Mathematics - 2017-07-04 (page 3 of 9)

Daminark

09:04

Oh, I never got far beyond just drawing cylinders

Balarka Sen

not a bad idea. you just have to figure out where to draw those cylinders

Daminark

That's the level of detail I could not quite hash out yet

Balarka Sen

no pressure; think about it and the $f^{-1}(x), f^{-1}(y)$ problem

Daminark

I mean the issue is as you say, you could sorta have a knot as your embedded copy of $S^1$, so I'm not really sure how to really see a cylinder from that

Balarka Sen

It won't literally be a cylinder.

Here's a thing: start with choosing a homotopy between $f, g$

Daminark

09:17

Sure

So, we should be able to choose a homotopy which fixes $p$, right?

Balarka Sen

$p$ is right there; it's a fixed point in the codomain. but cool idea

Daminark

Oh I mean whoops, what I wanted to say was something like

So the maps $h_t$ going from $h_0 = f$ and $h_1 = g$ all have $p$ as a regular value

Balarka Sen

:)

4

Las Vegas Raiders

Smiley's are not constructive.

Balarka Sen

I have a better suggestion; what if p is a regular value of $h$?

the homotopy M x I -> N itself i mean

Las Vegas Raiders

09:24

I'm going to flag your smiley as not constructive.

Balarka Sen

@lasvegas who's to decide?

lol

Las Vegas Raiders

acm

Astyx

Yo chat

Daminark

Hey @Astyx

Astyx

You can have either constructive smiley interference or destructive one

Balarka Sen

09:32

there are too many trolls in this chat. if i were Komrad Stalin i would have banned them from here

but i am not so

Astyx

What's up @Daminark ?

I wish I were Stalin

Balarka Sen

good ol' Stalin

SteamyRoot

Make a new chatroom called "Gulag" and invite people to it? :D

also hi

Balarka Sen

dictatorship 101

Daminark

Everything's alright, how about you?

Balarka Sen

09:35

@SteamyRoot like this?

encrypted-tbn0.gstatic.com/…

SteamyRoot

Perfect :D

Astyx

I'm good. I just underwent my first real maths oral exam

Well I had kinda one yesterday, so that's not 100% true

Balarka Sen

i mentally made up a russian accent for myself when typing the last couple messages

SteamyRoot

Oral exams (with written preparation) > Written exams

Astyx

No prep today

Yesterday was more about analysing a maths paper

So 2h prep

No exam > ...

Daminark

09:38

@Steamy on some pedagogical level I'm definitely down with that, though I hesitate somewhat if only because the stress of being watched can definitely harm performance

Balarka Sen

i'm lmaoing at the memes which came up in google images after i typed "gulag memes"

SteamyRoot

Well, if you don't want entrance exams for university, you can always move north a country, @Astyx

Daminark

May taught atop this last quarter and his exams were to write a lecture

Astyx

A bit too late to avoid exams I think :p

Balarka Sen

@Daminark Wow, cool idea

Astyx

09:39

And it ain't for university anyway

SteamyRoot

@Daminark Well, at least at our university, the standard exam form is like: 4 hours to solve 4 questions, and depending on the course 1 to 3 of those questions have to defended orally.

Astyx

I always found this terminology funny

Daminark

Midterm topic: “The long exact sequence of homotopy groups of a fibration”
Final topic: “Excision and the Mayer-Vietoris sequence”

@Balarka Yeah I'm totally gonna adopt that whenever applicable

SteamyRoot

The oral part ends up improving the grades of the majority of students, really. As a student, I really preferred having the oral part. But for professors, it takes a lot of extra work and time :/

Daminark

Perhaps, I guess if no one's undergoing the whole, basically stage fright in a lesser form, oral can help show that you know what's going on even if you're a bit stuck, in a way that's not possible for written exams

Astyx

09:47

@SteamyRoot Were you thinking of a country in particular ?

Balarka Sen

Ok, I have to stop myself from further meme-hunting and get some work done.

Daminark

Are they not synonymous? @Balarka

Balarka Sen

Don't. Massive amount of will power is needed to do this

over 9000

2

SteamyRoot

@Astyx Mine :P

Balarka Sen

ok i closed tab

Astyx

09:50

Legit

SteamyRoot

Well, hey, we get loads of Dutch students flocking to study here because of how cheap it is.

