Mathematics - 2016-11-02 (page 2 of 2)

Fargle

19:00

Specifically, let $x$ real and define $B(x)$ to be all $b^t$ for $t \leq x$. Prove that $b^r = \sup(B(r))$ for $r \in \Bbb Q$.

Danu

@TedShifrin You're leaving?

Ted Shifrin

Momentarily.

Balarka Sen

@MikeM: Your advisor has good taste for movies. I only just noticed.

Ted Shifrin

Not too bad, @Fargle. As I recall, there are steps.

Fargle

What I have down is:
Let $x = b^t \in B(r)$; then $t \leq r$, so that $b^{r-t} \geq 1$, giving $b^t \leq b^r$.

Then let $\alpha \neq b^r$ be an upper bound on $B(r)$--then $\alpha \geq b^t$ for all $t \leq r$; in particular, $\alpha \geq b^r$, and by assumption, $\alpha > b^r$, and we are done.

It just feels too easy, if you catch my drift.

Ted Shifrin

19:04

Just follows from increasing ($b>1$). Yup.

Fargle

Okay, phew. I was worried that I was missing something, or assuming too much.

Ted Shifrin

It gets harder, AIR.

Fargle

Now I have to prove $\sup(B(x+y)) = \sup(B(x)) \cdot \sup(B(y))$, which shouldn't be too bad.

Ted Shifrin

OK, heading to lunch. Bubye.

Fargle

'Bye!

73est

19:09

Is there anyway someone comfortable with optomizing a function using the steepest descent method could look at my question?

Tobias Kildetoft

What sort of distribution do the AMS fellows have of US vs non-US?

Hailbert

ams fellow is like a cheap version of FRS

Danu

Actually, that looks a bit rude. Sorry, I'm going to delete it.

Hailbert

what did you say?

Danu

Something along the lines of: What does that make us, hobos? But phrased in a seemingly more confrontational manner, which was not what I intended.

Hailbert

19:18

yeah these kinds of fellowships don't mean a lot

i think ramanujan was never elected a fellow of trinity college

because he was indian

s.harp

he was a fellow of trinity college

Hailbert

Ciprian Manolescu - don't think i have ever heard that name

Danu

He proved a big theorem not long ago.

saturatedexpo

My mathprof told me, that an id is always something that points stuff to itself. So f(g)=g would be id of g? (roughly) i mean g not as a value, rather some "stuff", like a set or something

Hailbert

what Theorem?

Danu

19:23

ams.org/journals/jams/2016-29-01/S0894-0347-2015-00829-3

Hailbert

big according to whom?

Danu

According to everyone in the field.

Non-existence of triangulation for some topological manifolds in any dimension $\geq 5$.

Or if you're more pop-science inclined: quantamagazine.org/…

Hailbert

i see

apparently this was proven 3 years ago though

Danu

That's pretty recent, lol.

Hailbert

but i think while the theorem is remarkable

i feel it would have been much better if every manifold was triangulizable

Danu

19:27

That's a pointless thing to say.

Hailbert

no

it would be much better

Danu

In any case, all smooth manifolds are triangulable. This just means that perhaps topological manifolds are not as nice as you'd hope.

Hailbert

because people could actually use it in further woork then

the theorem is like "quintics do not admit a General solution"

interesting result, but you can't use it

Danu

→ 10 messages moved to Trashcan

s.harp

how much sleep is recommended?

Fargle

19:33

@s.harp In general or for something specific?

I'd say 7-9 in general.

Hours that is.

teadawg1337

It varies from person to person

Danu

@s.harp Highly depends on the person, I'd say mostly.

teadawg1337

I personally function just fine off of 5-7 hours

Danu

I'm okay with 6-ish usually, but some people need like 9

Fargle

I need around 9 or 10, but I dunno. However much makes you feel best, honestly.

s.harp

19:35

Whenever I sleep less than 9 I get super tired, usually I do 10 at least

Fargle

@s.harp I'd suggest not much more than that. Anything over 12 is unhealthy AFAIK.

I say this as someone who slept for like 16 hours recently, so...

syzygy

too much sleep is unhealthy?

source?

teadawg1337

Too much of anything can be a bad thing

s.harp

recently I had an exercise for a tutorial that for some reason I couldn't solve.
If you have a vector space $V$ and look at the free vectorspace generated by $V$ (call it $F$) and the subset of $F$ given by the span of $\{\delta_{v+w}-\delta_v-\delta_w\mid v,w\in V\}\cup\{\delta_{\alpha v}-\alpha \delta_v\mid \alpha\in\mathbb K, v\in V\}$ where $\delta_v(x)=0$ if $x\neq v$ and $1$ if $x=v$ (calls this span $L$),

then the map $V\to F/L$ given by $[v]\mapsto[\delta_v]$ is an ismorphism.

linearity and surjectivity are trivial, but for some reason injectivity escapes me

syzygy

true dat, but i mean unhealthy? how does it damage your health

Fargle

19:40

@syzygy I've read in the past that it increases risk of heart disease, diabetes, stroke, and death, among other things

Grzegorz Baltissen

Does VarX <=1 && VarY <= 1 imply Cov(X,Y)<=1?

s.harp

@Fargle "it increases the risk of death"

XD

Danu

@s.harp Just check dimensions?

syzygy

i think too little sleep is much worse

Grzegorz Baltissen

As a side note: old people tend to sleep a lot => correlation of both death and sleep?

