12:01 AM
@TedShifrin If I run into trouble, I can ask someone at Warwick where I am; there are some good algebraic geometers here.

Yup, there sure are.

There is also an old paper in Astérisque that does compute this for $y^2+y = x^5$, and also some Chern classes, using Arakelov theory. Haven't been able to find a copy yet, but should be able to with inter-library loan.

I'm sure people have done these things. It's not the sort of algebraic geometry I worked on.

Out of curiosity, what were your main research interests?
(Within algebraic geo.)

I did some complex integral geometry and then a number of papers on singularities of Gauss maps in several dimensions.

12:15 AM
@TedShifrin complex integrals ? line integrals?

No, much higher-dimensional things, @Kasmir. Involving projections and averaging or intersecting and averaging.

hmm nice indeed :D
I want also to do some reseach on something in math
but i need 3 more years
at least ><

12:46 AM
hi chat

Hi
0

Consider the Koenigs function https://en.m.wikipedia.org/wiki/Koenigs_function Why is it analytic ? How to prove it is analytic ? I assume the Weierstrass M-test is used. https://en.m.wikipedia.org/wiki/Weierstrass_M-test In fact , i do not know another way to show it !? Maybe this relates...

1:02 AM
Hello @mick
I have an answer with 3 downvotes that I think are unreasonable, so if anyone could take a serious look at it.. thank you :-)
-2

By the Cauchy's repeated integral formula: $$\int_a^{x_1}\int_a^{x_2}\int_a^{x_3}\dots\int_a^{x_n}f(x_n)\ dx_n\dots\ dx_3\ dx_2\ dx_1=\frac1{(n-1)!}\int_a^{x_1}f(t)(x_1-t)^{n-1}\ dt$$ which I find analogous to the Cauchy integral formula from complex analysis for $n$th derivatives (instead of i...

hi @PVAL-inactive

I was discussing math today with a fellow grad student.
He was like I am sorry I don't discuss math with other people so I don't increase their grade.
...
very stupid

1:04 AM
I mean cmon grade school should be about discussion of math not some competition.

There's a lot of competition, but I don't think it has anything to do with grades.

yeah

Competition?
What's that?

in terms of getting research out and jobs

yeah

1:07 AM
Hm...
I should get a tutoring job
@N1ng Hello sir
Or mistress

@SimplyBeautifulArt Hello~

Just here for the random chat?

@SimplyBeautifulArt I just want to see what mathematics lovers focus on~

Hm...
That is a good question
As my username may suggest, I like to make math "simple, beautiful, and artful"

@SimplyBeautifulArt As we all like it to be :)

1:16 AM
Ok, cool
@N1ng But don't upvote just because. Your only allowed to upvote if you think its good

@SimplyBeautifulArt Okey, I will not touch it~ Just looking.

is a 1x1 matrix still a matrix?

It's like a box of chocolates with only one chocolate. Is it really a box of chocolate, or did you eat it all? Fundamentally, it is a box of chocolate, but it's not very satisfying. That is, unless you are the one who ate it. @Socrates

don't try to mimic Siri

1:24 AM
I'm sorry. I do not understand what you are asking.

is by definition a 1x1 matrix still a matrix

Yes. By definition, a "___" something is still that something.

beautiful ignore function

@Socrates Hello, I found that there is something en.wikipedia.org/wiki/Matrix_(mathematics)#Empty_matrices

Lol, zero vector space sounds fun

1:32 AM
mmh, 0 matrices are even wierder^^, thanks for pointing me to that @N1ng

@Socrates But you should be aware that I'm not pretty sure about it~ And I would ask the same if you encounter~ :)
if I encounter. Typo~

:)

@mick You good mate?

How surprising that @Ramanujan is here... Hello

...?
So?

1:40 AM
@SimplyBeautifulArt sup?

