Mathematics - 2015-01-18 (page 5 of 6)

masfenix

21:01

Hey guys, I am trying to show that an Algebra (closed under finite set operations) does not imply a $\sigma$-Algebra which is closed under countable set operations. The book gives the example of using $\mathbb Z$ as the space and letting the Algebra consists of sets $A \subset \mathbb Z$ so that either $A$ is finite or $A^c$ is finite.

but I don't see how this isn't a $\sigma$-Algebra.

evinda

@Axoren Could we prove like that the time complexity?
$D$ contains n elements, so the time complexity of Heapsort(D) is O(n*lgn).$E$ contains m elements, so the time complexity of Heapsort(E) is O(m*lgm). The while-loop is executed at most n times and the time-complexity of binary_search is O(lgm), so the time-complexity of the while-loop is O(n*lgm). Then the time complexity of the if and else statement is O(1).
In total, the time complexity of the algorith is $O(n \cdot lgn+m \cdot lgn+n \cdot lgm)=O((n+m) \cdot lgn+n\cdot lgn) \leq O((n+m) \cdot \lg(n+m)+(n+m) \cdoot \lg(n+m))=O((n+m) \c…

Axoren

Hold on, something's sketchy about your equations.

This part: $O(n \cdot lgn+m \cdot lgn+n \cdot lgm)=O((n+m) \cdot lgn+n\cdot lgn)$

$O((n+m)\log n+n \log m)$

Chris's sis

There will be a long night to me ... much work to do ... (hmmm, some coffee?)

Hippalectryon

@Chris'ssis Try to sleep a bit or you won't be able to be too productive tomorrow :)

Chris's sis

@Hippalectryon No sleep, but work, work, work ...

:D

Hippalectryon

21:11

plot twist : @Chris'ssis is a plant

Chris's sis

@Hippalectryon :-))))))

@Hippalectryon In my village there was a real math genius that went crazy after a while. So, some relatives often tell me to pay attention and work less, not to happen to me the same thing. :-)

Ted Shifrin

@Laters: I use "trivial" only in the technical sense (like "trivial solution" or "trivial linear combination"). I have been known to comment that some argument or computation is straightforward.

@Hippa: Tu te comportes bien ce soir? :)

As you can tell, @Chris'ssis, most of us are crazy.

2

Chris's sis

@TedShifrin Well said!3:-)))))

Hippalectryon

@TedShifrin >.> quelle question !

Ted Shifrin

Oui, enfin, quelle bêtise de ma part.

@Hippa: So it seems one of your French friends is concerned that we "have a problem"? :P

Hippalectryon

21:16

Who is that ?

teadawg1337

I went crazy before I realized I wanted to become a mathematician

Ted Shifrin

I don't remember his name, @Hippa. I just saw a comment when I was checking on a ping last night.

Hippalectryon

@teadawg1337 You became a mathematician because you were crazy :p

Ted Shifrin

Yes, @teadawg, you're a lost cause, too :D

Hippalectryon

@TedShifrin By any chance, Ramanewbie ?

Axoren

21:17

BizarroHippa

Ted Shifrin

Je ne me rappelle plus. Lui, il est français?

Axoren

Oh wait, the Z is backwards.

BisarroHippa

Hippalectryon

@TedShifrin That's the account that was once named TedShfrin

Ted Shifrin

Oh, would you come in pretending to be that person and ask why we don't get along?

Hippalectryon

Haha.The thing is, I gave him that account which had >50rep so that he could use the chat @TedShifrin. it's not mine anymore

Axoren

21:19

@TedShifrin The body snatcher cliché? Every one who is an imposter clone has to be evil?

Ted Shifrin

Je me méfie de toi, vraiment, @Hippa.

Axoren

I think he might just be a young kid or something.

evinda

Does someone of you have the second version of the book Introduction to Algorithm?

Hippalectryon

@TedShifrin Je me demande pourquoi ! regarde par la fenêtre d'un air innocent

Gato

Good evening chat-team!

