Mathematics - 2016-05-06 (page 1 of 3)

12:00 AM

So, we should forget about the smoothly varying condition? I am worried that might destroy smoothness of the volume form.

Mike Miller

Write down the Gram-Schmidt process explicitly. The fact that the metric is smoothly varying is the proof that your orthonormal vector fields are.

@Semiclassical for you

PVAL

I'm at the point where a calculation I am trying to do is so frustrating for me to do by hand, that I'm considering having to build a practical model a la Tait.

Semiclassical

hmm! @MikeMiller

Balarka Sen

I can see why I can choose orthonormal vector fields smoothly inside a chart, if that was your question. I am just not sure if choosing a locally smooth set of orthonormal vector fields is enough to conclude smoothness of the volume form $\omega$. I suppose it is: if I have $n$ many smooth vector fields, I can locally orthonormalize them by the metric. And then that orthonormalization is still (locally) smooth, so my form eats that and spits out something smooth.

So yeah, all I need is smoothness of the inner product.

Mike Miller

@BalarkaSen Your form, defined locally, is smooth. You need to verify that the form does not depend on the choice of orthonormal basis; that finishes the job (but you needed that in the first place to check everything made sense).

@Semiclassical I did not understand his comment to my comment, maybe you do.

Balarka Sen

12:08 AM

I was going to do that.

It's not actually hard. If I have another orthonormal basis I can just take it to there by a basechange matrix which is orthogonal. Plus it's det = 1 because everything is positively oriented.

Since by basechange my form gets multiplied by det of the basechange matrix, it's the same.

Mike Miller

Yes, the point is just that $SO(n) \subset SL(n)$.

When you learn some fancy geometry you'll see that that's true in a very literal sense. Another time.

Balarka Sen

Alright, I look forward to it.

What's next? I forgot, let me check.

Mike Miller

I've forgotten everything I asked, you'll need to tell me as you go.

Balarka Sen

I need to show that my volume form is unique given the Riemannian metric, and is $\sqrt{|g|} dx_1 \wedge \cdots \wedge dx_n$.

*given the Riemannian metric and the positive orientation

Erm.

Mike Miller

You just did the uniqueness thing (which actually was more a well-definedness thing).

Balarka Sen

12:15 AM

You're right, I proved if a form is 1 on some orthonormal basis, it's 1 on all orthonormal basis. That means any two such things must take the same value everywhere.

I am trying to remember what was my proof sketch that it was that expression in local coordinates. I think I tricked myself into thinking I had a proof.

Mike Miller

You probably had a correct proof.

Are you sick? YOu should sleep.

Balarka Sen

I am sick. I should indeed sleep.

But will I?

Mike Miller

If you die you can't do math.

Balarka Sen

Oh yeah I remember what my proof sketch was. Let $\omega$ eat $\partial/\partial x_1, \cdots, \partial/\partial x_n$, the tangent vectors coming from the parametrization. $\omega$ eats those and spits out $|A|$, the determinant of the basechange matrix w.r.t the orthonormal basis $\omega$ is $1$ on.

Mike Miller

Where does the square root come from?

Balarka Sen

12:23 AM

Yeah, I am a little confused now. Wait a second, I was sure I thought about it.

OK, $|A|$ is the same as $\sqrt{|A|^2}$. And $|A|^2 = |A A^T|$. If in that local coordinate my Riemannian metric was Euclidean, I'd be done right now. What am I missing?

I guess I need to do something tricky here.

Oh nah I don't. I am being silly, that's it.

$AA^T$ is precisely $g = g_{ij}$.

Mike Miller

:)

Semiclassical

@MikeMiller I haven't touched Zworski myself, so I'm in the dark as well

Mike Miller

Time to be in the air for about another 5 hours I think, soon.

Semiclassical

best i can think of is that some of the terms/concepts may be defined in there but not presented in context

Balarka Sen

@MikeMiller Heading for the airport? Have a productive flight.

Mike Miller

12:38 AM

in the plane

Balarka Sen

Ah. I'll quit badgering you then.

Mike Miller

why?

I haven't started working yet, though I have a paper to read and its author behind me

Balarka Sen

I thought you weren't allowed to use internet on air.

Mike Miller

I'm on the plane, not in the air. There's a boarding process and then a taxiing process.

Balarka Sen

OK, quick thing about the area 2-form then.

Mike Miller

12:39 AM

Also, we talked about this before, you can but you have to pay. I do not intend to pay.

