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11:28
Yep it was still here after last time :D
Know that feel, bro.
Really cozy place. Nice for solving discretized differential equations.
What are the characteristics for being a good place for doing that?
Good question. Having access to good affordable beer helps a lot for me though :D There is also some positivity atmosphere which is kind of inspiring. Don't know exactly which places you would get away with that within the cultural norms on daytime on non-weekends. Well the beer that is, hopefully being positive is still accepted ;)
@Faust not going so well , think am gonna take the re-exam next month >< to much stuff and I dont feel i got all of it
@Faust how is it going for you ?
11:41
Haha. My favorite place for doing algebra is the pub. I usually buy one of my favorite beers once I go there, and then sip on it for hours. If I drink to quickily my focus goes away too quickily. Then I'm vulnurable for friends passing by and chatting with me. When I do analysis I have to sit in a quiet place, I can't do those long computations otherwise. I really hate those...
I just do algebra whereever I can
e.g. in the analysis lecture
Of course, but if I were to choose one place.
Haha, exactly! I can't keep focus anyway.
@MatheiBoulomenos ich weiss, Sylowmeister
Hi @Mathei
@BalarkaSen hi
11:43
hey
Finally, some action around here.
Hi @BalarkaSen @LeakyNun and everyone else
@OskarTegby would you prefer group action?
You know I would.
11:44
@LeakyNun This is a Christian chatroom!
hehe, sind doch ein paar deutchern hier im Cafebar :D hmm, welcher genus hat das wort Cafebar?
Bar ist weiblich
@MatheiBoulomenos btw your way of saying that it normalizes only one subgroup and hence fixes no other subgroups provided for me a perfect demonstration of Sylow third theorem
@LeakyNun that's actually how you prove Sylow's thid theorem
@LeakyNun That is exactly how I showed the third Sylow theorem in my lectures
11:45
yes but it was just words for me
Ist doch ein sehr gutes atmosphar hier :P Vielleicht als es Freitag ist...
@MatheiBoulomenos I mean when I lass es auf meines Buch es wäre nur Wörter für mich
wäre es*
is it lass?
I’m conjugating verbs randomly now
When I'm done with probability I'm going to burn this book.
fyi
*Als ich es in meinem Buch las, waren es nur Wörter für mich if you want past tense
wäre is not indicative
11:49
las mit ein s. it is also an article in spanish. las und los, was ist los? lol.
major language confusion reporting in.
@MatheiBoulomenos waren?
@mathreadler entonces hablas tres idiomas
Well, is it's more than one word you're talking about, right?
es?
es waren?
es war
?
wir waren?
sie waren
estaban
11:51
but dont trust me, i tried learning hollandich after deutsch. it is seriously bad for grammar.
well, pronounciation too :D
it's kinda strange in German
But in this case, you have to use the verb in the plural
many words are very similar, but the pronounciation is really strange.
If you use "es" + a form of to "sein" and another noun in the plural, then that form of "sein" must be plural
11:53
i once had it described to me that listening to a dutchman is like listening to a seriously drunk Kölner.
it's kind of like this little "es" is not a real subject
If you want to emphasize a verb, you can always mix the word order up by introducing an "es" in the beginning. Like "Es sind viele Leute hier im Chat an Deutsch interessiert."
i see
please tell me when I've had one too many. I'd better log off then, lol.
is it okay to upload images on this chat here?
This is an attempt to make a function morpher, but as you can see one of the edges get some unnatural overshoot. any ideas of a differential equation constraint I can use to reduce it? The pink curved curves should all be below the box one.
12:11
o..o
the curved pinkies should be shorter so they mirror the violet and blue ones.
I think if you want to smooth a function, you would usually take the convolution with some kind of bump function
But I'm not analyst so idk
Yep and I have done that many times. Now I am trying to solve it implicitly with a discretized differential equation. Like an adaptive filter fulfilling some constraint expressed with norms of expressions with partial differentials. But I must have missed something because it should not grow completely uncalled for like that.
That is true
But it is Friday, maybe it is just happy. Lol!
12:23
In general one takes convolution with approximate identities to approximate a function smoothly by an arbitrary small amount, I think
Yeah, that's how you show that the space of compactly supported smooth functions is dense in $\operatorname{L}^p(\Bbb{R}^n)$ for $p \in [1,\infty)$
Yes I can do that already. I want to learn to do it in a new cooler more adaptive way.
But you have a very good point, I want the local average to not differ too much. I think I can express that with a convolution equation.
I don't know much about discretized differential equations so unfortunately I'm of no help. Sorry.
