00:00 - 20:0020:00 - 00:00

12:00 AM
Our specification only has these points listed for it i.imgur.com/ZjTnoRR.png and the answers to some of the past questions seem intuitive so I don't think it goes into much depth

None of those are so bad.

so $s\mid p-1$ means $p=ms+1$ for an $m$, and $p\mid s^n-1$ means that $s^n=rp+1=r(ms+1)+1$ so the question is about $s^n-rms-r-1=0$

Awesome, thanks for looking over it :)

finding integer solutions for $r$ and $m$ and $n$... hm

exponential Diophantine equations are really really hard
I don't agree that the questions are the same @Alex. just funding solutions to the last one doesn't mean that $m$ corresponds to a prime $p$

12:06 AM
Yeah that's not real easy
@MikeMiller that's true.

I don't see any reason to believe it should be true, but maybe it is

@MikeMiller I hope it is. I need to make a lot of $2$-Frobenius groups. :(

@AlexanderGruber You don't believe in infinite things; I don't believe in primes $p$ such that there are infinitely many primes $q$ with $p \equiv 1 \mod q, q \equiv 1 \mod p^n$ for some $n$

@MikeMiller I believe in countably infinite things. :)

@AlexanderGruber Then you believe in uncountably infinite things too, or else your notion of set theory sucks

12:14 AM
@MikeMiller how is that?

I slept for 8 hours and woke up, yay. But I am still having the flu.

@AlexanderGruber you should believe power sets exist

@MikeMiller well, the issue is really that I don't believe countably infinite sets are sets
which is precisely because i don't believe they have a power set

@AlexanderGruber that's fine, then. ZF {minus infinity} is a consistent theory.
@AlexanderGruber "does anybody know what the status of the whole π+e thing is? Is there a particular thing that is difficult about it, or do we just not really know what angle to attack the problem from"

Just morbid curiosity really
it's just past halloween, i still have a taste for the grotesque

12:18 AM
it's very hard in general to prove that numbers are irrational or transcendental unless they have a good reason to be. the reason we can do it for $\pi$ and $e$ is that they have some very nice functions that they matter to (trig functions for the first; the latter has a quickly converging power series, showing it's very well-approximable by rationals)
there's nothing like this, as far as we know, for $\pi + e$
so we don't really have any good techniques to attack them with.

I guess that means it's the 2nd thing then
i.e. there is nothing special about it being $\pi+e$, we probably don't know it for $\gamma + e$ and whatnot either

right
we know it for $e^{\text{algebraic}}$ because of the exponential function again, iirc

I wonder if we do know it for any two transcendental numbers, other than dumb stuff like $e+(1-e)$

no idea

Probably will get an earful about my $q\mid p^n-1$ thing too

12:23 AM
I deleted that because he told me recently that he thinks math is a solitary endeavor and does not need to be communicated, so I've decided that I should not encourage people to communicate with him
:)

Hi guys, I have a question from my old test that I need clarification on. Is this something suited for main? I'd post the question and post the solution and then ask for clarification. What would the title be of such a post?

sounds suitable to me, though you should TeX up the question and solution instead of posting a picture or a file or whatever.

Oh yeah ofcourse.

12:52 AM
Hi @Alex

@TedShifrin hi there

@TedShifrin Jacob went "eww" when I told him I was reading Kobayashi.
I believe he called me a dirty topologist.

1:14 AM
@Mike: That's absurd. There's way more geometry in K-N than in Peter's book. Perhaps a bit less analysis and definitely more algebra. The subject has changed in 50 years.

@Ted He wasn't actually hostile, of course. The joke was that it's about general affine connections, which he has so far, I think, found little use for in his work.
(This was friendly banter... I guess that wasn't clear.)

Nor have I. But I have done projective connections.

They're all the rage in topology, with Heegard Floer being Morse theory on the space of connections.
That's such a great idea... study the topology by, in some sense, studying all the possible geometry.

I'm not sure if you lose much doing just linear connections.

I don't know the answer to that, of course.

1:25 AM
Cartan thought about torsion in terms of the infinitesimal holonomy of the translation part of the affine connection.
torsion of the connection on the tangent bundle ...