Astyx

If I get the admission to the schools I want I'll actually get paid to study

SteamyRoot

Okay, yeah, we don't have that. Unless you go to the military

Astyx

There is a agreement between you and the state : you owe 10 years of work, but you get a quality education (which implies you getting paid, cause you need to eat as well)

There is a military aspect as a tradition in one of these schools

(It goes back to Napoléon)

(glorious days)

SteamyRoot

What does "you owe 10 years of work" imply?

In Belgium, the military will pay you to study and such, but after you graduate you have to work for them for at least 10 years. If you quit any sooner, you have to refund what they paid you during your studies :P

Balarka Sen

09:57

Speaking of will, here's a nice story I remember. So one time when testing a computer-based translator (I forget which; one of the very first?) they put in "The flesh is weak but the spirit is strong" (from the New Testament, I think?) in there.

And then translated it to Russian, and re-translated back to English.

What came up was: "The meat is soft but the vodka is strong".

Daminark

Could it maybe be that they don't want someone just running away to another country upon finishing? @Steamy

Astyx

Not too sure about my vocabulary on this matter. You have to work for the fonction publique for at least ten years (4 of which you spend studying anyway), for instance with trains, gas, electricity for the country. If you intend to do fundamental research it does not matter anyway cause in France it is "regulated" by the state. You can also pay to compensate for some years you don't want to do

Pretty much the same as what you said then

SteamyRoot

@Daminark Running away to any other job, really.

Balarka Sen

hutchinsweb.me.uk/MTNI-11-1995.pdf

Astyx

@SteamyRoot Yeah exactly the same

Which to be fair is understandable

Daminark

10:00

Wait I'm a bit confused now

This 10 year thing, is that for all schools in France (so that you have to work in some position that could be said to be public service) or a particular subset of them?

Astyx

@Daminark Only that particular subset

Daminark

I see. So is there any... categorization, I guess, of them? Like the military in one case

Astyx

And the fundamental research for the other case (the ENS)

but it's just schools that are regulated by the states really

Daminark

I see. What does "fundamental" research mean?

Astyx

That's just a term I use to tell it appart from applied research which is what private companies would do

Daminark

10:05

Ah

Astyx

However I know very little, this might be a total misconception

user84215

I am ready to do fundamental research.

SteamyRoot

Science for the sake of improving science, rather than "real-life application"-based.

Astyx

Kinda

Balarka Sen

The bourgeois science, one might say

Astyx

10:07

"Does your research have any relevant application ? If so, then it's useless"

A bit like philosophy

SteamyRoot

Ew no

Daminark

Philosophy is like, the dankest subject tho

Astyx

Never said it wasn't

Daminark

Yeah I was rebutting Steamy

(removed)

Balarka Sen

meh I'd rather take literature

SteamyRoot

10:12

Do you know the difference between a mathematician and a philosopher?

Daminark

braces What?

SteamyRoot

A mathematician only needs pen, paper, and a garbage can. A philosopher doesn't even need the latter.

8

Astyx

Litterature is philosophy to my narrow-minded brain

Las Vegas Raiders

The Reading Room

Welcome to chat for literature.stackexchange.com — Read any go...

4

user84215

You should first define mathematics and philosophy.

Astyx

10:15

Typical mathematician

Las Vegas Raiders

Define define

Daminark

Define " "

Astyx

waiting for Balarka to jump on the occasion

Balarka Sen

philosopher

user84215

please take a starting point.

Balarka Sen

10:16

mathematician

@SteamyRoot

Astyx

Gromov looks zen to me

Balarka Sen

Gromov is fantastic

Fruitful Approach

Can anyone please comment on this simply stated prime gap proposition having to do with polynomials.

0

Q: Basic consequence of Zhang, Maynard, Tao.

Let $S$ be the set of all polynomials in $\Bbb{Z}[X]$, such that $$ (1) \ \ \liminf(f(p_n - 2) - f(p_{n-1})) = 0 \\ $$. Since $f(X) = X - K_0 \in S$ for some $K_0 \leq 122$ by Zhang, Maynard, and Tao's results. Similarly, $g(X) = a(X - K_0) \in S$ as well as any $g(X) = a(X-K_0)^b$ for $a, b \...

