text:
## When our two-state PDA constructed from CFG is non-deterministic PDA?

We can always convert our GNF-CFG/CNF-CFG to a two-state PDA but i’m wondering when our PDA is non-deterministic? i’m sure we can not make DPDA for non-Deterministic-CFL , and i suspect that same rule which applied for differing between DCFG and non-Deterministic-CFG also applied here. i mean when we have non-Deterministic δ (delta) implied non-Deterministic edge in our PDA. if my suspicion is right , then for every DCFL exist at least one DPDA. Am i right?

``R ≝ Production Rules of CFG (x,y,"LBL") is a labeled-edge between x and y with “LBL” as a label  ∀r∊R: r= (A,aⱰ) ( A∊V ⋀ a∊T ∧ Ɒ∊V*) add (q,q,"a,A/Ɒ") to E Add (q,q,"ε,z/Sz′") to E Add (q,f,"ε,z′/z′") to E ``

## If anything can be verified efficiently, must it be solvable efficiently on a Non-Deterministic machine?

Suppose, I wanted to verify the solution to $$2$$^$$3$$. Which is $$8$$.

The $$powers~of~2$$ have only one 1-bit at the start of the binary-string.

## Verify Solution Efficently

``n = 8 N = 3 IF only ONE 1-bit at start of binary-string:   IF total_0-bits == N:    if n is a power_of_2:      OUTPUT solution verified, 2^3 == 8 ``

A solution will always be approximately $$2$$^$$N$$ digits. Its not possible for even a non-deterministic machine to arrive to a solution with $$2$$^$$N$$ digits faster than $$2$$^$$N$$ time.

Question

Can this problem be solved efficently in non-deterministic poly-time? Why not if the solutions can be verified efficently?

## Does this imply Hamiltonian path cannot be decided in nondeterministic logspace?

Suppose I nondeterministically walk around in a graph with n vertices.

When looking for a Hamiltonian path, at some point I’ve walked n/2 vertices.

There are (n choose n/2) different combinations of vertices I could have walked (meaning the unordered set, not the ordered walk).

Each of those states must be distinct from one another.

If not, then, depending on the remaining n/2 vertices, I would decide the wrong answer.

Therefore, midway through my processing, at n/2, I need (n choose n/2) different states. That is too big for logspace.

Therefore you cannot decide a Hamiltonian path by nondeterministically walking around.

Does this imply Hamiltonian path cannot be decided in nondeterministic logspace – at least by “walking around”?

## Why aren’t distributed computing and/or GPU considered non-deterministic Turing machines if they can run multiple jobs at once?

So we know a nondeterministic Turing machine (NTM) is just a theoretical model of computation. They are used in thought experiments to examine the abilities and limitations of computers. Commonly used to dicuss P vs NP, and how NP problems cannot be solved in polynomial time UNLESS the computation was done on the hypothetical NTM. We also know an NTM would use a set of rules to prescribe more than one action to be performed for any given situation. In other words, attempt many different options simultaneously.

Isn’t this what distributed computing does across commodity hardware? Run many different possible calculations in parallel? And the GPU, does this within a single machine. Why isn’t this considered an NTM?

## Statement TRUE or FALSE: there exist more non-deterministic TMs than deterministic TMs

Statement TRUE or FALSE:

There exist more non-deterministic TMs than deterministic TMs

For my point of view (I ask if it’s correct):

Under the assumption $$P \neq NP$$, the Statement is TRUE because (1) P is a subset on NP and (2) P is a special case of non-determinism.

Under the assumption $$P = NP$$, the Statement is FALSE, because for each deterministic TM we have a non-deterministic TM.

## Writing a non-deterministic Turing machine

I got an exercise in my college in which ive been asked to write a Non deterministic Turing Machine that accepts words of this form:

Ive manage to find some sort of solution but in a Deterministic way

## Can a Non-Deterministic Pushdown Automata recognize \$ \# a^nb^{2^n} \# \$ which a TM can?

$$\# a^nb^{2^n} \#$$

such that

• The alphabet of the machine is {, a, b, x}.

• The symbol x will never appear on the input a.

• The contents of the tape at completion may be anything.

• The head begins on the lefthand #.

• n ≥ 0.

I know that a Turing machine could recognize this language. But can a NPDA recognize this language too? I am thinking it can but I do not know how to start proving how/why?

## Converting non-deterministic algorithm to deterministic

I was thinking about an non-deterministic algorithm to generate all the subsets of the $$\{1..n\}$$ set.

``Subsets(n)     S' = null     for i = 1..n             u = choice(1..n)         if u in S'            fail         S' = S' + u     for_each u in S'          print u ``

How would I convert this algorithm into a deterministic form? I know that it is a lot more complicated to convert the `choice` statement in a deterministic manner but I want to get a general idea about maybe representing it in a deterministic manner.

## motivation and idea of defining non-deterministic Turing machine

This is a very basic question but I spent some time reading and find no answer. I am not computer science majored but have read some basic algorithm stuff, for example, some basic sorting algorithms and I also have some basic knowledge of how computers operate. However, I am really interested in the idea of a Turing machine, especially the non-deterministic one.

I have read Wiki about the definitions of a Turing machine (and watched some youtube videos) and I sort of accept that, although I really feel that this is a huge jump from an algorithm to an abstract machine. From my understanding (you are more than welcome to correct me):

1. A Turing machine is a machine performing works specified on a cookery book (algorithm).
2. The pages of the cookery book represent the “states” of your machine and each page contains a table saying that which state and which cell your machine will move to given the alphabet the machine read and your current state. (NB. This is not a function but a partial function because it is possible that the machine stops.)
3. So, to guess the idea and motivation of defining an abstract Turing machine, I imagine that the algorithm corresponds to the partial map, the memory of the computer corresponds to the (infinitely long) tape and what’s finally on the tape is the answer to the question you wanna solve.

So, Turing machine looks like a machine to realize any algorithm to solve problems. One just “translates” any algorithm to a set of mysterious simple rules (i.e. the partial function) and let the machine do the laboring job and then we get the solution.

In this respect, Turing machine is always deterministic, because algorithms are deterministic. It tells you what to do next precisely. This is no uncertainty. Turing machine is just a machine to realize any algorithm.

OK, This is very abstract and I sort of accept it. However, then I read something called non-deterministic Turing machine (NTM) and then I was knocked down. A NTM is pretty much similar to a Turing machine except that the partial function is now replaced by a “relation”. That is, it is a one-to-multiple map and it is no longer a (partial) function.

Could someone explain to me why we need such multiple options? I would never expect to encounter something uncertainty in the implementation of an algorithm. It is like telling the machine: you first do A, then if you find yourself in a state B and find the data is now B’ then you choose for yourself one of the 10 allowed next steps?

Are NTM’s corresponding to a set of algorithms that need uncertainty? for example the generation of random numbers? If no, why do we need to allow multiple choices for a Turing machine?

Any help will be appreciated!

