Mathematics - 2019-12-13 (page 1 of 2)

12:00 AM

In more generality a 2 dimensional surface is something you can locally parametrize by 2 coordinates

12:10 AM

For continuity, there's two definitions usually. A metric one and one with open sets. They're equivalent.

This boils down to the metric topology is generated by open balls so it suffices to check continuity on the open balls which is the definition of continuity.

Why is the equivalence of definitions important/what problems does it solve? It makes one's mind set up connections between metric space concepts and topology concepts so that one can work with both better. In particular abstract results about topology can walk through the bridge to give us results about metric spaces. But they g…

Ultradark

I met an economics major who knew about the banach tarski paradox

unrelated:

What if all ocean water left the planet at the speed of light?

but simultaneously all sharks turned into black holes

Ted E

"What if xyz" isn't a question

Rithaniel

12:29 AM

What if "What if xyz" were a question

Akiva Weinberger

What even is a Calabi–Yau manifold

@Rithaniel What if this question weren't a question?

Rithaniel

What if humans had never developed the word "what"

Akiva Weinberger

¿Qué?

Alternative gag: Scottish accent Whit?

Rithaniel

That is the abstract notion signified by the word "what," not specifically the English sound made to represent the notion

Ultradark

12:50 AM

Consider a sketch of a flow in the upper half plane. In fact there are two identical copies of the same flow, with one reflected across the middle.

If we let the flows interact then this amounts to adding the vectors in the box portion. There should be generally a vertical flow after this adding process.

I see how to add the vectors in the box but not the vectors in the top right corner and the top left corner.

[![enter image description here][1]][1]

[1]: https://i.stack.imgur.com/mY6la.jpg

Akiva Weinberger

Yet Another Sphere Eversion Video

Sphere Eversion with Transparency

►

Ultradark

How do you add vectors that don't overlap?

Akiva Weinberger

What's in the blank spaces? Zero vectors?

Ultradark

or maybe I should say how do you add

flows

yeah!

zero vectors!

maybe I'm not understanding something, but does that just mean you add the vectors in the box and the rest stays the same?

Akiva Weinberger

Yeah

Rithaniel

12:56 AM

Now that is a smooth sphere eversion

Ultradark

that's what I call

Semiclassical

testing:

Akiva Weinberger

I love sphere eversions because it was such a complete accidental discovery

Semiclassical

fail

nope, didn't work

Akiva Weinberger

It's a start

$\triangleright$

Semiclassical

1:02 AM

it's supposed to be an animation

working example: i.stack.imgur.com/uI9ha.gif

though right now it's cycling too fast

slowed-down example: i.stack.imgur.com/SxoME.gif

that's better

@AkivaWeinberger this is an animation of your trisection construction from earlier

evolving from a triangle to a hexagon towards something smooth

Akiva Weinberger

Neat

I still haven't tried to find an equation for it, but I think two possible approaches are:

1) write out coordinates for the vertices of P_n in barycentric coordinates (probably in base 3) and see if I find a pattern

Semiclassical

one nice thing about doing it in mathematica is that I can extract the areas of the progressive pictures

Akiva Weinberger

2) For each $\theta$, see if there's an equation for the tangent line at angle $\theta$

Semiclassical

and, relative to the area of the initial triangle, I get the following so far:

Akiva Weinberger

(like a sort of dual polar coordinates)

Semiclassical

1:09 AM

1, 2/3, 16/27, 140/243, 1252/2187, 11252/19683, 101236/177147, 911060/1594323, 8199412/14348907, 73794452/129140163, 664149556/1162261467

Akiva Weinberger

Decimals?

Ah

4/7

Semiclassical

1., 0.666667, 0.592593, 0.576132, 0.572474, 0.571661, 0.57148, 0.57144, 0.571431, 0.571429, 0.571429

oh, nice

Akiva Weinberger

4/7 is indeed the correct answer (but I don't know why)

(I previously said I knew the area but nothing else)

Semiclassical

well, I am a physicist, so obtaining it experimentally works for me :P

Akiva Weinberger

This also means 6/7 of the hexagon I think

Semiclassical

1:11 AM

hexagon is 2/3 of the triangle, so that checks out

I'm more interested in the sequence of fractions at this point

Akiva Weinberger

Do you have a static image of the final shape?

