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12:20 AM
@TedShifrin I'm honestly not sure if I'm happy I figured it out or sad that it took so long but.. When given $\int_{1}^{2}\int_{x}^{x^2}dydx$ to change the order you do $\int_{1}^{2}\int_{\sqrt{y}}^{y}dxdy + \int_{2}^{4}\int_{\sqrt{y}}^{2}dxdy$
I didn't consider you had to do it in two parts, was trying to figure out how to write it all in one integral
I gotta finish a lab report but thank you for the help!
 
Well done!
 
12:56 AM
"Type theory is not new! Homotopy type theory is just a version of Martin-Löf type theory, which was created in the 1980s."
heh, my definition of 'new' is totally skewed
hmmm, 1970 onwards? or 1975? is new
or maybe 1980 is the exact boundary of new
then, before that, going to 1945, i would say, 'relatively new'
then, all the way to 1912, i would probably say, 'modern'
 
Well, no.
 
1:59 AM
Stopping by to say hello
 
hello
 
Hello, Shmo
 
Hey Ted
What’s that about measure theory for 8th graders in Germany ?
That Midas guy is a piece of work
 
2:15 AM
Oh, did he respond?
 
He edited his answer to address your objection
And contradicted himself multiple times while he was at it. It was hysterical to read.
I pointed that out and he deleted his answer. So it’s not there anymore.. I had him at “the zero function”.
 
Some people shouldn’t be allowed learner’s permits, let alone driver’s licenses. I don’t usually downvote unless the OP ignores me, but this was too much of a travesty.
 
I hope this answer is correct:
0
A: If $(R:_{Q(R) } S)$ is non-zero, then does $(R:_{Q(R) } S)$ contain a non-zero-divisor?

Lukas HegerConsider $R=k[x,y]_{(x,y)}/(xy)$ and $S=R[\frac{x}{x+y},\frac{x}{x^2+y},\dots]$, then $(R:_{Q(R)}S)$ contains $y$, so it is non-zero. Suppose that $d \in (R:_{Q(R)}S)$. Then for all $n$, we have $d\frac{x}{x^n+y} \in R$, so $$dx \in \bigcap_{n \geq 1} (x^n+y) \subset \bigcap_{n \geq 1}(x^n,y)=(y)...

no idea what the geometric interpretation of that example is :D
I mean I understand the geometric interpretation of $R=k[x,y]_{(x,y)}/(xy)$
 
2:32 AM
What does that notation even mean?
 
it means those elements $d$ of $Q(R)$ such that $dS \subset R$
(I hope!)
 
Never seen anything like it.
 
for two ideals $I$ and $J$, the colon ideal $(I:J)$ is sometimes useful
 
And people complain diff geo has too much notation ….
 
I actually think that commutative algebra notation is way easier than diff geo notation
at least it's more consistent if you vary the author
if $X$ and $Y$ are algebraic subsets of affine space, then $I(X):I(Y)=I(X\setminus Y)$
so it's not a completely ungeometric notion
 
2:39 AM
Well, the big bugaboo in geometry is left-invariant versus right-invariant. And the idiocy in exterior algebra with factors of factorial or not.
Neither is really geometry. It’s algebra.
 
I mean the joke that differential geometry is the study of objects invariant under change of notation doesn't exist for no reason
 
Eh.
 
but I've gotten better with diffgeo
in the beginning it was intimidating
to me much more so than scheme theory :D
for some reason
maybe because it's not a bunch of algebra glued together
 
Or because you had no concrete experience with it beforehand.
 
yeah possible
maybe should've started with curves and surfaces
 
2:45 AM
Uh huh, and stuff like Guillemin and Pollack. Geometry pedagogy largely sucks.
 
I learned a lot from Taubes's book
 
I don’t know it.
 
I like his choice of topics, he does a lot of examples for everything and covers Lie groups, vector bundles, principal bundles, the classifcation of flat bundles, characeteristic classes, complex manifolds
 
3:02 AM
@Ted what's your favorite diff geo book?
among those that treat abstract manifolds
 
the best way to do math is to first do higher category theory. then you specialize after.
4
 
sure
I think elementary school should not do arithmetic with the natural number object in a specific topos
just work in a general topos
and of course carrying should be explained via group cohomology
 
4:10 AM
Why does a change of variables work here: $$\frac{1}{2} \Big[ \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac{m^2n}{3^m(m3^n + n3^m)} + \sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac{n^2m}{3^n(n3^m + m3^n)} \Big]$$
Further, why does it work in general?
 
4:22 AM
$dx_i(f(p)(\sum_j v_j D_jf_1(p), \sum_j v_j D_jf_2(p),\cdots, \sum_j v_j D_jf_m(p))=\sum_j v_i D_jf_i(p)$
Can anyone please explain why this is true?
 
user564465
5:08 AM
ok
 
5:33 AM
@Koro. Get rid of all that complication. What is $dx_i(p)(v)$ for any vector $v$?
 
