Mathematics - 2015-12-19 (page 5 of 6)

skill patrol

21:00

Are you a regular in this room?

Yes or no.

Balarka Sen

@skill How's that relevant?

OFFSHARING

Maybe I should become smaller and smaller, even decrease in size the text I put on channel.

Semiclassical

so much headdesk, this conversation is

OFFSHARING

@Semiclassical It's my fault only that I kept coming here. This is the only point.

Balarka Sen

He is a user, actively contributed to the math in this chat for more than a week, and is a mod.

I'd think that's enough.

skill patrol

21:02

It's called context.

OFFSHARING

@BalarkaSen Enough for what?

Anyway.

Balarka Sen

@skillpatrol That's utterly irrelevant. Expressing the opinion that huge messages create noise doesn't require much context.

Danu

@skillpatrol I don't see what you're trying to imply.

skill patrol

Knowing the various personalities in this room.

Danu

Also, you may or may not be aware that I've been a long-time spectator of this room.

@skillpatrol I think you underestimate my knowledge of this room.

skill patrol

21:04

You have to interact with people to get to know them pal.

::Enough said::

Danu

I don't think that's strictly true, and I think it isn't really relevant here.

OFFSHARING

@Danu What was that?

"@OFFSHARING No, I don't. I've seen your monologues before."

Balarka Sen

@TedShifrin Hi Ted.

Welcome to the drama.

Danu

Yeah, let's cut it out.

skill patrol

NOW please :-(

Ted Shifrin

21:07

Balarka: You have serious math to be working on.

Balarka Sen

I am working on it!

Ted Shifrin

@skull: @Danu is a friend. Stop.

Balarka Sen

@Semiclassical told me a cool theorem in dynamical systems.

OFFSHARING

Look, as long as I enter this chat I want to recommend some things: 1) if you don't know me, introduce yourself and greet me before saying anything else

2) don't talk about my mathematics if you have no idea about it and don't tell me what to do or not as long as I don't break the rules of the chat

3) if you wanna talk to me about my math, be polite and talk with me

Semiclassical

@BalarkaSen yep, period three implies chaos

Danu

21:10

Btw, @BalarkaSen how old are you? Someone said 16; in that case I must congratulate you with your superior knowledge of math :P

Balarka Sen

I don't really know what chaos means there, but I was referring to period 3 implies every other period.

Semiclassical

though i think i got at least one detail wrong. i forgot that, just as a fixed point can either be attracting or repelling, so too can a cycle. and the one that zero is nearby seems to be an attracting one

Danu

@TedShifrin D'aww, thanks!

Balarka Sen

@Danu: I am 51.

Danu

@BalarkaSen Really? That's a big difference from 16 :P

Ted Shifrin

21:11

smacks @Balarka and his foot-long nose

Semiclassical

so it wasn't actually a numerical issue on Mathematica's part. zero doesn't produce a 4-cycle, but it converges rapidly to it

Vrouvrou

please, what it mean: $I: W^{1,p}_0(\mathbb{R}^N})\rightarrow \mathbb{R}$ is rotationally invariant

Semiclassical

@TedShifrin I was explaining the things that could go wrong with Newton's method, and gave $f(x)=x^3-x-3$ as an example

Danu

So you're not comfortable disclosing your age? That's fine.

Huy

@Vrouvrou: Which part don't you understand?

Vrouvrou

21:12

@Huy: is rotationally invariant

Balarka Sen

No, I really am 51, an Australian computer scientist.

Semiclassical

and in doing so I realized that it gave a nice example of said result in dynamical systems

so i'm to blame on distracting him, I guess :P

Huy

@BalarkaSen: which is precisely the reason why people keep telling you to go bed earlier.

Balarka Sen

exactly!

skill patrol

:D

Semiclassical

21:14

another cool thing with newton's method in that vein, btw, is that you can use it to make fractals :)

Huy

BTW, I'm off to bed.

good night everyone.

