The h Bar

Pieter

00:15

@peterh Ça ira!

dmckee

00:26

Ice rime a couple of millimeters thick on the roads here and we're suppose to be having a home carer come over to sit up and look after a convalescing friend we're putting up.

I can salt and sand my patch, but I'm worried about their having to drive.

peterh

@EmilioPisanty There are around 8000 living languages on the world, but the majority of them will die out within a century. Wikipedia exists on 300 languages.

@EmilioPisanty With it, the collective knowledge of the humanity will get an irrecoverable loss. Most of these languages won't be ever written down, no recording of their spakers will exist, nothing. They will disappear, as if they hadn't ever existed.

Secret

ugh I hate my timezone, so many interesting things get missed above

Secret

01:00

3 hours ago, by Sir Cumference

Asking a question to post an answer

in The Factory Floor, 2 days ago, by Secret

Last night dream: There is a flashback involving Slereah and 0celo discussing about manifolds which they both left a comment saying that since you are gluing two surfaces together, the result is obviously not homeomorphic to a sphere. Later on in WSE (which has a background of MSE meta) Futurehistorician asked a question which had received two downvotes.

On closer inspection, found the whole question plus a 3 line paragraph in the comments by him are probably an answer to some question which the dream never give in detail what it is. His answer make use of the two manifolds discussed by Sl…

O wow, now we have a real life example

Emilio Pisanty

@peterh I'm glad you've expanded the reach of your list to more than four languages

given your comments above, I'm not sure you fully understand what that linguistic diversity entails, but hey

I'll just link here and here and here instead

something like this, say

Languages Without Verb Tenses?!

►

Secret

01:16

their are even languages where time notions goes backwards or even does not exist. How a language is structured does have a strong effect to the listener's actions and thinking

Emilio Pisanty

any claim that a substantial fraction of the world's languages shares any given feature (like, say the six-fold spread of pronouns of several European languages) needs to be backed by some pretty substantial evidence

Secret

meanwhile, I don't know how one can have gender neutral pronouns in the Chinese language

(there are no Chinese entry on the third person pronoun article)

slate.com/blogs/lexicon_valley/2013/12/26/…

> In recent years, however, there has been an attempt to get rid of the gender distinctions for the third person pronoun and go back to a genderless stage. What is most curious, though, is the manner in which this is being done, namely through Pinyin, the system by which Chinese characters are transcribed into the Roman alphabet. In other words, 他, 她, 它, 牠, and others— all pronounced tā—are now being replaced by the actual letters "ta"!

lol that's interesting

Semiclassical

@EmilioPisanty i was about to make a smart-a** comment about how all languages involve sound, but sign language is a thing

Secret

are there languages that cannot be written or coded?

Most languages I knew seemed to have some kind of structure

even sign languages and braille only make senses if you put the gestures and symbols in a certain order

Semiclassical

all languages involve syntax, is what I think you're after

Secret

01:29

Hmm I see, guess we cannot really call art an language then

peterh

@EmilioPisanty And the funny thing is... if we could find some measurement unit for the complexity of a language... and we would order the languages by this unit, I think English would be nearly at the end of this ordering.

Secret

Language complexity

Language complexity is a topic in linguistics which can be divided into several sub-topics such as phonological, morphological, syntactic, and semantic complexity. The subject also carries importance for language evolution. Although the concept of language complexity is an old one, the current interest has largely emerged since the beginning of the 21st century as it was previously considered problematic in terms of political correctness. Language complexity has been studied less than many other traditional fields of linguistics. While the consensus is turning towards recognizing that complexity...

peterh

01:55

@Secret I think somehow the information of the meaning should be measured per bits of entropy-normalized text.

Secret

But what will the microstates be?

peterh

03:01

@Secret I think it depends on the model of the meaning. The length of the entropy-normalized text is simpler, essentially we can get any long text and compress it with the best compressor. :-)

Dawood ibn Kareem

Are we going to reinvent the science of linguistics today? Let me get my hat and my coat.

peterh

@Secret Funny thing is, in the elementary school (!) I learned some linguistics. Essentially, a modelling of the sentences. It was a different classification of the words, as the verb/pronoun/etc . It classified them by their role in the sentence.