5:45 AM
$dx_i(p) = \phi_i(p)=p_i$, where $\phi_i$ is the elementary 1-tensor.
because $dx_i:=d\pi_i=\phi_i$
so I'm not sure how to compute it further (how to get $v_i$ as an answer to your question).
no. $dx_i(p)(v)= \tilde{\phi_i}(p, v)= v_i$, the second equality is by the notations I am using.
 
6:20 AM
So $\phi_i(e_j) = $?
 
 
1 hour later…
7:48 AM
$0$ for all j not equal to i.
 
8:10 AM
When solving the questions in 1st chapter Spivak, some questions can be solved using many theorems, some in the previous sections, some not. Is it really important that I stick to the axioms/theorems he has exposed, or can I use my other knowledge that i take for granted? Feels like some of the qs are supposed to be challenging but another idea eradicates this. For example, applying sine substitutions
 
 
3 hours later…
11:14 AM
peep
 
Isn't a redundant here? Why not just write s>0?
 
11:44 AM
@rschwieb sorry for spamming your ring theory database with suggestions
 
 
2 hours later…
1:59 PM
Hello! I come back to ask if you have any idea on that: math.stackexchange.com/questions/4586173/…. Thank you in advance!
 
 
2 hours later…
3:40 PM
peep
 
@onepotatotwopotato poop
 
4:07 PM
peekaboo
 
cock-a-doodle-doo
 
4:36 PM
@Xander Is this some form of greeting? duck-a-doodle done.
 
I have a question on policies. Does context always have to mean include some partial/failed attempts? or can it be also motivation, intuition or definitions
 
@LukasHeger Yes, we encourage motivation, intuition coupled with why, and definitions. Most of all, we appreciate specification of what the asker doesn't understand. Including failed attempts is not necessary; but motivation, specification of why stuck is then crucial.
 
I have a feeling that there are some people close-voting everything that doesn't include an answer to the question "what have you tried?", but maybe that's just my feeling
 
lukas, it's definitely possible to provide contexts that doesn't include 'failed attempts,' depending on the problem, sometimes getting stuck makes it impossible to make an attempt. stuff about motivation/intuition, or "if X were changed to Y, i'd know how to do this because Z, but here Z doesn't apply." that kind of thing, can be good context.
 
@LukasHeger Replying in a comment: "It's just an intuition", is not enough. I'm talking about information provided in the question, itself.
 
4:41 PM
lukas i have noticed that sometimes with more specialized/advanced questions, there is greater risk that people who don't know the subject matter will fail to recognize context, even if it is provided, because sometimes you need to actually know the subject matter to understand the context.
 
@amWhy that's not what I meant by providing intuition. I meant providing an intuitive reason for why something might hold
 
@leslietownes I doubt that. That's a typical cop out: blame the reviewer.
 
amwhy, i added in a 'sometimes' up above. people certainly accuse reviewers of doing that more often than they are actually doing that, but it does happen.
and when it happens, it tends to be not with calculus or whatever, but stuff at a level where it is more likely that most of the voters to close don't know the subject matter
in any case, i think a solution to avoiding that is providing enough english language, non jargon, non symbol discussion around the problem so that someone who doesn't know the area can still recognize that an attempt has been made to place the problem in context.
as opposed to, as you note, simply blaming the reviewer for not understanding something.
 
@LukasHeger Then that's fine. Reviewers here, at least a lot of us, fret more about posts simply telling us what they expect users to do for them. Including attempts is not the end-all and be-all.
 
lukas i certainly wouldn't put in 'throwaway' attempts, i have sometimes voted to close when it looks like someone is only complying with the letter but not the spirit of the guidelines.
 
4:46 PM
@leslietownes I get that. :) And thumbs up for your most recent comments.
 
thanks @leslie @amWhy for the discussion
 
5:06 PM
@leslietownes It happens, but it is rare. And a statement in the question like "I am a graduate student studying Lie algebras," can often be enough to provide the needed context for reviewers, who can say "Oh, I don't know anything about Lie algebras. I should probably skip this one."
Also, as I have said many, many times in the past: I regard "attempts" as the absolute worst kind of context. In my ideal world, an attempt would not count for context at all (as "problem statement + attempt" is typically a good way to use Math SE as a platform for getting someone else to do your homework for you).
I would generally discourage folk from posting attempts (throwaway or not), and encourage them to focus on other forms of context, e.g. some outline of what they are studying, or why the question is interesting, or some discussion of applicable theorems or definitions.
 
