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# How do I escape @ mentions in comments?

When commenting, I recently came across an issue when I needed to explain instance variables using the Ruby syntax @variablename; of course, I'm only allowed to mention one user per comment, so this proved tricky.

How do I stop comments from automatically creating user links when @ is used, and simply be interpreted as plain text?

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When I read this title, I thought you meant "escape" as in "run away from". Disappointed that you're not fleeing from your inbox =) –  jadarnel27 Jan 6 '12 at 0:24

Use \@ instead of @. That avoids the character is thought to be a @-reference, and allows it to be used.

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Simple, really should have thought of that. Cheers. –  Yule Jan 5 '12 at 23:28
Sadly the \ still shows up, but this does work –  Michael Mrozek Jan 5 '12 at 23:36

Use backticks  to make the @ mention into an inline code block.

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Reasonable suggestion, however not applicable in this case as wouldn't want the text to look like code in this particular scenario. Good to know you can do this though –  Yule Jan 5 '12 at 23:31
@Yule Admittedly, I like kiam's suggestion better, but isn't your @variable` representing something in the code anyway? Hehe. –  jonsca Jan 5 '12 at 23:33
It is representing something in the code, but it's not actual code. It's discussing the use of \@variables, rather than a particular \@variable in the question –  Yule Jan 5 '12 at 23:35
@Yule Okay, got it. –  jonsca Jan 5 '12 at 23:57
I think this approach looks more natural in most cases. +1 –  jadarnel27 Jan 6 '12 at 0:27

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The proof uses a certain procedure of randomized rounding of an orthogonal matrix to a permutation matrix. >if so, i am a type of computer with no creativity. Since J\ and J2 in (1.3) have the same inertia, J2 = PJ\PT for some permutation matrix P, and hence (QP)TJi(QP) = J\. Overview. Lv 7. An orthogonal matrix is the real specialization of a unitary matrix, and thus always a normal matrix.Although we consider only real matrices here, the definition can be used for matrices with entries from any field.However, orthogonal matrices arise naturally from dot products, and for matrices of complex numbers that leads instead to the unitary requirement. Eugene. Then the code generated by the rows of its block-point incidence matrix of D is contained in C(G,Î©). Does the first one below qualify as a proof? 7 years ago. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): Abstract. Symmetric Permutation Matrices Page 3 Madison Area Technical College 7/14/2014 Answering the Question: If P is a symmetric matrix, i.e. The proof uses a certain procedure of randomized rounding of an orthogonal matrix to a permutation matrix. Approximating orthogonal matrices by permutation matrices Item Preview remove-circle Share or Embed This Item. Their respective normalized eigenvectors are given in order as the columns of Q: Q= 1 3 0 @ 2 1 2 2 2 1 1 2 2 1 A: Problem 2 (6.4 ]10). Deï¬nition 4.1.3. This means that C 1 = Ï(C 1 ) = C 2 , a contradiction. The proof uses a certain procedure of randomized rounding of an orthogonal matrix to a permutation matrix.Comment: 18 page Topics: Mathematics - Functional Analysis, Mathematics - Metric Geometry, 05A05, 52A20, 52A21, 46B09, 15A48, 15A60 The first three arise by definition of norms. Comments: 18 pages: Subjects: Functional Analysis (math.FA); Metric Geometry (math.MG) MSC classes: 05A05, 52A20, 52A21, 46B09, 15A48, 15A60: Cite â¦ A necessary condition for such a linear combination to be an orthogonal matrix is that the sum of the coefficients in the linear combination be ±1. a50 Lemma 2.4. It is easy to observe that is orthogonal, so . (Cholesky Decomposition) If A is symmetric, and A = LU can be found without any row exchanges, then A = LLT (also called square root of a matrix). Why is it faster to reheat something than it is to cook it? Given a diagonal matrix whose diagonal entries are . In the case A = C, quantum magic squares become magic squares and quantum permutations become permutation matrices (that is, a matrix with exactly 1 in every row and column and 0 elsewhere). For some permutation s on {1,...,n}, P_ij = d_s(i)j, where d_kl = 1 if k = l â¦ All permutation, rotation, reflection matrix are orthogonal matrices. Since interchanging two rows is a self-reverse operation, every elementary permutation matrix is invertible and agrees with its inverse, P = P 1 or P2 = I: A general permutation matrix does not agree with its inverse. Its inverse equals its transpose, Pâ»¹ = Páµ. 4. How did Fremen produce and carry enough thumpers to use Sandworms as de facto Ubers? Since it is symmetric, it is diagonalizable (with real eigenvalues!). The idea is inspired by Barvinok in [2]: to round an orthogonal matrix Q to a permutation matrix P , consider its action on x â R n sampled from a Gaussian distribution. Comments: 18 pages: Subjects: Functional Analysis (math.FA); Metric Geometry (math.MG) MSC classes: 05A05, 52A20, 52A21, 46B09, 15A48, 15A60: Cite â¦ EMBED. Conditions when a permutation matrix is symmetric Getting prompted for verification code but where do I put it in? The same argument as in Lemma 2.1 shows that the product of two anti-good matrices is good, and that the product of a good and an anti-good matrix is anti-good. First, what's the _definition_ of the norm of a matrix A? Let P be an n x n permutation matrix. How to write capital alpha? Proof. That SO n is a group follows from the determinant equality det(AB)=detAdetB.There-fore it is a subgroup of O n. 4.1.2 Permutation matrices Another example of matrix groups comes from the idea of permutations of integers. (proof). There should be also lots of irreducible examples of these. Yes. ,n} such that PÏ(j),j =1 (i.e. $\endgroup$ â Padraig Ó Catháin May 10 at 19:14 if PPT = , then P is its own inverse and for every i and j in {1, 2, 3, â¦ n}, ,,(),,1(), T ij jipi j p j ipji The matrix P âM n(C)iscalledapermutationmatrix An Extreme Matrix Here is a larger example, when the uâ s and the vâs are just columns of the identity matrix. For orthogonal matrices the proof is essentially identical. $\begingroup$ Check out weighing matrices -- they are nxn orthogonal matrices with k non-zero entries in each row and column. Solution: The characteristic polynomial of the matrix is ( 1)( +1), so the eigenvalues are 0, 3 and 3. Since P is a permutation Prove that every permutation matrix is orthogonal. 1 Answer. Why weren't discrete x86 CPUs ever used in game hardware? The matrix A splits into a combinationof two rank-onematrices, columnstimes rows: Ï 1u1v T +Ï 2u2v T 2 = â 45 â 20 1 1 3 3 + â 5 â 20 3 â â1 1 = 3 0 4 5 = A. Permutations are âorthogonal Aram Harrow (UW) CIRM 13 Jan, 2012 12345 21345 54321 Friday, January 13, 12 Most bit efficient text communication method? Proof: Call an orthogonal matrix anti-good if it has the transformation law given in Lemma 2.2. Should a wizard buy fine inks every time he want to copy spells into his spellbook? The paper deals with those orthogonal matrices which can be expressed as linear combinations of permutation matrices. Favorite Answer. Relevance. >or is the "invariance to the 2-norm when multiplying it on a matrix >or vector" property all that ever needs to be known. Another property of permutation matrices is given below. It is a permutation matrix by just looking at it. And second, what does it mean to say that A is orthogonal? It is Markov since the columns add to 1 (just by looking at it), or alternatively because every permutation matrix is. $$\displaystyle P_\sigma$$ is an orthogonal matrix, that is, $$\displaystyle (P_\sigma)^{-1} = (P_\sigma)^T$$. Note that one can change the order of the singular values in by changing the orders of columns of the orthogonal matrices in SVD, we have , where is an arbitrary permutation of . J-orthogonal matrices as hypernormal matrices [2]. In the framework of unitary spaces, it is enough to notice that rows of a permutation matrix form an orthonormal set of vectors in the space $$\,K^m,\$$ where $$\,K=Q,\,R\$$ or $$\,C.\$$ This is just a necessary and sufficient condition for a matrix to be orthogonal. 1. It is not a projection since A2 = I 6= A. LU factorization. Proof. permutation matrix associated to the permutation of M, (ii 1,, n); that is to say, the permutation matrix in which the non-zero components are in columns ii1,, n. Equivalently, the permutation matrix in which the permutation applied to the rows of the identity matrix is (ii 1,, n ). A general permutation matrix is not symmetric. (Second non-singular case) There exist a permutation matrix P that reorders the rows, so that PA = LU. As discussed, steps in the Gaussian elimination can be formulated as matrix multiplications. No creativity required here, just two definitions. A product of permutation matrices is again a permutation matrix. 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2020 permutation matrix orthogonal proof


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### K - Know your Aliens

##### Languages: C, C++, Java, Python, ... (details)

Our world has been invaded by shapeshifting aliens that kidnap people and steal their identities. You are an inspector from a task force dedicated to detect and capture them. As such, you were given special tools to detect aliens and differentiate them from real humans. Your current mission is to visit a city that is suspected of have been invaded, secretly inspect every person there so as to know whose are aliens and whose aren't, and report it all to Headquarters. Then they can send forces to the city by surprise and capture all the aliens at once.

The aliens are aware of the work of inspectors like you, and are monitoring all radio channels to detect the transmission of such reports, in order to anticipate any retaliation. Therefore, there have been several efforts to encrypt the reports, and the most recent method uses polynomials.

The city you must visit has $N$ citizens, each identified by a distinct even integer from $2$ to $2N$. You want to find a polynomial $P$ such that, for every citizen $i$, $P(i) > 0$ if citizen $i$ is a human, and $P(i) < 0$ otherwise. This polynomial will be transmitted to the Headquarters. With the aim of minimizing bandwidth, the polynomial has some additional requirements: each root and coefficient must be an integer, the coefficient of its highest degree term must be either $1$ or $-1$, and its degree must be the lowest possible.

For each citizen, you know whether they're a human or not. Given this information, you must find a polynomial that satisfies the described constraints.

#### Input

A string $S$ of length $N$ ($1 \leq N \leq 10^4$), where $N$ is the population of the city. For $i=1,2,\cdots, N$, the $i$-th character of $S$ is either the uppercase letter "H" or the uppercase letter "A", indicating respectively that citizen $2i$ is a human or an alien.

#### Output

The first line must contain an integer $D$ indicating the degree of a polynomial that satisfies the described constraints. The second line must contain $D + 1$ integers representing the coefficients of the polynomial, in decreasing order of the corresponding terms. It's guaranteed that there exists at least one solution such that the absolute value of each coefficient is less than $2^{63}$.