Balarka Sen

Night.

Semiclassical

Newton fractal

The Newton fractal is a boundary set in the complex plane which is characterized by Newton's method applied to a fixed polynomial . It is the Julia set of the meromorphic function which is given by Newton's method. When there are no attractive cycles (of order greater than 1), it divides the complex plane into regions , each of which is associated with a root of the polynomial, . In this way the Newton fractal is similar to the Mandelbrot set, and like other fractals it exhibits an intricate appearance arising from a simple description. It is relevant to numerical analysis because it shows that...

Balarka Sen

Don't let mapping class groups bite.

Huy

I'm worried about mapping tori

skill patrol

21:14

Later pal

Balarka Sen

@Huy But those are friendly.

S^1-bundles. Friends.

Huy

we can discuss this some other time. next week.

Ted Shifrin

Ah, @Semiclassical ... of course, doing it over $\Bbb C$ gives all the famous chaos pictures with Julia sets.

Balarka Sen

sure.

Semiclassical

yep

Balarka Sen

21:16

Urk, I meant bundle over S^1. Oh well.

Semiclassical

i picked x^3+stuff because i wanted something whose second derivative vanishes at zero

and then decided on x^3-x-3 more or less arbitrarily

Ted Shifrin

Right, hence the need for a (positive) lower bound on $|f''|$ in what Balarka's supposed to figure out.

Semiclassical

right.

Ted Shifrin

FYI, @Semiclassic, this is Kantarovich's Theorem, which I learned from Hubbard & Hubbard's book.

Balarka Sen

@Danu I was lying earlier. I am 15, everyone here knows that. And don't know much math. Learning.

Semiclassical

21:17

mmkay

what i'm curious about myself now (though I suspect there's not a nice result) is the rate of convergence to attracting cycles

Ted Shifrin

Hubbard proves the Inverse Function Theorem using it (which I think is ridiculous), but it's interesting to get constructive bounds on sizes of neighborhoods, etc.

Semiclassical

it's geometric convergence if it's just the case of a fixed point, but i dunno about cycles

Ted Shifrin

I know not of this.

Danu

@BalarkaSen How'd you start studying math?

Semiclassical

it'd have to be pretty delicate

actually, wait

Balarka Sen

21:19

I hated everything else. Didn't know what to do.

The Substitute

Off topic. Is the conjugate $a-bi$ of $a+bi$ the same type of conjugation that arises in algebra (gxg^-1)? If so, how can I express $a-bi$ in this form?

Danu

How does one get started? Or do your parents/other mentors know a lot of math?

Semiclassical

if i want to consider an $n$-cycle, all I should need to do is take $g(x)=f^{(n)}(x)$ (nth composition, not derivative)

Khallil

Can anybody recommend a great book on measure theory?

Balarka Sen

@TheSubstitute Nope, they're different.

Very abusive terminology.

Semiclassical

21:20

and then the result for the fixed point case should apply once more.

Danu

@BalarkaSen I actually think I recently saw they're related.

In a HNQ on math.se.

Balarka Sen

I posted an answer there, yes.

Semiclassical

not sure that's easy to address, since one has replaced a map with its nth composition, but it's a rather well-defined question

Balarka Sen

I don't really agree with other answers.

Danu

What does "agree" mean when you have theorems? :P

Balarka Sen

21:21

here.

The Substitute

thanks

Danu

Your answer is the standard one one receives in physics.

Balarka Sen

@Danu I don't see any theorem relative the two.

Danu

C. Falcon's answer.

Balarka Sen

The one mentioned in the other ones just state Galois groups of different embeddings are conjugate.

Semiclassical

21:22

So, I guess here's a more precise question: What information on $f(x)$ is required to infer appropriate bounds on the first and second derivatives of $f^{(n)}(x)$?

Danu

It has at least a relation between conjugation and complex conjugation.