Semiclassical

I can get one

If I drop the triangle and rescale to the hexagon, then the sequence of fractions is 1, 8/9, 70/81, 626/729, 5626/6561, 50618/59049, 455530/531441, 4099706/4782969, 36897226/43046721, 332074778/387420489

converging to 6/7 evidently

the denominators are powers of 9

oeis doesn't recognize the sequence of numerators

ah, but I can instead write those as 1,1-1/9, 1-11/81, 1-103/729, 1-935/6561

and oeis does recognize 1,11,103,935...

specifically, as the coefficients in the expansion of 1/((1-2*x)*(1-9*x))

which suggests some recurrence relation

oh. better observation: the differences between successive fractions (i.e. the area lost at each step) are 1/9, 2/9^2, 2^2/9^3, 2^3/9^4, ...

so the area lost at each step decreases by 2/9

which is also true if I start from the triangle, so I guess that should go back

Akiva Weinberger

1:35 AM

So you end up with $1-\sum 2^n/9^{n+1}$? @Semiclassical

Semiclassical

looks right, yeah

Akiva Weinberger

Neat pic (Waterman polyhedron, convex hull of lattice points contained within a sphere)

Semiclassical

nice

Ultradark

1:52 AM

can the addition of two hyperbolic flows be hyperbolic?

Ted E

Do you even know the definitions of the words you are using?

5

Ultradark

Are you serious?

Akiva Weinberger

2:11 AM

I don't know the answer to that question

Soham Chowdhury

god that's so w i d e

Akiva Weinberger

Well - I don't fully understand the definition, but wouldn't a hyperbolic flow added to itself be hyperbolic?

But yeah try Math SE

Ultradark

okay

Akiva Weinberger

What subject would that be? Differential geometry?

Ultradark

dynamical systems

Akiva Weinberger

2:12 AM

Aha. I should learn that someday

So this is connected to double pendulums and Lorenz attractors somehow?

Soham Chowdhury

probably those systems have some kinda configuration spaces which are manifolds and then you ... study how a point on said manifold moves

and that forms whatever a "flow" is

Ultradark

akiva, I like your question. what about an Anosov flow added to a reflected copy of itself

Akiva Weinberger

Prof Ghrist (creator of the Calculus Blue series on YouTube) has some dynamical systems/chaos animation

Ultradark

then there would be a flow path that essentially remains invariant over the course of the evolution

the invariant would be the central vertical line dividing the flow into symmetric parts

the flow below splits everywhere except one path

vzn

2:45 AM

Mathematician Proves Huge Result on ‘Dangerous’ Problem (Tao on Collatz!) / quanta mag

Semiclassical

for reference, this Tao's blog post on the paper: terrytao.wordpress.com/2019/09/10/…

2

vzn

3:08 AM

anyone around here looked at the proof? :)

Akiva Weinberger

I continue to have mixed feelings about Quanta

vzn

have not met many experts who are fans of pop sci, quite to the contrary... :|

Leaky Nun

3:24 AM

@AkivaWeinberger it's a necessary evil

J. Doe

3:48 AM

Hello

If $R$ is a division ring, then the only ideals of $R$ are $R$ and $\left\{0 \right\}$

Suppose $I \subseteq R$ is an ideal of $R$. Assume $I \ne \left\{0 \right\}$. Suppose $a \ne 0 \in I$ and $b \in R$. Then there's $x \in R$ s.t. $ax =xa= 1$. Thus $1 \in I$. Hence $1 \cdot b = b \cdot 1 = b\in I$. So $R \subseteq I$. Therefore $I = R$.

This is what the book said: suppose $I \subseteq R$ is an ideal of $R$. Assume $I \ne \left\{0 \right\}$. Let $a \ne 0 \in I$ and $b \in R$. Then $ax=b$ is solved in $R$, hence $b \in I$, thus $I=R$.

