Mathematics - 2022-07-07

leslie townes

12:00 AM

i took a class on this crap once. far too long ago, apparently.

user193319

I am trying to determine $\widehat{Hom(S^1,S^1)}$.

leslie townes

you can tell a lot about a person by whether they write the circle group as S^1 or T.

user193319

Lol what does it say about me?

leslie townes

i dunno. i think harmonic analysts, and people contaminated by them, are more likely to use T.

user193319

Lol

leslie townes

12:06 AM

have you looked in fell and doran? i recall they had some results where they generalize pontryagin duality for various non commutative things. i forget if any of it was of independent interest in the commutative case.

Xander Henderson

@leslietownes >:(

user193319

No, I don't think so. What's the name of the book/paper?

Xander Henderson

But yes, $T$. Or $T^1$ (for the circle), and $T^2$ (for the surface), and so on.

The circle group is clearly $T$. :P

(Because "circle" starts with "T", just like "integer" starts with "Z".)

@leslietownes I feel like maybe Katznelson does things in a general enough way as to also deal with non-commutative BS (though it has been a while since I've looked at these things, and I am most interested in the Pontryagin dual of the $p$-adics, which are lovely and commutative).

leslie townes

user: something like 'representation of star-algebras and locally compact groups'

xander: you are weird.

Xander Henderson

Oh, no. I lied. Katznelson only generalizes to commutative Banach algebras.

leslie townes

12:13 AM

although, for the record i am also on team T. when i see S^1 i think of something loopy that might not have a group structure.

that section or appendix is very good though. on commutative banach algebras.

it is bested only by my advisor's treatment of the subject in his spectral theory book.

Xander Henderson

$S^1$ is a topological space. $T$ is a topological group.

@leslietownes Who was your advisor?

leslie townes

bill arveson.

the original papers on the gelfand theory are also really good.

Xander Henderson

@leslietownes Hrm... I feel like I know that name, but I don't recognize it immediately.

:/

leslie townes

well i've never heard of your advisor either. although somehow we're friends on linkedin.

:)

Xander Henderson

Ha!

leslie townes

12:17 AM

bill is the indie rock band where just knowing him makes you cooler.

he's not your favorite mathematician, he's your favorite mathematician's favorite mathematician.

Xander Henderson

Ha!

Oh, I have one of his books!

leslie townes

the C-star algebras one?

Xander Henderson

"A short course on spectral theory"

leslie townes

oh, his spectral theory one. that's the best.

Xander Henderson

Yeah. It isn't my area of expertise, but I have found it quite useful once or twice.

Bruce told me to get it.

leslie townes

12:19 AM

there are a few exercises in the first edition of that book that are broken. just wrong and broken. i used to get email about them.

but he fixed them in errata on his webpage, which is still up 10+ years after his untimely demise

Xander Henderson

Ugh... I've been driving all day, got out of a meeting 10 minutes ago, and very much need to shower and sleep. Later.

leslie townes

cheers.

and thanks for reminding me i have to drive to the office tomorrow. womp womp

AMDG

1:08 AM

Are there any relationships (explicit or implicit) between primes and irrationals?

Dr. Will Wood's recent videos on lattices and primes made me realize that primes sufficiently large act as irrationals. A line whose slope is irrational will never touch a single point on an integer lattice necessarily, however, arbitrarily large primes act like imperfect irrationals, never touching any point but itself.

Thorgott

@XanderHenderson $T$ is the topological group, $S^1$ is the pointed homotopy cogroup

leslie townes

thorgott: eww.

AMDG

The Riemann Zeta Function in the Integer Lattice

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Thorgott

i swear its very useful!

AMDG

All Primitive Triangles have an Area of 1/2

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Thorgott

1:16 AM

re the discussion, it's crucial to specify whether we're taking continuous homs or not

i have an inkling suspicion the set of group homomorphisms $S^1\rightarrow S^1$ is as badly behaved as you could imagine and sensitive to choice

leslie townes

good point. i was implicitly assuming continuity.

