Mathematics - 2019-12-27 (page 2 of 2)

Simply Beautiful Art

22:00

but it's kind of nasty doing that kind of computation

The definition is already numerically unstable to begin with

Maximilian Janisch

So I guess let’s not do that (?)

Simply Beautiful Art

but I wanna see what it looks like x.x

Shootforthemoon

@MaximilianJanisch ok, thanks!

Maximilian Janisch

@SimplyBeautifulArt You really are a special type of person 😄

well but cant you just increase precision to like 600 digits (apart from the symbolic computations)

Simply Beautiful Art

>:P and what's that supposed to mean

(but I don't wanna)

kek

Hawk

22:10

can any1 try to answer this...math.stackexchange.com/questions/3488809/…?

Maximilian Janisch

@SimplyBeautifulArt you really remind me of the quote „oh boy, that‘s a good sequence“ by Neil Sloane (see here youtu.be/etMJxB-igrc?t=167 )

Simply Beautiful Art

lmao

Maximilian Janisch

@Hawk If I understand correctly you need to prove that all eigenvalues of the $T$ have modulus $1$ ?

Nevermind that is garbage

Hawk

No I just don't understand why we are starting with the eigenvalues of T instead of \sqrt{T*T}.

I don't even feel like we need self-adjointness here

Maximilian Janisch

22:26

Okay why do we have $T^*Tv=\lambda_T\overline{\lambda_T}v$ @Hawk

Hawk

Because $\lambda$ are the eigenvalue of $T$

Maximilian Janisch

But I don't understand how you know a priori that $v$ is an Eigenvector of $\sqrt{T^*T}$ and also $T$

Hawk

So we start with \sort{T^*T} first

it has eigenvalue/vector pair (s,v)

then apply \sqrt{T^*T} to both sides of \sqrt{T^*T}v = sv

so s\sqrt{T^*T}v = s(sv) = s^2 v

Maximilian Janisch

Okay sure

Hawk

LHS is now T^*Tv = s^2v

Maximilian Janisch

22:29

Yes

Hello alternative @TedE

Hawk

Ah i think i see where ppl are confused and the downvote is for

cuz now I claim Tv = lambda v

Maximilian Janisch

Yes

that's why I am confused as well

Hawk

ah okay that's why we need Spectral theorem

Maximilian Janisch

But I am still confused

Nevermind actually I get the idea now

because if $T$ is self-adjoint then $T^*=T$

Leaky Nun

@Thorgott polynomial multiplication with x=10 is basically multiplication

Hawk

22:31

I think my book actually dropped the Self-Adjointness in the later edition

Maximilian Janisch

What book is it @Hawk

Nevermind

I think it is enough that $T$ is normal

Then by the spectral theorem we have the unitary diagonalization $T=U\Lambda U^*$

So $\sqrt{T^*T}=\sqrt{U\Lambda^2 U^*}=U\lvert T\rvert U^*$

Hawk

Its Axler's book

Maximilian Janisch

Actually I am still not sure if my argument works in infinite dimensions

Mike Miller

@AkivaWeinberger ok

Maximilian Janisch

How does the Hilbert space of your $T$ look like @Hawk ?

I think my argument works if it is separable

Thorgott

22:47

Multiplying a polynomial with a constant isn't all that convolve-y anymore

Mike Miller

@MaximilianJanisch you certainly don't have an orthonormal diagonalization of an arbitrary self adjoint operator on a Hilbert space unless you change what that means lol. think of the left-shift operator on ell^2, which has spectrum every complex number with absolute value less than 1.

this is clearly a finite dimensional question, confirmed by the fact that it's from axlers book

Maximilian Janisch

@MikeMiller You are right

there might even be a counter-example in infinite dimensional spaces

thats a keyboard slip :D

Here is what I meant to say: When $X$ is separable and $T:X\to X$ is self-adjoint, linear, then $T$ is unitarily equivalent to a bounded linear operator $M_\phi:L^2(Z)\to L^2(Z), f\mapsto \phi\cdot f$ for some measure space $(Z,\mu)$ and $\phi\in L^\infty(Z)$

So my $\Lambda$ gets replaced by the $M_\phi$

($X$ is a Hilbert space of course)

