Mathematics - 2019-01-29

user616128

01:50

@ChthonicOne Also, linguistically, this is nonsense. We need signs to communicate the signified right?

user76284

@VaneVoe What Thorgott said was exactly correct. You need to review some basic definitions of functions before accusing others of "lack of expertise" and "making themselves look smart".

Ted Shifrin

Yes, upon reviewing the transcript, I was shocked by @VaneVoe's rudeness and ignorance.

user616128

02:50

@VaneVoe I concur with above, the given answer was correct, and then you moved the goalposts, and then you moved them back without a worry in the world.

You ask for a function $f:\Bbb R\to \Bbb R$ that is surjective. For no reason you bring up bijective functions, possibly not understanding the directionality of implication? You then give a non-example of a function $\Bbb R\to \Bbb R$. You then start using R for the real numbers and some unknown other domain in the same sentence. You then write some comment that is nonsense.

That's my $.02

Vane Voe

03:14

The definition of function I gave was concurrent with bijective

they ignored that for no reason

they clearly have no idea what they're talking about

I don't need a function to be bijective in general

but I specifically asked about bijective functions and they specifically ignored what I asked, indicating they don't know how to address my question

Furthermore, I shouldn't have to define as conventional a termonology as function anymore than someone should have to prove 1+1=2. Are there other systems of math? Infinitely many, but there is one world-wide one taught in nearly every school within which 1+1=2, there is no reason to assume otherwise unless specified. Because if their lack of understanding of conventions, they definitely lack expertise.

The domain definitely isn't unknown, I said from R to R. I shouldn't have to tell someone with experience that a function that maps from all real numbers to all real numbers that is bijective has a domain of R and a range of R.

nCm

04:59

Does $\log^3_2 n$ mean$(\log_2 n)^3$? I don't know what this shorthand mean...

Mike Miller

05:38

Bad writing to not make it clear.

I think your guess is right.

Jacksoja

if X is a subset of R

to show that X is compact

we can just find a subcover corret?

@MikeMiller does it sound correct?

Mike Miller

05:56

R is the real line? A subcover of what? What is your definition of compactness?

Jacksoja

@MikeMiller Yes real line

I mean if X has finitly many elements

Mike Miller

If X has finitely many elements it is compact. I am sure you can show that with whatever definition you are using.

Jacksoja

okay thanks

@MikeMiller how would you prove it ?

Anubhab Ghosal

06:25

$\sum_{k=0}^{\infty}\frac{1}{3k+2}-\frac{1}{3k+4}}$. Has this been posted?

Silent

@MatheinBoulomenos Oh wow! Thank you very much

Laotse

08:24

Hello. Is it possible to obtain an approximate solution with method of multiple scales to a nonlinear ODE y''(x) + a - 1/\epsilon*y(x)^2 = 0? I can't get the scales "match".

nCm

09:03

Does complex symmetric matrix must have real eigenvalues?

Tobias Kildetoft

@nCm No, that fails even for diagonal ones

nCm

I only found answer about real symmetric has real eigenvalue...

@TobiasKildetoft Thanks! Wow now I understand it...

It's a test question (I don't like test) and it gives the answer true, now I think it implicitly assume real symmetric matrix...

Tobias Kildetoft

Yeah, usually we don't care about complex matrices being symmetric, but rather about them being hermitian

ninja hatori

10:34

@TobiasKildetoft do you know Hilbert theorem 90?

ÍgjøgnumMeg

12:48

@ninja what about it?

abenthy

13:36

I have a question. If $M$ be a connected $2$-dimensional manifold with a complete hyperbolic metric $g$, does the uniformization theorem state that such a $g$ is unique in some sense?

Perturbative

What does the highlighted part here mean?

Mike Miller

Haven't we talked about this?

@abenthy Unique in its conformal class for closed manifolds. I'm hesitant about saying anything in the noncompact case.