Professor @TedShifrin do you use the new app for the iPad?

Yeah, but I can't get the latest beta version to download, and I can't access chat on Safari. Screw the latest OS disaster.

1 hour later…
2:41 AM
hey Topologists
what is the term for when the $d$-skeleton of a simplicial complex is complete (i.e. has all possible faces) and the $d+1$-skeleton is empty?
e.g. {{1}, {2}, {3}, {1,2} {1,3} {2,3}}
is that just a $d$-simplex, or does a $d$-simplex have {1,2,3} too?

It's the boundary of the 2-simplex.

@TedShifrin there is no special name for that? like an empty simplex or something

Nope. Empty?

hm dang

Hello friends.

2:47 AM
Hi @AWertheim

Hello @TedShifrin. Long time no see :)

Where you been hiding?

Just been working hard, that's all. At least I like to think, anyway. :)
It's a busy time.

Well, so long as someone thinks so ...

Hehehe. Ain't that the truth. =P

2:53 AM
Fine, thanks ... Some awesome students and some frustrating (=failing) ones.

@AWertheim Hahaha, I like that dog in your picture

Always a dawg fan, @Zach :D

I don't think there's a name for that, @AlexanderGruber

Glad to hear it! Ah, that's a shame. It's always hard to see people struggle.
@ZachSaucier thanks! My friend insisted my old avatar was boring, and that I needed a cuter one. :)

Well, if you can't be cute, get a cute avatar :)

2:56 AM
A good strategy, I think. :D
How's your first year going, @MikeMiller?

It's good, @AWertheim

I have lost a few kg over the last few weeks, yay!

I don't know that I have much more to say about it than that. :P

Lol, no worries. I didn't mean to put you on the spot or anything. Nice to hear it's been good nonetheless.

Hi, I have problems with the following I'd appreciate any hint. Let $0\le f_n \uparow f$, so there is a sequence of simple function for each $n$ s.t., $0\le f_{n,\ell}\uparrow f$. If we define $g n=\text{max}\{f_{1,n}\ldots f_{n,n}\}$, it is clear that $g_n$ is simple, non-negative and nondecrescent, but I have problems to show that $g_n\uparrow f$, I thought it was simple but I find problems in the argument. I'd appreciate any suggestion. Thanks

2:59 AM
You didn't, I just don't have a whole lot to say. You graduated from Duke, huh? What're you up to now?

I am up to no good.
My runny nose turned into a sore throat, which turned into a cough.
This means it is about to end.

I don't know why I have problems to show the convergence, intuitively is clear that the $g_n\uparrow f$. But I have troubles to develop the argument formally.

@MikeMiller: Indeed I did! Working for the year while applying to grad school. :)

@AWertheim What're you working on in the interim / where're you applying?

@MikeMiller: working at a software company in the meanwhile, and trying to learn a lot math when I go home.
Many excellent places. To save the (excessively) long list, UCLA, Michigan, UC Berkeley, Brown, Northwestern, Yale, NYU, Columbia are some of the usual suspects.

3:17 AM
Noble aspirations (re: math at home). And good luck!

Thanks! Appreciate it. :)

OK i need another thing
trying to figure out a term, if I have [1],[2],[3],[4], [12],[13],[14], [23],[34], [123]
then I've got [24] missing, [134] missing
and everything above those
what do I call just those parts... like, the "tallest" things missing from the complex

3:51 AM
OK the complement is an upper set. What I was looking for are the "minimal elements" of the complement of the complex.

4:27 AM
What does it mean to "Show that the differential form is exact" ?
just that I can integrate it?

So mathy.
Byes.

4:46 AM
I suppose it does. But integrate it in a fast way and plug in end points

5:19 AM
anyone out there?
hello @TheArtist. Why do you call yourself that?