Fargle

19:41

@GrzegorzBaltissen Most likely.

Danu

lol

Sleep is the cousin of death [citation: gangsta rap]

Fargle

@syzygy It is, but that doesn't make too much sleep good...

syzygy

i am not saying that is good, i was just wondering

s.harp

@Danu in a sense that is just as difficult, well at least i dont see how to see the dimenion of $F/L$. And even then if the vector space is infintie dimensional that doesnt give you injectivity

syzygy

also when you sleep 16h this sucks, because you have only 8h to do something

^^

Grzegorz Baltissen

19:42

How can you possibly sleep too much? unless you count lying in bed awake as sleep?

s.harp

@GrzegorzBaltissen over-medication?

Grzegorz Baltissen

That's unhealthy.

Fargle

what have I wrought upon this room

s.harp

isn't this the sleep-health chat room?

Danu

@s.harp Right.

Expand in a basis, then.

Fargle

19:46

I was always under the impression that this was the pun and lowbrow humor chatroom, featuring math sometimes.

Danu

It's still an iso between finite-dimensional vector spaces.

s.harp

showing that tje map preserves the notion of linear independence may be workable

syzygy

or perhaps that it maps a basis to a basis :P?

Secret

[Division by zero] Division by zero semiring. It however seemed too ridgid to be expanded into a larger size

Meanwhile still scanning for signs of collapse

teadawg1337

I find it disappointing how simple questions gain more attention than challenging ones on MSE

s.harp

19:58

the only questions/answers of mine that got many upvotes were actually crappy

whereas if I think for long and write a text its 1 or 2 or zero

Secret

Scan complete, the semiring is stable. Now what to do next...

teadawg1337

I just posted a question that I've been stuck on for around 8 months now, and it's garnered 16 views and two upvotes. Meanwhile an elementary question regarding limits was posted after mine and currently has four upvotes and 27 views

s.harp

@Secret do you know about complete semi-rings?

or maybe called eilenberg complete

Secret

not aware of the term

s.harp

if where all sums $\sum_{i\in I}a_i$ are defined, even $I$ is super duper infinite

its a fnny thing to think about

if you look at examples you will see that they wouldn'T work if you had a ring, rather than a semi-ring. so if you dont have many negatives the sum is more like a thing that detects how much $a_i$ is in your term, and you cant take away from it

an example is $\mathbb R_{≥0}\cup\{\infty\}$

Secret

20:06

is that some kind of real projective line but throwing away all negatives?

saturatedexpo

is f(x)=x only a special Relation that also fullfills all whats needed to be a function?

Secret

@s.harp the way it works reminds of convergent series or even cauchy sequences, so I am guessing we can define a notion of limits on top of them?

s.harp

@secret no, its a semi-ring with finite addition and multiplication (on the $\mathbb R$ part) being the same as the normal $\mathbb R$ addition amd multiplication

Secret

But you said $I$ can be infinite, or do you mean we only need to pick finite number of $a_i$ to do the addition?

s.harp

no, consider ie $\sum_{n\in\mathbb N} n$ in that semi-ring, what do you think it evaluates to?

Secret

20:12

logically it has to be the element $\infty$?

s.harp

ye

but this is a countable sum and in a sense from analysis you already understand countable sums, what would $\sum_{r\in \mathbb R} a_r$ be if all the $a_r\neq0$?

Secret

If all $a_r \neq 0$ the sum has to = $\infty$ if only some $a_r\neq 0$ then the sum can be $< \infty$

so as you mentioned, it does roughly keep track on how many nonzero $a_r$ in the sum

s.harp

yeah the sum is only finite if a countable number are nonzero, and if a countable number is nonzero you know how to think about countable sums in $\mathbb R$

the example is a little boring in the sense that you dont really have interesting uncountable sums avaivable. I think $\mathfrak P(X)$ for some set $X$ is also a complete semi-ring with addition given by $\cup$ and zero $\emptyset$ and multiplication by $\cap$ with unit $X$.

