« first day (1318 days earlier)      last day (3705 days later) » 

6:00 AM
@alexander you should check that out too
 
@eXtremiity we're worldwide :)
 
It's pretty amazing.
..how we can all collaborate with mathematicians from all around the world. With ease too.
Never knew Pedro was from Argentina.
 
@eXtremiity internet's an amazing thing.
 
Indeed. I think they should change the sound chat makes when you get tagged in a message.
I find this sound rather annoying and uninviting.
 
did you guys like the story
 
6:19 AM
@Mike i haven't read it yet. i'm working.
 
ok.
 
a module of finite type is just a finitely generated module right?
 
@r9m Never.
 
i don't think so @alexander
 
@Mike what is the definition?
 
6:32 AM
@Mike Yeah, it's interesting.
 
i don't remember! which is why i'm not giving you a yes/no
ask karl. he's the man on this stuff.
 
@KarlKronenfeld
 
@AlexanderGruber The only possible synonyms I can think of are finitely generated or finitely presented. I have not seen the term before.
Based on a question at MSE, it seems to be the former.
 
i was sure i'd seen 'finite type' in A-M
 
@Mike i think it's a scheme theory term
my professor is using it in an assignment without defining it
 
6:36 AM
:I
 
@AlexanderGruber A completely different possibility here
 
i'm gonna get eviscerated tomorrow.
 
Hey guys, I'm finally enroled in a mathematics bachelor. My classes are going to start 31 this month. I want to thank for all your support. You've given me motivation and courage and inspiration (and also helped me a lot answering my stupid questions!). Without you nothing of this would be possible. Thanks.
7
 
@PristineKavalostka nice job. good luck.
 
@Mike You're right that "finite-type" is a term used in atiyah-macdonald. However, it is only defined for algebras over rings.
 
6:48 AM
yeah, i do remember that.
"oh... I'm sorry kid, we don't get much poor up here."
 
7:20 AM
@r9m Never ever, plum.
 
r9m
@Sawarnik then what are these pings for ?? (sorry for late reply .. had classes)
 
7:42 AM
@r9m About not talking to you, bcoz I m still angry.
@robjohn Did you get the 2048 btw?
 
@Sawarnik you acted very badly while Prof Ted was here, pal.
 
@skullpatrol Ah, I was having fun that time with Balarka. What has Prof Ted got to do with it?
 
Why didn't you take your "fun" to the other room @Sawarnik?
 
@skullpatrol Becuase the rules there are strict. At Nick's, you mean na? Where did I act 'badly'?
 
Can anyone familiar with Littlewood-Richardson coefficients tell me is this "duality" is well-known: Given partitions $\lambda = (\lambda_1,\lambda_2,\dots,\lambda_n)$, $\mu$ and $\nu$ and such that $\lambda_i - \lambda_{i+1}\geq \mu_1$ for all $1\leq i\leq n-1$ and such that $\nu = \lambda + \sigma$ for some partition $\sigma$, the Littlewood-Richardson coefficient $c_{\lambda,\mu}^{\nu}$ is equal to the number of semistandard young tableaux of shape $\mu$ and type $\sigma$
 
7:46 AM
Quick question: a hypersurface (eg vanishing set of some homogeneous polynomial) need not be connected in $P^1$, right? I can consider the vanishing locus of $X^2-Y^2$, which is [1:1] and [1:-1
], yes? I feel crazy and would like confirmation/denial
 
@r9m Did you get the 2048 monkey?
 
@KReiser Doesn't the definition of hypersurface entail the ideal is prime (i.e. in this case that the polynomial is irreducible)?
 
Hmm, good point. The homework I'm doing doesn't say hypersurface, it only says "Let $Y$ be a closed subscheme of $X=\PP^n_k$ defined by one homogenous equation $f$ of degree $d$."
 
@skullpatrol Is it wrong to have fun?
 
