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# Optical quorum cycles for efficient communication

Optical quorum cycles for efficient communication Many optical networks face heterogeneous communication requests requiring topologies to be efficient and fault tolerant. For efficiency and distributed control, it is common in distributed systems and algorithms to group nodes into intersecting sets referred to as quorum sets. We show efficiency and distributed control can also be accomplished in optical network routing by applying the same established quorum set theory. Cycle-based optical network routing, whether using SONET rings or p-cycles, provides the sufficient reliability in the network. Light-trails forming a cycle allow broadcasts within a cycle to be used for efficient multicasts. Cyclic quorum sets also have all pairs of nodes occurring in one or more quorums, so efficient, arbitrary unicast communication can occur between any two nodes. Efficient broadcasts to all network nodes are possible by a node broadcasting to all quorum cycles to which it belongs ( $$O(\sqrt{N})$$ O ( N ) ). In this paper, we propose applying the distributed efficiency of the quorum sets to routing optical cycles based on light-trails. With this new method of topology construction, unicast and multicast communication requests do not need to be known or even modeled a priori. Additionally, in the presence of network link faults, greater than 99 % average coverage enables the continued operation of nearly all arbitrary unicast and multicast requests in the network. Finally, to further improve the fault coverage, an augmentation to the ECBRA cycle finding algorithm is proposed. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Photonic Network Communications Springer Journals

# Optical quorum cycles for efficient communication

, Volume 31 (2) – Sep 4, 2015
10 pages

/lp/springer_journal/optical-quorum-cycles-for-efficient-communication-RbSe20rp60
Publisher
Springer US
Subject
Computer Science; Computer Communication Networks; Electrical Engineering; Characterization and Evaluation of Materials
ISSN
1387-974X
eISSN
1572-8188
D.O.I.
10.1007/s11107-015-0561-8
Publisher site
See Article on Publisher Site

### Abstract

Many optical networks face heterogeneous communication requests requiring topologies to be efficient and fault tolerant. For efficiency and distributed control, it is common in distributed systems and algorithms to group nodes into intersecting sets referred to as quorum sets. We show efficiency and distributed control can also be accomplished in optical network routing by applying the same established quorum set theory. Cycle-based optical network routing, whether using SONET rings or p-cycles, provides the sufficient reliability in the network. Light-trails forming a cycle allow broadcasts within a cycle to be used for efficient multicasts. Cyclic quorum sets also have all pairs of nodes occurring in one or more quorums, so efficient, arbitrary unicast communication can occur between any two nodes. Efficient broadcasts to all network nodes are possible by a node broadcasting to all quorum cycles to which it belongs ( $$O(\sqrt{N})$$ O ( N ) ). In this paper, we propose applying the distributed efficiency of the quorum sets to routing optical cycles based on light-trails. With this new method of topology construction, unicast and multicast communication requests do not need to be known or even modeled a priori. Additionally, in the presence of network link faults, greater than 99 % average coverage enables the continued operation of nearly all arbitrary unicast and multicast requests in the network. Finally, to further improve the fault coverage, an augmentation to the ECBRA cycle finding algorithm is proposed.

### Journal

Photonic Network CommunicationsSpringer Journals

Published: Sep 4, 2015

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# Re: python plotting with greek symbols within labels recommendations?

On 4 October 2005 at 19:48, Michael Gilbert wrote:
| i am a Debian user looking for a python plotting library that can support
| rendering greek symbols and other various characters on plot axes labels,
| etc. i would prefer something that adheres to tex formatting (as implemented
| in latex, matlab, etc and has the form $\alpha$ to represent the
| greekcharacter alpha for example).

You could consider R which has a lot of plotting functionality, including
almost arbitrary math symbol and various alphabets -- after 'apt-get install
r-base-core' try
\$ R
> demo(plotmath)
> demo(graphics)
> demo(image)
> demo(persp)
> demo(Hershey)
> q("no")

Moreover, R can be driven quite well from Python using the python-rpy
package(s), so try 'apt-get install python-rpy'.

(Note that unstable already has the last beta of R 2.2.0 which is to be
released on Thursday, and unstable's rpy has not yet been rebuilt for
it. Testing should work just fine.)

Hth, Dirk

--
Statistics: The (futile) attempt to offer certainty about uncertainty.
-- Roger Koenker, 'Dictionary of Received Ideas of Statistics'

format: plain


text:
Sibirskii Matematicheskii Zhurnal
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 Sibirsk. Mat. Zh.: Year: Volume: Issue: Page: Find

 Sibirsk. Mat. Zh., 2006, Volume 47, Number 6, Pages 1265–1274 (Mi smj932)

On the exact distributional asymptotics for the supremum of a random walk with increments in a class of light-tailed distributions

S. Zacharya, S. G. Fossb

a Heriot Watt University
b Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences

Abstract: We study the distribution of the maximum $M$ of a random walk whose increments have a distribution with negative mean which belongs for some $\gamma>0$ to a subclass of the class $\mathscr S_\gamma$ (for example, see Chover, Ney, and Wainger [5]). For this subclass we provide a probabilistic derivation of the asymptotic tail distribution of $M$ and show that the extreme values of $M$ are in general attained through some single large increment in the random walk near the beginning of its trajectory. We also give some results concerning the “spatially local” asymptotics of the distribution of $M$, the maximum of the stopped random walk for various stopping times, and various bounds.

Keywords: supremum of random walk, exact asymptotics, ruin probability

Full text: PDF file (230 kB)
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English version:
Siberian Mathematical Journal, 2006, 47:6, 1034–1041

Bibliographic databases:

UDC: 519.21

Citation: S. Zachary, S. G. Foss, “On the exact distributional asymptotics for the supremum of a random walk with increments in a class of light-tailed distributions”, Sibirsk. Mat. Zh., 47:6 (2006), 1265–1274; Siberian Math. J., 47:6 (2006), 1034–1041

Citation in format AMSBIB
\Bibitem{ZacFos06} \by S.~Zachary, S.~G.~Foss \paper On the exact distributional asymptotics for the supremum of a~random walk with increments in a~class of light-tailed distributions \jour Sibirsk. Mat. Zh. \yr 2006 \vol 47 \issue 6 \pages 1265--1274 \mathnet{http://mi.mathnet.ru/smj932} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=2302843} \zmath{https://zbmath.org/?q=an:1150.60380} \transl \jour Siberian Math. J. \yr 2006 \vol 47 \issue 6 \pages 1034--1041 \crossref{https://doi.org/10.1007/s11202-006-0112-8} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=000243454700005} \scopus{https://www.scopus.com/record/display.url?origin=inward&eid=2-s2.0-33845504874}

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Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. S. G. Foss, “On Exact Asymptotics for a Stationary Sojourn Time Distribution in a Tandem of Queues with Light-Tailed Service Times”, Problems Inform. Transmission, 43:4 (2007), 353–366
2. Foss S.G., Puhalskii A.A., “On the limit law of a random walk conditioned to reach a high level”, Stochastic Process Appl, 121:2 (2011), 288–313
3. Wang Yu., Wang K., “Random walks with non-convolution equivalent increments and their applications”, J Math Anal Appl, 374:1 (2011), 88–105
4. L. V. Rozovskii, “Ob asimptotike svertki raspredelenii s regulyarno eksponentsialno ubyvayuschimi khvostami”, Veroyatnost i statistika. 28, Zap. nauchn. sem. POMI, 486, POMI, SPb., 2019, 265–274
5. A. A. Borovkov, “Granichnye zadachi dlya obobschennykh protsessov vosstanovleniya”, Sib. matem. zhurn., 61:1 (2020), 29–59
6. Rozovsky V L., “on Large Deviations of a Sum of Independent Random Variables With Rapidly Decreasing Distribution Tails”, Dokl. Math., 101:2 (2020), 150–153
•  Number of views: This page: 245 Full text: 87 References: 50


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4.4: Examining the Central Limit Theorem

The normal model for the sample mean tends to be very good when the sample consists of at least 30 independent observations and the population data are not strongly skewed. The Central Limit Theorem provides the theory that allows us to make this assumption.

Central Limit Theorem - informal definition

The distribution of $$\bar {x}$$ is approximately normal. The approximation can be poor if the sample size is small, but it improves with larger sample sizes.

The Central Limit Theorem states that when the sample size is small, the normal approximation may not be very good. However, as the sample size becomes large, the normal approximation improves. We will investigate three cases to see roughly when the approximation is reasonable.

We consider three data sets: one from a uniform distribution, one from an exponential distribution, and the other from a log-normal distribution. These distributions are shown in the top panels of Figure 4.20. The uniform distribution is symmetric, the exponential distribution may be considered as having moderate skew since its right tail is relatively short (few outliers), and the log-normal distribution is strongly skewed and will tend to produce more apparent outliers.

The left panel in the n = 2 row represents the sampling distribution of $$\bar {x}$$ if it is the sample mean of two observations from the uniform distribution shown. The dashed line represents the closest approximation of the normal distribution. Similarly, the center and right panels of the n = 2 row represent the respective distributions of $$\bar {x}$$ for data from exponential and log-normal distributions.

31Here the null hypothesis is that the part is not broken, and the alternative is that it is broken. If we don't have sufficient evidence to reject H0, we would not replace the part. It sounds like failing to x the part if it is broken (H0 false, HA true) is not very problematic, and replacing the part is expensive. Thus, we should require very strong evidence against H0 before we replace the part. Choose a small significance level, such as $$\alpha = 0.01$$.

