7:00 PM
@Blue don't know
Is it just me or it happens to you too: I have posted questions or answers here in the past, but now that I read them I have almost no idea what I was talking about!!

@0celo7 Well, they're "regular" compared to higher vampires, but belong to the more powerful of the lesser vampires

@ACuriousMind so they're not higher vampires?
I guess I'm confused by what exactly vampires are in the witcher universe
are they not people who turn into bat-like things?
Geralt seemed surprised that the bruxa could talk, but not surprised that the innkeep's daughter was a bruxa
there seems to be a disconnect there

@0celo7 No, they are simply intelligent lesser vampires. I think the games' lore is a bit inconsistent on that point, though
For instance, bruxae should not be able to change shape based on the books

All that have been translated
I think there's still one or two that only exist in Polish

7:13 PM
@0celo7 You seem to take these too seriously...
it seems to be an action game. just go kill everybody, blow up everything and finish the game

That's simply wrong
@ACuriousMind Should I start with the short stories or go direct to the novels?
Or play TW2 first?
Or finish TW3 first :P

@Mostafa That...would be entirely missing the point of the game - the actiony parts are decent, but not extraordinary

@0celo7 Are they chronologically arranged? I mean, if you start TW3, wouldn't you miss out on the storyline?

@Sid too late now
I'm mostly done with TW3
just have one DLC and some side quests left
there are lots of parts of TW3 that reference earlier stuff and don't quite make sense

@0celo7 To get more background on what's happening in TW3? Probably TW2 has the highest density of relevant information. To learn how Geralt and Yen met, read the short stories, in particular The Last Wish. It's been a while since I've read the novels, but at least one of them should have more on Ciri
TW2 is also a very good game, just much shorter than TW3. Though one should play it twice, since the entire second act differs based on one decision :P

7:18 PM
@ACuriousMind Does this sentence make sense to you? "Let us denote $\theta$ to be normal angle for a convex closed curve $\gamma$ (sic)"

The..."normal angle"?
No, that doesn't make sense to me

Sigh, is Springer the only publishing house that actually has editors?
Anything written by foreigners that's not Springer is just terrible.

@0celo7 why hire editors? it's not like your library is going to cancel its subscription

@EmilioPisanty I guess it's funny because it's true? :P

@ACuriousMind oh, it very much is

7:21 PM
It's supposed to be a parameter for the curve.

I don't even think it's funny

It can't be the usual polar angle because that's not necessarily a parameter...or is it for a convex curve?

@EmilioPisanty Why did I snort when I read that, then?

If there's any steepest-descent aficionados around, I'm to decide whether what I'm thinking in a non-standard case makes any sense.

@0celo7 foreigners?

7:22 PM
@ACuriousMind because it's true? =P

But it must be such that $\gamma_\theta^1\cos\theta+\gamma_\theta^2\sin\theta=0$

Remember this? scottaaronson.com/writings/journal.html It's ten years old. How have things changed since then?
@Semiclassical ::raises hand::

@Mostafa Non-native speakers.

mmkay
So, for $p>0$ the following integral makes sense:

"I'm 21 but you know PDE my kinfolk" - 21 Savage
o.O
What does that even mean

7:24 PM
$\int_{-\infty}^\infty e^{\beta (px-xe^x)}\,dx$ where $\beta$ is large and positive

@Semiclassical ewwwwww

@EmilioPisanty I saw your paper in Nature (in your profile). Did you prepare the figure yourself?

@Mostafa paper in Nature?
I wish
you presumably mean my News & Views in Nature Photonics

Actually, lemme check if I should have a minus sign in the exponent of that (I always forget the signs)

7:25 PM
@Semiclassical wait, lemme hit zoom
$$\int_{-\infty}^\infty e^{\beta (px-xe^x)}\,dx$$

@EmilioPisanty News and Views
@EmilioPisanty Yeah

Okay, sign is fine.

Yeah, I made the figure myself
@Semiclassical is that... even...

@ACuriousMind The only thing i can think of is that it's the angle of the normal vector

The angle of the normal vector with what?

7:27 PM
As $x\to\infty$ the exponent is dominated by $-\beta xe^x$ so the integrand goes to zero rapidly

@ACuriousMind velocity

Wouldn't that be, like, 90° always?

oh, polar angle
@ACuriousMind :P

@Semiclassical is $p$ always positive?

As $x\to -\infty$, it instead behaves like $\beta px$ and the integrand again goes to zero.

7:27 PM
You pick an $x-y$ system, then measure the angle of the normal vector

for that integral to be well-defined, yes.

it's gonna be orthogonal to $\gamma_\theta$ by design
I don't know how to show that's a valid parametrization
I haven't done curve geometry in years

However, I need some not-entirely-absurd continuation to $p<0$.

@Semiclassical power series, clearly

7:29 PM
Obviously the moment I do that, though, the behavior at negative $x$ is going to make that integral baaadly divergent

@Semiclassical indeed it will

@EmilioPisanty Could you please point me to some MMA stuff to get me started on solving a system of singular, nonlinear ODE?