I maybe have a bias for higher-level question in my interest areas (which is kind of natural I guess), but I think a lot of higher-level question without attempts are actually fine, not all of them of course! but the percentage of questions with absolutely zero contexts seems to be much lower than in areas such as, say calculus
 
@LukasHeger Sure, there exist questions where "an attempt" likely provides sufficient context for interested parties. But I believe that this set of questions is vanishingly small, and that any such question at a higher level could easily be converted to something good without "an attempt".
 
@XanderHenderson oh I'm not disagreeing with you about attempts
sometimes it seems attempts are included to satisfy the policy requirements, not to make the question more informative
 
@LukasHeger Exactly.
And there are far too many "helpful" individuals who leave comments like "what have you tried?" or "show us an attempt".
Which rather misses the point.
@amWhy It is what a rooster says. :(
 
English-speaking roosters have sophisticated language
 
5:28 PM
@XanderHenderson I think I made your point pretty effectively.
 
@LukasHeger If I recall correctly Russian roosters go "дудл дудл дудл", which is, roughly, "doodle doodle doodle". Much less interesting. :D
@amWhy Yup.
I was just reinforcing it.
 
@XanderHenderson I responded to the sequence of peep, poop, peekaboo, cock-a-doodle-doo. They do not, collectively, lead to the conclusion: "It's about roosters". Consecutive, btw!
 
German roosters go kikeriki
 
@XanderHenderson Doing so is a good thing! :)
 
@amWhy I regarded those four comments a nonsense, and nothing more. I don't think that you should read into them too much. :D
 
5:33 PM
:P @Xander !! And I regarded your comment as nonsense. Mine was meant to be nonsense to.
 
@Alessandro what do Italian roosters sound like?
 
@amWhy got it! thumbs up emoji
 
apparantly Italian roosters sound like German roosters
just with different spelling
chicchirichì
 
My recollection is that Mexican roosters sound similar.
 
we should rename this room to rooster linguistics
 
@XanderHenderson No, more like "Uff."
@XanderHenderson That's Spanish for Ugh!
 
Ah, so it is like Welsh spelling, where the orthography makes no sense.
 
@XanderHenderson Indeed! Nice find, but so close to the Italian version. Latin roots, likely.
 
@amWhy I would imagine so.
Or possibly older Indo-European roots, as I seem to recall that Hebrew roosters are similar.
omniglot.com/language/animalsounds/cockerel.htm suggests an Indo-European root (the languages that stick out are mostly east Asian).
 
Maybe we need an etymology stack exchange site in the mix!
 
5:44 PM
@amWhy Maybe, though I suspect that it is too broad a topic with questions that are too narrow. At some point, you are just reproducing the OED.
 
@XanderHenderson Intriguing.
@XanderHenderson Indeed. :(
Time for me to get some work done!
 
5:59 PM
@LukasHeger chicchiricchì
 
so Italian and German rooster can communicate well
at least oral communication
 
6:18 PM
We've seen a couple of these on Math SE, as well. Please flag.
 
6:57 PM
@LukasHeger lol very important info
 
7:12 PM
Does anyone here know what a "jet bundle" is?
Could someone with expertise please look at the following answer:
3
A: How is $F(FM)$ related to 2-Jet bundle $F^{2} M$?

MidasThe 2-jet bundle $\mathcal{F}^2_M$ of a manifold $M$ is a bundle that encodes second order derivative information about smooth functions on $M$. This means that, for any point $p$ in $M$, the fiber over $p$ in $\mathcal{F}^2_M$ consists of the set of 2-jets at $p$ of smooth functions defined on a...

 
i don't know jets, but the row and column finite matrices answer from that user is suspect.
 
I kind of suspect the answerer of posting AI generated nonsense.
 
I can't even make sense of the question
what on earth should the frame bundle of a frame bundle be
 
it's a funny form of turing test. is this nonsense ai-generated, or could it be human-generated?
 
a frame bundle is a principal bundle associated to a vector bundle
 
7:25 PM
Heh. I have no clue. But I don't know what a jet bundle is.
My guess is that a jet bundle is how Delta buys planes.
@leslietownes Really, my question is: "Is it nonsense?"
Human generated or otherwise.
 
I know what a Jet bundle is, but both answer and question read like gibberish to me (which is not necessarily an insinuation that they're nonsense)
 
7:39 PM
Hello everyone! Maybe someone can give me a hint on this calculus problem: Calculate the integral: $\int\limits_{0}^{1} \dfrac{x^{\alpha} -1}{x\cdot \ln^2{x}} - \dfrac{\alpha}{\ln{x}}dx, \ where \alpha > 0$ using the fact that $\int\limits_{0}^{1} x^{\alpha -1}dx=\dfrac{1}{\alpha}$. I tried different representations of the under-integral function, but none of them seemed to me solvable
 
Also the supposed Wikipedia link is likely AI generated as well
 
@LukasHeger What wikipedia link?
 