32The normal approximation becomes better as larger samples are used.

Figure 4.20: Sampling distributions for the mean at different sample sizes and for three different distributions. The dashed red lines show normal distributions.

Exercise

Exercise 4.41 Examine the distributions in each row of Figure 4.20. What do you notice about the normal approximation for each sampling distribution as the sample size becomes larger?33

Exercise

Example 4.42 Would the normal approximation be good in all applications where the sample size is at least 30?

Not necessarily. For example, the normal approximation for the log-normal example is questionable for a sample size of 30. Generally, the more skewed a population distribution or the more common the frequency of outliers, the larger the sample required to guarantee the distribution of the sample mean is nearly normal.

TIP: With larger n, the sampling distribution of $$\bar {x}$$ becomes more normal

As the sample size increases, the normal model for $$\bar {x}$$ becomes more reasonable. We can also relax our condition on skew when the sample size is very large.

We discussed in Section 4.1.3 that the sample standard deviation, s, could be used as a substitute of the population standard deviation, $$\sigma$$, when computing the standard error. This estimate tends to be reasonable when $$n \le 30$$. We will encounter alternative distributions for smaller sample sizes in Chapters 5 and 6.

Example 4.43 Figure 4.21 shows a histogram of 50 observations. These represent winnings and losses from 50 consecutive days of a professional poker player. Can the normal approximation be applied to the sample mean, 90.69?

We should consider each of the required conditions.

1. These are referred to as time series data, because the data arrived in a particular sequence. If the player wins on one day, it may inuence how she plays the next. To make the assumption of independence we should perform careful checks on such data. While the supporting analysis is not shown, no evidence was found to indicate the observations are not independent.
2. The sample size is 50, satisfying the sample size condition.
3. There are two outliers, one very extreme, which suggests the data are very strongly skewed or very distant outliers may be common for this type of data. Outliers can play an important role and affect the distribution of the sample mean and the estimate of the standard error.

Since we should be skeptical of the independence of observations and the very extreme upper outlier poses a challenge, we should not use the normal model for the sample mean of these 50 observations. If we can obtain a much larger sample, perhaps several hundred observations, then the concerns about skew and outliers would no longer apply.

Caution: Examine data structure when considering independence

Some data sets are collected in such a way that they have a natural underlying structure between observations, e.g. when observations occur consecutively. Be especially cautious about independence assumptions regarding such data sets.

Figure 4.21: Sample distribution of poker winnings. These data include some very clear outliers. These are problematic when considering the normality of the sample mean. For example, outliers are often an indicator of very strong skew.

Caution: Watch out for strong skew and outliers

Strong skew is often identi ed by the presence of clear outliers. If a data set has prominent outliers, or such observations are somewhat common for the type of data under study, then it is useful to collect a sample with many more than 30 observations if the normal model will be used for $$\bar {x}$$. There are no simple guidelines for what sample size is big enough for all situations, so proceed with caution when working in the presence of strong skew or more extreme outliers.

You won’t be a pro at assessing skew by the end of this book, so just use your best judgement and continue learning. As you develop your statistics skills and encounter tough situations, also consider learning about better ways to analyze skewed data, such as the studentized bootstrap (bootstrap-t), or consult a more experienced statistician.

Contributors

David M Diez (Google/YouTube), Christopher D Barr (Harvard School of Public Health), Mine Çetinkaya-Rundel (Duke University)


text:
## scalar quantities vs. vector quantities

Is it correct to say that scalar quantities are abstract, idealized quantities with no precise counterparts in the physical world, whereas vector quantities properly represent reality because they take directions into account?


text:
# Square root of 5

 Binary 10.0011110001101111... Decimal 2.23606797749978969... Hexadecimal 2.3C6EF372FE94F82C... Continued fraction $2 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4 + \cfrac{1}{4 + \ddots}}}}$

The square root of 5 is the positive real number that, when multiplied by itself, gives the prime number 5. It is more precisely called the principal square root of 5, to distinguish it from the negative number with the same property. This number appears in the fractional expression for the golden ratio. It can be denoted in surd form as:

$\sqrt{5}. \,$

It is an irrational algebraic number.[1] The first sixty significant digits of its decimal expansion are:

2.23606 79774 99789 69640 91736 68731 27623 54406 18359 61152 57242 7089... (sequence A002163 in OEIS).

which can be rounded down to 2.236 to within 99.99% accuracy. As of December 2013, its numerical value in decimal has been computed to at least ten billion digits.[2]

## Proof of irrationality

This irrationality proof for the square root of 5 uses Fermat's method of infinite descent:

Suppose that √5 is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as $\frac{m}{n}$ for natural numbers m and n. Then √5 can be expressed in lower terms as $\frac{5n-2m}{m-2n}$, which is a contradiction.[3] (The two fractional expressions are equal because equating them, cross-multiplying, and canceling like additive terms gives $5n^2=m^2$ and hence $\tfrac{m}{n}=\sqrt{5}$, which is true by the premise. The second fractional expression for √5 is in lower terms since, comparing denominators, $m-2n since $m<3n$ since $\tfrac{m}{n}<3$ since $\sqrt{5}<3$. And both the numerator and the denominator of the second fractional expression are positive since $2<\sqrt{5}<\tfrac{5}{2}$ and $\tfrac{m}{n}=\sqrt{5}$.)

## Continued fraction

It can be expressed as the continued fraction [2; 4, 4, 4, 4, 4...] (sequence A040002 in OEIS). The sequence of best rational approximations is:

${\color{LimeGreen}{\frac{2}{1}}}, \frac{7}{3} , {\color{LimeGreen}{\frac{9}{4}}} , \frac{20}{9} , \frac{29}{13} , {\color{LimeGreen}{\frac{38}{17}}} , \frac{123}{55} , {\color{LimeGreen}{\frac{161}{72}}} , \frac{360}{161} , \frac{521}{233} , {\color{LimeGreen}{\frac{682}{305}}} , \frac{2207}{987} , {\color{LimeGreen}{\frac{2889}{1292}}}, \dots$

Convergents of the continued fraction are colored; their numerators are 2, 9, 38, 161, ... (sequence A001077 in OEIS), and their denominators are 1, 4, 17, 72, ... (sequence A001076 in OEIS). The other (non-colored) terms are semiconvergents.

## Babylonian method

When $\sqrt{5}$ is computed with the Babylonian method, starting with r0 = 2 and using rn+1 = (rn + 5/rn) / 2, the nth approximant rn is equal to the 2n-th convergent of the convergent sequence:

$\frac{2}{1} = 2.0,\quad \frac{9}{4} = 2.25,\quad \frac{161}{72} = 2.23611\dots,\quad \frac{51841}{23184} = 2.2360679779 \ldots$

## Relation to the golden ratio and Fibonacci numbers

The √52 diagonal of a half square forms the basis for the geometrical construction of a golden rectangle.

This golden ratio $\varphi$ is the arithmetic mean of 1 and the square root of 5.[4] The algebraic relationship between the square root of 5, the golden ratio and the conjugate of the golden ratio ($\Phi = 1 / \varphi = \varphi - 1$) are expressed in the following formulae:

$\sqrt{5} = \varphi + \Phi = 2\varphi - 1 = 2\Phi + 1$
$\varphi = \frac{\sqrt{5} + 1}{2}$
$\Phi = \frac{\sqrt{5} - 1}{2}.$

(See section below for their geometrical interpretation as decompositions of a root-5 rectangle.)

The square root of 5 then naturally figures in the closed form expression for the Fibonacci numbers, a formula which is usually written in terms of the golden ratio:

$F\left(n\right) = {{\varphi^n-(1-\varphi)^n} \over {\sqrt 5}}.$

The quotient of √5 and $\varphi$ (or the product of √5 and $\Phi$), and its reciprocal, provide an interesting pattern of continued fractions and are related to the ratios between the Fibonacci numbers and the Lucas numbers:[5]

$\frac{\sqrt{5}}{\varphi} = \Phi \cdot \sqrt{5} = \frac{5 - \sqrt{5}}{2} = 1.3819660112501051518\dots = [1; 2, 1, 1, 1, 1, 1, 1, 1, \dots]$
$\frac{\varphi}{\sqrt{5}} = \frac{1}{\Phi \cdot \sqrt{5}} = \frac{5 + \sqrt{5}}{10} = 0.72360679774997896964\dots = [0; 1, 2, 1, 1, 1, 1, 1, 1, \dots].$

The series of convergents to these values feature the series of Fibonacci numbers and the series of Lucas numbers as numerators and denominators, and viceversa, respectively:

${1, \frac{3}{2}, \frac{4}{3}, \frac{7}{5}, \frac{11}{8}, \frac{18}{13}, \frac{29}{21}, \frac{47}{34}, \frac{76}{55}, \frac{123}{89}}, \dots \dots [1; 2, 1, 1, 1, 1, 1, 1, 1, \dots]$
${1, \frac{2}{3}, \frac{3}{4}, \frac{5}{7}, \frac{8}{11}, \frac{13}{18}, \frac{21}{29}, \frac{34}{47}, \frac{55}{76}, \frac{89}{123}}, \dots \dots [0; 1, 2, 1, 1, 1, 1, 1, 1,\dots].$

## Geometry

Conway triangle decomposition into homothetic smaller triangles.