Thorgott

a good practice

leslie townes

as someone said, you never just assume the operations exist. you assume the compatibility with the other structure.

some wise guy said that.

Thorgott

sounds like a wise guy indeed :P

leslie townes

my dad is funny. he sent me like 20 texts today saying i have to watch this rare and unknown show called "better call saul." he's discovered binge-watching.

shintuku

1:23 AM

i was wondering what i should watch next

just finished ozark

leslie townes

my dad liked ozark too.

i asked my dad if he'd seen bob odenkirk in anything else, and he said no, so i sent him a long list of recommendations.

shintuku

i'm going to try better call saul

leslie townes

he couldn't watch breaking bad because the subject matter was too painful. i also had to peace out of it very early. we have a family member who is basically jesse pinkman. i don't want to know what happens to that guy.

very effective television writing though.

shintuku

seems like these shows all have a common theme running through

leslie townes

apparently BCS is less nihilistic and a lot of the worst stuff hasn't happened yet.

shintuku

1:26 AM

i started watching these crime dramas because netflix ran out of space fiction

leslie townes

as xander would tell us if he were here, breaking bad was originally written to take place in riverside, where he got his phd. they relocated it because of new mexican local tax incentives.

which are also why a ton of stuff is filmed in georgia right now. it's a race to the bottom, where local legislatures say "we will pay you to film here," and then do. until the next state pays even more.

shintuku

i wonder if they get tourism from that

it's probably why

leslie townes

they definitely do in new mexico. i dunno about georgia.

michigan was trying to get in on this when i lived there. they filmed some stuff near our house.

shintuku

huh, ozark is in georgia

leslie townes

the walking dead and stranger things also.

shintuku

1:29 AM

the more you know

leslie townes

my wife and i were watching a movie once where there's this meth-head who has a party in this trashed rural location, and we were like "uh, is that the house down the street?" and it was.

when we lived in saline, michigan.

the funniest story about anything being filmed anywhere was from my barber in cambridge. he was watching a movie where vanilla ice was playing himself, and the joke was he was living in this shithole of an apartment. they filmed it in the apartment where his ex-wife had taken his daughter after they divorced.

i still laugh about that.

shintuku

hahahahah, that's unfortunate

leslie townes

when your ex-wife raises your daughter in a place where a location scout says "yes, that will be an appropriately funny place for vanilla ice to live in our movie."

haha

how does it get funnier than that? it doesn't.

vanilla ice's film "cool as ice" is a really hilarious movie, if it's on some streaming services, i recommend it. the cinematographer cut his teeth on it before winning an oscar for cinematography for schindler's list.

Get With the Hero - Cool as Ice (2/10) Movie CLIP (1991) HD

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shintuku

the wikipedia page for cool as ice doesn't have a link to the director's page

i mean, the director doesn't have a page

signs of an obscure movie

leslie townes

the cinematographer does. he also did saving private ryan.

shintuku

1:36 AM

cool!

leslie townes

i think they filmed cool as ice somewhere in southern california too. maybe in the inland empire somewhere.

it's a great movie and i highly recommend it.

i also like really stupid movies, so i feel i should point that out.

the dumber the better.

i like movies where the protagonist is supposed to be 'cool' and desirable and yet every signal he gives off (it is always a "he") is in the opposite direction.

Akiva Weinberger

2:03 AM

Puzzle

Who wrote the following

leslie townes

Einstein

ok, not einstein

Kahan?

he would have been more traumatized by the widespread adoption of java

Akiva Weinberger

Wilkinson

Wilkinson's polynomial

In numerical analysis, Wilkinson's polynomial is a specific polynomial which was used by James H. Wilkinson in 1963 to illustrate a difficulty when finding the root of a polynomial: the location of the roots can be very sensitive to perturbations in the coefficients of the polynomial. The polynomial is w ( x ) = ∏ i = 1 20 ( x − i ) = ( x − 1...

From here

Finding roots of polynomials numerically is difficult because they're extremely sensitive

> If the coefficient of $x^{19}$ is decreased from −210 by $2^{-23}$ to −210.0000001192, then… the root at x = 20 grows to x ≈ 20.8.

