text:
# Fight Finance

#### CoursesTagsRandomAllRecentScores

A project has an internal rate of return (IRR) which is greater than its required return. Select the most correct statement.

A furniture distributor offers credit to its customers. Customers are given 25 days to pay for their goods, but if they pay immediately they will get a 1% discount.

What is the effective interest rate implicit in the discount being offered? Assume 365 days in a year and that all customers pay either immediately or on the 25th day. All rates given below are effective annual rates.

The following table shows a sample of historical total returns of shares in two different companies A and B.

 Stock Returns Total effective annual returns Year $r_A$ $r_B$ 2007 0.2 0.4 2008 0.04 -0.2 2009 -0.1 -0.3 2010 0.18 0.5

What is the historical sample covariance ($\hat{\sigma}_{A,B}$) and correlation ($\rho_{A,B}$) of stock A and B's total effective annual returns?

Two call options are exactly the same, but one has a low and the other has a high exercise price. Which option would you expect to have the higher price, the option with the or exercise price, or should they have the price?

Two put options are exactly the same, but one has a low and the other has a high exercise price. Which option would you expect to have the higher price, the option with the or exercise price, or should they have the price?

In Australia in the 1980's, inflation was around 8% pa, and residential mortgage loan interest rates were around 14%.

In 2013, inflation was around 2.5% pa, and residential mortgage loan interest rates were around 4.5%.

If a person can afford constant mortgage loan payments of $2,000 per month, how much more can they borrow when interest rates are 4.5% pa compared with 14.0% pa? Give your answer as a proportional increase over the amount you could borrow when interest rates were high $(V_\text{high rates})$, so: $$\text{Proportional increase} = \dfrac{V_\text{low rates}-V_\text{high rates}}{V_\text{high rates}}$$ Assume that: • Interest rates are expected to be constant over the life of the loan. • Loans are interest-only and have a life of 30 years. • Mortgage loan payments are made every month in arrears and all interest rates are given as annualised percentage rates (APR's) compounding per month. A company advertises an investment costing$1,000 which they say is underpriced. They say that it has an expected total return of 15% pa, but a required return of only 10% pa. Assume that there are no dividend payments so the entire 15% total return is all capital return.

Assuming that the company's statements are correct, what is the NPV of buying the investment if the 15% return lasts for the next 100 years (t=0 to 100), then reverts to 10% pa after that time? Also, what is the NPV of the investment if the 15% return lasts forever?

In both cases, assume that the required return of 10% remains constant. All returns are given as effective annual rates.

The answer choices below are given in the same order (15% for 100 years, and 15% forever):

A share currently worth $100 is expected to pay a constant dividend of$4 for the next 5 years with the first dividend in one year (t=1) and the last in 5 years (t=5).

The total required return is 10% pa.

What do you expected the share price to be in 5 years, just after the dividend at that time has been paid?

Which Australian institution is in charge of monetary policy?

Question 860  idiom, hedging, speculation, arbitrage, market making, insider trading, no explanation

Which class of derivatives market trader is NOT principally focused on ‘buying low and selling high’?

format: plain


text:
## Files in this item

FilesDescriptionFormat

application/pdf

QUANRUD-DISSERTATION-2019.pdf (2MB)
(no description provided)PDF

## Description

 Title: Fast approximations for combinatorial optimization via multiplicative weight updates Author(s): Quanrud, Kent Director of Research: Chekuri, Chandra Doctoral Committee Chair(s): Chekuri, Chandra Doctoral Committee Member(s): Har-Peled, Sariel; Erickson, Jeff; Young, Neal E; Blum, Avrim Department / Program: Computer Science Discipline: Computer Science Degree Granting Institution: University of Illinois at Urbana-Champaign Degree: Ph.D. Genre: Dissertation Subject(s): Approximation algorithms Linear programming Combinatorial optimization fast approximations traveling salesman problem Abstract: We develop fast approximations for several LP relaxations that arise in discrete and combinatorial optimization. New results include improved running times for explicit mixed packing and covering problems, nearly linear time approximations for tree packings, nearly linear time approximations for the Held Karp bound (leading to a significantly faster $(3/2 + \epsilon)$-approximation for metric TSP), faster approximations for covering LPs with knapsack covering constraints (the bottleneck for covering integer programs), and nearly linear time $(2+\epsilon)$-approximations for $k$-cut via the LP. Along the way we develop new techniques for the MWU framework and put forth two frameworks, "lazy MWU" for deterministic algorithms and "randomized MWU" for randomized algorithms, that algorithm designers can use to obtain nearly linear running times for their own problems of interest. This thesis has been organized as a user friendly guide, where we include basic background and analysis of the MWU framework, establish clean interfaces for the two frameworks, and use the applications as examples of the frameworks. Issue Date: 2019-09-30 Type: Text URI: http://hdl.handle.net/2142/106153 Rights Information: Copyright 2019 Kent Quanrud Date Available in IDEALS: 2020-03-02 Date Deposited: 2019-12


text:
How to reset one Bootstrap Multiselect when another Bootstrap Multiselect is used via jQuery

I am using the Bootstrap-Multiselect (http://davidstutz.github.io/bootstrap-multiselect/)

I have two dropdowns, when an item is selected in one dropdown I need to reset the other dropdown. I have been attempting to do this using the onchange jQuery but without success.

If an option is selected from dropdown1, then deselect/reset everything in dropdown2.

<form>
<select id="dropdown1" name="dropdown1" multiple="multiple">
<option value="1">One</option>
<option value="2">Two</option>
<option value="3">Three</option>
</select>

If an option is selected from dropdown2, then deselect/reset everything in dropdown1.

    <select id="dropdown2" name="dropdown2" multiple="multiple">
<option value="1">One</option>
<option value="2">Two</option>
<option value="3">Three</option>
</select>
</form>

My failed jQuery

$("#dropdown1").change(function () {$('#dropdown2').multiselect('refresh');
});

$("#dropdown2").change(function () {$('#dropdown1').multiselect('refresh');
});

How should this be accomplished?

I suspect you need to reset the select element and then do the refresh, like so:

$("#dropdown1").change(function () {$('#dropdown2 option:selected').each(function() {
$(this).prop('selected', false); })$('#dropdown2').multiselect('refresh');
});

$("#dropdown2").change(function () {$('#dropdown1 option:selected').each(function() {
$(this).prop('selected', false); })$('#dropdown1').multiselect('refresh');
});

That code is borrowed from the reset button code on the Bootstrap Multiselect site. See the further examples

• Worked perfectly, thankyou – Terry Jun 25 '16 at 19:47
• No problem. You were nearly there. – d3v_1 Jun 25 '16 at 19:50

Another way of deselecting all options. (Because multiselect('refresh') did not work for me.)

$('#ID').multiselect('deselectAll', false);$('#ID').multiselect('updateButtonText');


text:
# Math Help - Frustrating word problems

1. ## Frustrating word problems

Hello, I am new to these forums. After hitting a wall on my take-home quiz, I am on the hunt for some assistance. Our professor told us we can use any means necessary to get help, so I am doing just that! Always been terrible at word problems, algebra in general for that matter. Have two problems here that left me with nothing but a blank paper and a nasty migraine!

1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width.

2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be $200? Thanks in advance! 2. Originally Posted by Fails_at_Math Hello, I am new to these forums. After hitting a wall on my take-home quiz, I am on the hunt for some assistance. Our professor told us we can use any means necessary to get help, so I am doing just that! Always been terrible at word problems, algebra in general for that matter. Have two problems here that left me with nothing but a blank paper and a nasty migraine! 1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width. 2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be$200?

1. Solve 80 = x(2x + 6) => 2(x - 5)(x + 8) = 0 for x, where x is the width. Keep the answer that makes sense.

2. Solve 200 = x^2 - 15x - 50 => (x - 25)(x + 10) = 0 for x. Keep the answer for x that makes sense.

3. Originally Posted by Fails_at_Math
1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width.
First you need to write a formula that relates length, width, and area. Fortunately, this is a common formula: A=L*W

We know A since it is given, so plug that into the formula:

$80 = L*W$

Now we have two variables, but lets get it down to one so that we can actually solve for something. How do length and width relate to eachother? Well we were told that length is 6 more than twice the width, so L = 6+2W

Now we can substitute this value in for L

$80 = (6+2W)*W$

$80 = 6W +2W^2$

$40 = W^2 + 3W$

$0 = W^2 + 3W - 40$

Solve by factoring (Hint: -40 = -5*8)
This will give you two solutions (use the positive answer since you can't have negative width).
Then plug that answer into the equation again and solve for length.
Originally Posted by Fails_at_Math
2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be \$200?
p is profit = 200 so plug that in:

$200 = x^2 -15x - 50$

$0 = x^2 -15x - 250$

Now you just factor it (hint: -250 = -25 * 10)
This will give you two solutions, evaluate which makes the most sense (hint: it doesn't make sense to sell negative videos)

4. I see now how it relates back to the rest of the problems we have been doing in class. Just hit a wall trying to put the words into equation form, but I guess that's the real challenge of word problems. Thanks MHF!

5. Originally Posted by Fails_at_Math
Just hit a wall trying to put the words into equation form, but I guess that's the real challenge of word problems. Thanks MHF!
For me the real challenge of word problems is that I can't read :/

I don't even know what the question was, I just copied Mr. Fantastic's answer and changed the variable letters. I don't even know what you just posted, I'm just randomly hitting keys right now, I don't even know what I'm typing!

6. Originally Posted by angel.white
For me the real challenge of word problems is that I can't read :/

I don't even know what the question was, I just copied Mr. Fantastic's answer and changed the variable letters. I don't even know what you just posted, I'm just randomly hitting keys right now, I don't even know what I'm typing!

The resemblance is uncanny

7. Originally Posted by mr fantastic
The resemblance is uncanny
Why do you think I wear a hat and sunglasses?


text:
# Difference: 20151127_MSciSpectrocopyLab (1 vs. 3)

#### Revision 328 Nov 2015 - JosephBayley

Line: 1 to 1

 META TOPICPARENT name="StewartBoogertSpectroscopy2015"

# Additional Data Taking

Line: 8 to 8

• Using the Helium lamp, with micrometer set at 4.0mm.
• Starting at 0.05s exposure time, increasing in 0.05s increments, finishing at 1s.
>
>
-- AshleaKemp - 27 Nov 2015

 META FILEATTACHMENT attachment="latex36aba0fbae95c73643ac90e9c7dde91c.png" attr="h" comment="" date="1448624412" name="latex36aba0fbae95c73643ac90e9c7dde91c.png" user="zxap013" version="1"
>
>
 META FILEATTACHMENT attachment="20151127_JB_AK.zip" attr="h" comment="" date="1448739982" name="20151127_JB_AK.zip" path="20151127_JB_AK.zip" size="74029750" user="zxap044" version="1"

#### Revision 227 Nov 2015 - AshleaKemp

Line: 1 to 1

 META TOPICPARENT name="StewartBoogertSpectroscopy2015"

# Additional Data Taking

• Previously, plotted against exposure time for one spectral line from the Helium lamp
• Noticed this plot was not smooth, and fluctuated quite significantly for different exposure time
Changed:
<
<
• To try and smooth out this plot, we are taking 5 frames at each exposure time and finding the average error, rather than just analysing the one.
>
>
• To try and smooth out this plot, we are taking 10 frames at each exposure time and finding the average error, rather than just analysing the one.

• Using the Helium lamp, with micrometer set at 4.0mm.
Changed:
<
<
• Starting at 0.05s exposure time, increasing in 0.05s increments.
>
>
• Starting at 0.05s exposure time, increasing in 0.05s increments, finishing at 1s.
-- AshleaKemp - 27 Nov 2015

#### Revision 127 Nov 2015 - AshleaKemp

Line: 1 to 1
>
>
 META TOPICPARENT name="StewartBoogertSpectroscopy2015"

# Additional Data Taking

• Previously, plotted against exposure time for one spectral line from the Helium lamp
• Noticed this plot was not smooth, and fluctuated quite significantly for different exposure time
• To try and smooth out this plot, we are taking 5 frames at each exposure time and finding the average error, rather than just analysing the one.
• Using the Helium lamp, with micrometer set at 4.0mm.
• Starting at 0.05s exposure time, increasing in 0.05s increments.