Nevermind
 
Okie dokie.
 
remember when the only people who suspected that things around them were being generated by an AI were crazy people?
 
7:46 PM
@leslietownes I miss the good ol' days. :/
 
i honestly don't like it, haha. i was talking about this with my wife. everything now needs an extra layer of scrutiny.
 
8:31 PM
@Thorgott Well, $SO(3)$ is the frame bundle of $S^2$ and certainly it has a natural metric and its own frame bundle.
 
ohh
now I get it
so the frame bundle of the tangent bundle of the total space of the original manifolds frame bundle
now I feel silly
 
You're allowed to be silly from time to time :)
@MagnusAlexander It screams for integration by parts.
 
there are integrals that simply yield to integration by parts, and there are integrals that scream for integration by parts.
 
8:46 PM
This one was rather loud.
 
8:57 PM
@TedShifrin Well, that's what I did. Don't know what to do after. I have an integral of $\dfrac{x^{\alpha-1}-1}{\ln{x}}$
 
9:09 PM
Oh, I was thinking $\alpha^x$, but ….
Ah, have you differentiated the integral as a function of $\alpha$?
 
9:29 PM
that's a good trick. i remember the first time someone showed me that idea.
 
@TedShifrin I did just now, but I was hoping there is a simpler solution xd
I'm kinda bad in determining whether integral converges continiously or not: here we should consider $\int\limits_{0}^{1} x^{\alpha -1}dx$. I know it looks simple, but I'm a bit lost here answering the question whether it converges continiously on $(0, +\infty)$
 
No, definitely not a simpler solution. You have to be careful with limits and then solve an easy differential equation.
 
Well, we have here that $I'(y)=\dfrac{1}{\alpha}$
 
9:45 PM
Slow down. You mean $I'(\alpha)$? Where did you get that?
 
Oh, sorry, yeah, I'm fuzzy. $I'(\alpha)=\dfrac{1}{\alpha}$, because if we differentiate by parameter $\alpha$ our function $\dfrac{x^{\alpha-1}-1}{\ln{x}}$ we get $x^{\alpha-1}$
 
I'm not following.
Are you differentiating the original integral or the one you got integrating by parts?
Oh, I see, you're evaluating the integral we got integrating by parts by differentiating. I did the original. OK, cool. So what about the "constant" term in your integration by parts? What happened to it?
 
10:26 PM
I have flagged this question. A different user (with no rep) posted it a week ago, identically with the same picture and words. I gave a solution with discussion of envelopes. I can no longer find that post. I now suspect it's a competition question somewhere.
 
10:39 PM
for the space of normal distribution with mean 0 and variance sigma does it make sense to take the variance parameter to be discrete, like in the naturals? considering space of normal distributions with mean and variance as parameters, can identify the manifold to be the upper half plane given that the mean runs over the reals and the variance is positive
 
Sorry again @Ted, I am not quite acquainted with the good rules of public and private chats on this website.
Thank you for making me aware
 
Shoot, you can’t draft people to be interested in your particular question. You’re better off posting a well-written question on the main site, where it will get a much wider audience.
 
Can someone please help with the question, I have been struggling for quite long with this: quant.stackexchange.com/questions/73966/…
 
Maybe my English level is not adequate enough. I also bountied the question but it did not help
 
guess if it was discrete wouldn't be a connected manifold
 
10:44 PM
But you are right about not dragging people
I was abusing your known availability on the main chat
 
11:35 PM
chatgpt might be a good pedagogical tool, near infinite supply of "spot the error in this proof" type exercises
 
Ooh, I love the result!
 
note that the proof 'generalizes' to show any periodic group is abelian!
hey ted, guess who just tested positive for what
 
Given $f(k)=6\cdot\left(2^{k}-1\right)+8\cdot f(k-1)$ and $f(1)=1$, how do I find $f$?
 
You, for unmentionable
DogAteMy Guess/induct?
 
@leslietownes Corvids?
 
11:43 PM
akiva: caw caw caw
 
Is Munchkin sick with flu/respiratory yuck?
 
she had a runny nose but we thought it was "just" a sinus infection, so she's at day care right now
testing her is gonna be a PITA
akiva: OEIS doesn't know it
 
Got it
It's $\frac{6}{7}-2\cdot2^{x}+\frac{29}{56}\cdot8^{x}$
 
no its not, $f(7)$ is different
 
chatGPT's solution isn't funny enough for me to post (it can't even compute f(2), but it comes close)
 
11:59 PM
Correction
$\dfrac67-2^x+\dfrac{15}{56}8^x$
 
glad to help
 

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