@SimplyBeautifulArt That guy was a genius: youtube.com/watch?v=oXGm9Vlfx4w

mhm
he died too young
A moment of silence for Ramanujan

@SimplyBeautifulArt Teachers will be very angry if I do the same "How do you know" "I just do.". (Kicked out of the school)

Lmao, that sucks

1:45 AM
@N1ng I usually know how I know ;)
Holy [do not insert bad word here]
I just realized something!

What?

That I'm like one of top 5 users who gained the most rep so far this year
And the others are from Stackoverflow

But there's no way I can compete with them because that site gets 7.8k questions per day, but MSE gets 571 questions per day
~7800 questions PER DAY. Jesus, no wonder they get so much rep
:-/ and somehow I'm tailing closely behind them

@Socrates you are that user- Null ?

2:01 AM
@Ramanujan yes

Oh, so I do recognize you
:d
:D

2:14 AM
question, im learning basic derivatives at the moment but how would you ever be able to find the original function from a derivative given two different functions with different rules can have the same derivative rule. Is it simply not a reversible process?

It is called integration

@WDUK Practice
@Ramanujan Hehe, yup

so it is possible then ? i haven't got to integration yet.

@SimplyBeautifulArt :P

@WDUK Yes

2:16 AM
like i had x^2 and x^2 + 3 then realised they both are 2x so there would be infinite possible functions with a rule that has 2x derivative

The solution there is x^2+c
for constant c
;)

but c isn't solvable in integration ?

@WDUK When you learn Fundamental theorem of calculus, you throw c out the window
I.e., it cancels with itself when you apply integration

ah i see - guess ill find that out soon enough

:P
Anyways, goodnight

2:37 AM
Hello there, evening

2:51 AM
Hello all, i need help on identifying direction field of a separable ordinary differential equation. The ode is unknown but separable.
http://math.stackexchange.com/questions/2106520/identify-separable-ordinary-equation-with-directional-field?noredirect=1#comment4332496_2106520
im clueless on how to begin since the ODE is not known.

1 hour later…

@DHMO clever hehe

@Socrates did you?

I seriously don't know. But for what it's worth, my netbalance is <1k

I mean, did you know that it is not the real url?

Conspiracy Keanu: what if all phishing mails are serious, and all "serious" mails are in fact a phish?
@DHMO well, I looked in the linkdescription

4:27 AM
oh, then you won

@DHMO if you would have made the link not that obvious I would have clicked it, but what you then have apart from my IP?

