text:
7:00 PM
I wanna show that the characteristic map from interior of n ball is homeomorphisim to its image , can I give Xn the subspace topology to show the characteristic map is an open map onto its image ?

I saw an interesting article that day about how all languages call tea in essentially one of two ways: something that sounds like tea, or something that sounds like cha, and both ways essentially come from Chinese @LeakyNun, because teh is tea in Hokkien and cha is tea in Mandarin.

interesting!

Of course, we are just saying that the action of $\Bbb Z/3$ on $\Bbb R^3$ given by permuting the coordinates splits as the sum of a 2-dimensional irreducible representation (the subset of sum-0) and the trivial representation (the possible sums)

It's a quadratic equaton right? So one can complete the square and use the 2a+b+c=0 thing to work out where the roots will lie with the help of the discriminant?

@JasperLoy I see

7:00 PM
@TedShifrin As a HW problem for someone reviewing a lot of different math subjects, I find this problem artificial but not terrible. Being able to plot stuff in Mathematica helps

interessant!

As a test question, though, it's stooopid

@JasperLoy and of course, the sound in "tea" before the Great Vowel Shift would have been exactly like Hokkien

@Secret could, yeah

@MikeMiller heh...

7:02 PM
but the IVT approach is simpler

@Semiclassical Is there any other approach too?

@LeakyNun i got confused for a half-second about why an abelian group had a 2d irrep
but only a half-second

@Secret My favourite text editor for TeX is TeXworks. =)

@Abcd i dunno, probably

Torskun diert ba man comm goa

7:03 PM
@MikeMiller because R isn't algebraically closed :P

we found a method that worked and went with it

@JasperLoy Processing 3 is not a text editor, it is a programing console. But sure, I use texwork and texlive for my LaTex report typing

@LeakyNun whence half-second :)

@JasperLoy @LeakyNun The one from Chinese that really surprises me is "Ketchup" dictionary.com/browse/ketchup?s=t

that's Cantonese, my native tongue :P

7:04 PM
now let's make sure abcd writes the fact down on his homework that Z/3 has a 2d real irrep
truly crucial

@JasperLoy I tend to use TeXworks offline as well

@TedShifrin Apparently my TAs had a bit of trouble solving one of the exercises I had posed. It was showing that finite abelian groups satisfy converse Lagrange

Que whal sur jo?

@Alessandro: I told you so!

7:05 PM
but that's mostly because the other tex editor I downloaded ends up crashing pretty much immediately

@LeakyNun Are cantonese speakers really aware "ketchup" originates from a chinese language?

sure

Oh, that's some sort of induction problem, @Tobias. I've done it before.

@rschwieb If you are interested in etymologies, etymonline.com is a great free resource.

highly skeptical

7:05 PM
@TedShifrin Yeah, exactly.

@JasperLoy Yeah, I've seen it berore

I have solved the catalan conjecture

Really? Well, good for you.

Thanks

Wiktionary is actually very well-developed

7:06 PM
@JasperLoy @rschwieb vocaroo.com/i/s10V8449ltuA

I had even given a hint to use induction and the correspondence of subgroups with the quotient

welcome to the chat preda mihăilescu

@JasperLoy I always find it funny when there are familiar words whose etymology isn't known
like strawberry

I'm a bit surprised your TAs didn't get it, @Tobias — not surprised the students didn't.

@TobiasKildetoft oh, we aren't allowed to use classification?

7:07 PM
@MikeMiller Americans would like merriam-webster.com while British would like en.oxforddictionaries.com.

@LeakyNun No, that would make it trivial

no one really knows where the word strawberry comes from

well what if I prove it?

@TedShifrin Yeah. Possibly they just hadn't thought that much about it, since the exercises are usually much easier, so this harder one caught them off guard

7:08 PM
wasn't it just invented?
a berry from the straw fieldsd
that is red
strawberry.

ok I am off, this is not going to finish in 2 hours and it is already 5:07

secret
sleep is important

However, I find that the dictionary with the best definitions is the American Heritage Dictionary of the English Language, at ahdictionary.com.

@LeakyNun In this case, the students have also not seen group actions yet

7:10 PM
what is the etymology of conglomerate

Well at least I got 70% of the introduction done, so that will make me less likely to procrastinate on my thesis

Another big surprise is the surname Yuengling which looks and sounds chinese but is a weird anglicization

what is the origin of species

@Ultradark It is a book by Darwin.

7:12 PM
can we stop feeding the troll?
3

why does everyone put periods at the end of the sentence.

@LeakyNun I have a question about a proof.

@Ultradark Did you go by another username in the past? I don't remember you.

@JasperLoy he is geocalc33

I am the ultra darkest wizzi

7:13 PM
@LeakyNun Oh, I don't know that one either.

wizzi = wizard

wow what a dark wizard

I went by Geocalc before
but now I am not just dark. I am ultra dark
I wanted to turn a new leaf

I like dark chocolate.
And I want to turn into a new leaf.

I like it too. What's percent of darkness do you prefer your chocolate to be In terms of the amount of cocoa?

7:16 PM
@LeakyNun I haven't started yet, but it's a question about starting. It's a set theory proof: Prove that, for a function f:X→Y, we always have A⊂B ⇒f (A)⊂f(B). I presented it to a teacher today and he said it may be too abstract, to begin with.

I like 60%

I don't know. I am not an expert.

@CaptainAmerica16 high school math teacher is the worst

mercio

@LeakyNun I'm familiar with set theory and figured it was worth a shot, although it is a bit a more abstract than I'm used to. What do you think? Does it seem out of range for a high school student? It's a logical statement at first glance.

7:17 PM
@CaptainAmerica16 Why did you present it to your teacher?

I have been thinking about the problem you gave to me

@CaptainAmerica16 just ignore your teacher. this problem is fine.

@JasperLoy I don't really have any guidance with my math studies, so I go to him sometimes with questions.
@LeakyNun Ok, I'm going to start on it today.

@CaptainAmerica16 I see. This is a very simple question, can't be too abstract.

7:19 PM
@JasperLoy for a high school student studying "conventional" maths it might be too abstract
I think the teacher just doesn't know how to do it

@mercio I have been thinking about the problem of Navier Stokes. It seems like a very difficult problem. Do you have any suggestions. Where would you begin?

idk I'm not a specialist about ODEs

@LeakyNun Yes, but a high school teacher should not tell a student trying to write this proof that it is too abstract, if that is the situation here.

and if it were so easy maybe someone would have claimed the million dollar prize by now

4 mins ago, by Leaky Nun
@CaptainAmerica16 high school math teacher is the worst

7:22 PM
I'm looking at weak vector fields
that seems to be the best way forward

@Ultradark I suggest you begin by writing to Navier and Stokes.
2

okay
i believe they are not alive anymore
but I definitely would if they were still around

Seems I've stumbled upon some nerds
2
:0

@Daminark this is probably the most me_irl thing i've ever seen in chat

you are correct

7:25 PM
3

Gold

Diamond

that's one interpretation

I dunno if that was the point, more asking whether Borel and Langlands knew rep theory

I was just diverging

7:28 PM
Though I usually think of "Borel" as analysis
Like in Borel sets
Probably not the same guy, right? I feel the Borel of Borel sets was a bit too old to be in the presence of Langlands

@Daminark Emile is measure theory, Armand is representation theory
4
"He used to answer the question of whether he was related to Émile Borel alternately by saying he was a nephew, and no relation."

Heh, interesting

Emil and Michael Artin are father and son.

I am used to saying "Borel equivariant homology" on a daily basis so I am used to the idea of him as a G-action guy

How would you succinctly show that $\frac{z^3+i}{z^2-3z+2}$ is analytic except where z=1 and z=2?

7:32 PM
I have a question: Why do the pictures on the internet of lie groups look like graphs?

because they are not pictures of lie groups

what are they pictures of

I don't remember

they are lieing

In the mathematical field of Lie theory, a Dynkin diagram, named for Eugene Dynkin, is a type of graph with some edges doubled or tripled (drawn as a double or triple line). The multiple edges are, within certain constraints, directed. The main interest in Dynkin diagrams are as a means to classify semisimple Lie algebras over algebraically closed fields. This gives rise to Weyl groups, i.e. to many (although not all) finite reflection groups. Dynkin diagrams may also arise in other contexts. The term "Dynkin diagram" can be ambiguous. In some cases, Dynkin diagrams are assumed to be directed,...

