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02:14
$\tau$ of course...
I have a bad joke to share.
What do you call a cookie baked by an amateur?
A rookie cookie
@SoumikMukherjee I don't get it (English is not my first language)
@Derso An amateur is someone who is inexperienced. A rookie means someone who just started something, so also inexperienced. So a cookie made by an amateur is not good enough, so it's a rookie cookie.
@SoumikMukherjee that takes the biscuit
@leslietownes I made a pilgrimage to Moe's today. Unfortunately no Folland, in fact slim pickin's really. I went to the ASUC 'bookstore' to find that they no longer carry actual books, all online now. I suppose that explains the paucity. It also means I can no longer get a feeling for what is being taught by looking at the class books. Depressing really, the effective demise of hardcopy.
02:46
coppper: and the kind of material that used to be easily browsed on "course websites" is now often students-only, behind some walled garden of course management software that even students at the uni who aren't taking the class might not be able to see
03:02
yep. the notion of public seems to have disappeared...
03:26
People are also merging with the online it seems
 
2 hours later…
05:55
theres a cantor set like construction using more general sequences, of the type $a_0=1, 0<2a_n<a_{n-1}$ and so on. why would we do this, and not the usual case where interval length is taken to be $1/3^n$ at each juncture ?
 
6 hours later…
11:54
Hi
Hi everyone. I am getting $\frac{5\pi}{3}(1+5\sqrt6-5\sqrt2)$ for the question below, but that is in none of the options. The correct option according to the book is (C). Can anyone help please?
Here’s a Desmos plot for the same: desmos.com/calculator/cmfbjjli7y
12:13
@SoumikMukherjee Ok, so maybe I got it from the start, but it was really just a bad joke lol
@SineoftheTime $\int^{\infty}_{0} \frac{1}{(x^2+y^2)^{3/2}} dx$
Here after making the substitution $x = y \tan{\theta}$, I have to change the integration extremes, right?
So when $x \to 0 \quad \theta \to 0$
$x \to \infty \quad \theta \to \frac{\pi}{2}$
I didn't understand this last substitution
arctan($\infty$) is it defined as π/2?
Oh yes I think that's why, now I remember
12:55
@Pizza that's correct
as $x\to +\infty$, since $x=y\tan \theta$, then $\theta \to \pi/2$
13:25
how to group theory 101: given a theorem, barely check theorem for |G|=1, then assume theorem holds true for all groups of ord < G, find a proper normal subgroup, take quotient, apply the assumed theorem on the quotient, apply inverse of projection map on the quotient. Abuse lattice isomorphism theorem
@Derso hehe, I mentioned it's gonna be a bad joke :p
step 3- find proper normal subgroup--> not always easy, tho
13:59
@nickbros123 what are you referring to. Fat cantor sets?
 
2 hours later…
15:39
@Jakobian im not sure if this is the thing. My professor did something in class. Ill post it later on. I have a numerical analysis exam coming up :(
15:55
@nickbros123 If you take out less than 1/3 at each step, You can make "thick" Cantor sets which have measure greater than 0 but still completely disconnected
Bml
Bml
16:27
Hi everyone. I would like to ask whether you think it is reasonable to receive a downvote in this answer without any reason being given in comments section. Is there some intrinsic feature of the answer (length, articulation, etc.) that makes an answer subject to downvotes? Thanks to those who would like to reply.
I think you should not worry about this, period.
You have 8 upvotes and a bounty.
Bml
Bml
@BenSteffan I know, I am not worrying. It was just out of curiosity.
Unless somebody swings by to give a reason the downvote should be considered spurious.
@VladimirLysikov well, homeomorphic to the standard Cantor set
@Bml "reasonable to receive a downvote" what does that even mean
Bml
Bml
@Jakobian Well, it means that if a question is badly formatted, poorly written or partial, it is reasonable to expect a downvote.
16:35
an action can be reasonable. How can an event be reasonable. I don't understand
to expect a downvote, okay, that's different
Bml
Bml
@Jakobian Yes, sorry. English is not my mother tongue.