## What publication first introduced the concept of a non-deterministic Turing machine?

What publication first introduced the concept of a non-deterministic Turing machine?

Turing did not define the concept in his 1936 paper.


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# Translating sentence to FOL question

I am trying to translate the following English sentence to FOL, however, I am uncertain if my logic in the translation is correct. I will appreciate it very much if anyone can point out my errors.

Question: In addition to the existing predicates in FOL, we can use $Even(x)$, $Prime(x)$, the predicate "$<$", the variables $w,y,z$ and all truth-functional connectives. But we cannot use the predicate "$>$" and the predicate $Square(x)$, but we can express $Square(x)$ with the function symbol "$1x1$", for example the square of 2 is "$2x2$". The domain of discourse is in the set of natural numbers. Translate the following to FOL:

1) All squares are even.

2) No square is prime.

3) The square of some prime is prime.

4) The square of some prime other than 3 is odd.

5) The square of any number greater than 2 is greater than 3.

My Solution:

1) $\forall w Even(wxw)$

2) $\forall w \neg Prime(wxw)$

3) $\exists w (Prime(wxw) \wedge prime(w))$

4) $\exists w (Prime(w) \wedge Prime(wxw) \wedge w \neq 3)$

5) for this question, a more explicit interpretation is: "for any number $w$, if $w$ is greater than 2, then the square of $w$ is greater than 3", so:

$\forall w [(\neg(w < z) \wedge w \neq 2) \rightarrow (\neg ((wxw) < 3) \wedge (wxw) \neq3)]$

I only used one of the three given variables $w,y,z$, so I am not sure if there are any other ways to translate some of the above sentences with the rest of the variables in use. Perhaps I am missing something here?

Thanks

All correct, though I assume that for the last one you meant to write a $2$ rather than a $z$:
$\forall w [(\neg(w < \color{red}{2}) \wedge w \neq 2) \rightarrow (\neg ((w\times w) < 3) \wedge (w\times w) \neq3)]$
But yes, with a $2$ there that's correct.
$\forall w [2 < w \rightarrow 3 < w \times w]$

format: plain


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## Question 12.45

Acidity $K_{a}$
Basicity $K_{b}$
The Conjugate Seesaw $K_{a}\times K_{b}=K_{w}$

Steven Chau 1B
Posts: 21
Joined: Fri Sep 29, 2017 7:07 am

### Question 12.45

Arrange the following bases in order of increasing
strength on the basis of the pKa values of their conjugate acids,
which are given in parentheses: (a) ammonia (9.26);
(b) methylamine (10.56); (c) ethylamine (10.81); (d) aniline
(4.63).

After changing the pKa to pKb, how do we determine the strength of the bases? What is the connection between pKb to pOH?

Annie Lieu-1H
Posts: 68
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 1 time

### Re: Question 12.45

The smaller the pKB, the stronger the base is.

Rachel Formaker 1E
Posts: 86
Joined: Fri Sep 29, 2017 7:04 am
Been upvoted: 2 times

### Re: Question 12.45

The lower the pKb, the stronger the base.
Because pKa + pKb = 14 (so when pKa increases, pKb decreases), the larger the pKa value for an acid, the stronger the conjugate base.

Danah Albaaj 1I
Posts: 50
Joined: Fri Sep 29, 2017 7:07 am

### Re: Question 12.45

For this question, you use the given pKb to calculate the pKa by subtracting the pKb from 14. From there you are able to calculate the molecule's Ka using the equation: Ka=10^-pKa The resulting value should then help you determine the strength of each acid and rank them accordingly.

The whole process is very similar to that of finding the concentration of hydroxide/hydronium based on the pOH/pH given to you because the same rules from the autoprotolysis of water apply here.


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# Homework Help: HARD equation.

1. Jun 10, 2010

### Sirsh

Hi, i'm just wondering if anyone can help me with this equation: I need to determine the values of A,B and C from the given equation.

$$\frac{A}{(x-1)}$$ + $$\frac{B}{(x+1)}$$ + $$\frac{C}{x}$$ = $$\frac{5x^2+2x+1}{x^3-x}$$

I've tried to simplify it by factorising the denominator on the LHS to x^3 - x, and i am suspicious that the values of A,B and C are the 3 parts of the numerator of the RHS respectively to 5x^2,2x and 1.

Thanks

2. Jun 10, 2010

### Tedjn

No, A, B, and C are numbers. You need to combine terms on the LHS and then compare the numerators to get a system of equations for A, B, and C.

3. Jun 10, 2010

### qualal

You need to make the three "fractions" you have on the LHS into just one "fraction" (google "addition and subtraction of algebraic fractions" as a hint).

once you have that you can then equate the [STRIKE]denominators[/STRIKE] numerators in a separate equation. Then you can start to solve for A, B and C

Last edited: Jun 10, 2010
4. Jun 10, 2010

### Sirsh

How is that possible, i'd have to solve for 4 unknowns A,B,C and x?

(Ax^2+Ax+x^2B+x^2C-xB-C)/(x^3-x) is the LHS

5. Jun 10, 2010

### qualal

you leave x as an unknown. Instead you will need equate the coefficients of $$x^{0},x^{1},x^{2}$$

this will give you enough equations to solve for A,B and C

6. Jun 10, 2010

### qualal

apologies, I said to equate the denominators in an earlier post. I meant to say "equate numerators"

7. Jun 10, 2010

### Staff: Mentor

As already noted, the idea is not to solve for x. The original equation is supposed to be true for all values of x other than those that make the denominators zero (i.e., x = 0, 1, or -1).

8. Jun 10, 2010

### Sirsh

I am confused, I know that you can solve for three variables by doing an eliminiation process like a silmutaneous solving method, but i dont understand how to get it to the three equations..

9. Jun 10, 2010

### Staff: Mentor

Starting from this equation:
$$\frac{A}{x - 1} + \frac{B}{(x+1)} + \frac{C}{x} = \frac{5x^2 + 2x + 1}{x(x - 1)(x + 1)}$$

Multiply both sides of the equation by x(x - 1)(x + 1).
Solve for A, B, and C. The simplest way to get the three equations is to substitute specific values for x. I would use x = 0 to get one equation, x = 1 to get another, and x = -1 to get the third equation.

10. Jun 10, 2010

### Daft

Whether you choose to combine them into 1 fraction, and compare numerators, or multiply both sides by x(x-1)(x+1), it would be easier on you if you do not expand the brackets

eg. instead of writing "Ax2 + Ax", leave it as "A(x+1)(x)"

Since the values of A, B and C are the same for all values of x, you yourself can choose any value of x to substitute into the equation to obtain expressions for A, B and C

Choosing the right numbers to substitute is the key to making this simple

eg.: for the term "A(x+1)(x)", if I substitute the values x = -1 or x = 0, that term becomes 0, which means I eliminate the term containing A completely

11. Jun 11, 2010

### Shaybay92

Both sides have the variable x right? So you dont want to solve for x. What is missing is the A B C on the right hand side. You want to expand out the left hand side (through cross multiplication) and then say all x^2 terms must have some coefficient (in this case 5 on the right side) and so any terms on the left with x^2 must sum to give 5x^2. Similarly with 2x and the constant 1.