@EmilioPisanty Mathematica? I thought it's created using Inkscape

@Mostafa nope

My thinking was to recast the original integral as some kind of contour integral, picking a contour that continues to makes sense for $p<0$

7:31 PM
@Semiclassical have you tried to approximate it, see if it even converges for $p<0$?

@Semiclassical yeah, that could well work

@0celo7 As I said above, it doesn't converge as written for $p<0$.

@Semiclassical Missed that

mmkay
And I only care about the leading-order behavior, so a steepest-descent approximation is all I should need.
(the critical points can be determined in terms of the Lambert-W function, so numerically that's not a big deal)
Mostly I was trying to remember if the logic makes sense; it's been a while since I had to do steepest decent stuff beyond the trivial case ("everything with a narrow peak is Gaussian!")

... but on further examination looks unlikely

7:36 PM
in that plot, sure. but the bounds on im(x) aren't large enough (go out to pm 7)

or maybe you can deform your contour down to the imaginary axis instead of the negative real axis?
(duh)

e.g. this:
I'm trying to understand how they work. Specifically, how can they have unidirectional edge modes while the whole structure seems to be reciprocal (no external magnetic field, etc.)

@Mostafa very very very little

yeah.
It's not exactly nice :/

if you want to be able to handle $p<0$, your only hope is to relate it to a contour that comes in from $+\infty$ and then goes down through $\pm 2\pi i +\infty$
or actually
yeah, I know how to do it
come in from $+\infty$, then loop down and take the first valley on your left and go back to $-2\pi i +\infty$

7:43 PM
Yeah.
Though if all I care about is a steepest descent approximation, I may not need to worry about the fine details so much.
All I have to ensure is that there's some contour that makes sense and can be deformed to pass through the critical points.

i.e. come back in from $-2\pi i +\infty$, loop to your left, into the next valley, and then out to $-2\times 2\pi i +\infty$
and then do that again and again

Right.

lol

And each trajectory can be deformed to pass through a critical point.

the worst contour

7:44 PM
and then close off with a big loop from $-\infty$ to $-i\infty$

lol, yes.

so you've traded your big ugly negative-reals part for a sequence of valley loops

Well. I can either forget about it or it destroys the approach entirely :P

I'm no physicist, but how does that make the integral more convergent?

Well, those trajectories can be easily deformed to pass through the critical points.

7:46 PM
@0celo7 it doesn't change the convergence for $\mathrm{Re}(p)>0$
@Semiclassical no, it doesn't destroy the approach
you want an analytical continuation for $f(p,\beta) = \int_{C} e^{\beta(px-xe^x)}\mathrm dx$

Right.

what you've found is another contour $C'$ such that $f(p,\beta) = \int_{C'} e^{\beta(px-xe^x)}\mathrm dx$ for all $p>0$

@BalarkaSen Do you remember anything about the curvature of curves?

so you can forget about $C$ entirely and just work with $C'$

Yeah. And on that contour one can meaningfully continue to re(p)<0

7:49 PM
@Semiclassical exactly
so look what happens on the imaginary axis

Yeah, I'm a bit paranoid about that

Hrm.

and here it is with a negative real part

7:50 PM
@Semiclassical why? it's about the best you could hope for

yeah. doesn't mean it's fun to look at

dunno why you're complaining, it's about the best that it gets

that's fair.

don't focus on the big ugly spikes
you don't need to go there
focus on the nice deep valleys, where you can go

right. main thing to focus on is the saddle points in that graph

7:52 PM
yeah, exactly
so you get this nice series of contours
each of which passes a single saddle

Right.

and all of your saddles are nice and sharp, so you get good approximations there
and the second saddle is way below the first one
so it gets exponentially suppressed

Well, one has to be a bit careful about that last statement.

similarly with the third one w.r.t. the second one and so on

If $p$ is sufficiently negative, then all of the critical points are actually complex
The two on the real axis collide and move off axis.
So that complicates the picture a tad. But nothing major.
The main statement I wanted, though, was that if $p$ is negative, but small enough that there are two real critical points, then the contribution from that first saddle point is still the relevant one.
Which does seem to still be the case, so I'm happy with that.

7:56 PM
Christ, this isn't even a quasilinear PDE and they reference a book on quasilinear PDE for it

@Semiclassical negative (real) $p$ is the only place you can't go, though

Hrm.
There's no way to think of it as $im(p)=\epsilon>0$?