Astyx

Define $p_n$ explicitely

"Since" does not make sense in that second sentence

Finally make the . be inside the four $ so that it does not appear after the line break

Finally the last $f(X-K_0)$ doesn't make much sense to me

"or in general any polynomial." is explicit enough

Just my personnal opinion though, I have no clue on the actual question

(Don't have the will to think about it right now)

Secret

10:51

I have no idea how to write the line $P_{jk_1}$ etc. in a better way, but what is happening on that line is that when any matrix $A$ is multiplied to a block diagonal matrix whose blocks are column vectors of all 1s (thus there are y of these column vectors), then the resulting matrix is a matrix with y columns, where each column of the resulting matrix is the sum of the first k column vectors of $A$, then the next column is the sum of the k+1th to 2k of $A$ and so on all the way
up to yk-k to yk

$$SPS^T$$

Let $S^T=U$ and $(1)_{xy}$ be the components of the matrix of all 1s. Now write everything in terms of indices, i.e.:

$(S)_{ij}=S_{ij}$, $(P)_{kl}=P_{kl}$, $(U)_{mn}=U_{mn}$

Now

$(PU)_{kn}=\sum_{l}P_{kl}U_{ln}$, $(SP)_{il}=\sum_{j}S_{ij}P_{jl}$

$(SPU)_{in} = \sum_{j}S_{ij}(\sum_{l}P_{jl}U_{ln})$, $(SPU)_{in} = \sum_{l}(\sum_{j}S_{ij}P_{jl})U_{ln}$

Given $P$ is doubly stochastic, i.e.

$\sum_{k}P_{kl}=(1)_{1l},\sum_{l}P_{kl}=(1)_{k1}$

Now let $w,v$ be the row and column vectors of all 1s

Secret

@BrownNinja if $\vec{y}$ majorise $\vec{x}$, then I am guessing any "subvector" formed by taking the 1st kth elements of the original vector also satisfy the majorisation relation and hence the sums as well?

since majorisation means that two vectors of the same number of components have the same component sum and that the sum of each subvector from the smallest element are also $\geq$ (for majorise from below) or $\leq$ (for majorise from above) of the other

user84215

If f is a bijection between affine spaces of same finite dimension that takes any three collinear points into collinear points, prove that f is semiaffine.

Secret

so the sum of the first kth elements must be $\geq$ the sum of the next kth elements and so on

@SteamyRoot nowadays, computers as well

user84215

11:15

is there anyone awake?

Dattier

$$\text{Calculate }\lim\limits_{n\rightarrow \infty} P_n(x) \mod (x^2+1)^2 \text{ with }P_n(x)=\sum \limits_{k=0}^n \frac{x^k}{k!}$$

Hi

Secret

11:52

in The h Bar, 11 mins ago, by ACuriousMind

@Secret Yes, that is true, because multiplying a matrix with that column/row vector is the same as summing up all the rows/columns.

in The h Bar, 9 mins ago, by Secret

Hmm, this eigenvector also seemed quite special for these matrix. You can do a basis change on the original matrix and not only you still have the same eigenvalue, the eigenvector is also exactly the same, meaning this is one of the few eigenvectors that are left unchanged by a basis transformation

in The h Bar, 8 mins ago, by ACuriousMind

@Secret You'll have to restrict your notion of "basis transformation" considerably for that to hold true.

in The h Bar, 3 mins ago, by ACuriousMind

@Secret For instance, my example above can be diagonalized to $\mathrm{diag}(0,1)$, and diagonalization is a basis change and that matrix clearly does not have the vector $\begin{pmatrix}1\\1\end{pmatrix}$ as an eigenvector.

in The h Bar, 2 mins ago, by Secret

Hmm... Let $P$ be a row stochastic matrix and $\vec{1}$ be the column vector of all 1s. Let $S,S^{-1}$ be a pair of basis transformation matrices, and let primed variables be the matrix and vector under the new basis. Then:

$$P\vec{1}=\vec{1}$$

$$P'\vec{1}'=(S^{-1}PS)S^{-1}\vec{1}=S^{-1}P\vec{1}=S^{-1}\vec{1}$$

= fail unless $S^{-1}$ has $\vec{1}$ as eigenvector with eigenvalue $1$

Right, and yours is a nice counterexample of the general result here

and that's why I love formal manipulations, you frequently can get the set of all counterexamples in one sweep