Which cannot be said to be "wrong"

Balarka Sen

Yes, don't see why that means $a + bi$ and $a - bi$ are related to conjugation as in group theory.

robjohn

@OFFSHARING bad mood? They pass pretty fast.

Danu

@BalarkaSen "Remark. The complex conjugation is a ring automorphism of C. The field conjugation defined above is thus a generalization of the complex conjugation."

From the answer

Now you're claiming those conjugate fields have nothing to do with conjugation in the group sense?

Balarka Sen

How does that relate to group conjugation?

Danu

21:24

Is that really true?

That'd surprise me.

Balarka Sen

I am not claiming anything. I don't see a relation.

Ted Shifrin

hi @robjohn: Happy holidays

Balarka Sen

I mentioned other motivations in my answer below.

skill patrol

How's your Christmas so far @robjohn?

Ted Shifrin

@Danu @Balarka: The conjugate fields have invariant subgroups that are conjugate.

Danu

21:25

^woop

Ted Shifrin

By invariant subgroups I mean the subgroup of the Galois group leaving the subfield fixed.

Balarka Sen

What does conjugate field mean?

Galois groups of different embeddings are conjugate, yes

Danu

It's defined in C. Falcon's answer.

You didn't read it? :P

Balarka Sen

But not sure how that relates to $a + bi$ and $a - bi$

robjohn

@TedShifrin Happy Holidays!! I am taking the next two weeks off. Actually from Dec 24-Jan 1, we must take Vacation or LWOP

Balarka Sen

21:26

@Danu You're not getting my point.

Ted Shifrin

LWOP?

@robjohn: I'll be heading east to snow and ATL for about 3 weeks.

Balarka Sen

If $i, i' : L \to K$ are different embeddings, $Gal(i)$ and $Gal(i')$ are conjugate. This is standard fact (and what C.Falcon says)

But how in the world is it related to Galois conjugates?

Ted Shifrin

Well, @Balarka, that use of conjugate refers to field elements (in the orbit of the Galois group). :P

Semiclassical

too many conjugates :/

robjohn

@skillpatrol Christmas is a few days off

Ted Shifrin

21:28

Well, they are images of different embeddings of the quadratic extension?

Balarka Sen

@TedShifrin The questions asks whether they are related.

Ted Shifrin

Hmm, no, that's not right. It's the same subfield.

Never mind. I withdraw.

Go back to chapter 6 :)

Balarka Sen

:)

Ted Shifrin

@Semiclassic @Danu: I prefer to conjugate verbs.

Balarka Sen

@Danu told ya, there's no apparent relation except vague hints.

Semiclassical

21:29

Just to be clear, how many variations on 'conjugation' showed up in that conversation?

Balarka Sen

in particular, Slade mentions it's an abuse of terminology.

Ted Shifrin

And then there are conjugate visits ... oh wait ...

Semiclassical

i heard complex conjugation, galois conjugates, field conjugates

Ted Shifrin

complex conjugate is a special case of Galois conjugate

Balarka Sen

group conjugation. $gxg^{-1}$.

Danu

21:29

@BalarkaSen I think that's not true, but okay.

I don't know any field theory in the mathematical sense.

So I can't meaningfully contribute or investigate this question.

Ted Shifrin

sad that no one honored his bad joke with a groan

Balarka Sen

@Danu If you can give me a good reason, I'll happily stand corrected.

Ted Shifrin

@Balarka: You're taller when you sit :)

Semiclassical

i always find it funny how different in meaning the phrase "field theory" is for a mathematician versus a physicist

Balarka Sen

@TedShifrin What's a conjugate visit?

Ted Shifrin

21:31

Don't even start with the abuse of the word "normal" in mathematics.

Semiclassical

same word, but so very very very far apart in meaning

Bungo

in spanish, "con jugar" means "to play together", so "conjugate visit" is appropriately named

Ted Shifrin

@Balarka: it's really a conjugal visit

Balarka Sen

ah.