Might be "unique in the conformal class of complete metrics".

abenthy

Yeah, we already talked about this possibly. Maybe I'm just confused

Perturbative

Ohhh sorry @MikeMiller uploaded the wrong thing

What does the highlighted part mean here ^^

Specifically what does it mean to "identify all the summands to a single sphere"? Is it sort of a projection map for wedge sums?

Fargle

I think it's stipulating that the restriction of $p$ to each $S_n$ in the wedge sum is the identity map.

Alessandro Codenotti

13:48

It's a map that restricted to any of the sphere in the wedge looks like the identity

Fargle

pew pew

Alessandro Codenotti

Damn snipers

Mike Miller

@abenthy Sorry, that first line wasn't directed at you.

Perturbative

Thanks! @Fargle @AlessandroCodenotti

Fargle

No problem :)

Rithaniel

13:54

A manifold does not necessarily need to be orientable, correct?

abenthy

@MikeMiller No problem, and thanks for your answer!

Fargle

@Rithaniel I don't think any usual definition requires it, no.

Rithaniel

I suppose I should have just googled a definition, actually. Thank you replying.

Fargle

14:08

No problem

Anonymous

14:42

@VaneVoe Irrespective of whether you were right or wrong, it certainly was not okay to belittle other users. Take a day off.

Anonymous

@MikeMiller Do you mind if I remove your recent messages from the starboard, assuming the case is shut now?

Mike Miller

That's fine

Anonymous

Thanks.

Mike Miller

Feel free to do so anytime, I have no connection to those

On mobile I don't even see them :)

Anonymous

It seems the math chat is lacking active room owners. Let us know if you need more ROs. :)

Anonymous

14:50

Timeouts and kick-mutes are useful features in such cases.

ÍgjøgnumMeg

Ted!

lol

Anonymous

Yep, I know Ted is the only active RO (from what I've seen).

ÍgjøgnumMeg

I haven't seen any other italicised names

Anonymous

@ÍgjøgnumMeg The full list is here.

Anonymous

(bottom-left)

ÍgjøgnumMeg

14:52

Ah okay. I've occasionally seen robjohn and anon but not all that often

Mike Miller

15:12

The question is who would be an acceptable candidate :)

ÍgjøgnumMeg

VaneVoe

Rithaniel

I'd actually suggest you, Mike. Also Fargle and ÍgjøgnumMeg

(Though, I'm more a lurker and not active enough to really make such a call)

Mike Miller

15:32

If I'm irritable enough to post comments like I did then, is that not a disqualifier?

ÍgjøgnumMeg

there are too many logs in Analytic NT

Fargle

I'm definitely not active enough. Plus that just sounds like a lot of work

ÍgjøgnumMeg

anyway, I've just discovered "Bach's bound" which apparently gives a much tighter bound on the classes of primes that generate the class group of a number field, at the cost of assuming GRH

though it's asymptotic

Mike Miller

Eh but that's true so whatever

ÍgjøgnumMeg

Right

no more Minkowski's bound for me!

Once you go Bach, you never go back

$12\log^2 \Delta_K \leq N\mathfrak{p} \leq (4 + o(1))\log^2 \Delta_K$

Mike Miller

15:45

Too hard for me man

ÍgjøgnumMeg

the derivation of the bound is a bit ridiculous

Eran

16:06

if $U=\bigcup^{\infty}_{n=1} U_n $ and $C=\bigcap^{\infty}_{n=1} C_n $ is $U\setminus C = \bigcup^{\infty}_{n=1}(U_n\setminus C_n)$ ?

Mike Miller

16:26

Seems right. $$\bigcup (U_n \setminus C_n) = \bigcup (U_n \cap (X \setminus C_n)) = (\bigcup U_n) \cap (\bigcup (X \setminus C_n)) = (\bigcup U_n) \cap (X \setminus \bigcap C_n).$$

Anonymous

16:56

@MikeMiller The system generally chooses in order of room activity. You (84.9k) come just after Ted (165k). So if you don't have any issues, I could make you a RO now. Are you okay with that? Otherwise, this needs to be raised on math meta. (And I believe when given responsibility users tend to behave more maturely. So I don't think the slightly irritable comments you posted before should affect your behavior in the future.)

