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12:00 AM
i took a class on this crap once. far too long ago, apparently.
 
I am trying to determine $\widehat{Hom(S^1,S^1)}$.
 
you can tell a lot about a person by whether they write the circle group as S^1 or T.
 
Lol what does it say about me?
 
i dunno. i think harmonic analysts, and people contaminated by them, are more likely to use T.
 
Lol
 
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.
 
@leslietownes >:(
 
No, I don't think so. What's the name of the book/paper?
 
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).
 
user: something like 'representation of star-algebras and locally compact groups'
xander: you are weird.
 
Oh, no. I lied. Katznelson only generalizes to commutative Banach algebras.
 
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.
 
$S^1$ is a topological space. $T$ is a topological group.
@leslietownes Who was your advisor?
 
bill arveson.
the original papers on the gelfand theory are also really good.
 
@leslietownes Hrm... I feel like I know that name, but I don't recognize it immediately.
:/
 
well i've never heard of your advisor either. although somehow we're friends on linkedin.
:)
 
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.
 
Ha!
Oh, I have one of his books!
 
the C-star algebras one?
 
"A short course on spectral theory"
 
oh, his spectral theory one. that's the best.
 
Yeah. It isn't my area of expertise, but I have found it quite useful once or twice.
Bruce told me to get it.
 
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
 
Ugh... I've been driving all day, got out of a meeting 10 minutes ago, and very much need to shower and sleep. Later.
 
cheers.
and thanks for reminding me i have to drive to the office tomorrow. womp womp
 
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.
 
@XanderHenderson $T$ is the topological group, $S^1$ is the pointed homotopy cogroup
 
thorgott: eww.
 
i swear its very useful!
 
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
 
good point. i was implicitly assuming continuity.
 
a good practice
 
as someone said, you never just assume the operations exist. you assume the compatibility with the other structure.
some wise guy said that.
 
sounds like a wise guy indeed :P
 
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.
 
1:23 AM
i was wondering what i should watch next
just finished ozark
 
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.
 
i'm going to try better call saul
 
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.
 
seems like these shows all have a common theme running through
 
apparently BCS is less nihilistic and a lot of the worst stuff hasn't happened yet.
 
1:26 AM
i started watching these crime dramas because netflix ran out of space fiction
 
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.
 
i wonder if they get tourism from that
it's probably why
 
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.
 
huh, ozark is in georgia
 
the walking dead and stranger things also.
 
1:29 AM
the more you know
 
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.
 
hahahahah, that's unfortunate
 
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.
 
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
 
the cinematographer does. he also did saving private ryan.
 
1:36 AM
cool!
 
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.
 
2:03 AM
Puzzle
Who wrote the following
 
Einstein
ok, not einstein
Kahan?
he would have been more traumatized by the widespread adoption of java
 
Wilkinson
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.
 
 
2 hours later…
3:49 AM
@robjohn seems very complicated to prove decreasing.
 
the idea of doing numerical analysis in the 1960s is incomprehensible to me.
 
I got the difference as:
$-\sum_{r=0}^2 \frac 1{3n-r}+\frac 1{5n+1}+\frac 1{5n}+\sum_{r=1}^3\frac 1{5n-r}$
 
@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
 
Showing the above expression negative for all $n\in \mathbb N$ seems difficult. But I think I can try using differentiation here.
 
yes, at a high level i get it, but at the level of that. .000000000001192.
 
3:54 AM
Ah, sure
 
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.
 
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.
 
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.
 
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.
 
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)
 
haha, what was the line on the simpsons?
very few cartoons are broadcast live, homer. it's a terrible strain on the animators' wrists.
 
leslie, did you watch the new season of stranger things?
 
my daughter's reading one of her books to the cat. it's 9pm.
 
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
 
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
 
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
 
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.
 
@leslietownes yeah that subplot was unnecessary imo.
 
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.
 
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
 
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.
 
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
 
the german show 'dark' had some really good 80s stuff.
 
not that I've ever lived in 1986 Soviet Ukraine
 
yeah, chernobyl was amazing.
 
