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# Linear Algebra - Plane

A plane is a two dimensional vector space.

A plane has a dimension of two because two coordinates are needed to specify a point on it.

## 3 - Type

### 3.1 - Containing the origin

• Two-dimensional: All points in the plane: Span {[1, 2], [3, 4]}
• Span of two 3-vectors {[1, 0, 1.65], [0, 1, 1]} is a plane in three dimensions.
# A more familiar way to specify a plane
{(x, y, z) : ax + by + cz = 0}
# Using dot-product, the above equation becomes a set of vectors
# satisfying a linear equation with right-hand side zero
{[x, y, z] : [a, b, c] * [x, y, z] = 0}

#### 3.1.1 - Plane Intersection

The intersection of the two following plane:

• {[x, y, z] : [4,-1, 1] · [x, y, z] = 0}
• {[x, y, z] : [0, 1, 1] · [x, y, z] = 0}

is

• {[x, y, z] : [4,-1, 1] · [x, y, z] = 0, [0, 1, 1] · [x, y, z] = 0}

### 3.2 - Translation

The translation of a plane translate it in a way that it doesn't contain the origin.

You can express such plane as

• an affine hull
• a solution set of an equation

Vector addition is used to defined a set of points forming an plane that not necessarily go through the origin.

You translate the plane by adding a vector c [0.5, 1] to every point in the plane.

abbreviated: $c + nu$

The result is a plane through c instead of through origin.

#### 3.2.3 - Equation

The solution set of an linear equation:

• ax + by + cz = d
• In vector terms:{[x, y, z] : [a, b, c] · [x, y, z] = d}

Data Science
Data Analysis
Statistics
Data Science
Linear Algebra Mathematics
Trigonometry

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# Predicting major product of a reaction with nitrous acid

$\hspace{4.5cm}$

For the above reaction I am asked to predict $\ce{X}$, the major product.

Of the options given, I thought product (1) would be correct, seeing as how it has the most stable intermediate cation (tertiary) and a highly substituted double bond.

$\hspace{3.9cm}$

The answer however states that product (2) is the major product.

I can't understand why the second product would be favored over the first. There's no obvious stabilizing factor here in product (2).

• Oxonium cations are more stable than tertiary. – Mithoron May 4 '16 at 23:15
• Isn't a secondary carbocation the intermediate and not an oxonium ion for product 2?I mean, after N2 is eliminated and a hydride transer occurs from the carbonyl carbon to the carbocation, isnt a secondary carbocation formed next to electronegative oxygen? – Daipayan Mukherjee May 4 '16 at 23:35
• ...Which is a protonated carbonyl - oxonium cation. Somehow this powerful mesomeric stabilisation is often omitted – Mithoron May 4 '16 at 23:43
• (2) cannot be the major product as of the images you are supplying now unless a carbon source is added. The carbon atoms don’t add up. – Jan May 5 '16 at 12:05


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# Difference between revisions of "2006 AIME II Problems/Problem 7"

## Problem

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

## Solution

### Solution 1

There are $\left\lfloor\frac{999}{10}\right\rfloor = 99$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when $a$ or $b$ have a 0 in the tens digit, and since the equation is symmetric, we will just count when $a$ has a 0 in the tens digit and multiply by 2 (notice that the only time both $a$ and $b$ can have a 0 in the tens digit is when they are divisible by 100, which falls into the above category, so we do not have to worry about overcounting).

Excluding the numbers divisible by 100, which were counted already, there are $9$ numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling $9 \cdot 9 = 81$ such numbers; considering $b$ also and we have $81 \cdot 2 = 162$. Therefore, there are $999 - (99 + 162) = \boxed{738}$ such ordered pairs.

### Solution 2

Let $a = \overline{cde}$ and $b = \overline{fgh}$ be 3 digit numbers:

 cde
+fgh
----
1000

$e$ and $h$ must add up to $10$, $d$ and $g$ must add up to $9$, and $c$ and $f$ must add up to $9$. Since none of the digits can be 0, there are $9 \times 8 \times 8=576$ possibilites if both numbers are three digits.

There are two other scenarios. $a$ and $b$ can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are $9 \times 8 \times 2=144$ possibilities (the two accounting for whether $a$ or $b$ has three digits) and for the second case there are $9 \times 2=18$ possibilities. Thus, thus total possibilities for $(a,b)$ is $576+144+18=738$.

### Solution 3

We first must notice that we can find all the possible values of a between 1 and 500 and then double that result.

When 1 < a < 100 there are 9 X 9 = 81 possible solution for a so that neither a nor b has a zero in it, counting 1 through 9, 11 through 19 ... 81 through 89. When 100 < a < 200 there are 9 X 8 =72 possible solution for a so that neither a nor b has a zero in it, counting 111 through 119, 121 through 129 ... 181 through 189. This can clearly be extended to 100k < a < 100(k+1) where k is an integer and 0 < k < 9. Thus for 100 < a < 500 there are 72 X 4 = 288 possible values of a.

Thus when 1 < a < 500 there are 288 + 81 = 369 possible values of a and b.

Doubling this yields the 738.


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# Probability lollipops

Probability Level 1

There are 4 orange lollipops, 2 red lollipops and 3 green lollipops in a bag.

Martha takes and eats a lollipop at random. She then takes and eats another lollipop at random.

What is the probability that the lollipops are the same colour?

×


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# A question on modular arithmetic

Assume that \begin{align*} a &\equiv x \pmod p \\ b &\equiv y\pmod q.\end{align*}

Does this imply an equation involving the numbers $a,b,x,y$ modulo $pq$? One possible example would be $$ab \equiv xy\mod pq$$

-
Not quite. Take $a = 0$, $x = p$, $b = 1$, and $y = q+1$. Then your suggestion gives $$0 \equiv (q+1)p \bmod pq$$ which doesn't have to be true! However, if you avoid picking anything that reduces to 0..? (Also, I'm guessing $p$ and $q$ are primes?) –  Alex Feb 23 '12 at 14:42
I don't think you can do better than $$ab\equiv (x\mod q)(y\mod p)\mod pq$$ since the value of $x\mod p$ and $y\mod q$ don't tell you much about $x\mod q$ and $y\mod p$. –  Alex Becker Feb 23 '12 at 14:54
Well, you have $$(a-x)(b-y) \equiv 0 \mod pq$$ –  Joel Cohen Feb 23 '12 at 15:04

Here is an equivalent question: given that $(a - x)$ is a multiple of $p$ and $(b - y)$ is a multiple of $q$, can we conclude that something is a multiple of $pq$? The answer is yes: $(a - x)(b - y)$, $(a - x)q$, and $p(b - y)$ all must be multiples of $pq$. So, among other things, we have:

\begin{align*}(a - x)(b - y) &\equiv 0\phantom{0} \pmod{pq}\\ aq &\equiv xq \pmod {pq}\\ pb &\equiv py \pmod{pq}.\end{align*}

The first equation might look nicer if we multiply it out:

$$ab + xy \equiv ay + xb \pmod{pq}.$$

-
There is similiar law $$a \equiv b\pmod n \\c \equiv d\pmod n$$ imply $$ac \equiv bd\pmod n \\a+c \equiv b+d\pmod n$$


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# 2914. A+B(II)

### 输入格式

Input contains an integer N in the first line, and then N lines follow. Each line consists of a pair of integers a and b, separated by a space, one pair of integers per line.

### 输出格式

For each pair of input integers a and b you should output the sum of a and b in one line, and with one line of output for each line in input.

### 样例

Input
2
1 5
10 20

Output
6
30

381 人解决，400 人已尝试。

427 份提交通过，共有 611 份提交。

0.7 EMB 奖励。


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# Online MS in applied maths or physics

1. Jul 25, 2012

### steve0000

Hi all,
I am an engineer and I would like to take an online master in applied mathematics or physics - I am interested in several subjects in both field, so right now I am open to either choice.

To prevent common comments I saw in other post, let me point out I hold a PhD in Aerospace engineering, have several years of experience as researcher, and have a good background in maths. I was teaching assistant (in a top university) for several courses that were offered to Campus and online students, so I know that distance learning can work well, at least in engineering.

What I do not know, is which top universities offer online courses in maths or physics. I spent hours looking for options, but looks like most good universities only offer online courses in engineering schools, if any.

Only options I found so far are Columbia and Open University. Do you guys know more??

Steve

2. Jul 25, 2012

### chiro

Hey steve0000 and welcome to the forums.

One question I have is why you want a degree at all if you already have a PhD? Do you want to get the degree and the title of having the degree for reasons other than personal ones? Do you need to have an official degree from somewhere else? Are you looking perhaps for a career change?

The reason I ask is because since you have a PhD, you are probably by now an extremely independent learner, even moreso than a Masters student, so I'm wondering whether there is any real advantage for yourself to even take a formal course at all.

3. Jul 25, 2012

### steve0000

Chiro
Thank you for your interest. There are of course several reasons why I , as an engineer working in a narrow field, would like to improve my knowledge in maths and physics.

