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7:03 PM
@JasperLoy Whut? Still a long road for that. =D
@Complexanalysis Did you try anything?
 
@PedroTamaroff finished .
 
@PedroTamaroff Maybe we will be classmates, hehe.
 
@JasperLoy Where are you planning to study?
 
@PedroTamaroff US/Canada
 
@JasperLoy Nice.
 
7:04 PM
@JasperLoy UG ?
 
@Complexanalysis I finished my undergrad long ago, I am an old man now, lol.
 
@JasperLoy So MS or Phd ?
 
@Complexanalysis Going for PhD, because it is easier to get funding that way.
 
I am thinking of doing MS in Financial mathematics , is it a good idea ?
@JasperLoy cool .
 
I think it is always better to try to get a PhD.
If you are thinking of becoming a quant,
Mark Joshi has nice advice on that.
 
7:08 PM
@JasperLoy but its very hard to get a PhD position . and the thing is it just requires too much of things , GRE blah blah for US , Canada etc
 
@Complexanalysis So where are you gonna do your MS?
 
Germany i think .
@JasperLoy
 
Ah, good good. I wanna be born in Germany my next life.
 
you believe in Rebirth ? @JasperLoy
 
@Complexanalysis Yes, I do. In particular, I believe in Theravada Buddhism.
 
7:10 PM
@JasperLoy where are you from ?
 
@Complexanalysis I am from Singapore, a tiny dot on the map.
 
I did my undergrad locally, and it was a bad experience.
The courses really disappointed me. I think I wasted my time there.
 
After getting my degree, I realised that I had not done much real math.
So I will try to catch up by reading on my own.
 
7:13 PM
Is real math math with real numbers?
 
@Complexanalysis Are you from Germany?
 
Then you are lucky!
Nice place with good math it seems to me.
 
its good @JasperLoy i would say
 
@PedroTamaroff LOL@comments
 
7:16 PM
@Mike What comments?
 
The comments on my answer.
 
@mike So have you finally read my paper? lol
 
crickets
 
Wow, what happened to this chat?
 
@JasperLoy You happened to this chat.
 
7:22 PM
I was looking through Spivak and Apostol.
markjoshi.com @Complexanalysis if you want to be a quant read this site.
 
@JasperLoy what's that , is that your web page ?
 
@Complexanalysis Nope, it is that of Mark Joshi!
 
7:37 PM
Just answered a lhf!
 
@JasperLoy Wonderful! What a great job you did! I should announce to the whole world.
 
@Sawarnik Nobody appreciates my answer
 
@JasperLoyI never said that.
Its good enough, so +1 :)
 
@Sawarnik It's very good, you mean.
 
@JasperLoy No just good.
 
7:43 PM
@Sawarnik No, very good.
 
@JasperLoy You feel its extremely very very good. I feel its good.
 
@Sawarnik OK, whatever!
 
@JasperLoy Thanks! :)
@JasperLoy One is enough!
 
Has anybody an idea why $\TeX/\LaTeX$ chose to use $\Re$ and $\Im$ instead of $\operatorname{Re}$ and $\operatorname{Im}$ like basically the entire rest of the world?
 
@DanielFischer No, must ask the experts.
 
7:50 PM
@DanielFischer I wasn't even sure what it meant when I saw it the first time.
 
Jasper, enough.
 
@DanielFischer Looks waaay more badass.
I think I have a case.
 
@Sawarnik Not me
 
Another crazy thing: we have a code for \ker but not for \im
 
@PedroTamaroff You are a nutcase.
 
7:53 PM
@PedroTamaroff That also irritates me a lot.
 
@DanielFischer You need to relax.
 
@PedroTamaroff You should have written $\mathfrak{badass}$.
@JasperLoy In this case, by "a lot" I meant "often". Practically every time I write some linear algebra.
 
Hi, I have some tribble with latex, Can I make it clearer visibility for $a_{r+j}^{2^{j}}$, Unfortunately $2^{j}$ is very small. Thanks
 
@DanielFischer I think I know why I am not irritated. Because I don't write anything, lol.
 
trouble*
 
7:56 PM
$a_{r+j}^{\large 2^j}$
 
@DanielFischer Thanks!
 