More generally, it is much easier in abstract algebra to solve a problem by formal manipulations plus a few theorems compared to analysis where you need to worry about the specific detail related to the functions themselves (and the limit gymnasiums) and then I lost track because too many things are happening at the same time

Secret

Besides the underlying rules of close forms for a wide variety of integrals and series, I am also interested in the "closed form" of counterexamples to a given proposition

this is because, close forms often containing the "why", not just the "how"

One reason I like to seek the set of all counterexamples even though it is difficult is because of 2 reasons:

1. I am a perfectionist, thus I like to get all possible things that satisfy a given request

2. I frequently ran into cases involving accidental cancellations, mathematical coincidences and so on, thus I often have trouble finding counterexamples unless I knew the problem well enough

42 mins ago, by Dattier

$$\text{Calculate }\lim\limits_{n\rightarrow \infty} P_n(x) \mod (x^2+1)^2 \text{ with }P_n(x)=\sum \limits_{k=0}^n \frac{x^k}{k!}$$

mod a polynomial?

$$\frac{e^x}{(x^2+1)^2}$$??

RE60K

12:10

Is there any way to find triangles with integer sides and area multiple of $\sqrt3$

i.e. $$\Delta = \sqrt{s(s-a)(s-b)(s-c)} \in \mathbb Z\sqrt{3}$$

Simply Beautiful Art

@Secret Nah, that's not how it works lol

SteamyRoot

As it's written in the question, I doubt it's well defined. But if you place the limit outside the mod, it might be. Then it's just modding out an ideal in a polynomial ring.

Simply Beautiful Art

desmos.com/calculator/am3heyxkv7

Secret

ah ok

@SimplyBeautifulArt That looks really oscillatory

Simply Beautiful Art

It appears to approach $e^x$ everywhere except a small region $[0,0.58]$

Secret

12:20

As for that heron's formula question, that looks like a Diophantine equation of degree 4, 3 unknowns with mixed terms, and equal to $3k^2$, where $k$ is integer, I am not sure if that has an analytical solution

SteamyRoot

It's not like that Desmos plot is anywhere near correct.

Astyx

Depends on the topology

Simply Beautiful Art

Actually, I'm pretty sure the limit is $e^x\mod(x^2+1)^2$

SteamyRoot

Modding out $(x^2 + 1)^2$ means whatever you're left with is at most a degree $3$ polynomial...

Akiva Weinberger

@RE60K Doesn't an equilateral triangle with side length $2$ work?

Or do you want all of them?

I think you want $a\equiv b\equiv c\not\equiv0\pmod3$, as well as $a+b+c\equiv0\pmod2$. But I'm not sure.

(You also need to make sure that the triangle inequality is satisfied.)

(So like $(7,7,10)$ for example)

Oh, wait, that doesn't work. That has area $10\sqrt6$.

user84215

12:40

Continuing Berger's book is becomming hard for me. It assumes that you are familiar with measure theory, functional analysis and algebraic topology.

Akiva Weinberger

OK, yeah, I see where I messed up. Not sure how to fix it, though.

Secret

[Testing]
$\exists a,b \in \Bbb{Z},a^2=3b^2\implies \sqrt{3}\text{ is rational}\implies ⨳$

$\exists x,P(x)=0\implies P(x)\neq 0 \implies ⨳$

$a^2\equiv b^2 \mod 3 \implies \sqrt{3}\text{ is rational}?$

Define $\text{Root}(y)=\{y:P(y)=0\}$

if $P(x)=0$, then $\text{Root}(P(x))=x$

but $\text{Root}(0)=x$?