Dark, Chernobyl both were amazing : )
 
I haven't seen Dark. I'll have to check it out
 
4:21 AM
the plot is complete nonsense but the vibes are really cool.
 
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
 
there was a film version of the john le carre 'tinker tailor soldier spy' that had the aesthetic done perfectly.
 
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
 
people want to imagine that somewhere is less shitty than this timeline.
 
has anyone seen the movie Downfall (2002)?
 
4:23 AM
racacoonie!!!!!!
 
using the concept to explore nihilism and ADHD and stuff
@leslietownes :D
 
koro: i've seen it. and the million memes.
 
haha
 
my wife and i watched EEAAO two weeks ago. we loved racacoonie.
 
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
 
4:25 AM
i haven't seen any of those movies. the marvel stuff has completely passed me by.
 
(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
 
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)
 
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.
 
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"
 
My brother wrote a neat time-loop multiverse story
 
4:32 AM
is it out?
 
Nah
One day they'll make it
In the meantime he's working on other projects
 
does he do that full time?
 
did he go to tisch? i feel like i should know this. i've followed him for several years on twitter
 
He has a separate script called Precipice which has directors and producers attached, and now they're working on casting
@leslietownes Yeah
 
4:34 AM
wow
 
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
 
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
 
and there are travelers from our time with the mission to set up a "gate"
 
leslie, i also love apocalypse genre shows
like the walking dead
 
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
 
4:38 AM
koro have you seen 'the road'? it is based on a novel. it has viggo mortensen in it
 
not yet. I'll check it out. Does it have zombies in it?
The plot seems nice. I'll watch this one.
 
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'
 
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})
 
5:04 AM
huh.
 
 
2 hours later…
6:46 AM
What is the limit of $x_{n+1} = x_n - x_n^{n+1}$ with $x_1 = {3\over 4}$?
 
 
3 hours later…
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 $0$
 
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.)
 
11:17 AM
@MartinSleziak Thanks
@vidyarthi Seems not that simple
 
@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?
 
 
1 hour later…
12:27 PM
so, boris resigned
 
wait, really? is this the UK boris?
 
yes
time for that second scottish independence referendum
 
wow
didnt think he had it in him
 
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%
 
Is the monodromy theorem in complex analysis related to monodromy in covering space?
 
1:16 PM
What is the monodromy theorem in CA?
But yes, very closely
 
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
 
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
 
2:38 PM
@Thorgott If you restrict to a nbh it is, right?
 
I'm not sure
I was hoping it is, but then I couldn't convince myself anymore lol
 
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
 
 
6 hours later…
hahahahhahahahhahahahhahahahahhahahahah
hell that's funny
 
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
 
8:52 PM
And so it shall continue to do, in spades.
 
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 $a$?
 
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 $b=(1,x_1,\dots,x_{n-1})$
 
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
 
9:56 PM
How is the Hausdorff measure defined
 
as the infimum over all countable cover of the set, in this case the cube
 
10:08 PM
This makes no sense. What is $d$? The $n$-dimensional Hausdorff measure should just be the usual $n$-dimensional volume.
 
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?
 
$d$ is the usual euclidian distance $d(x, y)=\sqrt{\sum_{i=1}^{d}\left(x_{i}-y_{i}\right)^{2}}$
 
On the codimension $1$ symplectic subbundle, we should have $i_X d\alpha = \pm df$, so that much checks out.
 
@Alex, $H_n(Q)$ is a number.
 
@AlekMurt $d$ is a function, not a number. What is $(\sqrt{d})^n$?
 
10:13 PM
@AlekMurt So… $\sqrt d$ is a function? $d$ of what?
 
oh no
w8
 
Not everyone at once, lol
 
don't listen to me
 
EVERYBODY EVERYWHERE ALL AT ONCE!
 
I just realized what i am saying, srry boys
 
10:14 PM
Please don't assume that everyone here is male, or that they are happy being referred to as "boys".
 
As time goes on the more important I think typechecking is for math
 
* I meant $i_X d\alpha = df(R)\alpha - df$
 
Srry sorry now i feel bad.
 