5:11 AM
guys would you consider $75 spent on a pair of sports shoes to be too much? 5:29 AM math.stackexchange.com/questions/2106983/… can someone check my answer is correct(suitable because it looks odd to other answers)? Now it looks right because 3 upvotes :P but still it's oddest in all answers i upvoted u Oh,thanks Oh,its total 5 ups now bang! How the hell OP posted 2 consecutive questions within 2 minutes? that is wierd i cant post 2 only after 5 mins 5:37 AM Sometimes it asks me for 30 minutes maybe it depends on the activity or time =p 6:17 AM List of maths fields I have interest in: 1. Group theory in terms of orbits and actions 2. Zero term algebra and division by zero algebra 3. Integration in the language of abstract algebra and as a functional, symmetry of integrands 4. Optimising proofs given axiomatic systems 5. Category theory 6. Unnatural algebraic structures 7. Patterns in expanding multiplications of polynomials 8. Set of all counterexamples given a proposition 9. Tensor visualisation and intuitions 10. Numerical analysis methods to explore special regions or points of mathematical functions or systems of equations FAQ ans: There isn't much in the geometry, topology and number theory branches of maths. That is because my background is still not strong (and had been working on it) thus it is uncertain whether what I think I am interested in those branches is really what I interested in despite I am known to be a highly visual learner Trivia: Interestingly, one reason I am interested in 3, 7 and 8 is because I am lazy. I will let you figure out what that means 7:11 AM Why I am not able to post this question on main site If r and s are non zero constant vectors and the scalar b is chosen such that magnitude of$\vec r +b\vec s$is minimum, then we have to show that the value of$\vert{b\vec s}\vert^2+\vert{\vec r +b\vec s}\vert^2=\vert\vec r\vert$My thought We have given$\vert\vec r -(-b\vec s) \vert$is minimum so it should form right angle triangle . So as said above br.s=0 7:50 AM @MikeMiller Level 4/5 IIRC. I did it though Oh the first one. That's level 3 Ah right The next couple maps aren't too difficult. Did you just restart the campaign? yep I'd say just keep pushing forward and don't worry about rank. 8:03 AM Don't worry too much about nullity either Got it, that's what I was trying to improve Did you get any S-es? :) No :( Oh well. You'll get better by doing more. The hardest score is usually speed. Yeah. I finished Level 3 in 10 moves but I got a B :( 8:12 AM Did you lose points on technique or power? Maybe I lost some on power, I don't remember. Do you remember the formulas for perfect score? Not all of them; were they discussed in the training? Might have to check it out once again then No, I just told you at some point. Oh, no, I don't. 8:17 AM Power = kill Ceil(Total number of enemy units built / 10) units in one turn So for Eagle you'll certainly max that if you kill three things in a turn. Ahh, that makes sense. Technique = Lose at most Floor(total number of your units built/5) units You can always boost this by spam-building new units. Gotcha. I'll try these out eg for that grit map you're bound to lose a few units, so just build two recons per turn for the last 5 turns to compensate :p right. the grit map took me a lot of time (i think i ended it at 15 or 16 moves). it was essentially a sneak attack with mech units 8:31 AM Yeah, the optimal strategy there is probably to push up enough tanks to kill the Rocket at the base (along with an artillery to kill the mid tank). Once you've done that, either capture or rout. Of course one must pay close attention to where he is w/r/t co power yup 8:43 AM If three points are given which makes a triangle(say triangle ABC) then for ring angles in triangle we first convert them in Direction cosines of AB,BC,CA and cos (angle A)=mod(a1a2 +b1b2 +c1c2) where a1,a2,a3 are DC's of AB and b1,b2,b3, are DC's of CA My question is why we use mod? If any triangle is obtuse then it will show wrong angle… Heya friends $$I=\int x^2 \sin x e^x dx$$ Using Feymann's integration trick (Special case of Lebniz rule) and$\sin x= \mathcal{I}(e^{ix})\begin{align} I & =\left.\frac{d^2}{dt^2}\int -\sin (tx) e^x dx\right|_{t=1}\\ & = \left.