text:
## Marks 1

More
Consider a laminar flow at zero over a flat plate . The shear stress at the wall is denoted by $${\tau _w}.$$ The axial ...
GATE ME 2017 Set 2
For steady flow of a viscous incompressible fluid through a circular pipe of constant diameter, the average velocity in ...
GATE ME 2017 Set 1
Three parallel pipes connected at the two ends have flow-rates $$\,\,\,{Q_1},\,\,\,{Q_2},\,\,\,\,$$ and $${Q_3}$$ respec...
GATE ME 2015 Set 3
Couette flow is characterized by
GATE ME 2015 Set 3
Consider fully developed flow in a circular pipe with negligible entrance length effects. Assuming the mass flow rate, d...
GATE ME 2015 Set 1
Maximum velocity of a one-dimensional incompressible fully developed viscous flow, between two fixed parallel plates, is...
GATE ME 2010
The velocity profile of a fully developed laminar flow in a straight circular pipe, as shown in the figure, is given by ...
GATE ME 2009
In fully developed laminar flow in the circular pipe, the head loss due to friction is directly proportional to ....... ...
GATE ME 1995
For a fully developed laminar flow through a pipe, the ratio of the maximum velocity to the average velocity is ________...
GATE ME 1994

## Marks 2

More
Consider steady flow of an incompressible fluid through two long and straight pipes of diameters $${d_1}$$ and $${d_2}$$...
GATE ME 2017 Set 1
Consider a fully developed steady laminar flow of an incompressible fluid with viscosity $$\mu$$ through a circular pip...
GATE ME 2016 Set 3
For a fully developed laminar flow of water (dynamic viscosity $$0.001$$ $$Pa$$-s) through a pipe of radius $$5$$ $$cm,... GATE ME 2015 Set 2 The head loss for a laminar incompressible flow through a horizontal circular pipe is$${h_1}$$. Pipe length and fluid r... GATE ME 2015 Set 2 A fluid of dynamic viscosity$$2 \times {10^{ - 5}}\,\,kg/m.s$$and density$$1kg/{m^3}$$flows with an average velocity... GATE ME 2014 Set 3 Water flows through a$$10mm$$diameter and$$250m$$long smooth pipe at an average velocity of$$0.1m/s....
GATE ME 2014 Set 2
Consider laminar flow of water over a flat plate of length $$1$$ $$m.$$ If the boundary layer thickness at a distance of...
GATE ME 2014 Set 2
Water flows through a pipe having an inner radius of $$10$$ $$mm$$ at the rate of $$36$$ $$kg/hr$$ at $${25^ \circ }C$$....
GATE ME 2014 Set 1
For a fully developed flow of water in a pipe having diameter $$10$$ $$cm,$$ velocity $$0.1$$ $$m/s$$ and kinematic visc...
GATE ME 2014 Set 1
The velocity profile in fully developed laminar flow in a pipe of diameter $$D$$ is given by $$u = {u_0}\left( {1 - 4{r^... GATE ME 2006 For laminar flow through a long pipe, the pressure drop per unit length increases. GATE ME 1996 The discharge in$${m^3}/s$$for laminar flow through a pipe of diameter$$0.04m$$having a centre line velocity of... GATE ME 1988 ## Marks 5 More A$$0.20m$$diameter pipe$$20km$$long transport oil at a flow rate of$$0.01{m^3}/s. Calculate the power r...
GATE ME 1988

### EXAM MAP

#### Graduate Aptitude Test in Engineering

GATE ECE GATE CSE GATE CE GATE EE GATE ME GATE PI GATE IN


text:
# Weeknotes: 116

## Work

I spent much of this week writing documentation and thinking about things, sometimes in the other order. One of the things I’ve been thinking about is how cross-domain analytics will work with GOV.UK Accounts

GOV.UK uses cross-domain analytics (only if the user has consented to cookies on both domains, of course) to give a joined up view of how users interact with government. So naturally this has to work with GOV.UK Accounts too, or we’ll have an ever-growing blind spot in this joined up view.

Cross-domain analytics works by sticking a _ga query param onto cross-domain links, which the analytics code at the link destination picks up. The problem is that with accounts, most of the cross-domain stuff is in the form of OAuth journeys: we have cross-domain redirects. So we have non-cross-domain links which may do a cross-domain redirect; for example, navigating from an unauthenticated part of a service to an authenticated part, the link is a normal internal link, but it’ll redirect a logged-out user to the accounts domain.

So I spent some time thinking through all the cases:

• is the link from the accounts domain or the service domain?
• is the link to an authenticated or an unauthenticated page?
• does the user have an active session on the service domain?
• does the user have an active session on the accounts domain?

For each, figuring out which links a user will click on, which redirects will happen, and where a _ga param needs to be passed along for the cross-domain analytics to all fit together.

Geez, the web would be much simpler00And less secure. Ah, if only we could just trust everyone to not be malicious!

## Books

• Midnight Tides by <NAME>, the fifth of the Malazan Book of the Fallen.

This was an interesting one, as I had a lot of difficulty figuring out when it was set. Malazan can jump across continents and time when it changes from one book to the next, but this was particularly tricky. It’s the story of the Tiste Edur people, before <NAME> was abandoned in a warren for his crimes. So it’s clearly before the other books, but how long before?

The humans in this book use the Tiles of the Holds, not the Deck of Dragons, and even that’s considered a more modern alternative to the traditional Edur sorcery, so is it in the distant past, thousands of years before the other books?

There’s more evidence throughout the book which let me place it fairly accurately in the timeline by the end, but it took most of the book to get there.

format: plain


text:
“The law regarding the distinction between extortion and settling a legal claim is very unsettled—it’s not well defined”

GeneWilderFascination.jpg

A Mastodon instance for maths people. The kind of people who make $\pi z^2 \times a$ jokes.

Use $ and $ for inline LaTeX, and $ and $ for display mode.


text:
These are some notes about AES encryption in R, because it took me a little fiddling to produce a working example of my own.

library(digest)

First, the example from the digest vignette

msg <- as.raw(rep(1:16, 2)) # encrypt this
key <- as.raw(1:16)         # with this key

Now in CBC mode: each encoding is different

iv <- sample(0:255, 16, replace=TRUE)
aes <- AES(key, mode="CBC", iv)
code <- aes$encrypt(msg) code Need a new object for decryption in CBC mode aes <- AES(key, mode="CBC", iv) And here’s the original msg: aes$decrypt(code, raw=TRUE)

Next, my own example

encryptthis <- "There's no place like home."
stepone <- charToRaw(encryptthis)

Now append to the raw vector as many bytes as needed for its length to be a multiple of 16. What bytes you might append? Just pick the raw byte representation of the number needed. This is also known as PKCS5 padding:

rem <- 16L - length(stepone) %% 16L

This rule also ensures that your raw string is always padded. If by chance the length of the original string in raw format is a multiple of 16 already, then rem = 16 and you will pad it with as.raw(rep(16, 16)). This is nice because if the reader knows that this is your recipe for padding, they just need to check what number the last character is, move back as many elements in the decrypted raw vector, and drop them as padding.

steptwo <- c(stepone, as.raw(rep(rem, rem)))

Now set a key that will be shared with the recipient:

key <- as.raw(sample(0:255, 16, replace=TRUE))

And set an initial vector for encryption:

iv <- sample(0:255, 16, replace=TRUE)

Now encrypt, CBC mode:

aes <- AES(key, mode="CBC", iv)
decryptthis <- aes$encrypt(steptwo) Pre-pend the iv as raw vector, and send to recipient. decryptthis <- c(as.raw(iv), decryptthis) Now the recipient gets the message decryptthis and the key key. She knows that the first 16 raw bytes of the message are the IV, and expects that there’s padding at the end. So, with this key and this knowledge, here’s the decryption object: aes <- AES(key, mode="CBC", decryptthis[1:16]) And here’s the original message, in raw bytes: padded <- aes$decrypt(decryptthis[17:length(decryptthis)], raw = TRUE)

Now find the size of the padding, remove it, then show the original message in plain text:


text:
Page 1 of 1

### Chittagong-Higher secondary 2016

Posted: Mon Dec 24, 2018 12:56 pm
The lengths of the sides of the rectangle ABCD are 10 and 11. An equilateral triangle is drawn in such a way that no point is situated outside ABCD. The maximum area of the triangle can be expressed as (p√q)/r where p, q and r are positive integers and q is not divisible by any square number. p+q+r=?

