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# Questions tagged [mechanism-design]

Mechanism design is a field in economics and game theory that takes an engineering approach to designing economic mechanisms or incentives, toward desired objectives, in strategic settings, where players act rationally.

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### Auction theory reference request

I am interested in finding the literature or at least some references (books, journal papers etc.) for the following problem: Suppose there is an auction with one seller and one buyer. The seller ...
53 views

### Allocating chores using auctions

I live in a student dorm with 16 people and we share a kitchen. Every week someone has to clean the kitchen. Right now we just take turns in a way determined by the numbers of our respective rooms. ...
16 views

### Incentive compatibilty conditions for multi-item auctions

In this paper by Myerson on optimal single item auctions, the incentive compatibility condition is shown to be equivalent to the following simpler conditions on the auction - (i) the allocation ...
17 views

### Market mechanism design and decision rule based on a graph

I am doing a micro course right now. I came upon this question while studying and I was hoping for help/hints on how to solve it or what resources to consult to better understand the topic. the ...
29 views

### Is a resource allocation problem in mechanism design a direct or indirect mechanism?

From L. Hurwicz's work and book "Designing Economic Mechanisms," I cannot figure out whether a resource allocation problem in mechanism design is a direct or indirect mechanism. I think the answer is ...
21 views

### Renewal of multiple notions of value in contemporary (green, public good) economy and are there general game theories?

https://en.wikipedia.org/wiki/Exchange_value mentions 4 notions of value: economic value, exchange value, use value, price. But this article also mentions that these are Marxian concepts and the ...
50 views

### Cardinal Voting, Incentive Compatibility and Secrecy

Is there any available/feasible/practical way to make a Cardinal Voting both Incentive Compatible and Secret? A method to make a cardinal voting incentive compatible would be to force them to put ...
17 views

### Questions regarding 'Efficient Trade with Interdependent Values' by Marek Pycia and Peng Wang,2015

This question is with reference to the paper 'Efficient Trade with Interdependent Values' by Marek Pycia and Peng Wang,2015(See here). At page no. 4 of the paper, the authors describe $v_i$ as the ...
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### How to determine the first bid for a Double-Bounded Dichotomous-Choice Contingent Valuation model?

I'm very new to the survey format of double-bounded dichotomous-choice for contingent valuation, so I apologize for my naive question. Could you please let me know how a researcher decides the initial ...
11 views

### Designing the payment function in Mechanism Design problems

Suppose we have a network in which agents request access to its resources. Thus we have a resource allocation problem. Ideally, we want to incentivize agents to send social-welfare supporting ...
22 views

### Going from DRMs to Indirect Mechanisms

I have recently started studying mechanism design (after having taken a basic graduate micro course) through Borgers' 'Introduction to the theory of mechanism design'. I was wondering if people are ...
47 views

### Price of Anarchy when the cost is to society and not to the selfish agents

Let's say there is some scarce resource and several agents compete to access this resource. Each agent is able to convert the resources into products that are beneficial to society, but each agent has ...
22 views

### Is a proft tax not distorting unless we know the opportunity cost?

A Profit tax where we actually know the Entire Cost Function Including the Opportunity cost(Time, Incommodation, best alternative for the easily monetized resources etc) is not distorting. But the ...
71 views

### Direct revelation mechanism's sets of strategies and types

Mas-Colell, Whinston and Green in Microeconomic Theory describe the direct revelation mechanism as it follows: The employed notation is the following: $θ_i$: Player i's type $Θ_i$: Set of types for ...
17 views

### Difference between social choice function and mechanism outcome function

Mas-Colell, Whinston and Green's Microeconomic Theory (3rd edition) defines the social choice function as the following: Later, the mechanism outcome function is also defined: The relationship ...
45 views

### Bayesian incentive compatibility for a general distribution

Suppose there are $n$ bidders and a seller. Bidder $i$ observes a private signal $v_i$ from $[a,b]$. Let $\mathcal{X} = \times_{i=1}^n[a,b]$ Each bidder is represented by a random variable, that has a ...
35 views

### Non-Bayesian Mechanism Design

Suppose we have a mechanism where a finite number of agents possess private information that is not drawn from a probability distribution. The agents' types are given and fixed but agents only know ...
40 views

### Is a combination of Political economics and Game theory possible and beneficial?

By Political economics I do not mean the "economical" advice given by some people(see the Wealth of Nations by Adam Smith) but rather the heavily mathematized subfield of Economics studying and trying ...
92 views

### AGV mechanism and Individual rationality

I have the following question. I consider AGV mechanism which is as follows: $u_i(x,\theta_i) = v_i(k,\theta_i) + t_i$ is an utility function where $x = (k,t_1,...,t_n)$ vector of alternatives. There ...
52 views

### Monotonicity of second price auctions with bidder-specific reserve prices

I am having difficulty understanding the problem below: Given CDFs $F_1, ..., F_n$, prove that the allocation rule of a second-price auction with bidder-specific reserve prices is monotone. The item ...
53 views

### Are there any studies, academic papers or articles that defy “common” knowledge", i.e that patents give an incentive for R&D to private business?

In Economic Growth and particularly dynamic optimization. My professor taught us that an intertemporal allocation is a dynamic optimum(Pareto Optimum) only if(necessary condition) at any given ...
56 views

153 views

### Are direct mechanisms always truthful?

I am very new to mechanism design so I am a bit lost among the concepts, even the most basic ones. One thing that is not clear to me is the concept of direct mechanism and its relationship with the ...
24 views

### Synonyms for One-to-One Matching Mechanisms

I am trying to find a more elegant usage of wording for one-to-one matching and I am need of some thoughts from others. Bilateral matching? Any help would be respected.
441 views

### What is virtual valuation?

I’m reading through a textbook in auction design when it describes a term, virtual valuation $$\phi_i(v_i) = v_i - \frac{1 - F(v_i)}{f(v_i)}$$ where $f$ is the pdf of a bidders valuation and $F$ is ...
114 views

### Help needed to understand proof in Myerson and Satterthwaite (1983)

The following image is an excerpt taken from Myerson and Satterthwaite's 1983 paper Efficient Mechanisms for Bilateral Trading: I know what the theorem is all about, all I need is some help with the ...
33 views

### Efficient social allocation awards item to the highest valuer in a sponsored search auction. Whom is it optimal for?

In mechanism design, for example in the case of a single item auction, we often assume that a socially efficient outcome is to allocate the good to the bidder with the highest valuation of that good. ...
76 views

### Is Myerson's Lemma valid for piecewise linear allocation functions?

In Tim Roughgarden's fantastic course on Algorithmic Game Theory, he begins the course by introducing basic mechanism design, and Myerson's Lemma. He then proves that Myerson's Lemma is valid for ...
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### Do mechanisms always result in a series of strategic games being played out?

Mechanism design is an offshoot of game theory, and many call it "inverse game theory", since it involves the use of incentives to ensure that the end result of the game being played out is beneficial ...
85 views

### What are some examples of indirect mechanisms that are not auctions? [duplicate]

I'd like to know if there are any examples of indirect mechanisms that are not auctions. Auctions seem like a very controlled environment. I'd like to see some examples of indirect mechanisms "out in ...
40 views

### Is quadratic voting efficient in multiple winner elections?

Quadratic voting approximates efficiency for a large collective making a sequence of binary decisions. What if a group can take multiple actions? Say, for example, a series of decisions where 2 out of ...
271 views

### Question on uncertainity

Please imagine that Nicole is uncertain of her future wealth. Her wealth in the bad state of the world is zero. Her wealth in the good state is $w>0$. Each state is initially equally likely. ...
291 views

### Ex-post vs Ex-ante Budget Balance in Auctions

Have I understood this correctly? When I say my auction mechanism is ex-post budget balanced it means that in practice (I have done market simulations) and according to the results my mechanism is ...
113 views

### Auction question

I’m studying on auction. But I couldn’t understand some questions about this topic. And I cannot solve such type questions. For example, I know that a set of symmetric and risk neutral bidders ...
215 views

### Marginal Revenue Interpretation of Auctions

I'm reading Klemperer's survey on auction theory (pages 17-18) which discusses the relation between traditional microeconomic price theory and the revenue equivalence result. Firstly, I don't quite ...
65 views

### How to improve the underperforming construction productivity by correcting its market failures?

Globally, labor-productivity growth in construction has averaged only 1 percent a year over the past two decades, compared with growth of 2.8 percent for the total world economy and 3.6 percent in the ...
149 views

### Bayesian Mechanism vs Prior-free Mechanism

I have a double auction mechanism in which the valuations of the agents for the items are drawn from a known random distribution. To be precise, the valuations are the probability of each agent ...
82 views

### Incentive compatibility: Weakly dominant strategy versus Nash equilibrium?

When it comes to proving that a mechanism e.g. auction is incentive compatible this is the approach I'm using: I break down all the cases that might happen if the agent reports an untruthful value to ...
188 views

### What types of 'market mechanisms' are said to exist besides auctions?

The book Understanding Auctions states: The term auction covers a wide range of market mechanisms that are used to exchange products and services by determining who receives an item and how much is ...
88 views

### A manifold of agents?

In contract theory and mechanism design, if the agents have one-dimensional characteristics, then they are usually considered as a contiuum on $\mathbb R$; if they have n-dimensional type, then they ...
33 views

### VCG mechanism with a minimum payment threshold

Suppose that we want to find a socially-optimal allocation in the following setting: There are $n$ firms and $m$ workers. Each firm $i$ needs one worker and has value $v_i$ for having a worker (any ...
115 views

### Does the VCG algorithm work for any bundle valuation function?

Assume we have $n$ buyers of $|M|=m$ items in a combinatorial auction. So as we know, we can use the Vickrey-Clarke-Groves (VCG) algorithm to allocate bundles of items to buyers as best as possible. ...
113 views

### Incentive compatibility

I have a set of objects $A=\{a_1,a_2,a_3,\cdots,a_m\}$ and one player, who proposes a vector of valuations $t=(t_1~t_2~t_3~\cdots~t_m)$ for the $m$ objects. $T$ is the set of all such valuation types. ...
349 views

### Checking incentive compatibility of a mechanism

I was thrilled at seeing an interesting bidding system in an academic institutions. This bidding is for registration of courses. Each student is allocated some points based on her grade point(GPA). So,...
73 views

### Allocation of an Indivisible Good

Two kids at school are arguing over who owns a cool video game cartridge, and the schoolteacher, a ruthlessly practical social planner who knows that kids can only enforce so much AND that the game is ...
2k views

### Code finding all stable matchings in one-to-one problem

Do you know of any publicly available code in python or R (or any other free high level language) that returns all the stable matchings for any one-to-one matching problem? Note: This is related but ...
997 views

### What is the difference between contract theory and mechanism design?

Can someone explain the difference between contract theory and mechanism design?
Let $\theta_i$ be $i$'s private information about the state of the world, $\theta \in \Theta$, where $\theta = \times_i \theta_i$. Each agent reports $\hat{\theta_i}$ to a social planner, so that \$\...


