D. Bachman, Stabilizing and destabilizing Heegaard splittings of sufficiently complicated $3$-manifolds, Math. Ann. 355 (2013), 697–728.
• [3] J. Birman, “On the equivalence of Heegaard splittings of closed, orientable $3$-manifolds” in Knots, Groups, and $3$-Manifolds, Ann. of Math. Studies 84, Princeton Univ. Press, Princeton, 1975, 137–164.
• [4] F. Bonahon and J. Otal, Scindements de Heegaard des espaces lenticulaires, Ann. Sci. École Norm. Sup. (4) 16 (1983), 451–466.
• [5] W. Breslin, Curvature bounds for surfaces in hyperbolic $3$-manifolds, Canad. J. Math. 62 (2010), 994–1010.
• [6] A. Casson and C. Gordon, Reducing Heegaard splittings, Topol. Appl. 27 (1987), 275–283.
• [7] T. H. Colding and C. De Lellis, “The min-max construction of minimal surfaces” in Surveys in Differential Geometry, Vol. VIII, Boston, 2002, Surv. Differ. Geom. 8, Int. Press, Somerville, MA, 2013.
• [8] T. H. Colding and D. Gabai, Effective finiteness of irreducible Heegaard splittings of non-Haken $3$-manifolds, Duke Math J. 167 (2018), 2793–2832.
• [9] R. Engmann, Nicht-homöomorphe Heegaard-Zerlegungen vom Geschlecht $2$ der zusammenhängenden Summe zweier Linsenräume, Abh. Math. Sem. Univ. Hamburg 35 (1970), 33–38.
• [10] J. Hass, A. Thompson, and W. Thurston, Stabilization of Heegaard splittings, Geom. Topol. 13 (2009), 2029–2050.
• [11] K. Johannson, Topology and Combinatorics of $3$-Manifolds, Lecture Notes in Math. 1599 (1995), Springer, Berlin.
• [12] J. Johnson, Bounding the stable genera of Heegaard splittings from below, J. Topol. 3 (2010), 668–690.
• [13] J. Johnson, Calculating isotopy classes of Heegaard splittings, preprint, arXiv:1004.4669 [math.GT].
• [14] D. Ketover, Genus bounds for min-max minimal surfaces, to appear in J. Differential Geom., preprint, arXiv:1312.2666 [math.DG].
• [15] F. Laudenbach and S. Blank, Isotopie de formes fermées en dimension trois, Invent. Math. 54 (1979), 103–177.
• [16] T. Li, Heegaard surfaces and measured laminations, II: Non-Haken $3$-manifolds, J. Amer. Math. Soc. 19 (2006), 625–657.
• [17] T. Li, An algorithm to determine the Heegaard genus of a $3$-manifold, Geom. Topol. 15 (2011), 1029–1106.
• [18] Y. Moriah, Heegaard splittings of Seifert fibered spaces, Invent. Math. 91 (1988), 465–481.
• [19] Y. Moriah and M. Lustig, Nielsen equivalence in Fuchsian groups and Seifert fibered spaces, Topology 30 (1991), 191–204.
• [20] Y. Moriah and J. Schultens, Irreducible Heegaard splittings of Seifert fibered spaces are either vertical or horizontal, Topology 37 (1998), 1089–1112.
• [21] J. Pitts, Existence and Regularity of Minimal Surfaces on Riemannian Manifolds, Math. Notes 27, Princeton Univ. Press, Princeton, 1981.
• [22] J. Pitts and J. Rubinstein, “Applications of minimax to minimal surfaces and the topology of $3$-manifolds” in Microconference on Geometry and Partial Differential Equations, 2 (Canberra, 1986), Proc. Centre Math. Anal. Austral. Nat. Univ. 12, Austral. Nat. Univ., Canberra, 1987, 137–170.
• [23] K. Reidemeister, Zur dreidimensionalen Topologie, Abh. Math. Sem. Univ. Hamburg 9 (1933), 189–194.
• [24] H. Rubinstein and M. Scharlemann, Comparing Heegaard splittings of non-Haken $3$-manifolds, Topology 35 (1996), 1005–1026.
• [25] J Schultens, The stabilization problem for Heegaard splittings of Seifert fibered spaces, Topology Appl. 73 (1996) 133–139.
• [26] E. Sedgwick, The irreducibility of Heegaard splittings of Seifert fibered spaces, Pacific J. Math. 190 (1990), 173–199.
• [27] J. Singer, Three-dimensional manifolds and their Heegaard diagrams, Trans. Amer. Math. Soc. 35 (1933), 88–111.
• [28] F. Smith. On the existence of embedded minimal $2$-spheres in the $3$-sphere, endowed with an arbitrary Riemannian metric, Ph.D. thesis, University of Melbourne, Melbourne, Australia, 1982.
• [29] F. Waldhausen, Heegaard-Zerlegungen der $3$-Sphäre, Topology 7 (1968), 195–203.


text:
1. Jul 24, 2009

orange03

How do you convert radians to meters?? Is it even possible?

2. Jul 24, 2009

rock.freak667

That would be converting an angular displacement to a linear displacement. You'd need more data to find a length given an angle in radians.

