Mathematics - 2019-03-30

James Groon

00:09

Got it thanks

maths student

07:31

0

Q: tensor product map

$V\otimes_k k[t] \to L$ by $\ell_i\otimes p \mapsto p\ell_i$. Since the $\ell_i$ formed a $k[t]$ basis for $L$, this map is an isomorphism of $k[t]$ modules. Why this map is isomorphism of $ k[t]$ module? $V$ is $V=\{\sum_i c_i\ell_i : c_i\in k\}$ $k$-vector space. Attempt: Injectivity :if $p\el...

@LeakyNun can you have look at above question ?

maths student

08:03

k[t] basis are k basis for vector space over k?

Right or wrong?

Silent

08:47

I just learned that if field $F$ is characteristic zero, or a finite field, then if $f(x)$ is irreducible polynomial in $F[x]$,then $f(x)$ is separable. Hence I was trying to come up with counterexample of converse: I thought of this: $x^2-5x+6$ is separable but reducible in $\mathbb R[x]$, right?

So, it seems like irreducibility depends on underlying field, but separability is free of field.

Leaky Nun

10:19

@Silent x^2-5x+6 is separable and reducible in any field

and indeed irreducibility depends on the underlying field and separability is independent

Alessandro Codenotti

10:45

I think you need to be somewhat careful with what exactly you mean, $(x-1)(x-3)$ is separable over $\Bbb R$ but not over $\Bbb Z/(2)$

Larry Eppes

hello

Leaky Nun

11:09

@AlessandroCodenotti but that's (x-2)(x-3)

insert mathematicians can't count meme

Alessandro Codenotti

Sure, but I'm arguing that "separability is free of field" requires some care

Leaky Nun

oh

@Silent separability is independent of fields as long as the two fields are "connected" by some chain of morphisms

in the tree of fields

@AlessandroCodenotti how ridiculous that the present tense of "I have elected" is "I elect"

"I have fixed" is "I fix" etc

ahorn

12:16

If I have a series of conjunctions, can I use a big $\wedge$ operator with an index? E.g. $\wedge_{i=1}^n x_i < a$ instead of $x_1 < a\ \wedge\ x_2 < a\ \wedge\ x_3 < a\ \wedge\ \dots\ \ \wedge\ x_n < a$ ? What is the $\LaTeX$ code for that big symbol?

I think I'm right.

$\bigwedge$

Secret

Infinitary logic

An infinitary logic is a logic that allows infinitely long statements and/or infinitely long proofs. Some infinitary logics may have different properties from those of standard first-order logic. In particular, infinitary logics may fail to be compact or complete. Notions of compactness and completeness that are equivalent in finitary logic sometimes are not so in infinitary logics. Therefore for infinitary logics, notions of strong compactness and strong completeness are defined. This article addresses Hilbert-type infinitary logics, as these have been extensively studied and constitute the most...

$\bigvee \bigwedge$

Leaky Nun

14:01

let $X = \sum_{n=1} \zeta_n$ where $\zeta_n \sim U(0,2^{-n})$

then $X = \zeta_1 + \frac12 X$

Secret

$\int X$

Semiclassical

14:21

@LeakyNun that definitely suffices at the level of expectation values

Not so sure it works for computing probabilities

as in, suppose X and Y are identically distributed. It’s true that X and Y will have the same expectation values, but that doesn’t mean X=Y with probability one

Perturbative

14:55

If an author says let $X$ be an infinite dimensional normed vector space, do they refer to infinite dimensional with respect to Hamel or Schauder basis?

Daminark

Should be equivalent

Perturbative

Oh nice, I'll just refer to them interchangeably in that case

Alessandro Codenotti

Hamel if nothing else is specified since Schauder basis exist only in particular cases I'd say?

Daminark

The idea is that a Schauder basis just allows for possibly infinite sums, so if you have a finite Hamel basis that's trivially a Schauder basis, since in finite dim there's no infinite sums.

The reason you care about Schauder is that Banach spaces are always of uncountable dimension (Baire Category thm) but you might have a countable Schauder basis

Perturbative

Okay yeah that makes sense

Perhaps a silly question, but I'm assuming that it turns out every infinite dimensional normed vector space has a Shauder basis correct? (because of something similar to the proof that every vector space has a basis?)

