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## hyper-spherical coordinates in Minkowski space?

<jabberwocky>
<a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<TITLE>Usenet ASCII</TITLE>
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\nHi,\n\nis there any way to define global hyper-spherical coordinates in\nMinkowski space? Something compatible with the metric (+---) is\n\nx0 =3D r cosh xi\nx1 =3D r sinh xi sin theta cos phi\nx2 =3D r sinh xi sin theta sin phi\nx3 =3D r sinh xi cos theta\n\nBut there is trouble as the image of cosh is [1,oo) and not [-1,1] as\nneeded. Is there any other way or do I have to wick rotate to euclidean\nsignature first?\n\nRen=E9\n\n

');">   View this Usenet post in original ASCII form

</jabberwocky>Hi,

is there any way to define global hyper-spherical coordinates in
Minkowski space? Something compatible with the metric (+---) is

$$x0 =3D r[/itex] cosh \xi $x1 =3D r$ sinh $\xi sin \theta cos \phix2 =3D r$ sinh $\xi sin \theta sin \phix3 =3D r$ sinh $\xi cos \theta$$ But there is trouble as the image of cosh is [1,oo) and not [itex][-1,1]$ as
needed. Is there any other way or do I have to wick rotate to euclidean
signature first?

$$Ren=E9$$

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[Please post in ASCII and avoid MIME quoting such as "=3D".] meyr2@web.de wrote: > Hi, > > is there any way to define global hyper-spherical coordinates in > Minkowski space? Something compatible with the metric (+---) is > > $x0 =3D r$ cosh \xi > $x1 =3D r$ sinh $\xi sin \theta cos \phi$ > $x2 =3D r$ sinh $\xi sin \theta sin \phi$ > $x3 =3D r$ sinh $\xi cos \theta$ > > But there is trouble as the image of cosh is [1,oo) and not $[-1,1]$ as > needed. Is there any other way or do I have to wick rotate to euclidean > signature first? What you point out is not a problem. However, your equations are incomplete. If you fix r in your formulas, you'll notice that they parametrize the future hyperboloid distance r from the origin. If you vary r from to oo, you fill in the interior of the future half of the light cone. Varying r from to $-oo,$ you will fill in the interior of the past half of the light cone. However, you still have not parametrized the region at space-like separation from the origin, outside the light cone. To do that you need a second set of equations $$x0 = r[/itex] sinh \xi $x1 = r$ cosh $\xi sin \theta cos \phix2 = r$ cosh $\xi sin \theta sin \phix3 = r$ cosh $\xi cos \theta$$ Now you're done. In short, use the first set of equations when [itex]x0^2-x1^2-x2^2-x3^2 >$ and the second set of equations for $x0^2-x1^2-x2^2-x3^2 <$ . Hope this helps. Igor

 Similar discussions for: hyper-spherical coordinates in Minkowski space? Thread Forum Replies Advanced Physics Homework 7 Linear & Abstract Algebra 4 Advanced Physics Homework 1 Special & General Relativity 4 Special & General Relativity 2


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# (X2 + 1) Dy + (2y − 1) Dx = 0 - Mathematics

Sum

(x2 + 1) dy + (2y − 1) dx = 0

#### Solution

We have,

$\left( 1 + x^2 \right)dy + \left( 2y - 1 \right)dx = 0$

$\Rightarrow \left( 1 + x^2 \right)dy = \left( 1 - 2y \right)dx$

$\Rightarrow \frac{dy}{\left( 1 - 2y \right)} = \frac{1}{\left( 1 + x^2 \right)}dx$

Integrating both sides, we get

$\int\frac{1}{\left( 1 - 2y \right)}dy = \int\frac{1}{\left( 1 + x^2 \right)}dx$

$\Rightarrow - \frac{1}{2}\log\left| 1 - 2y \right| = \tan^{- 1} x - \log \sqrt{C}$

$\Rightarrow - \log\left| 1 - 2y \right| = 2 \tan^{- 1} x - 2\log \sqrt{C}$

$\Rightarrow - 2 \tan^{- 1} x = - \log C + \log\left| 1 - 2y \right|$

$\Rightarrow - 2 \tan^{- 1} x = \log \left| \frac{1 - 2y}{C} \right|$

$\Rightarrow e^{- 2 \tan^{- 1} x} = \frac{1 - 2y}{C}$

$\Rightarrow C e^{- 2 \tan^{- 1} x} = \left( 1 - 2y \right)$

$\Rightarrow 1 - C e^{- 2 \tan^{- 1} x} = 2y$

$\Rightarrow \frac{1}{2} - \frac{C}{2} e^{- 2 \tan^{- 1} x} = y$

$\Rightarrow y = \frac{1}{2} + K e^{- 2 \tan^{- 1} x},\text{ where }K = - \frac{C}{2}$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 45 | Page 146

format: plain


text:
# All Questions

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A sequence of points $x_1,x_2,\dots$ on the unit sphere $S^{D-1}$ is said to be uniformly distributed if \begin{align} \lim_{N \rightarrow \infty} \frac{1}{N} \sum_{j=1}^N f(x_j) = \int_{x \in ...
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### Reed-Muller-Codes

Let $F$ be the field with two elements, $V_m=F^{2^m}$.Let $R(r, m)\subset V_m$ be the binary Reed-Muller Code. Define $R_m:=R(1, m)$. Then the dimension of $R_m$ is $1+m$ and its minimal distance is ...
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### Disprove this Piece of Jensen's Inquality “Black Magic”

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### LQR solution when there are linear terms in the cost function?

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# Approximately what percentage of solar energy falling on a plant is assimilated in photosynthesis?

Feb 13, 2018

1%

#### Explanation:

Out of all the sunlight energy that falls on a plant, only 1% is absorbed by the photosynthetic pigments. This energy is then converted into chemical energy in order to form sugars in dark reactions of photosynthesis. The remaining 99% of solar energy is reflected or transmitted which is then used to evaporate water, heat up soil etc...

Hope it helps!


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# Water Elevation data file (*.csv)¶

Water Elevation data file contains the water surface elevation of river cross-sections in river survey data. List 19 shows an example of water elevation data file.

List 19 Example of Water elevation data file
Point,Normal,H27,H29
27.00,2,7,10.5
27.50,2.2,7.2,10.7
28.00,2.3,7.3,10.8
28.50,2.4,7.4,10.9
29.00,2.45,7.45,10.95
29.50,2.48,7.48,10.98
30.00,2.5,7.5,11

Water Elevation data file is a Comma-delimited text file. Each column has the following meaning:

• 1st column: The name of cross-section
• 2nd column and the followings: Water elevation

The first line is recognized as a header line.

Note

Importing multiple water elevation data

iRIC 3.0 and later versions can import multiple water elevation data from one file.

iRIC 2.x just skipped the header line, but iRIC 3.0 and later load the names of water elevation data from the 2nd column and the followings, from the header line. For example, in case of List 19, “Normal”, “H27”, “H29” are loaded as names.


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<< problem 265 - Binary Circles Billionaire - problem 267 >>

# Problem 266: Pseudo Square Root

The divisors of 12 are: 1,2,3,4,6 and 12.
The largest divisor of 12 that does not exceed the square root of 12 is 3.
We shall call the largest divisor of an integer n that does not exceed the square root of n the pseudo square root (PSR) of n.
It can be seen that PSR(3102)=47.

Let p be the product of the primes below 190.
Find PSR(p) mod 10^16.

# My Algorithm

The product of all primes below 190 is 2 * 3 * 5 * 7 * ... * 181 = 5397346292805549782720214077673687806275517530364350655459511599582614290.
That number is obviously too large for my poor C++ compiler ... but its logarithm isn't:
log(2) + log(3) + log(5) + ... + log(181) = log(53973462...2614290) approx 167.472

My program looks for an optimal subsets of those primes where
log(p_1) + log(p_2) + ... + log(p_n) approx 167.472 / 2 approx 83.736 = (because log(sqrt{x}) = log(x) / 2)

There are 2^42 approx 4.4 * 10^12 subsets. And again, I have a number that is too large for my computer ...
But there is hope - I split those 42 primes into two equally-sized sets:
the container left will be "responsible" for the first 21 primes, and right will control the next 21 primes.

Now there are 2^21 approx 2.1 million subsets of each left and right.
For each of those subsets I compute the logarithm of the product of its primes and a bitmask that indicates which primes were used.
If the n-th bit is set the n-th prime was used (that's the case for left) or the (n+21)-th prime was used (right).

For each value of left I need to find the largest value of right such that their sum doesn't exceed 83.736 (logRoot).
std::upper_bound can perform this task very fast if the input data is sorted (but it returns an iterator that is "one step too far").

Until now, all computations were based on logarithms. I could try e^{best sum} but rounding issues produce only garbage.
That's why I have to look at the bitmasks of the best combination of values from left and right.

I multiply all used primes according to leftMask and rightMask and repeatedly apply modulo 10^16.