3. Jul 25, 2009

tiny-tim

Hi orange03!

arc-length = rθ (with θ in radians),

so if r is in metres, then the arc-length will be in metres also.


text:
»  Home »  Bookstore »  Title index »  Books on Stata »  Stata for the Behavioral Sciences

## Stata for the Behavioral Sciences

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See the back cover

Inside preview

Print eBook Kindle

$65.00 Print Add to cart What are VitalSource eBooks? Your access code will be emailed upon purchase.$56.00 VitalSource eBook

$52.00 Amazon Kindle Buy from Amazon  Author: Michael N. Mitchell Publisher: Stata Press Copyright: 2015 ISBN-13: 978-1-59718-173-0 Pages: 646; paperback Price:$65.00
 Author: Michael N. Mitchell Publisher: Stata Press Copyright: 2015 ISBN-13: 978-1-59718-198-3 Pages: 646; eBook Price: $56.00  Author: Michael N. Mitchell Publisher: Stata Press Copyright: 2015 ISBN-13: 978-1-59718-199-0 Pages: 646; Kindle Price:$52.00

### Comment from the Stata technical group

Stata for the Behavioral Sciences, by Michael Mitchell, is the ideal reference for researchers using Stata to fit ANOVA models and other models commonly applied to behavioral science data. Drawing on his education in psychology and his experience in consulting, Mitchell uses terminology and examples familiar to the reader as he demonstrates how to fit a variety of models, how to interpret results, how to understand simple and interaction effects, and how to explore results graphically.

Although this book is not designed as an introduction to Stata, it is appealing even to Stata novices. Throughout the text, Mitchell thoughtfully addresses any features of Stata that are important to understand for the analysis at hand. He also is careful to point out additional resources such as related videos from Stata's YouTube channel.

The book is divided into five sections.

The first section contains a chapter that introduces Stata commands for descriptive statistics and another that covers basic inferential statistics such as one- and two-sample t tests.

The second section focuses on between-subjects ANOVA modeling. The discussion moves from one-way ANOVA models to ANCOVA models to two-way and three-way ANOVA models. In each case, special attention is given to the use of commands such as contrast and margins for testing specific hypotheses of interest. Mitchell also emphasizes the understanding of interactions through contrasts and graphs. Underscoring the importance of planning any experiment, he discusses power analysis for t tests, for one- and two-way ANOVA models, and for ANCOVA models.

Section three of the book extends the discussion in the previous section to models for repeated-measures data and for longitudinal data.

The fourth section of the book illustrates the use of the regress command for fitting multiple regression models. Mitchell then turns his attention to tools for formatting regression output, for testing assumptions, and for model building. This section ends with a discussion of power analysis for simple, multiple, and nested regression models.

The final section has a tone that differs from the first four. Rather than focusing on a particular type of analysis, Mitchell describes elements of Stata. He first discusses estimation commands and similarities in syntax from command to command. Then, he details a set of postestimation commands that are available after most estimation commands. Another chapter provides an overview of data management commands. This section ends with a chapter that will be of particular interest to anyone who has used IBM® SPSS®; it lists commonly used SPSS® commands and provides equivalent Stata syntax.

This book is an easy-to-follow guide to analyzing data using Stata for researchers in the behavioral sciences and a valuable addition to the bookshelf of anyone interested in applying ANOVA methods to a variety of experimental designs.

Michael Mitchell is a senior statistician working in the area of sleep research as well as the prevention of child maltreatment. He is the author of A Visual Guide to Stata Graphics, Data Management Using Stata, and Interpreting and Visualizing Regression Models Using Stata. Previously, he worked for 12 years as a statistical consultant and manager of the UCLA ATS Statistical Consulting Group. There he envisioned the UCLA Statistical Consulting Resources website and wrote hundreds of webpages about Stata.


text:
B. Covered Path
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The on-board computer on Polycarp's car measured that the car speed at the beginning of some section of the path equals v1 meters per second, and in the end it is v2 meters per second. We know that this section of the route took exactly t seconds to pass.

Assuming that at each of the seconds the speed is constant, and between seconds the speed can change at most by d meters per second in absolute value (i.e., the difference in the speed of any two adjacent seconds does not exceed d in absolute value), find the maximum possible length of the path section in meters.

Input

The first line contains two integers v1 and v2 (1 ≤ v1, v2 ≤ 100) — the speeds in meters per second at the beginning of the segment and at the end of the segment, respectively.

The second line contains two integers t (2 ≤ t ≤ 100) — the time when the car moves along the segment in seconds, d (0 ≤ d ≤ 10) — the maximum value of the speed change between adjacent seconds.

It is guaranteed that there is a way to complete the segment so that:

• the speed in the first second equals v1,
• the speed in the last second equals v2,
• the absolute value of difference of speeds between any two adjacent seconds doesn't exceed d.
Output

Print the maximum possible length of the path segment in meters.

Examples
Input
5 64 2
Output
26
Input
10 1010 0
Output
100
Note

In the first sample the sequence of speeds of Polycarpus' car can look as follows: 5, 7, 8, 6. Thus, the total path is 5 + 7 + 8 + 6 = 26 meters.

In the second sample, as d = 0, the car covers the whole segment at constant speed v = 10. In t = 10 seconds it covers the distance of 100 meters.


text:
# Not enough information!

