Mathematics - 2021-01-22 (page 1 of 2)

schn

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 $f(x)=\sum_{k=0}^m \frac{f^{(k)}(0)}{k!} x^k + \frac{O(x^{m})}{x^{m}}$?

Balarka Sen

2:42 AM

@Thorgott Of course there are many examples, right? Something like the Gabriel's horn

Anonymus 25- Reinstate Monica

Newbie here

Balarka Sen

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

Ted Shifrin

@BalarkaSen That seems wrong.

Oh, I do the horn from 1 to $\infty$.

It shouldn't matter.

Balarka Sen

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

Ted Shifrin

Yeah, I guess the profile curve has infinite length either way.

Balarka Sen

3:03 AM

Meh, this isn't a good idea

Ted Shifrin

I am skeptical!

user 85795

:D

Balarka Sen

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?

Ted Shifrin

Huh?

Just a mess.

Balarka Sen

I don't understand the question

In this language it's just choosing a function whose integral diverges

Semiclassical

3:12 AM

"Delete it then, and you'll end up limiting yourself and your calculus abilities to solve great mysteries of the universe"

4

Ted Shifrin

Oh, what you have is right with a $2\pi$. You need $|f’|<1$, I guess.

Balarka Sen

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...

123

Hi All..

user 85795

hi

Việt Nguyễn

hello, does anyone have a brilliant geometric math quiz? thnks

Ted Shifrin

3:29 AM

Balarka, you need arclength param, and the integral blowing up finitely.

User 873110

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.

Balarka Sen

Of course not. $\varphi$ has nothing to do with the Riemannian metric on $\Sigma$, why would it preserve the property of being shortest?

User 873110

ok

Karl Kroningfeld

"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."

Subhasis Biswas

6:48 AM

@BalarkaSen How are you doing? Do you, by however miniscule chances remember me?

copper.hat

7:53 AM

the world seems so quiet without all those tweets...

Balarka Sen

8:08 AM

@SubhasisBiswas Oh yeah I remember you. Hi

Subhasis Biswas

You've completed your BMath?

I'm in Belur now. Doing MSc.

Balarka Sen

I'm in my last year.

That's great!

Belur is a nice place

Head of the Department now is Adhikari, right?

Subhasis Biswas

Yessir!

Balarka Sen

I know him well.

Subhasis Biswas

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.

Balarka Sen

8:14 AM

Yeah he's really the one who modelled the math department and syllabi.

Subhasis Biswas

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?

Balarka Sen

No, not really.

Subhasis Biswas

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.

Balarka Sen

Thanks. It's great you're in Belur now; all the best!

Subhasis Biswas

<3

user586228

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.

Astyx

12:14 PM

What's your question ?

Edward Evans

Mornin'

Astyx

heya

No more exams for me

Edward Evans

how's it going?

Astyx

I can cry in peace

Edward Evans

jawohl

hahaha

Astyx

12:17 PM

wby?

Edward Evans

I think my interview went well yesterday

I think

Astyx

Good!

Alessandro Codenotti

Which interview?

Edward Evans

I had an interview for a position as a trainee software developer

Alessandro Codenotti

Ah nice

Edward Evans

12:26 PM

:D

Alessandro Codenotti

Are you looking for a job to work after your masters? Or concurrently with it?

Edward Evans

I'm taking an indefinite hiatus until I'm mentally stable enough to continue

lol

Alessandro Codenotti

I see, prioritizing health is a good plan

Edward Evans

Truth

Balarka Sen

12:44 PM

@EdwardEvans Nice!

Edward Evans

lol

Of course, it remains to be seen if it actually went well

Balarka Sen

Fingers crossed for you

Edward Evans

Please do

hahaha

LHC2012

1:21 PM

Is anyone here familiar with steffensen's method in numerical analysis?

Peter

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.

user2103480

Challenge: make me understand these goddamn n-gons for genus g surfaces

John K. N.

Morning everybody.

user2103480

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?

Edward Evans

Genus g represented the United Kingdom at the Eurovision Song Contest in Oslo 1996

user2103480

1:32 PM

no that was Tina T

Edward Evans

weyyy

John K. N.

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:

2

Q: cube and one example

if 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?

monoidaltransform

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?

user2103480

@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

Thorgott

@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

Balarka Sen

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.

monoidaltransform

@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 $(x^{k+1},......,x^n)$?

Thorgott

where do you get such an embedding from?

Balarka Sen

@monoidaltransform this is only locally true

Thorgott

locally it exists, of course, but globally?

monoidaltransform

oh right, yeah

Thorgott

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

John K. N.

@user2103480 Thanks.

monoidaltransform

what would be an example of a 2 dimensional embedded sub manifold that has trivial normal bundle in $\mathbb{R}^3$?

Thorgott

S^2

monoidaltransform

and an example of one that doesn't?

Thorgott

for codimension 1 submanifolds, trivial normal bundle is equivalent to orientability

Balarka Sen

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

monoidaltransform

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?

