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15:13
is the function f(t) = d[C*f(t)*e^(-t/k)]/dt a common differential equation to solve?
C, k constants
@Hippalectryon rob and chris.
@robjohn Are you here?
@Sawarnik I already asked Chris but she couldn't help with my problem :/
@Hippalectryon Sorry for the confusion .. try asking robjohn :D :P
@Sawarnik Wait... Chrissis's brother is here sometimes ? I thought you referred to her
Ok, I'll ask him when he's here
@Sawarnik @Hippalectryon lol, you both are so funny talking about brothers and sisters :-)))
15:19
:-)
:D
@Hippalectryon Do you have a sis, hippo?
@Sawarnik Two brothers
Oh, that's sad, sisters are so much fun.
@Hipp Prove that among five points inside a square of side 2, there are two apart by at most $\sqrt 2$.
@Sawarnik Isn't that obvious ?
@Hippalectryon How come? :O
15:26
I might be wrong, but here's how I's do that :
Cut the main square into four equal squares
At least two points are in the same square right ?
Done :P
Wow.
grr... and i was making a list of coordinates and trying pigeons in it.
Well since we know that sqrt(2) is the diagonal of a square it kind of occured naturally to me
we can actually extend the $\sqrt{2}$ to an almost twice smaller distance and we could still prove it.
Cool problem though
@Hippalectryon :'(
@cirpis How come?
15:29
@cirpis What is the limit distance ?
@cirpis I'm not so sure about 'twice smaller'. You can easily get a configuration where they are all distant by more than 1
$\frac{\sqrt{2}}{2}$ but we can arrange five points if it is exactly $\frac{\sqrt{2}}{2}$, soa bit more than that
@Hippalectryon Yup.
In fact, if you place it near the corners and one at the middle...
@Sawarnik exactly
well yes
@Hippalectryon Toujours là depuis hier ? :p
15:32
@Chris'ssis How about this...
Since $2^{-n}\binom{n}{k}$ has mean $\frac n2$ and variance $\frac n4$
$$
2^{-n}\binom{n}{k}\sim\frac1{\sqrt{\pi n/2}}e^{-2(k-n/2)^2/n}
$$
Therefore,
$$
\begin{align}
\frac1{n^p}\sum_{k=0}^n(n-2k)^{2p}2^{-n}\binom{n}{k}
&\sim\frac1{n^p}\frac1{\sqrt{\pi n/2}}\int_{-\infty}^\infty(2x)^{2p}e^{-2x^2/n}\,\mathrm{d}x\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty x^{2p}e^{-x^2/2}\,\mathrm{d}x\\
&=\frac2{\sqrt{2\pi}}\int_0^\infty x^{p-1/2}e^{-x/2}\cdot\tfrac12\mathrm{d}x\\
&=\frac{2^p}{\sqrt{\pi}}\int_0^\infty x^{p-1/2}e^{-x}\,\mathrm{d}x\\
3
@Gato Seuls les faibles dorment :P (je n'ai pas cours cette après midi, j'ai juste eu une colle de maths)
@hipp A square field with side 1 is divided among three people. Prove that at least one of the persons owns two points distant by strictly more than 1? :P
something something omelette du fromage
@robjohn wow, that looks really awesome! :-)
@Sawarnik I did this one
15:33
je suis très faible dans ta définition donc.
@Chris'ssis how did you approach it?
@Sawarnik Hint : draw circles
@Hippalectryon O o . its there on your top posts :O
@Hippalectryon Comment font-ils pour autant calculer d'intégrales ? Ils sont vraiment bons!
@Gato C'est la passion de @Chris'ssis
Intégrales, sommes, ...
15:35
Btw, @hipp Can I ask your what's your real name? :D
@Sawarnik No, it's forbidden by old pigeon laws :/
Amaury
Hello@Sawarnik
Hi @Rememberme I share your frustation with chem.
@Hippalectryon passion, il faut aussi du talent.
@Sawarnik I wouldn't have given my name before, but since I'm probably going to send a paper with my name on it for a contest soon, I don't care anymore :p
@Gato Ca ils en ont plein :D
15:37
@Hippalectryon What was teh problme in revealing your name?
