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## Found 10 Documents (Results 1–10)

100
MathJax

### Dynamics of rotating fluids. (English)Zbl 1167.76047

Dormy, Emmanuel (ed.) et al., Mathematical aspects of natural dynamos. Selected by Grenoble Sciences. Grenoble: Grenoble Sciences Université Joseph Fourier; Boca Raton, FL: CRC Press (ISBN 978-1-58488-954-0/hbk). The Fluid Mechanics of Astrophysics and Geophysics 13, 119-198 (2007).

### Formation of dynamical structures in axisymmetric fluid systems. (English)Zbl 0933.76025

Busse, Friedrich H. (ed.) et al., Evolution of spontaneous structures in dissipative continuous systems. Berlin: Springer. Lect. Notes Phys., New Ser. m: Monogr. m55, 86-126 (1998).

### Double-layer convection: A brief review and some recent experimental results. (English)Zbl 0875.76016

Renardy, Yuriko Y. (ed.) et al., Advances in multi-fluid flows. AMS-IMS-SIAM joint summer research conference, held in Seattle, WA (USA), 23–27, July 1995. Philadelphia, PA: SIAM. Proc. Appl. Math. 33-41 (1996).
MSC:  76-05 76E20 76E15

### An experimental and theoretical investigation of the onset of convection in rotating spherical shells. (English)Zbl 0526.76055

MSC:  76E15 76E20 76U05
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### Shear flow instabilities in a rotating system. (English)Zbl 0461.76032

MSC:  76E20 76U05 76F10
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MSC:  76E20
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### Thermal instabilities in rapidly rotating systems. (English)Zbl 0224.76041

MSC:  76E20 76E15
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all top 5


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Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

### Paper:

TR03-076 | 8th September 2003 00:00

#### Testing the independence number of hypergraphs

TR03-076
Authors: Michael Langberg
Publication: 16th October 2003 17:21
A $k$-uniform hypergraph $G$ of size $n$ is said to be $\varepsilon$-far from having an independent set of size $\rho n$ if one must remove at least $\varepsilon n^k$ edges of $G$ in order for the remaining hypergraph to have an independent set of size $\rho n$. In this work, we present a natural {\em property testing} algorithm that distinguishes between hypergraphs which have an independent set of size $\geq \rho n$ and hypergraphs which are $\varepsilon$-far from having an independent set of size $\rho n$. Our algorithm is natural in the sense that we sample $\simeq c(k)\frac{\rho^{2k}}{\varepsilon^3}$ random vertices of $G$, and according to the independence number of the hypergraph induced by this sample, we distinguish between the two cases above. Here $c(k)$ depends on $k$ alone ({\em e.g.} the sample size is independent of $n$).

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Showing:
Handle Score
@Ibrahim Nash 5765
@akhayrutdinov 5111
@mb1973 4993
@Quandray 4944
@saiujwal13083 4506
@sanjay05 3762
@marius_valentin_dragoi 3516
@sushant_a 3459
@verma_ji 3341
@KshamaGupta 3318
Infix to Postfix
Medium Accuracy: 23.7% Submissions: 2708 Points: 4

Given an infix expression in the form of string str. Convert this infix expression to postfix expression.

• Infix expression: The expression of the form a op b. When an operator is in-between every pair of operands.
• Postfix expression: The expression of the form a b op. When an operator is followed for every pair of operands.
â€‹Note: The order of precedence is: ^ greater than * equals to / greater than + equals to -.

Example 1:

Input: str = "a+b*(c^d-e)^(f+g*h)-i"
Output: abcd^e-fgh*+^*+i-
Explanation:
After converting the infix expression
into postfix expression, the resultant
expression will be abcd^e-fgh*+^*+i-

Example 2:

Input: str = "A*(B+C)/D"
Output: ABC+*D/
Explanation:
After converting the infix expression
into postfix expression, the resultant
expression will be ABC+*D/

This is a function problem. You only need to complete the function infixToPostfix() that takes a string(Infix Expression) as a parameter and returns a string(postfix expression). The printing is done automatically by the driver code.

Expected Time Complexity: O(|str|).
Expected Auxiliary Space: O(|str|).

Constraints:
1 ≤ |str| ≤ 105


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# What is the molecular geometry of the BCl_3 molecule?

Nov 21, 2016

$\text{Boron trichloride}$ is trigonal planar.
There are 3 boron-centred electrons, which form $3 \times B - C l$ bonds. The geometry around boron is thus trigonal planar with $\angle C l - B - C l = {120}^{\circ}$.
Now much of the chemistry of boron relates to the filling of its empty $\text{p-orbital}$; with extra halide ligand, $\text{tetrachloroborate anion}$ is formed:
$B C {l}_{3} + C {l}^{-} \rightarrow B C {l}_{4}^{-}$


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# Znanstveni kolokviji

## Isometries and isomorphisms

 Vrijeme: 30.1.201917:00 Predavaonica: 005 Predavač: Lajos Molnar, University of Szeged and Budapest University of Technology and Economics Naziv: Isometries and isomorphisms Opis: Surjective isometries are likely the most important "symmetries" of metric spaces. When the underlying space is also equipped with some compatible algebraic structure, it frequently turns out that the (surjective) isometries are closely related to corresponding algebraic isomorphisms. In this talk we are concerned with that phenomenon in areas of linear algebra and functional analysis. In the first part of the talk we present some classical results on linear isometries of matrix algebras. In the second part we consider non-linear problems on the positive definite cone in matrix algebras and explain how certain algebraic considerations help to characterize their isometries with respect to certain important metrics. If time permits, we will comment on infinite dimensional extensions, too.

<< Povratak na popis kolokvija


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On the deletion channel with small deletion probability
Yash Kanoria Andrea Montanari
Proceedings of the IEEE International Symposium on Information Theory, Austin, TX, USA, June 2010
##### Abstract

The  deletion   channel  is the simplest point-to-point communication  channel  that models lack of synchronization. Despite significant effort, little is known about its capacity, and even less about optimal coding schemes. In this paper we initiate a new systematic approach to this problem, by demonstrating that capacity can be computed in a series expansion for  small   deletion   probability .We compute two leading terms of this expansion, and show that capacity is achieved, up to this order, by i.i.d. uniform random distribution of the input. We think that this strategy can be useful in a number of capacity calculations.


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# Recent questions and answers in English Language

1
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2
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3
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6
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18
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20
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21
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23
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24
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25
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26
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27
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28
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1 vote
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How to find all irreducible polynomials in Z2 with degree 5? [duplicate]

I am totally lost on how to do this one. I am supposed to accomplish the following:

Find all irreducible polynomials in $\mathbb{Z}_2[x]$ with degree $5$. I may use the fact that x, $x+1$ and $x^2+x+1$ are the irreducibles of degree less than or equal to 2

If someone could provide a step by step explanation of how to do this, that would be amazing! Thanks in advance!

P.S.: This is not a homework, but is a question on a past exam that I couldn't answer and I want to know how to do it.

marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Oct 6 '18 at 14:59

Here is a rough procedure for finding all irreducible polynomials in $\mathbb{Z}_{2}[x]$. I will leave some of the details of computation to you.

If we let $f(x) = x^{5}+a_{4}x^{4}+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0} \in \mathbb{Z}_{2}[x]$, we see that each $a_{i}$ can be $0$ or $1$, yielding two choices for each of the five coefficients. Hence, we note that there are $32$ polynomials of degree $5$ in $\mathbb{Z}_{2}[x]$.

