Mathematics - 2019-12-16 (page 4 of 4)

@TedShifrin I figured out how to derive the functional equation for that one ? I'm on about. Or, at least, I know formally how to do so---I strongly suspect it can be made rigorous, but I definitely don't know how to justify it.

It comes out like this. I have a mapping $g:\mathbb{R}^2\to \mathbb{R}^2$. I want to find $(x,y)$ such that iterating $g$ converges to the (unstable) fixed point at the origin.

Kari

That reminds me of the contraction mapping theorem!

(I think Ted's gone out for some lunch btw)

Semiclassical

oh, I'm sure

It seems that, for the present problem, there's a one-to-one mapping between such $x$ and $y$. So let $y=f(x)$ be that mapping. The asker's method amounts to assuming that, if $g(x,y)=(x',y')$, then $y'=f(x')$

So $g$ preserves the functional relation between $x$ and $y$.

Kari

Now that reminds me of the Implicit Function Theorem

Wow, it's been a while since I've done any maths

Semiclassical

I don't think that's justified in general. But I think the requirement that $(x,y)$ iterate to zero might do it?

I'm sure I don't know how to make this rigorous tho

Astyx

8:15 PM

Everytime I come to this chat Semi is talking about me

Semiclassical

lol

not my fault this time

Attempt at formulating that better: Let $h(x)=(x,f(x))$. Then the claim is $g(x,f(x)) = (g\circ h)(x) = (x',f(x'))=h(x')$

still not happy with that

I feel like I need a function mapping x -> x' for this to make sense.

Thorgott

yeah, I don't quite follow that set-up

Semiclassical

I mean, I don't either. That's why I say I understand it formally

It feels like it's missing some ingredient to justify it

I also shouldn't be using $h$, since the author of the question I have in mind already uses it

I'll switch to $q$, because why not

Kari

Do you have a link to the original problem/question, @Semiclassical?

Thorgott

it sounds like you want some kind of uniqueness assumption

Semiclassical

8:24 PM

9

Q: Solve functional equation $ h(y)+h^{-1}(y)=2y+y^2 $

I was trying to solve a certain physics problem, and encountered the functional equation that contains a function $h$ and its inverse $h^{-1}$: \begin{equation} h(y)+h^{-1}(y)=2y+y^2.\tag{1} \end{equation} Q: Does $(1)$ have a unique solution and is it possible to find it in closed form? Eq...

@Thorgott yeah. I mean, the assumption that g iterates (x,y) to (0,0) is very strong in this case

so that is presumably a huge part of this

Thorgott

what is $g$ in this question?

Kari

Ew physics

Semiclassical

@Kari lol

i deal with all the physics at the top

@Thorgott $g(x,y) = (x-y,y-(x-y)^2)$

so nonlinear

Thorgott

ok, that does look un-nice

Semiclassical

So in this case we're assuming a solution of the form $y=f(x)$ and $y-(x-y)^2 = f(x-y)$.

from some numerical work, it actually looks like it works when you choose (x,y) converging to (0,0)---a rather delicate task, since any finite-precision input will ultimately not converge

If I could find $H(x,y)$ such that $H\circ g=H$, then I think this would be easy

Kari

8:36 PM

So you're looking for a $H$ such that $H(x-y, y-(x-y)^2) = H(x,y)$.

Hey, @MikeMiller!

Semiclassical

Right.

I'm rather sure it's not polynomial H, alas

Kari

Is there any restriction on the kinda $H$ you'd like to find?

Semiclassical

not really, no

I think I may see a physics-y approach, though

and as I think about the physics approach, it starts going down renormalization-group territory...so that's 'neat'

nm, physics approach recovers what I was already doing.

Leaky Nun

8:59 PM

@BalarkaSen is there a simply connected 4-manifold with $H_2 = \Bbb Z^3$? I couldn't do this on the exam

Semiclassical

Is double torus Z^4?

(as in, a sphere with two handles)

Leaky Nun

that's $H_0 = \Bbb Z$, $H_1 = \Bbb Z^3$, $H_2 = \Bbb Z$

Semiclassical

oh, second homology group

okay, I think see the physics justification of that question

Leaky Nun

what is it?

Semiclassical

the physics problem is linked earlier in the history ("solve functional equation...")

Leaky Nun

9:06 PM

oh I thought you were talking about my question

Semiclassical

and I think the interpretation should go like this: Suppose I draw a circle around every resistor and nonlinear element except for the first pair

then what I have should again be an infinite chain, just with a different voltage and current

but it's the same chain, so the relationship between voltage and current should be the same regardless

hence $j_k=f(u_k)$ regardless of $k$.