No, ugly.
Perhaps ${\displaystyle a_{r+j}^{2^j}}$?
No!
Nor that.
 
Ask the experts in the TeX room!
 
@DanielFischer $$\exp (2^j \log (a_{r+j}))$$
 
@DanielFischer Already your first response it's good.
 
8:01 PM
@DanielFischer Pronounced 'bathass'?
 
@mike So have you read my paper? lol
 
Nope.
 
@JasperLoy it's just to make my answer more enjoyable to read
 
@Julien $a_{r+j}^{\large2^{\small j}}$
$a_{r+j}^{\huge 2^{\small j}}$
Kiddin'
 
Holy wow, I answer totally the wrong questions. Look at these vote counts.
 
8:09 PM
now you know why jasper loves his lhf
 
lhf?
..means?
 
Low hanging fruit
 
Anybody here familiar with algorithms?
I'm looking for some ideas for data structures that keep things sorted. A heap is the first and only thing that comes to mind.
 
@agent154 Binary search trees? AVL, red-black, what have you?
 
@DanielFischer Extra ordinary vote counts .
 
8:33 PM
@DanielFischer I just found out about Skip Lists. This may be a nice thing
 
8:45 PM
@robjohn I think I obtained an interesting solution to a particular case of the Hadjicostas's formula. (no variable change, no integration by parts)
@robjohn are you around?
 
How can a measure have more than one support ?
 
@Complexanalysis WAT-.
 
@PedroTamaroff I just read that the support of a measure need not be unique .
 
@Complexanalysis What's the definition of support?
 
@PedroTamaroff In $(X, \mu , \Sigma)$ , supp$(\mu) := \{E \in \Sigma , \mu(E)>0\}$ , I guess we need the closure of it .
 
8:55 PM
Well, $\Sigma$ neednt have a topology.
How is that not unique?
 
how do i show that filtered inductive limits $\varinjlim_{i} A_i$ of coherent rings are coherent if $A_i$ is a flat $A_j$-algebra whenever $i\geq j$?
 
@PedroTamaroff Pedro, my apologies for showing ridicule towards your responses to my questions in this chat room, a while ago.
 
@MatsGranvik Oh...?
@AlexanderGruber Brain explodes.
 
@PedroTamaroff you're tellin me, man.
 
@PedroTamaroff ya , if $\Sigma$ had a topology then we would need closure . Its uniqueness is a problem . I don't know.
@AlexanderGruber I don't know some words .
 
8:58 PM
@Complexanalysis did adding that hyphen help?
 
@AlexanderGruber Brain explodes.
 
Quick question I'm trying to read some books on gamedevelopment but they easily drown you in mathematical jargon. Especially when they are talking about sets/subsets. I know i'm being vague but is there an actual term for these sorts of things? I don't always know how to read these forms/equations.
 
@AlexanderGruber Only as english language . =D
 
@Sidar What kind of symbols are you talking about?
@MatsGranvik I really don't know what you're talking about.
 
@PedroTamaroff what do you think about my question ?
 
9:01 PM
@PedroTamaroff Ok never mind then. It was that Fried Tucky Chicken for the explanation of the abbreviation of FTC (Fundamental Theorem of Calculus) and some other comment that I have forgotten. But both were ridicule.
 
@MatsGranvik Smart move.
=D
@Complexanalysis I cannot see why ${\rm supp}\; \mu$ is ill defined.
 
@PedroTamaroff a simple example is ( from the book ) Ac = {e,e',...}. But then it hits me hard with stuff like t:RAc --> 2E ( where Ac and E are superscript, I don't know how to achieve that in the chat ). I was wondering if there is an actual name for such things so I could find some material on how to read these things
My math background has been extremely poor in a sense and as a programmer these mathematical forms are everywhere on such topics
 
@AlexanderGruber Yo.
This one is easy.
I think.
@Sidar You need to learn a bit of TeX. =)
 
@PedroTamaroff yeah?
 
@PedroTamaroff sure but that's beside my issue =P
 
9:07 PM
@Sidar tex 101: just put dollars signs around all your math, and if something doesn't show up, try putting a \ next to it.
 
@AlexanderGruber Yes. Let $G$ be a group.
 
i'll allow it.
 