5:42 AM
0

So, I was given this really nasty problem to solve Suppose C is parametrized by $\mathbf{g}(t) = \left[\begin{array}{c}e^{t^{3}\cos\!\left(2\pi t^{25}\right)}\cr t^{6}+3t^{3}+3\cr e^{\sin\!\left(3t\right)}\!\left(t^{3}-t\right)\cr\end{array}\right] \ , \quad 0\le t\le 1 \ .$ Let $\ome... ^^ For any of you interested. There must be some trick I don't know to use @ZachSaucier coz I'm an artist? :p May I see some of your works? @ZachSaucier just kidding bro. It's just a random name that came to my mind when making my account. Hey guys Hello @KajHansen 5:50 AM @Zach, if that field is conservative, then choose a simpler path ;) Path independence can make lives much easier :D 0 While thinking about this post of Clin, I wanted to consider the polynomial:$P(z) = 1+x_1z+x_2z^2+\cdots+x_nz^n$, satisfying: (I)$1\geq x_{1}\geq x_2\geq\cdots\geq x_{n}\geq0$and$\sum\limits_{k=1}^{n}x_{k}=1$. Then,$P(1) = 2$and$P$has no root (real or complex) inside the disc$|z| < 1$... I asked this Q in main .. anyone has any ideas how to proceed ? :-) Oh wow @r9m. +1 for quality thread...we don't get enough of these. @KajHansen thank u :) 6:13 AM @KajHansen to get the end points I just plugin t for the original parameterization? Indeed @ZachSaucier that's still a pain, haha More of a pain would've been actually doing the integral with that though ;) true 6:25 AM cool, I should have thought of that, given that's what we're doing xD maybe I'm just too tired 2 hours later… 8:46 AM Greetings @r9m It looks like you didn't want to answer this one in an elementary way 6 What tools would you recommend me for computing the limit below? $$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$ As soon as any useful idea comes to mind I'll make the proper update with the new findings. @Chris'ssis If I could do it I would discuss it with you on chat first and then decide if its good enough to post as an answer :-) ... I couldn't :) @r9m My feeling is that it can be done at the high school level ... (just a feeling) @Chris'ssis okay :) That question looks like a rare gem. Well, it is a rare gem. @Chris'ssis do you know why the community is giving big bounties on some of the already answered questions on main ? math.stackexchange.com/users/-1/community?tab=bounties 8:55 AM @r9m What is the meaning here? 20 I'm trying to find a closed form for the following sum $$\sum_{n=1}^\infty\frac{H_n}{n^3\,2^n},$$ where$H_n=\displaystyle\sum_{k=1}^n\frac{1}{k}$is a harmonic number. Could you help me with it? "One or more of the answers is exemplary and worthy of an additional bounty." That guy even realized the answer of Tunk-Fey is wrong? @Chris'ssis community is supposed to be a bot right ? How is it choosing the questions for bounty ? @r9m Behind the bot there is a programmer, right? @Chris'ssis ya ,, but how is he choosing ? (based on the number of views ?) @r9m I couldn't answer this question. I don't know. @Chris'ssis I think robjohn's answer here is brilliant !! but the question didn't get a bounty ! 9:01 AM @r9m Yeah, I know. @TedShifrin Artin is a cool guy. 9:16 AM @Chris'ssis any ideas about my question above ? ^^ @r9m I didn't think of it yet. @Chris'ssis okay :) I've been helplessly thinking about it for a week now .. :( I couldn't get anywhere ... 9:30 AM @DanielFischer Okay but how do you know you reach$1$in finitely many steps? How do you know that$\sum\delta$converges in this case? 9:40 AM @DanielFischer Oh I see, it's because it is a fixed$\delta$value in this case so the sum will converge to$1$in$\frac{1}{\delta}$steps of the series. 1 hour later… 10:58 AM Hello everyone! Can someone explain me the sentence: "If R=K[x] the prime ideals are <f(x)> where f(x) is an irreducible polynomial in K[x] and <0>, and again <f(x)> f(x) is also maximal." 