@Secret All sentences can be ordered into a tree structure. For example, the sentence: "The quick brown fox jumps over the lazy dog" can have around this structure: "((quick brown) fox) (jumps over) (lazy dog)". I think, the meaning of the sentences could be encoded into such a tree structure, and then the complexity of this tree structure could be measured.

@Secret It should also encode all the bonus information what a sentence has. For example, in German, if you citate something, you use a little bit different endings of the verbs, if also you believe what you citate, or if you don't. This information is simply lost on English. Somehow this bonus information could be encoded into the tree structure by metadata objects.

(btw, Germans are now on the best way to lose all of these beautiful feature of their language)

Dawood ibn Kareem

03:41

What if you want to quote someone, but remain neutral on whether you agree?

peterh

03:54

@DawoodibnKareem Then it is the different ending.

@DawoodibnKareem On German, you can citate others on a way that you don't state it.

@DawoodibnKareem It is essentially a light-weight conditional.

@DawoodibnKareem Or, because there are multiple past and future tenses, talking about multiple things in the past and in the future, you can order them on a way, to make it also clear, which happened first and later. Also Italian has this feature.

Semiclassical

closest thing I can think of in english is how academics use the latin pace to reference an opinion you respect but don't agree with it

Sir Cumference

06:34

Microsoft Word and OneNote have been pretty generous this week. I've only been averaging 3 random crashes per day.

Sid

06:47

@SirCumference Perhaps you are opening too many documents at once or something...

Sir Cumference

@Sid Well it usually happens when I try to copy equations. I have a library of math notes and definitions in OneNote, but sometimes as soon as I try to copy one, it'll crash.

Word just crashes randomly though. Microsoft really needs to fix these things.

Mithrandir24601

09:42

@peterh It's a bit of a late addition to this conversation, but that seems similar to the way languages are thought of in CS

John Rennie

09:58

@JaimeGallego Hi Jaime. I've been away, but I've just got home and there was the letter. Thanks :-)

Balarka Sen

11:19

@EmilioPisanty Ignoring all the meta-conversation that circled around this topic, I do not like to use the singular they because it makes me feel like I am not addressing the person in question directly but putting the person in the category of a group of people of unspecified genders, which makes the conversation awkward for me. There is however a very simple alternative while talking over internet: Just use the fucking username.

I have no need to call a person him, her or zhe and zher when I can call the person roblox456 because it's what's in the profile description.

Unfortunate as that username may be!

Secret

@BalarkaSen I think his link have already covered that alternative, and indeed some famous transgender or gender unidentified personalities are often referred directly with their names

Balarka Sen

Sure I didn't read the link really. Just pointing out the obvious

ACuriousMind

@BalarkaSen What's unfortunate about roblox?

Balarka Sen

In real life it can be a little awkward, I do think.

@ACuriousMind Fine, what about "dicktingler3345"

ACuriousMind

@BalarkaSen Yeah, alright (although that might just be an oblivious person named Richard Tingler ;P )

Secret

11:57

historicalcrimedetective.com/msm-richard-lee-tingler-jr-1968

ouch

Emilio Pisanty

12:49

@BalarkaSen No. There are good reasons in this instance to not mention the username at all.

You could use [user] throughout

But if you think that that is an improvement then you've got some very strange tastes

Balarka Sen

The reasons being?

Tanuj

13:42

@Blue Hey !

@Blue Where should I prepare the error analysis and all the Vernier calipers and screw gauge questions from ?

Anonymous

@Tanuj I used the mechanics-I textbook by B.M. Sharma

Tanuj

@Blue Would you advice it to me since time is less ?

Anonymous

Yes

Tanuj

Alrighty then , does it have a pdf online ?

