The h Bar

rob

14:04

Another useful version (but too hardcore for me to use regularly) is with the trig functions.

Secret

14:18

rob, I have a 2 weeks ago EM conceptual question. Not sure if you can help

0

Q: Radial magnetisation cases and toroidal monopole?

Recently when re-reading Griffth electromagnetism, where he explained about how if the magnetic system is cylindrical, solenodial, planar or toroidally symmetric (4th edition, p.283; 1999 reprint, p.273), then $$\nabla \cdot \vec{M}=0$$ always and thus the $\vec{H}$ field can be obtained via th...

heather

anyone here interested in an HTML question?

MartianCactus

anyone free to watch this 3 min video? khanacademy.org/science/physics/geometric-optics/…

if u go in the newest question tab then u can find my question

can u plz help?

khan acaademy guys dont ususally respond a lot so..

Anonymous

@MartianCactus Which comment ? The one by apuri ?

MartianCactus

yes

o sorry i didnt specify it

Anonymous

See physics.stackexchange.com/questions/29583/…

MartianCactus

14:33

the one by apuri

ok

ACuriousMind

@Qmechanic I think it's a perfectly fine statistical mechanics question, and it should not be migrated.

MartianCactus

oh ok

thanks!

Anonymous

welcome :)

Anonymous

bye

ACuriousMind

It could do a better job of explaining the question to people not familiar with the jargon, but that's no reason to migrate it.

Danu

14:38

Hi @ACuriousMind

ACuriousMind

@Danu ::sigh::

Danu

Nice timing

ACuriousMind

Heh :)

Danu

Speaking of equations, I've finally reached the coolest equation in my book

The Hirzebruch-Riemann-Roch theorem: $$\chi(X,E)=\int_X \operatorname{ch}(E)\operatorname{td}(X)$$

I wish I'd understand the Atiyah-Singer index theorem, though.

ACuriousMind

It sure seems to be important, although I don't really understand what a Todd class is.

Danu

14:42

Well, do you know the method of defining cohomology classes through symmetric invariant polynomials?

ACuriousMind

Nope

Danu

Wanna hear it?

ACuriousMind

Or, well, I suspect it's similar to the way that invariant polynomials in the field strength of a gauge theory tend to give topological invariants, but I don't know it

Danu

That's it, exactly.

I can elaborate a bit.

ACuriousMind

Please do

Danu

14:45

So if you have some symmetric invariant polynomial, the information is equivalent to its polarized version (where you plug in the same element in every entry), this is just from the polarization identity

ACuriousMind

"Invariant" under what, actually?

Danu

Invariance of $P$ means that $P\circ \operatorname{Ad}_C=P$ for any $C\in GL(n,\Bbb C)$

where the adjoint action acts on every entry

I'm sorry, I hsould be a bit more precise

So consider a symmetric polynomial $P:\mathfrak{gl}(n,\Bbb C)\times\dots\times\mathfrak{gl}(n,\Bbb C)\to \Bbb C$

Then the above definition of invariance makes sense

heather

wow, this has got to be one of the first times I've opened the review queue and there weren't any close votes to go over.

ACuriousMind

@Danu Yes, alright, it's a polynomial I can plug matrices into

Danu

And you just require it to be Adjoint-invariant

ACuriousMind

14:47

Yep, that makes sense

@heather Sundays tend to be slow

Danu

This is equivalent to requiring, for any $a\in \mathfrak{gl}(n,\Bbb C)$, that $\sum P(v_1,\dots, v_{j-1}, [a,v_j],v_{j+1},\dots,v_k)=0$

One direction is shown by differentiation of the equation involving Ad, which also already does most of the work for the proof in the other direction

ACuriousMind

Yes, makes sense to me from a general Lie theoretic viewpoint

Danu

Okay

ACuriousMind

Invariance under the group action and invariance under the algebra action are equivalent.

Danu

Yeah, in that precise sense of the above equations.

ACuriousMind

14:52

To a physicist, it's just writing $C = 1 + \epsilon a + \mathcal{O}(\epsilon^2)$ ;)

Danu

So now of course you can induce a polynomial
$$P:\bigg(\bigwedge\nolimits^{j_1}M\otimes \operatorname{End} E\bigg)\times\dots\times
\bigg(\bigwedge\nolimits^{j_k}M\otimes \operatorname{End} E\bigg)
\to \bigwedge\nolimits^m_{\Bbb C} M$$

The way you do it is to just pull the form parts out, and evaluate on the endomorphism part.

