Mathematics - 2015-02-02 (page 3 of 4)

robjohn

17:02

@Chris'ssis cool.

@Chris'ssis I will look at it in a while. I have a lot of things to do for work this morning.

Chris's sis

@robjohn OK :-). Soon I'll upload a version slightly modified, I mean that I also added a dummy parameter on arctan, but this doesn't change the proof.

Sayan Chattopadhyay

Well @ABeautifulMind ur name is jasper

infinitesimal

The twitter reactions to the Superbowl are fantastic
-2nd and goal from the 1 yd line -1 timeout and BEASTMODE IN THE BACKFIELD... I would not have throw that ball to Jerry Rice! #fail
U have a running back named beast mode n u line up in shotgun formation n throw the ball on 1 yr line...the real MVP IS SEATTLE OC
Seahawks with the worst offensive call in #SuperBowl history. Why didnt you just run #BeastMode
If I were GM of the Seahawks, I'd be happy though: it makes their QB quite a lot cheaper that he blew the biggest game at the 1 yard line.

Daniel Fischer

Who won?

Sayan Chattopadhyay

What game?

infinitesimal

17:11

New England

American football

Daniel Fischer

So Brady has now won four times? Is that a record?

Sayan Chattopadhyay

Oh u r American @infinitesimal

infinitesimal

Yes, four Superbowl rings ties the record @DanielFischer

Sayan Chattopadhyay

Is American football similar to soccer

Daniel Fischer

No. Vaguely related to Rugby football, @Sayan.

@infinitesimal I guess he's a very happy man the next few weeks then.

Sayan Chattopadhyay

17:14

Oh...

I never played it

I play basketball more

Ted Shifrin

Surely all infinitesimal minds are American.

infinitesimal

:D

Sayan Chattopadhyay

Hi@TedShifrin

Ted Shifrin

Hi @Sayan

Mike Miller

Hope you weren't rooting for the pats, skull....

infinitesimal

17:16

Never

Sayan Chattopadhyay

Ted where r u from

Lucio D

Hi all

infinitesimal

Hi

Sayan Chattopadhyay

Hi@LucioD

Chris's sis

@robjohn I uploaded the slightly modified version (I added that dummy parameter). This wasn't an easy limit, at least not to me, I had to put some efforts on it.

Mike Miller

17:17

Phew :)

A Beautiful Mind

I really need to find the thing which has been biting me. It's biting me again.

Sayan Chattopadhyay

Hi@MikeMiller

Lucio D

If anyone has a chance please have a look at my post, any ideas would be great.

A Beautiful Mind

I think my sheets are clean, so I wonder what it is.

infinitesimal

Bite it back

A Beautiful Mind

17:18

I am now pretty sure it is not a mosquito.

Sayan Chattopadhyay

A rat

A Beautiful Mind

It could be a small bug hiding somewhere.

infinitesimal

Clean your room thoroughly

Sayan Chattopadhyay

Ant

Bedbugs I hate them

@ABeautifulMind can I call I Jasper

U*

A Beautiful Mind

@SayanChattopadhyay Yes.

Lucio D

17:21

Hi @DanielFischer if you have a chance please take a look at question it's of grave importance and been stuck on this for a while.

A Beautiful Mind

Grave...

Sayan Chattopadhyay

Grave?

A Beautiful Mind

He said grave importance.

Sayan Chattopadhyay

Oh

Hi@Ramanewbie

Lucio D

What's the problem?

A Beautiful Mind

17:23

No problem.

Ramanewbie

@sayan hi

A Beautiful Mind

Just that I wonder why it is so important.

Sayan Chattopadhyay

U were saying about something about pseudo

Ramanewbie

@sayan that's right...

Lucio D

Oh it's because, I am submitting something and been stuck on this prob with deadline approaching...hence grave.

Sayan Chattopadhyay

17:25

I didn't understand that @Ramanewbie

Ramanewbie

@sayan what's your pseudo ? is it sayan or sayan Chattopadhyay ?