But to answer your question, the reason I'd also like to get a formal degrees (rather than studying some textbooks or watching the OpenCourseWare videos) are

(1) to meet people and work "actively" in the field - if the graduate class is good, I could participate to small projects, perhaps leading to conference papers etc...

(2) to stay motivated - studying outside my field will not become my priority unless I force it to be. I know, I'm lazy, I need this kind of motivations sometimes :)

(3) to gain credibility in academia - an additional MS in maths/physics is a stronger statement than "I'm also interested in xyz" in a cover letter or CV

steve

4. Jul 25, 2012

### chiro

The only concern I have with an online course is the credibility issue: not just for the course itself, but how it is viewed by others including academics, employers, and anyone else.

I definitely know what you're saying about the motivation: pretty much all of us need an environment that keeps us plodding along and that usually is one where someone else gives the direction as a kind of stressor and a pressure scenario to make us do things.

The last thing that you would want to go through is to undertake a program only to have it not be taken seriously, even if the program itself was a decent program (it's possible this could happen), since a lot of the online education out there is plagued with the stigma of being a "degree mill" and this stigma is one of many things that tarnish credibility.

You've mentioned Columbia University which has a reputation to uphold which is a good thing: you know that if someone has a good reputation, they will probably do whatever they can to keep it.

The only other question I have is what kind of specialty or area are you considering? Different schools have different specialties depending on the staff or the department itself and each specialty will have an impact on the kinds of projects and networking opportunities that they have. Since you have an engineering degree, you will no doubt know the basic areas of mathematics, and probably have a really good idea of some of those sub-areas as well.

I'm also wondering whether you have your own networking opportunities to find people (like say through a colleague of a colleague) that can hook you up with some active applied mathematicians? It's not a gaurantee, but given your in engineering, it's definitely more plausible than simply possible.

5. Jul 25, 2012

### steve0000

I agree with you regarding the perceived credibility of online courses. This is why I consider Columbia as the only interesting option I found so far.

Regarding the area of interests: in physics, I am interested in space sciences and astrophysics. In maths, almost anything in applied math is a good exercise for my brain and will help my current work too. I have a few subjects in mind but I'm open to new as well.

Regarding the networking opportunities, I do have them in fact. That'd be my back-up plan if I cannot find an online master. The problem with that, I'd need to learn quite a lot by myself before I can actually propose a collaboration, which requires some time and much motivation (!).

6. Jul 25, 2012

### chiro

It sounds like doing an online masters and then using your networking opportunities is one way of attacking the situation.

The hardest part IMO is the credibility issue and the next thing is the networking issue. You have a lot of credibility capital behind you with your bachelors and PhD in engineering, so it's not like you are some unknown person from TimBukToo or anything and you have people who you know and/or can contact.

With regards to astrophysics, you want to PM twofish-quant if you can't get a reply from him since he is very interested in the kinds of situations you are in: i.e. situations of online learning and they go.

The other thing is that his PhD was in Astrophysics so he knows a lot about that area from personal experience and knows about the community, so he can give you specific advice that someone like myself would really have no clue about.

7. Jul 25, 2012

### steve0000

I'll try to get in touch with twofish-quant!

Best
Steve

8. Jul 25, 2012

### twofish-quant

The only two that I know of are

http://www.space.edu/ - online masters of space science with University of ND
http://www.bt.pa.msu.edu/index_vubeam.htm - online beam physics at University of Michigan

Also there are some interesting reason why there aren't any more online programs in physics. The basic problem is that it takes a lot of time and effort to put together a good online program, and there are basically no incentives to do it. With MBA's you have lots of people that are willing to pay the university  to start a new program. Engineering schools do distance education because they are funded by companies that want to improve employer skills.

For physics programs, the funding comes from the government, and there is no pressure to increase funding, in part because the economic system can't deal with all of the physics graduates that the traditional system is putting out.

9. Jul 25, 2012

### twofish-quant

I don't think that's true at all. For research positions, an extra Masters will not help you at all. For community college/adjunct positions, an extra Masters is overkill. If you want to teach physics or astronomy at the community college level, you already have the necessary credentials.

Among researchers, degrees don't provide any credibility at all. What does provide credibility are publications. If you get your Ph.D., your dissertation will be a publication, and you can usually get some journal articles out of it. If you have a Masters degree and no publications, it's completely useless.

10. Jul 25, 2012

### steve0000

twofish-quant
first of all, thank for your quick reply and for the links. I didn't know them so I am now looking into them.

This depends on the field. My field, for instance, is very narrow, and for this reason, finding an academic position in a good university is almost impossible. An extra masters is a proof that I have competence and interests beyond what I am doing (words are not enough I can tell you). Also, as in many other engineering field, being multidisciplinary helps my career. Just an example: my engineering research has a larger impact, if I understand the state of the art problems in space science. (For this reasons, physicists prefer to work with engineers with a strong background in physics)

I agree on this. But in top universities, master courses are good enough that students can produce publication-worth projects- which is what I wanted to do.


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# Midterm 1 guide

Answers on the bottom. Also, you can bring a notecard (5in x 7in max) with notes or whatnot.

## Turing test

True or false? Turing first described a test between a monkey and a human before describing a test between a human and a machine.

True or false? The goal of the machine is to convince the judge that the human is a machine.

True or false? To pass the Turing test, the machine should be expected to fool the judge (convince the judge that the machine is a human) nearly 100% of the time; or, maybe just 70% of the time.

True or false? Dennett thinks that winning a chess championship is just as strong a test for computer intelligence as the Turing test.

## Search

### General search

Give an initial state, describe possible actions and the transition model (i.e., how states connect), and specify a goal criterion for the following search problem:

A robot wants a path to exit a maze. The maze is represented as a square grid; open areas are represented by white grid cells, and walls are represented by black grid cells.

Repeat this exercise for the following search problem:

Recall from calculus that the derivative is the inverse of the integral. Finding the derivative of a formula is easy (no search needed). But finding the integral of a formula is often challenging, and may require a search procedure. Describe this search procedure.

Complete the following table for weighted (different branches have different costs), finite search graphs:

Search algorithm Always complete? Always optimal? Uses a heuristic?
Random search
Depth-first search   No
Best-first search
A* search     Yes

Recall the Goodale route search problem from Practice with searching notes. Answer the following questions for each of breadth-first search, depth-first search, best-first search, and A* search (using some heuristic like distance "as the crow flies"). Assume we start at Woodruff & Tuttle and the goal is Goodale parking lot. For each question, there may be more than one correct answer.

1. What are the first two states (beyond Woodruff & Tuttle) that will be checked?
2. What does the tocheck list look like after checking those states from question 1? Be sure to order the list by an appropriate sort order (breadth-first, best-first, and A* will take the next state from the front of the list; depth-first will take the next state from the end).

What is the singular requirement for a heuristic to be called "admissible?"

Describe a utility function for an adversarial search procedure (e.g., minimax) for playing chess.

True or false? Minimax is an appropriate search procedure for single-player games (like a solitaire card game or a Rubik's cube)?

True or false? Alpha-beta pruning, as applied to the minimax procedure, produces more optimal solutions (e.g., produces game moves that allow the computer to win in fewer moves).

True or false? Alpha-beta pruning (applied to minimax) uses different utility functions than regular minimax.

True or false? Alpha-beta pruning (applied to minimax) is the same algorithm as minimax, just faster because irrelevant computations are omitted.

What is the best move, according to a minimax procedure, for 'x' to make in the following tic-tac-toe board?

 o o x x x o

## Knowledge representation

### Boolean logic

Simplify the following expression using Boole's and De Morgan's laws: $$\neg(\neg x \vee \neg y) \wedge (x \wedge (y \vee \neg y))$$

### Propositional logic

Build a truth-table for: $$\neg A \vee (B \leftrightarrow A)$$

Given,

1. $$(S \vee T) \wedge (A \vee B)$$
2. $$S \rightarrow Q$$
3. $$\neg Q$$
4. $$\neg C \vee Q$$

Prove (and state the rules and premises you utilize):

1. $$\neg S$$
2. $$T$$
3. $$\neg C \vee W$$

### First-order logic

Rewrite the following statements in first-order logic:

1. Tony is a tiger.
2. All tigers have stripes.
3. Some tigers are white.
4. If any tiger is white, then we know it lives in the snow (but not necessarily vice versa).
5. A tiger is not both white and orange.

Rewrite the following without using $$\forall$$ (hint: use negation): $$(\forall x)(P(x) \leftrightarrow (A(x) \wedge B(x)))$$

Rewrite the following without using $$\exists$$: $$(\exists x)(W(x) \vee \neg Q(x))$$

### Turing test

True or false? Turing first described a test between a monkey and a human before describing a test between a human and a machine.

False — the original test was between a man and a woman; the man was supposed to impersonate the woman.

True or false? The goal of the machine is to convince the judge that the human is a machine.

True — or, False — my fault, both answers are true: the machine is supposed to convince the judge that it is a human, which is equivalent to convincing the judge that the human is a machine.