12. Jun 11, 2010

### gpax42

When I attempted the problem I managed to get C = -1, A = 4 and B = 2...

any idea if this is right?

I multiplied through to get a common denominator and then set the numerators equal to one another.

A(x+1)(x) + B(x-1)(x) + C(x2-1) = 5x2+2x+1

I distributed and found...

Ax2 + Ax + Bx2-Bx+Cx2-C = 5x2+2x+1

Then I simply isolated like terms, found C = -1; A - B = -2 and A + B + C = 5

Solving the system of equations yielded C = -1, A = 4, and B = 2

Might have messed up the arithmetic somewhere but i think that is the general premise to the solution

13. Jun 11, 2010

### Staff: Mentor

These kinds of problems are easy to check. Put the numbers in the equation in the OP and see if you get a true statement.


text:
Why don't ionic compounds have electrical conductivity as a solid but they do as a liquid?

Apr 22, 2014

Solid ionic compounds do not have electrical conductivity because the ions are not free to move. In a liquid, the ionic compound dissociates into its respective ions.

Explanation:

Electrical conductivity requires the movement of charged particles. The common unit for charge is the Coulomb ($\text{C}$).

In the solid state, electrostatic forces hold the ions together in a crystal lattice structure, which is, in short, a 3D interconnected ion network.

A solid ionic compound does not conduct electricity because the ions are not free to move.

This flow of ions depicted immediately above is an electric current. You can tell because if we use $\text{C}$ for the unit of charge, $\text{A}$, or ampere, is the unit of electric current, which is the same as $\text{C/s}$, coulombs per second, the electrical analog of velocity (velocity is mass-motion, just as ampere is charge-motion).

Therefore, charge-motion is required for there to be electric current.

When we heat an ionic compound to its melting point, the heat energy actually breaks the electrostatic forces between the ions, loosening the structure into something more like a liquid; the positive ions can move to the negative terminal, and the negative ions are free to move to the positive terminal, as depicted immediately above.

Since melting many ionic solids takes a lot of heat, we call this the molten state. Thus, ionic compounds can conduct electricity in the molten state.

They can also conduct electricity when dissolved in water; as they will dissociate into their ions, having the ability to conduct electricity (as they may move around freely, being electrolytes in solution).


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# Kerodon

$\Newextarrow{\xRightarrow}{5,5}{0x21D2}$ $\newcommand\empty{}$

## Comments on Definition 4.4.1.1

Go back to the page of Definition 4.4.1.1.

Comment #567 by Tim Holzschuh on

Typo: It should read "$F(\overline{u}) = u$", not "$F(\overline{u}) = C$".

Comment #592 by Kerodon on

Yep. Thanks!

There are also:

• 2 comment(s) on Chapter 4: The Homotopy Theory of $\infty$-Categories
• 2 comment(s) on Section 4.4: Isomorphisms and Isofibrations

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


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# How to go about solving this definite integral [closed]

How do I go about solving the following integral: $$\int_0^{\pi/2} (\sin x)^{\cos x} (\cos x \cot x - \log (\sin x)^{\sin x})\, dx$$

Giving a first try doesn't help as the starting point cannot be grasped that is to say it baffles that how and where to start. The hint given is substitute $u =(\sin x)^{\cos x}$ then what remains to be computed is $\frac {d}{dx} (\sin x)^{\cos x}$ that is a problem in itself as it is differentiating exponential function of a function. Any thoughts:

Hint:Take $u={\sin x}^{\cos x}$ and find $du$
Note that $$d\left( (\sin x)^{ \cos x } \right) =(\sin x)^{ \cos x }(\cos x\cot x-\log (\sin x)^{ \sin x })dx$$ (you can check it) so from Newton- Leibniz formula we can conclude that $$\\ \\ \\ \int _{ 0 }^{ \pi /2 } \left( \sin x \right) ^{ \cos x }\left( \cos x\cot x-\log \left( \sin x \right) ^{ \sin x } \right) dx=\left( \sin { \frac { \pi }{ 2 } } \right) ^{ \cos { \frac { \pi }{ 2 } } }-{ \left( \sin { 0 } \right) }^{ \cos { 0 } }=1-0=1\\$$
• There's an extra $dx$ in there. Either $df(x) = f'(x) \, dx$ or $\dfrac d{dx} f(x) = f'(x)$, but not $\dfrac d{dx} f(x) = f'(x) \, dx$.


text:
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# The Edexcel C3 (14/06/12 - AM) Revision Thread Tweet

Maths exam discussion - share revision tips in preparation for GCSE, A Level and other maths exams and discuss how they went afterwards.

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1. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)

Take the cos of both sides

We know

Hence,
Thanks .
Yup I understood that section, but what I was asking was why using sin(y+π/2) on a graph transformation would not give the correct answer. Sorry if I didn't phrase that clearly.
2. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)
You both are wrong, see my post.
That is so weird ..at 1:25 I edited...at 1:25 you posted this --

em how can we differentiate

3. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by arvin_infinity)
That is so weird ..at 1:25 I edited...at 1:25 you posted this --

em how can we differentiate

May be we both did it at the same time.

For these questions use the chain rule.

I will do one question for you.

Differentiating it gives,
4. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)
May be we both did it at the same time.

For these questions use the chain rule.

I will do one question for you.

Differentiating it gives,
I was meant to ask a totally different question..was gonna say what is the integration of those- might as well ask them later on for C4 page then
5. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by zincoff)
Thanks .
Yup I understood that section, but what I was asking was why using sin(y+π/2) on a graph transformation would not give the correct answer. Sorry if I didn't phrase that clearly.
We usually write it as

The results are usually quoted as 90-x in the books.
6. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by arvin_infinity)
I was meant to ask a totally different question..was gonna say what is the integration of those- might as well ask them later on for C4 page then

To integrate we differentiate

Hence,

Others will be integrated in a similar way.

Though to integrate higher powers of we use the reduction formula or de moivre theorem, however this stuff isn't in C4.
7. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)
We usually write it as

The results are usually quoted as 90-x in the books.
That's alright . Thank you!
8. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
Hey people.
Can someone help with Review Exercise 1, Q8 (Edexcel C3 Book) i know how to do be but i don't know how to use the graph to find the function of f. Thanks
9. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
For solving trigonometric equations which ask for answers in radians, is it ok to do all calculations in degrees and convert the answers to radians at the end ? Thanks !
10. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
help me with this question please!

Use the identities for (sinA + sinB) and (cosA + cosB) to prove that

sin 2x + sin 2y
_______________ = tan (x+y)
cos 2x + cos 2y
11. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by ViixKey)
Hey people.
Can someone help with Review Exercise 1, Q8 (Edexcel C3 Book) i know how to do be but i don't know how to use the graph to find the function of f. Thanks
you dont have to find an equation for f(x), find g(root 23) first which is = 5

Now to figure out f(5) you use the graph, find x=5 in the graph and i think you can carry on from here !
12. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by Puloot)
help me with this question please!