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# Showing $x^8\equiv 16 \pmod{p}$ is solvable for all primes $p$

I'm still making my way along in Niven's Intro to Number Theory, and the title problem is giving me a little trouble near the end, and I was hoping someone could help get me through it.

Now $x^8\equiv 16\pmod{2}$ is solvable with $x\equiv 0\pmod{2}$, so I assume $p$ is an odd prime. From a theorem earlier in the text,

If $p$ is a prime and $(a,p)=1$, then the congruence $x^n\equiv a\pmod{p}$ has $(n,p-1)$ solutions or no solution according as $a^{(p-1)/(n,p-1)}\equiv 1\pmod{p}$ or not.

So since $(16,p)=1$, the problem reduces to showing that $16^{(p-1)/(8,p-1)}\equiv 1\pmod{p}$ holds for all $p$. I note that $(8,p-1)$ can only take values $2,4,8$. For $2$, the above equivalence is then $4^{p-1}\equiv 1\pmod{p}$, which is true by Fermat's little Theorem. For $4$, it is then $2^{p-1}\equiv 1\pmod{p}$, which again holds by FlT. However, the case where $(8,p-1)=8$ is throwing me off. At best I see that $16^{(p-1)/8}\equiv 2^{(p-1)/2}\pmod{p}$, but I'm not sure how to show this is congruent to $1$ modulo $p$. Maybe there's a more elegant way to do it without looking at cases. Thanks for any insight.

• This problem naturally splits into cases. Don't be afraid to think casewise. – Qiaochu Yuan Oct 14 '10 at 9:56
• This problem played a role in the development of the Grunwald–Wang theorem; see en.wikipedia.org/wiki/Grunwald-Wang_theorem#History – lhf Oct 17 '10 at 23:46

One way is to use the Legendre symbol identity $$2^{(p-1)/2} \equiv (\frac{2}{p}) \equiv (-1)^{(p^2-1)/8} \pmod p$$ (for odd primes p), keeping in mind that if $$(8,p-1)=8$$ then $$p \equiv 1 \pmod 8$$.

• Thanks, I overlooked that identity. I had a small question to add. I know if $a$ is a quadratic residue modulo $p$, then the Legendre symbol $(a|p)=1$ . However, if by some calculation I find that $(a|p)=1$, is it also the case that $a$ is a quadratic residue modulo $p$? – yunone Oct 14 '10 at 10:13
• Yep that's right (unless if a=0 (mod p)). – Douglas S. Stones Oct 14 '10 at 21:22

HINT $\rm\ \ \ x^8 - 16\ =\ (x^2 - 2)\: (x^2 + 2)\: (x^4 + 4)\:.\ \:$ If the first two factors have no roots in $\rm\ \mathbb Z/p\$ then $\:2, -2\:$ are nonsquares so their product $-4\:$ is a square, so $\rm\: i = \sqrt{-1} \in \mathbb Z/p\:$. Thus the third factor has a root since $\rm\ x^4 + 4\$ has roots $\rm\: \pm 1\pm i\:$.

• I'm only learning elementary number theory, so I'm not too familiar with this. Could you please explain how $-4$ being a square implies $i=\sqrt{-1}\in\mathbb{Z}/p$? – yunone Oct 14 '10 at 23:27
• If $j^2 \equiv -4$ then $(j/2)^2 \equiv -1$. Note: since p is odd, $1/2 \in \mathbb Z/p$, in fact $1/2 \equiv (p+1)/2$ – Bill Dubuque Oct 14 '10 at 23:31
• The quartic factors even further: $x^8 - 16 = (x^2+2)(x^2-2)(x^2+2x+2)(x^2-2x+2)$. The last two factors both have discriminant $-4$, so if neither 2 nor $-2$ is a square mod $p$ then $-1$ is and thus the last two factors both have roots mod $p$. – KCd Oct 11 '11 at 2:35

I usually set this as an exercise when teaching Number Theory. My hint is to ask the students: what are the solutions of $z^8=16$ in the complex numbers?

Another solution, which doesn't need quadratic reciprocity but does use the elementary fact that the product of two quadratic nonresidues is a quadratic residue.

According to $p\bmod 8$,

• if $p=1\mod 8$, $\pm 2$ are both quadratic residues
• if $p=3\mod 8$, $-2$ is a quadratic residue but $2$ is not
• if $p=5\mod 8$, $\pm 2$ are not quadratic residues
• if $p=7\mod 8$, $2$ is a quadratic residue but $-2$ is not

If either $2$ or $-2$ is a quadratic residue, $4$ is a 4th power, so $16$ is an 8th power.

This accounts for all primes except those of the form $p=8k+5$, modulo which neither $2$ nor $-2$ is a quadratic residue. The product of two quadratic non-residues is a quadratic residue, so $-1=x^2$ for some $x$.

If $x=y^2$ for some $y$, $y^4=x^2=-1$ so $y^8=1$ so $y$'s order is 8. But the order of any element of a group divides the order of the group, which in this case is $\phi(p)=p-1=8k+4$, and $8\nmid 8k+4$.

So neither $2$ nor $x$ is a quadratic residue, so $2x$ is a quadratic residue $2x=y^2$ for some $y$, so $y^4=(2x)^2=4x^2=-4$ so $y^8=16$.

For the sake of contradiction, suppose there exists a prime $$p$$ such that $$x^8 \equiv 16 \pmod p$$ has no solutions. We can factor $$x^8 - 16 = (x^2 + 2)(x^2 - 2)(x^2 + 2x + 2)(x^2 - 2x + 2).$$

Since this is not divisible by $$p$$, we must have $$\left ( \frac{2}{p} \right ) = \left ( \frac{-2}{p} \right ) = -1 \implies \left ( \frac{-1}{p} \right ) = 1.$$

But then $$x^2 - 2x + 2 = (x-1)^2 + 1 \equiv 0 \pmod p$$ has a solution, which is a contradiction.

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Multitribe evolutionary search for stable Cu–Pd–Ag nanoparticles using neural network models
We present an approach based on two bio-inspired algorithms to accelerate the identification of nanoparticle ground states. We show that a symbiotic co-evolution of nanoclusters across a range of sizes improves the search efficiency considerably, while a neural network constructed with a recently introduced stratified training scheme delivers an accurate description of interactions in multielement systems. The method's performance has been examined in extensive searches for stable elemental (30–80 atoms), binary (50, 55, and 80 atoms), and ternary (50, 55, and 80 atoms) Cu–Pd–Ag clusters. The best candidate structures identified with the neural network model have consistently lower energy at the density functional theory level compared with those found with traditional interatomic potentials.
Authors:
; ; ;
Award ID(s):
Publication Date:
NSF-PAR ID:
10093502
Journal Name:
Physical Chemistry Chemical Physics
Volume:
21
Issue:
17
Page Range or eLocation-ID:
8729 to 8742
ISSN:
1463-9076
3. ABSTRACT We present the results of a proof-of-concept experiment that demonstrates that deep learning can successfully be used for production-scale classification of compact star clusters detected in Hubble Space Telescope(HST) ultraviolet-optical imaging of nearby spiral galaxies ($D\lesssim 20\, \textrm{Mpc}$) in the Physics at High Angular Resolution in Nearby GalaxieS (PHANGS)–HST survey. Given the relatively small nature of existing, human-labelled star cluster samples, we transfer the knowledge of state-of-the-art neural network models for real-object recognition to classify star clusters candidates into four morphological classes. We perform a series of experiments to determine the dependence of classification performance on neural network architecture (ResNet18 and VGG19-BN), training data sets curated by either a single expert or three astronomers, and the size of the images used for training. We find that the overall classification accuracies are not significantly affected by these choices. The networks are used to classify star cluster candidates in the PHANGS–HST galaxy NGC 1559, which was not included in the training samples. The resulting prediction accuracies are 70 per cent, 40 per cent, 40–50 per cent, and 50–70 per cent for class 1, 2, 3 star clusters, and class 4 non-clusters, respectively. This performance is competitive with consistency achieved in previously published human and automated quantitative classification of starmore »


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# two consecutive primes

• January 26th 2010, 06:09 AM
melissaanderson
two consecutive primes
Congruence cases.

(1) Let b be the sum of two consecutive odd numbers that are primes (for example, b = 11 + 13). Prove that b has at least three prime factors (not necessarily different).

(2) Let b be the sum of two consecutive primes p and p', where p' > p >= 3 (for example, b = 89 + 97). Prove that b has at least three prime factors (not necessarily different).

Up for the challenge?
• January 26th 2010, 07:32 AM
TwistedOne151
I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see $n\ge{1}$, and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
as $n\ge{2}$, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.
• January 26th 2010, 07:45 AM
Manx
one of two
Let's consider #1.
Consecutive odd integers that are primes implies consecutive primes, neither of which is 2. Hence, both primes are odd.
Therefore, b is even. It has a factor of 2.
b=p1 + p2 = 2k.
Then k = (p1+p2)/2, the average of p1 and p2.
Assuming p1<p2, k = p1/2 + p2/2 > p2/2 + p2/2 = p2.
Similarly, k > p1.
But if p1 < k < p2, then k is the even integer that lies between the two odd primes. Therefore k = 2n for some integer, n. n > 1 since n=1 implies k=2 implies p1=1 and p2=3 which is a contradiction.

I think you can take a similar approach in #2.
Good luck.
• January 26th 2010, 09:49 AM
Dinkydoe
Quote:

I don't know about (2), but as for (1), let the smaller odd be 2n+1, for some integer n; we see http://www.mathhelpforum.com/math-he...e4f02030-1.gif, and the other odd number is 2n+3. Thus, their sum is (2n+1)+(2n+3)=4n+4=4(n+1)=2*2*(n+1);
as http://www.mathhelpforum.com/math-he...0368e1f9-1.gif, it has one or more prime factors, and so the sum has at least three (not necessarily different) prime factors.

--Kevin C.
You're assuming that 2 different consecutive primes $p,p'$ can be written as $p=2n+1, p'=2n+3$ for the same $n$.
This implies $p'-p= 2$ for all consecutive $p,p'$

(p=13, p'= 17 are also consecutive primes!)
• January 26th 2010, 10:36 AM
Dinkydoe
This is how you do it:

Observe that for any 2 consecutive primes $p,p'$, we have $2|p+p'$ and

$p<\frac{p+p'}{2} < p'$ hence $\frac{p+p'}{2}$ is composite since $p,p'$ were consecutive primes.