7:32 PM
yes they are

You can't draw any of the interesting pictures of Lie groups. Instead you say "simple Lie groups are classified by these pictures, called root systems". Those are the graphs.

this?

yes

yeahh
pretty cool diagrams

@SohamChowdhury You seem to have aged a bit.

7:34 PM
Uh, thank you?
I don't remember what it was :/

You looked 10, now you look 20.

that kind of looks like a Dynkin diagram

the picture in the thumbnail is not a dynkin diagram

@MikeMiller the one I posted is not

That picture is of a root system. A Dynkin diagram is a lot simpler

7:37 PM
This is a picture I drew for a professor
yeah it's a root system.
only took me about half an hour to draw it

@Rithaniel differentiability?

Yes, but it seems like a lot of work to prove that one is differentiable. Does not look like fun.

Separately show that polynomials are differentiable and quotients of differentiable functions are differentiable when the denominator is not 0

Thank you. That does sound much easier than what I was going to attempt.

7:45 PM
I call that develop the theory general enough until it's trivial

Also you can show that polynomials are differentiable via product rule
Since $f(z) = z$ is easy
(You could also do binomial stuff I guess, but I feel doing that directly for an arbitrary polynomial will be unpleasant)
Hmm, the torus is the only compact 2-dim Lie group, right?

It's the only compact 2-d parallelizable manifold, so yes

Ah makes sense

are all parallelizable manifolds also Lie groups?

not at all
all compact orientable 3-manifolds are parallelizable, including eg $SU(2)/G$ for any finite subgroup of $SU(2)$ - there are infinitely many such
however the only compact connected 3-dimensional lie groups are $SU(2), SO(3)$, and $T^3$
there are very few lie groups.

7:54 PM
right, I guess you do have a full set of globally-define vector fields which is closed under the Lie bracket, but to get a Lie group you need the structure coefficients to be constants

i would anticipate that's still true for $SU(2)/G$, no?

yes, so there's a local criterion which I mentioned, and a topological one which you did :)

I don't understand what you mean. I'm trying to say there's probably more or less no criterion you can apply to compact manifolds to see if they are Lie groups other than "is it on the list of compact Lie groups?"

although of course the question is "Can I find a basis whose Lie brackets have construct structure coefficients?", and I'm sure the answer is topological
*constant

8:15 PM

8:33 PM
@vzn Riemann attack, now that's a band name

This might be a weird question to ask here, but here goes nothing; quantum computing is just linear algebra in disguise with some group theory (on the matrices) thrown in. Is there some nice write-up on this? Sort of like "quantum computing for mathematicians"
4
Most write-ups either assume that I'm a researcher in both quantum and mathematics, or assume that I am a physicist who has done some mathematics

9:10 PM
@Krijn I only know some books about quantum theory with a mathematical treatment

@MikeMiller indeed the issue with what I asked about the other day was that my foliation is defined by the ambient space, not by the surface alone.
So the vector field $(\phi_t)_*\frac{\partial}{\partial x}$ does not have the leaves as flow lines.
Which means back to square one with this continuity thing.

Hey guys, can someone help me understand why if I have k circles on the plane, that divide it into k^2-k+2 parts, and draw another circle, then the maximum numbers of divisions it will add is 2k? I can't seem to comprehend it.
I have seen proofs that just say:"It's obvious that this circle will intersect with each one of the others in at most 2 points, so we get 2k new parts from k circles". The fact that there are 2 intersections for each circles don't make it obvious to me, that 2k new divisions will be added.

@anakhro Yeah, by the end of that I was sure I wasn't saying what you wanted, but wasn't sure what the right thing to talk about was.
By the way, do you want to send me an email and kill your anonymity? :)

I dunno, foliations? Contact structures? World peace? World conquest?
I'm just curious. ;)

9:24 PM
@MikeMiller can I send you an email about 2k plane divisions? :D

You will be sorely disappointed when you find out I am literally a nobody, @MikeMiller :P

@anakhro Sure, but one day we may meet, no? I guess I would know closer to that time.
@Coder-Man My email is publicly available, but I don't know what that means, and I probably won't respond. <3

Maybe!

9:36 PM
@MikeMiller here's to hoping I sent it to the right Mike Miller.
Also, is a diffeotopy the same thing as a smooth isotopy.
I keep on forgetting to ask my supervisor.
Because I always see isotopy used interchangeably for continuous or smooth isotopies, but diffeotopy would really help out the resulting ambiguity.

@anakhro are you a graduate student?

No, I am a talented grade 8 student.

Balarka 2: Electric Boogaloo

@anakhro really? or are you trolling?

@Daminark where is balarka
I miss him

9:44 PM
in The h Bar, Jul 19 at 11:26, by Balarka Sen
It's all bullshit anyway. Not even sure why I am going to a university.

@anakhro I think it is but that you would serve to confuse your readers by using it.
Better to say smooth isotopy and then "all isotopies are smooth". If they are.

terminology is kind of the worst

Undergrad has taken him from us

Hi chat
Wassup ?

Yo @Astyx! How's it going?

9:50 PM
Good and you ?

Doing alright, thanks!

@Krijn old but this might work/ be close to what you want, ping me for more refs, have a bunch, need to write them up sometime... An Introduction to Quantum Computing for Non-Physicists Eleanor G. Rieffel, Wolfgang Polak arxiv.org/abs/quant-ph/9809016

10:03 PM
@Daminark I would have thought undergrad would have been the easiest thing in the world for him
Clearly he's in some dean's basement in chains.
Doing math on command.
Fourier transforms or something vicious like that.

That's what the GRE felt like
Except with ordinary integrals
So maybe they're training him

10:20 PM
@Daminark can you help me define some parametric curves

@vzn Thanks!

If $f$ is Riemann integrable, $g$ is monotonic, does it follow that $f \circ g$ is Riemann integrable?

Apologies if this is a naive question as I don't have a strong background in higher mathematics but does anyone know of a standardized formula for calculating permutations when using all possible values for k?

LOL, @user193319 doesn't believe me. :P
oh, I missed @Astyx.
@Phaeze: What do you mean, precisely?

Haha...I knew you were going to enter the room any moment!

10:31 PM
@user193319: Do you know the $\epsilon$ criterion for integrability in terms of upper and lower sums?

Yes, I have that criterion.