Anonymous

@Tanuj I don't think so

Anonymous

13:46

Photocopy that particular chapter from someone

Tanuj

Dude !

found it ! :)

@Blue Should I go for other practicals at this moment ? Do questions from other practicals and salt analysis etc . also come now ?

Emilio Pisanty

@BalarkaSen that this is a discussion of a suspension

Anonymous

@Tanuj Which "other" practicals?

Balarka Sen

@EmilioPisanty Wait, so the reason is we shouldn't mention the username of the suspended user?

I don't get it

0celo7

This site's policy on discussing suspended users is insane

Balarka Sen

13:56

I know the policy but that says we shouldn't talk about the reasons circling around the suspension.

Tanuj

@Blue That book by cengage doesn't have the screw gauge and Vernier calipers part

Balarka Sen

Why should that impose restriction on talking about the username? Or why is saying the username any worse than using a gender-neutral pronoun to indirectly refer to the user?

Am I misunderstanding??

Anonymous

@Tanuj Hmm, I had forgotten that. I learnt the working of screw gauge and vernier calipers from some online videos and just solved some relevant questions from the previous years.

Anonymous

Those are easy stuff anyway

Tanuj

@Blue Is there any online platform where I can solve previous years from ? Like previous 37 years ?

0celo7

13:58

lmao

@Tanuj are you preparing for some Indian exams?

Balarka Sen

JEE i bet

Tanuj

Yup I am

0celo7

@Tanuj my thoughts and prayers are with you

Tanuj

JEE it is

0celo7

may Jesus Christ bless you

Tanuj

14:00

@0celo7 Dude ! That's so nice of you ! Thanks man :)

0celo7

amen

Anonymous

@Tanuj There are some but mostly they have lot of errors. I'd prefer buying a good book having all the papers instead

0celo7

@Tanuj I am being mostly sarcastic

but good luck anyhow

Tanuj

@0celo7 Awww :')

@Blue for the time being could you name some online platforms , till I buy the book ?

Balarka Sen

I accidentally stumbled upon some #deepmaffs @0celo7

0celo7

14:01

@BalarkaSen me too

peterh

@Mithrandir24601 Yes :) As far I know, the originally not science-based languistics and the CS-lingustistics are overlapping things now.

Anonymous

@Tanuj Embibe etc.

0celo7

constructing the Dirichlet Green's function for coercive Schroedinger operators isn't as straightforward as I had hoped

it's no wonder no one has written down the proof

Tanuj

@Blue 37 years on embibe ? Really ?

Anonymous

@Tanuj No

0celo7

14:02

@Tanuj are you serious about 37 years?

Tanuj

Yea ! Didn't you do them ?

0celo7

@BalarkaSen so what did you stumble upon?

Tanuj

@Blue Didn't you do them ? To what extent did you solve past year papers

Balarka Sen

@0celo7 I was thinking about commuting polynomials because it came up in a olympiad style exercises I was doing. Turns out it's a deep theorem of Ritt that if $F$ and $G$ are deg > 2 complex polynomials in one variable such that $F \circ G = G \circ F$ then upto conjugation either (1) $F$ and $G$ are monomials (2) $F$ and $G$ are related as $F^{\circ n} = G^{\circ m}$ (3) $F$ and $G$ are Chebyshev polynomials.

And when I say conjugation I mean simultaneously conjugate

And it seems any proof invariably involves dynamics and topology

Which means I should learn the story

0celo7

why is this deep?

Anonymous

14:08

@Tanuj I'm too lazy XD Just 2-3 maybe

Balarka Sen

hard 2 prove

and not just an algebraic fact

Tanuj

@Blue seriously ? Didn't you join any test series ?

