Mathematics - 2017-05-04 (page 4 of 5)

Alessandro Codenotti

20:03

You should construct a counterexample for the continuous maps case, there are some simple ones

Random Variable

@DanielFischer On rare occasions, an asymptotic evaluation of a definite integral with a parameter will result in the exact evaluation of the integral. Is there any way to know beforehand that this will (or at least might) happen?

James Bond

Hello @ted! I hope you are well. I just returned to the site by creating a new account today, LOL. I am the 'librarian', in case you are wondering who I am, LOL.

Paul Plummer

In one of these math tutoring rooms, and one of my students is asking me about a problem (didn't come to class today btw), and I am trying to get them to actually think about it and do some attempts. He says he gives up, then walks over to the other side and basically gets another tutor to do the problem for him.

Balarka Sen

...

James Bond

Hello @balarka. Nice to see you again. I hope you are sleeping well these days.

Daminark

20:08

... What the actual... Good God...

Paul Plummer

@BalarkaSen Hey

Balarka Sen

hi @JamesBond. yeah, sleep's okay

@PaulPlummer Rehi. The only way I can describe your student is as an asshole. :(

James Bond

You don't want to get too emotionally involved with your students in this way.

Alessandro Codenotti

At this point not thinking is their choice and they'll eventually deal with the consequences, you did your job

James Bond

They are not elementary school kids. We don't need to tell them how to live their lives.

Paul Plummer

20:12

I am probably going to give the tutor a bit of shit though

Although I have already mentioned to them that they shouldn't be doing other peoples homework

so I guess it doesn't really matter

James Bond

How long should one work on a problem before asking for help? Any ideas?

Balarka Sen

as long as it should take. there's no generic number

Daminark

@JamesBond 2 hours and 86 minutes exactly :P

Lol jk

It's problem dependent, there are some which should take 10+ hours to figure out, and others for which if you're not getting it, this signifies a lack of understanding

And yeah @Paul I mean, yeah, that student will get creamed on the midterm/final so I if the lesson can't be learned one way, it will another

Paul Plummer

Yah, it is just annoying

Daminark

Fairly so

Balarka Sen

20:17

Can you refuse to tutor this student?

TheGreatDuck

if there is a policy on academic dishonesty and they are continuing to be academically dishonest... just flat out report them

if it isn't academic dishonesty, then butt out of their own business

though unless they are practice problems and not graded assignments, I see absolutely no reasoning that could lead to me believing it isn't academic dishonesty

Paul Plummer

Ehh, not really. I tend to say things like "try something" or ask things to make them get more specific (where they would need to think more about what is actually going on), and not really give them much until they show some actual effort @BalarkaSen

TheGreatDuck

no i mean them going to the other tutor and the other tutor doing it for them

if the other tutor is literally giving them the solution... then it is cheating.

Paul Plummer

I was reply to Balarka

TheGreatDuck

ah

James Bond

20:22

I just asked a question about complex algebraic/analytic/differential geometry on the main site. More of terminology, what is what.

TheGreatDuck

well I don't think refusing to tutor is the right way to go about doing this

James Bond

Hello @TheGreatDuck. What an interesting name!

Balarka Sen

I am not sure if this is actual homework though, in which case you probably can report it.

TheGreatDuck

^^

like I said, if it is nothing more than practice problems all you can really report them for is being annoying

or being difficult with you

Balarka Sen

I really think the student and the tutor he got his problem done from needs more shit than a shrug.

James Bond

20:24

Hello @MikeMiller, LOL. I hope you are doing well in grad school.

TheGreatDuck

@BalarkaSen if it's a practice problem and it is not for a grade than it is no more different than if one asked a question on here. So, I don't feel academic dishonesty comes into play as the student is still only turning in their own work.

Balarka Sen

That last message wasn't a reply to what you said. Of course you cannot report if it's just practice.

not for dishonesty at least

James Bond

I finally got all three books by Herbert Enderton.

The list of my math books is coming to 36 now.