This makes sense in a trivialization (since End is really gl there)

ACuriousMind

Yes, I'm fine with that

Danu

And our invariance equations are precisely what we need to patch together a global definition

Okay, now this induced polynomial is actually not symmetric anymore (just graded symmetric), but evaluated on only 2-forms it's still symmetric.

So now you should be thinking about symmetric invariant polynomials in the curvature, and since you want cohomology classes you should ask your self whether this well produce a closed form

In particular, you want to compute $\rm d P(\gamma_1,\dots,\gamma_k)$ for endomorphism-valued forms $\gamma_j\in \mathcal A^j(M,\operatorname{End}E)$

ACuriousMind

$\mathcal{A}$?

Is that your notation for "section"?

Danu

Yeah

Well, sections of form-bundles.

ACuriousMind

14:57

Yes, alright

Danu

It replaces $\Gamma(\bigwedge^j T^*M\otimes \cdot)$

So we work locally

Writing $P(\alpha_1\otimes t_1,\dots,\alpha_k\otimes t_k)=\alpha_1\wedge\dots\wedge \alpha_k P(t_1,\dots,t_k)$ we see that $$\mathrm{d}P(\gamma_1,\dots,\gamma_k)=\sum_{j=1}^k (-1)^{\sum_{l=1}^{j-1}}P(\gamma_1,\dots,\mathrm d \gamma_j,\dots, \gamma_k)$$

Now, locally $\mathrm d=\nabla-[A,\cdot]$ for an endomorphism-valued one-form $A\in \mathcal A^1(M,\operatorname{End}E)$

But then our invariance equation exactly shows that we may replace $\rm d$ with $\nabla$ on the right hand side of the equation.

Conclusion of this technical lemma:
$$\mathrm{d}P(\gamma_1,\dots,\gamma_k)=\sum_{j=1}^k (-1)^{\sum_{l=1}^{j-1}}P(\gamma_1,\dots,\nabla \gamma_j,\dots, \gamma_k)$$

Now recall the Bianchi identity, and conclude that the polarized polynomial $\tilde P$ induces a cohomology class $\tilde P(F_{\nabla})\in H^{2k}(M,\Bbb C)$ for any connection on $E$.

ACuriousMind

$\nabla$ is the gauge covariant derivative?

Danu

Still with me @ACuriousMind?

@ACuriousMind We're on a vector bundle

ACuriousMind

That's not an answer :P

Danu

Just a connection

on a vector bundle

ACuriousMind

15:04

Okay

Danu

I'm not thinking about principal bundles

So now symmetric, invariant polynomials produce cohomology classes.

ACuriousMind

Yes, I see that

Danu

In particular, the Chern classes are defined by $\det(\operatorname{id}+A)=1+\tilde P_1(A)+\dots+\tilde P_1(A)$, and plugging in $A=i/2\pi F_\nabla$

The invariant polynomials given by $\operatorname{tr} e^A=\tilde P_0(A)+\tilde P_1(A)+\dots$ yield the Chern characters in (after plugging in $i/2\pi F_\nabla$)

ACuriousMind

Yep

Danu

And finally the Todd classes are defined via the following polynomials:
$$\frac{\det (tA)}{\det(\operatorname{id}-e^{-tA})}=\sum \tilde P_k t^k$$

So there are your Todd classes for you :)

ACuriousMind

15:09

lol

Danu

(the formal variable $t$ can be used in all of the above definitions)

ACuriousMind

Thanks, that's more than I knew about them before, but now I wonder why the heck one would be looking at that expression.

Danu

It's related to the definitions of $c$ and $\operatorname{ch}$, clearly. But it's not super easy or anything

ACuriousMind

(I fear the answer is "because it appears in Hirzebuch-Riemann-Roch" :P)

Danu

Yeah, exactly.

