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Daminark
Daminark
9:04 AM
Oh, I never got far beyond just drawing cylinders
 
Balarka Sen
Balarka Sen
not a bad idea. you just have to figure out where to draw those cylinders
 
Daminark
Daminark
That's the level of detail I could not quite hash out yet
 
Balarka Sen
Balarka Sen
no pressure; think about it and the $f^{-1}(x), f^{-1}(y)$ problem
 
Daminark
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
Balarka Sen
It won't literally be a cylinder.
Here's a thing: start with choosing a homotopy between $f, g$
 
Daminark
Daminark
9:17 AM
Sure
So, we should be able to choose a homotopy which fixes $p$, right?
 
Balarka Sen
Balarka Sen
$p$ is right there; it's a fixed point in the codomain. but cool idea
 
Daminark
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
Balarka Sen
:)
4
 
Las Vegas Raiders
Las Vegas Raiders
Smiley's are not constructive.
 
Balarka Sen
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
Las Vegas Raiders
9:24 AM
I'm going to flag your smiley as not constructive.
 
Balarka Sen
Balarka Sen
@lasvegas who's to decide?
lol
 
Las Vegas Raiders
Las Vegas Raiders
acm
 
Astyx
Astyx
Yo chat
 
Daminark
Daminark
Hey @Astyx
 
Astyx
Astyx
You can have either constructive smiley interference or destructive one
 
Balarka Sen
Balarka Sen
9:32 AM
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
Astyx
What's up @Daminark ?
I wish I were Stalin
 
Balarka Sen
Balarka Sen
good ol' Stalin
 
SteamyRoot
SteamyRoot
Make a new chatroom called "Gulag" and invite people to it? :D
also hi
 
Balarka Sen
Balarka Sen
dictatorship 101
 
Daminark
Daminark
Everything's alright, how about you?
 
Balarka Sen
Balarka Sen
9:35 AM
@SteamyRoot like this?
 
SteamyRoot
SteamyRoot
Perfect :D
 
Astyx
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
Balarka Sen
i mentally made up a russian accent for myself when typing the last couple messages
 
SteamyRoot
SteamyRoot
Oral exams (with written preparation) > Written exams
 
Astyx
Astyx
No prep today
Yesterday was more about analysing a maths paper
So 2h prep
No exam > ...
 
Daminark
Daminark
9:38 AM
@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
Balarka Sen
i'm lmaoing at the memes which came up in google images after i typed "gulag memes"
 
SteamyRoot
SteamyRoot
Well, if you don't want entrance exams for university, you can always move north a country, @Astyx
 
Daminark
Daminark
May taught atop this last quarter and his exams were to write a lecture
 
Astyx
Astyx
A bit too late to avoid exams I think :p
 
Balarka Sen
Balarka Sen
@Daminark Wow, cool idea
 
Astyx
Astyx
9:39 AM
And it ain't for university anyway
 
SteamyRoot
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
Astyx
I always found this terminology funny
 
Daminark
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
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
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
Astyx
9:47 AM
@SteamyRoot Were you thinking of a country in particular ?
 
Balarka Sen
Balarka Sen
Ok, I have to stop myself from further meme-hunting and get some work done.
 
Daminark
Daminark
Are they not synonymous? @Balarka
 
Balarka Sen
Balarka Sen
Don't. Massive amount of will power is needed to do this
over 9000
2
 
SteamyRoot
SteamyRoot
@Astyx Mine :P
 
Balarka Sen
Balarka Sen
ok i closed tab
 
Astyx
Astyx
9:50 AM
Legit
 
SteamyRoot
SteamyRoot
Well, hey, we get loads of Dutch students flocking to study here because of how cheap it is.
 
Astyx
Astyx
If I get the admission to the schools I want I'll actually get paid to study
 
SteamyRoot
SteamyRoot
Okay, yeah, we don't have that. Unless you go to the military
 
Astyx
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
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
Balarka Sen
9:57 AM
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
Daminark
Could it maybe be that they don't want someone just running away to another country upon finishing? @Steamy
 
Astyx
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
SteamyRoot
@Daminark Running away to any other job, really.
 
Balarka Sen
Balarka Sen
 
Astyx
Astyx
@SteamyRoot Yeah exactly the same
Which to be fair is understandable
 
Daminark
Daminark
10:00 AM
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
Astyx
@Daminark Only that particular subset
 
Daminark
Daminark
I see. So is there any... categorization, I guess, of them? Like the military in one case
 
Astyx
Astyx
And the fundamental research for the other case (the ENS)
but it's just schools that are regulated by the states really
 
Daminark
Daminark
I see. What does "fundamental" research mean?
 
