Mathematics - 2012-08-07 (page 4 of 6)

Matt N.

19:00

No joking.

So it will look as if I knew nothing. Apart from the fact that I struggle to compute the image of multiplication by $n$ on integers mod $m$, which is high school algebra.

robjohn

@t.b. hey there :-)

Matt N.

So particularly futile is going to the exam.

It'll be so devastating.

robjohn

@BillDubuque: was there ever an orientation after the last election?

t.b.

@MattN. I already noticed that you had a knack for positive thinking...

Matt N.

@t.b. Picture this: I play classical guitar as a hobby and when I have a lesson my hands are too shaky to play. Now extrapolate to taking an oral exam where stuff matters.

t.b.

19:02

@robjohn hey, robjohn

Matt N.

And that's with a teacher I like hanging out with for fun.

Whereas lecturers already scare the shit out of me without having to be examined by them.

And yes, by thinking about this I am only making it worse.

Jonas Teuwen

Stop thinking, start drinking.

skullpatrol

^ good point

t.b.

@MattN. I understand. Look, there's about nothing you can do against this kind of anxiety but it most definitely doesn't help to spend your time imagining your possible failure.

N3buchadnezzar

Oh man loads of drinking next weekend =)

Matt N.

19:06

I know!

But I'm too realistic. : (

@JonasTeuwen Can't afford. Need tomorrow.

Jonas Teuwen

"too realistic", you're now bullshitting.

You're inventing situations which might happen and will surely not happen as you are seeing them now.

skullpatrol

^ better point

Jonas Teuwen

@MattN. Sure you can. Just drink one, or whatever. This does not help you either. Go to a bar with a book and drink something.

Matt N.

Ok. I will stop now. I will memorise the stupid fact about multiplication on Z mod m and use it if I have to.

At least I have an idea about inverse and direct limits...

skullpatrol

Memorizing helps relieve anxiety.

Matt N.

19:09

Bleh. Sorry. Said I'd stop.

whistles

@t.b. Do you have a minute to have a look at a diagram chase?

I only did one part out of three because running out of time.

t.b.

depends on what it is

Matt N.

It's exactness of direct limits.

But it's probably right. (Feels right, anyway.)

t.b.

okay

Matt N.

It's lurking here.

It's a bit inconvenient, isn't it : (

t.b.

@MattN. that's perfect

Matt N.

19:16

@JonasTeuwen I saw that. When I wrote that my hand were a bit shaky from coffee overdose in combination with stress : (

@t.b. Thanks for looking! xxx

Jonas Teuwen

Still looks good.

Matt N.

I like it better when it looks neat.

t.b.

and it's actually the non-obvious part of it. The other two parts follow from general nonsense.

Matt N.

But I'd rather have a nicer handwriting.

@t.b. Oops. I thought all three parts were the same.

I guess he won't have me do all three because not enough time during the exam.

t.b.

@MattN. well, not quite, you don't have injectivity to use anymore.

Matt N.

19:18

True but I meant, I can just do something similar with surjectivity.

t.b.

yes you can, but there's an easier way.

Matt N.

It seemed to me that all diagram chasing is the same: know one, know them all.

You're scary too. I'm quite sure you wouldn't pass me if this was your lecture.

Ragib Zaman

Hey guys.

Matt N.

Hi.

skullpatrol

hi

t.b.

19:20

You know that taking direct sums is exact and then you know that the three colimits are obtained from taking a certain cokernel. The snake lemma then tells you exactness in the middle and at the right hand end, and you've just done the left hand thing which you need to do "by hand".

Ragib Zaman

Anyone know anything about PDE's? I want to solve $\frac{ \partial p}{\partial x_1} + x_1 \frac{ \partial p}{\partial x_2} = 0.$

t.b.

I'll be back in a moment

Matt N.

Ok!

I did not know that taking direct sums was exact. It feels like we get only bits and blobs of the whole picture in this lecture.

skullpatrol

Maybe the textbook gives a more complete picture.

t.b.

19:35

@MattN. it's rather straightforward to prove.

(the first sentence, not the second!)

Matt N.

But I don't have time to learn something that will not be asked.

Well, assuming he will insist on seeing a diagram chase.

Clearly something that might be tested.

Thanks for being nice to me and not telling me that I should learn during term. Because I can't.

