Mathematics - 2017-10-26 (page 3 of 4)

Secret

06:26

[Random]

To be checked:

1. Hilbert space where the underlying set is an infinite dedekind finite set
2. Vector space where the underlying set is an infinite dedekind finite set
3. Poncaire lemma, does it hold in infinite dimensional manifolds

For 3:

6

Q: Poincaré lemma in infinite dimensions

Hi everyone, Is the Poincaré lemma true in infinite dimensions? Here's a precise statement: Let $X$ be a Banach (or maybe Hilbert) vector space, $U$ a simply connected open set in $X$. Is it true that every closed (smooth) $1$-form on $U$ is exact? Thanks!

some aside:

26

Q: Important results that use infinite-dimensional manifolds?

Are Banach manifolds (or other types of infinite-dimensional manifolds) just curiosities, or have they been utilized to prove some interesting/important results? Where do they turn up? Important examples?

manifolds dg.differential-geometry fa.functional-analysis

Brody

@copper.hat Eh, the professor in question had some valid points imo. Her phrasing of certain statements has me iffy... but to her defense, much of the backlash is made purely of straw men.

Tobias Kildetoft

@Brody Hmm, I haven't seen any of that. What is it about?

Secret

06:47

[Random]

Brody

@TobiasKildetoft (unless copper.hat meant something else) it's about a UIUC professor's statements made recently in an article. if i'm not mistaken, basically saying that the current state of math education disadvantages minority students and perpetuates white privilege

Secret

Suppose we want a predicative version of analysis (here predicative means all constructs must be provable to exist and that there exists no constructs which is not build from constructs that are already built) meaning that we can only use a subset of reals which is dense and consists of computable reals (and possibly a few irrationals that can be computed with one turing jump), is it always possible to use a subspace topology so that almost all theorems that relies on continuity will not be

affected? I think there should be a lot of topologies that can maintain the notion of continuity and some weak notion of smoothness

[More random]

Corollary chains:

(Note that whatevr I said next is not sure if it is known in the literature)
Def: A corolloary chain is a countable sequence of corollaries not linked by induction such that the proof of each one relies on the result of the previous one

Write: Corollary$^n$ = Corollary of corollary of ... of corollary

Brody

Hmm... the more I'd read, the sillier it all seemed. I'll leave that at that since the kind of language involved tends to be an automatic trigger for political shitstorms :/ @copper.hat

Secret

07:02

::Insert political shitstorm::

(NB Corollary chain is not my idea, but inspired from one our our lecturer writing Corollary of corollary of corollary of corollaroy

Pretty sure somewhere in categorical logic, this can be formalised

Tobias Kildetoft

07:53

Making good progress on my representation theory notes. Just finished up Schur's lemma and the fact that all irreps of abelian groups are $1$-dimensional

Daminark

@TobiasKildetoft nice

Tobias Kildetoft

But then, I should be, since they will be needed in a few weeks

Daminark

Ah, you'll be getting to them soon, I see. Where are you now in the class?

Tobias Kildetoft

Started rings this week

Then yesterday I skipped forward a chapter in the book in order to introduce polynomials so I could have a good source of examples

Next week, we will be considering unique factorization and related topics

Daminark

Oh so it's a one semester groups/rings class? That's nifty

Tobias Kildetoft

07:59

yeah, one semester. It covers some number theory (the stuff involved in RSA plus quadratic reciprocity), group theory (up to Sylow), ring theory (mainly related to unique factorization stuff), polynomials (partially also used for quadratic reciprocity)

And this year, it will end with some representation theory. It used to be Gröbner bases, but I know too little about those to feel comfortable lecturing about them when there are other good options

Daminark

Oh that's pretty good. Here it's like, groups for the first quarter (through Sylow), rings/modules second quarter, fields/Galois third

So far we've been going a bit slow in group theory though

Tobias Kildetoft

The lecturer before me also did representation theory, but I think he only ended up doing it one time, since last time the course ran, the lecturer was replaced about half-way through with a guy who preferred to do Gröbner bases (having written the course book which has a lot on them)

Daminark

This Friday we'll have finished 5th week, essentially halfway through the class, and we're only just defining A_n

Tobias Kildetoft

We had a total of about 5 weeks of group theory

Daminark

I think I'd prefer that

I think part of why I'm a bit eh on this whole situation is that the first couple weeks were very heavy on geometry

Like we did a bunch of rotation and dihedral groups, which took a bit, and I was just like eh

