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# Math Help - Triangle Inequality Problem

1. ## Triangle Inequality Problem

Triangle TUV has sides TU, UV, and VT. TU=6x - 11, UV=3x - 1, and VT=2x + 3. Describe the possible values of x.

Any help is appreciated. I've been trying to figure this one out for a while.

2. Originally Posted by jstark
Triangle TUV has sides TU, UV, and VT. TU=6x - 11, UV=3x - 1, and VT=2x + 3. Describe the possible values of x.

Any help is appreciated. I've been trying to figure this one out for a while.
The vertices of the triangle are the points where these lines intersect (ie, the points where their equations are equal). Does that help?

3. I don't think I understand how the equations are equal there.

4. I don't think I understand how the equations are equal there.

5. Originally Posted by jstark
I don't think I understand how the equations are equal there.
All three equations represent straight lines. Those lines intersect each other at the points T, U and V.

For example, your equation for TU and your equation for UV intersect at the point U(a,b).

So x = a, and y = b are solutions to BOTH equations, since that point lies on BOTH lines.

If you think of the equations being y = 6x-11 and y = 3x-1. The point at which these two lines intersect lies on both lines. And hence the y-coordinate of that point is equal for both lines.

Hence y = y

Hence 6x-11 = 3x-1

3x = 10
x = 10/3

This is the x-coordinate of the intersection (and hence the x coordinate of point U!). If you plug that value of x into either of the two equations, you'll get the y coordinated for U.

6. OK

7. OK. I think I understand. Thank you so much.

8. Originally Posted by jstark
OK. I think I understand. Thank you so much.
Good.

Now, if you find the three points T U and V using the method above, then you should find which one has the LOWEST x coordinate, and which one has the HIGHEST x coordinate. The range of possible x values for this triangle is the values in between those two coordinates, including the coordinates themselves!

Let me know what answer you get, I'll let you know if you're right or not.

9. Originally Posted by jstark
Triangle TUV has sides TU, UV, and VT. TU=6x - 11, UV=3x - 1, and VT=2x + 3. Describe the possible values of x.

Any help is appreciated. I've been trying to figure this one out for a while.
If and only if the given terms refer to lengths of the triangles sides you have to use the triangle inequality:

The triangle ABC has the sides a, b, c. Then

$a+b>c\ \wedge\ a+c>b\ \wedge\ b+c>a$

So you get:

$6x-11+3x-1>2x+3\ \wedge\ 6x-11+2x+3>3x-1\ \wedge\ 3x-1+2x+3>6x-11$

$x>\dfrac{15}7\ \wedge\ x>\dfrac95\ \wedge\ 13>x$

Therefore you know:

$\boxed{\dfrac{15}7 < x< 13}$


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Hardcover | $35.00 Short | £24.95 | ISBN: 9780262015868 | 178 pp. | 6 x 9 in | | July 2011 Ebook |$25.00 Short | ISBN: 9780262299541 | 178 pp. | | July 2011

## The Tarskian Turn

Deflationism and Axiomatic Truth

## Overview

In The Tarskian Turn, Leon Horsten investigates the relationship between formal theories of truth and contemporary philosophical approaches to truth. The work of mathematician and logician Alfred Tarski (1901–1983) marks the transition from substantial to deflationary views about truth. Deflationism--which holds that the notion of truth is light and insubstantial--can be and has been made more precise in multiple ways. Crucial in making the deflationary intuition precise is its relation to formal or logical aspects of the notion of truth. Allowing that semantical theories of truth may have heuristic value, in The Tarskian Turn Horsten focuses on axiomatic theories of truth developed since Tarski and their connection to deflationism.
Arguing that the insubstantiality of truth has been misunderstood in the literature, Horsten proposes and defends a new kind of deflationism, inferential deflationism, according to which truth is a concept without a nature or essence. He argues that this way of viewing the concept of truth, inspired by a formalization of Kripke’s theory of truth, flows naturally from the best formal theories of truth that are currently available. Alternating between logical and philosophical chapters, the book steadily progresses toward stronger theories of truth. Technicality cannot be altogether avoided in the subject under discussion, but Horsten attempts to strike a balance between the need for logical precision on the one hand and the need to make his argument accessible to philosophers.

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vector movement

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lets say i have a struct that has a 2d position vector and a 2d velocity vector, to project it to the screen, i would use the x, y coords from the position vector (right?), then how would i go about calculating it''s next position such that it only moves 1 pixel per frame? thanks

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If that is the case you take your 2d velocity vector, make it a unit vector (so it only moves your position vector 1 pixel/unit at a time) and add it to your position vector.
So:
V(velocity vector)=(x,y)
|U|(unit velocity vector)=|V|/a=1
a- some scalar

P(position vector)=(x1,y1)
P(new position)=P+U

EPHERE

[edited by - EPHERE on December 5, 2002 12:36:05 PM]

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Your velocity vector should have a magnitude of one. The magnitude is its length and given by sqrt(x*x+y*y). To set a vector a desired length you multiply by the desird length divided by the current length. Once your velocity vector has the right length you just add it to the position. Strictly speaking for linear motion at a constant velocity it is position+velocity*time, but here your velocity is in pixels per frame and your time is measured in frames.

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ok, so if my data is,
Pos(2,3)
Vel(2,3)

my |V| is 2*2+3*3 = sqrt(13)

then i get stuck..
what next?

thanks

ah., ok, so i would then go
U = V/|V| to give me the unit vector..?

[edited by - ligh on December 5, 2002 6:03:54 PM]

ah ha, would u look at that.,
thanks for all the help EPHERE and LilBudyWizer!!

[edited by - ligh on December 5, 2002 6:10:04 PM]

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Pixels are discrete units. If you direction is (1,1), normalizing it won''t make you move one pixel/frame... What do you want exactly?

Cédric

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if i have an object, with an x,y for position and x,y for velocity, it will move across the screen at the given velocity, but smooth pixel/pixel movement..

eg,
struct vector{   float x,y}struct obj{   vector pos   vector vel}obj p;p.pos.x = 100;p.pos.y = 100;p.vel.x = 10;p.vel.y = -1;

after a time it should end up at 110,99 but i want it to visit all the places on the line along the way

and.. if i have two objects, p1, p2 given
p1.pos.x = 100
p1.pos.y = 50
p1.vel.x = 10
p1.vel.y = 0
p2.pos.x = 100
p2.pos.y = 150
p2.vel.x = 5
p2.vel.y = 0
the p1 shoudl move faster then p2..

[edited by - ligh on December 7, 2002 3:24:59 AM]

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## Avoiding abelian powers in binary words with bounded abelian complexity.(English)Zbl 1223.68089

Two words are abelian equivalent if they contain the same letters with the same number of occurences. The abelian complexity of a word is the number of its subwords of length $$n$$ that are pairwise abelian inequivalent. A nonempty word is an abelian $$k$$-power, if it is a concatenation of $$k$$ pairwise abelian equivalent subwords.
The authors consider the problem of the frequency of abelian $$k$$-powers in words having bounded abelian complexity and prove the followings theorems:
Every suffix of the well-known Thue-Morse word begins in an abelian $$k$$-power for all positive integers $$k$$;
There exists a uniformly recurrent infinite word with bounded abelian complexity that does not begin in an abelian square;
There exists a uniformly recurrent infinite word with bounded abelian complexity in which there are infinitely many positions where no abelian square occurs.
The authors also consider the effect of morphisms on abelian complexity and show that the bounded property of abelian complexity is preserved by morphisms.
Finally, the authors present an open problem, namely whether there exists an infinite binary word that avoids abelian squares in positions that occur in bounded gaps, and prove that it is equivalent to a well-known open problem proposed independently by G. Pirillo and S. Varricchio [“On uniformly repetitive semigroups”, Semigroup Forum 49, No. 1, 125–129 (1994; Zbl 0805.20052)] and L. Halbeisen and N. Hungerbühler [“An application of van der Waerden’s theorem in additive number theory”, Integers 0, Paper A07, 5 p. (2000; Zbl 0962.11017)], asking whether there exists an infinite word over a finite set of integers such that no two consecutive blocks of the same length have the same sum.