Paul Plummer

@TheGreatDuck @BalarkaSen Maybe, I think it is sort of grey zone with the way he was being "helped". I honestly don't care that much, just wish the tutor had the student be more involved instead of this "lazy" explain the whole problem step by step and working out most of it himself. And it is online homework, pretty sure most of the class cheats. I didn't really mean for the conversation here to go this long anyways, probably should just stop thinking about it for now.

Daminark

It's not dishonesty so much as, kinda disrespectful, but also just detrimental to learning. I think at this point the student's lesson is probably just gonna have to be the grade from bombing the final, but the tutor definitely deserves some shit for this, especially if he/she saw what Paul was trying to do

@JamesBond Physical books? Oh snap

I've got like, 4 physical ones

James Bond

20:29

@Daminark Yup. I usually evaluate them online before deciding to buy or not.

Balarka Sen

@PaulPlummer Ah alright.

TheGreatDuck

@PaulPlummer next time you are on campus go straight to whoever your supervisor is and tell them everything you told us.

don't be acusatory

Daminark

Sally's analysis book, Spivak, Silverman's "Introductory Complex Analysis" (loaned from a friend), and this one I checked out from the library but never got around to reading called "From Calculus to Cohomology"

TheGreatDuck

just say you are very concerned about the student's well being as it appears like the tutor is handing the student answers.

Astyx

What ? Books can take other forms than pdf ?

Daminark

20:30

@Astyx Right??? I mean djvu but everyone just immediately converts those to pdfs anyway

TheGreatDuck

@PaulPlummer if your boss is reasonable, they'll take of the situation and the tutor will be fired and/or reported for academic dishonesty along with the student.

James Bond

@Daminark Very interesting that you chose these 4 and not others.

@Astyx DJVU is more compressed than PDF.

Daniel Fischer

@RandomVariable Umm, I think time travel would work. And as with all things, experience helps making an educated guess.

Daminark

Well, my Calc class used Spivak, and analysis class first quarter used Sally (interesting selection of topics, but not terribly well-written), Silverman is what my friend used in his complex analysis class as taught by my professor second quarter

And Calc to Cohomology was on a whim

Paul Plummer

Anyways got to go see a visitors talk. Later @BalarkaSen @Daminark @TheGreatDuck @AlessandroCodenotti

Balarka Sen

20:33

see ya

Daminark

See you @Paul!

Alessandro Codenotti

see you

Astyx

Bye Paul

James Bond

@Daminark I use Kaplan and Lewis: Calculus and Linear Algebra for calculus, and Protter and Morrey: First course in Real Analysis.

Daminark

I don't know either of those books

James Bond

20:35

I chose them for special reasons, after looking at over 20 books for calculus and analysis.

TheGreatDuck

@PaulPlummer dont forget what we told you to do. And cya.

Daminark

So analysis first quarter used some Rudin for suprema, topology, continuity, uniform convergence, and also generally for problems. Sally for a bit on countability and rational approximation, then multivariable differentiation and some material on topology not covered by Rudin, like absolute continuity, bounded variation, BCT. Toward the end we used Buck for curves and surfaces, and throughout we got linear algebra problems for Hoffman and Kunze. These were all chosen by our professor, though

I only purchased Sally since I couldn't find a pdf for it

James Bond

For linear algebra I use Peter Petersen: Linear Algebra.

Daminark

I do not know that one either

Do you like them though?

James Bond

After that, one can use Loehr: Advanced Linear Algebra.

Yes, I have not studied them from cover to cover but I like them. The latter book goes into modules and multilinear algebra, but I use the former for vector spaces and linear algebra.

@Daminark I would think that Sally is a girl at first, LOL.

Daminark

20:40

I've used a a bit of Axler, which was meh, Hoffman and Kunze which I liked (minus lack of quotients, and that first chapter on basically rref which should've been cut out)

Oh, lol it's Paul Sally

James Bond

I don't really like the latest edition of Axler having colour. It is really not necessary.