Not really my style

Tobias Kildetoft

08:06

When I learned algebra originally, the course I had was 7 weeks in total. It covered groups, rings, polynomials, quadratic extensions

But it did assume some familiarity with permutations, since that had been covered in a previous course on discrete math

Daminark

Ah, yeah no this doesn't assume any discrete, just calculus and linear algebra. And it seems like they aren't making much use of the linear algebra

On the second day of class when we were just listing off examples of groups, GL_n came up, and today we did SL_n and PGL_n

Tobias Kildetoft

linear algebra does not tend to come into the picture much in introductore abstract algebra, other than maybe as examples

Daminark

Yeah I guess in any event this class isn't supposed to cover much by way of representations

But yeah I think we might have a bit of time after Sylow. We've got 5 weeks left, we've already done basics/examples, actions, homomorphisms, quotients, like we could do Sylow next week

Balarka Sen

@Daminark So given a pair $(X, A)$ where $A \subset X$ and $x_0 \in A$ is a basepoint you can define the relative homotopy group $\pi_n(X, A, x_0)$ as follows: consider the class of all based maps $f : (D^n, p) \to (X, x_0)$ such that $f(\partial D^n) \subset A$, and define the equivalence relation $f \sim g$ if $f$ and $g$ are homotopic through a family of maps $f_t : (D^n, p) \to (X, x_0)$ such that $f_t(\partial D^n) \subset A$.

Tobias Kildetoft

@Daminark I don't think most places cover that topic at all until graduate level courses

@Daminark If it had not been the topic of choice for the previous lecturer, I might not have chosen it either

Balarka Sen

08:14

I should say that $p$ is a fixed choice of a point on the boundary of the disk $D^n$

If $A = x_0$, for example, then the relative homotopy group is the same as $\pi_n(X, x_0)$ because all of the boundary maps to $x_0$

Tobias Kildetoft

@Daminark Yeah, Sylow just needs some actions for the proofs

Balarka Sen

This gets a natural group structure. Namely, if $f, g : (D^n, \partial D^n, p) \to (X, A, x_0)$ are two such maps, consider $D^n \vee D^n$, wedge of two copies of $D^n$'s along a boundary point $p \in \partial D^n$ on each, and define $f \vee g : D^n \vee D^n \to X$ by setting it to be $f$ on the first copy of the wedge and $g$ on the second copy of the wedge (this is continuous because they agree on the only point of ambiguity $f(p) = g(p) = x_0$)

And finally consider $h : D^n \to D^n \vee D^n \to X$ where the first map simply quotients the equatorial $(n-1)$-dimensional subdisk of $D^n$ containing $p$. (Pinching the diameter of a 2-disk gives wedge of two 2-disks)

It's not hard to see that $h(\partial D^n) \subset A$ and $h(p) = x_0$. That means we have again a map $h : (D^n, \partial D^n, p) \to (X, A, x_0)$. The multiplication map is then defined as $[f]*[g] = [h]$

@Daminark Is this clear?

Daminark

I'm still in processing mode but yeah I think so

Balarka Sen

Feel free to ask if you're confused, it took me a while to digest relative homotopy groups (I still don't know how to compute them efficiently)

Daminark

Okay so just to be sure I get this right, the reason why if $A = x_0$, then the relative homotopy group is just $\pi_n(X,x_0)$ is because if the boundary maps to a point, we have a continuous map from $D^n/S^{n-1} = S^n$ into the space, yeah?

Balarka Sen

08:24

Yes, exactly.

Daminark

And since the basepoint $p$ is the point that we're wedging on, you know that since both $f$ and $g$ respect it, there's no ambiguity, and its restriction to each sphere is continuous, so you're good on that note

Balarka Sen

But in general, if $A$ is bigger than a point, the maps $D^n \to X$ are allowed to be nonconstant on $\partial D^n$, as long as $f(\partial D^n)$ stays inside $A \subset X$

@Daminark Truly

Daminark

And we see that $h(\partial D^n) = f\vee g(\partial D^n \vee \partial D^n)$?

Balarka Sen

\vee, not \wedge, but yes, exactly

Daminark

At which point we use the fact that $f$ and $g$ both send the boundary of the disk into $A$, okay I'm chill with that

Balarka Sen

08:29

Yuppers

Daminark

Okay so I see why we have the operation, can I try to guess how the axioms are gonna play out?

Balarka Sen

Feel free. It's not hard to check.

