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12:21 AM
@TedShifrin Going back to my notes from analysis:
So in my case I can only write $f(x)=T_{m-1}+O(x^m)$, since $f\in C^m$.
From this is then, how can one derive $f(x)=\sum_{k=0}^m \frac{f^{(k)}(0)}{k!} x^k + \frac{O(x^{m})}{x^{m}}$?
2 hours later…
2:42 AM
@Thorgott Of course there are many examples, right? Something like the Gabriel's horn
Gabriel's horn but surface of revolution of $y = 1/x$ from $x = 0$ to $x = 1$, let's say (the usual horn is between $x = 1$ to $x = \infty$). The surface area is $2\pi\int_0^1 1/x \cdot \sqrt{1 + 1/x^4} dx \geq 2\pi\int_0^1 dx/x$, blows up
But it's this incomplete surface trapped in some finite diameter, which forces I think that eg a geodesic ball of radius $100$ centered at any point will be all of the surface
@BalarkaSen That seems wrong.
Oh, I do the horn from 1 to $\infty$.
It shouldn't matter.
Yeah this isn't quite right because arclength of $1/x$ from $x = 0$ to $x = 1$ is still fairly infinite.
The apparent diameter is finite but not the actual diameter
Yeah, I guess the profile curve has infinite length either way.
3:03 AM
Meh, this isn't a good idea
I am skeptical!
Surface of revolution of an arclength parametrized curve has $ds^2 = dx^2 + f(x)^2 dy^2$, so the volume of the ball of radius $r$ around let's say the origin is $\int_0^r f(t) dt$. Of course you can arrange this is to blow up at a finite $r$, right?
Just a mess.
I don't understand the question
In this language it's just choosing a function whose integral diverges
Oh, what you have is right with a $2\pi$. You need $|f’|<1$, I guess.
Yeah, I forgot the $2\pi$. Also yes.
The picture is like the $0 \leq x \leq 1$ Garbiel's horn but I don't know why that exactly didn't work. Anyhow...
Hi All..
hello, does anyone have a brilliant geometric math quiz? thnks
3:29 AM
Balarka, you need arclength param, and the integral blowing up finitely.
1 hour later…
4:41 AM
A loop on an orientable surface $\Sigma$ equipped with a Riemannian metric is called shortest if it has the shortest length in its free homotopy class. Let $\varphi$ be a self-homotopy equivalence of $\Sigma$ and $f$ be a shortest loop on $\Sigma$. Is it true that $\varphi\circ f$ is again a shortest loop on $\Sigma$? My guess is no.
Of course not. $\varphi$ has nothing to do with the Riemannian metric on $\Sigma$, why would it preserve the property of being shortest?
"My Own Notation, I beg people to stop abusing the letter d. I no longer allow myself to diverge from 8-bit calculations. The Solar Wind will wreck my computers. 8 bit is my home when working on computers."
2 hours later…
6:48 AM
@BalarkaSen How are you doing? Do you, by however miniscule chances remember me?
1 hour later…
7:53 AM
the world seems so quiet without all those tweets...
8:08 AM
@SubhasisBiswas Oh yeah I remember you. Hi
You've completed your BMath?
I'm in Belur now. Doing MSc.
I'm in my last year.
That's great!
Belur is a nice place
Head of the Department now is Adhikari, right?
I know him well.
I'd like to ask him about you. He'd confirm whether or not you're a bot.
I wish Mahan Mj was still here.
8:14 AM
Yeah he's really the one who modelled the math department and syllabi.
Yeah. I've heard a lot about him. This place seems so silent now after hearing stories about him.
are you planning to get out of India?
No, not really.
Well, wish you the very best of luck for whatever you do. I actually learnt a lot from whatever interaction we had in the past.
Thanks. It's great you're in Belur now; all the best!
4 hours later…
11:54 AM
For one dimensional heat and wave equation....
I am unable to figure out how to solve this...
Typical out of hyperbolic and prabaolic equation is second derivative in x and t both
And one only in x
Any logic behind this is welcome.
12:14 PM
What's your question ?
No more exams for me
how's it going?
I can cry in peace
12:17 PM
I think my interview went well yesterday
I think
Which interview?
I had an interview for a position as a trainee software developer
12:26 PM
Are you looking for a job to work after your masters? Or concurrently with it?
I'm taking an indefinite hiatus until I'm mentally stable enough to continue
I see, prioritizing health is a good plan
12:44 PM
@EdwardEvans Nice!