-- AshleaKemp - 27 Nov 2015

 META FILEATTACHMENT attachment="latex36aba0fbae95c73643ac90e9c7dde91c.png" attr="h" comment="" date="1448624412" name="latex36aba0fbae95c73643ac90e9c7dde91c.png" user="zxap013" version="1"

Copyright © 2008-2022 by the contributing authors. All material on this collaboration platform is the property of the contributing authors.
Ideas, requests, problems regarding RHUL Physics Department TWiki? Send feedback


text:
# Chemical Kinetics: Mechanisms | General Chemistry 2

Mechanisms in chemical kinetics are studied in this chapter: elementary reactions, intermediates, activation energy, Arrhenius equation, rate-determining step, reversible reactions, catalysis

## Mechanisms

Many reactions involve more than one step. They can be divided in elementary reactions

Elementary reactions:
- occur in a single step
- have reaction orders equal to the stoichiometric coefficients for each reactant

Intermediate: species produced in one step and consumed in a subsequent step
⇒ does not appear in the overall equation

Step (1): 2 NO → N2O2
Step (2): N2O2 + O2 → 2 NO2
Overall reaction (1) + (2): 2 NO + O2 → 2 NO2

2 NO + O2 → 2 NO2 can be divided in 2 elementary steps (1) and (2) and N2O2 is an intermediate.

rate law of (1) = k1 [NO]2
rate law of (2) = k2 [N2O2] [O2]

## Activation Energy

Activation energy Ea (in J.mol-1):
energy that must be provided to compounds to result in a chemical reaction

## Arrhenius Equation

k = Ae-Ea/RT

ln k = - $\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{RT}}$ + ln A

k = reaction-rate constant
Ea = activation energy (in J.mol-1)
R = molar gas constant = 8.314 J.mol-1.K-1
T = temperature (in K)
A = constant which depends on the reaction

## Rate-Determining Step

Rate-determining step: step which is much slower than any of the other steps
⇒ controls the overall reaction rate

Step (1): 2 NO2 → NO + NO3 (slow)
Step (2): NO3 + CO → NO2 + CO2 (fast)
Overall: NO2 + CO → NO + CO2

NO2 + CO → NO + CO2 is a two-step reaction and (1) is the rate-determining step
Law rate of the overall reaction = law rate of (1) = k [NO2]2

## Reversible Reactions

Reversible reactions: proceed in both the forward direction and the reverse direction
At equilibrium: rate of forward reaction = rate of reverse reaction

2 NO $⇌$ N2O4 is a reversible reaction
At equilibrium: k1 [NO]2 = k-1 [N2O4]

## Catalysis

Catalyst: substance that increases the reaction rate but is not consumed in the reaction
Homogeneous catalyst: same phase at the reaction mixture
Heterogeneous catalyst: different phase at the reaction mixture

A catalyst lowers the activation energy barrier to the reaction and thereby increases the reaction rate:


text:
# Better Sampling

A couple of days ago, I compared the images my ambient occlusion integrator produced with those of Modo using similar settings. I noticed immediately how much ‘cleaner’ the render from Modo was. Clearly there was an issue with the way I was picking my samples, so I set about improving things.

My approach for generating the ambient occlusion rays was to generate uniform random samples over the hemisphere about the normal. Based on two random numbers in the range [0,1), I calculate the normalized sample direction using the following function:

Vector3 Sample::UniformSampleHemisphere(float u1, float u2)
{
const float r = Sqrt(1.0f - u1 * u1);
const float phi = 2 * kPi * u2;

return Vector3(Cos(phi) * r, Sin(phi) * r, u1);
}

This generates points on a hemisphere from uniform variables u1 and u2, where each point has equal probability of being selected. The following image was generated with 256 random uniform samples:

It looks pretty noisy, that’s for sure. Part of the trouble comes from the fact that there’s no way to ensure that there’s an even distribution of the rays. A common way to alleviate this problem is to do stratified sampling instead of fully random sampling. The idea of stratified sampling is to split up the domain into evenly sized segments, and then to pick a random point from within each of those segments. You still get some randomness, but the points are more evenly distributed, which in turn reduces the variance. Less variance means less noise. Here’s the scene again, using 256 rays, but this time using stratified sampling:

As expected, it’s much less noisy, and for the same amount of computation!

## Sampling for Diffuse Monte Carlo Estimator

The stratified sampler helps out with the indirect diffuse lighting calculation too, but one other thing you can do to reduce noise for the Monte Carlo estimator is to choose random values that have a similar ‘shape’ to the integral you are estimating. Looking at the integral for diffuse reflections, you will see the familiar cosine term inside the integral:

$L_o = \int \frac{c}{\pi} L_i \cos \theta ,d \omega$

Where c is the diffuse material color, Li is the incoming radiance, and pi is the energy conservation constant.

Rather than wasting samples on areas of the integral where they will get mulitiplied out by the cosine term, why not just choose proportionally fewer samples in those areas?

Recall that the Monte Carlo estimator for an the integral of the function f(x), with probability density function p(x) is:

$F_N = \frac{1}{N} \sum_{i=1}^{N}{\frac{f(x_i)}{p(x_i)}}$

The probability density function is just a function that returns the probability that a particular value will be chosen. For the uniform hemisphere sampling function above, the pdf is just a constant, (1 / (2 * pi)). This makes the Monte Carlo estimator for the diffuse integral:

$L_o \approx \frac{2c}{N} \sum_{i=1}^{N}{L_i \cos \theta}$

Rather than mutliply by the cosine term above, we just want to generate proportionally fewer rays at the bottom of the hemisphere. The integral of the pdf over the hemisphere must equal one, so by switching to a cosine-weighted sample distribution, the pdf becomes (cos(theta) / pi).

This makes the estimator:

$L_o \approx \frac{c}{\pi N} \sum_{i=1}^{N}{\frac{L_i \cos \theta}{\frac{\cos \theta}{\pi}}}$

Which cleans up rather nicely to:

$L_o \approx \frac{c}{N} \sum_{i=1}^{N}{L_i}$

Normally I would post a couple of images up for comparison’s sake, but in this case, the difference is pretty difficult to perceive without being able to compare one on top of the other. The difference is small, but it is definitely worth it!

The common way to generate a cosine weighted hemisphere sampler is to generate uniform points on a disk, and then project them up to the hemisphere. Here’s some code:

Vector3 Sample::CosineSampleHemisphere(float u1, float u2)
{
const float r = Sqrt(u1);
const float theta = 2 * kPi * u2;

const float x = r * Cos(theta);
const float y = r * Sin(theta);

return Vector3(x, y, Sqrt(Max(0.0f, 1 - u1)));
}

Just by doing these two small steps, I’ve been able to clean up my images significantly. Here’s the scene from above again, this time with single bounce final gather with 256 rays, stratified cosine-sampled:

Next on my list is to take a look at path tracing, followed by irradiance caching (wasn’t that the point of all this?). This should allow me to get fairly cheap multi-bounce diffuse lighting.

## 13 thoughts on “Better Sampling”

1. Freeman

Hi,

Nice article

I just have a question about the randomly generated point on the hemisphere.
In your code, you never use the normal to generate it. How do you know that the hemisphere is centered on the normal ?

2. rory

This function just generates the ray direction in tangent space. It assumes that you will rotate the direction into world space afterwards.

I just use the normal and tangent vectors from the mesh (generating the bitangent from the cross product) and generate a basis (3×3 matrix) which transforms the ray from tangent space into world space.

3. ShinZo

You should change:

return Vector3(x, y, Sqrt(Max(0.0f, 1 – x * x – y * y)));

to

return Vector3(x, y, Sqrt(Max(0.0f, 1 – u1)));

because (x * x – y * y) = u1 in your example

4. Boris

to

return Vector3(x, y, Sqrt(1.0f – u1));

actualy

5. […] same topic is discussed here: http://www.rorydriscoll.com/2009/01/07/better-sampling/ This entry was posted in Uncategorized. Bookmark the permalink. ← A simple […]

6. Anon

ShinZo, I’m pretty sure (x * x – y * y) is not u1, it’s u1*cos(2*theta), or 2*x*x-u1 (which is just as costly as the original formula).

7. Anon

Hmm, maybe I’m wrong… bah, I’m too tired to work it out.

8. flemming

Agree with Boris and others. it should be
return Vector3(x, y, Sqrt(1.0f – u1))
since 1-(r*cos(t))^2-(r*sin(t))^2 =>
1 – r^2 =>
1-u1

9. Anon (same as before - I finally got it ;))

To make it a bit more explicit:

1 – x * x – y * y

1 – (r*cos(t))^2 – (r*sin(t))^2

1 – r^2*cos^2(t) – r^2*sin^2(t)

1 – r^2*(cos^2(t) + sin^2(t)) // Took out -r^2 as a factor

1 – (r^2*(1)) // Trig identities!

1 – r^2

1 – u1

Yay!

10. rory

Ok, ok! Thanks for the clear explanation. I changed the post.

11. Charlie

Hi Rory, I am still a bit confused by the transformation for the ray from tangent space into world space. Does it mean that I generate a normal vector using cosineSample(), and get the normal direction for a point from the mesh, then cross product them to obtain bitangent, finally get the transformation matrix? and multiply the matrix with the normal in tangent space then I can have the normal vector in world space, is this correct?

12. […] we will use cosine-weighted sampling. More information on cosine-weighted sampling can be found here. Below u_1 and u_2 are uniform random variables. Ultimately, we will reorient the resultant vector […]

13. […] little better. Let’s try 25 samples, evenly distributed around the hemisphere (incidentally see here for information on how to choose uniform sampling points on a hemisphere, using cosine weighting). […]


text: Why the mathematician call identification topology "quotient topology"?from my viewpoint,identification topology bear an analogy with difference(minus),not quotient(divide)...BTW,Could anybody tell me the reason why the mathematicians who coined the term "ring","ideal","soul" called them like that,is there any anecdote about this terminology...tell me sth about them...Thx...


text:
Thank you for visiting nature.com. You are using a browser version with limited support for CSS. To obtain the best experience, we recommend you use a more up to date browser (or turn off compatibility mode in Internet Explorer). In the meantime, to ensure continued support, we are displaying the site without styles and JavaScript.

# Electron phase-space control in photonic chip-based particle acceleration

## Abstract

Particle accelerators are essential tools in science, hospitals and industry1,2,3,4,5,6. Yet their costs and large footprint, ranging in length from metres to several kilometres, limit their use. The recently demonstrated nanophotonics-based acceleration of charged particles can reduce the cost and size of these accelerators by orders of magnitude7,8,9. In this approach, a carefully designed nanostructure transfers energy from laser light to the particles in a phase-synchronous manner, accelerating them. To accelerate particles to the megaelectronvolt range and beyond, with minimal particle loss10,11, the particle beam needs to be confined over extended distances, but the necessary control of the electron beam’s phase space has been elusive. Here we demonstrate complex electron phase-space control at optical frequencies in the 225-nanometre narrow channel of a silicon-based photonic nanostructure that is 77.7 micrometres long. In particular, we experimentally show alternating phase focusing10,11,12,13, a particle propagation scheme for minimal-loss transport that could, in principle, be arbitrarily long. We expect this work to enable megaelectronvolt electron-beam generation on a photonic chip, with potential for applications in radiotherapy and compact light sources9, and other forms of electron phase-space control resulting in narrow energy or zeptosecond-bunched beams14,15,16.

## Access options

from\$8.99

All prices are NET prices.

## Data availability

Source data for Fig. 4a, c are provided with the paper. The data in Fig. 4b, d that support the findings of this study are available in Zenodo with the identifier https://doi.org/10.5281/zenodo.4446542Source data are provided with this paper.

## References

1. 1.