Axoren

21:20

@evinda I might.

user153330

@Gato good evening

Hippalectryon

@Gato o/

Gato

@Hippalectryon qu'est-ce dont?

Hippalectryon

@Gato High five :D

Ted Shifrin

Indeed, @Hippa, it's clear you're a sham.

Gato

21:22

@Hippalectryon ah! what's up?

Ted Shifrin

Bonsoir, M le @Gato

Axoren

@evinda I don't.

Hippalectryon

@TedShifrin He's not me, though. Don't you think that !! >:c

@Gato The sky :P

Axoren

@TedShifrin It's true, they were having a genuine argument that was happening too fast to be him and himself.

Ted Shifrin

Pas vrai. Le ciel est bleu.

JohnMerlino

21:23

I was reviewing some arithmetic. It states that a divisor is any number that doesn't leave a remainder. So 12,3,4,6,12 are divisors of 12, since when you divide them into 12, you will not have a remainder. if you divide 12 by 24, you will not have a remainder either. Why isn't 24 a divisor of 12?

evinda

@Axoren A ok.. I am looking at questions of a previous exam and one that was related to Depth-first search must be from this. I wnted to ask you if you could take a look at it and tell me what types of exercises there are that are related to Dijkstra's algorithm..

Ted Shifrin

They were arguing, @Axoren?

evinda

@Axoren No problem... :)

Axoren

@TedShifrin ... I think so...

Ted Shifrin

@JohnMerlino: You have to divide and get an integer !!

So to say $a$ is a divisor of $b$ means that $b=a\cdot c$ for some integer $c$ !!

JohnMerlino

21:24

@TedShifrin 12/24 is the integer 2.

Ted Shifrin

No, it's the non-integer $1/2$ !!

You're upside-down.

Axoren

@JohnMerlino you're flipped sideways.

Ted Shifrin

Now we'll confuse him, @Axoren: sideways and upside-down are not the same :D

Chris's sis

Guys, I'm curious to know if you ever met someone that published a math book without having a math background. I'm just curious, maybe you know some in your countries. I'd feel more comfortable I'm not alone ... :-) -- maybe @TedShifrin knows some.

user153330

@TedShifrin le ciel est bleu-ciel

JohnMerlino

21:25

You are right, that will result in a decimal number or fraction

Ted Shifrin

Certainement, @user153330

Axoren

@TedShifrin Technically you were wrong because if you flip it upside down you get 1\5.

Or something to that effect :P

Studentmath

@Ted!

Ted Shifrin

Flipping $\dfrac{24}{12}$ gives $\dfrac{12}{24}$, @Axoren, so shaddup :D

user153330

@Chris'ssis did you ever try to post it as a question in academia.SE?

Ramanewbie

21:26

@TedShifrin what did Hippa do to you ?

Ted Shifrin

@Studentmath !!

Chris's sis

@user153330 No

Axoren

@TedShifrin I was making a jok-- OH MY GOD, HE'S HERE

And I can't prove he's not Hippa!

Ted Shifrin

Uh huh, @Axoren, uh huh. :)

You should ask him, @Ramanewbie.

Axoren

You could say that he's potentially HippaCompliant.

user153330

21:27

@Chris'ssis well try, you'll surely get an answer there

Gato

@TedShifrin I need some help, I would like to prove that two complement subspaces in $E_1$ a subspace of $E$ are isomorphic ($F_1+F_2=E_1$). I took a projector restrained to $F_1$ noted $u$. So $u:S_1\rightarrow S_2$ is a morphism right?

user153330

@Axoren he's a hippi (or hippy?)

Ted Shifrin

Wait, @Gato. You mean $F_1+E=E_1$ and $F_2+E=E_1$?

Horrible notation, btw.

Axoren

The seeds of doubt are planted and they bore fruit in an instant.

Ted Shifrin

LOL @Axoren ... That is probably not universally understood, compliant or not.

@Gato: How do you take a projector? Are you working with a Hilbert space, or are you working with quotient spaces?

Gato

21:29

@TedShifrin sorry I don't know the latex code for the direct sum... $F_1+E_1=E$ and $F_2+E_1=E$.