I suppose the writer meant to add this condition, though. Hmm.
 
r9m
7:53 AM
@Sawarnik I'm banging my head with that .. It seems all sorts of luck has abandoned me .. made the same mistake twice :( :(
 
8 mins ago, by skullpatrol
Why didn't you take your "fun" to the other room @Sawarnik?
 
@skullpatrol Is it wrong to have fun here?
 
if other users find it obnoxious (they do), then yes
2
 
1 min ago, by skullpatrol
8 mins ago, by skullpatrol
Why didn't you take your "fun" to the other room @Sawarnik?
 
@r9m So get that game out of your head, and do something useful.
 
7:54 AM
@KReiser is it contained in the def'n of closed?
 
@skullpatrol Is it wrong to have fun here?
 
1 min ago, by skullpatrol
1 min ago, by skullpatrol
8 mins ago, by skullpatrol
Why didn't you take your "fun" to the other room @Sawarnik?
 
@Mike I don't think so. Closed just means closed (topologically).
 
@KReiser no, a closed subscheme is not just a closed subset of the topological space
but you are right that it does not require the corresponding ideal to be prime
 
if i recall open subscheme is the simple topological definition and closed was more complicated
 
7:56 AM
@TobiasKildetoft Okay, fine. Equivalence class of .... . But it doesn't require the ideal to be prime.
 
r9m
@Sawarnik difficult to get it outta my head .. I'm dis (._.) close !!
 
IE two distinct rational points in A^1 can be a closed subscheme.
 
k
then yeah they want it to be prime
 
@skullpatrol Is it wrong to have fun here?
 
13 mins ago, by Mike
if other users find it obnoxious (they do), then yes
1 min ago, by skullpatrol
13 mins ago, by Mike
if other users find it obnoxious (they do), then yes
 
8:08 AM
Get it in an infinite recursion!
 
54 secs ago, by skullpatrol
1 min ago, by skullpatrol
13 mins ago, by Mike
if other users find it obnoxious (they do), then yes
14 mins ago, by skullpatrol
1 min ago, by skullpatrol
1 min ago, by skullpatrol
8 mins ago, by skullpatrol
Why didn't you take your "fun" to the other room @Sawarnik?
 
Hmm...better, but still not infinte!
 
Not surprised that somebody flagged part of that convo.
 
mods appearing in 3...2...1...
 
What's going on here?
 
8:13 AM
@ManishEarth No, no no. The line is "what's all this then?"
 
I have just finished performing an attitude adjustment on a spoiled teenager.
 
i'm not convinced it worked
 
The proof is in the pudding ;-)
 
@RоryMcCune Hi. :)
 
@Alenanno hello, this isn't a corner of Chat.SE I've been to before..
 
8:20 AM
it's normally between slightly less unruly and slightly more so
 
@Mike nah
3
 
@Mike nah
 
Oh I just saw the Guidelines. Seem reasonable enough if you have a lot of traffic in this room.
 
i'd say roughly 70% of the chat is just people talking about whatever, the other 30% is questions
 
Those "Guidelines" were written by a dictator!
 
8:25 AM
they're good guidelines, bring back the old dictator
 
In fact, a self-admitted $\Huge\text{Troll}$.
It took a lot to get him to let go of his ownership of this room.
 
3 mods in 4 owners? Uhm...
 
idgi
 
@Alenanno What's your point?
 
I hold him personally responsible for one users nervous break-down in this room >8(
 
8:32 AM
nervous breakdowns are cool
as long as you're not the one breaking down nervously
 
@KarlKronenfeld Owners have some powers in a chat room, but mods have (more of) them across the chat network anyway, so it feels like pointless to me. But in the end, I don't care that much lol
 
@Mike they are tragic to watch up close, pal
 
@Alenanno Ah, thanks for clarifying. (I saw Alexander Gruber use some his chat powers but didn't think much of it)
 
@Alenanno the obvious solution is to make me a mod
 
@Mike That's not up to me. :D
 
8:35 AM
Ahh, the kid left
 
don't forget that ignore is a useful feature
one i use perhaps too sparingly
 
I've never used it or ever will :-)
 
I never used it either...
 
what for? If you can't stand the heat, stay out of the kitchen.
 
so that I don't say "shut the f*** up" on a too-regular basis :)
 
8:41 AM
yep, sinking to their level is the greatest danger, pal
 
8:56 AM
@Sawarnik 2048?
 