Conclusion: $p+p'$ has 3 not necessarily different prime-factors.
• January 26th 2010, 02:20 PM
DKman
Quote:

Originally Posted by Dinkydoe
You're assuming that 2 different consecutive primes $p,p'$ can be written as $p=2n+1, p'=2n+3$ for the same $n$.
This implies $p'-p= 2$ for all consecutive $p,p'$

(p=13, p'= 17 are also consecutive primes!)

The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly.

Your solution does apply to both cases though.
• January 26th 2010, 02:58 PM
Dinkydoe
Quote:

The first question asks for two consecutive odd numbers that are prime. Not two consecutive prime numbers that are odd. So I believe he is looking at the problem correctly
.

Ai, you're right, I'm sorry Kevin C. I misinterpreted the question (Giggle)
• January 27th 2010, 11:10 AM
melissaanderson
Good job guys


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# Extended Kalman filter (EKF)

I am working on the localization problem of an underwater vehicle. However the problem is still in very simple and it does not matter that it is about underwater.

How can I use the EKF when I have multiple measurements but with different sampling? Not all the measurements are available at the same time to perform the update step.

Can I perform the update for the measurements that I have at each time step? For example, I have a pressure sensor and a speed sensor with frequency $1$ Hz. But I have also another sensor (USBL) system with $0.01$ Hz.

What if I perform update step for pressure and speed sensor at each time step adjusting the measurement model to that case (like I have only these sensors) and once I receive the USBL measurement I augment the measurement vector including the USBL measurement?

• Any help? As far as I concerned, what do need is called: Data fusion (or Multiple sensors Kalman filter data fusion). – Gluttton Feb 15 '16 at 20:48
• This answer to a similar question might have some insight. Let me know if it does. – Peter K. Feb 15 '16 at 22:05
• thanks for the replies but still do not know how to deal with asynchronous observation. Assuming, for example, one has different sensors which provide the observation vector. However, each of these sensors provides measurements in different time steps. Then, who can I perform the update step of the EKF? – user19571 Feb 16 '16 at 13:06


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Week 9

Program: Exercise PM02

• Q/A
• Hidden Markov Models I.: belief updates, filtering by forward algorithm

Homework: In case you will not finish the exercise during the lab, your homework is to

• finish the exercise at home,
• identify the potential issues, and
• ask questions on the forum.
• Hand in the Python module hmm_inference.py via BRUTE, together with the solution to the next exercise, PM03. This task has an automatic evaluation, you should see if your functions in hmm_inference.py work fine almost immediately.
• If you need to submit more than one Python file, you can submit an archive with all the modules. The files must be placed in the root of the archive - the archive should not contain any directories.
courses/ui/exercises/pm02.txt · Last modified: 2020/12/30 00:54 (external edit)


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# Correction for attenuation

Correction for attenuation is a statistical procedure, due to Spearman (1904), to "rid a correlation coefficient from the weakening effect of measurement error" (Jensen, 1998), a phenomenon also known as regression dilution. In measurement and statistics, it is also called disattenuation. The correlation between two sets of parameters or measurements is estimated in a manner that accounts for measurement error contained within the estimates of those parameters.

## Background

Correlations between parameters are diluted or weakened by measurement error. Disattenuation provides for a more accurate estimate of the correlation between the parameters by accounting for this effect.

## Definition

The disattenuated estimate of the correlation between two sets of parameters or measures is therefore

${\displaystyle \rho ={\frac {{\mbox{corr}}({\hat {\beta }},{\hat {\theta }})}{\sqrt {R_{\beta }R_{\theta }}}}.}$

That is, the disattenuated correlation is obtained by dividing the correlation between the estimates by the geometric mean of the separation indices of the two sets of estimates. Expressed in terms of Classical test theory, the correlation is divided by the geometric mean of the reliability coefficients of two tests.

Given two random variables ${\displaystyle X}$ and ${\displaystyle Y}$, with correlation ${\displaystyle r_{xy}}$, and a known reliability for each variable, ${\displaystyle r_{xx}}$ and ${\displaystyle r_{yy}}$, the correlation between ${\displaystyle X}$ and ${\displaystyle Y}$ corrected for attenuation is ${\displaystyle r_{x'y'}={\frac {r_{xy}}{\sqrt {r_{xx}r_{yy}}}}}$.

How well the variables are measured affects the correlation of X and Y. The correction for attenuation tells you what the correlation would be if you could measure X and Y with perfect reliability.

If ${\displaystyle X}$ and ${\displaystyle Y}$ are taken to be imperfect measurements of underlying variables ${\displaystyle X'}$ and ${\displaystyle Y'}$ with independent errors, then ${\displaystyle r_{x'y'}}$ measures the true correlation between ${\displaystyle X'}$ and ${\displaystyle Y'}$.

### Derivation of the formula

Let ${\displaystyle \beta }$ and ${\displaystyle \theta }$ be the true values of two attributes of some person or statistical unit. These values are regarded as random variables by virtue of the statistical unit being selected randomly from some population. Let ${\displaystyle {\hat {\beta }}}$ and ${\displaystyle {\hat {\theta }}}$ be estimates of ${\displaystyle \beta }$ and ${\displaystyle \theta }$ derived either directly by observation-with-error or from application of a measurement model, such as the Rasch model. Also, let

${\displaystyle {\hat {\beta }}=\beta +\epsilon _{\beta },\quad \quad {\hat {\theta }}=\theta +\epsilon _{\theta },}$

where ${\displaystyle \epsilon _{\beta }}$ and ${\displaystyle \epsilon _{\theta }}$ are the measurement errors associated with the estimates ${\displaystyle {\hat {\beta }}}$ and ${\displaystyle {\hat {\theta }}}$.

The correlation between two sets of estimates is

${\displaystyle \operatorname {corr} ({\hat {\beta }},{\hat {\theta }})={\frac {\operatorname {cov} ({\hat {\beta }},{\hat {\theta }})}{{\sqrt {\operatorname {var} [{\hat {\beta }}]\operatorname {var} [{\hat {\theta }}}}]}}}$
${\displaystyle ={\frac {\operatorname {cov} (\beta +\epsilon _{\beta },\theta +\epsilon _{\theta })}{\sqrt {\operatorname {var} [\beta +\epsilon _{\beta }]\operatorname {var} [\theta +\epsilon _{\theta }]}}},}$

which, assuming the errors are uncorrelated with each other and with the estimates, gives

${\displaystyle \operatorname {corr} ({\hat {\beta }},{\hat {\theta }})={\frac {\operatorname {cov} (\beta ,\theta )}{\sqrt {(\operatorname {var} [\beta ]+\operatorname {var} [\epsilon _{\beta }])(\operatorname {var} [\theta ]+\operatorname {var} [\epsilon _{\theta }])}}}}$
${\displaystyle ={\frac {\operatorname {cov} (\beta ,\theta )}{\sqrt {(\operatorname {var} [\beta ]\operatorname {var} [\theta ])}}}.{\frac {\sqrt {\operatorname {var} [\beta ]\operatorname {var} [\theta ]}}{\sqrt {(\operatorname {var} [\beta ]+\operatorname {var} [\epsilon _{\beta }])(\operatorname {var} [\theta ]+\operatorname {var} [\epsilon _{\theta }])}}}}$
${\displaystyle =\rho {\sqrt {R_{\beta }R_{\theta }}},}$

where ${\displaystyle R_{\beta }}$ is the separation index of the set of estimates of ${\displaystyle \beta }$, which is analogous to Cronbach's alpha; that is, in terms of Classical test theory, ${\displaystyle R_{\beta }}$ is analogous to a reliability coefficient. Specifically, the separation index is given as follows:

${\displaystyle R_{\beta }={\frac {\operatorname {var} [\beta ]}{\operatorname {var} [\beta ]+\operatorname {var} [\epsilon _{\beta }]}}={\frac {\operatorname {var} [{\hat {\beta }}]-\operatorname {var} [\epsilon _{\beta }]}{\operatorname {var} [{\hat {\beta }}]}},}$

where the mean squared standard error of person estimate gives an estimate of the variance of the errors, ${\displaystyle \epsilon _{\beta }}$. The standard errors are normally produced as a by-product of the estimation process (see Rasch model estimation).


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# Proof that a sequence of continuous functions $(f_n)$ cannot converge pointwise to $1_\mathbb{Q}$ on $[0,1]$

As a homework question, we got asked the following:

Construct a function $f:[0,1] \rightarrow \mathbb{R}$ which is not the pointwise limit of any sequence of continuous functions

Thinking about it, a function which is nowhere continuous should be sufficient; thus I decided to try to use the indicator function on the rationals, $1_\mathbb{Q}$. Basically, I was just wondering if my proof was alright or not:

Suppose $(f_n)$ is a sequence of continuous functions converging pointwise to $1_\mathbb{Q}$; thus for any $\epsilon > 0$ we have that for all $x$ we can find a $N$ such that if $n \geq N$ then $| f_n(x) - 1_\mathbb{Q} (x) | < \epsilon$. In particular, we have for some $x_1 \in \mathbb{Q}, x_2 \notin \mathbb{Q}$ a $N_1$ and $N_2$ respectively such that

• if $n \geq N_1$ then $| f_n(x_1) - 1 | < \epsilon$
• if $n \geq N_2$ then $| f_n(x_2) | < \epsilon$

Now, as $f_k$ is continuous, for any $\epsilon > 0$ we have that for all $x$ there exists a $\delta > 0$ such that if $| x- y | < \delta$ then $|f_k(x) - f_k(y) | < \epsilon$. However, if we pick an arbitrary $x \in \mathbb{Q}$, for $n \geq \max\{N_1, N_2\}$, for any $\delta > 0$ we can pick a $y \notin \mathbb{Q}$ such that $| x- y| < \delta$ yet $| f_n(x) - f_n(y) - 1| < 2\epsilon$ (by the triangle inequality and using the pointwise convergence properties); so if we choose $\epsilon = 1/10$, say, then we have by continuity that $|f_n(x) - f_n(y)| < 1/10$, yet $| f_n(x) - f_n(y) - 1| < 1/5$, which (I believe) is absurd, so we have a contradiction.

I'm a bit iffy about whether this is alright or not (I feel I may be subtly abusing the definitions here), so any feedback would be much appreciated.

