NOOOO. B is between A and C, right?

The book has an actual picture.

which chapter?

oh duh

It's not in your edition. I added this problem in 3rd or 4th.

do you have a picture on hand

@Daminark I burst out laughing like a madman at 6:08

No, Meow. Read the words I just typed.

Remember that B was between A and C.

ok what do you mean by B is between A and C

because maybe theres a miscommunication

B is on the circle right?

oh geez ... short arc, not long arc.

...

in the interior of $\angle AOC$.

lets redraw this

wow $\alpha - x$ that was quite easy

@TedShifrin We never really talked even about what the definition of a path was. I assume because its covered in some intro topology course. So I assumed if was some kind of map $\gamma: [a,b] \to X$. But I never figured out how you define them for tings like a path over all of $S^1$ where you can't use an interval.

nods @Meow

@Kevin: Closed path here.

mximize $(\sin x + \sin (\alpha - x)) / 2 $?

Ya I guess a closed path has to be something like a map from $S^1 \to X$?

Yes, Meow.

Right, Kevin.

cool thats easy just factor out a $\sin$

braces for smack

smacks

since all functions are linear.

so we need $\cos x = \cos(\alpha -x)$ when we derive and simplify

since our $x \in [0, \pi]$ and $\cos$ is bijective on that domain we can simplify to $x = \alpha - x$ right?

Which tells you precisely what you knew.

Hey

cool, done

next part

Why are you beating MeowMix?

since I like triangles, you can subdivide the polygons into triangles all with a common vertex

He deserved it ... indeed, asked for it.

@Meow: You mean the center of the circle?

Well, carry on then.

yes

I think folks in here volunteer Ted smacks more than Ted gives them now.

and then you apply the same argument

Well, as long as they're not Ted karate chops, I think we'll all live.

f you ask for it, gives time to brace for impact I imagine

What do you mean by "apply the same argument"?

So, is the smooth image of a smooth manifold not necessarily a smooth manifold?

NO, not at all, Kevin.

Think about turning the circle into a figure 8.

Watch out. You're gonna get smacked again.

BTW, Daenerys, I removed your diff geo tag from your applied math post. Just because there are gradients there doesn't mean it's remotely diff geo.

Okay, yea that is troublesome then.

Oh, so that's how you edited it. I didn't notice. Thought maybe you fixed some of my crappy formatting.

Thanks.

Nope. I just figure when people post diff geo questions, I should know what they're talking about.

I wish you did know what I was talking about. Hell, I wish I knew what I were talking about.

next problem

Ok yea, this is the whole point of talking about embeddings. If the smooth image of a manifold was always a manifold it'd be trivial.

Right, Kevin.

posting here for mathjax capabilities

@Meow: When you say "same argument," do you have an elegant solution in mind as opposed to repeating something?

Wouldn't you need some kind of conditions on that mapping in order to preserve manifoldness? Something analogous to uniform continuity or something? Or an isomorphism?

No. I mean homeomorohism.

Like in topology, that preserves topological properties.

Embeddings are precisely homeomorphisms to the image.

not particularly

Embeddings are monomorphisms, right?

I never remember. They are injections.

actually how would I prove that lol

I guess that's what monomorphisms are

I would like you to think how to write a few sentences making a concise, rigorous argument, @Meow.

thats what I meant. They don't have to be bijective.

mono=one

Of course I'm just talking out my anus.

im not sure why one would do that

considering you have a mouth for that reason

Concentrate, Meow.

unless youre into that kind of stuff i guess

i know very little of which I speak, but finding conditions that preserve "nice" properties has always been a big interest

Getting distracted by shiny objects, though. Gotta go finish my homework before I turn into a pumpkin.

i mean a naive thought would be to make it a problem of $n-1$ variables but that's just disgusting

a least as far as i can see what the solution would be

Remember I try to train you to reduce to what you've already proved, Meow.

And we don't know multivariable calculus.

proven*

Both are acceptable.

really?

Proven tends to be used more for the passive voice than for the active past participle. sticks out tongue

Night night. Don't let the Ted bugs smack you too hard.

LOL, night Daenerys.

nobody can do the classic "Actually" move on ted

except for ted himself of course

huh?

you know

the "Technically"

@TedShifrin what have we already proved?