11:51 AM @robjohn I think this is the guy that downvotes me daily -> math.stackexchange.com/users/31254/donantonio 2 @Chris'ssis Impossible. He has not logged in since July. @JasperLoy Not from that account, of course. @Chris'ssis So which account is he using? @JasperLoy I don't know, but it's probably one with low reputation. @Chris'ssis What are your reasons for suspecting it is Don? 11:56 AM @JasperLoy The pattern of the downvotes and the fact he was the only serial downvoter I faced in the past. Both things match. @Chris'ssis You're turning into a detective ? :P @Hippalectryon I'm pushed to do that. :-) 12:12 PM Hi @DanielFischer how are you feeling these days? @JasperLoy Got a cold. It's autumn. @DanielFischer I am thinking of studying French, German, Italian, and Spanish in that order. Just thought I would share with you. @JasperLoy lol @DanielFischer If you want to talk about your secrets some day, feel free to email me. My email is jasperloy at outlook dot com =) 12:31 PM @Chris'ssis You can always ignore him :) He just downvoted my answer here 0 It's trivial to notice by Cauchy-D'Alembert criterion that $$\lim_{n\to\infty} \frac{n}{\sqrt[n]{n!}}=e \Rightarrow \frac{1}{\sqrt[n]{n!}}\approx \frac{e}{n}$$ and hence the conclusion. Q.E.D. @Chris'ssis I seriously doubt a 100k user would create another account to downvote you. This is a very brilliant way, not because is mine, but because the way itself is brilliant (and very elementary). Damn, enough people enroled in my optional course about game theory, so I'll have to study game theory after Christmas. @Huy You offered a course that you have not studied? That is very weird. 12:36 PM @JasperLoy: I know. @JasperLoy Yeah, the surprises come from people you never expect ... (it wasn't my case here since in the past when I quarreled with him, he began to downvote me - and he also has a downvoting pattern) @Chris'ssis I guess there are many lunatics here, like me, lol. @JasperLoy: It's just for high school pupils though, so I hope I'll manage in time. @Hippalectryon don't miss this one math.stackexchange.com/questions/1012455/… @Chris'ssis Thanks :D 12:39 PM (Might we do it elementarily?) @Hippalectryon It's a very precious marvellous gem! (created by me - oh, no - by my brother) @JasperLoy I don't need to downvote him, the life will downvote him for his attitude. :-) @JasperLoy The attitude is the success key in everything I think, and sometimes it's not that easy to adopt the proper attitude. Hello everyone! To show that an ideal I=<f(x)> (where f(x) is irreducible) is a maximal ideal of K[x] do we have to show that K is a field and that I=<f(x)> a prime ideal? @Huy high school game theory? @UserX: If you like to call it like that. We definitely don't learn GT in high school I doubt that my teachers have heard of the field @UserX: At the high school where I'm teaching, in the second last year (where they typically learn things like differentiation and integration), students have to choose from a range of optional courses. There are courses in all kinds of subjects, e.g. literature, economics, security, physics, chemistry etc, and I'm offering a course in mathematics. I was told usually only very few students enrol in maths courses so they won't take place (minimum amount of students). 12:53 PM Why did you choose game theory? Linear algebra should be your choice imho @UserX: Because linear algebra doesn't appeal to a high school student, at first sight. Sets and probabilities? @UserX: Every teacher is allowed to describe the course in a few lines of text, but nothing more. And usually, students will learn some basic probability in high school. I didn't apart from some really really elementary stuff @UserX: We did mostly basic stuff compared to what I learnt at university level, but I absolutely hated what we did at university level. 12:57 PM Well good luck with game theory. Introduce them prisoners dillema. @UserX: I used the prisoner's dilemma in my short description of the course. ;-) Anyway bbl later @Huy The Monty Hall Problem may interest them. @skullpatrol: I'd classify it as an exercise in probability rather than game theory. @Huy true, but it is easily understandable. just a suggestion :-) 1:38 PM @Chris'ssis He's been inactive since July 18. I don't think he is the one. @robjohn From a different account ... @Chris'ssis Looking into it. @Chris'ssis It does not appear so, but that is all I can say. @robjohn OK @Chris'ssis guess what I'm doing =D @robjohn how are you @TheArtist are you preparing an elementary proof to this one? 7 What tools would you recommend me for computing the limit below? $$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$ As soon as any useful idea comes to mind I'll make the proper update with the new findings. 