Geometrically, the square root of 5 corresponds to the diagonal of a rectangle whose sides are of length 1 and 2, as is evident from the Pythagorean theorem. Such a rectangle can be obtained by halving a square, or by placing two equal squares side by side. Together with the algebraic relationship between √5 and φ, this forms the basis for the geometrical construction of a golden rectangle from a square, and for the construction of a regular pentagon given its side (since the side-to-diagonal ratio in a regular pentagon is φ).

Forming a dihedral right angle with the two equal squares that halve a 1:2 rectangle, it can be seen that √5 corresponds also to the ratio between the length of a cube edge and the shortest distance from one of its vertices to the opposite one, when traversing the cube surface (the shortest distance when traversing through the inside of the cube corresponds to the length of the cube diagonal, which is the square root of three times the edge).[citation needed]

The number √5 can be algebraically and geometrically related to the square root of 2 and the square root of 3, as it is the length of the hypotenuse of a right triangle with catheti measuring √2 and √3 (again, the Pythagorean theorem proves this). Right triangles of such proportions can be found inside a cube: the sides of any triangle defined by the centre point of a cube, one of its vertices, and the middle point of a side located on one the faces containing that vertex and opposite to it, are in the ratio √2:√3:√5. This follows from the geometrical relationships between a cube and the quantities √2 (edge-to-face-diagonal ratio, or distance between opposite edges), √3 (edge-to-cube-diagonal ratio) and √5 (the relationship just mentioned above).

A rectangle with side proportions 1:√5 is called a root-five rectangle and is part of the series of root rectangles, a subset of dynamic rectangles, which are based on √1 (= 1), √2, √3, √4 (= 2), √5... and successively constructed using the diagonal of the previous root rectangle, starting from a square.[6] A root-5 rectangle is particularly notable in that it can be split into a square and two equal golden rectangles (of dimensions Φ × 1), or into two golden rectangles of different sizes (of dimensions Φ × 1 and 1 × φ).[7] It can also be decomposed as the union of two equal golden rectangles (of dimensions 1 × φ) whose intersection forms a square. All this is can be seen as the geometric interpretation of the algebraic relationships between √5, φ and Φ mentioned above. The root-5 rectangle can be constructed from a 1:2 rectangle (the root-4 rectangle), or directly from a square in a manner similar to the one for the golden rectangle shown in the illustration, but extending the arc of length √52 to both sides.

## Trigonometry

Like √2 and √3, the square root of 5 appears extensively in the formulae for exact trigonometric constants, including in the sines and cosines of every angle whose measure in degrees is divisible by 3 but not by 15.[8] The simplest of these are

$\sin\frac{\pi}{10}=\sin 18^\circ=\tfrac{1}{4}(\sqrt5-1)=\frac{1}{\sqrt5+1},\,$
$\sin\frac{\pi}{5}=\sin 36^\circ=\tfrac{1}{4}\sqrt{2(5-\sqrt5)},\,$
$\sin\frac{3\pi}{10}=\sin 54^\circ=\tfrac{1}{4}(\sqrt5+1)=\frac{1}{\sqrt5-1},\,$
$\sin\frac{2\pi}{5}=\sin 72^\circ=\tfrac{1}{4}\sqrt{2(5+\sqrt5)}\, .$

As such the computation of its value is important for generating trigonometric tables.[citation needed] Since √5 is geometrically linked to half-square rectangles and to pentagons, it also appears frequently in formulae for the geometric properties of figures derived from them, such as in the formula for the volume of a dodecahedron.[citation needed]

## Diophantine approximations

Hurwitz's theorem in Diophantine approximations states that every irrational number x can be approximated by infinitely many rational numbers m/n in lowest terms in such a way that

$\left|x - \frac{m}{n}\right| < \frac{1}{\sqrt{5}\,n^2}$

and that √5 is best possible, in the sense that for any larger constant than √5, there are some irrational numbers x for which only finitely many such approximations exist.[9]

Closely related to this is the theorem[10] that of any three consecutive convergents pi/qi, pi+1/qi+1, pi+2/qi+2, of a number α, at least one of the three inequalities holds:

$\left|\alpha - {p_i\over q_i}\right| < {1\over \sqrt5 q_i^2}, \qquad \left|\alpha - {p_{i+1}\over q_{i+1}}\right| < {1\over \sqrt5 q_{i+1}^2}, \qquad \left|\alpha - {p_{i+2}\over q_{i+2}}\right| < {1\over \sqrt5 q_{i+2}^2}.$

And the √5 in the denominator is the best bound possible since the convergents of the golden ratio make the difference on the left-hand side arbitrarily close to the value on the right-hand side. In particular, one cannot obtain a tighter bound by considering sequences of four or more consecutive convergents.[10]

## Algebra

The ring $\scriptstyle\mathbb{Z}\left[\,\sqrt{-5}\,\right]$ contains numbers of the form $\scriptstyle a\, +\, b\sqrt{-5}$, where a and b are integers and $\scriptstyle \sqrt{-5}$ is the imaginary number $\scriptstyle i\sqrt{5}$. This ring is a frequently cited example of an integral domain that is not a unique factorization domain.[citation needed] The number 6 has two inequivalent factorizations within this ring:

$6 = 2 \cdot 3 = (1 - \sqrt{-5})(1 + \sqrt{-5}). \,$

The field $\scriptstyle\mathbb{Q}\left[\,\sqrt{5}\,\right]$, like any other quadratic field, is an abelian extension of the rational numbers. The Kronecker–Weber theorem therefore guarantees that the square root of five can be written as a rational linear combination of roots of unity:

$\sqrt5 = e^{2\pi i/5} - e^{4\pi i/5} - e^{6\pi i/5} + e^{8\pi i/5}. \,$

## Identities of Ramanujan

The square root of 5 appears in various identities of Ramanujan involving continued fractions.[11][12]

For example, this case of the Rogers–Ramanujan continued fraction:

$\cfrac{1}{1 + \cfrac{e^{-2\pi}}{1 + \cfrac{e^{-4\pi}}{1 + \cfrac{e^{-6\pi}}{1 + \ddots}}}} = \left( \sqrt{\frac{5 + \sqrt{5}}{2}} - \frac{\sqrt{5} + 1}{2} \right)e^{2\pi/5} = e^{2\pi/5}\left( \sqrt{\varphi\sqrt{5}} - \varphi \right).$

$\cfrac{1}{1 + \cfrac{e^{-2\pi\sqrt{5}}}{1 + \cfrac{e^{-4\pi\sqrt{5}}}{1 + \cfrac{e^{-6\pi\sqrt{5}}}{1 + \ddots}}}} = \left( {\sqrt{5} \over 1 + \left[5^{3/4}(\varphi - 1)^{5/2} - 1\right]^{1/5}} - \varphi \right)e^{2\pi/\sqrt{5}}.$

$4\int_0^\infty\frac{xe^{-x\sqrt{5}}}{\cosh x}\,dx = \cfrac{1}{1 + \cfrac{1^2}{1 + \cfrac{1^2}{1 + \cfrac{2^2}{1 + \cfrac{2^2}{1 + \cfrac{3^2}{1 + \cfrac{3^2}{1 + \ddots}}}}}}}.$

## References

1. ^ Dauben, Joseph W. (June 1983) Scientific American Georg Cantor and the origins of transfinite set theory. Volume 248; Page 122.
2. ^ Lukasz Komsta: Computations page
3. ^ Grant, Mike, and Perella, Malcolm, "Descending to the irrational", Mathematical Gazette 83, July 1999, pp.263-267.
4. ^ Browne, Malcolm W. (July 30, 1985) New York Times Puzzling Crystals Plunge Scientists into Uncertainty. Section: C; Page 1. (Note - this is a widely cited article).
5. ^ Richard K. Guy: "The Strong Law of Small Numbers". American Mathematical Monthly, vol. 95, 1988, pp. 675–712
6. ^ Kimberly Elam (2001), Geometry of Design: Studies in Proportion and Composition, New York: Princeton Architectural Press, ISBN 1-56898-249-6
7. ^ Jay Hambidge (1967), The Elements of Dynamic Symmetry, Courier Dover Publications, ISBN 0-486-21776-0
8. ^ Julian D. A. Wiseman, "Sin and cos in surds" [1]
9. ^ LeVeque, William Judson (1956), Topics in number theory, Addison-Wesley Publishing Co., Inc., Reading, Mass., MR 0080682
10. ^ a b Khinchin, Aleksandr Yakovlevich (1964), Continued Fractions, University of Chicago Press, Chicago and London
11. ^ Ramanathan, K. G. (1984), "On the Rogers-Ramanujan continued fraction", Indian Academy of Sciences. Proceedings. Mathematical Sciences 93 (2): 67–77, doi:10.1007/BF02840651, ISSN 0253-4142, MR 813071
12. ^ Eric W. Weisstein, Ramanujan Continued Fractions at MathWorld


text:
# Nach langer Pause mal wieder aktiv

Nach drei Jahren mal wieder ein neuer Post. Eigentlich nur, um auszuprobieren, ob ich aus R heraus posten kann, und wie dass dann aussieht.

summary(cars)

##      speed           dist
##  Min.   : 4.0   Min.   :  2
##  1st Qu.:12.0   1st Qu.: 26
##  Median :15.0   Median : 36
##  Mean   :15.4   Mean   : 43
##  3rd Qu.:19.0   3rd Qu.: 56
##  Max.   :25.0   Max.   :120

You can also embed plots, for example:

plot(cars)

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## Rätsel – Auflösung

Auch wenn das Feeback überschaubar war, möchte ich doch das Rätsel auch auflösen, dass ich in Ein Rätsel gestellt habe: Bei dem Plot handelt es sich um die Tagesrenditen des DAX, die entsprechend ihrer Größe sortiert, anschließend in Absolutwerte umgewandelt und logarithmiert wurden.