Use the identities for (sinA + sinB) and (cosA + cosB) to prove that

sin 2x + sin 2y
_______________ = tan (x+y)
cos 2x + cos 2y

SinP + Sin Q = 2* sin[(P+Q)/2 ] * cos[(P-Q)/2 ]

and CosP + Cos Q =2 * cos[(P+Q)/2 ] * cos[(P-Q)/2 ]

P=2x , Q=2y

using this you get

2 sin(x+y) cos(x-y)
--------------------------
2 cos(x+y) cos (x-y)

sin(x+y)
-----------
cos(x+y)

tan(x+y)
13. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by shahruk)
SinP + Sin Q = 2* sin[(P+Q)/2 ] * cos[(P-Q)/2 ]

and CosP + Cos Q =2 * cos[(P+Q)/2 ] * cos[(P-Q)/2 ]

P=2x , Q=2y

using this you get

2 sin(x+y) cos(x-y)
--------------------------
2 cos(x+y) cos (x-y)

sin(x+y)
-----------
cos(x+y)

tan(x+y)

Thank youuuu!
14. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by zincoff)
That's alright . Thank you!

The range of is

So this method gives,

While the range of is

Hence, we can't do this.
15. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)

The range of is

So this method gives,

While the range of is

Hence, we can't do this.
Thanks .
16. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by raheem94)
There is no need to draw a graph.

We know the domain of is

So by looking at we can deduce that the range of is

The range of is the domain of

Hence the domain of is

Hope it makes sense
A useful method that doesn't required a sketch. A little slip at the end though...

The range of is the domain of

Hence the domain of is
17. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
Hey people. This things been bugging me.
It says in the C3 book, in the example find

f(x) = f-1 (x)

then it solves it for you, but it just says to make it easier solving it goes straight to f(x) = x

how does this work?
18. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by sajidmatin)
Hey people. This things been bugging me.
It says in the C3 book, in the example find

f(x) = f-1 (x)

then it solves it for you, but it just says to make it easier solving it goes straight to f(x) = x

how does this work?
Do you mean on p27 where is states...?

When

By definition of the inverse function, f-1(x) is a reflection of f(x) about the line y=x. Therefore f-1(x) and f(x) will intersect on the line y=x, so f(x)=x at the point of intersection.

I agree with you that the book should have expanded this to make it more clear. I've attached a graph of the functions used in their example.

Blue :

Red :

Green :
Attached Thumbnails

Last edited by grazie; 10-06-2012 at 12:03.
19. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by pwcroberts)
None as such but you will need to be able to show all of the double angle formulae and factor formulae from the trig expansions.

Also it is possible you may have to prove some of the standard derrivatives in the formula book by using the derrivatives of sin and cos
okay thanks, think I can do all that (hopefully)!
20. Re: The Edexcel C3 (14/06/12 - AM) Revision Thread
(Original post by grazie)
Because the range of f(x) is f(x) > 2

It's well worth doing a quick sketch rather than trying to 'plug in' numbers. As stated in in the previous post, I think changing the fraction makes it much more obvious to see where the asymptotes are. So for the inverse function

It's obvious that there's horizontal asymptote at y=1 and a vertical asymptote at x=2.
Thankyou! Makes sense now, I guess I'll just have to remember to long divide in the exam
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# Palette reduction to pre-defined palette

I'm writing a bunch of GLSL effects for fun, but I can't wrap my head around this. Basically, I want to reduce a texture's palette into a pre-defined set of colors. For example, a post-processing shader would take the FBO result texture and a 1D / 2D texture that would contain the reduced palette (say, 64 colors), and the output would then be sampled based on these.

Something like this in its core:

sampler2D source;
sampler2D palette;

vec4 source = texture2D(...); // Source texture to sample
vec4 palette = texture2D(...); // Palette

vec3 result = /* color from the source converted to the closest value available in palette */;

Any ideas how to do this? And to be more precise, I don't mean palette swapping - the source texture is full-colored result, and this shader would ultimately be a post-processing shader, reducing the amount of colors to the ones that are predefined in the palette texture.

Another way to do this is to make a texture that maps each RGB to a colour on the palette, an image like this (from the NES colours):

Then in a post processing shader you can the RGB colour from your regular image in a way like this:

uniform sampler2d paletteMapping;

vec3 mapColor( vec3 realColor ) {
vec3 mappedColors = floor( realColor * 16 );
int mappedI = mappedColors.r + mappedColors.g * 16 + mappedColors.b * 16 * 16;

return texture2D( paletteMapping, floor( vec2( mappedI / 64, mod( mappedI, 64 ) ) );
}

I've not tested the shader as I'm at work, but this should work on a texture that has interpolation disabled (i.e. it uses nearest neighbour sampling).

The map image I created with a little tool I wrote, from the NES palette image on wikipedia

• This is just awesome, and the fact you've done that tool... Pure gold. Thanks! Jul 22 '14 at 7:59
• If you can re-visit the code at some point, I'd be glad - I tried it out and I basically get only red, white and black with the sample palette you included. :) Jul 22 '14 at 8:33
• Bleh, sorry for the late reply. What code bit? The shader part might need some work, I've been unable to set something up to test it properly. You could try debugging it by skipping the sampling part to see if the colors are encoded directly. Does the output you get make sense compared to what you're expecting? (i.e. is it random pixels everywhere or do the colors roughly fit)
– Elva
Jul 31 '14 at 12:07
• I fiddled around with it, and got it working almost correctly. Some colors, however, were wrong (light gray was yellow etc). It might be a problem with the palette, though. Jul 31 '14 at 12:21
• Possibly, my tool is also pretty basic simply taking the colors as points in 3d space and taking linear distance, could be improved to take human perception into account. If you want, you could edit my reply to add the tweaks you made to the shader :)
– Elva
Jul 31 '14 at 12:41

This problem is called Color quantization. Basically for each pixel you want to find a closest color present in the palette. This can be done by partitioning the 3D color space into regions each of which containing exactly one palette color. The partitioning can be performed using e.g. Voronoi diagram, octree or any other spatial partitioning data structure. Then, for each pixel, a point location problem has to be solved, i.e. the region containing the pixel color has to be found. In case of octrees this can be done efficiently in a straightforward way. The unique palette color in the region will replace the original pixel color.

Because this process is computationally intensive, you should perform as many steps offline in a preprocessing step (e.g. in your asset pipeline) as possible.

Performing the entire quantization step on your textures offline is the easiest way. ImageMagick can perform this via the -remap option. ImageMagick can also perform dithering automatically, which improves the visual quality of quantized images.