[Random]
Going to add an extra axiom to the logic of a foundation of mathematics:
Axiom of You cannot remain silent
o..o
12:29
Given a predicate P, if P is undecidable or its existence or nonexistent are both consistent in the model then:
@mercio i lol'd irl
do you have axioms that depend on the model ?
In the model, there exists objects A that serves as examples and object B that serves as counter examples to the given predicate
also are you talking of "undecidable" with this very axiom included in the system ?
so if there is A and B such that P(A) and not P(B), then something ?
12:33
oh logic, i am not very good at it. i try and stick to my smooth models.
actually, I am not sure whether I can use intuitionist logic to partition undecidable predicates into examples that satisfy it and some examples that don't and some examples that are ne Ther. But this preliminary idea is I want to create a foundation such that some objects e.g. obey the axiom of choice and some obey the negation of it. Hopefully it will not lead to contradictions...
:/
... some examples that are neither ?
oh yeah "A or not(A)" is not an axiom ?
so when you say that you partition predicates into examples
are examples elements of predicates ?
or subsets ?
If P or not P are both consistent, then there are examples that satisfy P and examples that satisfy not P (and there are examples that satisfy something in between if intuitionist logic is used). If the foundation is set theory, then the examples will be sets
so basically you can have both nonmeasurable sets and infinite dedekind finite sets because axiom of choice becomes a predicate that can be attached to sets of you want
But yeah that's what I had in mind for that axiom of you cannot remain silent
youtube.com/watch?v=J0QlPfTmwcw , good german song by the way.
I'm not sure what completeness of first order logic becomes without the law of excluded middle
12:46
That I will have to investigate... I am not sure what will result since intuitionist logic is commonly used in constructive mathematics but ZF has non constructive objects
One thing that need to be account for is suddenly there are now countable number of truth values besides true, false for each predicate
Btw a side note. I am pretty sure the predicate "There exists A" cannot have excluded middle violated, otherwise what does it mean for an object that neither exists nor nonexistent?
Actually no, since both $\exists,\forall$ are needed in intuitionist logic, so there is actually something in between...
gah. I need to stop thinking about semantics and focus on the syntax
13:04
Is there a theorem or anything that says that a function R^3--> R^3 is an isometry if it preserves distances between points and the origin?
Yeah it is, nevermind.
Actually, I might be able to include nonmeasurable sets as follows: There exists sets that are well orderable with some non constructive choice function, and there exists sets which cannot be well ordered
This should result in a ZF set theory where the axiom of choice is replaced by the existence of non constructive well orderings that can apply to some sets
Therefore, if the choice functions are never invoked, you have ZF-C
If this works, I will be able to call this axiom the Axiom of Optional Choice
Man, the trains today are terrible. Waited in one for about an hour this morning because the one in front of it had broken down. And this one is now more than 30 minutes delayed and they are not sure when it can leave.
@anakhronizein define isometry
Ahh, never mind. It just started moving now.
bijective function T:R^3\to R^3 such that d(T(x),T(y)) = d(x,y) for all x,y in R^3
13:18
oh lol
I didn't notice the subtlety
@LeakyNun but you agree it is true, right? ;)
what if it doesn't preserve the origin ?
@anakhronizein I don't
If the metric is not translation invariant...?
you can mix points that are at the same distance from the origin
making it not an isometry
13:31
Is it not the same as saying <T(x),T(x)> = <x,x>?
i.e. ||T(x)|| = ||x||?
Oh it would have to be then linear.
You're equipping R^3 with the standard metric, then.
That's something important you should've mentioned
just define a function that is the identity on $\Bbb R^3\setminus S^2$ and some rotation on $S^2$
@TastyRomeo it is for a geometry course, not anything like analysis.
Norm-preserving linear maps on R^n are isometries, I believe
This should follow from the polarization identity
Sure, but I can't guess that through the internet :P
13:33
@BalarkaSen just mix the basis elements
@MatheiBoulomenos Did you look at that exercise yet?
@BalarkaSen yes, it needs the linear, I just overlooked that.
@BalarkaSen don't you need contuinity?
@TobiasKildetoft I have been busy Sylowing
@MatheiBoulomenos Linear maps are continuous...
I missed the linear thing
13:34
@LeakyNun Huh?
Well, that exercise is all about Sylowing
The point is the inner product on R^n is determined by the norm; so it can't happen that you'd preserve the latter but not the former
I hope you're not trying to contradict that
@TobiasKildetoft Well the first part is easy
If you know about nilpotent groups
Yeah, that is just there because it may be useful for a later part
@BalarkaSen what if I take the action of (12)(34) on the 4 basis elements of R^n?