## Note

Programming languages with proper support for large number probably don't need to deal with logarithms.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):
Note: Enter the upper limit for the prime numbers

This is equivalent to
echo 100 | ./266

Output:

Note: the original problem's input 190 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>

// all primes below 190
std::vector<unsigned int> primes =
{
2,   3,   5,   7,  11,  13,  17,  19,  23,  29,
31,  37,  41,  43,  47,  53,  59,  61,  67,  71,
73,  79,  83,  89,  97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181
};

// store a subsets logarithm and a bitmask of used primes
{
// logarithm of product of used primes
double       logarithm;
// for std::sort and std::upper_bound
{
return logarithm < other.logarithm;
}
};

int main()
{
unsigned int maxPrime = 190;
std::cin >> maxPrime;
// keep only relevant primes
while (primes.back() > maxPrime)
primes.pop_back();

// compute logarithm of all primes
double logProduct = 0;
std::vector<double> logPrimes;
for (auto p : primes)
{
auto current = log(p);
logPrimes.push_back(current);
logProduct += current;
}

// log(sqrt(x)) = log(x) / 2
double logRoot = logProduct / 2;

// subdivides primes into two equally-sized sets (21 elements each)
const auto half = primes.size() / 2;

// generate all subsets of the lower half of the primes (left)
// and      all subsets of the upper half of the primes (right)
// key is the logarithm of product of the used primes, value is the bitmask
// (if n-th bit is set, then the n-th prime is used)
{
// find logarithm for the upper half of primes
double logRight = 0;
for (unsigned int pos = 0; pos < half; pos++)
if ((bitmask & (1 << pos)) != 0)
logRight += logPrimes[pos + half];

// store logarithms and their generating bitmasks
if (logRight <= logRoot) // skip a few of the very large products
}

// sort container / required for fast binary search (the algorithm behind std::upper_bound)
std::sort(right.begin(), right.end());

// find "best" combination of left and right:
// for each value of "left" look for best match in "right"
// keep the highest sum which doesn't exceed logRoot
double best = 0;
{
// find logarithm for the same bitmask of the lower half of primes
double logLeft = 0;
for (unsigned int pos = 0; pos < half; pos++)
if ((bitmask & (1 << pos)) != 0)
logLeft += logPrimes[pos];

// how much should come from the bigger primes ?
LogAndBitmask missing = { logRoot - logLeft, 0 }; // bitmask doesn't matter, set to zero

// find best match
auto iteLogRight = std::upper_bound(right.begin(), right.end(), missing);
// ... which will slightly overshoot, therefore go back one step
iteLogRight--;
auto logRight = iteLogRight->logarithm;

// higher than before ?
if (best < logLeft + logRight)
{
best = logLeft + logRight;
}
}

// last 16 digits of the product
unsigned long long result = 1;
const auto Modulo = 10000000000000000ULL;
// index of current prime
unsigned int currentPrime = 0;

// the first 21 primes ...
for (; currentPrime < half; currentPrime++)
{
// use that prime ?
if ((leftMask & 1) != 0)
{
result *= primes[currentPrime];
result %= Modulo;
}
// next bit
}
// ... and the same for the next 21 primes
for (; currentPrime < primes.size(); currentPrime++)
{
// use that prime ?
if ((rightMask & 1) != 0)
{
result *= primes[currentPrime];
result %= Modulo;
}
// next bit
}

// print result
std::cout << result << std::endl;
return 0;
}

This solution contains 16 empty lines, 35 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 36 MByte.

(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

July 3, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 65% (out of 100%).

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

 << problem 265 - Binary Circles Billionaire - problem 267 >>
more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
thanks to the KaTeX team for their great typesetting library !


text:
Meta & resources
Machine Learning: concepts & procedures
Machine Learning: fundamental algorithms
Machine Learning: model assessment
Artificial neural networks
Natural language processing
The computer science appendix

# Voronoi tasselation

The Voronoi tasselation consists in partitioning the plane into regions based on the distance to points in a specific subset of the plane.

For each point, there is a region of all points closer to it than any other one (Voronoi cells).

# Simplex

A simplex is the generalisation of a triangle/tetrahedron to an arbitrary number of dimensions. The k simplex is a k-dimensional polytope, the convex hull of its k+1 vertices.

Given points$u_0, \ldots, u_k \in \mathbb{R}^k$, with$u_1 - u_0, \ldots, u_k - u_0$linearly independent, the simplex determined by them is the set of points

$C : {\theta_0 u_0 + \cdots + \theta_k u_k | \theta_i \geq 0, 0 \leq i \leq k, \sum_{i=0}^k \theta_i = 1}$
• the 2-simplex is a triangle;

• the 3-simplex is a tetrahedron;

• the 4-simplex is a 5-cell


text:
# Show that $6m \mid (2m+3)^n + 1$ if and only if $4m \mid 3^n + 1$

Let $m$ and $n$ be positive integers. How to prove that $$6m \mid (2m+3)^n + 1$$ if and only if $$4m \mid 3^n + 1$$

-
What have you tried? I would first try using induction, and a binomial expansion. –  Andrew Jul 1 '12 at 9:19
I would like to know what is the origin of this problem. (If it is from a textbook, it might be useful to know what you are supposed to know at this stage. E.g. whether a proof using only basic divisibility properties is expected.) –  Martin Sleziak Jul 1 '12 at 13:46
Its from Pre-Indonesian TSTST 2010,but your proof is understandable . –  Frank Jul 1 '12 at 20:23

$\newcommand{\jaco}[2]{{\left(\frac{#1}{#2}\right)}}$I have tried to solve this - I hope I did not make too many mistakes there. Maybe there is much easier solution than this one.

Let us start by a few easy observations.

Since $2m+3 \equiv 3 \pmod m$, we see that $$m\mid (2m+3)^n+1 \Leftrightarrow m\mid 3^n+1. \tag{1}$$

It is obvious that $$3\mid (2m+3)^n+1 \Rightarrow 3\nmid m. \tag{2}$$

Since $3\nmid x$ implies $x^2\equiv 1 \pmod 3$ (Little Fermat's theorem), we get (using this fact for $x=2m+3$) that $$3\mid (2m+3)^n+1 \Rightarrow n\text{ is odd.}\tag{3}$$

Now we get $$3\mid (2m+3)^{n}+1 \Rightarrow m\equiv1\pmod3, \tag{4}$$ since $m\equiv2\pmod3$ $\Rightarrow$ $2m+3\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n+1\equiv2\pmod3$.

It is easy to see that $$4\mid 3^n+1 \Leftrightarrow n\text{ is odd}\tag{5}$$ and $$m\mid 3^n+1 \Rightarrow 3\nmid m.\tag{6}$$

For odd $n=2k+1$ we have $3^{2k+1}+1=(3+1)(3^{2k}-3^{2k-1}+\dots+3^2-3+1)$, which shows that $$\frac{3^{2k+1}+1}4\equiv 1 \pmod 2.$$ Hence $$4m\mid 3^n+1 \Rightarrow m\text{ is odd.}\tag{7}$$

Now a few more facts with a little less elementary proof. We will use Legendre symbol and quadratic reciprocity.

We first show: $$4m\mid 3^n+1 \Rightarrow m\equiv1 \pmod 3.\tag{8}$$ We know that $n$ is odd, let $n=2k+1$. It suffices to show that each odd prime factor $p$ of $3^{2k+1}+1$ fulfills $p\equiv1\pmod 3$.

Suppose that $p\mid 3^{2k+1}+1$ and, consequently $p\mid 3^{2(k+1)}+3$. Therefore $-3$ is a quadratic residue modulo $p$ and $\jaco{-3}p=1$. We get: $$\jaco{-3}p=\jaco{-1}p\jaco3p$$ $$1=(-1)^{\frac{p-1}2}(-1)^{\frac{p-1}2}\jaco{p}3.$$ Hence $\jaco{p}3=1$, which implies $p\equiv1\pmod{3}.$

Now we are ready to show that $6m\mid (2m+3)^n+1$ $\Leftrightarrow$ $4m\mid 3^n+1$.

$\boxed{\Rightarrow}$ If $6m\mid (2m+3)^n+1$ then:

• $\gcd(m,4)=1$ by (7)
• $n$ is odd by (3)
• $4\mid 3^n+1$ by (5)
• $m\mid 3^n+1$ by (1)

Together we have $4m\mid 3^n+1$.

$\boxed{\Leftarrow}$ If $4m\mid 3^n+1$ then:

• $n$ is odd by (5)
• $\gcd(m,6)=1$ by (6), (7)
• $m\mid (2m+3)^n+1$ by (1)
• $2\mid (2m+3)^n+1$ since $2m+3$ is odd
• $3\mid (2m+3)^n+1$ since $m\equiv1 \pmod 3$ by (8), which yields $2m+3\equiv 2\pmod 3$ and $(2m+3)^n\equiv2\pmod3$ (using the fact that $n$ is odd).

Together we have $2\cdot 3 \cdot m \mid (2m+3)^n+1$.

-


text:
# Double Angle Identity

Printable View

• Feb 25th 2009, 06:33 AM
Beard
Double Angle Identity
Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$
• Feb 25th 2009, 06:41 AM
HallsofIvy
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

Sorry, I thought you were trying to prove an identity!
• Feb 25th 2009, 06:43 AM
Chop Suey
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

I'll start you off:

$2\cos^{2}x = 2\sin(x)\cos(x) + 1 \implies 2\cos^{2}x -1 = 2\sin(x)\cos(x) \implies \cos{2x} = \sin{2x}$

When does the cosine of an angle equals the sine of an angle?
• Feb 25th 2009, 06:44 AM
ADARSH
Quote:

Originally Posted by Beard
Stuck on this, don't know where to start

$2\cos^{2}x = 2\sin(x)\cos(x) + 1,\ for\ 0^{\circ} \leq x < 360^{\circ}$

First of all this is not an identity
Put x= 0
In case you need to solve the equation

Put $1 = sin^2(x) + cos^2(x)$

EDIT: Far too late :D
• Feb 25th 2009, 06:45 AM
Beard
at 45
• Feb 25th 2009, 06:47 AM
Chop Suey
I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.
• Feb 25th 2009, 06:53 AM
ADARSH
Read the initial lines
This is an equation which has solutions (Nod) and not identity(Shake)
• Feb 25th 2009, 06:53 AM
HallsofIvy
Quote:

Originally Posted by Beard
at 45

Sine and cosine are equal for 45 degrees and at 180+ 45= 225 degrees.