Ted Shifrin

:) @Bungo

Bungo

21:32

for that matter, conjugate group elements play together too

Daniel Fischer

@TedShifrin You mean when some people think a space can be normal without being Hausdorff?

Ted Shifrin

That's serious abuse, @DanielF :)

Semiclassical

when i hear 'normal' my first instinct is 'surface normal'

Balarka Sen

annoyed by abuses

Ted Shifrin

and vectors (two kinds of normal!) and probability distributions and ...

Balarka Sen

21:33

oh, and torsion.

Bungo

normal subgroups of course

Ted Shifrin

I think all the math haters are really right.

Bungo

and operators

Ted Shifrin

Yup @Bungo

Semiclassical

"normal modes" is a physics one, though that's related to the vectors meaning

Daniel Fischer

21:33

And families.

Balarka Sen

@TedShifrin You know what? Torsion makes sense. Torsion of curves measure how flat it is, locally. Torsion of groups are captured by Tor functor, which measures how much flat a module is.

Funny.

Bungo

i always thought it was a bit mean to call some series subnormal

Semiclassical

subharmonic would be a pretty harsh diss for a musician :P

Danu

haha

Daniel Fischer

@Semiclassical Let them play standing on their heads, then they're superharmonic.

Ted Shifrin

21:35

Ah yes, @DanielF and @Balarka

Semiclassical

there we go

Ted Shifrin

LOL

Bungo

oh inverted world

Ted Shifrin

Where's @Clarinet when we need him?

Balarka Sen

I should post my observation about torsion somewhere in MSE.

Ted Shifrin

21:37

@DanielF: I see you're tangling with her highness the empress of the room.

Bungo

"three things wrong with euler's formula?? math.stackexchange.com/questions/1582626/…

i have no idea what the OP is getting at

Ted Shifrin

@Balarka: Be careful with flat in differential geometry. Literally it means zero curvature. You mean planar, not flat.

Balarka Sen

The margins are too short to write the errors down.

Bungo

haha @BalarkaSen

Balarka Sen

@TedShifrin Fair enough.

skull petrol

21:39

@TedShifrin we used to call balarka his "highness" too ;-)

Ted Shifrin

Nor I, @Bungo. It's not the OP, it's his instructor. It actually was not proved by Euler, so maybe that's one thing.

Balarka Sen

@skullpetrol I have evolved.

Ted Shifrin

@skull: I don't remember that, but Balarka has turned into 80% human.

Balarka Sen

80% human? hah. HAH.

Bungo

@TedShifrin Interesting, not proved in the sense that his proof was not rigorous by modern standards, or not proved in the sense that someone else proved it first?

Ted Shifrin

21:40

80% human who has Chapter 6 to learn.

Daniel Fischer

@BalarkaSen Still 20% Teenager. You gotta watch out.

Ted Shifrin

Someone else proved it first, @Bungo. Let me check.

Ah, I no longer have that book. Hang on.

Bungo

of course the list of things that euler did discover first is so ridiculously large that it barely matters

Ted Shifrin

Cotes, I think, @Bungo.

Wiki says Cotes was 1714, Euler was 1740's (introducing the complex exponential).

Bungo

interesting, i'm not sure i had even heard of cotes before now

Ted Shifrin

21:43

Brit, I believe.

Cotes apparently was the first to discover Simpson's Rule, too. He was a disciple of Sir Isaac.

Balarka Sen

I have heard of Coates, but not Cotes.

Daniel Fischer

@TedShifrin Yes

Bungo

a student of newton's

Ted Shifrin

Yup, disciple of Sir Isaac. :)

Bungo

died at 33 of a fever, tough world back then

Ted Shifrin

21:45

well, every winter we worry that Balarka may do the same ...

Danu

Are you referencing Ramanujan? :P

Balarka Sen

No, he's referencing me being sick.

Danu

(Balarka is Indian, right?)

@BalarkaSen Close enough.

Balarka Sen

And we have enough integrators here.