Daminark

I mean, my problems with Axler are more content based really

Balarka Sen

I am going to have to write a long heavily TeXed message to Ted now. God no.

James Bond

Axler doesn't seem to cover some of the most elementary parts of linear algebra.

But I guess most people using the book has covered those parts elsewhere already.

Daminark

It completely sticks to $\mathbb{R}$ and $\mathbb{C}$, and rigid insistence on avoiding determinants apparently puts you in the mindset of triangularizing everything, when oftentimes you only care about the determinant or characteristic polynomial.

Also some gripes that I had even with Hoffman and Kunze carry over to it, like lack of quotients and all

James Bond

20:44

Petersen does use arbitrary fields, and covers the elementary parts as well.

Daminark

Now, I don't like how people define just a sum over permutations and just prove substantial things using the unmotivated definition, but you could define the determinant properly, in terms of multilinear stuff or whatnot, and derive this formula. Then the resulting theorems will feel like they have more content, you know?

James Bond

Petersen also mentions all three canonical forms, Jordan, Frobenius and Smith which is what I like.

And it also covers the spectral theorem for normal operators, not just self-adjoint ones.

Petersen does cover determinants in the last chapter but he doesn't really use them in the preceding chapters.

By the way it is Petersen from UCLA, because there are other authors of linear algebra books called Peterson/Petersen, lol.

Daminark

OK, found it

James Bond

If you need recommendations for math books on any topic, as long as it is not too exotic you can ask me.

I have spent a lot of time comparing books of the same topic for many topics, and look out for all kinds of idiosyncracies.

Daminark

Aight, I will keep that in mind, thanks!

And lol I tend to gather quite a lot of books, my pdf stache is insane, but there are a lot that I just have and don't know much about. Also my opinions are somewhat... specific I guess? I have weird preferences

@Akiva Did you figure out the name just now?

James Bond

20:51

@Daminark Same, I choose them based on very very weird things, lol.

Daminark

I think I definitely have an inclination to books that are short/terse for sure

I like some amount of chattiness and especially humor, but it's nice to just sorta move quickly through things. Plus it's good practice to read between the lines and actually be convinced of the proof of something for which the details were somewhat... skimped

Yo @Eric!

Eric

Yo

Balarka Sen

@TedShifrin From the proof of uniqueness of torsionfree Riemannian connections, $X\langle Y, Z \rangle + Y \langle X, Z \rangle - Z \langle X, Y \rangle = \langle [X, Z], Y\rangle + \langle [Y, Z], X \rangle + \langle [X, Y], Z\rangle + 2 \langle \nabla_Y X, Z\rangle$. Set $X = Y$; you boil down to $2\langle Z, \nabla_X X\rangle = 2X \langle X, Z \rangle + Z\langle X, X \rangle$.

If $X, Y, Z$ are all left-invariant, then those inner products are constant and we remain with $\langle Z, \nabla_X X \rangle = 0$ for all left-invariant $Z$'s. That means $\nabla_X X = 0$, and $X$ is parallel so is integral to a geodesic. This means every 1 parameter subgroup is a geodesic; but through any point and any vector runs a 1 parameter subgroup (pick that vector and left-translate it; take it's integral curve) so every geodesic is a 1 parameter subgroup.

Hi @Eric.

Eric

@Balarka is this problem 3.3 in do carmo

Daminark

@Balarka After hearing you talk about topology for all that time it's interesting to see the geometry more

Balarka Sen

21:01

@Eric Is this in doCarmo? Huh

I wanted to understand geodesics in SO(n) and Ted told me to prove that

James Bond

I never got any book by do Carmo. He is Portuguese, I think.

Eric

I think it is

Daminark

@JamesBond Brazilian, I thought

I mean language is the same but still

Eric

He's brazilian

Balarka Sen

@Daminark I'm terribad at it man

Eric

21:02

From alagoas I think

James Bond

@Daminark Oh, maybe you are right. Same language, different dialect, lol.