Sayan Chattopadhyay

Sayan

Studentmath

Sayan reminds me of something from Dragon Ball Z

Sayan Chattopadhyay

Super sayon

Studentmath

Yeah, that

That was a great show

Sayan Chattopadhyay

17:26

Goku,gihan,vegita

Amazing but stopped in between

Studentmath

I don't really recall it anymore, it was too long ago that I last saw it

almost 8 years or so

Sayan Chattopadhyay

I saw it5 Years ago

Ramanewbie

sayan what about that then ? gyazo.com/6cdf3e137d481b44fecda0b2a37c0984

Sayan Chattopadhyay

I don't know who is that

I just have an MAA account

How did u find that

@Ramanewbie

Ramanewbie

@SayanChattopadhyay are you kidding me ???

Sayan Chattopadhyay

17:34

Why

Ramanewbie

@sayan lol -_-

Ted Shifrin

hi @Studentmath, good night @Mike

Sayan Chattopadhyay

Lol why @Ramanewbie

Studentmath

Hey @Ted

Sayan Chattopadhyay

Tell me

Ramanewbie

17:35

@sayan you're just switching posting messages with an account or another !

Sayan Chattopadhyay

No only one account

I have been talking with the same account

Ramanewbie

@sayan wuuuut ??

Sayan Chattopadhyay

Well what is gyazo

Ramanewbie

@sayan look at that ! gyazo.com/5071950f5d5f0c007b9b9b0488ef9fe3

Sayan Chattopadhyay

Yes I have been........

I know but what Is it

Ramanewbie

17:38

@sayan it's an other one, go and see it !

Sayan Chattopadhyay

The pics of our chat are coming there how

Oh my god........is this quantum entanglement

Ramanewbie

@sayan it's just a screen

Sayan Chattopadhyay

But how can that happen and even if it is what and how

Ramanewbie

@sayan how can what happen ?

Sayan Chattopadhyay

Well I am using my phone and that gyazo is for Mac

Ramanewbie

17:42

@sayan gyaso isn't for mac, it's at least for windows, too

Sayan Chattopadhyay

Do it might be someone els

Ramanewbie

@sayan what ?? didn't get that...

Sayan Chattopadhyay

It might be someone else

Ramanewbie

@sayan don't you see him ?

Sayan Chattopadhyay

How does it even matter

Ramanewbie

17:44

@sayan it doesn't... it's just weird

Sayan Chattopadhyay

What see

Well two people of same names can be

Ramanewbie

@Sayan look at that gyazo.com/6def6d745b150cb13608170a0fd11efb

Sayan Chattopadhyay

Well ur name is like srinivas ramanujan @Ramanewbie

I have seen it

Ramanewbie

@sayan who's that ?

Sayan Chattopadhyay

Might be someone its not me

My phone is android

Ramanewbie

17:47

ok @sayan

@sayan but why don't you appear in this list then ??

Sayan Chattopadhyay

Well srinivas ramanujan is the great Indian mathematician

Which liag

List

Ramanewbie

@SayanChattopadhyay you again !

Sayan Chattopadhyay

Again

What happened @Ramanewbie

Ramanewbie

@SayanChattopadhyay @Sayan I think this is an entire joke since the beggining...

Sayan Chattopadhyay

No joke I do t know him

Ramanewbie

17:51

@sayan I can't believe it then...

Sayan Chattopadhyay

I don't password for that account

How can I log in

It only takes the pics of screens right

Ramanewbie

@SayanChattopadhyay who are you ?

Sayan Chattopadhyay

It might be my dad

I have no idea but about this

Ramanewbie

@sayan lol...

Sayan Chattopadhyay

Well night

Peace

N3buchadnezzar

18:08

1

Q: Definite Integrals problem

The question is to find the value of : $$\frac{\displaystyle29\int_0^1 (1-x^4)^7\,dx}{\displaystyle4\int_0^1 (1-x^4)^6\,dx}$$ without expanding. According to the book, the answer is 7. I tried taking $I=\displaystyle\int_0^1(1-x⁴)^7\,dx$ and integrating it by parts taking 1 as the function to be...