Leaky Nun

23:07

@Thorgott no, I mean the indeterminate variable x being substituted with 10

I said x=10, not x-10

if you've heard of the "Japanese multiplication method", it's basically that

23 x 12 = 276
corresponds to
(2x+3)(x+2) = 2x^2 + 7x + 6
when you substitute x=10

Thorgott

ah, that's what you meant

yeah, that makes a lot of sense

Maximilian Janisch

@MikeMiller I see where the confusion came from: The singular values are only defined for compact operators

Semiclassical

That works fine so long as none of the new coefficients exceed 10. But then one could do x=100 instead

Maximilian Janisch

So I can use the spectral Theorem in a very similar way to the finite dimensional case

Semiclassical

When it works, tho, it does give an elementary example of how polynomial multiplication / discrete convolution

What I wonder now is the simplest way to introduce convolution of continuous functions

Leaky Nun

23:19

@Semiclassical I mean the idea is the same

you shift and dot product

Semiclassical

I guess you could take a discrete convolution, view it as a Riemann sum, and then get an integral out of that

nbro

I don't think that the following question is a duplicate of the one it was marked a duplicate of, so I suggest you vote to re-open it, even though the questions are related.

2

Q: Does $f(x)\,dx$ denote multiplication of $f(x)$ by $dx$?

In the integral form $\int \! f(x) \, \mathrm{d}x$ does $f(x)\,\mathrm{d}x$ can be seen as a multiplication of $f(x)$ and $\mathrm{d}x$?

Thorgott

The usual motivation of the standard infinite-dimensional inner products is also as "continuization" of the finite-dimensional ones via Riemann sums, so that works out to be the same.

nbro

I think it would be more a duplicate of the following question (but I don't think they are strict duplicates, anyway).

49

Q: Is $dx\,dy$ really a multiplication of $dx$ and $dy$?

On the answers of the question Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? it was told that $\frac{dy}{dx}$ cannot be seen as a quotient, even though it looks like a fraction. My question is: does $dxdy$ in the double integral represent a multiplication of differentials? The problem than ca...

calculus real-analysis analysis nonstandard-analysis

Simply Beautiful Art

derp

Maximilian Janisch

23:21

@nbro Did you already post it in the CRUDE chatroom?

@Simply How is your tetration going

Simply Beautiful Art

my tetration plugs the height into an exponential function, which implies it's periodic in the imaginary directions

nbro

No, I was not aware of that chat and I didn't even check it, of course

Semiclassical

Another way is via integral transform theory , which would seem like overkill if not for how useful the convolution theorem is

nbro

People say that dx/dy is not a quotient, but then this quotient is used in many proofs, etc

Semiclassical

it's not.

Ted E

23:23

@nbro What?

Thorgott

Understanding integral transforms before convolutions seems like a daunting, if not circular, task

Semiclassical

umm, what?

Simply Beautiful Art

Tetration

In mathematics, tetration (or hyper-4) is an operation based on iterated, or repeated, exponentiation. It is the next hyperoperation after exponentiation, but before pentation. The word was coined by Reuben Louis Goodstein from tetra- (four) and iteration. Under the definition as repeated exponentiation, the notation n a {\displaystyle {^{n}a}} means a a...

It's somewhat similar to this

Semiclassical

i knew the definition of the Laplace transform before I knew about convolutions

hi @ted (s)

Simply Beautiful Art

hmmmm

Ted Shifrin

23:25

Hi @Semiclassic

Simply Beautiful Art

but the thing there claims to only work for bases greater than e^(1/e) which is precisely when mine does work

Maximilian Janisch

you mean doesn't work?

Ted Shifrin

Yeah, I learned Laplace transform freshman year in differential equations, and convolutions only in graduate real analysis later, as I recall.

Ted E

Hi @Semiclassical

Fargle

I'm seeing double.

Simply Beautiful Art

23:26

er yeah

Ted Shifrin

@Fargle !!

Simply Beautiful Art

mine works for bases less than e^(1/e)

Semiclassical

@Fargle Four Teds?!?

Ted Shifrin

Yes, there's an impostor alternative Ted now.

Ted E

Yeah, we should really deal with him.

skillpatrol

23:27

seeing double and feeling single?

Ted Shifrin

Where have you been, @Fargle?

Ted E

What is feeling, but a different way of seeing?

Fargle

I will not abide these personal attacks (yes)

Semiclassical

I'm seeing double here, four Krusty's!

