Mathematics - 2017-05-15 (page 4 of 6)

Jason Bourne

7:00 PM

A promposal is a new word considered for inclusion.

Mike Miller

@Astyx I'm not fond of more centrist-right economic policy, which is what EM promises

Jason Bourne

It means asking someone to go to prom with you.

Daminark

Academie francaise (or something) is just straight up the authority on the French language

Eric

I usually take the approach that if people are using the word on a social level, it's a word regardless of whether or not it's in any dictionary

Astyx

@Mike I can understand and agree with that

Jason Bourne

7:00 PM

Even if there is a regulatory body we can choose to ignore it.

Daminark

Or at least in France, not sure whether Quebec and Africa subscribe to it

Mike Miller

I'm glad I can abbreviate and criticise both at the same time :)

Akiva Weinberger

Is there a word for the graph consisting of lattice points and the one-unit edges between them?

Eric

@Daminark portuguese has one but most people ignore it because it's an attempt to regulate something which is basically impossible to regulate.

Astyx

@Daminark Which is why we don't have "lol" (or "mdr" more appropriately in french) in our dictionaries :p

Akiva Weinberger

7:02 PM

Well, I suppose "grid" is fine

Jason Bourne

@MikeMiller I think Americans spell it as criticize.

Daminark

I mean sure people can use it but say, in an academic or governmental context they better follow the rules

I'm waiting for Merriam-Webster to adopt the word "Kek"

3

Mike Miller

I don't really care as long as people know what I mean and Ted doesn't make a fuss.

Jason Bourne

@Astyx Well, partly because lol is an English word, not a French word.

Astyx

We use it nearly as often as you do in english @JasonBourne

Eric

7:04 PM

@Daminark It's not used that much though,

Jason Bourne

@Astyx You mean mdr?

Daminark

My mission is to change that

@Jason He said mdr...

Astyx

@Jason No I mean "lol". Do you use "mdr" ? That would really surprise me

Of course in "you" I mean "common people"

Jason Bourne

@Astyx No, I don't even know what mdr means.

Fargle

Revelation: the fact that the characteristic polynomial of a $2 \times 2$ matrix $A$ is $\lambda^2 - \mathrm{tr}(A)\lambda + \mathrm{det}(A)$ very nicely anticipates Vieta's formula.

Astyx

7:06 PM

It's an abbreviation for "mort de rire", ie "dead out of laughter"

Or "dying of laughter" I guess

Jason Bourne

There was this stupid lecturer who went on and on about how he knew how to solve the cubic and he was the only one in the department who could do it. He did not even mention the quartic.

Fargle

And apparently we were doing big words, so if I may be forgiven for a foray into German:

Rindfleischetikettierungsueberwachungsuebertragungsgesetz

Balarka Sen

At some point in an apocalyptic near-future the vocabulary would probably just boil down to a bunch of acronyms, with little to no vowels present. (T sm pt in a nr-ftr th vcblry wd pbly jst bl dwn t a bnch of acrnms)

Jason Bourne

@Fargle I hereby award you a star, though in German they join words every day.

Astyx

@Balarka I doubt so

Akiva Weinberger

7:08 PM

@BalarkaSen Welcome to Hebrew

Fargle

I think that might be possible in interpersonal speech, but within subjects and industries the nature of jargon would probably keep the language full.

Astyx

There's a reason it hasn't yet, partially because redundance is essential to communication

however there are obviously exceptions such as, as Akiva pointed out, Hebrew

Akiva Weinberger

Where "Welcome to Hebrew" is spelled "brvk hb’ l‘bryt" @BalarkaSen

Balarka Sen

@Astyx It's Hal Draper's visions, not mine! He hypothesized that this would be part of a futuristic information-control mania (trillian.mit.edu/~jc/humor/Ms_fnd_in_a_Lbry.html)

Jason Bourne

Biangbiang noodles

Biángbiáng noodles (simplified Chinese: 面; traditional Chinese: 麵; pinyin: biángbiángmiàn), also known as 油泼扯面; 油潑扯麵; yóupō chěmiàn, are a type of noodle popular in China's Shaanxi province. The noodles, touted as one of the "ten strange wonders of Shaanxi" (陕西十大怪), are described as being like a belt, owing to their thickness and length. == Description == The noodle is broad and hand-made. It was originally part of a poor man's meal in the countryside, but has recently become popular in fashionable restaurants due to the unique character used in its name. == Use in dishes == Dishes with...