That, of course, would be 2x. Now what values does x have?
• Feb 25th 2009, 07:44 AM
Beard
Quote:

Originally Posted by Chop Suey
I kind of automatically assumed that Beard mistitled this thread and meant it as a solve the trigonometric equation.

Sorry about this


text:
Journal article

### On the motion of an elastic solid inside of an incompressible viscous fluid

Abstract:

The motion of an elastic solid inside of an incompressible viscous fluid is ubiquitous in nature. Mathematically, such motion is described by a PDE system that couples the parabolic and hyperbolic phases, the latter inducing a loss of regularity which has left the basic question of existence open until now. In this paper, we prove the existence and uniqueness of such motions (locally in time), when the elastic solid is the linear Kirchhoff elastic material. The solution is found using a top...

### Access Document

Publisher copy:
10.1007/s00205-004-0340-7

### Authors

Coutand, D More by this author
More by this author
Institution:
University of Oxford
Department:
Oxford, MPLS, Mathematical Inst
Journal:
Archive for Rational Mechanics and Analysis
Volume:
176
Issue:
1
Pages:
25-102
Publication date:
2004-07-21
DOI:
EISSN:
1432-0673
ISSN:
0003-9527
URN:
Source identifiers:
404779
Local pid:
pubs:404779
Language:
English
Keywords:


text:
# Accounts receivable

## Related Questions in Accounting Concepts and Principles

• ### --acat-accounts and notes payable

(Solved) January 27, 2011

After the accounts are adjusted and closed at the end of the fiscal year , accounts receivable has a balance of $430,000 and allowance for doubtful accounts has a balance of$25,000. What is the net expected realizable value of the accounts receivable? a. $25,000 b.$405,000 c. $430,000 d.$455,000

• ### net realization value accounts receivable

(Solved) February 03, 2011

• ### After the accounts are adjusted and closed at the end of the

February 22, 2012

After the accounts are adjusted and closed at the end of the fiscal year , Accounts Receivable has a balance of $450,000 and Allowance for Doubtful Accounts has a balance of$25,000. What is the net realizable value of the accounts receivable? $25,000$425,000 $450,000$455,000

• ### acat-accounts and notes recieveable

(Solved) January 27, 2011

Wesser, Inc. has written off a $2,000 account as noncollectable. The balance in accounts receivable was$271,000 and the balance in the allowance for doubtful accounts was \$8,000 prior to the write-off. What is the net realizable value of accounts receivable after the receivable was written off?...

• ### The following information comes from the accounts of Jersey Company. Account Title Beginning Balance...

(Solved) October 11, 2015

? What was the net realizable value of receivables at the end of the accounting period? b. The note has a 6 percent interest rate and 24 months to maturity. What amount of interest revenue was recognized during the period? How much cash was collected for interest?


text:
# Quaternion limitations

This topic is 4614 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi everyone, I made this nice engine that uses quaternions for rotation math, I am also attempting (<- notice it's not working >.<) to use quats for setting basic simulation features like rotational speed and acceleration. However quat's can only represent +- pi rad's of rotation (since they use sin & cos representations for the angular magnitude), so what happens is that I set a ball (or any other object) to rotate at 1.5pi rad/sec. When a frame completes I get a delta T (in seconds) in an update function. The update function scales the quat's rotation by dT and performs the rotation. However when I get above pi rad/sec in either direction it naturally wraps around to the other sign (like 1.5pi rad has the same net effect as -.5pi) and so the update function scales -.5pi by dT instead of 1.5pi rad/sec and (not entirely unexpected) my rotation is all screwed. Is there any way to overcome this limitation of quats (without resorting to saying the rotation must be in rad/(.01*sec) and the likes which only strech the cap point)? I have been reading some on wiki and came across tensors, they sound mighty interresting. Can they help? (are they any use at all in computer graphics?) Any thoughts are welcomed Greets, Egelsoft

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When using quaternions for dynamics it is important to realize the relationship between quaternions, an axis-angle representation of an orientation, and a rotational torque.

Effectively, a Quaternion is just an axis to rotate about and an angular amount of how much to rotate. It is simply in a slightly different form that has nice properties when interpolating between them.

Generally a quaternion is used to express an orientation which can only be in the 360 degree range.

But you could have an angular velocity that is greater than 2 PI/frame. It won't look great if you object is rotating that fast every frame. It would lead to that weird strobing effect you sometimes see on video of rotating objects. But you can do it.

So if your angular velocity is something big, you then need to intergrate it with respect to your timestep to get your new orientation which can then be expressed as a quaternion.

I would suggest reading David Baraff's course notes on this, as he gives the formula and it is sort of drawn out to derive. www-2.cs.cmu.edu/~baraff/sigcourse/notesd1.pdf

Formula 4-2 should show you what you want.

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Quaternion's aren't "amount of rotation", they are "change in orientation".

If you spin around twice, your orientation did not change -- you are still facing exactly the same way -- even if you rotated alot.

If you want to be able to encode any rate of rotation, but use Quaternion, try the following.

Let Q by a Quaternion. Let Q be the "reverse" of that Quaternion. (ie, QQ = QQ = no change in orientation. Q is the inverse of Q.)

Let R(d) be a rotation by d degrees around a fixed axis (like the z axis). Note that this rotation can be converted into a Quaternion relatively easily, but the amount of rotation should not be stored as a Quaternion.

Then, multiplied as quaternions, the following:
QR(d)Q
is the orientation caused by a rotation by d degrees around some axis.

Which axis? Whatever axis the quaternion Q moves the z axis to.

If you want to see the effects of 1/100th of that rotation, you simply calculate the quaternion:
QR(d/100)Q

So your rate-of-rotations should store two values:
Q -- the Quaternion that represents the axis around which you are rotating
d -- the amount of rotation you want to do around that axis.

(I am assuming that your library has right-most quaternion acting on your point first -- it is possible my multiplication is backwards.)

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Yes, exactly as they say; quaternion is not what you need to use for angular velocities.
Angular velocity can be best represented as 3D vector. That's how it is usually(or always) done in physics. Direction of vector specifies axis of angular velocity, length specifies rotation speed itself. There's no more need to use quaternions there than to use quaternions to store linear velocities.[smile]

Now, how inertia tensors is used: the momentum of spinning body is inertia tensor * angular velocity . This formula is usually used other way to get angular velocity from momentum, tho.

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Quote:
 Original post by DmytryAngular velocity can be best represented as 3D vector.

wow.. I never thought of it that way. I was strugeling with the fact that vectors would require an aditional variable to indicate rotation about that axis, never thought about encoding them in the vector length. I'll look it up on google. This could mean seriously elegant code...
Thanx!

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# A rectangle is formed by increasing two opposite sides of a square of

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A rectangle is formed by increasing two opposite sides of a square of side length x by y units, and decreasing the two remaining sides of the square by y units. What is the area of the rectangle?

A. $$4x$$

B. $$4x - 2y$$

C. $$x^2 - 2y$$

D. $$x^2 + 2y$$

E. $$x^2 - y^2$$

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Re: A rectangle is formed by increasing two opposite sides of a square of  [#permalink]

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26 Jun 2018, 04:31

Solution

Given:
• A rectangle is formed by increasing two opposite sides of a square of side length x by y units, and decreasing the two remaining sides of the square by y units

To find:
• The area of the rectangle

Approach and Working:
• The length of the rectangle = x + y
• The breadth of the rectangle = x – y
• Therefore, the area of the rectangle = (x + y) (x – y) = $$x^2 – y^2$$

Hence, the correct answer is option E.

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Re: A rectangle is formed by increasing two opposite sides of a square of  [#permalink]

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26 Jun 2018, 08:17
Bunuel wrote:
A rectangle is formed by increasing two opposite sides of a square of side length x by y units, and decreasing the two remaining sides of the square by y units. What is the area of the rectangle?

A. $$4x$$

B. $$4x - 2y$$

C. $$x^2 - 2y$$

D. $$x^2 + 2y$$

E. $$x^2 - y^2$$

The 2 sides of the Rectangle are $$(x + y)$$ and $$(x - y)$$, thus area of the rectangle is $$x^2 - y^2$$, answer must be (E)
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Re: A rectangle is formed by increasing two opposite sides of a square of  [#permalink]

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27 Jun 2018, 17:12
Bunuel wrote:
A rectangle is formed by increasing two opposite sides of a square of side length x by y units, and decreasing the two remaining sides of the square by y units. What is the area of the rectangle?

A. $$4x$$

B. $$4x - 2y$$

C. $$x^2 - 2y$$

D. $$x^2 + 2y$$

E. $$x^2 - y^2$$

The area of the rectangle is (x + y)(x - y) = x^2 - y^2 square units.

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Re: A rectangle is formed by increasing two opposite sides of a square of &nbs [#permalink] 27 Jun 2018, 17:12
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# A rectangle is formed by increasing two opposite sides of a square of

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Why are gold mirrors yellow?

Why are golden mirrors yellow?

Do they add a yellow component to the spectrum or absorb non-yellow components?