Now, for $f(x)$ to be irreducible, it cannot be divisible by any polynomial in $\mathbb{Z}_{2}[x]$ of degree $<5$. If a product of polynomials $\prod_{i} g_{i}(x)$ has degree $5$, then it follows that the degree of at least one $g_{i}(x) < 3$ (another way of saying this is that the any partition of $5$ into a sum of strictly positive integers contains a summand $< 3$). Hence, it suffices to find all $f(x) \in \mathbb{Z}_{2}[x]$ of degree $5$ such that no irreducible polynomial of degree $1$ or $2$ divides $f(x)$.

It is straightforward to test for division by the polynomials of degree $1$. We must ensure that our $f(x)$ does not have any roots in $\mathbb{Z}_{2}$. Simply take a test $f(x)$ and verify that $f(0)$ and $f(1)$ are nonzero. This narrows down our set of $16$ polynomials to our remaining possible choices. Call this remaining set $P$.

Now, to test for division by irreducible polynomials of degree $2$, we must first compute the irreducible polynomials of degree $2$ in $\mathbb{Z}_{2}[x]$. I leave this to you - comment if you need further help in this step. Once we have computed the irreducible polynomials of degree $2$, we can use simple polynomial long division (since $\mathbb{Z}_{2}[x]$ is a Euclidean domain!) to test each element of $P$ for divisibility by an irreducible polynomial of degree $2$. If $f(x) \in P$ is not divisible by any irreducible polynomial of degree $2$, it is an irreducible polynomial of degree $5$ in $\mathbb{Z}_{2}[x]$. I leave the details of these actual computations to you. Feel free to comment if you need further hints!

• This is a fine answer, although paragraph 3's phrasing is kind of awkward, since $f$ can always decompose as a product of more than two irreducibles; it is true that any reducible $f$ has some decomposition into factors of degrees 1+4 or 2+3 (or more broadly, that it's divisible by some irreducible polynomial of degree 1 or 2), but I think there's a better way of making that argument than your current phrasing. – Steven Stadnicki Oct 30 '14 at 17:41
• @StevenStadnicki: great points, thanks! I'll try to edit it to something better. :) – Alex Wertheim Oct 30 '14 at 17:42
• @StevenStadnicki: I've updated paragraph 3 in response to your suggestions. Hopefully it reads a bit more accurately (and smoothly!) now. Thanks again! – Alex Wertheim Oct 30 '14 at 17:51
• A good, generic way of doing this. You can generalize this to higher degrees by building a list of irreducibles as you go. Not unlike the sieve of Eratosthenes. – Jyrki Lahtonen Oct 30 '14 at 18:36
• @TheD3ac: I don't believe your edits change anything about my answer. As far as how to compute irreducible polynomials of degree two in $\mathbb{Z}_{2}[x]$, note that any reducible polynomial of degree $2$ must be divisible by a linear polynomial in $\mathbb{Z}_{2}[x]$, hence must have a root in $\mathbb{Z}_{2}[x]$. The irreducibles of degree $2$ in $\mathbb{Z}_{2}[x]$ are therefore the degree $2$ polynomials with no roots in $\mathbb{Z}_{2}$. Hence, generate a list of the four possible degree $2$ polynomials, then plug in $0$ and $1$ to see if either is a root. – Alex Wertheim Oct 30 '14 at 20:44

Here's a different take. A bit ad hoc, but to an extent I expect that here. Also I will be using several bits and pieces of finite field theory.

Claim 1. $p_1(x)=p(x)=x^5+x^2+1$ is irreducible.

Proof. $p(0)=p(1)=1$ so it has no linear factors. That leaves the sole quadratic irreducible $x^2+x+1$ as a possibility. But $x^3+1=(x^2+x+1)(x+1)$, so $$p(x)=x^2(x^3+1)+1\equiv1\pmod{x^2+x+1}.$$ Thus $p(x)$ has no factors of degree $\le2$. QED

Corollary 1. $p_2(x)=p(x+1)=(x+1)^5+(x+1)^2+1=x^5+x^4+x^2+x+1$ is irreducible.

Proof. A linear substitution takes an irreducible to an irreducible. QED

Fun fact. If $q(x)$ is irreducible with non-zero constant term(over any field), then so is its reciprocal polynomial $$\tilde{q}(x)=x^{\deg q}q(\frac1x).$$

Proof. If we had $\tilde{q}(x)=g(x)h(x)$, it is an easy exercise to see that we would also have $q(x)=\tilde{g}(x)\tilde{h}(x)$. QED

Therefore $$p_3(x):=\tilde{p_1}(x)=x^5+x^3+1$$ and $$p_4(x):=\tilde{p_2}(x)=x^5+x^4+x^3+x+1$$ are also irreducible.

Going back to linear substitutions we see that $$p_5(x):=p_3(x+1)=x^5+x^4+x^3+x^2+1$$ is irreducible, and so is its reciprocal $$p_6(x):=\tilde{p_5}(x)=x^5+x^3+x^2+x+1$$ which incidentally coincides with $p_4(x+1)$.

At this point we need another bit of theory. All these six polynomials have five zeros in the field $\Bbb{F}_{32}$, so between them they are minimal polynomials of all the thirty elements that are not in the prime field. Thus there are no others, and the list is complete.

I don't recall having done this exercise ever before, so thanks for letting me do it. I knew in advance that $p_1(x)$ is irreducible, because that occurs frequently enough. I also knew that there would be exactly six irreducible quintics. I also knew that the substitutions $x\to 1/x$ and $x+1$ generate a group of six automorphisms of the rational function field. This time the stabilizer was trivial, so finding one irreducible produced all six.

• Wow. This is a beautiful answer, with some really nice techniques I haven't seen before. Thanks for sharing. +1 – Alex Wertheim Oct 30 '14 at 18:29
• Switching to CW, because I just found a dupe target. No more rep for me here. – Jyrki Lahtonen Oct 6 '18 at 15:02

A straightforward inductive way to do this is to list all reducible polynomials of degree 5 and taking the complement set.

The list of reducible polynomials can be obtained by multiplying polynomials of degrees $a$ and $b$ such that $a+b=5$ in all the possible ways.

Since a polynomial will always factor into irreducibles, some shortcuts can be taken in the described strategy once you know the irreducible polynomials of degree $<5$.

The following argument is specific to the question asked. Although I am by no means a trained professional, please do not attempt this at home with polynomials of larger degree.

The irreducible polynomial must be of the form $x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + 1$, that is, the degree-$5$ term as well as the constant term must be nonzero. Of the $16$ possible choices of $(a_4,a_3,a_2,a_1)$ we can eliminate the $8$ vectors of even Hamming weight because the corresponding degree-$5$ polynomials have $1$ as a root. So, we are down to $8$ polynomials of which $6$ are necessarily irreducible because the $30$ elements of $\mathbb F_{32}-\mathbb F_2$ all have order $31$ and come in sets of $5$ conjugates. Starting with the lexicographically first polynomial $x^5 + x + 1$, we readily discover via long division that $$x^5 + x + 1 = (x^2+x+1)(x^3+x^2+1)$$ and so $x^5 + x + 1$ and its reciprocal $x^5+x^4+1 = (x^2+x+1)(x^3+x+1)$ are reducible polynomials. Thus, the irreducible polynomials of degree $5$ are \begin{align} x^5+x^2+1 &\quad x^5+x^3+1\\x^5+x^3+x^2+x+1 &\quad x^5+x^4+x^3+x^2+1\\ x^5+x^4+x^2+x+1 &\quad x^5+x^4+x^3 + x + 1 \end{align}


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## Elementary Statistics: A Step-by-Step Approach with Formula Card 9th Edition

Order matters, it is a permutation: 5 out of 20 $_{20}P_5$=1860480


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## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.2 Combining Functions - 3.2 Exercises - Page 244: 36

#### Answer

$-4x+16$

#### Work Step by Step

The student did not change the sign of $-9$ when the parentheses were removed. The correct simplified expression of $(2x+7)-(6x-9) ,$ is \begin{array}{l}\require{cancel} 2x+7-6x+9 \\\\= 2x+7-6x+9 \\\\= (2x-6x)+(7+9) \\\\= -4x+16 .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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# Challenge: The Road to GM 2011

Thread in 'Competition' started by m.kevin, 14 Nov 2011.