@MeowMix?

that uhh

so if you have two points on the unit circle $A$ and $C$

and a point $B$ between them

with origin $O$

So my professor forgot to upload a worksheet for tomorrow. I may be delaying my email letting him know intentionally.

screw it ill copy the problem statement

Suppose points A and C on the unit circle subtend an angle \alpha. How should you choose B between A and C to maximize the sum of the areas of triangles AOB and BOC?

how is it even related to the polynomial problem

it's not

then what have we already proved that is relevant?

or is it????

no, its not

NOOO ... Two totally separate problems.

this is a diff problem, ignore that one

But the first had two parts.

I'm interested in the polynomial problem

ooh a diff problem?

ive proven that B bisects the arc (if thats proper terminology) to maximize the area

PVAL, I TA-ed only once in grad school (not counting teaching for Chern). The prof in charge was highly competent and organized, a sheer pleasure. Sadly, he was brilliant and died of an aneurysm at the age of 30. :(

diff = different, not diff category

Well, actually, it is in the diff category.

Yeah thanks for explaining my pun.

Really helps me out

You deserved it.

hmm so you want me to like make a generalization of my proof?

or just apply it?

No, Meow. I want you to apply that result to prove the general version. Without any new calculus.

alright

This will be the 8th (out of 9) semesters I've been a TA.

let me think

@PVAL: Is that because Bob doesn't have grant support to help you out? ... An increasingly serious problem.

so initial thought

Well I don't know. He might just not want to help me.

That doesn't seem right unless he's given you ultimata at some point.

I always thought he was a nice, straight-forward guy.

Admittedly I chose an adviser at the end of my 2nd year

That seems pretty usual.

So there were 4 semesters no one helped me out on.

so let's say $V_i$ is the $i$-th vertex of the polygon

ugh how can i phrase this

It's clear this university needs TA's and the department was willing to make me believe I'd only work half or so semesters in order to get more TA's.

So they're overtaxed with too many undergraduates compared to graduates, PVAL?

@MeowMix n must be even or roots of f are imaginary?

I'm quite unimpressed with the way UCSD handles size issues, too, PVAL.

yes

@Fawad, @Meow: No, that's not why.

At ucsd the ratio I experienced was maybe 40 to 1

i thought that was just a question

and we had TA's in standard upper division undergrad

@TedShifrin then? Also how that can be solved?

wait it doesnt have to do with roots

The honors calc class (which is now using my book) has 2 recitations with 45 each, or something, @PVAL. That's absurd.

i mean i guess it does but just to ensure $f(x) > 0$

The undergraduate analysis class has 180 students.

at the start of the semester here the ratio is at least 100 to 1.

you can never have that when $n$ is odd

@Meow: $f(x)\ge 0$, yes.

since one direction will always approach $+\infty$ and the other $-\infty$

The analysis course had 50 when I was there.

Did they merge 140 and 142?

Yeah, PVAL, things are out of control. No, I don't think so. It's the "weaker" major analysis that's so huge. You didn't take that one.

@MeowMix direction? What that is?

@Meow: Back to the polygon so I can go cook dinner.

Yeah I still know from talking to people, that that one was nowhere near 100.

ok ok

when I was there.

We're getting old, PVAL :D

The weaker one is 142.

or 143

or something

The ordinary one is 140.

i don't know how to phrase this elegantly

I had over 40 students one time for a graduate physics class as a TA. It was such a horrific experience that took up so much of my time to do properly, it became a departmental rule after that that TAs for grad classes are limited to a max ratio of 25 to 1 or something like that

My friend who's TAing the honors calc hasn't complained to me of late, so I hope things are going better than last year.

but you like apply it to each of the adjacent triangles

OK, Meow, good idea.

Well the dumpster fire that was this class has fixed my TA to student ratio.

You mean adjacent pairs, @Meow.

I'm not sure how thrilled I can be about that.

sorry, yes

How did it fix? all the students dropped, PVAL?

I counted 32 (out of an original 100) in lecture today.

Great lecturer, eh?

The question is whether faculty raises are affected by such things.

If we graded for competency instead of distribution, I think less than ten students would pass.

Is that the students' fault or the prof's?

Probably both ...

I dont know how you guys do departmental review. But it cna't look good on any university or outside audits that there are classes where 2/3rds of students drop

Well lets say this doesn't compare well with previous similar classes I have taught.