BTW, I meant g $continuous* and monotonic. We were discussing continuous functions. OK, then apply the criterion. Oh, that changes things a lot. Let me give the problem another go. Remember the context of our discussion — I backed down from general continuous$g$. (Do you have a counterexample to that, by the way?) @TedShifrin I am generating test data for an encoding function that operates on a list of values. Given the count of values in the list I'm trying to find a simple way to find to find the total number of test cases generated. 10:33 PM No, I don't actually. This isn't a homework problem, so I haven't had a whole lot of time to think about it; but I'll add that to my list. OK, I think I have one, when you're ready to discuss it. Heya chat. Don't ever talk to me if you don't want a longer list :P heya @Fargle Haha, okay! The function generates every possible subset and ordering of those values so for 5 items it would be 5! + (5!/4!) + (5!/3!) + (5!/2!) + (5!/1!) 10:36 PM @TedShifrin given an algebraic variety, can you give me a fairly easy problem to work on, or do you know of anyone else who can give me a problem to work on? The algebraic variety in question is a set of solutions to a system of polynomial equations and I was just wondering if you knew of any particular problem with regards to that @ultradark: These are very hard problems. It's a whole field of research mathematics that uses a lot of material. You could start by reading about algebraic curves, but you need either some complex analysis or a good bit of algebra. @Phaeze: So it's ordered lists, not just lists? With unordered lists, you'd just get$2^5$. Hmm ... Yeah order matters as it changes the output of the encoding I don't know if I've ever seen this. Ahh okay, it did seem like a pretty rare scenario in all the googling I did. Thanks for giving it a thought. I made a list of the first ten values on Mathematica. I don't see anything familiar. For$6$it's just a bit more than my birth year :P For large$n$, I can tell you an approximation for what you have. It's going to be roughly like$n^n/e^{n-1}$, which is huge. 10:47 PM Yeah for n=7 it was ~14000 and n=8 jumps to ~109000 Yeah, my approximation isn't very good until we get to large$n$. Ah, so I need better Stirling to nail it. It looks like $$\frac{n^n}{e^{n-1}}\sqrt{2\pi n}.$$ Very, very close. For$n\ge 70$it's right to two decimal places in exponential notation$x.yz... \times 10^N$. heya @Krijn @vzn This is exactly what I wanted Heya @Ted What are you up to today? is$\ln(x)$an algebraic curve No what is it 10:54 PM awesome, thanks @TedShifrin What is ln(X) or what is an algebraic curve? both is ln(x) a transcendental curve or something you mean$y=\ln(x)$, which is$x=e^y$? Does that look like polynomial equations to you? no so what is that equation is it just called non algebraic? It's a transcendental equation. The graph is a perfectly lovely$1$-dimensional manifold, but NOT an algebraic anything. 10:56 PM are there varieties with transcendental equations Not algebraic varieties, no. what if transcendental equations intersect and the solutiion space consists of a system of polynomial equations over the reals what is that called? You can talk about real analytic varieties or complex analytic varieties. real please. Real is worse than complex, it turns out. 10:59 PM what do you mean by worse? Harder to understand. The complex ones are easier to grasp, for me at least I don't know anyone who has studied the question you just asked. Other than trivial circumstances, I don't know that it ever happens. I mean it's true, it's possible to construct a solution space in which real analytic functions intersect and the intersection points are found by solving a system of quartics or whatever. Not sure if that would be enough to define an analytic variety Isn't then the system of quartics itself enough to define an affine variety? 11:11 PM or maybe the algebraic varieties would just be defined from the intersection points themselves yeah so basically, each point would represent an affine variety, okay that makes much more sense No, you can't define varieties just knowing the intersection points. Just think about the point$(0,0)$in the plane. You can give it in countably many ways as the intersection of a pair of lines. And, working over the real numbers, you can always make things of arbitrarily high degree because polynomials like$x^2+y^2+7$have no real roots. That's one reason that complex is so much better than real. Ah yes, I remember now that alg. geom. was always trickier than I thought LOL But hey, I finished that sooooo... and those beastly non-reduced points ... 11:22 PM can you define a real analytic variety with just one variable$x$or do you need it to be two variable Whatever came out of that article on reduction modulo p? Also, quantum entanglement is so mindblowingly weird in physics, but the linear algebra behind is just "oh ok yeah clear" @Ultradark: Lots of variables. 11:37 PM @TedShifrin okay. I 'm just kinda confused as to how to categorize the space of some special transcendental equations, where their intersections require you to solve nasty polynomial equations. I was discussing this with @Semiclassical I haven't even considered the complex case Whether you can solve the equations or not, polynomials still have zeroes ... in general, you cannot solve in any sense. okay, thanks for helping me. I really learned a lot you're welcome. also I found a cool thing where one of the solutions equals$\phi -1$so if you solve:$e^{2/\ln(x)}=e^{4/\ln(1-x)}$you get$1/\phi$11:54 PM I'm claiming this is an artificial transcendental equation. It really comes from a polynomial that you then transform into something transcendental. But I consider that artificial. Hi Ted Sir How should I get this extra condition? You need the half of the circle corresponding to the argument condition. In the other half, the argument will be$-\pi/2\$.

so I'm confused, it's neither algebraic nor transcendental?


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# Rook's graph

Rook's graph
8x8 Rook's graph
Vertices nm
Edges nm(n+m)/2 - nm
Diameter 2
Girth 3 (if max(n,m) ≥ 3)
Chromatic number max(n,m)
Properties regular,
vertex-transitive,
perfect
well-covered

In graph theory, a rook's graph is a graph that represents all legal moves of the rook chess piece on a chessboard: each vertex represents a square on a chessboard and each edge represents a legal move. Rook's graphs are highly symmetric perfect graphs; they may be characterized in terms of the number of triangles each edge belongs to and by the existence of a 4-cycle connecting each nonadjacent pair of vertices.

## Definitions

An n × m rook's graph represents the moves of a rook on an n × m chessboard. Its vertices may be given coordinates (x,y), where 1 ≤ x ≤ n and 1 ≤ y ≤ m. Two vertices (x1,y1) and (x2,y2) are adjacent if and only if either x1 = x2 or y1 = y2; that is, if they belong to the same rank or the same file of the chessboard.

For an n × m rook's graph the total number of vertices is simply nm. For an n × n rook's graph the total number of vertices is simply $n^2$ and the total number of edges is $n^3 - n^2$; in this case the graph is also known as a two-dimensional Hamming graph or a Latin square graph.

The rook's graph for an n × m chessboard may also be defined as the Cartesian product of two complete graphs Kn $\square$ Km. The complement graph of a 2 × n rook's graph is a crown graph.

## Symmetry

Rook's graphs are vertex-transitive and (n + m − 2)-regular; they are the only regular graphs formed from the moves of standard chess pieces in this way (Elkies). When m ≠ n, the symmetries of the rook's graph are formed by independently permuting the rows and columns of the graph. When n = m the graph has additional symmetries that swap the rows and columns; the rook's graph for a square chessboard is symmetric.

Any two vertices in a rook's graph are either at distance one or two from each other, according to whether they are adjacent or nonadjacent respectively. Any two nonadjacent vertices may be transformed into any other two nonadjacent vertices by a symmetry of the graph. When the rook's graph is not square, the pairs of adjacent vertices fall into two orbits of the symmetry group according to whether they are adjacent horizontally or vertically, but when the graph is square any two adjacent vertices may also be mapped into each other by a symmetry and the graph is therefore distance-transitive.

When m and n are relatively prime, the symmetry group Sm×Sn of the rook's graph contains as a subgroup the cyclic group Cmn = Cm×Cn that acts by cyclically permuting the mn vertices; therefore, in this case, the rook's graph is a circulant graph.

## Perfection

The 3×3 rook's graph, colored with three colors and showing a clique of three vertices. In this graph and each of its induced subgraphs the chromatic number equals the clique number, so it is a perfect graph.

A rook's graph can also be viewed as the line graph of a complete bipartite graph Kn,m — that is, it has one vertex for each edge of Kn,m, and two vertices of the rook's graph are adjacent if and only if the corresponding edges of the complete bipartite graph share a common endpoint. In this view, an edge in the complete bipartite graph from the ith vertex on one side of the bipartition to the jth vertex on the other side corresponds to a chessboard square with coordinates (i,j).

Any bipartite graph is a subgraph of a complete bipartite graph, and correspondingly any line graph of a bipartite graph is an induced subgraph of a rook's graph. The line graphs of bipartite graphs are perfect: in them, and in any of their induced subgraphs, the number of colors needed in any vertex coloring is the same as the number of vertices in the largest complete subgraph. Line graphs of bipartite graphs form an important family of perfect graphs: they are one of a small number of families used by Chudnovsky et al. (2006) to characterize the perfect graphs and to show that every graph with no odd hole and no odd antihole is perfect. In particular, rook's graphs are themselves perfect.

Because a rook's graph is perfect, the number of colors needed in any coloring of the graph is just the size of its largest clique. The cliques of a rook's graph are the subsets of its rows and columns, and the largest of these have size max(m,n), so this is also the chromatic number of the graph. An n-coloring of an n×n rook's graph may be interpreted as a Latin square: it describes a way of filling the rows and columns of an n×n grid with n different values in such a way that the same value does not appear twice in any row or column.

 a b c d e f g h 8 8 7 7 6 6 5 5 4 4 3 3 2 2 1 1 a b c d e f g h
A non-attacking placement of eight rooks on a chessboard, forming a maximum independent set in the corresponding rook's graph

An independent set in a rook's graph is a set of vertices, no two of which belong to the same row or column of the graph; in chess terms, it corresponds to a placement of rooks no two of which attack each other. Perfect graphs may also be described as the graphs in which, in every induced subgraph, the size of the largest independent set is equal to the number of cliques in a partition of the graph's vertices into a minimum number of cliques. In a rook's graph, the sets of rows or the sets of columns (whichever has fewer sets) form such an optimal partition. The size of the largest independent set in the graph is therefore min(m,n). Every color class in every optimal coloring of a rook's graph is a maximum independent set.

## Characterization

Moon (1963) characterizes the m × n rook's graph as the unique graph having the following properties:

• It has mn vertices, each of which is adjacent to n + m − 2 edges.
• mn(m −1)/2 of the edges belong to m − 2 triangles and the remaining mn(n −1)/2 edges belong to n − 2 triangles.
• Any two vertices that are not adjacent to each other belong to a unique 4-cycle.

When n = m, these conditions may be abbreviated by stating that an n×n rook's graph is a strongly regular graph with parameters srg(n2, 2n − 2, n − 2, 2), and that every strongly regular graph with these parameters must be an n×n rook's graph when n≠4. When n=4, there is another strongly regular graph, the Shrikhande graph, with the same parameters as the 4×4 rook's graph. The Shrikhande graph can be distinguished from the 4×4 rook's graph in that the neighborhood of any vertex in the Shrikhande graph is connected in a 6-cycle, while in the rook's graph it is connected into two triangles.