Balarka Sen

closed submanifolds just to be clear

but yes

monoidaltransform

sorry, which part are you trying to be clear about?

Balarka Sen

the part where you say "submanifold"

you should say an adjective, that being "closed"

Thorgott

otherwise, Möbius strip

Balarka Sen

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

Thorgott

huh? the map needn't be proper

monoidaltransform

@Thorgott every codimension 1 sub manifold is orientable in $\mathbb{R}^3$

Thorgott

false

monoidaltransform

right?

Thorgott

you're forgetting Balarka's adjective

Balarka Sen

2:04 PM

@Thorgott this is irrelevant

user 85795

a

Thorgott

the preimage needn't be closed man

Balarka Sen

i dont understand your comment

Thorgott

you're implying closedness is automatic, which it isn't, so I'm confused

Balarka Sen

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

Thorgott

2:13 PM

Why would that imply the embedding is proper?

Balarka Sen

Why don't you tell me if it doesn't?

Seems like a good exercise to think about for you

Thorgott

oh, I misunderstood, of course the inclusion of a closed (in the top sense) subset of R^n is gonna be proper

Balarka Sen

:P

Thorgott

the issue's that the Möbius strip isn't even closed (in the top sense)

so that's a counter-example for trivial reasons

Balarka Sen

It is a counterexample to what?

Thorgott

2:18 PM

to the claim that every embedded submanifold is a level set of some submersion

Balarka Sen

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.

Thorgott

well yeah, but I thought you meant closed in the manifold sense initially

Balarka Sen

I did.

Thorgott

but that wouldn't be automatic

Balarka Sen

Think about all of this again. You're so confused.

Thorgott

2:21 PM

the preimage of 0 under a submersion isn't always gonna be compact

Balarka Sen

When did I claim this?

Thorgott

never explicitly, but if you aren't claiming that, why would we assume M is closed

Balarka Sen

Because there's no non-closed example of a codimension 1 submanifold which is not a level set.

Ryan Unger

if only that were true

Thorgott

hmmmmm

Balarka Sen

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)

Thorgott

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?

Balarka Sen

Yes.

Ryan Unger

yeah, euclidean space

but the only interesting manifold is the torus

Balarka Sen

@RyanUnger this actually

we dont even understand the torus

Ryan Unger

do you know the overtorical band inequality

Balarka Sen

2:38 PM

i was just thinking about the systolic inequality

Ryan Unger

have you read guth's proof

Balarka Sen

nope

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

Ryan Unger

I'm trying to learn pde now

Balarka Sen

lol

user2103480

@BalarkaSen substitute "math" with "life"

Ryan Unger

2:45 PM

this is no laughing matter

user2103480

(not that we're not already doing that)

Ryan Unger

speaking of pde, do you know the equation of fire

Balarka Sen

whats that

Ryan Unger

Kuramoto–Sivashinsky equation

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...

Eminem

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?

Ryan Unger

2:46 PM

someone told me it's linearly unstable but nonlinearly stable

Balarka Sen

lmao thats so garbage

nbro

If someone is interested in reinforcement learning, here you have a recurrent question.

4

Q: What is the difference between Q-learning, Deep Q-learning and Deep Q-network?

Q-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...

reinforcement-learning q-learning dqn deep-rl function-approximation

Balarka Sen

Gregory Sivashinsky (also known as Grisha) is a fellow of the Combustion Institute @RyanUnger

Ryan Unger

sounds like a place where you can dispose of a body

Balarka Sen

loool

user2103480

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

Eminem

@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...

user2103480

That is generally a problem we let computers do, in any case

BigSocks

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:
https://ibb.co/jTZbprP

user2103480

Here this might actually be possible, but with a different approach. The moment generating function of a poisson RV is $e^{\lambda(e^t-1)}$

BigSocks

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

user2103480

2:58 PM

and if $X,Y$ are independent random variables, the MGF of their sum is the product of their MGF

BigSocks

the top arrow should probably read $\varphi(a,b)$ but yeah

user2103480

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

Eminem

Yea but the main problem here is that we have $e^3t$ at the exponent...

Oh wait..

Thorgott

@BigSocks just looks like post-composing $\varphi$ with the projections, no?

BigSocks

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

user2103480

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, $\Bbb E[e^{t\cdot 3X}] = e^{\lambda(e^{3t}-1)}$

BigSocks

@Thorgott hold on I just made sense of what you said

it makes a better diagram, but still two arrows look the same...

user2103480

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 $\Bbb P(Y = 1) = 1$

BigSocks

ok here ibb.co/549FHFj

Eminem

@user2103480 Ahhhh, i got it. $X \sim Pois(4) \Rightarrow M_{3X+3e}(t)=e^{3t}e^{4e^{3t}-4}$

user2103480

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.