True or false? To pass the Turing test, the machine should be expected to fool the judge (convince the judge that the machine is a human) nearly 100% of the time; or, maybe just 70% of the time.

False — the machine should not be expected to be "more human than human," so convincing the judge just 25% of the time seems sufficient. Over 50% may well change our definition of humanity…

True or false? Dennett thinks that winning a chess championship is just as strong a test for computer intelligence as the Turing test.

False — beating humans in chess is not sufficient because chess requires only a narrow kind of intelligence, more calculation than anything

### Search

#### General search

Give an initial state, describe possible actions and the transition model (i.e., how states connect), and specify a goal criterion for the following search problem:

A robot wants a path to exit a maze. The maze is represented as a square grid; open areas are represented by white grid cells, and walls are represented by black grid cells.

• initial state: the robot's starting location
• possible actions: move north, south, east, west (not all movements are always possible)
• transition model: look at the map, connect grid locations via north/south/east/west links
• goal criterion: finding the exit from the maze

Repeat this exercise for the following search problem:

Recall from calculus that the derivative is the inverse of the integral. Finding the derivative of a formula is easy (no search needed). But finding the integral of a formula is often challenging, and may require a search procedure. Describe this search procedure.

• initial state: the starting formula that needs a symbolic integral to be found
• possible actions / transition model: create a database like this:
• sin(x) --> -cos(x)
• cx --> cx^2/2
• x^n --> (x^(n+1)/(n+1))
• goal criterion: say you are looking at a new formula g, and the original formula is f, then you want dg/dx = f; every "state" (formula) is checked in this way

Complete the following table for weighted (different branches have different costs), finite search graphs:

Search algorithm Always complete? Always optimal? Uses a heuristic?
Random search Yes No No
Depth-first search Yes No No
Best-first search Yes No Yes
A* search Yes Yes Yes

Recall the Goodale route search problem from Practice with searching notes. Answer the following questions for each of breadth-first search, depth-first search, hill-climbing search, best-first search, and A* search (using some heuristic like distance "as the crow flies"). Assume we start at Woodruff & Tuttle and the goal is Goodale parking lot. For each question, there may be more than one correct answer.

1. What are the first two states (beyond Woodruff & Tuttle) that will be checked?
• breadth-first (one possible answer): Lane & Tuttle, High & Woodruff
• depth-first (one possible answer): Lane & Tuttle, SR-315 & Lane
• best-first (only answer): High & Woodruff, High & 15th
• A* (only answer): Lane & Tuttle, High & Woodruff
2. What does the tocheck list look like after checking those states from question 1? Be sure to order the list by an appropriate sort order (breadth-first, hill-climbing, best-first, and A* will take the next state from the front of the list; depth-first will take the next state from the end). Assume already-checked states are not added.
• breadth-first (one possible answer): [SR-315 & Lane, High & 15th]
• depth-first (one possible answer): [High & Woodruff, SR-315 I-670 offramp, SR-315 & King]
• best-first: [High & 11th, US-23 & 15th, Lane & Tuttle]
• A*: [High & 15th, SR-315 & Lane]

What is the singular requirement for a heuristic to be called "admissible?"

• The heuristic must always underestimate (or exactly match) the true cost from the current state to the goal state.

Describe a utility function for an adversarial search procedure (e.g., minimax) for playing chess.

• One possible answer: Number of captured pieces, maybe with a weight for each piece (so a queen will have a large weight, a pawn not so much)

True or false? Minimax is an appropriate search procedure for single-player games (like a solitaire card game or a Rubik's cube)?

False — there is no adversary to simulate and minimize, thus no minimax

True or false? Alpha-beta pruning, as applied to the minimax procedure, produces more optimal solutions (e.g., produces game moves that allow the computer to win in fewer moves).

False — alpha-beta does not change the solutions.

True or false? Alpha-beta pruning (applied to minimax) uses different utility functions than regular minimax.

False — the utility functions are not changed, just which parts of the search space are actually searched.

True or false? Alpha-beta pruning (applied to minimax) is the same algorithm as minimax, just faster because irrelevant computations are omitted.

True — that's all alpha-beta pruning is; it still does minimax.

What is the best move, according to a minimax procedure, for 'x' to make in the following tic-tac-toe board?

 o o x x x o

The answer is that 'x' should move in the exact middle, as proved by this minimax search tree:

### Knowledge representation

#### Boolean logic

Simplify the following expression using Boole's and De Morgan's laws: $$\neg(\neg x \vee \neg y) \wedge (x \wedge (y \vee \neg y))$$

• Change $$y \vee \neg y$$ to $$T$$, yielding: $$\neg(\neg x \vee \neg y) \wedge (x \wedge T)$$
• Change $$x \wedge T$$ to $$x$$, yielding: $$\neg(\neg x \vee \neg y) \wedge x$$
• Distribute the negative, yielding: $$(x \wedge y) \wedge x$$
• Get rid of parentheses, reduce redundancy, yielding $$x \wedge y$$

#### Propositional logic

Build a truth-table for: $$\neg A \vee (B \leftrightarrow A)$$

$$A$$ $$B$$ $$B \leftrightarrow A$$ $$\neg A \vee (B \leftrightarrow A)$$
T T T T
T F F F
F T F T
F F T T

Build a truth table to find this:

$$A$$ $$B$$ $$C$$ $$B \wedge C$$ $$A \rightarrow (B \wedge C)$$ $$\neg A \wedge (A \rightarrow (B \wedge C))$$
T T T T T F
T T F F F F
T F T F F F
T F F F F F
F T T T T T
F T F F T T
F F T F T T
F F F F T T

The last four rows will provide satisfying assignments.

Given,

1. $$(S \vee T) \wedge (A \vee B)$$
2. $$S \rightarrow Q$$
3. $$\neg Q$$
4. $$\neg C \vee Q$$

Prove (and state the rules and premises you utilize):

1. $$\neg S$$
1. From 2, 3 and modus tollens.
2. $$T$$
1. From $$\neg S$$ (above) and 1, simplification, and disjunctive syllogism.
3. $$\neg C \vee W$$
1. From 3, 4 and disjunctive syllogism (which gives us $$\neg C$$); now apply addition plus introduce the arbitrary symbol $$W$$ (which is allowed with the addition rule)

#### First-order logic

Rewrite the following statements in first-order logic:

1. Tony is a tiger.
• $$Tiger(Tony)$$
2. All tigers have stripes.
• $$(\forall x)(Tiger(x) \rightarrow Striped(x))$$
3. Some tigers are white.
• $$(\exists x)(Tiger(x) \wedge White(x))$$
4. If any tiger is white, then we know it lives in the snow (but not necessarily vice versa).
• $$(\forall x)((Tiger(x) \wedge White(x)) \rightarrow LivesInSnow(x))$$
5. A tiger is not both white and orange.
• $$(\forall x)(Tiger(x) \rightarrow (\neg White(x) \vee \neg Orange(x)))$$

Rewrite the following without using $$\forall$$ (hint: use negation): $$(\forall x)(P(x) \leftrightarrow (A(x) \wedge B(x)))$$

$$\neg(\exists x)((P(x) \vee (A(x) \wedge B(x))) \wedge (\neg P(x) \vee \neg(A(x) \wedge B(x))))$$

Rewrite the following without using $$\exists$$: $$(\exists x)(W(x) \vee \neg Q(x))$$

$$\neg (\forall x)(\neg W(x) \wedge Q(x))$$


text:
Does solution of $x=\sum_{n=0}^\infty e^{-A_n/x}$ exist?

Usually when working with indefinite sums, I want to work out the sum or whether it convergence. But now I encountered a problem they other way around and I'm clueless...

Is there even a general solution of $$x=\sum_{n=0}^\infty e^{-A_n/x}$$ for $A_n$, where $x$ is given and real, $A_n >0\space\forall n$ and $\frac{dA_n}{dx}=0\space\forall n$?

Thank you

EDIT:

To make my question clearer for the commenters and others, I'm searching for a systematic sequence $A_n$ which, when entered in the equation above, yields $x$ and this should hold for all (real) $x$.

