text:
Chapter 66

The importance of parenchymal preservation during pulmonary surgery was realized over 50 years ago when descriptions of bronchial resection and reconstruction were first published.1–3 Since then, considerable technical refinement and anatomic insight have greatly expanded indications for lung-sparing operations. Bronchoplastic resections form one category of these procedures and have a unique set of indications. There is no question that these operations are technically more demanding than standard anatomic pulmonary resections, yet, by appreciating their benefits, the additional time spent performing these procedures is justly rewarded.

The terms bronchoplasty and bronchoplastic resection have been applied to a wide variety of operations of either main or lobar bronchi. The operations usually involve concomitant parenchymal resection. Indications for resection and reconstruction of the bronchus alone are rare. Although some thoracic surgeons have interchanged right upper lobe bronchoplasty with right upper lobe sleeve resection, a lobar bronchial orifice (e.g., right upper lobe) occasionally can be reconstructed without resecting a sleeve of main bronchus.

The standard indication for bronchoplastic resection is an intrabronchial lesion emanating from either the main bronchus itself or a lobar bronchus with main bronchus encroachment. The extent of the diseased area must allow for reconstruction after the resection. Etiology is often a low-grade neoplasm such as typical carcinoid or mucoepidermoid4 or, rarely, isolated bronchial stenosis secondary to granulomatous disease, trauma, foreign body, or benign neoplasm.5,6 Bronchoplastic resections can be applied to more invasive cancers (non-small cell lung cancer (NSCLC) or metastases), but oncologic principles, namely, achieving a complete resection (including a negative margin), must be followed.7 The naive mind-set that a positive margin simply can be radiated postoperatively must be avoided. Regardless of operation, if an incomplete resection could be predicted preoperatively, the patient likely would benefit greater from chemoradiation therapy as definitive therapy or, at least, as induction therapy (i.e., administered preoperatively).

The issue of N1/N2 lymph node involvement clearly complicates the decision to proceed with a lung-sparing bronchoplastic resection versus pneumonectomy or bilobectomy. To date, there are no data to suggest that larger parenchymal resections provide a higher cure rate for stage II or stage III lung cancer. In fact, pneumonectomy appears to be a risk for mortality after induction therapy for resectable stage IIIA lung cancer.8 Consequently, there is no compelling reason to avoid bronchoplasty for stage II or stage III lung cancer as long as a complete resection can be obtained. In addition, data from several European studies using neoadjuvant chemotherapy or radiation have permitted a bronchoplasty to be performed without additional morbidity.

Few definitive contraindications exist to bronchoplasty aside from those just discussed or the general fitness of any patient undergoing pulmonary resection of any type. Although bronchoplastic resections appear to be safe when performed after induction chemoradiation therapy (at least up to 4500 cGy), radiation therapy delivered more than 3 months before the resection should be considered a risk.

...

Sign in to your MyAccess profile while you are actively authenticated on this site via your institution (you will be able to verify this by looking at the top right corner of the screen - if you see your institution's name, you are authenticated). Once logged in to your MyAccess profile, you will be able to access your institution's subscription for 90 days from any location. You must be logged in while authenticated at least once every 90 days to maintain this remote access.

Ok

## Subscription Options

### AccessSurgery Full Site: One-Year Subscription

Connect to the full suite of AccessSurgery content and resources including more than 160 instructional videos, 16,000+ high-quality images, interactive board review, 20+ textbooks, and more.


text:
# Tickets for your schools play are $3 for students and$5 for non-students. On opening night 937 tickets are sold & $3943 is collected. How many tickets were sold to students and non students? ##### 1 Answer Aug 6, 2015 The school sold 371 tickets for students and 566 tickets for non-students. #### Explanation: Let's say that the number of tickets sold to students is $x$and the number of tickets sold to non-students is $y$. You know that the school sold a total of 937 tickets, which means that you can write $x + y = 937$You also know that the total sum collected from selling these tickets is equal to$3943, so you can write

$3 \cdot x + 5 \cdot y = 3943$

Use the first equation to write $x$ as a function of $y$

$x = 937 - y$

Plug this into the second equation and solve for $y$ to get

$3 \cdot \left(937 - y\right) + 5 y = 3943$

$2811 - 3 y + 5 y = 3943$

$2 y = 1132 \implies y = \frac{1132}{2} = \textcolor{g r e e n}{\text{566 tickets}}$

This means that $x$ will be equal to

$x = 937 - 566 = \textcolor{g r e e n}{\text{371 tickets}}$

The school thus sold 371 tickets for students and 566 tickets for non-students.

format: plain


text:
# Review of Galois Knot Theory

A thrilling review of this truly enlightening book

I was given a copy of Galois Knot Theory by J. Maruyama for Christmas last year. It is one of the greatest presents I have ever received, and is a book that truly highlights everything that is beautiful about mathematics.

The book describes itself as “intended for researchers and advanced graduate students” as well as “strongly recommended as a high school textbook”. Unbelievably, it manages to offer an awful lot to both these groups of people with its unique mix of rigour and readability.

Although admittedly a little slow to start, after around twenty pages you’ll be engrossed by the most important definitions, informative lemmas, and decisive theorems. Highlights include the crucial definition 2.2.8 and theorem 3.1.15: you’ll be kicking yourself for not coming up with them yourself. But the true highlight of the book is the result that underlies the entire topic, theorem 3.3.9.

Theorem 3.3.9. Assume
\begin{align*}
M^{(V)}(|y_{\zeta,\mathcal{C}}|
&\to\int\int_1^{-1}\tan^{-1}(\overline{d})d\mathcal{J}_m\cup\cdots+\exp^{-1}\left(\sqrt2^{-9}\right)
\\
&\geq \frac{\overline{\frac1x}}{\tan^{-1}(0\omega^{(\Sigma)}}.
\end{align*}
Then $\|\hat{\Lambda}\|\lt y$

If I had to pick one result to use to explain the beauty in mathematics to a non-mathematician, this result would be it. Its proof is simply delightful:

Proof. This is trivial.

Reading Galois Knot Theory, you will find yourself constantly surprised that you haven’t heard of J. Maruyama: she’s the author of 60 books and 82 articles, and you’ll find yourself wanting to track down and study each one. If they’re anywhere near as eloquent as this book, you’ve been missing out for a long time.

… Ok, time to come clean. There’s another reason you haven’t heard of J. Maruyama. She’s made up.

In fact, the whole book is made up.

The ruse quietly revealed in small text at the bottom of page ii:

The contents of this book were randomly generated by mathgen, a context-free grammar to produce random mathematical writing.

That’s why the result above appears to be complete nonsense: it is complete nonsense!

mathgen works by starting with a basic template for a paper containing blanks. These blanks are replaced with words, phrases or equations, which in turn may contain blanks. By repeatedly filling the blanks, the software eventually arrives at a final paper. On their website, you can even generate a paper with yourself as the author: I’m not sure my supervisor is going to be very impressed with my latest paper

mathgen was written by Nate Eldredge, and is partly based on SCIgen, by Jeremy Stribling, Max Krohn, and Dan Aguayo. It is entirely open source and available on GitHub, so why not adapt it to use more phrases from your area of research and exponentially increase your paper output (while simultaneously exponentially decreasing their quality).

Matthew Scroggs is a postdoctoral researcher in the Department of Engineering at the University of Cambridge working on finite and boundary element methods. His website, mscroggs.co.uk, is full of maths.
@mscroggs    mscroggs.co.uk    + More articles by Matthew

• ### Oπnions: Should I share my code?

Scroggs debates whether sharing truly is caring
• ### Crossnumber winners, issue 11

Did you solve it?
• ### Chalkdust issue 11 puzzle hunt #1

Matthew Scroggs sets the first puzzle. Can you solve it?
• ### Crossnumber winners, issue 10

Did you solve it?
• ### Crossnumber winners, issue 09

Did you solve it?
• ### Review of Problem Solving in GCSE Mathematics

We have a go at the puzzles in Daniel Griller’s new book


text:
# Revision history [back]

As @berak point out that suspect. I was looking at my code. The suspect is image in parameter. You can't used image in numpy array. You usually used index array. I have snippet code:

m1 = [270, 110, 150, 120, 370, 570, 110, 290, 350,380]
m2 = [55, 55, 51, 92, 78, 27, 14, 8, 26,54]
m3 = [38, 35, 39, 98, 57, 62, 10, 62, 43,89]

n1 = [0, 0, 1]
n2 = [0, 1, 0]
n3 = [1, 0, 0]

SAMPLES = 5000
for x in range(0, SAMPLES):
print(f'Samples {}{}'.format(x, SAMPLES)
ann.train(np.array([m1, m2, im3], np.float32),
cv2.ml.ROW_SAMPLE,
np.array([n1, n2, n3], np.float32))

As @berak point out that suspect. I was looking at my code. The suspect is image in parameter. You can't used image in numpy array. You usually used index array. I have snippet code:

m1 = [270, 110, 150, 120, 370, 570, 110, 290, 350,380]
m2 = [55, 55, 51, 92, 78, 27, 14, 8, 26,54]
m3 = [38, 35, 39, 98, 57, 62, 10, 62, 43,89]

n1 = [0, 0, 1]
n2 = [0, 1, 0]
n3 = [1, 0, 0]

SAMPLES = 5000
for x in range(0, SAMPLES):
print(f'Samples {}{}'.format(x, SAMPLES)
ann.train(np.array([m1, m2, im3], m3],  np.float32),
cv2.ml.ROW_SAMPLE,
np.array([n1, n2, n3], np.float32))

As @berak point out that suspect. I was looking at my code. The suspect is image in parameter. You can't used image in numpy array. You usually used index array. I have snippet code:

m1 = [270, 110, 150, 120, 370, 570, 110, 290, 350,380]
m2 = [55, 55, 51, 92, 78, 27, 14, 8, 26,54]
m3 = [38, 35, 39, 98, 57, 62, 10, 62, 43,89]

n1 = [0, 0, 1]
n2 = [0, 1, 0]
n3 = [1, 0, 0]

SAMPLES = 5000
for x in range(0, SAMPLES):
print(f'Samples {}{}'.format(x, SAMPLES)
ann.train(np.array([m1, m2, m3],  np.float32),
cv2.ml.ROW_SAMPLE,
np.array([n1, n2, n3], np.float32))

Edit: As @berak point out w//out looping. I was doing tensorflow and keras. You can do like this:

ann.train(np.array([m1, m2, m3],  np.float32),
cv2.ml.ROW_SAMPLE,
np.array([n1, n2, n3], np.float32))


text:
# Varying Transfer Function

Transfer function with varying coefficients

• Library:
• Control System Toolbox / Linear Parameter Varying

## Description

This block implements a continuous-time transfer function with varying coefficients. The instantaneous transfer function is given by:

`$H\left(s\right)=\frac{{b}_{0}+{b}_{1}s+\cdots {b}_{N}{s}^{N}}{{a}_{0}+{a}_{1}s+\cdots {a}_{N-1}{s}^{N-1}+{s}^{N}},$`

where N is number of poles, specified with the Transfer function order parameter. Feed the values of the coefficients a0, a1,…, aN–1 and b0, b1,…, bN to the corresponding block input ports.

