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12:22 AM
heya @EricSilva: Welcome back to the land of cold and ice.
 
it is so cold
 
I'm happy to be in SD ... where half the mathematicians will be in a week
 
what's happening? is there a big conference or smth?
 
Joint Math Meetings ... big January meeting
 
ohhhh shit jmm
Neves is giving a talk at that i think
 
12:27 AM
guess we're still on holiday in here
 
there's cool stuff it seems
i have class here though so rip
 
Hi, what is the cotrapositive of this exactly: If $x\neq 0$ and if $x^2$ is rational, then

>>$(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$
 
I never really enjoyed the huge meetings, although I heard a handful of good plenary lectures. Surprisingly many were rather bad.
What's the negation of "for all $k$, something happens"? @Leyla
 
being a good mathematician is independent of being a good orator and expositor i suppose
 
yup, Eric
Babai was one of the great ones I heard
 
12:34 AM
lmao André's talk is called "Wow, so many minimal surfaces! "
Babai is a fantastic expositor
 
Witten was about the worst.
 
like ive sat in on a few of his lectures and he made me excited about things i never get excited about
huh really? that's interesting
 
@EricSilva he may give that exact talk in minimal surfaces
 
tbh true
 
Also hello
 
12:36 AM
hello, Demonark
welcome back to the arctic north
 
huzzah for subzero temps
 
Tru. Though I was sick in Texas and am in much better shape now so that's a plus?
 
@Daminark so this Calegari class is either gonna be some index theory for elliptic boiz and topology around that, rational homotopy theory, or Thurston stuff
im p hype
 
did you get your dental stuff arranged, Demonark?
 
@Semiclassical when i got here it was -13 with windchill and i just had a light sweater :'(
 
12:38 AM
data point: apparently the last week in December was the coldest ever (132 years) at MSP airport
 
I got permanent fillings for the two front teeth and will get crowns when I go back in Texas for the summer
 
@TedShifrin there exists k , st that something doesnt happen. @TedShifrin
 
I think taking only a light sweater was unfortunate planning, Eric.
 
@Eric sounds dank
 
@Semiclassical so wut ur telling me is that global warming is fake
 
12:38 AM
Sorry to interrupt your chatting btw
 
OK, @Leyla. And if the something is an "and" statement, what does it mean for it not to happen?
 
@EricSilva oof. I've been making sure that I have a good coat/cap/gloves
 
@TedShifrin yeah for real, it was bad
 
-23 wind chill is not something you want to mess with
 
I had to run between airport and cab and then cab to apartment
literally sprint
 
12:39 AM
And hope there was no waiting, Eric?
 
i didnt have to wait thankfully
was kind of far from the cabs tho so that sucked
it's all good i didnt lose any fingers or anything
 
I've been making sure to avoid being outside much
 
Lol my RH emailed us the day before like, yo it was -30 windchill today, please don't die
 
I dont think I get your question @TedShifrin
 
only half your brain cells, Eric ... luckily, you have plenty to spare.
@Leyla: Your original proposition was "for all $k$ ... $P(k)$ and $Q(k)$ ... So negating is ... ?
 
12:41 AM
pls dont die k thanks
 
If $x\neq 0$ and if $x^2$ is rational, then $(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$
 
So how're you negating the conclusion, @Leyla?
heya DogAteMy!
 
Hey up
School tomorrow closed
 
Yup, pretty drastic temperatures and storms
More time for math? :P
 
More time for essay writing, more like
I have two due really soon and I'm not very much prepared
 
12:45 AM
Oh, you're still applying?
I figured you were happy to quit with Yale.
 
@TedShifrin No no
 
@Daminark i looked at the description of the talk and i straight up heard this talk from him before lmao
 
I'm guessing it's essays for school
 
For History and English class
 
Ohhh ...
Already? You're just back after holidays.
I guess they were assigned a month ago.
 
12:46 AM
Yep
 
And somebody procrastinated. I can't imagine who.
 
@EricSilva kek
 
Hi folks; I took a look at Numberphile's Squared Squares video
 
There exists $k \in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)=0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}.$ @TedShifrin
 
12:49 AM
Is it possible to generalize the question "What is the minimum number of distinct squares needed to create a larger square", be generalized to "What is the minimum number of distinct regular n-gons needed to create a larger regular n-gon?"
 