Hier der R-Code:

?View Code RSPLUS
 1 2 3 4 5 6 7  library(quantmod)   getSymbols("^GDAXI", src="yahoo", from=1991-01-02)   sort.rets <- sort(as.vector(dailyReturn(GDAXI)))*100 l10.sort.rets <- log(abs(sort.rets),base=10) plot(l10.sort.rets,ylab="",type="l")
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## Außenhandel seit 1950

Als Antwort darauf, dass langfristig keine Exportüberschüsse erzielt werden können.

Einheit: Exportüberschuss in % der Importe. Quelle: UNCTAD

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## Quod licet Iovi, non licet bovi.

Die Mortgage Banks Association in den USA hat eine PR-Kampagen, die betont, das Immobilienbesitzer, deren Eigentum nicht mehr so viel wert ist wie die ausstehende Hypothek, eine moralische Verpflichtung hätten, auf die Option der strategischen Insolvenz zu verzichten. Jon Stewart zeigt, dass die MBA das selbst gar nicht so eng sieht.

The Daily Show With Jon StewartMon – Thurs 11p / 10c
Mortgage Bankers Association Strategic Default
www.thedailyshow.com
 Daily Show Full Episodes Political Humor Rally to Restore Sanity

Tja, soll man jetzt weinen oder lachen?

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## Ein Interview von 1999

Ich bin zufällig über ein Interview mit Martin Mayer gestolpert, das von 1999 ist, und in dem Mayer damals schon die Schwachstellen der “Finanzmarktinnovationen” beschreibt, und deren Versagen im Krisenfall vorhersagt. Hier ein paar Ausschnitte:

That’s the secret of the JP Morgan lawsuit with the Korean banks. It was Korean Central Bank money that these guys pissed away. You had this incredible device—a Malaysian corporation formed to write instruments that would pay off according to the relative values of the Thai baht and Japanese yen, and designed solely for the purpose of selling these instruments to the Korean banks because the Korean Central Bank wanted to increase the return on its reserve.

Once you start to get into this sort of nonsense, serious problems can occur for everybody, because you are dealing in a world of thoroughly artificial instruments that have nothing to do with real production, real consumption, real trade or real anything. They are simply counters in a game. But when the players have to pay off, they have to pay off in real resources, and that sort of thing is extremely dangerous over time.

In October 1997, when the Brazilian real came under dramatic attack, nobody could figure out why. It turned out that in September ’97, between 20 percent and 25 percent of all the Brazilian Brady bonds were held by South Korean banks. The Brazilian Brady bonds had a large, liquid market because people were making a lot of money trading them and trading derivatives off of them. So when the South Korean banks got in trouble, the easiest thing for them to do was to sell Bradys. The bonds took a hit and the real took a hit. And that’s crazy.

It’s one thing to talk about a spillover effect with countries that are closely related to each other. It’s another thing to talk about contagion. There have been artificial links created between currencies because of the statistical correlation in their path behavior. That has nothing to do with anything. Correlations and causes can get mixed up anytime, but the notion that you put hard money on correlations where there is no—and can be no—causation other than the correlation itself is bad news.

What’s really happening is that the guys selling the hedging instruments are writing a form of insurance. It’s catastrophe insurance, but they don’t know it’s catastrophe insurance, so they have no way to price it that gives them accurate premiums for the risks. Nobody knows. The question is whether there shouldn’t be some regulator making sure that when the insurance doesn’t pay off, important players don’t go absolutely insane and bust. That’s got to be addressed.

Banks are supposed to do information-intensive lending. The notion that the ratings agencies know what the risks are on loans better than the banks do—well, that’s really remarkable. It means that everybody has adjusted to a world in which banks are not going to hold loans in their portfolios. They are going to sell them off. And the ratings are important for the purpose of selling off the loans.

We’re moving from a world that was dominated by specific information in banks to a world dominated by much less profound but much more widespread information in markets. The clash between these two systems of valuing instruments is what makes for the instability that we can see all around us and that will increase.

There’s obviously a real value in credit derivatives because they allow the farm-belt bank and the rust-belt bank to swap exposures. By diversifying, both of them have less risk. The reduction of their returns is minor because the cost of these contracts is minor.

In some cases, you’re not swapping portfolios of loans. You’re gambling on the relative return of a risk-free instrument against an equity portfolio, a portfolio with junk bonds, a portfolio of foreign currencies or whatever, in which there is a larger risk.Nobody knows what the hell is going on in these private deals, and nobody knows the volume of what is out there, including regulators. So nobody can have an informed opinion about what the dangers are. What has happened over and over again with these instruments is that people have assumed that there is measurable risk where in fact there is immeasurable uncertainty.

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text:
### 2006-08

#### Using rectangula

Series: 2006-08, Preprints

MSC:
65N30 Finite elements, ~Rayleigh-Ritz and Galerkin methods, finite methods
65N15 Error bounds

Abstract:
The streamline diffusion finite element method (SDFEM; the method
is also known as SUPG) is applied to a convection-diffusion
problem posed on the unit square whose solution has exponential
boundary layers. A rectangular Shishkin mesh is used. The trial
functions in the SDFEM are piecewise polynomials that lie in the
space $Q_p$, i.e., are tensor products of polynomials of degree
$p$ in one variable, where $p>1$. The error bound $\|I_N u-u^N\|_{SD}\le C N^{-(p+1/2)}$ is proved; here $u^N$ is the
computed SDFEM solution, $I_N u$ is chosen in the finite element
space to be a special approximant of the true solution $u$, and
$\|\cdot\|_{SD}$ is the streamline-diffusion norm. This result is
compared with previously known results for the case $p=1$. The
error bound is a superclose result; $u^N$ can be enhanced using
local postprocessing to yield a modified solution $\tilde u^N$ for
which $\|u-\tilde u^N\|_{SD}\le C N^{-(p+1/2)}$.

Keywords:
convection-diffusion problems, streamline-diffusion method, finite elements, error estimates


text:
# Tag Info

-1

L(angular momentum) = (moment of Inertia) x (angular velocity)

1

The angular momentum $L_{A/B}$ of a rigid body $A/B$ about its center of mass is $$L_{A/B} = I_{A/B} \omega_{A/B},$$ where $I_{A/B}$ is the inertia matrix of $A/B$ about its center of mass in the world frame and $\omega_{A/B}$ is the angular velocity of $A/B$. The angular momentum $L_{A/B}^0$ of a rigid body $A/B$ about the origin of the world frame is ...

0

Some bits and pieces on angular momentum: Angular momentum is that which is conserved in rotationally invariant systems, just like energy is that which is conserved in time translation invariant systems and momentum is that which is conserved in space translation invariant system. This is the essence of Noether's theorem. The analogue in QFTs are ...

0

If you are looking for quantitative values, look to further than the equations of motion \begin{align} \sum_{i} \left( \vec{F_i} \right) & = m \vec{a}_{cm} \\ \sum_i \left( \vec{\tau}_i + (\vec{r}_i-\vec{r}_{cm}) \times \vec{F}_i \right) & = I_{cm} \vec{\alpha} + \vec{\omega} \times I_{cm} \vec{\omega}\end{align} where the left hand side is ...

0

The angular equivalent of the impulse-momentum theorem states The change in angular momentum of a system is equal to the product of the (average) external torque time the time it is applies In math that is (finite version): $$\langle \vec{\tau}_{ext} \rangle \,\Delta t = \Delta \vec{L} \,,$$ or (infinitesimal version): $$\vec{\tau} \, \mathrm{d}t = ... 1 Some conservation laws are related to conservation of angular momentum. There is a famous example (from Feynman if I recall correctly), where you assume an infinite flat space and conservation of angular momentum about any point, and then you get conservation of linear momentum for free. Intuitively, to get say the x component of linear momentum is ... 0 Answer to first question: At the instant being considered, the space and body axes are identical, so at that moment the matrix a that relates the two sets of axes is simply the identity matrix. dG' is a vector, so a_{ji}dG_j' = dG_i' is simply equivalent to the statement that with I the identity matrix and V an arbitrary vector, IV=V. Answer ... 0 Electromagnetic field of light has two kind of angular momentum first spin angular momentum (SAM) and secondly orbital angular momentum (OAM). former one represent the dynamical rotation of electric (or magnetic) field of around propagation direction and indicate the polarization of beam. Later one represent the rotation of light around beam axes. The ... 0 By no means not a complete answer, more a criticism of @luksen’s one. It is posted here because the text is too long to fit in the comment field. First of all, the spin is not a well-defined concept for composite particles. More precisely, whether the spin of a particle is defined depends on how the “particle” is defined. Look at an atom: it has the nucleus ... 1 There is no law of physics that requires an external force to be present when some double time derivative of a parameter in a system changes. What you were taught in high school, that is$$F = m \cdot \dot v$$and$$M = I \cdot \dot \omega$$are actually the simplified version of$$F = \frac{d}{dt}m \cdot v = m \cdot \dot v + \dot m \cdot v$$and$$M ...