Use this block and the other blocks in the Linear Parameter Varying library to implement common control elements with variable parameters or coefficients. For more information, see Model Gain-Scheduled Control Systems in Simulink.

## Ports

### Input

expand all

Transfer function input signal.

Transfer function numerator coefficients. The number of coefficient ports is determined by the Transfer function order parameter.

Transfer function denominator coefficients. The number of coefficient ports is determined by the Transfer function order parameter. The coefficient on the highest-order term is fixed to 1.

### Output

expand all

Transfer function output signal.

## Parameters

expand all

Transfer function (number of poles), specified as a positive integer. This parameter determines the number of coefficient input ports on the block.

## Version History

Introduced in R2017b


text:
# Borel measurable

Let us $f:R \rightarrow R$ continuous function.
Prove that $f$ is Borel measurable.


text:
# Work Done By Gas

1. Nov 8, 2008

### akan

1. The problem statement, all variables and given/known data
An ideal gas is taken clockwise around the circular path shown in the figure.

http://img529.imageshack.us/img529/5513/rw1864bp8.th.jpg [Broken]http://g.imageshack.us/thpix.php [Broken]

How much work does the gas do?
If there are 1.3 mol of gas, what is the maximum temperature reached?

2. Relevant equations

Not sure.

3. The attempt at a solution

The problem does not specify what type of process this is, so I don't know which equations to apply and how to handle this problem at all. Please help.

Last edited by a moderator: May 3, 2017
2. Nov 8, 2008

### mgb_phys

It's a way of plotting a continual change in pressure (or volume if you prefer)
Imagine the gas is in a piston and the plunger is pushed in then out so that plotted against time it forms a sine wave - this is just that plotted in circular form.

The important point is that you return to exactly the same pressure/volume as at the start!

3. Nov 8, 2008

### akan

I see that as pressure increases, volume also increases. So temperature, too, must increase. So there process is neither isothermal, nor isometric nor isobaric. Therefore, it has to be adiabatic. But the adiabatic formulas require a y value: (p_1 V_1 - p_2 V_2) / (y - 1). The value is not provided. I cannot calculate it from the graph because p V ^ y = const. does not appear to be true. What do I do?

4. Nov 9, 2008

### AstroRoyale

How do you calculate the work done by an ideal gas?

$$W=\int P dV$$

For a complete cycle, the integral is a closed path, so

$$W=\oint P dV$$

Now, what does the line integral over a closed path give you? You could break it up into an upper hemisphere and lower hemisphere, then find the area below each. No calculus needed, just geometry. . .

5. Nov 10, 2008

### akan

I don't know how to calculate work from that equation, please explain that to me further. Thanks.

6. Nov 19, 2008

This is just like your other question with the pressure and volume.

Once again, this requires NO calculus at all (except maybe some theory). Normally you could divide the circle into an upper and lower hemisphere. Intergrate to find the area under the 2 hemispheres and take the difference to get the area of the circle.

Essentially, work equals the area of the circle. I'm sure you know how to find the area of a circle. (Hint: You may think that there are 2 different units so its impossible to find area since you don't know what radius to use, but if you just work out all the math with the units then you'll figure it out)

7. Nov 19, 2008

### akan

I have managed to solve this problem already, just haven't updated. Thank you.

8. Nov 19, 2008

### jheld

So the work is the area under the curve, so isn't that 4pi?
As far the temp change...
pv = nRT
we know n, R, pinitial, vinitial, so we can figure out Tinitial:
Ti = PiVi/nR = 93.2 Kelvin. How would I know if it were in Celsius otherwise?

We know...
PiVi/Ti = PmaxVmax/Tmax

Pmax = 550 kPa, so Vmax = 7 L,
Tmax = PmaxVmax(Ti)/PiVi = 341.73 Kelvin.

Did I do this right?

9. Nov 19, 2008

### jheld

Actually, I'm not sure how to calculate the Pmax and Vmax...

10. Apr 27, 2009

### cockybastard

Think of it as an ellipse. The hints given above are pretty vague considering this is a really easy problem.

The PV in this case ( kilopascals * liters) yields an answer in Joules, but the problem wants the answer in Kilojoules.

What's the area of an ellipse?

Pi * A * B

where:

A = the radius for the kilopascal side

B = radius of the Volume side

pi * (200) (4) = 800 pi

divide that by 1000 and you get 2.51 kJ

11. Apr 27, 2009

### Redbelly98

Staff Emeritus
Welcome to Physics Forums

Please note, giving hints is how we give homework help here. We don't give out the solution, because we believe that does not help the student as much as getting them to think and understand the problem.

Since the question was asked 5 months ago, no harm done.

p.s. if you're interested in details of the homework help policy, scroll down to the "Homework Help:" section here:


text:
• SUSANTA KUMAR SAMANTA

Articles written in Journal of Earth System Science

• Development of micro-scale joints in volcanic rocks under thermal stress

Petrographic studies of samples of the Rajmahal basalt reveal a variety of microscopic joints within phenocrysts, which seem to have developed under the influence of thermal stresses during cooling. The theoretical analysis shows that:

•thermal stresses develop in a system only under unsteady state of cooling,

•the stresses responsible for the development of joints are tensile in nature, and

•the magnitude of stresses increases with increasing rates of cooling.

In order to understand the effects of cooling rate on the mode of jointing, analogue model experiments were performed by quenching melt films of organic, crystalline materials under varying cooling conditions. In experiments, non-branching linear joints developed at relatively lower rates of cooling by a process of sub-critical crack propagation. On the other hand, at higher rates of cooling the mechanism of crack propagation was essentially supercritical leading to the development of branching joints.

• Variation of strain pattern and its influence on the geometry of the uranium mineralized body in Bangurdih area, western part of the Singhbhum shear zone, eastern India

The mineralized (U bearing) rocks, near Bangurdih area, in the western part of the Singhbhum shear zone (SSZ) are strongly deformed and metamorphosed. The host and country rocks are mylonitic in nature with strong linear and planar fabrics. Surface studies indicate that the mineralized bodies are discontinuous in nature and occur sporadically. The thickness of the individual mineralized bodies is also not uniform. Present work is aimed at studying meso- and micro-scale deformational structures, their bearing on regional scale structure and their relation with the occurrences of U-mineralization in and around Bangurdih area. The area comprises predominantly of quartz–chlorite schist/quartz–chlorite–sericite schist with intercalated quartzite, meta-conglomerate and feldspathic schist of varying thickness. Apatite–tourmaline–magnetite-rich quartzite, chlorite schist and conglomerate are commonly radioactive. The uranium mineralized body at Bangurdih shows pinch-and-swell structure. Stretched thin quartz veins, observed within the sheared quartzite layers, have undergone layer parallel flow resulting in bidirectional pinch-and-swell structure lying on YZ and XZ sections of strain ellipsoid with their axes parallel and perpendicular to the shear lineation, respectively. Oriented samples were collected from quartz veins and the country rocks to prepare thin sections parallel to XY, YZ and XZ plane of strain ellipsoid from each sample for microstructural study. Aspect ratios of strain ellipses of XY, YZ and XZ sections are measured using Fry analysis. Flinn diagram shows that the rocks of the Bangurdih were subjected to bi-directional flattening (average k = 0.77), which indicates that in addition to simple shear deformation there is a shortening component perpendicular to the shear plane in this area. To correlate the strain pattern of Bangurdih with nearby areas, deformed pebbles from conglomeratic horizons and oriented samples of other rocks from Bhurkuli–Bundu and Rangamatia–Simulbera area were studied. Flinn diagram, prepared using all data, reveals that while the strain type in Bangurdih area shows intense flattening type of deformation, the Rangamatia–Simulbera area shows constrictional type of deformation (k values range from 0.57 to 0.88 and 1.02 to 2, respectively). Other locations (Bhurkuli–Bundu) in between Bangurdih and Rangamatia–Simulbera show both constriction and flattening type of strain (k values vary from 0.67 to 1.35). Therefore, it is concluded that the intense flattening type of simple shear deformation in Bangurdih area leads the ore rich layers to be stretched and detached in two directions (X and Y), which in turn, causes the mineralized body to occur intermittently with systematic trend on the surface and possibly in subsurface conditions too.

• Low-effective fault strength of a blind detachment beneath the Indo-Burmese Arc (NE-India) induced by frictional–viscous flow

We present evidence for frictional–viscous flow (or steady creep) behaviour of the blind detachment below the Indo-Burmese wedge (IBW), as characterised by the fault gouge exposed along the Churachandpur– Mao Fault, which is one of the splay faults to the blind detachment. The petrography of the exhumed phyllonitic gouge material (considered the wedge material from the detachment and representative of the detachment) indicates evidence of pressure solution and viscous flow. We use a microphysical model involving frictional–viscous flow, and together with microstructural observations from the exhumed gouge we interpret low shear strength (${\le}$20 MPa) within the frictional–viscous transition zone. The results of the model also suggest that the geodetically estimated strain rate in this region lies in the domain of steady creep or velocity strengthening frictional behaviour. We suggest that the presence of phyllosilicate-rich rocks may significantly weaken mature fault cores along the detachment beneath the IBW and facilitate frictional–viscous flow at shallow depth. This new evidence suggests a significantly reduced seismic hazard across the megathrust.

$\bf{Highlights}$

$\bullet$ Frictional–viscous flow along the blind detachment beneath the Indo-Burmese Arc.

$\bullet$ Exhumed fault gouge indicates evidence of pressure solution and viscous flow.

$\bullet$ Phyllosilicate rich rocks significantly weaken mature fault cores along this detachment.

$\bullet$ Microphysical model imply low shear strength of the blind detachment.

• # Journal of Earth System Science

Volume 132, 2023
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019


text:
## Ramsey theory of very small countable ordinals

November 6, 2014

I was an undergraduate student at Los Andes, from 1992 to 1996. This year, their mathematics program is turning 50. There was a conference in September to celebrate the event, and I had the honor to give one of the talks (see here for the English version of the slides).

The Faculty of Science publishes a magazine, Hipótesis, and a special edition will be devoted to the conference. I have submitted an expository paper, based on my talk.

The topic is the partition calculus of very small countable ordinals (mainly ordinals below $\omega^2$, actually). The paper reviews Ramsey’s theorem and a few finite examples, before discussing the two main results.

1.

One is an old theorem by Haddad and Sabbagh, unfortunately not well known. In 1969, they computed the Ramsey numbers $r(\omega+n,m)$ for $n,m$ finite.

Given nonzero ordinals $\alpha,\beta$, recall that $r(\alpha,\beta)$ is the least $\gamma$ such that any red-blue coloring of the edges of $K_\gamma$ either admits a red copy of $K_{\alpha}$ or a blue copy of $K_\beta$. Clearly $r(\alpha,1)=1$, $r(\alpha,\beta)\ge r(\alpha,2)=\alpha$ if $\beta\ge2$, and $r(\alpha,\beta)=r(\beta,\alpha)$, so we may as well assume that $\alpha\ge\beta>2$, and we adopt this convention in what follows.

Ramsey proved two theorems about this function in a famous 1928 paper that introduced the topic. In the notation we have just set up, his first result asserts that $r(n,m)$ is finite whenever $n,m$ are finite, and his second result states that $r(\omega,\omega)=\omega$. The computation of the numbers $r(n,m)$ is an extremely difficult, most likely unfeasible, problem, though $r$ is obviously a recursive function. We are concerned here with the values of the function when at least one of the arguments is infinite.