Ohh wait no, just read that there are separable Banach spaces that have no Schauder basis

Bummer

vzn

15:06

@Secret lol cutting edge math + burning man luv it :) have a bunch of awesome links on burning man, we should have a cyber meme party sometime :) ... another idea for an exhibit/ installation: fractals :)

Perturbative

So just to recap, say I have some infinite dimensional normed vector space $X$, then I can conclude it has an infinite dimensional Hamel basis, but I can't say it has a Schauder basis, by what I said above

Alessandro Codenotti

15:21

yeah

Daminark

~~Just allow an uncountable Schauder basis~~

Learner

15:43

Hi, would anyone here take a look at my question?

0

Q: Is this a sound proof of the fact that $a_n <C$ infinitely often?

Let $a_n, b_n, c_n$ be real positive sequences. Assume $a_n\to C>0$, and that $a_n$ is not monotone on any neighborhood of $+\infty$. Suppose also the following lemmata have been proven: Lemma 1. Whenever $a_n\ge a_{n-1}$ one has $b_n\ge c_n$. Lemma 2. For infinitely many $n$ one has $\dfrac{a_...

real-analysis sequences-and-series proof-verification inequality proof-writing

Newbie

@Daminark I want to discuss something small in measure theory

Peter Krauss

Hi @Somos and @DavidK, that help me in answers, I would like to invite you to sign (as co-authors) the article where the information provided was relevant.

Hello everyone, we are a group of software developers, we have finished the first draft of an article about set theory and bit string representation... But we are not experts, we would like to invite any expert from this chat room to be more a co-author, if you take a few minutes to evaluate or review something in the text.

https://docs.google.com/document/d/19_X_QXpY56-72Aw7voWoPXclcGuZi7fosCNDj6uOcQM/

Alessandro Codenotti

16:22

@Newbie just ask, someone will answer

James Groon

16:58

Let's suppose I am proving that some real number K is a supremum of some set S ( subset of the reals) . If I show that K is greater than all of the members and that for any 1>$\epsilon$>0 there is some member x in S such that x<K-$\epsilon$ , is that enough ? I have only restricted $\epsilon$ from the usual characterization.

Ted Shifrin

17:53

@JamesGroon First, you should say $K$ is greater than or equal to all the members. Second, you need to find an $x>K-\epsilon$, not $<$.

Semiclassical

hi @ted

Ted Shifrin

Hi @Semiclassic

James Groon

@TedShifrin Yeah I see I made some mistakes. But the main thing I was wondering if I can restrict $\epsilon$ as 0<$\epsilon$<p, where p>0 (?)

Ted Shifrin

Sure. All you care about is arbitrarily small $\epsilon$s.

James Groon

Nice, thanks.

Semiclassical

17:58

currently trying to make sure I fully understand how mixtures of probability distributions work

Ted Shifrin

Don't ask me nothing about that, @Semiclassic.

Semiclassical

lol

well

Érico Melo Silva

is mixture a formal term

Semiclassical

Yes, but it's a pretty obvious one

first line of the wiki article: "In probability and statistics, a mixture distribution is the probability distribution of a random variable that is derived from a collection of other random variables as follows: first, a random variable is selected by chance from the collection according to given probabilities of selection, and then the value of the selected random variable is realized."

Érico Melo Silva

oh interesting ive never encountered this

weird

Semiclassical

18:00

specifically I'm looking at a case where I start with a random variable $X$

and then I symmetrize it as $X^*$ by sampling $X$ with probability 1/2 and $-X$ with probability 1/2

Érico Melo Silva

oh so it's like ur taking some kind of convex combination i got u

Semiclassical

ya

Ted Shifrin

Heya MR Eric.

Semiclassical

specifically I'm looking at a case where $X$ is a random vector

Érico Melo Silva

hi @Ted

Semiclassical

18:04

and I'm trying to convince myself that, if the random components have a certain amount of symmetry to begin with, then nothing I'm interested changes if I symmetrize it

Alessandro Codenotti

Hi @Ted

Érico Melo Silva

im trying to get back into properly doing math that isnt baby stuff so maybe ill finally read the bryant paper @Ted lmao

Ted Shifrin

hi demonic @Alessandro

Semiclassical

the tricky bit is that I'm interested in stuff like $P(X_1=x)$ and $P(X_1=x,X_2=y)$, but I'm not interested so much in $P(X_1=x,X_2=y,X_3=z)$

Ted Shifrin

Or learn about Chern classes and divisors, @Eric.