►

Fargle

Going through the rigors of a finance master's program.

-_-

Maximilian Janisch

23:28

What is so rigorous there? :D

Fargle

:rimshot:

Semiclassical

rigor mortis

skillpatrol

bean counting

Fargle

Semi has it right. But most of my capital-R research has been on stochastic differential equations.

Semiclassical

nice

Thorgott

23:29

Fair enough then. I haven't even learned the Laplace transform. Convolutions popped up naturally in probability theory.

Semiclassical

laplace transform = characteristic function

Fargle

It's something to do. Pretty far afield from what I tend to enjoy, but, hey, gotta diversify. Especially in this economy

Semiclassical

$F(s)=\int_0^\infty e^{-s x}f(x)\,dx$ @Thorgott

Ted Shifrin

Oh ... just wait for the economy next year!

Well, anyhow, I've missed you, @Fargle.

skullpatrol

^

Fargle

23:30

It's basically just a stepping stone to doing a pure math Ph.D without a 2.*mumble* GPA

Missed you too, Ted. :)

nbro

d/dx can be seen as an operator and dy/dx as the application of such operator to y.

Fargle

Maybe some day I can do real math again.

Ted Shifrin

I have nothing against applied mathematics.

Semiclassical

so the laplace transform of a function $f(x)$ is literally just the characteristic function $E[e^{-s X}]$ with respect to $f(x)$ as the 'pdf'

Thorgott

lol, today I learned...

Maximilian Janisch

23:31

@TedShifrin +1

nbro

However, we often see the manipulation of dy and dx to move from one side of an equation to the other side, saying that dy/dx is not a ratio is at least misleading.

Ted Shifrin

One of my favorite teaching experiences (and popular with a number of the students) was a year-long applied mathematics course. I worked my butt off teaching that course.

Granted, that was 1986-7.

@nbro: You are, after all, the expert.

nbro

No, I am not an expert and, of course, you're being sarcastic (or trying to provoke me), because you know I am not a mathematican.

Thorgott

this makes the convolution theorem for the laplace transform rather obvious then, doesn't it

Semiclassical

tbf, when we write stuff like $y\,dx = x\,dy\implies dx/x=dy/y$, we are essentially dy/dx as though it's a ratio

but that's not really to be taken literally. it's a shorthand for manipulations at the level of the integral. (so long as one isn't talking differential forms, anyways)

Maximilian Janisch

23:33

But I always smash something in my room when I see this notation @Semiclassical

Ted Shifrin

In the context of nonstandard analysis, it's a bona fide ratio. Otherwise, it is not. Learn about chain rule and differential forms.

nbro

@Semiclassical You manipulate it like a ratio and many expect people not to be confused or maybe they expect it but they simply don't care, whoever they are.

Semiclassical

it's definitely an abuse of notation, but it's such a useful one :P

Maximilian Janisch

Fair enough

Ted Shifrin

Differential forms explain perfectly nicely why $df = \dfrac{df}{dx}\,dx$.

Balarka Sen

23:35

It's meaningful if you treat everything as a form

Maximilian Janisch

When substituting in integrals even I use it in some sense

Balarka Sen

Hi @TedShifrin

Ted Shifrin

That's the chain rule, @Maximil.

heya a @Balarka.

Balarka Sen

I slept for 18+ hours

skullpatrol

:O

Leaky Nun

23:35

@BalarkaSen nice

Ted Shifrin

Felicitations. (Put the é if you want it in French.)

Simply Beautiful Art

I prefer $y~\mathrm dx=x~\mathrm dy$ meaning $y\frac{\mathrm dx}{\mathrm dt}=x\frac{\mathrm dy}{\mathrm dt}$ for whatever $t$ is

Ted Shifrin

There need be no $t$.

Leaky Nun

or rather you can always use $x$

Maximilian Janisch

@TedShifrin True, I like the german word Transformationssatz for the generalization to multidimensional integrals

Simply Beautiful Art

23:36

:P

Leaky Nun

maybe not

Balarka Sen

Time to study some math

Ted Shifrin

I prefer understanding that (at the right level) as $\omega = y\,dx - x\,dy = 0$ giving a differential equation.