Nope, not quite, @Leyla. If I say "Today is Wednesday and it's summer," how do you show me I'm wrong?
 
@Daminark i have 3 courses back to back tomorrow from 9:30 to 2 and im not hype about it tbh
 
I'm just wondering if it's a "valid" mathematical question
 
Oh that doesn't sound fun
 
@DarkRunner I think you can't make another regular n-gon out of smaller n-gons except for the cases of n=3,4 (triangle and square)
 
12:51 AM
I ignore Numberphile.
 
I've got a class MWF 10:30, 11:30, and then TR 11:00, which isn't too bad
 
@Semiclassical is it due to diagonals, or,...?
 
I think the angles don't work out
 
But it would be regular, so that should fix that issue
 
i feel like if you could do this you could tesselate a plane with n-gons and you can't do that for almost all n
 
12:52 AM
Is it because "and statement"?
 
Right, @Leyla. How does an "and" statement fail?
 
Hey @ted!
 
@EricSilva: Indeed. That's not too hard. $n=3, 4, 6$ is it.
 
@Daminark my 9:30 am on MWF is my only class on MWF which is like pretty nice for me
 
heya @Cookie.
 
12:53 AM
yeah that's just some basic angle solving
 
I cant see now @TedShifrin
 
I like to do it with a bit of lattice-work (pun intended), Eric.
 
lol
 
Well, that's why I gave you the English sentence, @Leyla. Is my sentence up there true or false?
 
@TedShifrin Sorry, but I don't understand why?
 
12:54 AM
and the n=6 case doesn't help here because when you put them together you don't get flat sides
 
@DarkRunner you should try to see why as an exercise
 
It takes a proof, @DarkRunner. There are elementary arguments, I'm sure.
 
Its false @TedShifrin
 
I'm not paying any attention to the subject being discussed.
@Leyla. But half of it was true, right?
 
yes
 
12:55 AM
I see;
 
So I'm absolutely stuck on that integral @ted. After substitution, and taking the derivative wrt $y$, I'm left with some form of $- \int_{-\infty}^{\infty} \frac{ue^{u}e^{yu}}{(1+e^{2u})(1+e^{yu})^{2}} du$ Could you give me another hint?
 
@TedShifrin I hear a pun
 
So to negate an "and" statement, we need $P$ is false OR $Q$ is false.
 
sends @Daminark to the dungeon
or the pungeon i guess
now i belong there too
 
I agree, @Cookie, except isn't it $e^u$ instead of $e^{2u}$ in the numerator?
 
12:57 AM
Yeah I just saw!
 
note that the angles at the corner of a regular n-gon are 180-360/n
 
There exists $k \in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)=0\text{ or }\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}?$ @TedShifrin
 
OK ... Great. Split off the $\int_{-\infty}^0$ and make a substitution to rewrite it as an integral $\int_0^\infty$.
 
you mean this?
 
Yup, @Leyla.
This looks like a horrendous thing to be doing contrapositive of, but ... that's the negation of the conclusion. The negation of the hypothesis is easy.
 
12:58 AM
[REDACTED]
hmm
 
nooo
 
i don't like this argument
 
@Semi don't give it away
 
So, we are done? Okay thanks a lot@TedShifrin
 
(redacted is a fun word)
 
1:00 AM
@Leyla: Assuming you remember that the contrapositive of $P\implies Q$ is $\text{not }Q \implies \text{not }P$. :)
 
yeap, sure :)
 
@Semi [REDACTED] is a fun word
 
3
Q: Deriving the Center of Mass of A semi-circular disk with Cylindrical Coordinates

Better WorldProblem: Derive the Center of Mass of a semi-circular disk of mass $M$ and radius $R$. $$$$ My attempt: $$Y_{CM}=\int ydm$$ Now, $$dm=\sigma dA$$ where $\sigma$ is mass per unit area. Converting into Cylindrical Coordinates,$$dA=rdrd\theta$$. Also, $$y=r\sin\theta$$ Hence the integral...

2 answers both I am unable to understand :/
 
i guess ull never know what word i was talking about
top secret
 
1:01 AM
@Fawad: Your original equation is wrong. You need to divide by total mass.
 