0

A simple picture here would be that dissipation leads loss of mechanical energy. But angular momentum has to be conserved. The lowest possible mechanical energy with a given value of the angular momentum of fluid inside a container is that of rigid body rotating around the axis of largest moment of inertia.

0

You apparently mean "simulate" when you used the word "model". You'll need two things to accomplish this: A better rotational integrator than the one presented in this answer to your other question, and A physical model of a system that loses energy while conserving angular momentum. Regarding the first item, that integrator is not bad. It has the ...

0

For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum $$p_0^\mu=(M,0,0,0).$$ Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ ...

1

The best way to understand spin is actually to consider the Dirac Equation $$i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi$$ or more compactly: $$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$ The solutions to the Dirac equation are collections of complex valued fields called spinors. The spinor solution actually ...

0

For what it's worth, I've always had the same feeling that the spin should have some sort of reason behind it. It seems so unsatisfying to be told more or less that "it just came that way." Is there any deeper sort of explanation at all? I recall a paper in AJP from years ago called "What is spin?" by Ohanian, but I didn't put in the effort to follow it. I ...

2

Spin is a wave property. It exists in classical relativistic wave theories as well. A circularly polarized wave carries an angular momentum that's related to the spin of the field. A gravitational wave (spin-2) can carry twice the angular momentum of a classical electromagnetic wave (spin-1). Being "pointlike" is a particle property. You can think of the ...

0

You are correct. The total angular momentum remains constant. As the objects are not approaching head-on, there must be angular momentum. That momentum remains in the joined object as it spins around its COM. The speed of the spin for the joined object will be proportional to the perpendicular distance and speed of approach, and inversely proportional to ...

14

Spin is not about stuff spinning. (Confusing, I know, but physicists have never been great at naming things. Exhibit A: Quarks.) Spin is a purely quantum mechanical phenomenon, it cannot be understood with classical physics alone, and every analogy will break down. It has also, intrinsically, nothing to do with any kind of internal structure. ...

6

"The electron has no known internal structure", but since it does have a spin, does that mean that we know the electron has an internal structure but we just don't know what it is? An electron has no known internal structure simply means that nobody knows if the electron has an internal structure. So far they know none and therefore they suppose it ...

0

Answer expected by following author's hints. \begin{align} \mathbf{L}_{eg} &= \dfrac{1}{4\pi}\int \mathbf{r'} \times \left [\mathbf{E} \times \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left (\mathbf{B.r'} \right)\mathbf{E} - \left (\mathbf{E . r'} \right) \mathbf{B} \right] d^3r'\\ & = \dfrac{1}{4\pi} \int \left [ \left ...

0

One does not need to switch "gravity off" to make a system like that. Replace gravity with a simple string, and cut the string. Physically that's completely equivalent to your problem, as far as I can see. And while this may seem counterintuitive, the angular momentum in a system of a mass rotating at the end of a string is, indeed, conserved, when you cut ...

0

I see you have ticked Muphrid's answer but I still think a less abstract reply could be useful. 1. Yes, angular momentum is conserved. 2. Yes, They will move off at constant velocity at a tangent to their circular motion. 3. Yes, their angular momentum will be of equal magnitude. 4. "but opposite sign." - No - they are travelling in opposite directions, ...

1

The question is "Why are the possible outcomes the same for all directions?" It happens also for observables of classical physics! QM does not matter here, the truly relevant idea is the fact that in a inertial system physics appears to be isotropic. In practice, it is not possible to physically distinguish different directions with physical experiments. ...

1

Although I see already two answers, I’ll add mine. The situation, as original poster described, is such for a massive spin-½ particle and is not symmetric in this sense for a massless spin-½ particle (some people say this case is properly named “helicity”, but it is a question of terminology). Why does it matter? Because a massive particle has its ...

3

Short intro to ladders As you say, they're ladder operators. Let's get rid of the annoying $\hbar$ by setting it to one, and call them more systematically $L_{-1},L_0,L_1$ instead of $L_-,L_z,L_+$. Then, the commutation relations take the uniform form $$[L_n,L_m] = (m-n)L_{m+n}$$ If we had countably many of these, we'd have a Witt algebra, if there was ...

1

Just remember that a Black Hole doesn't have infinite gravity - it just has however much mass created it in the first place. Yes, anything that gets within the event horizon is trapped forever, but that event horizon will actually be smaller than the size of the equivalent amount of mass composed of ordinary matter. This is also why "microscopic black holes" ...

4

Consider the following diagram: This shows a mass $m$ moving past a point $P$ in a straight line. Note that the mass isn't connected to $P$ in any way - it's just moving past in a straight line. The angular momentum of $m$ about $P$ is given by: $$\vec{L} = \vec{r} \times m\vec{v}$$ So the direction of $\vec{L}$ is normal to the screen and the ...

0

Ball A has linear momentum pointing in the direction of its flight path. But it is not conserved, because you have a force acting on it (centripetal force), that as no counter part (unless you specify where the axle is mounted and allow for the mount to move as well). Observe, that you can ascribe an angular momentum to B as well even when it does not move ...

2

Just like $j=1/2$ (spin-half) particles have two-complex-component wave functions, spinors, particles with spin $j$ have $(2j+1)$-dimensional wave functions describing the spin degrees of freedom. The dimension is what it is because $j_z$ always goes from $-j$ to $+j$ with the spacing equal to one. All the transformation rules under rotations may be ...

0

As you know, any operation, which can be continuous or discrete, in quantum mechanics is represented by an operator say $\hat{O}$. Under the operation, a physical quantity $\hat{X}$ transforms as $\hat{X}' = \hat{O}\hat{X}\hat{O}^{-1}$. In case $\hat{O}$ is continuous (forming a Lie group), one can often write it as $\hat{O} = e^{ir\hat{T}}$ (there is more ...

5

You already received several answers. However the fundamental physical reason is elementary: In classical, quantum and relativistic physics the physical laws describing an isolated physical system in an inertial reference frame are the same (are invariant) if you rotate (with an element of $SO(3)$) the system (there are many other symmetries depending on the ...

2

One can make sense of the introduction of $\mathrm{SO}(3)$ into quantum mechanics as follows: Consider a physical system in three spatial dimensions which we'll think of as residing in a box sitting on a table, like a table-top experiment. Suppose that we prepare the system in a particular way, so that the system is measured to be in a (pure) state ...

0

The very important group in physics is the Lorentz group $SO(3,1)$. It is built from the $SO(3)$ group and Lorentz boosts. The algebra of the Lorentz group generators (boosts $L_{i}$ and 3-rotations $R_{i}$) doesn't have the separation on $SO(3)$ and boosts' parts. But by introducing the "new" generators $J^{i}_{\pm} = \frac{1}{2}(R_{i} \pm iL_{i})$ we may ...

1

Spin in (non-relativistic) QM is fairly ad hoc, it has no deep reason. The underlying reason is that, in a relativistic setting for QM/QFT our states/fields must transform in some representation of the Lorentz group $\mathrm{SO}(1,3)$ if we want the amplitudes/Lagrangians to be Lorentz invariant (since something cannot be invariant if there's no rule how its ...

1

Physically, $R$ is just a rotation. However, when you write $R \psi(x)$, the general result is $R \psi(x) = (R\psi) (R^{-1}x)$. The $R^{-1}x$ takes in account that the rotation has an effect on the space-time coordinates, and it corresponds to orbital angular momentum , but you have to take in account that $\psi$ itself may be not a scalar (an invariant) ...

Top 50 recent answers are included


text:
## Convexity, Continuity, and Jensen’s Inequality Revision

A comment to my post Convexity, Continuity, and Jensen’s Inequality asks how we know that $x_{n}\rightarrow x_{0}$, as $n\rightarrow\infty$, in the proof of Lemma 3. It’s actually false that the sequence converges to $x_{0}$. What I believe to be a correct proof follows below. I will replace the false proof in the original post soon.

Lemma 3. If $f$ is midpoint convex and has a point of discontinuity $x_{0}$, then $\limsup_{x\rightarrow x_{0}}f(x)=+\infty$.

Proof. $f$ is discontinuous at $x_{0}$, so there exists a $\varepsilon>0$ such that

$\displaystyle\forall\delta>0,\ \exists x\in B_{\delta}(x_{0})\text{ s.t. }\left|f(x)-f(x_{0})\right|>\varepsilon$

Let $\delta>0$ be given. For such an $x$, define $x':=2x_{0}-x$ (i.e. the reflection of $x$ by $x_{0}$). By midpoint convexity,

$\displaystyle\dfrac{f(x)+f(x')}{2}\geq f\left(\dfrac{x+x'}{2}\right)=f(x_{0})$

which implies that $\left[f(x)-f(x_{0})\right]+\left[f(x')-f(x_{0})\right]\geq0$. I claim that there is $y\in\left\{x,x'\right\}$ such that $f(y)-f(x_{0})>\varepsilon$. Observe that if $f(x)-f(x_{0})<-\varepsilon$, then the estimate obtained above implies that $f(x')-f(x_{0})>\varepsilon$. But

$\displaystyle\left|x'-x_{0}\right|=\left|(2x_{0}-x)-x_{0}\right|=\left|x_{0}-x\right|<\delta$

which proves the claim. Hence, there exists a point $x_{1}\in B_{\delta}(x_{0})$ such that $f(x_{1})-f(x_{0})>\varepsilon$.