Akiva Weinberger

7:10 PM

(Well, the "v" and the "y" are actually functioning as vowels in the above)

Balarka Sen

I also wasn't doing much for than joking.

Jason Bourne

Chinese character with the most number of strokes. ^

Eric

I feel like I used to text like that, but only because I had a flip phone, once things changed and full keyboards became a thing, I didn't really do that anymore.

Akiva Weinberger

I've heard of that, it's pretty cool @JasonBourne

Balarka Sen

@AkivaWeinberger Huh! Weird

Akiva Weinberger

7:10 PM

It's not even in Unicode

Astyx

@BalarkaSen I guessed that :p Just wanted to input my (unasked for) opinion

Balarka Sen

Haha, fair enough

Jason Bourne

@AkivaWeinberger Yeah it doesn't show up in the box, one must click on the link to see it.

Daminark

I anticipate language going a memetic route and everyone will say "whomst'd've'ff"

Jason Bourne

@BalarkaSen Will you show us the most complicated word in some Indian language?

Fargle

7:12 PM

Language was already memetic in the strictest sense. It's very easy to glom culture onto language by its nature.

Balarka Sen

Nah

Astyx

@Daminark Mdr

Fargle

That's part of why language lends itself so well to idiomatic expressions and figurativity.

Mike Miller

@Daminark You'd.

Daminark

I was shooting more for memetic bastardization but kek

@Mike Lolol

Jason Bourne

7:13 PM

What does kek mean?

Eric

@Daminark there's an episode of star trek where an alien speaks entirely in references to their cultural myths

Daminark

It's an internet lol, apparently came from World of Warcraft

Nice @Eric

Jason Bourne

I have seen lel too.

Astyx

@Daminark And Koreans

Balarka Sen

kek is like a russian lol

Eric

7:15 PM

Which is kind of like speaking through memes

Jason Bourne

Do you guys know about ASMR?

Eric

Brazilians also sometimes write kek

but not in reference to wow

Mike Miller

@BalarkaSen it's just a deformation of lel

top kek

Jason Bourne

I don't get ASMR. I don't feel any tingles when I watch.

Akiva Weinberger

@BalarkaSen Thought the Russian lol was "xaxaxa"

Eric

7:15 PM

topkek and kek are unrelated @MikeMiller

Jason Bourne

Deformation theory of complex manifolds.

Mike Miller

i don't believe you

Eric

topkek is some turkish brand of snack

Daminark

There's also a Turkish dessert called "Topkek", which has somehow figured into the meme

Fargle

@Eric Darmok and Jalad at Tanagra.

Balarka Sen

7:16 PM

yeah i have heard of the origins but it feels russian

Akiva Weinberger

(Two people swinging at the same ball… @Eric @Damin)

Mike Miller

yeah but that's obviously not the origin

Eric

kek was the way alliance players saw Horde people say lol in chat in WoW

Balarka Sen

@AkivaWeinberger Oh, sounds familiar to my favorite word, "axaxaxax mlo"

ml\"o I should say

SteamyRoot

there was the meme with the top gun hat

Jason Bourne

7:16 PM

There was a girl who always said hahax instead of haha.

Eric

@Fargle that's like my favorite episode

SteamyRoot

and then people made it into top lel, and later top kek

Daminark

@Akiva For real today is the day of sniping

SteamyRoot

and then they found out "topkek" is actually real etc etc

Eric

@Fargle Gilgamesh and Enkidu at Uruk :)

Fargle

7:17 PM

@Eric It's a very clever one.

Mike Miller

wow

sorry for doubting you Eric

Daminark

Wait I thought they had kek, found out about topkek, said "kek, let's incorporate it", and then decided to pull it back to lel to get top lel

Jason Bourne

Did you know that once someone did not cast Tom Cruise because they said he was not handsome enough?