@TedShifrin I saw. That’s not my question
 
For a second I thought you said "total mess" and was confused
 
So ask a specific question, @Fawad.
 
lmao that's what i read too
 
Demonark, you are a total mess.
 
1:02 AM
Ok :)
 
This is true, and I embrace that
 
You can't calculate things and you can't read. Great going, UC.
 
@ted So essentially I'm asking myself "What function $f$ do I know such that $ f \to \infty$ as $x \to -\infty$, so that my substitution has the $\int_{0}^{\infty} - \int_{0}^{\infty}$?
Which therefore results in $0$ proving symmetricality
 
@Cookie. Nothing so involved. Just let $v=-u$?
 
Oh my god.
 
1:04 AM
Then you want to recognize that you have the negative of the original integral, yes.
 
I overthink everything.
 
well, some things, yes. :)
 
@TedShifrin this is what a top 5 education looks like according to US news, we're all illiterate :)
2
 
We focused so much on strict computation during our sections on integration, I have to remind myself that you can do pretty cool things with an integral, just the way you can with the rest of math
 
It's sort of a lost art these days, @Cookie. One of my favorite things that's disappeared from the curriculum is the exercise that you can parametrize a circle by rational functions, and this allows you to do any trig integral as an integral of a rational function.
@EricSilva: Don't get me started on education ... without the wonderful influence of Trompolini and deVos.
 
1:06 AM
lol
 
back in a moment
 
@TedShifrin that's more dyslexia than UC. Can't blame the school alone for incompetence :shrug:
 
@Daminark i read it wrong too tho
and i also go to UC
 
Shhhh
I'm sponsored by Zimmer
 
and two data points determine a trend and correlation implies causation
therefore etc etc
zim zam the flim flam as i like to call him
 
1:14 AM
@EricSilva, Demonark: I used to tell many of my students that reading was a prerequisite for my courses.
 
If $x\neq 0$ and if $x^2$ is rational, then $(\forall k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)\neq0\text{ and }\frac{f_{k+1}(x)}{f_k(x)}\notin\mathbb{Q}.$

If $(\exists k\in\mathbb{Q}\setminus\{0,-1,-2,\ldots\}):f_k(x)=0\text{ or }\frac{f_{k+1}(x)}{f_k(x)}\in\mathbb{Q}$
, then $x=0$ or if $x^2$ is irrational @TedShifrin
 
i actually make a lot of dumb mistakes while reading but i read a lot
so idk what the deal is
 
is it ok now? @TedShifrin
 
Yes, @Leyla, except delete the "if" in front of $x^2$ is irrational.
 
ohh
that was typo sprry
 
1:16 AM
No problem :)
 
Given the handwriting of at leave some professors I know, that's near insurmountable
 
thanks again, goodnight to you all :)
 
You're welcome, Leyla.
what's insurmountable, Demonark? And my handwriting is pretty good :P
 
The reading prerequisite
And yeah that's helpful. The person who substituted my algebra class today... Did not have good handwriting
Though he's actually fantastic
 
who subbed
 
1:19 AM
In AoPS I have to write on the tablet screen. It's so awkward. My writing is horrid. My kingdom for a blackboard. Even the little whiteboard is better.
 
Today was just the axioms for a ring and all that other stuff and he actually made the lecture interesting by talking about some historical stuff
 
Rings really came before groups.
 
Not much but like, just to give an idea of how people think about stuff now vs how they did a long time ago. Also some etymology
Eric: Emerton
 
oh shit Emerton is like
great
 
If there are so many great lecturers/teachers, how is it that the curriculum is so f***ed?
 
1:22 AM
Yeah now I see why people like him so much, normally today would be the dullest lecture but it was actually quite good
 
I always found it difficult to substitute ....
 
@TedShifrin this makes sense, rings are like integers
 
Groups first showed up in the context of Galois groups.
 
Ted: this curriculum evolved from how Sally did things. Though for what it's worth, I think it's mostly concentrated around honors analysis anyway, the rest of it is probably generic for the most part
 
whereas with groups you really wanna think about their actions which is kinda hard to abstract to the algebraic distillation
at least the historical path is unclear to me
 
1:24 AM
And this year I think Silvestre intends to do the whole second quarter on multi/vector calc
 
yippee
 
Though I have also heard that he asked for Schlag's course so let's see how long this lasts
I'm not really keeping up as much with that class anymore
So don't take my word for much
 
my boi luis
 
Do I ever?
 
