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2:00 AM
Well, $8 \cdot 9 = 72$...
 
@TedShifrin I said a mobius strip in combination with a geometry that treats a surface as having two sides to contain points. That might lead to a contradiction.
 
Um, no, wayyyy to big, @ALannister. Is that what we did with $4$ and $6$?
 
since locally a mobius strip is two sided.
 
No, but then Ithought about finding the primary factorization of 72
 
But your question presupposes an orientable surface, Typhon.
 
2:01 AM
which is $2^{3}3^{2}$
 
OK, @ALannister. Go on.
 
@TedShifrin oh it does?
wasn't sure
 
You said "two sides of a surface." That assumed two-sidedness.
 
ah I see
so the illogicality of the mobius strip would come from violating that implied assumption.
oops
 
Right.
 
2:02 AM
hadn't thought of it that way before
thanks
 
Sure.
 
so what's new with you?
 
LOL
G'night @MikeM.
 
what?
 
Who the **** is starring everything I say? I'm gonna end up banned for that.
 
2:04 AM
I only starred the speeding ticket one.
 
LOL ... yeah, but someone else is being a troublemaker.
So what do you think the $n$ is for $8$ and $9$? :)
 
If I had to bet, I'd say it was Semi.
@TedShifrin still thinking
 
I doubt it.
 
doubt that Im thinking or doubt that it was Semi ;P
 
Well, I'm going to cook dinner soon, so I want us to settle this, @ALannister. If you understand this, you'll know how to do the general case.
LOL ... hush, you.
 
2:06 AM
With 4 and 6, we were able to write 1/4 as something over a power of 6
We can't do that here.
 
So what about $6$ made it all work?
 
@TedShifrin nah. You cannot get banned for being starred.
 
You know @TedShifrin I'm really not sure
 
and i think it is a mix of everyone
I starred one or two
someone else starred the others
 
I'm not sure, Typhon. Someone will complain about things I've said if they're all so evident.
 
2:08 AM
meh
 
You didn't say anything bad.
 
@ALannister: Your prime factorization thing was good. Think about it.
 
Anyway, what was it about 6?
 
if someone gets offended by geometry in a math chat, then by all means frogs jumping in a toad pond is an offense.
 
Oh, not geometry, Typhon.
 
2:09 AM
Well, the prime factorization of $6$ is $2 \cdot 3$...
 
Yup.
On the other hand, Typhon, I am a bit proud to be the top answerer in diff geo by a landslide. :)
 
The prime factorization of $8 \cdot 9$ is $2^3\cdot 3^2$
 
So, what should you pick for the $8$, $9$ case?
Go back to $4$ and $6$. Then $4\cdot 6 = 2^3\cdot 3$.
 
Well, $6$ is the product of the two bases
 
Aha.
Now finish your homework :P
 
2:11 AM
So, $6$ again?
 
Yup.
 
Spock eyebrow fascinating.
 
Figure out how to write $1/8$ and $1/9$ using the fact that $1/6$ makes sense.
love the Spock eyebrow
 
All right, Ted. Still not sure I'm going to come up with a solution tonight, but I will continue on this route, let it percolate, and see where it leads me...
 
You'll get it. Seriously, understanding examples (in an intelligent way) leads you to proofs.
And now you can do any $k$ :)
 
2:14 AM
Hello Ted! @TedShifrin
 
Just to make sure: How do we get $1/8\in\Bbb Z[1/6]$?
Hi @Kasmir
 
Hopefully to an isomorphism somehow... like a mapping of the $\frac{\text{something}}{n_{1}^{t_{1}}n_{2}^{t_{2}}} \mapsto \frac{\text{something else}}{n_{1}\cdot n_{2}}$
 
Don't worry about isomorphism. Start by showing the sets are the same.
 
Well, if the sets are the same, then you gots the iso
 
It's not $n_1\cdot n_2$ !!!
 
2:15 AM
no, n1 and n2 are the bases
 
Well, not quite, but almost. You really need to think about the sets, then let the algebra flow.
Yeah, but $n_1$ and $n_2$ might not be prime!!
Go back to our examples!
 
right! oh bugger.
Ok.
 
LOL
How do I get $1/8\in\Bbb Z[1/6]$?
 
Don't star me saying bugger
 
I ain't never the one doing the starring.
 