Daminark

Associativity is just, you're wedging 3 copies of $D^n$ when you sorta factor through that construction, but it doesn't matter which 2 you wedge first since you're gonna take the wedge along $p$ anyway, at which point everything is symmetric

Identity is constant map, which is good because if you take the intermediate wedged stage where it's constant on one copy and $f$ on the other, you'd reverse engineer it to the single disk before you modded out by the equator, at which point you'd have one hemidisk correspond to what $f$ will be, the other is constant, and they agree on the equator

But now you have this hemidisk which you nullhomotope the living shit out of back to $p$, so you should be good

Balarka Sen

Well, pretty much. But remember it took some work to show the group multiplication is associative in $\pi_1(X, x_0)$. The point is even though the picture is wedging three loops in $X$ based at $x_0$, the speed of $(f * g) * h$ and $f * (g * h)$ are different. Eg in the first $f$ is traced in $1/2 \cdot 1/2 = 1/4$ of the time and in the second $f$ is traced in $1/2$ the time

Daminark

Yeah, this is all up to homotopy

Balarka Sen

08:36

Yeah

@Daminark Quality proofs there

Daminark

The inverse should be, forgive my vagueness, "the symmetric map about the wedge point"

Some top-tier mathematical exposition, that was

Balarka Sen

Yep, sounds about right

Daminark

But yeah the idea being, if you sorta trace a point on one of the disks to the corresponding one on the other, you just sorta take that central point that maps to the basepoint and then spread it out until it swallows the whole disk

Or I guess this is also like, reverse engineered to see that half the disk looks like the map, but then you have a symmetry about the equator, so you should be able to nicely push it back up

Balarka Sen

Yeah I see what you mean. I think if $f : D^n \to X$ is your map, the inverse is $g : D^n \to D^n \to X$ where the first map is the antipodal map on $S^{n+1}$ taking upper hemisphere to lower hemisphere, and the second is $f$

Daminark

Okay I'm happy enough at this stage

Balarka Sen

08:41

Alright me too

So, now, the long exact sequence. This is where the #realmafs begin

I claim there is a long exact sequence $\cdots \to \pi_n(A, x_0) \to \pi_n(X, x_0) \to \pi_n(X, A, x_0) \to \pi_{n-1}(A, x_0) \to \cdots$

Daminark

Bold claim

Big if true

I'm loling IRL at this, but yeah don't mind me :P

I always throw in random comments like so

Balarka Sen

lolol

No I love this

Learn math through memes, is my motto

Daminark

Truly the best way to go about life

Balarka Sen

So, the first map $\pi_n(A, x_0) \to \pi_n(X, x_0)$ has an obvious candidate: $i_*$ where $i : A \to X$ is the inclusion map

Just push a map $(S^n, p) \to (A, x_0)$ to $(S^n, p) \to (A, x_0) \stackrel{i}{\to} (X, x_0)$

Daminark

True that

Balarka Sen

08:48

The second map $\pi_n(X, x_0) \to \pi_n(X, A, x_0)$ also has an obvious candidate: think of $\pi_n(X, x_0)$ as the relative group $\pi_n(X, x_0, x_0)$ and look at the map induced from inclusion $j: (X, x_0, x_0) \to (X, A, x_0)$ in the second component

Daminark

Oh that's slick, right yeah because of what we talked about earlier

Balarka Sen

Right

And the map $\pi_n(X, A, x_0) \to \pi_{n-1}(A, x_0)$ that I think works is, take a map $(D^n, S^{n-1}, p) \to (X, A, x_0)$, and look at the map $(S^{n-1}, p) \to (A, x_0)$ that you get from forgetting about the first component in the triplet

Namely, literally take $[f] \in \pi_n(X, A, x_0)$ and look at $[\partial f] \in \pi_{n-1}(A, x_0)$

Just $f$ restricted to the boundary of the disk $D^n$

Daminark

Okay so verifying that this is exact then

First two maps are p clear

Balarka Sen

Right...

ABcDexter

hey

Daminark

08:58

So the second and third map because if you map via the inclusion of x_0 in A, then even restricting the map to the sphere won't change that nullhomotopicity

Balarka Sen

True. And conversely, if $[\partial f] = 0$, then $\partial f$ is nullhomotopic as a map $S^{n-1} \to A$. That means $f$ is homotopic as a map of pairs to a map $(D^n, S^{n-1}, p) \to (X, A, x_0)$ which is constant on $S^{n-1}$. So $f$ is image of the map induced from $j$

ABcDexter

a quick question:

Let X be a geometric random variable with parameter p. Find the probability that X≥10 ?

Balarka Sen

That means kernel of the third map is image of the second, right?