No, Kevin, often Dean's offices get real pissed off at that, especially when parents complain.

That probably isn't mostly the students' fault.

honestly i have no idea how to do this without ugly garbage

On the other hand, department heads/assistant heads shouldn't put people they know are incompetent in such classes. But that rewards faculty for being incompetent. It's a really tough no-win situation. I've been there ... (for 8 years, officially).

You have the idea, @Meow.

So what must be true about adjacent pairs to get a maximum for that pair?

We had a departmental review of TA jobs the year I got there. It was universally agreed to be terrible.

All they did was fit the same amount of people in 2 section instead of 3

they must be equal

Meaning that TA's aren't properly trained/supported?

Meaning that they are overworked.

Yeah, seems clearly so to me, PVAL.

UGA handles that well, even though they pay sub-par.

@Meow: So how does maximizing adjacent-pair areas relate to maximizing the whole thing?

well like

if you add them all up

you get twice the whole thing

Awesome. So do you have a proof?

that was my original thought process but i dont have a proof for it

unless

ok so, heres my proof

Hey everybody!

subdivide the polygon into triangles from the origin

by the first part of this problem, to maximize the area between two adjacent triangles their angle must be equal

Well, we started with that.

applying this to all the triangles gives that each angle is equal to it's two adjacent angles, and so you just apply like transitive or whatever to get all the angles are equal

Hey @Daminark

If we say that "a map is multilinear over $C^{\infty} (M)$ " is that the short way of saying that linear combinations of inputs result in the corresponding linear combination of $C^{\infty} (M)$ outputs>

Right, @Meow. If we had some adjacent pair of angles not equal, we could increase the area.

And if they're not adjacent, make 'em adjacent.

cool

How's it going?

Hi Demonark.

ill think about 2 while you consume various foodstuffs

as part of the daily ritual known as "dinner"

OK, Meow. You can let me know about the polynomial problem later. :) It's pretty cool.

One of my brilliant high school students at UGA gave me that problem years ago.

to be honest im probably going to sleep soon

@KevinDriscoll What is the map?

bio exams

Where is it from/to?

@MeowMix same

so gn, thanks

have a swell night mathse

See you @Meow!

@BalarkaSen Let $\omega \in \Gamma(T^k M)$ then the map $\Psi: (\chi(M) \times . . . \times \chi(M) \to C^{\infty}(M): (v_1 , . . . , v_k) \mapsto \omega(v_1, . . . , v_k)$

Right, C^infty(M)-multilinear then means it's linear on each of those components

Though if you just said multilinear I'd assume they just meant multilinear wrt to $\Bbb R$ or $\Bbb C$.

which is why they write C^oo(M) multilinear.

[Random]

What if we have a countable alphabet (with the only restriction being that such string must be of order type $\omega$) to notate ordinals:

0,1,11,111,1111,11111,...,111... = $n,\omega$

(111...)+1,(111...)+11,(111...)+111,...,(111...)+(111...) = $\omega +n, \omega 2$

(111...)+(111...)+(111...)+... = $\omega^2$

(111...)(111...)(111...)... = $\omega^{\omega}$

$(111...)^{(111...)^{(111...)^{\cdots}}} = \epsilon_0 = \psi(0)$

actually no, try again

@BalarkaSen @PVAL-inactive $\mathbb{C}$ and $\mathbb{R}$ are the only fields. Got it.

Well, C^infty(M) is not a field

It's a ring, but you can talk about linearity, multilinearity etc over rings too

That's what modules are

Modules over rings are a generalization of the notion of vector spaces over fields

Close enough for memeing, imo

Alphabet: $\Bbb{N} = \{0,1,2,3,4,5,6,7,8,9,...\}$

Pick $n \in \Bbb{N}$, $n+n+n+...,nnn...,n^{n^{n^{\cdots}}}$ are all valid notations for $\omega$

where is the binomial here: $$\frac{2^n}{n^3} = \sum_{k=0}^{n}\frac{\binom{n}{k}}{n^3}$$

?