2:16 AM
I'm not sure what $\mathbb{Z}[1/6]$ is... isn't that a polynomial ring or something?
 
Hi, DogAteMy.
It's the thing you get with allowing all powers of $1/6$.
Yes, polynomial ring in variable $1/6$ :)
 
With integer coefficients
okay...
 
You enjoying your last evening of math presence, DogAteMy? :D
 
so, then $1/8 = 27/6^3$
 
There you go, @ALannister. :)
 
2:18 AM
Who is DogAteMy?
 
Akiva.
 
Too many names...
 
LOL ... I have a couple of pet names.
It's only friendly :)
 
while on four week vacation ,Herbert will play at least one set of tennis each day , but he wont play more than 40 sets in total during this time. prove that no matter how he distribute his sets during the four weeks , there is a span of days during which he will play exactly 15 sets
 
I have also been many names on here, but sometimes I think it's better if peeps don't remember me as those
 
2:19 AM
Kasmir: I liked your multivariable calculus better :P
 
I need only some ideas to push me a bit
@TedShifrin Haha =p I got an A thanks to your help on that =p
 
@ALannister: i now believe that you understand how to do it all and will complete your homework. Now just stay calm and go back through examples and do it :)
 
now doing discrete math
 
Do I get a partial credit when you graduate, @Kasmir? :D
 
So, Imma let this percolate, and I also have to get started putting together a 3 hour workshop where I'm going to try to give high school students a sneak peek into Calculus hopefully explaining things like limits to them in an intuititve way so that it doesn't confuse the bleep out of them the first time they see it like it did me.
 
2:21 AM
@TedShifrin Ofc :D you were the most helpfull person in my study =p
 
@Kasmir: That can't be right. What if he only plays a total of 12 sets the whole time?
 
he cant do that
because he has to play at least 1 game each day
 
I wouldn't do limits, @ALannister. I'd do more obviously conceptual things that aren't technical. But that's my two cents' worth. I'll be glad to chat about it as you think about it.
Aha, you got me, @Kasmir.
 
Thanks @TedShifrin I have to do it on Wednesday and I'm getting paid $400 to do it.
 
2:23 AM
That's awesome.
That kind of thing is not easy.
 
I just found out about it a couple of days ago b/c the person who was supposed to do something on compsci instead backed out at the last minute.
 
Well, if I can help, let me know.
 
Thanks, man. I appreciate that.
 
It's great that you're gonna do it.
 
Well, they have nobody else and I don't get any of my grad student stipend over the summer, so I kind of have to.
 
2:25 AM
@Kasmir: Unless I'm being stooopid, it's easy.
 
hmm its a pigeon hole principle question =p
 
Anyway, ciao for now.
 
@ALannister: Think of it as a great experience and an opportunity to get kids (particularly women :) ) excited about math.
 
but I cant find the pigeons nor the holes ><
2
 
Ciao. :)
 
2:25 AM
Okay, that I did star #sorrynotsorry
;P
 
LOL ... the hell with pigeons.
 
hi chat
 
Heya Eric.
You feeling better?
 
Hey Eric
 
yeah a bit
 
2:26 AM
BRB 2 mins ( Ted G
give me some hints please =p
 
I'm going to dinner ...
No hints.
 
i get chronic migraines and the pain usually lingers for a couple days at a time but im pretty ok now
 
oh okay untill next time ! bonne apetit :)
 
What did you cook for me, Eric?
Does a bit of caffeine or something else help?
 
I spent the day making Feijoada and mandioca
it was v worthwhile
 
2:28 AM
Hint @Kasmir: The pigeonholes are the days. The pigeons are the sets of tennis.
 
caffeine helps a looot
 
Translate for an ignoramus.
 
it's a brazilian black pean and pork stew over rice with tapioca flour
 
Yeah, I thought caffeine would help. It helps when I get headaches from mild caffeine withdrawal.
Tapioca flour ... interesting.
We have to cook for a week sometime :)
 
and the mandioca is a fried yucca
 
2:29 AM
OK, that I've had plenty.
 
the tapioca flour is in LOTS of brazilian dishes
esp the indigenous stuff i grew up eating
 
Very cool.
 