@TobiasKildetoft I heard Sylow
@MatheiBoulomenos I'm on a lecture now lol
but I'm reading Galois so I can't Sylow
13:39
@Leaky What if what? I don't see a question.
nvm
can a linear norm-preserving map reduce the dimension?
no, it's always injective
it's an isometry...
i always find the use of the word isometry occasionally confusing, when some authors also use it for isometric embedding
13:41
@MikeMiller oh yeah that is true
I find literally all words in algebra confusing, lol.
i prefer isometries to be surjective but meh
words in algebra are fine imo
no they are silly
sillier than thou
words in category theory
13:45
almosed fuxed it
*fixed
PROPOSITION 19.7 If $K$ is a subfield of $\Bbb R$, generated by the coordinates of points in a subset $P \subseteq \Bbb R^2$, and if $\alpha$ and $\beta$ lie in a normal extension $L$ of $K$ such that $L \subseteq \Bbb R$ and $[L:K] = 2^r$ for some integer $r$, then $(\alpha,\beta)$ is constructible from $P$.
Does this look right? (cc @TedShifrin)
dammit autotourettes
what if $P$ is empty ?
@mercio then it generates $\Bbb Q$
@Mathei Ayy, watch abstract nonsense getting rekt by MO here (check out Anton's answer + comments)
13:52
and what points are constructible from P ?
@mercio $(0.5,0.75)$, for one
@BalarkaSen I don't really see how how abstract nonsense is getting rekt
arent you reading Dylan Wilson's comments
I mean, you can go overkill with things over than category theory
oh you assume you have $(0,0)$ and $(1,0)$ available always ?
14:00
Well, true. But trolling with overkill category theoretical proofs seems to be a prevalent phenomenon in mathematics :P
and I'm afraid it scares off some students thinking mathematics is nonsense.
I think there's a post my J P May where he disses at somebody for calling a more concrete construction (model categories) replaceable by the infinity theory
but maybe some consider that more of a feature than a bug
So that's what I meant by "abstract nonsense". Generalizing everything over 9000x times is not the best approach in mathematics - according to the people who uses abstract arguments in their day to day life in mathematics, not me - and is generally frowned upon in a large section of the mathematical community
And that answer is a prime example of such a phenomenon
@Balarka I guess that perception is mostly based on antipathy towards category theory. If someone uses a lot of analysis or toplogy to prove an easy result while also giving deeper insight, would you describe this as trolling?
14:04
@MatheiBoulomenos You're trying to imply guys like Dylan Wilson or JP May himself perhaps has antipathy towards category theory? lmao
I was responding to "But trolling with overkill category theoretical proofs seems to be a prevalent phenomenon in mathematics :P"
Yes, and that is not my opinion, that is the opinion of the people I quoted.
Who uses category theory like fast foods in their daily livelihood in mathematics
categories are for pussies, real men use dogegories
or real dawgs even
@mercio sure
It's not like I ever said you should always use $\infty$-categories if you can
The other day, we were talking about category theory in Aluffi
14:07
The point being, ncatlab style category theory is frowned upon even in the homotopy theory community
which is like the basics of the basics
And I don't see anything wrong with Anton's answer as the OP explicitly asked for an abstract nonsense proof tbh
@MatheiBoulomenos apply abstract nonsense in Galois/Sylow theory?
I don't think it works for Sylow theory
Galois?
14:10
In mathematics, Grothendieck's Galois theory is a highly abstract approach to the Galois theory of fields, developed around 1960 to provide a way to study the fundamental group of algebraic topology in the setting of algebraic geometry. It provides, in the classical setting of field theory, an alternative perspective to that of Emil Artin based on linear algebra, which became standard from about the 1930s. The approach of Alexander Grothendieck is concerned with the category-theoretic properties that characterise the categories of finite G-sets for a fixed profinite group G. For example, G might...
thanks
Grothendieck's Galois theory isn't even the most general form
I once tried to read Borceux and Janelidze's book
are all these abstract nonsensers french?
imo claiming that the abstract nonsense in Aluffi is useless is ridiculous as well
14:14
a lot of them are but no not really
@Mathei I liked Aluffi personally, so :)
romantic french music starts playing oh great.
The thing I don't understand is where all the hate comes from
As someone doing research in higher representation theory, I too find that it is possible to go overboard with categorical stuff.
I mean, you can just ignore the higher category stuff if you don't like it
grate! ungrateable cheese even.
14:16
@mathreadler plays Russian nightclub music to counterattack
and by Russian nightclub I mean Vitas
@BalarkaSen lol wut?