Koro

3:49 AM

@robjohn seems very complicated to prove decreasing.

leslie townes

the idea of doing numerical analysis in the 1960s is incomprehensible to me.

Koro

I got the difference as:
$-\sum_{r=0}^2 \frac 1{3n-r}+\frac 1{5n+1}+\frac 1{5n}+\sum_{r=1}^3\frac 1{5n-r}$

Akiva Weinberger

@leslietownes I'd imagine it's one of the oldest branches of math, actually

For most cultures in most of history, math was more about calculating things than proving theorems

and finding more efficient ways to calculate things

Koro

Showing the above expression negative for all $n\in \mathbb N$ seems difficult. But I think I can try using differentiation here.

leslie townes

yes, at a high level i get it, but at the level of that. .000000000001192.

Akiva Weinberger

3:54 AM

Ah, sure

leslie townes

we've probably lost most of what we used to know about organizing calculations to preserve half that many digits.

you do make a good point, particularly in the early days of computing, the available disk space and memory requirements, on top of the processing time, more or less forced you to care about stuff that is an afterthought now.

i upset someone at work once. he wanted to write an article about how it was bad that there wasn't stronger patent protection for software. i said, no, software is the absolute worst use case for patent protection. someone can lazily describe an invention via some stupid setup that is entirely impractical until completely unrelated actual advances in storage and processing make it viable. nobody should be able to patent that.

everything that goes by the name of machine learning, AI, etc. is an example of this. you don't need to be smart anymore, you just need a dumb algorithm and a lot of CPU time.

Akiva Weinberger

You also find that sort of thing in video game design. Early game designers had to find creative ways to work around the tight limitations. (Famously, the clouds and bushes in Mario were the same sprite, just different colors, to save memory space)

With modern hardware, that level of creativity just isn't required anymore.

leslie townes

haha, i'm reminded of the films of edgar ulmer, who was famous (exaggeratedly so) for working on a budget. most notably in "detour" (1945) where a character driving back to las vegas was a flipped image of earlier footage of him driving away from las vegas. so he's on the wrong side of the road and the wrong side of his car.

Akiva Weinberger

Hah

leslie townes

he'd also do setups where people would enter a building and then suddenly be engaged in a conversation against a blank white wall.

great movies, though.

Akiva Weinberger

4:03 AM

I'm sure in animation you can find instances of objects conveniently covering up mouths - or otherwise extreme wide shots or extreme close-ups - so that they don't have to animate the lips moving with the dialogue

(Animation is expensive as hell though)

(An actor will just move automatically. You gotta draw dozens of pictures per second to make a picture move)

leslie townes

haha, what was the line on the simpsons?

very few cartoons are broadcast live, homer. it's a terrible strain on the animators' wrists.

Koro

leslie, did you watch the new season of stranger things?

leslie townes

my daughter's reading one of her books to the cat. it's 9pm.

Akiva Weinberger

Honestly, animatics (rough sketches for every beat rather than fluid animations) are pretty fun to watch

I'd probably be able to sit through a show that was just animatic-style

leslie townes

koro, i did. i love 1980s nostalgia. the show is like it was designed in a lab to appeal to my demographic, and it mostly works, but god, it dragged near the end.

too many characters, too many people reuniting. it was like that lord of the rings movie where it has 5-6 endings over 45 minutes. blegh

Akiva Weinberger

4:08 AM

On TikTok, people are doing "analogue animatics" (that's not a real phrase) where they draw each frame on a piece of paper and just hold the camera above each frame

I guess the difference between that and, say, a comic, is you don't have to worry about layout on the page. And also you can have audio

leslie townes

they should have cut the entire russia subplot and half of the california stuff, and everything in that lab in the desert.

i'll watch the fifth season, but it feels like it's overstaying its welcome. i like HBO's "barry," very short episodes. even if it's crazy and unhinged it's over in 30 minutes.