Balarka Sen

@LeakyNun What about $3\Bbb{CP}^2$

Leaky Nun

what is that

simply connected?

Balarka Sen

Connected sum of three CP^2's

Leaky Nun

does that work?

I don't know anything about connected sum

Balarka Sen

9:09 PM

Just Mayer-Vietoris it out

Leaky Nun

suspicious

I guess it works

Balarka Sen

$H_2(S^3) \to H_2(M) \oplus H_2(N) \to H_2(M \# N) \to H_1(S^3)$

No problemo

Leaky Nun

cool

are $S^2 \vee S^4$ and $\Bbb CP^2$ homotopy equivalent?

Balarka Sen

No

The Hopf map is not null

Leaky Nun

I guessed correctly

yeah but that's the wrong direction

Balarka Sen

9:11 PM

Essentially the same idea, cup square isn't 0

This is the Hopf invariant

Leaky Nun

oh, the cup product!

hey @loch

loch

Hi

Leaky Nun

@BalarkaSen by "cup square" did you mean the cup product?

Balarka Sen

Cup suqare of the unique degree 2 generator

That's zero in the former, nonzero in the latter

Leaky Nun

aha

understood

Balarka Sen

9:14 PM

Even $\Sigma \Bbb{CP}^2$ and $\Sigma (S^2 \vee S^4)$ are not homotopy equivalent I believe but this should be harder to see.

We'd have to use the Pontryagin square

Leaky Nun

hmm

Balarka Sen

@MikeMiller knows this cold

On the other hand, $S^1 \times S^2$ and $S^1 \vee S^2 \vee S^3$ are not homotopy equivalent but become homotopy equivalent once you suspend it once. The attatching map is just the lollipop map $S^2 \to S^2 \vee S^1$, which is nullhomotopic if you suspend it once.

This was pointed out to me by Dennis Sullivan once in his only appearance in MSE I think

Leaky Nun

maths is weird

@BalarkaSen have you played choker?

Balarka Sen

haha no I can't play poker

Leaky Nun

me neither

Balarka Sen

9:22 PM

i saw that video recommended to me

Leaky Nun

but it's fun

Semiclassical

well. i don't know how to answer the question that I've been on about, but I do finally understand -why- that's the question. so...progress?

Ultradark

$\frac{1}{f(x)} = f\left( \frac{1}{x} \right)$

Balarka Sen

@LeakyNun Regarding earlier example: It's exactly analogous to $S^1 \times S^1$ and $S^1 \vee S^1 \vee S^2$ not being homotopy equivalent, but after suspension being so. The commutator map $S^1 \to S^1 \vee S^1$ becomes nullhomotopic once you suspend it.

Leaky Nun

cool

Balarka Sen

9:36 PM

Essentially since $\Sigma(S^1 \vee S^1) \simeq S^2 \vee S^2$ which has abelian $\pi_2$. There is no such thing as a commutator

Leaky Nun

aha

Balarka Sen

I hadn't actually noticed $\Sigma T^2$ is $S^2 \vee S^2 \vee S^3$ before. Shame on me.

Leaky Nun

that's a nice decomposition

aha

so stable homotopy theory is all about decomposition into wedges :P

Balarka Sen

Yup

Stable homotopy class being zero would tell you an appropriate cell complex breaks up into a wedge of spheres after suspending for a long time

Leaky Nun

reasonable

Balarka Sen

9:43 PM

@BalarkaSen Oh I think Hatcher has a general exercise here which says if $X$ is union of $n$ contractible spaces then cup of any $n$ element dies.

Ultradark

Say I want to visualise a transformation matrix acting on some points in the plane. What's a good way to do this?

Balarka Sen

So $\Sigma X$ always has trivial ring structure, eh

That's why it's hard to see $\Sigma \Bbb{CP}^2$ and $\Sigma S^2 \vee S^4$ are distinct I think

Ring structure tells you nothing, you have to go to higher cohomology operations

Anyhow Mike told me something about Steenrod squares from time immemorial but I never read this story

Leaky Nun

maybe I'll study it next term

Balarka Sen

Teach it to me if you do

Leaky Nun

sure

Khallil

9:49 PM

@Ultradark map a shape that you already know under the matrix?

I think there's also a way of decomposing a matrix into a composition of basic operations (like translation, rotation and scaling)

Akiva Weinberger

Long rambly answer to what's really a simple question! Thoughts?