Just don't start now
 
1:25 AM
he's so salty
 
At any given time I minimize the risk of being taken seriously but in case I falter I must put a disclaimer
 
i was talking to a friend in the math building and he said "I don't understand how i got an A in (some math class)" as prof. Silvestre walked by, and Silvestre just said "pure luck"
it was great
 
LOL, sounds like me :P
Worthy of a @Hippa meme.
 
Was that in a Silvestre class?
 
it was just in the hallway next to his office
(he knew both me and my friend so we are no strangers to his saltiness)
 
1:28 AM
No I mean like, was the A in question from a Silvestre class?
 
no lol
 
Kek, that'd have been funny
 
In general, teachers with personality and humor beat out those without. :)
 
I wanna interact with the guy just to see the salt
Especially since I don't have Soug to roast me anymore
@TedShifrin 'tis true
 
A professor who actually cares — even if not the greatest lecturer — is still a good thing.
 
1:31 AM
Yeah, thankfully I don't think I've had any so far that seem like they don't care but... I know some who have and it's not a good time
 
Consider the IVP $$y''+\gamma y' + y = k\delta(t-1), y(0)=0, y'(0)=0$$
Let $\gamma=1/2$. Find the value of $k$ for which the response has a peak value of $2$
I got the solution $y(t) = ku_1 \dfrac{4}{\sqrt{15}}e^{-1/4(t-1)}\sin(\dfrac{4}{\sqrt{15}}(t-1))$
Should I find the value of $k$ s.t. $\max(y) = 2$?
 
One would say so.
Does that really satisfy the initial conditions, @Lozansky?
You don't have the right solution. You need a general solution of the homogeneous plus a particular solution of the inhomogeneous.
 
I'll try again
Laplace should do it
 
Oh, wait.
The initial conditions get rid of the general solution of the homogeneous, I guess.
I'm still not sure.
No, better double-check that.
 
@TedShifrin Laplacing both sides yields $s^2F(s)+\dfrac{1}{2}sF(s)+F(s) = ke^{-s}$
Agreed?
 
1:42 AM
I don't remember how Laplace plays with delta.
 
$\mathcal{L}\{\delta(t-T)\} = e^{-Ts}$
 
Oh, well, then OK. :)
 
OK
Thus $F(s) = k \dfrac{1}{(s+1/4)^2+(\dfrac{\sqrt{15}}{4})^2}e^{-s}$
 
Yeah, you be right.
 
Notice that $\mathcal{L}^{-1}\{\dfrac{1}{(s+a)^2+b^2}\} = \dfrac{1}{b}e^{-at}\sin bt$
 
1:47 AM
I don't remember this stuff.
 
It's from a Laplace table
 
So if you have the correct solution, then, yes, you need to find the maximum value.
 
That seems kinda cumbersome
Hmm
 
Well, it's just setting a derivative equal to 0 and solving. Slightly cumbersome in this case, but not horrendous. But, wait. Are we looking only at $t\ge 0$ or something?
Otherwise, that function has no maximum.
 
Yeah, because of $u_1$
 
1:49 AM
What's $u_1$?
 
Heaviside
 
Oh, cool. Then, yeah, think of the graph.
 
Yeah
So it's decaying exponentially
 
And it's 0 at 0.
 
And at 1
I think it would hit its maximum when $\dfrac{4}{\sqrt{15}}(t-1) = \pi/2$
But I haven't done any calculatios
I retract that statement
 
1:56 AM
No, you'll need a tangent in there.
 
I now think it's $\pi/4$
 
I don't.
But I'm not working it out.
 
Do I have to differentiate it?
 
You sure do.
 
Damn it
 
1:57 AM
shrugs
HNY, @MikeM.
 
@TedShifrin Ok I got $t=\sin^{-1}(\sqrt{\dfrac{16}{15} \cdot \dfrac{1}{1/16+16/15}}) \approx 2.29$
 
I have no idea, @Lozansky, but I would have had it in terms of $\tan^{-1}$.
 