Suppose that we have constructed $x_{1},\ldots,x_{n}$ as desired. Set $x_{n+1}:=2x_{n}-x_{0}$. Note that $\left|x_{n+1}-x_{0}\right|=2\left|x_{n}-x_{0}\right|$. By midpoint convexity,

$\displaystyle\dfrac{f(x_{n+1})+f(x_{0})}{2}\geq f\left(\dfrac{x_{n+1}+x_{0}}{2}\right)=f(x_{n})$

which implies that

$\begin{array}{lcl}\displaystyle f(x_{n+1})-f(x_{0})&\geq&\left[f(x_{n+1})-f(x_{n})\right]+\left[f(x_{n})-f(x_{0})\right]\\[2 em]\displaystyle&\geq&2\left[f(x_{n})-f(x_{0})\right]\\[2 em]&>&\displaystyle2^{n}\varepsilon\end{array}$

By induction, we obtain a sequence $(x_{n})_{n=1}^{\infty}$ with the property that $f(x_{n})-f(x_{0})>2^{n-1}\varepsilon$ and $\left|x_{n}-x_{0}\right|<2^{n}\delta$.

I claim that every open neighborhood of $x_{0}$ contains a point $z$ such that $f(z)>M$, for any $M>0$. Indeed, let a neighborhood $B_{\delta_{0}}(x_{0})$ and quantity $M>0$ be given. Let $i$ be a sufficiently large integer so that $2^{i-1}\varepsilon>M$. Run the argument above with $\delta=2^{-i-1}\delta_{0}$ to obtain a sequence $(x_{n})_{n=1}^{\infty}$. Observe that

$\displaystyle\left|x_{i}-x_{0}\right|<2^{i}\delta=2^{i}\left(2^{-i-1}\delta_{0}\right)=\dfrac{\delta_{0}}{2}$

and

$\displaystyle f(x_{i})-f(x_{0})>2^{i-1}\varepsilon>M$

$\Box$


text:
Properties of Frobenius matrix and decomposability

If $D$ is Frobenius matrix, is there any relationship of the matrix being Frobenius matrix with $D$ being $D = (A-I)B^{-1}$ where $A$ is non-negative square matrix?

Also, how is Frobenius matrix used?

-
You can read a lot about frobenius matrix in wikipedia en.wikipedia.org/wiki/Frobenius_matrix –  dineshdileep Dec 17 '12 at 11:14


text:
# Definition:Common Divisor/Integers

## Definition

Let $S$ be a finite set of integers, that is:

$S = \left\{{x_1, x_2, \ldots, x_n: \forall k \in \N^*_n: x_k \in \Z}\right\}$

Let $c \in \Z$ such that $c$ divides all the elements of $S$, that is:

$\forall x \in S: c \mathop \backslash x$

Then $c$ is a common divisor (or common factor) of all the elements in $S$.


text:
Convert Between ANOVA Effect Sizes

## Usage

eta2_to_f2(es)

eta2_to_f(es)

f2_to_eta2(f2)

f_to_eta2(f)

## Arguments

es

Any measure of variance explained such as Eta-, Epsilon-, Omega-, or R-Squared, partial or otherwise. See details.

f, f2

Cohen's f or f-squared.

## Details

Any measure of variance explained can be converted to a corresponding Cohen's f via:

$$f^2 = \frac{\eta^2}{1 - \eta^2}$$

$$\eta^2 = \frac{f^2}{1 + f^2}$$

If a partial Eta-Squared is used, the resulting Cohen's f is a partial-Cohen's f; If a less biased estimate of variance explained is used (such as Epsilon- or Omega-Squared), the resulting Cohen's f is likewise a less biased estimate of Cohen's f.

## References

• Cohen, J. (1988). Statistical power analysis for the behavioral sciences (2nd Ed.). New York: Routledge.

• Steiger, J. H. (2004). Beyond the F test: Effect size confidence intervals and tests of close fit in the analysis of variance and contrast analysis. Psychological Methods, 9, 164-182.

eta_squared() for more details.
Other convert between effect sizes: d_to_r(), diff_to_cles, odds_to_probs(), oddsratio_to_riskratio()


text:
# [OS X TeX] \makebox{} Ignorance

Charlie Ros5e ros5e at odorf.us
Sun Jan 29 08:31:19 EST 2012

On 2012 Jan 29, at 12:41 PM, Manfred Braun wrote:

> On 29.01.2012, at 14:23, Charlie Ros5e wrote:
>
>> Gentlefolk,
>>
>> I want to produce the following lines in a document:
>>
>> 4. Given the graph of the polynomial function f
>>   a. What is the minimum degree this polynomial can have?
>>
>>   b. Find a possible formula for f(x).
>>
>> But I also want to have a graphic of a polynomial placed to the right of this text. I've never used \makebox{} before but it seemed to be the thing I need. I thought I could put two \makebox{} commands on the same line (with their widths chosen wisely), put the code for the problem statement shown above into the first \makebox{} and the code to display the graphic in the second \makebox{}. Doesn't work. The text of the problem statement gets put into a single line, and I can't figure out how to break it into the multiple lines I need.
>>
>> Questions:
>>
>> 1. Is \makebox{} even the command I need for this application?
>
> no!
>
>>
>>       If no, what *should* I use?
>
> \begin{minipage}[t]{5cm}
> \end{minipage}\hfill
> \begin{minipage}[t]{7cm}
> \end{minipage}
>
> the option [t] makes the minipages top aligned, if this is wanted.

Manfred ... that does it! A little diddling with some negative vertical space put the graph just where I wanted it. Many *many* thanks!

___________________________________________
>
>>
>> Charlie Rose
>> Oberaudorf, Deutschland
>
> Manfred Braun
> Kempen, Deutschland
> Tallinn, Estonia
>
> ----------- Please Consult the Following Before Posting -----------
> TeX FAQ: http://www.tex.ac.uk/faq
> List Reminders and Etiquette: http://email.esm.psu.edu/mac-tex/
> List Archive: http://tug.org/pipermail/macostex-archives/
> TeX on Mac OS X Website: http://mactex-wiki.tug.org/
> List Info: http://email.esm.psu.edu/mailman/listinfo/macosx-tex
>

____________________________________________
Charlie Rose
Oberaudorf, Deutschland

Web Site:                    http://www.ros5e.com
My PGP Public Key: Temporarily Unavailable
-------------------------------------
When Matjes Herring is outlawed,
Only outlaws will have Matjes.

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text:
thanks.

2. Prove that between two zeros of $e^x\, \sin(x)=1$ there is a zeros of $e^x\,\cos(x)=-1$

Let: $f(x) = e^x\, \sin(x) - 1$

First observe that:

$e^x\,\cos(x)+1=f'(x)-f(x)$,

Now if $x_1$ and $x_2$ are consecutive zeros of $f(x)$ then the signs of $f'(x_1)$ and $f'(x_2)$ would be different (the other possibility that one or both of these is zero does not occur as that would imply that for one ,or both, of the zeros that $\cos(x)=\sin(x) \Rightarrow x=\pi/4+n\pi$, but such a point cannot be a zero of $f(x)$).

But the sign of $f'(x_1)$ and $f'(x_2)$ being different, implies that $e^x\,\cos(x)+1$ is positive at one root and negative at the other, so as $e^x\,\cos(x)+1$ is continuous, by the intermediate value theorem must be zero at some point between $x_1$ and $x_2$. which proves the result.

RonL

3. Find where the function:

$f(x)=\left( x+\frac{1}{x} \right) ^x$

is an increasing and where it is a decreasing function.

First we restrict attention to $x>0$, as $f(x)$ is complex or undefined otherwise.

Rewrite $f(x)$ as: $f(x)=e^{x\,\ln(x+1/x)}$

Now $f(x)$ is differentiable on $(0,\, \infty)$and its derivative is:

$\frac{df}{dx}=\left[\frac{x^2}{x^2+1}+\ln(x+1/x) \right] e^{x\,\ln(x+1/x)}$.

But the term in the square brackets is always positive, as is the term with the exponential. Therefore $f(x)$ is increasing on $(0, \infty)$.

RonL

4. First you probably mean,
$\lim_{x\to 0^+}x^{x^x}$
You can write,
$x^{\exp(x\ln x)}$
Thus,
$\exp(\ln x\exp(x\ln x))$
Let us look at the term inside,
$\ln x\exp(x\ln x)$
Now,
$\lim_{x\to 0^+}\ln x=-\infty$
And,
$\lim_{x\to 0^+}\exp (x\ln x)=1$
By using the L'Hopital rule inside and then applying the limit composite rule.
Thus,
$\lim_{x\to 0^+}\ln x\exp (x\ln x)=-\infty$ by the extended reals.

Thus,*)
$\lim_{x\to -\infty}\exp x=0$
If we subtract 1 from this limit we have -1 as the answer to the limit question.