Algebra Level 2

If $x$ and $y$ are real numbers such that $x+y=7$ and $x^3+y^3=133$, find the value of $xy$.

×


text:
# zbMATH — the first resource for mathematics

Homotopy perturbation method: a new nonlinear analytical technique. (English) Zbl 1030.34013
Summary: In this paper, a new perturbation method is proposed. In contrast to the traditional perturbation methods, this technique does not require a small parameter in an equation. In this method, according to the homotopy technique, a homotopy with an imbedding parameter $$p\in [0,1]$$ is constructed, and the imbedding parameter is considered as a “small parameter”, so the method is called the homotopy perturbation method, which can take the full advantages of the traditional perturbation methods and homotopy techniques. To illustrate its effectiveness and its convenience, a Duffing equation with high order of nonlinearity is used; the result reveals that its first order of approximation obtained by the proposed method is valid uniformly even for very large parameter, and is more accurate than the perturbation solutions.

##### MSC:
 34A45 Theoretical approximation of solutions to ordinary differential equations 65L99 Numerical methods for ordinary differential equations 34E99 Asymptotic theory for ordinary differential equations
Duffing equation
Full Text:
##### References:
 [1] G.L. Liu, New research directions in singular perturbation theory: artificial parameter approach and inverse-perturbation technique, in: Conference of 7th Modern Mathematics and Mechanics, Shanghai, 1997 [2] Liao, S.J., An approximate solution technique not depending on small parameters: a special example, Int. J. non-linear mech., 30, 3, 371-380, (1995) · Zbl 0837.76073 [3] Liao, S.J., Boundary element method for general nonlinear differential operators, Eng. anal. boundary elem., 20, 2, 91-99, (1997) [4] He, J.H., Approximate analytical solution for seepage flow with fractional derivatives in porous media, Comput. meth. appl. mech. eng., 167, 57-68, (1998) · Zbl 0942.76077 [5] He, J.H., Approximate solution for nonlinear differential equations with convolution product nonlinearities, Comput. meth. appl. mech. eng., 167, 69-73, (1998) · Zbl 0932.65143 [6] He, J.H., Variational iteration method: a kind of nonlinear analytical technique: some examples, Int. J. non-linear mech., 34, 4, 699-708, (1999) · Zbl 1342.34005 [7] He, J.H., A review on some new recently developed nonlinear analytical techniques, Int. J. nonlinear sci. numer. simul., 1, 1, 51-70, (2000) · Zbl 0966.65056 [8] He, J.H., Homotopy perturbation technique, Comput. meth. appl. mech. eng., 178, 257-262, (1999) · Zbl 0956.70017 [9] He, J.H., A coupling method of homotopy technique and perturbation technique for nonlinear problems, Int. J. non-linear mech., 35, 1, 37-43, (2000) · Zbl 1068.74618 [10] Nayfeh, A.H., Problems in perturbation, (1985), Wiley New York
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:
# Example and Counterexamples: Cyclic groups

Study of cyclic groups form a crucial part in group theory. Here are some important definitions, facts and examples of cyclic groups.

#### Cyclic group and its generator

Let $G$ be a group and $a \in G$. Then the set $=\{a^n : n\in \Bbb Z\}$ forms a subgroup of G and is called the subgroup generated by $a$.

If for some $a\in G, =G$, then G is called a cyclic group and $a$ is referred to as it’s generator.

#### Examples

1. $(\Bbb Z,+)$
2. Integer modulo n, $( \Bbb Z /n \Bbb Z)$
3. $\Bbb Z_m \times \Bbb Z_n$ whenever $Gcd(m,n)=1$
4. $A_n$, for $n\le 3$

#### Counterexamples

1. $(\Bbb Q,+)$
2. $(\Bbb R,+)$
3. $\Bbb Z_p \times \Bbb Z_p$, where p is any prime.
4. $\Bbb Q/ \Bbb Z$
5. Any non-abelian group.

#### Important facts

1. Any cyclic group is abelian.
2. Any cyclic group is either finite or countable.
3. Any subgroup of cyclic group is cyclic.
4. Any quotient group of cyclic group is cyclic.
5. Any group of prime order is cyclic.
6. Any group G, with $o(G)=pq$, $p, where p and q are distinct primes is cyclic.
7. Any non-abelian group is non-cyclic.
8. Any group which is uncountable is not cyclic.
9. If $a \in G$ is a generator of G, then $a^-1$ is also a generator of G.
10. Any infinite cyclic group has exactly two generators and is isomorphic to $(\Bbb Z,+)$.
11. Any finite cyclic group G is isomorphic to $\Bbb Z_n$, where $n=o(G)$.

#### Book Suggestions

Textbooks (Use the amazon link below to buy the product and support us!)


text:
# A randomized version of the Littlewood Conjecture – Part I – Henna Koivusalo (University of Vienna)

#### Venue

Sala Conferenze (Puteano, Centro De Giorgi).

#### Abstract

The Littlewood Conjecture in Diophantine approximation can be thought of as a problem about covering the plane by a union of hyperbolas centered at rational points. In this paper we consider the problem of translating the center of each hyperbola by a random amount which depends on the denominator of the corresponding rational. Using a randomized covering argument we prove that, not only is this randomized version of the Littlewood Conjecture true for almost all choices of centers, but an even stronger statement with an extra factor of a logarithm also holds. This work is joint with Alan Haynes (Houston).

