Mathematics - 2016-04-26 (page 2 of 5)

Evinda

1:04 PM

Hey!!!

How do we get that $\phi( \epsilon)-\phi(-\epsilon)=2 \epsilon \phi'(0)+O(\epsilon^2)$ ?

@EricStucky Hi.

ello

What's up?

I have 60 minutes to eat and be in the classroom :)

ideally not simultaneously

Balarka Sen

1 hour is a lot of time.

Eric Stucky

1:08 PM

yeah

it's weird being out of bed so early

I've been taking longer to get from awake to out of bed recently

Balarka Sen

1 hour = 60 minutes = 3600 seconds! 3600! That's a hell of a lot.

Eric Stucky

so many times

Evinda

Hey @DanielFischer
Do you have an idea?

Eric Stucky

llooks pretty Taylor-y to me evinda

did you try that?

Akiva Weinberger

@EricStucky Why not

Eric Stucky

1:12 PM

que?

Akiva Weinberger

eat in class

(Not 100% serious)

Balarka Sen

why eat at all?

Albas

Coz math requires mental stamina and food gives you the stamina

Akiva Weinberger

6.832, not 3600

An hour is 6.832! seconds

:P

@usukidoll If the domain is $(0,1)$, then those things you mentioned are all bounded. If the domain is $\Bbb R$, then they're unbounded.

Semiclassical

@TobiasKildetoft What I find strange about that question is that $[A,B]=\lambda I$ does show up in physics a lot in a form that amounts to $[\frac{d}{dx},x]=1$. But that's an exception that proves the rule, since these can't be represented by finite-dimensional matrices.

Akiva Weinberger

1:27 PM

(There's a theorem that says that every continuous function with domain $[0,1 ]$ — that is, the closed interval — is bounded. Thus, if you have an unbounded function on $(0,1)$ — the open interval — then it can't have a continuous extension to $[0, 1]$.)

Semiclassical

So the assumption that one has $n$-by-$n$ matrices is crucial. (Not a surprise, given that one can't take the trace of the identity matrix otherwise.)

Albas

I have a question. Given $a,b>2$ $a,b\in \Bbb{N}$ I have to prove that $2^a+1$ is not divisible by $2^b-1$. My idea is to take cases. The first being $b>a$. It becomes very easy because $2^b-1>2^a+1$ and hence $ 2^a+1$ is not divisible by $2^b-1$. But how should I proceed further?

Balarka Sen

Not only bounded but also have a maximum and a minimum. @Akiva.

Akiva Weinberger

(For example, take $\frac1x$, with domain $(0,1)$. It's unbounded, and it can't be continuously extended to $[0, 1]$ because it can't be defined at $0$.) @usukidoll

Evinda

Hey @Semiclassical @TobiasKildetoft
How do we get that $\phi( \epsilon)-\phi(-\epsilon)=2 \epsilon \phi'(0)+O(\epsilon^2)$ ?
Do you have an idea?

Semiclassical

1:32 PM

What's $\phi$?

Evinda

A test function @Semiclassical

Akiva Weinberger

Also, every ultrafilter on it with closed elements is principal @BalarkaSen

(I assume. I'd have to prove it.)

user 1618033

@robjohn hey

@robjohn these days I derived an amazing family of infinite series, and also some very interesting integrals. Thinking to propose some new stuff in more journals.

Balarka Sen

@AkivaWeinberger I believe you.

:)

user 1618033

For instance ...

$$\sum_{n=1}^{\infty}\frac{1}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots -\frac{1}{n^2}\right)\left(\zeta(2017)-1-\frac{1}{2^{2017}}-\cdots -\frac{1}{n^{2017}}\right)$$

Akiva Weinberger

1:39 PM

Wait, no, it's trivially true — no ultrafilter has only closed elements

user 1618033

@robjohn see above ^^^ I doubt anyone ever did such ones before. This is just an easy example I have. I have harder ones.

Akiva Weinberger

But, for every ultrafilter on $[0,1 ]$, there is an element of $[0, 1]$ that's in the boundary of every element of the ultrafilter

Semiclassical

@user1618033 i presume there's nothing particular about $2017$?

user 1618033

@Semiclassical no ... (btw, do you know who I am? chris's sis:-))

LeGrandDODOM

where is your book ?