Could someone explain what "$ax = b$ is solved in $R$" means?

loch

4:16 AM

the book means that there exists $x$ such that $ax=b$

J. Doe

4:32 AM

Thanks.

Akiva Weinberger

5:08 AM

youtube.com/playlist?list=PLPljbRce3kzE2xd6kMgNf5MuqFHwL8gFc

Put together a sphere eversion playlist

Any video it's missing?

Akiva Weinberger

5:26 AM

Balarka probably tried to explain this to me at one point. I still understand nothing

> A jet is really nothing more than the collection of all of the low-order
derivatives of a function up to a certain point. For instance, the
two-jet of f(x), a function of one variable, can be represented
by the triple (f, df/dx, d^2f/dx^2)

Hm OK

(from here)

Soham Chowdhury

5:45 AM

does it really "matter" whether "x is a limit point of A" is defined to mean "every open contains a point of A" with or without the "other than x" condition?

Hatcher in his general topology notes omits the "other than x", and calls attention to it: p7 here

isn't his definition what goes by (iirc) "adherent point"?

and like i said does it matter at all for any applications (say, you know, if i'm doing algebraic topology with one definition vs another)

Akiva Weinberger

They differ at isolated points, if I remember right

I think one important sequence (which probably has a name) is $A,A',A'',\dots$

where $'$ is the set of limit points

Soham Chowdhury

huh wonder if that always stabilises

Akiva Weinberger

and you can define $A^{(\omega)}=A\cap A'\cap A''\dotsb$ and continue up the ordinal hierarchy

Dunno if that's the right notation but then $A^{(\omega)} '=A^{(\omega+1)}$

Soham Chowdhury

to me the simplest difference is whether a the range of a(n eventually) constant sequence has the constant it tends to as a limit point, which it does with Hatcher's definition and not otherwise

Akiva Weinberger

(dunno why the LaTeX isn't working)

The Great Duck

5:51 AM

0

Q: Algorithm to get the list of triangles resulting from clipping a triangle against a convex polygon.

We have a 2D triangle. It has 3 points. We have a 2D convex polygon. It has N points. Normally when clipping, one wants to produce the polygon or list of triangles inside the convex polygon. I want to do the opposite. I want to get the list of triangles that do not intersect aside from borderin...

Soham Chowdhury

i don't see what's wrong with the latex wtf

Akiva Weinberger

And if I remember right this was Cantor's original motivation for defining the ordinal hierarchy in the first place

The Great Duck

@AkivaWeinberger do you have the plugin enabled in your web browser?

Soham Chowdhury

interesting

Akiva Weinberger

I don't remember the exact statement this is used to prove

Sayan Chattopadhyay

5:52 AM

@SohamChowdhury Well atleast in functional analysis it does matter. To show that l^infinity is not metrizable or seperable, you need to use that the neighborhood intersects the set at points other than center of the neighborhood

Akiva Weinberger

@TheGreatDuck Long time no see. It's a link on the top right, remember?

The Great Duck

@AkivaWeinberger doesnt answer my question

Akiva Weinberger

@SohamChowdhury A limit point of a sequence is different from the limit point of a set

In a sequence you can have the same point multiple (or infinitely many) times

@TheGreatDuck I have the LaTeX renderer thingy yeah

The Great Duck

anyway

please help

Soham Chowdhury

yeah i'm talking about the set definition

The Great Duck

5:54 AM

been banging my head all night

uuurgh

Soham Chowdhury

range of the sequence not the sequence itself

Akiva Weinberger

I dunno specific examples where it matters

Soham Chowdhury

Akiva Weinberger

All my point-set topology knowledge is used in service of other-topology

LaTeX test

Soham Chowdhury

it doesn't really matter for other-topology is what i'm hoping

Akiva Weinberger

5:56 AM

$A^{(\omega)} '$

Leaky Nun

yeah that's double superscript

' is actually a superscript

Akiva Weinberger

$a^b'$

Soham Chowdhury

huh

Akiva Weinberger

${a^b}'$

$a^{b\prime}$

Huh

Leaky Nun

how about $(a^b)'$

Akiva Weinberger

5:57 AM

$A^{(\omega)\prime}$

${A^{(\omega)}}'$

I guess I have a choice

$A^{(\omega)}\prime$

Leaky Nun

this sequence can be used to show that every closed set in $\Bbb R$ is a countable set union a perfect set