Koro

@leslietownes yeah that subplot was unnecessary imo.

leslie townes

i do love whoever does the set design for stranger things. they get it so right. i've recognized stuff like desk lamps and toys from my sister's room in the 1980s on that show.

they also did something where the kids in hawkins dressed slightly differently from the kids in california. it was subtle but i noticed. also accurate.

when i arrived in iowa in the late 2000s the students were wearing stuff that would have been outdated in the late 1990s in california.

Akiva Weinberger

I think one of the best things Avatar did was have an expiration date

It was clear from the start that it would be over in three seasons

leslie townes

the other cool thing about stranger things is that it's getting people into all of the tv and movies that they're blatantly ripping off.

Akiva Weinberger

4:18 AM

Also, y'know, funny comedy for the kids and nuanced themes for the adults (and thinkier kids)

(still re: Avatar)

Speaking of 80s set design, I thought HBO's Chernobyl was fantastic

leslie townes

the german show 'dark' had some really good 80s stuff.

Akiva Weinberger

not that I've ever lived in 1986 Soviet Ukraine

leslie townes

yeah, chernobyl was amazing.

Koro

Dark, Chernobyl both were amazing : )

Akiva Weinberger

I haven't seen Dark. I'll have to check it out

leslie townes

4:21 AM

the plot is complete nonsense but the vibes are really cool.

Akiva Weinberger

There's a great show called Counterpart that's set in modern times but is a Cold War vibes story

(I've only seen the first season)

There's a portal to an alternate universe under Berlin and there's spy shenanigans with the alternate universe

leslie townes

there was a film version of the john le carre 'tinker tailor soldier spy' that had the aesthetic done perfectly.

Akiva Weinberger

I wonder what it is about the zeitgeist that multiverse stories are so popular

Everything Everywhere All At Once (great film btw) tackled those themes directly

leslie townes

people want to imagine that somewhere is less shitty than this timeline.

Koro

has anyone seen the movie Downfall (2002)?

leslie townes

4:23 AM

racacoonie!!!!!!

Akiva Weinberger

using the concept to explore nihilism and ADHD and stuff

@leslietownes :D

leslie townes

koro: i've seen it. and the million memes.

Koro

haha

leslie townes

my wife and i watched EEAAO two weeks ago. we loved racacoonie.

Akiva Weinberger

The Marvel Dr Strange Multiverse thingy was a little disappointing in that it didn't tackle any of the depth there at all

Honestly, it felt like they just used the multiverse as an excuse to kill a trillion people

You know, stakes

leslie townes

4:25 AM

i haven't seen any of those movies. the marvel stuff has completely passed me by.

Akiva Weinberger

(In actuality the audience connects more with small stakes that are personal to an empathetic character than, like, "oh we gotta do blah to save the universe" stakes)

@leslietownes It's not all that good

leslie townes

this is something that season 4 of stranger things did pretty well. it was less "the entire world is in danger" (although maybe it was). the antagonist wanted to kill specific people. it was confined in time and space (except for the dumb russian subplot)

Koro

also strange to see how they had kept that monster but when the monster was loose, nobody was able to control it.

that subplot was unnecessary.

leslie townes

the show never explained what those things are. do they think independently? are they under the control of the main guy? why do we care about them? it seemed to hurt the main guy when they were harmed at the end.

it was incoherent.

the best scifi movie about a beast that cannot be stopped is the original "alien"

Akiva Weinberger

My brother wrote a neat time-loop multiverse story

leslie townes

4:32 AM

is it out?

Akiva Weinberger

Nah

One day they'll make it

In the meantime he's working on other projects

leslie townes

does he do that full time?