Daminark

This is interesting

@Jason I'm presuming this Tom Cruise is handsome?

Jason Bourne

Anyway, many actors look like shit when they are not filming...

Balarka Sen

7:18 PM

I don't believe you... You're a liar!!!

►

Jason Bourne

@Daminark He is the star in Top Gun.

Akiva Weinberger

Is there a name for Eulerian paths in grids that only intersect themselves transversally?

Eric

memes converging is weird

Akiva Weinberger

A grid being a finite rectangular lattice of points with the one-unit edges between them

Well, not even necessarily Eulerian, actually

Mike Miller

@AkivaWeinberger You could say "No tangent vertices"

or something

Jason Bourne

7:19 PM

@AkivaWeinberger When people say 3 by 5 grid do they mean there are 15 dots?

Daminark

@Eric Is the space of memes complete?

Akiva Weinberger

Actually, "Eulerian" is definitely not what I want

Mike Miller

absolutely not

Akiva Weinberger

I want it to hit every vertex, not edge

Mike Miller

hamiltonian

Eric

7:20 PM

It's discrete @Daminark

Astyx

@Daminark How do you define meme addition anyway ?

Jason Bourne

@Astyx LOL + KEK = LOLKEK

Daminark

Lel

Eric

So it's automatically always going to be complete

Akiva Weinberger

@MikeMiller But it can hit the same vertex twice, as long as it's transverse (like a cross rather than a ")(" shape)

Daminark

7:21 PM

@Astyx I think concatenation/composition could serve as our operation. Though completeness doesn't quite need that

Astyx

Then definitely not @JasonBourne That's like the free group

Akiva Weinberger

@JasonBourne Sure

@Daminark KEK+LOL=KOK?

Astyx

Except it's not a group

Akiva Weinberger

asks naïvely

Astyx

The free monoid

Fargle

7:22 PM

@AkivaWeinberger KEK + LOL = WTW

Daminark

Is our identity just the pure form of meme? If so maybe you could make a group out of it

Akiva Weinberger

♫ Lolkok, oriental setting and the city don't know what the city is getting ♫

Jason Bourne

@Daminark I am waiting for Merriam-Webster to print Webster's Fourth New International Dictionary, because the third was in 1961.

Balarka Sen

what's happening here

2

Mike Miller

@AkivaWeinberger I know what you meant

Balarka Sen

7:22 PM

i can't keep track of stuff

Astyx

Lots and lots of stars being given this evening

Daminark

"For every meme there exists another such that their composition makes it devoid of content and reduces it to the pure form of meme"

Akiva Weinberger

$\Large\star$

Jason Bourne

@Astyx Did you know that a conductor once told a soprano who said she is a star that there are only stars in the sky?

Daminark

Yeah I don't even know how the starboard works anymore

Though that sentence is rather relatable

Akiva Weinberger

7:24 PM

@Daminark It's the right side of the ship, yeah?

Astyx

@Jason Who said what to whom ?

Daminark

snaps eyy

Astyx

Was the conductor female ?

Jason Bourne

@Astyx I don't know. My pianist friend told me years ago.

Astyx

(I guess the soprano was)

Jason Bourne

7:25 PM

Yeah, though there are male sopranos.

They either sing in falsetto or their voices did not really crack.

Astyx

Sure, that's why it's a guess :)

Jason Bourne

I wonder whether there are female basses?

Astyx

Surely

Akiva Weinberger

@MikeMiller If manifolds with boundaries aren't manifolds, perhaps I can let transverse Hamiltonian paths not be Hamiltonian paths.

Jason Bourne

Anyway guys the thickest single volume book I have is my 2800 page Oxford Paravia Italian Dictionary.

Akiva Weinberger

7:27 PM

In any case: Conjecture: There is precisely one transverse Hamiltonian cycle through a $2^n\times2^n$ grid (counting dots, not edges) that always turns clockwise.

For $n=2$ it's that clover thing on Mac keyboards.