Astyx
Astyx
That's just a term I use to tell it appart from applied research which is what private companies would do
 
Daminark
Daminark
10:05 AM
Ah
 
Astyx
Astyx
However I know very little, this might be a total misconception
 
user84215
I am ready to do fundamental research.
 
SteamyRoot
SteamyRoot
Science for the sake of improving science, rather than "real-life application"-based.
 
Astyx
Astyx
Kinda
 
Balarka Sen
Balarka Sen
The bourgeois science, one might say
 
Astyx
Astyx
10:07 AM
"Does your research have any relevant application ? If so, then it's useless"
A bit like philosophy
 
SteamyRoot
SteamyRoot
Ew no
 
Daminark
Daminark
Philosophy is like, the dankest subject tho
 
Astyx
Astyx
Never said it wasn't
 
Daminark
Daminark
Yeah I was rebutting Steamy
(removed)
 
Balarka Sen
Balarka Sen
meh I'd rather take literature
 
SteamyRoot
SteamyRoot
10:12 AM
Do you know the difference between a mathematician and a philosopher?
 
Daminark
Daminark
braces What?
 
SteamyRoot
SteamyRoot
A mathematician only needs pen, paper, and a garbage can. A philosopher doesn't even need the latter.
8
 
Astyx
Astyx
Litterature is philosophy to my narrow-minded brain
 
Las Vegas Raiders
Las Vegas Raiders

 The Reading Room

Welcome to chat for literature.stackexchange.com — Read any go...
1
 
user84215
You should first define mathematics and philosophy.
 
Astyx
Astyx
10:15 AM
Typical mathematician
 
Las Vegas Raiders
Las Vegas Raiders
Define define
 
Daminark
Daminark
Define " "
 
Astyx
Astyx
waiting for Balarka to jump on the occasion
 
Balarka Sen
Balarka Sen
 
user84215
please take a starting point.
 
Balarka Sen
Balarka Sen
10:16 AM
@SteamyRoot
 
Astyx
Astyx
Gromov looks zen to me
 
Balarka Sen
Balarka Sen
Gromov is fantastic
 
Fruitful Approach
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.

Fruitful ApproachLet $S$ 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 \...

elementary-number-theory prime-numbers modules integers prime-gaps
 
Astyx
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
Secret
10:51 AM
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
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
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 AM
is there anyone awake?
 
Dattier
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
Secret
11:52 AM
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
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
RE60K
12:10 PM
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
Simply Beautiful Art
@Secret Nah, that's not how it works lol
 
SteamyRoot
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
Simply Beautiful Art
 
Secret
Secret
ah ok
@SimplyBeautifulArt That looks really oscillatory
 
Simply Beautiful Art
Simply Beautiful Art
It appears to approach $e^x$ everywhere except a small region $[0,0.58]$
 
Secret
Secret
12:20 PM
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
SteamyRoot
It's not like that Desmos plot is anywhere near correct.
 
Astyx
Astyx
Depends on the topology
 
Simply Beautiful Art
Simply Beautiful Art
Actually, I'm pretty sure the limit is $e^x\mod(x^2+1)^2$
 
SteamyRoot
SteamyRoot
Modding out $(x^2 + 1)^2$ means whatever you're left with is at most a degree $3$ polynomial...
 
Akiva Weinberger
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 PM
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
Akiva Weinberger
OK, yeah, I see where I messed up. Not sure how to fix it, though.
 
Secret
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
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
Secret
if $P(x)=0$, then $\text{Root}\circ (P)(x)=x$
 
SteamyRoot
SteamyRoot
I guess all you then have to do is calculate the limits for each coefficient. Can't be bothered.
 
user84215
12:53 PM
Is there anyone who wants to read Berger's book together ?
 
Secret
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
SteamyRoot
Pretty sure $0^2 \equiv 3^2 \mod 3$.
 
Secret
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
Daminark
 
SteamyRoot
SteamyRoot
Lol
 
Balarka Sen
Balarka Sen
1:01 PM
 
Secret
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
Secret
I have not read any books except munkres, definitely cannto help you on that
 
Mathmore
Mathmore
Hello!!
 
Secret
Secret
The issue is that polynomials are nonlinear maps wrt their arguments, which makes it very hard to decompose them
 
Mathmore
Mathmore
1:05 PM
Want a little help in understanding this proof of Gallian.
 