• It is quite a bit harder than that. See math.stackexchange.com/questions/77307/… – Rob Arthan Oct 27 '13 at 17:03
• @RobArthan Unfortunately, we've been asked this question in a course which doesn't involve measure theory/functional analysis or the likes. While I'm aware of how the result follows by the Baire categorization theorem because $1_\mathbb{Q}$ is discontinuous everywhere, the idea is to try and find a proof without using such a thing. The general idea (I presume) is that because the limit function is discontinuous everywhere, at some point $f_n$ has to be very close to two discontinuous points which seems to give a contradiction; I can't think of a good way of going about this though. – Andrew D Oct 27 '13 at 17:13
• I think you should post the answer then if you find or learn of a simpler proof. – Rob Arthan Oct 27 '13 at 19:08
• I should have said that the problem with your proposed answer is that $\delta$ depends on $x$, while you are going to need one $\delta$ that works for two distinct values of $x$, one rational and one irrational. The argument in the link I cited solves this problem by giving you a point where the limit is continuous. – Rob Arthan Oct 28 '13 at 0:12
• @RobArthan Ah, I see then; thanks! I'll post an answer giving a more "elementary" proof when I learn of one. – Andrew D Oct 28 '13 at 9:20

Lemma For all non-empty segment $S \subseteq [0,1]$, there exists an arbitrarily large $N$ and a non-empty segment $S' \subseteq S$ such that $f_N(S') = [\frac 1 4, \frac 3 4]$.
Proof: let $S$ such a segment, we can find $x$ and $y$ in $S$ such that $x$ is rational, $y$ is irrational and $x < y$. Since $(f_n)$ converges pointwise to $1_{\mathbb Q}$, there exists an arbitrarily large $N$ such that $f_N(x) \geq \frac 3 4, f_N(y) \leq \frac 1 4$. By the intermediate value theorem, there exists a non-empty segment $S' \subseteq[x,y] \subseteq S$ such that $f_N(S') = [\frac 1 4, \frac 3 4]$.
Using this fact, we can build a subsequence $f_{\phi(n)}$ and a non-increasing sequence of non-empty segments $(S_n)$ such that $f_{\phi(n)}(S_n) = [\frac 1 4, \frac 3 4]$. Using compactness, $\bigcap_{n} S_n \neq \emptyset$: let $x \in \bigcap_{n} S_n$. We have $f(x) = \lim_{n \to \infty} f_{\phi(n)}(x) \in [\frac 1 4, \frac 3 4]$, a contradiction.


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# In a solution consisting of 82% N2, 15% Ar, and the rest O2, which gas is the solvent? Type the name of the gas?

The solvent is that component of a solution which constitutes a larger proportion of the solution. In your question, ${\text{N}}_{2}$ constitutes 82%, $\text{Ar}$ constitutes 15%, and ${\text{O}}_{2}$ constitutes 3%.


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# How do you solve 2^(3x+5)=128?

$x = \frac{2}{3}$
Notice that $128 = {2}^{7}$ so
${2}^{3 x + 5} = {2}^{7}$
thus $3 x + 5 = 7$
$3 x = 2$
$x = \frac{2}{3}$


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# Understanding curvature tensor equation

1. Feb 6, 2012

### avery

hi,

I am trying to understand the meaning of the following equation in the simplest way possible

Last edited: Feb 6, 2012
2. Feb 6, 2012

### elfmotat

Your equation is incorrect because the free indices don't match. Anyway, the simplest way to think about it would probably be a matrix multiplication of the 1x4 row vector A with the 4x4 matrix g. I'm also not really sure what this has to do with the curvature tensor.

Last edited: Feb 6, 2012
3. Feb 6, 2012

### HallsofIvy

As elfmotat suggests, what you want is
$$A^\sigma= g^{\sigma\mu}A_{\mu}$$

And, as he says, this is not directly related to the curvature tensor, it is simply the relation between the covariant and contravariant components of a tensor. In fact, it can be considered the definition of "contravariant components".

4. Feb 6, 2012

Or it can be understood as: metric tensor gives us a particular linear map between cotangent and tangent spaces, the spaces of of forms and the spaces of tangent vectors. In this particular case the contravariant metric can be even degenerate, like it is the case for Newton-Galilei space-times. If it is nondegenerate, then we have an isomorphism, if it is degenerate, we gave only morphism (general linear map).

5. Feb 6, 2012

### avery

thank you Elfmotat and HallsofIvy and arkajad

from what I understand its a matrix that transform a 1xn row vector
if Aμ is a (1x2 row vector) in the two dimensional surface of the sphere.
how can I understand the following equation with a simple numerical example

(gμμ) has a fixed valued? or it depends on Aμ

Last edited: Feb 6, 2012
6. Feb 6, 2012

One possible meaning can be: suppose the metric is diagonal. Then you look at the general formula:

$$A^\mu= g^{\mu\sigma}A_{\sigma}$$

where you have to sum over sigma. But only one term is nonzero, where sigma=mu. This is your equation.

Or, if you prefer: if the matrix is diagonal, then each component of the transformed (row) vector is proportional to the corresponding component of the column vector.

Last edited: Feb 6, 2012
7. Feb 6, 2012

### elfmotat

Don't overload your indices. An index shouldn't appear more than twice in any term. Valid ways of writing your original equation would be:

$A^\mu =g^{\sigma \mu}A_{\sigma }$
$A^\rho =g^{\mu \rho}A_{\mu}$
$A^\nu =g^{\lambda \nu }A_{\lambda }$

etc.

Taking t=constant and r=constant, the metric of a two-sphere is:

$ds^2=r^2(d\theta ^2+sin^2\theta d\phi^2)$

Written as a matrix:

$$g_{\mu \sigma}=\begin{bmatrix} r^2 & 0\\ 0 & r^2sin^2\theta \end{bmatrix}$$

where μ,σ range from θ to ϕ.

The inverse metric is therefore:

$$g^{\mu \sigma}=\begin{bmatrix} r^{-2} & 0\\ 0 & r^{-2}csc^2\theta \end{bmatrix}$$

If the components of the one-form A are $A_\mu=\begin{bmatrix} a & b \end{bmatrix}$, then the components of the vector A are given by:

$$A^{\sigma}=g^{\mu \sigma}A_{\mu}= \begin{bmatrix} a & b \end{bmatrix} \begin{bmatrix} r^{-2} & 0\\ 0 & r^{-2}csc^2\theta \end{bmatrix} = \begin{bmatrix} ar^{-2} & br^{-2}csc^2\theta \end{bmatrix}$$

The components of the (inverse) metric tensor are independent of the components of any particular one-form.

Last edited by a moderator: May 5, 2017
8. Feb 6, 2012

Well, here it appears. That means it is no longer free! It is arbitrary, but fixed.

9. Feb 6, 2012

(gμμ) has a fixed value. It is mu-th diagonal element of the metric. It should not depend Aμ, thou mu in (gμμ) and in Aμ should be the same.

10. Feb 6, 2012

### HallsofIvy

I feel compelled to point out that a tensor can be represented by a matrix, in a given coordinate system, in the same sense that a velocity vector can be represented as, say $<v_x, v_y, v_z>$. But neither the tensor nor the vector is that representation.

Last edited by a moderator: May 5, 2017
11. Feb 6, 2012

### elfmotat

It doesn't matter. First of all, gμμ doesn't really make sense. It's also not clear which index is being summed over (first or second) which would matter if the metric wasn't symmetric.

12. Feb 8, 2012

### avery

thank you all for your help

13. Feb 8, 2012

Except for the case of $\mathbf{R}^n$, where we often define vector as an n-tuple of numbers $(x_1,...,x_n)$.


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# Hidden messages

Mapp and Lucia aficionados are clearly at an advantage here. This was part of a birthday treasure hunt for someone special: a poster hanging at the Barbican Library, and a transparency in a book at the Guildhall Gallery that, when overlapped, gave a cue.

The poster. Sesquiannual meetings!

The transparency. The parallel bars are a different puzzle.

A small academic cottage industry has improved methods for hiding information so that it is only revealed if innocuous-looking pictures are overlapped. Properly done visual cryptography can offer the strength of a one-time-pad and still be decoded by merely overlapping a pre-defined number of transparencies (or “shares”) with seemingly random or unrelated patterns.

No such pretensions here though, and a very simple method was used: I represented each grayscale pixel in the patent (i.e. obvious) images with a 2×2 black-and-white matrix.  0 – 1/4 – 1/2 – 3/4 – 4/4 of the matrix was black, depending on the darkness of the original pixel. When dithering an image this way, for middle tones there is a choice of different patterns with equal darkness. For example, there are six ways to set half the matrix black: ▚, ▞, ▌, ▐, ▀, ▄. Depending on the pattern combination, two overlapping 1/2 black matrices can be 1/2, 3/4, or 4/4 black:

▚ + ▚ = ▚   or   ▀ + ▀ = ▀

▚ + ▌ = ▙   or   ▀ + ▐ = ▜

▚ + ▞  = █   or  ▀ + ▄ = █

and so on.

Generally a 2×2 matrix in the hidden image can represent any grey that is at least as dark as the most dark of the overlapping two matrices. Continue reading

# Our daily bread, equivalent of 11-37 kg of batteries

A back of the envelope calculation on how many kilograms of lithium-ion batteries we would have to carry around to power us for a day — until the next nightly recharge.

Depending on age and sex, about 8-13 MJ energy are needed for a day’s existence, assuming light work. Lithium-ion batteries are widespread in smart phones, electric cars and other electronics not least because of their relatively large specific energy of 0.3-0.7 MJ/kg.

Given this, our battery pack would weigh between 11 and 37 kg for once-a-day recharge.

Having calculated this, how about body fat, our own kind of storage medium? Population average body fat content is around 20 %, and about 10 % body fat is essential. Assuming a body weight of about 80 kg, this leaves about 8 kg storage fat per person or 8 kg * 39.5 MJ/kg = 316 MJ stored energy.

Even in the best case, that are 451 kg batteries to carry.

# Newton stood on the shoulders of 14 m tall giants (or less)

If I have seen further it is by standing on the shoulders of giants.

(Newton in a letter to Robert Hooke)

From this we can calculate the giants to be at most 14.3 m tall, assuming they are human-shaped. This is because for seeing further Newton’s eyes must be higher than the eyes of the giant, i.e. his eye height, standing, must be larger than the shoulder-eye-distance of the giants. Sir Isaac is reported to have been five feet six inches (UK) which is about 167.6 cm. Using present-day median values for eye height and shoulder height (see below) for approximate proportions, his eyes were at 155.1 cm. Using this as the shoulder-eye-distance for the giant, by proportions it follows that the giant is at most about 9.198-times taller than Newton, that is about 14.3 meters tall.

FInishing, it is is worth considering that as giants are taller they are probably also proportionately wider and thicker than Newton, so that at maximum they are 778 times heavier than him. When proportionately scaling up body sizes the weight scales by cubes but bone cross section area, which determines their maximal load, only by squares. If giants are subject to biological limits of bone strength, then their bones have at worst only a tenth of the relative strength of Newton’s. Thus such giants can probably best bear their body weight (and Newton’s) when standing neck-deep under water. That however would defeat the purpose.