⌘

(I know that there's at least one, I don't know that there's at most one)

Daminark

@Akiva that sounds interesting

Jason Bourne

@AkivaWeinberger Should I get a Mac if I never tried one?

Akiva Weinberger

I haven't tried both so I can't say

Balarka Sen

Maybe I'll do some foliations before sleeping today

SteamyRoot

@JasonBourne install gentoo

Daminark

7:30 PM

Rip Riemannian geometry

Jason Bourne

@AkivaWeinberger Does your Mac freeze now and then?

Akiva Weinberger

Not very often

Jason Bourne

@SteamyRoot I have used Debian for a while. Gentoo is too difficult for me, lol.

Akiva Weinberger

Protip: Don't download "Clean Your Mac."

Jason Bourne

One thing about Mac is that though it has Office from Microsoft it is not supported for as long and not as updated.

Astyx

7:32 PM

Office isn't that good anyway (imho)

Jason Bourne

What's the Mac equivalent of Office? Is it really good?

SteamyRoot

Just get LibreOffice. It can take a while to get used to if you're coming from Microsoft Office; but it pretty much has all you need, and it's FLOSS

Astyx

^

Jason Bourne

@BalarkaSen Maybe you can take a look at Petersen's Riemannian Geometry and Sakai's Riemannian Geometry as well. I think those are very hard though.

Balarka Sen

I have enough trouble learning stuff from a single book

i don't collect books, i also try to learn stuff from them

Astyx

7:35 PM

You think you have enough

Jason Bourne

@SteamyRoot Well, I got Office because I don't like LibreOffice, especially the equation editor. I don't always use TeX for equations.

Daminark

recommends yet another Riemannian geo book to Balarka

Balarka Sen

noo

Jason Bourne

Of course, I just like Lee's Riemannian Manifolds, though it doesn't contain much material.

SteamyRoot

@JasonBourne If you attended my science communication course, you'd just have failed.

Eric

7:37 PM

@Balarka weren't you using that Gallot something something one

Jason Bourne

But I like his treatment of the Hopf's Umlaufsatz and the Gauss-Bonnet theorem.

SteamyRoot

Oooh, are we spamming Riemannian geometry? I can upload some course notes...

Balarka Sen

yeah, Hulin and Lafontaine being those something somethings :P

I also use doCarmo to side read

Eric

Ah yes yes

Neves had me read some of it and I find it really hard to read GHL

Balarka Sen

actually i thought that but i am getting the hang of it

Jason Bourne

7:38 PM

GHL seem to be students of Marcel Berger.

Mike Miller

It's the only intro Riemannian book I like

Eric

Because do Carmo doesn't do any forms stuff and he wanted to do some stuff about the laplacian on forms

Jason Bourne

@MikeMiller Why do you like GHL?

Mike Miller

Jost is my taste but doesn't really make you feel the geometry

Eric

@MikeM I was planning on picking up Jost at some point cause it gives me the impression that it's kind of analysis-y, which I like

Jason Bourne

7:40 PM

Jost also has a book on calculus of variations.

Balarka Sen

<- no love for symbolic calculations

Jason Bourne

Marcel Berger has a long book on A Panoramic View of Riemannian Geometry, which is like a survey and not a textbook.

Calculations are very important. Without them, there is no food.

Mike Miller

In working through the book you get a good concrete picture of the geometry of Riemannian manifolds; discussion of lots of explicit examples (and geodesics before curvature IIRC which I prefer as it's more visual)

the calculus of variations book is a conference proceedings

Daminark

@Balarka same tho

Jason Bourne

@MikeMiller Wait, I think we are talking about different books. Jost has a COV textbook, published by CUP, grey cover.

Mike Miller

7:43 PM

ok

@Eric You would like Jost. How much PDE do you know?

Eric

I've done some PDE from a functional analysis course and have read and done exercises for like half of Evans book

Jason Bourne

I don't know any PDE but I am using Folland's book for that.