## Domination

The domination number of a graph is the minimum cardinality among all dominating sets. On the rook's graph a set of vertices is a dominating set if and only if their corresponding squares either occupy, or are a rook's move away from, all squares on the m×n board. For the m×n board the domination number is min(m,n) (Yaglom & Yaglom 1987).

On the rook's graph a k-dominating set is a set of vertices whose corresponding squares attack all other squares (via a rook's move) at least k times. A k-tuple dominating set on the rook's graph is a set of vertices whose corresponding squares attack all other squares at least k times and are themselves attacked at least k − 1 times. The minimum cardinality among all k-dominating and k-tuple dominating sets are the k-domination number and the k-tuple domination number, respectively. On the square board, and for even k, the k-domination number is nk/2 when n ≥ (k2 − 2k)/4 and k < 2n. In a similar fashion, the k-tuple domination number is n(k + 1)/2 when k is odd and less than 2n (Burchett, Lane & Lachniet 2009).

format: plain


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# Community Promotion Ads - 2017 [duplicate]

It is a bit late into this new year, being that we're already in the second month, but we are now cycling the Community Promotion Ads for 2017!

### What are Community Promotion Ads?

Community Promotion Ads are community-vetted advertisements that will show up on the main site, in the right sidebar. The purpose of this question is the vetting process. Images of the advertisements are provided, and community voting will enable the advertisements to be shown.

### Why do we have Community Promotion Ads?

This is a method for the community to control what gets promoted to visitors on the site. For example, you might promote the following things:

• useful tools or resources for the mathematically inclined
• interesting articles or findings for the curious
• cool events or conferences
• anything else your community would genuinely be interested in

The goal is for future visitors to find out about the stuff your community deems important. This also serves as a way to promote information and resources that are relevant to your own community's interests, both for those already in the community and those yet to join.

### Why do we reset the ads every year?

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## marked as duplicate by GNUSupporter 8964民主女神 地下教會, YuiTo Cheng, JonMark Perry, mrtaurho, Arnaud D.May 13 at 12:05

• Glad to see this here! 3blue1brown is the 3blue1bomb! – Omnomnomnom Mar 29 '17 at 4:24
• I had no idea that this was being advertised, awesome! – Morgan Aug 29 '17 at 17:28
• This made me understand linear algebra after using it (with headache) for 8 years! – akraf Sep 27 '17 at 10:07
• I really enjoy watching your videos! Thank you for the effort you put it! – Cristian Em. Nov 25 '17 at 9:37

• Love it for those who don't use super advanced graphing tools but still want to be able to draw integrals, parametric/polar equations, and more! – Simply Beautiful Art May 24 '17 at 1:00

• – wythagoras Feb 4 '17 at 13:10

• The evergreen math helper! – Jaideep Khare Jun 3 '17 at 0:00
• @JaideepKhare: That slogan seems more appropriate for the Desmos commercial. :-) – Lucian Jun 3 '17 at 8:17

• Repost last year's version since I was surprised it wasn't already suggested. As a commenter mentioned on that post, Overleaf offers both free and paid services, in case that is a factor for anyone's decision. I enjoy the free service quite a lot... – rschwieb Feb 6 '17 at 15:46

• This image is 96×92 at the moment. Do you think you can get an official version at higher resolution? If not, it might make sense to create your own vector version of this (SVG or similar) and then export a suitably sized version from that. You also need to add a border, as far as I understand it. – MvG Feb 5 '17 at 9:31
• Logo fixed. Thanks to Alexander Bogomolny - CTK webmaster. – Jack D'Aurizio Feb 18 '17 at 16:29

• This is a demonstration post to indicate how this should look when an ad is posted. It also doubles as your twitter ad, but it's up to you if you wish to promote it by voting – Grace Note Feb 2 '17 at 14:04

• this needs more votes ! – Gabriel Romon Nov 28 '17 at 19:44

• – Alexander Konovalov Feb 9 '17 at 14:00
• link is not working for me? (took too long to respond) – MCMastery Sep 17 '17 at 1:50
• @MCMastery thanks for reporting - I've checked, works fine for me. Could you try again, please? – Alexander Konovalov Sep 17 '17 at 10:36
• now it works, guess server was just overloaded? – MCMastery Sep 17 '17 at 15:57
• @MCMastery could be some random network glitch - usually works fine. Was useful to look at the ad, prompted by this - I've also updated the URL, changing http to https. – Alexander Konovalov Sep 17 '17 at 16:53

• Reposted from yesteryear. Indispensable for evaluating many definite integrals. – Lucian Feb 26 '17 at 21:12
• That's pretty funny, given the reception the DLMF ad got the first time around =P. – E.P. Jun 21 '17 at 16:55

• I'm not sure if it makes much sense to put a link in the description. – punctured dusk Nov 9 '17 at 21:06

• I am writing here, on behalf of mathsteps' evykassirer to welcome developers to join mathsteps development on Github. Also, if those who votes down mathsteps can provide an suggestion to mathsteps' community, that would be much appreciated. – Wei Zhong Feb 7 '17 at 5:02

• Did you maybe notice the $300 \times 250$ dimensions requirement for ads? – J. M. is a poor mathematician Oct 22 '17 at 5:57
• @J.M.isnotamathematician No. Thanks for pointing out. How can I resize it? – tatan Oct 22 '17 at 8:01
• any good image editor should be able to do resizing or cropping as needed – J. M. is a poor mathematician Oct 22 '17 at 8:11
• @J.M.isnotamathematician I resized it... but the thing lost its beauty..;-( – tatan Oct 22 '17 at 14:40

• I am not sure that this picture can be used for community ads. The OP explicitly says: "The image that you create must be 300 x 250 pixels, or double that if high DPI." (I suppose the 2018 is about to start anyway, so you might prepare image with correct dimensions for the new thread.) – Martin Sleziak Jan 17 '18 at 15:51
• Note that the new thread has been posted here – user99914 Jan 29 '18 at 14:35


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# Are Regularity schema and $\in$-induction schema equivalent in intuitionistic logic?

the answer was that they are equivalent in classical first order logic with membership "$$\in$$".

But the answer relies on rule of excluded middle? hence the following question:

Question: Are Regularity (as presented in the above mentioned link) and $$\in$$-induction schemas equivalent in intuitionistic logic?

• This nLab article would seem to indicate no, they are not equivalent. Note also that regularity plus the existence of a non-empty set is enough to imply excluded middle. Nov 9, 2018 at 15:37
• There are several inequivalent non-classical set theories, which one do you mean? Nov 10, 2018 at 5:14

As discussed in the nLab link I posted in my comment, the existence of a non-trivial relation satisfying regularity implies the law of excluded middle. We can use this to show that regularity and $$\in$$-induction are not intuitionistically equivalent. I'll transcribe their proof into your context:

Assume that there are two elements $$a$$ and $$b$$ satisfying $$a\in b$$ and assume that $$\in$$ satisfies the regularity scheme. Let $$\varphi$$ be any sentence and define a formula $$\psi_\varphi(x)$$ given by $$x=b \vee \left( x \in b \wedge \varphi \right)$$. Since we have the regularity scheme this applies to the formula $$\psi_\varphi (x)$$, i.e. we have

$$\exists x\psi_\varphi(x) \rightarrow \exists x (\psi_\varphi (x)\wedge \lnot\exists z\in x(\psi_\varphi(z)))$$

Clearly we have that $$\psi_\varphi (b)$$ holds, so $$\exists x\psi_\varphi(x)$$ holds and by modus ponens $$\exists x (\psi_\varphi (x)\wedge \lnot\exists z\in x(\psi_\varphi(z)))$$ holds as well. Let $$c$$ be such that $$\psi_\varphi (c) \wedge \lnot \exists z\in c (\psi_\varphi(z))$$. Since $$\psi_\varphi (c)$$ holds we have two possibilities: Either $$c=b$$ or $$c \in b \wedge \varphi$$. Obviously in the second case $$\varphi$$ holds. In the first case we can show by contradiction (which is intuitionistically valid as long as we're proving a negative) that $$\neg \varphi$$ holds. Specifically, assume that $$\varphi$$ holds, then this would imply that $$a\in b \wedge \psi_\varphi (a)$$, which implies $$\exists z \in b (\psi_\varphi (z))$$ and hence $$\exists z \in c (\psi_\varphi (z))$$, contradicting the choice of $$c$$. Therefore in this case, $$\varphi \rightarrow \bot$$, or in other words, $$\neg \varphi$$. So by cases we have $$\varphi \vee \neg \varphi$$, i.e. the law of excluded middle holds for $$\varphi$$. Since $$\varphi$$ was arbitrary we have that it holds for every sentence.