we also make a fermented alcoholic drink out of it
 
I don't cook much Jewish/Russian stuff that's my heritage. I much prefer French, Italian, and Asian :)
 
very strong and very tasty
 
2:30 AM
Although I did make circassian chicken salad the other night ... still having leftovers for tonight.
cheers Eric :)
 
idk much about russian or jewish food
cheers
 
BTW, this might be a good exercise for you.
 
cheers
idk much about russian or jewish food
 
I saw that :)
 
whoa my internet broke for a second
 
2:32 AM
LOL
Ah.
 
i think it sent my messages multiple times
 
I'll make you stuffed cabbage in exchange for Brazilian :)
yup.
 
ooh sounds good
 
But I normally do other stuff :) ... OK, see ya later.
 
see ya Ted
 
2:33 AM
You saw my link to the exercise for you?
 
oh yes i see it now
 
OK :)
Back later.
 
oh this looks cute
 
hi chat
 
hi @Semi
 
2:44 AM
Hi
Long time no chat
 
hi
working on my 3rd paper. will try to have it published in AMS... It's very nice :)
 
I wanna go back to uni and study more math :p
 
Holy crap Brin and Stuck is the most terse book I have read in recent memory
I love it but like, it skips steps so aggressively
@ForeverMozart Awesome!
 
what abt rudin man
 
I like making beautiful pictures with autocad
 
2:53 AM
that book is like the e.g. in the dictionary definition of terse
 
I think this helps my chances
 
Nah, Rudin doesn't skip details the way this one does
Like it does a lot for sure
 
rudin skips loads of details
 
of course I have a killer lemma also
 
But I found reading Rudin a year ago to be easier than reading this now
 
2:54 AM
i havent been reading brin and stuck
i think i recally rudin being very hard but i read it when i was way younger
 
Like, tbf the things I've been getting caught up on were a bit stupid for the most part
Which was also kinda the case in Rudin
 
@Daminark have u been to harolds
 
And part of the difficulty is the typsetting errors, but still like yeesh
I've been there once some time ago, I remembered that I didn't particularly like it
You?
 
so i really like south side chicago sauces
like mild sauce
and i just made some from scratch
and hoooooooly fuck it's good man
 
Nice
How'd you do it?
 
3:04 AM
hello
 
@Daminark it's basically a mixture of hot sauce with barbecue and some other stuff
i made em all from scratch
it's good stuff
 
Dank
 
*Daminank
 
Lmao
Now time for topological entropy
Totally didn't hijack this term from the physicists
Not at all
 
suuuure
i'll believe you ;-) ;-)
 
3:13 AM
\('-')/
 
Did you know the Unsolved Pigeon Hole Problem is AKA Lengendre's Conjecture?
 
The one about a prime being between any two squares?
 
it is literally a question of counting composite numbers
@Daminark mmhm
wait
hold on a second
can we prove legendres conjecture for cubes?
 
Kek
And @Typhon I dunno
 
3:23 AM
@Daminark be interesting to ask. Reverse induction might be useful.
(for powers other than 1)
 
Hmm, I am suspicious if that if that would work. It might, but it feels like knocking down an exponent requires gaining extra resolution, I don't think there's this kinda "higher reduction"
 
what did i miss
 
@Daminark fair enough
plus... if we could do induction
that would imply all integers are prime
 
3:38 AM
@BalarkaSen I am preparing a beautiful paper
 
noice
 
ah the strong version of the twin prime conjecture
 
@EricSilva there are no numbers larger than 7?
@ForeverMozart wait btw it's on topology, right?
 
that's the strongest goldbach conjecture
 
Oh whoops, yeah
 
3:50 AM
you mean very strong
 
the strongest twin prime conjecture states that any pair of integers are twin primes
yeah that's the wording
 
@Daminark yes topology
I just discovered that the converse of my big theorem is almost trivial. I'm actually happy about that.
 
4:24 AM
Nice
(Also sorry I'm really bad at responding, I'm alternating between here and dynamics a lot (
 
5:24 AM
Hi@TedShifrin
long time :)
 
 
1 hour later…
6:32 AM
"On the Internet, nobody knows you're a dog" is an adage about Internet anonymity which began as a cartoon caption by Peter Steiner and published by The New Yorker on July 5, 1993. The cartoon features two dogs: one sitting on a chair in front of a computer, speaking the caption to a second dog sitting on the floor listening to the first. As of 2011, the panel was the most reproduced cartoon from The New Yorker, and Steiner has earned over US$50,000 from its reprinting. == History == Peter Steiner, a cartoonist and contributor to The New Yorker since 1979, said the cartoon initially did not get...
 