One cannot expect to not receive downvotes
A user can downvote you for any reason whatsoever, and they don't have to leave a comment either
I don't see which cohomology theory is suitable for $\mathcal I$
Cubical, floer, de Rham, simplicial - they don't seem quite right
It would take several pages to explain what $\mathcal I$ is
now its another thing to ask if your question attracts downvotes for some reason
given that you have been downvoted, perhaps another user was salty because you received the bounty
or someone who doesn't like you downvoted you because of the interactions you had
those are only speculations - either way its not a crime, because no one states their intentions
nor they are obliged to
as long as they don't downvote you in a systematic, targeted manner, there is nothing wrong with such behavior, according to the rules of the site
Bml
Bml
16:52
@Jakobian I understand. If you like, you can look at the answer and tell me what you think :-)
I am looking to buy a book on intersection cohomology. Can anyone recommend the easiest book on this topic?
I am thinking: "Stratified Morse Theory" by Mark Goresky and Robert MacPherson
@Jakobian yep. if you don't want to ever get downvoted on this site, don't post
I don't know if there even is another book on the subject
Goresky-MacPherson is like the bible
@Bml it looks like a long, exhausting read to an elementary question. Sorry but I'll pass
Bml
Bml
@Jakobian Don't worry :-)
17:11
Imagine if 1 downvote=-10 rep
I've thought for a while that downvotes should be heavier on the rep penalty
not -10, but maybe -5
I think in the past 1 upvote=+5
that 1 upvote balances 5 downvotes, reputation-wise, is strange to me
Bml
Bml
@BenSteffan I agree, but I would make it a constraint to explain the reason for the downvote, wouldn't you?
17:17
no
you don't want 5 more or less identical comments on some bad question explaining the same thing
Bml
Bml
@BenSteffan No, I have expressed myself badly. I simply mean that the downvote could be -5, but in my opinion one should distinguish between two cases. 1) Bad questions/answers, no comments needed to justify the downvote; 2) Good questions/answers that received several upvotes, need to justify why the downvote. I do not find it fair for a user to see -5 in their reputation for a well-written answer.
What's unfair about it?
Voting is not a reward dispenser
It's a democratic tool
@Bml This presupposes that there is some objective metric regarding what is a "good" question or anwer.
The whole point of voting is that there is not such an objective standard, and everyone with sufficient XP is allowed to express an option via their votes.
Overall the system works pretty alright
You write a decent answer, you almost always end up with a positive vote total
Is anyone here aware of a shape that has exactly 3 homotopy classes in its fundamental group?
17:27
@ModularMindset Any $\mathbb{Z} / 3$ lens space
@BenSteffan Even if you write a super marginal answer, the expected value is positive. People don't downvote nearly as much as they upvote, and downvotes are worth less than upvotes, vis-a-vis XP.
@ModularMindset There are pretty standard constructions for producing spaces with given homology / homotopy / whatever groups.
take a circle and attach a disk via a degree 3 map
@BenSteffan definitely
@BenSteffan it kind of is. The whole point of the system is gamification and instant gratification
17:39
there is some portion of down/upvotes that are misvotes due to the user hitting the wrong button and not noticing it. i don't know if the site design or any other UX considerations would bias this error rate up or down. i might expect people are less likely to notice mistaken upvotes because there is no effect on voter reputation associated with them. plus the bias on upvotes mattering more to the recipient's rep than downvotes...
But at the same time, I think downvoting should be the opposite of reward
people should really be interrogating those who upvote without comment. "ARE YOU SURE??? EXPLAIN THE UPVOTE"
5
@leslietownes you should ask: are you sure? explain the stars
exactly 3 homotopy classes in its fund. group s.t. the union of the 3 homotopy classes is path connected topological graph
and it's 4-regular (the top. graph)
@leslietownes Thank you for showing us very clearly the hypocrisy here
17:43
People who get 5+ upvotes should go to the mods and say ""I don't understand these upvotes?!
go to the mods is my main thesis here btw
because they are very powerful
@ModularMindset sure, if you allow loops in your graphs
I don't understand the upvotes on this answer of mine: math.stackexchange.com/questions/4964654/…
the graph will obviously not be 4-regular
@BenSteffan yeah true
the property $(xy)z=x(yz)$ is called associativity, what do you call $1x=x=x1$? unicity or unitality?
