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01:00
\wedge
It usually is, @Antonios.
no still nothing
$\wedge$
You need dollar signs, @JoeShmo.
be back in a little bit. walking home
01:00
$\wedge$
still nothing
ill figure it out later.
so the question is
can i consider ∂c as a (k-1)-cube?
but then funny stuff happens
@Ted how do you feel about korean food
@TedShifrin its basically this mapping
$(x^2,x^3,x^5 ,\dots) \to (x^{(2-|3-2|},x^{(3-|5-3|},\dots)$
Eric, like it a lot, but probably like Thai more.
No, @JoeShmo, $\partial c$ is a linear combination of $6$ $2$-cubes.
Also I've tried to further edit the question
ah! ok. i like that
let me snap a picture
01:04
i just went to this korean place in north chicago that blew my mind, had this bi bim bap with chicken hearts that i think fundamentally changed me as a person @Ted
if youre ever in chicago i highly recommend parachute
Ah, well, OK. ;)
@MoreAnonymous: I don't see that you've told me how to map a general sequence of polynomials. What would I do with $(x,x^4, x^2, x^2, x^3,\dots)$?
@TedShifrin Can I ask why the origin $(0,0,0)$ doesn't satisfy the equation $x+y+2z=2$ ?
@Leyla: Because that would mean $0=2$.
Yes, but that point is also in the plane
No, it's the other corner of the tetrahedron. The plane is just the TOP of the region.
01:07
@TedShifrin
https://imgur.com/a/NowRu
https://imgur.com/a/Xy45Y
but then i just get 0
and the rest of them are kind of symmetric
It shows up sideways and I can't read it.
tralala
let me try again
how's this -- imgur.com/a/GHtMy
So, the $1/4$ is correct. I find it easier to compute $c^*\alpha$ and then restrict to faces.
Three of the faces give 0, but the others don't necessarily.
$(x,x^4, x^2, x^2, x^3,\dots)$

$a_2 = 4=3+1 \implies a_1 = 1-3 =-2 $

$a_2 = 4=3+1 \implies a_1 = 1-3 =-2 $

$a_4 = 2+2= \implies a_1 = 3-1 =2 $
sorry was still typing accidently pressed enter
ignore that
Well, I don't really want to think about it, @MoreAnonymous.
01:14
Theorem: An abelian group $M$ admits a $\Bbb{Z}_n$-module structure if and only if $M$ has exponent $n$ as a group...Question: Is the theorem referring to AN exponent of $M$ or THE exponent? I ask because I don't see how to prove that $n$ is THE exponent, only that it is an exponent.
ohk ... :(
@TedShifrin yeah c_(i,0) give me 0 for all i
Right, @JoeShmo. But not for $c_{(i,1)}$.
@user193319: If you have a $\Bbb Z_4$-module structure, don't you also have a $\Bbb Z_2$-module structure?
Hmm, maybe that's not right.
01:17
did i screw up my integral over c_(1,1) or am i doing everything right?
MoreAnonymous: Decide if it's really linear.
I didn't see you do that integral, @JoeShmo.
you gotta scroll down
@TedShifrin I'm not sure...I just started teaching myself module theory.
I only see $c_{1,0}$, @JoeShmo.
Does that not only work the other way around?
01:19
to the right of it is c1,1
its not properly punctuated
I don't see it.
I think exponent $n$ means x^n =1 for all x
so if k|n if something has exponent k it should also have exponent n
Oh, it was there. You're integrating a $2$-form, silly!
01:21
you can call me an idiot.. i dont mind :)
@MoreAnonymous: It seems to me your vector space is infinite sequences of polynomials in $x$.
this whole mathematics experiment is an attempt to graduate idiocy.. :0)
You turned it into a $3$-form, @JoeShmo. It should be $t_2t_3 dt_3\wedge dt_2$ and then we have to decide sign.
so i'm integrating 2-forms... so.....?
01:22
If you're gonna try to integrate a $3$-form over a $2$-chain, you'll always get $0$.
@PVAL-inactive So the theorem is referring to an exponent? I was able to show that $x^n =1$ for all $x \in M$, but I was having trouble showing that $n$ was the smallest such value. Hence my reason for wondering if it's referring to an exponent and not the exponent.
right
If you simplify your expression, you get $0$. You're wedging the same guys with themselves.
But it should only be a $2$-form, as I wrote.
You need practice with basics.
right, which is why i decided not to proceed and ask instead
because i'll be getting 0's all over the place
@user193319 the exponent of a group is the least common multiple of the orders of the elements of the group, at least in a finite group I think
01:25
So if you set $x=t_2t_3$, $y=t_1t_3$, $z=t_1t_2$, what is the pullback of $x dy\wedge dz$ when you set $t_1=1$?
when i set t_1 to what?
oh
You were still doing $d\alpha$.
@ÍgjøgnumMeg Okay. So the theorem is referring to $n$ as THE exponent. How do I show this? As I said, I could only show that $nx=0$ for all $x \in M$ (using additive notation).
i was doing alpha
one sec
Oh, actually, I have no idea what you were doing.
01:27
im trying to follow the example he gave in class, and im starting to realize he didn't know what he was talking about
i computed lhs, and then i needed an example for rhs
My advice was the right advice. Compute $c^*\alpha$ once and for all. Then restrict to the various faces.
okay. how do i restrict? by setting t_i=0,1?
I don't do chains in my course, so I can't tell you to go watch a lecture. But I did a dozen examples with Stokes's Theorem in various forms.
Yup.
@TedShifrin