Ted Shifrin

Heya @JMoravitz

Axoren

$\oplus$

Ted Shifrin

\oplus

JMoravitz

g'afternoon ted

Ted Shifrin

smacks everyone

Axoren

21:30

Ted is right

Ted Shifrin

Imagine that :)

Studentmath

He had a doubt

Chris's sis

@user153330 there is a first time for all things. I don't insist on finding out an answer though.

Laters

@TedShifrin alright. a professor of mine says that in his opinion using the words "trivial" etc is like using a weapon

Axoren

Ted's been filling my head with doubts.

Ted Shifrin

21:31

LOL @Axoren ... clever guy.

Gato

@TedShifrin so $F_1\oplus E_1=E$ and $F_2\oplus E_1=E$. $E$ is just a vector space.

Ted Shifrin

Yes, @Laters, in general, making one's audience feel stupid is not well-advised.

Laters

he often says that one should not use them. nonetheless he uses it in the next sentence

Ted Shifrin

Finite or infinite dimensions, @Gato?

Ah, @Axoren, you're from my old home. I spent 10 years in Boston/Cambridge area.

Laters

btw does anyone know about the gato derivative

JohnMerlino

21:32

I was just watching a video on youtube where 2 in the panel argued for a multiverse and the two others believed in one universe. And the two who believed in universe said that they cannot believe that infinity is real.

Ted Shifrin

Gateaux

Gato

quelconque so the proof should works for any dimensions.

Axoren

@TedShifrin Been to Somerville?

Laters

I no

Axoren

It's full of Hipsters now.

Laters

21:32

actually there are quite a few people who spell it Gato

Ted Shifrin

Of course, @Axoren, many times. Tufts, Steve's Ice Cream, and my car repair shop was there :P

Laters

I am not joking

Ted Shifrin

It was slightly so even in the 70s, @Axoren.

Do you know about quotient vector spaces, @Gato?

Axoren

@TedShifrin Now, it's entirely. They replaced the McDonalds in Davis with a coffee shop. Then surrounded it with two more coffee shops.

Laters

is the weak derivative (Gato's derivative) of the functional i(Fi). The formula of finite increments for the functionals permit to estimate (by Gato's derivative) .

Ted Shifrin

21:33

If so, prove that any complement is $\cong E/E_1$.

I've never seen it spelled that way, @Laters, never.

Well, any time a McDonalds bites the dust, it makes me happy, @Axoren :P

Gato

@TedShifrin does it works as the same as group theory?

Ted Shifrin

Yes, @Gato.

Same homomorphism theorem :)

Axoren

@TedShifrin WingWorks is gone, replaced by a Shwarma Palace. That's where I have to draw the line.

The only place left for good chicken is Mike's now

Laters

apparently it is mainly used in translations of russian mathematical texts

somehow

Ted Shifrin

Well, I don't know what those are, @Axoren. I left in 1981, and have only been back to visit a few times, last time 1999. :)

Studentmath

21:35

@Ted won't the proof be identitcal if it wasn't a Vector space, but just a group? Or would it be false then?

Hippalectryon

@Axoren That's because chicken run

Ted Shifrin

@Laters: Translations of Russian notoriously misspell everything.

Axoren

@TedShifrin Ahh, I misunderstood, I thought you meant 10 years ago, not 10 years total

Gato

@TedShifrin okay, cool :). Je commencer à essayer, see you later!

Studentmath

Shwarma meant Shawarma?

Axoren

21:36

I never learned how to spell it.

Ted Shifrin

Yup, high school, college, and then postdoc ... 10 years, @Axoren.

Axoren

It's the vertical meat cycles

Studentmath

It's awesome. What do they give it with?

Axoren

That they carve and make food out of.

Studentmath

Yeah

Yesyes

Ted Shifrin

21:37

@STudentmath is from the Mideast :P

Axoren

I don't know, I've never been to it.

Studentmath

You should try it out

Axoren

The owners of the place are really smart

I knew the past owners.