-
You want to solve for $A_n$? That seems to be one equation in an infinite number of unknowns. To have convergence, you need the $A_n$ to increase fast enough, but once you do, you can bump $A_1$ up and $A_2$ down and maintain the equality. I'm not sure to make of the condition $\frac{dA_n}{dx}=0\space\forall n$ –  Ross Millikan Jul 10 '11 at 1:01
Does not make much sense to me. You could, for example set $a_n = b (n+1)$ –  leonbloy Jul 10 '11 at 1:15
@Ross, the condition $\frac{dA_n}{dx}=0$ only means $A_n$ does not depend on $x$. I state this for completeness, otherwise a trivial solution could be found, e.g $A_n(x)=-x(log(x)+e^{π2/6n^2})$. @leonbloy: That $A_n$ is not a solution of the equation. –  JBSnorro Jul 10 '11 at 2:08
But what does $A_n$ does not depend on $x$ mean? Given a certain solution sequence $A_n$, either it satisfies the equation for a fixed $x$ (and then it is implicitly a function of $x$ : different $x$ give different $A_n$), either it satisfies the equation for all $x$ (I'd like to see that) –  leonbloy Jul 10 '11 at 3:56
Clearly the right-hand side is always positive, so this can only happen for $x>0$. @Ross, you answer shows that this can always be arranged so that the series $\sum_{n=0}^\infty e^{-A_m/x}$ converges to a function $x$ for at least one point; I think the interesting question that's hiding in here is whether it's possible for the series to converge to $x$ for $0<a<x<b$ for some interval $(a,b)$ in the positive reals. –  Vladimir Sotirov Jul 10 '11 at 6:47

It cannot be done. For the proof write $x:={1\over y}$. Then we should have $${1\over y}\ \equiv\ \sum_{n=0}^\infty e^{-A_n y}\qquad(*)\ ,$$ say for all $y\geq1$. In particular $\sum_{n=0}^\infty e^{-A_n}=1$, so necessarily $\lim_{n\to\infty} A_n=\infty$. It follows that $\alpha:=\inf_n A_n>0$ and therefore $$\sum_{n=0}^\infty e^{-A_n y}=\sum_{n=0}^\infty e^{-A_n} \ e^{-A_n(y-1)} \leq e^{-\alpha(y-1)} \qquad (y\geq1)\ .$$ This shows that $(*)$ cannot hold for all $y\geq 1$.

-
Brilliant. This in fact shows that if $\frac 1y=\sum_{n=0}^\infty e^{-A_ny}$ and $\frac1x=\sum_{n=0}^\infty e^{-A_nx}$ with some $y>x$, then $\frac xy\leq e^{-\alpha(y-x)}$, limiting the intervals on which we could have convergence (because when $x/y\to0$ slower than $e^{-\alpha(y-x)}$ as $y\to\infty)$. Some numerical exploration with Mathematica suggests that for $\alpha\geq 1$, $x\geq 1$ the above system has no solution other than $x=y$, which seems to follow by computing derivatives $-x/y^2>-1>-\alpha e^{-\alpha(y-x)}$. –  Vladimir Sotirov Jul 10 '11 at 18:57

Certainly there are many solutions. As $x$ grows, the right side decreases monotonically, so for any series $A_n$ that is convergent there will be an $x$ that solves the equation. So pick any $x$ and series $A_n$ that solve the problem. Given a different $x$, just change your favorite $A_n$(s) to make it work.

For a specific example, take $x=1, A_n=\ln 2^{n+1}$. If you want a solution for $x=2$, just decrease any set of $A_n$'s to add enough to the RHS.

-
Two remarks on your post. First, a necessary condition for the RHS to be finite is that $A_n\to+\infty$. Second, more importantly, the RHS is nondecreasing. –  Did Jul 10 '11 at 3:12
@Didier: I remarked on the need for $A_n$ to increase in my comment on the original question. I was more focused on the fact that there are so many solutions the problem is not too interesting. –  Ross Millikan Jul 10 '11 at 3:17
But, of course, this implies that $A_n$ is implicitly a function of $x$ –  leonbloy Jul 10 '11 at 3:53
@Ross: My remarks addressed your post and your post only, and specifically your As $x$ grows, the right side decreases monotonically and for any series $A_n$ that is convergent. I can understand neither of them. First, as $x$ grows, the RHS increases. Second, for any sequence $(A_n)$ such that $A_n\not\to+\infty$ and any fixed positive $x$, $\mathrm{e}^{-A_n/x}\not\to0$ when $n\to+\infty$, hence the RHS is infinite. –  Did Jul 10 '11 at 9:26
@Ross: Not interested in clarifying the above? –  Did Jul 21 '11 at 21:41

Divide both sides by $x$ to get:
$$1 = \Sigma\frac{e^{-A_n/x}}{x},$$ As $x\to \infty$, the left hand side is 1 while the right hand side goes to zero for each $n$.

This might give a hint on what $A_n$ won't work.

-
It seems that what's going on here is that we're trying to write $1/x$ as a Fourier transform but where the transform is restricted to involve only a countable number of frequencies $A_n$. That's clearly impossible. –  Carl Brannen Jul 10 '11 at 5:42
you're right that what's happening is writing $-1/x$ as a series in $\{e^{mx}\}$ for positive real $m$. I agree that this is impossible when we require the series to converge absolutely, but I'm not sure that we have impossibility under conditional convergence (to a non-holomorphic function). –  Vladimir Sotirov Jul 10 '11 at 6:41


text:
# I forgot a resistor on my PCB prototype - how do I add one in?

I need to add a resistor into a data line on my project. The data line is a trace and carries low speed data. A through-hole resistor will be fine. The pins for both ends of the trace are exposed, but how do I add the resistor and remove the trace?

-
Why did you create a bodge tag? –  Kevin Vermeer Jun 25 '11 at 16:36
@Kevin - I didn't notice that. Maybe rework is more appropriate. That one at least exists already. –  stevenvh Jun 25 '11 at 16:41
@Kevin My error, but I am thinking it would be a bodge. The solution, of course, is to do another design revision. I've already ordered this, but it will be some time as I selected slow turn to lower costs. –  Thomas O Jun 25 '11 at 19:04

If your PCB has a solder mask you can fix this with an SMT resistor even when traces are close together. Cut the trace and scrape the solder mask away on both sides of the cut, over a few mm. Solder an 0603 or 0402 over the cut. If I patch like that I actually make two cuts a mm apart, and remove the trace(*) between both cuts to make sure there's a decent interruption of the trace.
Alternatively you can do the same and solder a PTH resistor, whose wires you first bent so that they lay flat on the trace. Won't probably work as well on narrow traces, like 0.5mm.

This type of knife allows for very accurate work (pingswept mentioned the brand X-Acto):

Success!

(*) davidcary uses a dental pick for this,

but the tip of the X-Acto is fine and sharp enough to do it with that.

-
To add to stevenvh: use an X-Acto knife to cut the trace. –  pingswept Jun 25 '11 at 16:15
@pingswept - Yes, that's the type of knife I'm using. Could be another brand, though. –  stevenvh Jun 25 '11 at 16:17
After cutting a gap in the trace with utility knife or X-Acto knife, I use a "dental pick" to scrape away the green soldermask upstream and downstream of the cut, revealing shiny scratched-up copper. –  davidcary Jun 25 '11 at 22:40
@David - I just use the same knife for that. Changing tools costs money! :-) –  stevenvh Jun 26 '11 at 10:33


text:
# A sum of $6,500 deposited in an account grows to$7,500 in 8 years. Assuming annual compounding, what interest rate is being paid?

$7500 = 6500 \cdot {x}^{8}$
${x}^{8} = \frac{7500}{6500}$
$x = \sqrt[8]{\frac{7500}{6500}}$ = 1.01804854685
ir = $\left(1 - 1.01804854685\right) \cdot 100$ = 1.804854685%


text:
# Magnet dropping through a loop

1. Apr 16, 2010

### semc

A permanent magnet of mass m is located on the vertical axis of a stationary, conducting , non-magnetic ring for a long time. The ring has a radius a, resistance R and inductance L.At time t=0 the magnet is released. Suppose the magnet has dipole moment $$\mu$$z. Write down the equation of motion.

Is the retarding force due magnetic flux generated by the induced current passing through the loop? If that is the case how do I evaluate the equation of motion of the magnet?

2. Apr 16, 2010

### chrisk

Consider Lenz' Law. Energy must be conserved, so the work done to create the current in the loop must equal the thermal energy generated in the loop.

Last edited: Apr 16, 2010


text:
# American Institute of Mathematical Sciences

2017, 24: 28-37. doi: 10.3934/era.2017.24.004

## Rigidity of the ${{L}^{p}}$-norm of the Poisson bracket on surfaces

 1 School of Mathematical Sciences, Faculty of Exact Sciences, Tel Aviv University 2 Department of Mathematics, Faculty of Natural Sciences, University of Haifa

Received  September 29, 2016 Published  May 2017

Fund Project: We wish to thank Lev Buhovsky and Leonid Polterovich for reading a preliminary version of the paper and making useful comments and for their interest. We thank the anonymous referee for reviewing the paper. KS is partially supported by the Israel Science Foundation grant number 178/13, and by the European Research Council Advanced grant number 338809. FZ is partially supported by grant number 1281 from the GIF, the German–Israeli Foundation for Scientific Research and Development, and by grant number 1825/14 from the Israel Science Foundation.

For a symplectic manifold $(M,ω)$, let $\{·,·\}$ be the corresponding Poisson bracket. In this note we prove that the functional $(F,G) \mapsto \|\{F,G\}\|_{L^p(M)}$ is lower-semicontinuous with respect to the $C^0$-norm on $C^∞_c(M)$ when $\dim M = 2$ and $p < ∞$, extending previous rigidity results for $p = ∞$ in arbitrary dimension.