Aad, G. et al. A particle consistent with the Higgs boson observed with the ATLAS detector at the large hadron collider. Science 338, 1576–1582 (2012).

2. 2.

Chatrchyan, S. et al. A new boson with a mass of 125 GeV observed with the CMS experiment at the large hadron collider. Science 338, 1569–1575 (2012).

3. 3.

Focus: Synchrotron techniques. Nat. Rev. Mater. https://www.nature.com/collections/vjzmtcbvzy (2018).

4. 4.

Bucksbaum, P., Möller, T. & Ueda, K. Frontiers of free-electron laser science. J. Phys. At. Mol. Opt. Phys. 46, 160201 (2013).

5. 5.

Karzmark, C. J. Advances in linear accelerator design for radiotherapy. Med. Phys. 11, 105–128 (1984).

6. 6.

Podgorsak, E. B. Radiation Oncology Physics: A Handbook for Teachers and Students (International Atomic Energy Agency, 2005).

7. 7.

Breuer, J. & Hommelhoff, P. Laser-based acceleration of nonrelativistic electrons at a dielectric structure. Phys. Rev. Lett. 111, 134803 (2013).

8. 8.

Peralta, E. A. et al. Demonstration of electron acceleration in a laser-driven dielectric microstructure. Nature 503, 91–94 (2013).

9. 9.

England, R. J. et al. Dielectric laser accelerators. Rev. Mod. Phys. 86, 1337–1389 (2014).

10. 10.

Wangler, T. P. RF Linear Accelerators 2nd edition (Wiley-VCH, 2008).

11. 11.

Chao, A. W., Mess, K. H., Tigner, M. & Zimmermann, F. Handbook of Accelerator Physics and Engineering 2nd edition (2013).

12. 12.

Courant, E. D. & Snyder, H. S. Theory of the alternating-gradient synchrotron. Ann. Phys. 3, 1–48 (1958).

13. 13.

Niedermayer, U., Egenolf, T., Boine-Frankenheim, O. & Hommelhoff, P. Alternating-phase focusing for dielectric-laser acceleration. Phys. Rev. Lett. 121, 214801 (2018).

14. 14.

Schönenberger, N. et al. Generation and characterization of attosecond microbunched electron pulse trains via dielectric laser acceleration. Phys. Rev. Lett. 123, 264803 (2019).

15. 15.

Black, D. S. et al. Net acceleration and direct measurement of attosecond electron pulses in a silicon dielectric laser accelerator. Phys. Rev. Lett. 123, 264802 (2019).

16. 16.

Niedermayer, U. et al. Low-energy-spread attosecond bunching and coherent electron acceleration in dielectric nanostructures. Phys. Rev. Appl. 15, L021002 (2021).

17. 17.

Panofsky, W. K. H. & Wenzel, W. A. Some considerations concerning the transverse deflection of charged particles in radio-frequency fields. Rev. Sci. Instrum. 27, 967 (1956).

18. 18.

Shimoda, K. Proposal for an electron accelerator using an optical maser. Appl. Opt. 1, 33–35 (1962).

19. 19.

Lohmann, A. Electron acceleration by light waves. IBM Tech. Note 5, 169–182 (1962).

20. 20.

Cesar, D. et al. High-field nonlinear optical response and phase control in a dielectric laser accelerator. Commun. Phys. 1, 46 (2018).

21. 21.

Leedle, K. J., Fabian Pease, R., Byer, R. L. & Harris, J. S. Laser acceleration and deflection of 96.3 keV electrons with a silicon dielectric structure. Optica 2, 158–161 (2015).

22. 22.

Kozák, M. et al. Optical gating and streaking of free electrons with sub-optical cycle precision. Nat. Commun. 8, 14342 (2017).

23. 23.

Leedle, K. J. et al. Phase-dependent laser acceleration of electrons with symmetrically driven silicon dual pillar gratings. Opt. Lett. 43, 2181 (2018).

24. 24.

McNeur, J. et al. Elements of a dielectric laser accelerator. Optica 5, 687–690 (2018).

25. 25.

Black, D. S. et al. Laser-driven electron lensing in silicon microstructures. Phys. Rev. Lett. 122, 104801 (2019).

26. 26.

Sapra, N. V. et al. On-chip integrated laser-driven particle accelerator. Science 367, 79–83 (2020).

27. 27.

Shiltsev, V. & Zimmermann, F. Modern and future colliders. Rev. Mod. Phys. 93, 015006 (2021).

28. 28.

Naranjo, B., Valloni, A., Putterman, S. & Rosenzweig, J. B. Stable charged-particle acceleration and focusing in a laser accelerator using spatial harmonics. Phys. Rev. Lett. 109, 164803 (2012).

29. 29.

Niedermayer, U., Egenolf, T. & Boine-Frankenheim, O. Three dimensional alternating-phase focusing for dielectric-laser electron accelerators. Phys. Rev. Lett. 125, 164801 (2020).

30. 30.

Kozák, M. et al. Ultrafast scanning electron microscope applied for studying the interaction between free electrons and optical near-fields of periodic nanostructures. J. Appl. Phys. 124, 023104 (2018).

31. 31.

Yousefi, P. et al. Dielectric laser electron acceleration in a dual pillar grating with a distributed Bragg reflector. Opt. Lett. 44, 1520 (2019).

32. 32.

Roques-Carmes, C. et al. Towards integrated tunable all-silicon free-electron light sources. Nat. Commun. 10, 3176 (2019).

33. 33.

Wiedemann, H. Particle Accelerator Physics (Springer, 2015).

34. 34.

Hirano, T. et al. A compact electron source for the dielectric laser accelerator. Appl. Phys. Lett. (2020).

35. 35.

Zhao, Z. et al. Design of a multichannel photonic crystal dielectric laser accelerator. Photon. Res. 8, 1586–1598 (2020).

36. 36.

Staude, I. et al. Waveguides in three-dimensional photonic bandgap materials for particle-accelerator on a chip architectures. Opt. Express 20, 5607–5612 (2012).

37. 37.

Egenolf, T., Niedermayer, U. & Boine-Frankenheim, O. Tracking with wakefields in dielectric laser acceleration grating structures. Phys. Rev. Accel. Beams 23, 054402 (2020). 0987654321

38. 38.

Kimura, W. D., Poaorelsky, I. V & Schächter, L. CO2-laser-driven dielectric laser accelerator. In 2018 IEEE Advanced Accelerator Concepts Workshop (AAC) https://doi.org/10.1109/AAC.2018.8659403 (2018).

39. 39.

Egerton, R. F. Outrun radiation damage with electrons? Adv. Struct. Chem. Imaging 1, 5 (2015).

40. 40.

Cros, B. & Muggli, P. Input to the European Particle Physics Strategy Update. in Advanced Linear Collider Study Group (ALEGRO collaboration) (2018).

41. 41.

Brüning, O. S. et al. LHC Design Report CERN-2004-003-V-1 (2004).

## Acknowledgements

We acknowledge discussions with the members of the Accelerator on a Chip International Program (ACHIP). We thank the clean-room facility staff at the Max Planck Institute for the Science of Light for continued assistance. We acknowledge funding by the Gordon and Betty Moore Foundation (#GBMF4744), ERC grants NearFieldAtto (#616823) and AccelOnChip (#884217) and BMBF projects 05K19WEB and 05K19RDE.

## Author information

Authors

### Contributions

T.C. and J.I. measured the data. R.S. and U.N. designed the structures and performed simulations. P.Y. fabricated the structures. J.I., R.S. and T.C. analysed the data. N.S. and A.M. inferred stringent tolerance requirements from initial measurements. J.I., R.S. and P.H. wrote the manuscript. P.H. supervised the experiment.

### Corresponding authors

Correspondence to R. Shiloh, J. Illmer or P. Hommelhoff.

## Ethics declarations

### Competing interests

The authors declare no competing interests.

Peer review information Nature thanks James Rosenzweig, Yelong Wei and the other, anonymous, reviewer(s) for their contribution to the peer review of this work.

Publisher’s note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

## Extended data figures and tables

### Extended Data Fig. 1 Photonic nanostructure on top of a mesa: over-focusing structure.

This SEM image shows a structure fabricated on top of a 60 μm-high mesa. The photonic structure for the over-focusing measurement is visible atop the mesa. The five gaps between the macro cells are also clearly visible. Here, 24 pillar pairs build one macro cell. Input (left) and output (right) apertures (thick blocks) are used for alignment during experiment. Additional pillars to the left of the input aperture act as markers that identify the specific structure during the experiment. The mesa allows us to focus the laser beam under 0° incidence angle from the side (see also Extended Data Fig. 1c).

### Extended Data Fig. 2 Schematic of the experimental set-up.

(a) Both IR (red) and UV (blue) laser pulses are generated with the help of an optical parametric amplifier (OPA). The UV pulses are focused onto the Schottky emitter of the SEM, where they release electron pulses. The electron pulses pass through the electron column and are focused into the nano-photonic channel. The IR pulses first pass a neutral-density (ND) filter for variable optical power attenuation before they traverse a delay stage, where the time delay between IR laser pulses and electron pulses is set. A bandpass filter limits the IR spectrum so that the laser pulses are stretched to a duration of 680 fs (FWHM). A cylindrical telescope is used to generate an elliptic laser beam with a 1:6 ratio. The beam is then split into two parts, where one part is used to monitor the power during measurements and the other is focused on the structure via an aspherical lens (ASL). The back reflection from the sample is used to align the laser beam to the structure. The electron energy is measured with a magnetic deflection spectrometer and an MCP detector with a phosphor screen, viewed from outside the vacuum chamber (indicated by the dashed line) with a CCD camera (not shown). For structures without a mesa (b) the incident angle of the laser beam is 5°, while for structures on top of the mesa (c) this angle is 0°.

### Extended Data Fig. 3 Photonic nanostructure on flat substrate: high contrast structure.

The scanning electron microscope (SEM) image shows the dual pillar transport structure for high contrast measurements. The dual pillar transport channel can be seen on the right as a colonnade structure. The four solid slabs left-above of the colonnades structure are a distributed Bragg mirror31. An alignment aperture is placed at the input of the structure (thick blocks). Electrons are focused into the colonnade structure, that is, into the channel between the rows of pillars. The laser beam impinges on the structure from the side, from bottom-right here, perpendicular to the pillars and with a 5° angle to the substrate (see also Extended Data Fig. 1b).

### Extended Data Fig. 4 Particle tracking simulation: One optical phase vs. all optical phases.

(a) Example particle trajectories for the optimal guiding field strength. The differently colored curves in between the pillars denote individual particle trajectories. The APF behaviour is clearly visible in the breathing of the envelope of the particle trajectories. For illustrative purposes the simulation was conducted with an electron pulse length of 0.001 fs. This way, the APF effect is apparent because electrons only sample fields of a small random fraction of the optical cycle, away from the crest. The lattice periodicity of 2 × (7.149 μm + 0.589 μm) = 15.47 μm (see Fig. 3), which is equal to the beta function periodicity, can be directly seen in the envelope of the trajectories, where the phase jump is manifested as a bending of the curve. Hence we can here directly assign F to macro cell 2, D to macro cell 3, and so on repeatedly (see numbering at the top of each macro cell, macro cell 1 consists of 2 half cells at the beginning and the end of the structure). (b) Same as (a) but with all optical phases uniformly sampled. The electron pulse (flat-hat here) is one optical period long. Evidently, the existence of the two fixed points in phase, π/2 and 3π/2, cannot be easily discerned close to the entrance of the structure, meaning that F and D cannot be uniquely assigned to each cell. Halfway through the structure, however, the lattice oscillations become visible again in the form of two “modes”, namely two trajectory classes, each directly linked to the lattice period again. The evolution into this two-mode structure is a consequence of the electrons accumulating around the two fixed points in phase space separated by π13. The result is an overlapping of two breathing motions in the trajectories shifted by one macro cell. (c) Same as (b) though the flat-hat electron pulse is here 400 fs long, the APF scheme still leads to particle propagation with hardly any loss, and the overall envelope nearly matches that of the single-cycle pulse in (b) (same as Fig. 4e of the main text). (d) Example particle trajectories with the over-focusing behaviour visible (same as Fig. 4f of the main text). Again, we only show particle trajectories of an electron pulse with a duration of 0.001 fs. Particles are obviously lost, where the lost particles’ deviation from the design axis exceeds the aperture of the structure. (e) Same as in (d) but with all phases sampled. A similar effect as depicted in (b) takes place, where the loss of particles at the structure boundary occurs at multiple locations. The reason is, again, that macro cells act differently depending on where the electron is with respect to the optical phase, and hence experience forces shifted in phase. Most importantly, the overall performance of the APF scheme is maintained, if not for the limitation of the beam size due to the structure aperture. The colours in panels a, c, and d were chosen so individual trajectories can be better discerned. The colours in b and e indicate the injection time of the electron (see colour bar).