0celo7

what's a test series

Anonymous

@0celo7 It's a tradition to appear for a mock test every month (or every alternate month) for 12 months before the real exam :P

Anonymous

Phew...I just want to forget those days

0celo7

14:11

@BalarkaSen here's a freaky fact for you

Anonymous

@Tanuj I had, but then I left

Anonymous

Too much stress

0celo7

if "difficult to prove" means "deep", then this is deeeep

like Deepack Chopra deep

@BalarkaSen Suppose $(M,g)$ is a complete, locally conformally flat Riemannian manifold with positive scalar curvature. Then $\pi_i(M)=0$ for $i=2,\dotsc, \lfloor n/2\rfloor$

Balarka Sen

I mean it's a completely dynamical statement. I think commuting complex polynomials => same Julia sets

@0celo7 coolio

that sounds kinda Lefschetz hyperplany

0celo7

@BalarkaSen I'm debating putting the proof in my thesis for the lulz

I'm doing some of the legwork that leads up to it

but GMT $\cap$ homotopy theory is a scary place

Balarka Sen

14:15

since both of those are scary, i imagine the intersection to be doubly so

0celo7

@BalarkaSen It has been said Yau wrote this paper and gave lectures on it while doing so

and the final writeup was by a student who really understood none of it

Balarka Sen

lmao

0celo7

and so here we are, with an impossible to read paper

Curio

Is CuO a covalent bond right?

But what is its Lewis formula then? A quadruple bond?!

0celo7

@BalarkaSen The proof is like "well the hypotheses establish some GMT consequences so you can just perturb the map from the cube and it's nullhomotopic"

like, really? Would more details have killed you?

Balarka Sen

14:18

rip

Oh I read a chunk of Milnor btw

0celo7

"since $i\le k$"

Curio

Because they only can reach 8 electrons sharing all ones

0celo7

idk, maybe that makes sense

@BalarkaSen which chunk?

Slereah

Riddle me this

Balarka Sen

i'd say transversality if i didn't know better. $\partial \Omega$ is a fucked set

Slereah

14:20

Take a $3$-manifold

Remove a line from it

Balarka Sen

what is a line in a 3-manifold

Slereah

Some $1$-submanifold

0celo7

@BalarkaSen do you know Bill Meeks

Balarka Sen

no

@Slereah closed?

Slereah

I think so in our case, yes

Balarka Sen

14:22

ok proceed

Slereah

what is the boundary, is it the original line, or two copies of that line?

Balarka Sen

there is no boundary

it's a no-boundary manifold if your 3-manifold was closed

Slereah

Well if you were to construct the boundary

0celo7

boundary of what

Slereah

By taking the Cauchy completion of it

Balarka Sen

14:23

it's not compact, but it doesn't have boundaries

Slereah

I'm trying to figure out how the Deutsch-Politzer spacetime is constructed

Balarka Sen

you can blow up that 1-submanifold to a tubular neighborhood and boundary of that dude will be your boundary

Slereah

Deutsch Politzer spacetime has the identification of a single line, though

I'm trying to figure out if it's a well defined manifold

DP is basically

Take a manifold

Remove two points

Balarka Sen

@0celo7 A little of chapter 1, a little of 2. You wanted to know the details of 3.5, right?

Slereah

Remove the line between the two points

0celo7

14:24

@BalarkaSen I did, yes

Slereah

Then identify the edges of that removed line

0celo7

in the noncompact case

he does this thing with an exhaustion I think?

and then applies that Combinatorial Homotopy paper

and a retract

fuck, it's been two years since I read that

Balarka Sen

Right, so somehow you want to say a direct limit of CW complexes is a CW complex.

0celo7

@BalarkaSen Yeah basically

(of course I've never needed Morse theory outside of the compact case so I didn't obsess over this)

(well, this Morse theory anyway. Morse theory on $\Bbb R^n$ is pretty useful)

Balarka Sen

It's all in the Hatcher's appendix A I'm certain

0celo7

14:27

@BalarkaSen Hatcher is no longer under my monitor

but I'm also not at school rn

@BalarkaSen @Slereah my seminar :')

GENERAL RELATIVITY SEMINAR
TITLE: Clifford algebras, spin groups, and their representations
SPEAKER: Carl Sundberg, University of Tennessee
TIME: 5:00 PM-6:00 PM
ROOM: Ayres 113
I will give a quick mini-course on the subject of the title and discuss conditions for a Euclidean vector bundle to admit a spin structure.