@Kirill $\displaystyle \sum_{k=0}^n \binom n k = \sum_{k=0}^n \binom n k 1^k 1^{n-k} = (1+1)^n = 2^n$

wow

thank you, @LeakyNun

Ooooo I just noticed I can pull the $n^3$ out

@MathematicsAminPhysics somebody flagged the "notice notice" banner as spam / offensive, any guidance for how that should be handled? it looks innocuous to me but if you have had heated debates about that I suppose the flag is warranted

@tripleee: Are you aware of who that user is?

@user21820 the flagger, or the apparent course announcement bot?

I guess not. Have fun reading the following:

thanks, I'll approve the flags then

@user21820 stop pulling Typhon to this business. He is not that user.

@LeakyNun Sorry wrong link then.

It gets messy tracking all these.

^ added that to the chat guidelines page / question linked from the room's description

@tripleee @LeakyNun: Since in my previous message I had one link for the wrong user, here are some correct ones to justify that point:

chat.stackexchange.com/search?q=troll&user=84215&room=

@LeakyNun: See last link; it seems it comes in bouts.

=P

oh god almighty

@LeakyNun When we have Math SE, we don't need reality TV.

quite the slippery one .. I feel I am missing something here ^^' (= total clueless :P )

@r9m Why should it be true, when it's false even for non-random variables? Hint: 1,2,0,3,0,0,...

Wait... 1/n of the max...

@r9m: What if X[n] is n with probability 1/n and 0 otherwise? Then we have max(X[n/2..n]) ≥ n/2 with probability at least 1/2. But I can't see why sum(X[1..n])/n → 1 in probability, and it may be false.

I feel the claim is not true because a sum can smooth the densities out more than a max, but that is my intuition and may be faulty.

@user21820 Thanks for documenting all this. I had seen the 'course' announcemnts and I actually thought it was some officially sanctioned reading club or study group or something. Its not an area Im interested in so I never went over there, but now I know to let anyone else know about the context

@KevinDriscoll Yes that fact (that many people think it is officially sanctioned) is what irks me (and many others who know) the most.

Hi I wonder about matrix operation notation. Is it different between writing by capital and small alphabet? like \lVert V-WH \rVert ^2 = \sum_{ij}(v_{ij} - (wh)_{ij})^2, should I use capital or normal letter in summation?

@user21820 it's actually an exercise in the lecture notes (by Amir Dembo) we are following .. but I am so beaten over it I have begun to suspect if its true or not :( .. the result seems to be true for independent RVs ..

@r9m Aha so that's the trick. We can arrange my example to be not independent to force the average to tend to one.

@user21820 I take it the user in question also once had 'admin' in their name and was forced to change it to avoid any misattribution of official standing as well?

The reference use the capital letter but I remember somewhere we use small letter when talking about element in matrix like v_{ij} in V matrix. Or this is not matter at all?

@jan forgot your dollar sins for your latex

@KevinDriscoll No it started off as aminliverpool and now is MathematicsAminPhysics.

Oh okay, maybe the admin then was just a typo I saw somewhere

@KevinDriscoll: I note that I have no problem with him setting up a personal chat-room and doing whatever he wants there, but to use the Math SE name for his pet ideas is sheer cheek.

@Jan Yo your actual question though, it doesnt make a bit of difference. As long as it is clear and consistent in your writing, theres no reason to prefer captial or lower-case letter for this

@user21820 Noted. Its pretty clear from the chat logs that no ones trying to run said user off of the site.

@KevinDriscoll ok! I'm writing a computational paper (but I'm bio person) so I just want to be very accurate for math stuffs. Thanks!!

@user21820 Actually I was kind of shocked at how much legitimate advice people have been trying to offer. Although I have this nagging doubt that the thing on the other side of the account is some kind of neural network chat AI or something. But maybe I just havent seen enough logs.

@KevinDriscoll I wasn't shocked at the great amount of advice, but I did feel that we had spent enough time and energy and ultimately there is a limit.

It's definitely not a bot though.

Certainly. My shock was that the limit wasnt reached about 30 seconds after the conversation started

Hahaha...