@robjohn Here
 
9:15 AM
Did the math teaching and learning site proposal get cancelled @Mike?
 
9:26 AM
@skullpatrol check the email you use for stackexchange. if you committed you have an invite to the private beta
 
@Mike thanks :D
 
I already benefited from one of the questions :)
 
cool
 
Greetings
 
The greatest one!
 
9:35 AM
@skullpatrol Hi :D
 
@Chris'ssis Hello, how are you?
 
@skullpatrol Like sh*t these days. How about you?
 
Just say shit, lol.
 
@Chris'ssis I'm sorry to hear that :(
@JasperLoy It's not polite to tell people what to say, pal
 
@JasperLoy :-)
 
9:39 AM
@Chris'ssis I'm ok, thanks for asking.
 
@skullpatrol That's great.
 
Why is it typical to replace the i with a symbol and not, e.g., h? s*it
 
So Vrabie's Differential Equations treats ODE, and a little PDE, integral equations, and calculus of variations. Perhaps I should get it.
 
usually they use vowels
 
@JasperLoy It's a good book.
 
9:40 AM
@KarlKronenfeld Usually the vowel becomes an asterisk.
 
@JasperLoy Why might that be usually the case?
 
@Chris'ssis You think so? I am deciding between it and another one, Teschl's ODE and Dynamical Systems.
@KarlKronenfeld I am not an expert, I dunno.
 
every word must have at least one
 
@JasperLoy I didn't read Teschl's ODE and Dynamical Systems.
 
@Chris'ssis Ah OK, Vrabie's book had a second edition in 2011, hope you know.
 
9:44 AM
@JasperLoy Work by Vrabie -> math.uaic.ro/~ivrabie/lucrari.html
@JasperLoy Vrabie, I. I. (2011) Differential Equations. An introduction to basic results, concepts and applications, Second Edition
 
i'm tired
 
sleep
 
need to work
I wish my phone wasn't dead, I could set an alarm and nap for a bit
 
Your phone was never alive, lol.
@Chris'ssis I didn't know he was so fat, lol.
 
@Mike Here, I'll set my phone alarm for you. Perfect plan
 
9:49 AM
:(
 
@JasperLoy hehe :-)
 
can't you set something on the internet?
Alarms.com
 
@robjohn Ah, sorry that was meant for r9m. I wanted to ask you wheter the answer I linked was right or not?
 
10:09 AM
If $\hookrightarrow$ represents injection. What represents surjection?
 
$\twoheadrightarrow$
 
I like these symbols. I don't think my school has ever used them.
Bijection?
 
They're not exactly standard.
 
Shame.
 
unless you're drawing a lot of diagrams (which is rare except in some parts of math) they're roughly useless
very little is saved by not saying "injective map" when it's done once or twice
 
10:13 AM
In the long term of my time doing mathematics.
2 seconds per comment
 
@eXtremiity I guess, $\cong$...
 
an alternative, which is sometimes more practical, is to write $0\to A\to B$ for injective and $A\to B\to 0$ for surjective.
and then assume all sequences to be exact
 
Oh wow, I've never seen that before, @TobiasKildetoft.
 
you'll see exact sequences eventually
 
but this only makes sense in abelian categories (ie, usually modules or similar)
 
10:15 AM
I see.
 
What do to if an Wolfram-Mathworld article contains an error?
 
he's working on sets and bijections and such right now @Tobias
 
@Mike ahh, ok
in that case, ignore those
 
That is correct.
 
the ones with the $0$ require more structure
 
10:16 AM
I'm really enjoying real analysis. Fell in love with complex analysis. But this stuff is beautiful too.
 