Semiclassical

1:42 PM

well, considering you're talking about writing your book on series and integrals that haven't been done before---yes, i can probably guess :p

user 1618033

@Semiclassical :D

@LeGrandDODOM It's in front of me, I'm working on it.

@LeGrandDODOM If you guess how many pages it will have you'll receive from me a copy for free (a good approximation, say).

Semiclassical

my instinct would be to find a nice integral representation of $\sum_{k=n+1}^\infty \frac{1}{k^p}$

user 1618033

@Semiclassical wait a second

Semiclassical

and then maybe use that to get a double integral? not sure

user 1618033

OK, back.

Akiva Weinberger

1:48 PM

@AkivaWeinberger *closure, not boundary

user 1618033

@Semiclassical That might be a good start but not sure what you will do then.

Balarka Sen

How do you self-ping??

Semiclassical

not sure, yeah

user 1618033

@LeGrandDODOM maybe it's not fair to ask you to guess the number of pages, this is a very difficult task, trust me, even to make a good approximation.

Balarka Sen

@BalarkaSen ping.

Ah.

Semiclassical

1:50 PM

if nothing else, one has (stealing from Wiki's page on the integral test) the bounds $$\int_N^\infty f(x)\,dx \leq \sum_{n=N}^\infty f(n)\leq f(N)+\int_N^\infty f(x)\,dx$$

user 1618033

@Semiclassical will it help?

Semiclassical

shrug

it might be informative

it may only lead to bounds on the result, though

user 1618033

@Semiclassical Sure, it's good to think of all variants. Then, I have under development a tool for calculating infinite series which has some little connections with what you propose. The idea is pretty sophisticated at this point, but as I said it's under development.

Semiclassical

gotcha

user 1618033

@Semiclassical one can also try this variant $$\sum_{n=1}^{\infty}\frac{H_n}{n}\left(\zeta(2)-1-\frac{1}{2^2}-\cdots -\frac{1}{n^2}\right)\left(\zeta(2017)-1-\frac{1}{2^{2017}}-\cdots -\frac{1}{n^{2017}}\right)$$

Semiclassical

1:57 PM

oof

LeGrandDODOM

@user1618033 300 pages

user 1618033

@LeGrandDODOM Much more.

LeGrandDODOM

less than 800 I'd say

user 1618033

@LeGrandDODOM In the actual form it has almost 700 pages.

LeGrandDODOM

@user1618033 it's all about computing series and integrals in closed form or you also ask for asymptotic estimates, inequalities and other real-analysis stuff ?

user 1618033

2:07 PM

@LeGrandDODOM There are also other interesting problems, not only computing series and integrals in closed form. I have some inequalities added and also limits that involve finding asymptotic estimates.

LeGrandDODOM

ok sounds good

will it be part of the Springer's problem books series ?

robjohn

@user1618033 it looks related to an identity I proved a while ago, but generalized a bit

user 1618033

@robjohn Yes, true. That 1/n in front make it much harder.

Semiclassical

do you have it for all pairs of powers $(p,q)$? (the above being $(p,q)=(2,2017)$)

user 1618033

@Semiclassical For a certain type of combinations only :D (***)

Semiclassical

2:15 PM

hrm.

any common thread for the ones that don't work?

user 1618033

@Semiclassical Yeap. Of course, one can do them even so, but going a different way, and that way might be far far harder.

Semiclassical

interesting.

user 1618033

You don't find such series calculated in any paper (they are crazy difficult).

Semiclassical

well, hard to say that it's not calculated in any paper. (could appear in some random physics paper, for example, and never be used again).

user 1618033

@Semiclassical ask the best mathematicians that are friend of yours, and see what they say. They require special techniques to calculate in closed form. I'm not aware of the existence of any such paper.

It might be a great suprise to me to find one though. Perhaps I would be glad to mention that paper in my book too.

Semiclassical

2:22 PM

eh, there's a lot of random stuff that shows up in, for example, field theory papers

user 1618033

@Semiclassical I know what you mean, sure.

@Semiclassical My point is that these series are insanely difficult. Well, you never know when some comes up with a super brilliant idea. It might be.

Mike Miller

@BalarkaSen What have you learned since yesterday?

Semiclassical

yeah. of course, my statement is entirely speculative---i don't know any reason myself why a physicist would run into such series.