Now to see the separation consider Heyting arithmetic and let $$<$$ take the role of your symbol $$\in$$. It's not hard to see that the induction scheme in Heyting arithmetic is literally $$\in$$-induction with $$<$$ taking the role of $$\in$$. It's well known that Heyting arithmetic does not prove the law of excluded middle for all sentences. On the other hand if we take Heyting arithmetic and replace the induction scheme with the regularity scheme, again with $$<$$ taking the role of $$\in$$, then by the argument above, using $$a=0$$ and $$b=1$$ we get that the law of excluded middle holds for every sentence, i.e. the system we get is Peano arithmetic, which of course is not intuitionistically equivalent to Heyting arithmetic.

Since we got that the regularity scheme and the $$\in$$-induction scheme are not logically equivalent in the presence of additional assumptions (i.e. the non-induction axioms of Heyting/Peano arithmetic), they cannot possibly be intuitionistically equivalent over any weaker set of assumptions, including no assumptions at all. So they are not intuitionistically equivalent.

• this shows that Regularity schema implies $\in$-induction, but the converse is not true, so the Regularity scheme is stronger than $\in$-induction schema, under intuitionistic logic. Nov 11, 2018 at 15:41

I just wanted to extend on the answer of James Hanson, that along his arguments, it is not only the case that $$\in$$-induction is strictly weaker than Regularity schema given the conditions of Heyting arthimetic or weaker grounds! Even if we hold all axioms of ZF-Regularity and add to it $$\in$$-induction [on top of intuitionistic first order set language], still the resultant system is a strict subtheory of the theory resulting from adding Regularity schema, instead of $$\in$$-induction, while keeping the rest of axioms. The reason is because intuitionistic ZF doesn't prove the law of excluded middle, while if Regularity is added then this would enforce excluded middle and thus get full (classical) ZF.


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# Step 4 in Map Projection: Properties of Map Projection (Preserving Metric Properties) (Especially for GATE-Geospatial 2022)

Glide to success with Doorsteptutor material for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.

Examrace Books on Mapping, GIS, and Remote Sensing prepares you throughly for a wide range of practical applications.

It is known that a globe is a true representation of the earth, which is divided into various sectors by the lines of latitude and longitude.

Theoretically map projection might be defined as “a systematic drawing of parallels of latitude and meridian of longitudes on a plane surface for the whole earth or a part of it on a certain scale so that any point on the earth surface may correspond to that on the drawing.”

Metric Properties of maps

Maps are flat, but the surfaces they represent are curved. Transforming, three-dimensional space onto a two-dimensional map is called “Projection.” Map projections are necessary for creating maps. This process preserves at least one of the following properties of map:

• Shape
• Area of Map
• Distance
• Direction

### 1. Angle in Projection

A projection in which the scale, although varying throughout the map, is the same in all directions at any point, so that very small areas are represented by correct shape and bearings are correct. A projection that retains this property of angular conformity is called a conformal or orthomorphic map projection.

### 2. Distance in Projection

The problem with angular conformity is variation in distance. If our purpose in projecting a map is to accurately measure distances, we must select a projection that preserves distance. Such projections are called “equidistant projections.” In an equidistant map projection, the length of particular lines in the map are the same as the length of the original lines on the curved reference surface.

### 3. Direction in Projection

Usually equidistant maps are equal direction maps.

### 4. Area in Projection

In an equal-area map projection the areas in the map are identical to the areas on the curved reference surface that means that areas are represented correctly on the map.

## Properties of Projection: Which Metric Properties to Preserve

All mapped areas have the same proportional relationship to the areas on the Earth that they represent. A map projection is a set of rules for transforming features from the three-dimensional earth onto a two-dimensional display. No flat representation of the earth can be completely accurate, so many different projections have been developed, and each suited to a particular purpose. A map projection would preserve some of the above metric properties while compromising others.

### Rules of Projection

Map projections differ in the way they handle four properties

• No projection can preserve all four simultaneously, although some combinations can be preserved, such as Area and Direction
• No projection can preserve both Area and Angles. The map-maker must decide which property is most important and choose a projection based on that.

Developed by:


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# Ex.1.1 Q4 Real Numbers Solution - NCERT Maths Class 10

Go back to  'Ex.1.1'

## Question

Use Euclid’s division lemma to show that the square of any positive integer is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m.}$$

[Hint: Let $$x$$ be any positive integer then it is of the form $$3q,\, 3q + 1$$ or $$3q + 2.$$ Now square each of these and show that they can be rewritten in the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1.$$]

Video Solution
Real Numbers
Ex 1.1 | Question 4

## Text Solution

To Prove:

The square of any positive integer is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m}$$ (using the Euclid’s division lemma).

Reasoning:

Suppose that there is a positive integer ‘$$a$$’. By Euclid’s lemma, we know that for positive integers $$a$$ and $$b,$$ there exist unique integers $$q$$ and $$r,$$ such that $$a=bq+r,\,0\le r\lt b$$

If we keep the value of $$b = 3,$$ then $$0 ≤ r < 3$$ i.e. $$r = 0$$ or $$1$$ or $$2$$ but it can’t be $$3$$ because $$r$$ is smaller than $$3.$$ So, the possible values for $$a = 3q$$ or $$3q + 1$$ or $$3q + 2.$$ Now, find the square of all the possible values of $$a.$$ If $$q$$ is any positive integer then its square (let’s call it as “$$\rm{m}$$”) will also be a positive integer. Now, observe carefully that the square of all the positive integers is either of the form $$3\,\rm{m}$$ or $$3\,\rm{m} + 1$$ for some integer $$\rm{m.}$$

Steps:

Let $${“a”}$$be any positive integer and $$b=3.$$

Then, $$a=3q+r$$ for some integer $$q\ge 0$$ and $$r = 0, 1 , 2$$ because $$0\le r <3.$$

Therefore,

\begin{align} a&=3q \;\text{ or }\;3q+1\;\text{ or }\;3q+2\text{ or}\;\\ {{a}^{2}}&={ }{{\left( 3q \right)}^{2}}\;\text{ or }\;{{\left( 3q{ }+{ }1 \right)}^{2}}\;\text{or }\;{{\left( 3q{ }+{ }2 \right)}^{2}} \\ {{a}^{2}}&={ }3{{(3q)}^{2}}\;\text {or }\;(9{{q}^{2}}+6q+{ }1{ })\;\text{or }\;(9{{q}^{2}}+{ }12q{ }+{{4}^{{}}}) \\{{a}^{2}}&={ }3{ }\left( 3{{q}^{2}} \right)\;\text{or }3\left( 3{{q}^{2}}+2q \right)+{ }1\\&\qquad \text{ or}\;3\left( 3{{q}^{2}}+{ }4q{ }+1 \right){ }+1 \\&=\rm{m}\; \text{or} \;3\rm{m}+1 \\\end{align}

Where $$\rm{m}$$ is any positive integer.

Hence it can be said that the square of any positive integer is either of the form \begin{align}{3\,\rm{m} \;\rm{or}\; 3\,\rm{m} + 1.} \end{align}

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text:
Documentation

### This is machine translation

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# dsp.PhaseExtractor

Extract the unwrapped phase of a complex input

## Description

The `dsp.PhaseExtractor` System object™ extracts the unwrapped phase of a real or a complex input.

To extract the unwrapped phase of a signal input:

1. Create the `dsp.PhaseExtractor` object and set its properties.

2. Call the object with arguments, as if it were a function.

## Creation

### Syntax

``phase = dsp.PhaseExtractor``
``phase = dsp.PhaseExtractor(Name,Value)``

### Description

example

````phase = dsp.PhaseExtractor` returns a phase extractor System object that extracts the unwrapped phase of an input signal.```

example

````phase = dsp.PhaseExtractor(Name,Value)` returns a phase extractor System object with the specified property name set to the specified value.```

## Properties

expand all

Unless otherwise indicated, properties are nontunable, which means you cannot change their values after calling the object. Objects lock when you call them, and the `release` function unlocks them.

If a property is tunable, you can change its value at any time.

Specify if the phase is to be unwrapped only within the frame, as a logical scalar.

When you set this property to:

• `false` –– The object returns the unwrapped phase while ignoring boundaries between input frames.