Daminark

Yup

Then finally, why the first map for the next group checks out with the third

Balarka Sen

Right. Ah, that's easy. Because suppose $[f] \in \pi_{n-1}(A, x_0)$ is in the kernel of the first map of the next group

Then $i_* f : S^{n-1} \to X$ is nullhomotopic

Daminark

09:04

Oh I misread lmao

Balarka Sen

Maybe I mistyped?

(Fixed)

I want to show $[f]$ is in the image of the third map

Daminark

Nah it was me, I forgot that this is just π_n and π_(n-1)

Balarka Sen

Ah OK

Daminark

Like for the same set A

So that is actually easy

Balarka Sen

Uh, no, I am not entirely sure what you mean

Daminark

09:06

As in, I forgot that we're just mapping one homotopy group of A to another, as opposed to relative business

Balarka Sen

I have this little piece of the LES $\pi_n(X, A, x_0) \to \pi_{n-1}(A, x_0) \to \pi_{n-1}(X, x_0)$ which I want to show is exact at the center, right?

Daminark

So it's more clean

Balarka Sen

Oh

Right. There's still a sneaky construction though

Do you want to do it?

Daminark

I'll try it out and see

ABcDexter

Did anyone try this? chat.stackexchange.com/transcript/message/40779770#40779770

Daminark

09:13

Okay so if you are in the kernel of the first map of the next segment, you have a nullhomotopic map from the sphere into X which you can then extend because this feels like a "degree 0" thing

Balarka Sen

You have the right intuition

Right, so if I have $[f] \in \pi_{n-1}(A, x_0)$ that's in the kernel, $[i_\star f] = 0$, so $i_\star f : S^n \to X$ is nullhomotopic

But choose a nullhomotopy $S^n \times I \to X$ that starts at $i_* f$ and ends at the constant map $S^n \to X$ mapping everything to $x_0$

Tobias Kildetoft

There, completed one of the more tricky parts given that I don't want to introduce the group algebra. I think I have managed to make it no harder than the corresponding proof using the group algebra

Balarka Sen

That gives a map $S^n \times I/S^n \times \{1\} \to X$

(Because the nullhomotopy is constant on $S^n \times \{1\}$)

Tobias Kildetoft

Essentially, I need the center of the group algebra, without the algebra.

Balarka Sen

But this thing is just a cylinder with the top pinched up, which is a disk $D^n$. So that gives a map $D^n \to X$

And the boundary is exactly $i_* f$

This is then the relevant element in $\pi_n(X, A, x_0)$ image of which under $\partial$ is $[f]$.

Bam, done

Daminark

09:18

Okay sick

We should continue this tomorrow because it's 4AM and tired

ABcDexter

okay, Summation and infinite GP done.

thx.

Balarka Sen

@Daminark For sure, I should sleep too lmao

Daminark

But yeah up to staring at this more and all I'm chill

Balarka Sen

It's pretty cool stuff

I literally know nothing about homotopy theory beyond this point upto quoting blackbox theorems, by the way

So you know everything I know now

Daminark

I mean you probably know like, what a fibration is

Balarka Sen

09:21

Oh right

I forgot about that

Yeah perhaps the fibration long exact sequence

BTW, I think $\pi_n(X, A, x_0)$ has something to do with $\pi_n(X/A, A/A)$ but I don't know the relevant theorem (I think this is the homotopy excision theorem). There is a map $\pi_n(X, A, x_0) \to \pi_n(X/A, A/A)$ by sending $(D^n, S^{n-1}, p) \to (X, A, x_0)$ to $(D^n/S^{n-1}, S^{n-1}/S^{n-1}) \to (X/A, A/A)$

I think for CW complexes $X$ and subcomplex $A \subset X$ under some dimension conditions this map becomes an isomorphism

Hopefully we'll eventually end up learning that too

Daminark

I see

Well, with that, good night!

Also @Tobias lmao

Balarka Sen

Night

Tobias Kildetoft

12:00

I wonder how few examples I can make do with to keep referring to in order to get examples of all the possible phenomena I cover in these notes. So far I seem to be able to make do with like 3 or 4.

BAYMAX

Hi

I had a probability question!