Input
HHH
Output
0 1
Input
AHHA
Output
2 -1 10 -21
Input
AHHHAH
Output
3 1 -23 159 -297

#### Hints

Original statement: https://matcomgrader.com/media/contests/6254/K.pdf


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Address 2465 E Semoran Blvd, Apopka, FL 32703 (321) 283-4335 https://www.pcgeeks2go.com

# newton cotes error Altamonte Springs, Florida

Thus n + 1 = 2 {\displaystyle n+1=2} . Mathispower4u 37.372 προβολές 7:01 Gauss Quadrature Rule: Two Point Rule - Διάρκεια: 8:44. Ordinary Differential Equations: An Elementary Textbook for Students of Mathematics, Engineering, and the Sciences. ISBN0-486-65241-6.

The text used in the course was "Numerical Methods for Engineers, 6th ed." by Steven Chapra and Raymond Canale. Κατηγορία Εκπαίδευση Άδεια Τυπική άδεια YouTube Εμφάνιση περισσότερων Εμφάνιση λιγότερων Φόρτωση... Αυτόματη If the first point is , then the other endpoint will be located at (1) and the Lagrange interpolating polynomial through the points and is (2) (3) (4) Integrating over the W. (1986). ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.8/ Connection to 0.0.0.8 failed.

Jacob Bishop 16.603 προβολές 7:29 Integración Numérica - Regla del trapecio - Διάρκεια: 10:07. A. See also Quadrature (mathematics) Interpolation Spline interpolation References ^ Pavel Holoborodko (2011-03-24). "Stable Newton-Cotes Formulas". Handbook of Mathematical Functions with Formulae, Graphs, and Mathematical Tables.

A 2-point open extended formula is (49) Single interval extrapolative rules estimate the integral in an interval based on the points around it. The closed Newton–Cotes formula of degree n is stated as ∫ a b f ( x ) d x ≈ ∑ i = 0 n w i f ( x i Given that n = x n − x 0 h {\displaystyle n={\frac {x_{n}-x_{0}}{h}}} , the global error is | ∑ i = 0 n − 1 − 1 12 h 3 Then find polynomials which approximate the tabulated function, and integrate them to approximate the area under the curve.

As can be seen in the following derivation the weights are derived from the Lagrange basis polynomials. Wolfram|Alpha» Explore anything with the first computational knowledge engine. In general, the -point rule is given by the analytic expression (25) where (26) (Whittaker and Robinson 1967, p.154). Learn more You're viewing YouTube in Greek.

It simply means that the cubic term is identically zero. Jacob Bishop 1.256 προβολές 2:51 Lagrange Error Bound - Διάρκεια: 4:56. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. f ( n + 1 ) ( ξ ) ∫ 0 n s ( s − 1 ) ⋯ ( s − n ) d s . {\displaystyle E_{\text{integrate}}=\int \limits _{x_{0}}^{x_{n}}E_{\text{interpolate}}(x)dx={\frac

To integrate a function over some interval , divide it into equal parts such that and . The error term can be obtained from the next term in the Newton polynomial, obtaining E integrate = h ∫ 0 2 s ( s − 1 ) ( s − Applying (1 ), we get E integrate = h ∫ 0 1 s ( s − 1 ) 2 h 2 f ″ ( ξ ) d s = − 1 Contents 1 Description 2 Instability for high degree 3 Closed Newton–Cotes formulae 4 Open Newton–Cotes formulae 5 Composite rules 6 See also 7 References 8 External links Description It is assumed

Classical Formulas for Equally Spaced Abscissas", Numerical Recipes: The Art of Scientific Computing (3rd ed.), New York: Cambridge University Press, ISBN978-0-521-88068-8 Josef Stoer and Roland Bulirsch. Malcolm, and Cleve B. patrickJMT 407.642 προβολές 7:21 Newton's Method - Διάρκεια: 7:30. MathWorld.

cetremo14 84.928 προβολές 10:07 27 βίντεο Αναπαραγωγή όλων Part 6: Numerical Methods: Integration and DifferentiationJacob Bishop Bisection Method: Example - Διάρκεια: 9:54. By picking the intervals at which to sample the function, this procedure produces more accurate approximations (but is significantly more complicated to implement). There are two types of Newton–Cotes formulae, the "closed" type which uses the function value at all points, and the "open" type which does not use the function values at the You can change this preference below. Κλείσιμο Ναι, θέλω να τη κρατήσω Αναίρεση Κλείσιμο Αυτό το βίντεο δεν είναι διαθέσιμο. Ουρά παρακολούθησηςΟυράΟυρά παρακολούθησηςΟυρά Κατάργηση όλωνΑποσύνδεση Φόρτωση... Ουρά παρακολούθησης Ουρά __count__/__total__ 6.1.8-Numerical

Marcel Derkker, INC. Fornberg, B. "Calculation of Weights in Finite Difference Formulas." SIAM Rev. 40, 685-691, 1998. Contact the MathWorld Team © 1999-2016 Wolfram Research, Inc. | Terms of Use THINGS TO TRY: 8_1 knot cusps 1+x-(x^2(1-sqrt(7)x^2)^2)^(1/3) grad sqrt(r) cos(theta) Newton-Cotes Quadrature Chris Maes Newton-Cotes Quadrature Formulas Eugenio Adaptive Quadrature is explained everywhere.

New York: Dover, 1972. (See Section 25.4.) George E. Generated Fri, 21 Oct 2016 11:56:48 GMT by s_wx1011 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.9/ Connection s ( s − 1 ) ( s − 2 ) … ( s − n ) h n + 1 f ( n + 1 ) ( ξ ) {\displaystyle This is called a composite rule, see Numerical integration.

Since d x = d ( x 0 + s h ) = h d s {\displaystyle dx=d(x_{0}+sh)=hds} , the error term of numerical integration is E integrate = ∫ x Closed Newton–Cotes Formulae Degree Common name Formula Error term 1 Trapezoid rule b − a 2 ( f 0 + f 1 ) {\displaystyle {\frac {b-a}{2}}(f_{0}+f_{1})} − ( b − a share|cite|improve this answer answered Jul 28 '13 at 19:40 Guido Kanschat 20627 thank you that is generally what I was looking for. –Boldwing Jul 29 '13 at 6:50 add Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization.

Open Newton–Cotes formulae This table lists some of the Newton–Cotes formulae of the open type. The 3-point rule is known as Simpson's rule. and Robinson, G. "The Newton-Cotes Formulae of Integration." §76 in The Calculus of Observations: A Treatise on Numerical Mathematics, 4th ed. Text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply.

The 4-point closed rule is Simpson's 3/8 rule, (17) (Ueberhuber 1997, p.100). Computer Methods for Mathematical Computations. In each interval, the integration uses two end points. Jacob Bishop 3.431 προβολές 6:30 NM7 4 Newton Cotes Integration - Διάρκεια: 27:15.


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# Problem of the Week Problem D Count on This

Determine the number of integer values of $$n$$ that satisfy the following inequality: $\dfrac{1}{9} \leq \dfrac{7}{n} \leq \dfrac{1}{5}$


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# Has there ever been an application of dividing by $0$?

Regarding the expression $a/0$, according to Wikipedia:

In ordinary arithmetic, the expression has no meaning, as there is no number which, multiplied by $0$, gives $a$ (assuming $a\not= 0$), and so division by zero is undefined.

Is there some other kind of mathematics that is not "ordinary", where the expression $a/0$ has meaning? Or is the word "ordinary" being used superfluously in the quoted statement?

Is there any abstract application of $a/0$?

• Sure. Look up projective geometry. $\frac{1}{0}$, or just $\infty$, is a convenient name for the "point at infinity" on the projective line. – Qiaochu Yuan May 3 '16 at 2:33
• Something that may be of interest: $$\lim_{x,y \to 0} x^y$$ – zerosofthezeta May 3 '16 at 2:36
• I don't know if the first comment covers this, since I'm not good at projective geometry, but if we compactify the real line using one point (i.e. "turn it into an infinite circle" - see Alexandroff compactification), then we can make the function $1\over x^p$ continuous on the ENTIRE real line, not just at $x \not= 0$ by defining $1/x=\infty$ in accordance with "intuition". – Chill2Macht May 3 '16 at 2:38
• @LorryLaurencemcLarry If you asked Euler, he would have said that $0/0$ can be anything; see Chapter 3 paragraphs 84 and 85 of imcs.dvfu.ru/lib.int/NEW/Math/MC_Calculus/… – user332714 May 3 '16 at 7:20
• There is no problem in defining the result of $a/0$ to be $\infty$. The problem is that $\infty$ won't have properties we expect from a number. – Curd May 6 '16 at 9:45

You said you wanted an application. Inspired by the example from Exceptional Floating Point, consider the parallel resistance formula:$$R_{total} = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2}}$$This formula tells you the effective electrical resistance of a path when the current can choose two routes to take.

Let's pretend that $R_1=0$. Then we have:$$R_{total}=\frac{1}{\frac{1}{0}+\frac{1}{R_2}}=\frac{1}{\infty+\frac{1}{R_2}}=\frac{1}{\infty}=0$$The resistance being zero is indeed the correct answer; all current flows along the single wire that has no resistance.

Naturally, you need to make appropriate definitions for arithmetic on $\infty$ (i.e., use the projective reals). For well-behaved applications like this, that's fairly straightforward.

• I don't think $0$ resistance is technically possible---should the wire itself not have some, albeit negligible, resistance? This is a somewhat idealized situation. – MathematicsStudent1122 May 3 '16 at 21:16
• @MathematicsStudent1122 Superconductors have exactly zero electrical resistance and have many practical applications. – In silico May 3 '16 at 21:28
• @MathematicsStudent1122 Wire has a resistance of 0 in the same way that a spherical cow falls in a vacuum. It's an abstraction used by practitioners in the field because it's Close Enough. – fluffy May 4 '16 at 4:20
• @Insilico, I find it hard to believe that anything has exactly zero resistance. Presumably, what's really going on is that our understanding of superconductivity is based on a model, an idealization, and in that model, the resistance is exactly $0$. But I bet you one day we'll have a better model, and the predicted resistance will be small-but-nonzero in that better model. – goblin GONE May 4 '16 at 5:26
• @goblin Our understanding of superconductivity is based on measurement, not models; when it was discovered, we had no clue how to explain it. Zero resistance was based on multiyear experiments. Maybe you're right, and it's not zero, but "I find it hard to believe" is not an appropriate justification for ignoring a large body of experiments on the subject. – prosfilaes May 4 '16 at 19:52

In complex analysis, we talk about the value of a function at infinity. To evaluate $f(z)$ at infinity, compute $f(1/z)$ then plug in $0$. This allows us to talk about things like the order of zeros and poles at infinity.

