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# Information algebra

Information algebra

Classical information theory goes back to Claude Shannon. It is a theory of information transmission, looking at communication and storage. However, it has not been considered so far that information comes from different sources and that it is therefore usually combined. It has furthermore been neglected in classical information theory that one wants to extract those parts out of a piece of information that are relevant to specific questions.

A mathematical phrasing of these operations leads to an algebra of information, describing basic modes of information processing. Such an algebra grasps a lot of formalisms of computer science, which seem to be different on the surface: relational databases, multiple systems of formal logic or numerical problems of linear algebra. It allows the development of generic procedures of information processing and thus a unification of basic methods of computer science, in particular of distributed information processing.

Information algebra

Information relates to precise questions, comes from different sources, must be aggregated and can be focused on questions of interest. Starting from these considerations, information algebras Harv|Kohlas|2003 are two sorted algebras $\left(Phi,D\right),$, where $Phi,$ is a semigroup, representing combination or aggregation of information, $D,$ is a lattice of domains (related to questions) whose partial order reflects the granularity of the domain or the question, and a mixed operation representing focusing or extraction of information.

Information and its operations

More precisely, in the two-sorted algebra $\left(Phi,D\right),$, the following operations are defined

Models of information algebras

Here follows an incomplete list of instances of information algebras:
*Relational algebra: The reduct of a relational algebra with natural join as combination and the usual projection is a labeled information algebra, see .
*Constraint systems: Constraints form an information algebra Harv|Jaffar|Maher|1994.
*Semiring valued algebras: C-Semirings induce information algebras Harv|Bistarelli|Montanari|Rossi1997;Harv|Bistarelli|Fargier|Montanari|Rossi|Schiex|Verfaillie|1999;Harv|Kohlas|Wilson|2006.
*Logic: Many logic systems induce information algebras Harv|Wilson|Mengin|1999. Reducts of cylindrical algebras Harv|Henkin|Monk|Tarski|1971 or polyadic algebras are information algebras related to predicate logic Harv|Halmos|2000.
*Module algebras: Harv|Bergstra|Heering|Klint|1990;Harv|de Lavalette|1992.
*Linear systems: Systems of linear equations or linear inequalities induce information algebras Harv|Kohlas|2003.

Worked-out Example: Relational Algebra

Let $\left\{mathcal A\right\},$ be a set of symbols, called attributes (or columnnames). For each $alphain\left\{mathcal A\right\},$ let $U_alpha,$ be a non-empty set, theset of all possible values of the attribute $alpha,$. For example, if $\left\{mathcal A\right\}= \left\{ exttt\left\{name\right\}, exttt\left\{age\right\}, exttt\left\{income\right\}\right\},$, then $U_\left\{ exttt\left\{name,$ couldbe the set of strings, whereas $U_\left\{ exttt\left\{age,$ and $U_\left\{ exttt\left\{income,$ are boththe set of non-negative integers.

Let $xsubseteq\left\{mathcal A\right\},$. An $x,$-tuple is a function $f,$ so that$hbox\left\{dom\right\}\left(f\right)=x,$ and $f\left(alpha\right)in U_alpha,$ for each $alphain x,$ The setof all $x,$-tuples is denoted by $E_x,$. For an $x,$-tuple $f,$ and a subset$ysubseteq x,$ the restriction $f \left[y\right] ,$ is defined to be the$y,$-tuple $g,$ so that $g\left(alpha\right)=f\left(alpha\right),$ for all $alphain y,$.

A relation $R,$ over $x,$ is a set of $x,$-tuples, i.e. a subset of $E_x,$.The set of attributes $x,$ is called the domain of $R,$ and denoted by$d\left(R\right),$. For $ysubseteq d\left(R\right),$ the projection of $R,$ onto $y,$ is definedas follows::$pi_y\left(R\right):=\left\{f \left[y\right] mid fin R\right\}.,$The join of a relation $R,$ over $x,$ and a relation $S,$ over $y,$ isdefined as follows::As an example, let $R,$ and $S,$ be the following relations::Then the join of $R,$ and $S,$ is::A relational database with natural join as combination and the usual projection $pi,$ is an information algebra.The operations are well defined since

• If $xsubseteq d\left(R\right),$, then $d\left(pi_x\left(R\right)\right)=x,$.
It is easy to see that relational databases satisfy the axioms of a labeledinformation algebra:; semigroup : and ; transitivity : If $xsubseteq ysubseteq d\left(R\right),$, then $pi_x\left(pi_y\left(R\right)\right)=pi_x\left(R\right),$.; combination : If $d\left(R\right)=x,$ and $d\left(S\right)=y,$, then .; idempotency : If $xsubseteq d\left(R\right),$, then .; support : If $x = d\left(R\right),$, then $pi_x\left(R\right)=R,$.

Connections

; Valuation Algebras : Dropping the idempotency axiom leads to Valuation Algebras. These axioms have been introduced by Harv|Shenoy|Shafer|1990 to generalize "local computation schemes" Harv|Lauritzen|Spiegelhalter|1988 from Bayesian networks to more general formalisms (including belief function, possibility potentials, etc.) Harv|Kohlas |Shenoy|2000.; Domains and Information Systems: "Compact Information Algebras" Harv|Kohlas|2003 are related to Scott domains and Scott information systems Harv|Scott|1970;Harv|Scott|1982;Harv|Larsen|Winskel|1984.; Uncertain Information : Random variables with values in information algebras represent "probabilistic argumentation systems" Harv|Haenni|Kohlas|Lehmann|2000.; Semantic Information : Information algebras introduce semantics by relating information to questions through focusing and combination Harv|Groenendijk|Stokhof|1984;Harv|Floridi|2004.; Information Flow : Information algebras are related to information flow, in particular classifications Harv|Barwise|Seligman|1997.; Tree decomposition : ...; Semigroup theory : ...

Historical Roots

The axioms for information algebras are derived from the axiom system proposed in (Shenoy and Shafer, 1990), see also (Shafer, 1991).

References

Harvard reference | Given=Gerard R. Renardel | Surname=de Lavalette | Chapter= Logical semantics of modularisation | Editor=Egon Börger, Gerhard Jäger, Hans Kleine Büning, and Michael M. Richter | Title=CSL: 5th Workshop on Computer Science Logic | Pages=306–315 | Publisher=Volume 626 of Lecture Notes in Computer Science, Springer | Year=1992 | ISBN=3-540-55789-X
URL=http://citeseer.ist.psu.edu/484529.html

Harvard reference | Given1=K. G. | Surname1=Larsen | Given2=G. |Surname2=Winskel | Chapter=Using information systems to solve recursive domain equations effectively | Editor=Gilles Kahn, David B. MacQueen, and Gordon D. Plotkin | Title=Semantics of Data Types, International Symposium, Sophia-Antipolis, France, June 27-29, 1984, Proceedings | Volume=173 of Lec- ture Notes in Computer Science | Pages=109–129 | Location=Berlin | Year= 1984 | Publisher=Springer

Harvard reference | Given1=S. L. | Surname1= Lauritzen |Given2=D. J.|Surname2=Spiegelhalter | Title=Local computations with probabilities on graphical structures and their application to expert systems | Journal= J. Royal Statis. Soc. B |Volume= 50 | Pages=157–224 | Year= 1988

Harvard reference | Given=Dana S. | Surname= Scott | Title= Outline of a mathematical theory of computation | Publisher=Technical Monograph PRG–2, Oxford University Computing Laboratory, Programming Research Group | Year=1970

Harvard reference | Given1=P. P. | Surname1=Shenoy | Given2=G. | Surname2=Shafer | Chapter=Axioms for probability and belief-function proagation | Editor=Ross D. Shachter, Tod S. Levitt, Laveen N. Kanal, and John F. Lemmer | Title= Uncertainty in Artificial Intelligence 4 | Volume= 9 | Journal= Machine intelligence and pattern recognition |Pages = 169–198 | Place=Amsterdam | Year= 1990 | Publisher=Elsevier | ISBN= 0-444-88650-8

Harvard reference | Given1=Nic | Surname1=Wilson |Given2= Jérôme | Surname2= Mengin | Chapter=Logical deduction using the local computation framework | Editor=Anthony Hunter and Simon Parsons, | Title=Symbolic and Quantitative Approaches to Reasoning and Uncertainty, European Confer- ence, ECSQARU’99, London, UK, July 5-9, 1999, Proceedings, volume 1638 of Lecture Notes in Computer Science | Pages= 386–396 | Publisher= Springer | Year= 1999 | ISBN = 3-540-66131-X | URL = http://springerlink.metapress.com/openurl.asp?genre=article&issn=0302-9743&volume=1638&spage=0386

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• Information theory — Not to be confused with Information science. Information theory is a branch of applied mathematics and electrical engineering involving the quantification of information. Information theory was developed by Claude E. Shannon to find fundamental… …   Wikipedia

• Information Integration Theory — was proposed by Norman H. Anderson to describe and model how a person integrates information from a number of sources in to make an overall judgment. The theory proposes three functions. [1] The valuation function V(S) is an empirically derived… …   Wikipedia

• algebra — /al jeuh breuh/, n. 1. the branch of mathematics that deals with general statements of relations, utilizing letters and other symbols to represent specific sets of numbers, values, vectors, etc., in the description of such relations. 2. any of… …   Universalium

• Algebra — This article is about the branch of mathematics. For other uses, see Algebra (disambiguation). Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from… …   Wikipedia

• Information theory and measure theory — Measures in information theory = Many of the formulas in information theory have separate versions for continuous and discrete cases, i.e. integrals for the continuous case and sums for the discrete case. These versions can often be generalized… …   Wikipedia

• Information science — Not to be confused with Information theory. Contents 1 Introduction 2 A multitude of information sciences? 3 Definitions of information science 4 History …   Wikipedia

• information theory — the mathematical theory concerned with the content, transmission, storage, and retrieval of information, usually in the form of messages or data, and esp. by means of computers. [1945 50] * * * ▪ mathematics Introduction       a mathematical… …   Universalium

• information processing — Acquisition, recording, organization, retrieval, display, and dissemination of information. Today the term usually refers to computer based operations. Information processing consists of locating and capturing information, using software to… …   Universalium

• Homological algebra — is the branch of mathematics which studies homology in a general algebraic setting. It is a relatively young discipline, whose origins can be traced to investigations in combinatorial topology (a precursor to algebraic topology) and abstract… …   Wikipedia

• History of algebra — Elementary algebra is the branch of mathematics that deals with solving for the operands of arithmetic equations. Modern or abstract algebra has its origins as an abstraction of elementary algebra. Historians know that the earliest mathematical… …   Wikipedia


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# Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1

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Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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Updated on: 12 Apr 2018, 01:57
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Is A positive?