If you want to apply a palette to your rendered scene while using your original textures, but the palette is fixed, the color space partitioning can still be done offline. E.g. you can build an octree using the palette colors, transfer the tree to the GPU, and use it in the shader to perform the quantization. ImageMagick uses this approach (on the CPU), the algorithm is described in the source code.

If you want to be able to switch between palettes, the partitioning has to be done on-the-fly, but reasonably on the CPU.

• I understand, that this is not that practical and immediately usable answer as Kevin's, so please accept that, if it solves your problem. As a sidenote: his solutions is a special case of the general framework I described. The spatial partitioning structure of the color space is a 16x16x16 grid, the palette is preprocessed to build that spatial partitioning structure, which is encoded in a texture, the texture is transferred to the GPU, and it is used to solve the point location (by texture sampling).
– zogi
Jul 22 '14 at 6:43
• Thanks for your answer. I pretty much need an online solution (fragment shader) because the original assets are affected by shaders and such. I want the end result to be limited to the palette, and it seems Kevin's answer is just what I was looking for. :) Jul 22 '14 at 8:01


text:
Read counts generated from featureCounts
1
0
Entering edit mode
8 months ago
basuanubhav ▴ 80

Hey all, I had a doubt regarding the output generated by featureCounts. So, what would be the difference in output between the following commands:

featureCounts -p -s 2 -F GTF -t exon -a annotation.gtf -o counts.txt sample.bam

featureCounts -p -s 2 -F GTF -t gene -a annotation.gtf -o counts.txt sample.bam

From what I understand, the second command will only read the lines in the GTF which have 'gene' in the 3rd column while the first command will count the exon lines. Also, the sum of the reads for all exons of a gene should be the same as the number of reads for the gene as a whole. So, should the output file have the same counts for both?

Thanks,

featureCounts • 405 views
ADD COMMENT
4
Entering edit mode
8 months ago

From what I understand, the second command will only read the lines in the GTF which have 'gene' in the 3rd column while the first command will count the exon lines.

This is correct. Both will be aggregated on the gene_id metafeature (that is the default for option -g) in a 1 - 1 relationship with -t gene or in a n - 1 relationship with -t exon.

Also, the sum of the reads for all exons of a gene should be the same as the number of reads for the gene as a whole.

This is not necessarily true. -t gene will also count reads mapped on introns. Usually this will account for low read count because introns are often spliced out but there are many cases of intron retention, and in some species (such as mammals) average intron length is much longer than average exon length. Also because of splicing dynamic, it is possible to sequence unspliced or partially spliced pre-mRNA. So in general, you should not expect the option -t exon and -t gene to produce the same counts.

ADD COMMENT
0
Entering edit mode

Ah, thanks for the nice explanation. I hadn't thought of the aspect of reads aligning to introns skewing results. So, ideally, for an RNAseq experiment, I would want to use -t exon, correct?

ADD REPLY
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Entering edit mode

Yes, definitely in most cases !

ADD REPLY
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Entering edit mode

Thanks a lot, cheers!

ADD REPLY
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Entering edit mode

PS: unrelated to your question, but usually we do not use the -o option because it is often better to discard read that can not be unambiguously assigned, at least for differential expression analysis.

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0
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Yes absolulety! Even I dont use the -O option. The -o I have used is just to define the output file.

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text:
# Triangular Form of a matrix

1. Nov 28, 2008

### Ioiô

I am confused on how to find a matrix B in triangular form for some linear transformation T over a basis $$\{v_1,v_2, v_3\}$$.

Suppose we are given a minimal polynomial $$m(x) = (x+1)^2 (x-2)$$.

Do I want to find a basis $$\{w_1,w_2\}$$ for $$null(T+1)^2$$ such that $$(T+1) w_1 = 0$$ and $$(T+1) w_2 \in S(w_1)$$? Is this because $$(x+1)^2$$ has degree two? This is the part I'm not sure about.

For $$w_3$$, should I just let it be a basis for $$null(T-2)$$?

I tried this for a specific transformation T and got the correct matrix B. (I checked the work by computing the matrix S that relates the old basis (v's) to the new basis (w's) and used the relation $$B = S^{-1} A S$$ where A is the matrix of T.)

Thanks for the help!

2. Nov 29, 2008

### morphism

What is S?

The general idea is as follows. If you take a basis for null(T+1)^2 and a basis for null(T-2), then their union will be a basis for your space V with respect to which T is block diagonal. In this case the first block will be 2x2 and the second one will be 1x1 (with just a '2' in it). So to make sure that T becomes upper triangular, we're going to have to see to it that the 2x2 block is upper triangular. One way to do this is to let w_1 be an eigenvector for T corresponding to -1 (i.e. pick a w_1 in null(T+1)) and then extend {w_1} to a basis {w_1, w_2} for null(T+1)^2. This choice of w_1 ensures that we get the 0s that we want on the first column.

Last edited: Nov 29, 2008
3. Nov 29, 2008

### Ioiô

In $$(T+1) w_2 \in S(w_1)$$ S means subspace. (I know it is confusing when S is also used for the matrix relating the two basis.)

The space V is a direct sum of V1 = null(T+1)^2 and V2 = null(T-2). Is this the reason for the 2x2 matrix block, the diagonals are all -1 and the 1x1 matrix block, the diagonal is 2?

To make the example concrete, suppose the matrix corresponding to the transformation T is
$$A = \begin{bmatrix} -1 & 0 & 2 \\ 3 & 2 & 1 \\ 0 & 0 & -1 \end{bmatrix}$$.

A vector in null(A + I) is x = v1 - v2 (just by solving the equation (A + I)x=0 for x). Likewise, a vector in null(A + I)^2 is y = v2-v3.

This gives me a new basis w1 = v1 - v2, w2 = v2 - v3, and w3 = v3. I checked this and it works. However, another vector in null(A + I)^2 is y'= v3-v2, so I'd get a different set of w1, w2, and w3 and I didn't get the matrix in block form as before. Shouldn't I get still get a matrix in diagonal form (but with different number in the upper triangular block)?

I don't see how choosing a vector w1 such that T(w1) = -w1 will also guarantee the matrix for the eigenvalue -2 will also be in block form.

4. Nov 30, 2008

### morphism

No. This is the reason that we can write T as a direct sum of a 2x2 block and a 1x1 block with respect to this decomposition of V. Of course the 1x1 block will have to be a '2', because Tx=2x for vector in V2.