7:31 AM
I read that a topological space (X, T) with T being the weak topology, is Hausdorff iff points are separated, that is, for every nonzero x in X, there exists a linear functional f such that f(x) is not equal to zero. But how does that show points are separated when we have only taken a single point x?! ..shouldn't we need to take two pionts x_1 and x_2 from X and do something to them wtih f to show that they are separated?
 
7:50 AM
@eurocoder What is the weak topology? And how does it make sense to have a linear functional in this generality? Did you mean for this to be a topological vector space?
 
8:13 AM
2
Q: How to decide $36^\text{th}$ smallest element in max-heap tree of $100$ elements?

Mithlesh UpadhyayConsider a max heap tree with $100$ elements and a node from the same level is chosen randomly among all valid balanced heap trees with $n$ nodes. What is the probability that it is the $36^\text{th}$ smallest element______ . My attempt: (Please, find definition of Binary Heap on Wiki: https...

Bounty Grace Period Started.
 
 
1 hour later…
9:15 AM
Is the notion of "Borel subgroup" (maximal closed connected subgroup) the same for a linear algebraic $\mathbb{C}$-group when considered as a complex Lie group?
 
@abenthy Do you mean whether we get the same subgroup?
 
9:30 AM
@TobiasKildetoft Yes. As I see it, every Zariski-closed subgroup of $G$ (linera algebraic $\mathbb{C}$-group) is also Euclidean-closed and hence is a Lie subgroup. But the converse is not clear to me.
So could there exists a subgroup of $G$ which is a Borel subgroup in the realm of Lie groups, but not in the realm of algebraic groups?
 
@abenthy Hmm, no idea. I am only used to the algebraic side.
 
10:36 AM
@abenthy I have always thought that "Borel subgroup" also includes solvability among the properties?
 
@JyrkiLahtonen Right, we need solvable in there as well
 
Welcome @JyrkiLahtonen
 
user84215
What do I do if I want to type multiple integral with more than five integral symbol? The command \iiiint is used at most for four integral symbols.
 
@aminliverpool What about \int\int?
 
Thanks @LasVegasRaiders. Just checking out the chat room. I should probably spend more time here. In addition to being fun it just might help me calibrate my moderator actions (as a bonus).
 
user84215
10:48 AM
it's good. but is there any simpler way?
 
@JyrkiLahtonen And hi btw
 
user84215
@TobiasKildetoft \iint is not the same as \int\int
 
@aminliverpool You could try inserting negative space between the \int signs. That would bring the signs closer to each other. Did you know that \! inserts a negative thin space? Anyway, that suggestion is probably highly non-kosher, and makes the TeX-gurus squirm.
So something like \int\!\!\!\int\!\!\!\int... experiment with the number of \! in between.
Hi @TobiasKildetoft :-)
 
user84215
Why is the \iiiint command restricted to at most four integral symbols?
 
@aminliverpool Because nobody made more
 
user84215
10:55 AM
Negative thin space is a good point.
 
I dunno @aminliverpool. May be the physicists never integrate over spaces of dimension more than four?
 
user84215
@JyrkiLahtonen In superstring theory we have at least 10 spatial dimensions.
 
Actually I don't remember the last time I would have needed four or more. Then again, I'm more of an algebra guy (even though I teach vector calculus as often as the rest of us)
Oh yeah! Good point about string theory! May be ask at TeX.SE?
 
Without our usual regulars robjohn and DanielFischer around, it's just not the same anymore.
 
Daniel works very hard handling the flags and all the other dirty moderator work. Arthur & Daniel know so much more about how this place works than I do. The moderator team would be lost without them. Robjohn would also know that stuff, but he has not been very active lately.
 
11:07 AM
$$\int \!\!\!\int \!\!\!\int \!\!\!\int \!\!\!\int=\left(\right)^{(-5)}$$
 
Thanks for info/update @JyrkiLahtonen
 
$$\iiiint$$
 
user84215
What is the advantage of \over command to frac{}{}?
 