@BalarkaSen : hehe, maybe later tonight unless I pass out before that :D
french cheese is well known to build a stable stomach for weekend activities... hmm, maybe not :D we'll see!
@BalarkaSen ive been to a few drinking establishments in russia and none of the looked like that
there's some serious cocaine shit going on there, so that's understandable
brlrlrlr ahaha
14:21
I had to turn it off the sound was grateing and the video more disconcerting than the actual sound
havent tried it yet, there's quite good music in this place already.
Hi @Alessandro
Morning
14:22
hi @AlessandroCodenotti
But there is less than some hundred meters to russian visa house. Never been there actually.
Hi @AlessandroCodenotti
it seems that I missed another user with good tastes in movies this night @Balarka
so much hw
so little brain power
@Alessandro yeah
he seconded two of your reco's
@mathreadler In this place, as in where you live? (Where are you from?)
@TobiasKildetoft I think I got (2.)
Have to write out to be sure though
14:29
No, where I am right now. In Prague. =) Really nice city. For many definitions of nice =)
@Balarka you said you think about the Hilbert basis theorem geometrically, can you elaborate on that?
@BalarkaSen : Where are you?
@Alessandro So do you know what an affine algebraic variety is?
@mathreadler Right here :)
zero set of a (system of) polynomial?
14:30
In Praha?
Or in the idea world of the internet.
In the world of internet indeed.
@Alessandro Right, if you denote the affine $n$-space over $k$ by $\Bbb A^n$ (this is as a set the same as $k^n$), then it's just intersection of the zero locus of a each polynomial in a subset $S \subset k[x_1, \cdots, x_n]$.
I.e., "common zero locus of a bunch of multivariable polynomials"
@TobiasKildetoft (1.) and (2.) down! this exercise is really cool
@Alessandro the H-dawg's basis theorem says even if $S$ is infinite, you can choose a finite subset $T$ of $S$ such that common zero locus of $T$ is exactly the same as the original affine alg variety
So every affine algebraic variety can be cut out by just finitely many polynomial equations
That is all
Ah, that's a cool perspective
14:35
Well, in the modern formulation Hilbert's basis theorem is not about fields at all
whats with the g:s turning to h:s in czech? geese usually don't like hay.
sure just replace everything i said by the affine scheme
The formulation I know is $A$ is Noetherian iff $A[X]$ is
@Alessandro Right, what I said is an application of $k[x_1, \cdots, x_n]$ being Noetherian
14:37
A special case of Hilbert's basis theorem if you will
@BalarkaSen if $R$ is not an algebraically closed field, then $\operatorname{Spec}(R[x_1, \dots, x_n])$ has more closed points than $R^n$
@Mathei I know, but don't care :)
I do algebraic geometry over algebraically closed fields, and no less
so restrictive
that's how i roll
let $A$ be a Noetherian ring and let $0\longrightarrow M_1\stackrel{f}{\longrightarrow} M_2\stackrel{g}{\longrightarrow} M_3\longrightarrow 0$ be a short exact sequence of $A$-modules, I want to show that $M_2$ is Noetherian iff $M_1$ and $M_3$ are
14:45
Do you know the analogous result if "Noetherian" is replaced with f.g.?
btw the hypothesis that $A$ is Noetherian is not needed
yes and I suppose I have to use it here
over a Noetherian ring f.g. is equivalent to Noetherian for modules
you could just use the ascending chain definition too
ah, right, I was assuming too much, $A$ is not Noetherian here
But I guess you need something similar to this exercise for that so nevermind
14:47
there's a correspondence between submodules of M containing N and submodules of M which you have to use iirc
if you have any submodule $N \leq M_2$, then $0 \to f^{-1}(N) \to N \to g(N) \to 0$ is again exact
now apply the result for f.g.
So I pick a submodule $N\subseteq M_2$ and I want to show it's f.g
you can also do the thing with chains
ah, I see, I didn't think about constructing another exact sequence with $N$ in the middle
$2^{37}-1 $ does anyone know how to show this isnt prime?
14:50
Lucas-Lehmer-Test? Not sure if you can do that by hand
it should follow from a trick involving mersenne primes
but i cant figure out which one
how do u do the lucas lehmer test i tried reading it and it looked like gibberish
well if $2^p-1$ is not prime, then every prime divisor of $2^p-1$ is of the form $1+2pk$
yeah but i cant stick $2^{37} -1 $ into m calculator n check if 75 then 75+37 etc divides it
can i mod 75 somehow?
No shit
well, 75 is not prime, so we don't need to check that one
14:58
why?