Balarka Sen

Sick bro

What are those conditions that you mention?

0celo7

it's not me

Slereah

Parallelizability?

0celo7

It's Carl

Balarka Sen

Oh right

0celo7

14:29

He's an operator K theory guy who has found a love of algebraic topology late in his career

He's gonna tell us all about Stiefel Whitney classes, etc.

Balarka Sen

@Slereah You can be non-parallelizable and still admit a spin structure...

0celo7

well, he's always liked algebraic topology

but anyway

@BalarkaSen I'm gonna give sick talk in two weeks

"Surgery of black hole horizons"

Balarka Sen

I never remember characteristic classes

@0celo7 ::ok-hand::

0celo7

(assuming I can actually prove what I think I can prove)

maybe 2 weeks is optimistic

@BalarkaSen looking

scrolling through chapter 4 right now, scary stuff

Balarka Sen

Let me summarize what Milnor does for ya

Slereah

14:32

@0celo7 are you gonna do Israel junction stuff

0celo7

@Slereah No, I'm going to glue together Riemannian black holes along a stable minimal surface and control the scalar curvature

Secret

in Mathematics, 43 mins ago, by philmcole

Can somebody explain to me why we need to take a derivative with respect to some independent variable? For example we take the derivative of a position vector $\vec{r}=(x(t),y(t),z(t))$ with respect to time, because time is not a component of the vector. What is the problem if time is a component of the vector like $\vec{r}=(t,x(t),y(t),z(t))$ with taking the time derivative? This is usually not dealt with in mathematics class, since time is independent in euclidian space.

Newtonian spacetime = not a good idea

Balarka Sen

You have an exhaustion $M^{a_1} \subset M^{a_2} \subset M^{a_3} \subset \cdots$ of your noncompact manifold $M$ where $a_1, a_2, a_3, \cdots$ are the heights on $M$ determined by the Morse function, so that there's a single critical point between $a_i$ and $a_{i+1}$.

0celo7

yes

look at prop A.11 in allen

Balarka Sen

By compact stuff, there's a homotopy equivalence $M^{a_k} \to K_k$ (Kek) for all $k$.

Where $K_k$ is a CW complex.

0celo7

14:35

why is $K_k$ kekworthy?

Balarka Sen

K_k

Slereah

@0celo7 Is it that one

gregegan.net/ORTHOGONAL/06/GRExtra.html#RS

Balarka Sen

just natural association

0celo7

have you been reading my thesis i.gyazo.com/0c6936c03d302a297655a2698828f873.png

@BalarkaSen yes.

Balarka Sen

@0celo7 Right, so the point is you can arrange these $K_k$'s in an chain of inclusions $K_1 \subset K_2 \subset K_3 \subset \cdots$ with those homotopy equivalences going between the exhaustion of $M$ and this dude. Take the direct limit of these complexes to be $K$. By the universal property of direct limits, this gives a map $M \to K$ which is a weak homotopy equivalence.

0celo7

14:38

weak homotopy equivalence?

Balarka Sen

weak because it's an isomorphism on $\pi_n$; given a map $S^n \to M$, you can homotope that to a map $S^n \to M^{a_k} \subset M$ for some $k$ by cellular approximation theorem.

And we know it's an isomorphism on $\pi_n$ on each $M^{a_k}$.

0celo7

what is a weak homotopy equivalence

Balarka Sen

A map which is an isomorphism on all homotopy groups

0celo7

isn't there a theorem that says if a space is an ENR or some shit then that implies homotopy equivalence?

Balarka Sen

Whitehead's theorem says this is the same as a homotopy equivalence if you have CW complexes as your domain/range

Right, it works for spaces dominated by CW complexes too

That's the theorem Milnor refers to

0celo7

14:40

right

is that in Hatcher?

Balarka Sen

Cor A.9

0celo7

so what

every compact manifold is an ENR?