Example: $$f(z) = \frac{az+b}{cz+d}$$ with $ad-bc \neq 0$ and say $a,c \neq 0$. $$f(1/z) = \frac{\frac{a}{z}+b}{\frac{c}{z}+d} = \frac{a+bz}{c+dz}$$ So $f(\infty) = \frac{a}{c}$.

• Couldn't you do the same thing with limits? Why do you need to define division by zero here? – jpmc26 May 3 '16 at 11:11
• @jpmc26: Why would you go through all the trouble with limits when you can just do arithmetic? – user14972 May 3 '16 at 17:07
• @jpmc26: It's difficult to use limits here since we are in the complex plane, and thus you would have to show that the limit of $f$ is independant of the path used to go to infinity... – Alexandre C. May 3 '16 at 18:36
• @Hurkyl Doesn't all the math here still hold if you use $\lim_{z\to0} f(\frac{1}{z})$? Alex: Isn't that important even if you try to define $\frac{1}{0}$? If the function has multiple values as you approach infinity, I'm not sure how you can handwave that away by declaring $\frac{1}{0} = \infty$. – jpmc26 May 3 '16 at 20:35
• @Hurkyl I'm asking the opposite question, though. Why redefine arithmetic when you can use limits? I assume that doing so leads to other complications. Also, how is $\frac{1}{x}$ continuous? It wouldn't be even if you do define $\frac{1}{0}$ since it hops between $-\infty$ and $\infty$ when you cross from $x < 0$ to $x > 0$. It doesn't seem like defining $\frac{1}{0} = \infty$ simplifies anything here. I feel like there's something major I'm missing, but I really don't know what. But that's my point: this answer doesn't point out where the knowledge gap is. – jpmc26 May 5 '16 at 7:21

The algebraic structure "wheel" is an algebra with division by zero. The one point compactification of the complex plane into the Riemann sphere almost produces a wheel (one still needs to adjoin the element $0/0$).

This may not be what you would consider an "application', but for computer floating point arithmetic, division by zero is useful for setting up three special values: 1.0/0.0 gives +inf, which is a valid floating point value satisfying the usual extended number line properties. -1.0/0.0 gives -inf, and my all-time favorite is 0.0/0.0, which gives NaN (not-a-number). In spite of the name, NaNs are also valid floating point numbers in the IEEE 754 standard, which almost all modern computers implement. A few languages like MATLAB let you directly specify those (e.g., a statement like x = inf is allowed) but others do not allow this - requiring the divide by zero syntax.

Possibly irrelevant to you but NaNs are useful in some numerical computing applications and highly useful in debugging some codes. The plural "NaNs" is valid because there are many of them: at least $2^{51}$ of them in 64-bit arithmetic. If you are interested, look for papers written by W. Kahan (father and grand inquisitor for the IEEE 754 standard).

• I would suggest that floating-point math doesn't really have positive and negative zeroes so much as it has positive and negative infinitesimals, but most implementations lack a true zero or unsigned infinitesimals and substitute a positive infinitesimal instead. If x is a positive number which is small enough that xx would underflow, then I would suggest that 1/(xx) should yield "+INF" [since x*x is clearly positive] but 1/(x-x) should yield "NaN" [because x-x should be neither positive nor negative]. – supercat May 4 '16 at 15:37
• supercat, Floating point implementation in computers has representation for zero. " if every bit is zero (the sign bit being irrelevant), then the number is considered zero." From here -> cprogramming.com/tutorial/floating_point/… – r12 May 5 '16 at 5:14
• @supercat This answer is basically saying that it's useful to define division by zero in discrete mathematical systems that attempt to approximate the real numbers. In such systems, zero can be used as an approximation for values too small for the system to represent, and so it's useful to have a definition for what happens when you divide by it. Whether that's a good idea or there are better ideas is, of course, up for debate, but it is a real world system that's widely used. So +1. – jpmc26 May 6 '16 at 0:21
• @r12: In trying to have the sign of 1/(x*y) remain constant as the product becomes too small to represent, IEEE-754 severely damaged other aspects of the structure of floating-point numbers. For example, x==y should imply 1/x==1/y, but in IEEE-754 it doesn't. x-y should be equivalent to -1*(y-x), but in IEEE-754, it doesn't, etc. – supercat May 6 '16 at 4:14

Yes, in projective geometry or hyperbolic geometry for example, you can see applications or geometric entities that are $\frac{a}{0}$ or just $\infty$ . Generally, non-euclidean spaces have such type of entities or applications.

Theory of relativity. $$m = \frac {m_0}{\sqrt{1 – (\frac vc)^2}}$$ This means that as your velocity (speed) increases, and gets closer and closer to the speed of light, your mass increases (therefore, mass is related to velocity). It also proves, that it is impossible to travel faster than the speed of light. If an object were to do that, its mass would reach infinity, and that is impossible, so travelling faster than the speed of light is impossible.

• no, actually you reach infinity when $v = c$, and there are particles that do what (they have $m_0 = 0$). Going faster then $c$ would mean that the square root becomes complex. I don't know about physics but my guess is that in that case this equation can no longer be used. (and the existence of this equation per se does not imply that $v > c$ is impossible, only that this equation makes sense only when $v < c$) – Ant May 4 '16 at 20:49
• Traveling faster than light is perfectly well-defined (if complex) in that formula. All it really says is that your velocity can't pass through the speed of light. – Mark May 4 '16 at 21:27
• This doesn't answer the question. The question is asking about forms of math that do allow for division by zero. The conclusions you mention require that division by zero be undefined. You reach these conclusions by taking a limit and showing that it approaches infinity (which is just a shorthand for "is unbounded" when discussing limits). If you define division by zero, then the mass becomes whatever value you defined $\frac{m_0}{0}$ to be (which may or may not be some kind of "legitimate" infinity). Bottom line: defining $\frac{1}{0}$ doesn't help here; it makes things harder. – jpmc26 May 6 '16 at 0:03
• @Mark If you allow $m$ to have complex values, then yes, it is well defined. This equation is only useful for particles with real positive mass, since other values don't let us draw any physical conclusions (as far as I know, anyway). So you can rule out $v > c$ on that basis. – jpmc26 May 6 '16 at 0:14

Though not actually correct, I've seen it used to "trick" people with a proof that 1=0

Consider two non-zero numbers x and y such that

$x = y$

Then $x^2 = xy$

Subtract the same thing from both sides:

$x^2 - y^2 = xy - y^2$

Dividing by $(x-y)$, obtain

$x + y = y$

Since $x = y$, we see that

$2 y = y$

Thus $2 = 1$, since we started with y nonzero.

Subtracting 1 from both sides,

$1 = 0$

Su, Francis E., et al. "One Equals Zero!." Math Fun Facts. http://www.math.hmc.edu/funfacts

• You just divided numbers by (x - y) which is zero because x and y are different from zero and x = y, you can't divide a number by zero, it's against the laws of mathematics. – GingerHead May 4 '16 at 9:16
• @GingerHead That's the point, and why it's a "trick" (that's explained in the link). Still, it is an application of dividing by zero – Xen2050 May 4 '16 at 9:23
• OK, I get your point, we used to do this at highschool – GingerHead May 4 '16 at 9:26

In two dimensional space, if we want a vector corresponding to a given angle $t$, we can do:

$(x, y) = ( \cos(t), \sin(t) )$

But what about the reverse? We have a vector, and want to know the angle. A common solution is to use arctan:

$t = \arctan( \frac{y}{x} )$

Here $\frac{y}{x}$ represents the gradient of the vector. So it has a meaningful value even when $x = 0$.

Trivia: Many program languages will fail to perform that calculation, because division by zero will produce an error. They usual address this by offering an alternative function which takes two arguments: atan2(y, x).

But a notable exception is Javascript, which has the concept of +Infinity and -Infinity as numbers, and can perform this calculation even when x = 0.

$node > Math.atan( 1 / 0 ) 1.5707963267948966 (Pi/2) > Math.atan( -1 / 0 ) -1.5707963267948966 (-Pi/2)  Another language which can perform this calculation is Haskell. Inspired by Qiaochu Yuan's comment. I gave an answer to a similar question some time ago here. To put it briefly, if you are ever solving an equation in one variable by just addition, multiplication, subtraction, and division, and you end up with a contradiction, then you can conclude with certainty that the one time you divided both sides by an expression involving a variable, the expression you divided by was actually$0$(because having multiple answers can only happen if you had a$\frac 00$). Also, when computing limits, if you get a$\frac a0$for$a\neq0\$, you can directly conclude the limit does not exist (as a finite number).


text:
# Sum-free sequence

In mathematics, a sum-free sequence is an increasing positive integer sequence

$\{n_k\}_{k\in\mathbb N}$

such that for each $k$, $n_k$ cannot be represented as a sum of any subset of the preceding elements of the same sequence.

The definition of sum-free sequence is different of that of sum-free set, because in a sum-free set only the sums of two elements must be avoided, while a sum-free sequence must avoid sums of larger sets of elements as well.

## Example

The powers of two,

1, 2, 4, 8, 16, ...

form a sum-free sequence: each term in the sequence is one more than the sum of all preceding terms, and so cannot be represented as a sum of preceding terms.

## Sums of reciprocals

A set of integers is said to be small if the sum of its reciprocals converges to a finite value. For instance, by the prime number theorem, the prime numbers are not small. Paul Erdős (1962) proved that every sum-free sequence is small, and asked how large the sum of reciprocals could be. For instance, the sum of the reciprocals of the powers of two (a geometric series) is two.

If $R$ denotes the maximum sum of reciprocals of a sum-free sequence, then through subsequent research it is known that $2.065 < R < 2.8570$.[1]

## Density

It follows from the fact that sum-free sequences are small that they have zero Schnirelmann density; that is, if $A(x)$ is defined to be the number of sequence elements that are less than or equal to $x$, then $A(x)=o(x)$. Erdős (1962) showed that for every sum-free sequence there exists an unbounded sequence of numbers $x_i$ for which $A(x_i)=O(x^{\varphi-1})$ where $\varphi$ is the golden ratio, and he exhibited a sum-free sequence for which, for all values of $x$, $A(x)=\Omega(x^{2/7})$, subsequently improved to $A(x)=\Omega(x^{1/3})$ by Deshouillers, Erdős and Melfi in 1999 and to $A(x)=\Omega(x^{1/2-\varepsilon})$ by Luczak and Schoen in 2000, who also proved that the exponent 1/2 cannot be furthermore improved.


text:
## Thursday, January 28, 2016 ... /////

### Gordon Kane's ideas at Dawid's testability workshop

An occasional TRF guest blogger Gordon Kane of Michigan has posted an edited, written version of his talk at the Munich workshop on the "testability of string theory etc." that was organized by Richard Dawid:

The 18-page PDF file is much more technical than other contributions but you find tons of easily comprehensible ideas in the paper, too.