Torna in cima


text:
# How do you solve  3(x - 5) + 8 = -14.8?

Use the distributive property to get $3 x - 15 + 8 = - 14.8$. Combine like terms $3 x - 7 = - 14.8$. Add 7 to each side to get $3 x = - 7.8$. Divide each side by 3 to get $x = - 2.6$.


text:
Lemma 15.88.4. Assume $\varphi : R \to S$ is a flat ring map and $I \subset R$ is a finitely generated ideal such that $R/I \to S/IS$ is an isomorphism. For any $f_1, \ldots , f_ r \in R$ such that $V(f_1, \ldots , f_ r) = V(I)$

1. the map of Koszul complexes $K(R, f_1, \ldots , f_ r) \to K(S, f_1, \ldots , f_ r)$ is a quasi-isomorphism, and

2. The map of extended alternating Čech complexes

$\xymatrix{ R \to \prod _{i_0} R_{f_{i_0}} \to \prod _{i_0 < i_1} R_{f_{i_0}f_{i_1}} \to \ldots \to R_{f_1\ldots f_ r} \ar[d] \\ S \to \prod _{i_0} S_{f_{i_0}} \to \prod _{i_0 < i_1} S_{f_{i_0}f_{i_1}} \to \ldots \to S_{f_1\ldots f_ r} }$

is a quasi-isomorphism.

Proof. In both cases we have a complex $K_\bullet$ of $R$ modules and we want to show that $K_\bullet \to K_\bullet \otimes _ R S$ is a quasi-isomorphism. By Lemma 15.88.2 and the flatness of $R \to S$ this will hold as soon as all homology groups of $K$ are $I$-power torsion. This is true for the Koszul complex by Lemma 15.28.6 and for the extended alternating Čech complex by Lemma 15.29.5. $\square$

There are also:

• 2 comment(s) on Section 15.88: Formal glueing of module categories

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Mon compte

connexion

inscription

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## définition - 0 k

voir la définition de Wikipedia

Wikipedia

# Absolute zero

(Redirected from 0 K)

Absolute zero is the temperature at which entropy reaches its minimum value. As implied by the laws of thermodynamics, absolute zero cannot be reached by artificial or natural means because this would require a system to be fully removed from the rest of the universe. A system at theoretical absolute zero possesses quantum mechanical zero-point energy. While all molecular motion does not cease at absolute zero, the system does not have enough energy for transference to other systems. It is therefore correct to say that molecular energy is minimal at absolute zero.

By international agreement, absolute zero is defined as precisely 0 K on the Kelvin scale and as −273.15° on the Celsius scale.[1] Absolute zero is also precisely equivalent to 0 R on the Rankine scale (same as Kelvin but measured in Fahrenheit intervals) and −459.67° on the Fahrenheit scale. Though it is not theoretically possible to cool any substance to 0 K,[2] scientists have made great advancements in achieving temperatures close to absolute zero, where matter exhibits quantum effects such as superconductivity and superfluidity.

For the kinematics of molecules at absolute zero on a larger scale, which is easier to understand, see kinetic energy.

## History

Robert Boyle pioneered the idea of an absolute zero.

One of the first to discuss the possibility of an absolute minimal temperature was Robert Boyle. His 1665 New Experiments and Observations touching Cold, articulated the dispute known as the primum frigidum. The concept was well known among naturalists of the time. Some contended an absolute minimum temperature occurred within earth (as one of the four so-called 'elements'), others within water, others air, and some more recently within nitre[citation needed]. But all of them seemed to agree that, "There is some body or other that is of its own nature supremely cold and by participation of which all other bodies obtain that quality."[3]

### Limit to the 'degree of cold'

The question whether there is a limit to the degree of cold possible, and, if so, where the zero must be placed, was first attacked by the French physicist Guillaume Amontons in 1702, in connection with his improvements in the air thermometer. In his instrument temperatures were indicated by the height at which a column of mercury was sustained by a certain mass of air, the volume or "spring" which varied with the heat to which it was exposed. Amontons therefore argued that the zero of his thermometer would be that temperature at which the spring of the air in it was reduced to nothing. On the scale he used, the boiling-point of water was marked at +73 and the melting-point of ice at 51, so that the zero of his scale was equivalent to about −240 on the Celsius scale.

This close approximation to the modern value of −273.15 °C[4] for the zero of the air-thermometer was further improved upon in 1779 by Johann Heinrich Lambert, who observed that −270 °C might be regarded as absolute cold.[5]

Values of this order for the absolute zero were not, however, universally accepted about this period. Pierre-Simon Laplace and Antoine Lavoisier, in their 1780 treatise on heat, arrived at values ranging from 1,500 to 3,000 below the freezing-point of water, and thought that in any case it must be at least 600 below. John Dalton in his Chemical Philosophy gave ten calculations of this value, and finally adopted −3000 °C as the natural zero of temperature.