Balarka Sen

Ah, right, forgot. In the planes I have been, they tell you to switch it off.

You asked me to prove the area form on a hypersurface is a volume form. let's try it with something simple: if I have a parametrized surface $S$ inside $\Bbb R^3$, it's area 2-form is in local coordinates $n_1 dy \wedge dz + n_2 dz \wedge dx + n_3 dx \wedge dz$. That eats two tangent vectors $v, w$ and barfs out $\text{det}(n, v, w)$. That's precisely area of the parallelogram spanned by $v, w$. So it's the volume form on $S$, because that's what it does.

What threw me off at first is that there are three variables $x, y, z$, but I realized that doesn't actually matter because since $S$ is immersed in $\Bbb R^3$ I can immersion theorem it so that locally it's $(x^1, x^2, x^3)$ with $S$ given by $x^3 = 0$. That's two variables. Anyway, this is a trivial note. (I haven't actually tried to immersion theorem it out to conclude explicitly that this is the same as the first fundamental form, but I suppose that can be done)

So, the exact same thing can be done for the general case. $(\iota\omega_n)(v_1, \cdots, v_{n-1}) = \omega(n, v_1, \cdots, v_{n-1})$. And if $v_i$ are orthonormal in $T_pS$, that's $1$. So it's the volume form.

Mike Miller

12:55 AM

@Balarka: The actual point is that you're pulling back that 2-form to the submanifold. But yes, it's easy to see this gives the volume form.

Adeek

1:20 AM

hi @BalarkaSen

Balarka Sen

1:33 AM

@MikeMiller I agree.

Sorry I got disconnected for a while.

@Adeek Hi.

Albas

3:54 AM

@BalarkaSen I still do not get how to proceed with the proof. According to your hint a number $x$ is prime iff $\pi(x+1)-\pi(x)=0$or $1$. But I still do bot understand how to use this idea in $\sum1/p$

Albas

4:18 AM

Okay Balarka . Nevermind. I think I might have got one more way

Forever Mozart

5:13 AM

@@@

Forever Mozart

5:48 AM

pikachu

usukidoll

rawr

Forever Mozart

6:09 AM

@Huy

Huy is an amazing Guy

Albas

6:19 AM

@ForeverMozart hello

Huy

thx

user147690

He sure is Mozart, he sure is

Albas

Hey@AlexClark

user147690

Hey @Albas

Albas

What are you studying?

It always makes me think that when you must have taken that photo of yours which is now your profile picture ,which would have been many many years back you must have had cameras which clicked coloured photos. That's pretty advanced for that time@Huy

Huy

6:35 AM

@Albas: that was actually taken on my iPhone 5, Asians generally look a bit younger than they are

user147690

@Albas Doing homework for birfurcation, my least enjoyable class (by an absurd margin)

Tobias Kildetoft

@AlexClark Hi

user147690

Hey there

user147690

Is there an easy way to make your code run parallel @TobiasKildetoft?

Tobias Kildetoft

@AlexClark No idea

I don't really know anything about parallel stuff

user147690

6:41 AM

Sure, maybe I'll investigate it out of curiosity. It seems I can't get it to resume, but I can easily restart it, so something weird has happened, or maybe it just stops running for some unknown reason after so many calculations. But if I can get it to run parallel, I can run it overnight on ~30 computers easily

Tobias Kildetoft

@AlexClark that would be cool

Mike Miller

hello

Huy

go to bed

Mike Miller

no

user147690

ok

usukidoll

7:21 AM

burrrrrrp

Mary Star

8:30 AM

Hello!!

Let $R$ a ring.
Suppose that $\phi : R[x]\rightarrow R$ is an epimorphism.
Does it follow then that $R[x]/\ker\phi \cong R$ ?

Hello @TobiasKildetoft !! Do you have an idea?

Balarka Sen

8:48 AM

Hi @iwriteonbananas

iwriteonbananas

Hey @BalarkaSen

How's it going?

Balarka Sen

Learning, @iwriteonbananas

Had a couple interesting realizations in the past few days.

iwriteonbananas

Nice, what did you realize?

Balarka Sen

I realized what (1) Stokes' theorem means (2) what forms really are (3) where does the $\sqrt{|g|}$ in the volume form on a Riemannian metric comes from.

Unrelated ponderings, but adds up a lot to the global understanding.