Which agrees with wolframalpha.com/input/?i=max(4%2Fsqrt(15)exp(-0.25*(t-1))*sin(4%2Fsq‌​rt(15)*(t-1))
 
Are you sure you differentiated correctly?
 
Yeah, pretty sure
 
2:07 AM
I mean, when I differentiate I get something like $\beta \sin\alpha x + \alpha \cos\alpha x = 0$.
 
Sure
 
So I naturally get $\tan\alpha x = -\alpha/\beta$.
 
To solve it, you square both sides
 
Um, no.
You're making it way hard, but I see.
 
Well, you move one trig term to the other side
But yeah
I get the same result as W|A
 
2:09 AM
OK, so then you never needed to bother us ...
 
Well, when I plug my $t$ and solve $f(t) = 2$ for $k$ I get $k \approx 2.75$
Whereas the answer sheet suggests $k\approx 2.8108$
 
Could be errors from using approximations instead of complete decimals in there
 
Idk... it really feels like there is some neat way to solve it
 
Beats me.
 
Really dull exercise otherwise
 
 
2 hours later…
4:39 AM
TACO
 
 
3 hours later…
7:16 AM
Hello nerds
 
Heyy
That's pretty good
 
How's it going?
 
7:49 AM
Morning
 
Yo
 
8:08 AM
Back
i NeEd To StUdY cHeMisTrY
2
@Daminark This meme is juicy
I give this a strong Smale/10
 
@BalarkaSen measure theory*
 
@AkivaWeinberger but $\bigcup_{r>0}\Omega_r=\mathbb{R}^2\setminus\{(1,-1)\}\notin \tau $
 
Hi all
 
Anyone on who can explain in brief how quantum groups (especially at roots of unity) are used in physics?
 
For a separable normed space $E$, is the space of continuous linear maps on $E$ also separable?
 
8:33 AM
I think this can fail even for Hilbert spaces
It can indeed, if $E$ is an infinite dimensional separable Hilbert space and $\mathcal L(E,E)$ is the space of continuous linear functions on $E$ there is an isometric embedding $\ell^\infty\to\mathcal L(E,E)$ and $\ell^\infty$ isn't separable
I can write out the details in about half an hour if you want
 
I think I get it, thanks :)
 
9:09 AM
Is there a belt trick to see $\pi_2(SO(3)) = 0$?
I suppose by thinking of it as $\pi_1(\Omega SO(3))$?
 
9:24 AM
The idea is that if $E$ is Hilbert and separable then it has a countable orthonormal basis $\{e_n\}_{n\in\Bbb N}$. For every $a\in\ell^\infty$ you can consider the linear function $T_a:E\to E$ defined by it's values in the basis: $T_a(e_n)=a_ne_n$, then $T_a\in\mathcal L(E,E)$ and $||T_a||=||a||_\infty$. Moreover $T_{\alpha a_1+\beta a_2}=\alpha T_{a_1}+\beta T_{a_2}$ so the function $\ell^\infty\to\mathcal L(E,E)$ that sends $a\mapsto T_a$ is the embedding
 
@TobiasKildetoft looks around apparently not
@Balarka $SO(3)$ is homeomorphic to $\mathbb{RP}^3$
I took a screenshot of what I wrote when I did the problem in difftop but I'm... less than happy with it, to say the least
Hmm
Okay so
You take some map $f:D^3 \to SO(3)$
Where a point is mapped to the rotation along its axis by the appropriate angle
That's pretty continuous
(Appropriate meaning like, $\pi\|x\|$)
Okay so I'm saying "along its axis" to specifically mean that if we take a point on the same axis but the other side, we rotate the opposite direction
Now, the only points that are mapped to the same thing are antipodal points on $S^3$
So the induced map on $\mathbb{RP}^3$ is a bijection
But by compactness that does it
So now we'd just need to compute $\pi_2(\mathbb{RP}^3)$
 
9:52 AM
Okay so I don't know much about CW complexes, I'm guessing it has something to do with, oh you just added a 3-cell to $\mathbb{RP}^2$ so RIP homotopy groups, but... that feels a bit vague for me atm
 