@Hippalectryon ajd j'ai calculé $\int_0^{+\infty}cos(t^2)dt$ j'étais content :D.
:21741462 is it easy to see that the $p^\text{th}$ derivative of $\cosh^n(x)$ at $x=0$ is $(2p-1)!!$?
@Sawarnik I just don't get the logic behind chemistry especially physical chemistry
@Sawarnik My parents don't want me (my indentity) to be too much on the internet or whatever
@robjohn I'll show you my proof when I put things on paper.
15:38
@Rememberme logic ? in what part ?
@Chris'ssis sounds good
@Gato @Gato laisse moi réfléchir :P
@Rememberme I don't see logic anywhere in chemistry :((((((((((((((
Chemistry is very logical though
Yes logic...in physical chemistry @Hippalectryon we don't have proofs in it :(
15:39
What's the difference between chemistry and physical chemistry ?
We just assume stuff which has no damn proof
@Hippalectryon that's so so odd. can revealing your name pinpoint you?
Like what ?
@robjohn you don't need the $p$th derivative but the $2p$th derivative. By induction you show that it holds for $2p$ and then prove it for $2p+2$. All flows naturally and easily.
@Sawarnik Ask my parents q_q
15:40
@Hippalectryon like everything.
@robjohn How did you approach $\int_{0}^{\infty}cos(t^2)dt$ ? We did this in exam today but a long way using a 'contour' with $\sqrt{\pi}/2+ir, -\sqrt{\pi}/2+ir,\sqrt{\pi}/2-ir,-\sqrt{\pi}/2-ir,$...
In physical Chem we talk about solutions concentration etc @hippa
@Sawarnik What kind of things are you doing in phys chem ? (chapter, whatever)
@Rememberme ? So.. what do you assume there ?
@Gato I have two proofs on the site, one using complex integration and one using real methods.
@robjohn Cool, I will search.
15:41
@Hippalectryon i am not doing anything in chem.. but till now what i have read there is no logic.
Btw, @hipp Can I ask whether you are in college or school?
@Gato I will get the links
@Sawarnik More like college (some kind of French equivalent)
We assume that whatever is the formula for calculating these stuff is the way it is my prof scolds me when I ask him about proofs @hippa
@Gato Urm... ce n'est pas évident :c
15:43
@Rememberme I don't really see what kind of formula you 'assume' though. Concentration is just mol/volume
@Hippalectryon You have a lot of free time unlike college like school.
@robjohn $n^p$ (not the coefficients as I mentioned :-)) from numerator and denominator vanish and you get the result. That's all - and you're done.
@Sawarnik Not a lot. We're in a mid-exams period right now, it's a bit special
@robjohn I don't think is that..
We have finished written exams and if we did well enough we'll do oral exams and hopefully get the school we want @Sawarnik
15:45
@Gato That's the one using real methods
@Hippalectryon C'était un exercice de l'examen de ce matin.
No logic .. just assume and do .. that is chem.
No, they can be proved.
Oh, Balarka.
Just because you don't know what the logic is doesn't mean it has no proof :P
15:46
@Sawarnik You fill the inner ones first, not the outer ones
Because they're lower energy orbitals
@BalarkaSen Its not proved to me.
So I have to assume. I won't wait til masters to get my proof.
Well, read it if you don't want to wait.
@Sawarnik You don't need proofs, as long as you have a good understanding of what's happening. It's phys and chem, not maths
Grab up a book on quantum physics and read if that makes you happy.
@BalarkaSen And really, when did you start studying these things? :O :O
15:47
(at your level at least)
@Hippalectryon No, physics is quite intuitive .. there is logic in most things.
@Sawarnik intuitive doesn't mean easy to prove
I am not saying proof.
I am saying what I am reading should make a bit of sense.
The point of physics and chemistry is that you not only have to prove things, but also have to set up a reasonable set of axioms.
chemistry is intuitive too actually, once explained properly
15:48
That's what makes it harder.
@robjohn I don't find the one with complex methods.
that's a stupid thing to say, really.
No one is done with QP
@BalarkaSen hmm, true.