Peter Tamaroff

Guys

t.b.

@MattN. that wouldn't be a very helpful thing to tell you at this point in the semester, right? :)

@PeterTamaroff guy?

Peter Tamaroff

@t.b. I have a quick question, since I'm off to work in a while.

t.b.

@MattN. I think spending ten minutes on that wouldn't hurt...

@PeterTamaroff sure, shoot

Peter Tamaroff

19:40

I want to prove that if $\Sigma$ is a basis for a topology $\mathfrak I$ on $X$, then $\bigcup \Sigma= X$.

Matt N.

@t.b. : ) But it would not work at any time.

Peter Tamaroff

Now, it seems pretty easy

Matt N.

@t.b. Ok. But let me finish finding a free resolution for a f.g. Z-module first.

t.b.

@PeterTamaroff isn't that part of the definition?

Peter Tamaroff

@t.b. I'm meddling with the two definitions, proving they are equivalent. =)

t.b.

19:42

then meddle away!

Peter Tamaroff

Since for every $x\in X$ we have a basis element $B$ such that $x\in B \subset X$. Thus $\bigcup_{x\in X} B_x=X$

Correct?

t.b.

(but tell me what the definitions are...)

Peter Tamaroff

@t.b. Oh, fair enough.

Matt N.

Ooooh! That was easy.

Peter Tamaroff

The first one is, "$\Sigma$ is a basis if every open set $O$ is a union of basis elements."

t.b.

19:43

@MattN. take $n$ generators map the standard basis $\mathbb{Z}^n$ to them. Take the kernel, continue.

yes, verrry easy

Peter Tamaroff

The other is "$\Sigma$ is a basis for $\mathfrak I$ provided, for each $O\in \mathfrak I$ and each $x\in O$ there is a basis element for which $x\in B\subset O$, and provided $x\in B_1\cap B_2$, there is a basis element $B_3$ such that $x\in B_3\subset B_1\cap B_2$"

And as I'm proving, both are equivalent.

Well, actually I got the secod implies the first form Munkres. I am proving the other direction

t.b.

right.

Peter Tamaroff

Note it is for each open set. So $X$ is there too.

t.b.

yup

so, where's the trouble?

Peter Tamaroff

@t.b. I thought you said something about the empty set being a basis, that's why. Just a glitch?

@t.b. Just asking if I was making sense, dunno.

t.b.

19:48

@PeterTamaroff yes, I mistook an I for a Sigma

(sheesh topologIes that's why there's an I)

Peter Tamaroff

@t.b. Yes, sorry. I'm too used to $\mathfrak I$. I usually use $\mathscr B$ for basis, but the Russians use $\Sigma$, and I think it is cool notation. Suggestive.

Jonas Teuwen

Annihilate existence.

Peter Tamaroff

OldJohn gave me a book full of exercises.

t.b.

Because in your dreams you think of $\Sigma$ases?

Peter Tamaroff

Actually, it is like a theory book, but you gotta prove the theory.

They are asking now "Riddle: What topological structures have exactly one base?"

Matt N.

19:51

Quickie: For $A$ f.g. $R$ module, $B$ any $R$-module, I can do the following free resolution (because every module is isomorphic to quotient of a free module):
$$ 0 \to M \to \bigoplus_{k=1}^n R \to A \to 0$$ and then from
$$ 0 \to M \otimes B \to \bigoplus R \otimes B \to 0$$
one sees that $\mathrm{Tor}_n (A,B) = 0$ for $n \geq 2$ and since the resolution is free hence in particular projective, we get the same for $\mathrm{Ext}^n$?

t.b.

@MattN. but why is $M$ free?

not in general

Matt N.

It's a submodule of the direct sum. Aren't submodules of free modules free?

t.b.

nope otherwise there wouldn't be any interesting projective modules.

Jonas Teuwen

@MattN. Those formulas look cool on a grave stone. With the words: "I don't know what those mean, but they looked cool. Cheers!"

Matt N.

@JonasTeuwen Yeah order me one please for Thursday.

Jonas Teuwen

19:53

@MattN. 8-).

Matt N.

@t.b. Then I assume that's why they set $R = \mathbb Z$ in the HW question?

I thought I could generalise.

t.b.