### MSC:

 68R15 Combinatorics on words

### Keywords:

avoidability in words; abelian power; abelian complexity

### Citations:

Zbl 0805.20052; Zbl 0962.11017
Full Text:

### References:

 [1] Richomme G., J. London Math. Soc. [2] Currie J., Adv. in Appl. Math. [3] DOI: 10.2307/2371431 · Zbl 0022.34003 [4] DOI: 10.1016/j.aam.2010.01.006 · Zbl 1203.68131 [5] DOI: 10.1051/ita/2010017 · Zbl 1211.68303 [6] DOI: 10.1007/BF02573477 · Zbl 0805.20052 [7] Halbeisen L., Integers 0 pp #A07– [8] DOI: 10.1016/0097-3165(74)90041-7 · Zbl 0279.05001 [9] DOI: 10.1016/0097-3165(79)90044-X · Zbl 0437.05011 [10] Shallit J., A Second Course in Formal Languages and Automata Theory (2009) · Zbl 1163.68025 [11] Allouche J.-P., Electron. J. Combin. 5 pp #R27– [12] DOI: 10.1017/CBO9781107341005 [13] DOI: 10.1016/S0304-3975(03)00092-6 · Zbl 1059.68083 [14] DOI: 10.1016/S0012-365X(97)00029-0 · Zbl 0895.68087 [15] DOI: 10.1016/j.tcs.2007.10.027 · Zbl 1133.68036 [16] Szemerédi E., Acta Arith. 27 pp 299–
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# A Diabolically Difficult Chemistry Puzzle

Originally posted 2022-07-24

Tagged: chemistry

Obligatory disclaimer: all opinions are mine and not of my employer

In 2012, I was a mentor at the USNCO Study Camp, which prepares and selects the top students in the U.S. to send to the International Chemistry Olympiad. I wrote a question for the final exam which only 2 of the 20 students completed. Both students admitted to brute-forcing the system of equation in 5 unknowns. I found it hilarious at the time that I’d stumped the entire camp with a seemingly trivial problem. The students were not as amused! Without further ado:

Balance the following equation.

$\ce{BH4- + BF3 -> B11H14- + BF4- + H2(g)}$

This 11-boron, 14-hydrogen species has $$C_{5v}$$ symmetry (dodecahedron with missing vertex), and its parent species $$\ce{B11H11^2-}$$ is apparently fluxional, rapidly interconverting with no one boron identifiable. If you’re wondering how I stumbled across such an obscure reaction, I decided at some point that boron was probably a simple enough element that I could learn everything there was to know about boron if I’d really tried. Boy, was I wrong! It turns out a lot of research went into obscure borohydride species as potential energy-dense rocket fuels during the Cold War.

A hint and solution for this equation comes after the spoiler break.

v v v v v v

## Hint

This is a redox equation. Find the two half-reactions.

v v v v v v

## Solution

The borohydride and fluoroborate species all have boron in their (+3) oxidation state, while the B11 species has a 15/11 oxidation state, which is about (+1.36) on average. The hydrogen, on the other hand, goes from hydride at the (-1) oxidation state to hydrogen gas at the (0) oxidation state. We’ll have to find the right ratios that lead to net electron transfer balance.

In a traditional redox reaction, typically what happens is that you first insert electrons to form the correct half-reaction, and then use $$\ce{H2O}$$, $$\ce{H+}$$ or $$\ce{OH-}$$ species to balance oxygens, hydrogens, and charges. As a simple example, for the reduction of Cr(+6) to Cr(+3):

$\ce{CrO4^2- -> Cr^3+}$

$\ce{CrO4^2- + 3e- -> Cr^3+}$

$\ce{CrO4^2- + 3e- -> Cr^3+ + 4H2O}$

$\ce{CrO4^2- + 3e- + 8H+ -> Cr^3+ + 4H2O}$

How do these steps apply to this problem? This isn’t an aqueous reaction - borohydrides are pyrophoric in air! Instead of using $$\ce{H+}$$ to balance our charges, we’ll use $$\ce{H-}$$.

$\ce{11BH4- + 20e- -> B11H14- + 30H-}$

$\ce{2H- -> 2e- + H2}$

Balanced reaction:

$\ce{11BH4- -> B11H14- + 10H- + 10H2}$

Free hydride is not in the final equation - this is where the fluoroborate species come in. You could see the role of $$\ce{BF3}$$ as a hydride sponge.

$\ce{4BF3 + 4H- -> 3BF4- + BH4-}$

Balancing and cancelling the hydrides, the final balanced equation is

$\ce{17BH4- + 20BF3 -> 2B11H14- + 15BF4- + 20H2}$


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# Expectation of inverse of normal RV, given that it is below a certain value

by dotpixel   Last Updated August 14, 2019 04:19 AM

I have a normal random variable $$X$$ with mean $$\mu$$ and variance $$\sigma^2$$. Any advice on how to compute the conditional expectation $$E[\frac{1}{X}|X \leq T]$$ where $$T$$ is a positive constant?

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## Related Questions

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SMS scnews item created by Hannah Bryant at Wed 25 May 2022 1227
Type: Seminar
Distribution: World
Expiry: 31 May 2022
Calendar1: 31 May 2022 1500-1600
CalLoc1: Quad S224 & via Zoom
CalTitle1: SMRI ’What is ...a virtual knot?’ Hans Boden (McMaster University)
Auth: hannahb@w1d4n6z2.staff.sydney.edu.au (hbry8683) in SMS-SAML

# SMRI ’What is ...?’: Boden -- ’What is ...a virtual knot?’

SMRI ’What is ...?’ seminar:
’What is a virtual knot?’
Hans Boden (McMaster University)

Tuesday, 31st May, 3:00pm - 4:00pm (AEST)
Quad S224 & Online via Zoom
Register for online attendance here:
https://uni-sydney.zoom.us/meeting/register/tZEqd-GpqTojEtay-NNhn95MOjOPMqvakwqB

Abstract: Virtual knots were introduced by Louis Kauffman in 1999 as a completion of
classical knot theory in which planarity is no longer required.  Virtual knots have been
studied using a variety of approaches, including algebra, combinatorics, and geometric
methods.  They also have strong connections to quantum topology and finite type
invariants.  This talk will survey some fascinating results that have been established
and present also open problems and directions for future research.

Biography: Dr.  Hans U.  Boden is a professor of mathematics at McMaster University in
Canada.  He is visiting the University of Sydney and SMRI from May 17 to June 11.  His
research interests are on the geometry and topology of manifolds, especially gauge
theory and low-dimensional topology.  In recent years, his work has focused on
developing geometric methods to understand knotting and linking in 3-dimensional
manifolds.  While in Sydney, he will be working closely with Dr.  Zsuzsi Dancso on a
collaborative project related to the Tait conjectures in knot theory.

Actions:


text:
# 需要常量表达式Constant expression is required

Const 语句没有正确初始化常量，或数组声明使用了变量来指定元素数。A Const statement does not properly initialize a constant, or an array declaration uses a variable to specify the number of elements.

## 更正此错误To correct this error

1. 如果该声明是 Const 语句，请检查以确保常量被初始化为文本、以前声明的常量、枚举成员，或结合了运算符的文本、常量和枚举成员的组合。If the declaration is a Const statement, check to make sure the constant is initialized with a literal, a previously declared constant, an enumeration member, or a combination of literals, constants, and enumeration members combined with operators.

2. 如果该声明指定数组，请检查是否正在使用变量指定元素的数量。If the declaration specifies an array, check to see if a variable is being used to specify the number of elements. 如果是，请用常量表达式替换该变量。If so, replace the variable with a constant expression.


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# nLab Poisson-Lie T-duality

Contents

### Context

#### Duality in string theory

duality in string theory

general mechanisms

string-fivebrane duality

string-string dualities

M-theory

F-theory

string-QFT duality

QFT-QFT duality:

# Contents

## Idea

What has come to be called nonabelian T-duality (Ossa-Quevedo 92) or Poisson-Lie T-Duality (due to Klimcik-Ševera 95, von Unge 02) is a generalization of T-duality from fiber bundles equipped with an abelian group of isometries (torus bundles) to those with a nonabelian group of isometries.

Poisson-Lie T-duality may also be made manifest at the level of type II supergravity in the framework of double field theory on group manifolds. Using this framework both the NS/NS sector and the R/R sector are captured, and this allows to derive the transformation of the RR fields for full Poisson-Lie T-duality (Hassler 17).

## As a partial duality of string theory

While ordinary abelian T-duality is supposedly a full duality in string theory, in particular in that it is an equivalence on the string perturbation series to all orders of the squared string length/Regge slope $\alpha'$ and the string coupling constant $g_s$, it has apparently been shown by Martin Roček (citation?) that there are topological obstructions at higher genus for non-abelian T-duality, letting it break down in higher orders of $g_s$; and already a genus-0 (tree level) it apparently breaks down for the open string (i.e. on punctured disks) at some order of $\alpha'$.

But see Hassler 20, Borsato-Wulff 20.