Eminem

3:20 PM

Oops it should be $X \sim Pois(4) \Rightarrow M_{3X+3}(t)=e^{3t}e^{4e^{3t}-4}$

user2103480

$3X + 3$, for some $X \sim Pois(4)$

Forget everything about almost surely constant random variables, I just meant a constant

Eminem

yea i got it, thanks :)

user2103480

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

Thorgott

$Z_g(\overline{k})$ 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

Balarka Sen

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

user2103480

@BalarkaSen if you like, I can tell you a few breakbeat-ish tracks that I like more than the burial stuff

Balarka Sen

Absolutely! Actually I am a fan of Venetian Snares

Rossz Csillag Alatt Született one of the best breakcore albums of all time

BigSocks

ibb.co/0Y2NwJ1

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 :/

Thorgott

$\varphi$ goes into $Z_g(\overline{k})$ which goes into $\overline{k}^2$ by inclusion

then your diagram is accurate

BigSocks

yeah I was just noticing I put $\varphi$ in the wrong spot...

but then what is that unique map from the universal property

$(\varphi_1 ,\varphi_2)$

?

user2103480

3:49 PM

@BalarkaSen whats with the hungarian hahahaha google says he's canadian

I'll listen to that album!

BigSocks

hmm or is it $i \circ \varphi$

$i$ is inclusion

user2103480

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

Balarka Sen

Thanks, I'll listen!

Thorgott

Ye

Balarka Sen

@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

BigSocks

3:53 PM

@Thorgott so like this ibb.co/JK4rQH7

fixed link^

@Thorgott thanks! I'm gunna try a couple more

Thorgott

Yes, precisely

BigSocks

nice, feelin good

so it is really 2 squares pasted

user2103480

@BalarkaSen lmao everybody who's been there reimagines a life inside that

Eminem

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 $E[XY]$?

123

4:37 PM

Hello guys..

EM4

Hello!

Thorgott

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?

user2103480

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

Astyx

What does MGF mean?

user2103480

Moment generating function

Thorgott

modularly generated field

Astyx

Metal Gear Folid

hi Ted

user2103480

The only thing I remember is that $M_{XY} = M_X M_Y$ if and only if $X,Y$ are independent

Thorgott

5:44 PM

metal gear foliation

Astyx

@user2103480 Isn't it $M_{X+Y}$ ?

user2103480

Yeh fucc I remembered too

The right answer is $\varphi_{XY}(t) = \Bbb E[\varphi_X({tY})]$

Argh

I shouldnt do math with a terrible headache

Astyx

Is it possible to do maths without a terrible headache?

user2103480

6:03 PM

It is easier if the headache isn't there when you start

Ted Shifrin

@Astyx I would say that doing math(s) with a terrible headache is nigh impossible.

Hello, @Astyx @user2103480.

anakhro

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?

Ted Shifrin

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.

user2103480

@TedShifrin hi!

anakhro

$\int_\gamma \frac{d}{dz}\log|f(z)|\,dz = 0$ by FTC?

user2103480

@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

anakhro

OH

Because it's a loop

Ted Shifrin

6:27 PM

Of course, @anakhro.

anakhro

Wow, I completely missed this.

@TedShifrin thanks for being there to show me all the easy stuff.

Ted Shifrin

Sure, of course.

Yes, we're doing closed curves if we want to enclose points :P

anakhro

Okay, so now I understand the reason they call it the argument principle.

Ted Shifrin

Yes. It all is the multivalued nature of log, of course.

anakhro

That's pretty neat. I am wondering why it wasn't expressed this way in my complex analysis course so many years ago.

robjohn

6:30 PM

testing dropbox: can people see this image?

anakhro

Yes.

Ted Shifrin

@robjohn Of course.

robjohn

Great... I will edit the post. the image is too big for imgur

Astyx

@TedShifrin Precisely

Ted Shifrin

@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

anakhro

6:35 PM

Will do! What have you been up to, @TedShifrin?

Astyx

@robjohn Very cool gif

Ted Shifrin

Nothing too exciting, although I did luck out and get vaccine #1 earlier this week.

robjohn

@Astyx It is a higher resolution animation for a post on physics.se

anakhro

Ted is near invincible now.

robjohn

@TedShifrin you get the second next week?

Ted Shifrin

6:37 PM

No, no, at least 28 days.

Hardly invincible.

robjohn

I thought it was 7-10 days

Ted Shifrin

No, not for any of 'em. Pfizer is 21. Moderna is 28.

robjohn

The people I know (medical workers) got their second Pfizer 7-10 days later. hmm

Ted Shifrin

Well, what do I know.

robjohn

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+

Ted Shifrin

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.

robjohn

@TedShifrin You have a double heart; I should have known ;-)

Ted Shifrin

Or negative.

@robjohn Do you have any idea what this person is blathering about?

Eminem

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...

robjohn

@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 $2\pi i\frac{f^{(k-1)}(z_0)}{(k-1)!}$

Ted Shifrin

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?

anakhro

6:58 PM

He is a filthy Riemannian geometer/physicist.

Ted Shifrin

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.

Semiclassical

"(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.)"

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$Mathematics$ Mathematics