### Lord Kelvin's work

After J.P. Joule had determined the mechanical equivalent of heat, Lord Kelvin approached the question from an entirely different point of view, and in 1848 devised a scale of absolute temperature which was independent of the properties of any particular substance and was based solely on the fundamental laws of thermodynamics. It followed from the principles on which this scale was constructed that its zero was placed at −273.15 °C, at almost precisely the same point as the zero of the air-thermometer.[6]

## Very low temperatures

The rapid expansion of gases leaving the Boomerang Nebula brings it to the lowest observed temperature outside a laboratory.

The average background temperature of the Universe today is 2.73 Kelvin.

Absolute zero cannot be achieved artificially, though it is possible to reach temperatures close to it through the use of cryocoolers. Laser cooling is a technique used to take temperatures to within a billionth of a degree of 0 K.[7] At very low temperatures in the vicinity of absolute zero, matter exhibits many unusual properties including superconductivity, superfluidity, and Bose-Einstein condensation. In order to study such phenomena, scientists have worked to obtain ever lower temperatures.

• The current world record was set in 1999 at 100 picokelvin by cooling a piece of rhodium metal.[8]
• In February 2003, the Boomerang Nebula was observed to have been releasing gases at a speed of 500,000 km/h (over 300,000 mph) for the last 1,500 years. This has cooled it down to approximately 1 K, as deduced by astronomical observation. This is the lowest natural temperature ever recorded.[11]
• In May 2006, the Institute of Quantum Optics at the University of Hanover gives details of technologies and benefits of femto-kelvin research in space.[13]

## Thermodynamics near absolute zero

At temperatures near 0 K, nearly all molecular motion ceases and, when entropy = S, ΔS = 0 for any adiabatic process. Pure substances can (ideally) form perfect crystals as T → 0. Max Planck's strong form of the third law of thermodynamics states the entropy of a perfect crystal vanishes at absolute zero. The original Nernst heat theorem makes the weaker and less controversial claim that the entropy change for any isothermal process approaches zero as T → 0:

$\lim_{T \to 0} \Delta S = 0$

The implication is that the entropy of a perfect crystal simply approaches a constant value.

The Nernst postulate identifies the isotherm T = 0 as coincident with the adiabat S = 0, although other isotherms and adiabats are distinct. As no two adiabats intersect, no other adiabat can intersect the T = 0 isotherm. Consequently no adiabatic process initiated at nonzero temperature can lead to zero temperature. (≈ Callen, pp. 189–190)

An even stronger assertion is that It is impossible by any procedure to reduce the temperature of a system to zero in a finite number of operations. (≈ Guggenheim, p. 157)

A perfect crystal is one in which the internal lattice structure extends uninterrupted in all directions. The perfect order can be represented by translational symmetry along three (not usually orthogonal) axes. Every lattice element of the structure is in its proper place, whether it is a single atom or a molecular grouping. For substances which have two (or more) stable crystalline forms, such as diamond and graphite for carbon, there is a kind of "chemical degeneracy". The question remains whether both can have zero entropy at T = 0 even though each is perfectly ordered.

Perfect crystals never occur in practice; imperfections, and even entire amorphous materials, simply get "frozen in" at low temperatures, so transitions to more stable states do not occur.

Using the Debye model, the specific heat and entropy of a pure crystal are proportional to T 3, while the enthalpy and chemical potential are proportional to T 4. (Guggenheim, p. 111) These quantities drop toward their T = 0 limiting values and approach with zero slopes. For the specific heats at least, the limiting value itself is definitely zero, as borne out by experiments to below 10 K. Even the less detailed Einstein model shows this curious drop in specific heats. In fact, all specific heats vanish at absolute zero, not just those of crystals. Likewise for the coefficient of thermal expansion. Maxwell's relations show that various other quantities also vanish. These phenomena were unanticipated.

Since the relation between changes in Gibbs free energy (G), the enthalpy (H) and the entropy is

$\Delta G = \Delta H - T \Delta S \,$

thus, as T decreases, ΔG and ΔH approach each other (so long as ΔS is bounded). Experimentally, it is found that all spontaneous processes (including chemical reactions) result in a decrease in G as they proceed toward equilbrium. If ΔS and/or T are small, the condition ΔG < 0 may imply that ΔH < 0, which would indicate an exothermic reaction. However, this is not required; endothermic reactions can proceed spontaneously if the TΔS term is large enough.

Moreover, the slopes of the derivatives of ΔG and ΔH converge and are equal to zero at T = 0. This ensures that ΔG and ΔH are nearly the same over a considerable range of temperatures and justifies the approximate empirical Principle of Thomsen and Berthelot, which says that the equilibrium state to which a system proceeds is the one which evolves the greatest amount of heat, i.e., an actual process is the most exothermic one. (Callen, pp. 186–187)

One model that estimates the properties of an electron gas at absolute zero is the fermi gas. But, interestingly, the fermi temperature of electrons, where the lattice temperature is zero, is non-zero. In fact, the electrons have very high velocities. For an isolated system this is probably a violation of conservation of momentum because—as the electrons are interacting with the lattice and the total momentum is zero—the lattice cannot be at zero velocity. Also, if the lattice was precisely at zero temperature its nuclei would have infinite de Broglie wavelength. (If the momentum goes to zero, the wavelength becomes infinite).