Akiva Weinberger

Perfect

@LeakyNun That's what it was

Soham Chowdhury

the last one looks hideous

Leaky Nun

which is used to prove continuum hypothesis for closed subsets of $\Bbb R$

Akiva Weinberger

Perfect is like, Cantor set-y things and interval-y things?

Soham Chowdhury

5:58 AM

every point is a limit point

Akiva Weinberger

@SohamChowdhury ${A^{(\omega)}}_{\huge\prime}$

Soham Chowdhury

ahh my eyes

Leaky Nun

@AkivaWeinberger yeah

every non-empty perfect set contains a Cantor set

so that's $2^{\aleph_0}$

Akiva Weinberger

$\stackrel\prime{A^{(\omega)}}$

Soham Chowdhury

looks like a diacritic

Akiva Weinberger

6:00 AM

$Á^{(\omega)}$

Leaky Nun

$\stackrel{\prime}\prime$

$\stackrel\prime{\prime^\prime_\prime}$

${\stackrel\prime\prime}^\prime_\prime$

Akiva Weinberger

^It's raining

upside-down rain

(Raindrops are actually shaped like pancakes and parachutes but whatever)

Akiva Weinberger

6:46 AM

Guess what?

Online interactive sphere eversion demo! homepages.loria.fr/BLevy/GEOGRAM/geogram_demo_Evert.html

Akiva Weinberger

7:03 AM

Theoretically this works too (www3.math.tu-berlin.de/geometrie/lab/vr.shtml#sphereEversion) but I haven't been able to get it to work

oliver

8:39 AM

Good morning, I have found a funky definition in a text book and on wikipedia that a complex square matrix is always unitary

Am I misunderstanding this ? that doesn't seem right

Unitary would mean A*A = I, but I can think of so many complex square matrices where this isn't the case

Soham Chowdhury

9:13 AM

that can't be right

for instance, for a unitary matrix clearly det(A)^2 = 1

er, i mean |det(A)| = 1

Balarka Sen

@AkivaWeinberger Someone's been thinking about sphere eversion I see

I dunno if I ever gave you this puzzle but: Can you find a deformation of the graph of $z = x^2 + y^2$ over $\Bbb D^2 \setminus (0, 0)$ to the graph of $z = -(x^2 + y^2)$ over $\Bbb D^2 \setminus (0, 0)$ such that at no step in the deformation the surface is horizontal, i.e., some point of the surface has unit normal vertically pointing up.

Essentially everting a punctured hemisphere without ever making any part of the surface horizontal

@Akiva The "simple sphere eversion" is great. I always thought there should be some way to just to it by doing calculus on the singularity but never managed to figure out what's happening.

Somehow you have introduce a circle's worth of double points, and apparently you can do that by some magic on the cancelling pair of fold lines meeting at two cusps

Balarka Sen

9:53 AM

Wonder what I should talk about in an upcoming student colloquium

user

10:33 AM

Is the stereoraphic projection orientation reversing or preserving (from the Riemann sphere to the complex plane)... I have two different sources ( Gamelin "Complex analysis" and my course) contradicting each other on this.

Balarka Sen

That depends on what your choice of orientation on the Riemann sphere and the complex plane are, respectively.

user

Gamelin: "s.p. is orientation reversing, as a map from the sphere with orientation determined by the outer unit normal vector to the complex plane with the usual orientation"

Balarka Sen

That's correct

user

So if the inner unit normal is taken the s.p. is orientation preserving?