1:55 PM @Chris'ssis Im learning the proof for : @Chris'ssis :D @TheArtist That one is easy. @Chris'ssis oh no :p obv Im a beginner @Chris'ssis Yep for you :) Im on this part of my fav book ;) @Chris'ssis maybe one day il be able to prepare an elementary proof for that ;) right now im still small in the field :) @TheArtist That one requires a crazy amount of experience. @Chris'ssis that's why I said one day (future) :) @Chris'ssis oh ok :/ then that's like far future @TheArtist for this you may also use the Fourier seires of the first periodized Bernoulli polynomial. This way also requires very goob abilities. 2:01 PM @Chris'ssis Don't forget me when you win the Fields medal. @JasperLoy lol :-) @Chris'ssis hmmmm :) @JasperLoy what's a field medal? :) @JasperLoy oh ok googled it =p Okay bye people @TheArtist after a while you reach the second formula here. Well, not the way you expect. The idea is to differentiate once with respect to$z$and then after you set$z=1$you get in the right side the integral you lastly obtain when using the way I told you about above. Have fun! @Chris'ssis (favorited and +1) .......will Check this out, thanks :) , U too , have fun. 2:47 PM How are limits than go to infinity(either positive or negative) called? Divergent limits sounds bad 3:15 PM I got a nice DE for anyone interested$y'=x^2-e^y$This little fella uses almost all elementary methods in its solution 3:26 PM Although I need help here; @Chris'ssis$\displaystyle\int e^{\frac{x^3}{3}}\mathrm{d}x$I should somehow get an incompete gamma function along with other stuff 3:37 PM I just posted a very nice question 0 In the following link http://integralsandseries.prophpbb.com/topic407-50.htmlI found the integral & the evaluation of $$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$ I'll also include a simpler version together with the question: is it possible to find some easy ways of computing bo... @Chris'ssis your link shows page not found ! @r9m Thanks! :-) @Chris'ssis lul ,.. those double euler sums are piece of cake ! (wonder why did S go through all the trouble of writing them like that!) what kind of alternative approach are you looking for ? :-) @r9m In the first moment I saw it I realized that one can do great things there. 3:54 PM @Chris'ssis poetic ! but I don't understand what is needed ! @r9m A neat proof, short & easy. :18564785 Let me ask you a different question: how would you compute the first integral? I wont .. I go jogging (back in 30-60 min) 4:18 PM @Chris'ssis I enjoy how humble you are, haha 6 @ZachSaucier What do you study? computer science and computer systems engineering Do you understand that? @TedShifrin That is an amazing quote ! @PedroTamaroff yep :) want me to explain? 4:27 PM @ZachSaucier Well, I think the solution the tutorial goes like this. The first is A. The second is C. That's all I could guess. @ZachSaucier Sure. Sorry. A, B, C. | is the "or" symbol, so if one is true the statement is true. It is also true if both A and B are true so the only condition when the statement is NOT true is when both A and B are false pedro you should try these instead By setting it up like they did, with one on top and one on the right, they are combining them, essentially an AND (A | B) & (A | Z) @ZachSaucier Oh, OK. I guess that' yes. So I should return A because? So the answer is A, because A being true would satisfy both requirements 4:30 PM OK. What do [ - ] stand for? @MikeMiller I am not going to pay to play crosswords. this is a weird tutorial, but they stand for range Range? OK. @PedroTamaroff Most of them are free. I think the daily one is free. essentially [ABC] would accept A OR B OR C 4:32 PM Also, I could just give you my password. ^ means "not"? you probably haven't learned this yet, but [ABC]* would mean A OR B OR C repeated 0 or more times, so it would accept ABCAABC for example @ZachSaucier I study math, not compsci. mmhm @MikeMiller I cannot solve those crosswords. 