This one should be solvable by parts, by defining a reqursive relation. Anyone agree? I just had some problems..

infinitesimal

@MikeMiller

Mike Miller

18:34

@infinitesimal :(

Ramanewbie

@hippa I just got a message on my computer screen ! but Tox isn't on !

evinda

Hi @DanielFischer!!!
How could we show that $\sqrt{1-p} \in \mathbb{Q}_p$ for each prime $p, p \neq 2$?

Ramanewbie

18:52

@hippa ??

Daniel Fischer

@LucioD Done.

@N3buchadnezzar Substituting $t = x^4$ and invoking the beta function is not fair game?

Lucio D

@DanielFischer Thanks a lot for your response, I'm busy looking at it now.

N3buchadnezzar

@DanielFischer NAH

Daniel Fischer

19:11

@N3buchadnezzar Then this?

Ilya_Gazman

hey @robjohn

N3buchadnezzar

@DanielFischer +1

Daniel Fischer

:D

N3buchadnezzar

@DanielFischer say you are a college student, then your solution is more understandable. You agree right?

Daniel Fischer

@N3buchadnezzar Depends. Are you a college student familiar with the beta function?

Lucio D

19:14

Hi @DanielFischer everything looks good except I don't follow how you get "$$\liminf_{n\to\infty} \bigl( \langle A(u_n), u_n-u\rangle + \langle A(u_n), u-v\rangle\bigr)
= \lim_{n\to\infty} \langle A(u_n),u_n-u\rangle + \liminf_{n\to\infty} \langle A(u_n), u-v\rangle$$"?

N3buchadnezzar

@DanielFischer I am a university student well endorsed with the function, but the students I have taught have never seen it before. On college level that is :p

Lucio D

Is this a basic idea for sequence that $$\liminf_{n \rightarrow \infty}(a_{n}+b_{n}) = \lim\limits_{n \to \infty}a_{n} + \liminf\limits_{n \rightarrow \infty} b_{n}$$

Daniel Fischer

@LucioD If you have two sequences of real numbers, $(x_n)$ convergent (to $x$), and $(y_n)$ arbitrary, then you have $\liminf\limits_{n\to\infty} (x_n + y_n) = \lim\limits_{n\to\infty} x_n + \liminf\limits_{n\to\infty} y_n$.

Lucio D

Okay understood.

robjohn

@DanielFischer: darn, you saw N3buchadnezzar's comment before I got there. I need to be more attentive.

@Ilya_Gazman hi there

Ilya_Gazman

19:20

@robjohn can you please take a look at my latest question. I want to find an efficient way guessing solutions for $n$

Lucio D

@DanielFischer What if you have sequence $x_{n} = -1,1,-1,1...$ and $y_{n} = 1,-1,1,-1,...$, then $\liminf\limits_{n \rightarrow \infty}(a_{n} + b_{n}) = 0$ but $\liminf\limits_{n \rightarrow \infty}a_{n} = \liminf\limits_{n \rightarrow}b_{n} = -1$.

Daniel Fischer

@LucioD But in this case, neither of the two sequences converges. To split, it is important that at least one of the two converges.

Lucio D

Oh yeah I see.

Daniel Fischer

@evinda You have $1^2 = 1 \equiv 1-p \pmod{p}$. And $$\left(1 + \tfrac{p-1}{2}p\right)^2 \equiv 1 + (p-1)p \equiv 1-p \pmod{p^2}.$$ So you have a square root modulo $p^2$. In a similar vein, you inductively show that if you have an $x_k$ such that $x_k^2 \equiv 1-p \pmod{p^k}$, then there is an $x_{k+1}$ such that $x_{k+1} \equiv x_k \pmod{p^k}$ and $x_{k+1}^2 \equiv 1-p \pmod{p^{k+1}}$.