Balarka Sen

Yup

user

10:38 AM

Btw is there a convention in / standard choice of outer/inner unit normal or is this entirely text dependent...?

Thanks

Balarka Sen

The standard orientation on the sphere is understood to be determined by the outward pointing normals.

user

Thanks!

Mike Miller

Huh, I never observed that stereographic projection is orientation reversing. I see it now though, if I draw a little circle just east of the North Pole and project it, it flips upside down.

It's because 1/r is orientation reversing on R^+ I suppose

Balarka Sen

Yeah, it goes the other way wrt the usual chart maps of CP^1

oliver

@SohamChowdhury yes. But it says that any complex square matrix is always unitary. So that's not possible right?

Balarka Sen

10:53 AM

As Soham pointed out, "of course not". Who says that?

Secret

oliver

I think wikipedia

If U is a square, complex matrix, then the following conditions are equivalent:

U is unitary.

it really confuses me

Unitary matrix

In mathematics, a complex square matrix U is unitary if its conjugate transpose U∗ is also its inverse—that is, if U ∗ U = U U ∗ = I , {\displaystyle U^{*}U=UU^{*}=I,} where I is the identity matrix. In physics, especially in quantum mechanics, the Hermitian conjugate of a matrix is denoted by a dagger (†) and the equation above becomes U...

Balarka Sen

You're misinterpreting what "the following conditions are equivalent" means

Secret

You knew the problem is not simple when it shows pseudosymmetry in its graph

oliver

There is a textbook where I found the same phrase

Balarka Sen

10:54 AM

Yes, the phrase is fine.

oliver

oh @balar

@BalarkaSen thank you

I was thinking I'm misunderstanding this phrase

what does it mean?

ahhh

that if it's unitary it's got to be that ....

?

Balarka Sen

The following conditions are equivalent doesn't mean the condition is equivalent to the hypothesis, it means the conditions (a), (b), (c), ... as stated are equivalent

oliver

ahhhhh

OMG thank you

I thought I was completely off somewhere

Balarka Sen

Also your gravatar is broken

oliver

so just means that if it's complex and square to say a, is the same as b and c

not sure how to fix it

guess I just have to upload a new one

Silent

10:58 AM

I am reading about regular curves, and definition of that is: A differentiable curve is said to be regular if its derivative never vanishes. Wikipedia has this explanation about it: "In words, a regular curve never slows to a stop or backtracks on itself." I can't understand what 'backtracks on itself' means.

Balarka Sen

@MikeMiller Sent you a message elsewhere

@Silent Consider $f : [-1, 1] \to \Bbb R^3$, $f(t) = (t^2, 0, 0)$. This is trajectory of a particle travelling from $(1, 0, 0)$ to $(0, 0, 0)$ and then back again to $(1, 0, 0)$.

This is what backtracking on itself means; note how $f'(0) = (0, 0, 0)$.

Silent

@BalarkaSen I suspect that backwarding will always give a point where the derivative of curve vanishes, because of Darboux theorem, am i right?

Thanks for that example!

Balarka Sen

No, nothing to do with Darboux.

This is just Rolle's theorem

maths student

11:25 AM

Let the relation R = {(n, m) | n, m ∈ ℤ, ⌊n/4⌋ = ⌊m/4⌋}. is any of the set below a correct equivalence class of R? A) {2, 4, 6, 8}. B) {1, 3, 5, 7}. C) {4, 5, 6, 7}. D) {1, 2, 3, 4}.

I think answer is C but can somebody explain it?

Silent

@BalarkaSen I quoted Darboux because it seems like backwarding will reverse the direction of tangent, hence, component-wise, we can apply Darboux. Where am I wrong? (Thanks for quoting Rolle's btw)

oliver

Another dumb question if I may, how can a complex matrix be hermitian if it isn't square?