@MattN. you can. It's true in PID's

Jonas Teuwen

@MattN. Existential crisis eh?

Crisis sounds too dramatic. Need a different word. NOW.

t.b.

@MattN. for an easy counterexample: Take $\mathbb{R} = \mathbb{Z}/(4)$. Now look at the submodule of $M = R$ generated by $2$.

Matt N.

Oh, that's of course not free.

Thank you.

Jonas Teuwen

19:58

Why does a productivity book has such a bloody long boring text in chapter 1 already? Oh man.

t.b.

It wants you to be productive, viz. it wants you not to read it.

Matt N.

Jonas, if you used all this time you spend reading productivity books... : )

Beat me to it : )

So. Now for the real deal. Claim: direct sum is an exact functor on R-Mod.

Jonas Teuwen

@MattN. Correct... the time I have used to actually try to read them.

robjohn

@PeterTamaroff $X$ is open, and must be a union of basic open sets. Does that not fit with your definitions?

Matt N.

I think I read this somewhere. I need to prove this by induction.

t.b.

20:03

well for finitely many it's not so hard....

it suffices to do it for two where it's pretty clear, no?

Matt N.

I get infinitely slow after like 8 or 9 hours of work. So: yes, it's not so easy.

@t.b. As always I'm not even sure what I want to prove so I'm still trying to figure out my preconditions.

I think by induction does the trick. I start with a sequence os SESs and then just apply the direct sum to the whole sequence.

t.b.

Given $M' \overset{f'}{\rightarrowtail} M \overset{f''}{\twoheadrightarrow} M''$. and similarly for $n$. You want to prove that the direct sum is exact.

Matt N.

I think given $0 \to A_i \xrightarrow{f_i} B_i \xrightarrow{g_i} C_i \to 0$ I want to prove that $0 \to \bigoplus A_i \xrightarrow{\oplus f_i} \bigoplus B_i \xrightarrow{\oplus g_i} \bigoplus C_i \to 0$ is exact.

t.b.

Meaning: $$M' \oplus N' \overset{(f',g')}{\rightarrowtail} M \oplus N \overset{(f'',g'')}{\twoheadrightarrow} M'' \oplus N''$$ is exact

@MattN. exactly.

Matt N.

So there is nothing to prove : )

Of course sticking stuff together like that won't hurt exactness.

t.b.

20:08

careful: it's not true with arbitrary products

Matt N.

Ah, those direct products have to be finite?

t.b.

For products you need finiteness or some esoteric conditions you don't want to know about right now.

For direct sums it's always true whether the family is finite or not.

But I agree for finitely many summands it's pretty obvious

Matt N.

Ok. So I guess I do have to prove it.

t.b.

Do it for two

Matt N.

But I thought we just agreed that it's obvious for finitely many! Then product and direct sum are the same on top of that.

Ok. Let's see.

But isn't that boring?

t.b.

20:12

it is.

Matt N.

I pick a thing in the image of $(f^\prime , g^\prime)$ say $(x,y)$ and then $x$ is in the image of $f^\prime$ and $y$ is in the image of $g^\prime$ then those are in the respective kernels ...

I take too much time to type.

t.b.

so that's exactness in the middle. right. the other two are just as easy

Matt N.

Yes, same.

t.b.

so go for infinite sums

user19161

@JonasTeuwen Well, reading most of these productivity books is not productive!

t.b.

20:14

@JasperLoy third

user19161

@t.b. I just understood what you meant by that. :-)

t.b.

@MattN. the important thing is that an element of an infinite sum only has finitely many nonzero entries

@JasperLoy I was deliberately cryptic :)

Matt N.

@t.b. I'm stumped.

t.b.

Did you understand that sentence?

Matt N.

Yes. But I don't see how to make that into a proof.

t.b.

20:18

Let's look at exactness in the middle

Matt N.

I mean yes, we know that it's exact for finite sums.

Can we just say an element is nonzero between indexes -N and N and then use the finite case?

t.b.

or just do it explicitly.

Matt N.

I thought that was explicit.

Pick one element $x$ in the image of $f$, only non-zero between -N and N. Then by the finite case it is in the kernel of $g$ which is contained in the infinite sum.

t.b.