• `true` –– The object treats each frame of input data independently, and resets the initial cumulative unwrapped phase value to zero each time a new input frame is received.

## Usage

For versions earlier than R2016b, use the `step` function to run the System object algorithm. The arguments to `step` are the object you created, followed by the arguments shown in this section.

For example, `y = step(obj,x)` and `y = obj(x)` perform equivalent operations.

### Syntax

``p = phase(input)``

### Description

example

````p = phase(input)` extracts the unwrapped phase, `p`, of the input signal. Each column of the `input` signal is treated as a separate channel. The System object unwraps the phase of each channel of the input signal independently over time.```

### Input Arguments

expand all

Data input, specified as a vector or a matrix. This object supports variable-size input signals. That is, you can change the input frame size (number of rows) even after calling the algorithm. However, the number of channels (number of columns) must remain constant.

Data Types: `single` | `double`
Complex Number Support: Yes

### Output Arguments

expand all

Unwrapped phase of the input, returned as a vector or a matrix. The size and data type of the unwrapped phase output match the size and data type of the input signal.

Data Types: `single` | `double`

## Object Functions

To use an object function, specify the System object as the first input argument. For example, to release system resources of a System object named `obj`, use this syntax:

`release(obj)`

expand all

 `step` Run System object algorithm `release` Release resources and allow changes to System object property values and input characteristics `reset` Reset internal states of System object

## Examples

expand all

Note: This example runs only in R2016b or later. If you are using an earlier release, replace each call to the function with the equivalent `step` syntax. For example, myObject(x) becomes step(myObject,x).

Create a `dsp.SineWave` System object?. Specify that the object generates an exponential output with a complex exponent.

```sine = dsp.SineWave('Frequency',10,... 'ComplexOutput',true,'SamplesPerFrame',128); ```

Create a `dsp.PhaseExtractor` System object?. Specify that the object ignores frame boundaries when returning the unwrapped phase.

```phase = dsp.PhaseExtractor('TreatFramesIndependently',false); ```

Extract the unwrapped phase of a sine wave. Plot the phase versus time using a `dsp.TimeScope` System object?.

```timeplot = dsp.TimeScope('PlotType','Line','SampleRate',1000,... 'TimeSpan',1.5,'YLimits',[0 80],... 'ShowGrid',true,... 'YLabel','Unwrapped Phase (rad)'); for ii = 1:10 sineOutput = sine(); phaseOutput = phase(sineOutput); timeplot(phaseOutput) end ```

Note: This example runs only in R2016b or later. If you are using an earlier release, replace each call to the function with the equivalent `step` syntax. For example, myObject(x) becomes step(myObject,x).

Create a `dsp.TransferFunctionEstimator` System object?.

```tfe = dsp.TransferFunctionEstimator('FrequencyRange','centered'); ```

Create a `dsp.PhaseExtractor` System object?. Specify that the object must treat each frame of data independently.

```phase = dsp.PhaseExtractor('TreatFramesIndependently',true); ```

Create a `dsp.IIRFilter` System object?. Compute the transfer function of a third-order IIR filter. Use the `butter` function to generate coefficients for the filter.

```[b,a] = butter(3,.3); iir = dsp.IIRFilter('Numerator',b,'Denominator',a); ```

Extract the phase response of the transfer function. Plot using a `dsp.ArrayPlot` System object?.

```sampleRate = 1e3; phaseplot = dsp.ArrayPlot('PlotType','Line','XOffset',-sampleRate/2,... 'YLimits',[-15 0],... 'YLabel','Phase Response (rad)',... 'XLabel','Frequency (Hz)',... 'Title','System Phase response'); for ii = 1:100 % Generate input input = 0.05*randn(1000,1); % Pass through IIR filter filterOutput = iir(input); % Estimate transfer function transferFunction = tfe(input,filterOutput); % Plot transfer function phase phaseOutput = phase(transferFunction); phaseplot(phaseOutput); end ```

## Algorithms

Consider an input frame of length N:

`$\left(\begin{array}{l}{x}_{1}\\ {x}_{2}\\ ⋮\\ {x}_{N}\end{array}\right)$`

The `step` method acts on this frame and produces this output:

`$\left(\begin{array}{l}{\Phi }_{1}\\ {\Phi }_{2}\\ ⋮\\ {\Phi }_{N}\end{array}\right)$`

where:

`${\Phi }_{i}={\Phi }_{i-1}+\text{angle}\left({x}_{i-1}^{*}{x}_{i}\right)$`

Here, i runs from 1 to N. The `angle` function returns the phase angle in radians.

If the input signal consists of multiple frames:

• If you set `TreatFramesIndependently` to `true`, the `step` method treats each frame independently. Therefore, in each frame, the `step` method calculates the phase using the preceding formula where:

• ${\Phi }_{0}$ is 0.

• ${x}_{0}$ is 1.

• If you set `TreatFramesIndependently` to `false`, the `step` method ignores boundaries between frames. Therefore, in each frame, the `step` method calculates the phase using the preceding formula where:

• ${\Phi }_{0}$ is the last unwrapped phase from the previous frame.

• ${x}_{0}$ is the last sample from the previous frame.


text:
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# What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?

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What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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26 Mar 2019, 17:34
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Question Stats:

75% (01:41) correct 25% (01:53) wrong based on 20 sessions

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What is the value of $$y$$ if $$y > 0$$ and $$4 - 3y(y-\frac{4}{y}) = 41 - 4y^2?$$

A) 25
B) 20
C) 12
D) 8
E) 5
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GPA: 3
Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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26 Mar 2019, 18:24
I am probably getting my algebra totally wrong, but I don't see how any of those options can be a correct answer, can you see whether there is a typo?
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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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26 Mar 2019, 18:42
By simplifying ,
the equation becomes 4y^2 -3y -25 = 0
This equation will have imaginary roots!!! Is there something wrong in the question

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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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26 Mar 2019, 19:34
The question has a typo. The denominator y is just below 4/y, not below entire (y-4)/y
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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?  [#permalink]

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26 Mar 2019, 20:26
Hi guys,
The problem should be 4-3y(y-(4/y))=41-4y^2

Simplifying LHS we get
4-3y(y^2-4)/y
I.e 4-3y^2+12=41-4y^2
Substraction yields
y^2=25.
Therefore answer is y=5 as y>0

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Re: What is the value of y if y > 0 and 4 - 3y(y-4/y) = 41 - 4y^2?   [#permalink] 26 Mar 2019, 20:26
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# Tag Info

## New answers tagged protocol-analysis

2

No. Yes, by choosing an authenticated encryption scheme with a known $\:I\hspace{.03 in}V\hspace{.04 in}||\hspace{.04 in}C\hspace{.04 in}||\hspace{.04 in}tag\:$ and $k_{\hspace{.02 in}0}$ and $k_1$ such that decrypting $\:I\hspace{.03 in}V\hspace{.04 in}||\hspace{.04 in}C\hspace{.04 in}||\hspace{.04 in}tag\:$ with $k_{\hspace{.02 in}0}$ and $k_1$ yields ...

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text:
Endre Tóth (SZTE): Solution sets of systems of equations over finite algebras

 Wednesday, 28. November 2018, 10:00 - 12:00
Abstract. Solution sets of systems of homogeneous linear equations over fields are characterized as being subspaces. But what can we say about the “shape” of the set of all solutions of other systems of equations? We study solution sets over arbitrary algebraic structures, and we give a necessary condition for a set of n-tuples to be the set of solutions of a system of equations in n unknowns over a given algebra. If the condition is sufficient as well, then we will talk about (the given algebra having) Property (SSCC). In the case of Boolean equations we obtain a complete characterization, and we also characterize lattices and semilattices having Property (SSCC).
Location : Bolyai Intézet, I. emelet, Riesz terem, Aradi Vértanúk tere 1., Szeged


text:
# infinite geometric series

Infinite Geometric Series

An infinite series that is geometric. An infinite geometric series converges if its common ratiorsatisfies –1 < r < 1. Otherwise it diverges.


text:
# Reference on a theorem of algebraic geometry

In the book GTM 52 by R.Hartshone, there is a theorem as following :

Every variety of dimension $r$ is birational to a hypersurface in $\mathbb{P}^{r+1}$

Could you please tell me, who is the author of this theorem ?

I have read somewhere that birational equivalence was considered in Italian school, so could you please show me what exactly they did found relating to classification of variety(w.r.t birational equivalence).

Thank!