No need for more.

Ted Shifrin

OK, I'm gone.

Balarka Sen

21:47

Bye.

skull petrol

Later

Danu

Cyas

Balarka Sen

Hi @KarlKronenfeld

skull petrol

Bye

Karl Kronenfeld

why hello

skill patrol

21:48

Hi

:-)

Balarka Sen

what're you upto?

skull petrol

You missed the fireworks ;-)

A lot of the juicy comments have been removed.

At least there was no suspensions, this time.

Balarka Sen

Chris'ssis got suspended.

Semiclassical

sigh

skull petrol

In the past yes.

Balarka Sen

21:55

(s)he got suspended today, I mean

skull petrol

Not today.

Balarka Sen

@Danu Something you might enjoy: recall the fact about $Gal(i)$ and $Gal(i')$ being conjugate above? does this remind you of the basepoint changing in $\pi_1$?

[wink]

skull petrol

(removed)

Balarka Sen

@skullpetrol I checked his/her profile during the suspension.

Semiclassical

just added a bounty on a question I put up

see math.stackexchange.com/q/1568212/137524

skull petrol

21:58

She doesn't get suspended that often @BalarkaSen but when she does, I can see it coming a mile away.

Semiclassical

that integral still bugs me. i don't feel like the antisymmetry should just be accidental, but there's not an obvious mechanism

Bungo

interesting question @Semiclassical, certainly it does not seem apparent from the integral expression

Semiclassical

yeah. and while proving the stated identity certainly establishes the symmetry, it doesn't explain it (to my satisfaction, anyways)

Agawa001

oh one of ur post-deleted lines

nvm

OFFSHARING

@robjohn I'm very sad because of that. It's not my top research, not at all, but still, it was some stuff that I liked. It's not stuff posted here, it happened in other circumstances.

Will Jagy

22:02

@TedShifrin , found this book preview with Roger Cotes books.google.com/…

skull petrol

Welcome @WillJagy :-)

Balarka Sen

@ForeverMozart Thinking about anything pathological?

Bungo

i just got a $150 amazon gift card for an early xmas gift. what math book to buy...

Forever Mozart

oh yes. I still have a nice example but it is not what I really wanted. still thinking

Semiclassical

hrm. it's still bugging me, so i guess i'll follow up on this

OFFSHARING

22:06

@DanielFischer I have nothing personally with you, just to know that. Nothing personally with anyone here, and it would be nonsense to have anything (like that - in a negative sense). I just don't like some things, no matter who does them.

Balarka Sen

@ForeverMozart Want to tell me a cool pathological problem?

Semiclassical

@ted How complicated is a 'proper' derivation of $\partial F^*/\partial c = \lambda?$ (where $G=G(x)=c$ is the constraint and $F^*$ is the objective function $F$ evaluated at its constrained maximum)

Daniel Fischer

@OFFSHARING That's okay. Just don't become aggressive when you don't like something.

Forever Mozart

can you take the knaster continuum and deform it in 3 dimensions so that its closure is decomposable?

OFFSHARING

@DanielFischer I'm just natural, and wrongly perceived as aggressive.

Balarka Sen

22:08

@Semiclassical Are you referring to the problem I and Ted were discussing?

Semiclassical

yeah

Balarka Sen

I solved it. I was being a dope.

Daniel Fischer

@OFFSHARING "I don't care your worthless opinions." is aggressive.

skill patrol

worthless is the wrong word.

OFFSHARING

@DanielFischer It was said in a certain context. It seemed to me he made fun of me talking about my monologues. That looked really like a worthless opinion. No generalization intended.

Semiclassical

22:11

my handwavey explanation was that $$\nabla F = \lambda \nabla G \implies \lambda \dfrac{|\nabla F|}{|\nabla G|}$$ since both gradients are necessarily parallel

Balarka Sen

@Semiclassical The problem, iirc, was if $a = \psi(c)$ is a local extremum of $f$ rel $g = c$, then $(f \circ \psi)'(c) = \lambda$. That doesn't seem to match with your problem.