If they absorb, then why are they used in telescopes being imperfect?

If they add a yellow component, then where do they take energy for it from?

JWST mirrors are coated with gold

Do they add some corrections in the on-board computer to compensate for the color of gold?

• Since the Webb telescope operates in the IR, the color in the visible is pretty irrelevant... Aug 17 '15 at 16:19
• On the color of metals: physics.stackexchange.com/q/72368 Aug 17 '15 at 16:23
• Just as a fun fact - visible light is actually a pretty crap approximation to other forms of light. Like you may think "close enough" and hope but wavelengths matter, for example infrared lenses are opaque, they're made out of ... uhh... a metal? Lens: i.imgur.com/ypYhXdb.jpg radiowaves are weirder still! i.imgur.com/rfVfT42.jpg (this is 22.5cm tall for reference) Aug 17 '15 at 23:34
• @Johannes that's something I addressed in my (first!) answer here. Aug 18 '15 at 2:00
• I'm surprised this question isn't merely: "Why is gold yellow?" since most gold things are yellow (plus metallic lustre). The answer is, of course, relativity. physics.stackexchange.com/questions/72368/… Aug 18 '15 at 19:31

I've made this into an answer because it's too long for a comment, and I really want to show the pictures.

It is tempting to think of visible light as "close enough" to (near by wavelengths) and to conclude that "yes, actually, the yellow does affect it. I want a mirror without an obvious tint"

However you are wrong, Physics will slap you down.

Exhibit A

(There's a book called Optics by Eugine Hecht that has a picture of some such lenses, but I couldn't find that picture. This is the best I found with quite a bit of googling)

This lens is made from "silicon or germanium" according to this link (where I found the picture) - I could have swore it was steel, anyway!

This lens is completely opaque to us, but to actual infrared cameras (the 1-$\mu$ sort of wavelengths) this is probably a good lens!

Let us go further.

Exhibit B

For reference this is 22.5cm tall

This is a lens for radiowaves. As you can just about see, it doesn't even "refract" our visible light, that light just goes right through the MASSIVE GAPS in the mesh. However to radiowaves, this is a lens!

Exhibit C

Finally, we come to visible light, I took this picture (bit of shameless "I'm proud of this", I know):

As you can probably tell from the soft background and such it was taken with a lens with a very large aperture (50mm f/1.4), and as you can (hopefully not) see the different wavelengths of light have actually gone different paths through the lens. Notice:

As you can see, "glass" isn't even very good at treating the visible wavelengths the same way. A picture is worth a thousand words.

I used a good lens, which is why the effect is so small, but this shows the principle behind it.

Exhibit D

This is a microwave door - it is opaque to microwaves, but as you can see, lets visible light through. (See Faraday Cage)

Exhibit E

WiFi. It can pass through walls and doors.

It should be clear now that light doesn't exactly behave like what our brain calls "light"

Finally

I hope this helps. As you can see - WAAAY to long for a comment.

• Great synopsis of the many varied interactions between electromagnetic radiation and matter. A bit off the original topic, but still. Incidentally, chromatic aberration on modern digital cameras is often mitigated with a lookup/scaling: the pixel maps for R,G and B are magnified by different factors and turned into the composite. For the price of half a pixel's worth of blurring, the aberration is almost completely eliminated. Aug 18 '15 at 12:33
• @Rafael I had hoped to find a picture of some of the lenses used for the massive radiowave telescopes, as they're literally just poles of different lengths! Unfortunately my Google-fu is weak. I am really glad that you and 5 other people at least found this fun. I was hoping to go for "cool related facts" Aug 18 '15 at 13:11
• I liked it enough that I leafed through my Hecht and Zajak to see if I could find the image you remembered. But it does not exist in the 1980 printing. I know, I am dating myself... Aug 18 '15 at 15:29
• @Floris just for reference, I read the 4th edition book (2001 edition) - it was a library copy because there's no way I could afford that! It is (sadly) a very good book though. I'd love an alternative if anyone has any books in mind! Also thanks. I think I'll like this SE site. Aug 18 '15 at 15:49
• Most confusing but "oh wow!" moment of college was when my chem professor taught me that radio waves are light. Then I realized you can shine that light through a house. Mind blown. Mar 4 '16 at 5:45

If you look at the reflectivity of gold (vs silver or aluminum) you can see a plateau at wavelengths below 500 nm source:

If blue wavelengths are not reflected as well as other colors, the resulting image will look "more yellow" - which is what you see.

At longer wavelengths, gold is a very good reflector (better than the other two above 600 nm). It also doesn't tarnish, so its reflectivity is less affected by atmospheric contamination.

If you need anything approaching accurate measurement, you have to calibrate your system at any rate - beside the mirrors and lenses, you need to consider the response of the detector, effects of the atmosphere, and pretty much everything in (or near) your optical path. Serious photometry needs serious calibration, as Chris White pointed out in the comment.


text:
# Can relativization change the direction of separation?

Are any $A$, $B$, and $O$ such that:

• $O$ is a set (for oracle),
• $A$ and $B$ are the names of two known complexity classes,
• $A^X$ and $B^X$ have well-defined accepted meanings,
• $A=A^\emptyset\subset B^\emptyset=B$, i.e. $A$ is strictly contained in $B$,
• $A^O\supset B^O$, i.e. $A^O$ strictly contains $B^O$.

In the case of $\mathsf{P}$ and $\mathsf{EXP}$, it's impossible to find an oracle $O$ relative to which $\mathsf{P}^O$ strictly contains $\mathsf{EXP}^O$ since a $\mathsf{EXP}^X$ can completely simulate every step of any turing machine in $\mathsf{P}^X$ (i.e. the time hierarchy theorem relativizes).

I'm wondering if there are complexity classes $A$, $B$ s.t. $A \subset B$, yet for some oracle $O$, $A^O \supset B^O$. In other words, can the direction of strict inclusion change when complexity classes are relativized?

• Since you wrote “in the most general case,” what is your definition of “complexity class”? – Tsuyoshi Ito Jan 29 '13 at 3:08
• Also note that there are more than one definition for space-bounded complexity classes in a relativized world, depending on whether you count the oracle tape as part of space or not. “Take a complexity class, and attach an oracle” is not a nice operation because we are usually using different oracle models e.g. when we talk about L^O (where the oracle tape usually does not count towards the space complexity) and when we talk about PSPACE^O (where the oracle tape usually counts towards the space complexity). I think that there was a recent question related to this point…. – Tsuyoshi Ito Jan 29 '13 at 3:54
• It is the other way around. P^HALT ⊊ EXP^HALT, and as I said, it is just the time hierarchy theorem. – Tsuyoshi Ito Jan 29 '13 at 12:46
• Not all classes on the complexity zoo admit oracles: what would $CFL^O$ mean? – Max Jan 29 '13 at 14:49
• The zoo used to display a warning sign reading “Please do not feed oracles to the complexity classes! These classes require a specially balanced diet to ensure proper relativization.” Unfortunately, it seems to be gone (or well hidden) in its Wiki reincarnation. – Emil Jeřábek Jan 29 '13 at 17:45


text:
# IBPS Simplification Questions with Answers Practice online test 30

Description: free IBPS Simplification Questions with Answers Practice test 30 for IBPS PO Preliminary and Main online test Prepare bank PO banking mock exams adda

1 . What value should come in place of question mark (?) in the following questions?
$\sqrt {(10648)^{1 \over 3} - (7776)^{1 \over 5}}$ = $\sqrt [6]{?}$
 46656 4096 16384 1296 1024
2 . What value should come in place of question mark (?) in the following questions?
$1224 \over ?$ = $? \over 306$
 524 612 728 772 848
3 . What value should come in place of question mark (?) in the following questions?
[$8 \over 15$ + $4 \over 25$] $\times$ ? = 780
 1125 1250 1280 1375 1420
4 . What approximate value should come in place of question mark(?) in the following questions?
127% of 75 + 28% of 277 = ?
 162 173 181 187 None of these
5 . What approximate value should come in place of question mark(?) in the following questions?
${(0.18)^2} + {(1.6)^2} \over 0.08$ = ?
 24 28 32 36 40
6 . What approximate value should come in place of question mark(?) in the following questions?
(59.842 $\div$ 1.982) $\times$ 6.97 - 17.77 $\times$ 3.2 = ?
 115 135 105 155 165
7 . What approximate value should come in place of question mark(?) in the following questions?
$\sqrt [3]{1330000}$ = ?
 103 110 117 120 128
8 . What approximate value should come in place of question mark(?) in the following questions?
{(7878 + 333 + 632) - ($11.9 \times 2.1 \times 7.09$)} $\times$ 2.532 = ?
 16900 20100 21700 22800 23000
$(17)^{8.8}$ $\times$ $(289)^{- 14}$ $\div$ $1 \over (17)^{- 1}$ = 17 $\times$ $(17)^?$


text:
## But please don't let it become sentient!

Posted on 05 April 2016

In this post we’re going to explore Markov Chains and how to use them to create a twitter 'AI' that can create new tweets based on the behavior of other twitter users.

This post comes with a corresponding repository that contains a working example of a Twitter AI based on Markov Chains.

So what are Markov Chains? Markov chains, named after Andrey Markov, are a way to model behavior. They work by modeling behavior as a series of transitions between nodes in a graph. They can also be seen as a state machines where movement between states is based on chance.

A very simple Markov Chain can be drawn like this:

This diagram only has two states, A and B. You can also see that each state has two transitions: one to itself and one to the other state. These transitions also have a chance attached and for each state the sum of the chances adds up to 100%.