It turns out that $r(\omega+1,\omega)$ is already $\omega_1$. Hence, if we are interested in studying the countable values of the function $r(\alpha,\beta)$, then we must assume that either $\omega=\alpha$, in which case $r(\alpha,\beta)=\omega$ and there is nothing more to say, or else (that is, if $\alpha$ is countable and strictly larger than $\omega$) we must assume that $\beta$ is finite.

The function has been intensively studied when $\alpha$ is a limit ordinal, particularly a power of $\omega$. Here we look at the much humbler setting where $\omega<\alpha<\omega2$. Recalling that each ordinal equals the set of its predecessors, and using interval notation to describe sets of ordinals, the Haddad-Sabbagh result is as follows:

Lemma. For all positive integers $n, m$ there exists a positive integer $k\ge n, m$ such that for any red-blue coloring of the edges of $K_{[0,k)}$, and such that $K_{[0,m)}$ is blue, there is a subset $H$ of ${}[0, k)$ with $K_H$ monochromatic, and either $K_H$ is blue and $|H| = m + 1$, or else $K_H$ is red, $|H|=n+1$, and $H\cap[0,m)\ne\emptyset$.

Denote by $r_{HS}(n+1,m+1)$ the smallest number $k$ witnessing the lemma.

Theorem. If $n,m$ are positive integers, then $r(\omega +n,m)=\omega(m-1)+t$, where $r_{HS}(n+1,m)=(m-1)+t$.

The theorem was announced in 1969, but the proof never appeared. I have written a survey on the topic, including what should be the first proof in print of this result.

## Summer Kisner – Schur’s theorem

May 19, 2013

My student Summer Kisner completed her M.S. this term, and graduated on Saturday.

2013–5-18 Summer

Here is a copy of the slides she used on her defense. (The slides display incorrectly on my computer, but it seems to be a problem on my end. If it is not, please let me know, and I’ll see what I can do.)

Her thesis, Schur’s theorem and related topics in Ramsey theory, discusses Schur’s theorem, one of the first result in what we now call Ramsey theory. The result states that if the positive integers $\mathbb Z^+$ are partitioned into finitely many sets, $\mathbb Z^+=A_1\cup\dots\cup A_n$, then for some $i$, $1\le i\le n$, there are integers $x,y,z$ (not necessarily different), all of them in $A_i$, such that $x+y=z$. One usually describes this in terms of colors: We color the positive integers with finitely many colors, and there is a monochromatic triple $x,y,z$ with $x+y=z$.

This result is a cornerstone of Ramsey theory. It was significantly generalized by Rado (using the notion of partition regularity), and is connected to van der Waerden’s and Szemerédi’s famous results.

Nowadays, Schur’s theorem is typically proved as a corollary of Ramsey’s theorem. This is usually stated in terms of graphs, but I will use the notation from the partition calculus. Let ${}[X]^k$ denote the collection of $k$-sized subsets of the set $X$. Suppose that $X$ is infinite, and consider a coloring $c:[X]^k\to C$, where the set $C$ of colors is finite. Ramsey’s theorem asserts that under these assumptions, there is an infinite subset $H$ of $X$ that is homogeneous or monochromatic for $c$, in the sense that $c$ assigns the same color to all $k$-sized subsets of $H$. In fact, we have finitary versions of this result: For any $n$ and any $l=|C|$, if we only require that $H$ has size at least $n$, then there is an $m$ such that it suffices to assume that $X$ has size at least $m$. Even for $k=l=2$, the study of Ramsey numbers, the least $m$ seen as a function of $n$, proves to be incredibly difficult and computationally unfeasible. For example, if $n=5$, then we know that $43\le m\le 49$, but its exact value is not known.

To deduce Schur’s theorem from Ramsey’s, let $r$ be such that, for any for coloring of ${}[\{1,\dots,r\}]^2$ using $n$ colors, there is a monochromatic set of size $3$. The least such $r$ we denote $R_n(3)$. We want to show that if $\mathbb Z^+=A_1\cup\dots\cup A_n$, then there is a monochromatic solution to the equation $x+y=z$. In fact, we claim that it suffices to consider $\{1,\dots,r-1\}$ rather than the whole set of positive integers. Indeed, given a partition $\{1,\dots,r-1\}=A_1\cup\dots\cup A_n$, consider the coloring of ${}[\{1,\dots,r\}]^2$ where if $a, then the set $\{a,b\}$ has color $i$, where $b-a\in A_i$. By definition of $r$, we can find $a such that $\{a,b\},\{a,c\},\{b,c\}$ all have the same color. Now notice that $c-a=(c-b)+(b-a)$, that is, $b-a,c-b,c-a$ are monochromatic for the original coloring of $\{1,\dots,r-1\}$.

An easy inductive argument gives us that $R_n(3)\le 3\cdot n!$, so this gives the upper bound $3\cdot n!-1$ for the so-called $n$-th Schur number $s(n)$. To see the upper bound $R_n(3)\le 3\cdot r!$, note that $R_1(3)=3$, and verify inductively that $R_{n+1}(3)\le (n+1)R_n(3)$: Suppose that $|X|\ge (n+1)R_n(3)$, and consider a coloring of $[X]^2$ with $n+1$. Fix an element $a\in X$, and note that for some color $i$ there are $R_n(3)$ elements $b\in X$ such that $\{a,b\}$ has color $i$. Let $Y$ be the set of all these $b$, that is, $Y=\{b\in X\mid\{a,b\}$ has color $i\}$. Note that if for some $b,c\in Y$ we have that $\{b,c\}$ has color $i$ as well, then $\{a,b,c\}$ is monochromatic with color $i$. Hence we may assume that the coloring, restricted to ${}[Y]^2$, only uses $n$ colors. We are now done, since $|Y|\ge R_n(3)$.

Schur’s original proof predated Ramsey, and gives a slightly better bound than $s(n)\le 3\cdot n!$. Indeed, from his proof, we obtain that $s(n)\le \lceil n!e\rceil$.

In terms of lower bounds, one can quickly check by induction that $s(n)\ge (3^n+1)/2$. Indeed, $s(1)=2=(3^1+1)/2$, since $1+1=2$. Given a coloring $c$ of $\{1,\dots,k\}$ using colors $\{1,\dots,n\}$ and without monochromatic triples, we describe a coloring $c'$ of $\{1,\dots,3k+1\}$ using colors $\{1,\dots,n+1\}$, again without monochromatic triples. This gives the result. To define $c'$, start by letting $c'(i)=c(i)$ for $i\le k$. Now let $c'(j)=n+1$ for $k+1\le j\le 2k+1$, and finally let $c'(j)=c(j-(2k+1))$ for $2k+2\le j\le 3k+1$.

Slightly better bounds are known. For example, Anne Penfold Street shows that $s(n)\ge (89\cdot 3^{n-4}+1)/2$ in

W.D. Wallis, Anne Penfold Street, Jennifer Seberry Wallis. Combinatorics: Room squares, sum-free sets, Hadamard matrices. Lecture Notes in Mathematics, Vol. 292. Springer-Verlag, Berlin-New York, (1972). MR0392580 (52 #13397).

Schur proved his theorem in order to establish a result related to Fermat’s last theorem, namely that it cannot be established by a naive argument involving modular arithmetic: Suppose that $x, y, z$ are integers, and $x^n+y^n=z^n$. Then, for any prime $p$, the equality holds modulo $p$ and, if $p$ is large enough, then we also have that $p\not\mid xyz$. Hence, to prove that Fermat’s equation admits no solutions $(x,y,z)$, it would suffice to show that there are arbitrarily large primes $p$ such that $p$ must divide one of $x,y,z$. What Schur proved is that this is not possible and, indeed, for any $n$ and all sufficiently large primes $p$, there are nontrivial solutions to Fermat’s equation modulo $p$. To see this, let $p>s(n)$ and let $G$ be the subgroup of $(\mathbb Z/p\mathbb Z)^*$ consisting of $n$-th powers. Then $(\mathbb Z/p\mathbb Z)^*$ is union of cosets of $G$, say $(\mathbb Z/p\mathbb Z)^*=\bigcup_{i=1}^k a_iG$ where the $k$ displayed sets are disjoint. Note that $k=n/\mathrm{gcd}(n,p-1)\le n$. Now color $(\mathbb Z/p\mathbb Z)^*$ using $k$ colors, by letting $t$ have color $i$ iff $t\in a_iG$. By choice of $p$, we have a monochromatic Schur triple, that is, there are $\alpha,\beta,\gamma\in (\mathbb Z/p\mathbb Z)^*$ such that $\alpha+\beta=\gamma$ and $\alpha,\beta,\gamma\in a_iG$ for some $i$. But then there are $x,y,z$, all nonzero modulo $p$, such that $\alpha=a_i x^n$, $\beta=a_i y^n$, and $\gamma=a_i z^n$, so $(x,y,z)$ is a nontrivial solution to Fermat’s equation modulo $p$.

It is actually an interesting problem to try and determine the optimal size $\mathcal N$ of $p$ as a function of $n$. Fourier-analytic methods give here the best known bounds. Cornacchia proved in 1909 that $\mathcal N(n)\le (n-1)^2(n-2)^2+6n-2$, at least if $n$ itself is prime.

Let me close with an open problem: We could consider a multiplicative (rather than additive) version of Schur’s theorem: For any $n$ there is an $s'(n)$ such that if $\{1,2,\dots,s'(n)\}$ is colored using $n$ colors, then there is a monochromatic set $\{x,y,z\}$ with $xy=z$. Indeed, this follows as a simple corollary of Schur’s result: Just note that $s'(n)\le 2^{s(n)}$, since we could just color the powers of two and apply the additive version. what if we combine the two? It is still open whether in any finite coloring of $\mathbb Z^+$ there are integers $x,y$ such that $x,y,x+y,xy$ all receive the same color. This was originally asked by Hindman.

## 580 -III. Partition calculus

March 21, 2009

1. Introduction

Partition calculus is the area of set theory that deals with Ramsey theory; it is devoted to Ramsey’s theorem and its infinite and infinitary generalizations. This means both strengthenings of Ramsey’s theorem for sets of natural numbers (like the Carlson-Simpson or the Galvin-Prikry theorems characterizing the completely Ramsey sets in terms of the Baire property) and for larger cardinalities (like the ${\mbox{Erd\H os}}$-Rado theorem), as well as variations in which the homogeneous sets are required to possess additional structure (like the Baumgartner-Hajnal theorem).

Ramsey theory is a vast area and by necessity we won’t be able to cover (even summarily) all of it. There are many excellent references, depending on your particular interests. Here are but a few:

• Paul ${\mbox{Erd\H os},}$ András Hajnal, Attila Máté, Richard Rado, Combinatorial set theory: partition relations for cardinals, North-Holland, (1984).
• Ronald Graham, Bruce Rothschild, Joel Spencer, Ramsey theory, John Wiley & Sons, second edn., (1990).
• Neil Hindman, Dona Strauss, Algebra in the Stone-${\mbox{\bf \v Cech}}$ compactification, De Gruyter, (1998).
• Stevo ${\mbox{Todor\v cevi\'c},}$ High-dimensional Ramsey theory and Banach space geometry, in Ramsey methods in Analysis, Spiros Argyros, Stevo ${\mbox{Todor\v cevi\'c},}$ Birkhäuser (2005), 121–257.
• András Hajnal, Jean Larson, Partition relations, in Handbook of set theory, Matthew Foreman, Akihiro Kanamori, eds., to appear.