@user1618033 this answer provides about the best starting point i can find for what i'd have in mind (not that it necessarily would work, of course)

Balarka Sen

@MikeMiller Nothing new, sorry to disappoint. I worked through a couple of concrete calculations from Ted's exercises. I feel like I am missing out the big picture still.

user 1618033

@Semiclassical I see. I would love to see a promising start using that result.

Balarka Sen

2:32 PM

I want to straighten it out before moving on to the next section (which is on surface integrals, i think)

iwriteonbananas

Howdy @MikeMiller, howdy @BalarkaSen

Balarka Sen

Hi @iwriteonbananas

iwriteonbananas

learning about forms right now, Balarka?

Balarka Sen

yep

iwriteonbananas

kewl

give me an example of a closed but non-exact form

Mike Miller

2:33 PM

OK. Should I say what I was talking about about why FTC is obvious?

iwriteonbananas

:P

Semiclassical

what seems especially nice there is that $e^{-N x}$ term in the integrand. if i write down such representations for both, that gives $e^{-N(x+y)}$ in the combined integral, which is readily resummed with $\sum_{n=1}^\infty \frac{1}{n}$

Balarka Sen

@iwriteonbananas On $\Bbb R^2 - 0$, $1/\|\vec{x}\|^2(-ydx + xdy)$

iwriteonbananas

yep

Mike Miller

that's more symbols than just writing $x^2+y^2$ :P

Balarka Sen

2:35 PM

yeah but without \displaystyle \frac, fractions like (huge expressions)/(huge expressions) look bad.

@MikeMiller I am not sure if you should. Shouldn't I think about it more?

Mike Miller

I dunno.

It's not deep, really.

Balarka Sen

OK, maybe say it then.

Mike Miller

The derivative is the infinitesimal change at the point; how much the function is going up at that point. If you think of this like a vector field, it's how much the function is moving to the right, infintesimally. Integrating across the interval is adding up all those infinitesimal changes, which is the total change.

(I said it wasn't deep!)

Balarka Sen

I am more interested in the first question, about why being swirly is related to $\partial X^1/\partial x_2 = \partial X^2/\partial X_1$. That I want to spend some time on at some point of time.

Mike Miller

All you need to do is figure out how this exact 'argument' generalizes to two dimensions.

Balarka Sen

2:39 PM

@MikeMiller I agree.

Hmm, ok.

Mike Miller

Which is why in 2D I suggested thinking about curl as being what you get when you draw a small vertical (or whatever) line and seeing how much of the vector field points left, and how much points right.

Then in 3D, then in 4D ...

Balarka Sen

ah, right.

Mike Miller

Do you see why such a generalization would give the above argument again?

Balarka Sen

That's a bit weird, because I was thinking of integrating $\partial X^1/\partial x_2 - \partial X^2/\partial x_1$ as sum of all the "infinitisimal swirls" of the vector field, because yesterday I figured it measures local swirls, not globals.

Mike Miller

Sure, but there's no difference :)

between what you're saying and I'm, I mean

Balarka Sen

2:45 PM

ok, so we're summing up local swirls of (the vector field corresponding to) $\omega$ which would give me $\int_{\partial S} \omega$ (which, I think, measures how much swirly - globally this time - the vector field is on $\partial S$?)

Mike Miller

Try to interpret curl in terms of the things going left and right of a(n infinitesimal) line.

That's what I was suggesting above.

Then your total integral is "The amount of energy leaving the surface".

Balarka Sen

let me try to understand an example. e.g. take $(0, -x^2)$ - that's like a river flowing. so take a small vertical line near the bank $x = 1$. then the curl measures how much of the current escaping through the right of it than the left?

that makes sense, because that's the force which makes things closer to the bank rotate.

Mike Miller

I'm being uncareful about signs (right? left?) but that's iltimatelt where the orientation of your curve and the orientstoon of the plane come in

Balarka Sen

yeah, in my previous case i think the force is the negative of (things flowing the right) - (things flowing the left). yep, we can make the rigorous using the orientation of the curve.

I am afraid it's not intuitively clear to me why the total integral is amount of energy leaving the surface though.

Semiclassical

the usual physics phrase for that 'swirliness' integral is the circulation, btw

though i tend to default to the magnetic field context rather than fluid dynamics

Balarka Sen

2:57 PM

no, i will say swirly to take revenge upon physicists who handwave mathematics. there is no way to handwave physics, so the next best is to abuse physical terms.