(1) x^2 - 2x + A is positive for all x
(2) Ax^2 + 1 is positive for all x

Originally posted by noboru on 15 Jul 2010, 12:42.
Last edited by Bunuel on 12 Apr 2018, 01:57, edited 2 times in total.
Edited the question
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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15 Jul 2010, 14:40
21
19
noboru wrote:
Is A positive?

x^2-2x+A is positive for all x
Ax^2+1 is positive for all x

OA is A

Is $$A>0$$?

(1) $$x^2-2x+A$$ is positive for all $$x$$:

Quadratic expression $$x^2-2x+A$$ is a function of of upward parabola (it's upward as coefficient of $$x^2$$ is positive). We are told that this expression is positive for all $$x$$ --> $$x^2-2x+A>0$$, which means that this parabola is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> $$D=2^2-4A=4-4A<0$$ --> $$1-A<0$$ --> $$A>1$$.

Sufficient.

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 08:01
21
7
AndreG wrote:
Hi,

I dont get it sorry... I mean I understand your equations Bunuel, but I tried first with picking numbers:

If I pick -0.5 for x --> x^2-2x+A>0 will hold for A > -1.25

...

Where is my mistake??

The point here is that $$x^2-2x+A>0$$ for all $$x-es$$.

Let's do this in another way:

We have $$(x^2-2x)+A>0$$ for all $$x-es$$. The sum of 2 quantities ($$x^2-2x$$ and $$A$$) is positive for all $$x-es$$. So for the least value of $$x^2-2x$$, $$A$$ must make the whole expression positive.

So what is the least value of $$x^2-2x$$? The least value of quadratic expression $$ax^2+bx+c$$ is when $$x=-\frac{b}{2a}$$, so in our case the least value of $$x^2-2x$$ is when $$x=-\frac{-2}{2}=1$$ --> $$x^2-2x=-1$$ --> $$-1+A>0$$ --> $$A>1$$.

OR:

$$x^2-2x+A>0$$ --> $$x^2-2x+1+A-1>0$$ --> $$(x-1)^2+A-1>0$$ --> least value of $$(x-1)^2$$ is zero thus $$A-1$$ must be positive ($$0+A-1>0$$)--> $$A-1>0$$ --> $$A>1$$.

Hope it's clear.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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11 Apr 2012, 10:42
4
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rongali wrote:
Is A positive?

1) X^2-2X+A is positive for all X
2) AX^2 + 1 is positive for all X

given answer as A...but i thought it should be E..
source: hard problems from gmatclub tests number properties I

1) X^2-2X+A is positive for all X

For all values of X,$$X^2-2X+A > 0$$
This means, for X = 0, $$X^2-2X+A > 0$$; for X = 1, $$X^2-2X+A > 0$$; for X = -2, $$X^2-2X+A > 0$$ etc etc etc

Let's put X = 0. $$0^2-2*0+A > 0$$ should hold. Therefore, A > 0 should hold.
Sufficient.

2) AX^2 + 1 is positive for all X

For all X, $$AX^2 + 1 > 0$$
Here, A could be positive or A could be 0 (since, when A = 0, we get 1 > 0 which holds no matter what the value of X.)
Since A can be 0, we cannot say whether A is positive. Not Sufficient.

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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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24 Sep 2013, 20:44
2
VeritasPrepKarishma wrote:
rongali wrote:
Is A positive?

1) X^2-2X+A is positive for all X
2) AX^2 + 1 is positive for all X

given answer as A...but i thought it should be E..
source: hard problems from gmatclub tests number properties I

1) X^2-2X+A is positive for all X

For all values of X,$$X^2-2X+A > 0$$
This means, for X = 0, $$X^2-2X+A > 0$$; for X = 1, $$X^2-2X+A > 0$$; for X = -2, $$X^2-2X+A > 0$$ etc etc etc

Let's put X = 0. $$0^2-2*0+A > 0$$ should hold. Therefore, A > 0 should hold.
Sufficient.

2) AX^2 + 1 is positive for all X

For all X, $$AX^2 + 1 > 0$$
Here, A could be positive or A could be 0 (since, when A = 0, we get 1 > 0 which holds no matter what the value of X.)
Since A can be 0, we cannot say whether A is positive. Not Sufficient.

Responding to a pm:
Quote:
I still did not understand your solution

x^2-2x+A>0
if we take the value 3 for example ,
9-6+A>0
3+A>0
which gives
A>(-)3
so A can assume -2,-1,0 and so on and we still get the overall value as +ve.
Can you help me understand what i am missing ?

Given that x^2-2x+A is always positive. No matter what the value of x, the value of A is such that this expression is always positive.
Whether x = ...-2, 0, 1, 4, 100..., the expression will always be positive.
So let's put a few values of x.

Put x = -2
(-2)^2-2(-2)+A > 0
A > -8

Put x = 0
0^2 - 2*0 + A > 0
A > 0

Put x = 1
1^2 - 2*1 + A > 0
A > 1

Put x = 3
3^2 - 2*3 + A > 0
A > -3

and so on...
So we see that A must be greater than -8, it should also be greater than -3, it should also be greater than 0 and it should also be greater than 1. So what values do you think A can take? Values which are greater than all these values i.e. values like 8, 10 etc. In any case, we are asked whether A is positive and we know that it must be greater than 1. Hence, we know that A must be positive. Sufficient.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 09:36
1
Hi Bunuel,

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

$$x^2 - 2x + A = 9 +6 +A = 15 + A >0$$

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers
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Joined: 02 Sep 2009
Posts: 60605
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 10:16
1
1
nitishmahajan wrote:
Bunuel wrote:
nitishmahajan wrote:
Hi Bunuel,

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

$$x^2 - 2x + A = 9 +6 +A = 15 + A >0$$

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all $$x-es$$ means that no matter what $$x$$ you pick $$x^2 - 2x + A$$ must be positive. So it must be positive even for the lowest value of $$x^2 - 2x$$ which is -1 --> so $$-1+A$$ must be positive hence A must be more than 1.

Now again: if A>1 then for any $$x$$ expression $$x^2 - 2x + A$$ is positive.

But if A=-15 (or any other number less than 1) we can find some $$x-es$$ for which expression $$x^2 - 2x + A$$ is not positive, so theese values of A (values of $$A\leq{1}$$) are not valid.

Hope it's clear.

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

I think you just don't understand one thing in statement (1): $$x^2-2x+A>0$$ FOR ALL $$x-es$$.

You say that if $$x=-3$$ then $$A$$ can be for example -10 (or any number more than -15) and $$x^2-2x+A$$ will be positive, $$but$$ if $$x=1$$ does $$A=-10$$ makes $$x^2-2x+A$$ positive? NO!

So you should find such value of $$A$$ (such range) for which $$x^2-2x+A$$ is positive no matter what value of $$x$$ you'll plug. And the way how to find this range is shown in my previous posts.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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22 Oct 2013, 23:39
1
bsahil wrote:
VeritasPrepKarishma wrote:

X = 10 is fine but it doesn't help. We know that this inequality holds for all x. We need to plug in a value for x which tells us something about A. If we put x = 0, we are left with just A and that will tell us something about A. Just plugging in any value may not work; you have to look for a smart value.

So for this question, no other value of x(except x=0,2) gives us any information about A. hence we take only that value which gives us explicit and fixed result for A. Am I correct??
Similarly we would do in other similar questions as well...??

No, look, we know that $$x^2 - 2x + A > 0$$ for all x. For every value of x, this inequality should be satisfied.

Put x = 0, you get A > 0
Put x = 1, you get $$1 - 2 + A > 0 i.e. A > 1$$
Put x = 10, you get A > -80

Now the point is that A should take a value such that all these conditions are satisfied. Say A can be 5. If A is 5, it is > 0, > 1 and > -80.

When I look at $$x^2 - 2x + A > 0$$ given x can take any value, the first value that pops in my head to get a sense of A is x = 0. That provides exactly what I need. Had the question been whether A > 2, x = 0 would not have helped. I would have had to search a little for a pattern to see how the value of x changes. This question is made in a way that x = 0 helps immediately. Anyway, it is a good idea to try the value 0 in many circumstances. It simplifies things immensely and usually helps you eliminate a couple of options at least.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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23 Oct 2013, 21:56
1
ratanpandit wrote:
VeritasPrepKarishma wrote:
rongali wrote:
Is A positive?

1) X^2-2X+A is positive for all X
2) AX^2 + 1 is positive for all X

given answer as A...but i thought it should be E..
source: hard problems from gmatclub tests number properties I

1) X^2-2X+A is positive for all X

For all values of X,$$X^2-2X+A > 0$$
This means, for X = 0, $$X^2-2X+A > 0$$; for X = 1, $$X^2-2X+A > 0$$; for X = -2, $$X^2-2X+A > 0$$ etc etc etc

Let's put X = 0. $$0^2-2*0+A > 0$$ should hold. Therefore, A > 0 should hold.
Sufficient.

2) AX^2 + 1 is positive for all X

For all X, $$AX^2 + 1 > 0$$
Here, A could be positive or A could be 0 (since, when A = 0, we get 1 > 0 which holds no matter what the value of X.)
Since A can be 0, we cannot say whether A is positive. Not Sufficient.

if we put x=-3 in 1, then A can have -1 or -2 value also ??