## Rights and permissions

Reprints and Permissions

Shiloh, R., Illmer, J., Chlouba, T. et al. Electron phase-space control in photonic chip-based particle acceleration. Nature 597, 498–502 (2021). https://doi.org/10.1038/s41586-021-03812-9

• Accepted:

• Published:

• Issue Date:


text:
Review question

# Can we factorise $x^4 + Ax^2 + B$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7809

## Solution

Suppose that the equation $x^4 + Ax^2 + B = (x^2 + ax + b)(x^2 - ax + b)$ holds for all values of $x$.

1. Find $A$ and $B$ in terms of $a$ and $b$.
Expanding the right-hand side of the equation gives $x^4 + Ax^2 + B = x^4 + (2b-a^2)x^2 + b^2,$ so by equating coefficients we get \begin{align*} A &= 2b-a^2, \\ B &= b^2. \end{align*}
In fact, we have a difference of two squares here: \begin{align*} (x^2 + ax + b)(x^2 - ax + b) &=\bigl((x^2+b)+ax\bigr)\bigl((x^2+b)-ax\bigr)\\ &=(x^2+b)^2-(ax)^2\\ &=x^4+2bx^2+b^2-a^2x^2\\ &=x^4+(2b-a^2)x^2+b^2. \end{align*}
1. Use this information to find a factorisation of the expression $x^4 - 20x^2 + 16$ as a product of two quadratics in $x$.

We can write $x^4 - 20x^2 + 16 = (x^2 + ax + b)(x^2 - ax + b)$ if we can find solutions to $2b-a^2=-20, \qquad b^2 = 16.$

These equations have solutions \begin{align*} b&=4, & a&=2\sqrt{7} \\ b&=-4,\quad & a&=2\sqrt{3}. \end{align*}

We can use either one of these (we were only asked to find one factorisation). From the first solution, we get $x^4 - 20x^2 + 16 = (x^2 + 2\sqrt{7}x + 4)(x^2 - 2\sqrt{7}x + 4),$ and from the second solution, we get $x^4 - 20x^2 + 16 = (x^2 + 2\sqrt{3}x - 4)(x^2 - 2\sqrt{3}x - 4).$

1. Show that the four solutions of the equation $x^4 - 20x^2 + 16 = 0$ can be written as $\pm \sqrt{7} \pm \sqrt{3}$.

Take $x^4 - 20x^2 + 16 = 0.$ From above, this is equivalent to $(x^2 + 2\sqrt{7}x + 4)(x^2 - 2\sqrt{7}x + 4) = 0,$ so $x^2 + 2\sqrt{7}x + 4 = 0$ or $x^2 - 2\sqrt{7}x + 4 = 0.$ Now the first option leads to $x = -\sqrt{7} \pm \sqrt{3}$ (using the quadratic formula or completing the square), while the second gives $x = +\sqrt{7} \pm \sqrt{3}.$ Combining these we get $x = \pm \sqrt{7} \pm \sqrt{3}.$

Alternatively, if we begin with the factorisation $(x^2 + 2\sqrt{3}x - 4)(x^2 - 2\sqrt{3}x - 4)=0,$ we also get the same solutions $\pm \sqrt{7} \pm \sqrt{3}.$


text:
## Loser66 one year ago $$\sqrt{(x+3)^2+(y-2)^2}=\sqrt{(x-3)^2+y^2}$$ In the xy plane, the set of points whose coordinates satisfy the equation above is?? 1) A line 2) A circle 3) An ellipse 4) A parabola 5) One branch of a hyperbola Please, help

1. ganeshie8

Notice the distance formulas on left and right hand sides, the given equation is same as :  distance between (x,y) and (-3, 2) = distance between (x, y) and (3, 0)

2. ganeshie8

so you need to find the locus of points that stay at same distance from two points : (-3, 2) and (3, 0)

3. ganeshie8

Thats exactly the definition of perpendicular bisector!

4. Loser66

Got you, thank you so much.

5. Loser66

Damn!! They are distance formulas!! how can I not see it!!

6. ganeshie8

Haha solving them is fun too, square both sides and all quadratic terms cancel out.. gives the equation of perpendicular bisector..

7. Loser66

I did that way but stop at square both sides and look at the square of x and y, I think there must be other way to find the answer out quickly. So that I stopped and post the question. hehehe... I got it from you. Thank you so much

8. ganeshie8

np


text:
# How prove this there is an infinite number of primes in $S$

Let $S$ be the set of all rational numbers expressible in the form $$\frac{(a_1^2+a_1-1)(a_2^2+a_2-1)\ldots (a_n^2+a_n-1)}{(b_1^2+b_1-1)(b_2^2+b_2-1)\ldots (b_n^2+b_n-1)}$$ for some positive integers $n, a_1, a_2 ,\ldots, a_n, b_1, b_2, \ldots, b_n$. Prove that there is an infinite number of primes in $S$

and my teacher tell me this link post solution is not true,so How prove it? Thank you

-
Why do you think the given proof is not valid? –  apnorton Feb 4 '14 at 16:57


text:
# Measuring the vacuum permittivity

So I was reading the EF experiment that's used at the MIT to measure the vacuum permittivity and I was thinking about trying it just to see how it works:

http://ocw.mit.edu/courses/physics/8-02x-physics-ii-electricity-magnetism-with-an-experimental-focus-spring-2005/labs/experiment_ef.pdf

I have some questions about it and I hope you can help me with them:

1. It says "To find the electric force on the foil, assume that the charge density, σ , on the foil, is the same as that of the lower washer". How good that assumption is?

2. It says "Charges on the foil feel only horizontal forces from other charges on the bottom plate, so the vertical force on the foil is due to the electric field of just the top charge sheet". What horizontal forces are they talking about? I thought both the bottom plate and the upper plate were all exerting a vertical force on the foil and that's why I don't understand why the force is due just to the electric field of the upper plate $(V/2d)$ and not from the one of the two plates $(V/d)$.

3. Finally, why washers? I think it really doesn't matter if they're just two discs, does it?

-

## 1 Answer

1. The experiment calls for noting when the foil just lifts off the bottom plate. Since the foil and the washer are both presumably good conductors, the charge density should be the same until they stop touching. So the "goodness" of this assumption depends on the variation in the voltage across the capacitor after the foil lifts off.

2. The foil should feel forces from both plates, so I think your source is wrong in this. Horizontal forces come from the fact the plates are not infinite, and therefore some of the field goes beyond the plates.

3. Washers are both readily available and fairly good conductors, so they're used in a lot of "do it at home" type experiments.

-


text:
further inspects on vector space [closed]

If you have learned anything about linear algebra, checking whether a space is a vector space is the most fundamental task we must do in the first class. However, it always makes me feel tedious to do so. Therefore, I am asking for some cool or bizarre examples or some further explanations about what actually a vector space is to help me understand deeper what I am actually doing with checking the eight axioms. For example, associativity under addition is always trivial. But I can not imagine some spaces without associativity, or why associativity is so important ? Thank you!

closed as off-topic by Henricus V., Adam Hughes, Daniel W. Farlow, David K, Alex ProvostMar 29 '16 at 2:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Henricus V., Adam Hughes, Daniel W. Farlow, Alex Provost
If this question can be reworded to fit the rules in the help center, please edit the question.

• If the associativity of addition fails for a space, then it is not a vector space. – Henricus V. Mar 28 '16 at 23:31
• So, can you give me an example such that the associativity fails? Also, I am asking why associativity is so important so that it is one of the eight axioms? – Ivy Mar 29 '16 at 0:48
• Have you tried doing linear algebra without ever using associativity? Try it and see if you think that property is important. It's hard to tell where you want to go with this. – David K Mar 29 '16 at 2:15

The axioms of vector space is related to definition of operation. So, if you define the addition as subtraction then it is not associative,$$a-(b-c)\neq (a-b)-c$$ there are many operation that are not associative. But it should be noted that if a vector space is given, all its subsets (note! subsets) have associativity and other properties, and you have to check only closeness to addition and scalar multiplication to be subspace.


text:
## A framework for multi-robot motion planning from temporal logic specifications.(English)Zbl 1245.68227

Summary: We propose a framework for the coordination of a network of robots with respect to formal requirement specifications expressed in temporal logics. A regular tessellation is used to partition the space of interest into a union of disjoint regular and equal cells with finite facets, and each cell can only be occupied by a robot or an obstacle. Each robot is assumed to be equipped with a finite collection of continuous-time nonlinear closed-loop dynamics to be operated in. The robot is then modeled as a hybrid automaton for capturing the finitely many modes of operation for either staying within the current cell or reaching an adjacent cell through the corresponding facet. By taking the motion capabilities into account, a bisimilar discrete abstraction of the hybrid automaton can be constructed. Having the two systems bisimilar, all properties that are expressible in temporal logics such as linear-time temporal logic, computation tree logic, and $$\mu$$-calculus can be preserved. Motion planning can then be performed at a discrete level by considering the parallel composition of discrete abstractions of the robots with a requirement specification given in a suitable temporal logic. The bisimilarity ensures that the discrete planning solutions are executable by the robots. For demonstration purposes, a finite automaton is used as the abstraction and the requirement specification is expressed in computation tree logic. The model checker Cadence SMV is used to generate coordinated verified motion planning solutions. Two autonomous aerial robots are used to demonstrate how the proposed framework may be applied to solve coordinated motion planning problems.

### MSC:

 68T40 Artificial intelligence for robotics 68T20 Problem solving in the context of artificial intelligence (heuristics, search strategies, etc.) 68T27 Logic in artificial intelligence 68Q60 Specification and verification (program logics, model checking, etc.)