\frac{d^2}{dt^2}\int \mathcal{I}(-e^{itx}) e^x dx\right|_{t=1}\\ & = \left.-\frac{d^2}{dt^2}\mathcal{I}\left(\int e^{(it+1)x} dx\right|_{t=1}\right)\\ & = \left.-\frac{d^2}{dt^2}\mathcal{I}\left(\frac{1}{it+1}e^{(it+1)x}\right|_{t=1}\right)\\ & = \left.-\mathcal{I}\left(\left(\frac{-2}{(it+1)^3}+\frac{-2i}{it+1}+(it+1)\right)e^{(it+1)x}\right|_{t=1}\right)\\ 2 9:38 AM When I see the recent activity on both MO and MSE around Banach-Tarski paradox, it always reminds me of this (from Conservapedia): Quote: "The only miracle described in all four Gospels is the multiplication of loaves and fish by Jesus, thought for thousands of years to be a mathematical and physical impossibility. But the 20th century discovery of the so-called Banach-Tarski Paradox proved a mathematical basis for creating identical copies of an object out of no extra material." 10:24 AM Hello, i have a question regarding interval of validity for an Initial value problem (Ordinary Differential Equation). Should i only pick the interval for the found solution based on the following criteria? -Solution Exist -Solution is differentiable on the interval and hence continuous on the interval as well. -Interval contains the initial value given. 1 hour later… 11:52 AM 3 hours ago, by Ramanujan If any triangle is obtuse then it will show wrong angle… 3 hours ago, by Ramanujan If three points are given which makes a triangle(say triangle ABC) then for ring angles in triangle we first convert them in Direction cosines of AB,BC,CA and cos (angle A)=mod(a1a2 +b1b2 +c1c2) where a1,a2,a3 are DC's of AB and b1,b2,b3, are DC's of CA 3 hours ago, by Ramanujan My question is why we use mod? @DHMO @Ramanujan it shouldn't have the mod If not than in some cases one can get 2 obtuse angles because of negative sign no, you will only get 1 obtuse angle at most\Huge{\text{:/ }}$12:30 PM Testing: $$\Huge{G^a_b\LARGE{c}}$$ Let$p:E\to B$be a covering map. What do the components components of$E\times _B E$look like? 12:59 PM hi @Danu hi how's it going? Meh. it can only get better :) 1:14 PM hi. i need to show that$f(x) = x \ ^ 3 -3x +1 $dont have a rational root. if$m/n $is a root of$f$we get an expression that im not sure how to get to a contradiction from . someone see a way to do so? @Liad rational root theorem @DHMO thanks but im not allowed to use it :/2 @Liad then prove it before using it @DHMO dont you see another way? can't you just substitute this polynomial into the proof of the theorem 1:18 PM You get$m^3 - 3mn^2 + n^3 =0$You can take$m$and$n$coprime And study their 2-adic valuation Wait, does that work ? your expression is$(m-n) \ ^ 3 $with some "fixing" maybe i can use it What I said works 2-adic valuation? If$m$is even so is$n$That's not possible Same goes if$n$is even And if both are odd, it's not possible since you have the sum of three odd numbers, which is odd thus not 0 2-adic valuation might be better known as parity though :p @Astyx No, it is not the same as parity 1:22 PM Meh, I'm tired You're right though they are related However here I'm only using parity hm. could you explain why it is possible that m and n is even? and why m even implies n even? Sure @Liad m is even, so$m^3-3mn^2$is even, so$n^3$is even, so$n$is even 1:27 PM$m$and$n$can't be even because they are coprime And see DHMO's answer for the second thing huh. ok . nice! [Divergent series] The partial sum of the divergent series $$S=\sum_{k=1}^{\infty}(-1)^{k+1} k$$ has an interesting pattern (NB all kth partial sums are finite by definition, thus they have a well defined value):$S_1=1S_2=-1S_3=2S_4=-2$...$S_{n(even)}=-\frac{n}{2}S_{(n-1)(odd)}=\frac{n}{2}$@Secret not interesting {1,-1,2,-2,3,-3,...} However I suck at series computations @Secret What computation? 