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12:15 AM
(sorry for the double message, my phone was drunk)
12:37 AM
just imagine what i said.
2 hours later…
2:25 AM
@copper.hat What I imagine is way too offensive!
I bet he said the vector space dual is canonical :O
My imagination is too limited: "(removed)"
2:42 AM
@AlessandroCodenotti yeah but don't have anything of value to add
Imagination dead imagine
(That's a short novella by Beckett where he demonstrates that imagination is not dead by demonstrating he can imagine that imagination is dead)
Such mirth I furnish!
imaginary numbers
a + (removed)i
(removed) + bi
::the ghosts of departed thoughts::
3:24 AM
@TedShifrin :-)
3:37 AM
"just imagine what I said" = (removed) + bi
"imagine what I could've said" = a + (removed)i
4:06 AM
Hi @feynhat
Want to help me with something?
Can try.
I have a wall separating two regions and a wave inside one of these regions
My goal is to push the wave across the wall to the other side
can it tunnel
4:09 AM
@Semiclassical It can go through the wall yes
so you want to transfer a localized wave in one region, to a localized wave in the other?
I can do this, but there's an additional constraint.
My wall is not a normal wall, it admits a foliation. The constraint is throughout this transfering, the height of the terrain doesn't change along this foliation as it moves through the wall
That is to say at every instance of time $t$, the height of the terrain (which is a function) is constant along each leaf of the foliation that foliates the wall
Idea: Consider the initial height of the terrain function, look at it's level sets. Draw a vector field along these level sets flowing along which will move the wave out of the wall
Seems sound right?
No clue bro.
jeez dude
imagination dead imagine
what kind of math have you been upto
4:23 AM
All I know is that two lines intersect at a point
*two lines in gEnErAl pOsItiOn
I learned a heuristic argument for Kontsevich's recursion.
T e l l
Sure. You can also give a talk if you want
4:27 AM
Bruh... I am writing something which is due very soon.
I mean whenever you're free!
You can also give a talk in my college math club if you want
Yeah if you give a talk, invite me as well!
I'll send a link for sure
You will learn this in your course, right?
Yeah doing Chow Rings right now, I don't know if we will do any Kontseivich stuff
4:36 AM
Ciao Rings
I tried to read Fulton's Intersection Theory once many moons ago
I read a chapter and then gave up
Bruh that book is not for reading. Try 3264 and all that, Eisenbud and Harris, much better
over 600 pages massive
screw this man
i only read CONCISE books
4:41 AM
J P Dey and K. Ponchanon
More Concise IITJEE
i dont know where this is going abandoning this bit
EXCLUSIVE The Most Important Message You Will Ever Read:
Is the message blank? I feel like it should be blank.
And there you have it.
4:59 AM
i prefer mine
5:16 AM
@BalarkaSen -upgrade- -upgrade- -shit go back-
5:37 AM
Regarding this question: math.stackexchange.com/questions/519161/… can't one just take the long exact sequence (this should be reduced but I'm on mobile) $$...\rightarrow H_n(\sqcup_\alpha x_\alpha) \rightarrow H_n(\sqcup_\alpha X_\alpha) \rightarrow H_n(\sqcup_\alpha X_\alpha, \sqcup_\alpha x_\alpha) \simeq H_n(\vee_\alpha X_\alpha) \rightarrow ...$$
Since that disjoint sum of points is a deformation retract of the disjoint sum of the spaces if I'm not mistaken
Ah lol thats exactly what the person asking the question does
Should've read that instead of the answer
Agh no it's not what they did but I get now why the answer is the quickest fix to the asker's attempt. The notation $H_n\left(\bigsqcup_\alpha (A_\alpha,B_\alpha)\right)$ is something to get used to
6:05 AM
Holy fu.... Lmfao trump pardoned lil wayne
I'm crying
6:21 AM
please let us return to a world where the onion is satire
I must say he goes out gloriously
do be like that
I hope all the commenters predicting that "trump will be tamed by the seriousness of the office" or some shit reread their articles
he just doesn't seem capable of accomplishing everything that he did.
6:30 AM
Finally, a radical Trump’s foreign policy would destabilise America’s alliances and escalate tensions with rivals. His protectionist stance could incite a global trade war, and his insistence that allies pay for their own defence could lead to dangerous nuclear proliferation, while diminishing American leadership on the world stage.