### Software:

LTLCon; Cadence SMV
Full Text:

### References:

 [1] Balch T, Arkin R C. Behavior-based formation control for multirobot teams. IEEE Trans Robot Autom, 1998, 14: 926–939 [2] Fierro R, Das A K, Kumar V, et al. Hybrid control of formations of robots. In: Proceedings of the 2001 IEEE International Conference on Robotics and Automation, Shanghai, 2001. 157–162 [3] Zachery R A, Sastry S S, Kumar V. Special issue on swarming in natural and engineered systems. Proc IEEE, 2011, 99: 1466–1469 [4] Klavins E, Koditschek D E. A formalism for the composition of concurrent robot rehaviors. In: Proceedings of the 2000 IEEE International Conference on Robotics and Automation, San Francisco, 2000. 3395–3402 [5] Alami R, Fleury S, Herrb M, et al. Multi-robot cooperation in the MARTHA project. IEEE Robot Autom Mag, 1998, 5: 36–47 [6] Gerkey B P, Mataric M J. Sold: Auction methods for multirobot coordination. IEEE Trans Robot Autom, 2002, 18: 758–768 [7] Egerstedt M, Hu X. A hybrid control approach to action coordination for mobile robots. Automatica, 2002, 38: 125–130 · Zbl 0997.93074 [8] Egerstedt M, Martin C F. Conflict resolution for autonomous vehicles: A case study in hierarchical control design. Int J Hybrid Syst, 2002, 2: 221–234 [9] Alur R, Esposito J, Kim M, et al. Formal modeling and analysis of hybrid systems: A case study in multi-robot coordination. In: World Congress on Formal Methods in the Development of Computing Systems, Toulouse, 1999. Goos G, Hartmanis J, Van Leeuwen J, eds. Lecture Notes in Computer Science, 1999, 1708: 212–232 [10] Koo T J, Sastry S. Bisimulation based hierarchical system architecture for single-agent multi-modal systems. In: Hybrid Systems: Computation and Control, Stanford, 2002. Lecture Notes in Computer Science, 2002, 2289: 281–293 · Zbl 1044.93502 [11] Antoniotti M, Mishra B. Discrete event models + temporal logic = supervisory controller: Automatic synthesis of locomotion controllers. In: Proceedings of the 1995 IEEE Conference on Robotics & Automation, Nagoya, 1995. 1441–1446 [12] Quottrup M M, Bak T, Izadi-Zamanabadi R. Multi-robot planning: A timed automata approach. In: Proceedings of the 2004 IEEE International Conference on Robotics & Automation, Barcelona, 2004. 4417–4422 [13] Fainekos G E, Kress-Gazit H, Pappas G J. Temporal logic motion planning for mobile robots. In: Proceedings of the 2005 IEEE International Conference on Robotics and Automation, Barcelona, 2005. 2020–2025 [14] Belta C, Isler V, Pappas G J. Discrete abstractions for robot planning and control in polygonal environments. IEEE Trans Robot, 2005, 21: 864–875 [15] Fainekos G E, Kress-Gazit H, Pappas G J. Hybrid controllers for path planning: A temporal logic approach. In: Proceedings of the 2005 IEEE Conference on Decision and Control and the European Control Conference, Seville, 2005. 4885–4890 [16] Kloetzer M, Belta C. A fully automated framework for Control of linear systems from LTL specifications. In: Hybrid Systems: Computation and Control, Santa Barbara, 2006. Lecture Notes in Computer Science, 2006, 3927: 333–347 · Zbl 1178.93067 [17] Kloetzer M, Belta C. LTL planning for groups of robots. In: Proceedings of the 2006 IEEE International Conference on Networking, Sensing, and Control, Ft. Lauderdale, 2006. 578–583 · Zbl 1178.93067 [18] Kloetzer M, Belta C. Automatic deployment of distributed teams of robots from temporal logic motion specifications. IEEE Trans Robot, 2010, 26: 48–61 [19] Chen Y, Ding X C, Stefanescu A, et al. A formal approach to deployment of robotic teams in an urban-like environment. In: Proceedings of the 10th International Symposium on Distributed Autonomous Robotics Systems (DARS), Lausanne, 2010 [20] Smith S L, Tumova J, Belta C, et al. Optimal path planning for surveillance with temporal-logic constraints. Int J Robot Res, 2011, 30: 1695–1708 [21] Fainekos G E, Girard A, Kress-Gazit H, et al. Temporal logic motion planning for dynamic robots. Automatica, 2009, 45: 343–352 · Zbl 1158.93369 [22] Clarke E M, Grumberg O, Peled D A. Model Checking. Cambridge: The MIT Press, 1999 [23] McMillan K L. Symbolic Model Checking. Norwell: Kluwer Academic Publishers, 1993 · Zbl 0784.68004 [24] McMillan K L. Getting started with SMV: User’s manual. Berkeley: Cadence Berkeley Laboratories, 1998 [25] Chutinan A, Krogh B H. Verification of infinite-state dynamic systems using approximate quotient transition systems. IEEE Trans Automat Contr, 2001, 46: 1401–1410 · Zbl 1031.93123 [26] Podelski A, Wagner S. Model checking of hybrid systems: from reachability towards stability. In: Hybrid Systems: Computation and Control, Santa Barbara, 2006. Lecture Notes in Computer Science, 2006, 3927: 507–521 · Zbl 1178.93077 [27] Pappas G J. Bisimilar linear systems. Automatica, 2003, 39: 2035–2047 · Zbl 1045.93033 [28] Koo T J, Clifton C A, Hemingway G. Cascaded control design for a quadrotor aerial robot. In: Proceedings of the 2006 Asian Control Conference, Bali, 2006. 989–993
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:
Contests Virtual Contests Problems Submit Runs Status Rank List Forum

2913.   Frequent values
Time Limit: 4.0 Seconds   Memory Limit: 65536K
Total Runs: 803   Accepted Runs: 255

You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are given several queries consisting of indices i and j (1 ≤ i ≤ j ≤ n). For each query, determine the most frequent value among the integers ai , ... , aj.

### Input

The input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the query.

The last test case is followed by a line containing a single 0.

### Output

For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.

### Sample Input

10 3
-1 -1 1 1 1 1 3 10 10 10
2 3
1 10
5 10
0

### Sample Output

1
4
3

Source: University of Ulm Local Contest 2007
Submit   List    Runs   Forum   Statistics

Tianjin University Online Judge v1.3.0
Maintance:Fxz. Developer: SuperHacker, G.D.Retop, Fxz


text:
## FANDOM

10,828 Pages

This blog post contains everything about Dollars Function, including FGH and numbers.

## What to solve first?

In order of importance:

• numbers with the lowest & level (Ultimate Array Notation III)

• numbers with least & symbols (Ultimate Array Notation II)

• many &'s solve on lowest to highest & order (Ultimate Array Notation I)

• brackets with the lowest arrow level (Extended Array Notation)

• brackets with the least number of entries (Array Notation)

• brackets with the lowest level (Extended Bracket Notation)

• least nested bracket with the smallest brackets in it (Bracket Notation)

2 _ * r halt
2 1 * r 2
2 [ * r 2
2 ] * l 3
3 * * l 3
3 _ _ r 4
4 1 _ l 5
4 _ _ l 7
4 [ 0 r 7
4 0 0 r 4
5 _ 1 r 6
5 1 1 l 5
6 1 1 r 6
6 _ 1 r 4
7 ] 0 r 9
8 * _ r 8
8 _ * r halt
9 _ * l 10
9 * * * 2
10 * _ l 10
10 1 1 l halt

## FGH Bracket Notation

Bracket Notation FGH Exact value
a$[0] $$f_1(a)$$ $$a2$$ a$[1] $$f_2(a)$$ $$2^aa$$
a$[2] $$f_3(a)$$ a$[3] $$f_4(a)$$
a$[[0]] $$f_{\omega}(a)$$ a$[0][[0]] = a$a[[0]] $$f_{\omega}(f_1(a))$$ a$[1][[0]] $$f_{\omega}(f_2(a))$$
a$[[0]][[0]] $$f_{\omega}(f_\omega(a))$$ a$[1[0]] $$f_{\omega+1}(a)$$
a$[[0][0]] $$f_{\omega*2}(a)$$ a$[[1]] $$f_{\omega^2}(a)$$
a$[[2]] $$f_{\omega^3}(a)$$ a$[[[0]]] $$f_{\omega^{\omega}}(a)$$
a$[[[[0]]]] $$f_{\omega^{\omega^{\omega}}}(a)$$ a$[[[...[[0]]...]] $$f_{\varepsilon_{0}}(a)$$

## Numbers Bracket Notation

Deal Group: 1000$[a] a means prefix, for example Glorious Deal = 1000$[100]

## FGH Extended Bracket Notation

Extended Bracket Notation FGH
$$[0]_2$$ $$\varepsilon_{0}$$
$$[[0]_2]$$ $$\varepsilon_{0}+1$$
$$[1[0]_2]$$ $$\varepsilon_{0}+2$$
$$[[0][0]_2]$$ $$\varepsilon_{0}+\omega$$
$$[[0]_2[0]_2]$$ $$\varepsilon_{0}*2$$
$$[[[0]_2]]$$ $$\omega^{\varepsilon_{0}+1}$$
$$[[[[0]_2]]]$$ $$\omega^{\omega^{\varepsilon_{0}+1}}$$
$$[1]_2$$ $$\varepsilon_{1}$$
$$[2]_2$$ $$\varepsilon_{2}$$
$$[[0]]_2$$ $$\varepsilon_{\omega}$$
$$[[0]_2]_2$$ $$\varepsilon_{\varepsilon_{0}}$$
$$[0]_3$$ $$\zeta_{0}$$
$$[[0]_3]$$ $$\zeta_{0}+1$$
$$[[[0]_3]]$$ $$\omega^{\zeta_{0}+1}$$
$$[[[[0]_3]]]$$ $$\omega^{\omega^{\zeta_{0}+1}}$$
$$[[0]_3]_2$$ $$\varepsilon_{\zeta_{0}+1}$$
$$[[[0]_3]_2]_2$$ $$\varepsilon_{\varepsilon_{\zeta_{0}+1}}$$
$$[1]_3$$ $$\zeta_{1}$$
$$[[0]_3]_3$$ $$\zeta_{\zeta_{0}}$$
$$[0]_4$$ $$\eta_{0}$$
$$[0]_{[0]}$$ $$\varphi(\omega,0)$$
$$[0]_{[0]_{[0]}}$$ $$\varphi(\varphi(\omega,0),0)$$
$$[0,1]$$

$$\varphi(1,0,0)$$

## Numbers Extended Bracket Notation

Dollars Group: 1000$[1000]_a No groups: BigDollars: 1000$[1000]_[1000]

Cash Group: 1000$$$[0 \rightarrow_{\{0\}\text{&}_a[0]} 1]$$ ## Ultimate Array Notation IV and higher ## Treasure Function It marks the ultimate limit of my notation. Current Definition: treasure(0) = 1 treasure(a) = a$$$[0 \rightarrow_{\{0\}\text{|0|}_{\{0\}\text{|0|}_{\{0\}\text{|0|}_{...}1}1}1} 1]$$

with treasure(a-1) nests

## Treasure Numbers

Bronze Treasure = treasure(2)

Silver Treasure = treasure(3)

Emerald Treasure = treasure(4)

Golden Treasure = treasure(5)

Diamond Treasure = treasure(1000)

Legend Treasure = treasure1000(1000)


text:
# Specifying plugins to local_costmap.yaml causes it to disappear

As an additional feature to my navigation stack, I'd like to add social_navigation_layers (http://wiki.ros.org/social_navigation...) to my local costmap, to distinguish between obstacles and humans. This should add extra inflation to the local costmap, such that my teb local planner better can navigate in the environment. For this I'm trying to specify layers of both the global and local costmap in the form of plugins as specified by this tutorial: http://wiki.ros.org/costmap_2d/Tutori....

The problem I'm facing is, that when specifying any plugin (including default) for my local costmap, the local costmap disappears from RViz and the topic /move_base/local_costmap/costmap returns only zero's. This ruins my navigation. If I comment out all plugins in the local costmap then everything works fine, besides that the social_navigation_layers won't be used. Specifying any and all plugins for the global costmap, then this seems to work fine. Using 'rosparam get' I can see, that the plugins are used for both the local and global costmap, despite it not working for the local costmap. If I do not specify any plugins for the local costmap then pre-Hydro parameters will be used, indicated by a warning in the terminal

Currently I'm running Melodic (Ubuntu 18.04), but I've also tested this on Kinetic (Ubuntu 16.04) with the same issue.

Picture of missing costmap. The small curved green line represents parts of an obstacle detected by the laser scanner and shown through teb markers.

UPDATE 1: Added .yaml files, and warning with pre-Hydro parameters.