Take something in the kernel of the right hand map. It has finitely many non-zero entries. For each nonzero entry, the entry is in the kernel of the map of the coordinate. Hence it's in the image of the injection of that coordinate. Do this for every non-zero coordinate. Get something in the left hand side that's mapped to the element you started with.

@MattN. the non-trivial inclusion is that $\ker{g} \subset \operatorname{im}f$

Matt N.

@t.b. The proof looks symmetric to me. If what I wrote above is correct.

Probably isn't.

It's too late here.

t.b.

20:24

if $gf =0$ then $\operatorname{im}f \subset \ker{g}$ is automatic, no?

Matt N.

Sure.

t.b.

so it's the other inclusion that needs an argument

Matt N.

But we don't have $gf = 0$.

t.b.

we surely do

robjohn

I can only think of smart-ass answers to this question

Matt N.

20:26

I'm getting too tired, sorry.

t.b.

@robjohn like $20^{23423432}$?

@MattN. no worries, I don't want to keep you awake with this. But spend a few minutes with that.

Matt N.

I assume you'll be off soon? That is, if I go afk to do a little bit of revising when I come back you'll have left so I better say good night now?

robjohn

@t.b. No, I was thinking of $4\arctan(1)$ uses only two digits and gives all the digits of $\pi$

among others

t.b.

@MattN. I'll check back

Matt N.

I'll spend some minutes tomorrow. Ok, I'll be back later.

t.b.

20:29

@robjohn not bad :) depending on where you are in the States, 4 would count, too, I guess.

robjohn

@t.b. or 3 if you believe in certain books...

t.b.

oh, right

well, I never understood this obsession with $\pi$.

Jonas Teuwen

Obsession with $\pi$? That's $3 \leqslant \pi \leqslant 4$, right?

t.b.

20:56

@robjohn congratulations for the modhood!

anon

modhood?

t.b.

@anon this

anon

cool. I don't yet see a diamond though!

t.b.

moderatorship would have been the word, I guess

Jonas Teuwen

@robjohn Oh no! 8-). Now I should be afraid of you.

robjohn

21:03

@t.b. not yet, as far as I can tell.

Jonas Teuwen

@anon Secret police?

Matt N.

Back.

Old John

@JonasTeuwen The king is dead - long live the king!!

Jonas Teuwen

@OldJohn Yes.

Old John

(mean to say that to Rob - but never mind)

Jonas Teuwen

21:05

Does anybody know this book Gettings things done by David Allen?

I guess I need some Avro Pärt.

Matt N.

Yep, my girlfriend read it.

Jonas Teuwen

Beats any book anytime.

@MattN. And?

Matt N.

@JonasTeuwen She thinks it's got some good ideas but is too extreme.

Old John

@JonasTeuwen Not heard of the book - yes time for more Arvo Pärt. (or Avro Pärt. - take your pick)

Jonas Teuwen

@MattN. Sure, but good ideas are good. You see?

Nimza

21:07

Hello everybody!

Matt N.

@JonasTeuwen Sure.

Jonas Teuwen

@OldJohn I always get confused because... Avro is a TV station in NL :-).

Old John

@Nimza zdrasvyutye

Jonas Teuwen

@MattN. But it stinks to read a whole book for one idea.

Old John

@JonasTeuwen and there was an AVRO chess tournament many years ago, I remember

Jonas Teuwen

21:08

@OldJohn Yes.

Matt N.

@JonasTeuwen It also stinks to write a whole book just for one idea.

Jonas Teuwen

@MattN. Not if it makes you a millionaire.

Old John

@JonasTeuwen Capablanca / Alekhine era, I recall

Jonas Teuwen

I would go into the professional trolling business if I am a millionaire.

@OldJohn Yes, the cool days.

Matt N.

Assuming "to stink" = "to suck monkeyballs"

mixedmath

21:09

alo all

Nimza

@OldJohn И Вам не хворать!

Old John

@JonasTeuwen back in the days when chess was played between humans :)

Jonas Teuwen

Michael Tal is my favorite. So aggressive. I like aggressive chess.

@MattN. Yep.

Matt N.

Hey teddy, going to sleep soon. Good night!

Jonas Teuwen

@mixedmath Go away... you're so scary now 8-). A mod without diamond. Gives a stasi appearance! 8-).

Matt N.