No, A can never have a value of -1 or -2. It must be greater than 1.
If we put x = -3, we get
X^2-2X+A > 0
A > -15
So this tells us that A must be greater than -15. Putting other values of x such as 0, 1, 2 etc tell us that A must be greater than 0 and A must be greater than 1 etc. Since this inequality holds for ALL values of x, A must be greater than 1 because a value greater than 1 will automatically be greater than -15 as well as 0. If we take a value of A such as -14, it will be greater than -15 but not greater than 0 or 1 hence the inequality will not hold for ALL value of x.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 05:21
2
Hi,

I dont get it sorry... I mean I understand your equations Bunuel, but I tried first with picking numbers:

If I pick -0.5 for x --> x^2-2x+A>0 will hold for A > -1.25

...

Where is my mistake??
Intern
Joined: 22 Jun 2010
Posts: 36
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 08:26
Wow u rock man!
That was very clear!

I especially like the +1 -1 trick

Posted from my mobile device
Math Expert
Joined: 02 Sep 2009
Posts: 60605
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 09:49
nitishmahajan wrote:
Hi Bunuel,

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

$$x^2 - 2x + A = 9 +6 +A = 15 + A >0$$

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all $$x-es$$ means that no matter what $$x$$ you pick $$x^2 - 2x + A$$ must be positive. So it must be positive even for the lowest value of $$x^2 - 2x$$ which is -1 --> so $$-1+A$$ must be positive hence A must be more than 1.

Now again: if A>1 then for any $$x$$ expression $$x^2 - 2x + A$$ is positive.

But if A=-15 (or any other number less than 1) we can find some $$x-es$$ for which expression $$x^2 - 2x + A$$ is not positive, so theese values of A (values of $$A\leq{1}$$) are not valid.

Hope it's clear.
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Posts: 216
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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16 Jul 2010, 09:56
Bunuel wrote:
nitishmahajan wrote:
Hi Bunuel,

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

$$x^2 - 2x + A = 9 +6 +A = 15 + A >0$$

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all $$x-es$$ means that no matter what $$x$$ you pick $$x^2 - 2x + A$$ must be positive. So it must be positive even for the lowest value of $$x^2 - 2x$$ which is -1 --> so $$-1+A$$ must be positive hence A must be more than 1.

Now again: if A>1 then for any $$x$$ expression $$x^2 - 2x + A$$ is positive.

But if A=-15 (or any other number less than 1) we can find some $$x-es$$ for which expression $$x^2 - 2x + A$$ is not positive, so theese values of A (values of $$A\leq{1}$$) are not valid.

Hope it's clear.

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

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Re: Is A positive? 1. x^2 - 2x + A is +ve for all x 2. (A*x^2) +  [#permalink]

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08 Apr 2011, 07:28
asmit123 wrote:
Is A positive?

1. x^2 - 2x + A is +ve for all x
2. (A*x^2) + 1 is +ve for all x

I've posted about this question a couple of times. Many people solve this backwards and arrive at the wrong answer (those test takers relying on 'number picking' strategies almost always answer this question incorrectly). The important word in each of the two statements is 'all'. In Statement 1, x^2 - 2x + A is positive not just for some value of x; it must be positive for EVERY value of x. In particular, it's positive when x=0, so substituting x=0, we learn instantly that A is positive and Statement 1 is sufficient.

Statement 2 is also almost sufficient. It is only insufficient because of a technicality: it's possible that A=0.

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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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04 Jul 2012, 03:55
Bunuel wrote:
noboru wrote:
Is A positive?

x^2-2x+A is positive for all x
Ax^2+1 is positive for all x

OA is A

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

Why didn't you use the discriminant formula to assess statement 2?

I tried the discriminant rule and got a>0. I had 0-4a<0 which turns to a>0.

What am I missing here?

Thanks,
Diana
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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04 Jul 2012, 03:59
dianamao wrote:
Bunuel wrote:
noboru wrote:
Is A positive?

x^2-2x+A is positive for all x
Ax^2+1 is positive for all x

OA is A

(2) $$Ax^2+1$$ is positive for all $$x$$:

$$Ax^2+1>0$$ --> when $$A\geq0$$ this expression is positive for all $$x$$. So $$A$$ can be zero too.

Not sufficient.

Why didn't you use the discriminant formula to assess statement 2?

I tried the discriminant rule and got a>0. I had 0-4a<0 which turns to a>0.

What am I missing here?

Thanks,
Diana

You are right: if we use the same approach for (2) then we'll get A>0 BUT if A=0 then Ax^2+1 won't be a quadratic function anymore. So this approach will work only if A doesn't equal to zero, but we can not eliminate this case and if A=0 then Ax^2+1=1 is also always positive. Hence Ax^2+1 is positive for A>0 (if we use quadratic function approach) as well as for A=0, so for $$A\geq0$$.

Hope it's clear.
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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21 Aug 2012, 16:51
best way to deal this problem is to bet on A more than X..
it wud b yes if A>0 Or No ,if A<0 ....
then first assume A>0 , then check whether statement 1 & 2 is true or not for all value of X....
then assume A<0 ,then check whether statement 1 & 2 is true or not for all value of X....
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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19 May 2013, 00:51
Bunuel wrote:
Quadratic expression $$x^2-2x+A$$ is a function of of upward parabola (it's upward as coefficient of $$x^2$$ is positive). We are told that this expression is positive for all $$x$$ --> $$x^2-2x+A>0$$, which means that this parabola is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Hi Bunuel,
How did you infer that the parabola would be above X axis by looking at the equation?? Pls explain.
Regards,
H
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Posts: 60605
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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19 May 2013, 04:30
imhimanshu wrote:
Bunuel wrote:
Quadratic expression $$x^2-2x+A$$ is a function of of upward parabola (it's upward as coefficient of $$x^2$$ is positive). We are told that this expression is positive for all $$x$$ --> $$x^2-2x+A>0$$, which means that this parabola is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation $$x^2-2x+A=0$$ has no real roots.

Hi Bunuel,
How did you infer that the parabola would be above X axis by looking at the equation?? Pls explain.
Regards,
H

We have $$x^2-2x+A>0$$ and told that this expression is positive for all x, which means that the parabola is above X-axis (otherwise it wouldn't be positive for all x).
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Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1  [#permalink]

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17 Oct 2013, 04:46
Dear Brunel

We have (x^2-2x)+A>0 for all x-es. The sum of 2 quantities (x^2-2x and A) is positive for all x-es. So for the least value of x^2-2x, A must make the whole expression positive.

So what is the least value of x^2-2x? The least value of quadratic expression ax^2+bx+c is when x=-\frac{b}{2a}, so in our case the least value of x^2-2x is when x=-\frac{-2}{2}=1 --> x^2-2x=-1 --> -1+A>0 --> A>1. ....Want to know how +1 changes to -1 .....please explain
Re: Is A positive? (1) x^2 - 2x + A is positive for all x (2) Ax^2 + 1   [#permalink] 17 Oct 2013, 04:46

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#### Tensor networks and efficient descriptions of classical data

##### Sirui Lu, Márton Kanász-Nagy, Ivan Kukuljan, J. Ignacio Cirac

We investigate the potential of tensor network based machine learning methods to scale to large image and text data sets. For that, we study how the mutual information between a subregion and its complement scales with the subsystem size $L$, similarly to how it is done in quantum many-body physics. We find that for text, the mutual information scales as a power law $L^\nu$ with a close to volume law exponent, indicating that text cannot be efficiently described by 1D tensor networks. For images, the scaling is close to an area law, hinting at 2D tensor networks such as PEPS could have an adequate expressibility. For the numerical analysis, we introduce a mutual information estimator based on autoregressive networks, and we also use convolutional neural networks in a neural estimator method.

arrow_drop_up


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1. ## curve tracing

how to trace the curve
x^3+y^3=3ax^2
^ stands for power

2. ## Re: curve tracing

Originally Posted by envyh
how to trace the curve
x^3+y^3=3ax^2
^ stands for power
\displaystyle \begin{align*}x^3 + y^3 &= 3ax^2 \\ y^3 &= 3ax^2 - x^3 \\ y &= \sqrt[3]{3ax^2 - x^3} \end{align*}

Now I suggest using a graphing calculator to graph this function for a few different values of a to see what effect changing it has...

3. ## Re: curve tracing

is there any software that i can use if dont have a graph calculator


text:
# Seminar: April 9

## From Irreducible Representations to Locally Decodable Codes

A q-query Locally Decodable Code (LDC) is an error-correcting code that allows to read any particular symbol of the message by reading only q symbols of the codeword.

In this talk we present a new approach for the construction of LDCs from the representation theory. We show that if there exists an irreducible representation (\rho, V) of the group G and q elements g_1,g_2,..., g_q in G such that there exists a linear combination of matrices\rho(g_i) that is of rank one, then we can construct a q-query Locally Decodable Code C:V \to F^G.

We will show that both matching vector codes and Reed-Muller codes fall in this framework.

No prior knowledge in representation theory will be assumed.


text:
## Algebra 1

$c^{15}$
We start with the given expression: $(c^3)^5(d^3)^0$ To raise a power to a power, we multiply the exponents: $c^{15}d^0$ The zero as an exponent rule states that for every nonzero number $a$, $a^0=1$ Therefore, we simplify to: $c^{15}\times1$ The identity property of multiplication states that the product of any real number and $1$ is the original number. Therefore, we can simplify to $c^{15}$.


text:
#### Vue (2.6.11) – object prop missing fields

I have a Laravel-Vue application that has the function of displaying "projects" stored in a MySQL db. These are complicated objects which are retrieved with the following query via Eloquent:

return $this->load([ 'projectVersions' => function($query) {
$query->with([ 'values' => function($valueQuery) {
$valueQuery->with([ 'grid' => function($gridQuery) {
$gridQuery->with(['gridCols', 'gridRows.gridCells' => function($cellQuery) {
$cellQuery->with(['gridCol', 'gridRow', 'grid' => function($subGridQuery) {
$subGridQuery->with(['reference', 'lastGridRow', 'lastGridCol']); }]); }]); } ]); } ]); } ]);  Although it looks horrible, it returns a straightforward data hierarchy that goes: $project
-> $projectVersion ->$values
-> $grids ->$gridCols, $gridRows ->$gridCells...

