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01:00 - 17:0017:00 - 00:00

01:34
@MatheinBoulomenos Can I do better at all than "field -> excellent"? I'm having trouble judging 'excellent' in comparison to more familiar properties
Everything in life is enjoyable except life itself
because... work
success = work + play + keep your mouth shut
@rschwieb you could do field -> complete local + Noetherian -> excellent
I don't know much about excellent rings, they're not mentioned in most textbooks (the only references I know are stacks, EGA and Matsumura "Commutative Algebra")
hi @Daminark
02:49
Yo, how's it going?
@Mathein
@MatheinBoulomenos I was just about to ask the same question about complete local rings :( I feel it's awkward to make "noetherian complete local" a new property in my database, but possibly it would be a valuable node in the graphs i'm generating
@rschwieb what database are you making?
that's really cool!
@user31415 well, I hope you find it useful! And also register and make some suggestions. That's what really improves the site.
02:52
sure
When I was in grad school, I produced some complex flowchart diagrams of ring types
and then when I got into computer programming this site was the next step in the evolution
now I'm going back to re-generate those graphs with the new database
because who doesn't like pictures?
I suppose criminals don't like pictures...
lolol
anyhow: good evening. I've got to get to bed. See everyone later!
see you!
See you @rschwieb
@Daminark I'm having semester break right now, so I'm mostly chilling and meeting people etc. How's your REU going?
03:02
Going pretty well, thanks. We have participant talks starting tomorrow
how long do they want your talk to be?
:^)
@Daminark awesome!
love your talk title
15 minutes
And yeah lmao, thanks for the idea
03:37
Hi
"How to multiply complex numbers"?
Hi @Akiva
it's a joke, he's doing a REU on something called complex multiplication
Hey Akiva!
And yeah it's that
04:11
nice @Daminark
hi @LeakyNun
hi @MatheinBoulomenos
is there really no ANT people in the entirety of this chat?
I refuse to accept ANT for analytic number theory
Why are you looking for ANT people? lol
Yeah it's algebraic
And there's a good reason why there are no analytic number theory people here
The shit's for nerds
04:13
why?
If anything it seems like the good way to use analysis in NT is via modular forms
I think that's extremely inaccurate
@MatheinBoulomenos also have you heard of the generalization of the prime number theorem?
even if you really like algebraic number theory
04:14
@LeakyNun yeah, do you know "abstract analytic number theory"?
Spectral methods (such as trace formula) are analytic in nature, and are also key tools in algebraic number theory
James Arthur considers himself an analyst for example
@user31415 we're mostly memeing lmao
@LeakyNun I have done some things that fall into the analytic side of number theory, like modular forms (including Eichler-Shimura trace stuff) and a bit on L-functions. But I can't help you with contrived integral estimates
i mean sure, meh
so the prime number theorem says that the number of primes less than $X$ is approximately $X/\ln X$
and the generalization is that the number of prime ideals with norm less than $X$ is approximately $X/\ln X$ in any number field
04:16
Personally I was less than fond of what I saw because I'm just not the biggest fan of computational calculus (even though I actually quite like stuff like functional analysis for some reason), for the most part if I'm dissing an area of math it's ironic
@LeakyNun I found that this sounds pretty cool, though I haven't looked further:
Abstract analytic number theory is a branch of mathematics which takes the ideas and techniques of classical analytic number theory and applies them to a variety of different mathematical fields. The classical prime number theorem serves as a prototypical example, and the emphasis is on abstract asymptotic distribution results. The theory was invented and developed by mathematicians such as John Knopfmacher and Arne Beurling in the twentieth century. == Arithmetic semigroups == The fundamental notion involved is that of an arithmetic semigroup, which is a commutative monoid G satisfying t...