Mike Miller

You can do it if nobody else around objects. I don't think this is really relevant enough to post on the math meta, since this room has (IMO) only a tenuous link to matters on the main site.

Feeds

Blue has added Mike Miller to the list of this room's owners.

Eran

Congratulations

Anonymous

@MikeMiller Okay, let's give it a try anyway, as I see at least one user supporting it. :) Please have read through A guide to moderating chat when you get time!

Anonymous

And congratulations!

Mike Miller

17:08

I'm not sure I buy that, since I bet you can find at least one user who objects, too.

I'm also not sure this merits congratulations. But I'll read the thing.

user629353

Can someone please tell is this (ibb.co/stGFSKs) figure a polygon. If yes then is middle line would count as a edge

Mike Miller

No, that is not a polygon. Start from definitions: what is a polygon?

Érico Melo Silva

@MikeMiller how does it feel to be a big shot

user629353

Closed loop formed from continuous line segment, and it forms the same

Mike Miller

@user629353 That is not a definition of polygon I have ever heard (nor does it sound like a correct one).

In any case, what you have drawn cannot be called a "simple closed loop". At those points with three edges coming out, you'll have drawn the curve touching that point at least twice.

Anonymous

17:12

@MikeMiller Well, we don't have any system to carry out a poll here. It's just a leap of faith we need to take. Nevertheless, room owners will be subjected to the same level of moderation as any other user, in case something goes awry.

Eran

@Blue I don't want to be rude but this channel is SO INACTIVE I can't see what's all the fuss about owning it :S

Anonymous

The main reason for choosing you was "you are the second most active user here" rather than "a majority supported you". That's the default algorithm the system uses to choose ROs anyway. :)

Anonymous

@Eran Well, this room (along with the other math chat rooms) generates a lot of flags. It's a statistical outlier on the network.

Anonymous

You guys need people to cool down the atmosphere when conversations start becoming hostile. And ROs are given tools for that. I don't know why they use the term "room owner" though. It does sound misleading.

Eran

What are those math chats? there are so many times I need help with math and I can't get it from here because no one speaks here most of the time

Anonymous

17:17

@Eran These.

Eran

Thank you sir

ninja hatori

0

Q: About dual of finitely generated projective module

Let say $x \in M$ and $x \neq 0$ then is it true that for finitely generated projective module $M$ there is $g \in M^*$ such that $gx \neq 0$. If yes how to prove it. For vector space dual this result is true what about projective module. Also if $f \in M^*$ $f \neq 0$ then $fy \neq 0$ is also ...

Mike Miller

@Eran You'll have less luck everywhere else.

Eran

#math chat in freenode is great.. the only bad thing about it is the chat, here you can use latex...

They always answer and the discussion is mature and nice

Anonymous

@Eran It's difficult to find all the good things in one place. :P

Eran

17:27

True.

Mike Miller

We have LaTeX but we don't answer questions and aren't mature.

Anonymous

I love the SE chats because of their great UI. Albeit it's true you won't find a lot of experts all the time, but if you stay around you'll make good friends and come across many knowledgeable folks. It takes time to build academic relations online, but some of them are really worth it.

Eran

I wish Ted could be here 24/7, he's the best

haha

Anonymous

17:47

@Eran I've known Ted, Semiclassical, Leaky Nun and Balarka from here. They're all excellent mathematicians. :)

Ted Shifrin

18:39

So I see that now @MikeM is my partner in crime. Yippee!! :)

Mike Miller

I see I can kick-mute you.

Should I try?

Érico Melo Silva

political power grows out of the barrel of a gun i see

Ted Shifrin

The Drumpfification of the world continues apace.

Perturbative

19:08

Hey everyone!