$X$ is a non-negative integer valued random variable, with $E(X) = 1$ and $E(X^2) = 3$ then $\sum_{i=1}^{\infty}iP(X \geq i) = ?$

so $\sum_{i=1}^{\infty}i P(X =i) = 1$

and also $\sum_{i=1}^{\infty} i^2 P(X =i) =3$

oh

let me try i got an idea

ok got that finally we will reach $\frac{E(X)+E(X^2)}{2}$

MatheiBoulomenos

12:32

@TobiasKildetoft it is true that we have $p-1\geq n$ in the last part of the exercise on Sylow theory gave me? I think I have a proof, but I'm not sure if it's right, as you would probably have asked to prove the stronger inequality if it works

Tobias Kildetoft

@MatheiBoulomenos Yes, that is correct (I don't recall right now how I meant for the original inequality to be shown)

So it might be that either can be proven in the same way

The inequality you have there is best possible though

Secret

[Random]
I need a mathematical joke such that an at least 3 word acronym of that mathematical concept can be arbitrarily permutated and the result is still a mathematical acronym, but different

Tobias Kildetoft

(well, in general. In fact, the possible value of $n$ is uniquely determined by $p$ and $q$)

or rather, for each pair of primes, there is one $n$ that works for order $pq^n$ and one that works for $qp^n$

MatheiBoulomenos

ah, that's really cool. I wonder if that $n$ is somehow related the factorization of the $q$-th cyclotomic polynomial modulo $p$. The proof I have makes me think it is

Tobias Kildetoft

@MatheiBoulomenos $n$ is the order of $p$ mod $q$ (or vice versa)

this is related to the existence of irreducible representations of degree $n$ of the cyclic group of order $p$ over the field with $q$ elements (again, or vice versa)

MatheiBoulomenos

12:41

@Tobias oh yeah, of course and that's also the smallest $n$ such that $\Phi_q$ splits over $\Bbb F_{p^n}$

Tobias Kildetoft

right

MatheiBoulomenos

My representation theory course worked only over $\Bbb C$, that's why I didn't really think about it like a rep theory problem, I formulated the argument in terms of $\Bbb F_q[x]$-modules. You're comment right now made me realize the connection between simple modules over $\Bbb K[x]$ that are annihalted by some cyclotomic polynomial and irreducible representations of cyclic groups, thanks!

Tobias Kildetoft

@MatheiBoulomenos Subgroups like in this problem is a common way for representations over finite fields to come up when studying finite groups

MatheiBoulomenos

I see. I really need to work more on my representation theory I guess

Tobias Kildetoft

One nice thing one can do if the group is $p$-solvable is use the Fong-Swan theorem

This essentially gives an ordinary characters corresponding to the Brauer character of any of these actions on such subgroups, with some control over how $p'$-elements act

MatheiBoulomenos

12:59

There's yet so much to learn about finite group theory ... but when I learn ring theory I know that it is useful for algebraic geometry and algebraic number theory, whereas I don't really know if advanced group theory is going to help me with that (it probably is, I just can't see how right now)

Tobias Kildetoft

@MatheiBoulomenos finite group theory in itself does tend to be a bit insular

but not entirely of course

I started out with finite groups, but I had to move to a new place when I did my PhD, and my advisor for that was more into algebraic groups, so that ended up being where I did most of my research for the PhD. Now I am splitting my time something like 50/50 between algebraic groups and higher representation theory, with a slight overlap

Akiva Weinberger

How would we even prove that $\int_0^\infty t^xe^{-t}\operatorname d\!t$ is increasing for $x>0$

MatheiBoulomenos

@TobiasKildetoft I think I read in the AMS review of Eisenbud's commutative algebra text that representation theory is relevant for algebraic geometry. But I don't know what kind of representation theory was meant exactly

Tobias Kildetoft

@MatheiBoulomenos depends on what one wants to do in algebraic geometry

Akiva Weinberger

Oh wait I think you could use Hölder

Secret

13:06

0

Q: Gamma function is strictly positive and increasing

Rudin defined Gamma function for $s>0$ as integral $\Gamma(s)=\int \limits_{0}^{\infty}x^{s-1}e^{-x}dx$. How to prove that $\Gamma(s)$ strictly positive function and monotone increasing? Can anyone show please a rigorous proof!

real-analysis

?

Tobias Kildetoft

13:27

Well, that's the outline for the proof of Maschke's theorem done. Now to write it up properly as well.