1. ### m.kevin

It finally happened. 2 months ago, KevinDDR made himself the first western Ti Grand Master, followed by Steve, only days after the first one.

I feel 2011 is going to be the Grand Master year for western players, so I made this thread just for that.

Here's the goal of the challenge : get the Grand Master grade before 2012.
- Kitaru and Nahuel are the most probable GMs of this year, I know you can do it.
- K, you're a Orange GM in TAP, I know you can do the Ti GM.
- Amnesia, you can... ok maybe not.

Here's the thing : it'll be a crowning moment of awesome if more than 3 players manage to get GM grade in less than 4 months. This grade is already really prestigious, it's the equivalent of winning 10 million dollars on Super Millionaire.

Code:
Rank--Name----------------Grade - Lvl @ Time     - Date     - ST AC CO SK - Rotation - Notes
======================================================================================================================
1--[URL="http://tetrisconcept.net/forum/attachment.html?attachmentid=118&d=1318743865"][-Steve-][/URL]--------------Gm - 999 @ 06:16:11 - 10/16/11 - SI       GO - World    - 2nd western GM
2--[URL="http://p.twimg.com/AZNNkZsCMAAVeV8.jpg"]KevinDDR[/URL]---------------Gm - 999 @ 06:30:55 - 09/13/11 - BR       GO - World    - 1st western GM

Last edited: 14 Nov 2011


text:
# zbMATH — the first resource for mathematics

Some inequalities for a multiphase system. (Russian) Zbl 0544.90026
The paper contains some estimates of a capacity of multiphase (tandem) queueing systems having limited intermediate queues. A demand served in a previous phase cannot be set free if there are no places in the following queue (a blocking effect). It is proved that a value T exists such that if the intensity of the demand flow is less than $$T^{-1}$$, a stationary regime is achieved and it is not achieved if the intensity is larger than $$T^{-1}$$. By definition, the value $$T^{-1}$$ is the system capacity and the value T is the mean effective service time. It is stated that T is a monotone non-decreasing function of the service time in each phase and the quantities of the places between the phases (the time $$\xi_ 2$$ is not less than the time $$\xi_ 1$$ if $\int^{\infty}_{x}(1-F_ 1(t))dt\leq \int^{\infty}_{x}(1-F_ 2(t))dt\quad for\quad all\quad x$ where $$F_ 1$$, $$F_ 2$$ are the distribution functions of $$\xi_ 1$$, $$\xi_ 2)$$. The main result gives an upper bound for T in a system having 3 phases, k places before the 2-nd phase and r places before the third one. The bound coincides with the correct value of T if $$r=0$$ and the distribution of service times is exponential.
Reviewer: A.A.Pervozvanskij

##### MSC:
 90B22 Queues and service in operations research 60J25 Continuous-time Markov processes on general state spaces


text:
# Thread: Forming an equation and differentiating it

1. ## Forming an equation and differentiating it

So you need to find an optimum price, so I just thought you had to find a stationary point, i.e. when the differentiation equals 0

But if you differentiate Q you just get -80. So you can't set this to 0, nor can the optimum price be -80. So I thought, ok it must be a bit harder than that.

So the cost per unit 'a' = 300,000 + 20a since there is a fixed cost of 300000 and a variable cost of 20 per unit a

But I just don't know where to go from here since cost per unit is not necessarily the price per unit, so we can't sub it into the quantity demanded equation. Even if we did, it wouldn't matter because we would still get 0=a number when we differentiate. I think that you have to somehow set up a quadratic in terms of price per unit and then differentiate THAT. But I just don't know how to do it. Any help?

Screenshot of the problem:

2. ## Re: Forming an equation and differentiating it

Multiply the profit per unit by the quantity demanded to get the total profit on the units sold. You can then differentiate the expression for total profit to find a maximum.

3. ## Re: Forming an equation and differentiating it

But I don't know the profit per unit, only the cost to produce one

Profit per unit would be P - 300000 + 20a i.e. Price minus cost.
But then this would give us two variables if we multiply that by Quantity demanded

4. ## Re: Forming an equation and differentiating it

What if I sub in quantity demanded per unit, into the formula I wrote for cost per unit.

So 300,000 + 20a = 300,000 + 20(15000-80P)
and then times profit per unit = P - 300,000 + 20(15,000 - 80P) by quantity demanded

This would give me P = 187.6 rounded, after differentiation. Is this correct?

But when I do differentiate again (to find the nature of this stationary point) it turns out it is a minimum, not a maximum so it must be wrong, given the question

5. ## Re: Forming an equation and differentiating it

Nevermind I got it

I got Profit per units produced = Pa - 300,000 - 20a where a is the number of units
then subbed in quantity demanded for a to get total profit

then I got P=103.75 by calculating and differentiating, which is a maximum


text:
# Regression Estimated variance and Chi-square distribution

#### saaagar

##### New Member
Hi all

* If i have a data set and I model it using linear regression I estimate intercept, slope and variance Ymodel=a+bx and the corresponding MSE to estimate variance StdevModle^2 .

* The next NEW set of data come which follows same model as data in step 1. i calculate Sum of (YObservedNEW-YExpectedfrom Model)^2/StdModel^2, will this still follows chi-square as the denominator is not the population variance but estimated from data set of step1.


text:
# How to customize a .bst bibliographic style created with custom-bib for dealing with entries with no year and/or no author

I am using natbib with author-year-like cites style and with a custom references style (a .bst file) that I created with custom-bib. That style file for references is also author-year oriented, of course.

I am not an expert on .bst files and so on. The style I created with custom-bib is perfect for me, but there are two cases in which it could (and should) be improved:

1. entries without year and
2. entries without author, and also a not so important improvement:
3. entries that have an URL (for the records) but are not online websites.

I copy here my MWE (maybe not so minimal), and a link to my .bst file:

\begin{filecontents}{mybib.bib}
@book{GAMSCPLEX,
publisher = {GAMS Development Corporation},
title = {{GAMS/CPLEX} 10 Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/cplex.pdf}
},
@book{GAMSSBB02,
publisher = {GAMS Development Corporation},
title = {{GAMS/SBB} Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/sbb.pdf},
year = {2002}
},
@book{GAMSCONOPT,
author = {Drud, A},
publisher = {{ARKI} Consulting and Development},
title = {{GAMS/CONOPT} Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/conopt.pdf}
},
@article{Chou2006,
author = {Chou, C. Y. and Chen, C. H.},
journal = {Expert Systems with Applications},
number = {2},
pages = {233--242},
title = {Economic design of variable sampling intervals {$T^2$} control charts using genetic algorithms},
volume = {30},
year = {2006}
},
@article{DRUZOVEC98,
author = {Dru\v{z}ovec, M and \v{S}ostar, A and Godina, A},
journal = {Strojni{\v{s}}ki Vestnik-Journal of Mechanical Engineering},
number = {1-2},
pages = {29--34},
title = {{MPC} process control procedure},
volume = {44},
year = {1998}
}
\end{filecontents}