Eric

In the Northeast of the country

James Bond

@Eric I see. Brazilian waxing, lol.

Balarka Sen

@Eric Ah! It seems it is

Eric

There are a bunch of different dialects of portuguese in Brazil

The country is huge

Balarka Sen

21:03

I didn't learn Ch 3 from do Carmo

WDUK

Hi, i am learning about partial fractions but am a bit confused with this snippet in my book: i.imgur.com/PZOMwjj.png, why is constant factor 2 in the denominator not taken into account here?

Daminark

@Balarka We all have various ways of thinking we're more or less compatible with. It's good to get a lot of practice in, you can still get better, but it's fine if one thing or another doesn't click so nicely

Eric

Yup @Balarka I thought I remembered doing this exercise relatively recently

James Bond

I like languages but I don't intend to study Portuguese. It seems though that Brazilian Portuguese is a major variety, since there are so many people in Brazil.

Balarka Sen

@Daminark Yah my main problem is I'm totally bad at calculation

I think pictures

James Bond

21:05

An equation is worth a thousand pictures.

Daminark

I mean I'm basically the same way

WDUK

So what about a picture of an equation ?

Eric

@JamesBond the majority of speakers of the language come from brazil, iirc it's either the most or second most spoken language in the southern hemisphere

Balarka Sen

Maybe I can cover that up by having a good library of knowledge in diffgeo

Eric

I love calculations @Balarka

James Bond

21:06

@Eric Yeah, the other one being Spanish.

Daminark

I'm better at pictures and "ideas", whatever that may entail. Occasionally in algebra I've also just come up with the right kinds of arguments

But I'm neither good at, nor really enjoy, calculating

Eric

Especially all those wild curvature calculations from Riemannian geo :P

Balarka Sen

@Eric Nice! Maybe I'm going to ask you stuff to understand how you think

studying how other people's brains work has been my secret weapon in doing math actually

Daminark

And yeah @Balarka I intend to basically do the same, just have a lot of theorems in an inventory. See, the way I tend to look at stuff is by sort of imagining various possibilities as a kind of tree and killing them off until there's one left. Having a lot of theorems floating around is good for that

Interesting

Balarka Sen

@Daminark Yep, that's a pretty good strategy.

Eric

21:09

It's interesting to see how different people have different cognitive models of math or different processes

James Bond

@Eric My favourite book on that is Lee's Riemannian Manifolds. I think the second edition would be out this year.

Eric

I've never read his Riemannian book, just his smooth manifolds

Daminark

@Eric I found this to be particularly helpful in the stuff on spectra we did at the end of 208

I'd sorta say OK, so it can't be this because this, can't be that because of blah, then boom boom boom this works

Hey @EricS!

Phase

Hey, what's the simplest way of proving the Cylical permutation property of traces?

Eric

@Daminark I think I've told you about my horrible problem solving process

Daminark

21:12

I wouldn't describe it as horrible, I'd say it could help in various contexts

Eric

Idk for some people it just doesn't work

Balarka Sen

(Also, @Ted, I used $\langle [X, Y], Z \rangle = \langle X, [Y, Z]\rangle$ at one point in my calculation. Forgot to mention)

Daminark

My way of thinking helps a lot sometimes but also I rarely think at all constructively or "calculationally" for that reason, it's always, what do I have on hand that forces this to be true/false?

Balarka Sen

I wonder if I know how to prove that actually. I saw that identity in doCarmo when reading about left-invariant metrics but I skimmed over the proof.

James Bond

@Eric I am getting all his three books on manifolds, LOL. I am in love with Lee!

Balarka Sen

21:15

@Daminark I think, fundamentally, my problem solving strategy relies on memory. Not detailed memory of formulas (or even theorems for that matter) etc but the patterns and pictures

Daminark

ODEs have not responded to me quite as well, possibly for that reason in part

Balarka Sen

which is possibly why i'm bad at calculations

James Bond

@Daminark ODE also stands for 'Oxford Dictionary of English', LOL.