10:29 AM
hey guys, quick question
For groups G and H, G embeds into H means G is isomorphic to a subgroup of H?
 
@TwiNightyes
 
ok thx
 
10:53 AM
@Sawarnik Adjust the world to conform with the article.
 
@DanielFischer That is a lot tougher.
 
@Sawarnik Then try to find a link for leaving feedback or so.
 
Hmmm, what's the code to do $\implies$ backwards?
 
Something like \Longleftarrow
 
@eXtremiity \Leftarrow
 
11:01 AM
Perfect. Ty !
Tail could be a bit longer, but it works.
 
Capitalizing the first letter is what makes it a double arrow btw
 
Yes, I realised that after I got a single arrow :p , thanks Karl.
 
11:13 AM
@DanielFischer That is a better advice, done. Lets see they respond or not. :)
 
r9m
11:30 AM
@Sawarnik yo man ... still mad @ me ?
 
@r9m Its good to taste defeat.
 
12:36 PM
What's $A$ ?
 
@eXtremiity Glaisher Kinkelin constant
 
Lol, so $A$ comprises of $H_{n}$. $H_{n}$ comprises of $K_{n}$. And finally $K_{n}$ is defined as $1^1 . 2^2 . 3^3 ....$
That's nasty.
 
If $\omega_1$ and $\omega_2$ are two roots of unity and $\arg(\omega_1)+\arg(\omega_2) = \pi$, then $\omega_1 + \omega_2 = i \cos \arg( \omega_1)$ ?
 
Wait a minute. The summation of their arguments is equal to $\pi$.
Does that not imply that $\omega_{1} + \omega_{2}$ lies on the real axis ?
 
12:43 PM
@eXtremiity no, that their product does
 
Ah ! yes.
 
@eXtremiity You mean the imaginary axis right?
 
@N3buchadnezzar . Ahh doesn't matter. I was going to say that it must lie on the real axis therefore your answer can't be purely imaginary. But I am wrong.
 
@Sawarnik I asked a couple of times for clarification, but I kept getting answers that didn't really clarify. I didn't want to pester.
 
@robjohn But the bounty time is expiring very soon, what do I do?
 
12:58 PM
@robjohn What is your goal for the future now ?
 
@Sawarnik you have two more days, right?
@N3buchadnezzar goal? to answer questions. I guess the Legendary badge is not too far off.
 
@robjohn No body is going to answer now, its too old. So in case that happens, that will happen, I do what?
 
@Sawarnik I am looking at the question to understand the question. I may come up with something.
 
Ok :)
 
@N3buchadnezzar . Are you sure your answer is correct?
 
1:03 PM
@eXtremiity somewhat?
 
I got $i2sin(arg \omega_{1})$.
I probably did a mistake somewhere.
 
@eXtremiity Yeah, of course it should be sine.
 
@N3buchadnezzar Oh ? You had $cos$.
 
@robjohn Somewhat impressive you have gotten so many points without the legendary badge, it is probably coming soon
 
r9m
@robjohn @Sawarnik I wanted to answer that bounty question too :D ... I can explain Jack's steps if you want :)
 
1:17 PM
@r9m Is that answer completely correct?
 
r9m
@Sawarnik yes :) atleast it looks clear to me :)
 
@r9m And is your solution a bit different?
 
r9m
@Sawarnik essentially the same .. but I started my argument from $R+\bar{OM_A}\ge m_a$ .. and worked backwards :)
 
guys take a look at this. unrelated I know, math.stackexchange.com/q/712102/127574
@Sawarnik my connection crapped out earlier
 
@r9m Well maybe....if you would to give a clearer solution you will get the bounty....or I should give him?
 
1:24 PM
@Sawarnik I suggest you wait for the 2 days to get over
you will get a 24 hour grace period as well.
 
@Sabyasachi What is a grace period?
 