Of course, it remains to be seen if it actually went well
Fingers crossed for you
Please do
1:21 PM
Is anyone here familiar with steffensen's method in numerical analysis?
Challenge : Find a completely factored number of the form $\Phi_n(n)$ (the $n$ th cyclotomic polynomial at $x=n$) with $\omega(\Phi_n(n))>12$ (more than $12$ distinct prime factors). $n=125$ is the current champion with $12$ distinct prime factors.
Challenge: make me understand these goddamn n-gons for genus g surfaces
Morning everybody.
If I cut one disk out of that it's homotopy equivalent to just the wedge of 2g circles, right? And for every other disk that I cut out I just add a circle to the wedge I suppose?
Genus g represented the United Kingdom at the Eurovision Song Contest in Oslo 1996
1:32 PM
no that was Tina T
A person over on DBA.SE has a question regarding mathematics in a sense and I'm no pro when it comes to explaining things. Her question is:
Q: cube and one example

Betty Anderssonif a data presented with 4-dimension in which each dimension is dependent to hierarchical 3-level aggregate like (country, city, street), then we can summarize it into 4096 ways ! we know for cube with n dimension in which none of dimension is hierarchical we have 2^n summarizing ways, but in th...

Would this be a question suitable for this site?
Or does anybody here have time to earn some points over on DBA.SE?
Hi, M is an embedded sub manifold of dimension $k$ in $\mathbb{R}^n$ iff there exists a a submersion $f:\mathbb{R}^n \rightarrow \mathbb{R}^{n-k}$ such that $f^{-1}(0)=M$, right?
@JohnK.N. this is probably not a hard problem, but it's not clearly understandable how all this works, partly due do the lack of context, partly due to language difficulties
I think asking computer scientists is more appropriate
@BalarkaSen Yeah, Gabriel's Horn doesn't do the job since it has infinite diameter, but I posted it as question on main and received an answer while I was asleep that suggests something similar-looking to what you wrote here
1:46 PM
@monoidaltransform This is not true. Only if $M$ has trivial normal bundle in $\Bbb R^n$.
@Thorgott Oh ok
@user2103480 This is correct.
@BalarkaSen if $M$ is embedded in $\mathbb{R}^n$ with vanishing of n-k coordinates $x^{k+1},.......,x^n$, why can't we just take $f$ to be $(x^{k+1},......,x^n)$?
where do you get such an embedding from?
@monoidaltransform this is only locally true
locally it exists, of course, but globally?
oh right, yeah
1:55 PM
but of course it's true that every embedded submanifold locally looks like a level set of some submersion, precisely for this reason
@user2103480 Thanks.
what would be an example of a 2 dimensional embedded sub manifold that has trivial normal bundle in $\mathbb{R}^3$?
and an example of one that doesn't?
for codimension 1 submanifolds, trivial normal bundle is equivalent to orientability
1:59 PM
@monoidaltransform there's no example in 3D. RP^2 in R^4 is an example
any embedding of RP^2 in R^4
so every 2 dimensional sub manifold in $\mathbb{R}^3$ has trivial normal bundle?
and so may be recognised as the zero set of some global submersion?
closed submanifolds just to be clear
but yes
sorry, which part are you trying to be clear about?
the part where you say "submanifold"
you should say an adjective, that being "closed"
otherwise, Möbius strip
2:01 PM
it's clear from context that you want closed because you want $f^{-1}(0) = M$
but you should spell it out
huh? the map needn't be proper
@Thorgott every codimension 1 sub manifold is orientable in $\mathbb{R}^3$
you're forgetting Balarka's adjective
2:04 PM
@Thorgott this is irrelevant
the preimage needn't be closed man
i dont understand your comment
you're implying closedness is automatic, which it isn't, so I'm confused
i dont know what your question is so i dont know what my response should be
$f^{-1}(0)$ is a closed subset of $\Bbb R^3$, so if it's noncompact it's gonna be something like properly embedded, but eg Mobius strip has no proper embedding in $\Bbb R^3$
this is why i said he's implicitly assuming closedness of the manifold
anyway, this is again some technical detour that is not going to help @monoidaltransform
if there's no question i'll leave it at that
2:13 PM
Why would that imply the embedding is proper?
Why don't you tell me if it doesn't?
Seems like a good exercise to think about for you
oh, I misunderstood, of course the inclusion of a closed (in the top sense) subset of R^n is gonna be proper
the issue's that the Möbius strip isn't even closed (in the top sense)
so that's a counter-example for trivial reasons
It is a counterexample to what?