## References

The original articles are

Review includes

• I. Petr, From Buscher Duality to Poisson‐Lie T‐Plurality on Supermanifolds, AIP Conference Proceedings 1307, 119 (2010) (doi:10.1063/1.3527407)

• Konstadinos Sfetsos, Recent developments in non-Abelian T-duality in string theory, Fortschr. Phys., Special Issue: Proceedings of the “Schools and Workshops on Elementary Particle Physics and Gravity” (CORFU 2010), 29 August – 12 September 2010, Corfu (Greece) Volume59, Issue11‐12 (arXiv:1105.0537)

Relation to T-folds:

Discussion of the duality at the level of type II supergravity equations of motion is (using Riemannian geometry of Courant algebroids) due to

and in relation to double field theory:

Discussion within a broader picture of dual higher gauge theories, including 4d electric-magnetic duality:

Discussion of non-abelian T-duality from a comprehensive picture of higher differential geometry, relating Kaluza-Klein compactification on principal ∞-bundles to double field theory, T-folds, type II geometry, exceptional geometry, etc.:

• Benjo Fraser, Dimitrios Manolopoulos, Konstantinos Sfetsos, Non-Abelian T-duality and Modular Invariance (arXiv:1805.03657)

• Francesco Bascone, Franco Pezzella, Patrizia Vitale, Poisson-Lie T-Duality of WZW Model via Current Algebra Deformation (arXiv:2004.12858)

• Falk Hassler, Thomas B. Rochais $O(D,D)$-covariant two-loop β-functions and Poisson-Lie T-duality (arXiv:2011.15130)

Discussion in cosmology:

• Ladislav Hlavatý, Ivo Petr, Poisson-Lie plurals of Bianchi cosmologies and Generalized Supergravity Equations (arxiv:1910.08436)

Generalization to U-duality in exceptional generalized geometry:

• Emanuel Malek, Daniel C. Thompson, Energy Physics - Theory Poisson-Lie U-duality in Exceptional Field Theory (arxiv:1911.07833)

• Chris D. A. Blair, Daniel C. Thompson, Sofia Zhidkova, Exploring Exceptional Drinfeld Geometries (arxiv:2006.12452)

In the context of the BMN matrix model:

Discussion of $\alpha'$-corrections:

• Falk Hassler, Thomas Rochais, $\alpha'$-corrected Poisson-Lie T-duality (arXiv:2007.07897)

• Riccardo Borsato, Linus Wulff, Quantum correction to Poisson-Lie and non-abelian T-duality (arXiv:2007.07902)

Last revised on November 30, 2020 at 21:28:01. See the history of this page for a list of all contributions to it.


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# 1X + 5 = -10(a)How do I isolate x?A Add by 5 on both sidesB Subtract by 5 on both sidesMultiply by 5 on both sidesD

###### Question:

1 X + 5 = -10
(a)
How do I isolate x?
A Add by 5 on both sides
B Subtract by 5 on both sides
Multiply by 5 on both sides
D Divide by 5 on both sides
(b)
X=

### Christian created a model of his truck using a scale of 1 inch : 36 inches. If the height of his actual truck is 75.8 inches,

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### This is the type of growth experienced by a population that increases at a rate proportional to its size. the larger the

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### Was this Arizona a republican or deocratic state during the 2020 presidential election Identify congressmen and senators of the state

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### Write and solve an equation to find the value of x and the missing angle measures.

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### Alveolar ventilation remains constant when the respiratory rate is increased while the volume of air

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### 3< -5n+2n how to solve the inequality?

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### Is it true that in to kill a mockingbird, Mr. Underwood likened Tom Robinson’s killing to the senseless slaughter of songbirds

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### If you hold a rubber band loosely between two fingers and let it rest against your lips, and then stretch the rubber band,

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### Zoran graphed the line shown below. Which of the following represents theequation and slope of the line?A. y = -7; slope = -7B.

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### It takes a cheetah just 3s to reach his speed of 30m/s what is the acceleration?

It takes a cheetah just 3s to reach his speed of 30m/s what is the acceleration?...


text:
Asaf Karagila
I don't have much choice...

Recursion: recursion. As the joke goes. But that's actually a misnomer, since that would be an ill-founded definition, which is exactly the point where you can't do a recursive definition. I'm not here to analyse that joke, though. I'm here to talk about something else.

Some time ago, I was looking for something, and I couldn't even tell you what, and I came across an algorithm for computing the factorial of a number by recursion. Let's review the standard way recursive algorithm first:

$\operatorname{factorial}_1(n)=\begin{cases} 1 & n=0\\ n\cdot\operatorname{factorial}_1(n-1) & n>0\end{cases}$

This is the naive, and most straightforward way to do it. This is the classical example of factorial in programming, and it's fine. It works, and if you're limited to "standard integers" in programming (so $$2^{64}$$ or less, what with the year being 2020) then you can't even go beyond $$20!$$ anyway, so that's efficient enough.

In some compilers or even languages, e.g. Steel Bank Common Lisp, where tail-end optimisation is available, we can slightly modify the algorithm to get better results:

$\operatorname{factorial}_2(n)=\operatorname{fac}(n,1)$

$\operatorname{fac}(n,k)=\begin{cases} k & n=0\\\operatorname{fac}(n-1,n\cdot k) & n>1\end{cases}$

The compiler will recognise this and turn that into a loop, making the run much more efficient. But what about languages without that sweet-sweet tail-end recursion? Well, nothing much changed from the first attempt.

So, what happens if you're using a language that has "native support" for arbitrarily large natural numbers, and we want to compute something like $$10{,}000!$$? It has $$35{,}660$$ decimal digits, so it's certainly not a problem to do that (memory-wise). But suddenly the number of iterations becomes more pertinent. And here is the algorithm I saw.

$\operatorname{factorial}_3(n)=\operatorname{IP}(1,n)$

$\operatorname{IP}(i,j)=\begin{cases} i & i=j\\\operatorname{IP}(i,k)\cdot\operatorname{IP}(k+1,j) & i\neq j\text{ and }k\text{ is the midpoint between them}\end{cases}$

The idea is to break the set $$\{1,\dots,n\}$$ break it in half, compute the product over each half, and combine, and do that recursively until the intervals are "small enough". The implementation I saw was actually mixed, and for short enough intervals resorted to a loop instead (or: pre-optimised tail-end recursion if you want). The idea here is to have at most $$\log_2 n$$ stack frames at any given moment. So instead of calling the same function $$10{,}000$$ times, we are never more than $$15$$ steps into the stack. Suddenly, this becomes a big deal.

So instead of defining the recursion linearly, we do it on a tree. And this is why we came here. We didn't come here to talk about algorithm efficiency, I'm a set theorist, if it runs in finite time you should be quiet and thank me. We came here to talk about recursion and induction. Let's turn our heads to induction for a moment, another confusing matter for first year students.

Axiom. Suppose that $$S$$ is a set of natural numbers such that $$0\in S$$ and if $$n\in S$$, then $$n+1\in S$$, then $$S=\Bbb N$$.

This is a standard formulation of induction, and then we can use it, and the proofs are at first kind of funny, since we need to define $$S$$ explicitly. For example:

Proposition. For every natural number $$n>3$$, $$n!>2^n$$.

Proof. Let $$S$$ denote the set $$\{n\in\Bbb N\mid (n+4)!>2^{n+4}\}$$. We check that $$0\in S$$: $(n+4)!=4!=24>16=2^4=2^{0+4}.$ Suppose now that $$n\in S$$, then $(n+1+4)!=(n+5)\cdot(n+4)!>2\cdot(n+4)!>2\cdot 2^{n+4}=2^{n+1+4},$ so $$n+1\in S$$ as well. Therefore by the induction axiom, $$S=\Bbb N$$ as wanted. $$\square$$

It gets better as one understands how this template fits the one we were taught in high school, and maybe even clear some misconception about how it works. But that's the easy part.

Axiom. Suppose that $$S$$ is a set of natural numbers such that if $$n$$ satisfies that for every $$k < n$$, $$k\in S$$, then $$n\in S$$ as well.

This is the standard formulation of strong or complete induction. This is utter magic for students. Is there a base case? Is there a successor case? How do you use it?

Proposition. Every natural number greater than $$1$$ is divisible by a prime number.

Proof. Define $$S$$ to be the set of natural numbers satisfying the above with $$0$$ and $$1$$ added in (by vacuity). Suppose that $$n$$ is any natural number and for every $$k < n$$, $$k\in S$$. If $$n=0$$ or $$n=1$$, then $$n\in S$$ by vacuity; if $$n$$ is a prime, then it is divisible by itself, a prime; and finally, if $$n$$ is not a prime, write it as $$p\cdot q$$ where $$1 < p,q < n$$, and since $$p$$ and $$q$$ are already in $$S$$ by the induction hypothesis, $$p$$ is divisible by some prime, and so must be $$n$$. $$\square$$

This proof is the first example of a complete induction, and it's a nice one, since it is really unclear how to prove this by the "usual induction". How does knowing anything about $$n$$ will help us with $$n+1$$ for this proof? It simply won't. But it's still not quite clear, if there's no basis, why do we separate the primes in a sort of trivial way, like what normally would be the base case in standard inductions? Well. The answer lies in generality.