A "grid" is just a table with cols, rows, and cells that are attached to those rows. In addition however, the cells are also loaded with some information via a with-clause: their column, their row, and their grid, to the latter of which we also attach some more information. These fields are important for handling a cell in various functions on the front-end without having to retrieve its associated grid/row/col etc.

The query works fine, in any case, and we end up with the following array (showing the gridCells of one value):

which is also received on the front-end like so (retrieved through Chrome dev tools):

However, once that object is loaded as a prop into Vue, we lose some of the information!

As you can see, we’ve lost the grid and grid_col fields, even though they were present in the response. I suspect it has something to do with the size of the object – if I cut down the query, the fields show up again – but I need them and there doesn’t seem to be any indication anywhere that this is the exact cause. I also cannot find any other instances of this issue online. So, why might Vue be mangling my objects, and is there a way to overcome this short of completely rewriting my app? Thanks!

Source: Laravel


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Due to the COVID-19 madness in Germany I was forced to perform all my penetration testing remotely.[1] This is quite challenging when you have to make sure that devices can be powercycled reliably. Additionally, the power consumption is often an indicator for a crash. Having on overview of the current consumption of my setup makes perfect sense.

The powercycle problem was trivial. I grabbed a R&S power supply which offers a programming interface. A few lines of python and I’m done. Hint: Use the raw tcp port 5025, it is a very simple ASCII line based protocol. Even `telnet` can be used to remote control the device.

The challanging thing was video. During my testing I need visual feedback for what I am doing, i.e. "hit enter, look at the power consumption. Anything changed?". A video of the power supply display would be nice… I had set up a cheap Logitech webcam in order to stream the video over the network (through VPN and who knows…). The problem was… 30 sec delay. Unusable.

After a lot of tinkering, I found out these settings:

on the server
`$cvlc v4l2:// :v4l2-vdev="/dev/video0" --sout '#transcode{vcodec=h264,venc=x264{preset=ultrafast,tune=zerolatency,intra-refresh,lookahead=10,keyint=15},scale=auto,ab=128}:std{access=http{mime=video/x-ms-wmv},dst=:8082/stream.wmv}' -v --no-sout-audio` on the client `$ mpv --profile=low-latency --stream-lavf-o=reconnect_streamed=1 HOST:8082/stream.wmv`

I works quite well. Due to the massive amount of profiles and settings (and not being a video pro) I simply got lost. Hope this help.

1. I break automotive ECU’s at `\$DAYJOB`


text:
# A technical detail in the construction of a covering space: Is the projection $p$ continuous without assuming semilocal simple connectedness?

I am studying the construction of a covering space with prescribed characteristic subgroup. For simplicity, I will outline the case where the characteristic subgroup is trivial (i.e. we're dealing with the universal cover). Note that the following is very similar, but not identical to the section called "The Classification of Covering Spaces" in Hatcher's book. In particular, Hatcher assumes semilocal simple connectedness from the get-go, while I attended a lecture where the professor explicitly postponed making this assumption as long as possible.

We construct a covering space $E$ of a fixed base space $X$ (locally path connected and connected) in terms of homotopy classes of paths in $X$. The covering projection $p$ sends $[\sigma]$ to $\sigma(1)\in X$. Consider an open and path connected neighborhood $U$ of the end point $\sigma(1)$ of some path $\sigma$. Then, we define an open set in $E$ as

$$U_{[\sigma]}=\{[\sigma*\tau]\mid \tau:(I,0)\to (U,\sigma(1))\text{ a path}\}$$

and claim that the all sets of this form constitute a basis of the topology of $E$ (which can be verified with no problem). Now, we consider the projection $p$. It is clear that $p$ is surjective since local path connectedness plus connnectedness implies path connectedness. $p$ is also an open map (by construction) since $p(U_{[\sigma]})=U$.

Now comes the crucial part (for me): continuity of $p$. In the notes from my lecture we simply find the following statement:

$p$ is continuous because $[\sigma]\in U_{[\sigma]}\subset p^{-1}(U)$ by construction.

This, to me, is definitely not clear. Hence, I tried to piece together a more detailed proof, proceeding along the same lines as Hatcher (last paragraph of page 64-first paragraph of page 65), but as pointed out before one has to take great care in following Hatcher because he assumes semilocal simple connectednes (and hence has e.g. local injectivity of $p$, which I am do not have!). Therefore, I'm having a hard time figuring out how to proceed. Even understanding whether Hatcher's proof of continuity really relies on semilocal simple connectedness has proved to be beyond me, so far.

My question, therefore, is: Is it true that $p$ is continuous, if one does not assume semilocal simple connectedness? If so, an answer would ideally outline the proof in more detail than what I found in my lecture notes (or explain to me why that brief statement suffices). If not, a counterexample would be much appreciated.

• A few comments. (1) The Hatcher paragraph in question does much more than prove continuity of $p$. It also proves that $p$ is a covering map, and that's where semilocal simple connectivity comes into play. (2) Your parenthetical remark on injectivity does not make much sense. Injectivity of $p$ is not at issue in this passage of Hatcher, and that's a good thing because it is generally false. – Lee Mosher Feb 23 '16 at 13:46
• @LeeMosher (1) I realize that that's what he's proving. In my notes, this is not the case because semilocal simple connectedness is not yet assumed. (2) My remark says exactly the same as your comment, I don't understand what the confusion is here. I agree with you that semilocal simple connectedness guarantees injectivity, and that it fails in general (which is why I do not have it "available" in my notes, and Hatcher does). – Danu Feb 23 '16 at 13:52
• In (1), my intent is to pin down more precisely where semilocal simple connectivity is needed: as you ask, it is not needed in proving continuity of $p$; instead, it is used in proving that $p$ is a covering map. My mathematical point in (2) is that semilocal simple connectivity does not imply injectivity of $p$. – Lee Mosher Feb 23 '16 at 13:56
• @LeeMosher I'm sorry, in (2) of course you are correct that $p$ is not injective (globally) but its restriction to appropriate open sets (like $U_{[\sigma]}$) in $E$ is, and this is guaranteed by the assumption of semilocal simple connectedness. Surely, you can agree to that? – Danu Feb 23 '16 at 14:16
• Yes, that sounds correct. – Lee Mosher Feb 23 '16 at 16:25

Yes, $p$ is continuous regardless of whether $X$ is semilocally simply connected or not:

Let $[\sigma] \in E$ so that $p([\sigma]) = \sigma(1) = x_0$ and $U$ be an open neighborhood of $x_0$. By local path connectedness, pick a path connected neighborhood $V$ of $x_0$ contained in $U$. As $p(V_{[\sigma]}) = V$, $V_{[\sigma]}$ is contained in $p^{-1}(V) \subset p^{-1}(U)$, hence is an open neighborhood of $[\sigma] \in p^{-1}(U)$.

Similarly, for any $[\gamma] \in p^{-1}(U)$, I can construct an open neighborhood of $[\gamma]$ contained in $p^{-1}(U)$. Thus, every point in $p^{-1}(U)$ is an interior point, proving $p^{-1}(U)$ is open.

• (Semilocal simply connectedness is assumed to ensure the sets defined are a basis for some topology on $E$. Continuity is ensured by local connectedness.) – Pedro Tamaroff Jul 31 '16 at 3:26

In my book Topology and Groupoids (T&G) I develop the classification of covering spaces using the notion of covering morphism $q: Q \to G$ of groupoids.

This is a morphism of groupoids such that for each $y \in Ob(Q)$ and element $g \in G$ starting at $q(y)$ there is a unique $h \in Q$ starting at $y$ and such that $q(h)=g$. (If one drops "uniqueness" one get the notion of fibration of groupoids.)

If $p: Y \to X$ is a covering map of spaces, one proves by standard methods that $\pi_1(p) : \pi_1(Y ) \to \pi_1(X)$ is a covering morphism of groupoids.

So one is interested in the reverse process: given a covering morphism $q$ as above but with $G=\pi_1(X)$, can we find a $p:Y \to X$ such that such that $q$ is essentially $\pi_1(p)$?

A basic result on covering morphisms one uses here is that if $f: F,z \to G,x$ is a morphism of groupoids which is pointed, i.e. $f(z)=x$, and $F$ is connected, then $f$ lifts to a morphism $\tilde{f}: Z,z \to Q,y$ such that $q\tilde{f}=f$ if and only if $f(F(z) ) \subseteq q(Q(y))$; and if this lift exists it is unique.

This is easy to prove by choosing, using connectivity of $F$, elements $\tau(w): z \to w$ in $F$ for each $w \in Ob(F)$ with $\tau(z)=1_z$. (The detailed proof is in 10.3.3 of T&G.)

Now given $G= \pi_1(X)$, we want to get a topology on $Y=Ob(Q)$. We want to choose a path connected neighbourhood $N$ of $x \in X$ such that if $F=\pi_1(N)$ then the pointed morphism $f: F,x \to G,x$ lifts to a morphism $F,x \to Q,y$ for any $y \in q^{-1}(x)$. Hence we must assume that for all $x \in X$ there is a path connected neighbourhood $N$ of $x$ such that the image of $\pi_1(N,x)$ in $G$ lies in the intersection of the $q(Q(y))$ for all $y \in q^{-1}(x)$. Assuming this, it is not hard to prove that the images of these neighbourhoods in $Y=Ob(Q)$ give a topology on $Y$ making $Q \cong \pi_1(Y)$. The details are in Section 10.5 of T&G.

So to be more specific to the question, without the semilocal condition, one can't construct the topology by this method; with the condition, we can construct the topology on $Y=Ob(Q)$ and prove $p=Ob(q)$ is continuous and a covering map.