Akiva Weinberger

Yeah

leslie townes

did he go to tisch? i feel like i should know this. i've followed him for several years on twitter

Akiva Weinberger

He has a separate script called Precipice which has directors and producers attached, and now they're working on casting

@leslietownes Yeah

leslie townes

4:34 AM

wow

Akiva Weinberger

Precipice is a post-apocalyptic time travel story

In the far future, the planet has healed (from an unknown catastrophe, probably climate change or whatever). And civilization has returned to stone age

leslie townes

i love the apocalypse as a dramatic concept, i just wish we weren't currently hurtling toward it as quickly as possible in our actual lives

Akiva Weinberger

and there are travelers from our time with the mission to set up a "gate"

Koro

leslie, i also love apocalypse genre shows

like the walking dead

Akiva Weinberger

The sort of idea is time travel is hard unless you have a gate set up at both ends. So they managed to send two people ahead to try to set up the gate at the far end. But one accidentally lands fourteen years before the other and starts a family

and has second thoughts about bringing people from our time into that new world

leslie townes

4:38 AM

koro have you seen 'the road'? it is based on a novel. it has viggo mortensen in it

Koro

not yet. I'll check it out. Does it have zombies in it?

The plot seems nice. I'll watch this one.

leslie townes

no, but it's very similar.

there's a man and his son trying to survive, and the people they meet might as well be zombies.

cormac mccarthy needs a therapist. he also wrote 'no country for old men'

Akiva Weinberger

That's a good movie

A fantastic small-stakes movie is Bad Education

A scandal at a Long Island high school, based on a true story

I showed it to my suitemates - they all thought it was a treat choice

Oh and it's got Hugh Jackman

Here's a neat math fact I learned recently

Choose a finite poset P. Let Ω_P(k) be the number of maps f:P→{1,…,k} where if p≤q then f(p)≤f(q). (Note that this includes constant maps, for example.)

Then the function Ω_P is a polynomial, for any finite poset P.

(That's the divisor poset of 18, with a random function to {1,2,3,4})

leslie townes

5:04 AM

huh.

one potato two potato

6:46 AM

What is the limit of $x_{n+1} = x_n - x_n^{n+1}$ with $x_1 = {3\over 4}$?

Calvin Khor

9:59 AM

The International Congress of Mathematicians 2022 (ICM 2022)

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vidyarthi

10:33 AM

@onepotatotwopotato Since $x_{n+1}/x_n<1$, limit at $n\to\infty$ exists and can be found by equating $x_{n+1}=x_n$, which come out to be $0$

Martin Sleziak

11:05 AM

It seems that there are some questions on the main site related to this: Limit of $(x_n)$ with $0<x_1<1$ and $x_{n + 1} = x_n - x_n^{n + 1}$ and Estimating the limit $x_{n+1} =x_n - x_{n}^{n+1}$. (And probably several other posts.)

one potato two potato

11:17 AM

@MartinSleziak Thanks

@vidyarthi Seems not that simple

Martin Sleziak

@onepotatotwopotato I have only put the equation into Approach0. That's one of the tools mentioned in the FAQ post: How to search on this site?

shintuku

12:27 PM

so, boris resigned

Calvin Khor

wait, really? is this the UK boris?

shintuku

yes

time for that second scottish independence referendum

Calvin Khor

wow

didnt think he had it in him

shintuku

you know, it's standard in some political parties to have the leader resign when a confidence vote yields more than 30% opposition

some have even stricter standards

homeslice here had 41%

one potato two potato

Is the monodromy theorem in complex analysis related to monodromy in covering space?

Astyx

1:16 PM

What is the monodromy theorem in CA?

But yes, very closely

Thorgott

1:43 PM

I think the answer is "almost"?

analytic continuations of a holomorphic germ along a curve are liftings of that curve through the projection from the espace etale (yes, I forgot some accents, no, I'm not sorry) of the sheaf of holomorphic functions to the Riemann surface

but that projection is not a covering in general, I think

it's just an open local homeomorphism on a Hausdorff space

however, homotopy invariance of lifting of curves only needs those hypotheses

the stronger hypothesis of being a covering is only necessary to make statements about the existence of lifts of curves

Alek Murt

2:16 PM

Hey guys, could someone please help me with the following, I'm trying to find the interior of this set $F_{\alpha}=\left\{x \in \mathbb{R}^{d}: \alpha^{t} x \geq \alpha^{t}(g x)\right.$, for all $\left.g \in G\right\}$ where $G$ is a symmetry group. Could someone guide me pls

Astyx

2:38 PM

@Thorgott If you restrict to a nbh it is, right?