Let $\mathscr M(G)$ be the collection of maximal subgroups of $G$.
Then $\Phi(G)=\bigcap \mathscr M(G)$. It is then clear that if $\mathscr M(G)=\varnothing$, $\Phi(G)=G$ by definition, is it not?
We're taking an empty intersection.
 
so...is there an actual name for these types of things?
 
@Sidar Let me read your question.
 
9:09 PM
@PedroTamaroff yeah i think so
 
@Sidar $t:RA^c\to 2^E$ means a function from the complement of $RA$ to the powerset of $E$.
@AlexanderGruber Because $\Phi(G)=\bigcap_{M\in\mathscr M(G)}M$.
 
there does not exist an $M\in \mathscr{M}(G)$ for which $g\notin M$, so $g\in M$ for all $M\in \mathscr{M}(G)$, so $g\in \Phi(G)$.
 
@PedroTamaroff sure, I'm sure you're right on that =P but I need some extensive material on how to read any of those forms. So is there an actual technical term I could google that explains how to read such forms?
 
@Sidar Read as in "read out loud"?
 
No read as in "learn"
 
9:12 PM
@PedroTamaroff something i've been thinking about about $\Phi(G)$
 
@AlexanderGruber Ah?
@Sidar Oh. You can pick a good book on set theory. For example, Halmo's "Naïve Set Theory" comes to mind.
 
there's a series, $\Phi^{(0)}(G)=G$, $\Phi^{(i)}(G)=\Phi\left(\Phi^{(i-1)}(G)\right)$
 
Halmos was a top notch expositor.
 
alright thanks. Ill have a look. Lates.
 
the lowest $r$ for which $\Phi^{(r)}(G)=1$ is the Frattini length of $G$ (which exists, the series does terminate)
i have been trying to figure out what the Frattini length means about a group
 
9:15 PM
@FernandoMartin Hey.
@AlexanderGruber All normal in $G$.
 
@PedroTamaroff characteristic in fact
 
Hey @Pedro, read about your test
 
@AlexanderGruber Right, characteristicness is transitive.
 
That sucks, did you get to uni and found out there?
 
@FernandoMartin Nah, mail.
=D
@AlexanderGruber $C_p$ has no maximal subgroups.
 
9:16 PM
@PedroTamaroff i asked about this a while ago and Jack Schmidt told me he doesn't think it means anything, but I think it does.
 
If $|G|$ is composite, does $G$ admit maximal subgroups?
 
@PedroTamaroff if $|G|$ is finite, it has maximal subgroups
 
@AlexanderGruber Do we regard as $1$ maximal in $C_p$?
I thought we want nontrivial subgroups.
 
(and i think infinite groups still have maximal subgroups too, from Zorn's lemma)
 
But that's fine to me then.
 
9:18 PM
Hi
 
@AlexanderGruber $\Bbb Q$ has none, no! =D
 
@PedroTamaroff i think $1$ can be maximal
@PedroTamaroff god dang infinite group theorists and their meddling with the rules of the world.
 
@AlexanderGruber Proof If $M\leqslant \Bbb Q$ was maximal, $\Bbb Q/M$ would be simple abelian, hence finite. Impossiburu.
 
$G=G/1$ is simple iff $1$ is maximal, still makes sense
 
@AlexanderGruber That's for abelian groups, right...? I guess one can have a finite group with a maximal nonnormal subgroup.
 
9:20 PM
I was wondering about math.stackexchange.com/questions/706848/…. . It looks really interesting to me
But not even a single comment
 
@felix This is like the third time you bring it up. =D
 
Is it just too hard?
 
The magic word is: bump.
There, bumped.
 
@PedroTamaroff yes, for abelian.
 
That's because nothing changes :)
 
9:22 PM
i think it still works though, modding out by a maximal subgroup should give a simple group, if the maximal subgroup isn't normal then i guess there's no problem.
 
I am pretty sure that if $G$ is finite with no nontrivial maximal subgroups then $\sim C_p$.
 
@PedroTamaroff yeah, that's true by Cauchy's.
 
Mike posed a problem some time ago.
What are the groups $H$ for which $C_p\simeq {\rm Aut}\, H$?
(For some prime $p$)
 
@PedroTamaroff hm
does he know the answer? i think that is difficult.
 