it's like the child of ANT and ANT
oh well, when I said inaccuracy I meant "If anything it seems like the good way to use analysis in NT is via modular forms"
dissing is fine lolol, I can see why some don't like that kind of analytic number theory
it's more like an axiomaticed version of some stuff in analytic number theory
Oh that was kinda continuing the trend, for the most part I don't see this stuff as "good" or bad, taste is purely personal
04:20
haha sure
I'm aiming toward trying out modular forms because I heard that if you play your cards right it doesn't even really need to have too much analysis (in the sense of verifying convergence or evaluating series/integrals)
@user31415 I really like modular forms, but I can really live without worrying about worrying about weird integral estimates. And I'm not that interested in a paper where the main result involves the term $\frac{\log X \log \log X \log \log \log \log X}{(\log \log \log X)^2}$, I like structural results
yeah I understand that; it's just that analytic number theory is not just that kind of stuff
more weird estimates :P
okay
for some reason I'm not that excited about the prime number theorem
04:24
these things have two flavors i guess
asking why there should be a zero free region is a fairly "structural" question imho
@MatheinBoulomenos how about this analytic proof of the infinitude of primes
(of course you can further to say why the zeros are all on half line)
but small increments on zero free region is somewhat meh
04:26
there are analytic results I like: Chebotarev density or the Eichler-Shimura trace formula or Hecke's converse theorem
that's interesting
Chebotarev density is "similar" to prime number theorem
but you like one not the other :P
Chebotarev density implies that a number field is uniquely characterize by how primes split in it
I don't see that the prime number theorem implies something along those lines
@MatheinBoulomenos do you know Dirichlet convolution? Does that belong to ENT or ANT or ANT?
and do you dislike it as well?
it's motivated by analytic NT, but you can also treat it in an elementary way
I don't "dislike" the prime number theorem
well I first learnt it in ENT
04:30
I'm just saying that I'm not that excited
I see that makes sense
have you heard of the function $\Lambda$?
$\Lambda(p^m) = \log p$, 0 elsewhere
@LeakyNun yeah I heard about it
and then $\Lambda \ast 1 = \log$
take dirichlet transform and obtain $\psi(s) \zeta(s) = -\zeta'(s)$
which is just so amazing
04:57
@Secret It's actually the type float. It's not very accurate, using double is better when doing such high close-ups
Now it's nice and crispy again
I see, I should try to zoom into the chaos again when I got back home
maybe this time I can actually single out all the dense orbits
Yeah, you can just modify the code you already have. You see what I replaced with doubles. Remember that map() requires a float, so convert it.
Probably a doesn't even need to be double
Just x probably
won't hurt I think, because a float a is going to have imprecision that interacts and magnifies each time x is looped
and giving jagged paths again and obscure the chaos
 
1 hour later…
06:26
Before this year it was in a 15-way tie with many countries including Germany and Italy
I don't think thay's very meaningful with only 1 or 2 medals per country!
yay, coincidences! much hype!
We have to make math popular guys.
@AlessandroCodenotti yup, neither of the Austrian winners are "native."
06:49
I'm surprised Germany doesn't have more people
I would've expected them to have more people than, say, the UK
Hi chat!!
If a map T has period p , then what can we say about the period of the map $T^n$?
Should be the gcd of n and p
(Assuming p is the smallest period)
Err, lcm
I don't think so. if n=3 then I say the period is p.
also n and p don't need to have gcd/lcm
I think its np
Like it will be like dominos
@Leaky what do you mean they don't have to have an lcm?
06:58
like 1 and sqrt(2)?
Hm, perhaps he was thinking of non integer case
Uh, what?
How do you have non-integer period?
Compose 2 and a half times?
Yup my bad..
I see we're interpreting the question in two completely different ways
If you're talking about a semigroup of operators sure but that's not the same thing as a map having a period, which is the case that Baymax was talking about
06:59
Like for n=2
maybe p=np and you're both right?
$ToT(x+2p) = ToT(x)$
@MatheinBoulomenos oooo thats the open prob thingy😁
Ohhhh
I was thinking period in the sense of dynamics lmao
@Daminark so what was your thought in terms of dynamics
I thought you meant the period of a map as a weird way of talking about an idempotent operator
07:03
@BAYMAX I don't there is an answer that doesn't depend on the map
Okay so I still buy that it's $p$
for example if you consdier the map $f(x)=[\sin(2\pi x)]$, then $f \circ f$ is constant
who is ToT?