The paper argues that the qualitative character of testability of string theory is exactly the same as it has been for all respectable theories in the history of physics. And it uses the predictions for particle physics from M-theory on $G_2$ holonomy manifolds as a detailed example of all these claims. It's impressive how this framework has been developed to say something about all aspects of physics – including the nature of dark matter, patterns of superpartner masses, solution to the CP-problem, and many others.

Gordon Kane admits that like others, this scenario has no solution for the cosmological constant problem – why the CC is so small – so he assumes that this problem is solved by something that is "orthogonal" to the mechanisms deciding about the rest of particle physics.

He has written down lots of details about the history of M-theory and its $G_2$ holonomy compactifications and the detailed predictions that follow from it – winos are above $600\GeV$, binos are LSPs, gauginos are light, gravitons and light moduli are close to $50\TeV$, squarks and sleptons just a little bit lighter, gluinos at $1.5\TeV$ plus minus 10-15 percent, with cross sections somewhat lower than what is assumed in most of literature, and so on. Before the Higgs discovery, the Higgs mass was calculated as $126.4\GeV$; unfortunately, Gordon couldn't quantify the error margin.

The setup allows you to meaningfully incorporate some cosmological epochs, inflation, non-thermal evolution in cosmology, dark matter, and other things.

It's being argued that string theory indeed seems to be the only game in town that shows the promise to unify gravity with other required mechanisms and interactions and particles of matter and there exist vacua that do that. They seem to do everything that is needed. No non-stringy attempt has ever come anywhere close to these achievements.

Even if the number of vacua is large, it doesn't mean in any way that string theory is untestable. What is to be compared with the experiments is a particular compactification to 4 dimensions, Kane stresses, but this disclaimer is in no way different than it was in the past. People always needed specific models to make the theory testable. His oldest example is Newton's $F=ma$ that is very deep but it's also testable only once you enrich it by additional formulae for the force (e.g. the inverse square gravitational force).

People who talk about "testing of string theory" in general – or even testing string/M-theory as a 10- or 11-dimensional theory – generally don't know what they are talking about. And the large number of solutions is not a problem. After all, I would add, the goal of science was always to locate the viable or right theory as a tiny subset of the a priori possible but actually wrong (and to-be-excluded) competing hypotheses. String theory is no different – it just organizes lots of the "wrong theories" as "physically irrelevant, other vacua/solutions of the same correct equations". But you still have to pick the right vacuum before you test things.

It may be useful to add that Kane's favorite M-theory on $G_2$ holonomy manifolds is something like "1/5" of the string/M-theoretical phenomenology – even though he and some of his collaborators would argue that this picture is so good that it covers more than "50% of the potential" to connect string theory with experiments. I could personally place (at least sometimes) heterotic $E_8\times E_8$ models, their strongly coupled Hořava-Witten limits, Vafa's F-theory models with localized standard model in the extra dimensions, and some other braneworlds "above" the $G_2$ holonomy manifolds. But sometimes, I do feel that that the $G_2$ scenario is the best one. It starts with the maximum number of non-infinitesimal spacetime dimensions and compactifies the excessive ones in the most "democratic" way, and this could be a virtue making these scenarios natural.

Off-topic: Dutch physicist Andreas Wahl shot himself with a gun. The only problem was that he was in the water. The principle he demonstrated, namely the friction in water, is a bit less deep than M-theory on 7-dimensional manifold of exceptional holonomy, however, despite the label "Viten" in the corner.

Sometimes, string theorists may find the right vacuum as well as the reasons why this vacuum had to be the right one – some cosmological vacuum selection mechanism. But even if the latter step (which would really bring us a big step closer to the "face of God") never materializes, string theory is still exactly as testable and scientific as older theories. Unrealized vacua may be considered on par with "wrong candidate theories" and the goal is simply to find the right one – the right vacuum of string theory. Old, recent, as well as future experimental discoveries will surely influence this search – just like they always have. The number of "wrong competing theories" or "wrong vacua" may be very large but it has always been very large. String theory doesn't make these matters any worse or "less scientific". I am happy that not everyone has lost his mind. Gordon Kane clearly agrees with me that this whole discussion about "string theory as an activity abandoning science" is wrong from the very beginning.

Aside from the visibility of many totally incompetent people who decided to speak on the issue that hugely transcends their intellectual capabilities, one less outrageous but still unfortunate reason why these fallacious statements about string theory were able to spread this much is, as Gordon also mentions, that most of string theorists are formal theorists, not particle physics phenomenologists. Within the broad string theory community, the string phenomenologists represent a very small fraction. And they also give a small fraction of the talks at recent pan-string-theory conferences. I do think that it's unfortunate. And it's even more counterproductive if and when the non-phenomenological string theorists sometimes talk as if they were assuming that string phenomenology doesn't exist.

It exists and it has achieved and still is achieving amazing things. Even the first sketch of the "real world of particle physics within string theory" that was outlined in the mid 1980s was stunning. But there's been lots of progress – even if the progress hasn't been "definitive" so far.

Let me tell you that when I was publishing papers in journals, they were papers about formal string theory. It was natural for me to focus on this subdiscipline – especially after I made some important enough contributions to Matrix theory etc. I am and I was greatly intrigued by the inner structure of string theory, by the conceptual questions of quantum gravity in string theory, and so on.

But it may be fair to say that even at the moment when Tom Banks invited me to the U.S., I may have been even more impressed by string phenomenology than by the formal theory. I think it's fascinating that the models work so well, I believe that there is no reason why one of the string vacua shouldn't be "perfect" and why we shouldn't be able to find the correct one increasingly accurately (or "perfectly"), and I would love to be one who demonstrates that this perfectly correct vacuum is included in string theory. Around 1995, I was particularly intrigued by the heterotic string models, especially those in the free-fermionic construction, and so on, but I don't want to say too many things that could be too idiosyncratic.

The kibitzers surrounding theoretical physics – and it's indeed a vastly more accurate phrase to describe the likes of Lee Smolin than "physicists" – frequently offer their conspiracy theories about the suppression of some ingenious researchers and their research directions. I hate these victimist clichés (especially because they're just plain lies in almost all cases that we actually hear about) and those who use them to serve their own, personal, egotist interests. And you probably don't remember me making similar statements, at least not often.

But let me tell you something. The whole string phenomenology subcommunity is one of the actual examples of these "real-world Cinderellas". One of the reasons why I have never spent too much time with writing papers about string phenomenology was indeed my feeling that people who do this amazing work and the work itself are heavily underappreciated.

Off-topic: Zika virus was said to have exploded. Flu-like symptoms, rashes, muscle weakness – and especially the shrunken heads (and retardation) of newborns – are the symptoms. No fatalities are occurring. But due to the dependence on mosquitoes, the virus is predicted to spread to Florida and the "hot wet third world".

Dozens of extremely active and brave people keep on working on string phenomenology, anyway. Some of them are partly phenomenologists, partly formal theorists etc. They are developing models that are qualitatively analogous to the models within effective quantum field theory that are being developed by more ordinary "model builders". The string phenomenologists could probably work on any of these "normal model building" activities. But they work on string phenomenology because their skills and knowledge are deeper than those of the "ordinary model builders". This "elite within elite" status is the only true reason why string phenomenologists represent a small percentage both among string theorists and among the model builders. Feel free to check that when I talk about string phenomenologists in this way, it's an opinion coming from an impartial person.

Within quantum field theory, one may discuss things like supersymmetry, supersymmetry breaking, dark matter, patterns in fermionic masses, inflation etc. But these individual aspects of beyond-the-Standard-Model physics are "mostly" independent of each other. String/M-theory basically unifies all of those. It forces you to discuss all these things at the same time. It requires you to know the effective field theory methods because much of the experimentally accessible physics following from string theory may be described by the approximation known as effective field theory. But it also requires you to know something more – string theory and the new relationships between physical phenomena (and new "geometrizations" of the old phenomena) that string theory has brought us and is still bringing us.

Sometimes it's being said that string phenomenologists don't use all the insights from "formal string theory" too much. I have the same feeling. But you must understand that it's not really the primary goal of string or particle physics phenomenology to maximally utilize all cute and seemingly fundamental ideas that have been found. The goal is to describe phenomena in particle physics.

If one looks at the things fairly but from any perspective, it's damn obvious that string phenomenology is at least as great and as scientific as other approaches to similar questions (or simpler questions). Even though I have never made living by full-fledged string phenomenology, I find it obvious that all the individuals who openly contaminate the public discourse with their "criticisms" of string phenomenology as a whole are piles of worthless pseudointellectual junk.


text:
# Vector 3 v.3.80 , transmission breaks on ANT+, on BLTE is ok

Recently, I had quite a few breaks from one to several seconds in displaying power and cadence on the Edge 1030. I received messages about low battery, so I thought that replacing it would solve the problem.
After replacing the battery, the problem persists.
At the same time, I'm using Zwift on an Android phone. Bluetooth transmissions, no interruptions. I made a switch, paired Vector after BLTE with Edge 1030, ANT + with the phone. The Edge power / cadence was displayed correctly and the breaks were now on the phone. Therefore, I exclude 1030, and the problem is on the Vector 3 side. It seems to me that it is not related to the battery, because during power outages the signal should disappear both after ANT + and Bluetooth.
Has anyone had such a problem?


text:
#### Term I - Model Question Paper

11th Standard

Reg.No. :
•
•
•
•
•
•

Chemistry

Time : 03:00:00 Hrs
Total Marks : 100

Part A

Multiple Choice Question

20 x 1 = 20
1. 40 ml of methane is completely burnt using 80 ml of oxygen at room temperature The volume of gas left. after cooling to room temperature is

(a)

40 ml CO2 gas

(b)

40 ml CO2 gas and 80 ml H2O gas

(c)

60 ml CO2 gas and 60 ml H2O gas

(d)

120 ml CO2 gas

2. An element X has the following isotopic Composition 200X = 90%, 199X = 8% and 202X = 2% The Weighted average atomic mass of the element X is closet to

(a)

201 u

(b)

202 u

(c)

199 u

(d)

200 u

3. The number of water molecules in a drop of water weighing 0.018 g is

(a)

6.022 x 1026

(b)

6.022 x1023

(c)

6.022 x 1020

(d)

9.9 x 1022

4. Which one of the following represents 180g of water?

(a)

5 Moles of water

(b)

90 moles of water

(c)