Semiclassical

pffff

Balarka Sen

:)

Semiclassical

one of the links from that page has a nice set of animations, btw

Vorticity

In continuum mechanics, the vorticity is a pseudovector field that describes the local spinning motion of a continuum near some point (the tendency of something to rotate ), as would be seen by an observer located at that point and traveling along with the flow. Conceptually, vorticity could be determined by marking the part of continuum in a small neighborhood of the point in question, and watching their relative displacements as they move along the flow. The vorticity vector would be twice the mean angular velocity vector of those particles relative to their center of mass, oriented according...

Mike Miller

@Balarka I'm confused now too. So I guess you'll have to figure it out and explain it to me.

Balarka Sen

I will do that. Thanks for explaining the idea to me.

Mike Miller

3:00 PM

Are our resident Lie theorists here?

Balarka Sen

None of the Lie algebraists are here. Daniel might know Lie groups though.

Oh, there's @anon.

Balarka Sen

3:34 PM

@MikeMiller Yeah, I am pretty sure this should be the picture: I take a bowl full of some liquid, I apply some force at each molecule of the liquid so it all starts to turn and twist. Now take some small enough particle and place it at random on the liquid. It will start to move, but focus on how much it "spins around it's own axis", and the observation should be done on an infintisimally small interval of time.

You could have placed this particle anywhere on the bowl and it would have given you some number ("it's angle of spin around it's axis"). Sum these up. Now forget about the interior of the bowl and focus on it's boundary. the liquid moves there, and if you place some particle there and make it not to enter the interior, it'll go round the boundary of the bowl by some angle. I think Green says these are the same.

I don't know how to make the "escaping of energy" thing work.

The reason this works out is, I guess, because when you sum up all those "infinitesimal angle of rotation of a particle around it's axis", many things cancel out.

Kaustabha Ray

4:06 PM

$\int_{}^{}\int_{R}^{}\frac{sin(x)}{x}dx\,dy$, To solve this integral, with the conditions, Let $R$ be the triangle in the $xy$-plane bounded by the $x$-axis, the line $y = x$,and the line $x = 1$, would the limits of the integral for be from 0 to 1 and for y be from 0 to x?

Akiva Weinberger

So my brother and my dad got their glasses stolen at the beach

Not, like, sunglasses — their actual glasses

My brother in particular is pretty much blind now

Balarka Sen

@KaustabhaRay Yes.

But you need to be more tricky than that to compute that integral.

@AkivaWeinberger Yikes.

Mike Miller

@Balarka The picture to me is much clearer with divergence in 3D. Maybe I'm wrong that it should work as well in 2D. But it seems strange that there should be a visible difference.

Akiva Weinberger

Yeah. You're essentially asking if $R$ is the set of points where $0\le x\le 1$ and $0\le y\le x$.

Since $R$ is the set of point where $0\le y\le x\le1$, it's true.

@KaustabhaRay

(where each of those three inequality signs represents a side of the triangle)

Mike Miller

My bus has more or less not moved today. :(

Balarka Sen

4:10 PM

@MikeMiller I'll have a look at what divergence is from the comment you made a few days back.

Akiva Weinberger

Is divergence $\partial\cdot$ or $\partial\times$? Or am I thinking of something else entirely?

Kaustabha Ray

@BalarkaSen So is there a trick to simplifying it?

Balarka Sen

@KaustabhaRay Yes.

Hint: Change of order of integrals.

Kaustabha Ray

@BalarkaSen Okay I'll try it out

Thanks

Balarka Sen

The limits $0 \leq x \leq 1$ and $0 \leq y \leq x$ are not too useful because $\sin(x)/x$ doesn't have an "elementary antiderivative".

Mike Miller

4:12 PM

@Balarka Just wait on it, it's coming.

Semiclassical

$\text{div }\vec{F}=\nabla\cdot \vec{F}=\sum_k \frac{\partial F_k}{\partial x_k}$

Akiva Weinberger

Oh, I wrote the wrong letter, then

(Is $\partial$ a "letter"?)

Balarka Sen

$\nabla \times X$ is curl.

Akiva Weinberger

(Same question for $\nabla$.)

Semiclassical

that it's $\nabla\cdot$ rather than $\partial\cdot$ is probably more historical contingency than anything else

Akiva Weinberger

4:15 PM

(They're both, like, a delta, but wrong.)