17:47
municpality
thorgot: i call it The Oneness
[case sensitive]
@Thorgott unitality
unitality
17:48
in combination with "$\exists 1$"
Noun: unitality (uncountable)
  1. The property of being unital.
seems like the term exists in the dictionary, so yeah
as someone who does not work in this area, unitality at least sounds like the right answer, unicity sounds like a student who is unfamiliar with math and/or english aiming for "uniqueness" and failing
@Thorgott neutrality of $1$
now someone can inform me that this is what grothendieck called it
"Grothendieck with glasses"
17:53
ok thanks, I'll go with unitality
Glad I could help
Is there a name for this generalization?
a pen, not a blue pen
what material is that made out of? is it metal, paint on some kind of 3d printed resin, ??
copper, don't answer with what the pens and notebook are made of
@psie what ed. of Folland do you have?
@ModularMindset A shape.
18:07
@ModularMindset Patrick Star
some Meissner bodies I sent to an academic friend recovering from an injury: imgur.com/a/cnsZtAa
@ModularMindset how did you make the shape?
Meme answers only
guys
@psie Folland seems to assume $\mu X < \infty$ a lot more than necessary.
the less selfish meme is youyou
copper: not sure if you are asking something specific to a psie question, but per your recent bookstore visit, ~ 25 years ago, the book on the shelf for 202a would have been whatever edition was out at that time (new then). i don't think it has been updated since except for typos
18:12
Adjoint operator $A^\ast$: $\langle Au,v\rangle = \langle u,A^\ast v\rangle$, right? Then, in terms of matrices we have $[A^\ast] = [A]^T$ (transpose). Right?
derso: depends a tiny bit on what "matrices" means. that would hold if the basis with respect to which the matrix is taken is orthonormal
(which is common when an inner product is around, often so much that "matrix" is taken to mean that definitionally in those specific settings, but this is not automatic, and obviously sometimes the difference matters)
It's not true in general?
@leslietownes thanks. i had Kobayashi, he didn't recommend a specific text (if i recall) and Chernoff has his own notes
Like, $\langle Au,v\rangle = \langle u,A^T v\rangle$?
some perverts have linearity in the first parameter.
18:14
I'm thinking of the most simple case $A:\mathbb{R}^3\to \mathbb{R}^3$...
Ooooh, if THE BASIS taken is orthonormal. Yeah, yeah
derso: math.stackexchange.com/questions/1490851/… contains some discussion of what the formulas for the matrix of the adjoint would generalize to if the bases involved are not orthonormal
No, no. My basis is the standard one lol hahaha
yeah. "basis means orthonormal basis" is very frequently built into the background of a discussion of inner product spaces, but doesn't have to be
and things only get goofy when it isn't
@leslietownes sorry, a bit slow this morning, on a long boring work call. i just visited the asuc store for general interest, since my trip to Moe's was disappointing.
Nice
18:17
copper: i'm also slow this morning, and not because i'm on a work call :)
@copper.hat Thanks
:-)
@Derso i didn't help at all, the opposite in fact :-)
derso: ^T needs to mean conjugate transpose if the vector space is complex. i didn't notice that above, but that's another thing
@leslie townes Exactly. I've found about Hermitian operators at Wikipedia, but not the real case, which is the one I need lol
Complex cojugation is also needed in that case
The thing is, there's also something called adjoint matrix (transposed of the cofactor matrix, right?) which has nothing to do with adjoint operator
So it's pretty confusing all these names
(or adjugate matrix, whatever)
adjugate is a better term
18:20
yes, although you really have to be trying to run into the adjugate these days
differentiate volume
i think of that as some clunky old thing from the 19th century whose primary point of interest was how it mildly messed up this point of terminology
@copper.hat I designed it and then someone helped me design it in a 3d modeling software program and then I sent it off to a 3d printer
@leslietownes silver
silver because it conducts heat
efficiently
well i don't know any term for it, but it looks cool :)
thanks!
and silver effciently conducts electricty
I've been doing tests
very safe though
18:37
So if $A:I\to O(3)$ is a smooth curve of orthogonal matrices, we can write $\langle A'u,Av \rangle=-\langle Au,A'v \rangle$, for every $u,v\in \mathbb{R}^3$. :0
Just use that $A^TA=I\implies A'^TA+A^TA'=0$ and do the stuff with the inner product
why do you need the curve to be smooth
Geometer laziness
smooth curves for a smooth brain
Everything is smooth
Yeah
continuous curves for a... continuous brain
hmm
18:40
C'mon I'm not even asking it to be analytical, just smooth
the whitney approximation theorem was a mistake
@copper.hat u just dissed Sheldon Axler!1!1!