Balarka Sen

If $M$ is a manifold (need not be smooth), embed that little shit inside some $\Bbb R^N$, take an $\epsilon$-neighborhood, triangulate that subset of $\Bbb R^N$

This neighborhood retracts to $M$

So any manifold is dominated by a simplicial complex even

Slereah

Does the Riemannian Schwarzschild solution have a benefit in topology

0celo7

fucking hell, how do you know that's triangulable

also $\epsilon$-nbhd is probably too hopeful if $M$ is not compact

you need a gauged neighborhood

variable width

@Slereah wot

Slereah

14:44

why are you gluing together Riemannian Schwarzschild solutions

0celo7

@Slereah I'm trying to generalize a recent thing

it's a secret

Slereah

Will you publish it afterward

0celo7

No, I'm sure experts know it already

I talked to one who told me it was true, but couldn't tell me why

Typical expert opinion in math

Slereah

Damn experts

Think they're so smart

0celo7

@BalarkaSen Don't you need the neighborhood to be an analytic set for it to be triangulable?

Is that always known?

Is there an analytic Whitney embedding theorem and an analytic tubular neighborhood theorem?

@BalarkaSen Riddle me this

is $\mathcal Op(p)$ some open set or any open set?

Or can it be used as both?

Curio

15:26

Is CuO a covalent bond right?
But what is its Lewis formula then? A quadruple bond?!

John Rennie

We urgently need this chap to review questions here

9

Slereah

@JohnRennie I'm not sure if this is for or against mainstream physics

Anonymous

@Slereah Can be used in both cases

Anonymous

It's a collection of rants ;)

Mithrandir24601

15:52

@JohnRennie It doesn't violate causality if you define the norm properly, for Pete's sake!

for reference, this is the paper I'm talking about

Slereah

16:17

arxiv.org/pdf/gr-qc/9608029.pdf

this seems interesting

John Rennie

16:32

There is a place in Hell reserved for people who link directly to the pdf instead of to the abstract

Anonymous

@JohnRennie What's wrong with linking the pdf directly ?

Anonymous

The abstract is almost always printed on the first page anyway

Balarka Sen

@0celo7 some appropriate

@0celo7 If $U_\epsilon$ is an $\epsilon$-neighborhood of $M$ in $\Bbb R^N$, you can always triangulate $U_\epsilon$; since $M$ is compact, take a large $n$-simplex that contains $M$. Barycentrically subdivide iteratively so that the simplices of that subdivided simplex which intersect $M$ have diameter less than $\epsilon$.

Well, actually, maybe it's better to reformulate. You can always choose a neighborhood $V_\epsilon$ of $M$ contained in $U_\epsilon$ which is triangulable.

That is enough for us.

Curio

16:55

Is this true? youtu.be/xDG8EQ2Fq8w Please can anyone answer me? I found only conflicting fonts

Balarka Sen

The retract $V_\epsilon \to M$ is the domination by a simplicial (hence CW) complex.

Semiclassical

Got a simple QM question which I want to see if I'm resolving right. (It's really elementary, but my brain lately seems to be compromised by my having a bad cold.)

Curio

@Semiclassical me?

Semiclassical

nah

to the room in general

Curio

:(

Semiclassical

17:00

First, a terminology question: I've got an operator $H_0$ whose spectrum is discrete, bounded below, and non-degenerate. (Think harmonic oscillator.) What's the name for that?

bolbteppa

17:41

compact? locally compact, some word like that...?

Mithrandir24601

@bolbteppa A compact operator apparently has a continuous spectrum though... (it's been a while since I've done this kind of thing, so I could be missing something quite easily)

bolbteppa

Not sure, don't find the need to go into this stuff yet tbh

This method is extremely powerful and awesome

phys.uconn.edu/~rozman/Courses/P2400_16S/downloads/…

Got the Hydrogen atom associated Laguerre from it really fast and quick, can get Harmonic oscillator Hermite's fast from it, no series matching terms then spending ages getting the Rodrigues form

dmckee

17:59

@Blue The abstrtact page does two things for you. (A) It always leads to the most up-to-date version of the paper and (B) it lets you decide if it is worth your time to download umptidy-ump pages of PDF.

loocsieulb

it also loads a lot faster if you don't have fast internet connection.