*)I am note sure about this. Because I employed the limit composite rule for infinite limits (limits that do not exists). Thus, it might not work.


text:
# Supporting both DirectX & OpenGL

## Recommended Posts

Spykam22    157

Hello!  I re-writing my game engine from scratch and was trying to decide whether I should support DirectX & OpenGL or just DirectX?  I understand that OpenGL is cross-platform while DirectX is restricted to Windows & the Xbox.  I also have a couple more questions:

1. I am developing on Windows using Visual Studio 2013 Ultimate.  If I were to implement OpenGL,  how would I build for different platforms? Example: If I wanted to build on maybe, Linux or Mac.  Would I have to build my project on that operating system?
2. If I do develop for both API's, how should I setup my project? Should I create two separate projects? Or keep them both separated in one project using enums, and interfaces? Example:
enum NXAPIChoice
{
NXAPI_DirectX,
NXAPI_OpenGL,
}

NXAPIChoice NXGame::DefaultAPI = NXAPIChoice::NXAPI_DirectX;

void NXGame::FinalInitialize()
{
int sWidth = 0, sHeight = 0;
InitializeWindows(sWidth, sHeight);
SetupGPU(sWidth, sHeight);

//TODO: [HIGH PRIORITY] Initialize Input
//TODO: [HIGH PRIORITY] Initialize Audio

Initialize();

//TODO: [HIGH PRIORITY] Create & Start GameTime
}

void NXGame::SetupGPU(int width, int height)
{
//stuff...
NXGPU::Initialize(DefaultAPI, width, height, ...other);
}

void NXGPU::Initialize(NXAPIChoice choice, int width, int height, ...other)
{
if (choice == NXAPIChoice::NXAPI_DirectX)
InitializeDirect3D(width, height, D3D_FEATURE_LEVEL_11_0, ...other);
else
InitializeOpenGL(width, height, GL_VERSION, ...other)
}

Also if I were to develop for both, and I do have to compile my code on other platforms (linux or mac), will I be able to skip my DirectX code (assuming it would cause errors because DX is not available) and just compile GL code? Is it even worth adding support for both? Thanks!

[EDIT: Or perhaps, shall I finish my engine with DirectX first (creating empty methods for GL in the process) and just start working on GL support (fill in the empty methods) after DirectX is fully implemented?]

Edited by KamRandle

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L. Spiro    25638

OpenGL is cross-platform

It aims to be.

Would I have to build my project on that operating system?

It is always best to do so.

Should I create two separate projects?

Yes. The days of a virtual interface and run-time selectable graphics API are long gone.

will I be able to skip my DirectX code (assuming it would cause errors because DX is not available) and just compile GL code?

If you work your macros properly.

shall I finish my engine with DirectX first

Doing them at the same time is preferred if possible given your experience with both of them.
You can’t really make good decisions for your interface without thinking about both of them at the same time, and implementing them at the same time tests your decisions.

L. Spiro

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rpiller    839

If I was new to this stuff I would finish the engine with one, then make it again with the other, then I feel like I'd truly know how to make it the 3rd time with both. This would be if you are trying to use this as a learning experience of course. If you already know how both work well enough then this wouldn't be the way to go about it.

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SeanMiddleditch    17565

Hello!  I re-writing my game engine from scratch and was trying to decide whether I should support DirectX & OpenGL or just DirectX?  I understand that OpenGL is cross-platform while DirectX is restricted to Windows & the Xbox.  I also have a couple more questions:

More accurately, OpenGL supports Linux, OSX, Windows, iOS, and Android. DirectX supports Windows and Windows Phone. Every console - XBox included - uses a custom API. The XBox one certainly resembles DirectX pretty closely, but it's not identical. The PS3 does not support OpenGL, though it does support an infrequently-used non-standard GLES1-derived API with Cg shaders that led many to think that the PS3 uses GL (most games used the entirely proprietary libgcm, though).

The point being, if you really want to be ported to all main platforms, you _must_ support multiple graphics API. OpenGL doesn't get you as far as some might think.

Even if you use OpenGL, there's several variants you need to support (OpenGL Core on OSX, OpenGL Compatibility for most other desktop OSes, GL|ES for mobiles).

I am developing on Windows using Visual Studio 2013 Ultimate.  If I were to implement OpenGL,  how would I build for different platforms? Example: If I wanted to build on maybe, Linux or Mac.  Would I have to build my project on that operating system?

You can do what's called a "cross compilation" (a compiler that runs on one OS/architecture but makes binaries for a different one). There are cross compilers available for Windows, even some commercial ones that integrate with Visual Studio. This is how mobile development is done, after all, since you don't actually compile on a phone.

Usually though you'd just port your build system and then build on an actual Linux/OSX machine. You can use something like CMake or SCons to handle builds on both Windows and other platforms. You'd continue using Visual C++'s cl.exe to compile on Windows and then use GCC or Clang when compiling on OSX or Linux.

If I do develop for both API's, how should I setup my project? Should I create two separate projects? Or keep them both separated in one project using enums, and interfaces?

Whichever you find easier. It doesn't matter much at all.

I personally prefer building a separate static lib for each underlying API and then only building/linking the one I am currently using. The public interface is a header, with the implementation files being separate for DX, GL, or so on.

You can take that a bit further and build them as DLLs (or whatever the platform's shared object format is) and allow runtime switchable system, though really Windows is the only platform that gives you multiple realistic choices but there's no good reason to use anything but DX on Windows if you support it. I stick with static polymorphism here and don't bother with virtual interfaces; they buy you nothing.

Also if I were to develop for both, and I do have to compile my code on other platforms (linux or mac), will I be able to skip my DirectX code (assuming it would cause errors because DX is not available) and just compile GL code?

Just don't compile the .cpp files that have the DX code and wrap any DX-using code in #ifdef's. Pretty simple to conditionally compile code or whole modules.

Is it even worth adding support for both?

Depends on your goals. If you are only planning to support Windows, there is absolutely no good reason to prefer OpenGL over DirectX. If you want to support the other desktop OSes they all only support OpenGL - which Windows also supports - so there's a strong argument to be made to only spend time implementing OpenGL support. If you want to support most mobiles and desktop, you're still going to need to support multiple renderers (even if GL and GL|ES are very similar, the kinds of things you can do on the desktop don't perform well on mobile) so you still need to deal with multiple APIs and conditional code. If you want Windows Phone, you must have DX.

Or perhaps, shall I finish my engine with DirectX first (creating empty methods for GL in the process) and just start working on GL support (fill in the empty methods) after DirectX is fully implemented?

This is the path I recommend, assuming you mean Direct3D 11. DX11 is both easier to use than modern OpenGL (though if you stick to old deprecated fixed-function stuff the GL API is definitely simpler) and it tends to force you to learn how to think the "right way" (which applies equally well to GL) while OpenGL can easily lead you to believe that very bad things are correct (partly because GL3+ was designed around some incorrect guesses about where hardware was going and GL is still struggling under some of those poor choices and it's not obvious in GL when you're doing something you shouldn't).

There is _zero_ reason to ever use D3D10 (the only OS that supported it exclusively is Vista which also supports 111 with a service pack) and I wouldn't recommend D3D9 for any new code; if you really need to support XP in the 2-5 years it'll take at minimum for your game to come out, OpenGL will cover you and support other platforms, too.

As one last reminder, note that your graphics code is _hardly_ the only bit of platform-specific code you'll have to write. The native, low-level APIs for threading, file I/O, memory management, window management, networking, input, audio, and so on all differ between every platform. If you think you'll just be writing OpenGL code on Windows and then compiling it for Linux or OSX you might be in for a surprise. There are libraries to consider using that paper over the problem. SDL for instance covers input, windows, simple file I/O, simple audio, and so on, though only on desktops and mobiles (not any consoles that I know of). FMOD or Wwyse cover more advanced audio needs on every major gaming platform. There are networking libraries that support a wide variety of platforms. You're still likely going to have a good deal of platform-specific code outside of graphics, however, and you're going to need to make a lot of good choices when writing you're Windows-only good to have any hope of compiling on Linux or OS without a very long time spent porting.

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richardurich    1352

I'd honestly recommend switching to OpenGL if you feel comfortable with the idea. Choosing DirectX means you either choose to limit your audience or you choose to limit your graphics options. Even if your game isn't coming out for 5 years, the same situation that exists today with Windows XP not supporting DX11 will exist with Windows 7 not supporting DX13 or whatever. OpenGL doesn't have that problem since driver manufacturers control support levels. And if there's any chance your game would sell in China or other countries that lag further behind on OS adoption, the impact of using DirectX will really, really hurt. Even in America, you're still talking about a larger group than all Mac and Linux users combined.

Is there a reason you're hesitant to use OpenGL? If you have tools that only work with DirectX and make your life insanely easier, obviously you should stick with DirectX. But if you're just worried OpenGL is slower or harder to use or something, just jump in and you'll find out all the FUD is unjustified.

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Satharis    2449

Yes. The days of a virtual interface and run-time selectable graphics API are long gone.

Maybe I'm just being stupid but what is the reasoning behind that?

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Vortez    2714

That's what i did in my 3d engine in my signature, have a look at it if you wish (Camera.h and Camera.cpp are good example). If i want to use opengl, i simply declalre

a macro COMPILE_FOR_OPENGL, if i want dx9, COMPILE_FOR_DIRECTX9 and so on. It's a bit of pain to write at first, but it had the advantage of being

extremly simple to switch between apis. The other solution would be separate interfaces in dlls i guess.

Both have advantages and inconveniances, it's a double edged sword kinda deal.