\documentclass[11pt]{book}
\usepackage[a4paper,showframe]{geometry}

\RequirePackage[latin1]{inputenc}

%\usepackage{amsmath}

\usepackage[spanish]{babel}

\usepackage[autostyle=true]{csquotes}

\usepackage[round,semicolon,authoryear,longnamesfirst]{natbib}

\usepackage[%implicit=false,
pdftex,hyperfootnotes=false,
\pdfcompresslevel=9

\hypersetup{hyperfootnotes=false,
filecolor=red, urlcolor=red,
pdfpagemode=UseOutlines, pdfpagelayout=SinglePage,%
pdfhighlight=/N, pageanchor=true,%

\begin{document}

Referencias:

\begin{itemize}
\item \cite{Chou2006}.
\item \cite{DRUZOVEC98}.
\item \cite{GAMSCPLEX}.
\item \cite{GAMSSBB02}.
\item \cite{GAMSCONOPT}.
\end{itemize}

\bibliographystyle{mycustom}
\bibliography{mybib}

\end{document}

This is the link for mycustom.bst: https://dl.dropbox.com/u/155485/mycustom.bst

These are the results I get for the reference list:

OK, so my questions are obviously the following:

1. How can I tweak my .bst file (or the preceding .dbj file) in order to hide those ugly empty round brackets in the case there is no year for the item??

2. How should my .bst file deal with items without author? Should it take the publisher info? I suppose the right order could be: authors, if not then editors, if not then publisher or institution. How can I get that done in my custom style?*

3. I don't know what exactly to do with URL information. In the cases I am showing in my example, they are in fact online documents, so I think it's not wrong to show the links to them. But in other cases you have an URL address in your .bib file just for your information... Anyway, my question in this case is easier: How can I get rid of that ugly "URL" label before the URL address?

EDIT: Although this is not the goal of this question, I would also like to ask you the following:

Do you know any (natbib or just bibtex) existing style that is the same or nearly the same as my custom bib style?? I was no able to locate such a style, but maybe I missed something.

• I have also seen "n. d." (for "no date") for publications without given publication date. Nov 5, 2012 at 19:12
• I have a style file made from custom-bib that does not show the problems you have. I guess you used option pub-date that causes all that. To make that all easier, could you please post your .dbj file, so I can compare?
– mafp
Dec 26, 2012 at 23:14

1. To start, there are several things wrong with your bib file. Following the BibTeX manual, the book entry requires year, and author or editor. So I suggest to turn the first three entries into manual, which just requires title.
2. For those manuals, in particular when the author is missing, one should provide a key entry that BibTeX can use to sort the entries, and that is sometimes shown instead of the author (depends on the style).
3. I would also change the publisher fields of the manuals into organization fields, as those are optional fields of manual, while publisher is not.

The resulting database would then look like:

@manual{GAMSCPLEX,
key = {GAMS/CPLEX},
organization = {GAMS Development Corporation},
title = {{GAMS/CPLEX} 10 Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/cplex.pdf}
},
@manual{GAMSSBB02,
key = {GAMS/SBB},
organization = {GAMS Development Corporation},
title = {{GAMS/SBB} Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/sbb.pdf},
year = {2002}
},
@manual{GAMSCONOPT,
key = {GAMS/CONOPT},
author = {Drud, A.},
organization = {{ARKI} Consulting and Development},
title = {{GAMS/CONOPT} Solver Manual},
url = {http://www.gams.com/dd/docs/solvers/conopt.pdf}
},
@article{Chou2006,
author = {Chou, C. Y. and Chen, C. H.},
journal = {Expert Systems with Applications},
number = {2},
pages = {233--242},
title = {Economic design of variable sampling intervals {$T^2$} control charts using genetic algorithms},
volume = {30},
year = {2006}
},
@article{DRUZOVEC98,
author = {Dru\v{z}ovec, M. and \v{S}ostar, A. and Godina, A.},
journal = {Strojni{\v{s}}ki Vestnik-Journal of Mechanical Engineering},
number = {1-2},
pages = {29--34},
title = {{MPC} process control procedure},
volume = {44},
year = {1998}
}

Regarding the empty parentheses for non-existing years I have three suggestions:

1. Remove the yr-par option from your dbj file to remove all parentheses around years. (maybe not an option for you)
2. Just provide a year! If you use a specific version of, say, GAMS/CPLEX, look up its release date, and use that. It is reasonable to assume that the manual was released alongside the software release. (maybe not an option for you)
3. bst hackery: Although year is an optional field of manual, your style file does not check properly if the year is present. This is a bug in custom-bib. To circumvent it, add the indicated five lines to your style file, in function format.date:

FUNCTION {format.date}
{ year "year" bibinfo.check duplicate$empty$
{
}
'skip$if$
extra.label *
before.all 'output.state :=
duplicate$empty$     %% added
{ pop$"{}" } %% added { %% added " (" swap$ * ")" *
if$%% added }  In principle, this could be added to merlin.mbs, to fix all derived styles. With this hack, and the proposed changes to your database, I get: I want to add an answer to a part of my own question. But there still are open questions. About how to get rid of "URL" label (question 3), I just opened the .bst file I created (that I linked to in my question) and then I realized that this could work: \providecommand{\urlprefix}{ }  This line should be added before calling the .bst style, I mean, not after the line where \bibliographystyle{mycustom} is, in my case. For example, it can be added before \begin{document}. And as for question 2, I must say that a possible workaround here is manually adding an editor field to those entries without author and without editor. This editor could have the same content as the publisher field, and in fact it should replace it. This way, I've checked that my bib style takes the editor field as a kind of author when author field is empty. But that's not a good solution. Please, I am not an expert in LaTeX or BibTeX coding, so any other answer will be appreciated. • Please provide your dbj file, without it it is hard to tell what needs to be changed. – mafp Jan 18, 2013 at 22:56 Bst files are written in postfix language that's not too hard to figure out, if you're a little motivated, so you could wade in and add code to handle missing authors gracefully (I believe it already sorts such records by title, so there's not much that needs fixing). But it's probably best to hack your generated file as little as possible. 1. Many citation styles use no date or n.d. in place of the date for documents that don't have one. If you do that in your bibliography, you won't have to adjust the style. 2. No good solution to the missing-author problem comes to mind at the moment, besides editing the generated bst. 3. The ugly "URL" label at the front is the content of the macro \urlprefix. You could get rid of it by redefining \urlprefix in the preamble of each document that uses it. For a permanent fix, edit the .bst you generated and look for this snippet (in FUNCTION {begin.bib}):  "\expandafter\ifx\csname urlprefix\endcsname\relax\def\urlprefix{URL }\fi" write$ newline\$
It sets the default \urlprefix if your document hasn't already defined it, so you can change it to another prefix (including \relax).
3a. If I recall correctly, whether to always show the url field is controlled by an option during the custom-bib run. I suggest re-running custom-bib to generate your style without the option; it might do exactly what you want. If this doesn't work, the easiest work-around is to use a custom field name like myurl for urls you don't want to typeset.


text:
If $R$ denotes the reaction of the body to some stimulus of strength $x$, the \ sensitivity $S$ is defined to be the rate of change of the reaction with respect to $x$. A particular example is that when the brightness $x$ of a light source is increased, the eye reacts by decreasing the area $R$ of the pupil. The experimental formula can be used to model the dependence of $R$ on $x$ when $R$ is measured in square millimeters and $x$ is measured in appropriate units of brightness.