Recall that a relation is well-founded if every non-empty subset has a minimal element. And it turns out that well-foundedness is exactly what we need in order to do recursion (and proofs by induction). So when you see induction or recursion, your first thought should always be well-foundedness. And in this proof we really utilise this fact. Of course we need to reformulate induction.

Axiom. Suppose that $$(P,\leq)$$ is a well-founded relation and $$S\subseteq P$$ such that for every $$p\in P$$, if $$\{q\in P\mid q < p\}\subseteq S$$, then $$p\in S$$, then $$S=P$$.

Theorem. $$(\Bbb N\setminus\{1\},\preceq)$$ is well-founded, where $$n\prec m\iff\exists k(n\cdot k=m)$$.

Proof. Note that if $$n\in\Bbb N$$ and $$k\preceq n$$, then $$k\leq n$$. Therefore below each natural numbers there are only finitely many points. Therefore if $$A$$ is a non-empty set, let $$n\in A$$ be any point, and consider $$A_n=\{k\in A\mid k\preceq n\}$$. Since $$A_n$$ is finite, it has a minimal element, $$k$$. If $$k$$ was not minimal in $$A$$, there would be some $$m\in A$$ such that $$m\prec k$$, but then $$m\preceq n$$, so $$m\in A_n$$, and therefore $$k$$ was not minimal in $$A_n$$. $$\square$$

Corollary. Every non-zero natural number greater than $$1$$ is divisible by a prime.

Proof. Note that the prime numbers are exactly the minimal elements of $$(\Bbb N\setminus\{1\},\preceq)$$. Let $$n$$ be any natural number greater than $$1$$, then $$D_n=\{k\in\Bbb N\setminus\{1\}\mid k\preceq n\}$$ is a downwards closed set, by well-foundedness it has a minimal element $$p$$, which is minimal in the partial order (not just that in $$D_n$$). So by definition, $$p$$ is a prime. $$\square$$

This corollary's proof can be even shorter if one observes that this is really a fact about well-founded orders: every element lies above a minimal element. This makes the proof ridiculously simple to the point you might be confused why that's even a proof.

So really, what we're doing here is an inductive proof on a partial order different from $$\Bbb N$$ with its usual ordering. People who taught basic logic courses also did similar things when proving something like unique readability, or proofs "by induction on the length/complexity of the term/formula" are really just proofs on the order of "sub-formula".

And this takes us back to the factorial algorithm. Instead of running the recursion on $$\{1,\dots,n\}$$ to compute $$n!$$, we instead run the recursion on $$2^{\lceil\log_2 n\rceil}$$, presented as a binary tree (growing downwards, of course). And what we gain is speed in overheads, but also understanding that recursion shouldn't always be linear. It is beneficial to look sideways, downwards, diagonally, and upside-down and find different structures on which we can perform recursion and induction more clearly.

If you tell that to someone who is teaching first year students a basic course where induction is covered they will likely say something like "Yes, you're right, but we just don't have time to dwell on that". But that's just wrong. Induction and recursion are so basic, and if you're training people to be able to solve abstract problems or even one day become developers, then they will only benefit from the ability to look at a problem from a different angle.

(For two years now I've been planning to write an introductory paper about induction and recursion. One step at a time, I suppose. One day I will do it.)

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text:
Back in July, we built a software module to “guarantee” that a quadcopter never leaves a rectangular box. We spent months proving this guarantee as a mathematical theorem. We were pretty proud of ourselves, until we actually bothered to run our code. As it turns out, when it comes to cyber-physical systems, there’s more to it then just proving theorems.

### The problem

The purpose of the software module, which we call a monitor, is to allow the pilot or existing control software to move the quadcopter freely within a rectangular region but prevent it from ever leaving that rectangular region.

We proved that our monitor guarantees that the quadcopter will never leave the safe rectangular box, assuming a simple model of the quadcopter dynamics. This is a key point. In particular, our model doesn’t include a number of forces like wind. So as it turns out, when you actually fly a quadcopter with our monitor, it does occasionally leave the rectangular safe region. Minor violations are ok; in fact, they are inevitable in this domain since a model of the physical dynamics can never include all the details. But what happens when the quadcopter leaves the rectangular safe region? As the following figure depicts, the quadcopter gets stuck:

A prior post discusses why the quadcopter gets stuck. To summarize, while running our rectangular bounding monitor, when the quadcopter was pushed outside any of the boundaries by wind or other unmodeled forces, the logic of the monitor prevented the quadcopter from moving back inside the boundaries, regardless of what the pilot did. In this post, I’m going to describe potential solutions to this issue.

### Potential solutions

This issue is not a violation of any property we verified. Within our model of the dynamics of the quadcopter, the monitor prevents the quadcopter from ever leaving the rectangular boundaries, so there is no need to prove anything about what happens when the quadcopter is outside. However, when working with cyber-physical systems, there is always a gap between model and reality. In order to build a practical verified monitor, we need to verify something about what happens when the quadcopter is outside the boundaries enforced by the monitor. What exactly should we verify?

Informally, we want the quadcopter to behave reasonably with respect the the safe region defined by the monitor’s boundaries. For example, we want the quadcopter to stay close to the safe region if it starts close and to eventually move back inside the safe region if it starts outside. Moreover, we want whatever property we verify to reduce to our original safety property when the quadcopter starts within the safe region (i.e. the quadcopter stays within the safe region if it starts within the safe region1).

These informal notions of reasonable behavior are very similar to several core notions in control theory called Lyapunov stability. Roughly speaking, Lyapunov stability characterizes the behavor of a system of differential equations around an equilibrium (a point $x_e$ such that if the system is initially at $x_e$, then it remains at $x_e$ for all time). There are three common notions of stability in the theory of Lyapunov:

1. A system is Lyapunov stable if it can remain within any distance $\epsilon$ of the equilibrium as long as it starts within some distance $\delta$ of the origin. In other words, a system will remain arbitrarily close to the equilbirium as long as it starts sufficiently close.
2. A system is asymptotically stable if it is Lyapunov stable and approaches the equilibrium as time goes to infinity. In other words, the system remains close to the equilibrium and eventually converges to it.
3. A system is exponentially stable if it is asymptotically stable and if the distance to the equilibrium over time is bounded by an exponential function. In other words, the system remains close to the equilbirium and converges exponentially quickly to it.

These three definitions of stability seem to capture the kind of reasonable behavior I described earlier, except that our system (a quadcopter equiped with our rectangular bounding monitor) does not have an equibrium point. In other words, there is no single point such that if the quadcopter reaches that point, it should remain there for all time. However, we do have a region such that if the system reaches that region, it should remain there for all time. This region is the safe region. We can easily adapt the three notions of stability to our setting by replacing “distance to the equilibrium” with “distance to the safe region”.

Now we have three definitions of stability that apply to our setting. Which one do we actually use? The first definition, Lyapunov stable, does not require that the system ever re-enter the safe region if it leaves the safe region, so it permits the system to remain stuck on the boundary. The second definition, asymptotic stability, does require the system to converge to the safe region, but there is no bound on how long this must take. For all practical purposes, this could still allow the system to remain stuck on the boundary. Thus, it seems that the final definition, exponential stability, is the appropriate one. It requires not only that the system converge to the safe region but that it do so quickly.

We have not yet verified or tested the appropriateness of stability for capturing reasonable behavior of the quadcopter outside of the safe region. In fact, stability is not the only potential solution to our problem. Here are a couple other potential solutions that we are considering:

1. We could treat the problem as an optimization problem in which the monitor seeks to minimize some notion of “badness”. One possible definition of badness is the distance from the safe region. There is a tremendous amount of work on optimial control theory that we could draw from.

2. The quadcopter only leaves the safe region if there are unmodeled forces such as wind. What happens when there is a constant unmodeled force like a constant wind? Even an exponentially stable system may not converge to the safe region in the presence of a constant wind. Thus, we may want to verify that the system quickly converges to the safe region in the presence of such an unmodeled force. Again, there are ideas from control theory that we can draw on, such as the use of the I term in a PID controller.

We haven’t yet experimentally evaluated any of these ideas, but as always, this is very important. As soon as we do so, we will certainly write another blog post on the results.

1. We actually need to replace safe region with inductively safe region. The inductively safe region is a subset of the safe region such that if the system starts in any part of the inductively safe region, it remains there. To see that this is not the entire safe region, suppose the quadcopter starts just inside the upper boundary traveling at near infinite upward velocity. Even though the system is within the safe region, it will soon exit the safe region. Thus, it is not in the inductively safe region.


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If |x|-|y|=|x+y|, then which of the following must be true?

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If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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If |x|-|y|=|x+y|, then which of the following must be true?