$\over{a}{b}$, $\frac{a}{b}$
 
0
Q: Asymptotic behavior of a Riemann sum (not really) by an integral

beckoI have a sum of the form: $$S_n = \frac{1}{n} \sum_{i=0}^n \mathrm{e}^{n f(i/n)} g(i/n)$$ where $f(x)$ and $g(x)$ are smooth functions defined for $0\le x \le 1$. I am interested in the Asymptotic behaviour of $S_n$ as $n\rightarrow +\infty$. What I have tried is to replace the sum by an integ...

I need some help with this.
 
user84215
11:11 AM
@Secret that is, you can not make the first by \frac command?
 
I suspect sums of this kind have popped up before, but I cannot find anything in the literature.
I am probably missing some keywords.
 
$\frac{}{ab}$,$\over{a}{b}$
 
user84215
I think they have the same size.
 
72
Q: What is the difference between \over and \frac?

ikdcWhat is the difference between {some-expression}\over{another-expression} and \frac{some-expression}{another-expression} ? Which is preferable?

 
@TobiasKildetoft Ah I think I got it! Let $G$ be an linear algebraic group. Then a Borel subgroup of $G$ can be defined as a maximal connected solvable subgroup. Since the Zariski-closure of any such group is also connected and solvable, it follows that every Borel subgroup is Zariski-closed, hence also closed in the Euclidean topology, and so it is both an algebraic subgroup and a Lie subgroup of $G$. So there are no two notions of Borel subgroups.
 
user84215
11:16 AM
Thanks.
 
If someone could confirm the above makes sense, I would be grateful.
... or does Zariski-connected not imply Euclidean-connected? Then one could have a different notion for Lie groups...
 
@abenthy Zariski-connected probably doesn't imply Euclidean-connected. For example, I think that the integers form a Zariski-connected set of the complex numbers because their closure is all of C. OTOH Zariski-closed and connected might imply Euclidean connected as well. But, I'm afraid I'm very rusty at this :-(
 
11:33 AM
How to solve this question?
 
@Abcd Not using math
 
@TobiasKildetoft I know that. I ask Chemistry questions here because Chemistry Stack Exchange is too inactive.
 
@Abcd That is a terrible reason
 
And if there are Chemistry experts here, I get the needed help @TobiasKildetoft
@TobiasKildetoft Why?
 
@Abcd The second part makes it make more sense. But the first part by itself is nonsense
 
11:37 AM
@TobiasKildetoft Okay
 
user84215
in the command \{x+y+z}^{1/2} the first brace does not work. why?
 
try it without the \
 
@aminliverpool What was it supposed to produce?
 
Is a $C^1(x_1,\ldots,x_n)$ integrand enough to switch the order of integration?
 
$(x+y+z)^{1/2}$
 
11:46 AM
then you need () not {}
 
@Secret question ^
 
user84215
I want to see that can we insert {} instead of () in the first?
 
@Abcd From observing your interactions with other chat users throughout h bar, maths and chemistry over the past 5 months, You are really starting to sound like a help vampire. What exactly have you tried on that question?
aminiverpool: $\{x+y+z\}^{1/2}$
 
@Secret Help vampire?
 
user84215
Thanks
 
11:50 AM
@Secret Calculated the Z/e ratio for the various species but realised that that method works only for isoelectronic species. Then thought about the periodicity of atomic size but couldn't arrive at any conclusion.
@Secret Alright, my last question on chat.SE (of physics,chemistry and maths)
 
then \{x+y+z\}^{1/2}
$\{x+y+z\}^{1/2}$
 
@Abcd hydride ion is pretty big because there is only one proton to attract two electrons. It can easily be as big as chloride. Fluoride meanwhile is smaller due to having more protons while having 2 shells of electrons. Iodide has 4 shells, thus no matter what it has to be bigger than a hydride. Bromide has 3 shells, thus it also has to be bigger
 
@Secret Okay, thanks. (but you used the concept of shells)
 
If you don't like shells, think about the principle quanutm number. While hydride ion has very extensive orbitals due to having just one proton but two electrons, Br- and I- have orbitals with n=3,4 thus these are more diffuse orbitals and they also expereince more shileding from the s orbitals with lower n
 
@Secret Okay. As promised, it was the last question. Bye. TC
 

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