But it is actually more likely that Trump will pursue pragmatic, centrist policies.
i mean policy-wise trump was no different than his predecessors
he's just a generic american politician but mean tweets/no filters
no, but he trampled over almost every norm
he didn't really have any policy other than nepotism or the like
good, the american idealism is now demonstrably a sham
Hmm no his foreign politics were pretty bad. In a very different way than bush who was significantly worse, but different
always has been
6:33 AM
at least iraq is free of wmd now. omg.
A lot of young folks looking a trump don't realise the gravity and depravity of what Bush did, honestly.
idealism is always that.
i think that falls on cheney's plate.
at least the world has putin & panda to fall back on.
Betraying the kurdish fighters that were allies in the fight against ISIS. To please a dictator buddy. As with Putin. Estranging european powers, waging trade war against the EU, pulling troops from germany and trying to separate from NATO
Climate deals
Oh, and he reincited the west sahara conflict recently? To make morocco open up to israel
His foreign policy was mostly just.... Stupid
it is surprising (not) how a certain biblical narrative drives popular support
he does not have a foreign policy.
it is extremely hard to imagine eg Hilary would have been any different in terms of foreign policy
@copper.hat yeah lol
6:40 AM
i knew i could never vote for trump. but i had to think long and hard before voting for hilary (i did).
@BalarkaSen why? Demanding more military spending from european countries would have been an overlap, but the rest?
All the north korea BS?
@user2103480 o yeah fair enough lmao
Sanctioning iran would have been on her list as well
i think more eu spending was a good thing to push, but at the risk of splitting nato. oh wait, i forgot, ukraine...
@user2103480 yup.
6:42 AM
read about the dulles brothers. you can see their bones in modern us foreign policy.
@BalarkaSen but the policy in syria wouldn't have been a hillary thing I suppose. Couldn't imagine her being fond of erdogan. She would've stayed more involved in libya as well, since the current state there is directly connected to her actions
(basically on anything related to EU foreign policy, trump made our lives harder, so I've been looking forward to a more sensible US policy for a while)
i s'pose you could make a case that his republican ghouls modified the courts forever. climate crisis is a real thing but it seems to be transcending political affairs extremely fast; soon any political action will either hasten it or do nothing.
everybody shouts about climate change but they are not willing to give up anything for it.
but i think the most significant thing that happened is crackpots are coming out of the woodworks to babble about their opinions. the pro-lifers, the climate-deniers, ...
ahh, the interweb, morons with microphones
6:49 AM
coz if the president can say whatever the f he wants why cant we
@copper.hat the problem is that it's easy for me/us to demand more action, stricter laws, etc
@user2103480 this makes sense
convert our oil usage to nuclear. because, hey, its safe.
I'll belong to the highly educated urban people who may lose a bit of comfort but not my job
our leaders are reflections of ourselves.
i know lots of people with hybrid suvs.
6:51 AM
buy tesla because electric
and help fund the genocide in yemen
and separate their garbage, but still take a few transoceanic trips a year.
2 birds in 1 stone
well, there may be a connection but i doubt that it is explicit
the us has plenty of poverty issues on it own soil
(The government, I'd rather say)
this really is quite a venting block :-)
6:54 AM
Better delete, that was too edgy lol
too late to prune my little outbursts
@copper.hat I often think about garbage separation, but not from the end. It's crazy how many different types of plastic are used in some supermarket products
i really think all we can do is slow things down
we cannot change people
There's no way we are able to separate those again and recycle the resources
the oceans are full of micro particles
6:57 AM
end of the world is more plausible than radical new socioeconomic systems
capitalist realism
well, i sortof agree with churchill
that democracy is the worst form of government