=fail

Define $\text{Root}(P)=\{y:P(y)=0\}$

SteamyRoot

Right... It seems that $$\sum_{k=0}^{2n-1} \frac{x^k}{k!} \equiv \sum_{l = 0}^{n-1} (-1)^{l+1} \left( \frac{l-1}{(2l)!} + \frac{l-1}{(2l+1)!}x + \frac{l}{(2l)!}x^2 + \frac{l}{(2l+1)!}x^3 \right) \mod (1+x^2)^2$$

Secret

if $P(x)=0$, then $\text{Root}\circ (P)(x)=x$

SteamyRoot

I guess all you then have to do is calculate the limits for each coefficient. Can't be bothered.

user84215

12:53

Is there anyone who wants to read Berger's book together ?

Secret

(bleh, not knowing enough how polynomial ring ideals work, will try again later)

The underlying idea:

Given
$a^2\equiv b^2 \mod 3 \implies \sqrt{3}\text{ is rational}\implies ⨳$

Want

$P(x),x\text{ is not root of P}\implies P(x) = (something) \mod (something)\implies ⨳$

Or in english:

SteamyRoot

Pretty sure $0^2 \equiv 3^2 \mod 3$.

Secret

nonzero perfect squares being non perfect square multiples of perfect squares will trigger the contradiction that surds of non perfect squares are rational

and I want a condition such that plugging a non root x into some given polynomial P will trigger some kind of contradiction without actually evaluating P(x), which hopefully involving taking mod somehow

Daminark

Idontbelieveyou

►

SteamyRoot

Lol

Balarka Sen

13:01

No direction home ending scene

►

Secret

suppose the idea works, then we can write the following which will capture all transcendental numbers:

$$\forall P \in P(\Bbb{C}),\exists Q,R: P(x) = Q \mod R \implies ⨳$$

user84215

I and my posts are ignored by you?

Secret

I have not read any books except munkres, definitely cannto help you on that

Mathmore

Hello!!

Secret

The issue is that polynomials are nonlinear maps wrt their arguments, which makes it very hard to decompose them

Mathmore

13:05

Want a little help in understanding this proof of Gallian.

0

A: Need help to understand this proof of the question number $51$ of the $7$th chapter given in Gallian's Contemporary abstract Algebra.

$((\neg B)\implies B)\implies B\implies ((\neg A)\lor B)\iff (A\implies B).$

Secret

and then, there are ireeducible polynomials...

Mathmore

I found he proves it like this : $\alpha \notin H \implies \alpha \in H$. Thus it seems paradoxical.

SoumyoB

happy Independence day to all Americans on here

3

Secret

Actually wait a sec. Based on the fundamental theorem of algebra, all polynomials must have roots, thus it can always be decomposed into product of monomials of the form (x-root), even if the root itself has no closed form by galois theory

That means $P(x)=\prod_i(x-x_i)$, but how will that help us...

Hmm.... what does $P(x) \mod (x-x_1)$ mean, need to check...

SteamyRoot

Maybe start by checking the definition of a polynomial ring?

You'd realise that, unless you're working over the complex numbers, you can't always split into those linear factors.

SoumyoB

13:11

hmm no Americans on here right now?

SteamyRoot

And if you mod out a product by one of its factors, it's obviously zero.

Secret

yup, I plan to start at the complex numbers case, because fundamental theorem of algebra will guarantee we will have linear factors

hmm... I think I really need to read that up first for the 2nd point, but I guess you are right, suppose I plug a non root x into $P$ and thus I get a product of nonzero numbers, then modding any of them out will be obviously zero

LegionMammal978

@Secret So what are you trying to do?

SteamyRoot

Stop thinking about "plugging in".

You'll end up making mistakes like that.

The polynomials $x, x^2, x^3, ...$ are all distinct over $F_2$, even though for any element of $F_2$ the evaluate to the same values.

Secret

@LegionMammal978 I am trying to create some criteria to detect transcendental number by forcing them to create a contradiction when they are evaluated by any polynomials, but my background on polynomial rings is too poor for the job, so I will deal with it later

Leaky Nun

13:20

@Secret by division algorithm, $P(x) = Q(x)d(x)+R(x)$, with $\deg r < \deg d$, then $P(x) \equiv R(x) \pmod{d(x)}$

Secret

Right then $P(x) \equiv 0 \mod d(x)$ is not helping us since all the information is in $Q(x)$

Leaky Nun

@Mathmore it isn't paradoxical. The paradoxical statement is $\alpha \notin H \iff \alpha \in H$

$\alpha \notin H \implies \alpha \in H$ can be right when $\alpha \in H$.