Similarly, the Chern character is mostly there because it behaves nicely under tensor product: $\operatorname{ch}(E\otimes F)=\operatorname{ch}(E)\cdot \operatorname{ch}(F)$

In general, the Todd classes & Chern characters are just linear combinations of Chern classes.

koolman

16:18

Someone please help me

In above question how they got the relation between transverse and longitudinal magnification

Anonymous

16:51

@koolman Image not loading

koolman

@S007 please try , or upload it again

koolman

17:04

@S007 try it again

koolman

17:37

ACuriousMind

How did this question get 11 upvotes? I can't tell what it is asking at all.

koolman

@ACuriousMind can you please help me out with the above question

vzn

17:55

Simulations Show Swirling Rings, Whirlpool-Like Structure in Subatomic ‘Soup’ / Berkeley lab

vzn

18:05

↑ finally a substantial simulation, & more of that new buzzword hydrodynamics vzn1.wordpress.com/2016/11/07/…

MartianCactus

18:33

i have a question

when we deal with parabolic mirrors

we take the object and take any arbotrary point on it

then after that why do we draw 2 rays going from that point to the morror instead of just 1 ray?

and what does those rays even mean? The object is not necessarily luminous

Anonymous

@MartianCactus One ray cannot form image of a point

Anonymous

Atleast two rays are needed

MartianCactus

2 rays are needed to form an image of one point?

why?

Anonymous

@MartianCactus See physics.stackexchange.com/questions/233603/…

Anonymous

3

Q: Why do you need at least two rays to form an image?

Why isn't enough one light beam to form an image in your retina for example?

optics geometric-optics lenses

MartianCactus

18:41

they say that you can use 20 but 2 are minimum

but why?

why are 2 minimum?

Anonymous

@MartianCactus An image is formed at intersection of 2 rays

Anonymous

1 ray cannot intersect with itself

MartianCactus

it is related to the way our brain percieves things

isnt it?

so out brain scans 2 rays

and then traces em back to where they meet

ACuriousMind

The rays are not real.

Anonymous

@MartianCactus No, it is not related to the brain...

MartianCactus

18:43

and then determines what the image is

@ACuriousMind what do you mean?

ACuriousMind

The light rays of geometric optics are tools we use to figure out where the image is formed, they do not represent something that's happening physically.

Anonymous

Even when taking a picture using a camera atleast 2 rays are needed to create image of a point

MartianCactus

but how do 2 rays create an image?

ACuriousMind

@MartianCactus They don't "create an image". The image is created by all light passing through the lens. But, if we want to know where the image is formed, we can just use two rays and see where they intersect

MartianCactus

i am not talking about lens..i am talking about a concave mirror

ACuriousMind

18:46

lens, mirror, it doesn't matter.

MartianCactus

why are 2 rays required then?

why not just one ray

the 2 rays will meet at 1 point anyway

ACuriousMind

How would you figure out from just 1 ray where it image is formed?

MartianCactus

why cant 1 ray do the same?

Anonymous

@MartianCactus Can you draw a picture as to how 1 ray forms an image ?

Anonymous

Try drawing on a paper

MartianCactus

18:47

ok

Anonymous

@MartianCactus Yes, try drawing it and then upload it here. Maybe then we can help you spot the mistake

MartianCactus

ok

imgur.com/a/AetUe

done

ACuriousMind

and...how did you decide where the point is at which the image is formed? Why is it not further back, or closer to the mirror?

Kyle Kanos

Attention: I'm a daddy again :D

Anonymous

@MartianCactus So which one is the image ? That dotted one on the right hand side ? Why isn't it more towards the left or more towards the right ?

ACuriousMind

18:54

@KyleKanos Congratulations! Boy or girl?

MartianCactus

oh

so thats why

you need at least 2 rays to tell

Kyle Kanos

Girl. Elizabeth. 9 lbs, 6 oz (4.25 kg)

10

MartianCactus

but why do people take one ray to be completely parallel?

@KyleKanos congrats!

Anonymous

@KyleKanos We are talking for the first time probably. But congratulations from my side too :)

Kyle Kanos

@MartianCactus Thanks

@ACuriousMind Thanks!

@S007 Thanks

ACuriousMind

18:55

@MartianCactus Because it's the easiest to draw.

Bernard Meurer

@KyleKanos!

Hey man!

@KyleKanos Congrats!

Kyle Kanos

@BernardMeurer Thanks!