@copper.hat estimated 150 million plastic bottles raining down in microplastics onto yellowstone each year. There's also microplastic in babies placentas now
except for all the others that have been tried
isnt the pacific garbage vortex the size of australia by now lol
see, its hopeless to fiught it
i may sound flip, but i really think it hopeless
6:59 AM
yeah no i buy it
@copper.hat but... Only 75% of insects died in the last 20 years :-)
eat more insects
Here in germany, at least
my mom was offered thalidomide when she was pregnant with me
The better formulation is that the population shrank this much
7:00 AM
what sort of caring world do we have
i dont see a butterfly which has any pigment on its wings... iirc this happens because of pollution
literally all of them are grey
future peoples may view grey as pretty, natural even
the world we are in today would appaul our forebears
excuse my wording, too tired/lazy to straighten it out
7:02 AM
SB-129 is fantastic
I don't trust any of these simpsons predictions
i wouldn't view them as such
just the fact that it was considered is impressive
i mean that trump might 'lead'
People say anything was predicted by the simpsons and often enough it's just not in the episode
7:05 AM
and its ilk
if you say enough you are bound to get something right
sort of like the digits of $\pi$.
10 hours to go
2 hours later…
8:51 AM
@user2103480 cOpRoDuCt iN tHe CatEgoRY oF pAiRS
9:07 AM
$f:D^n\to D^n$ has no fixed point, then $f$ can be thought of as a map from the northern hemisphere to itself. Here $D^n=\{x\in R^n:||x||\leq 1\}$ why this is true?
what does it mean to think of f as map from the northern hemisphere to itself
What does "can be thought of" mean here? $S^n$ is homeomorphic to the union of two $D^n$ with common intersection $S^{n-1}$, so if you want this is fulfilled for every map from a disk to a disk
(where each $D^n$ stands for one hemisphere)
@Thorgott grr
That actually elucidates it though. And makes sense with my earlier question about reduced relative homology
9:22 AM
you hate to admit it
not really tbh the only reason I'm taking this course is that if one day I will use algebra in my prospective research
it'll probably be slightly categorical and I can't think of anything more probable than (co)homology
Okay, there's some hopf-algebras stuff in SPDE. But I don't intend on doing anything explicitly geometrical so I don't expect Lie groups or other related stuff to show up
just do abstract homological algebra
lots of nice diagrams
but I for sure will never see splitting fields again
@Thorgott yes with 0 motivation
No thanks, I learned my lesson with simplicial sets
proving commutativity of various diagrams is a purpose in and of itself, not requiring any extrinsic motivation
If they actually proved commutativity only once instead of just positing it
9:31 AM
just a diagram chase
@user2103480 see here the third line of second paragraph of second answer math.stackexchange.com/questions/247933/…
Sorry first answer
yeah ok, that's just meaning the hemisphere (with equator) is homeomorphic to the disk
@Thorgott ah yes this I remember well
proof by "this is just a diagram chase"
using the axiom of this-diagram-commutes
nice nice
The worst thing when I look at it now, is that all the diagrams that are actually really bad to justify look totally innocent, while the diagrams I understood pretty well look terrifying
10:06 AM
Hi all, I have a question
Basically, a guy is pitching balls and I need to create a transition matrix for his pitches
I already have an answer, but it's super strange because I don't have some probabilities (I think)
Anyway, here's the problem:
Here's my attempted transition matrix:
My question is -- how can I create the full transition matrix T if I don't have the all the probabilities of going between the three states?
If anyone has ideas on how to solve this, that would be great
I've been struggling with this problem for quite awhile now;
I think it's actually missing information for the other probability states, but if that's the case -- how am I supposed to do Part B?
@Astyx Sorry to ping, but do you have any ideas on how to solve the above problem?
1 hour later…
11:32 AM
Hello!! Could someone of you take a look at my question in analysis?
Q: How do we get the formula of the sequence?