Mathmore

@LeakyNun Oh!

Brown Ninja

@Secret Thank you very much. I was away, so did not see your message earlier.

Mathmore

@LeakyNun I see. I had never seen proof of such type.

Leaky Nun

13:30

@Mathmore I didn't bother to look at the proof :p

Mathmore

@LeakyNun hahaha. You don't have to. I have summarized it in two lines. :D

@LeakyNun $\alpha \notin H \implies \alpha \in H$ can be right is a shocker to me though.

Leaky Nun

@Mathmore well then you need to learn logic properly :p

Mathmore

@LeakyNun any suggestion for the logic reference book? :/

Leaky Nun

@Mathmore just learn what each symbol means

e.g. what does $\implies$ mean?

@Mathmore I don't understand why $A_5 = H \cup \alpha H$.

Mathmore

@LeakyNun $\implies$ is implication. That is if $A$ is a statement written to the left of $\implies$, then we can deduce that $B$, which is written at the right side of $\implies$ is true. I mean it can be deduced from $A$.

Leaky Nun

13:36

@Mathmore that's not what it means at all

sure, it has some semantic relationship with implication, but its formal definition is simple and to the point and rigorous.

Mathmore

$A_5=H \cup \alpha H$ because $H$ is index two set in $A_5$. Thus for any element not in $H$, say $\beta \notin H$, we have $A_5=H \cup \beta H$.

Leaky Nun

@Mathmore oh, right

Lagrange theorem

Mathmore

@LeakyNun yes! :) So what understanding I have about $\implies$ is not accurate. :/

Leaky Nun

@Mathmore so look it up online, or ask me

Mathmore

@LeakyNun first I choose to ask you. :D

Leaky Nun

13:40

@Mathmore $A \implies B$ is equivalent to $\neg (A \land \neg B)$

that means, "it is not possible that A is true and B is false"

Secret

23 mins ago, by SteamyRoot

The polynomials $x, x^2, x^3, ...$ are all distinct over $F_2$, even though for any element of $F_2$ the evaluate to the same values.

hmm? so that means $\{x:P(x)\neq 0\}$ is not a sufficient condition for $x$ being transcendental?

Mathmore

@LeakyNun Okay.

Leaky Nun

@Secret yes. sufficient and necessary.

but the polynomials must have integer coefficients

Mathmore

This I know. Because I have done only this type of proofs.

Leaky Nun

@Mathmore so is the following true?

$(1=2) \implies (1+2=4)$

@Secret (or any other people here) how do I write "polynomials with integer coefficients over the complex numbers"?

Mathmore

13:44

But we know $1 \neq 2$ right?

Secret

For the field being complex numbers. If $P(x)\neq 0$ then since $P(x)=\prod_i(x-x_i)$ all linear factors also also nonzero

$\Bbb{Z}[\Bbb{C}]$ I suppose...

Leaky Nun

@Secret really?

@Mathmore yes, but that isn't what I asked

Semiclassical

hi chat

Secret

A polynomial ring is denoted $K[X]$ where $K$ is the ring and $X$ are the indeterminates, so if $X=\Bbb{C}$ then the ineterminates are taken from $\Bbb{C}$?

Brown Ninja

@Secret I have written my question in a better way here. I am sorry earlier it was poorly defined.
https://math.stackexchange.com/questions/2346189/does-this-vector-majorizes-all-other-similar-vectors

Mathmore

13:46

@LeakyNun Ah I see. I think that statement is true after substituting $1=2$. Then we get $2+2=4$. Right?

Leaky Nun

@Mathmore you're using your own definition again ^^

Secret

> I define an $\ell$ element vector $\vec{x}_{n,k}$ obtained from $\vec{x}$ (where elements are arbitrarily ordered) by summing every $k$ elements of $\vec{x}$

Mathmore

@LeakyNun Okay.

Secret

arbitrary ordered? in that case I don't see why the majorisation has to hold, unless its proof is not obvious

Leaky Nun

@Secret do you think this is legible to anyone?

@Mathmore so...?