$\frac { 6.022\times { 10 }^{ 23 } }{ 180 }$ molecules of water

(d)

6.022 x 1024molecules of water

5. The oxidation number of Cr in Cr2O72- __________ is

(a)

+6

(b)

-6

(c)

+7

(d)

-7

6. Which of the following contain same number of carbon atoms as in 6 g of carbon-12.

(a)

7.5 g ethane

(b)

8 g methane

(c)

both (a) and (b)

(d)

none of these

7. Which of the following compound(s) has /have a percentage of carbon same as that in ethylene (C2H4)

(a)

propene

(b)

ethyne

(c)

benzene

(d)

ethane

8. The change in the oxidation number of S in H2S and SO2,in the following industrial reaction:
2H2S(g) + SO2(g) $\longrightarrow$ 3S(s) + H2O(g)

(a)

-2 to 0, +4 to 0

(b)

-2 to 0, +4 to -1

(c)

-2 to -1, +4 to 0

(d)

-2 to -1, +4 to -2

9. In which of the following reactions, hydrogen peroxide acts as an oxidising agent?

(a)

I2+ H2O2 + 20H- $\longrightarrow$ 21- + 2H2O + O2

(b)

PbS + 4H2O2 $\longrightarrow$ PbSO4 + 4H2O

(c)

2MnO4-+ 3H2O2 $\longrightarrow$ 2MnO2 + 3O2 + 2H2O + 2OH-

(d)

HOCI + H2O2 $\longrightarrow$ H2O+ + Cl- + O2

10. A sample of ${ CaCO }_{ 3 }$ is 50% pure.  On heating 1.12 L of ${ CO }_{ 2 }$ ( at STP) is obtained.  Residue left (assuming non volatile Impurity is

(a)

7.8g

(b)

3.8g

(c)

2.8g

(d)

8.9g

11. Polyethylene can be produced from calcium carbide according to the following sequence of relations.

${ CaC }_{ 2 }+{ H }_{ 2 }O\rightarrow CaO+HC\equiv CH$

$n(HC\equiv CH)+{ nH }_{ 2 }\rightarrow { \left( { CH }_{ 2 }-{ CH }_{ 2 } \right) }_{ n }$

The mass of polyethylene, which can be produced from 20.0 kg of ${ CaC }_{ 2 }$ is

(a)

6.75 kg

(b)

7.75 kg

(c)

8.75 kg

(d)

9.57 kg

12. 1.0g of magnesium is burnt with 0.56g${ O }_{ 2 }$ in a closed vessel.  Which reactant is left in excess and how much? (At.wt.Mg=24; 0=16)

(a)

Mg. 0.16g

(b)

${ O }_{ 2 },0.16\quad g$

(c)

Mg,0.44 g

(d)

${ O }_{ 2 },0.28\quad g$

13. ${ 10 }^{ 21 }$ molecules are removed from 200 mg of ${ CO }_{ 2 }$.  The moles of ${ CO }_{ 2 }$ left are

(a)

$2.88\times { 10 }^{ -3 }$

(b)

$28.8\times { 10 }^{ -3 }$

(c)

$288\times { 10 }^{ -3 }$

(d)

$28.8\times { 10 }^{ -3 }$

14. 1+2 (17-15.1)=x. Hence, x is equal to

(a)

4.8

(b)

19.9

(c)

5

(d)

5.0

15. Number of significant figures in the answer of the following will be $\frac { (29.2-20.2)(1.79\times 105) }{ 1.39 }$

(a)

4

(b)

3

(c)

2

(d)

1

16. H2SO4 solution is 500 molal (d=1.8 g/mL) hence, its molarity is

(a)

18 M

(b)

36 M

(c)

9 M

(d)

4.50 M

17. Molality of sucrose solution changes if

(a)

temperature is increased

(b)

(c)

in both (a) and (b) cases

(d)

in none of the above cases

18. Equal moles of H2O and NaCI are present in a solution. Hence, molality of NaCI solution is

(a)

0.55

(b)

55.5

(c)

1.00

(d)

0.18

19. SI units of pressure is

(a)

pascal

(b)

atmosphere

(c)

metre

(d)

dyne/cm2

20. SI unit of emf is

(a)

joule

(b)

millivolt

(c)

volt

(d)

ampere

21. Part B

10 x 2 = 20
22. Calculate the molar mass of the following compounds. -Sulphurci  Acid [H2SO4]

23. Write a note on Thomson's plum pudding model of an atom.

24. What are the defects of Rutherford's model?

25. How many neutrons and protons are there in the Following nuclei?
$_{ 6 }^{ 13 }{ C }$,$_{ 2 }^{ 18 }{ O }$.$_{ 12}^{ 24}{ Mg }$,$_{26 }^{ 56}{ Fe }$,$_{ 88}^{ 38}{ Sr }$

26. Why Pauli exclusion principle is called exclusion principle?

27. Bring out the similarities and dissimilarities between a 1s and 2s orbital.

28. Calculate the wave length of an electron (mass = 9.1 x 10-31 kg) moving with 'a velocity of 10ms-1 (h = 6.6 x 10-34 kg m2 sec-1).

29. The kinetic energy of a subatomic particle is 5.58 x 10-25 J. Calculate the frequency of the particle wave. (Planck's constant h = 6.626 x 10-34 kg m2 s-1)

30. A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1$\overset { o }{ A }$  What is the uncertainty involved in the measurement of its velocity.

31. Explain how matter has dual character?

32. Part C

10 x 3 = 30
33. Matter is defined as anything that has mass and occupies space. All matter is composed of atoms

34. Write a note on the differences between elements and compounds

35. Write down the formulae for calculating the equivalent mass of an acid, base and oxidizing agent.

36. 1.05g of a metal gives on oxidation 1.5g of its oxide. Calculate its equivalent mass.

37. A piece of cut apple becomes brown. Why? Can you prevent it by a simple method

38. Balance the following reaction:
H2C2O4+KMnO4 + H2SO4 $\rightarrow$ H2SO4 + MnSO4 + CO2 + H2O

39. Calculate the molar volume of 146 g of HCI gas and the number of molecules present in it

40. Calculate the molar mass of 20 L of gas weighing 23.2 g at STP.

41. Describe in brief Lothar Meyer's classification of elements.

42. Explain classification of elements based on Newland's law of Octaves.

43. Part D

6 x 5 = 30
44. Write note on combination reaction.

45. Arrange the elements silver, Zinc and copper in the order of their decreasing electron releasing tendency and justify your arrangement with an appropriate experiment.

46. Balance the following equations by oxidation number method.
KBr + MnO2 + H2SO4 ⟶ KHSO4 + MnSO4 + H2O + Br2

47. Calculate the empirical and molecular formula of a compound containing 32% carbon, 4% hydrogen and rest oxygen. Its vapour density is 75.

48. Light of wavelength 12818 $\overset { o }{ A }$ is emitted when the electron of a hydrogen atom drops from 5th to 3rd orbit. Find the wavelength of a photon emitted when the electron falls from 3rd to 2nd orbit.

49. Explain about the factors that influence the ionization enthalpy.


text:
eLibrary of Mathematical Instituteof the Serbian Academy of Sciences and Arts
 > Home / All Journals / Journal /

 Kragujevac Journal of MathematicsPublisher: Prirodno-matematički fakultet Kragujevac, KragujevacISSN: 1450-9628Issue: 44_4Date: 2020Journal Homepage

 Results on Tauberian Theorem for Cesàro Summable Double Sequences of Fuzzy Numbers 495 - 508 B. B. Jena, S. K. Paikray, P. Parida and H. Dutta
AbstractKeywords: double Cesàro summability; slow oscillation; Tauberian condition; sequence of fuzzy numbersMSC: 40G05; 40E05 40A10; 03E72
 Sharp Bounds on the Augmented Zagreb Index of Graph Operations 509 - 522 N. Dehgardi and H. Aram
AbstractKeywords: augmented Zagreb index; general sum-connectivity index; general Randić index; graph operationsMSC: 05C12 05C07
 On Equienergetic, Hyperenergetic and Hypoenergetic Graphs 523 - 532 Samir K. Vaidya and Kalpesh M. Popat
AbstractKeywords: equienergetic; hyperenergetic; hypoenergeticMSC: 05C50; 05C76
 A Generalized Class of Close-to-Convex Functions 533 - 538 Pardeep Kaur and Sukhwinder Singh Billing
AbstractKeywords: analytic function; univalent function; close-to-convex functionMSC: 30C80 30C45
 A New Class of Integrals Involving Generalized Hypergeometric Function and Multivariable Aleph-Function 539 - 550 Dinesh Kumar, Frédéric Ayant and Devendra Kumar
AbstractKeywords: multivariable $I$-function; multivariable $H$-function; double finite integralsMSC: 33C60; 33C99 44A20
 Lower Bounds for Inverse Sum Indeg Index of Graphs 551 - 562 I. Gutman, M. Matejić, E. Milovanović and I. Milovanović
AbstractKeywords: degree (of vertex); degree (of edge); inverse sum indeg index; Zagreb indexMSC: 05C12 05C50
 A Note on the Definition of Bounded Variation of Higher Order for Double Sequences 563 - 570 Bhikha Lila Ghodadra and V. Fülöp
AbstractKeywords: double sequence of bounded variation; double sequence of bounded variation of order $p$ ($p\in \mathbb N$); double sequence of bounded variation of order $(p;0)$; double sequence of bounded variation of order $(0;p)$; double sequence of bounded variation of order $(p;p)$MSC: 40B05
 Some Grüss Type Inequalities for Fréchet Differentiable Mappings 571 - 579 T. Teimouri-Azadbakht and A. G Ghazanfari
AbstractKeywords: Fréchet differentiable mappings; $C^*$-modules; Grüss inequalityMSC: 26D10 46C05; 46L08
 The $\bar\partial$-Cauchy Problem on Weakly $q$-Convex Domains in $\Bbb{C}P^n$ 581 - 591 Sayed Saber
AbstractKeywords: $\bar\partial$; $\bar\partial$-Neumann operator; $q$-convex domainsMSC: 32F10 32W05
 Johnson Pseudo-Contractibility and Pseudo-Amenability of $\theta$-Lau Product 593 - 601 M. Askari-Sayah, A. Pourabbas and A. Sahami
AbstractKeywords: $\theta$-Lau product; Johnson pseudo-contractibility; pseudo-amenabilityMSC: 46H05; 46H20 43A20
Article page: 12>>