Anonymous

@dmckee Alright, yes, that makes sense. For papers on ArXiv I mostly do link to the abstract. But for papers behind paywalls.....(you know most people are gonna use SciHub or something similar :P)

Anonymous

But then, in a formal setting it's better to link to abstract

Anonymous

Yeah

loocsieulb

18:14

it's always better to link the abstract. it's not your responsibility to link sci-hub, you can just link them the paywall abstract and they can find it illegally if they want.

i guess the only time it's better to link the pdf is it someone asks lol

Jake Rose

Could anyone help me with a partial diffenetiation question?

xyz + x^3 + y^4 + z^5 = 0

How would one find (del x/del y) keeping z constant

Dawood ibn Kareem

18:57

Differentiate the whole thing, partially, by y. There'll be a couple of terms with del x / del y in. So you can then rearrange things.

Anonymous

The shortcut is to just find $$-\dfrac{\partial f/\partial y}{\partial f/\partial x}$$

Anonymous

Where $f$ is the left hand side...

Jake Rose

19:17

Where does that come from?

dmckee

smbc-comics.com/comic/get-me-a-scientist

Anonymous

@JakeRose Search for implicit differentiation

dmckee

The hover-over text is a riot

Balarka Sen

The extra minus sign always catches people off guard.

But it has a very good reason to be there

Jake Rose

Wont that just come up with the ordinary differential version?

Balarka Sen

19:20

Well it comes from the chain rule

Jake Rose

Anybody got a simple proof?

Anonymous

Hint: Your function $f$ satisfies a relation of the form $f(x(t),y(t))=0$. Now how do you represent the total derivative of $f$ w.r.t $t$?

Balarka Sen

You don't need a parametrization for this.

Anonymous

@BalarkaSen Sure

Anonymous

That's one of the ways.

Balarka Sen

19:30

Well, on introspection, being a graph is a parametrization. You're writing $f(x, y) = 0$ (locally) as $y = g(x)$; that's a parametrization of the level set as $(t, g(t))$.

Jake Rose

But once you have that formula how do you go about finding the y/x?

The derivative that is

Anonymous

@BalarkaSen Indeed, writing $y=g(x)$ is the alternative I had in mind. But it is equivalent to parameterizing.

Jake Rose

Because you have the del f/delx,y,z

Balarka Sen

@JakeRose Take the derivative of $f(t, g(t)) = 0$ with respect to $t$. What does the multivariable chain rule say?

Anonymous

$z$ is a constant here. Treat it like a number.

Jake Rose

19:34

= delf/delt * dx/dt + same for y

I dont know the the chat language for partial derivatives sorry

Anonymous

\partial

Balarka Sen

Well, x(t) = t and y(t) = g(t) in my case @Jake

So it's df/dt * dt/dt + df/dg * dg/dt, right?

dt/dt is 1, so df/dt + df/dg * dg/dt = 0 is what you are left with.

Cross-multiply. dg/dt = -(df/dg)/(dg/dt)

Jake Rose

Im confused could we start again?

Balarka Sen

Can you read LaTeX?

(ie do you have chatjax enabled?)

Jake Rose

yeah

Balarka Sen

19:46

Alright. So you start with a function $f(x, y) = 0$ and you write that (locally) as $y = g(x)$

Jake Rose

Okay

Balarka Sen

This means points of the form $\mathbf{x} = (x, g(x))$ satisfy $f(\mathbf{x}) = 0$. To be explicit, $f(x, g(x)) = 0$.

Jake Rose

lyup

Balarka Sen

So now you differentiate both sides of the equation $f(x, g(x)) = 0$ with respect to $x$. What does differentiating $f(x, g)$ with respect to $x$ give, using the multivariable chain rule? Well, just $$\frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial x} + \frac{\partial f}{\partial g} \cdot \frac{\partial g}{\partial x}$$

Is this ok?