Secret

13:49

though I did proved that the matrix $SPS^{T}$ is also doubly stochastic by fiddling with the indices, not sure how that helps

Mathmore

@LeakyNun I'm thinking and confused. But I would then say that we don't know what $+$ means and what the symbols $1,2,4$ indicate.

Brown Ninja

with the definition of 'Orthogonal ( :p )' $S$ I provided?

Secret

yeah, your $S$ is a very special matrix. It's a block diagonal matrix with each block being $\vec{1}$ the vector of all 1s

(My proof is also somewhere near your last reply above, it's a long wall of text)

Leaky Nun

@Mathmore eh, it's just a simple example...

Brown Ninja

Right. I will read it carefully.

Semiclassical

13:51

that'd mean $S$ can be thought of as the Kronecker product of the identity matrix and a matrix of ones (which is itself equivalent to the inner product $uu^T$ where $u$ is a matrix of ones.

(Note: I forget if it'd be Kronecker(identity,uu^T) or Kronecker(uu^T,identity). the ordering there is something I usually get wrong.)

Secret

well, except that the matrix of ones is a column vector or row vector

Leaky Nun

@Mathmore let's say we know what those symbols mean.

Mathmore

@LeakyNun Honestly I have never studied logic. But I clearly see how dumb I am too ignore such an important topic. ;(

Semiclassical

woops, yes.

should be a column vector of ones.

Secret

$$SPS^T$$

Let $S^T=U$ and $(1)_{xy}$ be the components of the matrix of all 1s. Now write everything in terms of indices, i.e.:

$(S)_{ij}=S_{ij}$, $(P)_{kl}=P_{kl}$, $(U)_{mn}=U_{mn}$

Now

$(PU)_{kn}=\sum_{l}P_{kl}U_{ln}$, $(SP)_{il}=\sum_{j}S_{ij}P_{jl}$

$(SPU)_{in} = \sum_{j}S_{ij}(\sum_{l}P_{jl}U_{ln})$, $(SPU)_{in} = \sum_{l}(\sum_{j}S_{ij}P_{jl})U_{ln}$

Given $P$ is doubly stochastic, i.e.

$\sum_{k}P_{kl}=(1)_{1l},\sum_{l}P_{kl}=(1)_{k1}$

Now let $w,v$ be the row and column vectors of all 1s

Leaky Nun

13:53

@Mathmore as I said, $A \implies B$ is equivalent to $\neg(A \land \neg B)$

or, using de Morgan's, $\neg A \lor B$

$\neg A$ is true, so $\neg A \lor B$ is true, so $A \implies B$ is true.

Mathmore

@LeakyNun Okay. Then $1+2=3 \neq4$. ?

Leaky Nun

@Mathmore yes

in other words, the former statement is false, so the implication must be true

2

Q: Why a false statement can imply anything?

According to the truth table, If $P$ is false,then $P->Q$ is true. if pigs fly, then $1+1=3$. Why is this implication true? How do you prove it?

logic

cut-the-knot.org/do_you_know/falsity.shtml

Mathmore

@LeakyNun So if the hypothesis is false, then it's implication is true?

Leaky Nun

@Mathmore when I say "the implication" I mean "the statement $A \implies B$"

I'm not sure what $B$ would be called

and yes, $A \implies B$ is a statement.

Mathmore

@LeakyNun I see. I remember studying in classroom that the statement "sun rises in west $\implies$ India is a country" is a true statement.

Oh false statement implies anything!!!

Going through your links...

Semiclassical

13:57

Another way to write S is as the direct sum of uu^T matrices

Secret

yup

Astyx

Any cool math today ?

Brown Ninja

@Secret Right. So I was using the matrix $S$ in the context of $k$ element summation of an $n$ element vector. $\vec{x}_{n,k} = \vec{x} S$. Right?

Leaky Nun

In informal speech, "if $A$ then $B$" and "$A$ implies $B$" are mainly used when it is believed there is a causal connection between $A$ and $B$. The truth-functional connective $\rightarrow$ does not capture this feature of "implies." — André Nicolas Apr 28 '12 at 3:24

@Mathmore ^

Brown Ninja

@Semiclassical nice to see you :)
You helped me a lot when I asked a question here a few days back.

@Semiclassical true

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