I have determined the maximum area of the triangle (221√3 - 400).But the matter of irony is that the area isn't in the form as mentioned in the question.

### Re: Chittagong-Higher secondary 2016

Posted: Mon Feb 18, 2019 4:28 pm
mizan_24 wrote:
Mon Dec 24, 2018 12:56 pm
The lengths of the sides of the rectangle $ABCD$ are $10$ and $11$. An equilateral triangle is drawn in such a way that no point is situated outside ABCD. The maximum area of the triangle can be expressed as $\dfrac{p\sqrt q}{r}$ [where $p, q$ and $r$ are positive integers and $q$ is not divisible by any square number. $p+q+r=?$

I have determined the maximum area of the triangle $(221\sqrt 3 - 400)$.But the matter of irony is that the area isn't in the form as mentioned in the question.
What is the question number?


text:
Open access peer-reviewed chapter

# Economics of Carbon Capture and Storage

By <NAME> and <NAME>

Submitted: April 25th 2016Reviewed: November 21st 2016Published: March 8th 2017

DOI: 10.5772/67000

## Abstract

Human-engineered capture of CO2 emissions at the point source and subsequent long-term storage of this CO2 underground represent a potential mitigation strategy for global warming. The so-called carbon capture and storage (CCS) projects are technically feasible but have not been well established from an economic efficiency perspective. This chapter uses economic theory to describe the costs, benefits, and economically efficient level of CCS provision. Achieving the economically efficient level of CCS provision requires consideration of both the private and public costs and benefits of CCS and will also likely require some degree of government intervention in the form of economic incentives and/or direct regulation.

• CO2
• emissions
• point source
• capture
• storage
• economics
• costs
• benefits

## 1. Introduction

Since the late twentieth century, a newly developed technology has become one of the tools that can help mitigate the negative impacts on climate change from rising levels of greenhouse gases, especially CO2. This technology is commonly known as the carbon capture and storage (CCS). CCS technology involves “capturing” CO2 emissions, say from a coal-fired power plant, and then depositing the captured CO2 gas in a storage site, such as an underground geological formation, where it will not enter the atmosphere. CCS projects are currently being tested and implemented throughout the world. However, economic feasibility of human-engineered CCS is not well established [14]. The purpose of this chapter is to discuss the economic benefits and costs of CCS projects from both private and public perspectives in order to shed light and provide insight on the potential for CCS technology to provide a viable mitigation strategy for helping to meet twenty-first century global CO2 emission reduction goals, such as set forth in the 2015 United Nations Climate Change Conference in Paris, France.

## 2. Carbon and oxygen cycles1

Carbon (C) is the basic building block for plant, animal and human life—all are “carbon-based” organisms. Plants, animals, and humans also depend on oxygen (O2) for survival. The cycling of carbon and oxygen in ecosystems is ultimately powered by solar energy. In photosynthesis, plants combine carbon dioxide (CO2), water (H2O), and solar energy to produce sugars, oxygen, and energy. In cellular respiration, animals and humans combine sugars and oxygen to produce carbon dioxide, water, and energy. Carbon-oxygen-hydrogen compounds (e.g., sugars) pass through the food chain or web in ecosystems via herbivores, carnivores, and omnivores. In the food chain, some of the carbon and oxygen stored in organic compounds are returned to the environment in the form of CO2 and H2O via cellular respiration. When a large organism such as a plant or an animal dies and is decomposed by microorganisms, more of the CO2 and H2O stored within the plant or animal is returned to the environment where it can be taken up again by plants to produce more carbon-oxygen-hydrogen compounds which can then be taken up again by animals and humans.

Not all carbon and oxygen are recycled in the relatively short-term cycle described above. Some carbon and oxygen from decomposing plants and animals are converted by relatively long-term geologic processes into rocks (e.g., carbonate rock formations such as limestone) and minerals (e.g., coal, oil, and natural gas) stored in the earth’s crust. When coal, oil, and natural gas enter economic systems, they are termed fossil fuels. The “fossil” part of this term derives from the fact that they come from fossilized remains of plants and animals. The “fuel” part is derived from the fact that coal, oil and natural gas, and their processed derivatives (e.g., gasoline) are burned as fuel in engines and other machinery found throughout our economic system (e.g., planes, trains, automobiles, electricity power plants, and home furnaces). When fossil fuels are burned, CO2 (and other emission gases—CH4, N2O) stored in these minerals is released back into the environment. The release of CO2 from burning fossil fuels is the focus of recent concern and debate over global climate change.

As indicated in the discussion above, human activities affect global climate change through impacts on the carbon and oxygen cycle. Burning of fossil fuels is a major contributor to releasing more CO2 into the atmosphere, primarily from terrestrial sources of stored carbon (e.g., coal deposits, oil deposits, and trees). Human activities can also help to remove CO2 from the atmosphere, with one of the primary means being increasing the storage of carbon in terrestrial plants. For example, taking actions to protect “green space” including farmland from development (and managing forests in a sustainable manner following an optimal harvest and replanting schedule) helps to remove CO2 in the atmosphere through carbon sequestration in plants via photosynthesis. Farms, forests, and other green space areas thus act as “carbon sinks” helping to counteract the greenhouse effect. Another means for storing carbon is through human-engineered carbon capture and storage projects.

## 5. Optimal CCS provision

### 5.1. Concept of economically efficient level of CSS size

According to economic efficiency, the optimal level of carbon capture and storage is where the marginal benefits and marginal costs of CO2 captured and stored are equal. In Figure 1, we show the marginal benefit curve for CCS (MBCCS), and the marginal cost curve for CCS (MCCCS). The marginal benefit curve is downward sloping because, following the law of diminishing returns, each additional unit of CO2 captured and stored provides less private and social benefits. The marginal cost curve is upward sloping because both the private and social costs of CCS go up with each additional unit of CO2 captured and stored. The upward-sloping nature of the marginal cost curve indicates that it would be very expensive (and likely cost prohibitive) to capture and store 100% of all CO2 found in emissions from a point source such as a coal-fired power plant or industrial factory.

The economically efficient level of CCS (Q*) is shown graphically in Figure 1 where the marginal benefit curve and marginal cost curve for CCS cross; at this point,

MCCCS =TCCCSQCO2=MBCCS =TBCCSQCO2 E5

If all private and social benefits and costs of CCS could be “internalized” into economic markets, transactions between buyers and sellers could lead automatically to an economically efficient level of CCS, given certain conditions (e.g., perfect competition). It is notoriously difficult, however, to “internalize” all social benefits and costs because of the public good (or “bad”) characteristics of these benefits and costs such as nonexclusiveness and nonrivalry. Thus, achieving an economically efficient level of CCS would most likely require some degree of government intervention into markets such as economic incentives (e.g., taxes and subsidies) and/or direct regulation ([5], Chapter 10).

### 5.2. Estimates of CSS optimal level

As previously described in this chapter, under the condition where marginal benefits and marginal costs of CO2 captured and stored are equal, there exists a relationship between the optimal carbon price and the optimal level of carbon capture and storage. For a given carbon price range of US $146–$257/tC (or US $40–$70/tCO2), the optimal level of CO2 captured and stored is in the estimated range of 0–8MtC (or 0–29.48MtCO2) per year [29, 30].

## 6. Summary and conclusions

From a public policy perspective, since the general public also benefits from carbon dioxide being captured, stored, and prevented from entering the atmosphere, there is economic justification for public policies targeted at providing economic incentives for private companies to invest in CCS technology, such as direct subsidies or tax breaks. Whether or not CCS technology will prove to be one of the “tools” in the global warming, mitigation “tool box” in the long run is yet to be seen.

In addition to the Petra Nova project in the United States, private companies in Canada, Germany, and China are investing in large-scale CCS projects, with mixed economic feasibility results from a private firm perspective. Scaling-up from the private firm level to the society level where public benefits from global warming mitigation are taken into account, the private and public economic benefits of CCS projects seem likely to outweigh the private costs. Thus, public polices, which help private companies to defray the high costs of large-scale CCS projects, may be justified from an overall benefit-cost analysis perspective.