Balarka Sen

9:16 AM

@Albas Did you figure out how to do it?

iwriteonbananas

Stokes's theorem is really cool

Balarka Sen

It's nice. It's hard to interpret what it actually means though.

At least in the abstract forms language.

iwriteonbananas

Which version of it do you know?

Balarka Sen

$\displaystyle \int_M d\omega = \int_{\partial M} \omega$, but I can't interpret what it means in that form. How do you think about it?

I.e., what is your intuition about Stokes' theorem?

iwriteonbananas

Not sure I have that much intuition for it. It's a significant generalization of the FTC.

Balarka Sen

9:24 AM

Eh, sure.

iwriteonbananas

What's your interpretation of it?

Hippalectryon

@BalarkaSen Isn't Stoke's theorem just integrating rotationals ? That's the version I know, and it's quite obvious intuitively (from Physics's point of view at least)

Balarka Sen

That ^

iwriteonbananas

Also, I'm pretty sure one can see Stokes's theorem in singular cohomology

Balarka Sen

For dimension 2 and 3 you can interpret it as "small swirlies add up to big swirly on the boundary".

iwriteonbananas

9:25 AM

if one translates to forms using de Rham, one gets something like Stokes's theorem

Hippalectryon

Yeah, what's different in dimension $n$ ?

Balarka Sen

Or, if you apply the Hodge star, "small divergences add up to big divergence on the boundary".

@Hippa There's no vector fields to interpret it into.

Vector fields can only get you till dimension 2 and 3.

Hippalectryon

Ah :(

Balarka Sen

@iwriteonbananas Sure. $\langle \sigma, \delta \psi \rangle = \langle \partial \sigma, \psi \rangle$.

iwriteonbananas

Right.

Balarka Sen

9:27 AM

But I wouldn't want to think of forms as just cochains. I don't feel like that's the whole story.

E.g., in that context, what does $k$-vector fields mean?

iwriteonbananas

Good point

Balarka Sen

I think Hippa's physical interpretation is the best I have - as Hippa said it's obvious why it's true from that point of view.

Especially the Hodge star-ed version, because divergence is easier to "see".

Hippalectryon

I don't know the Hodge star one :P I only use Stoke's theorem in physics

iwriteonbananas

I don't really understand the physics point of view very well

Balarka Sen

@Hippa There's two ways to write down Stokes theorem. $\displaystyle \int_{\partial D} \mathbf{F} \cdot \mathbf{T} ds = \int_D \text{curl}(\mathbf{F}) dS$, and $\displaystyle \int_{\partial D} \mathbf{F} \cdot \mathbf{n} ds = \int_D \text{div}(\mathbf{F}) dS$.

Hippalectryon

9:35 AM

@iwriteonbananas What part ?

iwriteonbananas

The whole thing

Balarka Sen

The latter is, in the context of forms, obtained from Hodge star-ing the first.

iwriteonbananas

never thought about it much

Hippalectryon

@BalarkaSen What's $D$ and $\partial D$ ? (I'm used to 3D at most, so I suppose $D$ is a surface and $\partial D$ a contour ?)

Balarka Sen

@iwriteonbananas Suppose I have a 1-form $\omega$ on $U \subset \Bbb R^2$. What's $d\omega$, in local coordinates?

@Hippa What could those possibly be? ;)

Yes.

iwriteonbananas

9:38 AM

@BalarkaSen Hold on, I was about to head out

Hippalectryon

I wish I had time to do more maths :P too many interesting things out there

Balarka Sen

Sure, bananas.

iwriteonbananas

I'm running late for something, I"ll be back tonight

Albas

With the help of robjohn @Balarka I managed to write the sum as a Riemann -Stieltjes integral and then use integration by parts. He had this answer so I basically did not figure everything out myself

Balarka Sen

ok...

Albas

9:40 AM

But I am trying to be better at this

It is a great book Nivan Zuckerman

Balarka Sen

@Hippalectryon Why can't you do math?

Hippalectryon

@BalarkaSen Because there are so many other interesting things :D and also because I have upcoming oral exams (in a month or so) that are very very important. Once I succeed in those and get into the school I want, I'll have more free time

Albas

@Hippalectryon I guess you also have something like physics troubles like me

Balarka Sen

@Hippa Great, good to hear.

Albas

Very true. Like ISI

Hippalectryon

9:42 AM

@BalarkaSen In the little free time I have I'm more into neural networks right now. Which, very interestingly, are deeply related to complex topology problems

Balarka Sen

No idea what those are. But sounds cool.