A. x-y>0
B. x-y<0
C. x+y>0
D. xy>0
E. xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 00:18
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If x & y are both equal to 0. Then none of the options are true. So if want to find which MUST be true then answer is none. Question should be missing some part i guess.

If the question states that x and y are non zero. Then we can see that x and y should be off opposite polarity to satisfy the equation.
Illustration :

x = 5, y = -1

1)true 2)false 3)true 4) false 5) true

x= -5, y = 1

1)false 2)true 3)false 4) false 5)true

xy<0

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 02:34
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Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 03:24
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Marcab wrote:
if |x|-|y|=|x+y|, then which of the following must be true?
1) x-y>0
2) x-y<0
3) x+y>0
4) xy>0
5) xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

The correct answer is E, but should be $$xy\leq{0}$$ and not $$xy<0$$. Otherwise, none of the answers is correct.
The given equality holds for $$x=y=0$$, for which none of the given answers is correct.

The given equality can be rewritten as $$|x| = |y| + |x + y|$$.
If $$y=0$$, the equality becomes $$|x|=|x|$$, obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of $$x$$. Corrected E holds.
If $$y>0$$, then necessarily $$x$$ must be negative, because if $$x>0$$, then $$|x+y|>|x|$$ ($$x+y>x$$), and the given equality cannot hold.
If $$y<0$$, then necessarily $$x$$ must be positive, because if $$x<0$$, then again $$|x+y|>|x|$$ ($$-x-y>-x$$) and the given equality cannot hold.
It follows that $$x$$ and $$y$$ must have opposite signs or $$y=0$$.

Answer corrected version of E $$\,\,xy\leq{0}$$.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 03:28
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Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

$$(|x| - |y|)^2 = (|x + y|)^2$$

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 03:57
Macfauz , the solution was perfect. Thanks.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 04:02
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EvaJager wrote:
Marcab wrote:
if |x|-|y|=|x+y|, then which of the following must be true?
1) x-y>0
2) x-y<0
3) x+y>0
4) xy>0
5) xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

The correct answer is E, but should be $$xy\leq{0}$$ and not $$xy<0$$. Otherwise, none of the answers is correct.
The given equality holds for $$x=y=0$$, for which none of the given answers is correct.

The given equality can be rewritten as $$|x| = |y| + |x + y|$$.
If $$y=0$$, the equality becomes $$|x|=|x|$$, obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of $$x$$. Corrected E holds.
If $$y>0$$, then necessarily $$x$$ must be negative, because if $$x>0$$, then $$|x+y|>|x|$$ ($$x+y>x$$), and the given equality cannot hold.
If $$y<0$$, then necessarily $$x$$ must be positive, because if $$x<0$$, then again $$|x+y|>|x|$$ ($$-x-y>-x$$) and the given equality cannot hold.
It follows that $$x$$ and $$y$$ must have opposite signs or $$y=0$$.

Answer corrected version of E $$\,\,xy\leq{0}$$.

Many thanks for the explanation.
It will be great if you elaborate on how to solve split modulus questions such as given above.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 04:26
MacFauz wrote:
Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

$$(|x| - |y|)^2 = (|x + y|)^2$$

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.

OR 0, that's why the correct answer should be $$xy\leq{0}$$.
Otherwise, very nice solution.
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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22 Oct 2012, 05:01
EvaJager wrote:
MacFauz wrote:
Marcab wrote:
Thanx for the reply Macfauz. Agree to your illustration but it will be great if you can go with the algebraic method.
Such modulus questions are painful if one doesn't the knows the correct approach.

Squaring both sides we get :

$$(|x| - |y|)^2 = (|x + y|)^2$$

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

So.,

|x||y| = -xy

So -xy is positive (since modulus cannot be negative) and hence xy should be negative.

OR 0, that's why the correct answer should be $$xy\leq{0}$$.
Otherwise, very nice solution.

I was solving on the basis of my previous comment where I had just added the phrase "where x and y are non zero" to the question.

But seeing as how it is much more probable to leave out a <= sign than an entire sentence, I guess the question frame is right and the answer should be xy <= 0
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19 May 2016, 09:57
Squaring both sides of equation:-

x^2+y^2 - 2 lxl lyl= x^2 +y^2 +2xy

-lxl lyl= xy

for this to hold true xy should be -ve i.e <0

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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08 Jun 2016, 20:54
EvaJager wrote:
Marcab wrote:
if |x|-|y|=|x+y|, then which of the following must be true?
1) x-y>0
2) x-y<0
3) x+y>0
4) xy>0
5) xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

The correct answer is E, but should be $$xy\leq{0}$$ and not $$xy<0$$. Otherwise, none of the answers is correct.
The given equality holds for $$x=y=0$$, for which none of the given answers is correct.

The given equality can be rewritten as $$|x| = |y| + |x + y|$$.
If $$y=0$$, the equality becomes $$|x|=|x|$$, obviously true.
From the given answers, D cannot hold, and A,B or C holds, depending on the value of $$x$$. Corrected E holds.
If $$y>0$$, then necessarily $$x$$ must be negative, because if $$x>0$$, then $$|x+y|>|x|$$ ($$x+y>x$$), and the given equality cannot hold.
If $$y<0$$, then necessarily $$x$$ must be positive, because if $$x<0$$, then again $$|x+y|>|x|$$ ($$-x-y>-x$$) and the given equality cannot hold.
It follows that $$x$$ and $$y$$ must have opposite signs or $$y=0$$.

Answer corrected version of E $$\,\,xy\leq{0}$$.

I think xy=0 wouldn't be the answer because when xy = 0 , either only x or only y can also be zero. Then if x=0 but y is not, then it won't hold true for xy=0. Let me know if I am wrong

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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08 Jun 2016, 22:46
Marcab wrote:
If |x|-|y|=|x+y|, then which of the following must be true?

A. x-y>0
B. x-y<0
C. x+y>0
D. xy>0
E. xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

say x = 4 and y = 2. then x+y = 6. |x|-|y| = 2. Not equal. Hence D is out. Also C is out.

Now say x = 4 and y = -2. then x+y = 2. |x|-|y| = 2. Equal. Hence E could be correct. Also B is out.

take Now say x = -4 and y = 2. then |x+y| = 2. |x|-|y| = 2. Equal. Hence A is out.

E is correct option.

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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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18 Jun 2017, 01:07
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Re: If |x|-|y|=|x+y|, then which of the following must be true? [#permalink]

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18 Jun 2017, 03:04
Marcab wrote:
If |x|-|y|=|x+y|, then which of the following must be true?

A. x-y>0
B. x-y<0
C. x+y>0
D. xy>0
E. xy<0

I was unable to find its answer. Hence after trying, I guess the answer is x+y>0.
Please correct me if I am wrong.
Source: Jamboree

If lxl - lyl = lx+yl and xy does not equal to o, which of the following must be true?

A. x-y > 0
B. x-y < 0
C. x+y >0
D. xy>0
E. xy<0

$$|x|-|y|=|x+y|$$ --> square both sides --> $$(|x|-|y|)^2=(|x+y|)^2$$ --> note that $$(|x+y|)^2=(x+y)^2$$ --> $$(|x|-|y|)^2=(x+y)^2$$ --> $$x^2-2|xy|+y^2=x^2+2xy+y^2$$ --> $$|xy|=-xy$$ --> $$xy\leq{0}$$, but as given that $$xy\neq{0}$$, then $$xy<0$$.

Another way:

Right hand side, $$|x+y|$$, is an absolute value, which is always non-negative, but as $$xy\neq{0}$$, then in this case it's positive --> $$RHS=|x+y|>0$$. So LHS must also be more than zero $$|x|-|y|>0$$, or $$|x|>|y|$$.

So we can have following 4 scenarios:
1. ------0--y----x--: $$0<y<x$$ --> $$|x|-|y|=x-y$$ and $$|x+y|=x+y$$ --> $$x-y\neq{x+y}$$. Not correct.
2. ----y--0------x--: $$y<0<x$$ --> $$|x|-|y|=x+y$$ and $$|x+y|=x+y$$ --> $$x+y={x+y}$$. Correct.
3. --x------0--y----: $$x<0<y$$ --> $$|x|-|y|=-x-y$$ and $$|x+y|=-x-y$$ --> $$-x-y={-x-y}$$. Correct.
4. --x----y--0------: $$x<y<0$$ --> $$|x|-|y|=-x+y$$ and $$|x+y|=-x-y$$ --> $$-x+y\neq{-x-y}$$. Not correct.