@nickbros123 It isn't @copper.hat's fault that Axler is a pervert.
18:57
@nickbros123 i'm anti anti linear in the second parameter
@ModularMindset that is pretty cool
@Derso no conjugate transpose. Transpose of an operator is a different operation
In linear algebra, the transpose of a linear map between two vector spaces, defined over the same field, is an induced map between the dual spaces of the two vector spaces. The transpose or algebraic adjoint of a linear map is often used to study the original linear map. This concept is generalised by adjoint functors. == Definition == Let X # {\displaystyle X^{\#}} denote the algebraic dual space of a vector space X . {\displaystyle X.} Let...
@Jakobian I see. Names are just confusing. The property I'm really interested in is this one: math.stackexchange.com/questions/2823831/…
@Derso the dot product has a formula in terms of matrices, $x\cdot y = xy^T$
Then $Ax \cdot Ax = Ax(Ax)^T = Axx^TA^T = xx^TAA^T = x(AA^Tx)^T = x \cdot (AA^Tx)$
i would write $x \cdot y = x^T y$.
The third equality follows because $xx^T$ is a real number
I've assumed the most common convention, that vectors are column vectors
19:08
$x x^T$ is a dyad, surely?
@Jakobian It looks like you assumed the opposite
@VladimirLysikov you're right, my bad
i think @Jakobian is working from behind his screen
man messes up order of matrix multiplication, more at 11
banned from conjugating visits
20:06
See arxiv.org/pdf/math/0309036 On page 3, it says that "the action is proper if it is metrically proper for the above given pseudo-metric (i.e. for any sequence $g_n$ of elements in $G$ tending to infinity, then $d(g_n(p),p)$ tends to infinity in $\mathbb{R}_+$ for any $p \in Y$)." What does it mean for a sequence of group elements to tend or converge to infinity? My initial thought was that for any finite subset $E \subseteq G$, there exists $n_0 \in \mathbb{N}$ such that
for any $n \ge n_0$...But that would mean every infinite sequence of distinct group elements converges to infinity...I guess that's okay...I just wasn't sure if that would make the notion trivial.
if $G$ is discrete, then that is what it means, yes
of course the cool MPI seminar I'd really like to attend is going to be in one of the two timeslots where I have non-negotiable obligations :((
what does MPI stand for?
max planck institute
it's a German research institute which has a branch here and whose events people from the university can attend
Yep I know it, it's famous for its phys department as well
20:20
and so they have all these cool research-level events, lectures, seminars, etc., and there's one recently announced that I would really like to go to
about ambidexterity in chromatic homotopy theory
@Thorgott Hmm...Okay. I guess my gut instinct was right. Thanks for the confirmation.l
but alas
lol that's quite a name
@BenSteffan ah yes of course
don't blame me, I'm not making up the terminology
homotopy theorists have been at war with the algebraic geometers over whose terminology is the coolest for the last few decades and I think we might be winning :)
20:24
ben: run the abstract against one of those AI checking tools just to be sure
highly suspicious
@BenSteffan I skimmed this in under 10 seconds, truly inspiring
jokes apart, mathematicians putting a seemigly unrelated historical image before unloading 90 pages of the most complicated things ever written in human history will never cease to be comical to me
the image is seemingly unrelated only to the outsider
hard to tell if it is physics, biology or chemistry
it's a reference to the chromatic telescope conjecture, recently disproved
at the time the paper was published it was still open
20:33
(d) None of the above.
@BenSteffan I see
the term "telescope", in turn, comes from the fact that it is an incarnation of a mapping telescope, which is named because there is a canonical way to draw a sketch of what a mapping telescope is which looks a bit like a telescope :)
copy of Smale's horseshoe
"homotopy sequential colimit" just doesn't have the same ring to it
20:38
hoseqco sounds like an interesting drink
@BenSteffan Is there an anecdote about this anecdote about the image as well? I feel like chances aren't that low
maybe there is, but I don't know it
I don't know who coined the term
math terminology is always fun
the more unhinged the name the easier the concept is to remember
it's all strategy
no one wanted to repeat the disaster that are the separation axioms in point-set topology, naming-wise
I love calling things regular or normal
20:57
@BenSteffan I was careful to not mess it up, it just didn't work
if you give me two matrices to multiply I will cry
matrix multiplication isn't so bad once you understand that it just represents the universally constructed maps between biproducts
@user193319 If your space is locally compact, then we can take its one-point compactification and say that a sequence $g_n$ converges to infinity if it converges to the infinity point. In practice it means that for any compact $K\subseteq G$ there is $N$ such that $g_n\notin K$ for $n\geq N$
@Thorgott I love calling things regular and normal.