Hippalectryon

@Albas physics troubles ?

@BalarkaSen Artificial intelligence thingy

Albas

Ya stuff like fluid dynamics

Hippalectryon

@Albas I love fluid dynamics :D it's very cool. I like pretty much everything I've studied so far in physics except induction and electronics.

Albas

It is very complicated but interesting

@BalarkaSen did you go and watch civil war??

@Hippalectryon You should do electronics and electromagnetism. If you like multivariable calc then you will love how it is applied in it

Balarka Sen

9:46 AM

No I am not that keen on watching that to go on theaters. I'll get a DVD.

Hippalectryon

@Albas I already have :-) I'm just not fond of "partical" electronics

Albas

Have you read the comics @Balarka they present a better picture of the whole plot

Ahh@Hippalectryon . What all have you done in physics?

Balarka Sen

no unfortunately I don't have the time to read a series of comics.

Hippalectryon

@Albas Hmm... Thermodynamics, kinetics, fluid dynamics, electromagnetism, quantum physics, diffusion (of matter and of heat), waves in solids. I might be forgetting minor chapters

Albas

that's correct@Balarka there is very less time after one spends 3/4 th of the whole day doing maths

@Hippalectryon my teacher destroyed thermodynamics for me

Hippalectryon

9:51 AM

@Albas O_o what did he do

Albas

She taught us thermodynamics as if she was teaching history.

I had this question that what is the reason that why is the carnot engine is the most efficient intuitively and her answer was that it is because that's what is given in the textbook

Hippalectryon

._.

"Hey teacher why can particles go through the barrier by the tunnel effect ?" - "Cuz it's written in your textbook"

Albas

Then came entropy. Very fascinating stuff. I asked her that if entropy meant things going to disorder then can we somehow create a mathematical model that would tell us how much time will something take to go to disorder

Hippalectryon

off for lunch, brb ~15 mins

Albas

Fine.

Hippalectryon

10:12 AM

back

Albas

Hello again

Hippalectryon

@Albas Yeah, there are plenty of models

Albas

Could you tell me one?@Hippalectryon

Hippalectryon

Well take for instance the diffusion of a gas in an infinite space, a simple diffusion model tells us everything we need

Albas

Hmm I haven't been taught diffusion stuff yet

Hippalectryon

10:17 AM

Ah that's a shame, it's fairly easy.

That being said, "going to disorder" doesn't make a lot of sense to me

There's not an exact time at which we are "at disorder"

Albas

It is a continuous process

Hippalectryon

Exactly (except at $t=0$ maybe)

Albas

My notes say entropy is the property of a system having a tendency to go to disorder

Hippalectryon

uh... that doesn't sound like a very clear definition to me :P

Whether it goes to disorder or not, it has an entropy

Albas

You are talking of entropy as if it is something that is inhibited by a system. I do not understand. What is the correct definition if entropy

@Hippalectryon I do not expect my teacher to give correct definitions. She doesn't herself know what is lateral displacement

Hippalectryon

10:23 AM

The "correct" definition of entropy isn't very interesting as far as having a physical sense of it is involved. That's why the definitions given are a bit vague (like the definition of energy)... but usually they make more sense that the one you've been given

Entropy, just like potentiel energy, is just a property that any thermodynamical system has. Overall (i.e. in a totally closed system), it can only increase. That property is closely linked to the "disorder" of the system.

Albas

So its like a ball being thrown up with a infinite amount of force such that it keeps on going up and potential energy keeps on increasing

Hippalectryon

Uh... the analogy to energy might be misleading, since energy is conserved whereas entropy is created

Also, "disorder" does not exactly coincide with the intuitive definition we might have. Since you've studied fluid mechanics, have you studied Stokes flows by any chance ?

Balarka Sen

@Hippa What's the definition of an entropy?

Albas

@Hippalectryon you would start crying if I tell you how much of stokes law we have been taught. There is just one line ( According to my teacher this is enough fir solving numericals) $F=6\pi\eta a v$ where a is the radius of a ball dropped into the liquid and v is its velocity

Hippalectryon

@BalarkaSen In physics it's "easily" defined by Boltzman's definition $S=k_B\ln\Omega$, but it givs little insight about what it really is. ($\Omega$ is basically the number of different possible configurations of the system

Albas

10:31 AM

That stuff involves things like microstates right?@Hippa

Balarka Sen

What is $S$, what is $k_B$ and what is $\Omega$?