UPDATE 2: Changed local_costmap.yaml to the way I want them to be, added output of 'rosparam get' and picture of missing costmap from RViz.

costmap_common.yaml

# Common settings for costmaps
# May be overwritten in specific global and local costmap files

# Robot Footprint Parameters
#footprint: [[0.55,0.5],[0.55,-0.5],[-0.55,-0.5],[-0.55,0.5]]   #Square robot

# Global Filtering Parameters
obstacle_range: 7
raytrace_range: 4.0

costmap_global.yaml

global_costmap:
# Coordinate frame and tf parameters
global_frame: map         # Test with /map as per documentation
transform_tolerance: 1.5

# Rate parameters
update_frequency: 0.5
# publish_frequency: 1.0

# Map management parameters
rolling_window: false
static_map: true

plugins:
- {name: static_layer,         type: "costmap_2d::StaticLayer"}
- {name: obstacles_layer,      type: "costmap_2d::VoxelLayer"}
- {name: inflation_layer,      type: "costmap_2d::InflationLayer"}

# Layer Definitions
static_layer:
map_topic: /map

inflation_layer:
# cost_scaling_factor: 10.0 # Default: 10.0

costmap_local.yaml

local_costmap:
# Coordinate frame and tf parameters
global_frame: odom
transform_tolerance: 1.0

# Rate parameters
update_frequency: 10.0
publish_frequency: 10.0

# Map management parameters
static_map: false
rolling_window: true
width: 10                 # The width of the map in meters. Default: 10
height: 10                # The height of the map in meters. Default: 10
resolution: 0.05          # The resolution of the map in meters/cell. Default: 0.05

plugins:
- {name: obstacles_layer,          type: "costmap_2d::ObstacleLayer"}
- {name: inflation_layer,          type: "costmap_2d::InflationLayer"}
- {name: social_pass_layer,        type: "social_navigation_layers::PassingLayer ...
edit retag close merge delete

I apparently don't have enough points to add pictures of my *_costmap_params.yaml files to this issue.

and that's a good thing, as .yaml files are simply text. There is no need to post screenshots of text.

Simply copy-paste it into your question, then select all the lines, press ctrl+k or the Preformatted Text button (ie: the one with 101010 on it).

As to your problem: YAML is very sensitive to indentation. Make sure to have consistent and correct indentation.

( 2020-04-08 04:10:53 -0500 )edit

Thank you for your response. I've updated the issue with the .yaml files. I've focused on the indentation, so I think it should be correct.

( 2020-04-08 04:45:39 -0500 )edit

Can you set the parameters the way you want them to be (i.e. uncomment the plugins) and then post the rosparam get / output.

( 2020-04-17 09:05:16 -0500 )edit

Thank you for your response. I've updated the issue with changes to the parameters, posted the output of rosparam get and inserted a picture of the missing obstacle.

( 2020-04-20 04:51:46 -0500 )edit

Can you change it to the output of rosparam get /move_base/local_costmap?

( 2020-04-20 12:41:11 -0500 )edit

Yes of course. I've updated the issue with both local and global rosparam get /move_base/*_costmap

( 2020-04-21 01:15:16 -0500 )edit

Sort by » oldest newest most voted

After comparing the result of rosparam get /move_base/local_costmap for the file costmap_local.yaml with specified plugins and without (pre-Hydro parameters) I noticed two issues:

1.The laser_scanner and observation_sources parameters were not correctly contained inside the obstacle_layer of the local costmap. To correct this I moved these two parameters from a separate .yaml file in which I specified them (loaded into the local_costmap ns inside my .launch file) and to the costmap_local.yaml specified in the following way:

Specified in costmap_local.yaml

  obstacle_layer:
observation_sources: laser_scanner
laser_scanner: {sensor_frame: scan_combined, data_type: LaserScan, topic: /laser_unified/scan, marking: true, clearing: true, expected_update_rate: 9.0}

Notice that this must be specified inside costmap_local.yaml and not costmap_common.yaml as I set static_map: true in the costmap_global.yaml. If placed inside costmap_common.yaml a warning similar to this will occur:

[ WARN] [1587471627.753143928, 471.157000000]: The origin for the sensor at (0.02, 0.02, -0.00) is out of map bounds. So, the costmap cannot raytrace for it.

2.Default or pre-Hydro layers call the obstacle layer in its singular form (without an s). Therefore, I changed the name of my obstacle layer plugin to be in the following form:

Specified in costmap_local.yaml

plugins:
- {name: obstacle_layer,          type: "costmap_2d::ObstacleLayer}

more

So you're all set?

( 2020-04-21 15:18:52 -0500 )edit

I’m set to the extend that I have the plugins working (including social_navigation_layers), and I can publish to /people (people_msgs/People) topic to insert humans and additional inflation into my local costmap. So the problem presented is this issue seems to be resolved. Next step/issue is to configure the inflation of this inserted human, such that my local planner (teb) will actually adjust its path, instead of going straight through it. I want a lethal/inscribed/high cost inflation to published humans to be approx 1 meter in radius. Currently, the inserted inflation of the human has no effect on the local trajectory, which can be seen here. Changing the amplitude parameter of social_navigation_layer only has minor effects to the inflation scaling. How would you change scaling of the inflated area, such that the robot avoids the published human?

( 2020-04-21 17:24:46 -0500 )edit

I wrote a whole paper about it: http://wustl.probablydavid.com/entry1... It's not easy.

( 2020-04-22 13:54:23 -0500 )edit

So you are confident that this is a tweaking issue, and not something incorrect in the way that I publish messages to the /people topic, or other potential errors in the way that I implement the social_navigation_layers? Published message can be seen here.

( 2020-04-27 02:48:08 -0500 )edit

After reading your paper: Given that we're using a standard algorithm (Dijkstra) for global planner and timed elastic band (TEB) as local planner, would you then argue that the method presented in the paper is still valid for our case, as we have no added constant value of P? Finally, are there any way of introducing lethal obstacles using social_navigation_layers, as I believe TEB only accounts for those?

( 2020-04-27 06:26:10 -0500 )edit


text:
# Smash product of pointed spaces is not associative

I've read many times that the smash product of pointed topological spaces is not associative, for example $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q})$ is not homeomorphic to $(\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$. How to prove this?

In general $\mathbb{N} \wedge X = \bigvee_{n>0} X$, so that $\mathbb{N} \wedge (X \wedge Y) = \bigvee_{n>0} (X \wedge Y)$ and $(\mathbb{N} \wedge X) \wedge Y = (\bigvee_{n>0} X) \wedge Y$. By the universal property of the wedge sum, we get a canonical continuous map $$\mathbb{N} \wedge (X \wedge Y) \to (\mathbb{N} \wedge X) \wedge Y,~(n,(x,y)) \mapsto ((n,x),y).$$ It is clearly bijective. For $X=Y=\mathbb{Q}$ it should not be open - why? Probably one has to take some open subset of $\mathbb{Q} \times \mathbb{Q}$ which approaches the base point $(0,0)$ in a weird way?

Can someone explain/supplement Joriki's proof?

-
If $X,Y,\mbox{ and }Z$ are all locally compact Hausdorff then the triple smash products agree. Note that the rationals based at $0$ are not locally compact. –  Sanath Devalapurkar Feb 28 '14 at 23:44
Have you seen the proof in section 1.5 of math.uiuc.edu/K-theory/0716/MaySig.pdf –  Niels Diepeveen Mar 2 '14 at 18:50
@Niels: Wow, that's exactly what I was looking for! When you post this as an answer, I will accept it. –  Martin Brandenburg Mar 2 '14 at 19:09

There appears to be a detailed proof due to Kathleen Lewis in section 1.5 of Parametrized Homotopy Theory by J.P. May and J. Sigurdsson.

-
Notice that they don't prove what they claim: They only prove that the canonical map $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is not a homeomorphism. But they claim that there is no homeomorphism at all. –  Martin Brandenburg Mar 2 '14 at 19:38
Good point. I wonder if there is an easy argument that there can be no other homeomorphism. –  Niels Diepeveen Mar 2 '14 at 19:42

Look at the following diagram (created using http://Presheaf.com. I didn't know how to do the
\Bbb-letters there)

Since both paths simply identify $(\Bbb N×\Bbb Q×\{0\})\cup(\Bbb N×\{0\}×\Bbb Q)∪(\{0\}×\Bbb Q×\Bbb Q)$ to a point, we have induced set maps $s$ and $\tilde s$, where the later is a bijection. Since the lower left arrow is a quotient map, $\tilde s$ is continuous if and only if $$s:(\Bbb N\wedge\Bbb Q)×\Bbb Q\to \Bbb N∧(\Bbb Q∧\Bbb Q)$$ is continuous. In order to show that it is not, we have to find an open set in $\Bbb N∧(\Bbb Q∧\Bbb Q)$ whose preimage is not open. Since the preimage of $V$ under $s$ is the same as the image of $q^{-1}(V)$ under $l×1$, this amounts to finding an open $q$-saturated set $U$ in $\Bbb N×\Bbb Q×\Bbb Q$ whose image is not open.
The product $\Bbb N×\Bbb Q×\Bbb Q$ is a disjoint union of $\Bbb Q×\Bbb Q$'s, one for each natural number, starting at $0$. A set in $\Bbb N×\Bbb Q×\Bbb Q$ is $q$-saturated if it contains the entire $(\Bbb Q×\Bbb Q)$-copy at $n=0$, and for each $n>0$ it contains the coordinate axes of $\Bbb Q×\Bbb Q$.
Consider the set $U:=(\{0\}×\Bbb Q×\Bbb Q)\cup\bigcup_{n>0} U_n$, where $$U_n=\{(n,x,y)\mid y<\pi/n\text{ or }y>x+\pi/n\}$$ Each $U_n$ is open and contains the axes, so $U$ is saturated and open. Its image $V=(l×1)(U)$ contains $(b,0)$, where $b$ denotes the base point of $\Bbb N∧\Bbb Q$. However, we can show that $(b,0)$ is not interior point of $V$:
For $(b,0)$ to be in $\text{int }V$, there had to exist an open neighborhood $B$ of $b$ and an $ϵ>0$ such that $B×[-ϵ,ϵ]⊆V$. One can think of $\Bbb N∧\Bbb Q$ as the wedge sum $\bigvee_{n>0}(\Bbb Q,0)$, countably many copies of $\Bbb Q,$ with all $0$'s identified. The set $B$ thus contains as a subspace the wedge $\bigvee_{n>0}((-\delta_n,δ_n),0)$, where all $δ_n>0$. Now for $k\in\Bbb N$ such that $\pi/k<ϵ$, the product $(-δ_k,δ_k)×[-ϵ,ϵ]$ is not contained in $U_k$.

This also shows that for a quotient map $q$ the product $q\times 1_Y$ need not be a quotient map if $Y$ is not locally compact. The reason why it fails here is that the neighborhood $K:=[-ϵ,ϵ]$ of $0$ in $\Bbb Q$ is not compact. Otherwise we could consider the set $W$ which is the largest subset of $\Bbb N×\Bbb Q$ such that $W×K⊆U$. This set turns out to be open and saturated. But here the openness would require that we can apply the Tube Lemma to $\{(k,0)\}×[-ϵ,ϵ]⊆U$, which is not possible as the interval isn't compact in $\Bbb Q$.

By the way, a smash product $X∧Y∧Z$ is associative if $X$ and $Z$ are locally compact, but also if two out of the three spaces are compact Hausdorff. The reason is that if $Y,Z$ are compact Hausdorff, then so is $Y∧Z$, which means that $p:Y×Z\to Y∧Z$ is perfect. For perfect $p$ a map $1_X×p$ is closed, hence a quotient map.

-
Thank you! This is very similar to the proof by Kathleen Lewis mentioned in the answer by Niels Diepeveen. But your construction of $U_n$ (there $V(\alpha/n)$) is simpler. –  Martin Brandenburg Mar 2 '14 at 20:36
I guess it is similar, from what I've seen in that document, though I haven't gone into the details, there. I was a bit disappointed to see that the bounty had already been awarded to the Niels' answer, which he posted when I was making up the proof. But that's the risk when you wait until the last hours of the grace period. –  Stefan Hamcke Mar 2 '14 at 20:47
@MartinBrandenburg: I suggest that you accept this answer instead of mine. –  Niels Diepeveen Mar 2 '14 at 21:58
@NielsDiepeveen: Thank you for passing on the bounty to me :-) –  Stefan Hamcke Mar 2 '14 at 23:16

The preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ of an open neighbourhood of the origin of $(\mathbb N\wedge\mathbb Q)\wedge\mathbb Q$ must contain a product of open neighbourhoods of the origins of $\mathbb N\wedge\mathbb Q$ and $\mathbb Q$, so its preimage in $\mathbb N\times\mathbb Q\times\mathbb Q$ must contain a set of the form $\{0\}\times\mathbb Q\times U$, where $U\subseteq\mathbb Q$ is a neighbourhood of $0$. However, $\{(n,x,y)\mid|xy|\lt1\}$ doesn't contain such a set, and its image in $\mathbb N\wedge(\mathbb Q\wedge\mathbb Q)$ is open.