$\psi$ here is differentiable.

Semiclassical

so many edits, ugh

skill patrol

What does "worthless" mean to you @OFFSHARING

Forever Mozart

relax people

3

OFFSHARING

@skillpatrol Of no importance to me in that context

skill patrol

22:13

That's not the usual definition

Balarka Sen

@ForeverMozart knaster continuum means the fan?

Forever Mozart

the buckethandle

OFFSHARING

@skillpatrol This is the meaning I know for it. You mean I'm wrong using it like that with this sense?

Semiclassical

with the handwaving being that one can replace that ratio with $\partial F/\partial c$ since $G=c$.

Balarka Sen

Ah.

Agawa001

22:14

worthless =/= costless

Balarka Sen

@Semiclassical See the right problem above.

Semiclassical

well, I had the version in terms of Lagrangian multipliers in my head

Agawa001

@OFFSHARING i do think that callin ur declarions (monologues) is sorta disdainin, but sometimes u have to just let it go ...

Balarka Sen

You can prove this quite easily. $(f \circ \psi)'(c) = Df(a) \circ D\psi (c) = \lambda Df(a) \circ D\psi (c) = \lambda (g \circ \psi)'(c)$. And $(g \circ \psi)(c) = c$, so that is simply $1$.

Agawa001

and u did well

Balarka Sen

22:16

So you're done.

Semiclassical

and i was mostly proceeding via geometric intuition. in that sense i do see the meaning of $\psi(c)$ as the line traced out by the critical point as $c$ is varied

skill patrol

it is hard to use that word "nicely" @OFFSHARING so yes, you were wrong, but I couldn't use "worthless" to mean something nice :-)

Bungo

maybe a dumb question about domain names: if math.stackexchange.com, why not math.stackoverflow.com instead of mathoverflow.net

OFFSHARING

@skillpatrol You mean that between of no importance and worthless is a big difference? Worthless sounds tougher?

Balarka Sen

@Semiclassical And if you pushforward each $\psi(c)$ by $g$, you land in $c$, don't you?

Semiclassical

22:18

mmm, true

evinda

@DanielFischer We have the linear mapping $T: \ell^2 (\mathbb{N}) \to \ell^2(\mathbb{N})$ with $T(x_1, x_2, x_3, \dots)=(x_2, x_3, \dots)$
I have shown that $||T^n||=1 \forall n=1,2, \dots$ as follows:

We notice that:
$T^2(x_1, x_2, \dots)=(x_2, x_3, \dots), T^3(x_1, x_2, \dots)=(x_3, x_4, \dots), \dots, T^n(x_1, x_2, \dots)=(x_n, x_{n+1}, \dots)$

Since $x \in \ell^2(\mathbb{N}), ||T^n||= \sup \{ ||T^n x||: ||x||_2=1\}=1$

But then I have to show that $||T^n x||_2 \to 0 \forall x \in \ell^2(\mathbb{N})$.

skill patrol

Yes, exactly @OFFSHARING

Semiclassical

given the kind of room we're in, the word 'orthogonal' probably best captures what you were saying @OFFSHARING

skill patrol

Worthless sounds rude. @OFFSHARING

Balarka Sen

@ForeverMozart what does decomposable mean here? what does deform mean?

OFFSHARING

22:20

@skillpatrol Well, I only meant of no importance. Wait a second. Why should it sound rude?

skill patrol

Say what you mean @OFFSHARING

Semiclassical

e.g. "Your opinions are orthogonal to my interests." but admittedly that sounds kind've weird

it's a connotation of the word worthless in english, i guess

Balarka Sen

{all the worthless opinions in $\Bbb R^n$} = $O(n)$.

OFFSHARING

@Semiclassical Ah, I see.