Daminark

Hmm, I find that pictures come up for me more when understanding definitions more. As for theorems, I'm alright/good with remembering those, but details in just about any context make things iffy for me @Balarka

Also hey @Ted, just saw you now

And ohi @Steamy

Eric

@Balarka if I recall correctly that identity comes using bi invariance and taking a derivative

SteamyRoot

21:18

ohi

Eric

I think it's a short calculation

PVAL-inactive

Is there anyone post-PhD here?

Eric

@jamesbond smooth manifolds was quite good if that's something to go by. I know nothing about the other two

PVAL-inactive

I need to ask someone an academic question.

Balarka Sen

@Eric Hm. Taking a derivative of what?

Eric

21:24

so take $x, u, v \in T_{e}G$ and define $x_{t}$ to be the flow of $x$. Then $\langle u, v \rangle = \langle dR_{x_{t}(e)}u, dR_{x_{t}(e)}v \rangle$ or something

if you differentiate that i think you get $0 = \langle [u, x] v \rangle + \langle u, [v, x] \rangle$

Balarka Sen

Ah! interesting

Daminark

gazes over the chat horizon and then shrugs @PVAL

Balarka Sen

That's a nice little trick

PVAL-inactive

Hi @Dami

I had to send a somewhat formal request to someone I've sent quite a few casual emails ,so I wasn't sure how to handle it.

Eric

you get lots of cute little formulas coming out of bi-invariant metrics @Balarka

Daminark

21:30

Ah, that's a tricky situation

Eric

The sectional curvature has a really cute one, it's $K(X, Y) = \frac{1}{4}\| [X, Y] \|^{2}$

trilolil

Hello everybody,

I have a question related to cholesky matrix decomposition.
My question consists of two parts:

http://imgur.com/a/JYHGh

1) Why would the author of this book use "for i=(n+1)...2n" on the third line instead of what he did on the second line ie "for i=1...n"? It looks to me like they will both totally give the same result.

2) Could someone help me understand how the matrix called sigma is actually constructed? This part is totally unclear to me.

$\Sigma = SS^T$ where S is computed using the cholesky decomposition.

Balarka Sen

So I am guessing the covariant derivative of $dR_{x_t} u$ is $[u, x]$. I guess that looks a lot like $[u, x] = \mathcal{L}_x u$

Yeah I guess that's how you'd prove the derivative is that

(modulo messed up signs :P)

@Eric Oh, really nice.

Ted Shifrin

@JamesBond: Welcome with your new name. That book of Lee's, unlike his others, has far less content. There's not enough Riemannian geometry (and none of the stuff I particularly like).

hi @Balarka — should I let Eric take over with you? :)

Balarka Sen

I posted a proof of what you claimed way above.

(And pinged you with it)

James Bond

21:36

@TedShifrin Hehe, yes, but he says he will add more stuff for the new edition. =)

Ted Shifrin

Are you claiming you're using biinvariance for all that, @Balarka?

Balarka Sen

Yep, crucially.

Ted Shifrin

How?

Balarka Sen

In the identity $\langle X, [Y, Z] \rangle = \langle [X, Y], Z \rangle$

trilolil

Does anybody have an idea about my question?

Ted Shifrin

21:37

OK. I think you're using it somewhere else in the argument about why 1-parameter subgroups are geodesics.

Hi, DogAteMy.

James Bond

I don't mean to upset anyone, but has anyone seen Chris's Sis in chat lately?

Balarka Sen

I used left invariance to prove $\langle X, Y \rangle$ is constant for left invariant vector fields $X, Y$. But not right

Ted Shifrin

@trilolil: I'm not looking carefully at what you wrote, but for a positive definite (symmetric) matrix, there is a universal convention on what square root means.

Akiva Weinberger

@TedShifrin Hi

Ted Shifrin

@JamesBond: Only once or twice. I have that person on ignore, so I only know he/she is here when people are talking to a blank wall (to me).