Balarka Sen

@AkivaWeinberger There was a guy in Ted's youtube lectures who asked what $\partial$ is called when partial derivatives were defined. Another guy googled while Ted was trying to remember, and announced that the official name is "curly dee" in the wikipedia article.

Mike Miller

@Semiclassical I think the gradient is a sufficiently different notion that it should be separated from the symbol for the partial derivative

Akiva Weinberger

I usually just call things by their LaTeX names, honestly.

Mike Miller

It's pronounced del

Akiva Weinberger

I thought that was $\nabla$.

Semiclassical

4:17 PM

i usually just call $\partial$ as "partial"

Balarka Sen

Del is $\nabla$.

Akiva Weinberger

^

Balarka Sen

@Semiclassical Me too.

Akiva Weinberger

(Where does the name "nabla" come from?)

Semiclassical

so $\partial f/\partial x$ as "partial-f partial-x"

Balarka Sen

4:18 PM

nah blah.

Semiclassical

i tend to use nabla instead of del. not sure why.

Mike Miller

I have never heard some say del for the gradient operator, because it's just "the gradient".

Semiclassical

the one i find annoying is $\Delta$ for the Laplacian

i've sometimes seen it as "grad f" for $\nabla f$

which i don't mind either

Mike Miller

But you appear to be right which is completely baffling

Semiclassical

it's funny. when writing stuff out, i always use $\nabla$

Akiva Weinberger

4:19 PM

Someday, someone will find a use for expressions such as $\partial\nabla(\Delta\delta)$

Semiclassical

but when actually saying stuff, i rely on div/grad/curl

Balarka Sen

I call it nabla though.

Kaustabha Ray

@BalarkaSen $\int_{0}^{1}\int_{y}^{1}\frac{sin(y)}{y}dy\,dx$ would be the integral with change of order?

Balarka Sen

@MikeMiller Did it come?

Semiclassical

well, $\delta$ has the interpretation of a codifferential

Akiva Weinberger

4:20 PM

$R$ was $0\le y\le x\le 1$, right?

Semiclassical

so it's not entirely absurd.

Akiva Weinberger

I think it's $\int_y^1\int_0^1\dots dy\ dx$ @KaustabhaRay

Balarka Sen

@KaustabhaRay That's not right, how did you get that?

Semiclassical

another one that's annoying in practice is that $\nabla^2$ is occasionally interpreted as acting on vectors component-by-component

tatan

Can a circle be represented as a function?I think it cannot be as it fails the vertical line test...Am I correct?

Akiva Weinberger

4:22 PM

@BalarkaSen I think he just wrote the int signs backwards by accident

@tatan No, the circle is not (the graph of) a function. A semicircle can be, though

Balarka Sen

What you said it ok, but it doesn't help computing the integral either, hence not what I was referring to.

Semiclassical

@tatan as a single function of one variable i.e. $y=y(x)$, you're correct. but you can do it by parametrizing $x,y$ by some $t$

Kaustabha Ray

No I didnt write it backwards, I actually saw this technique once from a YouTube video, can you give some some links to easy resources on change of order computation?

Mike Miller

I'm going to die on this street.

Akiva Weinberger

Wait, you wrote $\sin(y)/y$ instead of $\sin(x)/x$

Balarka Sen

4:23 PM

Ah, that confused me.

tatan

@Semiclassical Can you give an example of the parametrizing?

Mike Miller

I'm glad I'm not supposed to be teaching at 10 or something.

Semiclassical

$(x(t),y(t))=(\cos t,\sin t)$

Akiva Weinberger

It should be be $\int_y^1\int_0^1\frac{\sin x}x dydx\\$

Balarka Sen

@KaustabhaRay Here's how to do it. Our integral was $\int_0^1 \int_y^1 \sin(x)/x dx dy$, yep?

Akiva Weinberger

4:26 PM

(Balarka, was my above comment right?)

Balarka Sen

Because $y$ varies from $0$ to $1$, $x$ varies from $y$ to $1$. So our region is $0 \leq y \leq x \leq 1$.

Akiva Weinberger

(My above comment was wrong, actually. Sorry.)

Balarka Sen

@AkivaWeinberger Your integral is horribly wrong.

The limits are messed up. The first integral integrates to $\sin(x)/x$.

So you get something in terms of $y$. Not what was supposed to happen.