## Notes

• This section appears also in Ref. ([5], p. 16–18).
• As discussed in Section 2, natural chemical cycles covert carbon to hard rock and mineral deposits which further enhances long-term storage with minimal leakage.
• For consistency, in the chapter the units of MtC and US $/tC are being used to describe economic values for marginal costs and benefits on average, assuming MC ≈ AC in all long-run CCS operations. We will note where the units of MtCO2 and US$/CO2 are applied as alternative measures. All are equivalent: (1) US $27.3/tCO2 (= US$100/tC) [26]; (2) US $10/tCO2 is approximately equivalent to US$37/tC [15].
• Because the atomic weights of carbon are 12 atomic mass units and carbon dioxide is 44 atomic mass units, a ratio factor of 3.67, or 44/12, is used, meaning one ton of carbon equals 3.67 tons of carbon dioxide, which can also approximately equal 1 tC = 3.7 tCO2 (as computed from (US $37/tC)÷(US$10/tCO2), or 3.66 tCO2 = (US $100/tC) ÷ (US$27.3/tCO2)). However, only the factor of 3.67 is applied for computation of all estimates in this chapter.
• These costs can be easily and quickly observed in Anderson and Newell (2004) (please see Table 3 in Ref. [8]).
• The later measures may be slightly higher after having been adjusting for inflation over time. Assuming a gas price of US $3 per million Btu (MBtu), which was the average price over the past decade, transport and storage costs of$37/tC stored were reported in [8]. Moreover, one can apply the following formulas to see how adjusted/expected benefits and costs are affected by inflation rates over time, that is, adjusted benefits in current-year = dollars in base-year × (CPICurrent-year/CPIBase-year), and adjusted costs in current-year = dollars in base-year × (PPICurrent-year/PPIBase-year), where CPI is the consumer price index, and PPI is the producer price index.
• CCU is also called carbon capture and reuse or carbon capture and recycling (CCR) [17].
• Urea, also known as carbamide, is an organic compound with the chemical formula CO(NH2)2 and one of the most common forms of solid nitrogen fertilizer. Urea is produced by the reaction between ammonia and CO2. See ([24], Appendix B).
• Permanent and non-permanent potential performances are referred to permanent and non-permanent storage. According to the Global CCS institute, reuse technologies that permanently store CO2 are considered to be an alternative form of CCS and referred to as "alternative CCS." EOR, ECBM, EGS, carbonate mineralization, concrete curing, bauxite residue carbonation, and potentially algae cultivation (depending on the end product) are considered to be alternative forms of CCS. See ([24] Part I: Section 3.2).
• In recent work [25], it was estimated that EOR storage of CO2 could generate net benefits as high as $335/tC stored, or cost as much as$270/tC stored. In a base-case calculation, EOR generates average net benefits of about $45/tC stored [8]. • Median and 95th percentile estimates reported in [27]. • The estimated social cost of carbon reported by [28] including uncertainty, equity weighting, and risk aversion is$44 per ton of carbon (or $12 per ton CO2) in 2005 US$. Second, including uncertainty increases the expected value of the SCC by approximately 8%. Finally, equity weighting generally tends to reduce the SCC.
• For consistency, we assume there is also a 2.4% per year increase in the PMBCCS reported in [23]. Thus, for 2030 the estimated range for PMBCCS is between US $20/tC and$41/tC.

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© 2017 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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text:
# Atiyah-Macdonald Exercise 2.6

I'm working on Exercise 2.6 from Atiyah-Macdonald's "Introduction to Commutative Algebra" and I had a question about the definition of bilinear maps.

Exercise 2.6: For any $$A$$-module $$M$$, let $$M[x]$$ denote the set of all polynomials in $$x$$ with coefficients in $$M$$, that is to say expressions of the form $$m_0 + m_1x + \dots + m_r x^r \qquad (m_i\in M)$$ Defining the product of an element of $$A[x]$$ and an element of $$M[x]$$ in the obvious way, show that $$M[x]$$ is an $$A[x]$$-module. Show that $$M[x] \cong A[x]\otimes_A M$$ (as $$A[x]$$-modules).

So I begin by trying to find explicit $$A[x]$$-linear maps from $$M[x]\to A[x]\otimes_A M$$ and then from $$A[x]\otimes_A M \to M[x]$$. I got a little confused about the definition of a bilinear map when trying to define the second map.

What is an $$A[x]$$-bilinear map on $$A[x]\otimes_A M$$? In particular, let $$\phi: A[x]\times M \to M[x]$$ be a map. Suppose I fix a first coordinate $$f\in A[x]$$, then what does it mean for $$\phi(f,-) : M\to M[x]$$ to be $$A[x]$$-linear? A priori, there is no defined multiplication of $$A[x]$$ onto $$M$$ right?

• There is an $A$-bilinear map $A[x]\times M\to M[x]$ which is just $(p(x),m)\mapsto p(x).m$. This will factor through $A[x]\otimes_A M$. Check that this is an isomorphism. Jan 11 at 21:55
• But I thought this isomorphism is an $A$-isomorphism, not an $A[x]$-isomorphism? Jan 11 at 22:00
• well both modules are isomorphic as $A$-modules. What you need to show is that this is compatible with the $A[x]$ action on both sides. What is the $A[x]$ action on $A[x]\otimes_A M$? What is it on $M[x]$? Jan 11 at 22:09
• Thanks, I got it now. Jan 11 at 22:13

First note that you want an $$A[x]$$-linear map from $$A[x] \otimes_A M$$ into $$M[x]$$, not an $$A[x]$$-bilinear map. This is because in the category of modules, the morphisms are the linear maps. Now in order to exploit the universal property of the tensor product in order to construct such a map, it is sufficient to start with a map on $$A[x] \times M$$ which is merely $$A$$-bilinear, because $$A$$ is the ring we take the tensor product over.
Such a map is given by $$(a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0, m) \mapsto a_n m x^n + a_{n-1} m x^{n-1} + \cdots + a_1 m x + a_0 m.$$
Thus, the universal property of the tensor product implies that we get a map $$A[x] \otimes_A M \to M[x]$$, which a priori is only $$A$$-linear. But this map is in fact $$A[x]$$-linear because it's the inverse of a map which is $$A[x]$$-linear. Remember that the inverse of a module isomorphism is again a module isomorphism.
This $$A[x]$$-linear map is given by $$M[x] \to A[x] \otimes_A M, m_n x^n + \cdots + m_0 \mapsto x^n \otimes m_n + \cdots + 1 \otimes m_0.$$


text:
# Minkowski's Inequality?

1. Aug 2, 2011

### St41n

I don't understand how it is possible to show using the Minkowski's Inequality that
$(\sum x_i )^a \leq \sum x_i^a$ where $x_i \geq 0 \forall i$ and $0<a<1$.

I also tried to prove this without using Minkowski, but to no avail.

This is driving me crazy although it seems to be trivial in the literature.
I will appreciate any help

2. Aug 2, 2011

### micromass

Staff Emeritus
Hi St41n!

It seems that you must prove that $(x+y)^\alpha\leq x^\alpha+y^\alpha$ for $x,y\geq 0$ and $0<\alpha<1$.

For that, you must look at the function

$$f:\mathbb{R}^+\rightarrow \mathbb{R}:x\rightarrow 1+x^\alpha-(1+x)^\alpha$$

Try to show that f is increasing and has its minimum in 0. It follows that $f(x)\geq 0$.

3. Aug 2, 2011

### St41n

Thank you very much for the quick reply!


text:
# argmax of sum of two functions given constraints on argmax of each of them

Suppose we have two single variable functions $$f(x)$$ and $$g(x)$$. Suppose further that

$$argmax\Big[f\Big]\in[a,b]\quad \&\quad argmax\Big[g\Big]\in[a,b]$$

where $$a,b\in\mathbb{R}$$, moreover, assume $$\max_x\Big[f(x)\Big]=\max_x\Big[g(x)\Big]$$

and also suppose that:

$$\exists\ C\ \subset[a,b]\quad s.t.\quad \forall x\in C\ ,\ \ f(x)=g(x)$$

And the above subset is the subset that contains all the points in which $$f(x)=g(x)$$.