The algebraic theory of covering morphisms of groupoids then gives an equivalence of categories of covering morphisms of a groupoid $G$ and functors $G \to Sets$.


text:
# What would be the major product of the following reaction

$\begin {array} {1 1} a \\ b \\ c\\ d \end {array}$

Hence (c) is the correct answer.


text:
# rayoptics.util package¶

package supplying utility functions for math and numpy support

The util subpackage provides miscellaneous functions for geometric calculations, color calculations and anything else that doesn’t have an obvious home. These include:


text:
Thursday, December 12, 2013

The Importance of Angry Letters

How can international institutions foster cooperation given that they lack enforcement capability? One view, quite simply, is that they can't. This view is shared by realists and many outside the academy.

Many would argue this critique is unfair. It is too easy to jump from "can't control rogue states" to "completely worthless" or "false promise" or what have you. Even states that view one another as friends sometimes fail to reap all the possible benefits of international cooperation due to coordination problems, collaboration problems, etc, and institutions may help such states leave a little less money lying on the ground. There's also pretty strong evidence that UN peacekeeping works, particularly when it has the consent of all the parties involved. Sure, that's an important caveat, but we shouldn't trivialize the large number of lives that have likely been saved as a result of the UN's efforts.

But let's set those things aside. Is the best we can say about the UN that it helps those who want to be helped but is of no real consequence to the behavior of "rogue" states? I would argue that the answer is "sort of, but only if we adopt a fairly extreme definition of 'rogue'." If we don't define "rogue" states as those that do misbehave, but instead as those who would like to, then the answer is likely that the UN does not just allow the good guys to do a little bit better on the margins, but actually changes the intentions of those we might otherwise see as bad guys.

Look below the fold to see why.

Brief Synopsis

The argument I develop formally below is relatively straightforward, though I think the details are easier to see once you work through the math. (At least, they were for me. YMMV.) Angry letters don't themselves deter anyone. But they can serve as alarm bells. That is, international institutions do not need enforcement capability so long as they affect the likelihood of individual states taking punitive actions. Below, I analyze a model where they do so by providing new information. But another channel through which institutions like the UN can increase the likelihood of individual states like the US engaging in punitive action is through shoring up public support for such efforts.

How much does this matter? Clearly, some states are still willing to misbehave. North Korea and Iran have active nuclear programs, many states sponsor terrorist organizations, etc. What we don't know is how many states would do those things in the absence of the UN. In the model I analyze below, some subset of states would pursue objectionable policies in a world without institutions like the UN but would not do so in a world with them. Sure, some states would never misbehave, and some will misbehave no matter what, but institutions in this model do more than simply allow states with good intentions to move closer to the Pareto frontier. They cause some states who would not otherwise do so to behave like model citizens of the international community.

Details of the Argument

Suppose we've got two states, cleverly named 1 and 2. While 1 is an upstanding member of the international community, the world has its doubts about 2.

At the start of the game, 2 decides whether or not to pursue policy $$p$$ at cost $$\kappa>0$$, where $$p$$ is a policy that threatens the interests of the international community, or at least those of state 1. The obvious example here is developing nuclear weapons, but $$p$$ could also refer to sponsoring terror, violating human rights, engaging in currency manipulation, providing subsidies to exporters, or whatever.

Note that 2's choice is not observed by 1. However, it may be detected by some international institution, denoted I. Specifically, assume that if 2 pursues policy $$p$$, I catches them in the act with probability $$r$$. And when I does so, I subsequently writes an angry letter that the entire world will see. Otherwise, I gives 2 a pass.

For the sake of simplicity, I assume away bias by I, false positives, and other potentially interesting frictions. The goal here is simply to show that angry letters might matter, not to understand when, where, and why they would be written or withheld. In other words, I isn't a strategic actor here. I is, as I suggested above, an alarm bell. And one that doesn't even always ring when it should.

After 1 observes I's choice of whether to write an angry letter or give 2 a pass, 1 decides whether to pursue some punitive action. At the extreme, this might take the form of a ground war culminating in regime change. But it can also be thought of as limited strikes or non-military responses such as sanctions, diplomatic isolation, etc. Otherwise, 1 engages 2. That might not involve much more than keeping the channels of communication open, but it could involve substantial levels of trade and security cooperation. Let's keep things general.

If 1 chooses to engage and 2 wasn't pursuing policy $$p$$, then both receive some benefit, worth $$\beta>0$$.

If 1 chooses to engage and 2 was pursuing $$p$$, 1 receives $$-1$$ and 2 receives some temptation payoff, denoted $$\tau>0$$ (more on this below), but pays the cost $$\kappa$$ mentioned above.

If 1 chooses to punish and 2 wasn't pursuing the policy, 1 receives $$-c_1$$, where $$c_1\in(0,1)$$ and 2 receives $$-c_2$$, with $$c_2\in(0,1)$$.

If 1 chooses to punish and 2 was pursuing $$p$$, 1 again receives $$-c_1$$ while 2 receives $$-c_2 - \kappa$$.  Note that 1 does not receive any direct, consumption benefit from punishing 2 when doing so is appropriate. The payoff is the same whether 2 was pursuing policy $$p$$ or not. Realistically, 1 might suffer some domestic cost once it becomes clear that 2 wasn't pursuing policy $$p$$, but for our current purposes, such details are unimportant. This simple setup still ensures that 1 wants to punish if and only if 2 is pursuing policy $$p$$ since $$-1 < -c_1 < \beta$$. Of course, what makes this interesting is that 1 doesn't know whether is doing so or not.

That brings us back to $$\tau$$. Let's suppose that the world believes $$pr(\tau=\color{blue}{\underline{\tau}}) = \color{blue}{q}$$ and $$pr(\tau=\color{red}{\overline{\tau}}) = \color{red}{1-q}$$, where $$\color{blue}{\underline{\tau}} < \beta < \color{red}{\overline{\tau}}$$. In other words, the world knows that there is a $$\color{blue}{q}$$ probability that 2 is the good, blue type, which wouldn't want to pursue policy $$p$$ even if it could get away with it, but also knows that $$\color{red}{1-q}$$ of the time, 2 is the bad red type, which faces a real temptation to misbehave.

For notational convenience, let $$\color{blue}{q^p}$$ denote the updated belief that $$\tau=\color{blue}{\underline{\tau}}$$ held by 1 after 1 observes I give 2 a pass and let let $$\color{blue}{q^a}$$ denote the updated belief that $$\tau=\color{blue}{\underline{\tau}}$$ held by 1 after 1 observes I write an angry letter. Finally, let $$\color{blue}{q^\prime} \in \{\color{blue}{q^p},\color{blue}{q^a}\}$$ denote 1's generic updated belief. Naturally, given the information above, $$\color{blue}{q^a}$$ is simply 0. But it is straightforward to show that, for any given values of $$r$$ and $$\color{blue}{q}$$, $$\color{blue}{q} < \color{blue}{q^p} < 1$$. That is, when I gives 2 a pass, 1's trust in 2 grows, but 1 remains aware that it's possible that 2 simply escaped detection.

So. Will 1 engage? Only if 1 trusts 2 enough. More formally, iff $$\color{blue}{q^\prime}\geq\hat{q}$$, where $$\displaystyle \hat{q}\equiv \frac{1-c_1}{1-\beta}$$, which is simply what we get when we compare the expected utility of engaging to the utility of punishing and solve for $$\color{blue}{q^\prime}$$. As $$c_1$$ increases and $$\beta$$ decreases, 1 becomes less willing to engage at any given level of trust, but there's also some value of $$\color{blue}{q^p}$$ for which this inequality can be satisfied. Of course, the same is not true of $$\color{blue}{q^a}$$. That is, 1 will never engage after I writes an angry letter.

That's not surprising, of course. That 1 will always punish 2 after I catches 2 red-handed is important to the story, but where things get interesting, and where the value of formalizing the argument becomes clear, is when we consider whether 2 would pursue $$p$$ in the first place.

If 2 is blue, 2 never pursues the policy. Doesn't even want to. The behavior of 1 and I are irrelevant. Fine. Dandy. Let's move on.

If 2 is red (i.e., has a relatively high temptation to pursue $$p$$) and 2 knows that 1 will punish it even if I does not write an angry letter, then 2 will not pursue the policy because 2 won't reap the benefits thereof and will incur cost $$\kappa$$ for nothing. Again, not particularly interesting.

Ah, but what about when 2 is red and 2 knows that 1 will engage if I gives it a pass? Here, 2 might be willing to pursue the policy. But the operative phrase here is "might be", for reasons that will become clearer in the next paragraph. More formally, 2 pursues the policy iff $$\color{red}{\overline{\tau}} > \displaystyle \frac{\beta + r(c_2+\kappa)}{1-r}$$. In words, 2 pursues the policy if and only 2 is not just bad, but really bad. How bad, you ask? Well, that depends on a lot of things. Specifically, the size of that ugly fraction you just glossed right over, which itself is a function of how beneficial cooperation is, how likely 2 is to get caught if 2 pursues the policy, how costly it is to be punished by 1, and how costly it is to pursue the policy. But whatever. We don't need to get into all the different comparative statics, interesting though some of them are.

Here's the real important point. Suppose there was no institution. That would mean that $$r=0$$. In that case, 2 would pursue the policy so long as $$\color{red}{\overline{\tau}}>\beta$$, which I already stipulated. In other words, the red type is by definition interested in misbehaving. In a world without institutions, every red type would pursue policies contrary to the interests of the international community if they thought they could get away with it. But in a world with institutions that write angry letters, some of them won't. Perhaps a large number of them. How many well-behaved states in the world today would be pursuing nuclear weapons or otherwise working against the interests of the international community if not for the UN and other institutions of global governance? I don't know. I won't even venture a guess. But I feel pretty confident in saying that we're not talking about an empty set.

Angry letters matter.


text:
### Lifting of model structures to fibred categories

Abhishek Banerjee
Source: Homology Homotopy Appl. Volume 13, Number 2 (2011), 1-17.

