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12:05 AM
this is a mystery to me, despite having changed computers several times, despite the website rejecting the application, the very first sequence of numbers I entered into it's search window which returned the same prompt to submit them for publication appear every time, I mean ive got hundreds of them now, and it's still far too much rope to give a person like me sitting along in a bedroom the capacity to freely describe any such sequence and their meaning if there isn't any already there
my maturity levels are extremely variant in time, that's just way too much rope to give me considering its only me the pursuits matter to, who knows what kind of outlandish crap I might decide to spam in each of them
but still, the first one from well, almost a decade ago shows up as the default content in the search window
well, now there is a bunch of stuff about them pertaining to "trees" and "nodes" but that's what I mean by too much rope you cant just let a lunatic like me start inventing terminology as I go
oh well "what would cotton mathers do?" the chat room unanimously ponders lol
1:09 AM
remarkably, it still presents the same sequence when I visit from a Linux virtual machine, having connected to the tor network from behind a private proxy server
1:20 AM
and yet, the proxy and tor service are indeed doing their job, my IP address changes as expected, now that is truly incredible stuff, oeis.org is a truly remarkable website in this regard
1:32 AM
i see Secret had a comment to make, is it really a productive use of our time censoring something that is most likely not blatant hate speech? that's the only real thing that warrants censorship, even still, it has its value, in a civil society it will be ridiculed anyway?
or at least inform the room as to whom is the big brother doing the censoring? No? just suggestions trying to improve site functionality good sir relax im calm we are all calm
1:46 AM
fair enough.
2:20 AM
I found it interesting though...
2:46 AM
A104101 is a hilarious entry as a side note, I love that Neil had to chime in for the comment section after the big promotional message in the first part to point out the sequence is totally meaningless as far as mathematics is concerned just to save face for the websites integrity after plugging a tv series with a reference
2 hours later…
4:39 AM
@Adam Lmao that is brilliant
5:37 AM
Draft 1: A set A is infinite if it cannot be decomposed into finite sets using a finite number of bipartitions
6:23 AM
@Secret wait that is what got removed the first time? @BalarkaSen thanks I appreciate that it did have that little rant as an ace up the sleeve but choose to give it away compulsively to the people
choose and compulsively really doesn't work there in hindsight yet so accurately sums me up
again mother has not stocked the kitchen sink with a bottle of sodium hydroxide when will the world see this child-man neglect before them
6:51 AM
But seriously @BalarkaSen, some of the most arrogant of people will attempt to play the most innocent of roles and accuse you of arrogance yourself in the most diplomatic way imaginable, if you still feel that your point is not being heard, persist until they give up the farce please
very general advice for any number of topics for someone like yourself sir
assuming gender because you should hate text based adam long ago if you were female or etc
if its false then I apologise for the statistical approach to human interaction
7:13 AM
I want to determine the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$.

I found that this is a root of $x^6-3x^4+3x^2-3$. To show that this polynomial is minimal, I have done the following:

We have that $\mathbb{Q}\subseteq \mathbb{Q}(\sqrt[3]{2})\subseteq \mathbb{Q}(\sqrt{1+\sqrt[3]{2}})$.

It holds that $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=\deg (x^3-2)=2$ by Eisenstein criterium.

Then we have that $[\mathbb{Q}(\sqrt{1+\sqrt[3]{2}}):\mathbb{Q}(\sqrt[3]{2})]=2$, or not? How can we prove that?
@MaryStar why don't you apply Eisenstein to that degree 6 polymial directly?
So after having found the polynomial $x^6-3x^4+3x^2-3$we can just apply Eisenstein to show that this is irreducible over Q and since it is monic, it follwos that this is the minimal polynomial of $\sqrt{1+\sqrt[3]{2}}$ over $\mathbb{Q}$ ? @MatheinBoulomenos
yes, a monic irreducible polynomial is the minimal polynomial of its roots
Ahh ok!! Thank you!!
you're welcome
7:22 AM
So, in Galois fields, if you have two particular elements you are multiplying, can you necessarily discern the result of the product without knowing the monic irreducible polynomial that is being used the generate the field?