Edited by Vortez

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L. Spiro    25638

Yes. The days of a virtual interface and run-time selectable graphics API are long gone.

Maybe I'm just being stupid but what is the reasoning behind that?

• Performance: Reducing virtual calls is usually a micro-optimization but when it’s virtual for every SetTexture(), SetBlendMode(), etc. it can add up. Plus other locations in code that have to become run-time if/else’s instead of just macro-excluded lines of code. With other options readily available there is just no need for the overhead.
• Code maintenance: But you still need macros around the Direct3D parts for your ports for Linux and OS X, so the macro spaghetti still exists to some degree.  On some platforms it is both and on some platforms it is only one.  It is easier to just make it, “On some platforms it is this and on others that—no cross-overs.”
• OpenGL on Windows is generally just not a good idea.  OpenGL hasn’t had any form of consistency standards by which vendors are required to adhere until very recently, meaning it may run as expected on some cards and not on others.  Again from #2, it is just best to use Direct3D on Windows and OpenGL on everything else.
#3 is the main reason.  Direct3D is near 100% consistent on all Windows machines and OpenGL more like 80%.  On Mac OS X OpenGL is near 100% consistent.  Linux OpenGL is somewhere between.  So you always want to use Direct3D on Windows; there is never really any reason for overlapping builds.

L. Spiro Edited by L. Spiro

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Spykam22    157
Hello everyone! Thanks for the answers. I think I may just write my DirectX version first. It is being developed alongside a game that I am using it for. It will be my first game to be released. After I am finished, then I'll focus on GL versions. How many people do you guys think use Bootcamp on OS X? Thanks! :)

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## Found 36 Documents (Results 1–36)

100
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### Wave scattering in hyperbolic balance laws and the zero relaxation limit. (English)Zbl 1469.35142

MSC:  35L65 35B25 35L45
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### Zero relaxation limit in slow time scaling for first-order quasi-linear hyperbolic systems. (Chinese. English summary)Zbl 07494481

MSC:  35L60 35B65
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### Zero relaxation limit to rarefaction waves for general $$2 \times 2$$ hyperbolic systems with relaxation. (English)Zbl 1332.35215

MSC:  35L60 35B25
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### On the steady state relativistic Euler-Poisson equations. (English)Zbl 1267.35230

MSC:  35Q75 35B40 85A05
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### An asymptotic expansion in the multi-dimensional hydrodynamic model for two-carrier plasmas with small parameters. (English)Zbl 1274.35263

MSC:  35Q05 35B40 35C20

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### Zero relaxation limit to centered rarefaction waves for Jin-Xin relaxation system. (English)Zbl 1213.35056

MSC:  35B25 35L60 35L45
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### Existence and uniqueness of weak solutions for precipitation fronts: a novel hyperbolic free boundary problem in several space variables. (English)Zbl 1204.35068

MSC:  35D30 35R35
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### Existence and some limits of stationary solutions to a one-dimensional bipolar Euler-Poisson system. (English)Zbl 1160.35352

MSC:  35B40 35B45 82C31 35J55
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### Relaxation approximation of some initial-boundary value problem for $$p$$-systems. (English)Zbl 1157.35005

MSC:  35B25 35C20 35L50 35Q72
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### Modelling traffic flow with a time-dependent fundamental diagram. (English)Zbl 1041.35035

MSC:  35L45 90B20 35L65
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### Global solutions of nonconcave hyperbolic conservation laws with relaxation arising from traffic flow. (English)Zbl 1015.35066

MSC:  35L65 90B20
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### Well-posedness theory of an inhomogeneous traffic flow model. (English)Zbl 1162.35415

MSC:  35L65 90B20
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### Global solutions and zero relaxation limit for a traffic flow model. (English)Zbl 0979.35096

Reviewer: A.Doktor (Praha)
Full Text:

### The relaxation of the hydrodynamic model for semiconductors to the drift-diffusion equations. (English)Zbl 0970.35150

MSC:  35Q60 82D37 35Q35
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### A hierachy of hydrodynamic models for plasmas. Zero-electron-mass limits in the drift-diffusion equations. (English)Zbl 0956.35010

MSC:  35B25 78A35 35B40
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### Boundary conditions for hyperbolic systems with stiff source terms. (English)Zbl 0929.35080

MSC:  35L50 35B25
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### Materials with internal variables and relaxation to conservation laws. (English)Zbl 0973.74005

Reviewer: G.A.Maugin (Paris)
Full Text:

### Singular perturbations of first-order hyperbolic systems with stiff source terms. (English)Zbl 0942.35110

MSC:  35L60 35B25
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### The relaxation approximation to hyperbolic system of conservation laws. (English)Zbl 0931.35100

Fey, Michael (ed.) et al., Hyperbolic problems: Theory, numerics, applications. Proceedings of the 7th international conference, Zürich, Switzerland, February 1998. Vol. I. Basel: Birkhäuser. ISNM, Int. Ser. Numer. Math. 129, 485-491 (1999).
Reviewer: Weike Wang (Wuhan)
MSC:  35L65

### Existence and the singular relaxation limit for the inviscid hydrodynamic energy model. (English)Zbl 0948.76007

Jerome, Joseph W. (ed.), Modelling and computation for application in mathematics, science, and engineering. Workshop, Northwestern Univ., Evanston, IL, USA, May 3-4, 1996. Oxford: Clarendon Press. Numerical Mathematics and Scientific Computation. 189-215 (1998).
MSC:  76A99 82D15

### Convergence of approximate solutions for quasilinear hyperbolic conservation laws with relaxation. (English)Zbl 0919.35087

MSC:  35L65 35B20 35L60

### Quasilinear hyperbolic systems and dissipative mechanisms. (English)Zbl 0911.35003

Singapore: World Scientific Publishing Co. viii, 223 p. (1997).

MSC:  35L65
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### The dispersion relation at large frequencies for linear surface gravity waves on a Maxwell fluid exhibiting a spatial decay. (English)Zbl 0841.76005

MSC:  76A05 74J15
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### Zero relaxation and dissipation limits for hyperbolic conservation laws. (English)Zbl 0797.35113

MSC:  35L65 76N15 74Hxx
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# What is the value of x ?

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What is the value of x ? [#permalink]

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07 Jun 2012, 23:49
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What is the value of x ?

(1) x^3 = 6x − x^2

(2) 14x = x^3 + 5x^2
[Reveal] Spoiler: OA

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Re: What is the value of x ? [#permalink]

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07 Jun 2012, 23:57
1
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macjas wrote:
What is the value of x ?

(1) x^3 = 6x − x^2

(2)14x = x^3 + 5x^2

Statement (1)

x^3 = 6x - x^2

by factoring we have: x (x + 3) ( x - 2) = 0 so x = 0, x = - 3 and x = 2 (ins.)

Statement (2)

14x = x^3 + 5x^2

by factoring we have: x (x + 7) (x - 2) = 0 so x = 0, x = -7 and x = 2 (ins.)

Statement (1) AND (remember in Mathematics "and" means intersection, in layman's term: what is the common in both statement?) Statement (2)

x = 0 as well as x = 2

So still insufficient because we have two values of x

Hope I helped
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Re: What is the value of x ? [#permalink]

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08 Jun 2012, 00:00
2
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Hi,

to find the value of x,

using (1)
$$x^3 = 6x-x^2$$
or $$x^3 - 6x + x^2=0$$
or $$x(x^2+x-6)=0$$
or $$x(x+3)(x-2)=0$$
or $$x=-3, 0, 2$$, no unique value of x. Insufficient.

using (2)
$$14x = x^3 + 5x^2$$
or $$x^3 + 5x^2-14x=0$$
or $$x(x^2+5x-14)=0$$
or $$x(x+7)(x-2)=0$$
or $$x=-7, 0 , 2$$, no unique value of x. Insufficient.

even considering (1) & (2) together,
x=0,2.

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Re: What is the value of x ? [#permalink]

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29 Aug 2012, 04:28
Hi, was just wondering if we could solve it like this:
From 1) x^3 = 6x + x^2
From 2) x^3 = 14x – 5x^2
From 1) and 2), we get 6x – x^2 = 14x – 5x^2
Solving the above, we get, x = 2
Hence, C

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Re: What is the value of x ? [#permalink]

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27 Feb 2013, 19:10
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reetskaur wrote:
Hi, was just wondering if we could solve it like this:
From 1) x^3 = 6x + x^2
From 2) x^3 = 14x – 5x^2
From 1) and 2), we get 6x – x^2 = 14x – 5x^2
Solving the above, we get, x = 2
Hence, C

Hi,
Even if you solve by the above method you again get two values of x.
4x^2+6x-14x=0
x(4x-8) = 0
x = 0, 2 so Insufficient

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Re: What is the value of x ? [#permalink]

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27 Feb 2013, 20:25
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As stated in 1)

$$x^3 = 6x - x^2$$

$$x^3 + x^2 -6x = 0$$

$$x (x + 3) ( x - 2) = 0$$

=>$$x = 0, - 3, 2$$

From statement (2)

$$14x = x^3 + 5x^2$$

$$x^3 + 5x^2 - 14x=0$$

$$x (x + 7) (x - 2) = 0$$

=>$$x = 0, -7, 2$$

From Statement (1) AND Statement (2)

x = 0 and 2

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Re: What is the value of x ? [#permalink]

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