If they don't notice, someone might "tip them off" ...

@AlexJBest another interesting possibility: could a lecturer be posting questions here to help his students achieve better grades? :P

@OldJohn Possibly :D however the questions asked correspond to two entirely separate modules, so probably not I'd say!

@AlexJBest good point!

You can't get cooler than Obama, man.

@PeterTamaroff :)

@PeterTamaroff When will the election results be announced?

@Argon No clue. They are usually announced the very same day.

Depends on how close it is in certain states.

Since USA is pretty huge, maybe it takes longer.

I would really appreciate if someone found the slip in this

@Argon @OldJohn @peoplepower

I'm looking at it.

@PeterTamaroff had a quick look - but can't see immediately where the missing $\sqrt{2}$ factor is ...

@OldJohn Yeah. It is driving me nuts.

:))

@OldJohn Informally, this seems ok, right?

I actually proved (well, Mariano helped wonders) that if $f$ is Riemann integrable over $[a,b]$ it is continuous at infinitely many points of $[a,b]$-

@PeterTamaroff Assuming you get the missing factor sorted out, I think your answedr is better explained than the other one :)

@OldJohn Hm, thanks.

@PeterTamaroff I like that - but I suspect that the fact that such an $f$ is continuous at infinitely many points is not so well-known

@OldJohn Hmm.

What is $\int_0^{\frac{\pi}{2}}\cos^2x\text{d}x$? I might be having a brain-fart, but this might be part of the problem @PeterTamaroff

@peoplepower Brain fart sounds so disgusting.

@peoplepower That is $=\dfrac{\pi}{4}$

Then, no problems, I forgot to divide by $2$ somewhere.

Check your definition of $a_{2m},a_{2m+1}$ for a missing $2$.

I.e. you had a $2$ before the radical sign when you began, but lost it when you wrote down your sequences

Yes, I realized that. But now I'm too lazy even to change that.

Will Romney win?

Anyone near NYC? They say if Romney wins, the lights will turn red, and blue if Obama does.

Vote Appel/Haken 2016! en.wikipedia.org/wiki/Four_color_theorem

2nd best would be the Zermelo/Frankel ticket. if you are pro-choice that is...

:)

Robertson/Seymour have a good platform over all but it excludes too many minority interests.

okay somebody else's turn now...

as for Goedel/Rosser i'm undecided.

oh c'mon somebody.

Erdos has so many connections but to speculate on his running mate would be far from elementary.

Certainly he would do well in the PRIMary!

hi @anon care to contribute? i can't keep this up forever. Hilbert/Cantor probably could if they hadn't been busted for all that funny business in the hotel.

I think the Levi-Civita campaign has better connections than Erdos.

it would be a purely symbolic victory.

and the Banach-Tarski platform has doubled its share of the vote

Hardy is seen as too apologetic to be commander-in-chief and maintain currrent American foreign policy.

Lebesgue's effect on the result is likely to be nonmeasurable

i predict Martin's campaign finance reform legislation will put an end to Banach/Tarski's unethical gerrymandering.

Grothendieck's gaffes on primes have paradoxically made him a better-liked, personable candidate, but his $\ell$-adic coideology is perhaps not down-to-earth enough for the lowest common denominator.

I am running out of ideas as to how to explain to this guy

MacLane's slogans are so elitist that the masses will never entirely buy in.

i would say that Euclid/Pythagoras look the best on paper.

The Löwenheim–Skolem campaign has been a model of perfection

but the Lobachevsky affair threatens to derail the entire campaign.

But Liouville's share of the vote has been remarkably constant

@OldJohn except for the downward momentum

@OldJohn but depending on the poll it can be quite erratic at times en.wikipedia.org/wiki/Liouville_function

Did the Picard campaign omit one or two states?

Thanks, @anon for helping out :)

2:30 am here and running out of steam :(

I just woke up.

First time in forever I've woken up at 8pm

this is too good keep going. Mathamerica 2016!

And I was like "where the hell am I?"