Mike Miller

ah, thats good

I was going to suggest Taylor's series

Astyx

SoumyoB's starred message some hours ago kinda lost it's use now with all those new stars

Jason Bourne

@Astyx One should always update his computer anyway

Balarka Sen

7:46 PM

@Eric/@MikeM: How do I prove smooth manifolds are triangulable? Do I prove that a geodesically convex atlas is a good cover, and throw nerve theorem at it?

Astyx

Sure, just saying

Jason Bourne

@Astyx But sometimes, after an update, something breaks. =)

Eric

I've been recommended Taylor's books but I'm broke and don't like online copies of books :/

Astyx

I don't update my mac, I don't want to sell my soul to Apple

Mike Miller

chicago library?

Jason Bourne

7:47 PM

If you are broke, you can always use copyleft versions first.

Eric

Right I guess I could check there

I usually never check out library books

Jason Bourne

Also, you need to return library books in X days.

Maybe you need X+Y days to read them.

Another solution is try to look for international editions of expensive books on abebooks.com

Daminark

I've checked out a few, they tend to give you a reasonable amount of time. I checked out a book at the beginning of the quarter and they gave me until June 30th

Mike Miller

at least as a grad student at UCLA I have 180 days per check out and can just renew online at home

so I don't need to take them back unless somebody asks

Jason Bourne

180 days is very good.

I like UCLA.

Daminark

7:51 PM

+1 to UCLA because nice

Jason Bourne

Someone returned a library book after a few decades, a news article said. Some months ago.

Mike Miller

actually I just checked and I was supposed to return the two books I have out from the main library over six months ago and they now consider them lost

Eric

I might get my hands on them from the library when summer starts since i'll have a lot of free time in june before a summer school starts

Balarka Sen

@MikeM Oops

Jason Bourne

@MikeMiller LOL. I think you at most need to pay a small fine of a few cents.

Mike Miller

7:53 PM

i worked through various chunks of those at some point but i can't promise i remember all of it

Eric

I need to review/learn more PDE anyway or ill be completely lost when I attend analysis things this summer

Balarka Sen

Actually, nerve theorem won't do. It just proves it's homotopy equivalent to a simplicial complex.

Jason Bourne

I think in the third volume of Taylor he proves the Nash embedding theorem for compact Riemannian manifolds.

Eric

oh that's cool

Alessandro Codenotti

@MikeMiller I like how they consider them lost rather than sending an email to the last person who borrowed them

Balarka Sen

7:56 PM

Can I prove a smooth manifold is homemorphic to a simplicial complex using geodesic neighborhoods?

Mike Miller

I don't remember getting emails

Balarka Sen

Not sure how.

Mike Miller

@BalarkaSen Try like this. Enumerate the finite cover and start in the first $U_1$. Pick a small standard simplex centered at the origin of as large radius as possible to fit into the geodesic chart, and exponentiate it. There will be pieces of the chart not covered, but you can glue a simplex (of a different shape) to the boundary of your last one inside your geodesic neighborhood to cover more space. Continue doing this until every point of $U_1$ not covered by your simplices is...

...contained inside one of the other $U_i$. Then make sure your boundary faces are small enough that each of them fits inside some $U_i$.

Daminark

Got class so peace out for now frenz

Mike Miller

The crucial fact that lets you induct between the charts is that the transition maps here are well-behaved - the boundaries of your simplices are (well, should be designed to be, maybe I did a bad job) totally geodesic, which is a property preserved by isometries

Astyx

8:00 PM

See ya @Dami

Balarka Sen

I am really unconvinced about filling in the gaps. I guess you could take a point in a gap and exponentiate a polygon from it to fill it in (if the gaps are small enough, which is by constuction)

And then triangulate polygon

Fargle

How should I refer to the fact that a shear is area invariant? I remember there was some kind of name for it for polygons.

Balarka Sen

@MikeMiller Ah, true.

Astyx

I'll get going too, bye

Mike Miller

@BalarkaSen Well, I'm not worried about triangulating each of the charts. The problem that stops you from doing this inductively in general is that when you change between charts the transition map might totally mess up your previous simplices.