1:35 PM O, that I refer in general for anything to do with infinite series, divergent or not. I am very bad at finding close forms, patterns etc. in series. The only thing I am somewhat ok at is the residual knowledge from my year 1 on how to check whether a series converges by ratio test, integral test, p series test etc. Me suck at infinite series is one reason I tend to avoid infinite structures until later and why my analysis is also kinda suck > Me suck at infinite series @Secret I don't see what you are trying to do with this one. Hello!! We have the field extension$E/F$and we also have that$F(\theta)\leq E$. Why does hold that$|\text{Gal}(F(\theta)/F)|\geq |\text{Gal}(E/F)|$? @MaryStar It doesn't the inequality needs to be reversed That is one of simpleart's challenges. While I am well aware that divergent series have no well defined values (and I am **not** going to compute the so called 'value' in the fashion of 1+2+3+4... because that is nonrigorous cheating.) The challenge has a hint that somehow, some notion of manipulating it is actually rigorous and I suspect we are perhaps calculating some kind of limit of an infinite polynomial with coefficients from that series such that setting the polynomial at$x\to 1$will arbitrarily gave a limiting value that the series should be, analogous to computing limit of func That is, what I think the problem is as follows: Consider an infinite polynomial$p\in \mathbb{P}(\mathbb{R})$such that the coefficients of$p$is given by the series$S$. find $$\lim_{x\to 1}p(x)$$ 1:46 PM @Secret What is an infinite polynomial? Do you mean a formal power series? ah, yes fogot that polynomials must have a finite degree, yes, that's a better term well, we can equally say it is a power series where its coefficients$a_n$are given by the sequence {1,-2,3,-4,...} @Secret it doesn't converge @DHMO Well, as a complex power series it does have radius of convergence$1$well, it is 1/(1+x)^2 right (x+1)f(x) = xf(x) + f(x) = {1,-1,1,-1,...} (x+1)^2 f(x) = {1,0,0,0,...} = 1 so f(x) = 1/(x+1)^2 I am looking at the proof of the following theorem: Let$E/F$be a cyclic extension of degree$n$and$F$contains a$n$-th primitive root. Then$\exists a\in F$such that$E$is the splitting field of$f=x^n-a$and$f$is irreducible over$F$. The proof is the following: We have that$\text{Gal}(E/F)=\langle \sigma \rangle$,$\text{ord}(\sigma)=|\text{Gal}(E/F)|=[E:F]=n$. So, the$id, \sigma , \ldots , \sigma^{n-1}$are different. From a theorem we have that$\exists \beta\in E$: $$\theta:=\beta+\omega\sigma (\beta)+\ldots +\omega^{n-2}\sigma^{n-2}+\omega^{n-1}\sigma^{n-1}(\beta)\neq 1:54 PM Here's my attempt:$$\lim_{x\to 1}p(x)=\lim_{x\to 1}\sum_{k=1}^{\infty}(-1)^{k+1} \frac{k}{x^{k+1}}=\lim_{x\to 1}\sum_{k=1}^{\infty} \frac{S_k}{x^{k+1}}$$where$S_k$denote the partial sum of$S$Thus the limit becomes some kind of laurent looking series (because it only has negative powers of$x$starting from$2$) Ok, time to do math @Arrow It's just isomorphic to the trivial$\pi_1(B)$-bundle on$E$, isn't it? By what we discussed before @Secret look, f(x) = 1/(x+1)^2 what do you think it equals when x->1? @MaryStar By the argument that all of those automorphisms are distinct. I didn't really read the rest in detail to see how they arrived at that For the middle step, if we apply p series test to all the kth partial sums, it should converge absolutely, thus the limit will converge to some well defined value Evaluating however is a different story (which is why I suck at series, especially this one has no factorials to allow me to pattern match some exp or trigonometric functions) DHMO: If that's the closed form of the given series, then its limit will be 1/2, but... isn't 1/(x+1)^2 continuous at x=1 thus it is not really representing the neighbourhood of x=1 for the power series p(x)? @TobiasKildetoft$\tau$is$\sigma$restricted at$F(\theta)$. There are$n$distinct automorphisms in$E$,$id, \sigma , \ldots , \sigma^{n-1}$. Why can there be more$F$-automorphisms of$F(\theta)$,$\tau$, than$n$? 1:59 PM @Secret p(x) = 1/(x+1)^2 as long as |x|<1 in fact, replace x with z. the limit is implicitly understood as x->1^- ah I see Hi can somebody help me with a calculus question that is driving me insane? @MaryStar Who said that there could be more? Does$|\text{Gal}(F(\theta)/F)|\geq n$means that there could be more? @TobiasKildetoft @Kane not if you don't ask it 2:05 PM @MaryStar No, it means there cannot be fewer :) I'm trying to use a comparison test to prove 1/(lnk)^9 diverges (k = 2..infinity) i can show that 1/lnk diverges, but the ninth power is throwing me off oh, and that's a series by the way :) @Kane Just compare to 1/k @Kane how do you show that 1/lnk diverges? @TobiasKildetoft So, why do we not have