Or at most $p$
what is the period of a constant map?
it's perioid wrt to every number
but $\sin(2 \pi \sin(2 \pi x))$ is not constant
so according to wikipedia, constant functions don't even have a fundamental period
so "the" period of $T^n$ might not exist
if it becomes constant
07:06
Claim: if a continuous function $f:\Bbb R \to \Bbb R$ satisfies the condition that $[f(x)]^2$ be a constant, then $f$ is a constant.
07:28
@LeakyNun its T composed with T, I got it, its np, thanks for looking to my query
Thanks all!
08:24
Any help would be much appreciated.

https://math.stackexchange.com/questions/2875778/showing-existence-and-uniqueness-for-a-solution-to-a-homogeneous-fredholm-type-i
08:48
$$\left|\frac12\int_0^1\sin(xy)f(y)\mathrm{d}y\right|\leq\frac12\int_0^1\left|\s‌​in(xy)f(y)\right|\mathrm{d}y\leq\int_0^1 |f(y)|\mathrm{d}y\leq\frac12||f||_\infty$$
@Oskar the function $f(x)\mapsto \frac12\int_0^1\sin(xy)f(y)\mathrm{d}y$ looks like a contraction of $C([0,1])$ to me
 
1 hour later…
09:52
Thanks, Alessandro. I'll look into it and then accept your answer.
10:12
Hi. In MSE elections if i choose only 1 person and do not choose my second and third choice, will that be okay? the vote will still count or do i have to cast all three choices?
why not just make all the choices the same
cant do that
not allowed
hmm, good question
in The Reading Room, 58 secs ago, by Rand al'Thor
@user1732 Yes, the vote still counts regardless of how many choices you make.
from a mod^
10:30
@user1732 thanks
np, pal
10:51
@TobiasKildetoft Yesterday, we started out with saying that the sign of the group operations "should be a homomorphism to the group with two elements". I got a problem with that. Aren't there often different 1-dim non-trivial irreps? But only one specific should correspond to the "sign". How to cover that?
@Rudi_Birnbaum Well, there can be more than one "sign". But an important thing about the sign is that it should only be able to take two different values (else the word would be a strange choice). So all group elements need to either act as $1$ or $-1$ on a sign representation
@TobiasKildetoft Oh I see. That means my problem actually has nothing to do with the sign representation.
or at least it seems under this premise at the moment to me so...
Lets put it that way: I know of three different point groups which have order two.
And only one of them ($C_2$) is of interest for my problem.
The other ones would be $C_i$ ("inversion") and $C_s$ ("mirror plane").
homomrphisms to these latter are irrelevant for my problem. Moreover I kind of "know" that temsor squares of two-dimensional irreps from the non-isometric groups contain exactly the irrep corresponding to the $C_2$ normal-subgroup in the anti-symmetric part.
Which is kind of my first "theorem".
(of three)
The second one is that when restricting to a maximal subgroup such that the 2-D irreps split into 1D irreps (say $\rho_1$, $\rho_2$), then $\rho_1 \otimes \rho_2$ will still contain the restriction of same non-trivial 1-D irrep (corresponding to a $C_2$ subgroup).
Question: Is this already a trivial conclusion?
11:12
Is anyone over here good at Game Theory by any chance?
@TheArtist combinatorial or the regular kind?
@TobiasKildetoft regular kind.
then not me
shouldn't the combinatorial be built-upon the regular kind?
I like the third kind better
11:18
@user1732 No, they are essentially two completely different topics that just happen to share part of the name
oh? pardon me then :-)
@TobiasKildetoft Maybe could you explain to me what happens when do this tensor-product formation of ireps?