0

A: Solving $f''(x) = -f(x)$ without using Euler's formula

Say you've computed $f$ numerically (using, say, Euler's method). That is, you don't know a formula for $f$, but you know some values of $f$ to some decimal places. If you graph $f$ you'll see a wave, but let's say you don't recognize that graph. One thing we can do is plot the configuration spa...

D. Zack Garza

Q: For $F$ be a free $R$-module, is the functor $X \mapsto X \otimes_R F$ both left and right exact?

Balarka Sen

Yup

Semiclassical

@AkivaWeinberger the usual way I do this is sorta the reverse of what you do: Multiply $f''(x)+f(x)=0$ by $f'(x)$ and note that what you get is a total derivative

and therefore $f'(x)^2/2+f(x)^2/2$ is constant.

Akiva Weinberger

On a sort of philosophical level, this is similar to iterated distillation and amplification

Semiclassical

10:01 PM

Moreover, that's got a nice physics meaning: $ma=mx''(t)=-k x(t)$ is the equation of motion of a harmonic oscillator, whereas its first integral $\frac12 mv^2+\frac12 kx^2$ is just the total mechanical energy

which, is conserved

Akiva Weinberger

(Re: distillation/amplification) We learn from computation, which is slow but is better at revealing patterns

Those patterns then are "distilled" into faster tools

and then we can "amplify" again by using those tools in our new slow computations

Semiclassical

the main reason to do it this way, though, is that it works for other equations of motion where it's now anywhere near as obvious what to do

for instance, try 'guessing' the solution for $f''(x) = -\sin f(x)$

(good f'ing luck)

Akiva Weinberger

Is that actually doable?

Semiclassical

yep. you get elliptic functions iirc

in brief:

Akiva Weinberger

Oh so it's not, like, elementary

Semiclassical

10:04 PM

definitely not

If you take a first integral, you get $f'(x)^2/2 = \cos f(x)+C$

typically you'd pick $x=0$ so that $f'(x)=0$, so $C=-\cos f(0)$

therefore $f'(x)=\pm \sqrt{2\cos f(x)-2\cos f(0}}$

Akiva Weinberger

> The philosophy here is iterated distillation and amplification. Computation is slow, but it "amplifies" our powers because it lets us see new patterns. We then "distill" those patterns into new tools in our toolbox. Now we can iterate the process again, using those new tools in our computations. This is why mathematics has gotten so far.

Posted this as a comment

Semiclassical

which, annoying $\pm$ aside, is solved as $$x = \int_{f(0)}^{f(x)} \frac{df'}{\sqrt{2\cos f-2\cos f_0}$$

Shine On You Crazy Diamond

@Ultradark

Semiclassical

if you do a bunch of finagling, that's an elliptic integral

Ultradark

Is it okay to say that a transformation matrix "acts" on the space of points in the first quadrant of the two dimensional euclidean plane?

Akiva Weinberger

10:07 PM

(Re: What I just posted) 'Cause, like, I believe it. Euler wouldn't have solved the Basel problem if he didn't use tools like Bernoulli summation (IIRC), and he wouldn't have had tools like Bernoulli summation had no one done the annoying computations that derive it

Ted Shifrin

@Leaky: Are you at MIT for spring as well?

Akiva Weinberger

Sorry I'll read what you wrote now

Semiclassical

and if you invert it to get $f(x)$ in terms of $x$, you get an elliptic function solution

Akiva Weinberger

Your MathJax is broken

Ted Shifrin

It is?

Akiva Weinberger

10:08 PM

$$x = \int_{f(0)}^{f(x)} \frac{df'}{\sqrt{2\cos f-2\cos f_0}}$$

Fixed

Ted Shifrin

Oh, Semi's mathjax.

Semiclassical

derp

I'm on a computer where it'd be inconvenient to enable mathjax atm

Akiva Weinberger

Yeah I see it

Semiclassical

so I probably borked it somewhere

Akiva Weinberger

re: integral and also computer troubles

Although, since it's an elliptic integral, maybe you can geometrically related it to an ellipse somehow?

Ted Shifrin

10:10 PM

I'm glad to see your proselytizing for the energy trick, @Semiclassic.

Akiva Weinberger

I mean probably not

Leaky Nun

@TedShifrin yes

Ted Shifrin

@LeakyNun Very cool. I trust you've enjoyed it.

Leaky Nun

sure

Akiva Weinberger

In any case I kinda just wanted to put that philosophical idea out there (albeit in a place where no one will ever read it)

Ted Shifrin

10:11 PM

I don't know if you ever said hi for me to various people.