## Relation with Bose–Einstein condensates

A Bose-Einstein condensate is an unusual state of matter that only exists at temperatures lower than a few billionths of a degree above absolute zero.[14]

## Absolute temperature scales

Absolute, or thermodynamic, temperature is conventionally measured in kelvins (Celsius-scaled increments) and in the Rankine scale (Fahrenheit-scaled increments) with increasing rarity. Absolute temperature measurement is uniquely determined by a multiplicative constant which specifies the size of the "degree", so the ratios of two absolute temperatures, T2/T1, are the same in all scales. The most transparent definition of this standard comes from the Maxwell-Boltzmann distribution. It can also be found in Fermi-Dirac statistics (for particles of half-integer spin) and Bose-Einstein statistics (for particles of integer spin). All of these define the relative numbers of particles in a system as decreasing exponential functions of energy (at the particle level) over kT, with k representing the Boltzmann constant and T representing the temperature observed at the macroscopic level.

## Negative temperatures

Certain semi-isolated systems, such as a system of non-interacting spins in a magnetic field, can achieve negative temperatures; however, they are not actually colder than absolute zero. They can be however thought of as "hotter than T = ∞", as energy will flow from a negative temperature system to any other system with positive temperature upon contact.

## Notes

1. ^ "Unit of thermodynamic temperature (kelvin)". SI Brochure, 8th edition. Bureau International des Poids et Mesures. 1967. pp. Section 2.1.1.5. Retrieved 2008-02-11.
2. ^ Davies, Jeremy Dunning (1996). Concise Thermodynamics. Horwood Publishing. p. 43. ISBN 1898563152.
3. ^ Boyle, Robert (1665). New Experiments and Observations touching Cold.
4. ^
5. ^ Lambert, Johann Heinrich (1779). Pyrometrie. Berlin. OCLC 165756016.
6. ^ "Cold". Encyclopædia Britannica (Eleventh Edition ed.). The LoveToKnow Wiki. 1911. Retrieved 2008-02-11.
7. ^
8. ^ "World record in low temperatures". Retrieved 2009-05-05.
9. ^ Knuuttila, Tauno (2000). Nuclear Magnetism and Superconductivity in Rhodium. Espoo, Finland: Helsinki University of Technology. ISBN 9512252082. Retrieved 2008-02-11.
10. ^ Low Temperature Laboratory, Teknillinen Korkeakoulu (8 December 2000). "Low Temperature World Record". Press release. Retrieved 2008-02-11.
11. ^ Sahai, Raghvendra; Nyman, Lars-Åke (1997). [Expression error: Missing operand for > "The Boomerang Nebula: The Coldest Region of the Universe?"]. The Astrophysical Journal 487: L155–L159. doi:10.1086/310897.
12. ^
13. ^
14. ^ Elizabeth A. Donley et al. (2001). "Dynamics of collapsing and exploding Bose–Einstein condensates". Nature 412: 295-299.

## References

• Herbert B. Callen (1960). "Chapter 10". Thermodynamics. New York: John Wiley & Sons. OCLC 535083.
• Herbert B. Callen (1985). Thermodynamics and an Introduction to Thermostatistics (Second Edition ed.). New York: John Wiley & Sons. ISBN 0-471-86256-8.
• E.A. Guggenheim (1967). Thermodynamics: An Advanced Treatment for Chemists and Physicists (Fifth Edition ed.). Amsterdam: North Holland Publishing. OCLC 324553.
• George Stanley Rushbrooke (1949). Introduction to Statistical Mechanics. Oxford: Clarendon Press. OCLC 531928.

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Lemma 10.154.3. Let $R$ be a ring. Let $A = \mathop{\mathrm{colim}}\nolimits A_ i$ be a filtered colimit of $R$-algebras such that each $A_ i$ is a filtered colimit of étale $R$-algebras. Then $A$ is a filtered colimit of étale $R$-algebras.

Proof. Write $A_ i = \mathop{\mathrm{colim}}\nolimits _{j \in J_ i} A_ j$ where $J_ i$ is a directed set and $A_ j$ is an étale $R$-algebra. For each $i \leq i'$ and $j \in J_ i$ there exists an $j' \in J_{i'}$ and an $R$-algebra map $\varphi _{jj'} : A_ j \to A_{j'}$ making the diagram

$\xymatrix{ A_ i \ar[r] & A_{i'} \\ A_ j \ar[u] \ar[r]^{\varphi _{jj'}} & A_{j'} \ar[u] }$

commute. This is true because $R \to A_ j$ is of finite presentation so that Lemma 10.127.3 applies. Let $\mathcal{J}$ be the category with objects $\coprod _{i \in I} J_ i$ and morphisms triples $(j, j', \varphi _{jj'})$ as above (and obvious composition law). Then $\mathcal{J}$ is a filtered category and $A = \mathop{\mathrm{colim}}\nolimits _\mathcal {J} A_ j$. Details omitted. $\square$

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# Math Help - Functions Where the Extended Real Line gives (or does not give) Continuity

1. ## [SOLVED] Functions Where the Extended Real Line gives (or does not give) Continuity

I'm examining the final paragraph on page 75 from, Oden - Applied Functional Analysis 2nd ed., and am hoping for clarity.