(I will note that I might have my definitions incorrect. I am under the impression that a Galois field is a field of the form $\mathbb{Z}/p\mathbb{Z}[x]/(M(x))$ where $M(x)$ is a monic irreducible polynomial in $\mathbb{Z}/p\mathbb{Z}[x]$.)
I have also an other quesion.... To show that the polynomial $x^5-26x^4-13x+13$ is irreducible over $\mathbb{Z}[i][x]$ dowe have to show first the irreducibility over $\mathbb{Q}$ ? @MatheinBoulomenos
Buongiorno @Alessandro
@MaryStar not necessarily, no. Do you know a more general version of Eisenstein than over $\Bbb Z$ and do you know how prime elements in $\Bbb Z[i]$ look like?
Sto scrivendo per il mio blog un post di introduzione alla teoria delle categorie
7:26 AM
@Mathein that sounds useful
Lo leggerò, mandami un link quando lo pubblichi!
@AlessandroCodenotti lo faccierò
@ÍgjøgnumMeg Guten Morgen :)
Farò* that's an irregular verb
@MatheinBoulomenos The elements of Z[i] are in the form a+bi where a and b are integers, correct?
Grazie, coniugazione di fare è difficile
@MaryStar right, this ring is called the ring of Gaussian integers
I suggest you read up a bit on it for this question
7:29 AM
@Mathein Guten morgen :)
Do we have to reduce the polynomial modulo 5$ ? @MatheinBoulomenos
$5=(2+i)(2-i)$, that's not a prime element of $\Bbb Z[i]$
I realize that I don't personally know what prime elements necessarily look like in $\mathbb{Z}[i]$
maybe @ÍgjøgnumMeg can give @MaryStar and @Rithaniel a heads up on the ring of Gaussian integers and its primes? I'm a bit busy at the moment
I also realize that I have my volume up too loud. That ping was startling.
7:37 AM
@Mary do you know what the norm is $N : \Bbb Z[i] \to \Bbb Z$ ?
Not really @ÍgjøgnumMeg
Hey @Balarka
Hi @Mathein!
And @ÍgjøgnumMeg, @Alessandro, @Rithaniel
(Oh, right, the norm.)
Sup Balarka
@Mary so given any Gaussian integer $a + bi$ you have $N(a + bi) = a^2 + b^2$
7:39 AM
@ÍgjøgnumMeg Ahh yes,right!!
(which is just the product of the integer and its conjugate)
Note that $\alpha = a + bi$ is a unit iff $N\alpha = 1$
You might like to learn some of the properties of $N$ first, because this is useful for discussing divisibility in these kinds of rings
(Plus I'm at work and am pretending I'm doing my job)
Anyway, particularly useful is the fact that if $\pi \in \Bbb Z[i]$ is such that $N(\pi)$ is a rational prime then $\pi$ is a Gaussian prime (easily proved using the fact that $N$ is totally multiplicative) and so, for example $5 \in \Bbb Z$ is prime, but $5 \in \Bbb Z[i]$ is not prime because it is the norm of $1 + 2i$ and this is not a unit.
Hi @Balarka
Hi @Balarka @Alessandro
@Mary using these facts you can show, for instance, that rational primes of the form $4n + 3$ are prime in $\Bbb Z[i]$
@Rithaniel That leads to an interesting result about primes that are a sum of two squares
(in more technical language these are called inert primes, because they remain prime when you extend to $\Bbb Q(i)$)
7:48 AM
also note that the norm makes $\Bbb Z[i]$ into an Euclidean domain
If $p \equiv 1 \bmod 4$ then it splits in the extension, and so is no longer prime (and this answers the question of when $p$ is representable as a sum of two squares)
@Mary @Rithaniel various cool facts :P
also $2$ is the unique rational prime in $\Bbb Z[i]$ such that it is the square of a Gaussian prime (called a ramified prime)
anyway I better go and do some work, happy arithmetising
That is indeed some pretty cool stuff.
Actually, I was wondering how people figured out the norm in the first place. Could that process be generalized for any algebraic number field?
8:03 AM
@Rithaniel sure, in the case of a Galois number field you can just take the product of an element with all of its conjugates
err that's the case for any number field*
@MaryStar very impressive well done and keep it up!
it is simplified when the field is Galois because you can just take a product over all the elements of the Galois group
@Rithaniel I know right? it's like an electric shock all throughout my body sometimes since I moved my main pc to the bedside
So, in $\mathbb{Z}[\sqrt[3]{-1}]$ what would the list of conjugates be?