Find the sensitivity corresponding to $x=2$.

Sensitivity =


text:
100, rue des maths 38610 Gières / GPS : 45.193055, 5.772076 / Directeur : Thierry Gallay

# Sébastien Gontard

Curvature of invariant metrics in non compact complex manifolds
Wednesday, 29 May, 2019 - 17:00
Résumé :

Given a domain $D$ in $\mathbf{C}^n$ (or more generally a complex manifold without boundary) it is natural to study the relationships between quantities containing information about the complex geometry of $D$. For instance one can ask whether the curvature of the boundary of $D$ is related to the curvatures of objects defined on $D$, such as complete Kähler metrics. We expect that the behaviour of the curvatures of a complete Kähler metric is influenced by the geometry of the boundary $\partial D$ of $D$, at least when we look at points of $D$ close to $\partial D$. More formally, given a bounded pseudoconvex domain $D$ and a boundary point $p$ of finite D'Angelo 1-type we study the following question:
(Q): "Does there exists a complete Kähler metric on $D$ whose holomorphic bisectional curvatures are negatively pinched in a neighbourhood of $q$?"

Since we study this question for a general pseudoconvex domain with no explicit description it is reasonable to pick a canonical metric and study its curvatures. In our case we work with the Bergman metric and with the Kähler-Einstein metric of negative Ricci curvature.

The situation is well understood if $\partial D$ "looks like the boundary of a ball $B$" near a given point p of the boundary (for instance if $\partial D$ is strictly pseudoconvex at $p$): in this case D is metrically "curved like $B$" when we look at points in $D$ near $q$. Especially, the answer to question (Q) is affirmative.
Apart from the above situation, very few is known.

This talk is divided in four parts. The first part provides with basic ideas regarding metrics, curvatures (both geometric and metric) and motivations. In the second part we study the behaviour of the Kähler-Einstein metric and its curvatures at "ball-like" boundary points. Then we explain what problems we encounter when dealing with weakly pseudoconvex boundary points and what results we have for the moment. Finally we end this talk with some more general questions arising when we study the boundary behaviour of metrics and their curvatures.

Institution de l'orateur :
IF
Thème de recherche :
Compréhensible
Salle :
4


text:
## local flame temperature, $$T_{1}$$

in flame emission and absorption spectrometry
https://doi.org/10.1351/goldbook.L03603
The effective @[email protected] in the observation space as measured by a specific sensor for a specified element (in $$\text{K}$$). The temperature of a flame (or other plasma) is not homogeneous. It is usually lower at the borders of the flame. It is therefore appropriate to speak of an effective temperature which represents an average value of all temperatures throughout the observation space. The flame temperature depends on several factors such as: kind of plasma, kind of gas or gas mixture and @[email protected] of the @[email protected] species in the observation space.
Source:
PAC, 1986, 58, 1737. (Quantities and units in clinical chemistry: Nebulizer and flame properties in flame emission and absorption spectrometry (Recommendations 1986)) on page 1741 [Terms] [Paper]


text:
### Home > INT2 > Chapter 6 > Lesson 6.2.6 > Problem6-103

6-103.

Copy the diagram at right onto your paper.

1. What are the measures of all the angles in the diagram?

The triangles in the diagram are isosceles, therefore their base angles are congruent.

2. Make a flowchart showing that the triangles are similar.

3. Cheri and Roberta notice that their similarity statements for part (b) are not the same. Cheri stated $ΔABC$ ~ $ΔDEC$, while Roberta wrote $ΔABC$ ~ $ΔEDC$. Who is correct? Or are they both correct? Explain your reasoning.

They are both correct. Since both triangles are isosceles, we cannot tell if one is the reflection or the rotation of the other (after dilation).


text:
# Simplify $ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$

1. Feb 22, 2009

### walker242

1. The problem statement, all variables and given/known data
As in title, simplify $$ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1)$$, x > 0.

2. Relevant equations
-

3. The attempt at a solution
So far
$$ln(x(\sqrt{1+e^x}-\sqrt{e^x})) + ln(\sqrt{1+e^{-x}}+1) = ln(x(\sqrt{1+e^x}-\sqrt{e^x})(\sqrt{1+e^{-x}}+1)) = ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))$$.

Is it possible to get further?

2. Feb 22, 2009

### Staff: Mentor

I worked it through and ended with what you have. I don't see anything more you can do.

3. Feb 22, 2009

Thanks. :)

4. May 8, 2009

### walker242

Actually, it's possible to get further:

$\ln \left(x \left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1 \right) = \ln x + \ln\left(\sqrt{1+e^{x}}-\sqrt{e^{x}}\right) + \ln\left(\sqrt{1+e^{-x}}+1\right) =$

$= \ln x + \ln\left(\left(\sqrt{1+e^{x}}-\sqrt{e^x}\right) \left(\sqrt{1+e^{-x}} + 1\right)\right) = \ln x + \ln\left(\sqrt{1+e^{x}}\sqrt{1+e^{-x}}+\sqrt{1+e^{x}}-\sqrt{e^{x}}\sqrt{1+e^{-x}}-\sqrt{e^x}\right)$

$= \ln x + \ln\left(\left(\left(1+e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}}-\left(\left(e^{x}\right)\left(1+e^{-x}\right)\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right) = \ln x + \ln\left(\left(1+e^{-x}+e^{x}+e^{0}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+e^{0}\right)^{\frac{1}{2}}-\left(e^{x}\right)^{\frac{1}{2}}\right)$

$= \ln x + \ln\left(\left(2+e^{-x}+e^{x}\right)^{\frac{1}{2}} + \left(1+e^{x}\right)^{\frac{1}{2}} - \left(e^{x}+1\right)^{\frac{1}{2}} -\left(e^{x}\right)^{\frac{1}{2}}\right) = \ln x + \ln\left(\sqrt{\left(2 + e^{-x} + e^{x}\right)} - \sqrt{\left(e^{x}\right)}\right)$

$= \ln x + \ln\left(\sqrt{\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}}\right)^2} - \sqrt{e^x}\right) = \ln x + \ln\left(e^{\frac{x}{2}} + e^{-\frac{x}{2}} - e^\frac{x}{2}\right) = \ln x + \ln\left(e^{-\frac{x}{2}}\right) = \ln \left(\frac{x}{e^\frac{x}{2}}\right)$

5. May 8, 2009

### Staff: Mentor

You're right. It's also possible to get to your result with much less work.
Here's where the OP ended.
$$ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x}))$$
The expression in the first radical happens to be a perfect square trinomial, something I didn't notice earlier.

$$2+e^{-x} + e^{x} = e^{x} + 2 + e^{-x} = (e^{x/2} + e^{-x/2})^2$$
The expression being squared is always positive, so we don't have to worry about the negative square root. IOW,
$$\sqrt{e^{x} + 2+e^{-x} } = \sqrt{(e^{x/2} + e^{-x/2})^2} = e^{x/2} + e^{-x/2}$$

Here's the complete work.
$$ln(x(\sqrt{2+e^{-x} + e^{x}} - \sqrt{e^x})) = ln(x(e^{x/2} + e^{-x/2} - e^{x/2})) = ln(xe^{-x/2}) = ln(\frac{x}{e^{x/2}})$$

Last edited: May 8, 2009


text:
# PROBLEM 3: The drift diffusion model of decision-making.¶

In a two-alternative forced choice task (2AFC-task), subjects are asked to choose between two alternative actions. We will here consider the case where a subject receives a visual motion stimulus (a set of points on a screen that are moving in different directions) and then needs to indicate whether the points were moving upwards or downwards. If such a motion stimulus is ambiguous or "noisy", the task can be quite difficult. We will assume that the motion stimulus continues until the subject has made a choice. This scenario is well described by the "drift-diffusion-model", in which the subject compares the firing rate $m_A$ of an upward-motion sensitive neuron with the firing rate $m_B$ of a downward-motion sensitive neuron and integrates the difference between the two:

$$\dot{x} = m_A − m_B + σ η(t)$$

where $η(t)$ is a noise term (Gaussian white noise with unit standard deviation) that simulates the noisiness of real neurons. If the integration variable $x$ surpasses a threshold $μ$, then the subject decides for outcome $A$; if $x$ decreases below a threshold $−μ$, then the subject decides for outcome $B$.