Citation: Karina Samvelyan, Frol Zapolsky. Rigidity of the ${{L}^{p}}$-norm of the Poisson bracket on surfaces. Electronic Research Announcements, 2017, 24: 28-37. doi: 10.3934/era.2017.24.004
##### References:
 [1] L. Buhovsky, The $2/3$ -convergence rate for the Poisson bracket, Geom. Funct. Anal., 19 (2010), 1620-1649.  doi: 10.1007/s00039-010-0045-z.  Google Scholar [2] S. S. Cairns, A simple triangulation method for smooth manifolds, Bull. Amer. Math. Soc., 67 (1961), 389-390.  doi: 10.1090/S0002-9904-1961-10631-9.  Google Scholar [3] F. Cardin and C. Viterbo, Commuting Hamiltonians and Hamilton-Jacobi multi-time equations, Duke Math. J., 144 (2008), 235-284.  doi: 10.1215/00127094-2008-036.  Google Scholar [4] M. Entov and L. Polterovich, $C^0$-rigidity of Poisson brackets, in Symplectic Topology and Measure Preserving Dynamical Systems, Contemp. Math., 512, Amer. Math. Soc., Providence, RI, 2010, 25–32. doi: 10.1090/conm/512/10058.  Google Scholar [5] M. Entov, L. Polterovich and F. Zapolsky, Quasi-morphisms and the Poisson bracket, Pure Appl. Math. Q., 3 (2007), 1037-1055.  doi: 10.4310/PAMQ.2007.v3.n4.a9.  Google Scholar [6] H. Federer, Geometric Measure Theory Die Grundlehren der mathematischen Wissenschaften, Band 153, Springer-Verlag New York Inc., New York, 1969.  Google Scholar [7] K. Samvelyan, Rigidity Versus Flexibility of the Poisson Bracket with Respect to the ${L}_p$ -Norm Master's thesis, Tel Aviv University, 2015. Google Scholar [8] F. Zapolsky, Quasi-states and the Poisson bracket on surfaces, J. Mod. Dyn., 1 (2007), 465-475.  doi: 10.3934/jmd.2007.1.465.  Google Scholar

show all references

##### References:
 [1] L. Buhovsky, The $2/3$ -convergence rate for the Poisson bracket, Geom. Funct. Anal., 19 (2010), 1620-1649.  doi: 10.1007/s00039-010-0045-z.  Google Scholar [2] S. S. Cairns, A simple triangulation method for smooth manifolds, Bull. Amer. Math. Soc., 67 (1961), 389-390.  doi: 10.1090/S0002-9904-1961-10631-9.  Google Scholar [3] F. Cardin and C. Viterbo, Commuting Hamiltonians and Hamilton-Jacobi multi-time equations, Duke Math. J., 144 (2008), 235-284.  doi: 10.1215/00127094-2008-036.  Google Scholar [4] M. Entov and L. Polterovich, $C^0$-rigidity of Poisson brackets, in Symplectic Topology and Measure Preserving Dynamical Systems, Contemp. Math., 512, Amer. Math. Soc., Providence, RI, 2010, 25–32. doi: 10.1090/conm/512/10058.  Google Scholar [5] M. Entov, L. Polterovich and F. Zapolsky, Quasi-morphisms and the Poisson bracket, Pure Appl. Math. Q., 3 (2007), 1037-1055.  doi: 10.4310/PAMQ.2007.v3.n4.a9.  Google Scholar [6] H. Federer, Geometric Measure Theory Die Grundlehren der mathematischen Wissenschaften, Band 153, Springer-Verlag New York Inc., New York, 1969.  Google Scholar [7] K. Samvelyan, Rigidity Versus Flexibility of the Poisson Bracket with Respect to the ${L}_p$ -Norm Master's thesis, Tel Aviv University, 2015. Google Scholar [8] F. Zapolsky, Quasi-states and the Poisson bracket on surfaces, J. Mod. Dyn., 1 (2007), 465-475.  doi: 10.3934/jmd.2007.1.465.  Google Scholar
Illustrating $K_n\subseteq U$ and an element $\Phi (Q') = Q$ in its subdivision
Producing ${\tilde{F}}$ and ${\tilde{G}}$ (the dashed curves)
 [1] Frol Zapolsky. Quasi-states and the Poisson bracket on surfaces. Journal of Modern Dynamics, 2007, 1 (3) : 465-475. doi: 10.3934/jmd.2007.1.465 [2] Sobhan Seyfaddini. Spectral killers and Poisson bracket invariants. Journal of Modern Dynamics, 2015, 9: 51-66. doi: 10.3934/jmd.2015.9.51 [3] Mads R. Bisgaard. Mather theory and symplectic rigidity. Journal of Modern Dynamics, 2019, 15: 165-207. doi: 10.3934/jmd.2019018 [4] Ling Lin, Dong He, Zhiyi Tan. Bounds on delay start LPT algorithm for scheduling on two identical machines in the $l_p$ norm. Journal of Industrial & Management Optimization, 2008, 4 (4) : 817-826. doi: 10.3934/jimo.2008.4.817 [5] Peter Weidemaier. Maximal regularity for parabolic equations with inhomogeneous boundary conditions in Sobolev spaces with mixed $L_p$-norm. Electronic Research Announcements, 2002, 8: 47-51. [6] Guozhen Lu, Yunyan Yang. Sharp constant and extremal function for the improved Moser-Trudinger inequality involving $L^p$ norm in two dimension. Discrete & Continuous Dynamical Systems, 2009, 25 (3) : 963-979. doi: 10.3934/dcds.2009.25.963 [7] Francisco Crespo, Francisco Javier Molero, Sebastián Ferrer. Poisson and integrable systems through the Nambu bracket and its Jacobi multiplier. Journal of Geometric Mechanics, 2016, 8 (2) : 169-178. doi: 10.3934/jgm.2016002 [8] Michael Khanevsky. Hofer's length spectrum of symplectic surfaces. Journal of Modern Dynamics, 2015, 9: 219-235. doi: 10.3934/jmd.2015.9.219 [9] Michael Entov, Leonid Polterovich, Daniel Rosen. Poisson brackets, quasi-states and symplectic integrators. Discrete & Continuous Dynamical Systems, 2010, 28 (4) : 1455-1468. doi: 10.3934/dcds.2010.28.1455 [10] Fernando Alcalde Cuesta, Françoise Dal'Bo, Matilde Martínez, Alberto Verjovsky. Minimality of the horocycle flow on laminations by hyperbolic surfaces with non-trivial topology. Discrete & Continuous Dynamical Systems, 2016, 36 (9) : 4619-4635. doi: 10.3934/dcds.2016001 [11] Donglei Du, Tianping Shuai. Errata to:''Optimal preemptive online scheduling to minimize $l_{p}$ norm on two processors''[Journal of Industrial and Management Optimization, 1(3) (2005), 345-351.]. Journal of Industrial & Management Optimization, 2008, 4 (2) : 339-341. doi: 10.3934/jimo.2008.4.339 [12] Braxton Osting, Jérôme Darbon, Stanley Osher. Statistical ranking using the $l^{1}$-norm on graphs. Inverse Problems & Imaging, 2013, 7 (3) : 907-926. doi: 10.3934/ipi.2013.7.907 [13] Yeping Li, Jie Liao. Stability and $L^{p}$ convergence rates of planar diffusion waves for three-dimensional bipolar Euler-Poisson systems. Communications on Pure & Applied Analysis, 2019, 18 (3) : 1281-1302. doi: 10.3934/cpaa.2019062 [14] Daniel N. Dore, Andrew D. Hanlon. Area preserving maps on $\boldsymbol{S^2}$: A lower bound on the $\boldsymbol{C^0}$-norm using symplectic spectral invariants. Electronic Research Announcements, 2013, 20: 97-102. doi: 10.3934/era.2013.20.97 [15] Fernando Alcalde Cuesta, Françoise Dal'Bo, Matilde Martínez, Alberto Verjovsky. Corrigendum to "Minimality of the horocycle flow on laminations by hyperbolic surfaces with non-trivial topology". Discrete & Continuous Dynamical Systems, 2017, 37 (8) : 4585-4586. doi: 10.3934/dcds.2017196 [16] Zhenqi Jenny Wang. New cases of differentiable rigidity for partially hyperbolic actions: Symplectic groups and resonance directions. Journal of Modern Dynamics, 2010, 4 (4) : 585-608. doi: 10.3934/jmd.2010.4.585 [17] Duo Wang, Zheng-Fen Jin, Youlin Shang. A penalty decomposition method for nuclear norm minimization with l1 norm fidelity term. Evolution Equations & Control Theory, 2019, 8 (4) : 695-708. doi: 10.3934/eect.2019034 [18] Pia Heins, Michael Moeller, Martin Burger. Locally sparse reconstruction using the $l^{1,\infty}$-norm. Inverse Problems & Imaging, 2015, 9 (4) : 1093-1137. doi: 10.3934/ipi.2015.9.1093 [19] P. R. Zingano. Asymptotic behavior of the $L^1$ norm of solutions to nonlinear parabolic equations. Communications on Pure & Applied Analysis, 2004, 3 (1) : 151-159. doi: 10.3934/cpaa.2004.3.151 [20] Ahmad Mousavi, Zheming Gao, Lanshan Han, Alvin Lim. Quadratic surface support vector machine with L1 norm regularization. Journal of Industrial & Management Optimization, 2021  doi: 10.3934/jimo.2021046