Well this is a subring of $\Bbb Q(\sqrt[3]{-1})$ so you'll have three conjugates since this is a degree $3$ extension, they're the complex third roots of unity
8:11 AM
Is there a simple way to tell whether $\Bbb Z[\alpha]$ for some algebraic $\alpha$ is the ring of integers of a number field?
8:25 AM
I don't think so
That's something I haven't learned yet lol, but there are algorithms
I think the hard part is determining an integral basis
8:41 AM
@Adam actually no, what I removed is the following:
$\forall i\neq j : B_i \cup B_j \iff B_i \cap B_j =\varnothing$
which is basically the definition of disjoint union
I removed it because at that time I am only using the chat to check my latex render, before moving that latex sentence to the actual bookkeeping I was doing
For those who are aware of my dream diary sharing in the past: Somehow the disjoint union is the uniting thread of all the stories in last night's dream
from making better polls to how to get people to see social events notices
@Alessandro (and it's easy when $\alpha$ is a quadratic integer)
and I guess for a cyclotomic integer you always have a power basis
but I don't know about other integers
because they're for nerds
The contrapositive of this statement:
A set A is infinite if it cannot be decomposed into finite sets using a finite number of bipartitions
which is:
@AlessandroCodenotti it is always an order, you can compute the conductor
there are some tricks
if it can be decomposed into infinite sets using a infinite number of bipartitions, then a set A is finite
is very strange
you can compute the discriminant of $\Bbb Z[\alpha]$ if that's square-free, then it is the ring of integers of $\Bbb Q(\alpha)$
but that's not an iff
8:55 AM
@Mathein nice
so either I have screwed up the negations in $\neg Q \to \neg P$ somewhere, or it is really true that you can place infinite bipartitions into a finite set
@Alessandro in general if $\mathcal O_K$ is the ring of integers of $\Bbb Q(\alpha)$, then $\Delta(\mathcal O_K) [\mathcal O_K:\Bbb Z[\alpha]]^2=\Delta(\mathcal O_K)$, I'd suggest you read up on orders, the index of an order and discriminants for orders if you want to go into that rabbit hole
also note that if the minimal polynomial of $\alpha$ is $p$-Eisenstein, then $p$ doesn't divide $[\mathcal{O}_K:\Bbb Z[\alpha]]$
this together with the above formula is sometimes enough to show that $[\mathcal{O}_K:\Bbb Z[\alpha]]=1$, i.e. $\mathcal{O}_K=\Bbb Z[\alpha]$
Is this the ideal $\Delta(\mathcal{O}_K)$
$\Delta$ is supposed to denote the discriminant
Oh I see what you are getting at sorry
9:02 AM
yeah $\Delta(\mathcal O_K)$ is an integer
Milne gives an algorithm in his notes
Something like $\Bbb Z[\alpha] \subset \mathcal{O}_K \subset \Delta^{-1}\Bbb Z[\alpha]$
A right
that, too
but it's kind of a loose bound I think
so it's impractical for number fields of large degree
unless you have a PC
the proof of the $p$-Eisenstein thing even starts with taking a $p$-Sylow subgroup of $\mathcal{O}_K/\Bbb Z[\alpha]$
(just as a quotient of additive groups, that quotient group is finite)
in particular, from what I've said, if the minimal polynomial of $\alpha$ wrt every prime that divides the discriminant of $\Bbb Z[\alpha]$ at least twice, then $\Bbb Z[\alpha]$ is a ring of integers
that sounds oddly specific, I know, but you can also work with the minimal polynomial of something like $1+\alpha$
there's an interpretation of the $p$-Eisenstein results in terms of local fields, too. If the minimal polynomial of $f$ is $p$-Eisenstein, then it is irreducible over $\Bbb Q_p$ as well. Now you can apply the Führerdiskriminantenproduktformel (yes, that's an accepted English terminus technicus)
9:17 AM
That's cool, I'm very excited to finally be getting some formal pure mathematics education
@MatheinBoulomenos You once told me a group cohomology story that I forget, can you remind me again? Namely, suppose $P$ is a Sylow $p$-subgroup of a finite group $G$, then there's a covering map $BP \to BG$ which induces chain-level maps $p_\# : C_*(BP) \to C_*(BG)$ and $\tau_\# : C_*(BG) \to C_*(BP)$ (the transfer hom), with the corresponding maps in group cohomology $p : H^*(G) \to H^*(P)$ and $\tau : H^*(P) \to H^*(G)$, the restriction and corestriction respectively.