Ted Shifrin

Hi @Perturb

Perturbative

19:56

Hey @TedShifrin :)

I'm a bit confused here, so if I let $N$ and $S$ be the two components of $S^{k-1} \setminus S^{k-2}$, then I know if I pick any $y \in S^{k-1}$ that's not contained in the image of the collapsed subspace $\mathbb{RP}^{k-2}$ in $S^{k-1}$ (under a homeomorphism), then $(q \circ \varphi)^{-1}(y) = \{x_1, x_2\}$ with $x_1 \in N$ and $x_2 \in S$

without loss of generality, so $\operatorname{deg}(q \circ \varphi) = \operatorname{deg}(q \circ \varphi |_{x_1}) + \operatorname{deg}(q \circ \varphi|_{x_2})$, but I don't see why one is the identity and the other the antipodal map

Mike Miller

He doesn't say that one is the identity and the other is the antipodal map. If $\tau$ is the antipodal map, he says that $q \circ \varphi \big|_{S}$ is $q \circ \varphi\big|_N \circ \tau$.

Therefore, the local degree at a point differs by the local degree of $\tau$ at a point (but the antipodal map $\tau$ is a diffeomorphism, so this is the same as its global degree).

Perturbative

Okay yeah that's right and so $\operatorname{deg}(q \circ \varphi|_{x_1} \circ \tau) = \operatorname{deg}(q \circ \varphi|_{x_1})\operatorname{deg}(\tau) = 1\cdot (-1)^k = (-1)^k$ (assuming that $\operatorname{deg}(q \circ \varphi|_{x_1}) = 1$)

Leaky Nun

hi

FuzzyPixelz

"The trivial case of the empty family must be regarded as linearly independent for theorems to apply." This is what Wikipedia states about linearly independent sets, is this a convention ? If yes, what is an actual trivial linearly independent set.

Mike Miller

20:11

i do not understand the question. the empty set is a linearly independent set of vectors. this is called the trivial case.

FuzzyPixelz

Alright, my bad. But I failed to demonstrate that from the definition

Perturbative

Okay now I get the argument. Thanks! @MikeMiller

Mike Miller

@FuzzyPixelz Then what is the definition of a linearly independent set?

FuzzyPixelz

If a linear combination of its elements is zero, then all of the scalars are zero

Mike Miller

What are the linear combinations of elements of an empty set?

FuzzyPixelz

20:15

Well zero?

It's an empty sum

Mike Miller

I see your point - you object that "empty scalars" are not the same thing as "0 scalars".

But let's go back instead to the definition of linear dependence. I guess I would say a set of vectors is linearly dependent if there is a non-trivial finite linear combination of its elements which is zero.

The empty set has no non-trivial finite linear combinations of its elements.

FuzzyPixelz

Okay then, that's a better way to put it.. Thank you. Although all this feels very shallow as every existence statement on the empty set is false anyway.

Jasper

@MikeMiller When did you become a room owner?

Mike Miller

I'm not sure what you mean. I would be surprised about a theorem you could give me about linearly independent sets that it failed.

FuzzyPixelz

Not at all. Thanks again. I was trying to say that the argument isn't very satisfying. But it works.

Jasper

20:24

I always try to make arguments and definitions work for the empty case. It is an interesting exercise.

Mike Miller

@FuzzyPixelz It never is.

@Jasper An hour ago or something.

Jasper

Paul Halmos is very precise about the empty case in his books Naive Set Theory and Finite-Dimensional Vector Spaces.

But most authors are not that careful, and screw up sooner or later in the book.

Perturbative

@MikeMiller Just wanted to clarify, one of those maps in the question I asked earlier will be either the identity on $N$ or $S$ correct?

bolbteppa

In how many ways can I put the three variables $x_1,x_2,x_3$ into the function
$F = F(a,b,c,d,e;f)$, where $F$ is symmetric in the first five indices, such that one is always in the "$f$" position?

Mike Miller

No, that was never asserted, and depending on your set-theoretic definition of $q$ (which you never told me) may not be proved.

FuzzyPixelz

20:28

I guess you could also say that a family is linearly independent if none of its elements is a linear combination of the others.. But the empty set has no elements so it's always true.