11:33
ok , np!
$$x^{\frac1{\log x}} = e$$
does this mean to sum over all n divisible by c? $$\sum _{n mod c }^{}
$$
\sum _{n \,mod c }^{}
\sum _{n\, mod\, c }^{}
$$\sum _{n\, mod\, c }^{}$$
@LeakyNun $x\in$?
(for $1$ its not defined anyway ...)
@TobiasKildetoft maybe a simpler question: When I have an axis of rotation of order $\ge 3$ I get an 2D irrep. When I square that irrep it decomposes into something containing the irrep for the $C_2$ rotation (but along the same axis). Why is that?
11:56
@Rudi_Birnbaum I am not quite sure how that axis of rotation gives you an irrep
@TobiasKildetoft: I mean a group containing an ....
A group containing (exactly one) rotation of order $\ge 3$ contains exactly one 2D irrep.
(well over the reals)
@Daminark they have more Physics and Chemistry Nobel Laureates.
but yeah, i thought that way also
@TobiasKildetoft A group containing (exactly one) rotation of order $\ge 3$ contains 2D irrep(s) over the reals.
and the direct tensor square of these irreps contains a 1D irrep corresponding to a $C_2$ subgroup.
alternatively (talking over $\Bbb C$) you have to take the direct tensor product of the pair of complex conjugate irreps, to get that.
Addition: If such a $C_2$ is not contained in the group (if the axis is of odd order), then still the (anti-symmetric part of the) tensor square/product contains the trivial irrep, which in this case represents the rotation around this axis.
12:38
@AlexClark Soon
Does the second order derivative is bounded mean that the first order derivative is bounded as well?
@Rudi_Birnbaum $x > 1$
@OskarTegby no, consider $x^2$
@LeakyNun : Right.
12:44
So how's uni?
It's ok
I'm trying to prove that $|f''(x)|\le M$ implies that $f$ is Lipschitz. When the first derivative is bounded everything is easy. Then it's just the mean value theorem, but I'm a bit lost here. Any clues?
Learning math, but the rest of the time is spent sulking over stuff
What stuff?
Arbitrary things
12:46
Did you look at the hyperrationals construction?
The story of my life.
Yeah, I can explain it now if you want
Sure. @AlessandroCodenotti: Was that question asked to me?
@OskarTegby Nope, to Balarka
Okay.
I want to do integration, but $\int |f''(x)|dx$ doesn't equal $|f'(x)|$. Right?
I need to go from second order derivative to first order.
12:52
omg @BalarkaSen is back
@BalarkaSen Sure
@OskarTegby not equal
\o @BalarkaSen
:D
@Alessandro Hmm, I need to scurry off in a few minutes actually, but I can do it tonight I think
Hi @Leaky, skull.
(Also bye)
cya, pal
12:54
Okay, @Leaky. You guys are smart. If not integration, and no relationship between derivatives like that, then what?
@BalarkaSen Of course, no problem!
I've been reading some notes on differential forms to learn about the de Rham cohomology so I might have some questions concerning that as well for you :P
@OskarTegby $|\int \cdot| \le \int| \cdot|$ though
Okay. So $|f'(x)|=|\int f''(x)dx|\le\int|f''(x)|dx=\int Mdx=C$?
why $|f''(x)| = M$?
13:04
Oh! I meant $|f''(x)|\leq M$. It was given in the problem.
That computation can't be true as your counterexample showed.
that computation is true and my counterexample is irrelevant
because here the integrals are all definite
(otherwise they would mean nothing)
Okay. So there we have it!
no we don't have it
No?
Don't we just use the MVT from here?
put your limits in and see what you have derived
I'm very confused by our calculations
13:08
Okay.
I don't think your statement is correct
26 mins ago, by Oskar Tegby
I'm trying to prove that $|f''(x)|\le M$ implies that $f$ is Lipschitz. When the first derivative is bounded everything is easy. Then it's just the mean value theorem, but I'm a bit lost here. Any clues?
Consider still $f(x) = x^2$
It's from an exam.