$\tau \circ p$ is multiplication by $|G : P|$, so if I work with $\Bbb F_p$ coefficients that's an injection. So $H^*(G)$ injects into $H^*(P)$. I should be able to say more, right? If $P$ is normal abelian, it should be an isomorphism. There might be easier arguments, but this is what pops to mind first:
By Schur-Zassenhaus theorem, $G = P \rtimes G/P$ and $G/P$ acts trivially on $P$ (the action is by inner auts, and $P$ doesn't have any), there is a fibration $BP \to BG \to B(G/P)$ whose monodromy is exactly this action induced on $H^*(P)$, which is trivial, so we run the Lyndon-Hochschild-Serre spectral sequence with coefficients in $\Bbb F_p$.
The $E^2$ page is essentially zero except the bottom row since $H^*(G/P; M) = 0$ if $M$ is an $\Bbb F_p$-module by order reasons and the whole bottom row is $H^*(P; \Bbb F_p)$. This means the spectral sequence degenerates at $E^2$, which gets us $H^*(G; \Bbb F_p) \cong H^*(P; \Bbb F_p)$.
@Secret that's a very lazy habit you should create a chat room for every purpose you can imagine take full advantage of the websites functionality as I do and leave the general purpose room for recommending art related to mathematics
@BalarkaSen I'm busy and I've got an appointment in like 5 minutes, I'll think about it later, but the question interests me
for example an old film entitled "Rainman" for which a popular artist that are known by today's kids dedicated a song to which I will provide a link, youtu.be/od6Ml8iQvOE
@MatheinBoulomenos No worries, thanks in advance. Just to add the final punchline, what I wanted to ask is what's the general algorithm to recover $H^*(G)$ back from $H^*(P; \Bbb F_p)$'s where $P$ runs over Sylow $p$-subgroups of $G$?
9:23 AM
@Balarka $H^*(G)$ with $\Bbb Z$-coefficients injects into the product of $H^*(P)$ for all primes
Aha, OK.
@Adam It's not that I cannot do it. I actually have 7 rooms in the SE. The problem is whoever that SE crew decided to put the code that causes auto freeze rooms when they are inactive
meaning the effort to keeping them from not freezing increase exponentially with the number of rooms and the inverse of the number of regulars in those rooms
@Secret yeah cool talk to @Simplifire about he just has a loop putting "." in the ones he doesn't want to auto freeze
I tried loop scripts before using feeds, they still froze. That's what happens in the Rambles, causing me to need to open the Star Wars room
i mean me personally i don't care ill just make a new one it takes less than a second but each to their own logic
9:27 AM
Unless simplifire's script does not involve the feeds bot
look bro a fully functional desktop marco program of many varieties is available for download for free it will take you two seconds if you are really that sentimental about your rooms in cyberspace
Ok I never thought about macros before. Hmm...
In other news, something completely unrelated to maths:
Life is like walking towards a social version of heat death
You find new forums, gaming groups, communities etc.
Everything starts out very active, and then plateau, and then some time later, all activity died and communities scattered no matter what you do
Bacterial growth is the asexual reproduction, or cell division, of a bacterium into two daughter cells, in a process called binary fission. Providing no mutational event occurs, the resulting daughter cells are genetically identical to the original cell. Hence, bacterial growth occurs. Both daughter cells from the division do not necessarily survive. However, if the number surviving exceeds unity on average, the bacterial population undergoes exponential growth. The measurement of an exponential bacterial growth curve in batch culture was traditionally a part of the training of all microbiologists...