Mary StarA woman is walking home with distance $d$ and speed $v$. The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth. (i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet? (ii) Determine the total path length of the dog with the help of $(d_n)_{n\ge...

1 hour later…
12:41 PM
If $k$ is a field then $k[x_1,...,x_n]$ is a UFD?
that's Gauss' lemma, effectively
1:09 PM
As in the usual proof of 'every nonzero ring has a maximal ideal', can we always find a maximal ideal in a nonempty collection of ideals with certain property?
Using Zorn's lemma
@love_sodam If your collection of ideals satisfies the conditions for Zorn's lemma, then sure
the condition that any nonempty collection of ideals contains a maximal element is equivalent to being Noetherian
but ofc be wary, being maximal in a specific collection of ideals does not necessarily imply that the ideal is a maximal ideal in the ring
1:28 PM
Does the notion of algebraic independence make sense when you have a field $k$ and a ring $R\supset k$?
And do isomorphic $k$-algebras need to have the same transcendence degree?
Because I'm confused about the morphism $k[x]\to k^n$, which evaluates $f$ at $n$ fixed points
$k[x]$ has 2 algebraically independent elements, while $k^n$ has $n$
Now, if we mod out by the kernel $I$, we would get $k[x]/I\cong k^n$
But then the transcendence degrees don't seem to match anymore
@ShaVuklia which two?
Oh, I thought $1$ and $x$
1 satisfies t-1=0
1:33 PM
Right wait I'm not using the right definitions
that's usually always a thing to consider
Maybe like this: $k[x]$ has 2 generators as a $k$-algebra, while $k^n$ has at least $n$
Now, how does that add up
$k[x]$ has exactly 1 generator, which is $x$
Because of the empty product?
if you consider what $x$ maps to via $k[x] \to k[x]/I \cong k^n$ then you can also find a generator for $k^n$
@ShaVuklia what does generator mean to you?
1:37 PM
algebraic independence makses sense for overrings, but I'm not sure if transcendence degree does
it probably does, but I don't wanna think about it
@love_sodam as long as your collection is closed under unions of chains basically
@LeakyNun Well I thought that if you apply all $k$-algebra operations to your set, you get your $k$-algebra, but I wasn't sure if taking the empty product was part of it.
@LeakyNun This is fair, let me check in my ex
I think I was underestimating that taking products in $k^n$ would yield more stuff
@ShaVuklia in other words, what is the sub-algebra that $x$ generates?
@LeakyNun I'm not sure, I think the empty product isn't allowed after all, so in that case $x$ just generates $(x)$
that is not a sub-algebra
the problem with your definition of "generates" is that a subset does not generate a sub-algebra
1:42 PM
A subalgebra should contain 1 right
If a collection $A_i:i\in I$ of events indexed by a nonempty index set I is
mutually independent, then the collection is also pairwise independent. How can I prove that?
So the 1 is included automatically I suppose, or you could say 'empty product is allowed'
I mean then $x$ generatres $k[x]$
how about you do it logic-theoretically, and say that the operations of a ring is 01+-x
so any subset generates automatically a ring that contains 1
@ElizabethJ.Magee write down the definitions
1:46 PM
@LeakyNun Oof, I forgot the precise logic pov, but I guess I'm good
@Thorgott should I write here?
if writing them down for yourself doesn't resolve the issue, then write them down here
Hello @robjohn !! Could you take a look at my question about analysis? I saw that in the past you have answered a similar question. Do you have an idea also about my question?
Definition 6.3 (Independence of events). Let $I \neq \emptyset$ and $\left\{A_{i}: i \in I\right\} \subset \mathcal{F}$ be a collection of measurable events, indexed by $I$.
- The events in the collection $\left(A_{i}\right)_{i \in I}$ are pairwise independent if for any $i, j \in I,$
\mathbb{P}\left(A_{i} \cap A_{j}\right)=\mathbb{P}\left(A_{i}\right) \mathbb{P}\left(A_{j}\right)
- The events in the collection $\left(A_{i}\right)_{i \in I}$ are mutually independent if for any (i.e. for every) finite, nonempty subset $J \subset I$,
I could not send image. why?
1:55 PM
Do you see a similarity between those two definitions?
pairwise: P(A_i /cap A_j) =P(A_i)P(A_j)
mutual: P(A_j1 /cap A_j2 /cap A_j3) =P(A_j1)P(A_j2)P(A_j3)
@Thorgott i got this
2:16 PM
why did you write what you wrote after "mutual"?
Something slightly bothering me: for a field $k$, with closure $\overline{k}$, polynomial rings $k[x,y]$ and $\overline{k}[x,y]$, say we have $\overline{M}$ a maximal ideal of $\overline{k}[x,y]$ and $M := \overline{M} \cap k[x,y]$.