Mike Miller

What was said is that it's an orientation-preserving homeomorphism on one of $N$ or $S$.

mathsssislife

guys is there a field in mathematics that just focuses on parameters of functions?

bolbteppa

$5 \cdot 4$ ways to put two of the variables into the first five positions, right? Does it not matter that I can put one of the initial three variables into the $f$ position?

Perturbative

Okay understood, $q$ was the canonical quotient map $q : \mathbb{RP}^{k-1} \to \mathbb{RP}^{k-1}/\mathbb{RP}^{k-2}$

Mike Miller

That is not correct, because the codomain is incorrect.

It's the canonical quotient composed with an identification $\Bbb{RP}^{k-1}/\Bbb{RP}^{k-2} \cong S^{k-1}$.

I would do that by thinking of $\Bbb{RP}^{k-1}$ as a disc with antipodal points on the boundary identified, and $S^{k-1}$ as a disc with all points on the boundary identified. But when I say "by thinking", I mean "by using a homeomorphism from whatever my usual model is to this".

(Presumably your usual model is as a subset of Euclidean space.)

These sorts of identifications matter for orientation counts

Though honestly it would be just as easy for me to say that everywhere he says "2" he really means "plus or minus 2"

Perturbative

20:35

Wait really is it "plus or minus 2",if so then I'm done

Mike Miller

not sure how much of what I just said parsed. the point is that to pin down the sign you need to understand all of the conventions involved (because orientations are finicky like that). to calculate homology you do not at all need to know the signs.

Perturbative

Ohh that's right for calculating homology because somewhere down the line I'll end up with either multiplication by $2$ or multiplication by $-2$ and $Z/2Z$ is the same as $Z/-2Z$ (I hope that makes some sense)

Mike Miller

yes

Leaky Nun

what did the empty set do to deserve such bashing from y'all

bolbteppa

Seems like there are $5 \cdot 4 \cdot 3$ ways to do it :'\

Perturbative

20:38

Disregarding homology should I worry much about pinning down the signs?

Thanks again for your help! @MikeMiller

Mike Miller

@Perturbative no

not until you absolutely have to

Perturbative

Okay cool

Alessandro Codenotti

Potentially trivial question: can the wedge of two manifolds be homotopy equivalent to a manifold?

Mike Miller

Conditions on "manifold"?

Daminark

Wedge of two circles is homotopy equivalent to the double punctured plane

Alessandro Codenotti

20:46

Ah, right, are there compact examples?

Eric Wofsey

A wedge sum of two closed manifolds cannot be homotopy equivalent to a closed manifold.

Mike Miller

a compact manifold without boundary cannot be decomposed as a wedge of two manifolds up to homotopy, because top $\Bbb Z/2$ homology is not $\Bbb Z/2$. Any finite CW complex is homotopy equivalent to a compact manifold with boundary (and hence its interior, a noncompact manifold without boundary).

Eric Wofsey

If they have the same dimension, the top mod 2 homology of the wedge is 2-dimensional whichis not possible for a closed manifold.

If they have different dimension, the mod 2 fundamental class for the lower-dimensional one can't have a poincare dual.

Mike Miller

Ah, fair enough, I implicitly assumed same-dimensional.

It's not clear immediately to me to what degree "$X \simeq K_1 \vee K_2$" is impossible for $X$ a manifold and $K_i$ complexes. (For instance, I don't immediately see why $BG$ for some acyclic group $G$ cannot be a wedge summand of a manifold.)

It seems more promising if $X$ is simply connected, where a repeated application of Poincare duality implies that for any field $k$ either $H^*(X_1;k)$ or $H^*(X_2;k)$ is nontrivial. But then I don't know that the summand which has vanishing cohomology doesn't depend on $k$.

I guess in the non-simply-connected case there might be some argument involving the locally finite fundamental class of the universal cover.

Alessandro Codenotti

I see, thanks @Mike @EricWofsey

Hatshepsut

21:31

Can I get some help on improving my question?