21:10

Good night bros!

t.b.

@mixedmath hello, congratulations to you, too!

Good night, @MattN.

Old John

@Nimza gavaryu tolka nyemnogo po-pusski :(

Matt N.

I feel violated. By oral exams : ,(

Old John

@JonasTeuwen I was always a capablanca (and Petrosian) fan

Jonas Teuwen

boingboing.net/2012/08/07/what-do-christian-fundamentali.html < bizarre.

@MattN. Lol.

mixedmath

21:12

@t.b. thank you -

Old John

my favourite Venn diagram

mixedmath

King Robjohn, too, sometime

Jonas Teuwen

@OldJohn My favourite Venn diagram is no Venn diagram.

@mixedmath Your profile... the mid-June is that 2013?

mixedmath

oh, I suppose I should change that now that I'm back, hmm>

;p

Jonas Teuwen

Too little cockroaches.

MAOR ROACHES.

t.b.

21:18

^ is this a Whisky?

Jonas Teuwen

@tb No... sounds like a good idea though.

Old John

@JonasTeuwen I'll second that (not to mention its a birthday :))) )

Jonas Teuwen

@OldJohn I'll drink on your health.

Old John

@JonasTeuwen Thank you - have a large one!

t.b.

@OldJohn congratulations!

Old John

21:21

@t.b. Thanks

Jonas Teuwen

@tb It is Periodic Brain Cramp Existential Crisis Deterioration. (whatever that is)

robjohn

@t.b. I guess it will be coming soon. I see the meta notice, however :-)

Old John

Sometimes I wonder how I came to be this old - and then I tell myself that I did it by not dying (yet)

Jonas Teuwen

"By eating lots of carrots"

t.b.

@robjohn I'm very happy about that. You had my vote, so...

Jonas Teuwen

21:23

You had my vote too. Perfect.

You will be a fine dictator.

5

t.b.

moderator and room owner. Better not upset him here...

Jonas Teuwen

Not yet network admin...

robjohn

@t.b. perhaps I should modify OldJohn's Venn Diagram :-D

Jonas Teuwen

Send us off into oblivion!

Undo existence. Fix the mistakes.

Finlaggan.

Even after washing the glass it still smells like rat juice...

Old John

@JonasTeuwen That could be made into a Haiku :)

(well - the first 3 lines )

Jonas Teuwen

21:28

@OldJohn Maybe it needs some cockroaches.

What is the precise definition of a Haiku? I'll Haikufy it!

Michael Boratko

Hello everyone - I have a question (hope I'm not interrupting).

In Apostol's Calculus, he discusses "mildly" nonlinear differential equations of the form $y''+y+\alpha y^2=0$ where $\alpha$ is a small nonzero constant

Old John

@JonasTeuwen Not sure too complicated for me tonight :(

Jonas Teuwen

@OldJohn Too much Finlaggan or too little Arvo Pärt?

Old John

@JonasTeuwen both!

robjohn

@t.b.: i.sstatic.net/qdrJH.png

Michael Boratko

21:33

He begins by assuming the solution can be expressed by a power series of the form $y=\sum_{n=0}^\infty u_n (x) \alpha ^ n$ which is valid in some interval $0<\alpha<r$, and then suggests that we substitute this series in the differential equation and equate suitable powers of $\alpha$ to determine $u_0$ and $u_1$

Jonas Teuwen

@OldJohn Can be fixed. I bet the amount of Finlaggan in the blood is a one-point compactification.

@robjohn Perfect! 8-).

Michael Boratko

It's this last part that I'm confused about - in general, one can use the method of equating coefficients with power series of a variable, because the terms are linearly independent

Jonas Teuwen

@robjohn Sorry about the meta post.

Michael Boratko

here he seems to treat $\alpha$ as though it is a variable, when he previously states it is a constant, can someone clarify my misunderstanding?

Old John

@JonasTeuwen actually it is quarter cask Laphroaig - a present :)))

Jonas Teuwen

21:36

@OldJohn Fabulous! I also have that one. Three bottles. Would be a shame if I run out of QC.

robjohn

@JonasTeuwen I didn't notice it until you mentioned it.

Jonas Teuwen

@robjohn I can delete it if you want me to :-).