Detailed calculation:

AVERAGE HUMAN (50th percentiles, in cm)
eye height = 163.26
shoulder height = 144.18
shoulder-eye-distance = 19.08
total height = 175.49
EH:TH = 0.93031
SE:TH = 0.10872414

NEWTON (see here and more here)
with some likelihood five feet six inches  = 167.64 cm, then eye height by proportion = 155.9572

GIANT
shoulder-eye-distance < 155.9572 then total height by proportion <1434.43.

Footnote: The giants and shoulders metaphor has been used at least since scholasticism.

# Filter FASTA files by sequence id using a regular expression

Use a regular expression for filtering sequences by id from a FASTA file, e.g. just certain chromosomes from a genome. There are other tools as part of bigger packages to install (and no regex support), mostly awk-based awkward (sorry for the pun) bash solutions, and scripts using packages that one needs to install and with still no support for regular expressions. This however is a simple, straightforward little python script for a simple task. It doesn’t do anything else and doesn’t need anything but a stock python installation. Based on the FASTA reader snippetContinue reading

# A script for qPCR analysis in R

An R script for those who like to be close to their qPCR data and catch problems early. It takes export files (multicomponent data, text format, “across columns”) of Life Technologies StepOne machines.

A standard analysis can be done in less than 5 minutes. It consists of these steps:
– plotting of the raw signal (and saving of the result) to catch odd amplification and strong offsets
– baseline correction
– magnified plotting (and save) to check correction and drift in signal
– cycle estimation by using a threshold
– tabulation of the results

# If Nature had done the maths properly …

A Nature report about a new, enzymatic assay of Mycobacterium tuberculosis is (inadvertently) mostly a stark reminder that the false positive and the false negative rates are both important for evaluating an assay’s performance. Superficially, no doubt the assay has advantages: it does not require PCR, prolonged bacterial culture or microscopy, and delivers a result in half an hour unlike current standard methods. It is also sensitive: in a test it flagged all samples positive which microscopy found. Microscopy missed 50% of all positive samples, but even of those missed by microscopy the new method flagged 80% positive. Overall the assay recognises 90% of the Tb+ cases as such.

Despite of these advantages this is not yet a promising method. There is a 27 % false positive rate, i. e. the assay flags a quarter of all tested patients as Tb positive even though there are no Tb-causing bacteria in their samples. This is a problem because only 2-400 / 100 000 people get tuberculosis in any country of the world (World Bank). The new test flags about 27 000 positive out of those 99 600 healthy persons in the population. Continue reading

# Poetry in programming: a Quine, suitable for birthdays

A program that emits its own python source code when run, i.e. a Quine.

a="a= ;print a[0:2]+chr(34)+a+chr(34)+a[3:]#Another happy return!";print a[0:2]+chr(34)+a+chr(34)+a[3:]#Another happy return!

A version in particular for biologists with added emphasis on the cycle of life:

a="a= ;print a[0:2]+chr(34)+a+chr(34)+a[3:]#Keep studying the miracle of life!";print a[0:2]+chr(34)+a+chr(34)+a[3:]#Keep studying the miracle of life!

# Old microscope + £100 (+ 3D printer) = GFP fluorescent microscope

[News: Someone linked this text from the Open Source Toolkit: Hardware article collection and channel of PLoS. Thanks.]

I built my first fluorescent microscope. A Leitz Labovert became a surprisingly decent fluorescent microscope for DIY after spending about £100 and 3D printing a few custom parts. It is a simplified design with only a barrier and an excitation filter, LED illumination and no dichroic mirror.

The quality is sufficient for observing fluorescent yeast and in general fairly faint Drosophila embryos in low-to-midrange magnification.

Drosophila embryonic tracheal system, GFP-marked. Weak additional bright-field illumination for embryo outline.

Yeast with nuclear GFP marker.

Fluorescence is short wavelength light in, longer wavelength light out. In principle a fluorescent microscope needs only two parts: 1) a light source that emits light at the excitation, but not at the emission wavelengths of the sample, and 2) a barrier filter, which lets only the emitted longer wavelength light through, so it doesn’t drown out in the excitation light. There are a few details though, and I will start with the barrier filter.

# Setting the lower cutoff of a miniprep

It maybe a less known fact that the lower cutoff of PCR cleanup and other DNA minipreps may be dialed in by appropriately diluting the binding buffer with water. PCR cleanup kits usually don’t bind fragments below 50-100 bp, depending on the manufacturer. Using water, dilution of the chaotropic salts in the binding buffer sets this limit higher. DNA smaller than the new limit runs through the coloumn, while higher MW DNA adsorbs as before. This can sometimes save the effort of gel purification. Continue reading

# Javascript date calculation

A little contribution to make the Genetics Department fly facility media ordering system more user friendly. This snippet calculates the earliest possible delivery date to be used as default delivery time. The challenge was of course writing it without a single comparison or [if] clause. It ran for a while until Sysadmin Ian decided that he wanted one with more functionality.

var asap = new Date(new Date().getTime()+129600000);
asap.setTime(asap.getTime()+86400000*((asap.getDay()+1)%5)*(Math.exp(0, (asap.getDay()%6))));

(Tomorrow, if it is before 12 o’clock, otherwise the day after tomorrow, but either only if that day is not a day of the weekend. In that case the Monday after.)


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# Integral $\lim_{n \rightarrow \infty} \int_{0}^{1} \sqrt{\frac{1}{x}+n^2x^{2n}}dx$

Evaluate $$\lim_{n \rightarrow \infty} \int_{0}^{1} \sqrt{\frac{1}{x}+n^2x^{2n}}dx$$.

The integral is non asymptomatic. The convergence is non uniform at both $$x=0$$ and $$x=1$$. I'm not sure how to proceed.

We have \begin{align} I_n &= \int_0^1 \sqrt{\frac{1}{x} + n^2 x^{2n}} \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + nx^n\right) \mathrm{d} x \\ &\le \int_0^1 \left(\sqrt{\frac{1}{x}} + (n+1)x^n\right) \mathrm{d} x \\ &= \left(2\sqrt{x} + x^{n+1}\right)\Big\vert_0^1 \\ &= 3. \end{align} Also, we have \begin{align} I_n &= \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x + \int_{1-\frac{1}{\sqrt{n}}}^1 \sqrt{\frac{1}{x} + n^2 x^{2n}}\ \mathrm{d} x\\ &\ge \int_0^{1-\frac{1}{\sqrt{n}}} \sqrt{\frac{1}{x}}\ \mathrm{d} x + \int_{1-\frac{1}{\sqrt{n}}}^1 nx^n \mathrm{d} x \\ &= 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right]. \end{align} Combining the above results, we have $$2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] \le I_n \le 3.$$ Note that $$\lim_{n\to \infty} 2\sqrt{1 - \frac{1}{\sqrt{n}}} + \frac{n}{n+1} \left[1 - \left(1-\frac{1}{\sqrt{n}}\right)^{n+1}\right] = 3.$$ By the squeeze theorem, we have $$\lim_{n\to \infty} I_n = 3$$. We are done.

• Very simple approach. +1 May 8, 2020 at 14:08
• @ParamanandSingh Thank you! May 8, 2020 at 14:09

Let \begin{align} I = \int^1_0 \sqrt{\frac{1}{x}+n^2x^{2n}}\ dx. \end{align} Observe \begin{align} I_2=I-\int^1_0 \frac{dx}{\sqrt{x}} = \int^1_0 \frac{\sqrt{1+n^2x^{2n}}-1}{\sqrt{x}}\ dx = \int^1_0 \frac{n^2x^{2n}}{\sqrt{x}(\sqrt{1+n^2x^{2n}}+1)}\ dx. \end{align} To evaluate $$I_2$$, we will divided the interval. Fix $$\delta=\frac{1}{n^2}$$. Then we see that \begin{align} I_2 = \int^1_{1-\delta}+\int^{1-\delta}_\delta+ \int^\delta_0 \cdots\ dx =J_1+J_2+J_3. \end{align}

For $$J_2$$, the integrand converges uniformly to zero pointwise every where on the interval. Hence $$\lim_{n\rightarrow \infty} J_2 = 0$$.

For $$J_3$$, we see that for $$n$$ sufficiently large we have the estimate \begin{align} J_3 \le&\ \int^\delta_0 \frac{\delta}{\sqrt{x}(\sqrt{1+\delta}+1)}\ dx \leq\ \frac{2\delta\sqrt{\delta}}{\sqrt{1+\delta}+1}. \end{align}

Finally, for $$J_1$$. Observe that \begin{align} |J_1-1| \leq&\ \int^1_{1-\delta} dx\ \left|\frac{n^2}{\sqrt{1-\delta}(\sqrt{1+n^2x^{2n}}+1)}-\frac{1}{\delta}\right|\\ \leq& \left|\frac{n^2\delta}{\sqrt{1-\delta}(\sqrt{1+n^2(1-\delta)^{2n}}+1)}-1\right|. \end{align}

Hence, if you take $$n\rightarrow \infty$$, we see that $$J_1 \rightarrow 1$$, $$J_2\rightarrow 0$$ and $$J_3\rightarrow 0$$.

Thus $$\lim_{n\rightarrow \infty} I_2 =1$$ which means $$I = 3$$ since $$\int^1_0 dx/\sqrt{x} = 2$$.

Remark: Heuristically, we see that for $$f_n(x) \rightarrow \frac{1}{\sqrt{x}}$$ pointwise for any fixed point on $$(0, 1)$$. In fact, on $$[\delta, 1-\delta]$$ the convergence is uniform. Hence the integral of $$\lim_n \int f_n$$ coincides with the integral of $$1/\sqrt{x}$$ on that interval.

However, outside of the interval things get tricky. For the interval $$[0, \delta]$$, this is not a big problem since $$n$$ doesn't affect the integrand as long as $$\delta$$ is small. In particular, I chose $$\delta = n^{-2}$$ which is small enough for this to work (in fact, $$n^{-2}$$ was chosen so that the next part would work).

The most tricky part is the interval $$[1-\delta, 1]$$ because the area of $$f_n$$ concentrates around $$1$$ making the integral over the small interval not negligible. In fact, the main point is that $$f_n \rightarrow \frac{1}{\sqrt{x}}+\delta(x-1)$$ where $$\delta(x)$$ is the Dirac delta function. In fact, we see that \begin{align} \lim_{n\rightarrow \infty} \int^1_0 f_n(x)\ dx = \int^1_0 \frac{1}{\sqrt{x}}+\delta(x-1)\ dx = 2+1 = 3 \end{align}

• For J2 isn't it a problem that the integration region varies with n when you exploit uniform convergence? Could you clarify a bit this? Sep 19, 2021 at 20:12
• Further, how did you guess that J1 was tending to 1? Sep 19, 2021 at 20:14

Hint :

Let $$f_{n} := \sqrt{\frac{1}{x}+n^{2}{x^{2n}}}$$, when $$x$$ is fixed, since $$x \in (0,1)$$ we note that $$\lim\limits_{n \to +\infty} f_{n} = \frac{1}{\sqrt x}$$, which is integrable on $$(0,1)$$.