@AkivaWeinberger It has saved my bacon more than once. That, and unit checking.
 
@AlekMurt 'Tsall good
 
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
 
@BalarkaSen I missed an episode
 
(aka I don't know all the background)
Can't help
 
Oh ok
No worries, just thinking out loud
 
Sure
I'm learning about Möbius functions on posets
 
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.
 
it's a $\sqrt{5/8}$
 
I haven't tried working it, but make the numbers simpler.
 
@TedShifrin That is as simple as the numbers get without making the problem trivial.
 
As long as you give them the graph I'm OK with it
 
dont give the graph and do tangent lines at (0, 0) just to confuse them
 
10:21 PM
Rescale to make the point $(1,1)$ or something.
 
@BalarkaSen Heh.
 
anarchy
 
I'm getting $(x^2+y^2)(x+yy')=2(x-yy')$
 
@AkivaWeinberger Yup.
 
Where $x^2+y^2$ is $1$
That's actually really nice^
So $\sqrt5+\sqrt3y'=2(\sqrt5-\sqrt3y')$
 
10:24 PM
now work out the tangent lines at (0, 0) akiva
 
@BalarkaSen DNE.
 
@BalarkaSen $y^2=x^2$ duh
 
@XanderHenderson diagnostic nasal endoscopy?
 
@BalarkaSen Yes, but also "Does Not Exist".
 
@AkivaWeinberger proof?
 
10:25 PM
@BalarkaSen [looks at camera feed] yup, that sure is a nose
 
@AkivaWeinberger Italicized parentheses and braces make me sad. :(
 
@XanderHenderson i reject this propaganda
 
zariski would like to have a word
 
$y'=\frac{\sqrt5}{3\sqrt3}$?
 
lmao thorgott
do it by (m/m^2)^*
 
10:27 PM
@AkivaWeinberger I got $\sqrt{5/27}$, which seems to be what you got.
 
Don't forget to dualize.
 
jeeze, ted, corrected
 
Good evening smart people
 
@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.
 
Stupid question is incoming, brace yourself
 
10:29 PM
i dont like the zariski tangent space
 
Do Nash blowup.
 
@BalarkaSen That's okay. It doesn't like you, either.
 
I'm TAing analysis 2 atm and we've defined tangent lines for arbitrary subsets of $\mathbb{R}^n$ lol
 
@TedShifrin much better
 
@TedShifrin Sounds like a great topic for Calc I. :D
 
10:30 PM
Thorgott i thought you would be much more advanced in your studies based on your answers. only now doing Ana 2?
 
@MadSpaces He's the TA.
 
the sole purpose of Calc I should be to confuse all students and traumatize them
 
Not a student.
 
Reading is a prerequisite for this room.
 
10:30 PM
Okay i dont know whats a TA lol
 
@MadSpaces he's also a superior german mathematician
 
Teaching assistant.
 
Ah i see
 
A TA is typically a graduate student who is paid to teach undergraduate classes.
 
@XanderHenderson $y=\dfrac{\sqrt5}{3\sqrt3}x+\dfrac{\sqrt2}{3\sqrt3}$
 
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.
 
@AkivaWeinberger Yup. I'd take that answer.
 
too slow akiva
 
Yes, you are holding a tutorium :)
 
@BalarkaSen Except Thor actually is!
 
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
 
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.
 
@XanderHenderson I'd recommend these numbers instead
 
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.
 
@XanderHenderson Correction
 
@AkivaWeinberger I thought about it (and am still thinking about it). I am somewhat torn about where to put the "ugly" numbers.
 
Remember that this turns into more of an algebra test than a calculus test for today’s students.
 
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").
 
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
 
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…
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With all the graphs Akiva posted, my question has long been gone to oblivion, oh well :(
 
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
 
Its fine, i am enjoying the graphs more
 
10:48 PM
I want the Hausdorff measure question to get fixed up. That one is actually in my wheelhouse. :(
 
@MadSpaces All I can think of is try to get it from the $\operatorname{sign}\sigma$ somehow
 

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