Quote your exam question verbatim
"Assume that $f(x)$ is a twice differentiable function on the open interval $(a,b)$ and that there exists $M\geq 0$" such that $|f''(x)|\leq M$ for all $x\in(a,b)$. Prove that $f$ is Lipschitz in $(a,b)$. This means that you need to show that for all $x_1$ and $x_2$ in $(a,b)$, one has that $|f(x_1)-f(x_2)|\leq C|x_1-x_2|$ for some constant $C\geq0$."
I guess skipping the mention of the interval was crucial.
yes, it was
13:19
Mea culpa.
you can prove that the derivative is bounded as well
I'll give you a bound
$$|f'((a+b)/2)| + M(b-a)/2$$
Where did that come from?
try proving it lol
There's not even an inequality involved. What's it even a bound of?
the derivative, as I said
13:25
Oh, okay.
"you can prove that the derivative is bounded as well / I'll give you a bound [of the derivative]"
Is there any test to apply if a function is asymptotic to another function?
Right.
@MatsGranvik for example?
The derivative of Riemann Siegel theta and the riemann zeta function.
13:29
Take the limit of the ratio? That's just the definition, of course
Is it possible to write formulas down here via MathBot?
When one lets the parameter cc go toward zero one gets the Riemann zeta function for the blue function.
(*start*)
f[t_] = D[RiemannSiegelTheta[t], t];
nnn = 60
cc = 10^-10;
g1 = Plot[(f[t] + cc + EulerGamma), {t, 0, nnn},
PlotStyle -> {Thickness[0.004], Red}, PlotRange -> {-2, cc + 5}];
c = 1 + 1/cc;
g2 = Plot[
Re[Zeta[1/2 + I*t]*Zeta[c]/Zeta[1/2 + I*t + c - 1]], {t, 0, nnn},
PlotRange -> {-2, cc + 5}, PlotStyle -> Thickness[0.004]];
Show[g2, g1, ImageSize -> Large]
(*end*)
The same spectrum with one part of the analytic continuation removed, and divided by t.
13:37
Is there a proof for the "Hilberts Hotel" theory? Professor told us to take a look at it.
Couldn't find one.
@KeJie we can see latex here; if you want to see latex follow the room description
Hilbert's paradox of the Grand Hotel (colloquial: Infinite Hotel Paradox or Hilbert's Hotel) is a thought experiment which illustrates a counterintuitive property of infinite sets. It is demonstrated that a fully occupied hotel with infinitely many rooms may still accommodate additional guests, even infinitely many of them, and this process may be repeated infinitely often. The idea was introduced by David Hilbert in a 1924 lecture "Über das Unendliche", reprinted in (Hilbert 2013, p.730), and was popularized through George Gamow's 1947 book One Two Three... Infinity. == The paradox == Consider...
1+z/(1-z) = 1/(1-z)
In terms of generating functions, that's Hilbert's hotel right there :P
@Leaky Nun Thanks for the video.
And is there a general proof for that?
14:07
I don't fully understand that inequality, @Leaky. Why do you consider $|f((a+b)/2)|$, and what is $M(a-b)/2$ really? What's the idea behind it?
Call me stupid. I'm sure I am, but I want to understand this.
@OskarTegby you need to think of it graphically
Yeah. I get that. I've drawn my sketch. The point |a+b|/2 is the midpoint of the interval, but I don't get what's so special about that point.
I'm using it as a reference point
Is this room also suited for easier question? I have a question but im a bit afraid to ask it.
Okay.
@KeJie What's there to be afraid of? If someone thinks that your question is stupid then they're stupid.
14:12
$$\left|f'(x) - f'\left(\frac{a+b}2\right) \right| = \left|\int_{\frac{a+b}2}^x f''(t) \ \mathrm dt \right| \le M\left|x-\frac{a+b}2\right| \le M\frac{b-a}2$$
Ive done something but I wonder If I could have proceeded differently at that point.
I mean the point ive marked with red.
it's just an old telescope @KeJie
What is an old telescope?
2
Haha! :)
https://en.wikipedia.org/wiki/Telescoping_series

Right?