@Sawarnik you can wait 24 hours after the bounty period expires before awarding the bounty
 
r9m
@Sawarnik if I put my solution and you give me the bounty .. it will be doing injustice to Jack .. there's very little difference and he answered it before me :D .. you can ask him to add all the steps and details .. then decide if you want to give him the bounty :)
 
Ok, then, if no other solution comes until the next 2 days, the bounty goes to Jack.
@Sabyasachi Is your connection back now?
 
@Sawarnik considering the fact that i am on chat, it would seem so.
@r9m you can always post an answer, linking to jack's answer, and clarifying that you are only explaning his steps.
 
1:31 PM
@Sabyasachi Then my question? You tried.
 
@Sawarnik no. I solved the first and second part and told you.
can't do the third. didn't try though.
tell me.
>_>
 
@Sabyasachi How did the first go, tell?
 
i told you already.
trig.
 
you had an incomplete sol.
 
maybe you aren't seeing the trig solution, because you have something else in mind.
 
1:38 PM
did you read the question? It asked something else!
 
you told me "tell me the approach"
find FI right?
or FH
something like that.
you can construct a triangle
and where there be triangles, trig can help you.
any day.
 
Ok, 1 min...
And I have trig in mind too....
Now which length did you get??
 
What was the first part again?
 
Angle bisector, silly.
 
@robjohn Argument for why the integral over $\int_\Gamma P(x)/(Q(x))$ goes to zero along the half circle from $R$ to $-R$ where $\deg P + 2 \leq \deg Q$ ?
 
1:40 PM
right.
@N3buchadnezzar correct your $\color{red}\LaTeX$
 
\begin{align*}
\int_\Gamma \frac{P(x)}{Q(x)}\,\mathrm{d}x \leq \pi R \sup_{z \in \Gamma} \frac{P(x)}{Q(x)}
\leq \pi R \frac{C |z|^{n-2} }{D |z|^n}
\leq \pi R \frac{C}{D R^2}
\end{align*}
 
@Sawarnik okay so we are given the three sides, use them to get angle GFE
 
@Sabyasachi Tell me which length did you get?
 
then half angle formula over the cosine
I am pretty sure we can get both FI and FH depending on choice of final triangle
 
Final triangle?? You have the lengths.
 
1:44 PM
@Sawarnik you didn't get my course of action apparently
find angle GFI
then take triangle $\Delta$ GFH
cosine rule again.
and we're done.
I admit it isn't elegant.
but its quick
 
@robjohn Are you familiar with the calculus of variations?
 
its dirty, and it gets the job done.
 
@Sabyasachi Where, which side and angle, in the last step you are taking cosine rue?
 
sorry, meant sine.
over angle GFH and side HG
and angle FGH and FH
 
@Sabyasachi How do we get angle FGH?
 
1:51 PM
by getting angle fhg
which is equal to angle feg
:D
of course.
 
Ok.
@Sabyasachi Good! Now think for the height case.
 
i sense you have a another solution?'
 
Yeah.
 
r9m
@Sawarnik are we doing that length of angle bisector thing ? :) (don't be angry at me .. I apologize :) )
 
don't want to. it is practically the same.
trig again.
 
1:55 PM
@Sabyasachi Ok fine :)
@r9m Over with the 2048, first tell?
 
what 2048?
tell me tell me tell me.
 
@Sabyasachi Better not know it!
 
why? :'(
 
@Sabyasachi Because its very additictive. Ask r9m :)
 
hit me. :D
at least tell me what it is?
 
r9m
1:57 PM
huha stuff :P
 
powers of 2
so?
$2^{11}$ so?
 
Happy $\pi$ day everyone
 
r9m
@Sabyasachi gabrielecirulli.github.io/2048 try it :)
 
Dont try it!
 
@agent154 happy $\large\color{red}\pi$ day to you too
 
r9m
1:59 PM
@agent154 happy $\pi$ day to you too :) ..
 
too late i clicked on the link
 

« first day (1318 days earlier)      last day (3705 days later) »