Silent

@oliver As far as i know, Hermitian matrix is square matrix!

oliver

11:41 AM

I thought so, I'm asked to show that an mxn hermitian matrix has rank n

i found some references

to mxn hermitian matrices where m != n

@silen

@Silent

i'm probably misunderstanding the question, happens to me a lot :-p

Silent

@oliver oh! then i dont know

oliver

it's ok I'll figure it out

the journey to AHA is always nice ! :-D

Silent

yeah! :)

Balarka Sen

12:13 PM

@Silent That's too complicated. Backwarding means $f(a) = f(b)$ happens, the curve is not injective.

In particular, that is. Backtracking is stronger than being non-injective of course

Silent

@BalarkaSen I was typing exactly this :)

Balarka Sen

But that's precisely why you can project and apply Rolle's theorem.

@LeakyNun @ÍgjøgnumMeg Cyclotomic extension are too complicated for me.

Silent

@BalarkaSen What do you mean by 'project'?

ÍgjøgnumMeg

@Balarka sem. I'm planning on reading Washington's book in full over the next 2 years lol

Balarka Sen

@Silent Formulate the question appropriately and try to think about it. I am not going to write it down

Silent

12:21 PM

ok

Balarka Sen

@ÍgjøgnumMeg So much number theory, I can't begin to start

It hurts my eyes

ÍgjøgnumMeg

hehe I'm interested in learning Iwasawa theory and I think Washington's book is the most basic reference for that

very very cool

Alessandro Codenotti

@ÍgjøgnumMeg have you heard back from the professor you asked to supervise your master's thesis?

ÍgjøgnumMeg

@Alessandro not yet :/

Balarka Sen

How do I even begin to understand $\Bbb Q(\zeta_n)/\Bbb Q$? Start by understanding what the minimal polynomial for $\zeta_n$, maybe? The $n$-th cyclotomic polynomial of whatever it is called.

All the automorphisms should send $\zeta_n$ to $\zeta_n^m$ where $(n, m) = 1$, preservation of order, etc

ÍgjøgnumMeg

12:28 PM

ye

Balarka Sen

Doesn't it directly imply $\Phi_n(x) = \prod_{(m, n) = 1} (x - \zeta_n^m)$ is irreducible? It's a minimal polynomial fam

Why does Dummit-Foote rant about it so much

ÍgjøgnumMeg

no idea lel

cyclotomic extensions are the setting for a pretty cool partial proof of FLT

Balarka Sen

You'll end up asking when $\Bbb Z[\zeta_p] = \mathcal{O}_{\Bbb Q(\zeta_p)}$ is class number 1, I guess.

Kummer primes etc

ÍgjøgnumMeg

well, in that proof all that matters is that there's no $p$-torsion in the class group of the $p$-th cyclotomic field

Balarka Sen

Also the proof of that $\Bbb Z[\zeta_p] = \mathcal{O}_{\Bbb Q(\zeta_p)}$ thing is a series of magic tricks from what I gathered by attending a lecture on an ANT class

How do number theorists do it

ÍgjøgnumMeg

12:32 PM

I haven't seen it rofl

Balarka Sen

Its soul crushing

ÍgjøgnumMeg

i think Milne's notes have a proof

but I don't remember it

Balarka Sen

i 'followed' the proof when i was attending the class but thats about it

number theory is seriously corrupting

ÍgjøgnumMeg

hahaha it's pretty dutty

Balarka Sen

i hate number theory, fuck y'all

Alessandro Codenotti

12:35 PM

Go back to topology @Balarka you walked the dark path but you can still redeem yourself

Balarka Sen

but i have no topology friends

Alessandro Codenotti

Time to find new friends I guess

ÍgjøgnumMeg

lol Milne shows that $\mathcal{O}_K = \Bbb Z[\zeta] + p^m\mathcal{O}_K$ and that for large enough $m$, $p^m\mathcal{O}_K \subset \Bbb Z[\zeta]$

Balarka Sen

i dont want to be the lonely wolf

@ÍgjøgnumMeg yeah something like that, it's horrendous

ÍgjøgnumMeg

It is expected that 61% of primes are
regular and 39% are irregular

fun

Silent

12:41 PM

@BalarkaSen, it seems to me that the issue is that even though Rolle guarantees that there is a point at which derivative of each component vanishes, but they may be distinct points for different components.