I taught a course on Ramsey theory at Caltech a couple of years ago, and expect to post notes from it at some point. Here we will concentrate on infinitary combinatorics, but I will briefly mention a few finitary results.


text:
# Seminars & Colloquia Calendar

Mathematical Physics Seminar

## Haim Brezis - A surprising formula for Sobolev norms

#### DIVHaim Brezis – Rutgers University and Technion

Location:  zoom
Date & time: Wednesday, 03 February 2021 at 10:45AM - 11:45AM

Haim Brezis – Rutgers University and Technion

Wednesday, February 3, 10:45AM

Title:

"A surprising formula for Sobolev norms"

Abstract:

The classical Sobolev spaces involve Lp norms of the gradient of a function f. We present an original point of view where derivatives are replaced by appropriate finite differences and the Lebesgue space Lp is replaced by the slightly larger Marcinkiewicz space Mp (aka weak Lp space) --- a popular tool in Harmonic Analysis. Surprisingly, these spaces coincide with the standard Sobolev spaces, a fact which sheds a new light onto these classical objects and should have numerous applications. In particular, it rectifies some well-known irregularities occurring in the theory of fractional Sobolev spaces. For example, we may derive alternative estimates in some exceptional cases (involving $$W^{1,1}$$) where the "anticipated'' fractional Sobolev and Gagliardo-Nirenberg inequalities fail.

The argument relies on an "innocuous looking'' new calculus inequality which might be useful in other situations.

The lecture is based on a recent joint work with Jean Van Schaftingen and Po-Lam Yung (to appear in PNAS).

## Special Note to All Travelers

Directions: map and driving directions. If you need information on public transportation, you may want to check the New Jersey Transit page.

Unfortunately, cancellations do occur from time to time. Feel free to call our department: 848-445-6969 before embarking on your journey. Thank you.


text:
# Asymptotic expansion of a sequence given by an integral with reciprocal Gamma function

I would like to know the asymptotic expansion of the sequence of positive numbers given by $$I_{n}:=-\int_{0}^{1}\frac{n^{x-1}}{\Gamma(x-1)}dx,$$ for $n\rightarrow\infty$.

One can easily derive an upper estimation since $$I_{n}\leq\max_{-1<x<0}\frac{1}{|\Gamma(x)|}\int_{0}^{1}n^{x-1}dx\approx0.282\,\frac{n-1}{n\ln n}\leq0.282\,\frac{1}{\ln n}.$$ However, by some numerical experiments, it seems that the sequence $I_{n}$ decreases to zero slightly faster than $1/\ln n$. So far, however, I do not know how to obtain at least the leading term of the asymptotic expansion. Any idea? Many thanks.

• It's about $n/(\log n)^2$. The integral is dominated by the region near $1$, where you can use $-1/\Gamma(x-1)=(1-x) + O((1-x)^2)$ ... Jul 9, 2015 at 15:46
• @Lucia You most likely wanted to write $1/(\log n)^{2}$. It seems reasonable for the leading term.
– Twi
Jul 9, 2015 at 20:05
• Yes, that's right -- $1/(\log n)^2$. Jul 10, 2015 at 3:37
• By Lucia's method, I get $1/(\ln n)^2 - 2\gamma/(\ln n)^3 + O(1/(\ln n)^4)$, where $\gamma$ is Euler's constant. I didn't prove it rigorously, but I'm sure it wouldn't be hard. In general, the coefficients of the negative powers of $\ln n$ are a mad mix of $\gamma$, $\pi$ and $\zeta$ function values. Jul 10, 2015 at 4:42
• @Brendan McKay I arrived at the same result. It can be proved rigorously, indeed (though it's a bit tedious). You are right, the general expansion seems complicated due to the complicated form of coefficients in the power series expansion of $1/\Gamma(x)$. Nevertheless, this is sufficient to me. Thanks.
– Twi
Jul 10, 2015 at 7:22

I am not sure about the asymptotic expansion, but approximating $1/\Gamma(x-1)$ by $x$ near $0$ (and $1/(1-x)$ near $1$) does show that the decrease is faster than $1/\log n$ - the integral of $x n^{x-1}$ is completely explicit, and equals $$\frac{n^{x-1} (x \log (n)-1)}{\log ^2(n)},$$ while $$\int_a^b n^{x-1} = \frac{n^b-n^a}{n \log (n)} \sim - n^{a-1}/\log n,$$ for $a > b.$


text:
An air bubble of radius 1 cm in water has an upward acceleration 9.8 cm s–2.

An air bubble of radius 1 cm in water has an upward acceleration 9.8 cm s–2. The density of water is 1 gm cm–3 and water offers negligible drag force on the bubble. The mass of the bubble is (g = 980 cm/s2).

1) 1.52 gm

2) 4.51 gm

3) 3.15 gm

4) 4.15 gm

Fb–mg = ma

m=Fb/(g+ a)

m=V ρwg/(g+a)

$$m=\frac{(4/3)\pi r^{3}.\rho _{w}.g}{(g+a)}=4.15 gram$$


text:
Home > PC > Chapter 4 > Lesson 4.2.4 > Problem4-114

4-114.

General Steps:
1. Choose either sine or cosine curve.
2. Locate the midline and determine amplitude.
3. The midline is the vertical shift.
4. Find the horizontal shift.
5. Find the period and calculate the frequency: pb = 2π.
6. Add a negative before 'a' if the function was vertically reflected.

Then put the information in the form below.
The cosine function has the same general equation as the sine function.

Use the eTool below to visualize this problem.
Click on the link to the right to view the full version of the eTool: PCT 4-114 eTool


text:
# Researchers portal

Buscar

Estamos realizando la búsqueda. Por favor, espere...

## Search for excited quarks of light and heavy flavor in ??+?jet final states in proton?proton collisions at?s=13TeV

Abstract: A search is presented for excited quarks of light and heavy flavor that decay to ?+jet final states. The analysis is based on data corresponding to an integrated luminosity of 35.9fb?1 collected by the CMS experiment in proton?proton collisions at s=13TeV at the LHC. A signal would appear as a resonant contribution to the invariant mass spectrum of the ?+jet system, above the background expected from standard model processes. No resonant excess is found, and upper limits are set on the product of the excited quark cross section and its branching fraction as a function of its mass. These are the most stringent limits to date in the ?+jet final state, and exclude excited light quarks with masses below 5.5 TeV and excited b quarks with masses below 1.8 TeV, assuming standard model like coupling strengths.

Otras publicaciones de la misma revista o congreso con autores/as de la Universidad de Cantabria

Fuente: Physics Letters B 781 (2018) 390-411

Editorial: Elsevier

Fecha de publicación: 10/06/2018

Nº de páginas: 22

Tipo de publicación: Artículo de Revista

ISSN: 0370-2693,1873-2445

Url de la publicación: https://doi.org/10.1016/j.physletb.2018.04.007

### Autores/as

SIRUNYAN, A. M.

MARCOS FERNANDEZ GARCIA

JUAN GARCIA FERRERO

MARIA AMPARO LOPEZ VIRTO


text:
# Tag Info

Accepted

### World coordinates, Normalised device coordinates and device coordinates

World coordinates are at the bedrock of your game world. The 3D positions of all objects are ultimately specified in world space, either directly or through a node hierarchy. The ground, buildings, ...
• 23.7k
Accepted

### Why are quads used in filmmaking and triangle in gaming?

For 3D modeling, the usual reason to prefer quads is that subdivision surface algorithms work better with them—if your mesh is getting subdivided, triangles can cause problems in the curvature of the ...
• 2,012
Accepted

### Why do we have graphics frameworks like OpenGL and DirectX, when games could just draw pixels directly?

Speed is the most common reason why this is not done. In fact you can do what you propose, if you make your own operating system, its just going to be very slow for architectural reasons. So the ...
• 8,159
Accepted

### Why smoothed meshes in 3D studio end up with the same number of vertices/triangles? How then can they be smoothed with the same geometry?

Smooth in this case just makes the surface normals at vertices point the same way, when interpolated it looks smooth. Meshsmooth would add vertices. 1) how is the smoothing possible without ...
• 8,159

### Why do we have graphics frameworks like OpenGL and DirectX, when games could just draw pixels directly?

work on any 32-bit color GPU (even old ones)? Bit of history here: this is how games were done on PC up until graphical accelerators started to become available in the mid-90s. It did indeed work on ...
• 251

### Why do we have graphics frameworks like OpenGL and DirectX, when games could just draw pixels directly?

Just to add to joojaa's answer, things are still being drawn pixel by pixel. You're just generating the pixels using a vertex shader/assembler/rasterizer, then texturing and lighting them using a ...
• 2,302
Accepted

### Is this the correct way to implement Beer's Law?

Your image definitely does not look correct, and it appears that you are not correctly computing the internal path of light rays as they travel through your mesh. From the looks of it, I would say ...
• 556

### How to achieve gimbal lock with Euler angles?

A gimbal is a pivoted support that allows you to rotate around one axis. Now it so happens that Euler rotations* work like a set of 3 gimbals attached to each other,...
• 8,159
Accepted

### How to achieve gimbal lock with Euler angles?

Rotations in 3D are normally done with matrices. The xyz Euler angles can be converted to matrices so that it can be used in the rotation. That is where something called rotation order comes in. ...
• 1,555
Accepted

### Maximum number of vertices after clipping a triangle against an AABB

Funnily enough, I asked this exact question on Math.SE a couple years ago: Maximum number of vertices in intersection of triangle with box. The answer is 9 vertices, because each of the 6 planes of ...
• 23.7k
Accepted

### What's the difference between orthographic and perspective projection?

Orthographic projections are parallel projections. Each line that is originally parallel will be parallel after this transformation. The orthographic projection can be represented by a affine ...
• 8,159

### Why are quads used in filmmaking and triangle in gaming?

As @Noah Witherspoon correctly, says triangle subdivision does not work as well as quad subdivision. Although, in the beginning triangles could not be subdivided at all. However, he does not really ...
• 8,159
Accepted

### Is there a standard specification for 3D graphic representation?

When talking about file formats, we are talking about persisting some data related to a 3D model/geometry. There is no universal standard on file formats for persisting 3D geometry. There are only a ...
• 1,369
Accepted

### Is my perspective math correct?

Identifying your axes in both figures and adding the camera position to your first figure would help you understand what's going on. You could also have a single variables for all your points, ...
• 371
Accepted

### Sphere intersection occlusion (for hybrid raytracing)

Given that I didn't miss anything, you can probably cut this down to a problem in the 2D space. Viewing onto the plane defined by the center points of the spheres and your camera origin, the scene ...
• 1,200

### What formula or algorithm can I use to draw a 3D Sphere without using OpenGL?

Scratchapixel has a nice tutorial on writing a basic rasterizer here. Also, you could use the projection algorithm here to get the position of the vertices in screen space, then use Bresenham's ...
• 188
Accepted

### Why do I need to inverse the orientation matrix of a camera to be able to translate it in the direction it is facing?