Studentmath

If it's Turkish and with Yoghurt, they are doing it wrong (though it's still awesome)

Ted Shifrin

I was thinking Lebanese

Axoren

21:37

The part owners didn't want to sell, but they kept asking. They put a ridiculous number on the table and they took it.

So the past owners were like "Oh, well damn. Sure."

Studentmath

Lebanese do it like we do it, with Hummus

Axoren

It's in a perfect location where it gets lots if visage.

So everyone knows it's there and will eventually try it out.

Ted Shifrin

I cook lots of cuisines, @Studentmath, but never have done Mideastern.

Axoren

Where is Falafel from?

Pretend I spelled it right.

Ted Shifrin

@Axoren: You're clearly not a native English speaker :P

user153330

21:39

best book cover

Ted Shifrin

Same part of the world.

Studentmath

Yeah you did, well we would claim it's form here. I have no idea where it is really from

Axoren

I'm a native English speaker but I've only been spelling for the past 5 years.

Ted Shifrin

LOL @Axoren. Oh. "gets lots of visage"?

Laters

hahha

Ted Shifrin

21:40

visage = face or appearance, just as in French.

Studentmath

@Ted I don't know what would count as mideastern, really. We have here a mess of African/Asian/Iraqi/European food - I don't know which would count as true mideastern :P (besides Shwarma, and Flafel.. Kabab..)

Axoren

"The surface of an object presented to view"

Gato

@TedShifrin Using the first isomorphism theorem (celui avec le Ker) it works perfectly. Thanks again.

Ted Shifrin

don't forget gefilte fish, @Studentmath :D

Yes, @Gato, c'est ça.

Studentmath

That's so mideastern!

I love it. It's like a dessert as a main dish. I never understood those that hate it

Ted Shifrin

21:41

dessert? huh?

I've had lots of bad matzo ball soup, too.

I tend to make a French version if it's my choice :P

Studentmath

@Ted don't you have it sweet, too? It's usually sweet here.

Seems like you've been through few Passovers :P

Ted Shifrin

Oh, I hate sweet gefilte fish. Yuck.

Probably they put beets in it or something?

My last year in grad school, actually, I collaborated with my neighbor upstairs. She was also a gourmet cook, and we made the best Pesach dinner either of us had ever had :P

Studentmath

I heard few that actually put -sugar- in it. But yeah, that too.

That sounds awesome

Mike Miller

Hi, @Ted

Laters

good old times huh

listening to spanier all day long

Ted Shifrin

21:46

Good night, @Mike.

Mike Miller

Oh - forgot my good morning

Ramanewbie

Hey guys I've a problem :

isn't $20−5y+4x=0$ equal to $\frac{4}{5}x+4$ ???

teadawg1337

yes

Ted Shifrin

@teadawg: What?

teadawg1337

$y=\frac45x+4$

Ted Shifrin

21:47

Ah, that's better.

Studentmath

Also, equivalent

Ramanewbie

that's what I thought, tks

Ted Shifrin

No ... An equation is never "equal to" an expression.

Ramanewbie

@TedShifrin forget the $=0$

my bad

Studentmath

@Ramanewbie wait what?

Ted Shifrin

21:48

No, you need the $y=$.

Just talk to your buddy @Hippa.

Ramanewbie

@TedShifrin why should I need that ?

don't I have choice to put it or not ?

teadawg1337

Well, it's not actually equivalent to $y=\frac45x+4$, since $20-5y+4x=0$ is asking for solutions $(x,\,y)$ that result in zero

Ted Shifrin

NO. No choice.

teadawg1337

If we're going to be technical

Ramanewbie

ok fine

Mike Miller

21:50

@TedShifrin Kobayashi didn't have what we wanted. Do you mind walking me through something? :)

Studentmath

Well, they have the same set of $(x,y)$ soltuions @Teadawg

Ted Shifrin

Let's try to be correct, @teadawg, especially when helping people less advanced than you.

I might need my martini first, @Mike.

Mike Miller

Maybe, @Ted, I've gone through more than a few beers when trying to figure this out.