A statement is made that for a function such as $\frac{1}{x^2}$ using the extended real line where the value at x = 0 is $\infty$, is continuous. The author then goes on to say that this continuity is not true for $\frac{1}{x}$ at x=0.

I am looking for illumination as to why this is the case. Thank you.

I am not sure if this may be of help but the section is "Functions with values in Bar-R (Extended Real Line)." Previous discussions include point-wise sup/inf.

Chris

2. As x approaches zero from either side, 1/x^2 goes to + infinity. As x approaches zero from either side, 1/x goes to either +infinity or -infinity. The extended reals include both infinities, and they are not equal. So you get a jump discontinuity at 0.

3. Thanks

4. You're welcome!

5. Another thing would be if we compatify $\mathbb{R}$ with only one point i.e. $\overline{\mathbb{R}}=\mathbb{R}\cup \{\infty\}$. In this case, $\lim_{x\to 0} (1/x)=\infty$ .

Fernando Revilla


text:
# Electron-positron annihilation: Feynman diagram

Maybe my question is just a blunder. Consider Compton interaction: $e^{-}+\gamma \rightarrow e^{-}+\gamma$. There are two Feynman diagrams related to this process to the lowest order of $\alpha$.

Now consider $e^{-}+e^{+} \rightarrow \mu^{-}+\mu ^{+}$. I think here there are two different diagrams as well. First, connect electron and positron to a virtual photon (propagator) with two free points that can be connected to a fermion called $A$ and $B$. I guess one can connect $A$ to muon or to anti-muon so there are two different Feynman diagrams for this process to the lowest order of the electron charge. However, I checked this interaction on QFT reference books such as Peskin (Chapter 5) and they have just calculated one of this diagrams. What is the problem?

Note: There are other diagrams too.

First one:

Second one:

• Can you draw the diagrams you're talking about? Jul 15 '16 at 17:39
• @knzhou I edited my question Jul 15 '16 at 17:41
• Put the arrows on the fermion lines and you should see the problem with your first diagram ;) Jul 15 '16 at 17:57
• @annav done! seems there is only 1 diagram in the lowest order but I don't know why the other (maybe others) isn't allowed Jul 15 '16 at 18:15

The diagrams you are drawing are not allowed. In the last one you have an electron going into an muon, by the emission of a photon. Try to isolate that part. If it works one way, it should also work the other way - a muon should be able to decay into an electron by the emission of a photon. This cannot happen. Muons decay to electrons in the weak interaction, not QED.

In the first one, you have the same problem, plus an additional one: you don't obey charge conservation in your vertices. An electron (charge -1) cannot become an anti-muon (charge 1), no matter how many photons it emits.

If you're only considering QED, there is only one vertex for this interaction -- electron and positron annhilate to photon, photon pair-produces muon and antimuon. Any electron, photon, muon vertex will have flavor conservation issues.

If you're not JUST considering QED, there is a second diagram, identical to the first, with the photon replaced with a $Z^{0}$. For low energies, the contribution of this one is small, compared to the first, because it's suppressed with factors of $m_{Z}$

Basically, your proposed vertex of $e^{-} \rightarrow \gamma + \mu^{-}$ violates symmetries that you haven't yet encountered in Peskin and Schroder.

• I edited my question Jul 15 '16 at 17:41
• @KNP: yes, in your edit, you include an $e^{-}$ hitting a vertex, and becoming a $\mu^{-}$ and a $\gamma$. This is forbidden by flavor conservation. Jul 15 '16 at 20:02

The two diagrams you draw are all wrong, for the vortex is determined by Lagrangian.

• I suggest to add more details. Jul 15 '16 at 19:05
• Yes. None of them are true. I want to know why they are not allowed? (or why they are zero?) Jul 15 '16 at 19:15


text:
## The affine Weyl group

Last update: 25 September 2012

## The affine Weyl group

For $\lambda \in P,$ the translation in $\lambda$ is

$tλ: 𝔥ℝ* ⟶ 𝔥ℝ* x ⟼ x+λ. (1.10)$

The extended affine Weyl group $\stackrel{\sim }{W}$ is the group

$W∼= { wtλ|w∈W ,λ∈P } , (1.11)$

with multiplication determined by the relations

$tλtμ= tλ+μ,and twλw=w tλ, (1.12)$

for $\lambda ,\mu \in P$ and $w\in W,$ and so $\stackrel{\sim }{W}$ is a semidirect product of $W$ and the group of translations $\left\{{t}_{\lambda }\phantom{\rule{0.2em}{0ex}}\mid \lambda \in P\right\}\text{.}$ It is the group of transformations of ${𝔥}_{ℝ}^{*}$ generated by the ${s}_{\alpha },$ $\alpha \in {R}^{+},$ and ${t}_{\lambda },$ $\lambda \in P\text{.}$ The affine Weyl group ${W}_{\text{aff}}$ is the subgroup of $\stackrel{\sim }{W}$ generated by the reflections

$sα,k: 𝔥ℝ*⟶𝔥ℝ* in the hyperplanes Hα,k= { x∈𝔥ℝ*∣ ⟨x,α∨⟩=k } ,α∈R+,k∈ℤ. (1.13)$

The reflections ${s}_{\alpha ,k}$ can be written as elements of $\stackrel{\sim }{W}$ via the formula