Secret

[Random]

Ever since Akiva became a horoshpere, he rarely visits this dimension slice of the chatroom

MatheiBoulomenos

@Tobias the proof was just averaging over a projection right? Plus some general non-commutative algebra facts on equivalent definitions for semisimple modules

Tobias Kildetoft

@MatheiBoulomenos No, just the first part really

(well, depending on how one formulates the result of course)

I will not be using any non-commutative algebra for these notes, as the course does not cover non-commutative rings at all (nor does it cover modules)

So I have decided to completely avoid the group algebra

MatheiBoulomenos

I see. For me, the module point of view brought a lot of clarity, but I guess you can do without

Tobias Kildetoft

Yeah, it does do some good, but not at the cost of having to introduce so much new stuff (I won't have that much time for the topic)

MatheiBoulomenos

13:40

For me, induced representations make a lot more sense as a tensor product than if you define them elementarily

Tobias Kildetoft

@MatheiBoulomenos Sure, but I won't get that far (well, except inducing the trivial representation of a subgroup since that is just a permutation representation)

MatheiBoulomenos

So this is an general abstract algebra course? We didn't do representation theory in those at all. (Though we mentioned it when we were doing Artin-Wedderburn)

user1770201

14:13

Do we have that $L^1(R) \subset L^2(R)$?

Martin Sleziak

I have added a few links about inclusions between L_p and L_q in the functional analysis chat room.

Secret

Crisis in the Foundation of Mathematics | Infinite Series

►

anakhronizein

"Show that x in GF(27) has multiplicative order 26 by expressing x^3 ,x^4 ,... as quadratic polynomials in x." <-- do you think they mean by finding x^3, x^4, ... x^30 and showing they are all different, or what are they trying to ask the reader to do?

user1770201

@MartinSleziak I checked out that link but can't figure out how to use that to determine whether L^1(R) is a subset of L^2(R)

But I think it must so, since 1/x is in L^2(R) but not in L^1(R)

Secret

5

Q: What is the future of Set Theory if it is NOT the foundation of Mathematics?

Homotopy Type Theory has been receiving a lot of hype during the last couple of years and been hailed as the most effective foundation of all Mathematics compared with Set Theory. My question: If HOTT is the foundation of all mathematics, will that pronounce death of Set Theory research ? or els...

set-theory homotopy-type-theory

[Superrandom]

What if we consider the foundation of foundation systems. I am not sure if it is well defined

Imagine a universe where type theory, set theory and other foundation systems exists as constituents. How far can that push the limit of comprehension of the nature of mathematics

7

Q: Escaping Gödel's proof

Is there any way in which a reasonably strong foundation of mathematics can get around the hypotheses of the incompleteness theorems?

logic foundations

1

Q: Shortcomings of "The Continuum" by Hermann Weyl?

I just read that Hermann Weyl used a certain type of type theory as the foundation of mathematics in his book "The Continuum" and he was able to derive calculus without the use of infinite sets. Naturally, I'm going to read it ASAP but I wanted to ask right away what it's fundamental flaws are, s...

foundations

The concept of (potential) infinity kinda exists whenever something is iterated

But it might be possible to forget about actual infinity depending on the scope of mathematics the foundation want to cover

> Blass's theorem is a very strong one indeed. If the axiom of choice does not hold then there is a vector space without a basis. It is unusual to be able and tell which vector space it is (unless assuming more, or constructing the model directly).

math.stackexchange.com/questions/103743/….

Faust

14:49

whta wierd monring

anakhronizein

15:14

Hi faust, how have you been?

Faust

STresseded

Secret

The proportion of the weird population in maths chat is dropping in recent months, followed by the rise of non-weird populations. If this trend continues it is expected to affect the operation of The Labs

anakhronizein

Why are you stressed, Faust?

Evinda

15:32

Hello!!! Does it hold that if $\lim \frac{1}{n} \sum_{k=0}^n x_k=0$ that then $\lim x_n=0$ ?

Silent

Please someone explain why this is true here: The sum $\frac1n+\frac1m$, with least one of $\frac1n$ or $\frac1m$ less than $\epsilon/2$, must be at a distance of at least $\epsilon/2$ from $x$.

Secret

forbes.com/sites/startswithabang/2017/10/26/…

[Philosophy]

It seems interesting how rarely the unknown is hostile. A lot of unknowns that are uncovered by science often turned out to be beneficial

It is not even clear if the behaviour of the unknown follows certain rules, or even can be captured by mathematics

Leaky Nun

15:48

@Evinda try $x_n = (-1)^n$

Evinda

How do we calculate $\lim_{n \to +\infty} (-1)^n$ ? @LeakyNun

Liad

$\not a$

anakhronizein

That's the thing, it is just the sequence -1,1,-1,1, ..., @Evinda

Semiclassical

Look at the sequence of partial sums

Liad

$\neg a$

$\or$

anakhronizein

15:57

Liad, do you need assistance?