I have listened to my class lectures (which are very good) and that's about it.
15:50
i meant you are reading college level QP?
No, I am not. I have better ways to spend my time.
like maths ? :P
@BalarkaSen wait a minute .. you mean in class 10th .. you are being taught QP?
@Hippalectryon Yes, surely that's a better way to spend time.
@Sawarnik I'm telling you, phys and chem are totally logical, you just need to get a good understanding of it
That's also what SE is for
@Sawarnik They are teaching us scratches of it, yeah.
15:52
@Hippalectryon The books and teachers I have makes this worse.
Physics and chemistry are completely logical.
@Sawarnik Then ask me if you have questions I'll do my best to help :P
@BalarkaSen ... mind blown..
@Hippalectryon What courses are you taking in college?
@BalarkaSen Don't know why, but it feels odd to hear that from your mouth.
@Gato here is one...
People do change, @Sawarnik.
15:53
@Sawarnik In France it's not like in the US/Whatever, we're way less specialized and see a bit of everything. I have maths, physics, chemistry, English, French, Informatics with a strong science specialization (maths,phys,chem)
yikes, @Hippa.
@BalarkaSen ?
@Hippalectryon That's quite a lot to study :D
The classes préparatoires aux grandes écoles (CPGE) (English: "Higher School Preparatory Classes"), commonly called classes prépas or prépas, are part of the French post-secondary education system. They consist of two very intensive years (extendable to three or exceptionally four years) which act as a preparatory course (or cram school) with the main goal of training undergraduate students for enrollment in one of the grandes écoles. The workload is one of the highest in the world (between 35 and 45 contact hours a week, including written and oral exams). The students from CPGE have to ta...
That'd too much for US/whatever averages, I mean.
15:54
So if I ask which subject degree you would have, what would oyu say?
No idea, I'm not sure how it works outside of France
Wait a minute ..
that sounds like high school.
@Gato the first approach links to the one you cited, the second approach uses contour integration to show that the two integrals are equal.
'part of the French post-secondary education system. They consist of two very intensive years (extendable to three or exceptionally four years) which act as a preparatory course (or cram school) with the main goal of training undergraduate students for enrollment in one of the grandes écoles'
that's like the last two years of high school here?
Not exactly
15:56
@StanShunpike: is it going to be a bumpy ride?
It's after high school
@StanShunpike: sorry for all the HP references :-)
@Sawarnik After 'preparatory school' you can for instance (if you're among the best) get in en.wikipedia.org/wiki/%C3%89cole_Polytechnique
haha, @robjohn
@StanShunpike of course, you sort of asked for it :-p
15:58
@robjohn Nice one. I think I need to lear some result about Gamma function and Beta.
@BalarkaSen what for? :/
@Hippalectryon that looks like college to me? .. so are you in the first year or second?
@Gato they are useful in many situations.
@Sawarnik 'among the best is clearly necessary' for polytechnique.
@Sawarnik second.. and I might have to take that year twice q_q
@Sawarnik "stan shunpike" didn't ring the bell, then?
15:59
@BalarkaSen not at all.
then fuhget about it.
@Gato Are you in this two year thing as well?
@BalarkaSen Wasn't this the institution you aspired to go to?
@Sawarnik No, but I 'know' polytechnique.
@Sawarnik which?
@Gato Alright.
@BalarkaSen ecole polytechnique.
16:01
nope.
alright.
i don't wish to go in any university right now.
not at anytime.
i want to do math.
-_-
@Hippalectryon Well, they say the workload is highest .. but you have quite a lot of free time :D
@Sawarnik I'm working right now -_-
Huy
Huy
@BalarkaSen: Can you help me with a very basic geometry problem?
16:04
I'm bad with hard geometry, but let me see.
@Huy Just ask; don't ask to ask
@Hippalectryon Sorry :D
Huy
Huy
@BalarkaSen: imgur.com/HiBtStF $\beta = 50^\circ$, $AD = DE$. Find the remaining angles. I seem to be missing another condition to solve it. I have used Thales' theorem, sum of angles and also $\epsilon = \gamma + 90^\circ$ but I need another condition. Do you see it?