$sα,k= tkαsα= sαt-kα. (1.14)$

The highest short root of $R$ is the unique element $\phi \in {R}^{+}$ such that the fundamental alcove

$A=C∩ { x∈𝔥ℝ*∣ ⟨x,φ∨⟩<1 } (1.15)$

is a fundamental region for the action of ${W}_{\text{aff}}$ on ${𝔥}_{ℝ}^{*}\text{.}$ The various fundamental chambers for the action of ${W}_{\text{aff}}$ on ${𝔥}_{ℝ}^{*}$ are ${\stackrel{\sim }{w}}^{-1}A,$ $\stackrel{\sim }{w}\in {W}_{\text{aff}}\text{.}$ The inversion set of $\stackrel{\sim }{w}\in \stackrel{\sim }{W}$ is

$R(w∼)= { Hα,k∣ Hα,kis betweenA andw∼-1 A } and ℓ(w∼)=Card (R(w∼))$

is the length of $\stackrel{\sim }{w}\text{.}$ If $w\in W$ and $\lambda \in P$ then

$ℓ(wtλ)= ∑α∈R+ ∣ ⟨λ,α∨⟩+ χ(wα) ∣ , (1.16)$

where, for a root $\beta \in R,$ set $\chi \left(\beta \right)=0,$ if $\beta \in {R}^{+},$ and $\chi \left(\beta \right)=1,$ if $\beta \in {R}^{-}\text{.}$

In type ${C}_{2}$ we have the picture.

Let

$Hα0=Hφ,1 ands0= sφ,1=tϕ sϕ=sϕ t-ϕ, (1.17)$

and let ${H}_{{\alpha }_{1}},\dots ,{H}_{{\alpha }_{n}}$ and ${s}_{1},\dots ,{s}_{n}$ be as in (1.9). Then the walls of $A$ are the hyperplanes ${H}_{{\alpha }_{0}},{H}_{{\alpha }_{1}},\dots ,{H}_{{\alpha }_{n}}$ and the group ${W}_{\text{aff}}$ has a presentation by generators ${s}_{0},{s}_{1},\dots ,{s}_{n}$ and relations

$si2 = 1 for0≤i≤n, sisjsi… ⏟ mijfactors = sjsisj… ⏟ mijfactors , i≠j, (1.18)$

where $\pi /{m}_{ij}$ is the angle between the hyperplanes ${H}_{{\alpha }_{i}}$ and ${H}_{{\alpha }_{j}}\text{.}$

Let ${w}_{0}$ be the longest element of $W$ and let ${w}_{i}$ be the longest element of the subgroup ${W}_{{\omega }_{i}}=\left\{w\in W\phantom{\rule{0.2em}{0ex}}\mid \phantom{\rule{0.2em}{0ex}}w{\omega }_{i}={\omega }_{i}\right\}\text{.}$ Let ${\phi }^{\vee }={c}_{1}{\alpha }_{1}^{\vee }+\dots +{c}_{n}{\alpha }_{n}^{\vee }\text{.}$ Then let

$Ω= { g∈W∼∣ℓ (g)=0 } ={1}∪ {gi∣ci=1}, wheregi=tωi wiw0, (1.19)$

(see [Bou1981, VI § no. 3 Prop. 6]). Each element $g\in \Omega$ sends the alcove $A$ to itself and thus permutes the walls ${H}_{{\alpha }_{0}},{H}_{{\alpha }_{1}},\dots ,{H}_{{\alpha }_{n}}$ of $A\text{.}$ Denote the resulting permutation of $\left\{0,1,\dots ,n\right\}$ also by $g\text{.}$ Then

$gsig-1= sg(i), for0≤i≤n, (1.20)$

and the group $\stackrel{\sim }{W}$ is presented by the generators ${s}_{0},{s}_{1},\dots ,{s}_{n}$ and $g\in \Omega$ with the relations (1.18) and (1.20). In the setting of Example 1.1, ${W}_{{\omega }_{1}}=\left\{1,{s}_{2}\right\},$ ${W}_{{\omega }_{2}}=\left\{1,{s}_{1}\right\},$ ${w}_{1}={s}_{2},$ ${w}_{2}={s}_{1}$ and ${w}_{0}={s}_{1}{s}_{2}{s}_{1}{s}_{2},$ and ${\phi }^{\vee }=2{\alpha }_{1}^{\vee }+{\alpha }_{2}^{\vee }$ so that ${c}_{1}=2,$ ${c}_{2}=1$ and $\Omega =\left\{1,{g}_{2}\right\}\cong ℤ/2ℤ,$ where ${g}_{2}={t}_{{\omega }_{2}}{s}_{2}{s}_{1}{s}_{2}\text{.}$

## Acknowledgements

The research of A. Ram was partially supported by the National Science Foundation (DMS-0097977), the National Security Agency (MDA904-01-1-0032) and by EPSRC Grant GR K99015 at the Newton Institute for Mathematical Sciences. The research of K. Nelsen was partially supported by the National Science Foundation (DMS-0097977 and a VIGRE grant) and the National Security Agency (MDA904-01-1-0032).