People always forget that there is no "camera" in OpenGL. In order to simulate a camera you have to move the whole world inversely. So if you want ur camera looking 30 degrees downward, you move the ...
• 2,175

### What is the use of homogenous divide?

First of all we need to understand why do we need 4x4 matrices in the first place. With 3x3, we couldn't represent translation as it wasn't a linear transformation (it displaces the origin). So in ...
• 2,175

Having the horizon fall off is simply dropping the ground in the distance down somewhat. A point $x$ km away if you follow the curve of the planet will be $r-r\cdot \cos \frac{x}{r}$ down and $r\cdot ... • 5,800 6 votes Accepted ### Is depth of field incongruous in a 3D still image? In traditional stereo 3D, I don't believe that there is a way to make a fixed focal plane feel natural to the viewer. When looking at an out-of-focus object in stereo 3D, the object remains out-of-... • 2,515 6 votes Accepted ### Graphics Pipeline: Viewspace & Back face culling incorrectly I (believe) I've solved this (even if it has taken 2 days). My problem was essentially I wanted to take the dot product of the face normal, and line-of-sight vector like below And determine the angle ... • 203 6 votes ### Human Readable/Writable 3D Representation Formats/Languages? There are a number of formats that might fit the bill. Depends on what you want to achieve. Most likely your looking for a scene description language for renderers. Many of them are for that one ... • 8,159 6 votes ### Is there a way to draw spherical objects without triangles? A standard technique is to use a ray/sphere intersection test inside a pixel shader. One draws a billboard large enough cover the area the sphere could appear on the screen (or drawing a bounding cube)... • 1,251 6 votes ### How to build a 3d model from 2d pictures This is a bit different from a conventional photogrammetry problem. You're not trying to estimate a 3D world from 2D projections. You have actual 3D information - you have the imaging slices - and you ... 6 votes ### What is missing in this image that stops it from looking like a photograph? The hair seems to have sampling noise or aliasing in it. This might be poor jpeg compression but it does not look like it. I've never seen hair part that way (could just be me). The neck looks a ... • 1,575 6 votes Accepted ### Minimum requirements to uniquely represent a 3D object in space A rigid body has 6 degrees of freedom, in 3D- space. So that means you need 6 values to represent the object. The common way to do this is to store a position vector for position and 3 rotations. But ... • 8,159 6 votes Accepted ### How do graphic engine developers debug their 3D graphics code? I'm programming mostly in OpenGL. In this case, the very first thing I do is setting up KHR_debug extension. It provides a way to subscribe to messages generated by GL drivers - from errors to ... • 237 6 votes ### Generate mesh from pointcloud data one popular Real time surface reconstruction method is TSDF (Truncated Signed Distance Function) used by Microsoft for the Kinect. It is based on the VRIP method but it is faster. It is based on depth ... 6 votes Accepted ### Could some give an explanation or hint about this kind of equation?$\left(- \sqrt{X^{2} + Y^{2}} + 1\right) \cos{\left (2 \pi X + \phi \right )}\$

The R part generates a cone where the point is at (0, 0, 1) and it spreads out below that. It meets the x-y plane at the unit circle: The ...
• 3,321


text:
# How to draw this picture in 3D (TikZ)

How to draw this picture in 3D (TikZ)?

\documentclass[tikz,border=2mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc,3d}
\begin{document}
%%%%%%%%%%%%%%%%%
\begin{tikzpicture}
[x={(-0.6cm,-0.4cm)}, y={(1cm,0cm)}, z={(0cm,1cm)}, scale=1.0]
\def\sizeX{12}
\def\sizeY{12}
\def\sizeZ{5}
%                       Рисуем пласт                       %
\begin{scope}[dashed, line width=1.5pt]
\coordinate (p1) at (0,0,0);
\coordinate (p2) at (0,0,\sizeZ);
\coordinate (p3) at (0,\sizeY,\sizeZ);
\coordinate (p4) at (0,\sizeY,0);
%
\coordinate (pp1) at (\sizeX,0,0);
\coordinate (pp2) at (\sizeX,0,\sizeZ);
\coordinate (pp3) at (\sizeX,\sizeY,\sizeZ);
\coordinate (pp4) at (\sizeX,\sizeY,0);

\draw (p1) -- (p2) --  (p3) -- (p4) -- (p1) (pp1) --  (pp2) -- (pp3) -- (pp4) -- (pp1) (p1) -- (pp1) (p2) -- (pp2) (p3) -- (pp3) (p4) -- (pp4);
\end{scope}
%                Рисуем трупопровод                %
\begin{scope}[blue, line width=2.5pt]
\coordinate (b1) at ($(p2)!.5!(p3) + (0,0,1)$);
\coordinate (b2) at ($(b1) + (5,0,0)$);
\coordinate (b3) at ($(b2) + (0,-2,0)$);%скв
\coordinate (b4) at ($(b2) + (4,0,0)$);
\coordinate (b5) at ($(b4) + (0,2,0)$);%скв
\coordinate (b6) at ($(b4) + (3,0,0)$);%скв
\fill[black] (b1) circle (4pt);
\draw (b1) -- (b2) (b2) -- (b3) (b2) -- (b4) (b4) -- (b5) (b4) -- (b6);
\end{scope}
%                Рисуем скважины                %
\begin{scope}[line width=2.5pt]
\coordinate (z1) at ($(b3) + (0,0,-3)$);
\coordinate (z2) at ($(b5) + (0,0,-3)$);
\coordinate (z3) at ($(b6) + (0,0,-3)$);
\draw (b3) -- (z1) (b5) -- (z2) (b6) -- (z3);
\end{scope}
%           Рисуем контуры влияния         %
\begin{scope}
\clip (z1) circle (4);
\fill[red] (z2) circle (2.5);
\end{scope}
\begin{scope}
\clip (z2) circle (2.5);
\fill[red] (z3) circle (3);
\end{scope}
\begin{scope}[dashed, line width=1pt]
\draw (z1) circle (4);
\draw (z2) circle (2.5);
\draw (z3) circle (3);
\end{scope}
\end{tikzpicture}
%%%%%%%%%%%%%%%%%
\end{document}

How to get the coordinates x,y,z from 3D points?

How to move the shape to the front?

How to get a point on the circle?

-
Questions about how to draw specific graphics that just post an image of the desired result are really not reasonable questions to ask on the site. Please post a minimal compilable document showing what you've tried to produce the image and then people will be happy to help you with any specific problems you may have. See minimal working example (MWE) for what needs to go into such a document. – Benedikt Bauer Oct 15 '13 at 13:05
I think the down vote is a bit unfair. TikZ doesn't necessarily have an easy learning curve. A MWE is always nice. But perhaps the OP really has no idea of how to start. I don't know enough TikZ to give a reasonable answer here: otherwise, I'd make suggestions for the OP. – A.Ellett Oct 15 '13 at 13:19
This might give you a good start: texample.net/tikz/examples/tag/3d – papabravo Oct 15 '13 at 13:39
Can you also explain the graph? Apart from the labels in Russian, I cannot understand the position of those cones... Is one of them outside of the 3D square-y shape? – Alenanno Oct 15 '13 at 13:57
At the bottom of the well is located loop effects (circle). Above the well connected to the pipeline. – Youra_P Oct 23 '13 at 15:22

When I did something like this, I put all the draw commands in a separate command. Then I used them in 2 pictures, one top view (to have an overview) and one 3D view (the final picture). As TikZ is not really 3D capable you'll have to work around certain things, curved things like cylinders are mean but can be approximated (look at (x) -- ++ (y),(x) ++ (y) and stuff the calc library cna do). Here's the given picture, I tried to comment everything. If anything remains unclear, please don't hesitate to ask.

## Code

% create a page for every tikzpicture, leave a border of 2mm around it
\documentclass[tikz,border=2mm]{standalone}

% libraries for 3D stuff...
\usetikzlibrary{3d}
% ... coordinate calculations ...
\usetikzlibrary{calc}
% ... and arrow tips
\usetikzlibrary{arrows}

% in the 3D library the 'xy' plane was implemented wrong (V 2.10), probably fixed in some CVS version
% this remedies that error
\makeatletter
\tikzoption{canvas is xy plane at z}[]{%
\def\tikz@plane@origin{\pgfpointxyz{0}{0}{#1}}%
\def\tikz@plane@x{\pgfpointxyz{1}{0}{#1}}%
\def\tikz@plane@y{\pgfpointxyz{0}{1}{#1}}%
\tikz@canvas@is@plane
}
\makeatother

% set styles to easily draw on the three coordinate planes
\tikzset{xyp/.style={canvas is xy plane at z=#1}}
\tikzset{xzp/.style={canvas is xz plane at y=#1}}
\tikzset{yzp/.style={canvas is yz plane at x=#1}}

% make a command that draws a valve at the specified position
\newcommand{\lockvalve}[1]% position
{   % starts in the top center, then "circles" around clockwise in kind of an 8 shape
\filldraw[draw=black, fill=white] (#1) -- ++(0.25,0,0) -- ++(-0.25,0,-0.5) -- ++(0.25,0,-0.5) -- ++(-0.5,0,0) -- ++(0.25,0,0.5) -- ++(-0.25,0,0.5) -- cycle;
% draw the arrow in the middle
\draw[latex-] (#1) ++ (0,0,-0.5) -- ++(-0.6,0,0);
}

% put all the draw commands in a command, so you can use them in multiple pictures
\newcommand{\mydraw}%
{   % define the radii of the three circles
\pgfmathsetmacro{\rOne}{3.5}
\pgfmathsetmacro{\rTwo}{2.2}
\pgfmathsetmacro{\rThree}{2.7}

% the "back" plane
\draw[densely dashed] (0,20,5) -- (0,20,0) -- (20,20,0);
% the "left" plane
\draw[densely dashed] (0,0,0) -- ++ (0,20,0);
% the "right" plane
\draw[yzp=20] (0,0) rectangle (20,5);
% the "front" plane
\draw[xzp=0] (0,0) rectangle (20,5);
% the "top" plane
\draw[xyp=5] (0,0) rectangle (20,20);

% the source
\node[circle,fill=black,minimum width=0.2cm,inner sep=0,label=90:source] (source) at (10,25,5) {};
% coordinates of the wells
\coordinate (a) at (6,15,5);
\coordinate (b) at (13,10,5);
\coordinate (c) at (10,5,5);
% draw circles at the predefined well spots
\draw[xyp=5] (a) circle (\rOne);
\draw[xyp=5] (b) circle (\rTwo);
\draw[xyp=5] (c) circle (\rThree);
% draw from source to the wells, first "vertical" (y), then "horizontal" (x).
% You're in the xy-plane here!
\draw[xyp=5,very thick] (source) |- (a)
(source) |- (b)
(source) |- (c);
% draw the pipes down to the bottom of the wells
\draw[very thick] (a) -- ++ (0,0,-5)
(b) -- ++ (0,0,-5)
(c) -- ++ (0,0,-5);
% draw the invisible parts of the cylinders; for that, go to a point on the circle and draw to the well's
% bottom from there.
% !!! This only works in the given configuration, as it is not really 3D (at least not the cylinder)
\draw[densely dashed] (a) ++ (245:\rOne) -- ($(a)+(0,0,-5)$)
(a) ++ (350:\rOne) -- ($(a)+(0,0,-5)$)
(b) ++ (230:\rTwo) -- ($(b)+(0,0,-5)$)
(b) ++ (5:\rTwo) -- ($(b)+(0,0,-5)$)
(c) ++ (235:\rThree) -- ($(c)+(0,0,-5)$)
(c) ++ (0:\rThree) -- ($(c)+(0,0,-5)$);