As a result, there does not exists a single group which lived long enough to belong to, and hence one continue to search for new group and activity
eventually, a social heat death occurred, where no groups will generate creativity and other activity anymore
Had this kind of thought when I noticed how many forums etc. have a golden age, and then died away, and at the more personal level, all people who first knew me generate a lot of activity, and then destined to die away and distant roughly every 3 years
9:43 AM
Well i guess the lesson you need to learn here champ is online interaction isn't something that was inbuilt into the human emotional psyche in any natural sense, and maybe it's time you saw the value in saying hello to your next door neighbour
(Actually the exact same curve applied for all my real life interactions except for close friends and neighbours)
infact i beg you to question in the future if anything responding in a typical manner is even sentient
in the most true sense of the word i mean
Well, if there is a chance of meeting sentient AIs, I actually will expect the same autonomous behaviour will still result, except in the standard of these machines.
Otherwise, I actually expect the further my stage of life is away from high school, the less likely I will meet genuine people who do not interact with scripts and then distant in finite time
well naturally their use in covert operations would be lost by making their development public in real time, since you like secrets i would assume you to have understood this
Or more likely, we will need to start recognising machines as a new species and interact with them accordingly
so covert operations AI may still exists, even as domestic AIs continue to become widespread
It seems more likely sentient AI will take similar roles as humans, and then humans will need to either keep up with them with cybernetics, or be eliminated by evolutionary forces
But neuroscientists and AI researchers speculate it is more likely that the two types of races are so different we end up complementing each other
that is, until their processing power become so strong that they can outdo human thinking
But, I am not worried of that scenario, because if the next step is a sentient AI evolution, then humans would know they will have to give way
However, the major issue right now in the AI industry is not we will be replaced by machines, but that we are making machines quite widespread without really understanding how they work, and they are still not reliable enough given the mistakes they still make by them and their human owners
That is, we have became over reliant on AI, and not putting enough attention on whether they have interpret the instructions correctly
10:15 AM
That's an extraordinary amount of unreferenced rhetoric statements i could find anywhere on the internet! When my mother disapproves of my proposals for subjects of discussion, she prefers to simply hold up her hand in the air in my direction
for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many females i have intercourse with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise
i feel as if its an injustice to all child mans that have a compulsive need to lie to shallow women they meet and keep up a farce that they are either fully grown men (if sober) or an incredibly wealthy trust fund kid (if drunk) that's an important binary class dismissed
10:31 AM
If we have a positive definite symmetric matrix in R, does the Jacobi method always converge?
how many times have you asked yourself dear?
Are you a troll @Adam
I have not done anything related to matrices or more than one dimension in quite a while but @TedShifrin would be much better versed to answer your question @MaryStar
@F.White explain to me what one is precisely and I can finally provide a precise answer to that question
If im troubling you that much I will oblige if you politely ask me to leave

Chatroom troll: A person who types messages in a chatroom with the sole purpose to confuse or annoy.
I was just genuinely curious
How does a message like this come from someone who isn't trolling:

"for example i tried to explain to her that my inner heart chakras tell me that my spirit guide suggests that many ... with are easily replaceable and this can be proven from historical statistical data, but she wont even let my spirit guide elaborate on that premise"
Anyway feel free to continue, it just seems strange @Adam
I'm genuinely curious what makes you annoyed or confused yes I was joking in the line that you referenced but surely you cant assume me to be a simpleton of one definitive purpose that drives me each time I interact with another person? Does your mood or experiences vary from day to day? Mine too! so there may be particular moments that I fit your declared description, but only a simpleton would assume that to be the one and only facet of another's character wouldn't you agree?
10:44 AM
Jun 7 at 1:02, by Balarka Sen
Adam is well-versed in making no sense, if you're not acquainted with him
Adam speaks in a different language
where only people who knew him enough will understand
Well no he is just being nice there I don't think I deserve that much credit
but yes what secret says is correct
@Adam riggggggggggggggght
There's actually some guideline on determining who is a troll in this chatroom: If they annoy Ted Shifrin like hell, then consensus said they are a troll
How on earth Ted found me more annoying than homework help vampires who show no effort, or people who multiping other people and make fake SE announcement in the past still eludes me
But basically the consensus for the normals in this chat room is:
I'm not familiar enough with the lore I guess
Normals think I am the top chat troll
whereas weirds think I am contributing interesting discussions and ideas
10:50 AM
Do you know much about manifolds secret?
are you a differential topology guy?
or differential geometry?
nah, the differential guys are balarka, Danu and a few others
Secret is more of a logician (and poet).