Érico Melo Silva

18:04

or that lol

Semiclassical

so I only care about symmetry up to a certain point

Érico Melo Silva

im only taking a few blow off classes and algebraic number theory this quarter so i should have tons of free time since it's my last term

Semiclassical

e.g. I want $(X_1,X_2)$ and $(-X_1,-X_2)$ to be identically distributed but I don't particularly care if $(X_1,X_2,X_3)$ and $(-X_1,X_2,-X_3)$ are as well

Daminark

Hey everyone!

Ted Shifrin

Yeah, @Eric, but you are a senior and get lazy :P

Hi Demonark.

Semiclassical

18:07

So I'm having to be careful

Alessandro Codenotti

Hi @Dami

Érico Melo Silva

im only a little lazy i think

sup @daminark

Daminark

How's everything going?

Érico Melo Silva

i officially committed finally @Daminark so that feels good, now im just reading the book nori posted to refresh my decaying algebrain

Daminark

West or East coast?

Érico Melo Silva

18:13

east

NoOneSpecial

Lets say you feed 2D images together with a human written textual description of each image to a system which analyse it all in some unknown way and creates some unknown internal data-structure for each image together with "fuzzy" weighted many-to-many relationships of "fuzzy" relationships types. Eache new image and text combination can affect all existing relations and data structures, The system can also on it's own create entirely new data structures, which is not a representation of neither an image nor a text.

Daminark

Nice, congrats! I haven't quite committed yet but in my mind I basically know where I'm going

Semiclassical

Were you thinking Wisconsin? I'm forgetting

Daminark

Yup

Semiclassical

cool

swing by Minneapolis some time, lol :P

(I know, not wisconsin, but still pretty close)

Érico Melo Silva

18:17

@Daminark ull go to the same school sougie went to!

Daminark

Yeah I think there's gonna be a bus that goes from Madison right to Minneapolis! And yeah I remember I asked him about it not too long ago and he was like yeah it's a good place but my experience there was irrelevant since it was so long ago

Semiclassical

So here's a question. Suppose I have a random real variable X which is equal in distribution to $-X$. Let the random real $Z$ be a 1:1 mixture of $X$ and $-X$. Is $Z$ identical in distribution to $X$?

My intuition would be yes but

y'know, assumptions

Érico Melo Silva

r they still a good place for analysis? i remember he told me i could apply if i wanted to but i was like nah i want out of the midwest lol

Daminark

They do seem to have quite a few folk in analysis

Érico Melo Silva

@Semiclassical isnt this obvious? if they have the same distributions isnt the formula for the distro of Z just 1/2 p(x) + 1/2 p(x) = p(x)

Semiclassical

18:24

Yeah, I think that case is indeed obvious

it's less obvious to me when I start doing stuff like $(X_1,X_2)$ and $(Y_1,Y_2)$ be equal in distribution

but all that should really be changing there is that I'm changing the domain of my random variable

and I don't think that should care a whit about the mixing

Érico Melo Silva

if the formula for the distribution of the mixed guy is just a convex combination of things that are identically distributed that's just taking a convex combination of 1 guy

which is just the original guy

Semiclassical

right

Daminark

It seems like they're mostly in harmonic analysis

Semiclassical

doesn't matter what the domain is. could be a single real, could be a vector etc

Daminark

Also a number of PDE people

Érico Melo Silva

18:27

ok that's why he told me to apply then lol

@Semiclassical yeah u only get something up to distro anyway right? so it shouldnt matter where the og rvs live

Semiclassical

right

Daminark

Yeah so it's mostly harmonic analysis, a few PDE (they seem to be listed as a separate group), a couple people do approximation and/or scattering theory, one person is analytic NT and he seems real fun to me (he works a bunch with arithmetic manifolds), one guy is a Webster student and does SCV