#### Abstract

A fibred category consists of a functor $p: \mathbf{N} \to \mathbf{M}$ between categories $\mathbf{N}$ and $\mathbf{M}$ such that objects of $\mathbf{N}$ may be "pulled back along any arrow of $\mathbf{M}$". Given a fibred category $p: \mathbf{N} \to \mathbf{M}$ and a model structure on the "base category" $\mathbf{M}$, we show that there exists a lifting of the model structure on $\mathbf{M}$ to a model structure on $\mathbf{N}$. We will refer to such a system as a "fibred model category" and give several examples of such structures. We show that, under certain conditions, right homotopies of maps in the base category $\mathbf{M}$ may be lifted to right homotopic maps in the fibred category. Further, we show that these lifted model structures are well behaved with respect to Quillen adjunctions and Quillen equivalences. Finally, we show that if $\mathbf{N}$ and $\mathbf{M}$ carry compatible closed monoidal structures and the functor $p$ commutes with colimits, then a Quillen pair on $\mathbf{M}$ lifts to a Quillen pair on $\mathbf{N}$.

First Page:
Primary Subjects: 55U35


text:
The Matching Problem (probability)

1. Sep 3, 2015

TheMathNoob

1. The problem statement, all variables and given/known data
Suppose that all the cards in a deck of n different cards are placed in a row, and that the cards in another similar deck are then shufﬂed and placed in a row on top of the cards in the original deck. It is desired to determine the probability Pn that there will be at least one match between the corresponding cards from the two decks

2. Relevant equations
P(AUB)= P(A)+P(B)-P(AnB)

3. The attempt at a solution
I already posted this problem before, but it was deleted because of the format. People tell me that my logic is wrong and I think that they are right, but I don't know why I am wrong.

This is how I see it

A B C D E.........
C B A D E..........

we shall determine the probability that at least one pair will match.

Simpler case

A B C D
A
A good way to do it would be to find the prob that no pair matches. In this case there 3 ways how this can happen so it would be 3/4 the prob that no pair matches in this case.

Similarly

A B C D
A B

The number of outcomes in this case would be 3*3

So for n cards it would be (n-1)(n-1)......... n times

so the prob would be

1-π(n-1)/n^2

I am still learning how to use the system in this forum. The multiplicative iterator goes from i=1 to n

2. Sep 4, 2015

BvU

Hi there, mathwiz !

Did you consider reversing this problem too ?

3. Sep 4, 2015

TheMathNoob

Hi BvU, I already read the other method for solving this problem and it made sense to me. They just set Ai as the event where some card i matches up and then they compute the prob of A1 U A2 U A3 U A4 all they way down to n cards. This makes sense because in sets, union can be looked as or statements so they are basically saying that one of them shall work in order to satisfy the statement, but I am just wondering why mine doesn't work.

4. Sep 4, 2015

BvU

My inroad was that for the first card you can pick 51 out of 52, for the second 50 out of 51, etcetera.... to get no matches at all. A bit of normalization and then 1 minus that is your answer. I'm not giving the answer on a plate (i have no idea, to be honest), just suggesting a way to look at this problem. Good luck.

 it's not right anyway ....

Last edited: Sep 4, 2015
5. Sep 4, 2015

haruspex

This is known as the derangement problem.
You describe a method in terms of A1 U A2 U A3 U ... , but without seeing more details I cannot comment whether it works. It is not straightforward since the events are not independent. Certainly that looks like a flaw in your solution, but I'm not sure I understood what your argument was.

An approach which does work uses the principle of inclusion and exclusion. Are you familiar with that?
Another sets up a recurrence relation on the probability of n objects being deranged, as a function of n. Are you accustomed to recurrence relations?

6. Sep 4, 2015

Ray Vickson

Do you really mean $1 - \prod_{i=1}^n (n-1)/n$---exactly as you wrote? That equals $1 - [(n-1)/n]^n$. I have no idea how you arrive at this, because you do not show your logic.

However, it is incorrect.

Let's look at your simple case of 4 objects:
$$\begin{array}{ccccc} \Box & \Box & \Box & \Box& \text{Row 1} \\ A&B & C & D&\text{Row 2} \end{array}$$
We want to put letters A, B, C, D into the boxes in Row 1 and look at the matches with Row 2. Your claim above is that the probability of no match equals $\text{claimed-}P0 = 1- (3/4)^4= 175/256 \doteq 0.6836$, and that the probability of $\geq 1$ match is $\text{claimed-}P_{\geq 1} =81/256 \doteq 0.3164$. In fact, the actual probabilities of no match and $\geq 1$ matches in this case are $\text{true-}P0 = 3/8 = 0.375$ and $\text{true-}P_{\geq 1} = 5/8 = 0.625$. These come from careful and detailed application of the inclusion-exclusion principle.

As I said: I cannot tell where you went wrong because you do not explain your method or your logic.


text:
# What is the point behind $(-1)^{n-1}$ in the general term

The question is:
How many terms of the series $$\frac{1}{(2)(3)(4)}-\frac{5}{(6)(7)(8)}+\frac{9}{(10)(11)(12)}-\frac{13}{(14)(15)(16)}+...$$ must be used to ensure that the error in the approximation of the limit of the sum does not exceed 0.00001?
The solution then states
The general term of the series is $$\frac{(-1)^{n-1}(4n-3)}{(4n-2)(4n-1)4n}$$
1. What is the point behind the $(-1)^{n-1}$ part? Is this not always equal to -1 since:
$(-1)^{1-1}=-1$
$(-1)^{2-1}=-1$
$(-1)^{3-1}=-1$
$(-1)^{4-1}=-1$
$(-1)^{1000-1}=-1$?

Correction The error I made here was not trusting what I knew and then testing it on my calculator. The calculator however ran its calculation as $-(1^{n-1})$ instead of $(-1)^{n-1}$.

So that would then yeald the series $$\frac{1}{(2)(3)(4)}-\frac{5}{(6)(7)(8)}\color{red}{+}\frac{9}{(10)(11)(12)}-\frac{13}{(14)(15)(16)}\color{red}{+}...$$

The solution then continues stating:
Let $u_n=\frac{(4n-3)}{(4n-2)(4n-1)4n}$. Since the series alternates, $u_{n+1}<u_n$, and $\lim_{n\rightarrow\infty}u_n=0$,....
2. How can $\lim_{n\rightarrow\infty}u_n=0$. Surely $\lim_{n\rightarrow\infty}u_n=1$, no?

-
$$(-1)^n=1\,\,\,\forall\,\,\text{even}\,\,n\in\Bbb Z$$ –  DonAntonio Nov 18 '12 at 12:12
You should know that an alternating series (one with alternation signs) in which the terms tend to zero is convergent, and the error in the partial sum is bounded by the absolute value of the first term omitted. If you don't know this you should find out how to prove it before you use it. –  Mark Bennet Nov 18 '12 at 13:35

$$(1)\;\;\;\;\;\;\;\;\;\;(-1)^n=1\,\,\,\forall\,\,\text{even}\,\,n\in\Bbb Z$$

$$(2)\;\;\;\;\;\;\;\;\;\;\;\frac{(-1)^{n-1}(4n-3)}{(4n-2)(4n-1)4n}\xrightarrow [n\to\infty]{} 0$$

since the denominator's degree in $\,n\,$ is higher than the numerator's

-
What happend to the -1 in $(-1)^{n\color{red}{-1}}$? –  Gineer Nov 18 '12 at 12:23
What's the diffence?! As $\,n\,$ runs over the naturals, or even the integers, so does $\,n-1\,$. Fo example, $\,n-1\,$ is even for $\,n=1,3,5,7,9,...\,$ –  DonAntonio Nov 18 '12 at 12:26
@Gineer how did you put the red color? –  doniyor Nov 18 '12 at 15:29

$(-1)^{n-1}=1 \ \forall$ ODD $n$.

-


text:
# Electric flux through a surface

## Homework Statement

A point charge q is situated a distance d/2 above a square surface of side d as shown (see attached).
a) It cannot be determined
b) 2q/ε0d
c) q/4$\pi$ε0
d) q/$\pi$ε0d2
e) q/6ε0

## Homework Equations

$\Phi$=qenc0=$\oint$E*dA

## The Attempt at a Solution

Shouldn't the answer be zero because whatever enters the area also leave the area? If not I know I should use the integral of E*dA but I don't know what E would be.

#### Attachments

• 5 KB Views: 545
Last edited:

Related Introductory Physics Homework Help News on Phys.org
rude man
Homework Helper
Gold Member
If the area were a thin volume you'd be correct, the net flux into the volume = 0, but flux thru a surface A = the integral over the surface of the dot-product of the electric field E and an element of surface area dA: flux = ∫∫ E*dA.

So yes, you need to integrate E*dA over the surface. dA is normal to the surface at every point on the surface.

You can manage without integration.
Imagine the charge is in the middle of a cube with side d.
Due to symmetry, the flux through each face of the cube is the same.
And the total flux through the cubic, closed surface can be found from Gauss' law.

rude man
Homework Helper
Gold Member
Nasu has come up with the right and simple way to solve this.


text:
# I.14.1 Small potential typo

In the solution for I.14.1 (in the repo, it is found in I.14.2/challengeDescription.md):

There may be some variation as well; for example, to create

\frac{1}{\sqrt{2}}(\vert 00\rangle + \vert 11\rangle)

the Z can be applied to either qubit to obtain the same result.

I believe the + sign should be a -. I tried to reason about it, and I wasn’t able to see how we can reach |+> by applying a Z to either qubit.

Hey @Kamal_Mohamed! You are correct! There’s a typo there. I will fix it!


text:
# The Number Line

## Greater than or Less than

Putting numbers on a number line makes it easier to decide which one is larger and which one is smaller:

Numbers on the left are less than numbers on the right.

Numbers on the right are greater than numbers on the left.