11:13 AM
$K \otimes_{\Bbb Q} \Bbb R$ for a number field $K$ is just.. extending scalars to $\Bbb R$ right?
what does $\otimes_{\Bbb{Q}}$ mean?
tensor over $\Bbb Q$
as a $\Bbb Q$-vector space
"$\Bbb Q$-bilinearity" becomes $\Bbb Q$-linearity in the tensor product i guess
I F.Black personally much like ol mate Prince Harry you might say well no megan is more what I think they call a quadroon
I feel like you need more punctuation
damn it
11:28 AM
Fancy "No" right there.
@ÍgjøgnumMeg it's an algebra, too
not just a vector space
ah ye?
it factors as a product of $\Bbb R$s (one for each real embedding) and a product of $\Bbb C$s (one for each pair of complex embeddings)
Ohhh I remember seeing that in the section on Dirichlet's unit theorem
11:34 AM
@ÍgjøgnumMeg if $A$ is a commutative ring and $B_1$ and $B_2$ are commutative $A$-algebras, then $B_1 \otimes_A B_2$ is a commutative $A$-algebra as well
Right this is just the tensor product of algebras
$K$ and $\Bbb R$ are $\Bbb Q$-algebras yeah?
Yeah ofc they are
I'm so rusty :(
12:20 PM
What up
chillin, you?
I am currently thinking about nested horospheres
or maybe, to make things simpler, horocircles
Heya Akiva
Suppose I have a pair of concentric horocircles, and I am an ant located in the "smaller" one, how can I go from the smaller horocircle to the big one when its boundary is located at infinity?
As in, they share the same idealized point at infinity?
I guess you can't get from one to the other without crossing the gap between them
which has the same width everywhere I think
12:26 PM
I see
In the half-plane model
the lines parallel to the boundary are horocircles
@Secret It's easiest to see why horospheres are Euclidean in that model
because they are really, lines in Euclidean space?
The horosphere would be a plane
but the metric in that model depends only on your distance from the boundary
so on the planes parallel to the boundary, the metric is a constant times the identity
12:44 PM
make sense
1:39 PM
So, there are some weakened forms of associativity. Such as flexibility ($(xy)x=x(yx)$) or "alternativity" ($(xy)x=x(yy)$, iirc). Tough, is there a place a person could look for an exploration of the way these properties inform the nature of the operation? (In particular, I'm trying to get a sense of how a "strictly flexible" operation would behave. Ie $a(bc)=(ab)c\iff a=c$)
1:54 PM
@Semiclassical conjecture: let $\phi_k$ be a sequence of spherical harmonics with eigenvalue $\to\infty$. then $area(\phi_k>0)/area(\phi_k<0)\to 1$
I usually explore non associative systems with homomorphisms on associators, but otherwise I don't know how they are done in the literature since I have not read much on those except lie groups
@RyanUnger You're the guy to ask for this sort of thing I think:

If I want to, by hand, compute $\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle$, then I just want to expand out $R(\partial_1,\partial_2)\partial_2$ in terms of the connection, then use linearity of $\langle -,-\rangle$ and then use Koszul's formula? Or there is a smarter way?
$$\langle R(\partial_1,\partial_2)\partial_2,\partial_1\rangle = \langle \nabla_1\nabla_2\partial_2 - \nabla_2\nabla_1\partial_2 ,\partial_1\rangle = \langle \nabla_1\nabla_2\partial_2 ,\partial_1\rangle -\langle \nabla_2\nabla_1\partial_2 ,\partial_1\rangle $$
and then apply Koszul's formula to both of these
Why am I the guy
@Semiclassical actually that one might be solved
For general manifolds it’s very much open
Or it'd be better to just compute the Christoffel symbols to determine $\nabla_1\partial_1,\nabla_1\partial_2$ etc
Balarka said you'd be the guy to ask something like this before :P
2:09 PM
I realized today that the possible x inputs to Round(x^(1/2)) covers x^(1/2+epsilon). In other words we can always find an epsilon (small enough) such that x^(1/2) <> x^(1/2+epsilon) but at the same time have Round(x^(1/2))=Round(x^(1/2+epsilon)).
Am I right?