 Remote Address: 3.221.159.255 • Server: elib.mi.sanu.ac.rsHTTP User Agent: CCBot/2.0 (https://commoncrawl.org/faq/)


text:
# Netzschleuder network catalogue, repository and centrifuge

Problems with this dataset? Open an issue.
You may also take a look at the source code.
The network in this dataset can be loaded directly from graph-tool with:
import graph_tool.all as gt
g = gt.collection.ns["cs_department"]

# cs_department — Aarhus Computer Science department relationships

Description

Multiplex network consisting of 5 edge types corresponding to online and offline relationships (Facebook, leisure, work, co-authorship, lunch) between employees of the Computer Science department at Aarhus. Data hosted by Manlio De Domenico.1

1. Description obtained from the ICON project.

Tags
Social Relationships Multilayer Unweighted
Citation
• Matteo Magnani, Barbora Micenkova, and Luca Ross, "Combinatorial Analysis of Multiple Networks." arXiv:1303.4986 (2013), http://arxiv.org/abs/1303.4986
Upstream URL OK
https://manliodedomenico.com/data.php
Networks
Tip: hover your mouse over a table header to obtain a legend.
Name Nodes Edges $\left<k\right>$ $\sigma_k$ $\lambda_h$ $\tau$ $r$ $c$ $\oslash$ $S$ Kind Mode NPs EPs gt GraphML GML csv
cs_department 61 620 20.33 11.22 13.78 4.46 0.03 0.52 4 1.00 Undirected Unipartite name nodeLabel weight layer 2 KiB 4 KiB 3 KiB 3 KiB
Ridiculograms


text:
# Probing the softest region of nuclear equation-of-state

Bao-An Li11Address after Aug. 10: Dept. of Chemistry and Physics, Arkansas State University, P.O. Box 419, State University, AR 72467-0419 and C. M. Ko Cyclotron Institute and Physics Department,
Texas A&M University, College Station, TX 77843
preprint: Phys. Rev. C (1998) in press.

An attractive, energy-dependent mean-field potential for baryons is introduced in order to generate a soft region in the nuclear equation-of-state, as suggested by recent lattice QCD calculations of baryon free matter at finite temperature. Based on a hadronic transport model, we find that although this equation-of-state has negligible effects on the inclusive hadronic spectra, it leads to a minimum in the energy dependence of the transverse collective flow and a delayed expansion of the compressed matter. In particular, the transverse flow changes its direction as the colliding system passes through the softest region in the equation-of-state.
PACS number(s): 25.75.+r

The primary goal of experiments on relativistic heavy-ion collisions is to create and study the Quark-Gluon-Plasma (QGP) (1). Lattice QCD calculations of baryon-free matter at finite temperature (2)have shown that although both the energy and entropy densities increase appreciably at about 150 MeV where the phase transition from the hadronic matter to the QGP occurs, the change in pressure is considerably weaker, leading thus to a small sound velocity which is given by the pressure gradient with respect to energy density, i.e., $V_{s}^{2}\equiv dP/de$ . This soft region in the nuclear equation-of-state is expected to have significant influence on the collective dynamics of the hot and dense matter formed in relativistic heavy ion collisions. In particular, a small sound velocity delays the expansion of the compressed matter and leads also to a reduced transverse collective flow. Indeed, recent studies based on the hydrodynamical model have demonstrated that as one varies the incident energy of a heavy ion collision, the system may reach the softest region in the equation-of-state and forms a long-lived fireball, which then results in the appearance of a minimum in the energy dependence of the transverse collective flow (3); (4); (5); (6). Although the hydrodynamical model provides a convenient framework to study the effects of nuclear equation-of-state on the collective behavior of colliding nuclei (5); (7), it requires the introduction of an ad hoc freeze-out procedure to compare the calculated results with the experimental measurements.

In this Rapid Communication, we take a different approach to explore the effect of a softened equation-of-state on experimental observables by using the hadronic transport model ART (11), which treats consistently the final freeze out. The possibility of describing the main features of lattice QCD phase transition in a hadronic model has recently been demonstrated by allowing hadron masses to decrease in medium, which has been argued to be related to the restoration of chiral symmetry (8). Although neither QGP phase transition nor chiral symmetry restoration actually happens in our approach as in the hydrodynamical models, results from our studies are useful for understanding the signals for a real phase transition. Indeed, we find that the softened equation-of-state leads to both a minimum in the incident energy dependence of the transverse collective flow and a delayed expansion of the compressed matter as in the hydrodynamical model. Our results further show that such an equation-of-state has almost no effect on the inclusive hadron spectra.

To introduce a soft region in the equation-of-state, we consider for the baryon sector of the reaction system the following relation between the pressure $P$ and energy density $e$ in the mean-field approach (9); (10):

 $\displaystyle P$ $\displaystyle=$ $\displaystyle\frac{2}{3}E_{k}\rho+\rho\frac{dW}{d\rho}-W,$ (1) $\displaystyle e$ $\displaystyle=$ $\displaystyle(m+E_{k})\rho+W,$ (2)

where $E_{k}$ , $W$ , $\rho$ , and $m$ are the average kinetic energy per baryon, potential energy density, baryon density, and nucleon mass, respectively. It was shown in ref. (10)that $W$ generally depends on both $\rho$ and $E_{k}$ . We require that at low energy densities, it coincides with the normal soft equation-of-state obtained from the well-known Skyrme interaction, i.e.,

 $P=\frac{2}{5}E_{f}\rho-179\rho^{2}/\rho_{0}+164(\frac{\rho}{\rho_{0}})^{7/6}% \rho~{}~{}~{}~{}({\rm MeV/fm^{3}}),$ (3)

where $E_{f}$ and $\rho_{0}$ are, respectively, the Fermi energy and normal nuclear matter density. Above the soft region, the pressure is taken to be

 $P=200(\frac{\rho}{\rho_{0}})^{1/3}\rho-153~{}~{}~{}~{}({\rm MeV/fm^{3}}),$ (4)

which is similar to that of a noninteracting quark gas confined in a MIT bag.

To satisfy the condition at high energy densities, the following potential energy density is introduced,

 $W=\rho[600(\frac{\rho}{\rho_{0}})^{1/3}-\frac{3}{5}E_{f}(\frac{\rho}{\rho_{0}}% )^{2/3}+153/\rho+C_{1}],~{}~{}~{}~{}({\rm MeV/fm^{3}})$ (5)

where $C_{1}$ is a constant depending on the energy density at which Eq. ( 4) is reached, i.e., the upper boundary of the soft region.

To illustrate the effects of the soft region of nuclear equation-of-state, we consider the extreme case of a vanishing sound velocity. This can be achieved by keeping a constant pressure at $P_{c}$ while changing the energy density via the following potential energy density in Eq. ( 1):

 $W=\rho(C_{2}-\frac{P_{c}}{\rho}-E_{k}),~{}~{}~{}~{}({\rm MeV/fm^{3}})$ (6)

where $C_{2}$ is again a constant determined from the energy density of the lower boundary of transition region. In deriving the above expression we have assumed that $E_{k}$ scales with $\rho^{2/3}$ as in an adiabatic process.

The resulting equation-of-state is shown in Fig. 1as super-soft eos1 and eos2, with the softest region in the energy density range of $1 and $1.5 , respectively. For comparison, we have also included the kinetic pressure, corresponding to a pure cascade model, as a function of energy density. The mean-field potential to be used in the transport model can be obtained from the relation $U\equiv\partial W/\partial\rho$ , which is repulsive outside but attractive inside the softest region. In particular, the mean-field potential in the softest region is energy-dependent in order to balance the increasing kinetic pressure as the energy density increases. We note that both parameters $C_{1}$ and $C_{2}$ have no effect on the dynamics as only the second derivative of the potential energy density appears in the equation of motion.

Effects of the softened equation-of-state can be studied using the relativistic transport model ART. We refer the reader to Ref. (11)for the detail of the model and its applications in studying various aspects of relativistic heavy-ion collisions at AGS energies. It has been found in these studies that the transverse collective flow, defined by

 $_{t}=\frac{1}{N}\int(y)\cdot\frac{dN}{dy}\cdot{\rm sgn}(y)% \cdot dy,$ (7)

is sensitive to the nuclear equation-of-state. We expect that the super-soft equation-of-state we introduced here will also affect the transverse collective flow if the energy density achieved in the collisions reaches the softest region, as in the recent study based on the hydrodynamical model (5); (7). In the latter study, a minimum in the excitation function of $_{t}$ at a beam energy of about 6 GeV/nucleon has been found when a softened equation-of-state due to the QCD phase transition is used.

In Fig. 2, we show the excitation function for the average transverse momentum predicted by the ART model using the four equation-of-states shown in Fig. 1. For the case of either the cascade or the normal soft equation-of-state, details can be found in our previous publications (11). It is interesting to see that the super-soft equation-of-state with the softest region leads to a minimum in the excitation function of the transverse flow, as in hydrodynamical calculations. Furthermore, the incident energy at which the minimum occurs depends on the location of the softest region in the equation-of-state. For both super-soft eos1 and eos2, the average transverse momenta near the minima are even smaller than those in the cascade calculations as attractive mean-field potentials have been introduced in these equation-of-states to counterbalance the positive gradient of the thermal pressure. This is more clearly seen in Fig. 3, where we compare the average transverse momentum distribution using the cascade model and the super-soft eos1 for the Au+Au reaction at $P_{\rm beam}/A=12$ GeV/c. It is seen that the flow parameter, defined as the slope of the transverse momentum distribution at mid-rapidity, changes from positive to negative sign as a result of the attractive mean-field potential in the softest region. Our results also show that the flow parameter changes gradually from positive to negative then to positive as the beam energy increases. A change in the sign of flow parameter as the beam energy increases is another clear indication of the softening of the nuclear equation-of-state. To observe this effect experimentally, one needs to measure the absolute direction of the transverse flow besides its strength. This can be done by studying the shadowing effect on pion flow (12); (13); (14); (15)and the circular polarization of the $\gamma$ ray emitted from the reaction (16), as in heavy ion collisions at low energies.

A softened equation-of-state is expected not only to slow down the expansion of the system but also to affect the amount of compression in the collision. Since the pressure decreases from the normal soft, super-soft to the cascade at the same energy density, one expects to reach the highest compression in the cascade model. Shown in Fig. 4are the radial density distributions of nucleons, pions and kaons at 20 fm/c from the Au+Au reaction at an impact parameter of 2 fm and a beam momentum of 12 GeV/c. At this already rather late time of the reaction, the normal soft equation-of-state gives the most extend particle distributions, indicating thus the fastest compression and expansion. The particle distributions from calculations using the super-soft eos1 are more compact, implying thus a slower compression and expansion.

It is worth mentioning that we have also studied other inclusive hadronic observables using the four equations of state and have found that it is almost impossible to distinguish the predictions using the softened equation-of-state from others.