If we can find an integrable function $$g$$, such that $$|f_{n}(x)|\leq g(x)$$, we can conclude by Dominated convergence Theorem the $$\lim\limits_{n\to +\infty} \int_{0}^{1} f_{n} = \int_{0}^{1} \lim\limits_{n\to +\infty} f_{n} = \int_{0}^{1} \frac{1}{\sqrt x} = 2$$.

• Here dominated convergence does not apply since $x \mapsto nx^n$ is not (a.s) dominated by any integrable function on $[0,1]$, for $\int^1_0nx^n\,dx=\frac{n}{n+1}\xrightarrow{n\rightarrow\infty}1$ whereas $nx^n\xrightarrow{n\rightarrow\infty}0$ a.s. It was proven by River Li that the desired limit is in fact 3 (not 2 ) May 8, 2020 at 15:12


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## G = (C2×C30)⋊D4order 480 = 25·3·5

### 4th semidirect product of C2×C30 and D4 acting via D4/C2=C22

Series: Derived Chief Lower central Upper central

 Derived series C1 — C2×C30 — (C2×C30)⋊D4
 Chief series C1 — C5 — C15 — C30 — C2×C30 — D5×C2×C6 — C2×C15⋊D4 — (C2×C30)⋊D4
 Lower central C15 — C2×C30 — (C2×C30)⋊D4
 Upper central C1 — C22 — C23

Generators and relations for (C2×C30)⋊D4
G = < a,b,c,d | a2=b30=c4=d2=1, ab=ba, cac-1=ab15, ad=da, cbc-1=b-1, dbd=b19, dcd=c-1 >

Subgroups: 1308 in 260 conjugacy classes, 60 normal (26 characteristic)
C1, C2, C2, C2, C3, C4, C22, C22, C22, C5, S3, C6, C6, C6, C2×C4, D4, C23, C23, D5, C10, C10, C10, Dic3, D6, C2×C6, C2×C6, C2×C6, C15, C22⋊C4, C2×D4, C24, Dic5, C20, D10, D10, C2×C10, C2×C10, C2×C10, C2×Dic3, C2×Dic3, C3⋊D4, C22×S3, C22×C6, C22×C6, C5×S3, C3×D5, C30, C30, C30, C22≀C2, C2×Dic5, C5⋊D4, C2×C20, C5×D4, C22×D5, C22×D5, C22×C10, C22×C10, C6.D4, C2×C3⋊D4, C2×C3⋊D4, C23×C6, C5×Dic3, Dic15, C6×D5, C6×D5, S3×C10, C2×C30, C2×C30, C2×C30, D10⋊C4, C23.D5, C2×C5⋊D4, D4×C10, C23×D5, C244S3, C15⋊D4, C10×Dic3, C5×C3⋊D4, C2×Dic15, D5×C2×C6, D5×C2×C6, S3×C2×C10, C22×C30, C23⋊D10, D10⋊Dic3, C30.38D4, C2×C15⋊D4, C10×C3⋊D4, D5×C22×C6, (C2×C30)⋊D4
Quotients: C1, C2, C22, S3, D4, C23, D5, D6, C2×D4, D10, C3⋊D4, C22×S3, C22≀C2, C5⋊D4, C22×D5, C2×C3⋊D4, S3×D5, D4×D5, C2×C5⋊D4, C244S3, C15⋊D4, C2×S3×D5, C23⋊D10, C2×C15⋊D4, D5×C3⋊D4, (C2×C30)⋊D4

Smallest permutation representation of (C2×C30)⋊D4
On 120 points
Generators in S120
(1 52)(2 53)(3 54)(4 55)(5 56)(6 57)(7 58)(8 59)(9 60)(10 31)(11 32)(12 33)(13 34)(14 35)(15 36)(16 37)(17 38)(18 39)(19 40)(20 41)(21 42)(22 43)(23 44)(24 45)(25 46)(26 47)(27 48)(28 49)(29 50)(30 51)(61 112)(62 113)(63 114)(64 115)(65 116)(66 117)(67 118)(68 119)(69 120)(70 91)(71 92)(72 93)(73 94)(74 95)(75 96)(76 97)(77 98)(78 99)(79 100)(80 101)(81 102)(82 103)(83 104)(84 105)(85 106)(86 107)(87 108)(88 109)(89 110)(90 111)
(1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30)(31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60)(61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90)(91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120)
(1 68 52 104)(2 67 53 103)(3 66 54 102)(4 65 55 101)(5 64 56 100)(6 63 57 99)(7 62 58 98)(8 61 59 97)(9 90 60 96)(10 89 31 95)(11 88 32 94)(12 87 33 93)(13 86 34 92)(14 85 35 91)(15 84 36 120)(16 83 37 119)(17 82 38 118)(18 81 39 117)(19 80 40 116)(20 79 41 115)(21 78 42 114)(22 77 43 113)(23 76 44 112)(24 75 45 111)(25 74 46 110)(26 73 47 109)(27 72 48 108)(28 71 49 107)(29 70 50 106)(30 69 51 105)
(1 16)(2 5)(3 24)(4 13)(6 21)(7 10)(8 29)(9 18)(11 26)(12 15)(14 23)(17 20)(19 28)(22 25)(27 30)(31 58)(32 47)(33 36)(34 55)(35 44)(37 52)(38 41)(39 60)(40 49)(42 57)(43 46)(45 54)(48 51)(50 59)(53 56)(61 106)(62 95)(63 114)(64 103)(65 92)(66 111)(67 100)(68 119)(69 108)(70 97)(71 116)(72 105)(73 94)(74 113)(75 102)(76 91)(77 110)(78 99)(79 118)(80 107)(81 96)(82 115)(83 104)(84 93)(85 112)(86 101)(87 120)(88 109)(89 98)(90 117)

G:=sub<Sym(120)| (1,52)(2,53)(3,54)(4,55)(5,56)(6,57)(7,58)(8,59)(9,60)(10,31)(11,32)(12,33)(13,34)(14,35)(15,36)(16,37)(17,38)(18,39)(19,40)(20,41)(21,42)(22,43)(23,44)(24,45)(25,46)(26,47)(27,48)(28,49)(29,50)(30,51)(61,112)(62,113)(63,114)(64,115)(65,116)(66,117)(67,118)(68,119)(69,120)(70,91)(71,92)(72,93)(73,94)(74,95)(75,96)(76,97)(77,98)(78,99)(79,100)(80,101)(81,102)(82,103)(83,104)(84,105)(85,106)(86,107)(87,108)(88,109)(89,110)(90,111), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120), (1,68,52,104)(2,67,53,103)(3,66,54,102)(4,65,55,101)(5,64,56,100)(6,63,57,99)(7,62,58,98)(8,61,59,97)(9,90,60,96)(10,89,31,95)(11,88,32,94)(12,87,33,93)(13,86,34,92)(14,85,35,91)(15,84,36,120)(16,83,37,119)(17,82,38,118)(18,81,39,117)(19,80,40,116)(20,79,41,115)(21,78,42,114)(22,77,43,113)(23,76,44,112)(24,75,45,111)(25,74,46,110)(26,73,47,109)(27,72,48,108)(28,71,49,107)(29,70,50,106)(30,69,51,105), (1,16)(2,5)(3,24)(4,13)(6,21)(7,10)(8,29)(9,18)(11,26)(12,15)(14,23)(17,20)(19,28)(22,25)(27,30)(31,58)(32,47)(33,36)(34,55)(35,44)(37,52)(38,41)(39,60)(40,49)(42,57)(43,46)(45,54)(48,51)(50,59)(53,56)(61,106)(62,95)(63,114)(64,103)(65,92)(66,111)(67,100)(68,119)(69,108)(70,97)(71,116)(72,105)(73,94)(74,113)(75,102)(76,91)(77,110)(78,99)(79,118)(80,107)(81,96)(82,115)(83,104)(84,93)(85,112)(86,101)(87,120)(88,109)(89,98)(90,117)>;

G:=Group( (1,52)(2,53)(3,54)(4,55)(5,56)(6,57)(7,58)(8,59)(9,60)(10,31)(11,32)(12,33)(13,34)(14,35)(15,36)(16,37)(17,38)(18,39)(19,40)(20,41)(21,42)(22,43)(23,44)(24,45)(25,46)(26,47)(27,48)(28,49)(29,50)(30,51)(61,112)(62,113)(63,114)(64,115)(65,116)(66,117)(67,118)(68,119)(69,120)(70,91)(71,92)(72,93)(73,94)(74,95)(75,96)(76,97)(77,98)(78,99)(79,100)(80,101)(81,102)(82,103)(83,104)(84,105)(85,106)(86,107)(87,108)(88,109)(89,110)(90,111), (1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30)(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60)(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90)(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120), (1,68,52,104)(2,67,53,103)(3,66,54,102)(4,65,55,101)(5,64,56,100)(6,63,57,99)(7,62,58,98)(8,61,59,97)(9,90,60,96)(10,89,31,95)(11,88,32,94)(12,87,33,93)(13,86,34,92)(14,85,35,91)(15,84,36,120)(16,83,37,119)(17,82,38,118)(18,81,39,117)(19,80,40,116)(20,79,41,115)(21,78,42,114)(22,77,43,113)(23,76,44,112)(24,75,45,111)(25,74,46,110)(26,73,47,109)(27,72,48,108)(28,71,49,107)(29,70,50,106)(30,69,51,105), (1,16)(2,5)(3,24)(4,13)(6,21)(7,10)(8,29)(9,18)(11,26)(12,15)(14,23)(17,20)(19,28)(22,25)(27,30)(31,58)(32,47)(33,36)(34,55)(35,44)(37,52)(38,41)(39,60)(40,49)(42,57)(43,46)(45,54)(48,51)(50,59)(53,56)(61,106)(62,95)(63,114)(64,103)(65,92)(66,111)(67,100)(68,119)(69,108)(70,97)(71,116)(72,105)(73,94)(74,113)(75,102)(76,91)(77,110)(78,99)(79,118)(80,107)(81,96)(82,115)(83,104)(84,93)(85,112)(86,101)(87,120)(88,109)(89,98)(90,117) );