I don't really follow what you're doing in the rest of the post. If the minimal polynomial of A is m(x)=(x+1)^2(x-1) (I haven't checked), then A won't be diagonalizable. But it will be upper triangular if you choose a good basis for null(A+1)^2. Think about why picking an eigenvector is a good idea.


text:
# Factorisation

## Exercise 14.3 (Part)

(e) 5pq(p^2 – q^2) ÷ 2p (p + q)

Answer: 5pq(p^2 – q^2) ÷ 2p (p + q)
Using (a +b) (a – b) = a^2 – b^2; the equation can be written as follows:
= 5pq (p + q) (p – q) ÷ 2p (p + q)
= 5q(p – q) ÷ 2

(f) 12xy (9x^2 – 16y^2) ÷ 4xy (3x + 4y)

Answer: 12xy (9x^2 – 16y^2) ÷ 4xy (3x + 4y)
This can be solved as previous question;
= 3 (3x + 4y) (3x – 4y) ÷ (3x + 4y)
= 3 (3x – 4y)

(g) 39y^3(50y^2 – 98) ÷ 26y^2 (5y + 7)

Answer: 39y^3(50y^2 – 98) ÷ 26y^2 (5y + 7)
= 2 xx 39y^3(25y^2 – 49) ÷ 26y^2 (5y + 7)
= 3y (25y^2 – 49) ÷ (5y + 7)
= 3y (5y + 7) (5y – 7) ÷ (5y + 7)
= 3y (5y – 7)

## Exercise 14.4

Find and correct the errors in the following mathematical statements:

Question 1: 4(x – 5) = 4x – 5

Answer: 4(x – 5) = 4x – 20

Question 2: x(3x + 2) = 3x^2 + 2

Answer: = 3x^2 + 2x

Question 3: 2x + 3y = 5xy

Here; x and y are as different as chalk and cheese and hence cannot be added together.

Question 4: x + 2x + 3x = 5x

It is like adding one apple, two apples and three apples.

Question 5: 5y + 2y + y – 7y = 0

Answer: = y

Question 6: 3x + 2x = 5x^2

Answer: = 5x

Question 7: (2x)^2 + 4(2x) + 7 = 2x^2 + 8x + 7

Answer: = 4x^2 + 8x + 7

Question 8: (2x)^2 + 5x = 4x + 5x = 9x

Answer: 4x^2 + 5x

Question 9: (3x + 2)^2 = 3x^2 + 6x + 4

Answer: Using (a + b)^2 = a^2 + 2ab + b^2;
= 9x^2 + 12x + 4

Question 10: Substituting x = - 3 in

(a) x^2 + 5x + 4 gives (-3)^2 + 5 ( - 3) + 4 = 9 + 2 + 4 = 15

Answer: = (-3)^2 + 5( - 3) + 4
= 9 – 15 + 4 = - 2

(b) x^2 – 5x + 4 gives (-3)^2 – 5 (- 3) + 4 = 9 – 15 + 4 = - 2

Answer: (- 3)^2 – 5 ( - 3) + 4
= 9 + 15 + 4 = 28

(c) x^2 + 5x gives (-3)^2 + 6( -3) = - 9 – 15 = - 24

Answer: ( - 3)^2 + 5 ( - 3)
= 9 – 15 = - 6

Question 11: (y – 3)^2 = y^2 – 9

Answer: = y^2 – 6y + 9

Question 12: (z + 5)^2 = z^2 + 25

Answer: = z^2 + 10z + 25

Question 13: (2a + 3b) ( a – b) = 2a^2 – 3b^2

Answer: = 2a^2 – 2ab + 3ab – 3b^2
= 2a^2 + ab – 3b^2

Question 14: (a + 4)(a + 2) = a^2 + 8

Answer: = a^2 + 2a + 4a + 8 = a^2 + 6a + 8

Question 15: (a – 4)(a – 2) = a^2 – 8

Answer: a^2 – 2a – 4a + 8
= a^2 – 6a + 8

Question 16: 3x^2 ÷ 3x^2 = 0

Question 17: (3x^2 + 1) ÷ 3x^2 = 1 + 1 = 2

Answer: = 1 + (1)/(3x^2)

Question 18: 3x ÷ (3x + 2) = ½

Answer: = 3x ÷ (3x + 2)

Question 19: 3 ÷ (4x + 3) = 1/4x

Answer: = 3 ÷ (4x + 3)

Question 20: (4x + 5) ÷4x = 5

Answer: = 1 + (5)/(4x)

Question 21: (7x + 5) ÷ 5 = 7x

Answer: = (7x)/(5) + 1


text:
# How Can I Prevent R Markdown From Escaping Characters in Custom Blocks

Let's say I want to recreate this LaTeX code into a custom block:

\begin{tabularx}{\linewidth}{p{0.5\linewidth} X}
\toprule
Column 1 & Column 2 \\
\midrule
\textbf{Lorem} & Ipsum \\
\bottomrule
\end{tabularx}

I tried to do this:

::: {.tabularx data-latex="{\linewidth}{p{0.5\linewidth} X}"}
\toprule
Column 1 & Column2 \\
\midrule
**Lorem** & Ipsum \\
\bottomrule
:::

But it automatically escapes all of the & and \\. Here is the LaTeX output:

\begin{tabularx}{\linewidth}{p{0.5\linewidth} X}
Column 1 \& Column2 \textbackslash{}
\midrule
\textbf{Lorem} \& Ipsum \textbackslash{}
\bottomrule
\end{tabularx}

How do I prevent them from being escaped? The main reason I want to do this is I've been in a situation where I want to use Markdown syntax in a LaTeX table but there's no good R package for creating that LaTeX table.

You are writing here some Markdown syntax. So Pandoc will convert anything as Markdow to LaTeX, unless you tell Pandoc not to do it. Or you need to write raw latex and mark the block as raw latex

Several solutions

::: {.tabularx latex="{\\linewidth}{p{0.5\\linewidth} X}"}
\toprule\
Column 1 & Column2\
\midrule\
**Lorem** & Ipsum\
\bottomrule\
:::

• Write raw latex using {=latex} block
::: {.tabularx latex="{\\linewidth}{p{0.5\\linewidth} X}"}
{=latex}
\toprule
Column 1 & Column2\\
\midrule
\textbf{Lorem} & Ipsum\\
\bottomrule

:::

or even using no custom block

So in conclusion, either you write raw LateX or you write Markdown syntax that would be transformed to LaTeX. Mixing both does not always work well.

This is almost a perfect solution but there is one problem: the & is still escaped. I don't know if there is a way to prevent it from being escaped.

That's what I did as a workaround, but I prefer not to do this because I want to use markdown syntax in the table. For example, if I want to cite a source [@source], it will not work in the LaTeX block.

If I cannot properly make a table in a custom block, I will probably fork kable from the knitr package and make my own table package.

Just look are documentation: we mentioned several time that writing Markdown requires escaping sometimes depending on your output. If you don't want a & but a special LaTeX character you need to escape:

So just use \& instead of & if you don't want it to be considered as Markdown.

Markdown in LaTeX tables is not something that is supported by Pandoc yet (A markdown_in_tex_blocks extension · Issue #2453 · jgm/pandoc · GitHub) that is why mixing both even in table package does not always works. Outputing a Markdown Table with knitr will work though. What does not work for you with this ?

Other table packages like gt or flextable are also to consider.

1 Like

Using \& instead of & doesn't fix it. The all_symbols_escapable Markdown extension unfortunately doesn't either. For now, I will continue to use =latex code chunks. None of the table packages listed in the R Markdown Cookbook do what I want.