2:18 PM
to the claim that every embedded submanifold is a level set of some submersion
Oh that's a dumb counterexample, of course you want the manifold to be a closed subset, aka, properly embedded.
The point is Mobius doesn't even properly embed in R^3.
well yeah, but I thought you meant closed in the manifold sense initially
I did.
but that wouldn't be automatic
Think about all of this again. You're so confused.
2:21 PM
the preimage of 0 under a submersion isn't always gonna be compact
When did I claim this?
never explicitly, but if you aren't claiming that, why would we assume M is closed
Because there's no non-closed example of a codimension 1 submanifold which is not a level set.
if only that were true
2:26 PM
It is true, every properly embedded codimension 1 submanifold is orientable, has trivial normal bundle.
Nonclosed topologically closed submanifold necessarily is properly embedded (If not you can find a sequence going off to infinity in the submanifold which converges to something finite in the ambient Euclidean space.... that point certainly cannot be a manifold point but it also cannot be a topological boundary point)
all top closed subsets are properly embedded
so you're just saying there aren't any codim 1 counter-examples at all (except for trivial non-top closed ones), right?
yeah, euclidean space
but the only interesting manifold is the torus
@RyanUnger this actually
we dont even understand the torus
do you know the overtorical band inequality
2:38 PM
i was just thinking about the systolic inequality
have you read guth's proof
i have seen the probability proof lol
there's too damn much to learn
math is a hopeless endeavor until they make mind uploading feasible
no point in voluntarily being Sisyphus
I'm trying to learn pde now
@BalarkaSen substitute "math" with "life"
2:45 PM
this is no laughing matter
(not that we're not already doing that)
speaking of pde, do you know the equation of fire
whats that
In mathematics, the Kuramoto–Sivashinsky equation (also called the KS equation or flame equation) is a fourth-order nonlinear partial differential equation, named after Yoshiki Kuramoto and Gregory Sivashinsky, who derived the equation to model the diffusive instabilities in a laminar flame front in the late 1970s. The equation reads as u t + Δ 2 u + Δ u + 1 2...
Say i know the generating function of $X$ to be $M_X(s)=e^{3s+4e^{3s}-4}$ what can I know about $X$ himself? I know that $E[X]=15$ from taking the derivative and assigninig $s=0$ but what now?
2:46 PM
someone told me it's linearly unstable but nonlinearly stable
lmao thats so garbage
If someone is interested in reinforcement learning, here you have a recurrent question.
Q: What is the difference between Q-learning, Deep Q-learning and Deep Q-network?

datdinhquocQ-learning uses a table to store all state-action pairs. Q-learning is a model-free RL algorithm, so how could there be the one called Deep Q-learning, as deep means using DNN; or maybe the state-action table (Q-table) is still there but the DNN is only for input reception (e.g. turning images in...

Gregory Sivashinsky (also known as Grisha) is a fellow of the Combustion Institute @RyanUnger
sounds like a place where you can dispose of a body
2:52 PM
@Eminem You can find out all the moments there, I think. Also, if X itself has a probability density on $\Bbb R$, you can transform this back (with the inverse laplace transform) to obtain the density
@RyanUnger edgy
@user2103480 Havent learned the laplace transform...I could calculate the variance as well, but as far as finding out $f_X(x)$ there is no option for me i think...
That is generally a problem we let computers do, in any case
So I am struggling with some notation in this book. I tried to draw a diagram to clear up where things are being taken, but the relevant arrow shows up twice, which to me is confusing. How is one supposed to reconcile this? Perhaps I am drawing it wrong. My guess is that we consider the elements $a, b$ as zeroes of $f$ in the first arrow going down, and then as field elements in the arrow going up to the right. Does this make things better?

The relevant image:
Here this might actually be possible, but with a different approach. The moment generating function of a poisson RV is $e^{\lambda(e^t-1)}$
ah, the diagram got a bit cut out, but the arrow going down has the same thing as the arrow going up to the right, like I said
2:58 PM
and if $X,Y$ are independent random variables, the MGF of their sum is the product of their MGF
the top arrow should probably read $\varphi(a,b)$ but yeah
so to solve this, you might try to rewrite the MGF into a product, find the MGF of a Poisson random variable in one factor of the product, and find the distribution that the other factor of the product corresponds to
Yea but the main problem here is that we have $e^3t$ at the exponent...
Oh wait..