10:28 AM
@Balarka here's a fully formal formulation (which is much more restrictive than what one does in algebraic geometry I think but anyway). Let $\sf{ACF}_0$ be the theory of algebraically closed fields of characteristic zero, i.e. the (necessarily infinitely many) axioms written in the first order language of fields describing algebraically closed fields of characteristic zero and everything that is provable from them.
Then for every sentence $\varphi$ in the first-order language of fields, $\sf{ACF}_0$ proves $\varphi$ if and only if $\varphi$ holds in $\Bbb C$
 
11:09 AM
@Daminark It's not hard to compute $\pi_2(\Bbb{RP}^3)$ (which is what the homotopy group is, like you said). For $n > 1$, $\pi_n(X)$ is isomorphic to $\pi_n(\tilde{X})$ where $\tilde{X}$ is the universal cover of $X$.
I was looking specifically for a belt trick technique
There are also non-explicit ways to establish $SO(3) \cong \Bbb{RP}^3$. One way to do it is to identify $SO(3)$ with the unit tangent bundle $T_1 S^2$, which is total space of an $S^1$-bundle of $S^2$, which you can identify to be the $\Bbb Z/2$-quotient of the Hopf bundle $S^1 \hookrightarrow S^3 \to S^2$.
So it is indeed $S^3/\Bbb Z_2 \cong \Bbb{RP}^3$.
@AlessandroCodenotti That's pretty cool!
 
The reason this holds is that $\sf{ACF}_0$ is $\kappa-$categorical in every uncountable cardinal $\kappa$ (meaning that any two uncountable algebraically closed fields of characteristic zero with the same uncountable cardinality are isomorphic), so by a theorem of Vaught $\sf{ACF}_0$ is complete (for every sentence $\varphi$ it can prove either $\varphi$ or $\neg\varphi$) and for complete theories all models are elementarily equivalent (the same assertions hold in any of them)
 
I did not at all know that there was a fully foundational statement for the Lefschetz principle.
Very cool
 
11:25 AM
It appears that people actually doing algebraic geometry ignore the formal version anyway because it's too restrictive, what I wrote above should be the same as the "classical result of Tarski" from the first answer
 
12:22 PM
Let $x_n \in H$ be a sequence with $(x_n, x_m) = \delta_{n,m}$. Then does $(x_n,y) \to 0$ (scalar product) for every $y \in H$ ?
 
0
Q: Conjectured values for infinite products?

mickWhen I look at simple open problems in calculus ( not including analytic Number theory mainly ) most conjectures are , or are equivalent to , infinite sums or integrals being equal to a closed form real number. So I wonder about famous open problems where we have the conjecture that an infinite...

Any ideas ?
 
1:07 PM
@blat I think so, such a family $\{x_n\}$ is orthonormal, so the Bessel inequality holds, $\sum\limits_{n=1}^\infty|(y,x_n)|^2\le||y||^2$ which should give the desired result since the series converges
 
1:45 PM
@AlessandroCodenotti Thanks
 
 
1 hour later…
 
2 hours later…
4:41 PM
Hey guys, I'm a bit rusty and have been asked about this integral. Any ideas? \int_R sin((pi-z)^3) dV, where R = {(x,y,z) | 0<=z<=x<=y<=pi}.
As far as I know, sin(x^3) doesn't have an anti-derivative expressible in terms of elementary functions.
 
4:55 PM
In the same way that $\Bbb R^n/ \Bbb R^m \cong \Bbb R^{n-m}$ for vector spaces, if I have the modules $\Bbb Z^n$ and $\Bbb Z^m$, is their quotient isomorphic to $\Bbb Z^{n-m}$?
 
Yes, dimension of quotient of free modules is dimension of the ambient module minus dimension of the submodule.
 
You can make the integrand a bit nicer by substituting $(x’,y’,z’)=(\pi-x,\pi-y,\pi-z).$ @Mankind
In those coordinates, I then suggest considering how you’d write this as an iterated integral
 
@ÍgjøgnumMeg it depends on how you embed $\Bbb Z^m$ into $\Bbb Z^n$
 
Hooowso?
 
For example take $m=n=1$ and embed $\Bbb Z$ into $\Bbb Z$ via $x\mapsto 2x$, then the quotient is $\Bbb Z/2\Bbb Z$
 
4:59 PM
Ok, so it becomes \int_R' sin((z')^3) dV, where R' = {(x',y',z') | 0<=y'<=x'<=z'<=pi}
 
Good point, I didn't think about that.
 

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