14:15
right
Ok, so my question is If there was another way to finish things off?
The part that is marked in red.
Okay. So, obviously it was $|f'(x)|\leq |f'((a+b)/2)|+|f'(x)-f'((a+b)/2)|$ that was being used.
what do you mean exactly by "finish things off"
@OskarTegby right
I want to know if there is a alternative way to solve it.
and which steps do we need to use?
14:24
@LeakyNun What do you mean?
it's a telescoping series
Yes.
so just telescope it
Ive already solved it but that was not what I actually meant.
Is there another way to express: $\frac{n}{n+1} + \frac{1}{(2n+3)(2n+1)}$ ?
Other then what I did.
Might seem like a dumb question but I can't think of any other way.
Stop saying that it's a dumb question.
14:33
I said it might seem like one for the people on this site because most here are master students etc
Hello, if I roll a die twice, are these events dependent or independent? (Not a homework! I just read in a website that these are independent and in my textbook a similar example exists and according to that, these are dependent, so I'm confused!) A: Getting two sixes B: Getting at least one six
I believe they are dependent because if I get B, the probability of A is higher, and if I don't the probability of A would be 0.
One tip: Never mention anything about homework in any way no matter what. Ive once asked a question on this site where I mentioned the word homework and I got spammed with downvotes.
It's not even my university course :( Thanks for the tip.
It's because it's kind of like cheating to ask people for help when you're being graded on it. If you can't do it while in the exam hall, then you shouldn't be able to do it when at home.
I don't think it's cheating. I have asked some questions related to "homework" ( not anymore) but not because I wanted solutions but because I was seeking for some hints or verifications on my solutions.
14:47
Would you get that when in an exam hall?
If you're told that you may cooperate, then it's more of a fuzzy line.
Just do whatever you want, but that's the reason for the downvotes.
Come one. Don't tell me you never worked in a group with people to solve a task or something.
I honestly don't know what to say, I'm not cheating or on any exams I swear! :D
Of course I have! I'm just trying to answer your question.
You can't tall me that you don't use the internet. Everybody has to start somewhere.
Bro, I'm on your side. I'm just said why people don't like it.
14:48
And I have no study group in real life that's why im using this site sometimes/talking to people on discord. How can you consider that to be cheating?
Easy now. Listen to what I said.
WoWoWo, I apologize guys asking my question the wrong way, could you please take a look at my question?
I have im just saying that not every homework question has the intention to cheat.
I all ears.
I'm*
nvm, I thought you guys were fighting ;-D
14:52
No, haha.
Haha! No. I wasn't at least. It's hard to tell when just having text to go on. I mean, we don't know each other or anything.
Sorry to interrupt.
Don't be.
Sorry for troubling you guys
You didn't do anything wrong in my opinion.
14:54
Don't be! I would've answered if I didn't have to study for this stupid exam.
\o @MikeMiller
Just ask a question on the site and that non-chat users will see it too!
Alright, thanks
No sweat.
Yeah, the main site is the way to go first.
15:09
I added an answer to a question which is 6 years old...
0
A: How to construct a dense subset of $\mathbb R$ other than rationals.

MasonFor any fixed $s$, $$\Bigg\{\sum_{n=1}^\infty\frac{a_n}{n^s}: \{a_n\}_{n=1}^\infty \text{ is a periodic sequence of integers.} \Bigg\}$$ Is a countable set which is dense in the reals. To demonstrate this we can apply checkmath's answer.

My answer seems fine right? No issues?
Just a thing I was thinking about anyway and I figured I'd type it up.
15:42
How do I convert mathML into latex?
user131753
Does anyone know whether the category of topological spaces is a reflective subcategory of the category of closure spaces?
16:25
0
Q: About invariant of qudratic space

sscool One associates "in-variants" with the quadratic space so that the space is determined byits invariants as completely as possible. What is meaning of this statement? Also I know that dimension is invariant property for vector space and it is equal or greater than 0 so we can construct grothen...

01:00 - 17:0017:00 - 00:00

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