Lucio D

19:52

Hi @DanielFischer, are you famaliar with the following result: Every convergent sequence $u_{n} \rightarrow u$ in $L^p(\Omega)$ has a pointwise convergent subsequence $u_{n_{k}}(x) \rightarrow u(x)$ a.e. $x \in \Omega$. Further, there exists an $h \in L^{p}(\Omega)$, where $h(x) \geq 0$ and $|u_{n_{k}}(x)| \leq h(x)$ a.e. $x \in \Omega$?

Daniel Fischer

@LucioD Yes. The first one is often used, and the second is an easy consequence of the proof. You pick a subsequence such that $\sum \lVert u_{n_{k+1}} - u_{n_k}\rVert < \infty$. Then take $h = \lvert u_{n_0} \rvert + \sum \lvert u_{n_{k+1}} - u_{n_k}\rvert$.

Lucio D

@DanielFischer Yeah I see how you obtain $h$. (the first one I know to be true.)

Ilya_Gazman

@robjohn Do you have any ideas that I can investigate for potential solution?

N3buchadnezzar

are two curves equal if they fill the same region?

Balarka Sen

20:08

not nessesarily, @N3buchadnezzar

robjohn

@Ilya_Gazman not at this time. Factoring is not that easy and your question seems to have a lot of background that is not apparent.

Balarka Sen

@N3buchadnezzar that's an easy example of two curves that bounds the same region, but are not equal.

:P

Studentmath

Hey @Balarka

Balarka Sen

hi

Hippalectryon

@BalarkaSen Beautiful

Ramanewbie

20:14

@hippa ?

Balarka Sen

Saying the guy who makes sonic unrecognizable, @Hippa

Mike Miller

@N3buchadnezzar Define 'curve'

Ramanewbie

@BalarkaSen I can reconize Sanic pretty well on his avatar...

Balarka Sen

sanic \neq sonic

Ramanewbie

@BalarkaSen I know that

N3buchadnezzar

20:19

@MikeMiller okay

$$
r_1(t) = ( \sin \pi t / 2 , \sin \pi t / 2) \\
r_2(t) = (t^3 , t^3)
$$

@MikeMiller Does $r_1$ and $r_2$ describe the same curve?

anon

@N3bu in order for your question to be nontrivial, perhaps you're asking if two paths are "equal" if (for instance) two paths are not the same function out of [0,1] but have the same image?

Mike Miller

@N3buchadnezzar I don't understand. Those don't even have the same image.

anon

@N3buchadnezzar well sin doesn't take values outside of [-1,1] but t^3 does...

N3buchadnezzar

@anon The domain should of course be $t \in [-1,1]$ otherwise the question does not make sense.

anon

k

Mike Miller

20:22

anyway, to me, and to most modern differential geometers, a curve is a map $I \to M$, where $I$ is an interval of real numbers, and $M$ is whatever your curve is on.

N3buchadnezzar

So the path is irrelevant, k

anon

http://en.wikipedia.org/wiki/Curve#Conventions_and_terminology:

Terminology is also not uniform. Often, topologists use the term "path" for what we are calling a curve, and "curve" for what we are calling the image of a curve. The term "curve" is more common in vector calculus and differential geometry.

Ilya_Gazman

@robjohn nop, all is there. I been looking at the mod function of $c$, and this is the behavior that I saw, can't explain it, can't understand it.

Mike Miller

what do you mean the path is irrelevant?

N3buchadnezzar

@MikeMiller Well both curves map $I$ to $M$

anon

20:25

and are different maps

one might as well say (0,t) and (t,0) "both map I to R^2"

Mike Miller

@N3buchadnezzar they're different curves because they're different maps, like anon said

Hippalectryon

@Chris'ssis Hello

@Chris'ssis If $f,g$ are continuous, integrable, how would you evaluate $\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+(x+g(t))^‌2}dtdx$ ?

Chris's sis

@Hippalectryon Hello :-) How are u doing?

Ramanewbie

@hippa you're here !