-
See mathoverflow.net/questions/4324/…, which is quite related. –  Mariano Suárez-Alvarez Jan 30 '13 at 7:15

The proof is very simple (let me work over an algebraically closed field $k$ of characteristic zero) If $K(X)$ is the field of rational functions on a variety $X$ of dimension $r$, then the trascendence degree of $K(X)$ over $k$ is $r$ and there are functions $x_1,\dots,x_r\in K(X)$ algebraically independent over $k$ such that the extension $K(X)\supset k(x_1,\dots,x_r)$ is finite and separable. It is therefore simple, so there exists an $x_{r+1}\in K(X)$ such that $K(X)=k(x_1,\dots,x_{r+1})$. Moreover, since the extension is finite, there is a polynomial $f$ such that $f(x_1,\dots,x_{r+1})=0$.
Now $f$ defines a hypersurface in the affine space $\mathbb A^{r+1}$, and its projective closure is a hypersurface in $\mathbb P^{r+1}$. It is easy to see that the field of rational functions on the latter is isomorphic to $K(X)$.
Could you please explain for me why the field of rational function of the latter is isomorphic to $K(X)$ ? –  Arsenaler Jan 30 '13 at 16:22


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This demonstration shows how to use the PmodAD2.

The PmodAD2, and an analog signal generator are required for this demonstration.

In this demonstration, an analog waveform is generated using the Digilent Analog Discovery 2, and the Waveforms software.:

For the waveform to be displayed, we collect multiple points in each period. However, according to the Nyquist theorem, the sample rate only has to be $2\times$ the frequency of the signal.

The following block of code is just an example. For the PMOD ADC, the minimum delay between two samples is around $0.3\,$ms (corresponding to a sampling period of $3\,$kHz). So the maximum frequency of the input signal can be $1.5\,$kHz.

In [1]:

from time import sleep
from pynq import Overlay

ol = Overlay("base.bit")
adc_id = int(input("Type in the PMOD ID of the ADC (1 or 2): "))
freq = int(input("Type in the frequency/Hz of the waveform: "))
period = 1/freq
log_interval_us = 0

# Assume Channel 0 is connected to the waveform generator
sleep(3*period)

# Draw the figure
%matplotlib inline
import matplotlib.pyplot as plt
plt.plot(range(len(log)), log, 'ro')
plt.axis([0, len(log), min(log), max(log)])
plt.show()

Type in the PMOD ID of the ADC (1 or 2): 2
Type in the frequency/Hz of the waveform: 200


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# Work done in inserting capacitor

We apply force to insert a dielectric slowly between capacitor .

While inserting , we are assuming charge is constant.

Now my sir told that

Work done by external agent $= Q^2 /2C[(1/k)-1].$

I could not why it is negative as according to me this should be the work done by force to insert it. And the direction slab displaces and force is same , so it should be positive m


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# Digital modulation and RF bandwidth

1. Jun 7, 2012

### dnyberg2

Why isn't it possible to invent a custom digital modulation scheme that would use much less RF bandwidth than all the schemes available? What are the limiting factors that do not allow for high data rates to be coded so that the occupied bandwidth is much less than all the others like ASK, PSK, MSK OOK etc...

2. Jun 7, 2012

### the_emi_guy

Look up "channel capacity" on the Internet.
Roughly summarizing you are trading between three things:
1 - information rate
2 - channel bandwidth
3 - noise

You can reduce channel bandwidth if you reduce noise, accept lower information rate, or both.
Note that "information rate" is not the same as bit rate.
There are also practical considerations. It is rarely possible to actually achieve reliable operation right at the channel capacity because it would require too much coding complexity.

3. Jun 7, 2012

### marcusl

Shannon showed that the capacity of a band-limited communication channel with AWGN is
$$C=Wlog_2(1+\frac{P}{N_0W})$$
where W is the channel bandwidth, P is the signal power and N0 is noise power spectral density. Reducing the bandwidth requires increasing transmit power to keep capacity the same, but the logarithmic function makes it very painful to do this.


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Differential graded Lie algebra

In mathematics, in particular abstract algebra and topology, a differential graded Lie algebra (or dg Lie algebra, or dgla) is a graded vector space with added Lie algebra and chain complex structures that are compatible. Such objects have applications in deformation theory[1] and rational homotopy theory.

Definition

A differential graded Lie algebra is a graded vector space ${\displaystyle L=\bigoplus L_{i}}$ over a field of characteristic zero together with a bilinear map ${\displaystyle [\cdot ,\cdot ]\colon L_{i}\otimes L_{j}\to L_{i+j}}$ and a differential ${\displaystyle d:L_{i}\to L_{i-1}}$ satisfying

${\displaystyle [x,y]=(-1)^{|x||y|+1}[y,x],}$

the graded Jacobi identity:

${\displaystyle (-1)^{|x||z|}[x,[y,z]]+(-1)^{|y||x|}[y,[z,x]]+(-1)^{|z||y|}[z,[x,y]]=0,}$

and the graded Leibniz rule:

${\displaystyle d[x,y]=[dx,y]+(-1)^{|x|}[x,dy]}$

for any homogeneous elements x, y and z in L. Notice here that the differential lowers the degree and so this differential graded Lie algebra is considered to be homologically graded. If instead the differential raised degree the differential graded Lie algebra is said to be cohomologically graded (usually to reinforce this point the grading is written in superscript: ${\displaystyle L^{i}}$). The choice of (co)homological grading usually depends upon personal preference or the situation as they are equivalent: a homologically graded space can be made into a cohomological one via setting ${\displaystyle L^{i}=L_{-i}}$.

Alternative equivalent definitions of a differential graded Lie algebra include:

1. a Lie algebra object internal to the category of chain complexes;
2. a strict ${\displaystyle L_{\infty }}$-algebra.

A morphism of differential graded Lie algebras is a graded linear map ${\displaystyle f:L\to L^{\prime }}$ that commutes with the bracket and the differential, i.e., ${\displaystyle f[x,y]_{L}=[f(x),f(y)]_{L^{\prime }}}$ and ${\displaystyle f(d_{L}x)=d_{L^{\prime }}f(x)}$. Differential graded Lie algebras and their morphisms define a category.

Products and coproducts

The product of two differential graded Lie algebras, ${\displaystyle L\times L^{\prime }}$, is defined as follows: take the direct sum of the two graded vector spaces ${\displaystyle L\oplus L^{\prime }}$, now equip it with the bracket ${\displaystyle [(x,x^{\prime }),(y,y^{\prime })]=([x,y],[x^{\prime },y^{\prime }])}$ and differential ${\displaystyle D(x,x^{\prime })=(dx,d^{\prime }x^{\prime })}$.

The coproduct of two differential graded Lie algebras, ${\displaystyle L*L^{\prime }}$, is often called the free product. It is defined as the free graded Lie algebra on the two underlying vector spaces with the unique differential extending the two original ones.

Connection to deformation theory

The main application is to the deformation theory over fields of characteristic zero (in particular over the complex numbers.) The idea goes back to Daniel Quillen's work on rational homotopy theory. One way to formulate this thesis (due to Vladimir Drinfeld, Boris Feigin, Pierre Deligne, Maxim Kontsevich, and others) might be:[1]

Any reasonable formal deformation problem in characteristic zero can be described by Maurer–Cartan elements of an appropriate differential graded Lie algebra.

A Maurer-Cartan element is a degree ${\displaystyle -1}$ element, ${\displaystyle x\in L_{-1}}$, that is a solution to the Maurer–Cartan equation:

${\displaystyle dx+{\frac {1}{2}}[x,x]=0.}$


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# Math Help - Prove s is in A or is an accumulation point of A..

1. ## Prove s is in A or is an accumulation point of A..

Let A be a nonempty set of real numbers that is bounded from above, and let s = sup(A). Prove that either s ∈ A or s is an accumulation point of A. Hence, in either case, s is in the closure.

For this, would my conclusion be that s is an accumulation point or both s is in A and is an accumulation point. If I know which direction I am aiming for, I think I can prove it.

2. A finite subset of the reals is closed. It is bounded above by its greatest element, which is s. In this case s is in A, but is not an accumulation point of A.

3. If $s\in A$ we are done. Right? So suppose that $s\notin A$.
Because $s-1 we know that $s-1$ is not an upper bound for $A$.
So that implies $\left( {\exists a_1 \in A} \right)\left[ {s - 1 < a_1 < s} \right]$. WHY!

Next let $\delta_2=\max\left\{a_1,s-\frac{1}{2}\right\}.$
Can you say $\delta_2 How do we select $a_2~?$

Carry on.