Somebody's downvoted it but I'm not sure where I can make it clearer

I haven't linked it because I think that's frowned upon?

Leaky Nun

how well do manifolds with corner behave? e.g. stokes theorem?

Seth Taddiken

If I'm told to "write the equation in y" does that mean isolate y on one side or does it just mean get rid of any other variables by subbing them out for functions in terms of y?

More context, I have a messy equation with x's and y's but I can sub in x=sqrt(L^2+y^2) for x and have a messy equation with y's on both sides. Is that "writing the equation in y" or do you think I'm expected to isolate y on one side

Mike Miller

21:51

@LeakyNun fine

Leaky Nun

great

mathsssislife

Guys, my definition of closed is that if its complement is open, how can I prove that a closed rectangle is indeed a closed set?

Leaky Nun

prove that its complement is open :P

mathsssislife

LOL

I haven't thought of that

My definition of open is that: A subset U $\subset \mathbb{R^n}$ is open if $\forall x \in U$ $\exists$ an open rectangle in A such that x$\in A \subseteq U$

Guys, is calculus on manifolds a good book in your opinion?

for self study

Fargle

I found that I couldn't really get through it personally. Other people here have also at other times said they don't care for the book, but I won't speak for them

CaptainAmerica16

22:07

@Fargle Spivak's? Can I ask what you found wrong with it?

Fargle

I just found it generally confusing in exposition and notations. Granted I didn't go too deeply into it.

mathsssislife

@Fargle what book do you suggest for an introduction to manifolds?

@CaptainAmerica16 nothing, i'm just curious about peoples opinion on the book

Fargle

I'm personally a fan of Guillemin and Pollack's diff top book.

CaptainAmerica16

@Fargle Ah, ok. That seems to be the problem with a lot of textbooks, lol.

Fargle

There are places where they're a bit "sloppy" with details, but with careful reading that wouldn't be too hard to deal with

Perturbative

22:35

@CaptainAmerica16 The Spivak textbook you're reading is different to the Calculus on Manifolds book (if you were thinking it's the same book)

Also +1 for G&P

CaptainAmerica16

@Perturbative Yeah, I know.

Ted Shifrin

Greeting, @CaptainAMerica.

Fargle

Heya @Ted.

Ted Shifrin

Heya @Fargle

Fargle

Here to call me out for opining on G-P despite only being up through transversality? :P

CaptainAmerica16

22:49

@TedShifrin Howdy

Ted Shifrin

You have a ways to go!

user193319

Dumb question: is $A \cap E^c = A \setminus E$ true, where $A$ and $E$ are sets? I think I proved that it is true.

Ted Shifrin

Of course. Think of the distributive property.

Mike Miller

@TedShifrin How much do you object to someone calling the theorem "if $M$ is a 1-manifold, then the signed count $\# \partial M = 0$" by the name "Stokes' theorem"?

CaptainAmerica16

Does E^c mean complement?

Ted Shifrin

22:56

I object strenuously, @MikeM.

CaptainAmerica16

One thing I don't like about set theory is that some stuff has like a million ways of being represented.

Érico Melo Silva

lmao what a weird complaint

Ted Shifrin

The notation that some books use that I will NOT accept is $\overline E$ for the complement of $E$.

Throw them in the trash.

Érico Melo Silva

who does that

Ted Shifrin

Various intro to higher math books, apparently.

Érico Melo Silva

23:00

ive never seen $\bar{E}$ not mean closure

Ted Shifrin

I don't recall.

Well, if you're doing just set theory and not point-set topology, who cares about closure ...

Érico Melo Silva

oh word

i didnt even think about that

Ted Shifrin

Still, I'm on your side, as students using such a book will be woefully confused if they go on.

CaptainAmerica16

I prefer E'

Ted Shifrin

But that also stands for the set of limit points of $E$. Nope.

$^c$ is safest.

Or $X-E$.

CaptainAmerica16

23:06

>:\

Ted Shifrin

So, @CaptainAmerica, are we talking about chapters 4,5,6 ... yet ?