One last question. In this post: https://math.stackexchange.com/a/70265/94106

Matt E's answer says:
"We may write $f(x)=g(x^{p^d}),f(x)=g(x^{p^d})$, where $g$ is irreducible and separable and $d\geq 0$.

Why is this true?
lets try that :)
i truly can't tell you enough how much i can't stand that class
01:30
That doesn't help, @JoeShmo :)
I like it when errors in papers make their way into books written by other people.
im putting in tons of work, and there's very little output. and it's not just me. the whole class is full of deer-in-the-headlights looks
no, i guess not. im venting.
Makes me confident that people actually check computations when ripping them off.
and the best part is that this stuff shouldn't take longer than 15 minutes
@user2154420: Better to ask @Mathein ... I haven't thought about this stuff in too long.
@JoeShmo: You need some simple practice, yes.
01:33
1
Q: prove $\int_{-C}\vec{F}d\vec{R}=-\int_{C} \vec{F}d\vec{R}$

Maneesh Narayanan We need to prove $\int_{-C}\vec{F}d\vec{R}=-\int_{C} \vec{F}d\vec{R}$ Usual attempt in Textbook: Suppose $C:\vec{R(t)},a\leq t\leq b$. Then $-C:\vec{R(-t)},-b\leq t\leq -a$. I am able to verify the result. Can I define as $-C:\vec{R(a+b-t)},a\leq t\leq b?$ $\int_{-C}\vec{F}d\vec{R}= \int_{a...

Please help me to find my mistake.
@TedShifrin yes, thats on me.
@Maneesh: You dropped a minus sign. There's one in the $dR/d\tau$ and another one with $d\tau = -dt$.
anyway, are you going anywhere interesting anytime soon ted? im making small talk while computing c*(alpha)
stay tuned
@TedShifrin in order to compensate the minus, I have reversed the limit of the integration. Am I wrong here?.
I wrote an answer. You have to reverse the limits when you change variables, because $\tau$ goes from $b$ to $a$. But there are the two compensating negative signs (you lost one of them).
01:40
@user2154420 are you familiar with the fact that if $f$ is irreducible and not separable, then $f'=0$?
@TedShifrin ... it is linear! For your sequence:
$$(x,x^4, x^2, x^2, x^3,\dots)$$

We write this as:
$$ K_0 =sx + s^2 x^4 + s^3 x^2 + s^4 x^3 + \dots \equiv A_0$$

Multiplying $s$

$$ s K_0 = 0+s^2 x + s^3 x^4 + s^4 x^2 \dots $$
Subtracting the equations:

$$ K_0(s-1) = -sx + s^2x(1-x^3) + s^3 x^2 (x^2 -1) + s^4 x^2(1-x)$$

Hence using $x^\lambda + y^\lambda = 2$

$$ K_0(s-1) = -sx + s^2x(y^3-1) + s^3 x^2 (1-y^2) + s^4 x^2(y-1)$$

Now using $xy=1$

$$ K_0(s-1) + sx + s^2x + s^3 +s^4 x^2 = s^2 x^{-2} + s^3 x^2 + s^4 x \equiv A_1$$
for t_1 = 1, is dt_1 = 0?
@MatheinBoulomenos Yes
Sure, @JoeShmo.
Officially, if you have $\iota(t_2,t_3) = (1,t_2,t_3)$, then $\iota^*(dx) = d(\iota^*x) = d(1) = 0$, @JoeShmo.
AAAH!!! an additional pullback :)
hence "degenerate"
01:44
@MoreAnonymous: It isn't linear point by point. You have to think about what happens if you add things component by component. More basically, if you know what $K$ does to the vector with $x^j$ in the $k$th entry and all other $0$, can you reconstruct what happens to the whole infinite sequence? And does it work with sums component by component?
@user2154420 okay, lets think about what this means. Write $f= \sum_{k=0}^n a_kx^k$, then $f'=\sum_{k=1}^n ka_kx^{k-1}$. If we look at this formula closely, then we see that we only way that $f'=0$ is that the only coefficients $a_k$ that are not $0$ are (possibly) those with $p \mid k$, because if $p \not \mid k$, then $k \neq 0$, so $ka_k \neq 0$ if $a_k \neq 0$
@JoeShmo: I definitely said that at some point in my lectures, but then you just start doing it automatically.
@MatheinBoulomenos Right...
@Mathein: Thanks for coming to help.
@TedShifrin chain rule and change of variable. right?
01:45
Right.
Thank you very much.
i strategically missed the computational stoke's stuff, which inadvertently was a mistake.
Sure. ;)
Yeah, computational expertise really does lead to better understanding of theory.
So we can write $f=\sum_{k=0}^{n'} a_k x^{pk}$, where $n'=n/p$ and thus if we define $h=\sum_{k=0}^{n'} a_k x^k$, then $h(x^p)=f(x)$
@MatheinBoulomenos so we would have to have $f$ be some power of $p^k$
01:47
@user2154420 right, exactly!
We just iterate this process until we hit a separable polynomial
as long as the polynomial is not seraparble, the degree is divided by $p$ each time, so this terminates
@TedShifrin do you know a better way to do this, my answer and the posted one are rather unsatisfying
1
A: Prove that T is diagonalizable if and only if the minimal polynomial of T has no repeated roots.

mechanodroidClearly $T$ is diagonalizable if and only if we can decompose $V$ into a direct sum of eigenspaces $$V = \ker (T-\lambda_1I) \dot+ \ker(T - \lambda_2 I) \dot+ \cdots \dot+\ker(T - \lambda_k I)$$ since we can then take a basis of the form $$(\text{basis for }T-\lambda_1I, \text{basis for }T-\lamb...