Semiclassical

i.e. not just "not of relevance to my goals/interests" but "not of purpose to anyone"

skill patrol

22:22

@OFFSHARING are you insulting my self worth by calling my comments worthless?

Daniel Fischer

@Bungo One thing is that Math Overflow was originally not part of the Stack Exchange network. It's still not completely part of it, the Math Overflow owners can leave again if they so wish.

OFFSHARING

@skillpatrol no no no. Wait.

skill patrol

You said worthless

OFFSHARING

@skillpatrol Not sure what you mean now. I said nothing bad to you.

Daniel Fischer

@evinda Basically, that's the difference between pointwise and uniform convergence.

OFFSHARING

22:24

@Semiclassical This is clear now. That means that worthless takes any kind of value of a thing for anyone.

evinda

@DanielFischer It holds that $||T^n x||_2=\sqrt{x_n^2+ x_{n+1}^2+ \dots}$. How can we deduce that this converges to 0?

skill patrol

@OFFSHARING just avoid such "tough" sounding words :-)

Forever Mozart

relaxxx

Daniel Fischer

@evinda What do you know about $$\sum_{k = n}^\infty x_k^2\,?$$

Forever Mozart

you people need to relax

OFFSHARING

22:27

@skillpatrol @Semiclassical It's like a worthless thing means that that thing is of no help, useless to anyone, right?

Something you might like to throw away or anything like that?

Semiclassical

right, or that said worthlessness is attributed to the subject rather than just their opinions (or to the subject's opinions in general rather than just on that topic)

skill patrol

It's more negative than that @OFFSHARING

OFFSHARING

hmmm

Forever Mozart

its ok relax

Bungo

@DanielFischer I wasn't aware of that history, I knew MathOverflow came before MSE but not that it was separate from StackExchange originally. I am reading the Wikipedia article now. Amusing quote from one of the founders: "Mathematicians as a whole are surprisingly skeptical of many aspects of the modern Internet... In particular, things like Facebook, Twitter, etc. are viewed as enormous wastes of time."

Semiclassical

22:28

i'd say that's about right, actually. the phrase 'worthless trash' sounds right to my ear

evinda

@DanielFischer That it converges to 0, while $n \to +\infty$ and that's why $||T^n x|| \to 0, \forall x \in \ell^2(\mathbb{N})$ ?

Semiclassical

well, in truth, they pretty much are :P

Bungo

as if no one ever wasted time on stackexchange :-)

Forever Mozart

LOL, I never had a facebook or twitter account

Daniel Fischer

@evinda Well, do you know that?

Semiclassical

22:29

i touch facebook like...once a month, if that

Forever Mozart

but mathematics is not a waste of time

Bungo

@ForeverMozart Me either, but I figured that's because I am old

Balarka Sen

the point is, it's a productive waste of time @Bungo

being in MSE, I mean

Semiclassical

i could see myself doing twitter for a specific purpose, but not as a social thing

Bungo

@ForeverMozart I of course agree, but "waste" is in the eye of the beholder I suppose. I'm sure people who watch cat videos all day don't think it is a waste of time either.

@BalarkaSen It can certainly sharpen one's mathematical skills, but it doesn't earn us any money, so it is not productive in the economic sense of the word.

Karl Kronenfeld

22:32

@OFFSHARING (Jumps in without looking back) It's an expression of one's own feeling that one suggests others should share. Calling person X's thoughts worthless suggests that in particular that person X should view these thoughts as worthless too.

Semiclassical

ehh, that depends. there's a lot of stuff I do on the internet which i do consider to be an objective waste of time, and which i do mostly b/c my procrastination is pretty compulsive :/

compulsive procrastination is a neat little hell sometimes

Bungo

@Semiclassical The internet is such a double-edged sword in that respect

Semiclassical

eh, i'm not sure i can blame it on the internet. if it wasn't there, i've no reason to think i'd be doing something else

OFFSHARING

@KarlKronenfeld I see, interesting. I used that with the meaning of no importance.