I can imagine - years ago I did some shift work and never knew what time of day it was

saved to post to my fb if no objections imgur.com/a/uFKrD excellent work guys

@DanBrumleve no problem

Holy Monkey, people are imaginative

@PeterTamaroff mwuahahaahaha.

Obama is up 162 to 153

@Will, I saw Lee's post...I deleted the pdf (copyright issues), replace with a link to Google books...

@Will Pretty cool, hey? (Lee commenting...)

@Will thanks...I was wondering about that...I noticed :-)

I get the feeling sometimes that people might get annoyed...since I do tend to elaborate...and I'm a perfectionist, always wanting to polish up my answers!

@WillHunting Hola.

ok we wrote 16 jokes. transcribed and anonymized with minor edits: pastebin.com/0ihmMAMb

also i annotated

was remarking to some friends earlier today that math education is the ultimate way to fix our broken political process. maybe with annotation they will be funnier to more people and this is a step in that direction.

First Class!!

favorite is my own lobachevsky/lewinsky riff :)

also the banach tarski vote fraud

who won?

CNN projects an obama victory.

is there something missing?

@leo first part of your answer seems to settle the issue.

what does a "cooked" proof mean anyway?

@DanBrumleve nothing. The OP was asking cook up a proof

oic i thought perhaps it had been deliberated in a sauna.

@DanBrumleve lol

either that or it implies some independent and speculative mischief. if i were to cook up a proof of P=NP right now it would be as valuable as cake

sorry i don't have enough background to easily verify the rest of your answer

@DanBrumleve maybe as valuable as a cheesecake even

haha i suppose i could bake a palatable bertrand's postulate at least.

night night ++obama/biden ++appel/haken

Hey guys

I've been reading an algebraic topology text in my free time (Rotman), and I'm in the process of reading about singular homology. To my understanding, intuitively, the nth homology group is the free abelian group of continuous maps from the n-simplex into the space, modulo all boundary curves.

I was a bit surprised to find that the nth homology group of the n-sphere is Z, and not trivial, as this implies that the sphere has a continuous image of the n-simplex in it that is not a boundary curve. Can somebody help to demonstrate for me a cycle on the sphere that is not a boundary curve?

@WillHunting From your nap? Hmm.

I am having trouble understanding the Perron-Frobenius theorem.

Interesting

Hi all

I spotted this last night on the BBC website:

How will he pull that off

@OldJohn ahh, the mystery of the negative votes. :-)

@JayeshBadwaik I just wonder how much they paid the guy who coded that script :)

A constitutional amendment is in order.

@peoplepower :))

@WillHunting Clearly, you are a star

@OldJohn You should not wonder too much, much more serious software has been much more badly written.

@WillHunting Yes, You're gonna be star, oh please let me drive your car.

@JayeshBadwaik Indeed it has - wonder if anyone has written a book of coding horrors ...

@OldJohn There are many!!

@OldJohn This is a start.

@JayeshBadwaik thanks

@WillHunting Not had breakfast yet! I stayed up to watch the US results

@WillHunting :) See the screenshot I posted above ...

They still haven't decided Florida yet.

@peoplepower It might be the rare time when elections have been decided before florida has been counted.

Anyone know if I can get wolfram to convert function equations from cartesian to polar?

Ah, yup, I can. Sort of.

Can I use the words "elements of X can be added closely" to mean that the addition is closed in X?

It makes for an interesting perspective, but I do not think it would be interpreted the way you want.

hmm. I thought so. Have to use some longer explanation. =)

Hi. I am looking for a one line answer for why "[a, b] is complete where a < b in R." Here is my answer "[a, b] is a closed subset of the complete metric space R, hence it is complete as completeness is closed-hereditary"

@sonicboom Full marks!

Wooo!!

{o,t,t,f,f,s,s,e,n}

Is there a way to directly convert this parabola equation to polar coordinates? mathbin.net/111900

@JayeshBadwaik Yo

My message got 10 stars and pinned! That must mean I am popular and have friends.