Thorgott

I'm not sure

I was hoping it is, but then I couldn't convince myself anymore lol

Astyx

I'm pretty sure the singularities of the function are precisely critical points of the projection of the Riemann surface

And analytic continuation only makes sense away from these singularities

leslie townes

8:27 PM

@CalvinKhor, @shintuku: twitter.com/Fritschner/status/1545077380527673344

shintuku

hahahahhahahahhahahahhahahahahhahahahah

hell that's funny

leslie townes

it really gets me around the 15 second mark when it starts up again with increased vigor

that's so british. i love it

it's also nice to see another country's government be a shitshow for once :D

the US was hogging the spotlight for quite a while

Ted Shifrin

8:52 PM

And so it shall continue to do, in spades.

Akiva Weinberger

9:26 PM

So legitimate question

If I have a finite vector space $\Bbb F_q^n$, that is, an $n$-dimensional vector spaces over a finite field, and if I also have a line (1D subspace) $a\subseteq\Bbb F_q^n$, how do I count the number of hyperplanes (dim $n-1$) not containing $a$?

Akiva Weinberger

9:39 PM

I guess it's $\dfrac{q^n-1}{q-1}-\dfrac{q^{n-1}-1}{q-1}=q^{n-1}$

using the fact that there are as many $n-1$-subspaces as $1$-subspaces

Is there a way to see it as $n-1$ choices of $q$ options?

Oh, you know what it is

We can assume $a=(1,0,\dots,0)$

and then if $b^\perp$ is the hyperspace not containing $a$, up to constant multiple, we have $b=(1,x_1,\dots,x_{n-1})$

Alek Murt

Hey guys any hints on how can i show $H_{n}(Q) \leq(\sqrt{d})^{n}$, where $H_n$ is Hausdorff measure and $d$ is the euclidian distance. Also, $Q$ is the n-dimensional unit cube

Akiva Weinberger

9:56 PM

How is the Hausdorff measure defined

Alek Murt

as the infimum over all countable cover of the set, in this case the cube

Ted Shifrin

10:08 PM

This makes no sense. What is $d$? The $n$-dimensional Hausdorff measure should just be the usual $n$-dimensional volume.

Balarka Sen

Huh, given a function $f : (M, \alpha) \to \Bbb R$ on a contact manfold, the contact Hamiltonian is apparently defined as the unique $X$ such that $\alpha(X) = f$ and $i_X d\alpha = df(R) - df$ where $R$ is the Reeb vector field. Why is this the correct definition?

Alek Murt

$d$ is the usual euclidian distance $d(x, y)=\sqrt{\sum_{i=1}^{d}\left(x_{i}-y_{i}\right)^{2}}$

Balarka Sen

On the codimension $1$ symplectic subbundle, we should have $i_X d\alpha = \pm df$, so that much checks out.

Ted Shifrin

@Alex, $H_n(Q)$ is a number.

Xander Henderson

@AlekMurt $d$ is a function, not a number. What is $(\sqrt{d})^n$?

Akiva Weinberger

10:13 PM

@AlekMurt So… $\sqrt d$ is a function? $d$ of what?

Alek Murt

oh no

w8

Akiva Weinberger

Not everyone at once, lol

Alek Murt

don't listen to me

Xander Henderson

EVERYBODY EVERYWHERE ALL AT ONCE!

Alek Murt

I just realized what i am saying, srry boys

Xander Henderson

10:14 PM

Please don't assume that everyone here is male, or that they are happy being referred to as "boys".

Akiva Weinberger

As time goes on the more important I think typechecking is for math

Balarka Sen

* I meant $i_X d\alpha = df(R)\alpha - df$

Alek Murt

Srry sorry now i feel bad.

Xander Henderson

@AkivaWeinberger It has saved my bacon more than once. That, and unit checking.