Hippalectryon

@Albas yeah

@BalarkaSen $S$ is the entropy, $k_B$ is a constant

Albas

$\Omega$ if I am not wrong is the number of microstates.

Balarka Sen

Entropy of what? What's the context?

Hippalectryon

@BalarkaSen thermodynamics. It's not the same definition as in information theory (in case you knew another one)

@Albas yeah

Balarka Sen

10:33 AM

I don't know any definition of entropy, thus I am asking for one. I don't know if "entropy of a thermodynamics" is a thing I can make sense of.

Hippalectryon

@Albas Uh that's not what I meant :P ok basically at very low Reynolds numbers, the quasi-static Navier Stokes equation is time reversible. So basically its entropy change is $0$

Albas

I just know Navier stokes equation comes in the millennium problems

I am in high school so I do not know what Navier Stokes says

Hippalectryon

@BalarkaSen Ah. Thermodynamics is (very roughly) the study of the different energies involved in the transformations of fluids. And one of the properties widely used is entropy, which (for the layman) measures the "disorder" of the system

@Albas Oh, ok my bad. Anyhow it's not really important. What matters is that the flow is time reversible, so the entropy stays constant

Albas

Its like how much chaos is there in the system

Hippalectryon

@Albas Now watch this youtube.com/watch?v=p08_KlTKP50 It's an example of such a time reversible flow. Entropy-wise, the "disorder" stays the same during the whole experiment, whether we'd intuitively says it increases a lot in the first half !

Balarka Sen

10:36 AM

@Hippalectryon I have heard that. But I am asking for a mathematical definition.

Hippalectryon

@BalarkaSen Well $S=k_b\ln\Omega$ then, where $\Omega$ is the number of possible arrangements of the system

Balarka Sen

What's a system?

Hippalectryon

@BalarkaSen A macroscopic set of particles (i.e. lots of particles)

Balarka Sen

Mhm. I mean a system as in an amount of gas inside some container? So you have homogeneous collision between those particles?

Hippalectryon

@BalarkaSen yeah. But you could also chose a subsystem composed of only some particles of the broader system.

Balarka Sen

10:40 AM

OK, what's an arrangement of a system?

Albas

@Hippalectryon you forgot the boundary. @BalarkaSen it is a set of particles inside a certain closed area marked by a boundary

Hippalectryon

@Albas But the boundary can change over time so it's fine

Albas

Hmm.. Yes expansion of a gas

@Hippa what did you mean by time reversible?

Hippalectryon

@BalarkaSen A configuration. For instance if your system is composed of 3 particles and there are exactly 5 "places" where they can belong (and two particles can't be in the same hole), then you have 5*4*3=60 possible configurations

Albas

Its like basic combinations and permutations

Hippalectryon

10:43 AM

@Albas Suppose you are in a state A. Now, you do a series of operations to the system an you end up in a state B. If it's time reversible, then by doing the series of operations backwards, you'll end up back in the state A.

Albas

Ahh... Like a cyclic process?

Hippalectryon

Kind of

Albas

Carnot cycle is then time reversible?

Balarka Sen

@Hippa So your system is moreover finite? That's a new assumption.

So your system is not a realistic thing.

Hippalectryon

@Albas Not necessarily. A cycle is rarely symmetrical.

Albas

10:46 AM

Symmetrical?

Hippalectryon

Let me explain another way with some examples. If you open a bottle and turn it upside down, the water will fall on the ground. Now suppose you record it, and play it backwards : you'd see the water rising from the flow back into the bottle. That's absurd ! Then it's not time reversible.

Balarka Sen

So if you have $n$ many slots and $k$ many particles, then $\Omega = \binom{n}{k}$?

Albas

If the k particles are not distinct

Hippalectryon

@BalarkaSen Not necessarily. For instance, now suppose that your particles have some given speed, then a configuration in which particle 1 has a speed +V is different from a configuration in which particle 1 has a speed -V

Albas

So then it would be better to take them as distinct

Balarka Sen

10:50 AM

Er... those were not in the definition of a system a couple messages ago. But that makes sense.

Alright, I can live with that.

So, what does $k_B \log(\Omega)$ really means?

Albas

Yes I get the meaning of time reversible @Hippa

Balarka Sen

I.e., how does it relate to "disorder"?