@BigSocks just looks like post-composing $\varphi$ with the projections, no?
but $\varphi$ doesn't get composed with anything? $\varphi(a,b) := (\varphi_1(a,b), \varphi_2(a,b))$, which is again kind of confusing since, yeah, they're the same point, but $\varphi : Z_f(\overline{k}) \to Z_g(\overline{k})$ whereas $\varphi_1, \varphi_2 : Z_f(\overline{k}) \to \overline{k}$
so the LHS is technically something in $Z_g(\overline{k})$ while the right is something in $\overline{k}^2$
as sets they are the same, sure
but idk, the maps are giving me a twist
3:11 PM
@Eminem Good point, there probably is some smarter substitution argument but I just checked the calculation and it seems that for a poisson-distributed RV X, $\Bbb E[e^{t\cdot 3X}] = e^{\lambda(e^{3t}-1)}$
@Thorgott hold on I just made sense of what you said
it makes a better diagram, but still two arrows look the same...
and for $Y$ being the distribution with $\Bbb P(Y = 3) = 1$, $\Bbb E[e^{tY}] = e^{3t}$
To make this more aesthetic, you can remark that $Y$ is the same as $3Z$ for some $Z$ with $\Bbb P(Y = 1) = 1$
@user2103480 Ahhhh, i got it. $X \sim Pois(4) \Rightarrow M_{3X+3e}(t)=e^{3t}e^{4e^{3t}-4}$
exactly! I assume the $e$ refers to the "random" variable with deterministic outcome 1
ah shit got lost in abstraction again. Yes. Just a constant.
3:20 PM
Oops it should be $X \sim Pois(4) \Rightarrow M_{3X+3}(t)=e^{3t}e^{4e^{3t}-4}$
$3X + 3$, for some $X \sim Pois(4)$
Forget everything about almost surely constant random variables, I just meant a constant
yea i got it, thanks :)
no probs, sorry for making it slightly more convoluted than it should be :P
In the literal sense lmao, if I would have followed my stupid ways I'd have calculated the distribution of $X + Y$ with a convolution
Constants? Sorry we only have delta-distributions here
$Z_g(\overline{k})$ is a subset of $\overline{k}^2$
project onto each factor $\overline{k}$ from the latter
that's precisely what $\varphi=(\varphi_1,\varphi_2)$ says
it's the universal property of the product, but we restricted to a subset first
3:36 PM
@user2103480 Oh by the way I forgot to respond to your comments about Burial. I haven't actually heard any proper album by him, just isolated tracks, and I am also not familiar with this genre very much, so I have no real opinions. Interesting to know yours though
@BalarkaSen if you like, I can tell you a few breakbeat-ish tracks that I like more than the burial stuff
Absolutely! Actually I am a fan of Venetian Snares
Rossz Csillag Alatt Született one of the best breakcore albums of all time
hmm but where does $Z_g(\overline{k})$ go...
@Thorgott yeah I drew the universal property, that was good to clear out
I know this is really dumb, but I just wanna get it right :/
$\varphi$ goes into $Z_g(\overline{k})$ which goes into $\overline{k}^2$ by inclusion
then your diagram is accurate
yeah I was just noticing I put $\varphi$ in the wrong spot...
but then what is that unique map from the universal property
$(\varphi_1 ,\varphi_2)$
3:49 PM
@BalarkaSen whats with the hungarian hahahaha google says he's canadian
I'll listen to that album!
hmm or is it $i \circ \varphi$
$i$ is inclusion
I quite like "You Are Going to Love Me and Scream" by Against All Logic (an alias of Nicolas Jaar)
"Pylon" by Lakker is also great, though on a less groovy, darker side than burial
Thanks, I'll listen!
@user2103480 It's because it's a concept album where he reimagines himself as a pigeon in Budapest's Royal Palace and eventually gets electrocuted
3:53 PM
@Thorgott so like this ibb.co/JK4rQH7
fixed link^
@Thorgott thanks! I'm gunna try a couple more
Yes, precisely
nice, feelin good
so it is really 2 squares pasted
@BalarkaSen lmao everybody who's been there reimagines a life inside that
4:23 PM
@user2103480 Hey, got another question. If i have two MGF $M_X(t),M_Y(t)$, how can i use them to calculate $Cov(X,Y)$?
I can calculate $E[X]\cdot E[Y]$ but what about $E[XY]$?
4:37 PM
Hello guys..
5:22 PM
if $(r,\theta)$ are polar coordinates on the plane, how am I supposed to make sense of $dr^2$ at the origin?
5:37 PM
@Eminem I think you can't say much per se
X and Y could be completely dependent of each other, something inbetween, or independent
What does MGF mean?