If you fix $\mathrm{Re}(p)<0$ and you take $\mathrm{Im}(p)$ to zero, then the first few nontrivial saddles will be bigger than the first one

ah. yeah, that's no good.

so you'll need to go to several valley runs before it starts converging
and as $\mathrm{Im}(p)$ decreases, you'll need to go further and further to get convergence

7:59 PM
that's unfortunate. what I was after really only makes sense if I only need to worry about the first one.

in the limit $\mathrm{Im}(p)\to 0$ you will likely have an asymptotic series

So that's disappointing.

and I really suspect that you'll have a branch cut on the real axis

wouldn't shock me.

mostly because you need to change which way you send your contour depending on the sign of $\mathrm{Im}(p)$

8:00 PM
that's unpleasant but not shocking.
I mean, I'm trying to extract a meaningful analytic continuation from something that behaves terribad when $p<0$.
Though, I do have that $\beta$ out front, and I can have that as large as I want...

well, you get almost all of that half plane

@dmckee thank you very much =)

@Semiclassical and actually, come to think of it, your integral isn't really that pathological, and it's actually, structurally, extremely close to the ones I use

Neat.
Mine is coming from some large deviations theory stuff

(fig. 2.2 in my thesis)

8:03 PM
Though this is more a toy version of the full case (because the full case is awful)
nice.

those valleys are exponential in the same sense as yours

What systems do you study?

we just like to see them as trigonometrics, instead
This is a semiclassical trajectory-based approximation to optical tunnelling, i.e. ionization of atoms by a strong, long-wavelength field

Nice.
The stuff for mine is from the KPZ equation.

good lord, that is a mouthful

8:05 PM
and my attempts to figure out wth the authors were actually doing :P

anyways, if you do figure out whether there's a branch cut or not, drop me a line here ;-)

mmkay

and good luck

thanks

how do I go about thanking the Russian server gods

8:09 PM

@EmilioPisanty that's likely illegal, and I didn't mean literally
I was thinking more of sacrificing someone
old-school

send a telegram message saying thank you
it'll reply saying there's no papers titled "thank you" in the database, but still

Well, I'm doubtful this book actually has what I need

ah, that server, not that server

The PDE is clearly not parabolic, I don't know why I should find info about it in here

8:13 PM
@EmilioPisanty Just to give a sense of what I'm after: For $p>0$, it's enough to look at the original convergent integral along the real line and worry only about the real saddle.

@0celo7 got distracted
what was the question?

@EmilioPisanty I need to numerically solve a system of nonlinear ODE
initial value problem

And that gives an answer which in some sense depends only on $W_0(p)$ (or some trivial variation thereof)

@0celo7 what's wrong with NDSolve?

where that's the principal branch of the Lambert-W function

8:14 PM
@EmilioPisanty I don't know, I was asking for any starting point whatsoever

@0celo7 NDSolve

cool
@EmilioPisanty damn Russian servers, the djvu won't convert

@Semiclassical well, those relationships are likely to hold

probably got a virus

But $W_0(p)$ isn't limited to $p>0$
It extends as an analytic function all the way down to $p=-1/e$.

8:15 PM
but you're likely to need more than one branch of that Lambert W as $p$ approaches the negative real axis

Hmm.

@Semiclassical is it really?
huh

Sure. It's (one of the) the inverse function of $p=we^w$.

(that last comment was a bit of a wild shot, based on the need for multiple saddles in that limit)
so?

And the minimum value of that for real $w$ is $-1/e$, achieved at $w=-1$.
So if one only worries about $p>-1/e,w>-1$ then that inverse function is perfectly well behaved.

8:18 PM

@0celo7 Do you want to convert a djvu to pdf?

@0celo7 Then what was the problem?

The blue part is the plot of $W_0(x)$ for $-1/e<x<0$, the orange part is $W_{-1}(x)$.

8:20 PM
@Mostafa the first djvu I tried didn't work
second one did

@Semiclassical that's for real $w$, though, right?

Yep.

so make it complex

Hmm.

@0celo7 DjVuLibre DjView is open source and can convert to pdf.

8:23 PM
@Mostafa I just had a problem with the .djvu for some reason
I know how this all works

While I'm plotting that for real $p$, here's a fun graph which I wish I understood better

Why are supplementary documents of APS journals not free??? :(

these are now the real parts of $W_0(x),W_{-1}(x),\frac12 W_0(x)+\frac12 W_{-1}(x)$ (blue, orange, green) for $x\in(-2/e,0)$
there we go
Apparently all three functions have the same real part for $x<-1/e$, and the average of the two branches is a smooth continuation of that real part.

Can I use SciHub to get the supplemental material? @0celo7

@Mostafa Why would I know?

8:28 PM
(all three branches are purely real $x\in(-1/e,0)$, and the two branches have opposite imaginary parts for $x<-1/e$. So the green function is purely real.)

I never use SciHub

you were talking about Russian servers and scihub above

I was not talking about SciHub

@Mostafa extremely unlikely

@Mostafa i don't think so.
not 100% sure, of course.

8:32 PM
(9︵9)

or #icanhazpdf
or BPPF@FB

So this equation isn't parabolic, and the book he references does not talk about such equations
@Mostafa I have the pdf

@0celo7 (ʘ‿ʘ)

8:44 PM
@0celo7 trying to figure out how to get it using University's subscription.
courtesy of scihub , I've never used it before as these APS supplemental materials are the only thing not supported by SciHub. (last time that I needed one, used my friend's account)

9:23 PM
Finally got it!!! ʘ̚ل͜ʘ̚
2
(after 39 mins trying)