Hmm....

You wanna have another friend?

@GregRos I want a friend =)

hmm

hi

Is there a visual editor for mathjax?

codecogs has a wysiwyg editor (which is what I assume you want)

Do they work on mac?

both codecogs and the mse previews occur inside an internet browser...

I haven't been going to class for awhile. Apparently we're behind a few sections and the test will be on monday.

ergo the next class I go to will be monday

@anon Are you not a teacher, or have I completely missed something?

I am whatever I am. The label designed for me is "freshman undergraduate."

@N3buchadnezzar Yo!!

Sup dawg

@WillHunting I thought field medallist was a binary function. :P

@WillHunting Low fruits, Low fruits everywhere

@N3buchadnezzar Good good, decided to organize my LA knowledge till now, writing up notes on it. It is fun.

Los Angeles knowledge?

Linear Algebra. :-/

@WillHunting Hmm. I will read that, but currently, I am now starting on my own. Writing my own axioms, developing my own theorems. Most of the stuff will be same, but it will be my effort.

Also, I finally got the Perron-Frobenius thing.

@WillHunting Yup. I actually have already done some of the parts of it. Just arranging and putting it down in one place.

Peter Peterson? Funny name

This book you mean?

math.ucla.edu/~petersen

@WillHunting Contradiction, Mr Watson.

@WillHunting Ahh, okay.

Did you buy the book?

@WillHunting Yup, must be.

@WillHunting Did you read that book?

@WillHunting How did you get that book? Peter Petersen?

@WillHunting I am your shadow -:)

hi...

hiiii

forever alone

I was refering to Garbage, nobody responded to his message.

In adititon to having friends and being popular, I also have a relation ship

hello

@N3buchadnezzar :D :D Nice.

@jdoe Hello.

I learned a cool theorem today

Which one?

If you have a monic irreducible polynomial $$x^n + a_{n-1} x^{n-1} + \ldots + a_0$$ in a CDVF then $v(a_0) \ge 0$ implies $v(a_i) \ge 0$ for all $i$.

"CDVF" just means things like the completion of the p-adic numbers

so basically if 1/p occurs in any of the coefficients other than $a_0$ the polynomial factors

...wow. I just had a major brain fart.

I think it's related to eisentein criteria, but I don't know how exactly

@GregRos I think you should let a doctor look at you

@jdoe hmm nice.

Going out now .Later.

bye

Hi @GarbageCollector

How does $pq = p + q$ imply $(p-1)(q-1)=1$?

(p-1)(q-1) = pq - (p+q) + 1

@WillHunting Don't let your rep go to your head.

@anon Hey :D

I read through most of the first version of the pdf while at work last night.

@anon thankx :d

I can send you my completed one.

I have cleared up some definitions

Checking now, I see you finished section 5.

Including that for the defn of a partition

Because I realised we need to allow trailing zeros in the definition

otherwise you get into all sorts of sticky situations

@anon would you like the completed one?

Yeah.

There's no need to cite Serre for Maschke's theorem when you don't bother citing anybody for Schur's lemma or Artin-Wedderburn or Frobenius' formula in my opinion.

I skipped most of the calculations in Lemma 3.6

Why use $\bf i$'s to tabulate conjugacy classes of sym groups when you can just tabulate them with partitions?

Also the numbering of everything is strange.

Like, (4.4) and (4.5) come before (4.3), for instance.

As I don't know much lie theory, perhaps you can clear this up: even in view of $\mathfrak{sl}_n$ being a complex lie algebra, when we decompose elements of $\frak h^*$ into linear combinations of $L_i$'s, how can we say when differences between coefficients are positive if the coefficients are (I presume) complex numbers? Or will they necessarily be real when working the roots associated to the ad rep?