equality? @MaryStar The whole point is to show equality, isn't it? 2:07 PM Yes. @TobiasKildetoft @MaryStar And the reverse inequality is already clear, so once the current one is established, we are done @TobiasKildetoft well i went lnk < k => 1/k < 1/lnk.... because 1/k diverges so does 1/lnk @Kane Right, now do the same here is this the way: lnk < k => (lnk)^9 < k^9 => 1/k^9 < 1/(lnk)^9 => 1/k < k^8/(lnk)^9? Why do we know that we have$n$distinct automorphisms? We have that$\tau$is$\sigma$restricted to$F(\theta)$. We have that there are$n$district$F$-automorphisms of$E\sigma$. How do we know that these are also$F$-automorphisms of$F(\theta)$? Or do the automosphisms$\tau$not have to be the automorphisms$\sigma$? @TobiasKildetoft 2:11 PM No, use that k is greater than any power of the log for large values of k @TobiasKildetoft omg that's the fact I've being trying to track down... why is that true?? @MaryStar As I said I didn't read it thoroughly, and I don't really feel like doing so right now @Kane this is a basic thing about comparison between exponentials and polynomials (or in this case the inverse) @TobiasKildetoft yes a tutor gave me a hint to think about the inverse. I've been searching on the net to study this x > log power thing but couldn't find anything. @Kane Do you know about derivatives? @TobiasKildetoft so I can legitimately say that k > (lnk)^power Yep, I know about derivatives 2:15 PM Then one can show the inequality fairly easily by induction on the power by noting that the derivetive of any power of ln is decreasing, and from some point it is strictly smaller than$1$so far enough out it will be overtaken by any polynomial @TobiasKildetoft oh I see! Ah ok. Thanks!! @TobiasKildetoft so for my question I can say that for sufficiently large k, k > (lnk)^9? @TobiasKildetoft champion thanks mate 2:31 PM Hi @AliCaglayan Hi @BalarkaSen Hi @Balarka @Balarka have you ever done hyperbolic geometry? Hi @raman also hi @Tobias 2:33 PM Hi @AliCaglayan @Ali I know a thing or two but that's it Hi, @Alessandro Hi @SteamyRoot Ohi @Balarka I am trying to understand what it means to move in hyperbolic space 2:37 PM Hi @Alessandro and @Balarka Hi @Astyx If I imagine hyperbolic plane to be the surface of a hyperboloid which is above a disk @AliCaglayan That can mean a lot of things depending on what you mean. Move a point along a geodesic? Pick an isometry taking one point to another? then the disk is the poincare model of the plane @BalarkaSen It makes sense to move in the plane I just don't understand what it means geometrically Let me draw a picture quickly 2:39 PM In the hyperbolic maze one uses a translational isometry to get from one point to another So the hyperbolic plane can be thought of as the surface of a hyperboloid right? So if I move in the plane what is happening to my disc? Maybe there's a way to do that, but I am not familiar with it. Literally, however, the hyperboloid is not locally isometric to$\Bbb H^2$because of curvature reasons. ohh I see So it makes sense as long as nothing moves In general it's hard to visualize what goes on in$\Bbb H^2$by thinking of a surface in$\Bbb R^3$because of Hilbert's theorem. Because it's complete, so cannot have a smooth isometric embedding in R^3. Though there exists (badly not compact) surfaces of constant negative curvature in R^3- eg pseudosphere. is someone here familiar with Ostrowski's fixed point theorem by any chance? 2:50 PM never heard of it take$\phi:\Bbb R\to\Bbb R$with$\phi\in C^1(\Bbb R)$and$\alpha$a fixed point of$\phi$, it says that if$|\phi'(\alpha)|<1$then there exist a neighbourhood of$\alpha$in which the sequence$x_{k+1}=\phi(x_{k})$converges to$\alpha$for every$x_0$in this neighbourhood ah. Yeah, that should be a straightforward consequence of Banach fixed point theorem and mean value theorem for$C^1$maps hello all, anyone here familiar with d-separation? ive posted a question on this link and just wish to confirm if my answer is correct that F is independent of H given G. math.stackexchange.com/questions/2107222/… what I'm wondering about is what's the relation between the neighbourhood of$\alpha$in which convergence is guaranteed by this theorem and the neighbourhood of$\alpha$in which$|\phi'(x)|<1\$