Albas

For that we should understand what k_B means@Balarka

Balarka Sen

I don't know what $k_B$ is, so sure.

Hippalectryon

I'm not sure what you wanna say about $k_B$ so go ahead :-) @Albas

Albas

10:53 AM

I do not know @Hippa I just said we would have to know about what$k_B$ to understand that expression

Hippalectryon

@Albas It's just a constant

Albas

But why is it there

Balarka Sen

@Hippa So, want to tell me how does it relate to "disorder"?

Hippalectryon

@Albas For historical reasons

@BalarkaSen From a classical point of view, a system at $T=0K$ is perfectly ordered. It has only one accessible state. When you heat it up, more states become accessible, and so $S$ increases.

ramsay

why a cubic equation with 3 real and distinct roots have to acquire this shape?

Hippalectryon

10:57 AM

@ramsay What other shape do you envision ?

Albas

Use maxima and minima of a function @ramsay

Balarka Sen

@Hippalectryon That makes sense.

Why the $\log(\Omega)$? Why not just $\Omega$, then...? :P

Albas

For historical reason :P

Hippalectryon

@BalarkaSen Beause $S$ is used in many other formulas, and we're not gonna change all those formulas, I guess

Balarka Sen

I mean I don't see the difference between entropy and # of states/configurations from the example you gave.

Hippalectryon

10:58 AM

@BalarkaSen Entropy has a complicated history. The concept was introduced before Boltzman's formula

Balarka Sen

@Hippalectryon OK, I am obviously not convinced with that reason, but that's fine.

Tobias Kildetoft

@BalarkaSen often when a log shows up such places it has something to do with how we perceive things (like in the decibel scale)

Hippalectryon

It's like asking why we define the audio intensity to be $I_{DB}=10\log(I/I_0)$. Why not simply use $I$ ? Because in the long run it's more convenient

Tobias Kildetoft

@Hippalectryon there is actually a good reason for that as I just mentioned

it is is because in our perception, doubling the intensity of sounds twice sounds like increasing by the same amount twice

Hippalectryon

:29486741 Entropy is not a process though, it's a property

Balarka Sen

11:01 AM

@Tobias @Hippa Fair, I haven't seen it coming up in many places, so I wouldn't know.

Hippalectryon

@TobiasKildetoft Yeah sure. I was just trying to give some mild analogy :-) I'm definitely not an expert in either field though

Balarka Sen

Thanks though, that's an interesting intuition, Tobias.

Albas

Really nice way to think @TobiasKildetoft

Tobias Kildetoft

I know they actually used the same principle when designing the grading scale used in the '70s in Denmark

Balarka Sen

Hah

Tobias Kildetoft

11:03 AM

which meant that the underlying point-scores for calculating averages looked really absurd

Hippalectryon

Well, we use functions like shifted $\tanh$ for our exams ._.

Albas

@Hippa what is stokes law actually except the formula given to me?

Hippalectryon

@Albas Stokes' law is what you know. Stokes flows are quasi-static flows at very low Reynolds numbers

Albas

Ahh.. Okay

Balarka Sen

*Stokes'

You said "Stoke's law".

Hippalectryon

11:12 AM

Ah my bad

Albas

@Hippa Want to indulge in an integral ?

Balarka Sen

[troll grin]

Hippalectryon

@Albas uh... sure ?

Albas

$\int_{\frac{1}{a}}^a \frac{\arctan(x)}{x}$ You have to solve this without invoking $Li$

Balarka Sen

Hi @Akiva.

Albas

11:17 AM

I had decided to solve it this afternoon but entropy came in between :p

user147690

@Albas Why are you working on integrals? For class?

Albas

A friend gave me this problem. Liked it a lot

Balarka Sen

Classic.

Albas

Haha .. What I just said was sheer nonsense

user147690

@BalarkaSen A great one.

user147690

11:26 AM

@Albas There was never a point in which I actually enjoyed integrals :S.

Albas

Did you give something like an entrance exam @AlexClark for your uni?

user147690

Definitely not.

Balarka Sen

How's life, @Alex?

Albas

Ahh.. Or else you would have faced such integrals in the entrance exam.paper

Hippalectryon

@Albas $a>0$ ?

user147690

11:28 AM

@BalarkaSen Varying right now, but probably averaging to neutral :P.

Albas

Yes Hippa. Sorry for not mentioning it.

user147690

I am just doing bifurcation homework right now, very boring stuff. Really looking forward to more AG and stuff for my research paper.