About this, I misunderstood as you did not shared what results you had. I though the & was converted to something else.

I believe the & is escaped as it is seen as raw Markdown, then converted to LaTeX and I believe & in LaTeX needs to be escaped, unless in table environment maybe ?
It seems Pandoc will at least escape it because not in a LaTeX block. Pandoc is doing the conversion from & to \&. If you use raw LaTeX in the markdown though, it will keep it as is

\begin{tabularx}{\linewidth}{p{0.5\linewidth} X}
\toprule
Column 1 & Column 2 \\
\midrule
\textbf{Lorem} & Ipsum \\
\bottomrule
\end{tabularx}

without escaping.

As I said, it is a limitation of Pandoc : Markdown syntax can't be mixed within a LaTeX block.

This would be interesting to share what you are looking for. R offers a lot of table framework and it would be interesting for them to know that you have a use case not handled by them.

I don't know what is missing really from what you described, but maybe there is a way you did not find. kable() can take table.envir argument among other to customise the LaTeX written. xtable has a tabular.envir and example in its doc with tabularx.

I am sorry I can't help more. I believe you are encountering the limitation mentioned above. If you can't use a R package for that, you need to write Markdown Table or Raw LaTeX directly in a way that Pandoc can convert them correctly.

I think my table is too complex to be able to make it in any available R package. Here is an example of a table I would make:

\begin{xltabular}{\linewidth}{p{0.1\linewidth} Y Y Y Y Y Y}
\toprule
Activity & \multicolumn{2}{l}{February 10, 2022} & \multicolumn{2}{l}{April 26, 2022} & \multicolumn{2}{l}{June 6, 2022}\\
\cmidrule(lr){2-3}
\cmidrule(lr){4-5}
\cmidrule(l){6-7}
& Score & Goal & Score & Goal & Score & Goal\\
\toprule
Activity & \multicolumn{2}{l}{February 10, 2022} & \multicolumn{2}{l}{April 26, 2022} & \multicolumn{2}{l}{June 6, 2022}\\
Push-ups & 20\newline Level 1 & 25\newline Level 2 & 29\newline Level 2 & 40\newline Level 4 & 30\newline Level 3 & 40\newline Level 4\\
Crunches & 24\newline Level 1 & 30\newline Level 2 & 40\newline Level 2 & 30\newline Level 3 & 26\newline Level 3 & 40\newline Level 3\\
Plank & 1:00\newline Level 1 & 1:30\newline Level 1 & 1:13\newline Level 1 & 1:30\newline Level 2 & 2:01\newline Level 3 & 3:00\newline Level 3\\
Vertical jump & 40 & 50 & 36 & 50 & 40 & 50\\
Sit, reach, and hold & 16 & 17 & 16 & 17 & 17 & 18\\
\bottomrule
\end{xltabular}

Here is the output:

Here are the things making it seem impossible to make this table in kable:

1. The table.envir argument replaces the table environment instead of the tabular environment. I do not want my table to be in a floating table environment. I want it to be in xltabular alone.
2. It will not work with the xltabular or tabularx environment because they expect two arguments, as opposed to tabular which just expects one argument.
3. I'm not sure how to add cmidrules in any R package. In LaTeX it's \cmidrule.
4. I want \addlinespace to be added between every column.

Yes it is possibly. You should included it in your document using Raw LaTeX then with

{=latex}

this way it will be preserved during Pandoc conversion.

Doing complex table in LaTeX without writing LaTeX directly is not that easy so it makes sense that you don't find what you want. gt or flextable can do complex layout, but not great (yet) for LaTeX unfortunately (unless shared in LaTeX as image not table probably).

Yeah, right now I just use the =latex code chunk. If I wanted to make my own R package for tables, is it better to start from scratch or fork kable from knitr?

You cannot just fork kable, you would need to fork knitr which is a bit too much.
you could definitely take inspiration from kable() and other package. Also, take care of the Licence when forking a piece of code : knitr is GPL for example:

One scenario is also submitting a PR to one of the existing package so that your use case is better supported. Maybe better than to start from scratch ?

By the way a few answer I found

This should probably be improved in kable() to allow modifying tabular environment ? You can submit a feature request or a PR.

Probably the same ?

This is definitely not supported currently in kable() and would require adding such feature

linesep in knitr::kable() allows to add \addlinespace

huxtable is another package that has complex table support - don't know for sure what is really supported or not. Looking at the code could help.

And I see you got it working \addlinespace is not added in kable(booktabs = TRUE) · Issue #2133 · yihui/knitr · GitHub

You're right. kable is almost perfect for me except for a few things that are missing. Should I create a fork on GitHub and submit a PR?

I will add a feature request or a PR.

As for tabularx and xltabular, this can probably be fixed by adding an option called tabularx or something like that to create a table like this:

\begin{tabularx}{\linewidth}{X X X}
\end{tabularx}

\begin{tabular}{lll}
\end{tabular}

For tabularx, the first argument is for the length of the table. The second argument is for the alignment or the length of each column. X X X` for three columns means it will divide the three columns evenly (1/3 each). This should be the default. I think there should be an argument to change this too.

Yeah, there should be an argument to put cmidrules.

I will submit pull requests if I can. If I cannot, I will do feature requests instead. Thank you.

This topic was automatically closed 7 days after the last reply. New replies are no longer allowed.

If you have a query related to it or one of the replies, start a new topic and refer back with a link.


text:
# Finding conjugate subgroups

This is a survey article related to:conjugacy
View other survey articles about conjugacy

This article describes techniques to solve problems of the following nature:

• Prove that two subgroups of a group are conjugate. In other words, given a group $G$ and two subgroups $H$ and $K$ of $G$, we need to show the existence of a $g \in G$ suchthat $gHg^{-1} = K$
• Prove a domination condition: Show that a given subgroup is conjugate to some subgroup of another given subgroup. In other words, given a group $G$ and subgroups $H$ and $K$, we want to show that there exists $g \in G$ such that $gHg^{-1} \le K$
• Prove the existence of another conjugate subgroup, with somewhat nicer properties

## Move the big, not the small

It is helpful to keep in mind that the following are equivalent for subgroups $H,K$ of $G$:

1. A conjugate of $H$ is contained in $K$
2. $H$ is contained in a conjugate of $K$

It is usually the second form that is more useful. In other words, it is usually more helpful to try to find a conjugate of $K$ containing $H$, rather than try to find a conjugate of $H$ contained in $K$. In some sense, the bigger groups are easier to move around than the smaller groups.

A formulation equivalent to (2) is:

• The action of $H$ on the coset space $G/K$ has a fixed point

### Proving the domination in Sylow's theorem

Further information: Sylow's theorem#Proof of domination

One part of Sylow's theorem states if for for a finite group $G$, a $p$-Sylow subgroup $S$ of $G$ and a $p$-subgroup $H$ of $G$, $H$ is conjugate to a subgroup of $S$.