Balarka Sen

@Silent but you can "zero in" on a common point where derivatives of all three component vanishes because the value of $f$ matches arbitrarily close to the point where the curve turns and starts backtracking on itself. that's why i told you to make it mathematically precise what you want "backtracking" to mean

Silent

ok.

Balarka Sen

@Alessandro

dsm

12:58 PM

can you guys make sense out of that problem?

I mean, it seems like $\tilde{g}^{\mu\nu} \equiv \tilde{g}(e^\mu,e^\nu) = g(e_\mu,e_\nu) = g_{\mu\nu}$, but that's not right.

self-study, not homework

Silent

$\phi:[a,b]\to\Bbb R^n$ with $c,d\in(a,b)$ and $\phi(d)\in\phi[a,c)$. So $\phi(d)=\phi(x)$ for some $x\in [a,c)$, and $\phi$ has this property that $\phi(t(s))=\phi(c-s)$ for $s\in[c,d]$ where $t:[c,d]\to\Bbb R$ is a continuous function. Is that correct way of defining backtracking?

@BalarkaSen

Adam

1:27 PM

@TedE no I'm not offended, I just prefer criticism to be constructive, and yours was not

if it does have substance, then I apologise and will expect you to answer one of the questions you deem basic

ÍgjøgnumMeg

you sound kinda salty tbf

Adam

what does salty mean?

anakhro

Almost contemptuous.

But to be fair, TedE was constructive. He did make a suggestion for how you could improve.

Adam

No it's just that if any of the questions ive posted that have not been answered are seen as basic by someone then obviously I want that person to answer them, that's only fair,right?

anakhro

I don't know if that is "fair".

When a student asks me a question which I think they have not really thought about, I usually try to encourage them to look into answering it themselves.

I think that is fairest in those particular cases.

It would be unfair to deprive them of the learning experience that comes with struggling with their own questions.

Adam

1:43 PM

Based on your logic no question posted on stack exchange should ever be answered

anakhro

I think you misunderstand if you think that's the logic that follows.

But I do agree with the statement that majority of answers on stackexchange exhibit poor pedagogy.

Often times users merely spoonfeed answers to people.

Instead, they should be giving hints and motivation for the questioners to continue working on it on their own.

Adam

you suggested that Ted is able to easily answer my questions, but does not do so because he doesn't want to deprive me of the learning experience, yes?

anakhro

No, I am saying that it doesn't follow that spoonfeeding answers is the "fair" thing to do.

Whether TedE did or did not do this, I am not sure. I only scrolled up so far.

What TedE might want to do is more clearly exhibit this fact (if he agrees with it) by giving you appropriate hints.

ÍgjøgnumMeg

"I am salty because the conservatives won a huge majority in the election" is a possible use

anakhro

And to compliment this process, you should perhaps ask for hints if you know the question has a known solution, or ask how you could approach a question that you are unsure has an answer.

This way you exhibit a willingness to learn properly (through working yourself, perhaps with some help from others) and TedE would exhibit a willingness to teach properly (through encouraging independence and helping when you actually need it).

Balarka Sen

1:49 PM

@Silent Looks a little clunky to me. Would it be more accurate to say it's a curve $\phi : I \to \Bbb R^n$ such that you can restrict to a subinterval $[a, b] \subset I$ with a point $c \in (a, b)$ such that exist reparametrizations $\gamma_1, \gamma_2 : [0, 1] \to [a, c], [c, b]$ such that $\phi(\gamma_1(t)) = \phi(\gamma_2(1-t))$?

@ÍgjøgnumMeg "I feel salty because I had too much salt with my morning sandwich"

I'm So Meta, Even This Acronym

(read the first letter in each word)

ÍgjøgnumMeg

yis

I am going to be salty soon because I just ordered a fucking giant döner to deal with my horrendous hangover

Adam

I don't get them

anakhro

@BalarkaSen I am going to start looking at the Mishachev + Eliashberg book

Balarka Sen

Awyeah