In summary, we have studied effects of the softened nuclear equation-of-state on heavy ion collisions at relativistic energies within a hadronic transport model. To keep a zero pressure gradient in the softest region, an attractive mean field for baryons has been introduced. The softened equation-of-state is found to lead to a minimum in the excitation function of transverse flow and a delayed expansion but has almost no effect on the inclusive hadronic observables. Although it is questionable to use the hadronic degrees of freedom at very high energy densities, our results do indicate that a super-soft equation-of-state can produce the effects expected from the phase transition. Experiments on the measurement of the transverse collective flow at different incident energies will be extremely useful in our search for the signatures of chiral and deconfinement transitions.

This work was supported in part by NSF Grant No. PHY-9509266, the Robert A Welch foundation under Grant A-1358, and the Texas Advanced Research Program.

## References

• (1) C. Y. Wong, Introduction to High Energy Heavy-Ion Collisions, (World Scientific, Singapore, 1994).
• (2) E. Laermann, Nucl. Phys. A610 , 1c (1996).
• (3) L. Bravina, L. P. Csernai, P. Levai and D. Strottman, Phys. Rev. C50 , 2161 (1994); Nucl. Phys. A566 , 461c (1994).
• (4) C. M. Hung and E. V. Shuryak, Phys. Rev. Lett. 75 , 4003 (1995).
• (5) D. H. Rischke, Y. Pürsün, J. A. Maruhn, H. Stöcker and W. Greiner, Heavy Ion Phys., 1 , 309 (1995).
• (6) M. Ploszajczak, A.A. Shanenko and V.D. Toneev, Preprint, Ganil-95.
• (7) D. H. Rischke and M. Gyulassy, Nucl. Phys. A597 , 701 (1996); ibid , A608 , 479 (1995).
• (8) V. Koch and G. E. Brown, Nucl. Phys. A560 , 345 (1993) and references therein.
• (9) R.B. Clare and D.D. Strottman, Phys. Rep. 141 , 177 (1986).
• (10) C. Gale, G.F. Bertsch and S. Das Gupta, Phys. Rev. C35 , (1987) 1666.
• (11) B. A. Li and C. M. Ko, Phys. Rev. C52 , 2037 (1995); ibid C53 , R22 (1996); C54 , 3283 (1996); C54 , 844 (1996); Nucl. Phys. A601 , 457 (1996).
• (12) J. Gosset, O. Valette, J. P. Alard, J. Augerat, R. Babinet et al. , Phys. Rev. Lett. 62 1251 (1989).
• (13) B. A. Li, W. Bauer and G. F. Bertsch, Phys. Rev. C 44 2095 (1991).
• (14) B. A. Li, Nucl. Phys. A570 , 797 (1994).
• (15) J. C. Kintner et al. for the EOS collaboration, Phys. Rev. Lett. 78 , 4165 (1997).
• (16) R. C. Lemmon et al., preprint MSUCL-1084, submitted to Phys. Rev. Lett.


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnlilne Submission ㆍMy Manuscript - For Reviewers - For Editors
 The number of representations of a positive integer by triangular, square and decagonal numbers Bull. Korean Math. Soc. 2019 Vol. 56, No. 5, 1143-1157 https://doi.org/10.4134/BKMS.b180914Published online September 30, 2019 Uha Isnaini, Ray Melham, Pee Choon Toh Nanyang Technological University; University of Technology, Sydney; Nanyang Technological University Abstract : Let $T_aD_b(n)$ and $T_aD'_b(n)$ denote respectively the number of representations of a positive integer $n$ by $a(x^2-x)/2 +b(4y^2-3y)$ and $a(x^2-x)/2 +b(4y^2-y)$. Similarly, let $S_aD_b(n)$ and $S_aD'_b(n)$ denote respectively the number of representations of $n$ by $ax^2 +b(4y^2-3y)$ and $ax^2 +b(4y^2-y)$. In this paper, we prove 162 formulas for these functions. Keywords : representations by binary quadratic forms MSC numbers : Primary 11E25, 11E16 Downloads: Full-text PDF   Full-text HTML


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# A simple but ambiguous expression [duplicate]

Possible Duplicates:
Implicit multiplication order of operation
What is 48÷2(9+3)?

i am so confused after knowing different answer of this ques
48/2(9+3)
though google cal comes up with ans 288 but i have seen its answer 2 on some websites

-

## marked as duplicate by Henry, t.b., Arturo Magidin, Andres Caicedo, Willie WongApr 30 '11 at 20:37

A/BC can mean either (A/B)*C or A/(BC). Without explicit bracketing or without conventions establishing operator precedence it is ambiguous. –  Bill Dubuque Apr 30 '11 at 20:20
Why would different people want to know the answer to this banal expression? –  Uticensis Apr 30 '11 at 20:23
@flutty: The problem here is not with the math, but with the notation. Many people seem to think that this question does not have a clear answer because the mathematics hasn't been figured out. Really, the problem is that it isn't clear what is being asked, and the question could be interpreted as one of several different math problems. It's like saying "I have a red house and I have a blue house. What color is my house?"--the problem is that it's unclear which house is being talked about, not that people haven't figured out how to determine the color of a house. –  Alex Becker Apr 30 '11 at 20:24
@Theo: This didn't actually start on Facebook, that's just one of the ways it's spread. It's an example of a meme, a quirky expression which spreads like wildfire on the internet. Information on it's origin can be found at knowyourmeme.com/memes/48293. –  Alex Becker Apr 30 '11 at 20:30
There must be a way to use this confusion in order to get huge discounts in shops..." What does this cost? 48$\$ $for 2 pieces? Give me 9! No wait, I want 3 more!" --- "That will be 2$, sir, have a nice day." –  Myself Apr 30 '11 at 20:35

By standard convention, 48/2(9+3) is 288. Anyone who wants to be well understood should write either $\frac{48}{2} (9+3)$ or (48/2)(9+3).


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# Math Help - Prime numbers

1. ## Prime numbers

Let $n \ge 2$ be a number that is not a prime.
Show that there is a divisor $p$ of $n$, with $p \ge 2$, so that $n \ge p^2$

On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.

2. Originally Posted by Joel24
Let $n \ge 2$ be a number that is not a prime.
Show that there is a divisor $p$ of $n$, with $p \ge 2$, so that $n \ge p^2$

On my paper, I've written down n=rs where n and s are any natural numbers >1. Not sure if that's needed, though.
If n>2 is not prime, then it has a factor p>=2, so there also exists a q>=2 such that pq=n. Now if both n<p^2 and n<q^2, then pq<n a contradiction.

Hence every composite number >2 has a factor whose square is less than or equal to the number.

RonL

3. Ah, a proof (or sorts) by contradiction. I would've never have thought of that.

Excellent. Thank you so much!


text:
## martingale posteriors

Posted in Books, Statistics, University life with tags , , , , , , , , , , , , on November 7, 2022 by xi'an

A new Royal Statistical Society Read Paper featuring Edwin Fong, Chris Holmes, and Steve Walker. Starting from the predictive

$p(y_{n+1:+\infty}|y_{1:n})\ \ \ (1)$

rather than from the posterior distribution on the parameter is a fairly novel idea, also pursued by Sonia Petrone and some of her coauthors. It thus adopts a de Finetti’s perspective while adding some substance to the rather metaphysical nature of the original. It however relies on the “existence” of an infinite sample in (1) that assumes a form of underlying model à la von Mises or at least an infinite population. The representation of a parameter θ as a function of an infinite sequence comes as a shock first but starts making sense when considering it as a functional of the underlying distribution. Of course, trading (modelling) a random “opaque” parameter θ for (envisioning) an infinite sequence of random (un)observations may sound like a sure loss rather than as a great deal, but it gives substance to the epistemic uncertainty about a distributional parameter, even when a model is assumed, as in Example 1, which defines θ in the usual parametric way (i.e., the mean of the iid variables). Furthermore, the link with bootstrap and even more Bayesian bootstrap becomes clear when θ is seen this way.

Always a fan of minimal loss approaches, but (2.4) defines either a moment or a true parameter value that depends on the parametric family indexed by θ. Hence does not exist outside the primary definition of said parametric family. The following construct of the empirical cdf based on the infinite sequence as providing the θ function is elegant but what is its Bayesian justification? (I did not read Appendix C.2. in full detail but could not spot the prior on F.)

“The resemblance of the martingale posterior to a bootstrap estimator should not have gone unnoticed”

I am always fan of minimal loss approaches, but I wonder at (2.4), as it defines either a moment or a true parameter value that depends on the parametric family indexed by θ. Hence it does not exist outside the primary definition of said parametric family, which limits its appeal. The following construct of the empirical cdf based on the infinite sequence as providing the θ function is elegant and connect with bootstrap, but I wonder at its Bayesian justification. (I did not read Appendix C.2. in full detail but could not spot a prior on F.)

While I completely missed the resemblance, it is indeed the case that, if the predictive at each step is build from the earlier “sample”, the support is not going to evolve. However, this is not particularly exciting as the Bayesian non-parametric estimator is most rudimentary. This seems to bring us back to Rubin (1981) ?! A Dirichlet prior is mentioned with no further detail. And I am getting confused at the complete lack of structure, prior, &tc. It seems to contradict the next section:

“While the prescription of (3.1) remains a subjective task, we find it to be no more subjective than the selection of a likelihood function”

Copulas!!! Again, I am very glad to see copulas involved in the analysis. However, I remain unclear as to why Corollary 1 implies that any sequence of copulas could do the job. Further, why does the Gaussian copula appear as the default choice? What is the computing cost of the update (4.4) after k steps? Similarly (4.7) is using a very special form of copula, with independent-across-dimension increments. I am also missing a guided tour on the implementation, as it sounds explosive in book-keeping and multiplying, while relying on a single hyperparameter in (4.5.2)?

In the illustration section, the use of the galaxy dataset may fail to appeal to Radford Neal, in a spirit similar to Chopin’s & Ridgway’s call to leave the Pima Indians alone, since he delivered a passionate lecture on the inappropriateness of a mixture model for this dataset (at ICMS in 2001). I am unclear as to where the number of modes is extracted from the infinite predictive. What is $\theta$ in this case?

Copulas!!! Although I am unclear why Corollary 1 implies that any sequence of copulas does the job. And why the Gaussian copula appears as the default choice. What is the computing cost of the update (4.4) after k steps? Similarly (4.7) is using a very special form of copula, with independent-across-dimension increments. Missing a guided tour on the implementation, as it sounds explosive in book-keeping and multiplying. A single hyperparameter (4.5.2)?