Érico Melo Silva

yeah harmonic is its own game

i like PDE but i literally know no harmonic analysis

Daminark

I'd like to learn some harmonic tbh, hopefully I can chat with some of these folk

Érico Melo Silva

doesnt harmonic stuff come up a lot in the analytic side of nt n stuff

Daminark

18:42

Yeah, on the one hand there's the whole business with harmonic analysis that's secretly representation theory of locally compact groups, and Tate's thesis seems to do quite a lot with that. Also the more modern "Langlands" stuff apparently has a good bit of it

I'm less familiar with how it figures in to the more "What sound does an analytic number theorist make when drowning? log log log log" stuff

Érico Melo Silva

lol

Semiclassical

i've yet to run into nested logs in any problem I've actually had to work on

I think I may have run into powers of logs once

Thorgott

The standard proof that the sum of reciprocals of primes diverges has a $\log\log$ in it iirc

Semiclassical

on that note, here's something interesting I randomly found while googling

"A property of prime powers used frequently in analytic number theory is that the set of prime powers which are not prime is a small set in the sense that the infinite sum of their reciprocals converges, although the primes are a large set."

So what makes the set of prime powers "large" is that it includes the primes, and what's left over is "small"

Daminark

18:59

Oh snap

Thorgott

19:10

That's very interesting. Do you know how that is proven?

Semiclassical

newp

user10478

Is it known generally how much of a 50 rep bounty one can expect to get back due to increased upvotes?

In other words, the net cost of a 50 rep bounty.

Semiclassical

19:27

It's an interesting question, though not one I think there's data for

The only way I know to track the number of upvotes after the bounty is to count how many there were before then

so unless someone were specifically tracking that, i think it's not easy to get that

Thorgott

Actually, $\sum_{k\ge2}\frac{1}{p^k}=\frac{1}{p^2}\sum_{k\ge0}\frac{1}{p^k}=\frac{1}{p^2}\frac{1}{1-1/p}=\frac{1}{p^2-p}=\frac{1}{p(p-1)}=\frac{1}{p-1}-\frac{1}{p}$ and $\sum_{p\text{ prime}}\left(\frac{1}{p-1}-\frac{1}{p}\right)<\sum_{n\ge2}\left(\frac{1}{n-1}-\frac{1}{n}\right)=1$, establishing the convergence. Am I missing something or does that do the job?

Semiclassical

@Thorgott to clarify, that's a sentence on the wikipedia article for prime powers

and they don't cite a source

user10478

Yeah, unless someone has heavily used bounties and has a rough idea from experience.

Semiclassical

@Thorgott nice, that seems to make sense

Daminark

@Thorgott that looks right to me

Semiclassical

19:31

What it does leave me wondering is what $\sum_{p\text{ prime}} \frac{1}{p(p-1)}$ actually converges to

Mathematica is smart enough to know that $\sum_{p\text{ prime}} p^{-k}$ is called a prime zeta function

it does not, by contrast, seem to know about $\sum_{p\text{ prime}} p^{-1}(p-1)^{-1}$

piece_and_love

hello all

Thorgott

Yeah, that's an interesting question, but if there is a satisfying answer, it probably requires a lot more machinery to demonstrate.

Leaky Nun

20:32

@Semiclassical IIRC you actually use that property to show that sum of reciprocal of prime diverge

Semiclassical

20:55

@Thorgott yeah, i don't doubt it

@LeakyNun neat

user76284

@Semiclassical Do you expect a closed form of some kind?

Semiclassical

frankly, no

but for that matter there's no known elementary closed form for the sum of reciprocal cubes, for instance.

nevertheless mathematica still does give an "answer" for that, in the form of it being just $\zeta(3)$

so my curiosity is to whether there's anything known about how to express the sum of $1/p-1/(p-1)$ for prime $p$

wouldn't be surprised if there isn't

Leaky Nun

@Semiclassical can't find any match on WA for N=1e7

Semiclassical

yeah

Leaky Nun

21:10

0.7731566.. btw

Semiclassical

that agrees with mathematica

Leaky Nun

I'm using Python btw

Semiclassical

nice

Semiclassical

21:29

Here is a question which I had on the main site recently. I got a good answer but I'm curious to see how 'obvious' their construction was

Leaky Nun

21:47

legend has it Semiclassical is still asking his question

Daminark

Okay I laughed at that

Ted Shifrin

ignores legends

Daminark

Rehi Ted, how's it going?