I don't want to read all that.
lol
fair enough nvm
@Faust I personally think that all this stuff becomes clearer if you think about it in terms of modules (but others may argue that's overkill or just a reformulation)
But Faust doesn't know that, Mathein.
01:49
Oh okay
faust doesnt know anything
nevermind, then
@TedShifrin I don't think one can do that unless it there is an obvious pattern (, if you know what $K$ does to the vector with $x^j$ in the $k$th entry and all other $0$, can you reconstruct what happens to the whole infinite sequence) ... Also yes it works with sums component by component
I didn't say that!
? no i did lol
01:50
If those answers are yes, @MoreAnonymous, then — yes — you have constructed a linear map from the space of sequences of polynomials to itself.
@MatheinBoulomenos Wait, does the degree of $f$ have to divide $p$, or just the degrees of some of the terms of $f$?
Faust: Think about one single eigenvalue. If the minimal polynomial isn't linear, then it won't be diagonalizable. That is clear?
@user193319 I think you can just define the $\Bbb Z/(n)$-action on $M$ as $km = \bar{k}m$ with $\bar{k}$ the remainder when dividing $k$ by $n$, since you have $k = na + \bar{k}$, so if you take $km = (na + \bar{k})m$ for all $m \in M$ then the exponent of $M$ ensures that the $na$ term dies, and then you can do a bunch of stuff to check this gives you a $\Bbb Z/(n)$-module
@TedShifrin ... Then why can't I find the matrix representation of this polynomial ... Also how do I improve m question ... since it's been put on hold
@Faust: The key thing is that diagonalizable is equivalent to basis of eigenvectors.
01:52
@user2154420 if $f$ is irreducible and inseparable, then $p$ has to divide the degree of $f$, since the leading coefficient is non-zero by definition, but we have shown that the only possible terms with nonzero coefficient are those with $p \mid k$
@TedShifrin that how i proved it
its too long
i want a better proof
@MoreAnonymous: Well, I made you say a lot of stuff that I couldn't figure out in all that question. You need to give domain and range. You need to say how the damn map is defined!
@MatheinBoulomenos Okay, this is clear now. Thanks!
@Faust: Unless you know fancier stuff, that's the proof.
@user2154420 you're welcome
01:53
:'(
@user193319 at least for one of the directions :P
I love the partial fraction argument the guy gave. I put an exercise like that in the second edition of my book. It's basically doing the module stuff without the fancy language.
@TedShifrin hrm. So, here's something not encouraging
its like a page in latex
when i learn about modules?
How should I know?
01:54
one of the ideas I had to explore what was going on was to generate a lot of random U,V matrices
Look at Artin's book.
@TedShifrin if you don't know how am i supposed to know?
I don't know the curriculum at your school, Faust. We didn't do modules in our undergraduate algebra at UGA.
i dont think we have a 4th year linear algebra course at my uni
MIT does it in its "advanced" algebra track, not in its easier track.
01:55
From there, I did 3D scatter plots of subsets of the four matrix elements e.g. $(m_{11},m_{12},m_{22})$
It's a standard topic in graduate algebra.
:(
how come all the intresting stuff is for graduates
basic modules over a PID is first year stuff here
And no matter which of the four coordinates I omit, I find that the outputs cover all of the [-1,1]^3 cube.
Ha ha @ first year stuff.
01:56
Henlo
Well, that's interesting, Semiclassic.
@MatheinBoulomenos Are you able to confirm what I just wrote above? Lol
@Daminark henlo u stinky thonktato
go thonk ugly
How does every German school teach math at an American graduate level to every first year undergraduate
01:57
@TedShifrin yes, 1/4
German high school isn't that amazing
So if the image is a proper subset of [-1,1]^4, it's one whose projections are all [-1,1]^3.
@0celo7 you don't have to do 40 other subjects in European universities, as I think is the case in US unis
Universities are very specialized in Europe, 0celo. No general studies.
Suppose $u(s)$ be the unit tangent vector to the curve $C$ at $P$. How |u'(s)| be the curvature($\kappa$)? suppose $\vec{b}=u(s)\times \frac{1}{\kappa}u'(s)$. what is torsion?How torsion($\tau(s)$) is |$\vec{b'(s)$|? I know the fact that curvature is the measure of how curvy is the curve. please help me.
01:57
I don't know we have a lot of people failing and dropping in their first year
I don't believe that gen eds have anything to do with it
That's not impossible, of course, but it means one doesn't get much insight into the 'shape' of the image from those projections.
You know you want to be a mathematician if you're going Mathein's route.
How do no gen eds make math easier?
You go to different universities.
01:58
it lets you do a lot more math
I couldn't have been a math and French major in Europe ... not easily.
Presumably you're doing all 4 math classes from day 1
They're not doing more math, they're doing harder math immediately
4 math classes is a bit rough in the first year
01:58
Well, look at where Akiva is starting in math next year!!
@Daminark not every school has max 4 classes dawg
dont you europeans specialize a bit earlier anyway
Well I'm guessing the module over PID stuff you were talking about earlier isn't literally day 1, but like, they have one full year calculus, one full year linear algebra, etc
In the UK we pretty much specialise at 16 lol
yeah, we have one full year linear algebra
module stuff is the second semester
@TedShifrin ... The domain and range are powers of $x$. The map is defined algorithmic-ally as I have illustrated
02:00
I took the Artin algebra course and learned modules in second semester of my second year of college.
@More: No, not powers. Polynomials.
@Eric wait really? Maybe semester schools do more?
max 4 is a uchicagoism
shrug
Nah, Demonark. We actually covered a bit less under semesters at UGA than we had under quarters.
I'm taking 6-ish classes
02:01
Oh I just meant more classes