@Sawarnik awful equations with free surface of fluid in interaction with a magnet
yikes, that's too hard for me.
i donwanna think about it.
Huy
Huy
@BalarkaSen: I figured so.
16:06
@Huy Surely there is some missing conditions.
Huy
Huy
@Sawarnik: How so?
@Huy With those two conditions, you can create a lot of different quads.
What is ϵ though??
Huy
Huy
@Sawarnik: The angle at the point $E$.
the center $M$ is passing through that line, right?
Aha.
16:10
then the quadrilateral is a parallelogram
@Huy the figure is very misleading though. AD and AE are nowhere near the same length on it
Huy
Huy
@BalarkaSen: Why?
@Hippalectryon Exactly.
draw $DE$.
Huy
Huy
I did.
16:11
the angle that line makes with $DE$ is twice $50$ degrees.
furthermore, $MB = ME$.
Huy
Huy
I'm not sure which angle you mean. If we draw $DE$, let's split $\delta = \delta_1 + \delta_2$, where the upper part $\delta_1 = 40^\circ$ due to Thales' theorem. Which angle is twice $50^\circ$?
@robjohn it's the knight bus, what do you expect? And I enjoy it! You're one of the few with a sense of humor. :D
there's not enough marking.
Well, we have 1) the sum of angles 2) the isocele triangle -> we need one more equation and we'll have the three angles, right ?
give me a fully marked picture, then I'll explain
Huy
Huy
16:13
@BalarkaSen: Just describe which angle you mean in your first statement.
Also, if we are going by how things looks, then is AE tangent to the circle?
Huy
Huy
Definitely.
call $B$ the point where $\beta$ is made. $DME$ is twice $DBE$
@robjohn my favorite part about the night bus was how beds magically appeared at night yet the bus continued to move so violently no one would be able to sleep without some sort of magic.
@Huy Then it can be done.
Oof.
16:15
@robjohn btw, I sent the solution to the Knuth's problem to AMM. They accepted it (I corrected the problem first). :-)
Huy
Huy
@BalarkaSen: I agree.
ok, good.
now $MB = ME$
@Chris'ssis which one was that?
Huy
Huy
Yes.
If AE is tangent, then y=140-90 .. the rest are easy then.
16:16
so $MEB = 50^\circ$
@Hippalectryon Well done. Though you missed the center.
Huy
Huy
@Sawarnik: Why?
oh, I'm sorry. it's not a parallelogram, but it still can be done.
'cause you know $BEA$.
16:18
@Huy Because, ∠EAD= ∠DEA, right?
@Chris'ssis Ah, that looks quite related to problems we've worked on here.
And DEB=90?
@robjohn That one is quite advanced. :-) Yeah, we attended in the past a series called harmonic-binomial or something like that.
Huy
Huy
@Sawarnik: Yes, so?
Also, MEB=50 and DEA=90, so BEA=140.
16:21
@robjohn Be careful, don't work on it since it's wrong. If you wanna attend it, work only on the right side of the equality.
Huy
Huy
@Sawarnik: Yes.
Huy
Huy
Why is MEA = 100?
$DEA$ is 50.
it's not 100. it's 90.
Huy
Huy
DEA? Do you mean $\gamma$?
16:22
@Chris'ssis You mean it's not $\log(C(z))^2$?
@robjohn No, it's not, and it's not that $C(z)$ as it is defined there. I wonder how they missed that since in general people need to send a solution together with the problem. Maybe since he is a celebrity they didn't ask him for a solution.
@BalarkaSen yeah, sure.
no, the angle $DEA$
Huy
Huy
I don't know which other angle you would mean.
no, no, no
MEA is 90
16:23
oh, I think I have it.
because AE is tangent
Huy
Huy
Oh, right, @Sawarnik.
angle $\gamma$ is $50^\circ$
Huy
Huy
So, $\gamma = 40^\circ$.
So, BEA= MEA+ MEB= 90 + 50 = 140
16:24
@Huy no, it's not.