but normal implies regular!
to be honest, we should start calling T_5 mundane or something
21:01
no it doesn't
$T_4$ implies $T_3$
There are $T_0$ normal spaces which aren't $T_1$, hence not regular
for example, Sierpiński space
17 mins ago, by Ben Steffan
no one wanted to repeat the disaster that are the separation axioms in point-set topology, naming-wise
@SineoftheTime I have kakis :)
Incredible
I finished to eat one in this second
I think for most people the separation axioms are such that $T_s$ implies $T_r$ for $s > r$, and the named properties are reserved for the ones where we don't require $T_0$ or $T_1$ assumption
unless you are assuming beforehand things such as "every space is Hausdorff", this is what is most common that I see
I've never seen people use $T_4$ to mean closed sets can be separated. And if someone used that, I would cringe
I went to buy at the store but there's still unripe
@SineoftheTime yah, it's a pain
21:07
they're still yellowish
I bought them because they're on sale
actually Sierpiński space is perfectly normal $T_0$ space which isn't regular
Does anyone here have a Nintendo Switch?
@Binky I don't know about anyone here, but I'm sure someone elsewhere does.
21:24
@Binky I dont
I know they have to announce the switch 2 though, if you're interested
Okay!
@Pizza I'm interested
you can search on Google
@Pizza But are you taking tests?
no he's playing on Nintendo switch :D
21:40
Ah!
@Pizza Can you give me the nikname?
@Pizza
em, I don't have the switch :(
@SineoftheTime ?
@Binky Anyway, I have to do them in January
@Pizza Today my computer ate an apple and decided to become a microwave oven
@SineoftheTime This is why the toast continues to hide the socks!
21:51
@Binky huh?
I don't understand anything :(
Ok, enough about me… what were we talking about before?
Ah, kakis... mysterious fruit! You never know if they're sweet or if they leave that weird feeling in your mouth.
@Binky what are you studying in this period?
21:57
kakis
Interesting question
I'm currently on break
You?
I'm planning to restudy some complex analysis topics
Nice!
22:20
A positive set $E$ is one for which $\nu(F)\geq0$ for all $F\subset E$, where $\nu$ is a signed measure. Why in the proof is there a sequence $\{P_j\}$ of positive sets such that $\nu(P_j)\to m$? At that point, we don't know yet that $m$ is finite, so we can't use the characterization of the supremum that says there exists a positive set $P_\epsilon$ such that $$m-\epsilon<\nu(P_\epsilon)\leq m$$to construct a sequence of positive sets $P_j$ such that $\nu(P_j)$ converges to $m$.
22:30
This exercise asks me whether the level set $L_0 = \{F = 0\}, F = x-y-e^{x+y}$ implicitly defines a function $y = g(x)$ on all of $\mathbb{R}$. I was wondering how to approach this? I know Dini's theorem (e.g. the Implicit FT) works locally on an open interval $I = ]x_0-a, x_0+a[, a>0$ when $F_y(x_0,y_0) \ne 0$, but in this case $F_y(x,y) \ne 0 \forall (x,y) \in \mathbb{R}^2$. Thus, I can extend $I \longrightarrow \mathbb{R}$?
@copper.hat I have the 2nd edition, 6th or later printings (here's the errata for that particular version by the way, useful at times). In one theorem in the earlier printings of the 2nd edition, I believe there was a bit of a flaw, which the 6th or later printings don't have.
@psie also, here's a typo. It should say $\nu$ does not assume the value $\infty$ in the first line of the proof.