1. can we conclude that we have the following? $$argmax\Big[f+g\Big]\in[a,b]$$

2. If the above conclusion is not generally correct, what can we say about $$argmax\Big[f+g\Big]$$?

Let $$f$$ be $$1+\frac{\epsilon}{2}$$ everywhere except $$f(a) = 1+7\epsilon$$ and $$f(C)=1$$
Let $$g$$ be $$1-\frac{\epsilon}{2}$$ everywhere except $$g(b) = 1+8\epsilon$$ and $$g(C)=1$$
Now add an extra peak somewhere outside the interval, for instance $$f(c) = 1+5\epsilon$$ and $$g(c) = 1+6\epsilon$$. $$f+g$$ is now maximized at $$c$$ which is arbitrary.


text:
# OpenMP Support¶

If installing pygram11 from conda-forge or via a binary wheel from PyPI, OpenMP acceleration should be available. If you don’t want to use conda-forge or you can’t use a binary wheel, keep reading.

If you are building from source, the setup.py script tests to see if OpenMP is available during installation. You can look at the builds.sr.ht continuous integration configuration files to see how builds are defined (Linux only) to ensure OpenMP acceleration is available. For macOS, take a look at the GitHub actions configuration files. We rely on libomp from Homebrew.

To check if OpenMP was detected and used while compiling the extension module pygram11._core, try the following:

>>> import pygram11
>>> pygram11.omp_available()
True

Needless to say, if you see False OpenMP acceleration isn’t available.

The histogramming functions use a named argument (omp) for requesting OpenMP usage. If pygram11.omp_available() is False the omp function argument is ignored.

You can use pygram11.omp_max_threads() to check the number of threads that OpenMP has determined are available on your CPU.


text:
# How do you use a graph to determine limits?

The limit of a function $f \left(x\right)$ at a given point $x = a$ is, essentially, the value one would expect the function $f \left(x\right)$ to take on at $x = a$ if one were going solely by the graph. For example, if given a graph which resembles the function $f \left(x\right) = x - 1$, one might expect the function to take on the value $f \left(x\right) = 0$ at $x = 1$. However, the function $f \left(x\right) = {\left(x - 1\right)}^{2} / \left(x - 1\right)$ would also be graphed like $f \left(x\right) = x - 1$, but would be undefined at $x = 1$.
In the case listed above, one would analyze the situation by examining the function's behavior in the graph for $x$-values slightly above and slightly below the desired point. For this case, suppose one examines the graph at the points $x = 0 , x = 0.5 , x = 0.75 , x = 1.25 , x = 1.5 , x = 2$. Doing this, we determine that as $x \to 1$ from both the right and the left, $f \left(x\right) \to 0$. Thus, the two-sided limit of the function $f \left(x\right) = {\left(x - 1\right)}^{2} / \left(x - 1\right)$ at $x = 1$ is 0, though $f \left(1\right)$ itself is undefined (as it takes on the form $\frac{0}{0}$)


text:
Abstract

FastGaussQuadrature.jl is a Julia package to compute n-point Gauss quadrature nodes and weights to 16-digit accuracy and in O(n) time. So far the package includes gausschebyshev(), gausslegendre(), gaussjacobi(), gaussradau(), gausslobatto(), gausslaguerre(), and gausshermite(). This package is heavily influenced by Chebfun.

An introduction to Gauss quadrature can be found here. For a quirky account on the history of computing Gauss-Legendre quadrature, see [6].

Our Aims

• The fastest Julia code for Gauss quadrature nodes and weights (without tabulation).
• Change the perception that Gauss quadrature rules are expensive to compute.

First example

To check an integral

$$$\int_{-1}^{1} x^4 dx = \frac{2}{5}$$$

by numerically, try following code.

julia> using FastGaussQuadrature, LinearAlgebrajulia> x, w = gausslegendre(3)([-0.7745966692414834, 0.0, 0.7745966692414834], [0.5555555555555556, 0.8888888888888888, 0.5555555555555556])julia> f(x) = x^4f (generic function with 1 method)julia> I = dot(w, f.(x))0.40000000000000013julia> I ≈ 2/5true


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# Reverse Polish Notation

Reverse Polish notation, or RPN, is an alternative to the familiar infix notation for mathematical expressions in which operations appear after their arguments rather than before or between them.
For example, the infix expression
$x - y$
is written in RPN as
$x \; y \; -$
One of the advantages of RPN is that it can unambiguously represent expressions without the need for brackets.
For example
$(x + 1) \times (y - 1)$
can be written as
$x \; 1 \;+ \; y \; 1 \; - \times$

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# Each of the 45 boxes on shelf J weighs less than each of the

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Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?

(1) The heaviest box on shelf J weighs 15 pounds.
(2) The lightest box on shelf K weighs 20 pounds.
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udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is
the median weight of the 89 boxes on these shelves?
(1) The heaviest box on shelf J weighs 15 pounds.
(2) The lightest box on shelf K weighs 20 pounds.

If a set has odd number of terms the median of a set is the middle number when arranged in ascending or descending order;
If a set has even number of terms the median of a set is the average of the two middle terms when arranged in ascending or descending order.

Set J+K has 89 terms, which means that the median would be the 45th term, so the heaviest box on shelf J (as each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K, when arranged in ascending order we'll have j1, j2, ..., j45, k1, k2, ..., k44).

(1) The heaviest box on shelf J weighs 15 pounds --> $$j_{45}=15$$--> directly gives us the answer. Sufficient.

(2) The lightest box on shelf K weighs 20 pounds --> $$k_1=20$$ --> just tells us that median is less than 20. Not sufficient.

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Good question.....and an amazing answer from the master....Kudos!

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Re: Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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03 Sep 2010, 00:36
Option 1 alone is enough to solve this question but option 2 can't alone resolve this question
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Re: Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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31 Dec 2014, 00:53
But if the boxes of shelf J is same and each of the boxes in other shelf is different ... will the median still follow the same logic.

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31 Dec 2014, 04:38
jhasac wrote:
But if the boxes of shelf J is same and each of the boxes in other shelf is different ... will the median still follow the same logic.

Yes.

{15, 15, 15, 15, ..., 15, j45 = 15, k1=20, k2, = 21, ..., k44 = 63}
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Re: Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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Hi jhasac,

DS questions on the Official GMAT are often based on patterns/concepts that you know, but are presented in ways that you're not used to thinking about.

If you were told you that there were 89 values in a group and you were asked to find the median of the group, what would you do?

You'd...
1) Put the numbers in order from least to greatest.
2) Look at the 45th number - even if there are duplicates (since that's the one in "the middle")

Those rules don't change, even though the question is worded in a quirky way. Maybe there's a way to figure out the 45th term WITHOUT putting ALL the numbers in a row first though....

We're told that EACH of the 45 boxes on shelf J weigh LESS than EACH of the 44 boxes on shelf K. This means that the 44 boxes on shelf K would be the "heaviest 44" numbers in the group; by extension, the "heaviest" value on shelf J MUST be the median. With that realization, the question is really asking "what is the heaviest value on shelf J?" From here, getting to the correct answer doesn't require much work.

The "take-away" from all of this is that by spending a little bit of time thinking about the "setup" in the prompt and doing some work to "rewrite" the question can make answering the question considerably easier (and save you some time).

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Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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15 Jul 2016, 10:57
udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?

(1) The heaviest box on shelf J weighs 15 pounds.
(2) The lightest box on shelf K weighs 20 pounds.

The important terms to note in this questions are "MEDIAN" and "89 box"
The median of a set containing odd numbers of elements is always the mid point
in this case it will be the $$45^{th}$$ term

(1) The heaviest box on shelf J weighs 15 pounds.
Sufficient :- shelf J has 45 box and the 45$$^{th}$$ term is 15 pound
Therefore we found the median==>sufficient

(2) The lightest box on shelf K weighs 20 pounds.
This is the $$46^{th}$$ term and thus not helpful in finding median ==> Insufficient

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Re: Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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14 Sep 2017, 10:45
udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?