Ted Shifrin

Rehi, Demonark.

I am struggling to type, having cut off the tip of my left index fingers whilst cooking the other night. Luckily, I didn't do any irreparable harm ... just have to wait for a month while it heals.

Leaky Nun

:o take care

Daminark

21:58

Ouch, sorry about that

Ted Shifrin

I did something similar 40 years ago, in grad school. That was the left thumb instead.

Daminark

Make sure to avoid the letters f, r, t, g, v, or c

But yeah really hopefully it isn't too painful still

Ted Shifrin

Well, I have it bandaged, so it's not so painful — just inaccurate.

Alessandro Codenotti

Do you want to think about some set theory? @Leaky

Leaky Nun

sure

Alessandro Codenotti

22:12

Do you happen to have a copy of Kanamori just lying around?

Leaky Nun

leider nein

Alessandro Codenotti

Ok, I'm looking at a proof that for every cardinal $\lambda$ there is a function $\omega$-Jónsson for $\lambda$ and I'm stuck on a step

Leaky Nun

what is Jonsson?

Alessandro Codenotti

A function $f:[x]^\omega\to x$ is said to be $\omega$-Jónsson for $x$ if for all $y\subseteq x$ with $|y|=|x|$, $f"[y]^\omega=x$

Leaky Nun

and what step?

Alessandro Codenotti

22:17

So the proof goes like this: let $\lambda$ be a fixed cardinal and consider the equivalence relation on $[\lambda]^\omega$ given by "having identical tails", that is for $x,y\in [\lambda]^\omega$ we have $x\sim y$ iff there is $\alpha<\bigcup x$ such that $x\setminus\alpha=y\setminus\alpha$

For every equivalence class $E$ fix a representative $x_E$ (this uses AC) and consider the function $g:[\lambda]^\omega\to\lambda$ sending $x\in[\lambda]^\omega$ to the least $\alpha$ such that $x\setminus(\alpha+1)=x_E\setminus(\alpha+1)$, where $E$ is the equivalence class of $x$

Now we don't really need to find a function $\omega$-Jónsson for $\lambda$, it's clearly enough to find one for an $A\in[\lambda]^\lambda$. In particular we want to find such an $A$ with the property that for every $B\subset A$ with $|B|=|A|=\lambda$ we have $g"[B]^\omega\supseteq A$, the book here claims that for such an $A$ a function $\omega$-Jónsson for $A$ can be obtained from $g$, but I'm not seeing how exactly

The next step is also unclear to me, it says: Suppose by contradiction that no such $A$ exists. Then for $n\in\omega$ there are $A_n\in[\lambda]^\lambda$ and $a_n\in\lambda$ with $A_n\supseteq A_{n+1}$ and $a_{n+1}\in A_n\setminus(a_n+1)$ such that $a_{n+1}\not\in g"[A_{n+1}]^\omega$ but it's not clear to me how to construct those sequences

For the $A_i$ I think I'm supposed to take $A_0\in[\lambda]^\lambda$ arbitrary and then pick $A_{n+1}\subseteq A_n$ with $g"[A_{n+1}]^\omega\not\supseteq A_n$, but then I don't see how to fix the $a_i$

usukidoll

22:43

How was this integrated? I split the exponential on my paper and I ended up with extra junk. I am able to get everything else though but this integration part is driving me nuts. (Uploaded pictures to follow)

This problem was done before and I understand most of it but after applying the minimum density formula I have to take the expected value which I do get what's posted but how the heck does the integration work because I tried splitting the exponential and then taking the antiderivative. When I intergrate I have extra terms instead of $\frac{\theta}{n-1}$. Nerghh. https://ibb.co/7y2kYCF
https://ibb.co/0nHTxC2

I could just solve for $E[Y^2]$ but that would mean doing this $Var(Y)+(E[Y])^2=E[Y^2]$

I used a lot of scratch paper on this before :/

Transcript for

$Mathematics$ Mathematics