Liad

what is the expression for "or" and "and" ?

yea :P

anakhronizein

\vee and \wedge

Liad

$a \vee b$

thanks

Evinda

Ah the limit does not exist since $(-1)^{2k}=1$ and $(-1)^{2k+1}=-1$, right? @LeakyNun @anakhronizein

Liad

and for iff?

anakhronizein

15:57

\leftrightarrow

Yes, @Evinda

Liad

$\rightarrow$

alright, i need to prove that i cant express $\neg$ with only $\vee , \wedge , \rightarrow , \leftrightarrow$ , someone can help?

Evinda

if it would exist, it would have to hold that $(-1)^{2k}=(-1)^{2k+1}=l$ for some l, for every $k \in \mathbb{N}$, right? @anakhronizein

Silent

yes

Semiclassical

You can also look at the partial sums

anakhronizein

At the very least the subsequences have to converge to the same thing.

Evinda

16:03

And since $x_{2k}=1$ and $x_{2k+1}=-1$, the sequence $x_n$ does not converge, right? @anakhronizein

anakhronizein

Yes.

Evinda

Okay

Leaky Nun

@Evinda it’s a counter example

Evinda

Yes, I see... @LeakyNun

And how do we show that $\sum_{k=0}^n (-1)^k=\frac{(-1)^n+1}{2}$ ?

anakhronizein

By cases.

n is either odd or even

Evinda

16:11

It holds that $\sum_{k=0}^n (-1)^k= \left\{\begin{matrix}
(-1)^0+(-1)+\dots+(-1)^{2u+1}=0 &, \text{ if } n=2u+1 \\
(-1)^0+(-1)+\dots+(-1)^{2u}=1 &, \text{ if } n=2u
\end{matrix}\right.$, right? @anakhronizein

anakhronizein

16:23

Yes, but that is not saying much so far.

user193319

17:16

Problem I am working on: $S_3$ is not the direct product of any family of its proper subgroups.

When they say "is not the direct product," may I presume they mean "not isomorphic to any direct product..."?

anakhronizein

Yes.

user193319

@anakhronizein Thanks!

Leaky Nun

Best proof of $\sqrt5 \notin \Bbb Q(\sqrt2,\sqrt3)$?

MatheiBoulomenos

@LeakyNun discriminants

or Kummer theory

@LeakyNun discriminants involves some computations and some theory

Evinda

17:40

@anakhronizein How else do we justify it?

anakhronizein

Evaluate the partial sums.

For n even, what do you find?

For n odd, what do you find?

Evinda

For $n=2u$, $\sum_{k=0}^{2u} (-1)^k=(-1)^0+(-1)+(-1)^2+(-1)^3+\dots+ (-1)^{2u-1}+(-1)^{2u}=0$.

For $n=2u+1$, $\sum_{k=0}^{2u-1} (-1)^k=(-1)^0+(-1)+(-1)^2+(-1)^3+\dots+ (-1)^{2u-1}=1$ @anakhronizein

Right?

anakhronizein

Yes.

Errr

No.

Wrong way around.

even ---> =1

odd ---> =0

Evinda

18:01

I tried n=4 and I got that the result is 0 @anakhronizein

anakhronizein

1-1+1-1+1 = 0?

Kevin Driscoll

@anakhronizein @Evinda forgetting to start at 0?

Evinda

Oh yes, you are right

Kevin Driscoll

clearly not a computer science major, heh

Evinda

:)

Liad

18:04

i need to prove that $\neg$ cant be expressed with only $\vee , \wedge , \rightarrow , \leftrightarrow$ , someone can help?

Evinda

And so it follows immediately that $\sum_{k=0}^n (-1)^k=\frac{(-1)^n+1}{2}$, right? @anakhronizein

anakhronizein

Yes.

Shahar Shalev

Hey, A little question, I want to know if i look from above the plane or not , how do i check it?
Lets assume that Z= (2x-3y)/5 and the point is (0,0,4)

Balarka Sen

18:16

Hi chat

MatheiBoulomenos

Hi @BalarkaSen

Kevin Driscoll

@Liad So presumably you can write the truth table for $p$ and $\neg p$

Liad

write \negate

alright, what now?

Kevin Driscoll

Then we can write the truth tables for $p \vee p$, $p\wedge p$, $p \rightarrow p$, $p \leftrightarrow p$.