Hippalectryon

Fine, what about you ? @Chris'ssis

@Ramanewbie ..... and ?

Chris's sis

20:33

@Hippalectryon I'm trying to finish a long proof to one of my questions.

Hippalectryon

@Chris'ssis Oh ok.

Ramanewbie

@hippa the gif didn't work ?

anon

@Mike can you verify if the $\pi$-coordinate of the element $g\mapsto f(g)$ under the isomorphism $L^2(G)\xrightarrow{\sim}\bigoplus_{\pi\in\widehat{G}}{\rm End}(V_\pi)$ is $\int_G f(g)\pi(g)dg$?

Hippalectryon

@Ramanewbie For god sake, when the hell will you ask clear questions ????????

Ramanewbie

@hippa but... I sincerely thought it was !

Hippalectryon

20:34

@Ramanewbie "the gif didn't work ?" how is that supposed to mean anything at all ????

Ramanewbie

@hippa I mean, maybe it was like frozen ? it didn't move ?

Mike Miller

@anon trying to compile your thing crashes my page, bizarrely. (PS: I don't know a single topologist that uses 'curve' to mean the image of a map.)

Hippalectryon

@Ramanewbie I don't care what you mean. What matters is what you say.

Ramanewbie

ok, don't get mad @hippa !

Hippalectryon

I'm always mad.

4

anon

20:36

@MikeMiller I know topologists sometimes choose to use path instead, but IIRC when people use curve to mean image it's usually in some kind of really basic geometry / vector calc context.

Ramanewbie

@hippa I had many occasions to see it... -___-

Mike Miller

yeah, I haven't seen it ever, but definitely not in the context of modern stuff. (robjohn told me once that when he was a student, curves were just images, and you took derivatives by attaching these images with tangent vectors, apparently)

I don't know fourier analysis on groups, @anon, which is what I assume you're doing?

my officemate has a copy of Rudin's book laying around somewhere if you'd like me to look at the isomorphism in question, if I can find it

anon

more properly called representation theory or abstract harmonic analysis, but yes

Mike Miller

sure, sure

anon

I have good heuristic reason to believe my thingie but I am trying to understand Peter-Weyl with as little functional analysis as possible (as I already have some slick stuff developed for finite groups, I want to move onto compact now). perhaps doing without functional analysis is ill-fated though.

Mike Miller

20:39

you can recover compact with very little functional analysis

it's LCA that one needs to work for

anon

that's what I was optimistic about

Mike Miller

and you're replacing $\Bbb CG$ with $L^2(G)$, I see

ok I'm fairly confident what you wrote down is correct now

anon

I think of an element of $L^2(G)$ as a "formal integral" $\int_G f(g)g dg$ (like a formal linear sum of elements of $G$, but infinite and with $L^2$ regularity), and applying $\pi:G\to{\rm End}(V)$ linearly should yield $\int_G f(g)\pi(g)dg$ within ${\rm End}(V)$

Mike Miller

not sure whence the g in your first integral there.

anon

I also had the idea yesterday that Peter-Weyl generalizes the idea that the Fourier transform turns convolution in the time domain into multiplication in the frequency domain. the operation in $L^2(G)$ is convolution of functions, and the operation on the direct sum on the other side of the isomorphism is coordinatewise, where each coordinate is like a Fourier mode.

Mike Miller

20:43

ah, nevermind, I see

anon

@MikeMiller I am replacing $\frac{1}{|G|}\sum_{g\in G}$ with $\int_G$ in the "formal sum" thingie

Chris's sis

@Hippalectryon did you try to change the integration order? All should flow pretty easily and you're done in one line.

Mike Miller

right

Hippalectryon

@Chris'ssis I don't wanna use Fubini. Only HS maths.

Chris's sis

@Hippalectryon btw, yesterday I also asked you a question. ;)

Hippalectryon

20:44

@Chris'ssis Yesterday ?

Mike Miller

@anon: I think you should be able to write down the proof that this is indeed an isomorphism just by following the same argument as the finite case, now that you've figured out the right formula

Chris's sis

@Hippalectryon Yeap.