Balarka Sen

Have to decide what to study

Thorgott

If $X$, $Y$ are independent with pdfs $f,g$, then $X+Y$ has pdf $f\ast g$. So $\mathcal{L}\{f\ast g\}(s)=\mathbb{E}[e^{-s(X+Y)}]=\mathbb{E}[e^{-sX}]\mathbb{E}[e^{-sY}]=(\mathcal{L}\{f\}\cdot\mathcal{L}\{g\})(s)$.

Maximilian Janisch

23:37

Actually I don't even know what to call it in english

Leaky Nun

that is not Lapl... whatever

Semiclassical

one way to avoid the issue: $$\frac{dy}{dx}=\frac{y}{x} \implies \frac{1}{y(x)}\frac{dy}{dx}=\frac{1}{x}\implies \int \frac{1}{x}dx=\int \frac{1}{y(x)}\frac{dy}{dx}\,dx = \int \frac{1}{y}dy$$

Thorgott

I'm going off of what Semi said :p

Ted Shifrin

@Maximil: Change of Variables Theorem.

Maximilian Janisch

@TedShifrin Thanks 🙂

Semiclassical

23:39

So one can avoid using differentials if one finds it distasteful,

nbro

In measure theory (particularly, in the context of the Lebesgue integral), $dx$ in $\int f(x)dx$ is a measure. Now, this clearly explains why you can take an integral with respect to a probability measure. However, some people claim that $dx$ is just notation or means infinitesimally small, but this fails to explain why $dx$ is a measure.

Semiclassical

but meh.

Thorgott

I remember being confused before learning the theorems that there are two rather different theorems which are both called Change of Variables/Transformationssatz

Maximilian Janisch

@nbro If you want you can write $\mathrm d\lambda(x)$ or something like that where $\lambda$ is your Lebesgue measure

nbro

@MaximilianJanisch Are you saying that $dx$ is a shorthand for $d\lambda(x)$?

Maximilian Janisch

23:40

Yes basically

Semiclassical

I was mis-remembering a bit earlier: I should've said the moment-generating function, not the characteristic function

Maximilian Janisch

I mean some people like to write $\lambda(\mathrm dx)$

Leaky Nun

hi @ÍgjøgnumMeg

Maximilian Janisch

and other stuff

Semiclassical

(and there's an extra minus sign in the exponent because of that)

ÍgjøgnumMeg

23:41

Hey @Leaky

nbro

@MaximilianJanisch Yes, I noticed that

ÍgjøgnumMeg

I have exercises to prove "Kronecker-Weber for quadratic fields" rofl

Leaky Nun

@ÍgjøgnumMeg that's a fun exercise

skullpatrol

@TedShifrin what do you expect the economy to be like next year?

nbro

Of course, this Lebesgue integral was developed after the Riemann integral, so I guess that the notation dx does not always mean that is is a measure. It can also be "just notation"

Balarka Sen

23:42

Characteristic function is the Fourier transform, which is $\Bbb E[e^{itX}]$.

ÍgjøgnumMeg

@Leaky yeah? I'll have a go in a bit

Leaky Nun

@ÍgjøgnumMeg that could lead to quadratic reciprocity

ÍgjøgnumMeg

can't be bothered atm rofl

Leaky Nun

but maybe you know it already

ÍgjøgnumMeg

oh yeah

Mike Miller

23:42

@BalarkaSen Learn surgery theory

Leaky Nun

@nbro it's a notation that was given alternative meanings

Ted Shifrin

@skull: Of course I am as far from an expert as one can be. But for various reasons, I believe (and hope fervently) that it'll have a serious downturn for months preceding November.

Balarka Sen

@MikeMiller That might be fun. What's a good reference?

Mike Miller

Wall's book

I think

Balarka Sen

That's huge tho

Mike Miller

23:43

I don't know surgery theory

Ted Shifrin

I was 99% sure he'd say Wall.

Leaky Nun

@BalarkaSen Oxford Handbook of Clinical Surgery, by Greg McLatchie

Ted Shifrin

smacks Leaky

2

skullpatrol

@TedShifrin I tend to agree

Mike Miller

It's the only text I know, lol

nbro

23:44

In any case, you can always say that dx is a measure, for every integral, because the Lebesgue integral is a generalization of Riemann's one (and I believe that the Lebesgue integral is the most general definition of an integral that we have developed, or am I wrong?)