Well, you can also do that yourself, but... 8-).

robjohn

@MichaelBoratko $\alpha$ can be changed, but it does not depend on $x$, so it is constant as far as the DE is concerned, but it can be varied.

t.b.

@robjohn heh, that's the General Audiences version :)

Michael Boratko

@robjohn so the terms in the series are linearly independent, and therefore each coefficient of $\alpha^n$ in my final power series representation must be zero? (treating it this way does give the correct answer, it just seemed sort of hand-wavy)

robjohn

21:41

@t.b. well, this is a PG forum, so I thought it appropriate.

t.b.

@robjohn interestingly your ping did ping me, but it remains visible in the starred board

that's spam, right?

robjohn

@t.b. I am confused. why is any of that surprising?

t.b.

@robjohn well, pings used to be removed in the starred board.

robjohn

@t.b. No, it seems that is a demonstration of visual noise.

t.b.

Okay, but rather borderline, it seems given the link in the bio

robjohn

21:45

@t.b. You mean the @t.b. part? some seem to remain and some seem to be stripped.

t.b.

@robjohn It probably depends on whether they were entered manually or using Jasper's favorite reply arrow feature thingie

user19161

Wait, can someone update me on the ping status? So now @t.b. pings tb? How about @tb?

Jonas Teuwen

Damn... the name says check your head but the link says shake. I am disappointed now :-(.

t.b.

@JasperLoy the former did, the latter didn't

user19161

I see. OK, very interesting.

robjohn

21:47

@t.b. yes, I entered that one by hand. Had it linked to a previous comment, it would probably have been removed.

user19161

Yes, I like to use the right arrow or simply manually typing in the first three letters. I never use tab complete.

t.b.

@robjohn I also see the @t.b. in the code block highlighted

@t.b.

those two didn't ping me

@t.b. but this one does

but isn't highlighted.

confusing.

user19161

I will try to ping myself.

t.b.

</playing around>

user19161

@jas

user19161

21:49

Self-pings don't work.

t.b.

I'm always a bit annoyed by downvotes and votes for closure in such questions

robjohn

@JasperLoy watch it! this is a PG forum ;-)

t.b.

Not the greatest question ever, but sort of feels legitimate, doesn't it?

user19161

@robjohn Oh, I really was not aware of any alternative meaning, now I need to go figure it out...

Old John

@t.b. agreed

user19161

21:52

@t.b. And the answer got 5 votes. Damn! I should have answered it!

t.b.

well, others get 50 votes for not proving that $\sqrt{2}$ is irrational in 3 different ways.

Michael Boratko

@robjohn Thanks for the help!

robjohn

@t.b. it is a fine question of operator precedence. If there is a problem, either point the OP to a previous answer, or answer the question.

Jonas Teuwen

@t.b. Mmm... is there any consensus about the minimum level? We don't want the site to get overrun by such questions, I suppose.

@tb If they prove the irrationality in ${\sqrt{2} \choose 1}$ ways they would get +1 from me.

t.b.

@JonasTeuwen from the faq: Mathematics - Stack Exchange is for people studying mathematics at any level and professionals in related fields.

robjohn

21:54

@MichaelBoratko I hope I did answer your question. I just have been getting a shower of coments. :-) If there are things left unanswered, just ask.

@JonasTeuwen It would be inappropriate for SO, but not here.

user19161

At most, it could be closed as too localized.

Jonas Teuwen

@tb Yep... but... then the question just shifts to "what is mathematics?".

robjohn

@JasperLoy That could be argued as valid.

ooh, I have a diamond!

t.b.

@JonasTeuwen I don't know, it'd be very hard to justify any kind of lower cut-off

user19161

@robjohn Of course! You think I talk crap here? :-)

user19161

21:56

@robjohn Really?

t.b.

@robjohn a shiny new mod diamond! glitter

Jonas Teuwen

@tb Yep. I agree. I would not vote to close unless he talks about Kalle numbers.

robjohn

It hasn't filtered to chat yet, though.

Jonas Teuwen

@robjohn As a friend... can you now ban people for me? Please? 8-).

user19161

The ELU election ended. Two new mods have been elected, both women.

t.b.

21:59

@robjohn did you notice that both user names are robjohn, according to the markdown source in the question on meta?

Transcript for

$Mathematics$ Mathematics