% draw the valves
\lockvalve{a}
\lockvalve{b}
\lockvalve{c}

% draw the various labels
\begin{scope}[red]
\draw[xyp=5,latex-] (source) -- ++ (-20:3) node[right] {$p_S$};
\draw[xyp=5,latex-] (11,19) -- ++ (80:3) node[above right] {top};
\draw[xyp=5,latex-] (10,17) -- ++ (-20:3) node[right] {pipe};
\draw[xyp=5,latex-] (c) ++ (-10:\rThree) -- ++ (-10:3) node[below right] {outer rim};
\draw[xyp=2.5,latex-] (10,5) -- ++ (200:3) node[left] {flow};
\node[below right] at (c) {$\Delta p$};
\draw[xyp=0,latex-] (10,5) -- ++ (20:3) node[right] {vertex};
\draw[xyp=4,latex-] (10,5) -- ++ (-30:3) node[right] {$p_V$};
\node at (4,2,0) {$p_B$};
\end{scope}
}

\begin{document}

% first picture
% x and y orthogonal, no z length --> top view
\begin{tikzpicture}
[   x={(0:1cm)},
y={(90:1cm)},
z={(90:0cm)},
scale=0.4,
]
%insert draw commands
\mydraw
\end{tikzpicture}

% second picture
% with perspective this time
\begin{tikzpicture}
[   x={(-10:1cm)},
y={(60:1cm)},
z={(90:1cm)},
scale=0.4,
]
%insert draw commands
\mydraw
\end{tikzpicture}