Bernard Meurer

Damn, that's a big baby

MartianCactus

imgur.com/a/Dkr51

here i used 3 rays

so we can literally take as many rays as we want and because of physics they will all end up back at the same point!!!

whoa

Kyle Kanos

@BernardMeurer Yes she is. Biggest of the lot

Previous owner of that was 1st born boy @ 9 lbs (4.08 kg)

Bernard Meurer

19:00

@KyleKanos Generally I'd dispute the use of 'lot' here, but I feel in your case it's the appropriate wording :P

@KyleKanos I think I was 4.5Kg when I was born, but I'm fat so maybe that's why :D

Kyle Kanos

I was 4.68 kg when I was born

The heaviest of my parents' 6 kids

MartianCactus

am i right?

@CuriousMind

Bernard Meurer

@KyleKanos Jesus, that's a lot of uncles and aunts for your kids

MartianCactus

@ACuriousMind

Anonymous

@MartianCactus yes you are

Anonymous

19:02

right

MartianCactus

oh ok :D

Kyle Kanos

@BernardMeurer Yes.

My parents have 10 grand kids now. Half of them are mine

Bernard Meurer

@KyleKanos Lol, you have a ~~world~~ family domination thing going on?

Kyle Kanos

@BernardMeurer I plead the 5th

MartianCactus

and also in a concave mirror, we take 2 rays from the tip of the arrow. One which goes through the focus and one which goes parallel to it. We chose these 2 rays just to make it simple right? We can literally chose any ray and the result would be the same

Bernard Meurer

19:03

@KyleKanos When they are all teenagers I hope God watches over you

Anonymous

@MartianCactus yes

Kyle Kanos

I hope He watches over me for more than just their teenage years...

MartianCactus

oh ok cool!

Bernard Meurer

@KyleKanos Lol, you're my hero

Anonymous

@MartianCactus At what level are you studying btw ?

MartianCactus

19:09

im in 10th grade

Anonymous

@MartianCactus Country ?

MartianCactus

India

/Mars

however u wanna take it :P

Anonymous

@MartianCactus Damn XD....almost every second person I meet on stack exchange is from India :-P

MartianCactus

lol

where ya from?

Anonymous

India....:-P

Anonymous

19:11

12 th grade

MartianCactus

o yo

which city?

Anonymous

KOL

Anonymous

KA

Anonymous

TA

Anonymous

:-D

MartianCactus

19:12

o gewd

Anonymous

you ?

MartianCactus

im frm Jaipur

Anonymous

Ow! The Pink City :-D

MartianCactus

i dont live in the pink city

i live outside it in the modern jaipur

Anonymous

Oh...I did'nt know that there is modern jaipur too

Anonymous

19:14

gd to knw :-)

MartianCactus

yea

Anonymous

@MartianCactus Okay. Goodnight :-D! Bye!

MartianCactus

same

bye!

heather

@KyleKanos, congrats!

koolman

Can someone help me out with a question please

heather

19:27

@koolman, go ahead and ask, you don't need to ask if you can ask =)

koolman

Ok

In the above question how they get the relation between transverse and lateral magnification

Anonymous

@koolman See youtube.com/watch?v=3buVroSPrpY

Anonymous

and also youtube.com/watch?v=iQuHVUMM9Yg

Anonymous

You will get it :-)

Anonymous

Sorry, I was away, so I could'nt answer your question before

koolman

19:31

Anyway , thankyou

Anonymous

@koolman Did you get it now ?

koolman

I will watch the videos

Anonymous

@koolman Okay. I hope you will get it. It is very easy.

koolman

Yeah@S007

Aritro Pathak

20:00

Is it safe to stay in the US now, as an Indian person?

Physics Meta

0

Q: Why can't I set a bounty on my question that I've answered myself?

My reputation is 76 and therefore I'm eligible for setting bounties, I can set a bounty for eligible questions and even on my own previous eligible questions, but when I want to set a bounty on my own question that I've given an answer myself (which is incomplete and I'm looking for other answers...

bug bounty

Anonymous

@AritroPathak Why not ? :-P

Anonymous

There are over 2 million indians living there :-D

dmckee

@AritroPathak Well, I suppose the odds of getting harassed in public are slightly higher, but frankly Trump has largely energized an existing base of discontent and made it more visible without really making a lot of new jerks.