Very pretty visual representations with also a more in-depth explanation can be found in this blog post.

## Modeling language behavior

So how does this knowledge apply to creating a twitter AI? Like many other forms of behavior the language patterns of a group of persons or a person can be expressed as a Markov chain. We model the behavior by feeding nodes to the graph 'strengthening' bond between certain words while also weakening bonds between others.

We do this by, in essence, counting occurrences of word combinations. We split the sentence (square in orange below) into a sequence of tokens (fat arrows in purple). From that we build a graph (circles) where the words are connected via edges leading to another word (arrows). The process is represented visually here:

So we have a sentence "foo bar baz", it’s split into individual words and then turned into a graph You might notice the {S} and {E} nodes. These are 'special' nodes that represent the {s}tart and {e}nd of the sentence. We need a start node so we know where to enter the graph and we also need an end node so we don’t generate sentences of infinite length.

The numbers above the arrows are the transition counts. Since we only fed the chain the sentence "foo bar baz" this is now also the only possible outcome: from the start the only possible transition is to "foo", from "foo" the only possible transition is "bar", then "baz" and then the only option is to finish the sentence. Since this is obviously quite boring we now feed it the next sentence:

We have now also inserted "foo baz bar" and as you can see three new connections have been created.

The possible results from this graph immediately have grown substantially. Not only can "foo baz bar" now be a result from this chain, if you trace through the graph you can see it is also possible for, for example, "foo baz" or "foo baz bar baz" to be generated. In fact: there is now a cycle between "baz" and "baz" so in theory it is possible to generate a string starting with "foo" and then repeating "bar baz" infinitely.

So why won’t this be an issue? Because each transition has a chance that is simply the transition count divided by the sum of all transitions out of that state. So in the case of {S} to "foo" this is 2/2 = 100%. In the case of transitions out of "baz" there is a 50% chance (1/2) of it transitioning to "bar" and a 50% chance (1/2) of it transitioning to {E}.

We’ve added "foo foo bar" and you notice that we now have a transition from "foo" to "foo" added while also incrementing the counts on "foo" → "bar" and "bar" → {E}.

We now have a beautiful stochastic model of our little textual world that can now be used to generate completely new text. It will often generate sentences we’ve fed into the chain (like "foo foo bar") but also sentences we didn’t ("foo foo baz bar") while still using the same type of language. It will never generate something like "bar bar foo" because the model doesn’t allow it to: we never allowed it to make a connection from the {S} node to "bar".

## Building our AI

After learning this I hope it becomes clear that we can apply this to any sequence of 'things'. It doesn’t really matter if it’s words from a bible text, tweets or even individual letters in english words: they can all be modeled this way as transitions between nodes in a graph.

So we’ll do just that: the most important class in our solution is the MarkovChain class. The add method allows you to add sentences to the chain while the generate() method generates sentences.

A very quick explanation of the 'add' process is that you split the input text up in pairs of words, with the first pair having a "from" that is empty and the last pair having a "to" that is empty. So "foo bar baz" gets split into (null, foo), (foo, bar), (bar, baz) and (baz, null). Each of these then gets passed to the word nodes (which get created on the fly if they don’t exist yet) and the counts get incremented.

Generating a sentence is also fairly straightforward: we grab the start node and from there select a random transition based on the transition chances. The chance for a transition is simply the count of an edge divided by the sum for the node (which we store with the node). The sentence 'ends' when the randomly selected node is the end node.

## Running the example

The repository contains a runnable Spring Boot application that showcases the 'bot'. To run the example you will need to provide a configuration.properties file with your Twitter OAuth2 credentials:

twitter4j.oauth.consumerKey=...
twitter4j.oauth.accessTokenSecret=...

So first fork and clone the example application and assuming you created the properties file (anywhere on the classpath works) correctly you can either start the Application class or use maven to build a runnable jar and run it from your command line.

After it’s started successfully you should be able to point your browser to http://localhost:8080 and see a small AngularJS application that automatically refreshes

## Playing Around

The main view shows the state of the 'main' Markov Chain that gets constantly fed from a sample of the twitter feed. Quite soon you’ll see that some words (The, I) occur more often than others. You should also see a constant feed (1 per second) of generated tweets based on the chain.

If you switch to the user view you can input a user name (or click one of the provided ones) and do the same thing. But instead of a random selection of all english tweets it’ll 'mimic' a real used based on their language. So by inputting 'realdonaldtrump' for example you will create tweets similar to those of DeepDrumpf.

## Conclusion

I hope I at least entertained you and perhaps even inspired you to build your own 'AI'. Feel free to contact me if anything is unclear either via the contactdetails in my about page or through issue’s in the repository.


text:
# Differentials on the arc space

-
Roi Docampo, University of Oklahoma
Fine Hall 322

We study the sheaf of Kähler differentials on the arc space of an algebraic variety. We obtain explicit formulas that can be used effectively to understand the local structure of the arc space. The approach leads to new results as well as simpler and more direct proofs of some of the fundamental theorems in the literature. The main applications include: an interpretation of Mather discrepancies as embedding dimensions of certain points in the arc space, a new proof of a version of the birational transformation rule in motivic integration, a new proof of the curve selection lemma for arc spaces, and a description of Nash blow-ups of jet schemes. This is joint work with Tommaso de Fernex.


text:
Logic at Columbia University

## Category: News

### A Course on Epistemic Game Theory

From May 12 – May 23, 2014, the EPICENTER will give a two-week intensive course on Epistemic Game Theory at Maastricht University.

Audience: This course is primarily meant for master and PhD students at all universities in the Netherlands and abroad. But researchers are also very welcome.

Instructors: The members of the EPICENTER: Christian Bach, Angie Mounir, Christian Nauerz, Andrés Perea and Elias Tsakas.

Register? Please send an E-mail to Andrés Perea at: course@epicenter.name

Master and PhD students at Maastricht University should also register via the course Capita Selecta: Econometrics, Mathematical Economics and Operations Research (EBC4011). Officially, this course is part of Capita Selecta.

Book:  The first seven days of the course will be based on the book Epistemic Game Theory: Reasoning and Choice, by Andrés Perea.

### Turing meeting at Boston University

Turing 100
Sunday, November 11 and Monday, November 12
Photonics Center, 9th floor Colloquium Room
8 St. Mary’s Street, Room 906

Sunday, 10:00am-12:00pm
I. Turing’s Philosophical and Logical Foundations

“On Formalism Freeness: A Meditation on Gödel’s 1946 Princeton Bicentennial Lecture”
Juliette Kennedy Mathematics and Statistics, University of Helsinki

“Turing, Church, Gödel, a personal perspective”
Michael Rabin Computer Science, Harvard University

“Turing and Wittgenstein”
Juliet Floyd Philosophy, Boston University

Sunday, 1:45pm-3:45pm
II. Turing and Mathematics: Computability and Definability

“Universality is Ubiquitous”
Martin Davis Courant Institute, NYU; Mathematics, UC Berkeley

“Collapsing Sentences”
Gerald Sacks Mathematics, Harvard University and MIT

“The Hierarchy of Definability: An Extended Thesis”
Theodore Slaman Mathematics, UC Berkeley

Sunday, 4:00pm-6:00pm
III. Turing and Cryptography

“Rational Proofs”
Silvio Micali Computer Science, MIT

“Turing and the Growth of Cryptography”
Ronald Rivest Computer Science, MIT

“Alan Turing and Voice Encryption”
Craig Bauer Mathematics, York College of Pennsylvania

Monday, 9:30am-12:15pm
IV. Turing and AI

Title TBA
Marvin Minsky Media Arts and Sciences, MIT

“Why Neanderthals Couldn’t Pass Turing’s Test and When Computers Will”
Patrick Henry Winston Computer Science, MIT

“What’s Wrong with the Moral Turing Test?”
Matthias Scheutz Computer Science, Tufts University

“Embodying Computation at Higher Types”
S. Barry Cooper Mathematics, University of Leeds

Monday, 2:00pm-4:00pm
V. The Church-Turing Thesis

“Normal Forms for Puzzles: an Enigmatic Variant of Turing’s Thesis”
Wilfrid Sieg Philosophy, Carnegie Mellon University

Title TBA
Stephen Wolfram Wolfram Research

“Is there a Church-Turing Thesis for Social Algorithms?”
Rohit Parikh Computer Science, Mathematics, Philosophy, CUNY

Monday, 4:15pm-6:30pm
VI. Turing, Physics, and Probability

“Algorithmic Randomness and Turing’s Work on Normality”
Rod Downey Mathematics, Victoria University of Wellington

“Spacetime Physics and Non-Turing Computers”
Mark Hogarth Philosophy, Cambridge University

“The Mysterious Thesis”
Leonid Levin Computer Science, Boston University

### Karby: How not to attack luck egalitarianism

COLUMBIA PHILOSOPHY COLLOQUIUM
How not to attack luck egalitarianism
Derrick Darby (Kansas Univ.)
Tuesday, May 8th, 2012, 4:10 – 6:00 PM
716 Philosophy Hall, Columbia University

Reception to follow

### CFP: The Constructive in Logic and Applications

THE CONSTRUCTIVE IN LOGIC AND APPLICATIONS.
A 60TH BIRTHDAY CELEBRATION HONORING SERGEI ARTEMOV
May 23-25, 2012, detailed program here.
CUNY Graduate Center, 365 Fifth Avenue, New York City.