So we have that either $$y<0<x$$ (case 2) or $$x<0<y$$ (case 3) --> $$x$$ and $$y$$ have opposite signs --> $$xy<0$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-x-y-x-y-a ... 83728.html
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Re: If |x|-|y|=|x+y|, then which of the following must be true?   [#permalink] 18 Jun 2017, 03:04
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text:
# Lirik Lagu Rohani Epic No Less - Come to the Cross

There is a cross, there is a crown, there is a hill
Where love's poured out and lost are found
There is a Man, so full of peace
And it's etched into His hands and on His feet

Come, come to the cross, come to the light
Come to the one who gave His life
Come just as you are, out of the night
Come to the one who gave His life, owohoo, ooh

There is a love that cannot be contained
No, He did not lose His life, He gave it away for you
Yeah, He's holding you, yeah, He's holding you
Yeah, He's holding you, yeah, He's holding you

Come, come to the cross, come to the light
Come to the one who gave His life
Come just as you are, out of the night
Come to the one who gave His life

Come, come to the cross, come to the light
Come to the one who gave His life
Come just as you are, out of the night
Come to the one who gave His life, owohoo, ooh

Come, come, come to the cross
Come, come, just as you are
Come, come, come to the cross
Come, come, just as you are

Come, come, come to the cross
Come, come just as you are
Come, come, come to the cross
Come, come just as you are

Come, come to the cross, come to the light
Come to the one who gave His life
Come just as you are, out of the night
Come to the one who gave His life

Come, come to the cross, come to the light
Come to the one who gave His life
Come just as you are, out of the night
Come to the one who gave His life

Come, come, come to the cross
Come, come just as you are


text:
Topology questions

1. Mar 1, 2008

jacobrhcp

a) Is there an algorithm for the number of topologies on finite sets?

b) If two spaces are homeomorphic, are intersections of opens sent to intersections of opens? Are unions of opens sent to unions of opens?

I tried to find an algorithm in the first part, and tried to prove the second part, both cases I failed. Anyone knows if there is a way to do this? The lecture notes I use does not seem to speak of it.

For the first one, I convinced myself that if it exists, it was very hard to do... so it probably doesn't exist or isn't found yet, otherwise I'd expect my notes to mention it.

For the second one, I suspect it's very easy (I'm just reading the chapter on homeomorphisms) because homeomorphisms are called 'the isomorphisms of topology'

Last edited: Mar 1, 2008
2. Mar 1, 2008

Hurkyl

Staff Emeritus
Brute force doesn't work? (Or did you forget to consider it?)

I assume you mean finite intersections?

How did you try to prove it?

Last edited: Mar 1, 2008
3. Mar 1, 2008

jacobrhcp

by brute force you mean writing out all topologies for sets with 1,2,3 and 4 elements? because I did, numer of labeled topologies on a set with:

1 element: 1
2 elements: 4
3 elements: 29
4: elements: complicated

the point is I don't really see any pattern here just yet. I must say I did not count how many nonhomeomorphic ones there were. *starts counting* for 1 it's 1, for 2 it's 3, for 3 I'd have to count a bit longer.... but since it's bigger than 5 at least, it's not an easy way to do get an algorithm (EDIT: considering symmetry of labeled topologies I suspect it's 8).... and then still, how to prove it?

and no I did not mean finite, but I think you mean that it's only true for finite intersections =). Although I'd expect infinite intersections of opens to be sent to infinite intersections of opens, only you wouldn't be sure they were open after taking the infinite intersection.

I tried writing out the definitions and then starting like this:

'suppose f:X->Y is a homeomorphism. then if $$U_1 -> U_1'$$, $$U_2 -> U_2'$$ this means $$U_1 \cap U_2$$is sent to an open subset of Y, because f and $$f^{-1}$$ are continuous.

Now I wanted to show this is $$U_1' \cap U_2'$$. I imagine this has to do with continuity as well... because the 'bijective' part of homeomorphisms is certainly not enough for this to work.

So admittedly I did not get very far, but I don't know how to continue.

I feel you suspect this is a homework question, but I assure you it's not: I'm trying to learn myself topology from lecture notes and internet, since I am not able to attend the exercise classes I don't even know what the homework is, nor is there any exercise that looks remotely like these.

Last edited: Mar 1, 2008
4. Mar 1, 2008

Hurkyl

Staff Emeritus
Yes -- but without stopping at 4. If you want to know how many topologies there are on a particular finite set, you could simply write out all topologies for it. That qualifies as an algorithm! Asking for an 'efficient' algorithm, or a 'simple' arithmetic expression is another question entirely.

Well, to state the algorithm more precisely, you would:
1. Write down every subset of P(X).
2. For each subset of P(X), test if it's a topology.
3. Count how many subsets of P(X) were, in fact, topologies.

If you so desired, you could include additional steps to sort the topologies into homeomorphism classes.

It may seem like a dumb algorithm, but it is an algorithm -- and dumb algorithms like these often offer easy answers to interesting theoretical questions.

Actually, I meant the latter. I read something into your original post that wasn't there.

The notions of intersection and union are set-theoretic -- you already know that the homeomorphism maps open to open (and back), so are you sure 'bijective' isn't enough to finish the question?

Have you thought about how your conjecture specializes to the case where each space has the discrete topology?

Anyways, the main thing you seem not to have done is invoke the definition of "intersection".

I hadn't really thought about it; I'm very much in the habit of trying to point people in the right direction -- they learn more when they solve the problem themselves! Ideally I won't even do that much; it's better to teach a person how to find paths and evaluates them than it is to show them the right path!

5. Mar 1, 2008

jacobrhcp

haha, thanks a lot =)

so there is no simple algorithm found (which was what I actually was looking for)? =(... that's a shame.

anyways, for the proof of 'my' conjecture ;)

the definition of intersection is: $$U_1' \cap U_2':=$$x in X| x in $$U_1'$$and x in $$U_2'$$

and in the case of the discrete topology, the 'continuous' part is a bit of an overkill in a homeorphism, because bijections between discrete topologies are automatically continuous, right? So in that case, 'bijection' should be enough.

In general,

$$f(U_1 \cap U_2$$) = f({x in X| x in $$U_1'$$ and x in $$U_2'$$})={y in Y| y=f(x) in $$U_1'$$ and y=f(x) in $$U_2'$$}=$$U_1' \cap U_2'$$, right?... but even now I'm a bit lost as to why the second equality holds, and why we needed openness again for this to work (or did we?)?

Last edited: Mar 1, 2008
6. Mar 1, 2008

jacobrhcp

btw, I like the xkcd comic in your signature =)... what does xkcd stand for anyway?


text:
Converts the rectangular components of a complex number into its polar components.

This node converts the rectangular components to polar components using the following equations:

$r=\sqrt{{x}^{2}+{y}^{2}}$
$theta=\mathrm{arctan}2\left(y,x\right)\text{\hspace{0.17em}}radians$

x

The x rectangular component of x + yi. This input supports scalar numbers, arrays or clusters of numbers, and arrays of clusters of numbers.

Data Type Changes on FPGA

y

The y rectangular component of x + yi. This input supports scalar numbers, arrays or clusters of numbers, and arrays of clusters of numbers.

Data Type Changes on FPGA

r

The distance from the origin to the point. r has the same data type structure as x and y.

Data Type Changes on FPGA

theta

The angle for the line, r, from the origin to the point in radians. theta has the same data type structure as x and y.

Data Type Changes on FPGA


text:
# Gaze-directed Text Entry in VR Using Quikwrite

Text entry in virtual environments is one of those old problems that never seem to get solved. The core issue, of course, is that users in VR either don’t have keyboards (because they are in a CAVE, say), or can’t effectively use the keyboard they do have (because they are wearing an HMD that obstructs their vision). To the latter point: I consider myself a decent touch typist (my main keyboard doesn’t even have key labels), but the moment I put on an HMD, that goes out the window. There’s an interesting research question right there — do typists need to see their keyboards in their peripheral vision to use them, even when they never look at them directly? — but that’s a topic for another post.

Until speech recognition becomes powerful and reliable enough to use as an exclusive method (and even then, imagining having to dictate “for(int i=0;i<numEntries&&entries[i].key!=searchKey;++i)” already gives me a headache), and until brain/computer interfaces are developed and we plug our computers directly into our heads, we’re stuck with other approaches.

Unsurprisingly, the go-to method for developers who don’t want to write a research paper on text entry, but just need text entry in their VR applications right now, and don’t have good middleware to back them up, is a virtual 3D QWERTY keyboard controlled by a 2D or 3D input device (see Figure 1). It’s familiar, straightforward to implement, and it can even be used to enter text.

Figure 1: Guilty as charged — a virtual keyboard in the Vrui toolkit, implemented as a GLMotif pop-up window with rows and columns of buttons.