Bungo

@Semiclassical Without the internet, we might all be so bored that even work would seem interesting :-)

Semiclassical

22:34

it occurs to me that another version of that ambiguity is "insignificant" versus "not significant"

at one level they're exactly the same, but they don't quite sound the same to my ear

Karl Kronenfeld

@Semiclassical Procrastination isn't innately bad: you will often have fresh thoughts on nagging questions upon returning to them from some distraction. That's just not how people typically use the technique of procrastination!

Semiclassical

ehhh

not to exaggerate it, but that way of understanding it strikes me as like comparing feeling sad to clinical depression, or being nervous with chronic anxiety

Bungo

@KarlKronenfeld I should send those thoughts to the manager who emailed this to several teams on Friday: "Finally I would request everyone involved to be available to be responsive (phone/emails) and work over the weekend in order to get a consistent & cleaner build and also try those out in the lab. We are in an extreme crunch situation so weekend work is very much needed. Thanks in advance."

OFFSHARING

@skullpetrol are there some resources where the negativity of the English words are measured, assessed?

skill patrol

context and practice @OFFSHARING are your only guides :-)

Karl Kronenfeld

22:37

@Bungo :D

Semiclassical

namely that said comparisons, to me, pretty much discredit the speaker as not having experienced how profoundly disabling those can be

i guess there's a bit of a 'no true scotsman' paradox there, though

evinda

Yes.

For each $m,n$ with $m>n-1$ we have $\sum_{k=n}^m x_k^2= \sum_{k=1}^m x_k^2- \sum_{k=1}^{n-1} x_k^2$.
We fix n and let m tend to $+\infty$: $\sum_{k=n}^{\infty} x_k^2= \sum_{k=1}^{\infty}x_k^2- \sum_{k=1}^{n-1} x_k^2$.

While $n \to +\infty$: $\lim_{n \to +\infty} \sum_{k=n}^{\infty} x_k^2= \sum_{k=1}^{\infty} x_k^2- \sum_{k=1}^{\infty} x_k^2=0$. Right?

OFFSHARING

@skillpatrol lol, now I have practice after a real situation. However one might like to know some in advance about the negativity power of the words.

skill patrol

Yep :D

OFFSHARING

Anyway.

skill patrol

22:40

That comes from practice

OFFSHARING

@skillpatrol Well, you hardly know what the other feels when you tell some stuff (if you're not a native English speaker).

skill patrol

True

That's why familiarity with the personality of that person is important

Daniel Fischer

@evinda A bit roundabout. But yes, the convergence of a series means that the sequence of remainders tends to $0$. So you have your pointwise convergence $T^n x \to 0$ for all $x\in \ell^2$.

Danu

@BalarkaSen You mean inner automorphisms?

skull petrol

Welcome back pal.

Karl Kronenfeld

22:48

Who might that pal be?

skull petrol

Danu et al

:-)

Karl Kronenfeld

til that pal is sometimes plural.

skull petrol

Pal et al

Karl Kronenfeld

So you wouldn't qualify the others as pals?

At least relative to the Danu

Danu

@OFFSHARING Googling "define [term]" does it.

In the case of "worthless" it returns, for instance:

1. Having no real value or use.
2. (of a person) having no good qualities; deserving contempt.

OFFSHARING

22:52

@Danu In the conext with you I meant of no importance to me.

Danu

Then you should have used those exact words.

OFFSHARING

Well, I'm not sure 100% they cannot be replaced by each other. To me they sound approximatively the same.

Danu

They are not synonymous.

Agawa001

now i think my last answer is worth an attention since i backed it with a c++ program

Danu

You can look up synonyms too, on the internet.

OFFSHARING

22:56

@skullpetrol are you sure there is absolutely no situation when you can replace worthless by of no importance?

skull petrol

What do you mean?

OFFSHARING

@skullpetrol I mean a context where you can replace worthless by of no importance.

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$Mathematics$ Mathematics