Akiva Weinberger

8:03 PM

@Fargle Cavalieri's principle

Mike Miller

So you're not sure how to continue the triangulation.

Fargle

@AkivaWeinberger Thanks! Was blanking on it.

Balarka Sen

@MikeMiller Yeah, I didn't notice you were doing it for a chart first and then inducting. That's good intuition.

Akiva Weinberger

@Fargle It's more than just linear shears.

Fargle

@AkivaWeinberger Right, I just couldn't express exactly what I meant succinctly. >_>

Balarka Sen

8:04 PM

The "adjacentness" of the geodesic triangles are preserved here, that's all. Gotcha.

Akiva Weinberger

It's interesting how they related a sphere to the complement of a double cone in a cylinder.

Fargle

Yeah, that's damn clever.

4/3 just falls right out

Balarka Sen

Yeah Cavelieri's principle is really strange

Fargle

Lucky thing it's true, though.

(If an inevitability can be lucky...)

Mike Miller

I think classical results like triangulation of surfaces follow from Schoenflies theorem, whose main content is that continuous curves in surfaces are locally flat.

Balarka Sen

8:07 PM

True that.

Mike Miller

So when you change charts the local flatness shows you can rectify the old triangulation to one that looks nice in your new chart.

Balarka Sen

Ah, I see

I haven't carefully read the TOP proof of triangulability of surfaces

Mike Miller

I mean, that's it

Do these authors somewhere claim you can triangulate manifolds w Riemannian geometry?

Balarka Sen

Nope, I just heard it somewhere, and I thought I saw an easy proof upto homotopy equivalence by nerve theorem. But I couldn't immediately come up with a proof upto homeomorphism.

(I think all manifolds are homotopy equivalent to simplicial complexes?)

Mike Miller

sure, embed them in high-dim'l euclidean space and take a normal neighborhood

Balarka Sen

8:11 PM

It should also be easy to do it smoothly without nerve. Take an embedding in R^n, take a tubular neighborhood, triangulate it carefully

and use that spaces dominated by simplicial complexes is simplicial complex

Mike Miller

OK, I'm not 100% convinced about what I said, since the only thing I can make sure are totally geodesic by exponentiating are the bits coming radially from the origin, but not the 'edges' of the simplices

Jing Weng

Do any of you know some techniques to get over regret?

Mike Miller

That's a hell of a question

Jing Weng

or reduce regret. I dont know

Balarka Sen

Well, in Riemannian manifolds, you admit neighborhoods which are totally geodesically convex. Any two points can be joined by geodesics

Ted Shifrin

8:13 PM

@Balarka: Better put in "complete"

Balarka Sen

@Ted Hm, why would I need that if I only care about local neighborhoods? Am I being dumb?

Mike Miller

Yeah but what I want is to fill in a totally geodesic $\partial \Delta^k$ with a totally geodesic $\Delta^k$. Can I do that?

Ted Shifrin

Just criticizing your glib "any two points can be joined by geodesics"

Balarka Sen

@MikeMiller Just join the vertices with geodesics, right?

Mike Miller

"in the totally geodesically convex nbhds"

Balarka Sen

8:15 PM

@Ted I meant inside the neighborhood

Ted Shifrin

OK ... I'll remember to be a mind-reader :)

Balarka Sen

lol :P

Fargle

Hi @Ted!

Mike Miller

@BalarkaSen I'm worried that those run into, say, some other part of the triangulation

Ted Shifrin

Hi @Fargle

Fargle

8:16 PM

You'll never guess what I'm doing.

Mike Miller

But you know this stuff better than I do, you should figure out whether this works

Ted Shifrin

Procrastinating?

Jing Weng

What do you all regret?

Fargle

The exact opposite!

Balarka Sen

I regret my existence

Ted Shifrin

8:17 PM

takes Fargle's temperature

Fargle

@JingWeng Procrastination. And what Balarka said.

Ted Shifrin

stop that, you guys

Jing Weng

@BalarkaSen did you tell your parents?

Mike Miller

Cover most of $U_1$ with a totally geodesic triangulation, then most of $U_2$ extending the previous, etc.