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text:
1
AIEEE 2010
+4
-1
Statement-1: The point $$A(3, 1, 6)$$ is the mirror image of the point $$B(1, 3, 4)$$ in the plane $$x-y+z=5.$$

Statement-2: The plane $$x-y+z=5$$ bisects the line segment joining $$A(3, 1, 6)$$ and $$B(1, 3, 4).$$
A
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
B
Statement - 1 is true, Statement - 2 is false.
C
Statement - 1 is false , Statement - 2 is true.
D
Statement - 1 is true, Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
2
AIEEE 2010
+4
-1
A line $$AB$$ in three-dimensional space makes angles $${45^ \circ }$$ and $${120^ \circ }$$ with the positive $$x$$-axis and the positive $$y$$-axis respectively. If $$AB$$ makes an acute angle $$\theta$$ with the positive $$z$$-axis, then $$\theta$$ equals
A
$${45^ \circ }$$
B
$${60^ \circ }$$
C
$${75^ \circ }$$
D
$${30^ \circ }$$
3
AIEEE 2009
+4
-1
Let the line $$\,\,\,\,\,$$ $${{x - 2} \over 3} = {{y - 1} \over { - 5}} = {{z + 2} \over 2}$$ lie in the plane $$\,\,\,\,\,$$ $$x + 3y - \alpha z + \beta = 0.$$ Then $$\left( {\alpha ,\beta } \right)$$ equals
A
$$(-6,7)$$
B
$$(5,-15)$$
C
$$(-5,5)$$
D
$$(6, -17)$$
4
AIEEE 2008
+4
-1
The line passing through the points $$(5,1,a)$$ and $$(3, b, 1)$$ crosses the $$yz$$-plane at the point $$\left( {0,{{17} \over 2}, - {{ - 13} \over 2}} \right)$$ . Then
A
$$a=2,$$ $$b=8$$
B
$$a=4,$$ $$b=6$$
C
$$a=6,$$ $$b=4$$
D
$$a=8,$$ $$b=2$$
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text:
# First Measurement of the Z -> mu(+) mu(-) Angular Coefficients in the Forward Region of pp Collisions at root s=13 TeV

Aaij, R, Abdelmotteleb, ASW, Beteta, C Abellan, Abudinen, F, Ackernley, T, Adeva, B, Adinolfi, M, Afsharnia, H, Agapopoulou, C, Aidala, CA
et al (show 981 more authors) (2022) First Measurement of the Z -> mu(+) mu(-) Angular Coefficients in the Forward Region of pp Collisions at root s=13 TeV. PHYSICAL REVIEW LETTERS, 129 (9). 091801-.

Access the full-text of this item by clicking on the Open Access link.

## Abstract

The first study of the angular distribution of μ^{+}μ^{-} pairs produced in the forward rapidity region via the Drell-Yan reaction pp→γ^{*}/Z+X→ℓ^{+}ℓ^{-}+X is presented, using data collected with the LHCb detector at a center-of-mass energy of 13 TeV, corresponding to an integrated luminosity of 5.1  fb^{-1}. The coefficients of the five leading terms in the angular distribution are determined as a function of the dimuon transverse momentum and rapidity. The results are compared to various theoretical predictions of the Z-boson production mechanism and can also be used to probe transverse-momentum-dependent parton distributions within the proton.

Item Type: Article LHCb Collaboration Symplectic Admin 10 Oct 2022 09:57 21 Jan 2023 06:08 10.1103/PhysRevLett.129.091801 https://arxiv.org/pdf/2203.01602v2.pdf https://livrepository.liverpool.ac.uk/id/eprint/3165102


text:
## An observation about how the OGG Opus CODEC may do Stereo.

One of the subjects which I’ve written about before, is the fact that the developers of the original OGG Vorbis form of music compression, have more recently developed the OGG Opus CODEC, which is partially the CELT CODEC. And, in studying the manpage on how to use the ‘opusenc’ command (under Linux), I ran across the following detail:

--no-phase-inv
Disable use of phase inversion for intensity stereo. This trades
some stereo quality for a higher quality mono  downmix,  and  is
useful when encoding stereo audio that is likely to be downmixed
to mono after decoding.

What does this mean? Let me explain.

I should first preface that with an admission, of the fact that an idea which was true for the original version of the Modified Discrete Cosine Transform, as introduced by MP3 compression and then reused frequently by other CODECs, may not always be the case. That idea was that, when defining monaural sound, each frequency coefficient needed to be signed. Because CELT uses a form of the Type 4 Discrete Cosine Transform which is only partially lapped, it may be that all the coefficients are assumed to be positive.

This will work as long as there is no destructive interference between the same coefficient, in the overlapping region, from one frame to the next, in spite of the half-sample shift of each frequency-value. Also, a hypotenuse function should be avoided, as that would present itself as distortion. One explicit way to achieve this could be, to rotate the reference-waves (n)·90° + 45° for coefficient (n):

Where ‘FN‘ refers to the current Frame-Number.

In general, modern compressed schemes will subdivide the audible spectrum into sub-bands, which in the case of CELT are referred to as its Critical Bands. And for each frame, the way stereo is encoded for each critical band, switches back and forth between X/Y intensity stereo, and Mid/Side stereo, which also just referred to as M/S stereo. What will happen with M/S stereo is, that the (L-R) channel has its own spectral shape, independent of the (L+R) channel’s, while with X/Y stereo, there is only one spectral pattern, which is reproduced by a linear factor, as both the (L+R) component, and the (L-R) component.

Even if the (L+R) is only being recorded as having positive DCT coefficients, with M/S stereo, the need persists for the (L-R) channel to be signed. Yet, even if M/S stereo is not taking place, implying that X/Y stereo is taking place, what can happen is that:

|L-R| > (L+R)

This would cause phase-inversion to take place between the two channels, (L) and (R). Apparently, a setting will prevent this from happening.

Further, because CELT has as its main feature, that it first states the amplitude of the critical band, and then a Code-Word which identifies the actual non-zero coefficients, which may only number 4, the setting may also affect critical bands for which M/S stereo is being used during any one frame. I’m not really sure if it does. But if it does, it will also make sure that the amplitude of the (L+R) critical band exceeds or equals that of the (L-R) critical band.

The way in which the CODEC decides, whether to encode the critical band using X/Y or M/S, for any one frame, is to detect the extent to which the non-zero coefficients coincide. If the majority of them do, encoding automatically switches to X/Y… Having said that, my own ideas on stereo perception are such that, if none of the coefficients coincide, it should not make any difference whether the specific coefficients belonging to the (L-R) channel are positive or negative. And finally, a feature which CELT could have enabled constantly, is to compute whether the (L-R) critical band correlates positively or negatively with the (L+R), independently of what the two amplitudes are. And this last observation suggests that even when encoding in M/S mode, the individual coefficients may not be signed.

(Update 10/03/2019, 9h30 … )

## Different types of music, for testing Audio Codecs – Or Maybe Not.

One of my recent activities has been, to start with Audio CDs from the 1980s and 1990s, the encoding of which was limited only by the 44.1kHz sample rate, and the bit-depth, as well as by whatever type of Sinc Filter was once used to master them, but not limited by any sort of lossy compression, and to “rip” those, but into different types of lossy compression, in order to evaluate that. The two types of compression I recently played with were ‘AAC’ (plain), and ‘OGG Opus’, at 128kbps both times.

One of the apparent facts which I learned was that Phil Collins music is not the best type to try this with. The reason is the fact that much of his music was recorded using electronic instruments of his era, the main function of which was, to emulate standard acoustical instruments, but in a way that was ‘acoustically pure’. The fact that Phil Collins started his career as a drummer, did not prevent him from releasing later, solo albums.

If somebody is listening to an entire string section of an orchestra, or to a brass section, then one factor which contributes to the nature of the sound is, that every Violin is minutely off-pitch, as would be every French Horn. But what that also means is that the sound which results, is “Thick”. Its spectral energy is spread in subtle ways, whereas if somebody mixes two sine-waves that have exactly the same frequency, then he obtains another sine-wave. If one mixes 10 sine-waves that have exactly the same frequency, then he still obtains one sine-wave.

Having sound to start from which is ‘Thick’, is a good basis to test Codecs. Phil Collins does not provide that. Therefore, if the acoustical nature of the recording is boring, I have no way to know, whether it’s the Codec that failed to bring out greater depth, or whether that was the fault of Phil Collins.

(Update 8/03/2019, 12h55 : )

Since the last time I edited this posting, I learned, that Debian / Stretch, the Linux version I have installed on the computer I name ‘Phosphene’, only ships with ‘libopus v1.2~alpha2-1′ from the package repositories. Apparently, when using this version, the best one can hope for is an equivalent, to 128kbps MP3 -quality. This was the true reason, for which I was obtaining inferior results, along with the fact that I had given the command to encode my Opus Files using a version of ‘ffmpeg’, that just happened to include marginal support for ‘Opus’, instead of using the actual ‘opus-toolkit’ package.