Any ordinary differential equation can be solved numerically using the Euler method, i.e. using the approximation:

$$x(t + ∆t) = x(t) + \dot{x} ∆t \qquad (6)$$

For stochastic differential equations, i.e. those that have a noise-term, the random part grows with the square root of the time step, a technical issue that you can ignore for now. It leads to the following discrete approximation of the drift-diffusion-model:

$$x(t+∆t) = x(t) + (m_A - m_B) ∆t + σ η(t) \sqrt{∆t} \qquad (7)$$

In [339]:
# /!\ I'm using Python 3

import plotly
import numpy as np
import plotly.plotly as py
import plotly.graph_objs as go

import matplotlib.pyplot as plt
import scipy.io as sio
from scipy import stats
%matplotlib inline

plotly.offline.init_notebook_mode(connected=True)
from IPython.core.display import display, HTML, Markdown
# The polling here is to ensure that plotly.js has already been loaded before
# setting display alignment in order to avoid a race condition.
display(HTML(
'<script>'
'var waitForPlotly = setInterval( function() {'
'if( typeof(window.Plotly) !== "undefined" ){'
'MathJax.Hub.Config({ SVG: { font: "STIX-Web" }, displayAlign: "center" });'
'MathJax.Hub.Queue(["setRenderer", MathJax.Hub, "SVG"]);'
'clearInterval(waitForPlotly);'
'}}, 250 );'
'</script>'
))

# Colorscales
def colorscale_list(cmap, number_colors, return_rgb_only=False):
cm = plt.get_cmap(cmap)
colors = [np.array(cm(i/number_colors)) for i in range(1, number_colors+1)]
rgb_colors_plotly = []
rgb_colors_only = []
for i, c in enumerate(colors):
col = 'rgb{}'.format(tuple(255*c[:-1]))
rgb_colors_only.append(col)
rgb_colors_plotly.append([i/number_colors, col])
rgb_colors_plotly.append([(i+1)/number_colors, col])
return rgb_colors_only if return_rgb_only else rgb_colors_plotly

## (a) Assume $m_A = 1$ and $m_B = 0.95$. Plot several runs of the drift-diffusion-model, always starting with $x(0) = 0$. Choose a stepwidth of $∆t = 0.1 \, \mathrm{ ms}$, a noise level $σ = 1/2$ and iterate the Euler method over $10000$ time steps up to time $t = 1 \, \mathrm{ s}$.¶

In [607]:
delta_t = 0.0001
t_max = 1

m_A, m_B = 1, 0.95

sigma = 0.5

x_0 = 0

In [440]:
def drift_diffusion(m_A=m_A, m_B=m_B, sigma=sigma, delta_t=delta_t, t_max=t_max, x_0=x_0):
x = np.zeros(len(np.arange(0, t_max, delta_t)))
x[0], sqrt_delta_t = x_0, np.sqrt(delta_t)

term1, term2 = delta_t*(m_A-m_B), sigma*sqrt_delta_t

for i in range(1, len(x)):
x[i] = x[i-1] + term1 + term2*np.random.standard_normal()

return np.array(x)

layout = go.Layout(
title= '$\\text{Runs of drift-diffusion model, for }'+'σ = {}, \, m_A = {}, \, m_B = {}$'.format(sigma, m_A, m_B),
hovermode= 'closest',
xaxis= dict(
title= 'Time (in seconds)',
ticklen= 5,
zeroline= False,
gridwidth= 2,
),
yaxis=dict(
title= '$\\text{Integration variable } x$',
ticklen= 5,
gridwidth= 2,
),
showlegend= True
)

legend_every = 1

# Plotting the evolution of the free parameter w
traces_dd = []

for i in range(10):
traces_dd.append(
go.Scatter(
x = np.arange(0, t_max, delta_t),
y = drift_diffusion(),
mode = 'lines',
name = 'Run {}'.format(i+1),
line = dict(
width = 2,
dash = 'solid'
),
hoverlabel = dict(
namelength = -1
),
showlegend = (i % legend_every == 0)
)
)

plotly.offline.iplot(go.Figure(data=traces_dd, layout=layout))


text:
# The coupon collector's most collected coupon

Suppose a coupon collector is collecting a set of $$n$$ coupons that he receives one-by-one uniformly at random.

If the collector stops exactly when the collection is complete, we know the expected number of coupons in his collection is $$n*H[n]$$. What is the expected number of copies, $$M$$, of his most collected coupon?

When $$n = 2$$, then $$M = 2$$ because after the first coupon he will collect the other coupon with probability p ~ Geom(1/2). So he will collect the same coupon one additional time on average, and that coupon is certain to be his most collected coupon.

I don't know the exact value for any $$n$$ larger than 2, but found some approximate values by simulation:

n M
3 2.8415
4 3.4992
5 4.0259
6 4.4633
7 4.8377
8 5.1649
9 5.4560

EDIT 1:

For small n enumerating the small possibilities converges faster than simulation, and from this approach I hypothesize that $$M[3] = (15 + 6 \sqrt{5})/10$$ but don't have any real argument to support that claim.

EDIT 2:

With some more thought, I find that M has an explicit sum formula. The probability the collection terminates with a given distribution of coupons $$v = \{c_1, ..., c_{n-1}\}$$ other than the final collected coupon is just the multinomial coefficient $$(c_1; ...; c_{n-1})$$ over $$n^\text{total # of coupons in the collection}$$. This is an infinite sum over n-1 variables. Here is Mathematica code the computes the sum for terms where the collection has no more than k copies of any coupon:

M[n_, k_] := Sum[Max[v]*(Multinomial @@ v)/n^Total[v], {v, Tuples[Range[1, k], n-1]}]

Mathematica can't actually evaluate this though for $$k \rightarrow \infty$$ though.

• You might be interested in Exercise 2.20 of Boucheron, Lugosi, and Massart's ""Concentration Inequalities: A Nonasymptotic Theory of Independence", which asks to prove an upper bound of $$e\log n \cdot ce^{W((1-c)/(ce))}$$for this expectation, when collecting $m = cn\log n$ coupons. ($W$ being the Lambert function.) May 21 at 6:52

Value of $$M_3$$. I will denote $$M$$ as $$M_n$$ to emphasize the dependence on $$n$$. Then by using A230137, we get

\begin{align*} M_3 &= \sum_{n=2}^{\infty} \frac{\sum_{k=1}^{n-1} \binom{n}{k} \max\{k,n-k\}}{3^n} \\ &= \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)n - 2(2n)}{3^{2n}} + \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)(2n+1) - 2(2n+1)}{3^{2n+1}} \\ &= 3\left(\frac{1}{2} + \frac{1}{\sqrt{5}}\right) \\ &\approx 2.84164, \end{align*}

which is the same as what OP conjectured.