Akiva Weinberger

@AlekMurt 'Tsall good

Xander Henderson

10:16 PM

@AlekMurt Don't feel bad---it takes time to learn the culture of a group. Just take the input on board, and don't do it again. Please. :D

Akiva Weinberger

@BalarkaSen I missed an episode

Balarka Sen

?

Akiva Weinberger

(aka I don't know all the background)

Can't help

Balarka Sen

Oh ok

No worries, just thinking out loud

Akiva Weinberger

Sure

I'm learning about Möbius functions on posets

Xander Henderson

10:18 PM

Okay, on a scale from 1 to "Xander, your students are going to murder you", how bad is the following question:

> Determine an equation for the line tangent to the curve $$(x^2+y^2)^2 = 4(x^2-y^2)$$ at the point $(\sqrt{5/8}, \sqrt{3/8})$.

This is an implicit differentiation problem for an exam.

Balarka Sen

it's a $\sqrt{5/8}$

Ted Shifrin

I haven't tried working it, but make the numbers simpler.

Xander Henderson

@TedShifrin That is as simple as the numbers get without making the problem trivial.

Akiva Weinberger

As long as you give them the graph I'm OK with it

Balarka Sen

dont give the graph and do tangent lines at (0, 0) just to confuse them

Ted Shifrin

10:21 PM

Rescale to make the point $(1,1)$ or something.

Xander Henderson

@BalarkaSen Heh.

Balarka Sen

anarchy

Akiva Weinberger

I'm getting $(x^2+y^2)(x+yy')=2(x-yy')$

Xander Henderson

@AkivaWeinberger Yup.

Akiva Weinberger

Where $x^2+y^2$ is $1$

That's actually really nice^

So $\sqrt5+\sqrt3y'=2(\sqrt5-\sqrt3y')$

Balarka Sen

10:24 PM

now work out the tangent lines at (0, 0) akiva

Xander Henderson

@BalarkaSen DNE.

Akiva Weinberger

@BalarkaSen $y^2=x^2$ duh

Balarka Sen

@XanderHenderson diagnostic nasal endoscopy?

Xander Henderson

@BalarkaSen Yes, but also "Does Not Exist".

Balarka Sen

@AkivaWeinberger proof?

Akiva Weinberger

10:25 PM

@BalarkaSen [looks at camera feed] yup, that sure is a nose

Xander Henderson

@AkivaWeinberger Italicized parentheses and braces make me sad. :(

Balarka Sen

@XanderHenderson i reject this propaganda

Thorgott

zariski would like to have a word

Akiva Weinberger

$y'=\frac{\sqrt5}{3\sqrt3}$?

Balarka Sen

lmao thorgott

do it by (m/m^2)^*

Xander Henderson

10:27 PM

@AkivaWeinberger I got $\sqrt{5/27}$, which seems to be what you got.

Ted Shifrin

Don't forget to dualize.

Balarka Sen

jeeze, ted, corrected

Mad Spaces

Good evening smart people

Xander Henderson

@Thorgott I'm sure he would. I should be more precise: for the purposes of introductory calculus, where varieties are a bit ouf of scope, we have no definition for a line tangent to a curve at a cusp, corner, or point of self-intersection. Within the scope of Calc I, such a tangent line does not exist.

Mad Spaces

Stupid question is incoming, brace yourself

Balarka Sen

10:29 PM

i dont like the zariski tangent space

Ted Shifrin

Do Nash blowup.

Xander Henderson

@BalarkaSen That's okay. It doesn't like you, either.

Thorgott

I'm TAing analysis 2 atm and we've defined tangent lines for arbitrary subsets of $\mathbb{R}^n$ lol

Balarka Sen

@TedShifrin much better

Xander Henderson

@TedShifrin Sounds like a great topic for Calc I. :D

Mad Spaces

10:30 PM

Thorgott i thought you would be much more advanced in your studies based on your answers. only now doing Ana 2?

Ted Shifrin

TA

Xander Henderson

@MadSpaces He's the TA.