James Bond

21:39

@TedShifrin When I was in middle school, I never knew what root 4 means, 2 or plus minus 2.

trilolil

@TedShifrin in this context the square root is defined as: $\Sigma = SS^T$

Ted Shifrin

That confuses people to this day, @James.

trilolil

according to the author.

Ted Shifrin

I don't know what all the letters mean. I was just answering your question about square root.

trilolil

@TedShifrin but my question was not about the square root at all. What makes you think that?

James Bond

21:41

@robjohn You should come back to this chat! LOL

trilolil

all the letters are scalars except $\Sigma$ which is a matrix

Akiva Weinberger

Wolfram Alpha believes $\sqrt[\Large3]{-1}=e^{i\pi/3}$.

Ted Shifrin

If $A$ is symmetric and positive definite (for example, anything of the form $SS^\top$ or $S^\top S$ for square $S$), then $A=Q\Lambda Q^{\top}$ for a diagonal matrix $\Lambda$ with positive entries. We define $\sqrt A = Q\sqrt{\Lambda}Q^\top$, where all the entries of $\sqrt\Lambda$ are positive.

robjohn is here occasionally, @James, but super busy.

That's one of 'em, DogAteMy. If you read their documentation, they probably explain such things.

James Bond

@TedShifrin Are you a James Bond fan? I just watched 19 Bond movies, 7 more to go...

trilolil

@TedShifrin thanks for this information, but I don't think this answers any of my two questions...

Akiva Weinberger

21:42

Yeah. It's just kinda strange, as the real cube root function seems like a more obvious choice.

Makes it easier for them to plot $(-1)^x$, though, I guess.

Ted Shifrin

@trilolil: If you want to take the time to explain to us what your question is, rather than explaining us to read something else, go ahead. Otherwise, don't be impudent.

Sylent Nyte

guys, weird question (i might be going crazy) $a^2 = 6^2 + c^2$ can simplify to $a = 6 + c$, can it?

trilolil

Ted Shifrin

@Balarka: Here's something important for you to think about. Suppose $X$ is a left-invariant vector field on $G$. What is the flow of $X$?

trilolil

Why would the author of this book use "for i=(n+1)...2n" on the third line instead of what he did on the second line ie "for i=1...n"? It looks to me like they will both totally give the same result. @TedShifrin I don't know what else I should add to clarify my question...

Balarka Sen

21:44

Isn't it a 1-parameter subgroup?

Ted Shifrin

@Balarka: Start at some point other than the identity. How do you give the flow?

Balarka Sen

Sorry, un-delete my previous comment. Integral curve of $X$ is the 1-parameter subgroup, I was confusing "flow" with that.

Daminark

First un-sleep, then un-delete, what next??

Balarka Sen

un-live

Ted Shifrin

@trilolil: I have no idea what any of this means. But did you notice he changed a sign in the definition?

Daminark

21:46

gulp

Ted Shifrin

@Balarka: Integral curves are flows (starting at a particular point), but I want to know a formula for the integral curve starting at $g\in G$.

Balarka Sen

Ah, well, $\gamma'(t) = X(\gamma(t))$ with initial condition $\gamma(0) = g$.

Ted Shifrin

I want a formula for $\gamma(t)$ (you may use exponential map). And I'm fine if you work just with a matrix group, where things are more concrete.

Balarka Sen

(I really think of flow as a 1-parameter family of diffeomorphisms, but of course I understand it's integral curve starting at a point)

@TedShifrin Ok, got it.

trilolil

@TedShifrin yes I did, but my question is about the part on the right side of that sign you are referring to. That part will, I think always give the same result, no matter what subscripts are used ('i' or 'i-n'). Which makes me question why the author wrote 'for i=1..n' and a little later 'for i=(n+1)...2n' as they will yield in the same result.

Balarka Sen

21:51

How would the formula look like? Just want to make sure what I am going to look for

Ted Shifrin

How does $i$ give the same result as $i-n$ when $i=n+5$?