CaptainAmerica16

Uuuummmm

I haven't quite reached 4 yet.

Ted Shifrin

And yet you continue wasting time ...

hi Demonark

Daminark

Hey there!

Ted Shifrin

(speaking of wasting time)

CaptainAmerica16

23:17

Oh, come on.

Daminark

:0

Érico Melo Silva

called out

CaptainAmerica16

Nevermind, that sounded really bad once I typed it out.

Daminark

Assuming $x\le y$ implies $y$ is big is a common and dangerous mistake

Ted Shifrin

Like the kid in my AoPS class who wants to do an independent study of my multivariable course with me next year, but he has yet to follow through on a single promise to turn in this year's homework assignments to me.

LOL, Demonark. Well, if $x$ is big, it seems logical.

CaptainAmerica16

23:19

@TedShifrin What, you would do that?

Ted Shifrin

Not looking like it.

Daminark

Turns out we have no class tomorrow because of the polar vortex

Érico Melo Silva

ya it's too fuckin cold

Ted Shifrin

Doesn't global warming know about projective duality, Demonark?

I wouldn't think Chicago would close anything for cold.

Daminark

It's cold enough that there's apparently legitimate frostbite risk

Érico Melo Silva

23:21

it's -40 w windchill

Ted Shifrin

that's pretty chilly

CaptainAmerica16

I don't know how cold it is where I am, but it's snowing.

Daminark

Originally I expected the university to just say "Toughen up kids" but there's probably a liability concern if people start getting hospitalized

CaptainAmerica16

And apparently, I have to help my older sister move into her new apartment.

;-;

Jake Rose

@TedShifrin hows your sturm louiville?

Ted Shifrin

23:23

Don't you have school to go to?

I haven't thought about it in 40+ years, Jake.

Jake Rose

Still probably more advanced than my knowledge

Ill ask and if you get it you get it if you dont you dont

Whats the Latex for differential operator?

Daminark

Oh I remember at the very end of analysis Schlag wrote a handout talking about some "further directions" we didn't have the chance to pursue

Ted Shifrin

you talking about a script L or something?

Daminark

Sturm-Liouville was one of them

Jake Rose

@TedShifrin

@Daminark Its a super interesting subject

About to dominate my quantum mechanics course

Daminark

23:25

Nice

Jake Rose

Actually

I dont fully know what Im askin

Not gonna waste peoples time

Let me think for a bit

CaptainAmerica16

Ted, just so I have a reference: How quickly would you expect someone to move through Spivak?

Assuming time isn't being wasted and said person isn't a slacker.

Ted Shifrin

That's a silly question.

CaptainAmerica16

-_-

Ted Shifrin

I know how fast my course moved. I can't speak for an individual who's doing it occasionally in spare time.

Érico Melo Silva

23:29

i feel like rate someone moves at for self-studying is so wildly variant that no answer could be good

CaptainAmerica16

Yeah, alright.

Mike Miller

parsed

Ted Shifrin

The moral of the story is: I'm losing patience with people who have intentions and then basically do not follow up on them reasonably. Stop pretending ...

Mike can kick me out of the room now.

Salut, @Astyx!

Astyx

Hi

CaptainAmerica16

@TedShifrin :(

I do plan on finishing Spivak.

Astyx

23:34

Actually I need to go to bed

Seeya another time ! :)

Ted Shifrin

Bonne nuit, @Astyx.

CaptainAmerica16

But I get what you're saying...I haven't taken it as seriously as I should.

And I mean that according to my own intentions.

Ted Shifrin

I'm not here to be your conscience or your nagging mom.

CaptainAmerica16

Of course not.

Daminark

@Eric so when I emailed the places about yesterday, 2 got back, Madison and Minnesota. The latter response seemed... mildly annoyed

Érico Melo Silva

23:38

RIP dude

Ted Shifrin

I guess it gets annoying to be pestered .... but I'm sure that's not so unusual for the graduate coordinator or his/her secretary.