Mike Miller

One should so the second edition, edited by Ranicki

Balarka Sen

Ranicki will have a copy in his site I am sure

Maximilian Janisch

@nbro If $f:\mathbb R\to\mathbb R$ is Riemann integrable on a compact interval then it also Lebesgue integrable on that interval and the both integrals coincide

Ted Shifrin

a @Balarka: Maybe this will be germane.

Balarka Sen

Ok its in Schmuel Weinberger's site

Thorgott

23:45

d-anything is never a measure. You write, say, $\mathrm{d}\mu$ if you integrate with respect to a measure $\mu$. The notation $\mathrm{d}x$ gets an explanation in differential geometry as far as I know, but I'm not qualified to talk about that.

Mike Miller

Actually the suggestion to read Kervaire Milnor and its predecessors is good @BalarkaSen

Thorgott

Also, in Euclidean spaces, the Henstock-Kurzweil integral supercedes the Lebesgue integral

Maximilian Janisch

@Thorgott True

But I like Lebesgue more :D

Ted Shifrin

I went through my entire mathematical career/life and never heard of Henstock-Kurzwell until about a year ago here.

3

nbro

@Thorgott Well, as someone wrote above, dx can be thought of a shorthand for $d\mu(x)$

Balarka Sen

23:47

@TedShifrin I think I first mentioned it to you

Ted Shifrin

I think that's quite likely.

Did you see the MO link on your surgery reference query?

Maximilian Janisch

I said that people write $$\int f(x) \,\mathrm dx$$ sometimes when they mean the Lebesgue integral $$\int f(x)\,\mathrm d\lambda(x)$$

Balarka Sen

Yeah, I did, Mike suggested their Kervaire-Milnor suggestion is better

nbro

Yes, I was referring to your comment then (I am lazy enough not to check it again above)

Leaky Nun

@nbro maybe you shouldn't get lost in abstractions (i.e. notations) and you should understand the intuition behind the concepts and connect them

Thorgott

23:48

Lebesgue integration is the richer theory overall, because the spaces are nicer.

Mike Miller

Wall's book is difficult

Maximilian Janisch

@TedShifrin In my opinion the Henstock-Kurzweil idea is very creative

nbro

@LeakyNun lol

Thorgott

Henstock-Kurzweil has the cool property of being able to integrate any derivative though

Mike Miller

To be honest I forget the various references KM will use (lol, KM) so you'll have to glance through to see what you need to read first

Balarka Sen

23:49

There are so many fucking integrals

The McShane integral

The Pfeffer integral

Mike Miller

damn if theres anything ive ever wanted to do its integrate any derivative!

Balarka Sen

The nobody-gives-a-rats-ass-about integral

nbro

@MaximilianJanisch Are your parents mathematicians or something? How did you manage to start a master's in math at only 16 years old?

Semiclassical

Feynman path integral, which is really another can of worms entirely

Thorgott

Also, dx can be thought of as a shorthand for dmu(x), sure. It's a symbol that even works without any inherent meaning in Riemann integration. I think the proper understanding, which, in the cases you are looking at, will be consistent with the Lebesgue theory is explored in differential geometry, but, again, I can't say anything on that matter.

Isn't McShane equivalent to Lebesgue?

Fargle

23:50

@TedShifrin You may not, but I tire of finance in particular.

Thorgott

Or was that Daniell?

Ted Shifrin

In my mathematical career, the Riemann integral (in the context of integrating on manifolds or real analytic spaces) has been sufficient, although in terms of reading papers and some teaching, of course, I have needed the Lebesgue integral, for distributions and currents, etc.

Mike Miller

predictive policing has its problems too

Ted Shifrin

@Fargle: I've never studied mathematical finance, tbh.

Maximilian Janisch

@nbro It is quite a long story. Basically I started studying mathematics while still in high school so at the end of high school I was almost done with the Bachelor

Fargle

23:51

A lot of it is just stats and stochastic dynamics.

Balarka Sen

@Thorgott Shrug, I only know Henstock-Kurzweil because I read it in Folland's analysis book once

Leaky Nun

ooh, Folland

Semiclassical

Fokker-Planck and all that

Ted Shifrin

I know a number of pure math PhDs who went to work on Wall Street or other avenues of math finance.

Mike Miller

im sure

Thorgott

23:52

@BalarkaSen have you ever looked at the historical definition of the Henstock-Kurzweil integral?