Asymptotic Formula of $$M_n$$. A heuristic seems suggesting that

$$M_n \sim e \log n \qquad \text{as} \quad n \to \infty,$$

The figure below compares simulated values of $$M_n$$ and the values of the asymptotic formula.

Heuristic argument. First, we consider the Poissonized version of the problem:

1. Let $$N^{(1)}, \ldots, N^{(n)}$$ be independent Poisson processes with rate $$\frac{1}{n}$$. Then the arrivals in each $$N^{(i)}$$ model the times when the collector receives coupons of type $$i$$.

2. Let $$T$$ be the first time a complete set of coupons is collected. Then we know that $$T$$ has the same distribution as the sum of $$n$$ independent exponential RVs with respective rates $$1$$, $$\frac{n-1}{n}$$, $$\ldots$$, $$\frac{1}{n}$$. (This is an example of the exponential race problem.) For large $$n$$, this is asymptotically normal with mean $$\sum_{i=1}^{n}\frac{n}{i} \sim n \log n$$ and variance $$\sum_{i=1}^{n}\frac{n^2}{i^2} \sim \zeta(2)n^2$$. This tells that $$T \sim n \log n$$ in probability, and so, it is not unreasonable to expect that

$$M_n = \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_T \Bigr] \sim \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \Bigr]. \tag{1}$$

holds as well.

3. Note that each $$N^{(i)}_{n\log n}$$ has a Poisson distribution with rate $$\log n$$. Now, we fix $$\theta > 1$$ and choose an integer sequence $$(a_n)$$ such that $$a_n > \log n$$ and $$a_n/\log n \to \theta$$ as $$n \to \infty$$. Then

\begin{align*} \mathbf{P}\Bigl( N^{(i)}_{n\log n} > a_n \Bigr) &\asymp \frac{(\log n)^{a_n}}{a_n!}e^{-\log n} \\ &\asymp \frac{(\log n)^{a_n}}{(a_n)^{a_n} e^{-a_n+\log n} \sqrt{a_n}} \\ &\asymp \frac{1}{n^{\psi(\theta) + 1 + o(1)}}, \qquad \psi(\theta) = \theta \log \theta - \theta, \end{align*}

where $$f(n) \asymp g(n)$$ means that $$f(n)/g(n)$$ is bounded away from both $$0$$ and $$\infty$$ as $$n\to\infty$$. This shows that

$$\mathbf{P}\Bigl( \max_{1\leq i\leq n} N^{(i)}_{n\log n} \leq a_n \Bigr) = \biggl[ 1 - \frac{1}{n^{\psi(\theta) + 1 + o(1)}} \biggr]^n \xrightarrow[n\to\infty]{} \begin{cases} 0, & \psi(\theta) < 0, \\[0.5em] 1, & \psi(\theta) > 0. \end{cases}$$

So, by noting that $$\psi(e) = 0$$, we get

$$\max_{1\leq i\leq n} N^{(i)}_{n\log n} \sim e \log n \qquad \text{in probability}. \tag{2}$$

This suggests that $$M_n \sim e \log n$$ as well.

4. I believe that this heuristic argument can be turned into an actual proof. To this, we need the following:

1. We have to justify the relation $$\text{(1)}$$ in an appropriate sense. We may perhaps study the inequality

$$\max_{1\leq i\leq n} N^{(i)}_{(1-\varepsilon)n\log n} \leq \max_{1\leq i\leq n} N^{(i)}_T \leq \max_{1\leq i\leq n} N^{(i)}_{(1+\varepsilon)n\log n} \qquad \text{w.h.p.}$$

and show that the probability of the exceptional event is small enough to bound the difference of both sides of $$\text{(1)}$$ by a vanishing qantity.

2. Note that $$\text{(2)}$$ alone is not enough to show this. So, we need to elaborate the argument in step 3 so that it can actually prove the relation $$\mathbf{E}\bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \bigr] \sim e \log n$$.


text:
Volume 358 - 36th International Cosmic Ray Conference (ICRC2019) - CRI - Cosmic Ray Indirect
Particle Identification in Smartphone Camera Images Using the Distributed Electronic Cosmic-ray Observatory
J. Vandenbroucke, M. Meehan, J. Peacock, J. Bourbeau, F. Campos, C. Schneider, M. Winter, A. Pizzuto*, R. Morgan, B. Gallay and A.L. Simons
Full text: pdf
Pre-published on: September 02, 2019
Published on: July 02, 2021
Abstract
The Distributed Electronic Cosmic-ray Observatory (DECO) is a global network of smartphones that searches images for evidence of cosmic rays and other ionizing charged particles. DECO was released to the public in 2014 and has citizen scientists from 80 countries on all seven continents participating in the project. We previously demonstrated that tracks seen in the DECO data set are caused by cosmic-ray muons by comparing the track length distribution of candidate events in the data set to the expected distribution from a cosmic-ray flux. However, robust particle identification necessary to separate cosmic rays from the radioactive background on an event-by-event basis had not previously been identified. We present a deep learning, computer vision algorithm for identifying and classifying charged particles in camera image sensors. The convolutional neural network was trained using images from the DECO data set and achieves classification performance comparable to human quality across four distinct DECO event classifications. We apply our model to the entire DECO data set and determine a selection that achieves a purity of $95\%$ when applied to cosmic-ray muons and $\ge 90\%$ for all event types. The automated classification is run on the public DECO data set in real time in order to provide classified particle interaction images to users of the app and other interested members of the public. The model and techniques used to develop it are applicable to other smartphone-based cosmic-ray detectors and data sets consisting of images of charged particles from solid-state camera sensors.
DOI: https://doi.org/10.22323/1.358.0390
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
2 corrected name

Birkhoff's HSP theorem.

Given a class of algebraic structures which satisfies a set of identities (universally quantified equations in a purely functional language with equality as the only relation; this is known as an equational class), homomorphic images, images of subalgebras, and cartesian products also satisfy the same set of identities (e.g. by inspection) and their consequences in equational logic. Conversely, given a class of algebraic structures of the the same type closed under products, subalgebras, and homomorphisms, one can realize every such algebra as an image of some free algebra with respect to the class; constructing the appropriate image of the term algebra gives a free algebra that lives in the closed class; the (fully invariant) congruences on the term algebra gives the set of equations that the closed class must satisfy; thus a class is an equational class iff it is closed under the class operator HSP.

Garret

Garrett Birkhoff's original paper on this theorem (called a preservation theorem when it came out in the 1930's) is easy to read. There may be within it a cleaner synopsis than the one above.


text:
# What is the distance between the following polar coordinates?:  (2,(23pi)/12), (3,(7pi)/8)

Nov 24, 2016

The distance is $\approx 4.9897$

#### Explanation:

Let's begin by rotating both points an additional $\frac{\pi}{12}$, because that will make the first point go around a full rotation and its angle will be zero:

$\left(2 , 0\right) , \left(3 , \frac{23 \pi}{24}\right)$

These points form a triangle with the origin.

side $a = 2$

side $b = 3$

The $\angle C \text{ between the sides} = \frac{23 \pi}{24}$

We can use the Law of Cosines to find the length of side c, which is, also, the distance between the two points:

c = sqrt(a^2 + b^2 - 2(a)(b)cos(C)

c = sqrt(2^2 + 3^2 - 2(2)(3)cos((23pi)/24)

$c \approx 4.9897$


text:
# Integration by parts, why is this allowed?

#### find_the_fun

##### Active member
I'm following this example where it is asked to integrate $$\displaystyle \int \ln{x} dx$$ using integration by parts. I don't understand how it's legal to set v=x since the only x in the equation is the argument of ln and that's already accounted for by u.