Jake Rose

so does that equal $\partial f$

oops

$\partialf / \partialx

$\partialf / \partialx$

damn

Balarka Sen

19:50

Put a space between \partial and f (and x likewise). Also notice that you can edit messages

Jake Rose

\partialf / \partialx$

$$\partial f / \partial x$

$\partial f / \partial x$

Got there in the end

Balarka Sen

Well the first bit equals $\partial f/\partial x$. You have $\partial f/\partial g \cdot \partial g/\partial x$ remaining on the second bit.

The full thing equals $\partial f/\partial x + \partial f/\partial g \cdot \partial g/\partial x$

Jake Rose

No I mean that generl expression

Balarka Sen

Oh, I see. No, it's $\dfrac{d}{dx} f(x, g(x))$

Jake Rose

Why is it not partial?

Balarka Sen

19:53

$f(x, g(x))$ is a single variable function of $x$! The partial comes in because $f$ itself is a multivariable function.

Jake Rose

Ohhhhh I see

Yes continue

Anonymous

If you're writing for an exam, then don't write $\partial g(x)/\partial x$ :P (as $g$ depends on only one variable $x$) The partial symbol is reserved for when the function whose derivative you're taking depends on two or more variables.

Balarka Sen

True ^ I shouldn't have put the partial there.

But it doesn't matter, really.

Also you're going to be faced with examples where $g$ depends on stuff other than $x$ so might as well write than anyway!

(implicit differentiation works when you're implitizing one single variable out of many)

@JakeRose So anyway, you have $$\frac{d}{dx} f(x, g(x)) = \frac{\partial f}{\partial x} \cdot \frac{d x}{d x} + \frac{\partial f}{\partial g} \cdot \frac{d g}{d x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial g} \cdot \frac{dg}{dx} = 0$$

Focus on the last equality

Jake Rose

OKAY

*okay

Balarka Sen

If you cross-multiply, you get $dg/dx = -\dfrac{\partial f/\partial x}{\partial f/\partial g}$, don't you?

Jake Rose

19:59

Yeah I see

Could we go through the same thing for 3 dimensions?

Anonymous

Then you'd have to take into account the Jacobians, but yes

Balarka Sen

Sure, you could have $f(x, y, z) = 0$ and you could implitize like $z = g(x, z)$ and go through the procedure to find $\partial z/\partial x$ and $\partial z/\partial y$.

Anonymous

It can be generalized

Jake Rose

Could I attempt to do it and you help me when I get stuck?

We havent done matrices and stuff yet sadly

Balarka Sen

I'm going to fall asleep in a few minutes but surely @Blue could help. If not, there's a mathematics chat room where some people roam around

Anonymous

20:02

I'm going to fall asleep too. The math chat room is the place to go :)

Jake Rose

Maths chat was silent when I asked earlier sadly

Anonymous

Or just ask here. I'll answer tomorrow morning

Anonymous

Or Balarka will

Anonymous

I wonder if there is a geometric way to visualize that formula though. For the $f(x,y)$ case. Should be...

bolbteppa

22:20

$$y(x) = \frac{A 2 \pi i(-1)}{(\frac{\eta}{2}-l-1)!} \frac{e^x}{x^{2l + 1}} \frac{(\frac{\eta}{2}-l-1)!}{2 \pi i} \int_C d{t} \dfrac{ t^{l+\frac{\zeta}{2}} }{(t - x)^{\frac{\zeta}{2} - l}} e^{- t} \sim e^x \text{as} \ x \to \infty $$
(except when $\zeta = 2n$) right?

Slereah

22:31

the hell is this

bolbteppa

Associated Laguerre up to a constant in integral form :(

bolbteppa

22:59

This may be crazy, but is there any way to solve the eigenvalue PDE $\hat{L}^2 Y(\theta,\phi) = - l(l+1) Y(\theta,\phi)$ just by solving Laplace's equation $\nabla^2 Y(\theta,\phi) = 0$?

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