You are not communicating clearly, @trilolil. Or maybe I just cannot understand you.

Waiting

I don't see @ShaVuklia which I wanted to greet ...

Balarka Sen

I guess I know how it looks like for matrix groups, because I "know" what exp is for SO(n) say

Ted Shifrin

Sure, @Balarka. I'm happy for you to write things like $e^{tX}$ where $X\in\mathfrak g$.

Sylent Nyte

guys, weird question (i might be going crazy) $a^2=6^2+c^2$ can simplify to $a=6+c$ right?

Ted Shifrin

21:53

WHAT????

Daminark

angery reacts only

Balarka Sen

You have gone crazy, yes

Sylent Nyte

lmao

Ted Shifrin

Let's try 1+1=2 ... so $1^2+1^2=2^2$. Yeah, right.

Next you'll tell me that $\dfrac 1{a+b} = \dfrac1a+\dfrac1b$.

Sylent Nyte

it isnt?

Ted Shifrin

21:54

sighs and goes to the corner

Sylent Nyte

$log(a) + log(b) = log(a+b)$ no?

Eric

@Daminark sad reacts

Ted Shifrin

No, @Sylent. Nor is it $2^{a+b}=2^a+2^b$.

Balarka Sen

@Daminark Angery, indeed.

Daminark

@Sylent you almost committed an even worse sin than saying that, which was screwing up the TeX

Ted Shifrin

21:55

One of my old students/friends was just ranting on Facebook about how his calculus students in a Georgia college were doing all these things on their final exam.

LOL, Demonark.

Daminark

@Eric But wait it was supposed to be angery!

Sylent Nyte

lmao just kidding about the logs guys dw, im not TOTALLY crazy

Balarka Sen

@Ted All of what you listed at once!

Ted Shifrin

We cannot believe you, @Sylent.

Sylent Nyte

I was just having a brainfart moment

Ted Shifrin

21:56

@Balarka: apparently different students did different ones.

Balarka Sen

Well that's a little better.

Eric

$\frac{\sin x}{n} = \text{six} = 6$ @Ted Just cancel the $n$s :)

trilolil

@TedShifrin ok, let's apply this. Imagine the first iteration. The first formula will obtain as subscript 1, while the second formula gets subscript i=(n+1)-n , where n=4. And there you have subscript 1 as well.

Ted Shifrin

@Eric: That one's famous, too.

Sylent Nyte

tho it just means that I have a messy equation eurgh

wishful thoughts

Daminark

21:57

@Eric That's one of my favorites

Ted Shifrin

@trilolil: He intends that. But the columns will not be the same because he's put $-$ where there was $+$ in the formula.

Daminark

Also 16/64 = 1/4 by cancelling 6

Balarka Sen

1/100

Sylent Nyte

ahhhh

Eric

lol @Daminark I've seen someone do this on a problem set

Danu

21:58

@TedShifrin heyo

Daminark

??????????????????

trilolil

@TedShifrin You sure? this sign will only influence the actual calculated result and not the position in the matrix I think, right? Between square brackets you will always obtain the same result for both equations.

Ted Shifrin

@Danu: Did you catch up on all our unit tangent bundle stuff from before? And on my quadric response?

Danu

@MikeMiller So my quadric is naturally homogeneous ($G_2/U(2)$ or $SO(7)/SO(2)\times SO(5)$)

Daminark

I mean eh it was probably a roflproof or something

No one would actually do that

Danu

21:59

@TedShifrin I saw that you concluded that there is torsion and relayed that information to my friend. I clearly didn't understand the details :P

Eric

I think they were just going to fast for their own good @Daminark, they generally made a lot of mistakes based around their handwriting being sloppy

Daminark

But everything is alright so long as it's ironic

Ted Shifrin

You're not interpreting the formula correctly, @trilolil. The 1st column is something $X+Y$ and the 4th column is something $X-Y$.

Danu

The quadric response---I'm reading

Daminark

Oh... Kek

Transcript for

$Mathematics$ Mathematics