# Someone at Oculus is Reading my Blog

I am getting the feeling that Big Brother is watching me. When I released the inital version of the Vrui VR toolkit with native Oculus Rift support, it had magnetic yaw drift correction, which the official Oculus SDK didn’t have at that point (Vrui doesn’t use the Oculus SDK at all to talk to the Rift; it has its own tracking driver that talks to the Rift’s inertial movement unit directly via USB, and does its own sensor fusion, and also does its own projection setup and lens distortion correction). A week or so later, Oculus released an updated SDK with magnetic drift correction.

A little more than a month ago, I wrote a pair of articles investigating and explaining the internals of the Rift’s display, and how small deviations in calibration have a large effect on the perceived size of the virtual world, and the degree of “solidity” (for lack of a better word) of the virtual objects therein. In those posts, I pointed out that a single lens distortion correction formula doesn’t suffice, because lens distortion parameters depend on the position of the viewers’ eyes relative to the lenses, particularly the eye/lens distance, otherwise known as “eye relief.” And guess what: I just got an email via the Oculus developer mailing list announcing the (preview) release of SDK version 0.3.1, which lists eye relief-dependent lens correction as one of its major features.

Maybe I should keep writing articles on the virtues of 3D pupil tracking, and the obvious benefits of adding an inertially/optically tracked 6-DOF input device to the consumer-level Rift’s basic package, and those things will happen as well. 🙂

# What is Presence?

Disclaimer: Presence research is not my area of expertise. I’m basically speaking as an interested layperson, and just writing down some vaguely related observations that have re-occurred to me recently.

So, presence. What is presence, and why should we care? Libraries full of papers have been written about it, and there’s even a long-running journal of that title. I guess one could say that presence is the sensation of bodily being in a place or environment where one knows one is not. And why is it important in the discussion of virtual reality? Because it is often trotted out as the distinguishing feature between the medium of VR (yes, VR is the medium, not the content) and other media, such as film or interactive 3D graphics; in other words, it is often a feature that’s used to sell the idea of VR (not that there’s anything wrong with that).

But how does one actually measure presence, and know that one has achieved it? Some researchers do it by putting users into fMRI machines, but that’s not really something you can do at home. So here are a few things I’ve observed over sixteen years of working in VR, and showing 3D display environments and 3D software to probably more than 1,000 people, both experts and members of the general public:

# Fighting Motion Sickness due to Explicit Viewpoint Rotation

Here is an interesting innovation: the developers at Cloudhead Games, who are working on The Gallery: Six Elements, a game/experience created for HMDs from the ground up, encountered motion sickness problems due to explicit viewpoint rotation when using analog sticks on game controllers, and came up with a creative approach to mitigate it: instead of rotating the view smoothly, as conventional wisdom would suggest, they rotate the view discretely, in relatively large increments (around 30°). And apparently, it works. What do you know. In their explanation, they refer to the way dancers keep themselves from getting dizzy during pirouettes by fixing their head in one direction while their bodies spin, and then rapidly whipping their heads around back to the original direction. But watch them explain and demonstrate it themselves. Funny thing is, I knew that thing about ice dancers, but never thought to apply it to viewpoint rotation in VR.

Figure 1: A still from the video showing the initial implementation of “VR Comfort Mode” in Vrui.

This is very timely, because I have recently been involved in an ongoing discussion about input devices for VR, and how they should be handled by software, and how there should not be a hardware standard but a middleware standard, and yadda yadda yadda. So I have been talking up Vrui‘s input model quite a bit, and now is the time to put up or shut up, and show how it can handle some new idea like this.

# More on Desktop Embedding via VNC

I started regretting uploading my “Embedding 2D Desktops into VR” video, and the post describing it, pretty much right after I did it, because there was such an obvious thing to do, and I didn’t think of it.

Figure 1: Screenshot from video showing VR ProtoShop run simultaneously in a 3D environment created by an Oculus Rift and a Razer Hydra, and in a 2D environment using mouse and keyboard, brought into the 3D environment via the VNC remote desktop protocol.

# VR Movies

There has been a lot of discussion about VR movies in the blogosphere and forosphere (just to pick two random examples), and even on Wired, recently, with the tenor being that VR movies will be the killer application for VR. There are even downloadable prototypes and start-up companies.

But will VR movies actually ever work?

This is a tricky question, and we have to be precise. So let’s first define some terms.

When talking about “VR movies,” people are generally referring to live-action movies, i.e., the kind that is captured with physical cameras and shows real people (well, actors, anyway) and environments. But for the sake of this discussion, live-action and pre-rendered computer-generated movies are identical.

We’ll also have to define what we mean by “work.” There are several things that people might expect from “VR movies,” but not everybody might expect the same things. The first big component, probably expected by all, is panoramic view, meaning that a VR movie does not only show a small section of the viewer’s field of view, but the entire sphere surrounding the viewer — primarily so that viewers wearing a head-mounted display can freely look around. Most people refer to this as “360° movies,” but since we’re all thinking 3D now instead of 2D, let’s use the proper 3D term and call them “4π sr movies” (sr: steradian), or “full solid angle movies” if that’s easier.

The second component, at least as important, is “3D,” which is of course a very fuzzy term itself. What “normal” people mean by 3D is that there is some depth to the movie, in other words, that different objects in the movie appear at different distances from the viewer, just like in reality. And here is where expectations will vary widely. Today’s “3D” movies (let’s call them “stereo movies” to be precise) treat depth as an independent dimension from width and height, due to the realities of stereo filming and projection. To present filmed objects at true depth and with undistorted proportions, every single viewer would have to have the same interpupillary distance, all movie screens would have to be the exact same size, and all viewers would have to sit in the same position relative the the screen. This previous post and video talks in great detail about what happens when that’s not the case (it is about head-mounted displays, but the principle and effects are the same). As a result, most viewers today would probably not complain about the depth in a VR movie being off and objects being distorted, but — and it’s a big but — as VR becomes mainstream, and more people experience proper VR, where objects are at 1:1 scale and undistorted, expectations will rise. Let me posit that in the long term, audiences will not accept VR movies with distorted depth.

# 2D Desktop Embedding via VNC

There have been several discussions on the Oculus subreddit recently about how to integrate 2D desktops or 2D applications with 3D VR environments; for example, how to check your Facebook status while playing a game in the Oculus Rift without having to take off the headset.

This is just one aspect of the larger issue of integrating 2D and 3D applications, and it reminded me that it was about time to revive the old VR VNC client that Ed Puckett, an external contractor, had developed for the CAVE a long time ago. There have been several important changes in Vrui since the VNC client was written, especially in how Vrui handles text input, which means that a completely rewritten client could use the new Vrui APIs instead of having to implement everything ad-hoc.

Here is a video showing the new VNC client in action, embedded into LiDAR Viewer and displayed in a desktop VR environment using an Oculus Rift HMD, mouse and keyboard, and a Razer Hydra 6-DOF input device:

# Small Correction to Rift’s Projection Matrix

In a previous post, I looked at the Oculus Rift’s internal projection in detail, and did some analysis of how stereo rendering setup is explained in the Rift SDK’s documentation. Looking at that again, I noticed something strange.

In the other post, I simplified the Rift’s projection matrix as presented in the SDK documentation to

$P = \begin{pmatrix} \frac{2 \cdot \mathrm{EyeToScreenDistance}}{\mathrm{HScreenSize} / 2} & 0 & 0 & 0 \\ 0 & \frac{2 \cdot \mathrm{EyeToScreenDistance}}{\mathrm{VScreenSize}} & 0 & 0 \\ 0 & 0 & \frac{z_\mathrm{far}}{z_\mathrm{near} - z_\mathrm{far}} & \frac{z_\mathrm{far} \cdot z_\mathrm{near}}{z_\mathrm{near} - z_\mathrm{far}} \\ 0 & 0 & -1 & 0 \end{pmatrix}$

which, to those in the know, doesn’t look like a regular OpenGL projection matrix, such as created by glFrustum(…). More precisely, the third row of P is off. The third-column entry should be $\frac{z_\mathrm{near} + z_\mathrm{far}}{z_\mathrm{near} - z_\mathrm{far}}$ instead of $\frac{z_\mathrm{far}}{z_\mathrm{near} - z_\mathrm{far}}$, and the fourth-column entry should be $2 \cdot \frac{z_\mathrm{far} \cdot z_\mathrm{near}}{z_\mathrm{near} - z_\mathrm{far}}$ instead of $\frac{z_\mathrm{far} \cdot z_\mathrm{near}}{z_\mathrm{near} - z_\mathrm{far}}$. To clarify, I didn’t make a mistake in the derivation; the matrix’s third row is the same in the SDK documentation.

What’s the difference? It’s subtle. Changing the third row of the projection matrix doesn’t change where pixels end up on the screen (that’s the good news). It only changes the z, or depth, value assigned to those pixels. In a standard OpenGL frustum matrix, 3D points on the near plane get a depth value of 1.0, and those on the far plane get a depth value of -1.0. The 3D clipping operation that’s applied to any triangle after projection uses those depth values to cut off geometry outside the view frustum, and the viewport projection after that will map the [-1.0, 1.0] depth range to [0, 1] for z-buffer hidden surface removal.