Constructivity is fundamental to the application of mathematics to the real world, whether in the exact sciences, the social sciences, or engineering. Constructive logics supporting this tradition have flourished over the last century. This conference is a celebration of the past and future of the constructive tradition. Distinguished logician Sergei Artemov is 60. He has made fundamental contributions exploring constructivity in logic, computer science, epistemology, game theory, and other areas. With this conference we honor his work, and by extension a subject whose content he has elucidated and whose boundaries he has expanded.

Scientific Committee: Anil Nerode (Cornell) and Melvin Fitting (CUNY)
Web page: ConstructivityAtCUNY.com.
Participants are invited to register, the registration is free.
Contact email: Info@ConstructivityAtCUNY.com.

The confirmed speakers are:
Arnon Avron (Tel Aviv University)
Sam Buss (University of California San Diego)
Robert Constable (Cornell University)
Solomon Feferman (Stanford University)
Melvin Fitting (City University of New York)
Harvey Friedman (Ohio State University)
Haim Gaifman (Columbia University)
Dexter Kozen (Cornell University)
Victor Marek (University of Kentucky)
Robert Milnikel (Kenyon College)
Joan Moschivakis (University of California Los Angeles)
Yiannis Moschovakis (University of California Los Angeles)
Pavel Naumov (McDaniel College)
Anil Nerode (Cornell University)
Rohit Parikh (City University of New York)
Vaughan Pratt (Stanford University)
Jeff Remmel (University of California San Diego)
Bryan Renne (University of Amsterdam)
Gerald Sacks (Harvard University /Massachusetts Institute of Technology)
Andre Scedrov (University of Pennsylvania)
Johan van Benthem (University of Amsterdam/Stanford University)

### Kehlmann’s new play about the life of Gödel

GOETHE-INSTITUT NEW YORK
English reading of Daniel Kehlmann’s new play
Tuesday, May 1, 2012, 6:00pm
Center for Jewish History
15 West 16th St. New York, NY 10011
Free

The Leo Baeck Institute and the Goethe-Institut New York are proud to present a special reading of Ghosts in Princeton, the first play by Daniel Kehlmann, in an English translation by Carol Brown Janeway. A montage of facts, fiction, and philosophy, the play follows stages in the life of the brilliant Viennese logician Kurt Gödel (1906-1978), who, by the age of only 24, had revolutionized the logic of mathematics. A deeply complex person, Gödel did not believe in the existence of time and, with age, suffered from delusions of persecution and paranoia. After being both falsely identified as a Jew in post-Anschluss Austria and at the same time declared fit for active service, Gödel, together with his wife Adele, set out on an arduous journey to Princeton to join their friend Albert Einstein at the Institute for Advanced Study.

Daniel Kehlmann is one of Germany’s most critically acclaimed young authors. His 2006 international best-seller Measuring the World (Die Vermessung der Welt) has been translated from the original German into more than 20 languages and awarded some of the most prestigious prizes in literature.

The author himself will introduce the reading, which will be followed by a brief Q&A and refreshments. The event is free of charge, but reservations are strongly encouraged as seating is limited. To reserve a seat, please call +1 (212) 744-6400 or send an email to Maryann Legaspi at mlegaspi@lbi.cjh.org.

Leo Baeck Institute and Goethe-Institut New York are extremely grateful to Deutsche Telekom for its support of this program. Special thanks also to H.E. Mr. Peter Wittig and Mrs. Huberta von Voss-Wittig.

Arancha San Ginés provides the info above.

### Baldwin: Axiomatic Set Theory and L_{omega_1,omega}

CUNY MODEL THEORY SEMINAR
Axiomatic Set Theory and $L_{\omega_1,\omega}$
John T. Baldwin (Mathematics, UIC)
Friday, April 20, 2012, 12:30 PM
Abstract. In the late 1960’s model theory and axiomatic set theory seemed to be inevitably intertwined. The fundamental notions of first order stability theory are absolute. We describe the role of this fact in the development of first order model theory independent from set theory since the 1970’s. The role of extensions of ZFC in infinitary logic is muddled. Important results are proved using weak extensions of ZFC; the use is not in general proved essential. We expound the following proof-scheme: (1) Prove an infinitary sentence is consistent with ZFC. (2) Prove there is a model of set theory for which this sentence is absolute. (3) Deduce the property it expresses is provable in ZFC. We will describe how this technique implies the following recent result: Theorem (Shelah) Let $\phi$ be a sentence of $L_{\omega_1, \omega}$. (a) If ‘algebraic closure’ fails exchange on models of phi then $\phi$ has many models in $\aleph_1$. (b) If $\phi$ is pseudo-minimal then it has model in the continuum. Here algebraic closure’ and pseudo-minimal’ are modifications of classical notions appropriate for the context.