Not sure about how much content is covered
i have a friend at a state school who takes like 6-7
Shit
Well, more classes total? Nah.
6-7 is absurd.
there's no limit to taking classes. A friend of mine took 7 classes before while also TAing
Suppose $u(s)$ be the unit tangent vector to the curve $C$ at $P$. How |u'(s)| be the curvature($\kappa$)? suppose $\vec{b}=u(s)\times \frac{1}{\kappa}u'(s)$. what is torsion?How torsion($\tau(s)$) is |$\vec{b'(s)}$|? I know the fact that curvature is the measure of how curvy is the curve. please help me. sorry for posting again. there was an error in the tex code. I am not able to edit it.
02:02
4 is usual, 5 is tough for the average student if you're taking hard maths.
this is an absurd friend to be fair
ah ... I see the domain and range are polynomials with coefficients of powers of $x$ ?
@TedShifrin
I don't think I've had a semester under 18 hours
Infinite sequences of those, @MoreAnonymous.
Remember, each one goes with a power of $s$.
Classes are free past 12 hours, so might as well take more
02:03
If you don't start flunking, sure.
but they can in turn be mapped to a row vector with an $x^\lambda$ entry right?
I dunno if I want to be taking more than 4 classes anyway. Maybe in certain quarters that'd work out alright but for the most part that'd risk being overwhelming
the only ppl i know who've taken 5 are dying @Daminark
I don't know what that means, @MoreAnonymous.
so dont
02:03
Not necessarily throughout the quarter but like, midterm week
20 hour weeks are the norm here I think? Perhaps the system is slightly different
Seems like 2-3 classes is good. I've had 4 this semester and last and not much enjoyed that.
I kind of took 5 this quarter and am dying
Also there's no way in hell I could fork the money for that
Is this correct?
$$ \underbrace{(x^a_1, x^a_2, \dots)}_{\text{domain}} \to \underbrace{(x^b_1, x^b_2, \dots)}_{\text{range}} $$
@TedShifrin
02:04
The average student here takes 3-4 classes per semester
If I finish this thesis my workload gets cut in half. Then I'd enjoy life
@MoreAnonymous: Remember, each term can be an entire polynomial. Otherwise, addition won't work at all.
ah ... so its not linear!
makes sense now!
such a silly mistake :P
Right now, and last winter, I had 3 classes, since fall this year and last year I got a bit burnt out, and this year there wasn't much that I was interested in. I would've taken linguistics but it conflicted with analysis
linguistics is really cool
02:05
Plus, you need different polynomials in the different slots ... you shouldn't have subscripts on $x_1$, $x_2$, etc. You should have polynomials $f_i(x)$.
@Daminark Cool that you can take electives outside of mathematics
quantum conflicts with grad complex and i wanna die @Daminark
I almost went into languages/linguistics.
@EricSilva: Is the grad complex a top notch teacher? You can save some stuff for grad school, seriously.
I've been wanting to take more linguistics classes (took only an intro class so far, and well all my ancient greek classes, but they were more on philology), but there's always overlap
Why not explore non-math stuff while you have the chance?
02:07
i mean ive had her before and i like her
Or no actually it conflicted with algebra
But, if you're into quantum, do it.
i just dont know which id enjoy more
id probably enjoy complex more just cause it's math
I'm just saying you'll probably take grad complex in grad school anyway.
@ÍgjøgnumMeg ah sorry, I forgot, yeah that's correct
02:08
@EricSilva who is teaching QM
@MatheinBoulomenos No problem, cheers :)
Lesson for myself: Convex sets may be easier to visualize than non-convex sets, but 4D sets are still harder to visualize than 3D ones.
Turns out it conflicted with both. And next quarter there are again two sections of intro to linguistics. One conflicts with algebra, the other with analysis
sigh
(captain obvious)
02:08
LOL, Semiclassic.
OK ... I think I'm liberated to leave now.
chat.stackexchange.com/transcript/message/43206218#43206218 can you please tell me at least. which text is good other than kreyzig advanced engineering mathematics?
See you @Ted
bye
See you
02:09
bye
@Semiclassical Ok time to do this damn fluids midterm
godspeed man
@Semiclassical I messed up and didn't bring my fluids 1 notes to the exam
so I had to do a bunch of vector calculus because I didn't remember stuff
turns out I can't into determinants
02:10
@Eric lol wait so two of your classes conflict and another is gone, yikes
yeah no my plans are in utter ruin
No core left aside from sosc?
the only saving grace is Neves swoopin in to give me interesting shit to do
i have to finish civ
but that'll be fall and winter next year
@Semiclassical it also confused me because you have to do perturbation theory on a nondynamical equation
I realized too late that it couples to the heat equation and that somehow gives it dynamics
Well, hopefully Neves comes through
02:17
yeah he's notoriously bad at following through with things he says he's going to do unfortunately
Oh really? Yikes
Also I didn't realize he said he was gonna do anything lmao, I think you vaguely mentioned something you were interested but I can't remember if that was a hope or if it was on firm grounding
i mean he offered me a job as his RA officially
he just needs to talk to people to get my payroll going
Oh shit nice
he's disorganized as hell but he's honestly such a nice dude that i cant mind
03:04
the last exercise im working on here is deducing Green, Gauss, Stoke and the gradient formulas from Stoke's theorem on chains.
it's pretty neat actually, reinterpreting as forms, and hitting it with Stoke's theorem.
04:02
anyone know peano's axioms? i want to see if i correctly proved commutative addition
@user537069 There exists a natural number, called 1.
Every natural number a has a successor, called a'.
1 is not the successor of any natural number.
@nitsua60 I wrote an answer on MSE but I feel like it got kind of hijacked with a non-answer
2
Q: Prove addition is commutative using axioms, definitions, and induction