Yeah this totally works. I was confusing myself by thinking about the exponential map.

You should just work entirely geometrically.

Balarka Sen

I don't think my existence has much to do with my parents. Certainly the source of my existence, but that's not what I regret.

@MikeMiller Right, I think it should work too.

Mike Miller

8:20 PM

The hardest thing is making the first simplex, in which you have to pick a bunch of points not on a plane...

but then you just pick an already-existing face and a new vertex, prove nothing else could be contained in this, and add a new simplex

Very cool.

Balarka Sen

Right, exactly

Vrouvrou

Hello, please iff $f$ is continuous over [0,1], and $\int_0^1 x |f(x)| dx=0$ how to deduce that $f\equiv 0$ ?

someone here ?

Alessandro Codenotti

Rehi chat

SteamyRoot

@Vrouvrou Start by proving that for $g$ continuous and $g(x) \geq 0$, $\int_0^1 g(x) dx = 0 \implies g \equiv 0$

Fargle

Hi @Alessandro

SteamyRoot

8:30 PM

ohi

Alessandro Codenotti

Can I bug you with a pde question? @Ted

lol he ran away as I was asking, was that a coincidence? maybe :P

Mike Miller

what was the question?

Vrouvrou

@SteamyRoot i know this then i have that $x=0$ or $f(x)=0$

SteamyRoot

But $x = 0$ only when, well, $x = 0$. So for all other $x \in [0,1]$ ...?

Alessandro Codenotti

We showed that if $f\in C^2_0(\Bbb R^n)$ then the identity $f(x)=\int_{\Bbb R^n}G_n(x,y)\Delta_yf(y)\text{d}x$ holds. Now we want to use this to construct solutions to the Poisson equation $\Delta f=\phi$ with $f(x)\to 0$ uniformly in all directions and $\phi\in C^2_0(\Bbb R^n)$

We asserted that there is a unique solution of the form $f(x)=\int_{\Bbb R^n}G_n(x,y)f(y)\text{d}x$ and we argue that it is a solution based on the fact above. I don't see why $f$ necessarily has compact support here though

($G_n(x,y)$ is a fundamental solution to the Laplace equation here)

ah, wait I wrote it wrong. The solution should be $f(x)=\int_{\Bbb R^n}G_n(x,y)\phi(y)\text{d}y$ and the identity we use is $f(x)=\Delta_x\int_{\Bbb R^n}G_n(x,y)f(y)\text{d}y$ (which is equivalent to the one above)

and I also got it now, I don't need $f$ with compact support, it's $\phi$ that needs to have compact support and that's given. That was dumb

SteamyRoot

8:49 PM

Hrmf. Random question: is there any general strategy/approach to determine if a $2$-parameter family (over $\mathbb{Z}$) can be written as a finite union of $1$-parameter families?

Vrouvrou

when $x\in]0,1]$ then $f(x)=0$ but in $x=0$ ? @SteamyRoot

SteamyRoot

@Vrouvrou $f$ is continuous

Jason Bourne

A continuous function that is positive somewhere is bounded below by a positive number over an interval. Meditate on this and you will find the way @Vrouvrou. And do remember what I told you the last time.

Akiva Weinberger

@Vrouvrou x\in{]0,1]} $x\in{]0,1]}$ has better spacing, in my opinion

Jason Bourne

@AkivaWeinberger IMAO spacing questions are for TeX SE, lol.

Akiva Weinberger

8:56 PM

(It stops the $\in$ from sticking to the $]$)

Alessandro Codenotti

Are there $C^2$ solutions $\phi:\Bbb R^n\to\Bbb R$ to $\Delta\phi=\phi$ apart from those of the form $\sum\limits_{i=1}^n (a_ie^{x_i}+b_ie^{-x_i})$?

Jason Bourne

The other thing @Vrouvrou is that a lot of these questions have been asked in books and online in many places, and the solutions are everywhere.

But if one does many questions without thinking about what one is doing, then one gets nowhere, except a bunch of solutions to a bunch of problems.

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$Mathematics$ Mathematics