Moment generating function
modularly generated field
Metal Gear Folid
hi Ted
The only thing I remember is that $M_{XY} = M_X M_Y$ if and only if $X,Y$ are independent
5:44 PM
metal gear foliation
@user2103480 Isn't it $M_{X+Y}$ ?
Yeh fucc I remembered too
The right answer is $\varphi_{XY}(t) = \Bbb E[\varphi_X({tY})]$
I shouldnt do math with a terrible headache
Is it possible to do maths without a terrible headache?
6:03 PM
It is easier if the headache isn't there when you start
@Astyx I would say that doing math(s) with a terrible headache is nigh impossible.
Hello, @Astyx @user2103480.
6:21 PM
@TedShifrin someone explained something to me about the argument principle but I didn't quite understand a part. They basically reduced $\int_\gamma f'/f dz$ down to $\int \frac{d}{dz}\log |f|\,dz + i\int d(\arg f)$.
ick, hold on as I fix this
They then claimed that the first integral is zero.
But I don't follow why exactly this is. How should this first part be interpreted?
Yes, assuming of course that $f$ is nowhere zero.
The first is just the fundamental theorem of calculus.
So you understand that you're looking at the winding number of the image $f(\gamma)$, so you understand it geometrically.
@TedShifrin hi!
$\int_\gamma \frac{d}{dz}\log|f(z)|\,dz = 0$ by FTC?
@TedShifrin it's making my weekend learning session impossible, I can't even choose on what to eat since I have no appetite at all. Developed a really diffuse set of symptoms in the last few days, although I haven't seen many people in the new year. Covid test result comes tomorrow
Because it's a loop
6:27 PM
Of course, @anakhro.
Wow, I completely missed this.
@TedShifrin thanks for being there to show me all the easy stuff.
Sure, of course.
Yes, we're doing closed curves if we want to enclose points :P
Okay, so now I understand the reason they call it the argument principle.
Yes. It all is the multivalued nature of log, of course.
That's pretty neat. I am wondering why it wasn't expressed this way in my complex analysis course so many years ago.
6:30 PM
testing dropbox: can people see this image?
@robjohn Of course.
Great... I will edit the post. the image is too big for imgur
@TedShifrin Precisely
@anakhro: I can't imagine not discussing the winding number of $f(\gamma)$ around the origin.
BTW, you should for practice calculate $f^*(\frac{dz}z)$ :P
6:35 PM
Will do! What have you been up to, @TedShifrin?
@robjohn Very cool gif
Nothing too exciting, although I did luck out and get vaccine #1 earlier this week.
@Astyx It is a higher resolution animation for a post on physics.se
Ted is near invincible now.
@TedShifrin you get the second next week?
6:37 PM
No, no, at least 28 days.
Hardly invincible.
I thought it was 7-10 days
No, not for any of 'em. Pfizer is 21. Moderna is 28.
The people I know (medical workers) got their second Pfizer 7-10 days later. hmm
Well, what do I know.
I'm not in line to get a vaccine any time soon. Not a medical worker, not in the right age range.
Right now, around here at least, it is 65+
6:41 PM
Well, I was fortunate in several respects. Got in with UCSD because I have a UCSD health account, and on Monday (MLK Day) they were undersubscribed at Petco Park with the duly appointed medical personnel, so over 65's were allowed in with the UCSD connection.
My double heart surgery and cancer never factored in.
@TedShifrin You have a double heart; I should have known ;-)
Or negative.
@robjohn Do you have any idea what this person is blathering about?
Given $X \sim Poi(a), Y \sim Poi(b)$, calculate $Cov(X,Y)$. How would I approach such problem?
I would have to calculate $E[XY]$, but im not sure how...
@TedShifrin I think they are trying to handle this the way you can with a simple pole. However, without a closed contour, the integral usually blows up for order $\gt1$. with a closed contour, the integral is $2\pi i\frac{f^{(k-1)}(z_0)}{(k-1)!}$
I know all that. I did say the first part of what you said. But he seems interested in area inside a closed curve. I think he's totally confused.
What you said is only true if the all the rest of the derivatives vanish (so that you have a simple pole of the function $\phi(z)/(z-z_0)^k$).
Check out the putdown from the lofty high school student, @robjohn. I am inclined to flag his wit.
@RyanUnger I thought you'd already studied a good deal of PDE?
6:58 PM
He is a filthy Riemannian geometer/physicist.
Now you're sounding as obnoxious as the high school engineering stud who just wittily insulted me.
Gotta love it when a high school kid who writes garbage gets on his high horse.
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