If $2^p-1$ is not prime, it must have some smaller prime dividing it
oh
and there's a lemma that says it must be of the form $1+2pk$
@Leaky: NO, it's wrong. The converse is correct. $\alpha$ constructible implies $\deg(\alpha) = 2^n$, but NOT conversely.
yeah i proved it
14:59
Hi @Ted
so 149 is the first prime
Hi @Alessandro et al ... Just had to reply to Leaky's ping. Back later.
so do i write mod 149 and see if i get remainder 0?
@LeakyNun what Leaky wrote was correct
I mean @TedShifrin lol
@Faust
yeah you can do that
15:01
@MatheiBoulomenos
you can also write $37$ in base two and use fast exponentiation
I see ill try the 149 first
i dont know base 2 so well
$37=32+4+1$
So $2^{37}=2^{2^5}\times 2^{2^2} \times 2$
this makes the mod calculations easier
you just need to square
i dont really understand what u have wrote...
it's just a trick that helps you reduce calculations
you don't need to use it
15:06
223
divides it
intresting
does this method only work for mersenne primes?
is that why there is so many base 2 big primes we have found?
I don't think it's due to this method specifically, but yes mersenne primes have some special properties that makes them easier to detect
u mean easier? :S
15:09
ok thank you1
whatcha do anyway nothing in your profile?
u a grad student or something?
no I'm an undergrad
ic
most of my classes i did 5-10 years ago so i feel really stupid with most things
I think math makes everyone feel stupid
4+ years of no math makes it really hard to come back +p
15:11
[Random]
@MatheiBoulomenos but its so intresting =)
$\sqrt{2}^{\sqrt{2}^{\sqrt{2}}}$ has an interesting property:
hi @parvin
If you square the topmost term, you get $2$, regardless of how large this exponential tower is
More generally:
Let ${}^m(n^{\frac{1}{n}})=(n^{\frac{1}{n}})^{(n^{\frac{1}{n}})^{\cdots^{(n^{\frac{1‌​}{n}})}}}$
Then powering the topmost term by $n$ gives $n$ for any $m \in \Bbb{N}$
can some one help me with this?
I don't understand it:
_____________
If x>=4 then x^2 <= 2^x :
15:15
@MatheiBoulomenos so when im testing primes to divide the larger number can i use the fact that they must have a remainder of 7 or 1 mod 8 for them to be divisors?
@Faust sure, if you're allowed to use that lemma
intresting
Morning Daimark
:s
@Daminark
@parvin instead of a growth argument, you can also use induction
Hi @Daminark
@TedShifrin even if it is normal?
I don't understand why it's mentioned $ x+1/x $
@MatheiBoulomenos
15:21
Well, you take both sides and you look at the ratio for consecuitive values for $x$
You still need to prove that $2 \geq (\frac{x+1}{x})^2$
@MatheiBoulomenos even if it is normal?
@LeakyNun No, your proposition is correct
@MatheiBoulomenos then what happens with the root of the irreudicble polynomial of degree 4 that isn’t constructible?
Ted is right that it's not enough that $[K(\alpha):K]$ is a power of $2$
oh, K(a) wouldn’t be normal
15:24
wait, I confused solvable and constructible
Well, either $K(a)$ is not normal or the degree of $K(a)$ is not a power of $2$ or both
but it’s irreducible
Right, I'm dumb
yes, then it won't be normal
nein du bist nicht dumm Sylowmeister
15:37
wer ist der Sylowmeister?
I want to use a logistic function (or any S-shaped function) for weighting. So let's say for a time interval [0,3], I'd want f(0)=0 and f(3)=1 , basically placing the curve between the lower and upper end of my given time interval - any idea how I could do this?
16:27
First thing I’d say is that there’s not going to be a unique answer
That said, this WP has some options: en.m.wikipedia.org/wiki/Sigmoid_function#Hard_sigmoid
Actually, those examples have asymptotes at infinity. If you want a finite interval, this page is more relevant: en.m.wikipedia.org/wiki/Smoothstep
thanks, the latter link looks exactly like what I'm looking for, at least on first glance
will give that a read after dinner
can you tell why the growth rate of x^2 is the ratio (x+1/x)^2 ?
--------------------------------
If x>=4 then x^2 <= 2^x :
As x grows larger than 4, the left side 2^x doubles each time x increases by
1.
However, the right side x^2 grows by the ratio (x+1/x)^2
If x >=4 then
(x+1/x) cannot be greater than 1.25 .
therefore {x+1/x)^2 can not be bigger than 1.56 .
----------------------
16:53
@mathreadler Mathei ist der Sylowmeister

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