(1) The heaviest box on shelf J weighs 15 pounds.
(2) The lightest box on shelf K weighs 20 pounds.

We are given that each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K and need to determine the median weight of the 89 boxes.

We may recall that when the boxes are ordered from least weight to greatest, the box with the median weight is in the (89 + 1)/2 = 90/2 = 45th position. Thus, the box with the median weight is on shelf J and it is the heaviest box on shelf J.

Statement One Alone:

The heaviest box on shelf J weighs 15 pounds.

Since the heaviest box on shelf J weighs 15 pounds, the 45th box weighs 15 pounds. Since the 45th box is the median of the boxes, the median weight is 15 pounds. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The lightest box on shelf K weighs 20 pounds.

Since the lightest box on shelf K weighs 20 pounds, when the boxes on both shelves are ordered from least to greatest weight, the 46th box weighs 20 pounds. However, we do not have enough information to determine the weight of the 45th box, i.e., the median weight.

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Re: Each of the 45 boxes on shelf J weighs less than each of the [#permalink]

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udaymathapati wrote:
Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on shelf K. What is the median weight of the 89 boxes on these shelves?

(1) The heaviest box on shelf J weighs 15 pounds.
(2) The lightest box on shelf K weighs 20 pounds.

Target question: What is the median weight of 89 boxes on these shelves?

Given: Each of the 45 boxes on shelf J weighs less than each of the 44 boxes on the shelf K.
Let J1 be the weight of the lightest box on shelf J.
Let J2 be the weight of the 2nd lightest box on shelf J.
.
.
.
Let J45 be the weight of the heaviest box on shelf J.
Let K1 be the weight of the lightest box on shelf K.
Let K2 be the weight of the 2nd lightest box on shelf K.
etc.

So, the given information tells us that J1 < J2 < J3 < ... < J45 < K1 < K2 < ...< K44
Since the 89 boxes are now arranged in ascending order (according to weight), the median weight will be the weight of the middle box.
That is, the median weight, will be the weight of the 45th box.
So, we can REPHRASE the target question....
REPHRASED target question: What is the weight of box J45?

Statement 1: The heaviest box on shelf J weighs 15 pounds.
PERFECT! Box J45 is the heaviest box on shelf J
So, box J45 weighs 15 pounds.
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: The lightest box on shelf K weighs 20 pounds.
This tells us that box K1 weighs 20 pounds.
This does not help us find the weight of box J45
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

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# Synopsis: Bulk carriers cannot take the strain

Mercury telluride—a semimetal in bulk—shows signs of becoming a topological insulator under strain.

Mercury Telluride ($\text{HgTe}$) has an “inverted” band structure, where the valence band rises above the conduction band. As such, it was the first material to be identified as a candidate topological insulator—an exotic insulator that has conducting states along its surface or edge that are unusually robust against defects and scattering.

Since bulk $\text{HgTe}$ is a semimetal, the properties of a topological insulator can only be observed by opening up a gap in its band structure. Accordingly, the conducting surface states were first found along the edges of two-dimensional $\text{HgTe}$ quantum wells, where the reduced symmetry of the well lifted the semimetallic character of the band structure. Analogously, three-dimensional (3D) topological insulating behavior should be observable in the surface states of bulk $\text{HgTe}$, provided a gap in the bulk electronic structure can be induced.

Now, <NAME> and colleagues at Universität Würzburg, Germany, and Stanford University, US, report in Physical Review Letters that they have observed quantized Hall conductance on the surface of a $70$-$\text{nm}$-thick sample of $\text{HgTe}$—a signature that this sample is a 3D topological insulator. When deposited on $\text{CdTe}$, the lattice mismatch between the $\text{HgTe}$ film and substrate creates a strain that opens the necessary band gap. In contrast to more-studied 3D topological insulators, such as ${\text{Bi}}_{2}{\text{Se}}_{3}$, ${\text{Sb}}_{2}{\text{Te}}_{3}$, and ${\text{Bi}}_{2}{\text{Te}}_{3}$, bulk $\text{HgTe}$ has a very low background doping, so that the strained bulk layer corresponds to an intrinsic topological insulator.

The next challenge for the group is to observe the exotic effects predicted to show up in the topologiocal surface states, such as quantized topological magnetoelectricity, the existence of magnetic monopoles, and the creation of Majorana bound states. – <NAME>

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# ELECTRICAL MEASURING INSTRUMENTS

By | December 5, 2020

It is understood that moving electrons will produce a magnetic field which of course can be attracted or repelled by other magnetic sources. This condition is used as the basis for making electric motors and simple electric meters to measure currents and voltages. The basic construction of an electric meter is shown in Figure 1.

Figure 1. Basic construction of electric meter

This basic meter consists of a permanent magnet in the form of a horseshoe with round poles. A coil with a core of malleable iron is positioned between the U and S poles so that it rotates freely. A pointer is attached to the coil and will move when the coil rotates.

The electric current to be measured is passed to the coil so that the coil will produce a magnetic field (electro-magnetic). The electro-magnetic poles of will interact with the permanent magnetic poles so that the coil rotates according to the amount of current in it.

## Basic Meter

The most important use is as a current measuring tool and a voltage measuring device. In use as an ampere meter (ammeter), striving for all current at a branch point that is measured can be through the ammeter. The goal is that at the branch point it appears as if a short circuit occurs, that is, it has a low resistance and a low voltage drop. For use as a voltmeter (installed between two points), try to keep the current passing through the meter (voltmeter) as small as possible. The goal is that the two connection points appear as if they were open circuits, that is, they have a very large resistance or very little current passes. Figure 2 shows how the two electricity meters are installed in the circuit. A basic meter typically requires a current of 1 mA (and about 0.1 V) to create afull-scale deflection.

Figure 2.  Installation of a voltmeter and ammeter in the circuit.

Base Meters as Ammeters

We can make a meter with a full-scale current indication (measuring limit) greater than its basic capability (but with the same full-scale voltage indication capability), by installing aresistance shunt in parallel with the meter.

Figure 3.3 Full-scale representation of the base meter:
a) ammeter and
b) voltmeter.

Figure 3 (a) shows a meter with a full scale designation (measuring limit) of 1 mA which will be converted to 1 A. Using the current dividing principle, the shunt resistance value is obtained:

$R_{p}=\frac{{R_{m}}}{(n-1)}$                                (1)

where n shows the magnification of the meter limit. For the above case, n is about 1,000 times and thus  Rp = 25 $\Omega$/ 999 = 0,025 $\Omega$

A multimeter usually have several scales measuring limit by connecting with corresponding terminals. In this case the shunt resistance is installed in the meter circuit. Figure 3.4 shows meters with measuring limits of 2 and 10 A made using the above principle.

Figure 4 Installation of a shunt to change the meter measurement limit.

Basic Meters as Voltmeters

We can also increase the measuring limit of a voltmeter by n times its basic measurement limit (with the same full scale current), that is, by installing an external resistance in series. For the circuit in Figure 3.3-b shows a basic meter with a maximum current measuring limit of 1 mA will be used to measure a voltage of 2 V. The total resistance (outer resistor + meter resistor) is

2 V / 1 mA = 2000 Ω

thus resistance The outside that must be installed is

RS = (2000 – 25) Ω = 1975 Ω

On a voltmeter with several measuring limits it is usually equipped with a switch to select the appropriate series resistor.

Figure 5 Setting the meter limit by installing a resistor.

Example

Suppose that a meter base 50$\mu$A has a resistance of 3000 Ω. Try to design a multimeter that can be used for measurements up to the measuring limits of 100 $\mu$A, 1 mA, 1 V and 10 V. A suitable circuit is shown in Figure 5.

At the limit measure 100A, a current of 50$\mu$A must flow through the meter and the resistance  (R1 + R2). Thus (R1 + R2)= 3000 Ω

At the 1 mA measurement limit, a current of 50 $\mu$A flows through ( R2 + 3000 Ω) and the balance of 950 $\mu$A through R1. Thus,

At the limit measure 1 V, drain current of 100 $\mu$A through meter and 50 $\mu$A via (R1 + R2). On the meter there is a voltage of

50$\mu$A x  3000 Ω =  0.15 V

thus the voltage at R3 is, or

R3 = 0.85 V / 100 $\mu$A = 8500 Ω

In the same way,  R3 = 9.85 / 100$\mu$A = 98.5  kΩ


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# FFT plot in GNU Radio

i got similar question as in FFT scaling in GNU Radio and quite not clear about the answer given so far. I understood that the FFT plot is a relative value in which we have to use measurement instrument to get the actual value.