I'm just asking what your approach would be I guess, since the computations seem tedious if I want to find, say, the sectional curvature for some surface with metric $$g=\frac{1}{h^2(x,y)}dx^2+dy^2$$
@LeakyNun thanks for your help from previous one year
I’m in a lecture so I can only take a look later
Sure thing
Basic functional analysis question: is the topological dual of a Banach space the space of all bounded linear functionals?
2:14 PM
can someone help me with a statistics problem? nobody in the cv room
2:26 PM
@mathsstudent you are welcome
2:50 PM
@RyanUnger not a perfect analogy but that sounds awfully close to oscillation theory of ODEs
@anakhro Thanks!
Would like to solve a two body problem but the other body lives on another continent and I have no money. If you said the probability of success is low, you'd be right
is that a bad pickup line
3:07 PM
@BalarkaSen I can't think of a way to prove that result without spectral sequences
how is an energy function defined on a multigraph
I set up the adjacency matrix, and need to find the eigenvalues. The energy function of a 'simple graph' is the sum of the absolute value of the eigenvalues
3:28 PM
What is an energy function, @Ultradark
@Rithaniel Depends on if you only have a multiplication operation, or also have an addition operation
@ÍgjøgnumMeg @Alessandro ho finito la prima parte della introduzione alla teoria delle categorie: wlou.blog/2019/06/18/a-brief-introduction-to-categories-part-1
If I recall correctly (and I might be mistaken), if an algebra has $x(yy)=(xy)y$ and $x(xy)=(xx)y$ then $x(yx)=(xy)x$, but you need to use addition/subtraction in the proof
so it holds for algebras but not necessarily nonassociative… groups? There's a word
@AkivaWeinberger magma is the word
Hey @Secret. I wrote a bit about categories on my blog. You might be interested (for the link see above)
3:45 PM
sounds good, checking now
Yeah, these are just single-operation objects.
Like, with the octonions @Rithaniel
The algebra generated by any two elements is isomorphic to the quaternions
Yes, but more generally than that example
meaning that the algebra generated by any two elements is associative
which is why the octonions have $x(xy)=(xx)y$
That intriguing.
3:49 PM
$(rp)s = ps = s$, but $r(ps) = rs = r$.
Why would one study magmas
With the sedenions, that doesn't happen
Not quite sure what sorts of things you get generated by two elements, but they're not in general associative
@RyanUnger you mean study magmas that aren't better described as something else?
Because all groups are magmas.
So studying groups is transitively studying magmas.
@RyanUnger nobody does study magmas without additional properties
@anakhro every group is a set, so group theory is set theory?
@MatheinBoulomenos at least keep my grammar: "a group theorist studies sets",
It's pedantry, of course.
But hey, it's not wrong.
4:00 PM
of course, technically correct is the best kind of correct
@Mathein Typo: "A category C consists of the following data"
Loving the prose
nicely written :P
thanks, glad to hear that from an Englishman
I think there are a couple of instances of "consits" instead of "consists"
not that it matters lol
4:04 PM
@MatheinBoulomenos do you mean "technically correct", or "accurate"?
@ÍgjøgnumMeg there were two, both are fixed now, thanks
Nise one
@Mathein I look forward to the next post :P
I'll ping you when I publish it
Cool :)
actually I'm supposed to be working through a book on galois groups and fundamental groups and a section is reserved for category theory
so maybe I'll use your blog
4:11 PM
We have the following Simpson method $$y^{n+2}-y^n=\frac{h}{3}\left (f^{n+2}+4f^{n+1}+f^n\right ), n=0, \ldots , N-2 \\ y^0, y^1 \text{ given } $$ Show that the method is implicit and state the stability definition of that method.

How can we show that the method is implicit? Do we have to try to solve $y^{n+2}$ as a function of $y^{n+1}$ ?
4:28 PM
Interesting, so category theory is basically morphisms obeying associativity and there is an identity, thus making them behave like a very abstract version of monoids
and the way categories works is really similar to the usual way I think when trying to find commonalities between dissimilar objects
Looking forward to the next blog post
4:56 PM
@anakhro an energy function of a graph is something studied in spectral graph theory. You set up an adjacency matrix for the graph, find the corresponding eigenvalues of the matrix and then sum the absolute values of the eigenvalues. The energy function of the graph is defined for simple graphs by this summation of the absolute values of the eigenvalues
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