@Huy its 50? you must have made an error.
note that angle $BDE$ is $40^\circ$, and $MD = ME$
Huy
Huy
@BalarkaSen: BEM is 40, so MED is 50. Then, using that MEA is 90, $\gamma$ should be 90-50, so 40, no?
Oh.
That's odd.
BEM is 50.
you're messing up.
Basically, gamma = AED = BEA - DEB= 140-90 = 50.
Huy
Huy
16:27
Ok, I think I got it. I just need to draw it again to make sure I got it.
Yay :D
I am leaving.
Hello leaving, I'm hippa :3
@Chris'ssis UCLA does not get a free copy of the full document. I can see the problem, but the whole document will cost 12 USD
@BalarkaSen Bye .. teach me chem soon.
@Hippalectryon Are hippas and hippos alike?
16:28
Not at all
mumbles something about olympiadish geometry being destructive to the mental faculty
@robjohn but you see the problem from that page, don't you? That part is visible.
Hippos have four legs, Hippa have 2
@BalarkaSen wRoNg.
oh, morning, @Mike.
16:28
Hippos don't have wings
@Chris'ssis yes.
@robjohn OK
@hipp Uh oh.
A hippalectryon (or hippalektryon, from Greek ἱππαλεκτρυών) is a type of fantastic hybrid creature of Ancient Greek folklore, half-horse (front) and half-rooster (hind), including the tail, wings and hind legs. Its colour varies between yellow and reddish. No myths related to it are currently known. The oldest representation currently known dates back to the 9th century BCE, and the motif grows most common in the 6th century, notably in vase painting and sometimes as statues, often shown with a rider. It is also featured on some pieces of currency. A few literary works of the 5th century mention...
morning
16:30
@robjohn You can see the whole problems on this page mat.uniroma2.it/~tauraso/AMM/amm.html
@Chris'ssis I recognize the last one :-)
@Hippalectryon That one is too easy though. ;)
@Chris'ssis All the PDFs give me a not found error though
@Hippalectryon yup.
@Sawarnik You keep deleting some of your messages.. you know you can edit them, right ?
16:38
@Hippalectryon No, actually, what i wrote was useless. :D
@Chris'ssis I know I've done 11821 and 11819 on chat at some point
@hipp You live in the Paris area?
@Chris'ssis 11810 looks very familiar, too
16:44
@Hippalectryon Do you prefer this system? Or the American/whatever where you have a bit more freedom?
Huy
Huy
@Sawarnik: So $\epsilon = 140^\circ, \gamma = 50^\circ$ and $\delta = 120^\circ$, right?
@Sawarnik Both have their advantages, but I'm very pleased with the French one.
@Hippalectryon that's odd.
Why ? (Don't get me wrong, the education system up to high school included is horrible. But prep school is great)
@Hippalectryon You are pleased with more and more workload and lesser freedom.
16:46
@Sawarnik Well, I chose that path :-)
I could have chosen others with more freedom and less work
@Huy Oh, sorry. You are totally correct :D
@Hippalectryon You are brave :D
@Sawarnik I'm not brave, I just like what I do (same as @BalarkaSen I guess)
@Sawarnik Anyway i'll say it once again, if you feel that some elements (or all of them xD) of chem/physics are obscure/don't make sense/need more explanations, feel free to ask. I'm not a genius but I can probably help
@hipp Do you get time for movies and fun?
16:49
yeah, I do. not a lot, but it's okay
@Hippalectryon Tell me one thing .. why are the arrangement of electrons in differnt orbitals are that way? Does it make sense?
@Sawarnik describe "that way". Do you mean their shape ? Or the order in which they are filled ?
The order in which they are filled.
And shape too maybe, but I don't know about that.
They're filled from the ones of lower energy from the ones of higher energy
Because the system always tends to chose the lowest energy state when it can
Why $2n^2$ (the max no. of electrons in nth shell)?
Why the outermost is 8?
16:52
@Sawarnik 2n^2 ? What's the context ? (we might not have the same notations etc)
Morning, @Ted
Well anyways.... nothing is comparable to topology.... :)
Good night @MikeM
Let $Y$ be a closed, oriented 3-manifold. So I want to make $\Bbb Z[\Bbb R]$ into a $\Bbb Z[H^1(Y;\Bbb Z)]$-module. I've got a 2-form $\omega$ on $Y$. Can you make sense of the statement "The action is defined by $\omega$"?