But, without the measurement instrument, in general, what value will be shown on the plot if the given input is for example complex cosine waveform with amplitude 1?

I've tried the following flowgraph: signal_source - throttle - qt_frequency_sink

and vary the cosine waveform amplitude from 0 to 100. other than amplitude, all blocks use default setting. I expect to get 0dB when the amplitude is 1, but the plot shows ~ -10dB instead,and ~32dB when amplitude is 100 (i'm not sure if this value make sense). SO, how actually the frequency sink calculates these value?

\begin{align} Y[f] &= \sum_{k=0}^{N_{DFT}-1} y[k] e^{j2\pi f \frac kN},\\ \text{for }f&=0\text{, it follows that } e^{j2\pi 0 \frac kN}\equiv 1,\text{ so}\\ Y[0] &= \sum_{k=0}^{N_{DFT}-1} y[k] &y\text{ being constant}\\ &= N_{DFT}\cdot y[0] \end{align}
Because it's handy for visualization, GR's approach to scaling of these values is to divide them by the DFT length $N_{DFT}$; otherwise, the same constant signal would have more power in a longer DFT.


text:
# Math Help - Infinite Dimensional Vector Spaces

1. ## Infinite Dimensional Vector Spaces

Suppose you have a infinite dimensional vector space $V$. Suppose $T \in \mathcal{L}(V,W)$. Can we have $\dim V = \dim \text{null} \ T + \dim \text{range} \ T$?

E.g. can we partition an infinite dimensional vector space into finite dimensional ones?

2. I'm not sure how your two questions are related. The dimension theorem holds for infinite-dimensional vector spaces; but all that this tells us is that if $\dim V = \infty$ then either the range or the kernel of $T$ must be infinite-dimensional.

It is impossible to partition a vector space into subspaces. If $W_1, W_2$ are subspaces and $W_1 \cup W_2 = V$ then we must have $W_1 \subset V$ or $W_2 \subset V$. On top of that, the zero vector would have to be in every part.


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The size Ramsey number of graphs with bounded treewidth

# The size Ramsey number of graphs with bounded treewidth

<NAME> School of Mathematics, Monash University, Melbourne, Australia (<EMAIL>)    <NAME> School of Mathematics and Statistics, UNSW Sydney, NSW 2052, Australia. Email: <EMAIL>. Supported by the Australian research council (DE170100789 and DP180103684).    <NAME> School of Mathematics, Monash University, Melbourne, Australia (<EMAIL>). Research supported by the Australian Research Council.    Liana Yepremyan Mathematical Institute, University of Oxford, Oxford, UK. Email: <EMAIL>. Supported by ERC Consolidator Grant 647678 and by a Robert Bartnik Fellowship of the School of Mathematics, Monash University. The author would like to thank for the hospitality the School of Mathematics, Monash University, and the School of Mathematics and Statistics, UNSW, where this work was partially carried out.
###### Abstract

A graph is Ramsey for a graph if every 2-colouring of the edges of contains a monochromatic copy of . We consider the following question: if has bounded treewidth, is there a ‘sparse’ graph that is Ramsey for ? Two notions of sparsity are considered. Firstly, we show that if the maximum degree and treewidth of are bounded, then there is a graph with edges that is Ramsey for . This was previously only known for the smaller class of graphs with bounded bandwidth. On the other hand, we prove that the treewidth of a graph that is Ramsey for cannot be bounded in terms of the treewidth of alone. In fact, the latter statement is true even if the treewidth is replaced by the degeneracy and is a tree.

## 1 Introduction

A graph is Ramsey for a graph , denoted by , if every -colouring of the edges of contains a monochromatic copy of . In this paper we are interested in how sparse can be in terms of if . The two measures of sparsity that we consider are the number of edges in and the treewidth of .

The size Ramsey number of a graph , denoted by , is the minimum number of edges in a graph that is Ramsey for . The notion was introduced by Erdős, Faudree, Rousseau and Schelp [16]. Beck [3] proved , answering a question of Erdős [15]. The constant 900 was subsequently improved by Bollobás [6], and by Dudek and Prałat [14]. In these proofs the host graph is random. Alon and Chung [2] provided an explicit construction of a graph with edges that is Ramsey for .

Beck [3] also conjectured that the size Ramsey number of bounded-degree trees is linear in the number of vertices, and noticed that there are trees (for instance, double stars) for which it is quadratic. Friedman and Pippenger [22] proved Beck’s conjecture. The implicit constant was subsequently improved by Ke [29] and by Haxell and Kohayakawa [25]. Finally, Dellamonica Jr [11] proved that the size Ramsey number of a tree is determined by a simple structural parameter up to a constant factor, thus establishing another conjecture of Beck [4].

In the same paper, Beck asked whether all bounded degree graphs have a linear size Ramsey number, but this was disproved by Rödl and Szemerédi [34]. They constructed a family of graphs of maximum degree 3 with superlinear size Ramsey number.

In 1995, Haxell, Kohayakawa and Łuczak showed that cycles have linear size Ramsey number [26]. Conlon [9] asked whether, more generally, the -th power of the path has size Ramsey number at most , where the constant only depends on . Here the -th power of a graph is obtained by adding an edge between every pair of vertices at distance at most in . Conlon’s question was recently answered in the affirmative by Clemens, Jenssen, Kohayakawa, Morrison, Mota, Reding and Roberts [8].

Their result is equivalent to saying that graphs of bounded bandwidth have linear size Ramsey number. We show that the same conclusion holds in the following more general setting. The treewidth of a graph , denoted by , can be defined to be the minimum integer such that is a subgraph of a chordal graph with no -clique. While this definition is not particularly illuminating, the intuition is that the treewidth of measures how ‘tree-like’ is. For example, trees have treewidth 1. Treewidth is of fundamental importance in the graph minor theory of Robertson and Seymour and in algorithmic graph theory; see [5, 24, 33] for surveys on treewidth.

###### Theorem 1.

For all integers there exists such that if is a graph of maximum degree and treewidth at most , then

 ˆr(H)⩽c|V(H)|.

?THM? LABEL:thm:tw implies the above bounds on the size Ramsey number from [8], since powers of paths have bounded treewidth and bounded degree. Powers of complete binary trees are examples of graphs covered by our theorem, but not covered by any previous results in the literature. (Note that complete binary trees have bounded treewidth but unbounded bandwidth, that is, they are not contained in a bounded power of a path.) Note that the assumption of bounded degree in ?THM? LABEL:thm:tw cannot be dropped in general since, as mentioned above, there are trees of superlinear size Ramsey number [4]. Furthermore, the lower bound from [34] implies that an additional assumption on the structure of , such as bounded treewidth, is also necessary. We prove ?THM? LABEL:thm:tw in Section 3.

We actually prove an off-diagonal strengthening of ?THM? LABEL:thm:tw. For graphs and , the size Ramsey number is the minimum number of edges in a graph such that every red/blue-colouring of the edges of contains a red copy of or a blue copy of . We prove that if and both have vertices, bounded degree and bounded treewidth, then . Moreover, we show that the host graph works simultaneously for all such pairs and and has bounded degree.

###### Theorem 2.

For all integers there exists such that for every integer there is a graph with vertices and maximum degree , such that for all graphs and with vertices, maximum degree and treewidth , every red/blue-colouring of the edges of contains a red copy of or a blue copy of .

The second contribution of this paper fits into the framework of parameter Ramsey numbers: for any monotone graph parameter , one may ask whether can be bounded in terms of . This line of research was conceived in the 1970s by <NAME> and Lovász [7]. The usual Ramsey number and the size Ramsey number (where and respectively) are classical topics. Furthermore, the problem has been studied when is the clique number [18, 32], chromatic number [7, 37], maximum degree [27, 28] and minimum degree [7, 19, 20, 35] (the latter requires the additional assumption that the host graph is minimal with respect to subgraph inclusion, otherwise the problem is trivial).