I dont know many examples, and im not finding super interesting examples on the web, but I found the structure funny. I heard about it in a talk about topological field theory, where speaker wanted to find invariants of topological spaces and the "lack of negatives" in the semi-ring sort of helpd the theory "accumulate whats there"

and the fact that you can have arbitrary sums also helped the bookkeeping in a way

Secret

Usually completeness is often linked to analysis. I wonder if I can made that into a module of some sort to explore it further some time...

s.harp

here you dont have a "completeness structure" in that if you ahve a sequence of elements $a_n$ you can find out whether or not they converge

the completeness is more algebraic in that it says every kind of sum exists

Alessandro

20:24

Aren't you just talking about a complete lattice in the last example where arbitrary sums are the joins?

s.harp

I guess yeah, that covers pretty much the way I was talking about it, and also the examples I was giving

Secret

I haven't gone into lattice yet, as I am currently still in group theory

teadawg1337

So posting my own question in here would be frowned upon, correct?

s.harp

anyway Im going to put the discussion about sleep into practice

teadawg1337

That's what I seem to remember

s.harp

20:26

@teadawg1337 no, I dont think so

teadawg1337

As in linking a question I posted on MSE?

Alessandro

A lot of people (me included) ask here all the time

s.harp

yes people do that, I have done it too, but odnt spam it :P

Secret

btw s.harp, for that semiring structure I linked, is there any structure related to it. I noticed there's a null semigroup in there

teadawg1337

Okay, here it is. I've spent eight months on that and I feel like it's getting buried. Link here

user243301

20:29

@Axoren I hope that your are right. I was thinking from your profile that perhaps if you are interesting in mathematics and computer science you could be interested in some problems concerning it. I say for example aliquot sequences, Markov conjecture, Collatz 3x+1 problem, abc conjecture... Also there are journals about mathematics and computations like as Mathematics of Computation and Experimental mathematics.

s.harp

@Secret I dont know

I never looked at a multiplication table seriously since kindergarden

Secret

ok

s.harp

so I cant really get information from them :)

saturatedexpo

does $R\circ R^{-1}$ contain the hole set of M, if R is a relation on M?

*whole set

Adeek

hi

s.harp

20:33

$R\circ R^{-1} = \{ (x,y) \mid \exists z: xRz \text{ and } yRz\}$?

Adeek

anyone want to discuss functional analysis ?

I just want to see any critique to a proof I did

teadawg1337

Hmmm, I'll focus on something else for a while. It's not like it'll get any more attention from me staring at it. This isn't me trying to get attention, either

saturatedexpo

or does $R\circ R^{-1}$ only contain those elements that R deals with in the beginning?

Secret

If I recall, one of the cool things about modules is that you can put torsion into what is effectively a vector space

Alessandro

I'll take the occasion to repost an old and unanswered question of mine as well mathoverflow.net/questions/211283/…

PVAL-inactive

20:36

There's probably a high correlation between sleeping a lot, and a generally inactive lifestyle. That doesn't mean the first implies the second

saturatedexpo

@s.harp can you point to an explanation source to how to derive that? i fail to do so :((mainly because i dont know what to search after)

teadawg1337

@PVAL-inactive Does doing independent math research count as an inactive lifestyle?

Secret

Not really, cause it can be done with your peers at a pub

All you need is a laptop, pen and paper

and heaps of discussion

PVAL-inactive

I don't think drinking at a pub is what I would consider "active"

teadawg1337

What if I have no peers that I know IRL?

I'm partially joking. I do have no social life, though :P

Secret

20:42

If your maths research have something to do with biology, it can involve being in the field

PVAL-inactive

@Secret I doubt it.

saturatedexpo

it also could involve being in something else then a field ;)

(mixed with drinking haha)

teadawg1337

My research has everything to do with sitting on my bum and writing in a notebook

I think I have an inactive lifestyle

Alessandro

I'm often impressed at some of my classmate's ability to do math while drunk

Alex

Hi @DanielFischer

Secret

20:46

about.illinoisstate.edu/biomath/pages/default.aspx

Adeek

0

Q: Suppose X is normed space such that $dim(X) = \infty$, then $dim(B(X,\mathbb{K})) = \infty$

I want to see if my proof is correct. Here $\mathbb{K}$ is the ground field. Suppose that $dim(B(X,\mathbb{K})) = n$ where n is any arbitrarily integer we shall construct $n + 1$ linearly independent vectors in $B(X,\mathbb{K})$. Since X is infinite-dimensional it has n + 1 linearly independent v...

Jake1234

23:19

[title]books.google.cz/…

would someone please help me out with the proof starting at page 133 ? The claim is that the derived subgroup of $G$ satisfies ACC (no infinite series of increasing subgroups), then union of two subnormal groups is subnormal.

Or more precisely, I think the claim halfway through about $\langle H, H^a, H^b \rangle$ is wrong - I don't see how it's equal to what it's claimed to be. Also, I think induction should start at s>0, not s>1.

Either way, if I understand it right, it should be fine just to take $\langle H,H^a \rangle$, which is subnormal in $H_{(1)}$, and then since $H^b$ is subnormal and satisfies the inductive hypothesis, I can say that their union, which is $\langle H, H^a, H^b \rangle$ is also subnormal in $H_{(1)}$

Shimmy

23:39

When is the $\ell$ used?

What's the purpose of this character a small script-like l

ok got here: en.wikipedia.org/wiki/Latin_letters_used_in_mathematics#Ll

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$Mathematics$ Mathematics