@AlexanderGruber I showed first that $H$ must be abelian.
Refresh the page, I wrote $H\simeq {\rm Aut}\, H$ first, but then changed to $C_p\simeq {\rm Aut}\, H$.
 
9:29 PM
@PedroTamaroff is this for nec finite groups?
 
@AlexanderGruber Finite unless told otherwise? ;)
 
@PedroTamaroff ok i agree then :p
 
Then the FTFGAG gives the result.
@AlexanderGruber This is a pwetty silly question. Suppose $\langle y,Y\rangle <\langle x,Y\rangle=G$, $x\notin Y$. Then necessarily $y\in Y$.
 
@PedroTamaroff yeah, that's pretty easy
kind of neat, that statement is symmetric with its contrapositive
 
@AlexanderGruber Nevermind it then.
I did something rather silly.
Tried to avoid a proof by contradiction.
Ended up having to write a more complicated one.
=D
I want to prove directly that if $x\in G$ is not a nongenerator, so there exists $Y\subseteq G$ for which $\langle Y\rangle <\langle x,Y\rangle$, then $\langle Y\rangle$ is maximal in $G$.
@AlexanderGruber Oh, just realized! =D
 
9:57 PM
@robjohn hehe, I just create a new family of multiple integrals where I've already conjectured the general case.
That's a good night since I'm pretty productive. :-)
 
@Chris'ssis cool
 
@robjohn anyway, you can see it there.
 
Mathematics Educators (beta) started yesterday matheducators.stackexchange.com
29
Q: How to assign homework when answers are freely available or attainable online?

mixedmathI find that making homework meaningful is becoming increasingly challenging. Let us suppose that I am planning for next semester's first-semester or second-semester calculus course at my university. In all likelihood, we will be using one of the common calculus texts because the university expec...

 
@AméricoTavares Cool! =D
@AméricoTavares ARGH, I cannot access it yet.
 
@PedroTamaroff I've just signed up today, but I will not ask or answer there. Just for reading.
 
10:11 PM
"Warning: this site is currently in private beta for at least 5 more days. To log in, you must have commited to the Area 51 site proposal and received the invitation email. Click the invitation link in the email to log in!"
 
@PedroTamaroff No, that was the message I've got yesterday, not today. The commitment phase ended, as far as I understand.
I've received no invitation email.
 
10:26 PM
@AlexanderGruber !!!
I think I just did something cool.
 
@PedroTamaroff Anyhow thanks for the warning! I will have to investigate if my understanding is correct. Or if it's a kind of bug with the site.
 
10:47 PM
No, nobody can now 'sign up'. People can be invited, but they're expected to contribute in some way.
 
@Mike You're in?
 
Yeah
I asked a question but then deleted it because it wasn't as good as the rest of the stuff getting asked
 
I am proud I could come with that argument myself.
Rotman provides quite a weak proof, IMO.
 
That seems like a bizarre definition of nongenerator
 
@Mike Eh?
 
10:52 PM
Shouldn't a nongenerator be $\langle g \rangle \subsetneq G$?
 
@Mike No man.
That's the effing definition. =P
 
It's a bad choice of word.
 
@Mike Do you approve?
@Mike Debatable.
 
Radiohead is good, yes.
 
If I had the energy, I would argue with you.
 
10:53 PM
Kid A is classic.
@PedroTamaroff "nongenerator" = "something that does not generate the group"
I've moved on with my life though, I understand where it comes from. :P
 
@Mike Not doing this.
 
Is this so upsetting to you?
 
I am not upset.
 
I certainly don't intend to upset.
Oh.
Okay, well then.
 
You will learn in time I am very hard to upset.
 
10:55 PM
That sounds like a challenge.
OK. @Pedro you are... smelly
I bet I got you with that one.
 
@Mike At the moment... sniff sniff... I am.
 
Me too. I need to shower.
 
I need to find the first integer $k$so $\sum_{n=0}^{1013}\binom{2n}{n}k^n$ is a multiple of 2027.
any hints?
 
gross
 
who here is good at number theory or combinatorics?
 
10:59 PM
@user4140 I would use a generating function, I guess.
Oh, wait.
 
well
i assume you want positive integer :P
 
Oh, so you would try to close it?
 

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