What I have now done was, to download and custom-compile ‘libopus v1.3.1′, as well as its associated toolkit, and just to make sure that the programs work. Rumours have it, that when this version is used at a bit-rate of 96kbps, virtual transparency will result.

And I’ve written quite a long synopsis, as to why this might be so.

(Update 8/03/2019, 15h50 : )

I have now run an altered experiment, by encoding my Opus Files at 96kbps, and discovered to my amazement, that the sound I obtained seemed better, than what I had already obtained above, using 128kbps AAC-encoded Files.

(Update 10/02/2019, 12h25 : )

When I use the command ‘opusenc’, which I’ve custom-compiled as written above, it defaults to a frame-size of 20 milliseconds. Given a sampling rate of 48kHz, this amounts to a frame-size – or granule – of 960 samples. This is very different from what the developers were suggesting in their article in 2010 (see posting linked to above). With that sampling interval, the tonal accuracy will be approximately twice as good as it was with MP3 encoding, or with AAC encoding, without requiring that any “Hadamard Transforms” be used. With the default setting, Opus will generally be able to distinguish frequencies that are 25Hz apart, while it was in the nature of MP3 only to be able to distinguish between frequencies that are 40Hz apart, except for the fact that the lowest distinguishable frequency was near 20Hz.

If Hadamard Transforms are in fact used, then it now strikes me as more likely that they will be used to increase temporal resolution at the expense of spectral resolution, not the reverse. And the reason I think this is the following paragraph in the manpage, for the ‘opusenc’ command-line:

--framesize N
Set maximum frame size in milliseconds (2.5, 5, 10, 20, 40,  60,
default: 20)
Smaller  framesizes  achieve lower latency but less quality at a
given bitrate.
Sizes greater than 20ms  are  only  interesting  at  fairly  low
bitrates.

(Update 8/12/2019, 6h00 : )

I have now listened very carefully to the Phil Collins Music encoded with AAC at 128kbps, and the exact same songs, encoded with Opus at 96kbps. And what I’ve come to find, is that Opus seems to preserve spectral complexity better than AAC. However, the AAC encoded versions of the same music, seem to provide slightly better perception of positioning all around the listener, of instruments and voices in the lower-mid-range frequencies, as a result of Stereo. And this would be, when the listener is using a good set of headphones.

Dirk

## A Gap in My Understanding of Surround-Sound Filled: Separate Surround Channel when Compressed

In This earlier posting of mine, I had written about certain concepts in surround-sound, which were based on Pro Logic and the analog days. But I had gone on to write, that in the case of the AC3 or the AAC audio CODEC, the actual surround channel could be encoded separately, from the stereo. The purpose in doing so would have been, that if decoded on the appropriate hardware, the surround channel could be sent directly to the rear speakers – thus giving 6-channel output.

While writing what I just linked to above, I had not yet realized, that either channel of the compressed stream, could contain phase information conserved. This had caused me some confusion. Now that I realize, that the phase information could be correct, and not based on the sampling windows themselves, a conclusion comes to mind:

Such a separate, compressed surround-channel, would already be 90⁰ phase-shifted with respect to the panned stereo. And what this means could be, that if the software recognizes that only 2 output channels are to be decoded, the CODEC might just mix the surround channel directly into the stereo. The resulting stereo would then also be prepped, for Pro Logic decoding.

Dirk


text:
## Stream: new members

### Topic: TC resolution

#### Andre Knispel (Aug 12 2020 at 07:50):

I'm trying to define and experiment with graded rings and I have a few issues with type classes. Here's my code:

import data.equiv.algebra
import data.polynomial
import algebra.module
import algebra.direct_sum

universes u v
variables (α : Type u) (M : Type v)
variables [add_monoid M] [decidable_eq M]

class graded_ring extends ring α :=
(pieces : M → Type u) (groups : ∀ m : M, add_comm_group (pieces m))
(iso : α ≃+ direct_sum M pieces)
(of : Π m : M, pieces m → α := λ (m : M) (r : pieces m), iso.inv_fun (direct_sum.of pieces m r))
(mul_respects_grading : ∀ (m m' : M) (r : pieces m) (r' : pieces m'),
∃ s : pieces (m + m'), of m r * of m' r' = of (m + m') s)

variables [comm_ring α] [r : ring α]

noncomputable instance : graded_ring (polynomial α) ℕ :=
@graded_ring.mk (polynomial α) ℕ _ _ (comm_ring.to_ring (polynomial α))
(λ _ : ℕ, α) (λ _ : ℕ, @ring.to_add_comm_group α r) _ _ _

First, I don't understand why the ring instance of polynomial isn't found. If I make it implicit, the search tries increasingly complicated versions of dfinsupp.add_monoid before giving up. Is there a nicer way to do this, ideally so that I can use the record syntax? Second, is there a nicer way to write the commutative group instance of α?

#### Scott Morrison (Aug 12 2020 at 08:38):

There's a rule that you shouldn't use extend when adding more data to a class, if you need to have more parameters than the thing you're extending.

#### Scott Morrison (Aug 12 2020 at 08:38):

So I'd suggest you want to substantially reformulate this.

#### Kevin Buzzard (Aug 12 2020 at 08:42):

These rules are extremely hard to find unless you're a regular here.

#### Kevin Buzzard (Aug 12 2020 at 08:42):

You just get people like you, Reid or Mario saying "oh this is an antipattern" occasionally

#### Kevin Buzzard (Aug 12 2020 at 08:44):

What would be really nice would be a resource with some simple bad classes and perhaps not an explanation of why they're bad but at least a description of the rules of thumb to follow. I don't know these rules at all. I am still really bad at making definitions in Lean which is why I'm making so much fuss on another thread about exactly what the definition of yet another concept of finiteness / cardinality should be.

#### Scott Morrison (Aug 12 2020 at 08:45):

I think the problem is just that it introduces a "bad instance", because now whenever Lean wants to check if something is a ring, is has to check if it is a graded ring, and suddenly there's a metavariable for the type M introduced into the typeclass search problem.

#### Kevin Buzzard (Aug 12 2020 at 08:46):

Yeah I'm sure that if you actually understand the typeclass system (which I certainly don't) then it's all obvious. The problem is that if a mathematician beginner comes along they can write something which looks completely mathematically reasonable but is somehow impossible to use because of some issue regarding [foo X] v extends foo X

#### Kevin Buzzard (Aug 12 2020 at 08:46):

or some issue regarding universe parameters in the definition

#### Kevin Buzzard (Aug 12 2020 at 08:47):

or some issue regarding something having a different number of variables to something else

#### Kevin Buzzard (Aug 12 2020 at 08:48):

or some other issue -- these things are too numerous to mention. Classes are extremely well understood now, and the "gotchas" are all known to the experts, but whenever I want to make a new definition in Lean I run it past the Zulip chat before I do anything with it at all because I don't know all the pitfalls

#### Andre Knispel (Aug 12 2020 at 11:48):

Ah, now that I know what I'm looking for, I found the loop ring -> graded_ring -> monoid (for M) -> ... -> ring. I've looked at modules, because they should have the same issue, but the only thing I see is set_option default_priority 100. Does this make the priority for the classes there very low?

#### Scott Morrison (Aug 12 2020 at 11:51):

I don't think module is comparable, because it is not using extends like you are.

#### Scott Morrison (Aug 12 2020 at 11:51):

I would guess that you just want an additional typeclass, so you write [ring R] [graded_ring R] to summon one.

#### Scott Morrison (Aug 12 2020 at 11:52):

You might also look at src#category_theory.graded_object for the bundled version.

#### Andre Knispel (Aug 12 2020 at 12:04):

Thanks, that helped!

#### Jalex Stark (Aug 12 2020 at 22:36):

Kevin Buzzard said:

What would be really nice would be a resource with some simple bad classes and perhaps not an explanation of why they're bad but at least a description of the rules of thumb to follow. I don't know these rules at all. I am still really bad at making definitions in Lean which is why I'm making so much fuss on another thread about exactly what the definition of yet another concept of finiteness / cardinality should be.

I guess some of these rules have been implemented in code. Maybe the document you want starts out as docs for the linter?

Last updated: May 08 2021 at 19:11 UTC


text:
# Lyapunov exponent

In mathematics the Lyapunov exponent or Lyapunov characteristic exponent of a dynamical system is a quantity that characterizes the rate of separation of infinitesimally close trajectories. Quantitatively, two trajectories in phase space with initial separation ${\displaystyle \delta \mathbf {Z} _{0}}$ diverge (provided that the divergence can be treated within the linearized approximation) at a rate given by

${\displaystyle |\delta \mathbf {Z} (t)|\approx e^{\lambda t}|\delta \mathbf {Z} _{0}|\,}$

where ${\displaystyle \lambda }$ is the Lyapunov exponent.