4:33 PM stuff like this is helpful for both though So [^AB] means AB are not in the range' in this case, yes, haha @PedroTamaroff You should learn. They're fun. in my experience it's more common to use ! for not Last Wednesday had "YOLO" as one of the answers/ 4:36 PM @ZachSaucier So what should that stand for? The answer to [^AB] and [ABC] is C. @PedroTamaroff you mean ^? We don't use it much @PedroTamaroff correct :) @ZachSaucier I don't know why, though. Regex is a language or something? @PedroTamaroff ya, it parses text any reason why you're learning regex? I would think something like Python would be more useful for a math guy @ZachSaucier I am not learning regex, I just found that game and was curious. I should learn how to use Mathematica or SAGE, and PARI/GAP. PARI is a subset of Sage. 4:43 PM @MikeMiller Mike is a subset of bother. 5:10 PM @ZachSaucier What do you mean? :-) (back from jogging) Greetings @rehband 3 In the following link here I found the integral & the evaluation of $$\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$$ I'll also include a simpler version together with the question: is it possible to find some easy ways of computing both integrals without using complicated sums that r... @rehband I'm sure you can finish the first one. @Chris'ssis can you help me solve that integral? In fact nvm, I'll post it on main site @UserX use the incomplete gamma function ... @Chris'ssis just the fact that you call your question "very nice" :P 5:19 PM @ZachSaucier hehe, I see. Well, I really think my questions (major part of them) are unbelievably awesome, but that's what I sincerely think. :-) 5:37 PM $$\Large{\text{I have so fun finding brilliant solutions!!! Just found another one!}:-)}$$ 5:49 PM @Chris'ssis can you please check and see if I've mistaken somewhere $$\int e^{x^3/3}dx$$ @Integrator that integral requires the use of the incomplete gamma function. I'll look at it a bit later. @Chris'ssis I've done it using normal inddefinite integration rules! @Chris'ssis Am I wrong then? @Integrator Ah, it works without that. Nice. I didn't have a close look at it. Right now I'm working on some crazy stuff. @Chris'ssis Could you please take a look! Please! @Chris'ssis Fine, Now I'll just hope that I won't get down-votes this night! 6:10 PM I'm working on this one $$\int_0^1 \frac{\log^2(1-x)\log(x) \operatorname{Li_3}(x)}{x}\ dx$$ that is very nice. ^^ $$\frac{\log 5}{\log 2} = 1 + \frac{\log (1 + 1/4)}{\log 2} \sim 1 + \frac{1}{4 \log 2}$$ And also this one is very nice $$\int_0^1 \frac{\log^3(1-x)\log(x) \operatorname{Li_3}(x)}{x}\ dx$$ It is fun with elementary algebra ^^ ^_^ And also this one is great $$\int_0^1 \frac{\log^2(1-x)\log^2(x) \operatorname{Li_3}(x)}{x}\ dx$$ @r9m ^^^ 6:24 PM @skullpatrol Howdy. @ParthKohli Hello. 7:04 PM Blah Any neat way to show$|x/(1+|x|)-y/(1+|y|)|\le |x/(1+|x|)-z/(1+|z|)|+|y/(1+|y|)-z/(1+|z|)|$besides opening it up with all possible options? 7:26 PM @Studentmath You have$\lvert f(x) - f(y)\rvert \leqslant \lvert f(x) - f(z)\rvert + \lvert f(z) - f(y)\rvert$for all functions$f$(and all$x,y,z$in its domain). That's just the triangle inequality of$\lvert\,\cdot\,\rvert$. Hello!!! How could we show that$\mathbb{C}[X,Y]/<X-1,Y+X^2-1>$is isomorphic to$\matbb{C}[Y]/<Y>$? @evinda Can you see that$\langle X-1, Y + X^2-1\rangle = \langle X-1,Y\rangle$? @DanielFischer Why is it like that? @evinda Because$X^2 - 1 = (X-1)(X+1)$is a multiple of$X-1$. hi guys what's the history of mollifiers? 7:38 PM @DanielFischer So, can I just say that , since$X^2 - 1 = (X-1)(X+1)$is a multiple of$X-1$, we have that:$\langle X-1, Y + X^2-1\rangle = \langle X-1,Y\rangle$? Or do I have to explain it further? @evinda For the time being, you have to explain it further. 7:53 PM @DanielFischer Could we prove it like that? We have that the ideal$I=\langle x-1, y+x^2-1 \rangle $is generated by two polynomials$x-1$and$y+x^2-1$. Since$y+x^2-1=y+(x+1)(x-1)$we have that$y+x^2-1=y+ \langle x-1 \rangle $. Do we conclude from that, that$I=\langle x-1, y+x^2-1 \rangle =\langle x-1, y \rangle\$ ?

Have I done something wrong? Is there a better way to prove it?

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