It is therefore interesting to ask whether can be bounded in terms of . Our next theorem shows that the answer is negative, even when replacing treewidth by the weaker notion of degeneracy. For an integer , a graph is -degenerate if every subgraph of has minimum degree at most . The degeneracy of is the minimum integer such that is -degenerate. It is well known and easily proved that every graph with treewidth is -degenerate, but treewidth cannot be bounded in terms of degeneracy (for example, the 1-subdivision of is 2-degenerate, but has treewidth ).

###### Theorem 3.

For every there is a tree such that if is -degenerate then .

A positive restatement of ?THM? LABEL:thm:d-degenerate is that the edges of every -degenerate graph can be 2-coloured with no monochromatic copy of a specific tree (depending on ). This is a significant strengthening of a theorem by <NAME>, Sanders and Vertigan [13, Theorem 3.9], who proved that the edges of every graph with treewidth at most can be -coloured with no monochromatic copy of a certain tree . We also note that a statement similar to Theorem LABEL:thm:d-degenerate does not hold in the online Ramsey setting, see Section 4 in [10] for more details.

Furthermore, ?THM? LABEL:thm:d-degenerate is tight in the following sense. If is a monotone graph class with unbounded degeneracy, then for every tree , there is a graph such that . Indeed, for a given tree , let be a graph in with average degree at least , which exists since is monotone with unbounded degeneracy. In any 2-colouring of , one colour class has average degree at least . Thus there is a monochromatic subgraph of with minimum degree at least , which contains as a subgraph by a folklore greedy algorithm.

## 2 Tools

Our proof of ?THM? LABEL:thm:offdiagonal relies on the following characterisation of graphs with bounded treewidth and bounded degree. The strong product of graphs and , denoted by , is the graph with vertex set , where is adjacent to in if and , or and , or and . Note that is obtained from by replacing each vertex by a clique and replacing each edge by a complete bipartite graph.

###### Lemma 4 ([12, 36]).

Every graph with treewidth and maximum degree is a subgraph of for some tree of maximum degree at most .

Our host graph will be constructed from a bounded-degree graph with certain expansion properties. An -graph is a -regular -vertex graph in which every eigenvalue except the largest one is at most in absolute value. The existence of graphs with is shown, for instance, by considering a random -regular graph on vertices, denoted by .

###### Lemma 5 ([21]).

Let be an integer and be even. With probability tending to 1 as , every eigenvalue of except the largest one is at most in absolute value.

For a graph and sets , let be the number of edges with one endpoint in and the other one in . Each edge with both endpoints in is counted twice. We will use the following well-known estimate on the edge distribution of a graph in terms of its eigenvalues, see, e.g., [31] for a proof.

###### Lemma 6 ([31]).

For every -graph and for all sets ,

 ∣∣∣e(S,T)−D|S||T|N∣∣∣⩽ λ ⎷|S||T|(1−|S|N)(1−|T|N).

The key tool that we use is the following implicit result of Friedman and Pippenger [22], which shows that every -graph with the appropriate parameters is ‘robustly universal’ for bounded-degree trees. Let be the set of all trees with vertices and maximum degree at most . The following lemma follows implicitly from the proofs of Theorems 2 and 3 in [22].

###### Lemma 7 ([22]).

Let and be integers. Let and be integers such that and , and let be an -graph with . Then every induced subgraph of on at least  vertices contains every tree in .

We summarise the above results in the following lemma.

###### Lemma 8.

For all integers , every and for all even there exists such that if , then there exists an -vertex -regular graph with the following properties:

• For every pair of disjoint sets with we have .

• Every induced subgraph of on at least vertices contains every tree in .

###### Proof.

Let be an even integer and . Let be an -graph where , which exists by ?THM? LABEL:thm:friedman. Property (2) follows from ?THM? LABEL:thm:fp:vertex. Moreover, for all sets with we have , which implies by ?THM? LABEL:thm:ks:edgedist, as desired. ∎

We also need the following lemma of Friedman and Pippenger [22]. For a graph and , let be the set of vertices in adjacent to some vertex in .

###### Lemma 9 ([22]).

If is a non-empty graph such that for each with ,

 |ΓH(X)|⩾(d+1)|X|

then contains every tree in .

Finally, we need the following standard tools.

###### Lemma 10 (Kövari, <NAME> [30]).

Every graph with vertices and no subgraph has at most edges.

###### Lemma 11 (Lovász Local Lemma [17]).

Let be a set of events in a probability space, each with probability at most , and each mutually independent of all except at most other events in . If then with positive probability no event in occurs.

## 3 Proof of ?THM? LABEL:thm:offdiagonal

We start with the following lemma that states that if a graph does not contain all trees in then its complement contains a complete multipartite subgraph where the parts have ‘large’ size.

###### Lemma 12.

Fix integers and let . In every red/blue-colouring of there is either a blue copy of every tree in , or a red copy of a complete -partite graph in which every part has size at least .

###### Proof.

Let be the spanning subgraph of consisting of all the blue edges. We may assume that does not contain every tree in . By Lemma 9, for every non-empty set , there exists such that and . Note that for such and , all the edges of between and must be red. Let and be sets of vertices in such that and , and for :

• with and , and

• .

We stop when . Note that are pairwise disjoint. Since all the edges of between and are red, all the edges between distinct and are red. Let . Note that

 N=m∑i=1|Xi∪ΓG[Si](Xi)|

Thus . We now combine the parts to reach the required size. Let , where is the minimal index such that . Since , we have the upper bound, . Repeating the same argument starting at , and noting that , we construct parts satisfying . ∎

Let be a rooted tree with root . For each vertex of , let denote the parent of , where for convenience we let . Let denote the grandparent of ; that is, . Moreover, we denote the set of children of by , and define as the set of children and grandchildren of . Let be the distance between and , that is, the number of edges on the path from to . For each integer , let be the set of vertices with . In the above definitions, we may omit the subscript if is clear from the context.

Given a tree rooted at , define another tree rooted at as follows. The vertex set of is defined to be . A pair with is an edge of if or . In particular, the -neighbourhood of is . We call the truncation of . Note that if has maximum degree , then has maximum degree at most .

Let and be integers and suppose we are given a graph , a vertex partition of , and an edge-colouring . Define an auxiliary colouring of the complete graph with vertex set as follows. For distinct , colour the edge blue if there is a blue between and in , and red otherwise. We call this edge-colouring the -colouring of . This auxiliary coloring also appears in [1], and subsequently in [8].

###### Lemma 13.

Fix integers , , , . Let be a tree in rooted at , and let be the truncation of . Let . Suppose we are given a graph , a vertex partition of , and an edge-colouring such that, for all , all the edges of are present and are blue, and . If there exists a blue copy of in the -colouring of then there exists a blue copy of in .

###### Proof.

Let be the -colouring of and suppose is an embedding of in the blue subgraph of . Let be the vertices of ordered by their distance from the root in . We will find a blue copy of whose vertices are in for . We warn the reader that in this proof we often use notation to denote the image of , for some subset , under some embedding mapping into , without precisely defining how acts on each vertex of , but rather claiming that such an embedding exists. This is done for brevity, and to keep the proof intuitive.

We define a collection of subsets of as follows. Let and for each , let . We call the bag of vertex . Observe that the bags are pairwise disjoint, and they partition the entire vertex set . They will help us keep track of the embedding of in .

We will proceed iteratively, starting from the root and following the order of the vertices we fixed earlier. At each step , we will have a partial embedding of in , where is the subtree . Our final embedding will be . At step we will embed in some way; this will define . At step , will be defined as an extension of , and the extension will be defined only on so that the image of the latter ‘links’ back appropriately to the embedding of . Before specifying our embedding, we list the properties that will satisfy:

• ,

• ,

• for every , ,

• every edge of will be coloured blue.

Note that (P2) implies that at most vertices are embedded in , and every other (with ) will contain at most embedded vertices by (P3). Moreover, (P4) will be satisfied for edges of embedded inside one partition class . To guarantee that those edges of that go between distinct partition classes and are blue, we will make use of the properties of the auxiliary colouring . Finally, we define our iterative embedding scheme from which properties (P1)–(P4) can be easily read out, thus completing the proof.