-
Thank you. But I don't understand why the set must contain $\{0\} \times \mathbb{Q} \times U$; why not $\{0\} \times V \times U$ for some $V$? Also, I argued that $\mathbb{N} \wedge (\mathbb{Q} \wedge \mathbb{Q}) \to (\mathbb{N} \wedge \mathbb{Q}) \wedge \mathbb{Q}$ is continuous, but your proof seems to produce an open subset of whose preimage is not open. Therefore I am confused. –  Martin Brandenburg Jan 9 '13 at 23:32
@Martin: The preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ must contain $V\times U$ with $V\subseteq\mathbb N\wedge\mathbb Q$ containing the origin of $\mathbb N\wedge\mathbb Q$, and the preimage of that origin in $\mathbb N\times\mathbb Q$ contains $\{0\}\times\mathbb Q$, so the preimage in $\mathbb N\times\mathbb Q\times\mathbb Q$ of the preimage in $(\mathbb N\wedge\mathbb Q)\times\mathbb Q$ must contain $\{0\}\times\mathbb Q\times U$. Regarding the continuity, you not so much argued as claimed this :-) Why do you believe that map is continuous? –  joriki Jan 10 '13 at 0:17
@Martin: Actually I think I didn't really understand what you're doing with that map. Does $\bigvee$ denote the disjoint union? Then I don't think that's equivalent, since in $\mathbb N\wedge\mathbb Q$ all of $\{0\}\times\mathbb Q$ is collapsed in the origin whereas in the disjoint union all copies look the same and are uncollapsed. –  joriki Jan 10 '13 at 0:27
If $X$ is locally compact, then $- \times X$ preserves quotient maps, right? And this would then imply that it also preserves wedges. –  Martin Brandenburg Jan 16 '13 at 13:25
There is also a result by Puppe (in Math. Zeitschrift, Vol. 69) which says that the smash product is associative for well-pointed spaces (where the inclusion of the base point is a cofibration). –  Martin Brandenburg Jan 16 '13 at 13:54


text:
# Rotational motion (confused about the variables)

we have many accelerations in rotational motion ,and I don't know the difference between them. first the linear acceleration which as I understand is the vector acceleration which has the same direction as the change of velocity and it affects the speed and direction then it's two components are centripetal acceleration which changes the angle and is towards the center and it doesnt change the magnitue of the velocity. Then there is the tangential acceleration which is perpendicular to the centripetal acceleration and it increases the magnitude of the velocity.
First is my understanding of the acceleration correct ? second I also see that we have the angular acceleration and it changes the the the angular velocity but this confuses me what is the difference between it and the centripetal acceleration? they both change the angle.

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
Here's how I'd put it. Something moving in a circle has a linear (or translational) acceleration. That acceleration has two components: a centripetal component and a tangential component.

What is usually called the angular acceleration is the rate of change of angular speed. The angular speed relates to the tangential speed via: ω = r*vt. In turn, the angular acceleration relates to the tangential acceleration via: α = r*at.

Does that help?

but what I don't understand is whether the centripetal acceleration and the angular acceleration are measuring the same thing or not? I mean they all affect the change of the direction of the velocity right?

Doc Al
Mentor
but what I don't understand is whether the centripetal acceleration and the angular acceleration are measuring the same thing or not?
No. Centripetal acceleration is the rate of change in linear velocity (for constant speed). Angular acceleration is the rate of change of the angular speed. For something moving with constant speed, there will be a non-zero centripetal acceleration while the angular acceleration will be zero.

No. Centripetal acceleration is the rate of change in linear velocity (for constant speed). Angular acceleration is the rate of change of the angular speed. For something moving with constant speed, there will be a non-zero centripetal acceleration while the angular acceleration will be zero.
oh thanks I see it now but if angular acceleration is the rate of change of angular speed then what is the rate of change of angular velocity?

Doc Al
Mentor
oh thanks I see it now but if angular acceleration is the rate of change of angular speed then what is the rate of change of angular velocity?
As long as the axis of rotation doesn't change, as when you stick to two dimensions, angular speed and velocity are the same (for practical purposes). When you deal in three dimensions, and the axis of rotation can change, things get complicated fast. Angular velocity is a vector (pseudovector, really) whose magnitude is the angular speed and whose direction is along the axis of rotation. See: http://en.wikipedia.org/wiki/Angular_velocity" [Broken]

Last edited by a moderator:
but isnt the rotation negative if it is clockwise?

Doc Al
Mentor
but isnt the rotation negative if it is clockwise?
That's the convention, sure.


text:
# Pluripotential Monge-Amp{\e}re flows in big cohomology class

Abstract : We study pluripotential complex Monge-Amp\ere flows in big cohomology classes on compact K{\"a}hler manifolds. We use the Perron method, considering pluripotential subsolutions to the Cauchy problem. We prove that, under natural assumptions on the data, the upper envelope of all subsolutions is continuous in space and semi-concave in time, and provides a unique pluripotential solution with such regularity. We apply this theory to study pluripotential K{\"a}hler-Ricci flows on compact K{\"a}hler manifolds of general type as well as on K{\"a}hler varieties with semi-log canonical singularities.
Document type :
Preprints, Working Papers, ...
Domain :

https://hal-univ-tlse2.archives-ouvertes.fr/hal-03466537
Contributor : Quang-Tuan Dang Connect in order to contact the contributor
Submitted on : Monday, December 6, 2021 - 12:55:34 AM
Last modification on : Tuesday, January 4, 2022 - 5:58:32 AM

### Identifiers

• HAL Id : hal-03466537, version 1
• ARXIV : 2102.05189

### Citation

Quang-Tuan Dang. Pluripotential Monge-Amp{\`e}re flows in big cohomology class. 2021. ⟨hal-03466537⟩

### Metrics

Les métriques sont temporairement indisponibles


text:
# State of DData

Carl Witty cwitty at newtonlabs.com
Fri Mar 26 12:46:11 EST 2004

On Fri, 2004-03-26 at 11:15, John Meacham wrote:
> A specified bias is very important for reasoning about time/space
> efficiency. I think it is quite important.
>
> plus, even though (==) means equality, every now and again it comes in
> handy to cheat a little :)

Given a Map type with a specified value bias, you can construct a Set
type with this bias with something like:

type Set t = Map t t
set_insert t set = map_insert t t set
set_iterate f set = map_iterate (\key value -> f value) set

Of course, it would be simpler to just document the bias on an existing
Set implementation. :)

Carl Witty


text:
# Electric field using dielectric constant

Tags:
1. May 24, 2015

### Issy

1. The problem statement, all variables and given/known data
What is the magnitude of the electric field due to a 6.0 x 10-9 C charge at a point located 0.025m from the charge? The charge and the point in question are located underwater (κ(water) = 80).

2. Relevant equations
I know that E = magnitude of electric field = kq/r2, where k = Coulomb's constant of 9 x 109, and I also know that κ = kappa = dielectric constant aka relative permittivity of a dielectric material, which we use in the equation C = capacitance = κεoA/d.

However, I can't workout how to use κ to find electric field strength, because I don't see how a capacitor is involved in this situation? And with the three variables I have been given (the size and electric polarity of charge, the distance, and the dielectric constant of water), I also don't know of any formulas I can use that don't require any other variables.

The practise test I'm working on is multichoice, and the answers state that the final answer is equal to 1.1 x 103 N/C, but I'm not sure how to work backwards from that either. Any help would be greatly appreciated! We have a test later this week, and I'm sure Murphy's Law will make sure that unless I get my head around it, a question like this will no doubt be in the exam, so any and all advice will be welcomed!

2. May 24, 2015

### ash64449

use this equation E=Kq/kr^2,where

K= 9 * 10^9
k=dielectric constant
q=charge
r=distance in question

Edit: Answer is 1.08 * 10^3 N/C

Equation reference: H C Verma Concepts Of Physics

Last edited: May 24, 2015
3. May 24, 2015

### ash64449

Do you need any proof of this equation?

4. May 24, 2015

### Issy

Thank you very much! So, is it fair to say that dielectric constant (ie Kappa) is actually always used in the E=kq/r2 equation when the surrounding environment is not specified? Which would mean it's assumed that it's just air, which has a dielectric constant of 1, so doesn't actually affect the equation?
...I guess what I'm asking, in a roundabout way, is should I use that equation for any question where the surrounding environment of the charge is not air?
Thank you again for your help, I had no idea how to approach it, you've saved me a lot of confusion!

5. May 24, 2015

### ash64449

Yes.. You are right! Thank you! you are welcome!

6. May 24, 2015

### ash64449

There is one thing that you should know... Once you use kappa in the equation.. you should never use the bound charge appearing due to polarization of the dielectric.

only free charge should be used..

7. May 25, 2015

### rude man

Use a gaussian surface around your charge, and Gauss's law.
Now you can handle any combination of charge and observation point being surrounded by various dielectric material.
.

8. May 25, 2015

### Issy

I'm going to bookmark this thread, because while I currently have no idea about Gauss' Law, no doubt I'll find out about it next semester, or soon enough. Thank you for your advice! Sometimes it seems like every time I learn a new physics concept, I become aware of half a dozen others that I also need to learn - but it makes my degree more entertaining if I think of it as fighting a Hydra! Thank you again

9. May 25, 2015

### mooncrater

The constant K we use in finding the electric field is= $$\frac {1}{4\pi \epsilon_0}$$
So , when you are talking about an dielectric just replace the $\epsilon_0$ part with $k\epsilon_0$ where k is the dielectric constant , just the way you do for the capacitors. Thats why you get that k in the denomimator of the formula.

10. May 25, 2015

### rude man

Har! Are you going for a physics major or something like "physical sciences" where calculus-level physics is not covered? My school had that option & most others do too I'm sure. So you may or may not run into Dr. Gauss.


text:
# X Terminates in What Quadrant?

• April 29th 2009, 02:26 PM
ThaBeretta
If $(sec x - 2)(2 sec x - 1) = 0$ , then x terminates in:

a) Quadrants I and II, only

b) Quadrants I and IV, only

c) Quadrants I, II, III, and IV

• April 29th 2009, 02:28 PM
VonNemo19
hello
Solve for x, and determine the domain my son.(Wink)
• April 29th 2009, 02:34 PM
ThaBeretta
Hmm...I'm really not sure how to go about solving for x. Also, how would I determine the domain once I have x?
• April 29th 2009, 02:41 PM
VonNemo19
When you solve a quadratic equation, in some cases you factor. Right.

Well your problem looks a lot like something like this: (X+a)(X+b)=0

and the solutions are: x=-a, x=-b

Can you see my meaning?
• April 30th 2009, 03:31 AM
ThaBeretta
Quote:

Originally Posted by VonNemo19
When you solve a quadratic equation, in some cases you factor. Right.

Well your problem looks a lot like something like this: (X+a)(X+b)=0

and the solutions are: x=-a, x=-b

Can you see my meaning?

So the domain would be the point (2,1), right? That point is located in the first quadrant, giving me the answer of "d".
• April 30th 2009, 03:43 AM
Sundae
Let secx = y
therfore, y = 2 & y = 1/2
Because both values are positive they are in the 1st and 4th quadrant.

in the first quadrant everything is positive, in second quadrant only sin/cosec is positive, in the third quadrant only tan/cot is positive and in 4th quadrant cos/sec is positive.

Is that what u were askin?
• April 30th 2009, 09:25 AM
ThaBeretta
So the correct answer is choice "b"?
• April 30th 2009, 09:30 AM
VonNemo19
yes
yes
• April 30th 2009, 09:58 AM
ThaBeretta
Okay, thank you.


text:
0

When the GCF of the numerator and the denominator of a fraction is 1?

Wiki User

2012-05-24 17:42:36

The fraction is in its simplest form.