Another name for the symmetric polynomials, perhaps more standard (especially with invariant theory), might be $\Bbb Z[x_1,\cdots,x_n]^{S_n}$ (as $X^G$ denotes the fixed points of a $G$-action on $X$) instead of $\Bbb Z_{\rm sym}[x_1,\cdots,x_n]$

@anon They are all integers :D

@anon I have corrected the numbering

@anon By $\mathfrak{sl}_n$ theory all the $a_i's$ are integers :D

did you know you can click the gray arrow on the side of posts in order to get pings that respond directly to specific posts?

@anon Just sent it in.

@anon The $a_i's$ are all integers

That's the beautiful thing about the theory.

is that sort of thing particular to sl?

I know it's true for $\mathfrak{sl}_n$

Because first of all you only need to know the weight of the highest weight vector.

By finding a suitable subalgebra isomorphic to $\mathfrak{sl}_2$

you get that the highest weight is an integer

now because it is a highest weight rep

any another vector is just the negative root spaces applied to the highest weight vector

I am sort-of learning lie theory from Krillov at the moment

@anon That's why they call it the weight lattice :D

If they weren't integers it would not be a lattice :D

@anon did you receive it?

@BenjaLim eh? how so?

yes I received it

@anon Well a lattice is a $\Bbb{Z}$ - linear combination of some basis vectors

in this case the $L_i's$

sure, but even if a set of eigenvectors have corresponding irrational eigenvalues, you can make a lattice out of those vectors, no?

I don't see how the eigenvalues matter in the process of forming a lattice out of the eigenvectors.

oh, you want the algebra to act on the lattice

or some kind of constraint resulting from wanting an action

@anon It boils down to $\mathfrak{sl}_2$ theory

every weight is an integer

Fulton and Harris lecture 11

and now in sl_n

what do those comments have to do with what I said?

by restricting to a suitable subalgebra isomorphic to $\mathfrak{sl}_2$

sigh, I keep forgetting your sentences are seven, eight, nine ... lines long

:-D

@anon Sorry.

Let me say this again.

We want to know why all the weights of $\mathfrak{sl}_n$ are integers.

Now the point is that we know from $\mathfrak{sl}_2$ that the weights are positive integers.

Consider $v$ the highest weight vector of an irreducible representation of $\mathfrak{sl}_n$.

My question is why the weights need to be integers in the first place in order for the $\Bbb Z$-linear span of the weight vectors to be a lattice.

@anon No you don't I was wrong there.

Okay.

But should we a priori "want" them to be integers so that we get something nice?

Well they have to be integers.

if they're not integers you'd have to scale them (rationals) up by a common denominator?

A priori they are complex numbers.

I'm talking motivation here bro.

Ah no.

No a priori a weight is just an eigenvalue

or more precisely a linear functional from the cartan subalgebra to $\Bbb{C}$

is there a correspondence between a "second highest weight vector" of L and the h.w.v. of the quotient of L by the root space associated to L's hwv?

iow, can you use a method of finding highest weights inductively to find all of the weights in decreasing order

@anon What is $L$?

some nice lie algebra

ah ok.

h.w.v.?

oh

context clues man

highest weight vector

@anon No I don't know about "second highest weight vectors"

@anon I don't think it is at all obvious how to find the weights in decreasing order.

All we know is that any vector in the representation is generated by linear combinations of elements of the negative root spaces applied to the hwv

this is of course assuming the representation is irreducible

more generally semisimple theory

if you learn sl_n theory

highest weight representation iff irreducible.

so if you have a root space decomposition, and you quotient out the highest weight factor, the resulting direct sum will no longer be accurate as a root space decomposition? if it were accurate, then its (the quotient's) highest weight factor would correspond to the second highest weight factor of the original space, right?

Hmmmm

I actually don't know the answer to this.

I don't know what you mean by "quotient of L by the root space associated to L's hwv"

I mean a root is in particular a weight of the adjoint representation.

lie algebras have decompositions into direct sums of subspaces associated to the weights or roots right?

well to the root spaces yes.