Step 0: Let and embed into , by picking any vertices in ; this determines . Recall that all edges inside are blue, hence indeed this is a valid embedding of .

Step : Having defined , we now show how to extend it to . Recall that . Let be the grandparent of . Since there is an edge in and was a blue embedding of in , there is a blue between and . Let be any such copy of . Define on to be a set of any vertices in disjoint from the image of . Define on to be any set of vertices in disjoint from the image of . This is possible since , and the total number of vertices embedded into during the procedure is at most . ∎

The next lemma is a well-known application of the Lovász Local Lemma. Given a graph let denote the blowup of where each vertex is replaced by an independent set of size , and each edge is replaced by a complete bipartite graph between and .

###### Lemma 14.

Fix . Let be a graph with maximum degree . Let be a spanning subgraph of such that for every edge there are at least edges in between and . Then .

###### Proof.

For each vertex of , independently choose a random vertex in . For each edge of , let be the event that is not an edge of . Since there are at least edges between and , the probability of is at most . Each event is mutually independent of all other events, except for the at most events corresponding to edges incident to or . Since , by Lemma 11, the probability that some event occurs is strictly less than 1. Thus, there exist choices for for all , such that is an edge of for every edge of . This yields a subgraph of isomorphic to . ∎

?THM? LABEL:thm:tw and ?THM? LABEL:thm:offdiagonal are implied by Lemma 4 and the following result.

###### Theorem 15.

For all integers there exists such that for all there is a graph with vertices and maximum degree , such that for all trees and with vertices and maximum degree , every red/blue-colouring of contains a red copy of or a blue copy of .

###### Proof.

Let . Let be the smallest even number larger than . Let be derived from Lemma 8 applied with this choice of , and . Choose and let be any -vertex -regular graph derived from Lemma 8. Set and . Denote the Ramsey number of by . Recall that is a graph on the same vertex set as and is an edge in whenever and are of distance at most three in . Let .

Since is -regular, has maximum degree at most , and has maximum degree at most . Let denote the copy of corresponding to . Let be any edge-colouring of . We will show that it must contain either a red copy of or a blue copy of .

By definition of , for each vertex , contains a monochromatic copy of , say on vertex set . Let be the set of vertices for which induces a blue . By symmetry between and , we may assume that . Let .

Let and be the -colouring of . Root at an arbitrary vertex. Let be the truncation of . If there is a blue copy of in , then ?THM? LABEL:lem:chopping implies that contains a blue copy of , so we are done.

We henceforth assume that there is no blue copy of in . Since has maximum degree at most and , by Lemma 12, there are sets of size at least such that all the edges in between two distinct parts and are red.

For , define an -matching to be a matching of edges in with one endpoint in and the other in . (Note that we are now considering the original graph not .) We will find a set satisfying , and a collection of -matchings such that each covers . We proceed by induction on . Assume at the end of step we have found a set with and a collection of -matchings , where each covers . At step , let be a maximum matching between and . If consists of fewer than edges, then, by Kőnig’s theorem, the bipartite graph between and has a vertex cover of order at most . But then we can find sets and with and . This contradicts property (1) from Lemma 8. Hence covers at least vertices of . We set and proceed. After steps, we reach the desired set , which we call .

For each vertex , for , let be the unique neighbour of in . Since , contains a copy of of on some vertex set by property (2) from Lemma 8. Next we show that there is a red copy of in contained in the vertex set of and use this copy to find a red copy of in via ?THM? LABEL:lem:lifting.

Root at any vertex . For every vertex let if is at even distance from and , otherwise. Note that for every , induces a red clique in because the vertices of are elements of distinct partition classes . Moreover, if and are adjacent in , then also edges between and are red in since all the vertices of lie in distinct partition classes . So this shows that the vertex set induces a red copy of in . It remains to ‘lift’ this copy to the graph with the colouring . First we observe that every edge in is in fact an edge of . Indeed, for any , and any , are of distance at most two in , hence is an edge in . Now if and are adjacent in , then for any and , the distance between and in is at most , so and are also adjacent in .

Recall that if is an edge of and is red in , then the complete bipartite graph between and in contains no blue copy of . Lemma 10 implies that has at most blue edges. Note that . Let and be the subgraph of consisting of all the red edges of over all . It is now easy to see that and satisfy the assumptions of ?THM? LABEL:lem:lifting. Therefore contains a red copy of which finishes the proof. ∎

## 4 Proof of ?THM? LABEL:thm:d-degenerate

Let be the complete -ary tree of height with a root vertex ; that is, every non-leaf vertex has exactly children and every leaf is at distance from . ?THM? LABEL:thm:d-degenerate is implied by the following. Recall that, for a rooted tree , denotes the number of edges of the path from the root to in .

###### Theorem 16.

For every integer , every -degenerate graph is not Ramsey for the tree .

###### Proof.

We proceed by induction on . For , is a tree, so fix an arbitrary vertex to be the root of and colour the edges of by their distance to the root modulo 2 (where the distance of an edge to the root in the tree is . There is no monochromatic path of length 3, and in particular no monochromatic copy of .

Now let and set and for brevity. Let be a -degenerate graph. It follows from the definition of degeneracy that has a vertex-ordering , such that each vertex has at most neighbours with . Form an oriented graph by choosing the orientation for an edge if . Then each vertex has in-degree at most .

We now partition into sets and such that both and are -degenerate. Start by assigning to . For , assume that have been assigned to or . Add to if contains at most of the in-neighbours of . Otherwise add it to . Note that in the latter case, contains at most of the in-neighbours of , since has at most in-neighbours in . Clearly, this does not affect the in-degree of in and .

By induction, there is a red/blue-colouring of the edges in not containing a monochromatic copy of . We extend to a red/blue-colouring of in the following way. For an edge assume without loss of generality that it is directed from to in , that is . Then colour red if , and blue if . In other words, the edge ‘inherits’ the colour from its source vertex in .

We claim that there is no monochromatic copy of in this colouring of . Assume the opposite and let be a monochromatic copy of in G. For each vertex in , we denote its copy in by . Without loss of generality we may assume that is red.

###### Claim 1.

If is a non-leaf vertex of that lies in , then there are least children of in such that for all .

###### Proof.

The in-degree of in is at most . Furthermore, each edge with is coloured blue in , by definition of and since . That is, those edges cannot be part of . It follows that at least neighbours of in are out-neighbours of that are elements of , and the claim follows. ∎

Let be the root of .

###### Claim 2.

For every vertex of distance at most from in we have that .

###### Proof.

Assume that and has distance at most in from . Apply the previous claim iteratively to and all of its descendants that lie in . In iterations (before reaching the leaves of ), we construct a red copy of whose vertices all lie in , that is, a red copy of in the colouring . This contradicts the property of . ∎

It follows that all vertices in of distance at most from must lie in , forming a red copy of in , which again contradicts the property of . ∎

## 5 Concluding Remarks

We have shown that for a graph of bounded maximum degree and treewidth, there is a graph with edges that is Ramsey for . We propose a multicolour extension of this result.

###### Question 17.

Given positive integers and an -vertex graph of maximum degree and treewidth , is there a graph with edges such that every -colouring of the edges of contains a monochromatic copy of , where depends on and only?

When is a bounded-degree tree, a positive answer (and a stronger density analogue) follows from the work of Friedman and Pippinger [22]. Han, Jenssen, Kohayakawa, Mota and Roberts [23] have recently shown that the above extension holds for graphs of bounded bandwidth (or, equivalently, for any fixed power of a path).

Our second result is that every -degenerate graph can be partitioned into two subgraphs, neither of which contains a copy of a tree . It would be interesting to show that cannot be replaced by a tree whose maximum degree is bounded by an absolute constant (independent of ).

On the positive side, Ding, Oporowski, Sanders and Vertigan [13] showed that for every tree , there is a graph of treewidth 2 such that every red/blue-colouring of the edges of contains a red copy of or a blue copy of a subdivision of .

###### Question 18.

Is there a function with the following property: for every graph of treewidth , there is a graph of treewidth such that every red/blue-colouring of the edges of contains a monochromatic copy of a subdivision of ?

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