@psie given the definition of m, if m = +oo, then for any n, there is some positive E_n with nu(E_n) >= n, and any sequence (E_n) thus chosen would have the property that nu(E_n) converges to +oo
22:46
ah right, ok, thanks leslie
it seems like you could even make the sequence E_n increasing (having chosen E_j as above, for any n the set F_n := E_1 union ... union E_n would also be positive by a short argument, and F_1 subset F_2 subset ... with nu(F_n) >= nu(E_n) >= n
im not sure if you would want that, but, you could easily arrange a kind of convergence of the sets themselves in that way and not just have them maybe arbitrarily skipping around your measure space
and sometimes in other arguments like this, you will see authors like folland omit this kind of simplification (hopefully expressly, by saying they are doing it, but not always)
\begin{align}
\begin{bmatrix}
u_x & u_y & u_z & 0 \\
v_x & v_y & v_z & 0 \\
n_x & n_y & n_z & 0\\
0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & 0 & -x_0 \\
0 & 1 & 0 & -y_0 \\
0 & 0 & 1 & -z_0\\
0 & 0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}
=
\begin{bmatrix}
u \\
v \\
n \\
w
\end{bmatrix}
\end{align}
Hello, I have this transformation. Without the homogeneous coordinate I understand that this is a change of basis vectors and that the rows are the basis vectors. If they are chosen to be orthonormal then the new coordinate system is also orthonormal etc. My question is what happens with the homogeneous coordinate. The combined transformation gives basis vectors that are not unit vectors. I want to know what are the basis vectors of the combined transformation in the previous coordinate system.
maybe I can restrict myself to $F(x_0,y), x_0 \in \mathbb{R} \text{ fixed}$. Now I can take, for instance, $x_0 =0$ and I have $F(x,y)|_{\{x=0\}} = \phi(y) = -y-e^y$
Maybe as a first more specific question, is it true that
$$\hat{u}=\langle 1,0,0,0 \rangle $$
$$\hat{v}=\langle 0,1,0,0 \rangle $$
$$\hat{n}=\langle 0,0,1,0 \rangle $$
$$\hat{w}=\langle 0,0,0,1 \rangle $$
in the new coordinate system?
23:04
$O(3)$ acts as isometries on $\mathbb{S}^2$. What about $\mathbb{H}^2$? In the halfplane model they're essentially Möbius transformations, but I wished I could work with the hyperboloid (embedded in the Lorentzian space) model instead.
Do we have a "nice" representation for isometries in this model?
wait maybe I can work with a generic $x_0$ instead of setting it to 0. Now, if $\phi(y) = 0$ for some value $y_0 \in \mathbb{R}$. But this means $F(x_0,y_0)=0$ which is good because the point $(x_0,y_0)$ works fine for the local version of Dini's theorem. Now since $x_0$ is arbitrary I can span the whole real line
So all I need to verify is that $\exists y_0 \in \mathbb{R} : \phi(y_0) = 0$, namely that $0 \in Im(\phi)$...
23:42
Hi can someone help me with how the matrix of a passie rotation looks like for this case:
So I want the x' axis to point in the direction of R
Becuase I don't know how the passive will look like I thought of the following
I am doing an active rotation around the z axis, so that I can bring R in the x axis
$\vec R(Rcos\phi,Rsin\phi,0)^T$ and it goes to $\vec R'=(R,0,0)$
So the matrix will look like: $[[cos\phi,sin\phi,0],[-sin\phi,cos\phi,0],[0,0,1]]$
Is this a valid matrix for active rotation?
If yes, then the passive should be the inverse of this, i suppose
if anyone is willing to have a look at my question, it would help me out a lot:

https://math.stackexchange.com/questions/4997704/flux-of-constant-vector-field-across-surface
@ShaVuklia uh, did they make you repeat undergrad or something? :D non jokingly those seem like "bad" boundaries for that problem, all but guaranteed to make the problem worse
even if it somehow is 16
lol yea i see one mistake now
i didn't normalise
i'm ta-ing a vector calc course, but it's been a while
i went through this about 100 times when i had to teach multivariable calculus twice during a postdoc
vector calc is tough because a lot of textbooks are drafted against a background reality that even "easy to work out by hand" problems might take more than "the usual amount" of computation
o wait, there is no mistake regarding normalisation
23:54
so you can't use "uh oh, this is taking more than the usual amount of computation" as a proxy for the existence of a simpler approach
found it :)
i multiplied by the norm of that cross product i took, but that's not correct
without it, it's very doable

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