#### MarkFL

Staff member
Re: integration by parts, why is this allowed?

We are actually setting:

$$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

$$\displaystyle dv=dx\,\therefore\,v=x$$

#### find_the_fun

##### Active member
Re: integration by parts, why is this allowed?

We are actually setting:

$$\displaystyle u=\ln(x)\,\therefore\,du=\frac{1}{x}\,dx$$

$$\displaystyle dv=dx\,\therefore\,v=x$$
Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?

#### MarkFL

Staff member
Re: integration by parts, why is this allowed?

Ok I think I get it. So with integration by parts you choose dv before you calculate v? In other words dv has to be part of the original equation but v does not?
Correct...integration by parts states:

$$\displaystyle \int u\,dv=uv-\int v\,du$$

So, you want to choose a $u$ and a $dv$, and then from these compute $du$ and $v$.

#### HallsofIvy

##### Well-known member
MHB Math Helper
Any time you use "integration by parts",
$$\int udv= uv- \int v du$$, there is NO "v" in the original integral. You get it by your choice of "dv".

Here, the choice is dv= dx, u= ln(x).

#### SuperSonic4

##### Well-known member
MHB Math Helper
If it helps you can think of

$$\displaystyle \int \ln(x) dx = \int \left(1 \times \ln(x) \right)dx$$

Then you can set $$\displaystyle \frac{dv}{dx} = 1$$ and $$\displaystyle u = \ln(x)$$


text:
Plzzzzzzzzzzzzzzoooooooooooooooooooolkoolook solve question number 6.......

Arun
25757 Points
3 years ago

charges be -q, +q and -q.
Distance between two charges = a
Energy of the system $U =\frac{ 1q_{i}q_{j}}{4\pi \varepsilon r_{ij}} ; i=1,2,3; j=1,2,3 ;i\neq j$
putting the values we get Uinitial = $-\frac{3q^{2}}{8\pi \varepsilon a}$
Final energy of the system, Ufinal = $-\frac{q^{2}}{8\pi \varepsilon a}$
Therefore, work required = Ufinal -  Ufinal
= $-\frac{q^{2}}{4\pi \varepsilon a}$
However, the system will become unstable and eventually collapse.


text:
Financial Analysis

The economic efficiency calculation in PV*SOL® is based on the net present value method.

For the capital value of the total investment $KW_{\text{Gesamtinvestition}}$ is as follows:

$$KW_\text{Gesamtinvestition} = \sum BW_\text{dynamisch} - i + f$$

• $\sum BW_{dynamisch}$: Total present value over $T$ years
• $i$: Investments
• $f$: Funding

Positive capital values mean investments that can be assessed positively from an economic point of view. The payback period is the period that the plant must run in order to generate a net present value of the total investment of zero. Amortization periods greater than 30 years are not issued by PV*SOL®.

The present value $BW_\text{dynamisch}$ of a price dynamic payment sequence $Z$ over $T$ years (lifetime) is valid according to VDI 6025:

$$BW_\text{dynamisch} = Z_\text{dynamisch} \cdot b(T,q,r)$$

with

b(T,q,r) = \left\{ \begin{aligned} \frac {1-(r/q)^T}{q-r} \quad\text{für }r\neq q \newline {} \newline \frac {T}{q} \quad\text{für }r= q\end{aligned} \right.

• $b$ - Present value factor
• $q$ - Capital return factor
• $r$ - Price change factor

and the price dynamic payment sequence $Z_{t} = Z \cdot r^{t-1}$ as a sequence of payment sequence $Z \cdot r, Z \cdot r^{2}, \dots$ periodically increasing by the price change factor $r$ beginning with the first payment.

If the price change factor $r = 1$ the price dynamic payment sequence can be converted into a constant payment sequence $Z_\text{konstant}$.It is:

$$BW_\text{konstant} = \frac{Z_\text{konstant}}{a(q,T)}$$ $$a(q,T) = \frac{1}{ b(T,q,r = 1)}$$

• $a(q,T)$ - Annuity factor

For the electricity production costs $k$ applies:

$$k = Z / E$$

• $E$ - Quantity of energy and electricity generated during the period under review


text:
Special Note
This Case is a modification of Case 16 by Dr. David Willis from Case Studies in Fisheries Conservation & Management: Applied Critical Thinking & Problem Solving. Much of the text is exactly as it appeared in the original Case chapter, though slightly rearranged.

## Background

#### Motivation

Students (and biologists for that matter!) have a tendency to accept sampling data at face value. If a gear type primarily collects small fish then they assume the population is dominated by small fish. If a gear type captures big fish and lots of them then they assume that the population is dominated by large fish. In reality, many biases are possible and are actually very common. To truly understand sampling data, biologists must first understand the biases associated with each gear, and only then will the true nature of the population’s structure (e.g., size or age structure) and dynamics (i.e., recruitment, growth, and mortality) be revealed.

Various sampling gears may be differentially effective for different species, and even differentially effective for different sizes of the same species. For example, largemouth bass are commonly sampled with electrofishing gear. The numbers and sizes of largemouth bass collected can vary widely across seasons. During the spring and fall, more and larger largemouth bass tend to be nearshore and vulnerable to the electrofishing gear, which is used in that shallow-water habitat. During midsummer, fewer largemouth bass would be sampled at the same locations because many of the larger bass will have moved offshore to deeper water as a result of the warm summer water temperatures.

In this case study, you will explore the differential size structure of Bluegills captured by two common sampling gears – electrofishing and trap nets (also known as modified fyke nets).

#### Data

Lake Louise is a 45-ha impoundment located in Hand County, South Dakota. The maximum depth is 6.5 m with a mean depth of 2.7 m. Fish were sampled with trap nets that had 1.2- X 1.5-m frames, dual throats, and 19-mm bar mesh. Night electrofishing was undertaken with pulsed DC electricity at approximately 250 V and 8 A. Samples were collected in late May at a water temperature of 23$$^{o}$$C.

The total length (TL; mm) of each Bluegill was recorded for fish captured in both gears in the LenData sheet of the LakeLouiseBG.xlsx Excel 2007 file (available from here). The variables in this data frame are defined as follows,

• len: The TL (mm) of the sampled fish.
• gear: A factor indicating the gear used to capture the fish. Choices are A and B (described later).

## Preparation

Prior to beginning this case study you should:

• Read Laarman and Ryckman (1982) and Schultz and Haines (2005). Both of these papers deal with length-related biases for sampling Bluegills (the latter article deals with other species as well). As you read these articles, concentrate on the gear bias for 8-cm and longer Bluegills (Bluegills smaller than this length are not reliably sampled by these gears).
• Be familiar with methods of comparing distributions of fish length (e.g., see Section 6.4.1 of the IFAR book).
• You should open the Excel file described above and save LenData as a CSV file. Then create a script that reads the CSV file into an object in R and displays the structure and a few rows of the data.frame.

## Analysis Questions

Construct length-frequency histograms and summary statistics separately for each gear using the same x-axis limits on each graph. Additionally, plot the cumulative distribution function for each gear and compute the Kolmogorov-Smirnov test of the differences between the two distributions.

Use thes results to answer the following questions.

1. Is there a significant difference in the distribution of lengths between the two gears? If so, describe the difference. Use two different types of graphics to present your results.
2. Is there a significant difference in the mean or median lengths between the two gears?
3. Identify which gear – trap nets or electrofishing – that corresponds to gear A and gear B, respectively. Explain your reasoning for your choices.
4. Explain at least two different conclusions you might have made if you had used one gear exclusively over the other. How might this have impacted a management decision.