G=PermutationGroup([[(1,52),(2,53),(3,54),(4,55),(5,56),(6,57),(7,58),(8,59),(9,60),(10,31),(11,32),(12,33),(13,34),(14,35),(15,36),(16,37),(17,38),(18,39),(19,40),(20,41),(21,42),(22,43),(23,44),(24,45),(25,46),(26,47),(27,48),(28,49),(29,50),(30,51),(61,112),(62,113),(63,114),(64,115),(65,116),(66,117),(67,118),(68,119),(69,120),(70,91),(71,92),(72,93),(73,94),(74,95),(75,96),(76,97),(77,98),(78,99),(79,100),(80,101),(81,102),(82,103),(83,104),(84,105),(85,106),(86,107),(87,108),(88,109),(89,110),(90,111)], [(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30),(31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60),(61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90),(91,92,93,94,95,96,97,98,99,100,101,102,103,104,105,106,107,108,109,110,111,112,113,114,115,116,117,118,119,120)], [(1,68,52,104),(2,67,53,103),(3,66,54,102),(4,65,55,101),(5,64,56,100),(6,63,57,99),(7,62,58,98),(8,61,59,97),(9,90,60,96),(10,89,31,95),(11,88,32,94),(12,87,33,93),(13,86,34,92),(14,85,35,91),(15,84,36,120),(16,83,37,119),(17,82,38,118),(18,81,39,117),(19,80,40,116),(20,79,41,115),(21,78,42,114),(22,77,43,113),(23,76,44,112),(24,75,45,111),(25,74,46,110),(26,73,47,109),(27,72,48,108),(28,71,49,107),(29,70,50,106),(30,69,51,105)], [(1,16),(2,5),(3,24),(4,13),(6,21),(7,10),(8,29),(9,18),(11,26),(12,15),(14,23),(17,20),(19,28),(22,25),(27,30),(31,58),(32,47),(33,36),(34,55),(35,44),(37,52),(38,41),(39,60),(40,49),(42,57),(43,46),(45,54),(48,51),(50,59),(53,56),(61,106),(62,95),(63,114),(64,103),(65,92),(66,111),(67,100),(68,119),(69,108),(70,97),(71,116),(72,105),(73,94),(74,113),(75,102),(76,91),(77,110),(78,99),(79,118),(80,107),(81,96),(82,115),(83,104),(84,93),(85,112),(86,101),(87,120),(88,109),(89,98),(90,117)]])

66 conjugacy classes

 class 1 2A 2B 2C 2D 2E 2F 2G 2H 2I 2J 3 4A 4B 4C 5A 5B 6A ··· 6G 6H ··· 6O 10A ··· 10F 10G 10H 10I 10J 10K 10L 10M 10N 15A 15B 20A 20B 20C 20D 30A ··· 30N order 1 2 2 2 2 2 2 2 2 2 2 3 4 4 4 5 5 6 ··· 6 6 ··· 6 10 ··· 10 10 10 10 10 10 10 10 10 15 15 20 20 20 20 30 ··· 30 size 1 1 1 1 2 2 10 10 10 10 12 2 12 60 60 2 2 2 ··· 2 10 ··· 10 2 ··· 2 4 4 4 4 12 12 12 12 4 4 12 12 12 12 4 ··· 4

66 irreducible representations

 dim 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 4 4 4 4 4 type + + + + + + + + + + + + + + + + + - + image C1 C2 C2 C2 C2 C2 S3 D4 D4 D5 D6 D6 D10 D10 D10 C3⋊D4 C3⋊D4 C5⋊D4 S3×D5 D4×D5 C15⋊D4 C2×S3×D5 D5×C3⋊D4 kernel (C2×C30)⋊D4 D10⋊Dic3 C30.38D4 C2×C15⋊D4 C10×C3⋊D4 D5×C22×C6 C23×D5 C6×D5 C2×C30 C2×C3⋊D4 C22×D5 C22×C10 C2×Dic3 C22×S3 C22×C6 D10 C2×C10 C2×C6 C23 C6 C22 C22 C2 # reps 1 2 1 2 1 1 1 4 2 2 2 1 2 2 2 8 4 8 2 4 4 2 8

Matrix representation of (C2×C30)⋊D4 in GL4(𝔽61) generated by

 60 0 0 0 0 60 0 0 0 0 60 0 0 0 17 1
,
 44 1 0 0 16 60 0 0 0 0 14 0 0 0 45 48
,
 39 8 0 0 8 22 0 0 0 0 48 20 0 0 22 13
,
 17 18 0 0 45 44 0 0 0 0 60 0 0 0 17 1
G:=sub<GL(4,GF(61))| [60,0,0,0,0,60,0,0,0,0,60,17,0,0,0,1],[44,16,0,0,1,60,0,0,0,0,14,45,0,0,0,48],[39,8,0,0,8,22,0,0,0,0,48,22,0,0,20,13],[17,45,0,0,18,44,0,0,0,0,60,17,0,0,0,1] >;

(C2×C30)⋊D4 in GAP, Magma, Sage, TeX

(C_2\times C_{30})\rtimes D_4
% in TeX

G:=Group("(C2xC30):D4");
// GroupNames label

G:=SmallGroup(480,639);
// by ID

G=gap.SmallGroup(480,639);
# by ID

G:=PCGroup([7,-2,-2,-2,-2,-2,-3,-5,141,219,1356,18822]);
// Polycyclic

G:=Group<a,b,c,d|a^2=b^30=c^4=d^2=1,a*b=b*a,c*a*c^-1=a*b^15,a*d=d*a,c*b*c^-1=b^-1,d*b*d=b^19,d*c*d=c^-1>;
// generators/relations

׿
×
𝔽


text:
# The Azimuth Project Blog - network theory (part 24) (Rev #4)

This page is a blog article in progress, written by John Baez. To see discussions of this article while it is being written, go to the Azimuth Forum.

### Review

The storm is brewing for the final climax, so let’s remember where we are. We start with a stochastic reaction network:

This consists of:

• finite sets of transitions $T$, complexes $K$ and species $S$,

• a map $r: T \to (0,\infty)$ giving a rate constant for each transition,

source and target maps $s,t : T \to K$ saying where each transition starts and ends,

• a map $Y : K \to \mathbb{N}^S$ saying how each complex is made of species.

Then we extend $s, t$ and $Y$ to linear maps:

Then we put inner products on these vector spaces as described last time, which lets us ‘turn around’ the maps $s$ and $t$ by taking their adjoints:

$s^\dagger, t^\dagger : \mathbb{R}^K \to \mathbb{R}^T$

More surprisingly, we can ‘turn around’ $Y$ and get a nonlinear map using ‘matrix exponentiation’:

$\begin{array}{ccc} \mathbb{R}^S &\to& \mathbb{R}^K \\ x &\mapsto& x^Y \end{array}$

This is most easily understood by thinking of $x$ as a row vector and $Y$ as a matrix:

$\begin{array}{ccl} x^Y &=& {\left( \begin{array}{cccc} x_1 , & x_2 , & \dots, & x_k \end{array} \right)}^{ \left( \begin{array}{cccc} Y_{11} & Y_{12} & \cdots & Y_{1 \ell} \\ Y_{21} & Y_{22} & \cdots & Y_{2 \ell} \\ \vdots & \vdots & \ddots & \vdots \\ Y_{k1} & Y_{k2} & \cdots & Y_{k \ell} \end{array} \right)} \\ \\ &=& \left( x_1^{Y_{11}} \cdots x_k^{Y_{k1}} ,\; \dots, \; x_1^{Y_{1 \ell}} \cdots x_k^{Y_{k \ell}} \right) \end{array}$

Remember, complexes are made out of species. The matrix entry $Y_{i j}$ says how many things of the $j$th species there are in a complex of the $i$th kind. If $\psi \in \mathbb{R}^K$ says how many complexes there are of each kind, $Y \psi \in \mathbb{R}^S$ says how many things there are of each species. Conversely, if $x \in mathbb{R}^S$ says how many things there are of each species, $Y^x \in \mathbb{R}^K$ says how many ways we can build each kind of complex from them.

So, we get these maps:

Next, the boundary operator

$\partial : \mathbb{R}^T \to \mathbb{R}^K$

describes how each transition causes a change in complexes:

$\partial = t - s$

As we saw last time, there is a Hamiltonian

$H : \mathbb{R}^K \to \mathbb{R}^K$

describing a Markov processes on the set of complexes, given by

$H = \partial s^\dagger$

But the star of the show is the rate equation. This describes how the number of things of each species changes with time. We write these numbers in a list and get a vector $x \in \mathbb{R}^S$ with nonnegative components. The rate equation says:

$\displaystyle{ \frac{d x}{d t} = Y H x^Y }$

We can read this as follows:

$x$ says how many things of each species we have now.

$x^Y$ says how many complexes of each kind we can build from these species.

$s^\dagger x^Y$ says how many transitions of each kind can originate starting from these complexes, with each transition weighted by its rate.

$H x^Y = \partial s^\dagger x^Y$ is the rate of change of the number of complexes of each kind, due to these transitions.

$Y H x^Y$ is the rate of change of the number of things of each species.

### The zero deficiency theorem

We are looking for equilibrium solutions of the rate equation, where the number of things of each species doesn’t change at all:

$Y H x^Y = 0$

In fact we will find complex balanced equilibrium solutions, where even the number of complexes of each kind don’t change:

$H x^Y = 0$

More precisely, we have:

Deficiency Zero Theorem (Child’s Version). Suppose we have a reaction network obeying these two conditions:

1. It is weakly reversible, meaning that whenever there’s a transition from one complex $\kappa$ to another $\kappa'$, there’s a directed path of transitions going back from $\kappa'$ to $\kappa$.

2. It has deficiency zero, meaning $\mathrm{im} \partial \cap \mathrm{ker} Y = \{ 0 \}$.

Then for any choice of rate constants there exists a complex balanced equilibrium solution of the rate equation where all species are present in nonzero amounts. In other words, there exists $x \in \mathbb{R}^S$ with all components positive and such that:

$H x^Y = 0$

Proof. Because our reaction network is weakly reversible, the theorems in Part 23 show there exists $\psi \in (0,\infty)^K$ with

$H \psi = 0$

This $\psi$ may not be of the form $x^Y$, but we shall adjust $\psi$ so that it becomes of this form, while still remaining a solution of $H \psi = 0$. To do this, we need a couple of lemmas:

Lemma 1. $\ker \partial^\dagger + \im Y^\dagger = \mathbb{R}^K$.