Like $\mathfrak{sl}_n = \mathfrak{h} \oplus \mathfrak{g}_+ \oplus \mathfrak{g}_-$

where each elementary matrix $E_{ij}$ with $1$ in (i,j) and zeros elsewhere

is a root vector with root $L_i - L_j$

So polynomials are not complete metric spaces, since there exists sequences that converge to functions that are not polynomials

@anon I have to go to bed.

I am really tired

Nite man!

nice talking

thanks as always!

later

I wonder why no one answered my question about dual basis

@jdoe is "trace" a word for a bilinear form?

trace is a particular one

lol @ {x1,...,yn} btw

oh no

if $\{y_i\}$ is a basis for the dual space $V^*$ then there must be an isomorphism $V\cong V^*$ you're implicitly using to make sense of $t(x_i,y_j)$, right?

as $t:V\times V\to K$

I suppose you are using the isomorphism given by $V\to V^*:x_i\mapsto (\sum a_jx_j\mapsto a_i)$

I suppose that is true, but I don't like it

so then $\{x_i^*\}$ as I just described is precisely the dual basis you want no?

@anon, well that does make sense - I really don't like how he used the dual while still calling it V though

that's so confusing

the lecturer

I should probably just drop this class and focus more on the others, but I do want to learn the stuff

@WillHunting Okay!! I got the lecture notes by the way.

newcriterion.com/articles.cfm/…

The Fifth problem: math and anti-Semitism in the Soviet Union

o.O it is real?

Can anyone recommend me a good book on proofs?

@GustavoBandeira I found that reading and emulating proofs in books by good authors is a good way to improve proof-writing

@OldJohn But aren't there books for proof writing?

Do you think it's a good idea?

@GustavoBandeira I am sure there are - but I have never used one, I'm afraid - just preferred to "learn from the masters"

@GustavoBandeira Proof writing does not stand in vacuum, it is always in context of some or the other field. If you want to go to hard core proofs, I believe, a course in logic might be a good option. (I am taking one right now on coursera.)

@OldJohn I hope what I said is okay?

Yep - I think so

Do you guys know "proofs from THE BOOK"?

IMHO too many people write proofs usimg too few words - I can find some examples if needed.

@GustavoBandeira I have it, yes

@GustavoBandeira I have it, not read it yet.

I've selected a book on logic

A First Course in Logic: An Introduction to Model Theory, Proof Theory, Computability, and Complexity - Hedman

Okay. I am taking the logic course for the first time. I am basically reading the notes on coursera currently. Don't know much about the books.

I did some courses on mathematical logic many years ago - but I am not sure they really helped me with the art of writing out proofs, though

@OldJohn Ohh.

I remember writing and reading lots of logic proofs with very few words in them, and found that was not very helpful when it came to writing "real" maths :)

@OldJohn Hehe. Yeah. :-) Writing proofs well is one thing and writing proofs correctly is another. My opinion was from the point of view of writing them correctly, avoiding handwaiving etc.

@JayeshBadwaik Yep - I think we are both giving advice about different aspects of the topic :)

@OldJohn Coming from engineering background, I have become so much accustomed to hand-waiving, that when asked to prove stuff like countable number of discontinuities on a closed interval for a monotonic function, I had to for the first few times, draw and/or/imply logic to convince myself I have formally proved stuff, and have not hand-waived parts of it.

@JayeshBadwaik Yep - I remember being in similar situations many years ago

anyway, dinner calls, see you guys later.

@OldJohn hmm. :-)

@JayeshBadwaik later!

Funky analysis makes my brain hurt, but it is really cool aswell

+1 for the plane all 8 of them are in

But the plane is defined by 3 points so these 8 together don´t give me one plane, do they?

Oh I just have got it! Thanks!

:D

sap?

@N3buchadnezzar maple syrup?

@N3buchadnezzar rubber?

@N3buchadnezzar amber?

@robjohn Functional analysis?

@N3buchadnezzar Functional analysis flows from a tree when you bruise it?

@robjohn I really wish it did!

@N3buchadnezzar Is $(\cdot,\cdot)$ an inner product?

looks kinda dirty...

@robjohn Yeah