@Hippalectryon Anyway, forget that. ;)

Hippalectryon

@Ramanewbie About 10 days ago, I asked you not to turn that chatroom into some skype clone.

Ramanewbie

@hippa but you're not on tox

Hippalectryon

@Chris'ssis What was it ? The only one I remember is the huge equation with roots.

@Ramanewbie That is irrelevant.

Chris's sis

20:46

@Hippalectryon By the way, did you find the solution to that nested radical? :D

Hippalectryon

Unfortunately I didn't :/

Don't tell me though

Chris's sis

@Hippalectryon OK

anon

@MikeMiller The biggest problem I have is that my argument for the finite case uses the fact an injection between spaces of the same dimension is an isomorphism ... if the spaces are finite-dimensional.

Mike Miller

how many accounts do you have?

anon

heh

Hippalectryon

20:48

@Chris'ssis You've made me curious now >:c what was it you asked me yesterday ?

Chris's sis

@Hippalectryon wait

Hippalectryon

Saved

Chris's sis

@Hippalectryon It's amazing! Believe me! Just give it a try! ;)

Hippalectryon

Ok I'll take a look then :-)

Chris's sis

OK :D

Hippalectryon

20:52

If you keep adding new problems to your book, it's never gonna end :P

Chris's sis

@Hippalectryon There will be 300 only. :-)

Mike Miller

@anon fuzzy memory, sorry: aren't there only finitely many irreps of a compact Lie group?

Hippalectryon

@Chris'ssis So 1-2 page / problem ?

anon

@MikeMiller I think generally there are countably infinitely many irreps all of which are finite-dimensional

Chris's sis

@Hippalectryon Yeap, something like that.

Hippalectryon

20:54

@Chris'ssis :O a 300-600 pages book then great :D

Mike Miller

by generally I guess you should mean always, because otherwise your sum doesn't work, dimension-wise? (on that note: shouldn't your RHS have a sort of closure above it?)

closure/completion

Chris's sis

@Hippalectryon There is much work to do! I hope the final result will be a nice one. :-)

Mike Miller

in that case, what's the deal with $S^1$? what are its interesting irreps? there's the obvious one on $\Bbb{R}^2$... what else?

Hippalectryon

Yep :DD @Chris'ssis

anon

@MikeMiller yes it's a hilbert space direct sum (the subspace of the direct product comprised of element with finite $L^2$-norm), which is bigger than the algebraic direct sum

Mike Miller

20:56

just checkin'

anon

@MikeMiller I am doing things over $\Bbb C$ so things are a bit nicer than over $\Bbb R$.

the irreps of $S^1$ of course are just the one-dimensional ones with power maps

Mike Miller

ahhh yes

Chris's sis

@Hippalectryon Only let me know how you would start there. Curious about the way you look at it.

Hippalectryon

@Chris'ssis By 'there', you mean the last one you showed me ?

Chris's sis

@Hippalectryon Yes, the limit I mean.

Hippalectryon

20:59

@Chris'ssis I'll do that in the room then :D

Chris's sis

@Hippalectryon OK

Don Larynx

21:10

@ABeautifulMind: Have faith brother.

robjohn

22:16

@Chris'ssis: Have you seen $$\sum_{n=1}^\infty\frac{\zeta(2n)-1}{n+1}=\frac32-\log(\pi)$$ before?

Chris's sis

@robjohn Possibly. Where do you have it from?

robjohn

I am trying to see how Mathematica evaluates it

@Chris'ssis It is from evaluating a fairly difficult product.

@Chris'ssis I have posted what I have up to that point here

Karlo

Why isn't 2^A finite set?

Chris's sis

22:34

@robjohn It should flow nicely by adding there an integral and using a generating function for $\zeta(2n)$. You can easily derive it by using the case for $\zeta(n) x^n$ generating function.