Why does $[\overline{M} \neq \overline{k}[x,y]] \Rightarrow [M \neq k[x,y]]$ hold?
I guess because it not being equal and being maximal means modding by it is $\overline{k}$, so it must not contain $1$ (or any constant really) so its intersection with $k[x,y]$ will not contain a constant and thus cannot equal $k[x,y]$ either?
a true statement implies a true statement
t r o l l
but yeah, the math response would be "then why is the consequent always true"
cause a proper ideal can't contain a unit
and $M$ a proper ideal because of what I said or na
2:29 PM
I mean $\overline{M}$, which is proper by definition
"(or any constant really)" $\mapsto$ "(or any constant other than $0$ really)"
it's true that $\overline{k}[x,y]/\overline{M}\cong\overline{k}$, but not entirely trivial
@Thorgott but that's in the anteceeedent
@Thorgott I thought "modding by a maximal ideal gives you a field" was ancient scripture
yes, but it's true by definition
yeah, yes
but still, I don't think it finishes answering the question
about the consequent
$[M \neq k[x,y]]$
2:32 PM
@BigSocks yes, but that that field is $\overline{k}$ is not entirely trivial
Here's the general scenario: if $f\colon R\rightarrow S$ is a ring homomorphism and $I\subseteq S$ is a proper ideal, then $f^{-1}(R)$ is a proper ideal in $R$. Prove this.
oh yeah, ok I buy that. there is also a piece of info I am omitting, that $\overline{M} = (x-a, y-b)$ for $(a,b) \in \overline{k}^2)$, but I don't think it should matter for the question (and I think it makes it clearer that you get $\overline{k}$ after modding, for this instance
yeah, it's obvious you get $\overline{k}$ then, but that the maximal ideals look like that is the non-trivial part
it's precisely the weak Nullstellensatz
and to prove Nullstellensatz, you use the Zariski lemma, which is precisely what you need to argue the quotient is $\overline{k}$ on a priori grounds
mmm I must've looked at this stuff so many times I internalized it, which is not so great
thanks for the reminder. good to always remember nontrivial stuff
nothing wrong with inernalizing the Nullstellensatz, arguably it's the right thing to do
cool cool well do you think the implication I mentioned depends on weak Nullstellensatz?
idk why I feel like maybe not, but that it's still a good intuition to have
2:37 PM
no, it depends solely on the general fact I mentioned above
right, ok, this is what I expected
preimages of proper ideals are proper, applied to the inclusion $k[x,y]\hookrightarrow\overline{k}[x,y]$
should it say $f^{-1}(I)$?
uhhhhhh, yes
why did I write $f^{-1}(R)$ lol
hahaha, no stress, probably just writing fast
hmm but what if $f$ is onto $I$?
like $f: \Bbb Z \to \Bbb Z$ with $1 \mapsto n$
so $f : \Bbb Z \twoheadrightarrow n \Bbb Z$
$f^{-1}(n) = \Bbb Z$
not proper
2:45 PM
$n\mathbb{Z}$ is not a ring
oh yeah yeah
my rings have units and my homomorphisms are unital, to be clear
I remember having an argument about this with someone once
for non-unital rings, this is indeed a counter-example (I never thought about that, huh)
about proper ideals not being rings since they didn't have $1$
but yeah, I guess if you map $1_R$ to $1_S$ you gotta be right huh
2:47 PM
yeah, precisely
3:06 PM
If you are interested in RL and EAs, maybe you will find the following question and answer useful and interesting.
Q: What is the difference between a fitness function and a reward function?

nbroIn reinforcement learning (RL), the reward function (RF), which can be denoted as $r(s)$, $r(s, a)$, $r(s, a, s')$, $r(s, s')$ depending on its specific definition, provides the learning signal, which guides the RL algorithm/agent towards the desirable behaviour. As opposed to a loss (or cost) fu...

hello. how can we express matrix exponential of $A_{n\times n}$ in terms of $n-1$ powers of $A$
do you know the Cayley-Hamilton theorem?
so what does that give you?
yes that gives me that but so long
ok I think its the only way
I have lot of questions and less time per question less than one minute
3:56 PM
if you want to compute matrix exponentials fast, you'll likely need some information about the underlying matrix @jeea
If we have a non symmetric and non diagonally dominant matrix,do we know that Jacobi methods diverges for a system of equations?
if you have the eigendecomposition, then that quickly gives the matrix exponential. if you have the jordan normal form (or more generally a decomposition into diagonal and nilpotent matrices) then that also gives a speedy result
if you have an arbitrary matrix, though, then that's going to be tough
so uh, not sure if i could ask this on Math.SE but, does anyone know an algorithm or formula to reference the position of specific characters in a string in the smallest notation possible?

Feel free to tell me if there any such thing. Already know of hamming distance and also using BCD to know the position of specific character but wondering if there others method i could use.
4:26 PM
I don't like attending online course.
thanks to Wikipedia- inverse function, I know what are left inverse and right inverse now.
I am so hungry and want to skip the Polish class to forage.
4:49 PM
So say you have a unique monic irreducible element $f(y) \in K[y]$ that generates $K[y]$. I guess that means you have a primitive polynomial $f_0(y) \in A[y]$, $A$ the ring from which $K$ is the field of fractions.

$1.$ $f$ is monic, but might it have some other coefficients that are not elements of $A$, say, fractions of elements of $A$
$2.$ if this is true, wouldn't you get $f_0$ not monic?
that first sentence is a very weird way of saying $f(y)=y$
I assume that's not what you mean
yeah I meant to say it generates a maximal (prime, of course) ideal $M$
and there's an assumption (bwoc) that $M \cap A = (0)$
hi chat
ok, so what is your set-up
do you start with $A$?
4:58 PM
$A$ a factorial domain (pretty sure that's just a ufd), $K$ its field of fractions, $M$ assumed to be a maximal ideal of $A[y]$ that is not principal
damn, so, again I made a mistake
because $f$ is actually said to generate *a non-trivial prime ideal of K[y] that contains M$
-part of the proof

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