ÍgjøgnumMeg

@BalarkaSen Yo

Kevin Driscoll

18:24

And we see that none of those are the truth table for $\neg p$. So if there's a relation its more complicated. Treat this as the base case and see if you can construct a kind of proof by induction. @Liad

@liad If you include only $\wedge$, $\vee$, and $\leftrightarrow$ then its very straightforward. $\rightarrow$ seems to be the hard part.

Silent

19:05

Please someone explain why this is true here: The sum $\frac1n+\frac1m$, with least one of $\frac1n$ or $\frac1m$ less than $\epsilon/2$, must be at a distance of at least $\epsilon/2$ from $x$.

Abdullah UYU

19:45

About the question math.stackexchange.com/questions/2491265/…, Can I take the $2\sqrt{6}^{th}$ power of both sides and say: $8^\sqrt{2}<9^\sqrt{3}$?

Abdullah UYU

19:55

I think somebody has wrote an answer based on this idea 8 minutes ago :)

Leaky Nun

@MatheiBoulomenos the most elementary proof?

MatheiBoulomenos

20:54

@LeakyNun well, the most elementary proof would be writing $\sqrt{5}$ as a rational linear combination of a $\Bbb Q$-basis of $\Bbb Q(\sqrt{2},\sqrt{3})$ and deriving a contradiction (which would involve squaring things and writing rational numbers as quotient of integers with gcd 1)

But if you use a little theory, you can reduce the computation. Note that if $\sqrt{5} \subset \Bbb Q(\sqrt{2},\sqrt{3})$, then $\Bbb Q(\sqrt{5}) \subset \Bbb Q(\sqrt{2},\sqrt{3})$ is a subfield of degree 2. Now use the Galois correspondence to compute the subfields of $\Bbb Q(\sqrt{2},\sqrt{3})$ and you can reduce the problem to showing that $\Bbb Q(\sqrt{5}) \neq \Bbb Q(\sqrt{m})$ for $m=2,3,6$

That may sound like it's more computation than the other approach, but it's actually less because the bases for $\Bbb Q (\sqrt{m})$ are smaller and it's easier to get to the contradiction

user193319

I am working on a problem in the Linear and Abstract Algebra chatroom. Anyone interested in taking a look at what I have written? chat.stackexchange.com/rooms/13473/linear-abstract-algebra

Daminark

21:12

Hey @Mathei!

MatheiBoulomenos

Hi

Daminark

How's it going?

MatheiBoulomenos

Well, my seminar talk didn't went that well, but other than that, pretty good

and for you?

Daminark

Aw, what went wrong?

And lol today I'm gonna mostly working on bio which... isn't that fun

MatheiBoulomenos

I made a computional error that I couldn't fix on the spot and I didn't get to the interesting stuff because there were too much things to talk about

Daminark

21:16

Oh that's unfortunate

Also do you know of a way to prove that matrices operating on $\mathbb{C}^n$ have an eigenvalue without resorting to the characteristic polynomial/determinant?

MatheiBoulomenos

yes

Daminark

How does that argument go?

Alessandro Codenotti

Hi chat

Daminark

Oi @Alessandro

Alessandro Codenotti

I have sinned @Daminark

I need someone for a few basic category theory doubts :P

Daminark

21:25

What've you done?

Oh, let's hear it

Category theory is the light tho

Alessandro Codenotti

Nah, I just want examples of categories without products or coproducts

I suspect something like measurable spaces or maybe even fields works

Daminark

I think finite groups might work as well

You can't take infinite or free products

Alessandro Codenotti

Hmm, true

MatheiBoulomenos

@Daminark Let $f$ be a linear endomorphism on a finitely generated vector space $V$ over an algebraically closed field $K$. Consider the evaluation homomorphism $K[X] \to \operatorname{End}(V)$ that takes $X \mapsto f$. As $K[X]$ is infinite-dimensional, but $\operatorname{End}(V)$ is finite-dimensional, this map has nonzero kernel. Let $P$ be a polynomial in the kernel, then it has a factor of the form $(X-\lambda)$, now this $\lambda$ is an eigenvalue of $f$.

Alessandro Codenotti

Fair, restricting the cardinality of the object messes up anthing with too big families

Daminark

21:31

Also does the category of ordered simplicial complexes have them?

@Mathei okay that makes sense actually

Semiclassical

22:30

@ACuriousMind since I think it’s be unhelpful for me to say it in h-bar, I’ll say it here: Thaaaaank youuuu

Leaky Nun

22:45

@MatheiBoulomenos nice

Faust

hows my favorite chat?4

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$Mathematics$ Mathematics