Ted Shifrin

hi @robjohn @Chris'sssis

robjohn

@Chris'ssis I am trying the generating function of $\zeta(2n)$ actually.

Mike Miller

Morning @Ted

Ted Shifrin

good night, @Mike

Ramanewbie

@MikeMiller "morning" ??

@MikeMiller isn't it 17h37 in the US ?

Chris's sis

22:37

@robjohn You mean you already tried that and it's ugly, right?

user112495

How do I determine whether a particular property of a metric space is a topological property?

Chris's sis

@robjohn $$\pi\;x\;\cot(\pi\;x)=-2\sum_{n=0}^\infty \zeta(2n)\;x^{2n}$$

Ted Shifrin

@Ramanewb: @Mike is in the far, far, far west corner of the country

Good question, @user112495. What does that mean?

Ramanewbie

@ted ooh, I see... In France we only have one unique hour

user112495

@TedShifrin We've been told that a property is topological if it makes sense for every metric space.

Chris's sis

22:40

@robjohn however this one is not a nice way. I'd try something else.

Ted Shifrin

whispers sotto voce to @Ramanewb: @Mike says good morning no matter what the hour.

robjohn

@Chris'ssis or I can cite this answer

user112495

@tedshifrin Ignore that

Ted Shifrin

is good at ignoring

Ramanewbie

@TedShifrin Is it usual ? Would anyone understand ?

Ted Shifrin

22:42

@Ramanewb: He's almost as insane as your brother.

anon

@user112495 Here's a heuristic answer: If it's a topological property, then there exists a condition on the topology of a topological space which for metric spaces is equivalent to the original condition. If it's not a topological property, then there should be two different metrics on a set that induce the same topology yet the original condition is satisfied under one metric but not the other.

Ted Shifrin

hi @anon

Chris's sis

@robjohn Yes. However, it's interesting that one needs so ugly operations to get the right answer.

anon

hi

Ted Shifrin

What are you teaching this term?

anon

22:42

same as last term

Ramanewbie

@ted That's quite difficult ! -_- hippa would agree himself...

user112495

@TedShifrin We've been told that a property (that makes sense for every metric) is topological if whenever $M$ has said property, so has every space homeomorphic to it.

Ted Shifrin

There you go, @user112495. That sounds good.

Mike Miller

right

anon

intermediate and college algebra. can't really teach anything else as an undergrad officially.

Ted Shifrin

22:43

So, is compactness a topological property?

Did your students end up doing better than average last term, @anon?

Chris's sis

@robjohn I have a different idea.

user112495

@TedShifrin We haven't got to compactness yet.

Alessandro

@TedShifrin is "sotto voce" an expression commonly used in english?!

Ted Shifrin

Ah, so what words are legal to use, @user112495?

No, @Alessandro: I threw those in just for you :D

Mike Miller

@Ted Do you know if people still use yang-mills/Donaldson theory?

Ted Shifrin

22:44

"use"? @Mike

anon

@TedShifrin my classes' averages were about median amongst all the averages

Ted Shifrin

Floer seems to have taken priority, but people still talk about 'em, yeah, @Mike.

ah, ok, @anon. That's still quite good for someone teaching the first time and probably not training the students like puppets for the tests :P Well done!

anon

but our system does pretty well anyway; about 80-85% averages, and there's no curves or extra credit anywhere. although the grades did get a bit higher when we removed conic sections a two or three semesters ago.

Mike Miller

Yes, @Ted, use. Probably I'd be vetter off asking Ciprian this, but I don't really know if it's worth my time to learn the technical details of Donaldson theory as opposed to, say, Seiberg-Witten, if the former is antiquated.

user112495

@TedShifrin Well we hadn't actually defined a topological space at the point of defining a topological property.

But the examples our lecturer gives are:
M is open in M; M is closed in M
M is finite; countable; uncountable
M has an isolated point;
M has no isolated points
Every subset of M is open

Ted Shifrin

22:46

Well, @Mike, should we not bother learning Chern classes because things have gotten more advanced?

user112495

@TedShifrin I just don't really see how to see that any of these are topological properties from the definition.