text: Example 3.4 The catenoid. An interior gradient bound for classical solutions of the minimal surface equation in n variables was established by Bombieri, De Giorgi, and Miranda in 1968. This not only stimulated new work on using the old parametric methods, but also demonstrated the importance of computer graphics to visualise the studied surfaces and numerical methods to solve the "period problem" (when using the conjugate surface method to determine surface patches that can be assembled into a larger symmetric surface, certain parameters need to be numerically matched to produce an embedded surface). Mathém. In general, We provide a new and simpler derivation of this estimate and partly develop in the process some new techniques applicable to the study of hypersurfaces in general. Show that the Euler{Lagrange equation for E[v] = Z 1 2 jrvj 2 vf dx (v : !R) is Poisson’s equation u = f: Problem 2. Definition 3.2 A smooth surface with vanishing mean curvature is called a minimal surface. 92. Oxford Mathematical Monographs. par div. In Fig. One way to think of this "minimal energy; is that to imagine the surface as an elastic rubber membrane: the minimal shape is the one that in which the rubber membrane is the most relaxed. Structures with minimal surfaces can be used as tents. uis minimal. The partial differential equation in this definition was originally found in 1762 by Lagrange,[2] and Jean Baptiste Meusnier discovered in 1776 that it implied a vanishing mean curvature.[3]. We provide a new and simpler derivation of this estimate and partly develop in the process some new techniques applicable to the study of hypersurfaces in general. (1 + jr j 2) 1 = = 0: (2) This quasi-linear … Minimal surfaces can be defined in several equivalent ways in R3. By viewing a function whose graph was a minimal surface as a minimizing function for a certain area Then the Jacobi equation says that. However, the term is used for more general surfaces that may self-intersect or do not have constraints. If the projected Gauss map obeys the Cauchy–Riemann equations then either the trace vanishes or every point of M is umbilic, in which case it is a piece of a sphere. The complete solution of the Plateau problem by Jesse Douglas and Tibor Radó was a major milestone. In mathematics, a minimal surface is a surface that locally minimizes its area. If the soap film does not enclose a region, then this will make its mean curvature zero. Savans, 10:477–510, 1785. The local least area and variational definitions allow extending minimal surfaces to other Riemannian manifolds than R3. The above equation is called the minimal surface equation. Miscellanea Taurinensia 2, 325(1):173{199, 1760. Brownian motion on a minimal surface leads to probabilistic proofs of several theorems on minimal surfaces. Minimal surface theory originates with Lagrange who in 1762 considered the variational problem of finding the surface z = z(x, y) of least area stretched across a given closed contour. Show that the Euler{Lagrange equation for the functional L W[v] = 1 2 Z R Z Rd jv We prove several results in these directions. Question. Yvonne Choquet-Bruhat. h�b"Kv�" ���,�260�X�}_�xևG���J�s�U��a�����������@�������������/ (\$,"*&.! Essai d'une nouvelle methode pour determiner les maxima et les minima des formules integrales indefinies. Thus, we are led to Laplace’s equation divDu= 0. Minimal surfaces are part of the generative design toolbox used by modern designers. Using Monge's notations: p := ∂ f ∂ x; q := ∂ f ∂ y; r := ∂ 2 f ∂ x 2; s := ∂ 2 f ∂ x ∂ y; t := ∂ 2 f ∂ y 2; Where f ∈ C 2 ( Δ ⊂ R 2, R) is the minimal surface (any other function with the same values on the border of Δ has a bigger surface over it). Another cause was the verification by H. Karcher that the triply periodic minimal surfaces originally described empirically by Alan Schoen in 1970 actually exist. The loss of strong convexityor convexity causes non-solvability, or non Triply Periodic Minimal Surfaces A minimal surface is a surface that is locally area-minimizing, that is, a small piece has the smallest possible area for a surface spanning the boundary of that piece. Sci. This has led to a rich menagerie of surface families and methods of deriving new surfaces from old, for example by adding handles or distorting them. For a given constraint there may also exist several minimal surfaces with different areas (for example, see minimal surface of revolution): the standard definitions only relate to a local optimum, not a global optimum. 317 0 obj <>stream He derived the Euler–Lagrange equation for the solution In this paper, we consider the existence of self-similar solution for a class of zero mean curvature equations including the Born–Infeld equation, the membrane equation and maximal surface equation. 189 0 obj <> endobj The surface of revolution of least area. Example 3.3 Let be the graph of , a smooth function on . A classical result from the calculus of ariations v asserts that if u is a minimiser of A (u) in U g, then it satis es the Euler{Lagrange equation r u. Oxford University Press, Oxford, 2009. xxvi+785 pp. [7] In contrast to the event horizon, they represent a curvature-based approach to understanding black hole boundaries. DIFFERENTIAL EQUATION DEFINITION •A surface M ⊂R3 is minimal if and only if it can be locally expressed as the graph of a solution of •(1+ u x 2) u yy - 2 u x u y u xy + (1+ u y 2) u xx = 0 •Originally found in 1762 by Lagrange •In 1776, Jean Baptiste Meusnier discovered that it … Then is a minimal surface if by Example 2.20. Other important contributions came from Beltrami, Bonnet, Darboux, Lie, Riemann, Serret and Weingarten. The term "minimal surface" is used because these surfaces originally arose as surfaces that minimized total surface area subject to some constraint. Derivation of the formula for area of a surface of revolution. Schwarz found the solution of the Plateau problem for a regular quadrilateral in 1865 and for a general quadrilateral in 1867 (allowing the construction of his periodic surface families) using complex methods. 9.2 Numerical Results 155. ) if and only if f satisﬁes the minimal surface equation in divergence form: div grad(f) p 1 + jgrad(f)j2! While these were successfully used by Heinrich Scherk in 1830 to derive his surfaces, they were generally regarded as practically unusable. Presented in 1776. Phys. This property is local: there might exist regions in a minimal surface, together with other surfaces of smaller area which have the same boundary. Mémoire sur la courbure des surfaces. This property establishes a connection with soap films; a soap film deformed to have a wire frame as boundary will minimize area. A minimal surface is a surface each point of which has a neighborhood that is a surface of minimal area among the surfaces with the same boundary as the boundary of the neighborhood. Minimal surfaces necessarily have zero mean curvature, i.e. One might think that if the minimal surface equation had a solution on a smooth domain D ⊂ R n with boundary values φ, it would have a solution with boundary values tφ for all 0 ≤ t ≤ 1. 8 Minimal Surface and MembraneWave Equations 137. Jung and Torquato [20] studied Stokes slow through triply porous media, whose interfaces are the triply periodic minimal surfaces, and explored whether the minimal surfaces are optimal for flow characteristics. The minimal surface equation is nonlinear, and unfortunately rather hard to analyze. By contrast, a spherical soap bubble encloses a region which has a different pressure from the exterior region, and as such does not have zero mean curvature. A surface in three dimensional space generated by revolving a plane curve about an axis in its plane. General relativity and the Einstein equations. the Smith conjecture, the Poincaré conjecture, the Thurston Geometrization Conjecture). a renewed interest in the theory of minimal surfaces [7]. Tobias Holck Colding and William P. Minicozzi, II. Minimal surface theory originates with Lagrange who in 1762 considered the variational problem of finding the surface z = z(x, y) of least area stretched across a given closed contour. A famous example is the Olympiapark in Münich by Frei Otto, inspired by soap surfaces. The definition of minimal surfaces can be generalized/extended to cover constant-mean-curvature surfaces: surfaces with a constant mean curvature, which need not equal zero. Ulrich Dierkes, Stefan Hildebrandt, and Friedrich Sauvigny. Additionally, this makes minimal surfaces into the static solutions of mean curvature flow. Derivation of the Partial Differential Equation Given a parametric surface X(u,v) = hx(u,v),y(u,v),z(u,v)i with parameter domain D, ... For a minimal surface, the eigenvalues of the matrix S are opposites of one another, and thus %%EOF Paris, prés. "The classical theory of minimal surfaces", "Computing Discrete Minimal Surfaces and Their Conjugates", "Stacked endoplasmic reticulum sheets are connected by helicoidal membrane motifs", "Touching Soap Films - An introduction to minimal surfaces", 3D-XplorMath-J Homepage — Java program and applets for interactive mathematical visualisation, WebGL-based Gallery of rotatable/zoomable minimal surfaces, https://en.wikipedia.org/w/index.php?title=Minimal_surface&oldid=1009225491, Articles with unsourced statements from March 2019, Creative Commons Attribution-ShareAlike License. 2. Between 1925 and 1950 minimal surface theory revived, now mainly aimed at nonparametric minimal surfaces. Initiated by the work of Uhlenbeck in late 1970s, we study questions about the existence, multiplicity and asymptotic behavior for minimal immersions of closed surface in some hyperbolic three-manifold, with prescribed conformal structure on the surface and second fundamental form of the immersion. o T do this, e w consider the set U g all tly (su cien smo oth) functions de ned on that are equal to g @. My question is the following: since a geodesic is just a special case of a minimal surface, is there some analogous equation for the deviation vector field between two "infinitesimally nearby" minimal (or more generally, extremal) surfaces? The "first golden age" of minimal surfaces began. Get the full course herehttps://www.udemy.com/course/calculus-of-variations/?referralCode=DCDA4C6662157C098CE5 We give a counterexample in R 2. )%-#+'����������������ohdlbjfnaiemckg�����������������8�xeQa����͙=k��ӦN�. So we get the minimal surface equation (MSE): div(ru p 1 + jruj2) We call the solution to this equation is minimal surface. Jn J1 + IY'ul2. This definition uses that the mean curvature is half of the trace of the shape operator, which is linked to the derivatives of the Gauss map. ]�[�2UU���%,CR�-qT�4 mY.-����m���Cn�������u���;һm���C�j��+,W��e��{�aO�\C�t�R�Y^�I��\��Fw�+|N�Eaa��|/�����/�6=� �6� Over surface meshes, a sixth-order geometric evolution equation was performed to obtain the minimal surface . This is equivalent to having zero mean curvature (see definitions below). 2 f 11f 2! 0 = 0 Inthiscasewealsosaythat isaminimalsurface. But the integrand F (p) = q 1+|p|2 is not strongly convex, that is D2F δI, only D2F > 0. Hence the catenoid is a minimal surface. hޜѽK�Q��so"d��M�A���m����DS���H��� NJhsP�bK����[-J4�����Z>��s�{Ϲ�c�Ŋ��!Ys�2@*���֠W�S�='}A&�3���+�@�!������2�0�����*��! If u is twice differentiable then integration by parts yields (2.2) or, equivalently, (2.3) div (a(\i'u)) = 0 This partial differential equation is known as the minimal surface equation. He derived the Euler–Lagrange equation for the solution. One cause was the discovery in 1982 by Celso Costa of a surface that disproved the conjecture that the plane, the catenoid, and the helicoid are the only complete embedded minimal surfaces in R3 of finite topological type. [5], Minimal surfaces have become an area of intense scientific study, especially in the areas of molecular engineering and materials science, due to their anticipated applications in self-assembly of complex materials. 2 the surface M is generated by revolving about the x axis the curve segment y = f(x) joining P 1 - P 2. Appendix A: Formulas from Multivariate Calculus 161. Fix ˚: @!R, and introduce L(;˚) := fu2C0;1(); uj @ = ˚g; (1.1) the set of Lipschitz functions on whose restriction to @ is ˚. An equivalent statement is that a surface SˆR3is Minimal if and only if every point p2Shas a neighbourhood with least-area relative to its boundary. Initiated by … minimal e surfac oblem pr is the problem of minimising A (u) sub ject to a prescrib ed b oundary condition u = g on the @ of. Bernstein's problem and Robert Osserman's work on complete minimal surfaces of finite total curvature were also important. Progress had been fairly slow until the middle of the century when the Björling problem was solved using complex methods. In discrete differential geometry discrete minimal surfaces are studied: simplicial complexes of triangles that minimize their area under small perturbations of their vertex positions. 8.2 Derivation of MembraneWave Equation 138. Weierstrass and Enneper developed more useful representation formulas, firmly linking minimal surfaces to complex analysis and harmonic functions. Physical models of area-minimizing minimal surfaces can be made by dipping a wire frame into a soap solution, forming a soap film, which is a minimal surface whose boundary is the wire frame. [4] Such discretizations are often used to approximate minimal surfaces numerically, even if no closed form expressions are known. 303 0 obj <>/Filter/FlateDecode/ID[<9905AF4C536B704FAAAE36E66E929823>]/Index[189 129]/Info 188 0 R/Length 287/Prev 1231586/Root 190 0 R/Size 318/Type/XRef/W[1 2 1]>>stream Show that the Euler{Lagrange equation for the ‘surface area’ functional A[v] = Z p 1 + jrvj2 dx (v : !R) is the minimal surface equation div ru p 1 + jruj2 = 0: Problem 3. In 1776 Jean Baptiste Marie Meusnier discovered that the helicoid and catenoid satisfy the equation and that the differential expression corresponds to twice the mean curvature of the surface, concluding that surfaces with zero mean curvature are area-minimizing. … Expanding the minimal surface equation, and multiplying through by the factor (1 + jgrad(f)j2)3=2 weobtaintheequation (1 + f2 y)f xx+ (1 + f 2 x)f yy 2f xf yf xy= 0 J. L. Lagrange. 2 Abstract. the positive mass conjecture, the Penrose conjecture) and three-manifold geometry (e.g. A simpler version of the equation is obtained by lineariza-tion: we assume that |Du|2 ˝ 1 and neglect it in the denominator. 8.1 Derivation of Minimal Surface Equation 137. Acad. [citation needed] The endoplasmic reticulum, an important structure in cell biology, is proposed to be under evolutionary pressure to conform to a nontrivial minimal surface.[6]. 9 The KPIWave Equation 149. Lecture 7 Minimal Surface equations non-solvability strongly convex functional further regularity Consider minimal surface equation div √Du 1+|Du|2 = 0 in Ω u = ϕ on ∂Ω. %PDF-1.5 %���� The thin membrane that spans the wire boundary is a minimal surface; of all possible surfaces that span the boundary, it is the one with minimal energy. Another revival began in the 1980s. The minimal surface equation is the Euler equation for Plateau's problem in restricted, or nonparametric, form, which can be stated as follows [3, §18.9]: Let fix, y), a single-valued function defined on the boundary C of a simply connected region R in the x — y plane, represent the … By the Young–Laplace equation, the mean curvature of a soap film is proportional to the difference in pressure between the sides. Exercise: (i) Verify the above derivation of the minimal surface equation. ¼ >A7Y>hz á â ã ä Ï B6>AG6\8XY>/W XY:6>)i87958B`AG X \d^ XY:6>m^bZ6G6cAXn


text:
# [luatex] luatex Digest, Vol 34, Issue 12

Mohamed Bana mohamed at bana.org.uk
Mon Oct 10 13:25:32 CEST 2011

I am having problems unsubscribing to this list.  Could someone please do
it.