Laters

how does one prove that the tangent bundle of the sphere is trivial iff n=1,3,7 ?

Axoren

Who here has some good experience with finite group theory?

Ted Shifrin

21:51

Adams spectral sequence stuff, @Laters. Hard.

Mike Miller

K-theory and spectral sequences. Hatcher does it in his vector bundles book.

teadawg1337

@Ted Still super tired, not paying as much attention as I should be :/

Ted Shifrin

It's ok, @teadawg. I still like ya :)

Studentmath

You drink when trying to solve? I find myself consuming tons of water but never alcohol

Mike Miller

@Studentmath The trying to solve leads me to drink.

Ted Shifrin

21:52

Wait 'til you're my age, @Studentmath. Grading exams is much more fun after a few martinis :P I guess I'm retiring before I become an alcoholic :D (just joking)

Axoren

@Studentmath They say water is the alcohol of devils and demons.

I drink a lot of water, too.

Studentmath

@Axoren what are you hinting?

@Mike @Ted lol

Laters

@TedShifrin I thought the proof was hard, but I heard usiing K theory it becomes sort of understandable

Mike Miller

@TedShifrin: I have $a \in \Omega^1(\mathfrak g_E)$. You tried to explain the other day what $a$ does as a map $\Omega^1(E) \to \Omega^2(E)$. I didn't understand it. Could you explain again and be explicit?

Laters

at some point I will understand it

Axoren

21:53

@Studentmath We're math demons.

Laters

I use spectral sequences like other people use matrices

Studentmath

@Axoren that would mean I actually have some skill in Math, so I am afraid I have to disagree

Axoren

@Studentmath There are plenty of demons from antiquity that had lackluster skills.

Mike Miller

Actually, the proof I am best aware of doesn't use spectral sequences, now that I think of it

Axoren

I'm pretty sure there was one known for just sitting behind people's eye lids and itching at them in the mornings.

Mike Miller

21:55

uses certain automorphisms of the K-theory functor and Bott periodicity

Axoren

But yeah, does anyone know a simple solution to this problem?

Ted Shifrin

I didn't explain, and I didn't think it through thoroughly, @Mike. But given a connection $D\colon \Omega^0(E)\to\Omega^1(E)$, I gave you the rule for the induced connection $D^{(k)}\colon \Omega^k(E)\to\Omega^{k+1}(E)$. I claim you can sort it out in the terms you want.

Mike Miller

@TedShifrin This is not what I asked about. $a$ is not a connection. I'm fine with $D$.

Ramanewbie

@Hippalectryon how do you say "repère" again please

Ted Shifrin

frame, @Raman

Ramanewbie

21:57

frame ?

Ted Shifrin

And wtf are you doing with that word when you can't do eighth grade algebra?

Ramanewbie

@TedShifrin I thought it was sth like coordinate xxx

Hippalectryon

@TedShifrin He's talking about basic frames

Ramanewbie

yes I bet I am !

Hippalectryon

Like $(O,\vec{i},\vec{j})$

Laters

21:58

@MikeMiller could you give me a link/reference for that proof?

Ramanewbie

exactly that

Ted Shifrin

I'll leave you two to chat with yourself.

Studentmath

French Education system is odd

Axoren

For a member $g$ of a monoid with presentation $\langle s, t\ |\ s^2 = 1,\ t^n = 1 \rangle$. How can I get the minimal string of $s$ and $t$ whose product produces that element $g$?

Hippalectryon

@TedShifrin I'm kind of busy atm sorry :/ I just checked the chat at that moment at random

Mike Miller

21:59

One wants to derive a formula for the curvature of $D+a$ in terms of $D$. The formula I've seen is $(D+a)^2 = D^2 + aD + a \wedge a$. But to interpret the series of formulas correctly, one needs to know how $a$ acts as a map $\Omega^1(E) \to \Omega^2(E)$.

@Laters Hatcher's vector bundles book

If you're fine with Bott periodicity just jump to section 2.3

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$Mathematics$ Mathematics