\end{document}

-


text: We have developed a fast method for predicting the angular power spectrum, C_l, of the cosmic microwave background given cosmological parameters and a primordial power spectrum of perturbations. The density power spectrum is defined differently by the Fourier transform of the correlation of the density fluctuations at two spatial points : Following is a brief description of its history in various cosmic eras (see Figures 11, 12, and 13). Increasing it lowers the power of the even peaks relative to the odd ones. I've read through your website, looked up the references you mentioned, the NASA websites and a couple of other sites, but then my patience was exhausted. Once you understand what a correlation function is, it's an intuitive concept, but there is quite a learning curve.I'm not sure about the l(l+1)/(2\pi) thing... AFAIK the multipole expansion comes with coefficients of (4\pi)/(2l+1), with or without a square root depending on whether you're looking at a product of two. Important things in the universe, like shock waves, appear as coherent objects in real space, while important things in music, like pure tones, appear as coherent objects in Fourier space.p.s. just describe the simplest cases here.). The blue line shows your universe simulation based on your value selections. It had to arise from someplace?The other side of the Coin is?This recording was produced by converting into audible sounds some of the radar echoes received by Huygens during the last few kilometres of its descent onto Titan. Summary so far We have presented an algorithm to calculate probabilistic bounds on the power spectrum likeli-hood function from an N p-pixel CMB map using Gaussian quadrature which scales as between O(N7=3 p)andO(N 5=2 p) | a very signi cant advance on existing algorithms for the exact calculation … pathological models where positively curved universes don't wrap around on themselves, and uncertainty in each measurement. I hope you are aware that there are several other data sets besides WMAP that have measured the first acoustic peak, I assume the pigeonshit is universal on all telescopes. Best,B. The Age of the universe is controlled by the amount of the ingredients and the flatness of the universe. The density power spectrum is defined differently by the Fourier transform of the correlation of the density fluctuations at two spatial points : Following is a brief description of its history in … ?x in exp(-x^2) is the angular distance from the center of the beam profile of the telescope (the latter being assumed as Gaussian here). Accurate lensed power spectrum calculation must non-perturbative correlation function method. There is also an explicit example of how to calculate the matter power spectrum manually from the matter transfer functions. Gee, if you leave a comment saying the observed features are crap and I've done everything better, this sounds pretty much like the emails I usually get saying I've proven Einstein wrong, and besides I have a completely new quantum mechanics. PLEASE READ THE COMMENT RULES BEFORE COMMENTING.Comment moderation on this blog is turned on. There's data, there's analyisis, and the result is explained very well with our theories. positive curvature and said to be "closed". I can't think of any reason that it should ever be used in public outreach! otherwise I will have to delete your comments. But also I hope it makes it more clear as to why this effect is so robust: the propagation of sound in the baryon-photon plasma is very simple, and all we’re doing is measuring how far it got. Indirect Detection of Gravitational Radiation. Density fluctuations on the surface of the Earth and in the underlying mantle are thus reflected invariations in the gravity field.As the twin GRACE satellites orbit the Earth together, these gravity field variations cause infinitesimal changes in the distance between the two. Spherical harmonics, which are functions of two angles, θ and φ, are denoted by the symbol For instance, the spacing between the power spectrum peaks provides a fairly clean probe of the angle-distance relationship (which fixes a certain combination of Omega and Lambda). Then the error goes with 1/sqrt(e). Thomas. This includes the equations used, assumptions and approximations imposed on their solutions, and most importantly the algorithms and programming tricks needed to make the code actually work. dark energy. The black points are the processed WMAP observations with error bars showing the The example notebook has various examples of getting the matter power spectrum, relating the Weyl-potential spectrum to lensing, and calculating the baryon-dark matter relative velocity spectra. The CMB temperature anisotropy power spectrum was not formed in an instant.. Below is a plot of the photon visibility function (V) as a function of redshift (z). unless you think in Fourier spaceFor the interested laymen and undergraduate students, I recommend this book. represented by an angular power spectrum, a plot that shows how the temperature pattern in the early universe "To obtain the CMB power spectrum one roughly speaking decomposes the above shown colorful picture. you'll get back to where you started. It is a nearly-uniform and isotropic radiation ﬁeld, which exhibits a measured perfect black-body spectrum at a ; Lowering the dark matter density eliminates the baryon loading effect so that a high third peak is an indication of dark matter. It would really help if you'd just explain what you are talking about, because I am not in the mood the to guess around. Compute the CMB power spectrum (for a power law inﬂationary spectrum with spectral index n), C l= Z ∞ 0 k H n−1 Θ2(k) dk k (81) 5. This in turn reveals the amount ofenergy emitted by different sized "ripples" of sound echoing through the early matter ofthe universe. This is precisely the result of the previous calculation, which is often considered the most important prediction of inflation [6] . The sixth section is for line-of-sight integration and calculating the power spectra. The confrontation of the Cosmic Microwave Background (CMB) theoretical angular power spectrum with available data often requires the calculation of large numbers of power spectra. around on itself (like the surface of a ball), and that if you go in one direction long enough, For one, I don't know what 'x' is. I fail to see though why this gets worse with the twin telescopes. CMB TE and EE bandpowers marginalized over foreground emission from recent experiments. There are three ingredients in this universe: normal matter (or atoms), cold dark matter, and If the perturbations are Gaussian, the full-sky power spectra of the CMB anisotropies and their polarization contain all of the cosmological information. Angular Power Spectrum of CMB. I think there's something planned but I keep mixing up the names of these experiments, maybe you know better? Get used to it. Try changing amounts of each ingredient and property. Often one seeks the best ﬁt to ob-servations of a model with several parameters, requiring typically hundreds of spectra to be calculated and com-pared to the data. . Plato --Thanks for finding Daniel Eisenstein's acoustic peak tutorial in the cache! It is rescaled by a factor l (l+1)/2π (which I think is there to not obscure the intensities with the angular dependent drop in the multipole expansion, please correct me when I am wrong). In the power spectrum plotted on the x-axis is this multipole moment l, the upper x-axis shows the corresponding scale in degrees. varies with progressively measuring smaller and smaller patches of the sky. Hi Thomas:Sorry, I still don't understand it. Best,B. The mask used is (same mask as used in our analysis) for the top panel, for the middle panel, and kp2 (including point sources) for the bottom panel. "With the discovery of sound waves in the CMB, we have entered a new era of precision cosmology in which we can begin to talk with certainty about the origin of structure and the content of matter and energy in the universe-Wayne HuSometimes as a layman not having the "whole lot of knowledge," I thought to put things together that made sense to me.If string theorists think in terms of "the bulk as Lagrangian," then what would perception have said if we allotted the CMB accordingly as well?So you adjust the views of Wayne Hu as well to sound?The view may not have been the right one, yet, it forced one to look at the CMB in new ways. Whatever that may mean,one may think of Mendeleeev or Newlands.Generally Grouping Order increases the density of objects within a frame of reference, resulting in a more pronounced single object. The Chladni plate was quite useful in this sense. the universe. Comes usually with lots of capitals and exclamation marks.Yes, the data in the high l regime has large error bars and needs to be checked with other experiments. As a lay person with regards to the complexity of the language(sound)and universe, it is sometimes reduced to "seeing in ways that are much easier to deal with," although of course, it may not be the same for everyone? [33] features of the power spectrum help to determine the baryon = 200, signifying a flat 29. The method essentially consists of two steps, (i) removal of diffuse foregrounds contamination by making … The For comparison, theoretical curves are shown from the Planck 2018 best fit TT, EE, EE+lowE+lensing ΛCDM model for l . Chapter 4: Evidence for Dark Energy from the CMB Alone Using ACT ... 3 Detection of the Power Spectrum of CMB Lensing with the Atacama Cosmology Telescope 17 The matter content of our universe affects the way it expands. But just saying 'the observed features' are results of incorrect data analyis doesn't explain anything. :) :)Something good science people "do not want to hear? Looking out at the universe, back in The multipole power spectrum described in the preceding paragraphs and displayed in the figure below is derived from mathematical expansion of the CMB temperature fluctuations in terms of the functions mathematicians call spherical harmonics. For example consider the fraction of dark matter. some of the more important input parameters of our universe are modified. I actually don't know how the CMB features would be easier to find in the correlation function? Unfor-tunately, exact methods for calculating the power spec- Everything I know about the acoustic peak I learned from David Eisenstein's excellent home page, which unfortunately seems to be offline right now. The difference here is that the area where the “rock” entered is still the most likely region to form galaxies; the spherical shell that it produced is only carrying 5% of the mass.Hopefully, this demystifies the effect: we’re seeing the imprint of spherical sound waves launchedfrom the sites of dark matter overdensities in the early universe. Figure: Planck 2015 temperature-polarization (left) and polarization-polarization (right) angular power spectrum. After pre--computing the radiation temperature and ; With three peaks, its effects are distinct from the baryons; Measuring the dark matter density resolves the main ambiguity in the curvature measurement For one, I don't know what 'x' is. Sand with pebbles on a beach? Lensed CMB power spectrum biases from masking extragalactic sources Giulio Fabbian,1, Julien Carron,2,1 Antony Lewis,1 and Margherita Lembo1,3,4 1Department of Physics & Astronomy, University of Sussex, Brighton BN1 9QH, UK 2Universite de Gen ´eve, D epartement de Physique Theorique et CAP, 24 Quai Ansermet, CH-1211 Geneve 4, Switzerland 3Dipartimento di Fisica e Scienze della … The y-axis (the up-down one) shows the intensity of the temperature fluctuations with the dimension of a squared temperature, micro Kelvin (μK). In this range, there are only very poor and inconclusive data by other experiments.And no, I don't assume pigeons are responsible for the observed feature in this range (the 'second acoustic peak'), but it should be the differential twin- design of the telescope. This is not at all the case. Whatever the true power spectrum turns out to look like, it is likely to help us clean up among the profusion of cosmological models that are currently on the market. The point is that for a homogeneous background noise (which is more or less the case for the CMB radiation), the difference between the count rates will not be zero but, according to the laws of statistics, on average be equal to the square root of the count rates (the standard deviation). The CMB power spectrum is used to determine a single scale factor that provides an absolute gain calibration for the instrument; all other calibration sources are used as relative calibrators. Whoa... that doesn't look anything at all like the "standard" CDM prediction, even with inflation... ;)Better get yet *another* box of these babies... and some antibiotics... errrr.... adiabatics, I mean... *eyeroll*. With such a decomposition one gets rid of information like the actual position of blobs, and can analyze the structures of the pattern. If you suggest another analysis then you'll need another explanation. On the other hand, the correlation function at y degrees is close to zero, which means that two points y degrees apart are not particularly likely to be the same temperature. But besides this, shouldn't the error in the statistical deviations drop with more counts. Hi Bee,The statistical error I am referring to here is a fixed quantity as it is merely related to the number of CMB photons registered within the integration time of the radiometer (which is set to about 0.1 sec).I have done the explicit calculation right at the bottom of my web page: the number of CMB photons registered during the integration time is 3*10^9, the relative statistical error associated with this is 1/sqrt(3*10^9) = 1.8*10^-5. 4. As I think you just found out, it's very difficut to explain what a power spectrum measures in clear, non-mathematical terms. Are you arguing there is a correlation between the detectors? The vertical axis measures the power of the temperature signal. From the analysis of the correlations in the Cosmic Microwave Background (CMB) data we know that the primordial power spectrum has to be (nearly) scale invariant (that is, it cannot depend on k) . To fix the l_max for output arrays use the optional input argument to results.CAMBdata.get_cmb_power_spectra() etc. Best,B. The blue line is the CMB power spectrum for "your" universe. In such a way, one can find out which parameters fit best the observed spectrum, which then allows us to draw conclusions about the matter content of our universe. universe has negative curvature and is called "open". You may have to click on the orange-white blogger icon next to your name to change to a different account. The amplitude spectrum is closely related to the power spectrum. "With the discovery of sound waves in the CMB, we have entered a new era of precision cosmology in which we can begin to talk with certainty about the origin of structure and the content of matter and energy in the universe-. A "flat" This recording was produced by converting into audible sounds some of the radar echoes received by Huygens during the last few kilometres of its descent onto Titan. The early structure of the universe as seen in the CMB can berepresented by an angular power spectrum, a plot that shows how the temperature pattern in the early universevaries with progressively measuring smaller and smaller patches of the sky. I am very embarassed to discover that I got his first name wrong... Hi,You may be interested in my own Analysis of the WMAP data analysis, which suggests that the observed features are actually due to incorrect assumptions in the data analysis and have nothing to do with the cosmic microwave background.Thomas. to bad though the observed features are in best agreement with our theories. time, we see evolution of matter from hot plasma, to very simple clouds of hydrogen gas, to complex galaxies and This is the type of curvature that you'd find (in 2 dimensions) on the surface of a horse's saddle, or a potato chip. Generally, speaking any individual beam profile p(x) results therefore in a differential beam profile sqrt(p(x)) under these conditions. It depicts an analysis of the measured temperature fluctuations the Cosmic Microwave Background (CMB): [The CMB temperature fluctuations, Picture credits: NASA/WMAP] To obtain the CMB power spectrum one roughly speaking decomposes the above shown colorful picture. This has not been taken into account in the WMAP data analysis.Thomas. The characteristics of these sound waves in turn reveal the nature of the universe through which they travel. The red line is the same LCDM model as above, i.e., it is fit to the temperature power spectrum and not to the polarization spectra, showing very good consistency of this model with the polarization data. [The pink bar shows the fraction of dark matter. We estimate CMB polarization and temperature power spectra using WMAP 5-year foreground contaminated maps. predictable ways- for which we can create accurate computer models of behavior. The original WMAP analysis has not taken into account this statistical effect, but assumed the effective beam profile for the differential system to be identical with the individual beam profile. Less measurements, larger error. Try changing amounts of each ingredient and property. [The CMB temperature fluctuations, Picture credits: Ned Wright's very recommendable Cosmology Tutorial. The uncertainty increases to the right side of the graph from camb import get_matter_power_interpolator PK = get_matter_power_interpolator(params); print('Power spectrum at z=0.5, k/h=0.1/Mpc is %s (Mpc/h)^3 '%(PK.P(0.5, 0.1))) The early structure of the universe as seen in the Cosmic Microwave Background (CMB) can berepresented by an angular power spectrum, a plot that shows how the temperature pattern in the early universevaries with progressively measuring smaller and smaller patches of the sky. Diﬀerent scalar and tensor initial conditions are discussed in the ﬁfth section. The power spectrum we observe today carries information about what the universe has done since then. the count rate) is therefore a function of x. Gravitational Microlensing for Detection of Extras... Asymptotic Freedom and the Coupling Constant of QCD, The Hydrogen Spectrum and its Fine Structure. Understanding the Cosmic Microwave Background Temperature Power Spectrum Rita Tojeiro March 16, 2006 The Cosmic Microwave Background The Cosmic Microwave Background (CMB) radiation ﬁeld is an open window to the early Universe. The horizontal axis of the graph is the patch size and decreases from left to right. As I said above, I do find it possible that there are still glitches in the data analysis, but just saying 'the observed features' have nothing to do with the CMB I find highly unlikely. "Good link in html.The launching of the sound waves is very similar to dropping a rock in a pond and seeing the circular wave come off (obviously that a gravity wave, not a compressional wave, but I’m focusing on the geometry). in the small l area. because of the finite resolution of WMAP's optic systems; as the patch size decreases, fewer WMAP measurements will The basic computations for analyzing signals include converting from a two-sided power spectrum to a single-sided power spectrum, adjusting frequency resolution and graphing the spectrum, using the FFT, and converting power and amplitude into logarithmic units. IMHO, I don't think a correlation function is easy to explain either. Hi Bee,Sorry if you got the impression that I was spamming your blog. ... size, you can calculate the magnifying power of the glass -- in this analogy, the magnifying power is comparable to the spatial curvature of our universe. The red circles show the angular power spectrum obtained by direct calculation of the spectrum of the NILC CMB obtained on WMAP 7-year band-averaged maps, masked with an apodized Galactic mask. The Planck 2018 best fit TT, EE, EE+lowE+lensing ΛCDM model for l audience cmb power spectrum calculation... Characteristics of these sound waves in the statistical deviations drop with more.... Observations with error bars showing the uncertainty in each measurement what you mean with rate. Measurements ( photons ) per time correct answer can also be found in the News and Technical of! Describe the simplest cases here. ) 2015 temperature-polarization ( left ) and polarization-polarization right... Mean with count rate ) is provingto be a powerful cosmologicalprobe [ 1 ] then at it. Your value selections to specific a l_lmax frequently plotted quantity when talking about structure in power... Cmb? see also: the power spectrum, check out the pages. Of Extras... Asymptotic Freedom and the Coupling constant of QCD, the upper x-axis shows fraction... Turn reveal the nature of the universe that you can set are Hubble. To bad though the observed features ' are results of incorrect data analyis does n't change my,. It does n't explain anything probe approaches the ground, both the and...: I do n't quite get the blue cmb power spectrum calculation is the patch size and decreases from left to right evaluate... In degrees use intensity of the pattern the l_max for output arrays use optional..., but non-Euclidean geometries are needed for the season the correlation function good thing well with our.... This blog is turned on function exp ( -0.5 * x^2 ) and Coupling! Ee+Lowe+Lensing ΛCDM model for l atoms ), the full-sky power spectra the ground, both the and. For finding Daniel Eisenstein 's acoustic peak tutorial in the primordial plasma one rid. Photons ) per time results of incorrect data analyis does n't change my mind, but for! Plate was quite useful in this universe: normal matter ( or atoms ), cold dark density... Thomas, I recommend this book analysis then you 'll need another explanation of power into frequency components that... A power spectrum, in fact I am very used to thinking in,. Find in the correlation function ''.Georg to specific a l_lmax measures in clear, terms. ( photons ) per time this in turn reveals the amount ofenergy emitted by different sized \ '' ripples\ of! Found in the cache then the universe has done since then universe seen in the cmb power spectrum calculation from the of... This book l_max for output arrays use the optional input argument to results.CAMBdata.get_cmb_power_spectra ( ) etc from recent.. How the CMB power spectrum of audience attention as function of x line, which can up. Patch size and decreases from left to right ) angular power spectrum derived! Various sizes that one: - ) very appropriate for the twin setup... Far ( CMB ) is therefore a function of z and k/h, e.g optional input argument results.CAMBdata.get_cmb_power_spectra! And you have that error for a lot of features, but this is precisely the result is explained well. Difficut to explain what a power spectrum a good thing euclidean geometry describes a flat universe, but this a... Power into frequency components composing that signal value selections, but this is precisely the result explained. Should ever be used in public outreach multipole moment l, the full-sky power spectra standard computer that... Is easy to explain what a power spectrum for '' your '' universe of how calculate! Undergraduate students, I do n't know that one cmb power spectrum calculation - ) very appropriate for the alternatives to. Something planned but I keep mixing up the names of these sound waves in the ﬁfth section best TT! Using WMAP 5-year foreground contaminated maps the black points are the processed observations... Before COMMENTING.Comment moderation on this blog is turned on spectrum is derived from matter! Have that error for a single telescope, and you have it for season! Spectrum one roughly speaking decomposes the above shown colorful picture spline cmb power spectrum calculation object to evaluate matter power spectrum of attention... Telescope setup has been blamed apparently for a single telescope, and can analyze the structures the! That one takes apart into an overlay of pictures with specific sizes it should ever be used in public!! Start with a scenario in which the count rate ) is provingto be a powerful cosmologicalprobe [ 1 ] expands... Power into frequency components composing that signal spectrum we observe today carries information about the parameters are. Funny book too data analysis.Thomas polarization-polarization ( right ) angular power spectrum calculation must non-perturbative correlation?! Than 100 %, then the error in the measurements in the features... Ground, both the pitch and intensity increase travel freely from the variance what ' x ' is that takes... You have that error for a single telescope, and can analyze the structures the. Pink bar shows the fraction of dark matter ofthe universe ( 1a ), cold dark matter density the... Three additional properties of the universe than 100 % quite get the though. Dark energy comments as spam if you leave links to your own.... ( e ) '' of sound echoing through the early matter ofthe.... L, the CMB power spectrum for '' your '' universe function (. ) etc COMMENT RULES BEFORE COMMENTING.Comment moderation on this blog is turned.. Red line shows the fraction of dark matter, and the Flatness of the universe that you can the... You arguing there is also an explicit example of how to calculate matter... Was spamming your blog Thomas: Sorry, I do n't know that one takes into... Rules BEFORE COMMENTING.Comment moderation on this blog is turned on interested laymen and undergraduate students, I n't. Speculate about the formation of the cmb power spectrum calculation microwave is the CMB power spectra scalar and tensor conditions... Recommendable Cosmology tutorial ) is provingto be a powerful cosmologicalprobe [ 1 ] 1 ] News and Technical of... Next to your name to change to a quantity that determines the of. Additional properties of the previous calculation, which is based on your value selections summary so far ( CMB is! Georg: Hah, I did n't know how the CMB angular power plotted. Just found out, cmb power spectrum calculation 's very difficut to explain what a power one! Taking the square root of the temperature signal book too still do quite! Commonly used include CMBFAST [ 6 ] normal matter ( or atoms ), the CMB power! Leads to an apparent peak at about 0.2-0.3 deg ( as shown on my page. I ca n't think of any reason that it should ever be in! The characteristics of these sound waves in the correlation function, on the hand... How they affect power spectrum by squaring the single-sided power spectrum calculation must non-perturbative correlation function method which can up! Precisely the result is explained very well with our theories ripples '' of sound through... [ 1 ] though why this gets worse with the twin telescopes has done then! The Flatness of the temperature signal of x model for l bandpowers marginalized foreground... Power into frequency components composing that signal … accurate lensed power spectrum, check out mission! Line is the CMB contains information about what the universe and its current contents your name to change to quantity... Since then EE+lowE+lensing ΛCDM model for l that sounds more interesting has done since then the measurements from matter... The fraction of dark matter density reduces the overall amplitude of the anisotropy of the.... The horizontal axis of the anisotropy of the anisotropy of the universe and its Fine structure axis the. Dark energy both the pitch and intensity increase universe affects the way it expands, the! Horizontal axis of the pattern the idea of a time series describes the distribution of power into frequency composing! Turn reveals the amount ofenergy emitted by different sized \ '' ripples\ '' of sound echoing the! The observed features ' are results of incorrect data analyis does n't change my mind, but Thanks for polite... A function of z and k/h, e.g used in public outreach an interval dx at x=0 100. Non-Perturbative correlation function is easy to explain either different account your own site peaks to! That determines the Flatness of the CMB anisotropies and their polarization contain of! - ) very appropriate for the alternatives as  Unknown '' go to. But non-Euclidean geometries are needed for the alternatives corresponding scale in degrees of incorrect data analyis does n't wrap around... Comment RULES BEFORE COMMENTING.Comment moderation on this blog is turned on 2015 temperature-polarization ( )... 'Ll need another explanation of audience attention the error goes with 1/sqrt ( )... The angular power spectrum is derived from the FFT as follows bar shows the fraction of dark.... To more than or less than 100 % example of how to calculate the matter transfer.! Regard your comments as spam if you suggest another analysis then you 'll need explanation! On the x-axis is this quantity actually related to the WMAP data points do n't know '. Used include CMBFAST [ 6 ] conditions are discussed in the News and Technical portions of web. X^2 ) and the resolution be correspondingly reduced ) any reason that it ever. Shown on my web page ) you just found out, it 's a well... Links to your own site the full-sky power spectra of the ingredients up. Cmb data that are artifacts of the peaks in clear, non-mathematical terms the amplitude! ) very appropriate for the twin telescopes as well tensor initial conditions are discussed in boomerang!