### Examples

#### 1. Less Than

$2$ is to the left of $5$, so $2$ is less than $5$

$-2$ is to the left of $2$, so $-2$ is less than $2$

$-4$ is to the left of $-1$, so $-4$ is less than $-1$

#### 2. Greater Than

$5$ is to the right of $2$, so $5$ is greater than $2$

$2$ is to the right of $-2$, so $2$ is greater than $-2$

$-1$ is to the right of $-4$, so $-1$ is greater than $-4$

#### 3. Comparing More than 2 Numbers

Jasmin, Sam and Al's mum is trying to get them to eat brussels sprouts. She decides to give them $50 \text{ c}$ for each brussels sprout they eat, and fine them $50 \text{ c}$ for each brussels sprout they give to their dog, Lucy. Lucy loves brussels sprouts! None of them have any money before dinner, and they are all served up the same number of brussels sprouts.

After dinner, Sam owes his mum $3$, Al has $\ 0$ and Jasmin has earned $\2$. Who ate the most brussels sprouts? Who gave the most brussels sprouts to Lucy?

The person who ate the most brussels sprouts will be the one with the most money after dinner. The person who gave the most brussels sprouts to Lucy will be the person who has the least money after dinner.

Place these amounts on the number line to see who has the greatest amount of money and who has the least amount of money.

1. Negative numbers mean that money is owed.
2. Positive numbers mean that someone has money.

Sam's marker is furthest to the left. So, he has the least money, and he must have fed the most brussels sprouts to Lucy.

Jasmin's marker is furthest to the right. So, she has the most money, and she must have eaten the most brussels sprouts.

Al's marker is somewhere in between: he has less money than Jasmin, but more money that Sam. He probably sat at the dinner table going "one for me", "one for Lucy" with his brussels sprouts.

## Using the Number Line

We can also use the number line to help us to do addition and subtraction. When we add, we move to the right along the number line. When we subtract, we move to the left.

### Examples

When we are adding, we move to the right on the number line.

1. Let's use the number line to find out what $2 + 5$ is:

We start out at $2$, and then move $5$ places to the right to add on 5. The answer is $7$.

2. Let's use the number line to find out what $-4 + 3$ is:

We start out at $-4$, and then move $3$ places to the right to add on $3$. The answer is $-1$.

3. Let's use the number line to find out what $-3 + 5$ is:

We start out at $-3$, and then move $5$ places to the right to add on 5. The answer is $2$.

#### 2. Subtracting

When we are subtracting, we move to the left on the number line.

1. Let's use the number line to find out what $7 - 6$ is:

We start out at $7$, and then move $6$ places to the left to subtract $6$. The answer is $1$.

2. Let's use the number line to find out what $-1 - 4$ is:

We start out at $-1$, and then move $4$ places to the left to subtract $4$. The answer is $-5$.

## Measuring the distance from zero

We can also use the number line to measure how far a number is from zero. This is called the absolute value of the number. To find it, you just count the number of spaces on the number line between the number and zero.

For example, "5" is 5 spaces from zero, and so is "-5".

We say that the absolute value of "5" is 5. We also say that the absolute value of "-5" is 5.

Absolute values are always positive numbers because they measure distances.

What do you think would be the absolute value of "4"? How about "-4"?

### Description

This mini book covers the core of Math for Foundation, Grade 1 and Grade 2 mathematics including

1. Numbers
3. Subtraction
4. Division
5. Algebra
6. Geometry
7. Data
8. Estimation
9. Probability/Chance
10. Measurement
11. Time
12. Money
13. and much more

This material is provided free of cost for Parent looking for some tricks for their Prekinder, Kinder, Prep, Year 1 and Year 2 children

### Audience

You must be logged in as Student to ask a Question.


text:
# BE Notes

Nyeste elementer:
2017-08-31
09:59
 Simulation of transverse beam splitting using time-dependent dipolar or quadrupolar kicks / Capoani, Federico Two simple systems with high relevance for accelerator physics have been studied in detail in the context of this Summer Student Project. [...] CERN-STUDENTS-Note-2017-142. - 2017. Full text

2017-08-25
11:37
 Characterization and simulation of a pillbox cavity / Moran Guizan, Carla A cylindrical pillbox cavity is characterized and its principal parameters, such as the cut-off frequencies of the modes, the quality factor and R/Q figure, are calculated [...] CERN-STUDENTS-Note-2017-106. - 2017. Access to fulltext

2017-08-18
19:38
 Upgrade of CERN OP Webtools IRRAD Page / Vik, Magnus Bjerke CERN Beams Department maintains a website with various tools for the Operations Group, with one of them being specific for the Proton Irradiation Facility (IRRAD). [...] CERN-STUDENTS-Note-2017-088. - 2017. Access to fulltext

2017-08-18
15:02
 Study of Nb-Cu 1.3 GHz SRF cavity resonators for future particle accelerators / Amelin, Kirill Niobium-coated superconducting radio-frequency cavities have a number of advantages over cavities made from bulk niobium [...] CERN-STUDENTS-Note-2017-079. - 2017. Access to fulltext

2016-09-30
14:11
 Beam Impedance Studies of the PS Beam Gas Ionization Monitor / Avgidis, Fotios The Beam Gas Ionization monitor (BGI) is a device for continuous beam size monitoring that is intended to be installed in the CERN Proton Synchrotron (PS) during the extended year-end technical stop from December 2016 to April 2017. [...] CERN-STUDENTS-Note-2016-252. - 2016. Access to fulltext

2016-09-23
23:28
 PERFORMANCE ANALYSIS OF MULTI-TURN EXTRACTION FROM THE PROTON SYNCHROTRON TO THE SUPER PROTON SYNCHROTRON / Abernethy, Samuel Within CERN's accelerator complex, the extraction from the Proton Synchrotron to the Super Proton Synchrotron has been done using the so-called Continuous Transfer" (CT) method since the 1970's. [...] CERN-STUDENTS-Note-2016-246. - 2016. Access to fulltext

2016-09-07
10:58
 Summer Student Report / Gavanas, Chris This document contains my summer student report for every application I've developed or upgraded in the period of 3 months (13/06- 12/09) CERN-STUDENTS-Note-2016-189. - 2016. Access to fulltext

2016-09-02
14:15
 Schottky signal analysis: tune and chromaticity computation / Chanon, Ondine Schottky monitors are used to determine important beam parameters in a non-destructive way. [...] CERN-STUDENTS-Note-2016-177. - 2016. Access to fulltext

2016-09-01
17:02
 Uncertainties and noise in coupling measurements at the Large Hadron Collider / Wierichs, David Alexander This report demonstrates the influence of noise in turn-by-turn data on the analysis of the coupling. [...] CERN-STUDENTS-Note-2016-168. - 2016. Access to fulltext

2016-08-25
14:04
 Fibre Optic Notch Filter For The Antiproton Decelerator Stochastic Cooling System / Simmonds, Max Vincent John The project scope included reverse engineering, upgrading, and recovering the operational conditions of an existing fibre optic notch filter. [...] CERN-STUDENTS-Note-2016-127. - 2016. Access to fulltext


text:
# MAC or Windows laptop for MFE students?

#### AlexandreH

I just ordered a Lenovo laptop (x220t). I haven't taken delivery yet but it is small, versatile and has the latest cpu (sandy bridge i5) and awesome screen (IPS screen with multitouch and digitizer for note-taking). It is Win7 so not for Mac fans.
Where did you buy it from? And, maybe you can share with the community your views on it once you use it.

#### Tsotne

I'm in need of high memory laptop. I have one application made in SQL and the local server should host the data which will be on my laptop. Speed is of less concern. Trying to differentiate in apple products. 500GB is a bit expensive...

#### SYau

##### Ting Ting
Where did you buy it from? And, maybe you can share with the community your views on it once you use it.

I haven't gotten it yet Seems like Lenovo has not shipped any of the x220 tablets yet. I ordered from Costcentral with this configuration. With no tax, it is actually much cheaper than ordering direct from Lenovo, even after a 20 or so percent discount.

#### Andy Nguyen

Nice job with using CostCentral. I bought my T60 from them.

#### AlexandreH

I haven't gotten it yet Seems like Lenovo has not shipped any of the x220 tablets yet. I ordered from Costcentral with this configuration. With no tax, it is actually much cheaper than ordering direct from Lenovo, even after a 20 or so percent discount.
Why does Lenovo/IBM have such a small HD????

#### AlexandreH

I haven't gotten it yet Seems like Lenovo has not shipped any of the x220 tablets yet. I ordered from Costcentral with this configuration. With no tax, it is actually much cheaper than ordering direct from Lenovo, even after a 20 or so percent discount.
When do you expect to receive it by? A friend of mine is visiting from the States and I was wondering if I can get it to him in time. He's leaving on June 10th.

#### menovive

##### Columbia MAFN
I haven't gotten it yet Seems like Lenovo has not shipped any of the x220 tablets yet. I ordered from Costcentral with this configuration. With no tax, it is actually much cheaper than ordering direct from Lenovo, even after a 20 or so percent discount.

When u place an order It gets configured in china and has to be shipped from there. May be thats why the time..!!

#### SYau

##### Ting Ting
When do you expect to receive it by? A friend of mine is visiting from the States and I was wondering if I can get it to him in time. He's leaving on June 10th.

Lenovo has delayed shipment to CC again this morning, to 5/12. I would hesitate ordering from CC right now unless Lenovo delivers to CC this week.

#### Lyosha

##### Psychic in Training
I used one of these http://www.anandtech.com/show/2585 for a while that was a hangover from before. Eventually I ordered a newer computer http://www.sonystyle.com/webapp/wcs...0151&langId=-1&categoryId=8198552921644784015

I got it because it was the cheapest out there that would give me a fermi video card and 1920 pixels on a horizontal line. Consequently a few days after I ordered it my previous laptop started to violently crash due to some hardware issue with the video card (nice timing on the order!)

My suggestion is as follows - my laptop is nowhere near "too heavy" to carry. You will be doing a significant amount coding. If you don't have eyesight problems and can deal with high resolutions, get a laptop with a 1920 horizontal resolution (i.e. 1080P resolution, or 1920x1200, although those are almost a dead breed... and hence horribly overpriced). You want all the horizontal space you can get. Vertical space is also nice but not as crucial.