4. Originally Posted by DrSteve
A finite subset of the reals is closed. It is bounded above by its greatest element, which is s. In this case s is in A, but is not an accumulation point of A.
Oh, I've seen a few proofs of this but I was trying to work it out myself but one proof that I've seen showed that s is not the least upper bound. Then I've seen a contradiction that showed that s being an accumulation point and then ANOTHER proof i've seen is a proof by contraposition showing that s is neither a point of A or an accumulation point of A. So im not sure which way to direct my proof.


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1 . . Some authors (for example Willard, in General Topology) use the term frontier instead of boundary in an attempt to avoid confusion with a different definition used in algebraic topology and the theory of manifolds. {\displaystyle (-\infty ,a)} If the inequality is or >, draw the boundary line dotted. If you were to look down at your property from a bird's eye view, you would see a geometric shape. + The boundary of a set is the boundary of the complement of the set: The interior of the boundary of a closed set is the empty set. 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For example, the function y = 1/x converges to zero as x increases. − The straight line in the graph of an inequality that defines the half-plane containing the solutions to the inequality. 2 Despite widespread acceptance of the meaning of the terms boundary and frontier, they have sometimes been used to refer to other sets. R Boundary Lines Law and Legal Definition Adjoining landowners can find themselves in disputes over fences, overhanging branches, water rights, subjacent and lateral support and party walls. ∂ One side of the boundary line contains all solutions to the inequality. ∂ Boundary definition, something that indicates bounds or limits; a limiting or bounding line. 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If the disk is viewed as a set in , where a is irrational, is empty. The explanation for the apparent incongruity is that the topological boundary (the subject of this article) is a slightly different concept from the boundary of a manifold or of a simplicial complex. Mathematics. } Felix Hausdorff named the intersection of S with its boundary the border of S (the term boundary is used to refer to this set in Metric Spaces by E. T. Copson). Definition Of Boundary. translation and definition "boundary", English-Tagalog Dictionary online. ∂ , then the boundary of the disk is the disk itself: Boundary definition: The boundary of an area of land is an imaginary line that separates it from other areas. ∂ {\displaystyle \mathbb {Q} } Q {\displaystyle \Omega =\{(x,y,0)|x^{2}+y^{2}\leq 1\}} ) a hit in which the ball reaches or crosses the boundary line of the field on one or more bounces, counting four runs for the batsman.Compare six(def 5). 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S The boundary line indicating an edge of something. Meaning of boundary line. x bounded - WordReference English dictionary, questions, discussion and forums. See more. ∂ The Boundary line defines the space or area. a S In discussing boundaries of manifolds or simplexes and their simplicial complexes, one often meets the assertion that the boundary of the boundary is always empty. the collection of all points of a given set having the property that every neighborhood of each point contains points in the set and in the complement of the set. It is not to be confused with, Learn how and when to remove this template message, https://en.wikipedia.org/w/index.php?title=Boundary_(topology)&oldid=989046165, Articles lacking in-text citations from March 2013, Articles with unsourced statements from May 2018, Creative Commons Attribution-ShareAlike License. In topology and mathematics in general, the boundary of a subset S of a topological space X is the set of points which can be approached both from S and from the outside of S. ... Rate this definition: boundary. IPA : ... An edge or line marking an edge of the playing field. A boundary line is the outline of an entire shape or area. Here you can see that one side is colored grey and the other side is colored white, to determined which side that represent y â¤ 2x - 4, test a point. with the usual topology (i.e. In the space of rational numbers with the usual topology (the subspace topology of Ω = Ω If the boundary line is solid, then the linear inequality must be either â¥ or â¤. Definition of boundary line in the Definitions.net dictionary. 2 ), then the boundary of the disk is empty. Conversely, the boundary of a closed disk viewed as a manifold is the bounding circle, as is its topological boundary viewed as a subset of the real plane, while its topological boundary viewed as a subset of itself is empty. [citation needed] Felix Hausdorff[1] named the intersection of S with its boundary the border of S (the term boundary is used to refer to this set in Metric Spaces by E. T. Copson). Cricket. = ) First, we need to graph the boundary line. Learn more. {\displaystyle \mathbb {R} ^{2}} Example. In math, we use the term perimeter to indicate the distance around the outer edge of a shape. Many properties are rectangular, but not all are. The limits of an area can be determined by the boundary line. Information and translations of boundary line in the most comprehensive dictionary definitions resource on the web. x R , | (In particular, the topological boundary depends on the ambient space, while the boundary of a manifold is invariant. 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Boundary definition is - something that indicates or fixes a limit or extent. For K-12 kids, teachers and parents. ... (a four) or 6 (a six) runs respectively for the batting team. Except where noted, content and user contributions on this site are licensed under CC BY-SA 4.0 with attribution required. = ( , Indeed, the construction of the singular homology rests critically on this fact. 1 Cricket. One has. This page was last edited on 16 November 2020, at 19:18. How to use boundary in a sentence. The boundary line is dashed for > and < and solid for â¥ and â¤. Notations used for boundary of a set S include bd(S), fr(S), and{\displaystyle \partial S}. Vertical Line Test: If any vertical line intersects the graph of a relation at more than one point, then the relation graphed is not a function. ( of a set) The set of points in the closure of a set $S$, not belonging to the interior of that set. ( {\displaystyle \mathbb {R} } Math explained in easy language, plus puzzles, games, quizzes, videos and worksheets. Class boundary is the midpoint of the upper class limit of one class and the lower class limit of the subsequent class. ( Boundary line: The line itself is called the boundary line. Notations used for boundary of a set S include bd(S), fr(S), and ... Boundary A line or border around the outside of a shape. ≤ Each class thus has an upper and a lower class boundary. A set is closed if and only if it contains its boundary, and. 30 synonyms of boundary from the Merriam-Webster Thesaurus, plus 40 related words, definitions, and antonyms. The interior of the boundary of the closure of a set is the empty set. For any set S, ∂S ⊇ ∂∂S, with equality holding if and only if the boundary of S has no interior points, which will be the case for example if S is either closed or open. All Free. ∂ y , } } Despite widespread acceptance of the meaning of the terms boundary and frontier, they have sometimes been used to refer to other sets. 2 | ≤ Ω | Visit to learn Simple Maths Definitions. x ∂ Linear Inequality Terminology: Half-plane: A straight line separates a plane into two half-planes. The straight line in the graph of an inequality that defines the half-plane containing the solutions to the inequality. 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## boundary line math definition

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[<< wikipedia] ProPhoto RGB color space
The ProPhoto RGB color space, also known as ROMM RGB (Reference Output Medium Metric), is an output referred RGB color space developed by Kodak. It offers an especially large gamut designed for use with photographic output in mind. The ProPhoto RGB color space encompasses over 90% of possible surface colors in the CIE L*a*b* color space, and 100% of likely occurring real-world surface colors documented by Pointer in 1980, making ProPhoto even larger than the Wide-gamut RGB color space. The ProPhoto RGB primaries were also chosen in order to minimize hue rotations associated with non-linear tone scale operations. One of the downsides to this color space is that approximately 13% of the representable colors are imaginary colors that do not exist and are not visible colors.
When working in color spaces with such a large gamut, it is recommended to work in 16-bit color depth to avoid posterization effects. This will occur more frequently in 8-bit modes as the gradient steps are much larger.
There are two corresponding scene space color encodings known as RIMM RGB intended to encode standard dynamic range scene space images, and ERIMM RGB intended to encode extended dynamic-range scene space images.

== Development ==
The development of ProPhoto RGB and other color spaces is documented in an article summarizing a presentation by one of its developers Dr. Geoff Wolfe at Kodak, now Senior Research Manager at Canon Information Systems Research Australia, at the IS&T/SPIE Color Imaging Conference in 2011.

== ProPhoto RGB (ROMM RGB) Encoding Primaries ==

== Viewing environment ==
Luminance level is in the range of 160–640 cd/m2.
Viewing surround is average.
There is 0.5–1.0% viewing flare.
The adaptive white point is specified by the chromaticity values for CIE Standard illuminant D50 (x = 0.345704, y = 0.358540).
The image color values are assumed to be encoded using flareless (or flare corrected) colorimetric measurements based on the CIE 1931 Standard Colorimetric Observer.

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{\displaystyle X'_{\mathrm {ROMM} }=I_{\mathrm {MAX} }\cdot {\begin{cases}0;&X_{\mathrm {ROMM} }<0.0\\16\cdot X_{\mathrm {ROMM} };&0.0\leq X_{\mathrm {ROMM} }