Using a projection matrix as presented in the previous post, or in the SDK documentation, will still assign a depth value of -1.0 to points on the far plane, but a depth value of 0.0 to points on the (nominal) near plane. Meaning that the near plane distance given as parameter to the matrix is not the actual near plane distance used by clipping and z buffering, which might lead to some geometry appearing in the view that shouldn’t, and a loss of resolution in the z buffer because only half the value range is used.

I’m assuming that this is just a typo in the Oculus SDK documentation, and that the library code does the right thing (I haven’t looked).

Oh, right, so the fixed projection matrix, for those working along, is

$P = \begin{pmatrix} \frac{2 \cdot \mathrm{EyeToScreenDistance}}{\mathrm{HScreenSize} / 2} & 0 & 0 & 0 \\ 0 & \frac{2 \cdot \mathrm{EyeToScreenDistance}}{\mathrm{VScreenSize}} & 0 & 0 \\ 0 & 0 & \frac{z_\mathrm{near} + z_\mathrm{far}}{z_\mathrm{near} - z_\mathrm{far}} & 2 \cdot \frac{z_\mathrm{far} \cdot z_\mathrm{near}}{z_\mathrm{near} - z_\mathrm{far}} \\ 0 & 0 & -1 & 0 \end{pmatrix}$

# Game Engines and Positional Head Tracking

Oculus recently presented the “Crystal Cove,” a version of the Rift head-mounted display with built-in optical tracking, which is combined with the existing inertial tracker to provide a full 6-DOF (position and orientation) tracking solution at low latency, and it is rumored that the Crystal Cove will be released as development kit mark 2 after this year’s Game Developers Conference.

This is great news. I’ve been saying for a long time that Oculus cannot afford to drop positional head tracking on developers at the last minute, because it will break several assumptions built into game engines and other VR software (but let’s talk about game engines here). I’m also happy because the Crystal Cove uses precisely the tracking technology that I predicted: active markers (LEDs) on the headset, and an external camera placed at a fixed position in the environment. I am also sad because I didn’t manage to finish my own after-market optical tracking add-on before Oculus demonstrated their new integrated technology, but that’s life.

So why does positional head tracking break existing games? Because for the first time, the virtual camera used to render a game world is no longer under sufficient control of the software. Let’s take a step back. In a standard, desktop, 3D game, the camera is entirely controlled by the software. The software sets it to some position and orientation determined by the game logic, the 3D engine renders the virtual world for that camera setup, and the result is the displayed image.

# A Follow-up on Eye Tracking

Now this is why I run a blog. In my video and post on the Oculus Rift’s internals, I talked about distortions in 3D perception when the programmed-in camera positions for the left and right stereo views don’t match the current left and right pupil positions, and how a “perfect” HMD would therefore need a built-in eye tracker. That’s still correct, but it turns out that I could have done a much better job approximating proper 3D rendering when there is no eye tracking.

This improvement was pointed out by a commenter on the previous post. TiagoTiago asked if it wouldn’t be better if the virtual camera were located at the centers of the viewer’s eyeballs instead of at the pupils, because then light rays entering the eye straight on would be represented correctly, independently of eye vergence angle. Spoiler alert: he was right. But I was skeptical at first, because, after all, that’s just plain wrong. All light rays entering the eye converge at the pupil, and therefore that’s the only correct position for the virtual camera.

Well, that’s true, but if the current pupil position is unknown due to lack of eye tracking, then the only correct thing turns into just another approximation, and who’s to say which approximation is better. My hunch was that the distortion effects from having the camera in the center of the eyeballs would be worse, but given that projection in an HMD involving a lens is counter-intuitive, I still had to test it. Fortunately, adding an interactive foveating mechanism to my lens simulation application was simple.

Turns out that I was wrong, and that in the presence of a collimating lens, i.e., a lens that is positioned such that the HMD display screen is in the lens’ focal plane, distortion from placing the camera in the center of the eyeball is significantly less pronounced than in my approach. Just don’t ask me to explain it for now — it’s due to the “special properties of the collimated light.” 🙂


text:
# Cross-referencing page numbers between two documents

I have two documents which I will subsequently merge (i.e. I will merge .pdf’s). In order to do that I have two issues to resolve, both have to do with cross-referencing.

First, I use firstpage and LastPage labels in headings and footing to indicate the first and the last pages of an article in the journal. The lines that allow me to do that are located in the preamble of the first document. Since the article consists of two parts (the files I want to merge), I need the lastpage label to refer to the last page of the second document. The same holds for the second document: I need the firstpage label to refer to the first page of the first document.

The second issue has the similar nature as the first one and, perhaps, may have the same solution. It goes without saying that the first page of the second document should have the number greater than the last page of the first document by one. So, I suppose the solution here is to put in the argument of the setcounter command in the beginning of the second document some label that refers to the last page of the first document and +1 (i.e. label+1). Sounds like a simple task but I don’t know how to cope with it.

Important note. The page numbers in the headings and footings should be hyperlinks, whereas (obviously) the first page number in the second document shouldn’t.

Edit. This solution doesn't work, because after implementing it I merged pdf's but clicking the last page number in the header or footer only leads to jumping to the first page of the merged document.

Here is the example of the first document (file1.tex) code:

\documentclass[12pt,a4paper]{article}

\usepackage{cmap}
\usepackage{mathtext}
\usepackage[T2A]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english,russian]{babel}
\usepackage{indentfirst}
\usepackage{letltxmacro}
\usepackage[blocks]{authblk}
\usepackage{xr-hyper}
\externaldocument{file2}
\usepackage{hyperref}
\frenchspacing
\usepackage{lipsum}

\usepackage{fancyhdr}
\pagestyle{fancy}
\lhead{\pageref{firstpage1} --- \pageref{lastpage2}}
\fancypagestyle{firststyle}
{
\fancyhf{}
\fancyfootoffset[R]{-4cm}
\fancyfoot[L]{\pageref{firstpage1} --- \pageref{lastpage2}}
}

\begin{document}
\setcounter{page}{1}
\author{И.\,О.~Фамилия1}
\title{Название}
\date{}
\maketitle
\thispagestyle{firststyle}
\label{firstpage1}
\lipsum
\label{lastpage1}
\end{document}

And the second one (file2.tex):

\documentclass[12pt,a4paper]{article}

\usepackage{cmap}
\usepackage{mathtext}
\usepackage[T2A]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage[english,russian]{babel}
\usepackage{indentfirst}
\usepackage{letltxmacro}
\usepackage[blocks]{authblk}
\usepackage{xr-hyper}
\externaldocument{file1}
\usepackage{hyperref}
\frenchspacing
\usepackage{lipsum}

\usepackage{fancyhdr}
\pagestyle{fancy}
\lhead{\pageref{firstpage1} --- \pageref{lastpage2}}
\fancypagestyle{firststyle}
{
\fancyhf{}
\fancyfootoffset[R]{-4cm}
\fancyfoot[L]{\pageref{firstpage1} --- \pageref{lastpage2}}
}

\begin{document}
\setcounter{page}{3}
\author{И.\,О.~Фамилия1}
\title{Название}
\date{}
\maketitle
\label{firstpage2}
\lipsum
\label{lastpage2}
\end{document}

• When you merge pdf's with e.g. pdfpages you will loose the links. You can get external references with the xr-package. – Ulrike Fischer Jun 10 '16 at 9:38
• I've tried to implement this solution tex.stackexchange.com/a/41542/101051, but when I click the last page number in the header or footer my pdf reader jumps to the first page of the document/ – David Jun 10 '16 at 10:42
• Well I won't spent my time to guess what you are doing or to build a complete test suite. – Ulrike Fischer Jun 10 '16 at 10:56
• @UlrikeFischer Sorry, I've edited the question – David Jun 10 '16 at 11:17


text:
## Precalculus (6th Edition) Blitzer

The required solution is $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=\frac{3}{4}$ .
The permutation formula is: ${}_{n}{{P}_{r}}=\frac{n!}{\left( n-r \right)!}$ Applying these formulas into the provided expression, we get: $1-\frac{{}_{3}{{P}_{2}}}{{}_{4}{{P}_{3}}}=1-\frac{\frac{3!}{\left( 3-2 \right)!}}{\frac{4!}{\left( 4-3 \right)!}}=1-\frac{3!}{1!}\times \frac{1!}{4!}$ $=1-\frac{3\times 2\times 1}{4\times 3\times 2\times 1}$ \begin{align} & =1-\frac{1}{4} \\ & =\frac{3}{4} \\ \end{align}