Keep in mind that many Trump supporters don't hold the more extreme views highlighted in the media, and that most people are pretty decent most of the time.

That said, there have always been places where a dark skinned person would be at more risk than a light-skinned one.

And places where dressing nattily puts you at more risk than jeans and t-shirts.

I'd like to be able to write a wholly encouraging message on the matter, but it only takes one nutjob working himself up to a frenzy to mess up your whole day if you are in the wrong place at the wrong time.

Aritro Pathak

20:18

yea, I'm aware the Trump supporters were always there. Hope they are not very emboldened now. Hoping to go through unscathed for the next few days.

dmckee

I live in a small, city on the border of a rural region with a reputation for backwardness, and I have two colleagues from near India. They've not been harrassed so far.

So things don't seem to be spiraling out of hand.

heather

i have a python program that takes in an equation that the user inputs. how can I "sanitize" this, so I can make it an expression python can manipulate?

Aritro Pathak

Hopefully nothing bad happens. There are enough decent and welcoming people around.

ACuriousMind

@heather You'll want a parser

dmckee

@heather Write a parser that only accepts safe expressions and test all input before evaling it.

If you only want arithmetic expressions such a parser is not very difficult, either.

heather

20:23

okay, thank you, I'll give it a shot

dmckee

Remember when I was talking about Jack Crenshaw's compiler tutorial? This would be a possible time to get a first taste of that.

As a parser is one of several elements that go into building a compiler.

heather

okay, one other quick question, an example input would be 5x+2=12, would it be especially difficult to add in variables to the normal arithmetic parser?

dmckee

The canonical link for Crenshaw's tutorial is compilers.iecc.com/crenshaw

heather

thank you

ACuriousMind

@heather Fixed number of variables with fixed names wouldn't be difficult

dmckee

20:25

@heather It's not hard, but you might consider requiring a syntax like 5*x+2=12 or equivalent. Otherwise you'll need a more restrictive lexer.

heather

okay, good to know

this project is going to be even more difficult than I thought, which is saying something =)

dmckee

(Lexing is the step where you decide what parts of a string represent a logical unit of input (called a token)).

Typical simple languages use [A-Za-z_]+ for keywords and variables [0-9]+(.[0.9]*)? for number and a list of opperators +, *, etc.

But it is worth reading up on what various languages you know do before deciding.

heather

21:03

darn

@BernardMeurer, when you have a moment, could you provide some linux assistance?

Bernard Meurer

@heather What up miss?

heather

@BernardMeurer, actually, I fixed the problem

Bernard Meurer

@dmckee What are you going on about? :P

@heather Did someone pervert you with regular expressions?

heather

@BernardMeurer, what are regular expressions?

Bernard Meurer

@heather Please don't look :P

dmckee

21:07

@heather You mentioned that this was going to be a bigger project than you knew? You were right. Welcome to the rabbit hole.

Kyle Kanos

@heather An amazing tool for finding text in large amounts of strings

Bernard Meurer

@KyleKanos That is a regular expression, is a way to define matching patterns in strings, basically

Emilio Pisanty

@heather they're a pretty neat tool for e.g. parsing html files and whatnot

dmckee

@BernardMeurer I'd argue that specifying the basic lexigraphical tokes of a simple language is one of the few places where regular expressions actually belong.

Bernard Meurer

You use it when you want to filter some input or file for a particular kind of data a lot

heather

21:08

oh, okay, that makes sense.

Bernard Meurer

@dmckee True

@heather They're also kind of the devil

Kyle Kanos

Or when you have a 2k line file and you need to change all instances of variable x with y

Bernard Meurer

Because it's hard to debug them

rob

@EmilioPisanty We have impressionable minds here.

Kyle Kanos

(where x and y are somewhat complicated)

Bernard Meurer

21:09

Also some monster made this

heather

@BernardMeurer, that's going to be annoying, because I think I'm going to need them for the proof-writing project.=D

Bernard Meurer

(?:(?:\r\n)?[ \t])*(?:(?:(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(?:\r\n)?[ \t]
)+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ \t]))*"(?:(?:
\r\n)?[ \t])*)(?:\.(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(
?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[
\t]))*"(?:(?:\r\n)?[ \t])*))*@(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \000-\0
31]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|\[([^\[\]\r\\]|\\.)*\
](?:(?:\r\n)?[ \t])*)(?:\.(?:(?:\r\n)?[ \t])*(?:[^()<>@,;:\\".\[\] \000-\031]+

Please god make it stop

Emilio Pisanty

@rob I was kind of hoping that poking the hive would produce a bigger uproar

Kyle Kanos

Good regex: .?|(..+?)\\1+

Bernard Meurer

@KyleKanos Take a look at that perversity

Kyle Kanos

21:10

What does it do?