user537069I wanted to try to prove the commutative property of addition before reading too much about it and "spoiling" things for myself. So I am curious how close I got. First, some axioms (statements/relationships we take to be true): $$\begin{align} 0 &\in \mathbb{N} \tag{$0$ is a natural number} \\ ...

For two natural numbers a and b, a=b iff a'=b'.
And then the PMI (suitably stated in terms of successorship) is the fifth.
@user537069 I don't see an answer you wrote.
I just linked it
oh, sorry, I meant I wrote a question
Oh, ok.
It would seem the high-rep user is telling you that checking your own work with well-known references is a better use of time than asking someone else to check your work. And, by extension, telling future readers that checking one's own work is a better way to come to a deep understanding of mathematics than by asking stackizens to check their work.
I'm not around math.se a lot (as you can see from my profile) so I don't really have an educated opinion on whether that's the sort of frame-challenging answer that the stack likes.
I will say that in the question post it seems unnecessary to me to state symmetry, reflexivity, transitivity. Once you use the "=" sign unless you define it otherwise every reader will assume it's an equivalence relation, which means it has those properties. But that's just my style/preference--take that advice as you will from someone who's answered exactly one question around here.
04:39
(You start at 0 where I start at 1, but that's no big deal.)
In mathematics, triality is a relationship among three vector spaces, analogous to the duality relation between dual vector spaces. Most commonly, it describes those special features of the Dynkin diagram D4 and the associated Lie group Spin(8), the double cover of 8-dimensional rotation group SO(8), arising because the group has an outer automorphism of order three. There is a geometrical version of triality, analogous to duality in projective geometry. Of all simple Lie groups, Spin(8) has the most symmetrical Dynkin diagram, D4. The diagram has four nodes with one node located at the center...
This is insane
04:57
10 hours ago, by Rick
2
Q: Sum involving binomial coefficients.

user326254 Prove that $${^{404}\mathrm C_4}-{^4\mathrm C_1}\cdot{^{303}\mathrm C_4}+{^4\mathrm C_2}\cdot{^{202}\mathrm C_4}-{^4\mathrm C_3}\cdot{^{101}\mathrm C_4} =(101)^4$$ I tried writing $101=102-1$, but couldn't move forward. Sorry for the inconvenience, I am new here and couldn't type it. '.' Me...

Can someone prove this algebraically?
05:24
can any one explain how they simplified these steps i.imgur.com/rJwxK2A.png i just don't know what they did to get it to that result
@WDUK Which step are you asking about?
the second = line becoming the third = line
They used the induction hypothesis that the formula is what they wrote in
ohh so they substituted k for k+1
05:31
@WDUK No, they used the assumption that the formula holds for $k$
oh i see it now
because using K+1 is equal to p(k) + (k+1)^3
still though, it really should communicate each step better when its trying to explain it for the first time lol
@WDUK those steps look very self-explanatory to me
The only one that is not a simple algebraic manipulation is labelled with what was done
hmm im not seeing it very easily at the moment
ok i figured out the steps now
took me enough tries !
05:45
I'm writing a book on solutions to NT olympiad problems, would anyone be willing to fix errors, see if there's something wrong with my solutions, or even willing to just take a look at the book? It'll be released ~13 months from now
06:05
@Semiclassical fuark my notes from semester 1 seem to be gone
that's not good
@Semiclassical I found half of them. I remember this. I was working on the final and only wanted to carry around half of the notes
so I made two piles
stupid fucking idea in hindsight
@Semiclassical found them
I really need to clean up this place :P
oh
I'm stupid
I really just forgot how curls work during the exam
or...not. Hmm.
06:50
https://math.stackexchange.com/a/2677393/322355
You're welcome!
07:03
@TheTestosteroneFanatic Thank you!
07:25
@Rick np, your problem was kinda the only thing I have genuinely enjoyed the past three days
I have been grading students' lab assignments this whole weekend
 
3 hours later…
09:56
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