We now consider the action of $H$ on the quotient space $G/S$, by left multiplication. Since the order of $G/S$ is relatively prime to $p$, there is some orbit whose order is not a multiple of $p$, but since $H$ is a $p$-group, this orbit must have size one. Thus, the action of $H$ on the quotient space $G/S$ has a fixed point, so $H$ is contained in the conjugate of $S$ that fixes that point.

### Proving that any finite subgroup is contained in the orthogonal group

Further information: Orthogonal group is finite-dominating in general linear group

Another example is the following: given a finite subgroup of $GL(n,\R)$ (the general linear group), there exists a conjugate of it contained in $O(n,\R)$ (the orthogonal group). In other words, the orthogonal group dominates all finite groups.

Again, it is easier instead to prove the following: given any finite subgroup of $GL(n,\R)$, there exists a conjugate of the orthogonal group containing it. Saying that a group is contained in a conjugate of the orthogonal group, is equivalent to saying that there exists a positive definite symmetric bilinear form invariant under the action of the group. Thus, the problem reduces to the following: given any finite subgroup of $GL(n,\R)$, find a positive-definite symmetric bilinear form invariant under the group.

This is achieved by taking any positive-definite symmetric bilinear form, and then averaging it over the group.

### Proving that any finite subgroup of diffeomorphism group has an invariant Riemannian metric

We can take this somewhat further: given any finite subgroup of the self-diffeomorphism group of a differential manifold, there exists a Riemannian metric on the differential manifold that is invariant under the finite group. In other words, there is a Riemannian metric such that the elements of the finite group are isometries for that Riemannian metric.

The proof is exactly the same as the previous one: we start with any Riemannian metric, and average under the action of the finite group. The key difference from the preceding case is that in the previous case, all the isometry groups for positive-define symmetric bilinear forms were conjugate. Now, it is no longer true that isometry groups of Riemannian metrics are all conjugate, so we cannot obtain a single group that is finite-dominating.

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]

## Move the big, and Sylow's theorem

We often need to combine the basic results of Sylow theory, with the idea that it's the bigger subgroup that needs to be moved around. Below are some examples.

### Moving to nice p-subgroups

The setup is like this: $G$ is a finite group, $S$ is a $p$-Sylow subgroup, and $T$ is a subgroup of $G$ contained inside $S$. Then, we need to find a subgroup $U$ of $G$ such that $T$ and $U$ are conjugate inside $G$, and such that $N_S(U)$ is a $p$-Sylow subgroup inside $N_G(U)$.

The key trick here is again, to focus attention on trying to move the big rather than the small. So, instead of chasing $U$, we look at the subgroup $N_G(T)$. Let $V$ be a $p$-Sylow subgroup of $N_G(T)$. Now, by the domination part of Sylow's theorem, $V$ is contained in some conjugate $S_1$ of $S$ (i.e. in another Sylow subgroup). Let us try to determine what $N_{S_1}(T)$ looks like.

Clearly, $N_{S_1}(T) = N_G(T) \cap S_1$. This contains $V$, because $V$ is contained in both subgroups. Also, it cannot be strictly bigger than $V$, because $V$ is Sylow inside $N_G(T)$.

So what we've done is: instead of moving $T$ to a good subgroup of $S$, we've found a Sylow subgroup $S_1$ in which $T$ is good. Now, of course, we can do a conjugation taking $S_1$ to $S$ and the image of $T$ under that conjugation is the required $U$.


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# Thread: Sin, sec and cosine.

1. ## Sin, sec and cosine.

If sec^2(x) – 1 = sin(x) , which one of the following is a possible value for x?

how do you do this?

Thanks in advance. I don't get the sin,cosine rules so please help.! =)

The choices are

a) - $\frac{\pi}{2}$

b)= $\frac{\pi}{4}$

c) 0

d) $\frac{\pi}{2}$

e) $\frac{\pi}{4}$

thanks in advance!

2. Originally Posted by fabxx
The choices are

a) - $\frac{\pi}{2}$

b)= $\frac{\pi}{4}$

c) 0

d) $\frac{\pi}{2}$

e) $\frac{\pi}{4}$

thanks in advance!
see, by inspection, 0 works. $\sec^2 0 - 1 = 1 - 1 = 0 = \sin 0$

note that the left hand side (which is actually $\tan^2 x$) is an even function, so if (a) and (b) don't work, (d) and (e) would give the same answer and hence not work either. (c) being 0 catches your eye though, and i would test that first, and what do you know, it worked!

doing this with the algebra would be a bit more complicated and time consuming. so just do the check. you have to check at most 3 values, which you can do pretty quickly if you memorized the trig values for the special angles.


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EN

## Economically Vulnerable Youth in OECD and Russia

PIIS013122270002973-1-1
DOI10.20542/0131-2227-2018-62-11-99-107
Publication type Article
Status Published
Authors
Affiliation: National Research University Higher School of Economics
Affiliation: National Research University Higher School of Economics
Address: Russian Federation, Moscow, Russian Federation
Journal nameMirovaia ekonomika i mezhdunarodnye otnosheniia
EditionVolume 62 Issue 11
Pages99-107
Abstract

The article is dedicated to the analysis of economically vulnerable categories of youth in OECD and Russia with particular focus on the comparative analysis of various characteristics of NEETs. Labour force participation of young people in the developed countries decreased during 2000–2015 which was associated not only with the cyclical crisis, but also with an increase in the proportion of young people who continue their education. The new statistical indicator – NEET-reflects the ratio of those young people who are not in employment, education or training. Thus, it should identify economically vulnerable groups of young people who experience difficulties with the transition from school to work. The article states that NEET should not be regarded as the only indicator of economic deprivation of young people. The reasons for the drop out from education and employment can be diverse. The category of NEET unites young people with different experience in employment, personal characteristics and life goals. For some of them economic inactivity is a voluntary choice. That is why one should always consider differences in NEET types. However, the majority of NEET group consists of economically vulnerable young people who definitely need support from the state. In OECD countries about 60% of all NEETs got in this state due to some disfunction of the labor market, and the remaining 40%  – due to some social or medical reasons. Countries of Southern Europe are characterized by the highest proportion of long-term unemployed and desperate NEETs, reflecting the tough character of youth labor market adjustment to the crisis. The risk of falling into the NEET group as a whole is higher among young women and those who have low levels of education. Effective youth policy is impossible without the consideration of all aspects of youth labor market. This is relevant both to developed and transition economies, including Russia. In Russia the share of NEET youth has been declining over the past twenty years and now is at the level of European mean values. At the same time the structure of NEET youth in Russia is dominated not by the unemployed, i.e. those actively seeking work, but by economically inactive young people who are outside the labor market. One of the most disturbing features of the structure of Russian NEET youth is an increase in the proportion of young graduates with higher education. It points to the difficulties of matching between education and labor market which can come as a result of so-called massovization of higher education.

Keywordsyouth, economic activity, unemployment, NEET, labor force, Russia, OECD
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