Ted Shifrin

So, if you put two different metrics on the same space (say $\Bbb R$), @user112495, which properties won't depend on the metric?

Chris's sis

@robjohn Can't you avoid the use of that series in $\zeta(2n)$? I'm sure it would be nicer to get rid of that form.

anon

@TedShifrin well, we'd need the metrics to induce the same topology, which might be begging the question for user****** here

Ted Shifrin

But, @Chris'ssis, you love series upon series upon series upon series!

Chris's sis

22:48

@robjohn better use this form $$\sum_{k=2}^\infty\left(-1-k^2\log\left(1-\frac1{k^2}\right)\right)$$

@TedShifrin Hello @TedShifrin! Sorry to answer you with delay! :-) I'm so caught into this stuff! :D

Ted Shifrin

right, @anon, but that's one step further.

LOL, no problem, @Chris'ssis

Mike Miller

@Ted That's not a fair comparison, as chern classes are still used, still important, still fundamental. If Donaldson theory has been completely eclipsed by something less technical, isn't it better to learn that?

robjohn

@Chris'ssis and how would you use that?

Ted Shifrin

I suspect it's still foundationally important, @Mike, but ask Ciprian.

Mike Miller

More apt, maybe, is the slant product: you can find it in loads of papers and books from the 60s, but nobody uses it anymore.

robjohn

22:50

@Chris'ssis That is essentially the product we are looking for.

Mike Miller

Thanks. I'm not trying to convince you its unimportant, just that it's not sacrilege to suggest it might be. :)

Chris's sis

@robjohn I'd try to get some telescoping sums and then probably the use of Stirling and Taylor series. Let me see.

Ted Shifrin

Well, @Mike, it's not exactly super-odd. It's just a contraction.

Mike Miller

You didn't see the groans when someone mentioned it in the topology seminar last quarter, @Ted

Alessandro

@TedShifrin I had never seen it used in english before (unlike other latin or italian words, viceversa, a priori, a posteriori and so on)... well, the more you know!

Ted Shifrin

22:51

well, @Alessandro, I am a pompous American, so I like to show off words in foreign languages. Don't judge by me :P

does one see groans, @Mike?

Mike Miller

One sees groans.

(Of course, the slant product - and chern classes - are probably a bit easier to pick up than the definition of Donaldson invariants.)

Ted Shifrin

Me guess you sees groans.

user112495

@TedShifrin Wait, but if you are looking at boundedness for example, then you can show that that isn't a topological property by looking at the sets $(0, 1)$ and $(0, \infty)$. How would you look at this in terms of decidng whether or not it's dependent on the metric?

Mike Miller

I wasn't one of the groaners, @Ted, I didn't even know what it was at the time.

Ted Shifrin

Awesome start, @user112495!

Well, with the usual metrics you know, @user112495, those spaces are homeomorphic, and one is bounded and the other isn't.

Studentmath

22:54

@Ted @Mike!

Ted Shifrin

Hi @Studentmath :)

Mike Miller

Hi, @Studentmath

Ted Shifrin

BTW, @user112495, can you put a different metric on $(0,\infty)$ so that it is bounded? :) But we want the same topology: i.e., we want close points with one metric to be close in the other.

Studentmath

@Ted I am latexing and flashing out my results now, it might actually come to something :) I still doubt it very much, but still it's a great practice

Ted Shifrin

Welcome to almost-adulthood, @Studentmath (army notwithstanding).

Studentmath

22:58

Hope that will go well too.. How're this semester students, @Ted @MikeM?

Mike Miller

They're good, @Studentmath. A few of mine keep surprising me with good questions.

Ted Shifrin

Given exams this week and next, @Studentmath, so we'll see. But I'm starting to have withdrawal symptoms ... 11 weeks left and I'm done ...

I hope they flummox you, @Mike.

Mike Miller

They've done so, sometimes.

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$Mathematics$ Mathematics