Secret
Secret
and then, there are ireeducible polynomials...
 
Mathmore
Mathmore
I found he proves it like this : $\alpha \notin H \implies \alpha \in H$. Thus it seems paradoxical.
 
SoumyoB
SoumyoB
happy Independence day to all Americans on here
3
 
Secret
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
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
SoumyoB
1:11 PM
hmm no Americans on here right now?
 
SteamyRoot
SteamyRoot
And if you mod out a product by one of its factors, it's obviously zero.
 
Secret
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
LegionMammal978
@Secret So what are you trying to do?
 
SteamyRoot
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
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
Leaky Nun
1:20 PM
@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
Secret
Right then $P(x) \equiv 0 \mod d(x)$ is not helping us since all the information is in $Q(x)$
 
Leaky Nun
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
Mathmore
@LeakyNun Oh!
 
Brown Ninja
Brown Ninja
@Secret Thank you very much. I was away, so did not see your message earlier.
 
Mathmore
Mathmore
@LeakyNun I see. I had never seen proof of such type.
 
Leaky Nun
Leaky Nun
1:30 PM
@Mathmore I didn't bother to look at the proof :p
 
Mathmore
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
Leaky Nun
@Mathmore well then you need to learn logic properly :p
 
Mathmore
Mathmore
@LeakyNun any suggestion for the logic reference book? :/
 
Leaky Nun
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
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
Leaky Nun
1:36 PM
@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
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
Leaky Nun
@Mathmore oh, right
Lagrange theorem
 
Mathmore
Mathmore
@LeakyNun yes! :) So what understanding I have about $\implies$ is not accurate. :/
 
Leaky Nun
Leaky Nun
@Mathmore so look it up online, or ask me
 
Mathmore
Mathmore
@LeakyNun first I choose to ask you. :D
 
Leaky Nun
Leaky Nun
1:40 PM
@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
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
Mathmore
@LeakyNun Okay.
 
Leaky Nun
Leaky Nun
@Secret yes. sufficient and necessary.
but the polynomials must have integer coefficients
 
Mathmore
Mathmore
This I know. Because I have done only this type of proofs.
 
Leaky Nun
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
Mathmore
1:44 PM
But we know $1 \neq 2$ right?
 
Secret
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
Leaky Nun
@Secret really?
@Mathmore yes, but that isn't what I asked
 
Semiclassical
Semiclassical
hi chat
 
Secret
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
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
Mathmore
1:46 PM
@LeakyNun Ah I see. I think that statement is true after substituting $1=2$. Then we get $2+2=4$. Right?
 
Leaky Nun
Leaky Nun
@Mathmore you're using your own definition again ^^
 
Secret
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
Mathmore
@LeakyNun Okay.
 
Secret
Secret
arbitrary ordered? in that case I don't see why the majorisation has to hold, unless its proof is not obvious
 
Leaky Nun
Leaky Nun
@Secret do you think this is legible to anyone?
@Mathmore so...?
 
Secret
Secret
1:49 PM
though I did proved that the matrix $SPS^{T}$ is also doubly stochastic by fiddling with the indices, not sure how that helps
 
Mathmore
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
Brown Ninja
with the definition of 'Orthogonal ( :p )' $S$ I provided?
 
Secret
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
Leaky Nun
@Mathmore eh, it's just a simple example...
 
Brown Ninja
Brown Ninja
Right. I will read it carefully.
 
Semiclassical
Semiclassical
1:51 PM
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
Secret
well, except that the matrix of ones is a column vector or row vector
 
Leaky Nun
Leaky Nun
@Mathmore let's say we know what those symbols mean.
 
Mathmore
Mathmore
@LeakyNun Honestly I have never studied logic. But I clearly see how dumb I am too ignore such an important topic. ;(
 
Semiclassical
Semiclassical
woops, yes.
should be a column vector of ones.
 
Secret
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
Leaky Nun
1:53 PM
@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
Mathmore
@LeakyNun Okay. Then $1+2=3 \neq4$. ?
 
Leaky Nun
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?

Gavin Z.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
 
Mathmore
Mathmore
@LeakyNun So if the hypothesis is false, then it's implication is true?
 
Leaky Nun
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
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
Semiclassical
1:57 PM
Another way to write S is as the direct sum of uu^T matrices
 
Secret
Secret
yup
 
Astyx
Astyx
Any cool math today ?
 
Brown Ninja
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
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
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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