The rate of separation can be different for different orientations of initial separation vector. Thus, there is a spectrum of Lyapunov exponents— equal in number to the dimensionality of the phase space. It is common to refer to the largest one as the Maximal Lyapunov exponent (MLE), because it determines a notion of predictability for a dynamical system. A positive MLE is usually taken as an indication that the system is chaotic (provided some other conditions are met, e.g., phase space compactness). Note that an arbitrary initial separation vector will typically contain some component in the direction associated with the MLE, and because of the exponential growth rate, the effect of the other exponents will be obliterated over time.

The exponent is named after Aleksandr Lyapunov.

## Definition of the maximal Lyapunov exponent

The maximal Lyapunov exponent can be defined as follows:

${\displaystyle \lambda =\lim _{t\to \infty }\lim _{\delta \mathbf {Z} _{0}\to 0}{\frac {1}{t}}\ln {\frac {|\delta \mathbf {Z} (t)|}{|\delta \mathbf {Z} _{0}|}}.}$

The limit ${\displaystyle \delta \mathbf {Z} _{0}\to 0}$ ensures the validity of the linear approximation at any time.[1]

For discrete time system (maps or fixed point iterations) ${\displaystyle x_{n+1}=f(x_{n})}$ , for an orbit starting with ${\displaystyle x_{0}}$ this translates into:

${\displaystyle \lambda (x_{0})=\lim _{n\to \infty }{\frac {1}{n}}\sum _{i=0}^{n-1}\ln |f'(x_{i})|}$

## Definition of the Lyapunov spectrum

For a dynamical system with evolution equation ${\displaystyle f^{t}}$ in an n–dimensional phase space, the spectrum of Lyapunov exponents

${\displaystyle \{\lambda _{1},\lambda _{2},\ldots ,\lambda _{n}\}\,,}$

in general, depends on the starting point ${\displaystyle x_{0}}$. (However, we will usually be interested in the attractor (or attractors) of a dynamical system, and there will normally be one set of exponents associated with each attractor. The choice of starting point may determine which attractor the system ends up on, if there is more than one. Note: Hamiltonian systems do not have attractors, so this particular discussion does not apply to them.) The Lyapunov exponents describe the behavior of vectors in the tangent space of the phase space and are defined from the Jacobian matrix

${\displaystyle J^{t}(x_{0})=\left.{\frac {df^{t}(x)}{dx}}\right|_{x_{0}}}$

The ${\displaystyle J^{t}}$ matrix describes how a small change at the point ${\displaystyle x_{0}}$ propagates to the final point ${\displaystyle f^{t}(x_{0})}$. The limit

${\displaystyle L(x_{0})=\lim _{t\rightarrow \infty }(J^{t}\cdot \mathrm {Transpose} (J^{t}))^{1/2t}}$

defines a matrix ${\displaystyle L(x_{0})}$ (the conditions for the existence of the limit are given by the Oseledec theorem). If ${\displaystyle \Lambda _{i}(x_{0})}$ are the eigenvalues of ${\displaystyle L(x_{0})}$, then the Lyapunov exponents ${\displaystyle \lambda _{i}}$ are defined by

${\displaystyle \lambda _{i}(x_{0})=\log \Lambda _{i}(x_{0})}$

The set of Lyapunov exponents will be the same for almost all starting points of an ergodic component of the dynamical system.

## Lyapunov exponent for time-varying linearization

To introduce Lyapunov exponent let us consider a fundamental matrix ${\displaystyle X(t)}$ (e.g., for linearization along stationary solution ${\displaystyle x_{0}}$ in continuous system the fundamental matrix is ${\displaystyle \exp \left(\left.{\frac {df^{t}(x)}{dx}}\right|_{x_{0}}t\right)}$), consisting of the linear-independent solutions of the first approximation system. The singular values ${\displaystyle \{\alpha _{j}{\big (}X(t){\big )}\}_{1}^{n}}$ of the matrix ${\displaystyle X(t)}$ are the square roots of the eigenvalues of the matrix ${\displaystyle X(t)^{*}X(t)}$. The largest Lyapunov exponent ${\displaystyle \lambda _{max}}$ is as follows [2]

${\displaystyle \lambda _{max}=\max \limits _{j}\limsup _{t\rightarrow \infty }{\frac {1}{t}}\ln \alpha _{j}{\big (}X(t){\big )}.}$

A.M. Lyapunov proved that if the system of the first approximation is regular (e.g., all systems with constant and periodic coefficients are regular) and its largest Lyapunov exponent is negative, then the solution of the original system is asymptotically Lyapunov stable. Later, it was stated by O. Perron that the requirement of regularity of the first approximation is substantial.

In 1930 O. Perron constructed an example of the second-order system, the first approximation of which has negative Lyapunov exponents along a zero solution of the original system but, at the same time, this zero solution of original nonlinear system is Lyapunov unstable. Furthermore, in a certain neighborhood of this zero solution almost all solutions of original system have positive Lyapunov exponents. Also it is possible to construct reverse example when first approximation has positive Lyapunov exponents along a zero solution of the original system but, at the same time, this zero solution of original nonlinear system is Lyapunov stable.[3][4] The effect of sign inversion of Lyapunov exponents of solutions of the original system and the system of first approximation with the same initial data was subsequently called the Perron effect.[3][4]

Perron's counterexample shows that negative largest Lyapunov exponent does not, in general, indicate stability, and that positive largest Lyapunov exponent does not, in general, indicate chaos.

Therefore, time-varying linearization requires additional justification.[4]

## Basic properties

If the system is conservative (i.e. there is no dissipation), a volume element of the phase space will stay the same along a trajectory. Thus the sum of all Lyapunov exponents must be zero. If the system is dissipative, the sum of Lyapunov exponents is negative.

If the system is a flow and the trajectory does not converge to a single point, one exponent is always zero—the Lyapunov exponent corresponding to the eigenvalue of ${\displaystyle L}$ with an eigenvector in the direction of the flow.

## Significance of the Lyapunov spectrum

The Lyapunov spectrum can be used to give an estimate of the rate of entropy production and of the fractal dimension of the considered dynamical system. In particular from the knowledge of the Lyapunov spectrum it is possible to obtain the so-called Kaplan–Yorke dimension ${\displaystyle D_{KY}}$, which is defined as follows:

${\displaystyle D_{KY}=k+\sum _{i=1}^{k}{\frac {\lambda _{i}}{|\lambda _{k+1}|}},}$

where ${\displaystyle k}$ is the maximum integer such that the sum of the ${\displaystyle k}$ largest exponents is still non-negative. ${\displaystyle D_{KY}}$ represents an upper bound for the information dimension of the system.[5] Moreover, the sum of all the positive Lyapunov exponents gives an estimate of the Kolmogorov–Sinai entropy accordingly to Pesin's theorem.[6]

The multiplicative inverse of the largest Lyapunov exponent is sometimes referred in literature as Lyapunov time, and defines the characteristic e-folding time. For chaotic orbits, the Lyapunov time will be finite, whereas for regular orbits it will be infinite.

## Numerical calculation

Generally the calculation of Lyapunov exponents, as defined above, cannot be carried out analytically, and in most cases one must resort to numerical techniques. An early example, which also constituted the first demonstration of the exponential divergence of chaotic trajectories, was carried out by R. H. Miller in 1964.[7] Currently, the most commonly used numerical procedure estimates the ${\displaystyle L}$ matrix based on averaging several finite time approximations of the limit defining ${\displaystyle L}$.

One of the most used and effective numerical techniques to calculate the Lyapunov spectrum for a smooth dynamical system relies on periodic Gram–Schmidt orthonormalization of the Lyapunov vectors to avoid a misalignment of all the vectors along the direction of maximal expansion.[8][9][10][11]

For the calculation of Lyapunov exponents from limited experimental data, various methods have been proposed. However, there are many difficulties with applying these methods and such problems should be approached with care. The main difficulty is that the data does not fully explore the phase space, rather it is confined to the attractor which has very limited (if any) extension along certain directions. These thinner or more singular directions within the data set are the ones associated with the more negative exponents. The use of nonlinear mappings to model the evolution of small displacements from the attractor has been shown to dramatically improve the ability to recover the Lyapunov spectrum,[12][13] provided the data has a very low level of noise. The singular nature of the data and its connection to the more negative exponents has also been explored.[14]

## Local Lyapunov exponent

Whereas the (global) Lyapunov exponent gives a measure for the total predictability of a system, it is sometimes interesting to estimate the local predictability around a point x0 in phase space. This may be done through the eigenvalues of the Jacobian matrix J 0(x0). These eigenvalues are also called local Lyapunov exponents.[15] (A word of caution: unlike the global exponents, these local exponents are not invariant under a nonlinear change of coordinates.)

## Conditional Lyapunov exponent

This term is normally used in regards to the synchronization of chaos, in which there are two systems that are coupled, usually in a unidirectional manner so that there is a drive (or master) system and a response (or slave) system. The conditional exponents are those of the response system with the drive system treated as simply the source of a (chaotic) drive signal. Synchronization occurs when all of the conditional exponents are negative.[16]

## References

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