Balarka Sen

the sole purpose of Calc I should be to confuse all students and traumatize them

Xander Henderson

Not a student.

Ted Shifrin

Reading is a prerequisite for this room.

Mad Spaces

10:30 PM

Okay i dont know whats a TA lol

Balarka Sen

@MadSpaces he's also a superior german mathematician

Xander Henderson

Teaching assistant.

Mad Spaces

Ah i see

Xander Henderson

A TA is typically a graduate student who is paid to teach undergraduate classes.

Akiva Weinberger

@XanderHenderson $y=\dfrac{\sqrt5}{3\sqrt3}x+\dfrac{\sqrt2}{3\sqrt3}$

Thorgott

10:32 PM

I'm not sure how different the systems are across different countries. I have a subset of students whose homework I grade and with whom I have a weekly session to discuss the previous weeks homework.

Xander Henderson

@AkivaWeinberger Yup. I'd take that answer.

Balarka Sen

too slow akiva

Mad Spaces

Yes, you are holding a tutorium :)

Ted Shifrin

@BalarkaSen Except Thor actually is!

Mad Spaces

anyway, my kvstchn:
I am in the middle of some Calculation with alternating products and stuff in multilinear algebra. and i am trying to show that this result i am getting $\sum_{\sigma \in S_k} sgn(\sigma) * \Pi_i Det A_{\sigma(i),i}$ is equal to the determinant of the Adjunct matrix, the problem is, the adjunct matrix has these $-1^{ij}$, thing going on for each element. How can i insert this in my result since its dependent on the permutation : so i can use the formel of Leibniz

Xander Henderson

10:34 PM

@Thorgott That is pretty typical. As a masters student, I was given nearly complete autonomy over precalc (there was a common final, but I could otherwise do whatever I wanted), and total autonomy over calculus; for other classes, I essentially ran review sessions (e.g. I'd go over homework problems with the students during a one hour "discussion" section, and grade homework).

The TA duties at my PhD institution were mostly a similar style of helping out with homework and grading.

Akiva Weinberger

@XanderHenderson I'd recommend these numbers instead

Mad Spaces

It is part of the proof of showing that for a linear function on two vector spaces, the outer product of degree K is given by the determenant of the minors of the matrix. i am trying to show if it is zero then the function does not have full rank, but i must show that multiplication with that equal to multiplying with the determinant, hence the problem arise.

Akiva Weinberger

@XanderHenderson Correction

Xander Henderson

@AkivaWeinberger I thought about it (and am still thinking about it). I am somewhat torn about where to put the "ugly" numbers.

Ted Shifrin

Remember that this turns into more of an algebra test than a calculus test for today’s students.

Xander Henderson

10:42 PM

I am also a fan of having problems where the solution isn't something which looks completely trivial ($y=x+1/2$ feels a little "too nice").

Mad Spaces

This looks a very nice path of a planet with two sun system

equal mass and position at 1 -1 or else that would look different. Turn it into a physics problem :D

Akiva Weinberger

This is an interesting problem

Jun 30 at 6:38, by Akiva Weinberger

For what value of $c$ does the spiral $r=e^{c\theta}$ have the property that, the points with vertical tangent are on the same y-coordinate as a point with horizontal tangent?

Its approximate form is 0.27441…

--

Mad Spaces

With all the graphs Akiva posted, my question has long been gone to oblivion, oh well :(

Akiva Weinberger

Turns out it's the solution to $1=xe^{x\cdot3\pi/2}$, which is $W(\frac{3\pi}2)/(\frac{3\pi}2)$ where $W$ is the Lambert W function (inverse of $xe^x$)

@MadSpaces sorry

Mad Spaces

Its fine, i am enjoying the graphs more

Xander Henderson

10:48 PM

I want the Hausdorff measure question to get fixed up. That one is actually in my wheelhouse. :(

Akiva Weinberger

@MadSpaces All I can think of is try to get it from the $\operatorname{sign}\sigma$ somehow

Transcript for

$Mathematics$ Mathematics