Balarka Sen

@AlexClark You know what's strange? I listened to that song and thought Bowie was mimicking Dylan's voice. And then I realized Dylan also wrote a song about time and changes.

Probably just a funny coincidence.

user147690

@BalarkaSen That's another great song though for sure.

user147690

My ex used to play that one.

Balarka Sen

11:32 AM

@AlexClark There's no doubt about it.

Cody

11:48 AM

Is a x b = HCF OF (a,b) x LCM of (a,b) hold true for more than 2 numbers ?

Akiva Weinberger

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Q: Property similar to connectedness

Recall that $X$ is connected if $X$ cannot be written as the union of nonempty open sets with empty intersection. Consider the following similar property: $X$ is good if $X$ cannot be written as the union of connected open sets with disconnected intersection. In other words, $X$ is good if...

Anyone have any idea for that?

Balarka Sen

@Akiva I doubt if there is a good space with $H_1(X; G) \neq 0$ for all $G$.

Cody

12:04 PM

@BalarkaSen Is a x b = HCF OF (a,b) x LCM of (a,b) hold true for more than 2 numbers ?

Akiva Weinberger

That's why I was asking about $H_1(P^2,\Bbb Z_3)$ yesterday, by the way

Balarka Sen

There is an analogue for 3 numbers, yes.

Note that gcd(a, b, c) = gcd(a, gcd(b, c)).

Cody

@BalarkaSen so a x b x c = HCF OF (a,b,c) x LCM of (a,b,c) is correct

Balarka Sen

No.

Akiva Weinberger

But it's not obvious what the right analogue is, right?

Balarka Sen

12:07 PM

Depends on what right or left means.

user147690

12:20 PM

@Cody a*b=gcd(a,b)*lcm(a,b)

Mambo

@AkivaWeinberger Where did the motivation of good spaces come from?

user147690

@Cody gcd(a,b,c)lcm(ab,ac,bc)=abc

Hippalectryon

And then, very good spaces : $X$ is very good if $X$ cannot be written as the union of good open sets with bad intersection.

And so on :D

user147690

and with the latter, exchanging gcd and lcm, it still works

Albas

Just an addition gcd(a,gcd(b,c))=gcd(gcd(a,b),c)=gcd(gcd(a,c),b)

@Hippa Did you get the integral?

Balarka Sen

12:26 PM

expanding on what i was saying earlier: gcd(a, gcd(b, c)) = gcd(a, b, c) gives abc = gcd(a, b, c)lcm(a, gcd(b, c))lcm(b, c).

That's the most general thing I know of.

Hippalectryon

@Albas no :(

Balarka Sen

@Hippalectryon $X$ is excellent if it cannot be written as any union of open sets at all.

2

Hippalectryon

(◔_◔)

Tobias Kildetoft

@BalarkaSen Well, that is close to the definition of irreducible :)

Albas

Sad. I do not know how to solve that integral.

Hippalectryon

12:35 PM

@Albas Have you seen any promising way though ?

Albas

Nope.

What did you try@Hippa

Hippalectryon

@Albas Ibp, substitution, power series

Mambo

@Hippalectryon How did you do that

millipede

Balarka Sen

I wanted to see if starring that would destroy the starboard. Apparently not. Oh well.

Hippalectryon

@BalarkaSen Nope. Use this if you want to destroy it

Balarka Sen

12:48 PM

No I won't.

Hippalectryon

Can't you star it and unstar it 20 secs later ? I'm actually interested in whever it works

Balarka Sen

Yeah that's ugly.

Hippalectryon

It works ._.

Balarka Sen

@Hippalectryon I did that to the previous message but apparently someone starred it again.

And apparently they do not intend to unstar it.

Hippalectryon

Bah a mod will unstar it

@BalarkaSen If I flag it, is there an option to add a comment like "some starred it but it's ugly on the starboard, it's my message" ? I've never flagged chat comments before so I don't know

Balarka Sen

12:54 PM

You shouldn't flag it. Other mods from other places will come and snarl at you for flagging non-offensive message.

Hippalectryon

Uh ok

Balarka Sen

@DanielFischer For when you come here: Hippa wants some mod to delete/unstar his message that was starred because it clutters the starboard. Would you kindly do it? Thanks.

Done, @Hippa.

Hippalectryon

Thanks

Mambo

unstar it

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$Mathematics$ Mathematics