@AkivaWeinberger hello, i have a proposition which says that if $cl(A)\cap B=\emptyset=cl(B)\cap A$ then $Int(A\cup B)=Int(A)\cup Int(B)$ in my case $cl(Z)=Z$ $X\cap ]0,3]=\{1,2,3\}\neq \emptyset $

so how to make the proof that $Int(Z\cup ]0,3])=]0,3[$ ?

cl is closure?

i know that $]0,3[\subset Int(Z\cup]0,3])$ if i take $x\in Int(Z\cup ]0,3]$ then there exists an open O such that $x\in O\subset (Z\cup ]0,3])$ how to find that $O=]0,3[$

@KevinDriscoll yes

Is there a difference in performance between using a sigmoid activation function and using the ReLU activation function in a shallow (3-4 layers) neural network?

@user76284 That probably depends on how your language is calculating the sigmoid output

What do you mean?

In the ReLU case you're making 1 check and then passing different outputs based on the result

In the sigmoid case you're just computing the output of the sigmoid function

My guess is the ReLU will always be slightly faster, though probably not by much. But if your language is using a very inefficient way to calculate the sigmoid output, that could slow things down.

Oh, calculation is no problem. I meant in terms of the general performance of the network itself.

For example, speaking of hidden layers, tanh (which goes from -1 to 1 and is symmetric about 0) is known to perform better than the logistic sigmoid (which goes from 0 to 1 and is not symmetric about 0). ReLU in turn is known to perform better than tanh for deep neural networks since it avoids tanh's gradient saturation problem.

Oh I see when you say 'performance' you mean how long the network takes to learn and such. Not like the algorithm performorance that would be measured in number of cycles to go through one loop or something like that.

Unfortunately I dont know. I havent ever looked at running ReLU

Hey everyone!

@Daminark Hi

How's it going?

Good. Starting to put together an exam

Now to figure out if I will be benevolent :)

Oh that's always a fun time. Is it a midterm?

no, final

we don't have midterms here

I see

I've had an interesting ride wrt finals

Though I've had just one class whose final I'd describe as unreasonable

I am looking through old exams for this course now. Man, they have varied a lot in difficulty, especially when you go back 10 or more years

On the whole, the ones which were harder have been more fun so I guess I'd push in that direction?

Is difficulty more about not having time or, problems are just hard to figure out?

One from 2004 had an exercise to show that the rings $\mathbb{Z}[\sqrt{-p}]$ where $p\geq 3$ is a prime were never PIDs

Oof

But I think the syllabus may have been different back then, so they may have had better tools than they get now

the current course has almost nothing on quadratic extensions in general

Oh I guess that makes sense

(one can emulate the proof that the Gaussian integers are an ED to see how to get elements for which unique factorization fails in those rings, and then that implies that the ring is not a PID)

but I am not sure how the students would have been able to realize that

I haven't quite seen much wrt Gaussian integers and the like, but I definitely buy that it at least feels like a jump

Especially since this is jumping from a detail in a proof that the Gaussian integers really do form a PID to showing some other ring is not a PID

It is an important general idea, but one that is probably more advanced than this course (it is the sort of thing that shows up in force when doing algebraic number theory)

Yeah, if I remember right you mostly focused on classical number theory, right?

Plus some quadratic reciprocity if I remember right?

yeah, and not very much of it

right. We still haven't proven that though (that will be done soon)

I see

We are almost done with stuff related to unique factorization. Just need to finally use it to count integer solutions to $x^2 + y^2 = n$

Then we will be going into polynomials, and use those to prove quadratic reciprocity

Ah, okay that makes sense. How large is the class btw?

about 100 students registered

probably about 50-ish show up to lectures (I don't count precisely)

Whoa

We've got two sections of algebra

Well, two classes, each with two sections

And each section probably has ~30 students

Though I think there's at least one algebra class that starts in the winter

I should ask my TAs for an estimate of how many from each of their classes they expect to qualify for the exam

What does it mean to qualify for an exam?

there are hand-ins during the course. They need enough of those approved

Hand-ins like problem sets?

Also hey @Alessandro!

yeah

Morning

True art

This is a tricky balance, trying to make sure the problems are hard enough to test them, but not so hard that they will just give up

Yeah, I imagine. Any problems you've got in mind?

Well I dunno if you're at liberty to disclose them here

I need to make sure not to write them anywhere yet

Makes sense

Would be a disaster if one of the students happened to see one of the problems here beforehand

hey

Yo @Balarka

Turns out my algebra prof won't be around on Wednesday and we'll have a guest lecturer

Hey, I need help with probability

i have this picture

how is defined $\phi(t)$ please