Proof. We need to use a few facts from linear algebra. If $V$ is a finite-dimensional vector space with inner product, the orthogonal complement $L^\perp$ of a subspace $L \subseteq V$ consists of vectors that are orthogonal to everything in $L$:

$L^\perp = \{ v \in V : \quad \forall w \in L \quad \langle v, w \rangle = 0 \}$

We have

$(L \cap M)^\perp = L^\perp + M^\perp$

where $L$ and $M$ are subspaces of $V$ and $+$ denotes the sum of subspaces. Also, if $T: V \to W$ is a linear map between finite-dimensional vector spaces with inner product, we have

$(\mathrm{ker} T)^\perp = \mathrm{im} T^\dagger$

and

$(\mathrm{im} T)^\perp = \mathrm{ker} T^\dagger$

Now, because our reaction network has deficiency zero, we know that

$\mathrm{im} \partial \cap \mathrm{ker} Y = \{ 0 \}$

Taking the orthogonal complement of this subspace of $\mathbb{R}^K$, we get

$(\mathrm{im} \partial \cap \mathrm{ker} Y)^\perp = \mathbb{R}^K$

and using the rules we mentioned, we obtain

$\ker \partial^\dagger + \im Y^\dagger = \mathbb{R}^K$

as desired.   █

Now, given a vector $\phi$ in $\mathbb{R}^K$ or $\mathbb{R}^S$ with all positive components, we can define the logarithm of such a vector, componentwise:

$(\ln \phi)_i = \ln (\phi_i)$

Similarly, for any vector $\phi$ in either of these spaces, we can define its exponential in a componentwise way:

$(\exp \phi)_i = \exp(\phi_i)$

These operations are inverse to each other. Moreover:

Lemma 2. The nonlinear operator

$\begin{array}{ccc} \mathbb{R}^S &\to& \mathbb{R}^K \\ x &\mapsto& x^Y \end{array}$

is related to the linear operator

$\begin{array}{ccc} \mathbb{R}^S &\to& \mathbb{R}^K \\ x &\mapsto& Y^\dagger x \end{array}$

by the formula

$x^Y = \exp(Y^\dagger \ln x )$

which holds for all $x \in (0,\infty)^S$.

Proof. A straightforward calculation. By the way, this formula would look a bit nicer if we treated $\ln x$ as a row vector and multiplied it on the right by $Y$: then we would have

$x^Y = \exp((\ln x) Y)$

The problem is that we are following the usual convention of multiplying vectors by matrices on the left, yet writing the matrix on the right in $x^Y$; in a sense the transpose $Y^\dagger$ of the matrix $Y$ serves to compensate for this.   █

Now, given our vector $\psi \in (0,\infty)^K$ with $H \psi = 0$, we obtain $\ln \psi \in \mathbb{R}^K$. Lemma 1 says that

$\mathbb{R}^K \in \ker \partial^\dagger + \im Y^\dagger$

so we can write

$\ln \psi = \alpha + Y^\dagger \beta$

for some $\alpha \in \ker \partial^\dagger$ and some $\beta \in \mathbb{R}^S$. Moreover, we can write

$\beta = \ln x$

for some $x \in (0,\infty)^S$, so that

$\ln \psi = \alpha + Y^\dagger (\ln x)$

Exponentiating both sides componentwise, we get

$\psi = \exp(\alpha) \; \exp(Y^\dagger (\ln x))$

where at right we are taking the componentwise product of vectors. Thanks to Lemma 2, we conclude that

$\psi = \exp(\alpha) x^Y$

So, we have taken $\psi$ and almost written it in the form $x^Y$—but not quite! We can adjust $\psi$ to make it be of this form:

$\exp(-\alpha) \psi = x^Y$

Clearly all the components of $\exp(-\alpha) \psi$ are positive, since the same is true for both $\psi$ and $\exp(-\alpha)$. So, the only remaining task is to check that $\exp(-\alpha) \psi$ is still a complex balanced equilibrium solution of the rate equation:

Lemma 3. If $H \psi = 0$ and $\alpha \in \ker \partial^\dagger$, then $H(\exp(-\alpha) \psi) = 0$.

Proof. It is enough to check that multiplication by $\exp(-\alpha)$ commutes with the Hamiltonian $H$, since then

$H(\exp(-\alpha) \psi) = \exp(-\alpha) H \psi = 0$

Recall from Part 23 that $H$ is the Hamiltonian of a Markov process associated to this ‘graph with rates’:

As noted here:

• John Baez and Brendan Fong, A Noether theorem for Markov processes.

multiplication by a function on $K$ commutes with $H$ if and only if that function is constant on each connected component of this graph. So, it suffices to show that $\exp(-\alpha)$ is constant on each connected component. For this, it is enough to show that $\alpha$ itself is constant on each connected component. But this follows from the next lemma, since $\alpha \in \ker \partial^\dagger$.   █

Lemma 4. A function $\alpha \in \mathbb{R}^K$ is constant on each connected component of the graph $s, t: T \to K$ iff $\partial^\dagger \alpha = 0$

Proof. Suppose $\partial^\dagger \alpha = 0$, or in other words, $\alpha \in \mathrm{ker} \partial^\dagger$, or in still other words, $\alpha \in (\mathrm{im} \partial)^\perp$. To show that $\alpha$ is constant on each connected component, it suffices to show that whenever we have two complexes connected by a transition, like this:

$\tau: \kappa \to \kappa'$

then $\alpha$ takes the same value at both these complexes:

$\alpha_\kappa = \alpha_{\kappa'}$

To see this, note

$\partial \tau = t(\tau) - s(\tau) = \kappa' - \kappa$

and since $\alpha \in (\mathrm{im} \partial)^\perp$, we conclude

$\langle \alpha, \kappa' - \kappa \rangle = 0$

But calculating this inner product, we see

$\alpha_{\kappa'} - \alpha_{\kappa} = 0$

as desired.

For the converse, we simply turn the argument around: if $\alpha$ is constant on each connected component, we see $\langle \alpha, \kappa' - \kappa \rangle = 0$ whenever there is a transition $\tau : \kappa \to \kappa'$. It follows that $\langle \alpha, \partial \tau \rangle = 0$ for every transition $\tau$, so $\alpha \in (\im \partial)^\perp$.

And thus concludes the proof of the lemma!   █

And thus concludes the proof of the theorem!   █

And thus concludes this post!

category: blog


text:
IMI Interdisciplinary Mathematics InstituteCollege of Arts and Sciences

• Dec. 5, 2017
• 4:15 p.m.
• LeConte 312

Abstract

We discuss some of the basic elements of Maximal functions and their connection to Hardy spaces, both in the classical and the multiparameter settings on $R^d$. We also present some resent results on Hardy spaces in the presence of a non-negative self-adjoint operator whose heat kernel has Gaussian localization.

© Interdisciplinary Mathematics Institute | The University of South Carolina Board of Trustees | Webmaster


text:
# Typesetting “C#” in LaTeX

How do you refer to the C# programming language in LaTeX? Simply typing C# doesn’t work because # is a special character in LaTeX. You could type C#. That works, but it looks a little odd. The number sign is too big and too low.

What about using a musical sharp sign, i.e. C$\sharp$? That also looks a little odd. Even though the language is pronounced “C sharp,” it’s usually written with a number sign, not a sharp.

Let’s look at recommended ways of typesetting C++ to see whether that helps. The top answer to this question on TeX Stack Exchange is to define a new command as follows:

\newcommand{\CC}{C\nolinebreak\hspace{-.05em}\raisebox{.4ex}{\tiny\bf +}\nolinebreak\hspace{-.10em}\raisebox{.4ex}{\tiny\bf +}}

This does several things. First, it prevents line breaks between the constituent characters. It also does several things to the plus signs:

• Draws them in closer
• Makes them smaller
• Raises them
• Makes them bold

The result is what we’re subconsciously accustomed to seeing in print.

Here’s an analogous command for C#.

\newcommand{\CS}{C\nolinebreak\hspace{-.05em}\raisebox{.6ex}{\tiny\bf #}}

And here’s the output. The number sign is a little too small.

To make a little larger number sign, replace \tiny with \scriptsize.

Related posts:

## 7 thoughts on “Typesetting “C#” in LaTeX”

1. JohnH says:

The C# symbol you last give is too high… essentially, the positioning you want, to match most renderings of C# I’ve seen, is to make the top and bottom stroke of the octothorpe line up with the edges of the letter C’s terminals; the top stroke with the top terminal, and the bottom stroke with the bottom terminal. This may prove difficult with LaTeX as the octothorpe in the typeface shown (or how the ‘special char’ is drawn in stead of rendered fromt he font) has the spacing between the top and bottom strokes much narrower than most typefaces.

This might be more correctable by using XeTeX for yout LaTeX processor with truetype fonts, set with setmainfont{Arial} (or some other better typeface).

2. alfC says:

why not ” C$^sharp$ ” ?

3. why not texttt{C#} ?

4. abc says:

The best looking one I’ve found is simply Ctexttt{#}

5. Thomas says:

If you use the relsize package this lines up the octothorpe nicely:

newcommand{CS}{Cnolinebreak[4]hspace{-.05em}raisebox{.4ex}{relsize{-2}{textbf{#}}}}

Relsize is more robust anyway. tiny and scriptsize are not going to work well if you put the macro in a section title.


text:
## Mello Group Title If (a/b)=(c/d), prove (a+b)/(a-b)=(c+d)/(c-d). one year ago one year ago

1. hartnn Group Title

hint : 1.Add 1 on both sides---->(1) 2.subtract 1 from both sides ---->(2) 3. divide : (1)/(2)

2. hartnn Group Title

@Mello did u get it or should i show detailed steps??

3. Mello Group Title

Yes, please. My brain isn't getting it.

4. Mello Group Title

I got step 1...

5. Mello Group Title

Im a bit lost from there.

6. hartnn Group Title

so u must have got : $$\frac{a+b}{b}=\frac{c+d}{d}$$------->(1) right? Step 2: subtract 1 from both sides: $$\frac{a}{b}-1=\frac{c}{d}-1$$ $$\frac{a-b}{b}=\frac{c-d}{d}$$------->(2) got step 2 ??

7. Mello Group Title

@hartnn Right. Oh subtract from the original. OK.

8. hartnn Group Title

sorry i didn't mention that, now divide: (1) / (2) did u see how u get your final result after dividing??

9. Mello Group Title

How would you divide an equation by an equation?

10. hartnn Group Title

ofcourse we can: if x=y and w=z after dividing we get (x/w)=(y/z) try dividing here. if u don't get, i'll show that step.

11. Mello Group Title

Oh, right. I remember learning something like that. I think I should be able to manage from here

12. hartnn Group Title

good,let me know if u still have doubts or could not get final answer.

13. Mello Group Title

Alrighty, Thank you so much for the help!

14. hartnn Group Title

welcome :)

15. Mello Group Title

Got it! Thanks again.