Balarka Sen

@LeakyNun It's a good one right?

No @Thorgott

Fargle

Lot of faffing about with Itou's lemma and so on.

Leaky Nun

@BalarkaSen the harmonic analysis one right

Thorgott

Wikipedia says it was done with transfinite induction over the number of singularities

I'm not sure what to think of that

Balarka Sen

@LeakyNun Oh no the "Advanced Analysis" thingy

nbro

23:52

@MaximilianJanisch Well, I've read in your dedicated Wikipedia page that your father was a mathematian. Anyway, this is very impressive!

Leaky Nun

oh

Balarka Sen

which is basically a compression of everything a graduate student needs to know about analysis

Semiclassical

Ito’s lemma is firmly in the nope-I-dunno category for me

Maximilian Janisch

@nbro Yes, he is still a math professor (now emeritus)

Mike Miller

its cool to try to make sense of all variations of the same notation as if they're all commensurable and appeared in the same historical context imo

Balarka Sen

23:53

I should also eventually read Falconer's little book on fractals

skullpatrol

tip: only read tough books after 18+ hours of sleep :P

Balarka Sen

supposed to be a good entry drug to GMT

Fargle

It's basically just the statement that "you must account for second-order variation in the chain rule when dealing with stochastic stuff".

Semiclassical

Ah

nbro

@MaximilianJanisch Well, I guess you're the only person I've ever encountered to which I don't have to say "good luck", so I won't say it :)

Thorgott

23:54

(It was McShane, which is equivalent to Lebesgue)

Maximilian Janisch

@nbro Haha 😄

Fargle

It falls right out of a Taylor approximation but I don't have the stuff in front of me. It being break and all

Leaky Nun

@BalarkaSen how about read Dummit and Foote

Ted Shifrin

Lazy bum, break and all @Fargle :D

Balarka Sen

I could

Semiclassical

23:55

I’ll note now that my experience with SDEs is to use the WKB approximation to get a PDE

Balarka Sen

I could just go back to Galois theory

Leaky Nun

do it

Ted Shifrin

Balarka needs an algebra-free zone.

Fargle

Most of my experience is numerical, if that makes you feel any better, Semi. I'm not out here trying to get analytic PDFs.

Leaky Nun

no he doesn't

Mike Miller

23:56

@BalarkaSen if you do algebra ill be forced to

Fargle

Just trying to work on better integration methods.

Balarka Sen

I promised I would read a substantial amount of Galois theory before the break gets over so I guess I should do Galois theory

Mike Miller

Lame

Maximilian Janisch

Like which ones? @Fargle

Semiclassical

@Fargle well, so was mine

Thorgott

23:56

good decision

Ted Shifrin

Whom did you promise, a @Balarka?

Balarka Sen

@MikeMiller Let's hop in the algebra wagon

Mike Miller

Sigh

skullpatrol

1 min ago, by Leaky Nun

@BalarkaSen how about read Dummit and Foote

Balarka Sen

@TedShifrin Nobody, that's the trick :P

Ted Shifrin

23:57

Precisely.

Semiclassical

Mostly because the KPZ equation, in the weak noise approximation, gives you a pair of coupled nonlinear pdes

Balarka Sen

OK, I will spend the next 3 hours finishing all of Galois theory

Leaky Nun

nice

I guess I'll be here answering your questions

Ted Shifrin

How nice of Leaky to nominate himself as the expert-on-high.

Semiclassical

Not holding out much hope of solving those analytically, especially when the boundary conditions are as miserable as they were

Thorgott

23:58

I guess I'll be here to lurk and learn

nbro

@MaximilianJanisch Btw, I've been to Luzern once. It's pretty nice there (apart from the train station and the surroundings). However, when I went there, it was a bad situation and moment, so maybe this is a biased impression. Anyway, I will never forget that place

Mike Miller

@BalarkaSen So long as you're done afterwards

Leaky Nun

@TedShifrin I didn't say I'm the expert

Fargle

@MaximilianJanisch Stochastic stuff doesn't generally do well with stuff like implicit methods---take, say, Adams-Bashforth-Moulton, properly generalized---because they derive future and current estimates simultaneously. But there are some nice matricial formulations of some high-order explicit methods that end up not being terribly computationally expensive.

Semiclassical

The stuff I did was in this paper: arxiv.org/pdf/1606.08738.pdf

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