Daminark

"Per your request, your updated statement of purpose has been uploaded to your online application.

Please refrain from sending any further materials." :(

Mike Miller

rekt

Ted Shifrin

Oh ... you weren't pestering about admission.

Daminark

Oh right I don't think I told you. There was a TeXnical error with my statements of purpose that I discovered yesterday. I think when I pasted the edited SOP over the previous document, it didn't actually delete the existing contents like I thought it did

Érico Melo Silva

23:40

ull probably be fine

best to just not worry too much

Daminark

So there was a Notre Dame SOP, \end{document}, and then the actual SOP I wanted to send to the school in question. The pdf only showed the former

Ted Shifrin

Didn't you tell me you couldn't update?

I'm confuzled.

The actually SOP followed the \end{document}?

Oh, you're using the letter format.

You have to be super careful with that.

Érico Melo Silva

i had separate files for each school so that i could keep track

Ted Shifrin

Yeah, I think that's advisable.

Daminark

I didn't know about the formats, I just use whatever the overleaf default was. But yeah I don't know how many places took updates but I was like eh you know what I'll just go for it, if they refuse to accept any updates I'm done anyway

Mike Miller

23:43

This really probably wasn't the deciding factor anywhere.

Daminark

Hopefully

Ted Shifrin

I last served on the graduate committee one year eons before the electronic age, so I can't help.

Érico Melo Silva

@TedShifrin how much did y'all care about these statements when u did serve tho

Ted Shifrin

Unless they were addressing something serious in the record, probably less important by far than letters of recommendation. But if the SOP stood out as unusual, that could be a plus.

Daminark

Like a good pun? :P

CaptainAmerica16

23:51

Anyone know what the "fundamental forces" are?

Érico Melo Silva

that would be an auto-reject if it were me

Leaky Nun

I keep seeing on the internet that English is a stress-timed language (where you spend roughly the same time per stress) whereas French and Spanish are syllable-timed language (where you spend roughly the same time per syllable). I think the latter is very accurate, and certainly English is far from being syllable-timed, but I'm heavily doubting the fact that it's stress-timed

Ted Shifrin

I tended to take undergraduate application essays more seriously than some other folks when I served on UGA's admissions committee. But then when it was clear that someone else had written the essay, I was in the position of saying, "well, this totally sucks."

I think American English is reasonably stress-timed, although I haven't thought about it carefully.

in what context, @CaptainAmerica?

Leaky Nun

I asked my friend to say "the dog walked in the park" as an experiment (without telling him the intention), and he spent much less time between "dog" and "walked" than between "walked" and "park"

CaptainAmerica16

@TedShifrin Physics

Ted Shifrin

23:53

I"m confused. There are two more words in the latter span.

Is this a native Brit or a furner?

Leaky Nun

hmm maybe the difference is not as much as I initially perceived

Ted Shifrin

gravitational, electrostatic, magnetic ? @CaptainAmerica

Leaky Nun

but those three syllables are the only syllables to receive stress in that sentence

CaptainAmerica16

@TedShifrin Maybe...there are four blanks on the assignment though.

Ted Shifrin

I was thinking more about multisyllabic words unto themselves.

Guess you should read the chapter, @CaptainAmerica.

Leaky Nun

23:55

I'm thinking about a sentence as a whole

as that's what English language teachers on youtube do when they demonstrate the stress-timed-ness of English

CaptainAmerica16

@TedShifrin Dang it

Érico Melo Silva

@TedShifrin electromagnetic is 1 force, then there's weak nuclear, strong nuclear, then gravity

Ted Shifrin

aha

I don't see why electric field and magnetic field get combined, though.

Leaky Nun

inb4 gravity is not a force

Karl Kronenfeld

wait who's saying gravity is not a force?

Érico Melo Silva

23:57

it's not in GR that's true

Leaky Nun

but I guess we're going quantum so we ignore relativity

until someone can unify those two theories together

Transcript for

$Mathematics$ Mathematics