Wiki User

2012-05-24 17:42:36
Study guides

20 cards

➡
See all cards
3.81
2278 Reviews


text:
对于初次参加SAT的考生来说，备考阶段的规划是很重要的，然而有些学生反应不知道该如何规划时间进行有效的复习，那么为了帮助大家更好的备战SAT考试，智课网小编为大家带来了关于SAT写作的相关内容，更多备考的精彩内容，欢迎关注智课网SAT频道。

Walter P. Chrysler(1875-1940)

Walter Percy Chrysler is the founder of what is today known as the third largest North American automaker: the Chrysler Corporation. Walter Chrysler was born on April 2, 1875, in Wamego, Kansas. Several years later the Chrysler family moved to the western Ellis, Kansas. His father worked for the railroad industry as an engineer. It was at the shops of the Kansas Pacific Railroad, where he spent much time with his father as a young boy, that he developed an acute interest in the workings of a locomotive. Walter attended the public school system of Ellis where he finished high school. He then became a machinist's apprentice at the Union Pacific Railroad.

At the age of eighteen he designed and built a miniature steam locomotive that ran on an eighth of a mile track that he also built. He made many of his own possessions because it was not enough to just know how to use his tools but how to make them as well. Mechanical engineering became a way of life not just a profession. The young machinist was well on his way to making a name for himself.

Walter P. Chrysler's automotive career started in 1911, as a Works Manager for Buick in Flint, Michigan. Buick was the cornerstone on which William C. Durant built General Motors in 1908.

W. C. Durant had lost the presidency of General Motors in 1910. Mr. Chrysler went to work resolving many engineering and production problems at Buick. His reputation quickly included a tremendous ability in automaking and sound management. In 1916, W.C. Durant regained the presidency of General Motors and fired Charles W. Nash who became the president three years earlier. Walter Chrysler was made President and General Manager of Buick. Durant found out that Mr. Chrysler was considering an offer to take over Packard Motors. Durant offered him a raise from his salary of $50,000 a year to$500,000 a year to stay at Buick. He accepted.

He was also named General Motors' Vice President in charge of production. Now the other divisions within GM had the fortune of Mr. Chrysler's expertise. His next promotion put him in the office of Executive Vice President of General Motors. He now worked closely with W. C. Durant. He became troubled with Durant's style of business: constant over-expansion. Mr. Chrysler felt this was a path to imminent corporate disaster. It was this disagreement in business philosophy that lead Mr. Chrysler to retire from General Motors in 1920. His reputation in the automotive industry now was immense. He had regained Buick's name as a synonym for soundness and quality.

He had increased production from forty cars a day to over five hundred cars a day. This all helped GM reach a profit of almost fifty million dollars annually. Later, William C. Durant was forced to resign by the bankers that held General Motors' eighty million-dollar debt. Walter P. Chrysler retired from the automotive industry in 1920, at the age of forty-five, financially independent and since then his great career began.

克莱斯勒(1875-1940)

克莱斯勒从小就表现出了在机械方面的天赋， 18 岁时，他制造了一辆微型蒸汽车，虽然这辆车更像是一件玩具，但却“五脏俱全”，可以在后院里他专门铺设的轨道上行驶，20 岁那年，他被一家工厂聘为机械师，可以拿到一份令人羡慕的薪金。但是，克莱斯靳具有一种对任何事情都十分好奇的正确性，不愿意始终呆在一个岗位上，总想寻找其他发展自己的机会。

年轻的克莱斯勒到底换过多少次工作，恐怕连他自己都无法记得清楚，直到33 岁那年，才相对稳定地受聘担任了芝加哥西部铁路的动力总负责人。

1910 年，克莱斯勒辞掉了年薪12000 美元的工作，受聘担任了通用汽车公司别克分部中一家工厂的技术经理，年薪只有6000 美元。由于他精通机械、技术超群，在通用公司的作用越来越重要，1912 年年薪增加到2.5 万美元，1915 年增加到5 万美元。“通用”一心一意想留下他为公司效力，但克莱斯勒本人却产生了离开“通用”，独自去干一番事业的想法。

正在此时，杜兰特重返“通用”，他为了振兴“通用”，急于招揽一批人才时，对其竭力挽留，不仅委任他担任了别克部的主要负责人和公司第一副总经理，而且还将其年薪一下子提高到50万美元。然而，由于克莱斯勒与杜兰特难以合作，他还是于1920 年3 月25 日离开了“通用”。

经过辗转，克莱斯勒终于促成立了自己的公司。汽车公司成立心以后，发展极其迅速。相继推出的“克莱斯勒4 号”和“系列58”两种新车为其发展做出了贡献。公司在1925 年的国内排名只有27 位，1926 年末升至第5 位，1927 年则又上升至第4 位。1928 年通过股票交易的方式买下了道奇公司和普利茅斯汽车公司。道奇公司当时在美国排名第三，有良好的商誉和可靠的销售网，买下它之后，克来斯勒在1929 年即跃升为美国三大汽车公司之一，后来还曾有过超过福特位居第二位的辉煌。1930-1934 年间，克莱斯勒相继成立了安普莱部和爱尔坦普部，从事轴承和散热器的生产。

以上内容由智课小编为大家整理的关于“SAT写作素材积累克莱斯勒”，希望可以帮助到大家，小编在此预祝大家在考试中取得好成绩!


text:
The category of finite locally-free commutative group schemes

I'm trying to understand the properties of the category $FL/S$ of finite locally-free commutative group schemes over an arbitrary base-scheme $S$. I know it is not in general an abelian category: Over the integers the morphism $\mathbb{Z}/2\mathbb{Z}\to\mu_2$ given by the ring homomorphism $\mathbb{Z}/(T^2-1)\to\mathbb{Z}\times\mathbb{Z}$ with $T\mapsto (1,-1)$ is a monomorphism and an epimorphism in $FL/\mathbb{Z}$ but it is clearly not an isomorphism.

What I don't know: Does every morphism in $FL/S$ have a kernel and/or a cokernel?

One has a notion of short exact sequences in $FL/S$: If $f:G'\to G$ and $g:G\to G''$ are morphisms in $FL/S$ we say that the sequence $0\to G'\to G\to G''\to 0$ is exact if its image under the embedding of $FL/S$ into the category of abelian fppf-sheaves is exact. How can we detect in $FL/S$ whether a sequence is exact? Specifically: If $f$ is a kernel of $g$ and if $g$ is a cokernel of $f$ (both in $FL/S$), is the corresponding sequence exact?

I'd appreciate (references for) answers to any of these questions.

-
What does locally free mean in this context? –  Mariano Suárez-Alvarez Dec 3 '09 at 19:23
We consider commutative group schemes G over S such that the structural morphism f:G->S is finite locally free, i.e. it is affine and the direct image f_* O_G is a finitely generated locally free O_S module. –  fls Dec 3 '09 at 19:34

If the base $S$ is Spec of a Dedekind domain, say $A$, and one restricts to finite flat groups schemes with etale generic fibres (which is no restriction if the fraction field of $A$ has char. 0), then one can form kernels and cokernels in the category of finite flat group schemes by taking scheme-theoretic closures of the corresponding notions on the generic fibre, and this can be useful. (Of course, one doesnt' get an abelian category: the map ${\mathbb Z}/2 {\mathbb Z} \rightarrow \mu_2$ has trivial kernel and cokernel, but is not an isomorhpism.)

An illustration of how this method can be used is given in e.g. in my joint paper with Calegari On the ramification of Hecke algebras at Einstein primes.

(The point of the restriction to a Dedekind base is that in this case, torsion free, which is what you get with scheme-theoretic closure from the generic fibre, coincides with flat.)

-

Voici tout ce qu'il y a à savoir sur cette catégorie FL/S.

En toutes généralités, c'est une catégorie exacte au sens de Quillen. Plus précisément, le plongement de FL/S dans la catégorie des faisceaux de groupes abéliens fppf sur $S$ font de FL/S une sous-catégorie stable par extensions dans cette catégorie abélienne de faisceaux fppf. Dans une catégorie exacte on dispose d'une notion de monomorphismes et épimorphismes stricts: ce sont ceux qui peuvent s'insérer dans une suite exacte. Alors, si $f:G\rightarrow H$ est un morphisme de schémas en groupes, $f$ est un monomorphisme strict si et seulement si c'est une immersion fermée. De plus, $f$ est un épimorphisme strict si et seulement si c'est un morphisme fidèlement plat.

En général cette catégorie exacte n'est pas abélienne. Comme rappelé précédemment c'est cepdendant le cas si $S$ est le spectre d'un corps.

Maintenant supposons que $S$ soit le spectre d'un anneau de valuation d'inégales caractéristiques que je note $\mathcal{O}_K$.

Lorsque $e_{K/\mathbb{Q}_p} < p-1$ Raynaud a montré que c'est une catégorie abélienne. De plus, le foncteur fibre générique $G\mapsto G\otimes K$ est pleinement fidèle et identifie FL/S à une sous-catégorie abélienne de FL/$spec(K)$ (i.e., après un choix d'une clôture algébrique $\overline{K}$ de $K$, une sous-catégorie abélienne de la catégorie des $Gal(\overline{K}|K)$-modules discrets finis en tant que groupe abélien).

Lorsque $e \geq p-1$ le résultat précédent est faux. Néanmoins on a le résultat suivant: dans FL/S tout morphisme possède un noyau et un conoyau. Plus précisément, si $f$ est un morphisme dans FL/S alors le platifié de $\ker f$ ($\ker f =$ noyau usuel dans la catégorie des schémas en groupes non-nécessairement plats) (platifié= on tue la $p$-torsion) est un noyau dans $FL/S$ du morphisme $f$. On construit de même l'image de $f$ comme adhérence schématique de l'image en fibre générique. Cependant, la catégorie précédente n'est pas abélienne. Soit en effet $K=\mathbb{Q}_p (\zeta_p)$ et $$f: \mathbb{Z}/p\mathbb{Z} \longrightarrow \mu_p$$ le morphisme qui à $\bar{1}\in \mathbb{Z}/p\mathbb{Z}$ associe $\zeta_p\in \mu_p$. Alors, $f$ est un isomorphisme en fibres génériques. On en déduit que dans FL/S les noyaux et conoyaux de $f$ sont nuls. Ce n'est cependant pas un isomorphisme !

-

Even if kernels (etc.) exist in the category of finite flat commutative group schemes over a scheme, they won't have good properties, and so probably won't be interesting. For example, they won't commute with base change: the kernel of $Z/2Z\to\mu_2$ over Z is zero, but over $F_2$ is is $Z/2Z$.

On the other hand, for a homomorphism whose kernel is flat, everything works as expected: the kernel, cokernel, image, and co-image exist, and the map from the co-image to the image is an isomorphism.

In the category of group schemes of finite type over a field, the isomorphism theorems from group theory hold. In particular, kernels exist, quotients by normal subgroups exist, and the the map from the co-image to the image is an isomorphism (this fails when you don't allow nilpotents in the coordinate rings, which is why you won't find it in any of the standard texts Borel, Humphreys, Springer). From this it follows immediately that the category of commutative group schemes of finite type over a field is abelian.

-
I suspect that there are some kernels and cokernels in $FL/S$. Moreover, the kernels should correspond to those in the category of fppf-sheaves (since you consider only FLAT schemes over S). The situation with cokernels is worse.
For your second question the answer is probably "Yes" (though maybe one needs some restrictions on S). I would suggest you to look at the generic fibre of your exact sequence. There is a paper Raunad M. Schemas en groupes de type $(p,\dots,p)$} Bull. Soc. Math. France. 1974. v. 102. P. 241--280. Raynaud proves that (under some restrictions) all group subschemes and factor-schemes of the generic fibre of G could be lifted to subschemes and factor-schemes of G itself in a unique way. You should obtain all G' and G'' coming from all possible short exact sequences (with G in the middle) this way.


text:
4

# The unit of equilibrium constant of a chemical reaction is the same as that of

Molar concentration

Temperature

Internal energy

None of these