I am quite happy with my current laptop. I dual boot W7 and almost stock Ubuntu 10.10 (needed a few tweaks for vid card and sound card, 11.04 should be fine once I upgrade), and this has been a very good setup, at least for me.

Don't get a mac. They look pretty, and a 17 inch macbook pro is one of the computers that offers 1920x1200 resolution, but you pay a disturbingly horrid premium for the shiny plastic. You actually pay a ton to get outdated hardware components and to be forced to put up with apple's monopolistic inefficiencies (how's the reception on that iPhone 4?). Not to mention that there's no advantage to having one, aside of course from looking damn cool.

#### AlexandreH

I used one of these http://www.anandtech.com/show/2585 for a while that was a hangover from before. Eventually I ordered a newer computer http://www.sonystyle.com/webapp/wcs...0151&langId=-1&categoryId=8198552921644784015

I got it because it was the cheapest out there that would give me a fermi video card and 1920 pixels on a horizontal line. Consequently a few days after I ordered it my previous laptop started to violently crash due to some hardware issue with the video card (nice timing on the order!)

My suggestion is as follows - my laptop is nowhere near "too heavy" to carry. You will be doing a significant amount coding. If you don't have eyesight problems and can deal with high resolutions, get a laptop with a 1920 horizontal resolution (i.e. 1080P resolution, or 1920x1200, although those are almost a dead breed... and hence horribly overpriced). You want all the horizontal space you can get. Vertical space is also nice but not as crucial.

I am quite happy with my current laptop. I dual boot W7 and almost stock Ubuntu 10.10 (needed a few tweaks for vid card and sound card, 11.04 should be fine once I upgrade), and this has been a very good setup, at least for me.

Don't get a mac. They look pretty, and a 17 inch macbook pro is one of the computers that offers 1920x1200 resolution, but you pay a disturbingly horrid premium for the shiny plastic. You actually pay a ton to get outdated hardware components and to be forced to put up with apple's monopolistic inefficiencies (how's the reception on that iPhone 4?). Not to mention that there's no advantage to having one, aside of course from looking damn cool.
I already have a 17 in laptop: A G73JW Asus with the quad-core i7-740 and 12 GB RAM, True HD res and Nvidia GeForce GTX 460M. That is top of the line for anything (including programming). I was thinking about getting the ThinkPad x200 Tablet but now am reconsidering given your feedback. Don't know actually...

#### SYau

##### Ting Ting
I know resolution will be an issue for the x220 tablet, so I plan to have a separate monitor/docking when I program at home. I think the weight/small form factor makes up for the screen. Plus, it has IPS screen so despite the resolution, it is quite beautiful!

#### Lyosha

##### Psychic in Training
Umm... that is overkill. I see absolutely no reason for you to want to upgrade.

...Just make sure you can carry the thing around. You'll be bringing your laptop with you every day.

#### alain

##### Older and Wiser
but you pay a disturbingly horrid premium for the shiny plastic.
shiny aluminum

#### Andy Nguyen

My Thinkpad T60 has served me well during my MFE education and anything in between. It's been a 5 years journey so "well" is a very understatement.
From time to time, I'm tempted with getting a better, faster, cooler machine. I confessed that I even looked at MBP. In any case, this beast does not die or "violently crash" so I'm getting no excuse to plunge for the guilty pleasure of having new machine.
For the foreseeable future, I am content with getting this T60 a new life by installing W7 on it.

#### amanda.jayne

Im running an x201 (laptop, not tablet). The screen size has been a limiting factor at times but Ive gotten used to it (for the most part anyhow). Apart from my Lenovo service incidents it's been great.

Weight was a big deal for me - a 17"er is hardly portable... Tho Im a fraction of Alexei's size

Unf. felt the same way about the 15"+ MBPs. My laptop comes *everywhere*

#### roni

##### Cornell FE
Here is a good online store located in CA:
http://www.cyberpowerpc.com
I ordered a desktop computer from them in 2009... I really love it, perfect quality...
Soon, I will be ordering a laptop for grad school.
The only thing is that they have 15/17 in.
I tried to compare prices and you can get it for $200-$400 cheaper than from Sony/Dell (definitely cheaper than a Mac).
If you live outside CA, you don't pay taxes. And they always have one of those promotion codes...

#### Devdeep

shiny aluminum

True!

I used one of these http://www.anandtech.com/show/2585 for a while that was a hangover from before. Eventually I ordered a newer computer http://www.sonystyle.com/webapp/wcs...0151&langId=-1&categoryId=8198552921644784015

I got it because it was the cheapest out there that would give me a fermi video card and 1920 pixels on a horizontal line. Consequently a few days after I ordered it my previous laptop started to violently crash due to some hardware issue with the video card (nice timing on the order!)

My suggestion is as follows - my laptop is nowhere near "too heavy" to carry. You will be doing a significant amount coding. If you don't have eyesight problems and can deal with high resolutions, get a laptop with a 1920 horizontal resolution (i.e. 1080P resolution, or 1920x1200, although those are almost a dead breed... and hence horribly overpriced). You want all the horizontal space you can get. Vertical space is also nice but not as crucial.

I am quite happy with my current laptop. I dual boot W7 and almost stock Ubuntu 10.10 (needed a few tweaks for vid card and sound card, 11.04 should be fine once I upgrade), and this has been a very good setup, at least for me.

Don't get a mac. They look pretty, and a 17 inch macbook pro is one of the computers that offers 1920x1200 resolution, but you pay a disturbingly horrid premium for the shiny plastic. You actually pay a ton to get outdated hardware components and to be forced to put up with apple's monopolistic inefficiencies (how's the reception on that iPhone 4?). Not to mention that there's no advantage to having one, aside of course from looking damn cool.

I like the VPCF2190X CTO. but the VPCF2190X CTO with Intel® Core™ i7-2720QM quad-core processor (2.20GHz), 4GB (2GBx2) DDR3-SDRAM-1333, NVIDIA® GeForce® GT 540M (1GB VRAM) dedicated graphics, 500GB Hard Drive (7200rpm), Blu-ray burner, and Large Capacity Battery is $2040. The 17" MacBook Pro with almost the same configuration is$2300. So ~ \$250 premium (not disturbingly horrid, right?) Plus you do get to work on Unix environment throughout and will not have to deal with Securities etc. for the W7 part... I am missing something?

@alain, If I have the 17" MacBook Pro, do I really need the windows boot?

Thanks!

#### amanda.jayne

@Devdeep , there are only a few instances I can think of where I really *needed* windows (vs. Ubuntu)

Excel/VBA - I don't know how nicely the new version coming out will play. There are VBA support rumors hovering

Visual Studio -most of the class uses this. Unless you code well enough for it to be cross compatible with G++ this can be annoying at times when working in a group

Rare vendor software - RITC competition comes to mind

Other than that, you shouldnt have any problems running mac.

There are plenty of windows machines on campus if you ever feel the urge

#### Devdeep

@Devdeep , there are only a few instances I can think of where I really *needed* windows (vs. Ubuntu)

Excel/VBA - I don't know how nicely the new version coming out will play. There are VBA support rumors hovering

Visual Studio -most of the class uses this. Unless you code well enough for it to be cross compatible with G++ this can be annoying at times when working in a group

Rare vendor software - RITC competition comes to mind

Other than that, you shouldnt have any problems running mac.

There are plenty of windows machines on campus if you ever feel the urge

Thanks, Amanda!

#### alain

##### Older and Wiser
@alain, If I have the 17" MacBook Pro, do I really need the windows boot?

Windows? Are you out of your mind?

When I use excel, it is because I'm forced to do it and I refuse to do VBA.

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text:
# Law of gravitation and net gravitational force

[SOLVED] Law of gravitation

In Figure 13-34, a square of edge length 15.0 cm is formed by four spheres of masses m1 = 5.00 g, m2 = 4.00 g, m3 = 1.50 g, and m4 = 5.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m5 = 2.10 g?

I know that since the mass of 1 and 4 are equal they cancel out and i know that the radius is .075*Square root(2). which gives me .106m. I use the equation Gm2m5/r25^2+Gm3m5/r35^2 and i get the answer 6.85e-14 which is not the right answer. I know that the masses are .004, .0015, and .0021 since they are in grams. Can some one help me

## Answers and Replies

Doc Al
Mentor
(1) You didn't provide the figure, but I assume that m1 and m4 are at adjacent corners not opposite corners. So they don't cancel out.
(2) The forces are vectors and must be added as such. Direction counts!

sorry Doc Al M1 and M\$ are on opposite corners here is the picture given

#### Attachments

• 13-34.gif
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Doc Al
Mentor
OK. Realize that the forces due to m2 and m3 point in opposite directions.

that should mean i should subtract their forces but i tried that and that did not work i got a value of 3.11e-14 which is also incorrect

Doc Al
Mentor
I get the same answer. But that's the magnitude of the force. They asked for the force in unit vector notation, which means they want x and y components.

But how would i do that I know how to go from unit vector notation but i do not understand how to do the opposite

Doc Al
Mentor
Just find the x and y components of the net force. What direction is that force? What angle does it make with the x-axis?

It is moving towards mass 2 and the angle should be 45

Doc Al
Mentor
It is moving towards mass 2 and the angle should be 45
Right. So what are the x and y components of a vector at an angle of 45 degrees to the horizontal?

so i set up the sin 45=o/h and solve so i get hsin 45=0 and i get 2.64e-14 but that is not the correct answer. i do the same thing with cos and get 1.63e-14. where should i go from here

Doc Al
Mentor
If a force F makes an angle $\theta$ with the x-axis, its components are:

$$F_x = F \cos \theta$$

$$F_y = F \sin \theta$$

Isn't that what i did, i just wrote it in different terms

Doc Al
Mentor
Isn't that what i did, i just wrote it in different terms
How did you get different answers? ($\sin 45 = \cos 45$.)

ooo i am in radian mode

Thank you Doc Al for your help