Emilio Pisanty

maybe a link to this one

4426

A: RegEx match open tags except XHTML self-contained tags

You can't parse [X]HTML with regex. Because HTML can't be parsed by regex. Regex is not a tool that can be used to correctly parse HTML. As I have answered in HTML-and-regex questions here so many times before, the use of regex will not allow you to consume HTML. Regular expressions are a tool th...

Kyle Kanos

Because mine finds primes

Bernard Meurer

@KyleKanos validate email addresses

Kyle Kanos

Ouch

103

Q: How to determine if a number is a prime with regex?

I found the following code example for Java on RosettaCode: public static boolean prime(int n) { return !new String(new char[n]).matches(".?|(..+?)\\1+"); } I don't know Java in particular but understand all aspects of this snippet except for the regex itself I have basic to basic-advanced ...

java regex primes

Bernard Meurer

@KyleKanos badass

Kyle Kanos

21:11

Horrifically expensive method, of course

Bernard Meurer

Regex is p. slow though, which is a shame

dmckee

@heather You should note that there was a time when "regular expression" had a precise meaning in terms of "regular languages", but that perl and other fonts of evil have perverted that and the now call non-regular pattern matching tools "regular expressions" despite the obvious absurdity of the nomenclature.

Bernard Meurer

Very useful indeed

heather

lol

xkcd is awesome

Bernard Meurer

dmckee

21:14

Link to the original, already: xkcd.com/208

Bernard Meurer

Another good mention

(xkcd.com/1313)

@dmckee Thanks, I always forget that I can just paste the whole link here

dmckee

And someone should mention "two problems": en.wikiquote.org/wiki/Jamie_Zawinski#Attributed and blog.codinghorror.com/…

Bernard Meurer

@dmckee As a beginner project you should try write a kernel in regular expressions

I was going to direct that to @heather, but hey it's just as ludicrous to dmckee :P

dmckee

@BernardMeurer Well, it'll keep you from bothering people, anyway.

Bernard Meurer

@dmckee :(

heather

21:21

a kernel in regular expressions...maybe I should learn what a regular expression is first =)

and how to write one

dmckee

I once read a operating systems textbook, and the prologue talked a bit about the development of computer hardware. The author said of the 'powerful' time-sharing systems of the seventies that they encouraged the serious programs to stay at work late when the machines were busy so that "we developed a generation of pasty, caffine-adicted insomniacs' or something smilar to it.

Bernard Meurer

@heather It was a joke, please don't ever try to do that

@heather I think if someone even thinks too hard about doing that they might just die

2

@dmckee Lol, No systems like that anymore and I'm still insomniac

dmckee

The big advantage of staying late these days is that there is no one to bother you: joelonsoftware.com/articles/fog0000000022.html

heather

So, I can write a regular expression that will check the input and make sure it is alright?

Kyle Kanos

The disadvantage is that you probably will stay single and childless...

dmckee

21:50

@heather No. Read the link @EmilioPisanty provided above (stackoverflow.com/questions/1732348/…). You can't keep nested parenthesis straight with real regular expressions, and you don't want to do it in the extended versions.

I really is better to write a recursive or stack-based parser.

heather

@dmckee, okay...that post is hilarious, btw.

dmckee

But you might use regular expressions to implement the lexer if the language is simple enough.

Danu

Leonard Cohen - You Want It Darker (Audio)

►

dmckee

22:19

I've just spent about 15 minutes trying to find a comic I read on the web once.

It features a couple of characters representing stack machines (AKA context free languages) trading strings of matching or un-matched parentheses and responding correctly while a character representing a finite state machine (AKA a regular language) looks on in confusion.

Can anyone help me out here?

Bernard Meurer

$$a_{n+1}=\frac{1}{3}(\frac{a_n^2}{2}+4)$$

Transcript for