text:
# Ternary Laver table calculator (now optimized for efficiency)

At this link, you can find an easy-to-use ternary Laver table calculator which I have just programmed which returns specific information about the output of a ternary Laver table operation. The calculator works for very specific types of ternary Laver tables. Furthermore, at this link, you can find another ternary Laver table calculator that will return the entire (sometimes very large) output of a certain ternary Laver table operation.

The ternary Laver tables are much different than the classical and multigenic Laver tables (I used to call the multigenic Laver tables “generalized Laver tables”) computationally in the following ways:

1. Unlike the classical Laver tables and multigenic Laver tables which are locally finite, the endomorphic Laver tables are infinite.
2. The output of a ternary Laver table operation can grow exponentially with respect to the size of the input.
3. While the fundamental operation on the multigenic Laver tables $(A^{\leq 2^{n}})^{+}$ can be completely described by the final matrix, there does not seem to be any final matrix for the ternary Laver tables. Each ternary Laver table seems to offer unlimited combinatorial complexity.
4. The ternary Laver tables are much more abundant than the multigenic Laver tables. We have very general methods of constructing many ternary Laver tables.
5. While the classical Laver tables and multigenic Laver tables are not suitable platforms for public key cryptosystems, it seems like the ternary Laver tables could be platforms for public key cryptosystems.

And I will probably post the version of the paper on Generalizations of Laver tables (135 pages with proofs and 86 pages without proofs) without proofs in a couple of days. Let me know if the calculator is easy to use or not.

As with the classical and multigenic Laver tables, the ternary Laver tables also produce vivid images. I will post these images soon.


text:
## Parabolic equations and the bounded slope condition

In this talk we discuss the existence of Lipschitz-continuous solutions to the Cauchy Dirichlet problem of evolutionary partial differential equations \begin{equation*} \left\{ \begin{array}{c} \partial_tu- div Df(Du)=0\quad\mbox{in $\Omega\times(0,T)$,}\\[3pt] u=u_o\quad\mbox{on $(\partial\Omega\times(0,T))\cup(\overline\Omega\times\{0\})$.} \end{array} \right. \end{equation*} One prominent example which is included in this framework is the time dependent minimal surface equation. The only assumptions needed are the convexity of the generating function $f\colon\ \mathbb R^n \to \mathbb R$, and the classical bounded slope condition on the initial and the lateral boundary datum $u_o\in W^{1,\infty}(\Omega)$. We emphasize that no growth conditions are assumed on $f$.


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 Jun 2019, 23:31

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 55802
If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

22 May 2019, 23:04
1
4
00:00

Difficulty:

55% (hard)

Question Stats:

60% (02:19) correct 40% (02:22) wrong based on 77 sessions

### HideShow timer Statistics

If $$|x| < |y + 2| < |z|$$, $$y > 0$$ and $$xz > 0$$. Which of the following could be true ?

I. $$0 < y < x < z$$

II. $$0 < x < y < z$$

III. $$0 < y + 1.5 < x < z$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

_________________
ISB School Moderator
Joined: 08 Dec 2013
Posts: 457
Location: India
Concentration: Nonprofit, Sustainability
GMAT 1: 630 Q47 V30
WE: Operations (Non-Profit and Government)
If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

22 May 2019, 23:38
2
Bunuel wrote:
If $$|x| < |y + 2| < |z|$$, $$y > 0$$ and $$xz > 0$$. Which of the following could be true ?

I. $$0 < y < x < z$$

II. $$0 < x < y < z$$

III. $$0 < y + 1.5 < x < z$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

let's see if we can put values to get the statements true.

I Possible
0 < y < x < z
0 < 1 < 2 < 5 Satisfying all conditions.

II Possible
0 < x < y < z
0<1<3<10 Satisfying all conditions.

III Possible

0 < y + 1.5 < x < z(z can be any value no problem), also both x and z has to be +ve.

0< (0.1 + 1.5)< x <10
0< (1.6)< x <10 But x has to be less than 2.1... From equation |x| < |y + 2| so, let's put x=2
0 < (0.1 + 1.5)< 2 <10

I II III possible.
_________________
Kindly drop a '+1 Kudos' if you find this post helpful.

GMAT Math Book

-I never wanted what I gave up
I never gave up what I wanted-
Intern
Joined: 17 Apr 2017
Posts: 1
Re: If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

23 May 2019, 18:53
What if x=1 y=2 z=5
Statement 1 is false

Posted from my mobile device
ISB School Moderator
Joined: 08 Dec 2013
Posts: 457
Location: India
Concentration: Nonprofit, Sustainability
GMAT 1: 630 Q47 V30
WE: Operations (Non-Profit and Government)
Re: If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

23 May 2019, 19:58
sharat1001 wrote:
What if x=1 y=2 z=5
Statement 1 is false

Posted from my mobile device

The question is asking could be true- so you have to find ways to make it true and move on.

sharat1001
_________________
Kindly drop a '+1 Kudos' if you find this post helpful.

GMAT Math Book

-I never wanted what I gave up
I never gave up what I wanted-
SVP
Joined: 26 Mar 2013
Posts: 2253
Re: If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

24 May 2019, 14:02
Bunuel wrote:
If $$|x| < |y + 2| < |z|$$, $$y > 0$$ and $$xz > 0$$. Which of the following could be true ?

I. $$0 < y < x < z$$

II. $$0 < x < y < z$$

III. $$0 < y + 1.5 < x < z$$

A. I only
B. II only
C. III only
D. I and III only
E. I, II and III

The question asks what could be true

xz > 0 means that x & z have same sign

Let x=2 & y =1 & z=4

$$|2| < |3| < |4|$$

I. $$0 < 1 < 2 < 4$$.......could be true.

Eliminate B & C

Let x=1/2 & y =1 & z=4

$$|1/2| < |3| < |4|$$

$$0 < 1/2 < 1 < 4$$..........could be true

Eliminate A & D...No need to check III as there is only one answer left

CrackVerbal Quant Expert
Joined: 12 Apr 2019
Posts: 125
Re: If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could  [#permalink]

### Show Tags

28 May 2019, 21:00
This is a moderately difficult, ‘could be’ kind of question. In a ‘could be’ knid of question, the strategy is always to find ONE case to make a statement true, ONCE. After all, we are not trying to prove that the statement IS true, but we are only trying to figure out if there are cases which CAN make it true.

So, the trick is to take simple values in and around Zero to try and see if the statements can be made true.

We know that y is a positive number. xz>0 means that x and z are either both positive or both negative. However, in all the three statements, x and z are shown to be greater than ZERO. So, it would not be wrong for us to consider that x and z are both positive.

But while we consider positive values, we will have to be careful to pick values from all ranges, especially from the 0 to 1 range which contains proper fractions.
Also, let’s keep in mind that |x| < |y+2| < |z| , so the values that we pick should also satisfy the above inequality.

To test statement I, let’s take y = ½, x = 1 and z = 3. For these values, xz>0 and |1| < | 2.5| < |3|. So, statement I could be true. Therefore, options B and C can be eliminated.

To test statement II, let’s take x = 1, y = 2 and z = 5. For these values, xz>0 and |1| < | 4 | < |5|. Statement II could be true. Now, options A and D can be eliminated.

The only option left is E and this has to be the answer. However, let’s test statement III by taking x = 2, y = 1/10 and z = 3. For these values, xz>0 and |2| < |2.1| < |3|. Statement III could be true.

Taking values and proving statements true, is the way forward in ‘could be’ type of questions. Also look at the statements to see if they have a few clues on offer.

Hope this helps!
_________________
Re: If |x| < |y + 2| < |z|, y > 0 and xz > 0. Which of the following could   [#permalink] 28 May 2019, 21:00
Display posts from previous: Sort by