Hello @TedShifrin how was your day
16:55
Still in the middle, Remember :D
It might be that it only depends on $[\omega]$.
Closed 2-form? Maybe they're using Poincaré duality to get to $H^1$?
It's not entirely clear to me whether or not it should be closed. They write $[\omega]$ later, so I think it must be. Can you make your suggested action more explicit?
@robjohn Yeah, that one was mentioned by @r9m. I think you can use some work of yours to finish it completely elementarily by series manipulations only.
(The action should also be trivial if $\omega$ is exact, I think)
17:00
@Chris'ssis It is possible. I may work on it later.
I guess you pair your 1-cohomology class with $\omega$ (wedge and integrate)?
@robjohn I tell you for sure, 100%.
I never remember group ring stuff ...
that was my guess... but how does that give me an action on $\Bbb Z[\Bbb R]$? if $r \in \Bbb R$, then $\alpha \cdot t^r = t^{r+\int \alpha \wedge \omega}$?
@Sawarnik Say your principal quantum number is $n$, then the secondary quantum number ($l$) can be anywhere in $[-(n-1),(n-1)]$. Each orbital slot can hold maximum two electrons (one up and one down). Thus, the maximal number of electrons in one shell is $\sum_{l=1}^n2(2l-1)=2n^2$.
17:01
I guess that would give me the stuff I'm asking for... defined if $\omega$ is closed, trivial if $\omega$ is exact...
@Sawarnik (gtg eating, bbl in about 20 minutes)
As I said, I don't remember what an action on $\Bbb Z[\Bbb R]$ is ....
@TedShifrin: $\Bbb Z[\Bbb R]$ is a ring (but more importantly an abelian group), freely generated by $t^r$, where $r \in \Bbb R$
an action of $\Bbb Z[H^1(Y;\Bbb Z)]$ on $\Bbb Z[\Bbb R]$ is just a fancier way of saying an action of $H^1(Y;\Bbb Z)$ (as a group) on the abelian group $\Bbb Z[\Bbb R]$
so we just need to specify what $\alpha \cdot t^r$ is for all $\alpha, r$
Ok ... So basically you need an additive action on $\Bbb R$.
essentially.
17:07
If $\omega$ isn't closed, is this well-defined?
no, $\int \alpha \wedge \omega$ doesn't make sense then... but I'm satisfied with only defining it for closed forms
in the case I care about, $Y$ is the boundary of a symplectic manifold and $\omega$ the restriction of the symplectic form
Ok, that's what I wanted.
What does it mean when a question was marked as a duplicate by Community♦?
Topology is too hard, @Ted
@MartinSleziak Hovering over Community, I see "This question's author approved a pending duplicate vote."
17:20
Oh, I did no hover... My bad!
BTW it's the same user, who asked on meta about question ban.
I thought that there were some changes and question ban can only be temproray now. But maybe I did not remember it correctly.
@MartinSleziak closing old questions won't lift the ban
@MartinSleziak if it is just the too many questions in a certain period ban, then it is temporary.
@robjohn That's not what I was saying. Anyway, this questions is marked as duplicate correctly, in my opinion.
@MartinSleziak That is true, but I don't see a connection between that vote and the meta question. Perhaps I am missing something.
@Sawarnik back
The only connection between that vote and the meta question was that I went to look to the list of questions of the particular user.
17:24
Ah
Which is how I stumbled upon the closure, which I asked about.
@Sawarnik Typo above, it's $1\le l\le n$, but it doesn't change anything
I went to look to their profile page to see whether they might have reached question quota 6/d, 50/m. (It seems not, but there might be also some deleted questions, which I do not see.)
@Hippalectryon Back. Lemme see. :-)
@Hippalectryon nice, nice, nice :-)
Hello!! We have that $\Omega$ is open and bounded $\subseteq \mathbb{R}$.

What is $\overline{\Omega}$ ?
@robjohn do you have an idea?

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