I tried using the online interface but that did not work.  Someone please
help.

—Mohamed

On 10 October 2011 11:00, <luatex-request at tug.org> wrote:

> Send luatex mailing list submissions to
>        luatex at tug.org
>
> To subscribe or unsubscribe via the World Wide Web, visit
>        http://tug.org/mailman/listinfo/luatex
> or, via email, send a message with subject or body 'help' to
>        luatex-request at tug.org
>
> You can reach the person managing the list at
>        luatex-owner at tug.org
>
> than "Re: Contents of luatex digest..."
>
>
> Today's Topics:
>
>   1. Re: ligatures with luaotfload (Javier M?gica de Rivera)
>   2. Re: ligatures with luaotfload (Khaled Hosny)
>   3. Re: Building LuaTeX (again). (Taco Hoekwater)
>
>
> ----------------------------------------------------------------------
>
> Message: 1
> Date: Sun, 9 Oct 2011 12:58:37 +0200
> From: Javier M?gica de Rivera  <javieraritz.ribadeo at gmail.com>
> To: luatex at tug.org
> Subject: Re: [luatex] ligatures with luaotfload
> Message-ID:
>        <CABCShTg3Oe25L1bP8V-g_FqW1D1T+OMcPQBKoEUM3Bib4vXEbg at mail.gmail.com
> >
> Content-Type: text/plain; charset=windows-1252
>
> > you have to explicitly set kern and ligatures: [...] +liga;+kern
>
> > if you don't select a script no feature will be enabled by default
>
> Thank you two, both of
>
> \def\otffont#1=#2 at{\font#1={#2: +kern;+liga} at}
> \def\otffont#1=#2 at{\font#1={#2: script=latn} at}
>
> work.
>
> Now that I know the answer I have looked again into the luaoftload
> manual and the OpenType Feature File Specification, and see that all
> the nedeed pieces are present, but they appear scatered here and
> there. You don't need to know anything about .tfm fonts in order to
> use them. I would expect the same for otf fonts as long as you don't
> ask for complicate operations. That is, I would expect
>
> \font\tenrm=LMRoman10-Regular
>
> to be the same as
>
> \font\tenrm=ec-lmr10
>
>
>
> "script
> OpenType script string, default value is dflt. Some fonts don?t assign
> features to
> the dflt script, in which case the script need to be set explicitly."
>
> but since Latin Modern fonts are designed just for the Latin script I
> could not imagine that keeping the default would miss features.
> And it is true that at the section of "Troubleshooting" there appears
> the example
>
> \font\myfont = MyFont.otf:script=latn;+liga;
>
> but here the key words latn and liga just happen to appear. That is,
> if you want kerning you still have to find the appropriate word for
> that somewhere else, as well as if you want some other script. And
> after a good deal of browsing through the Internet to find the right
> answer (the name for the kerning and ligature features) I had simply
> need to put ten different pieces of information together in order to
> make something work.
>
> To summarize, in my opinion
>
>    a) The package should try hard to make its default behavior what a
>
> and
>
>    b) It should provide a list of script names when it describes the
> keyword script, and a list of feature names when it mentions that font
> features are enabled/disabled "in the form of +feat/-feat" (there is
> not a single example).
>
>   The solution for point a) may just be to always activate kerning,
> ligatures and final or other alternatives shapes for glyphs (is this
> what is called contextual ligatures?)
>
>
> Finally, Khaled, thank you very much for the package! I had asked some
> two or three years ago how I could use an OpenType font in Luatex and
> the answer was essentially "program it yourself"
>
> Regards
> Javier A. M?gica
>
>
>
> ------------------------------
>
> Message: 2
> Date: Mon, 10 Oct 2011 01:56:15 +0200
> From: Khaled Hosny <khaledhosny at eglug.org>
> To: "General discussion of LuaTeX." <luatex at tug.org>
> Subject: Re: [luatex] ligatures with luaotfload
> Message-ID: <20111009235615.GA13725 at khaled-laptop>
> Content-Type: text/plain; charset=utf-8
>
> On Sun, Oct 09, 2011 at 12:58:37PM +0200, Javier M?gica de Rivera wrote:
> > > you have to explicitly set kern and ligatures: [...] +liga;+kern
> >
> > > if you don't select a script no feature will be enabled by default
> >
> > Thank you two, both of
> >
> > \def\otffont#1=#2 at{\font#1={#2: +kern;+liga} at}
> > \def\otffont#1=#2 at{\font#1={#2: script=latn} at}
> >
> > work.
> >
> > Now that I know the answer I have looked again into the luaoftload
> > manual and the OpenType Feature File Specification, and see that all
> > the nedeed pieces are present, but they appear scatered here and
> > there. You don't need to know anything about .tfm fonts in order to
> > use them. I would expect the same for otf fonts as long as you don't
> > ask for complicate operations. That is, I would expect
> >
> > \font\tenrm=LMRoman10-Regular
> >
> > to be the same as
> >
> > \font\tenrm=ec-lmr10
> >
> >
> >
> > "script
> > OpenType script string, default value is dflt. Some fonts don?t assign
> > features to
> > the dflt script, in which case the script need to be set explicitly."
> >
> > but since Latin Modern fonts are designed just for the Latin script I
> > could not imagine that keeping the default would miss features.
> > And it is true that at the section of "Troubleshooting" there appears
> > the example
> >
> > \font\myfont = MyFont.otf:script=latn;+liga;
> >
> > but here the key words latn and liga just happen to appear. That is,
> > if you want kerning you still have to find the appropriate word for
> > that somewhere else, as well as if you want some other script. And
> > after a good deal of browsing through the Internet to find the right
> > answer (the name for the kerning and ligature features) I had simply
> > need to put ten different pieces of information together in order to
> > make something work.
> >
> > To summarize, in my opinion
> >
> >     a) The package should try hard to make its default behavior what a
>
> OpenType is a complicated beast and will never be as simple as old TFM
> fonts (feature activation wise, else I see nothing simple about TFM
> fonts; you had to do a good deal of complicated setup to merely install
> them). However, \font\tenrm=LMRoman10-Regular was supposed to such work,
> but the way it was actually implemented requires selecting a script for
> things to work. This have been fixed in the unstable branch, so you may
> want to use that until there is a release.
>
> > and
> >
> >     b) It should provide a list of script names when it describes the
> > keyword script, and a list of feature names when it mentions that font
> > features are enabled/disabled "in the form of +feat/-feat" (there is
> > not a single example).
>
> I'm not going to duplicate OpenType layout tag[1] registry in the
> package documentation, sorry. However, improvements to make the
> documentation more clear to people not familiar with OpenType are very
> welcomed.
>
> >    The solution for point a) may just be to always activate kerning,
> > ligatures and final or other alternatives shapes for glyphs (is this
> > what is called contextual ligatures?)
>
> Things are more complicated than that since different scripts have
> different set of features that need to be activated by default, this why
> the old code required setting a script, now if no script is specified I
> just assume it is DFLT and activates the respective set of feature
> (which is the same as that of Latin script).
>
> > Finally, Khaled, thank you very much for the package! I had asked some
> > two or three years ago how I could use an OpenType font in Luatex and
> > the answer was essentially "program it yourself"
>
> Thanks all goes to Hans who is doing all the real work, Elie and others
> who wrote the plain tex/latex package, I'm merely glueing things
> together.
>
> Regards,
>  Khaled
>
> [1] http://www.microsoft.com/typography/otspec/ttoreg.htm
>
> --
>  Khaled Hosny
>  Egyptian
>  Arab
>
>
> ------------------------------
>
> Message: 3
> Date: Mon, 10 Oct 2011 10:05:32 +0200
> From: Taco Hoekwater <taco at elvenkind.com>
> To: "General discussion of LuaTeX." <luatex at tug.org>
> Subject: Re: [luatex] Building LuaTeX (again).
> Message-ID: <4E92A74C.4050203 at elvenkind.com>
> Content-Type: text/plain; charset=ISO-8859-1; format=flowed
>
>
> Hi,
>
> On 10/04/2011 11:10 AM, Keith J. Schultz wrote:
> > Hi Hartmut,
> >
> > I am not know the system you are using, but I can other you some
> thoughts.
>
> Same failure during crosscompile here
>
> > 2) the is not finding the reference for EnumFontsA at 16
>
> EnumFonts() is in gdi32.dll, which is not linked in luatex. That bit of
> code in GlobalParams.cc should have been preprocessed out (as it was in
> earlier versions, I think, thanks to PDF_PARSER_ONLY). Hartmut, is that
> enough information for you to fix it?
>
> Best wishes,
> Taco
>
>
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> End of luatex Digest, Vol 34, Issue 12
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