Kaustabha Ray

@BalarkaSen I had said "would the limits of the integral for be from 0 to 1 and for y be from 0 to x?" which you said to be correct , Now you said "Our integral was $\int_0^1 \int_y^1 \sin(x)/x dx dy$, yep?"

Fargle

G'morning, chat.

Balarka Sen

4:30 PM

They are the same thing. $0 \leq x \leq 1$ and $0 \leq y \leq x$ implies $0 \leq y \leq x \leq 1$.

That's our region.

Kaustabha Ray

Ah yes I see after sketching it out

Balarka Sen

Yes, draw something.

Anyway so $\int_0^1 \int_y^1 \sin(x)/x dx dy$ is problematic because the first integral in the iterated integral is not elementary.

So I want to do something tricky with the limits.

So, our region is $0 \leq y \leq x \leq 1$. How about I let $x$ vary from $0$ to $1$ and $y$ vary from $0$ to $x$?

Kaustabha Ray

Yes that makes sense

Balarka Sen

That's the integral $\int_0^1 \int_0^x \sin(x)/x dy dx$, right?

Kaustabha Ray

Yup

Balarka Sen

4:34 PM

And $\int_0^x \sin(x)/x dy$ is... what?

(You should note carefully that the order of the integral just got changed due to setting the independent variable in the limit to be something different. This is the strategy of changing order of integrals)

Kaustabha Ray

Not sure actually, just going through basics

Balarka Sen

Well, $\sin(x)/x$ has no $y$ term in it. So what should $\int \sin(x)/x dy$ be?

I mean, $\sin(x)/x$ is a constant relative to $y$, no?

Kaustabha Ray

oh, so that's a constant

so it would be the constant times y

Balarka Sen

Yes, but you have limits. $\int_0^x \sin(x)/x dy = ?$

Kaustabha Ray

sin(x)/x * x = sin(x)

Balarka Sen

4:39 PM

Yes. So your integral is $\int_0^1 \sin(x) dx$. That should be pretty easy to do.

Kaustabha Ray

Yes thanks

Balarka Sen

No problem. Always draw the regions when you solve things like this.

Mike Miller

morning @Fargle

Balarka Sen

It's worth noting why Akiva's thing here does not make sense, as Akiva pointed out. You need the outer integral to be independent of any other variables. The variable in the outermost $d(\text{variable})$ should be independent of everything else.

Kaustabha Ray

I'm still not dully clear on the change of variable technique...any references?

Balarka Sen

4:42 PM

@KaustabhaRay Maybe look at Shifrin's, "Multivariable Mathematics".

Hubbard & Hubbard is standard reference to multivariable calculus but I haven't read it.

Kaustabha Ray

ok

That integral I knew actually, I got confused with the dy part, I was caught up in the moment and was thinking its sin(x)/x dx

Balarka Sen

@MikeMiller Did the bus move anymore?

Mike Miller

Yes, I'm now halfway there.

Semiclassical

it seems worth mentioning that while there's really nothing nice about the indefinite integral $\int \frac{\sin x}{x}\,dx$, the definite integral over the entire real line is just $\pi$

Kaustabha Ray

@BalarkaSen Is there any trick to solving $\int_{0}^{1}x^n ln(1+x)dx$ types of integrals, or is it always integration by parts?

Balarka Sen

4:50 PM

Taylor series for the integrand maybe. I just do integration by parts.

Not really a very good integrator here. I just know a few things.

Kaustabha Ray

So what are some of the must know "other techniques", like the change of variable one you said, anything else comes in hnady?

Balarka Sen

Polar, spherical and cylindrical coordinates are worth knowing. Change of variables theorem in general.

For multivariable integrations at least.

Differentating under integral sign wrt dummy variable also comes handy time to time.

@MikeMiller Do you usually get too much traffic at LA?

Akiva Weinberger

@BalarkaSen Yeah. Sorry. I haven't done this in a while

Mike Miller

No, this is unnatural. There was construction on one of the bus routes, reducing the street from 3 lanes to 1.

Balarka Sen

No problem, change of order of integrals is confusing.

Mike Miller

4:57 PM

I mean, yes, there's usually traffic. But this was a lot of traffic

Balarka Sen

Ah, I see.

Akiva Weinberger

So, the divergence thing makes a lot more sense for 1D. And it's not so surprising that adding them is the way to go

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$Mathematics$ Mathematics