## Finite mixture models do not reliably learn the number of components

Posted in Books, Statistics, University life with tags , , , , , , , , , , , , , on October 15, 2022 by xi'an

When preparing my talk for Padova, I found that Diana Cai, Trevor Campbell, and Tamara Broderick wrote this ICML / PLMR paper last year on the impossible estimation of the number of components in a mixture.

“A natural check on a Bayesian mixture analysis is to establish that the Bayesian posterior on the number of components increasingly concentrates near the truth as the number of data points becomes arbitrarily large.” Cai, Campbell & Broderick (2021)

Which seems to contradict [my formerly-Glaswegian friend] Agostino Nobile  who showed in his thesis that the posterior on the number of components does concentrate at the true number of components, provided the prior contains that number in its support. As well as numerous papers on the consistency of the Bayes factor, including the one against an infinite mixture alternative, as we discussed in our recent paper with Adrien and Judith. And reminded me of the rebuke I got in 2001 from the late David McKay when mentioning that I did not believe in estimating the number of components, both because of the impact of the prior modelling and of the tendency of the data to push for more clusters as the sample size increased. (This was a most lively workshop Mike Titterington and I organised at ICMS in Edinburgh, where Radford Neal also delivered an impromptu talk to argue against using the Galaxy dataset as a benchmark!)

“In principle, the Bayes factor for the MFM versus the DPM could be used as an empirical criterion for choosing between the two models, and in fact, it is quite easy to compute an approximation to the Bayes factor using importance sampling” Miller & Harrison (2018)

This is however a point made in Miller & Harrison (2018) that the estimation of k logically goes south if the data is not from the assumed mixture model. In this paper, Cai et al. demonstrate that the posterior diverges, even when it depends on the sample size. Or even the sample as in empirical Bayes solutions.

## a journal of the [less] plague and [more] pestilence year

Posted in Books, Kids, Mountains, pictures, Travel, University life with tags , , , , , , , , , , , , , , , , , , , on July 10, 2022 by xi'an

Read Rankin’s last Rebus, A song for the dark times, which takes place between Edinburgh and the Far North (of Scotland). I reasonably enjoyed it, by which I mean I was not expecting novelty, but rather reuniting with a few characters, including the Teflon villain, Big Ger Cafferty, still around at his craft. Rebus is getting older, cannot climb stairs any longer, and cannot deliver a proper punch in a fight! Still enjoyable, with a dig into Second World War internment camps for German prisoners… While not yet into the COVID era, the spirit is definitely post-Brexit, with a general resentment of what it brought (and did not bring). The character of Inspector Fox escaped me, mostly, but otherwise, an enjoyable read.

Made a light (no baking) chocolate tart, with home raspberries on top (of course) that did not last long.

Watched two Japanese shows: Any crybabies around?! by Takuma Satô which revolves around the Namahage tradition in Northern Japan (to terrify children into being obedient and no crybabies!) and the immaturity of a young father acting as such a character until disaster strikes. With a lot of cringe moments, until the utter hopelessness of this man crybaby, more straw-like than his traditional costume made me stop caring. And the mini-series Switched. Which explores a (paranormal) body switch between two teenager girls to school pressure, bullying, and depression, but in a rather perturbing manner as the girl who initiated and forced the exchange does not come out nicely, despite her overweight issues, her abusive single mother, and the attitude of the rest of the school.  The most interesting character is the other schoolgirl who has to adapt to this situation without changing her (inner) personality, but the story is slow-motioned, predictable, and heavy-handed, esp. in the sobbing department. (Plus bordering at fat-shaming at some point.)

## ICMS supports mirrors

Posted in Travel, University life with tags , , , , on December 6, 2021 by xi'an

Received an announcement from the International Centre for Mathematical Sciences (ICMS) in Edinburgh that they will support mirror meetings (albeit in the United Kindgom) as well as other inclusive initiatives:

• ICMS@: this programme allows organisers to hold ‘satellite events’, ICMS-funded activity at venues elsewhere in the UK with logistic support by the ICMS staff. The aim will be to enable more activities at a high level distributed throughout the country and to facilitate participation by those who cannot easily travel. There will be an additional emphasis on regions that have found it difficult to fund local events in the past.

• Funds to allow participants at ICMS workshops to extend their visits in the UK, especially for the purpose of visiting other institutions and engaging in extended research interaction.

• A visitor programme for researchers from low- and middle-income countries to come to the UK to attend workshops and fund their stay for up to three months.

• Workshops or schools for postgraduate students and early career researchers which can be of varying lengths and intensities.

• A fund to help people with caring responsibilities attend our events.

## northern gannets rock

Posted in Kids, pictures, Travel with tags , , , , , , , , , , , on May 9, 2021 by xi'an


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LP-Swap

# How slippage is calculated

The intuition behind slippage is the following: if you try to buy big quantities in a market that has no liquidity, then you will have to pay more than the expected price (see this article for more details). Here we focus on the slippage occurring when swapping tokens on decentralised exchanges, that is when using liquidity pools.

So let us say that you want to swap $q_{input}$ BUSD tokens for WBNB tokens. For this you will use a BUSD-WBNB pool containing quantities $x^{\mathrm{BUSD}}$ and $x^{\mathrm{WBNB}}$ of both tokens before the swap. Recall from the swap formula that the real quantity $q_{output}^{real}$of WBNB tokens you get from the pool is given by

$q_{output}^{real}=\frac{(1-r)\times q_{input} \times x^{\mathrm{WBNB}} }{x^{\mathrm{BUSD}}+(1-r)\times q_{input}},$

where $0\leq r \leq 1$ is the fee rate of the pool used to pay to liquidity providers (typically it equals 0.002). However for very small input $q_{input}$ of BUSD tokens compared to the sizes of the pools, we obtain an ouput of

$q_{output}^{ideal}=(1-r)\times q_{input}\times x^{\mathrm{WBNB}}/x^{\mathrm{BUSD}} \text{ WBNB tokens.}$

The slippage of the transaction is the amount of WBNB tokens that you loose compared to buying all the WBNB tokens at the ideal price:

$q^{real}_{output}-q_{output}^{ideal}=-\frac{(1-r)\times q_{input}^2\times x^{\mathrm{WBNB}}}{ (x^{\mathrm{BUSD}}\times [x^{\mathrm{BUSD}}+q_{input}])}<0.$

The result is quite intuitive: we see for instance that the bigger the quantity $q_{input}$ of BUSD you swap, the higher the slippage. Besides, the bigger the quantity $x^{\mathrm{BUSD}}$ of BUSD already in the pool, the less significative your contribution to this pool and the impact of your swap, and so the lower the slippage. Similarly, the bigger the quantity $x^{\mathrm{WBNB}}$ of WBNB already in the pool, or rather the bigger the pool ratio $x^{\mathrm{WBNB}}/x^{\mathrm{BUSD}}$ of a BUSD token expressed in WBNB, the more WBNB tokens are withdrawn from the pool because of your swap, and so the higher the slippage.

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The top width and the top depth of flow in a triangular channel were measured as $4$ m and $1$ m, respectively. The measured velocities on the centre line at the water surface, $0.2$ m and $0.8$ m below the surface are $0.7$ m/s, $0.6$ m/s and $0.4$ m/s, respectively. Using two-point method of velocity measurement, the discharge (in $m^3/s$) in the channel is

1. $1.4$
2. $1.2$
3. $1.0$
4. $0.8$
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Check out our latest project: Algebrarules.com. The Rules of Algebra, Sweet 'n Simple. X

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By PiThagoras

Tags: Long Division, Division, Quotient, Remainder, Divisor,...

1

Posted: 2013-04-09
Long Division Procedure
Long Division Procedure

1. When dividing two numbers, for example, n divided by m, n is the dividend and m is the divisor; the answer is the quotient.
2. Find the location of all decimal points in the dividend and divisor.
3. If necessary, simplify the long division problem by moving the decimals of the divisor and dividend by the same number of decimal places, to the right, (or to the left) so that the decimal of the divisor is to the right of the last digit.
4. When doing long division, keep the numbers lined up straight from top to bottom under the tableau.
5. After each step, be sure the remainder for that step is less than the divisor. If it is not, there are three possible problems: the multiplication is wrong, the subtraction is wrong, or a greater quotient is needed.
6. In the end, the remainder, r, is added to the growing quotient as a fraction, $r/m$

More info can be found on Wikipedia: http://en.wikipedia.org/wiki/Long_division


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Introduction to CFD Questions & Answers
 Question by Student 201227147 Professor, in the section 2 of the table (Euler Equation in Generalized Coordinates), it says that $Q \equiv \Omega U$. But I guess it should be $Q \equiv \Omega \Gamma U$.
 06.06.17
True, but $\Gamma=1$, so it doesnt' matter in this case. 1 point bonus.
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# probability that 3rd head occurred on 9th trial

If it is known that in a series of coin tosses, the 4th head occurred on the 12th trial, what is the probability that the 3rd head occurred on the 9th trial?

Here's what I did, but I don't know if it's right:

$$\text{total outcomes with 12^\text{th} trial being 4^\text{th} head} = {11 \choose 3} = 165$$

$$\text{total outcomes with 9^\text{th} trial being 3^{\text{rd}} head} = {8 \choose 2}= 28$$

$$P(\text{3^\text{rd} head on 9^\text{th} trial \mid 4^\text{th} head on 12^\text{th} trial}) = \frac{28}{165}$$

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There were 2 heads in the first 8 trials; the 9th trial was a head; the 10th and 11th trails were tails. Analyzing it that way, can you work out the probability? – Gerry Myerson Jan 23 '13 at 5:39

Let us use the ordinary conditional probability formalism. Let $A$ be the event the fourth head occurred on the $12$th trial, and let $B$ be the event the third head occurred on the $9$-th trial. We want $\Pr(B|A)$. We have as usual $$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$ Now compute the probabilities on the right.

The $4$-th head occurs on the $12$th trial if there are $3$ heads in the first $11$, and then a head. The probability of this is $\dbinom{11}{3}(1/2)^{11}(1/2)$.

Now we find $\Pr(A\cap B)$. The probability of $B$ is $\dbinom{8}{2}(1/2)^9$. Given that $B$ happened, the probability of $A$ is the probability that in coin tossing, the first head occurs on the third trial. This is just the probability of TTH, which is $(1/2)^3$. Thus $\Pr(A\cap B)=\dbinom{8}{2}(1/2)^{12}$.

Remark: When we do the division, the powers of $1/2$ cancel, and we get precisely your answer. That answer was obtained in a combinatorially nicer way, but there is a certain amount of security in using standard conditional probability machinery.

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