Semiclassical

@TedShifrin it's an oldie but a goody

Balarka Sen

@Leaky These Naka vs Duda games are next level

Semiclassical

notably, the use of elliptic functions in this way goes all the way back to Legendre

so it's definitely classical math

Leaky Nun

@BalarkaSen I haven't watched that

Semiclassical

it also links up to Gauss's arithmetic-geometric mean, which is all kinds of great

Leaky Nun

10:15 PM

@BalarkaSen have you watched Hansen vs Carlsen?

Ted Shifrin

a @Balarka: Does this interest you?

Balarka Sen

@Leaky Yeah those are good

Eric is actually surprisingly good

Leaky Nun

does PL interest Balarka?

Balarka Sen

@TedShifrin Hm, yes

I would be surprised if the stated result in the title is true

Ted Shifrin

I think the OP would be, too, based on his last sentence.

Akiva Weinberger

10:21 PM

@Semiclassical Is that where you iterate $(a,b)\mapsto({\rm AM}(a,b),{\rm GM}(a,b))$ 'til they converge

Leaky Nun

why should "maximal number of points in a fiber" be the degree?

Semiclassical

Yep

Akiva Weinberger

and through magic find an integral that tells you what the limit is?

Semiclassical

Pretty much

Balarka Sen

@LeakyNun Because it's going to be constant along the top stratum

Akiva Weinberger

10:22 PM

That was actually an extra credit on one of the P-sets we had a while back

Balarka Sen

It's a PL map

Akiva Weinberger

And by "extra credit" I mean "for fun" - I don't think he gave actual points for them

I didn't think very hard about it

No wait

There was an actual question (not "for fun") which was, show it converges

Semiclassical

In the physics context, the agm gives you a fast way to compute the frequency of a mathematical pendulum for a given amplitude

Akiva Weinberger

and the "for fun" was to find what it converges to

@Semiclassical Oh so it's a neat algorithm for that

Cool

Semiclassical

For small amplitudes, you have agm of 1 and 1-epsilon, so basically just 1

Hence you get small amplitudes having the same oscillation frequency

Akiva Weinberger

10:26 PM

I see

Balarka Sen

@TedShifrin I was thinking that maybe you could compose the map $T^4 \to \Bbb{RP}^4$ with the 2-sheeted covering map $T^4 \to T^4$. Then the composition $T^4 \to T^4 \to \Bbb{RP}^4$ is certainly zero on $\pi_1$ because $\pi_1 \Bbb{RP}^4$ is $2$-torsion, which then lifts to $T^4 \to S^4$.

Akiva Weinberger

A more sophisticated version of the $\sin\theta\approx\theta$ argument

Semiclassical

If you do large amplitudes, you take agm of 1 and a number closer to 0

Leaky Nun

$\sin \theta = \theta = \tan \theta$

Semiclassical

So you get a smaller frequency (longer period) at higher amplitudes

Balarka Sen

10:27 PM

Oh no, it won't be zero on $\pi_1$.

ÍgjøgnumMeg

@Leaky fact

Akiva Weinberger

@LeakyNun Those are all odd functions

Semiclassical

If you invert a pendulum perfectly, that’s technically agm of 0 and 1 which converges to zero

Akiva Weinberger

which means they all equal 1 mod 2

Ted Shifrin

@Balarka: I get it on $H_4$, not $\pi_1$.

Akiva Weinberger

10:28 PM

thus they're almost equal QED

Balarka Sen

If I took the $16$-sheeted cover $T^4 \to T^4$ corresponding to the subgroup $(2\Bbb Z)^4$ then we can guarantee that.

ÍgjøgnumMeg

fields medal awarded

Semiclassical

Corresponding to the fact that an inverted pendulum is an (unstable) equilibrium

Akiva Weinberger

@Semiclassical I see

johnny09

0

Q: How to show that $\lim_{n\to\infty}\mathbb{P}\left(\bigg|\frac{1}{n}S_n-f(n)\bigg|>\varepsilon\right)=0,\quad\forall\varepsilon>0$

Consider a collection of independent random variables $(X_i)$ with $\mathbb{I}_{X_i}$ being the indicator random variable for $X_i$. Let $$f(n)=\frac{1}{n}\sum_{i=1}^{n}\mathbb{P}(X_i)\quad\text{and}\quad S_n=\sum_{i=1}^{n}\mathbb{I}_{X_i}.$$ I'm interested in showing that $$\lim_{n\to\infty}\m...

probability probability-theory solution-verification

Ted Shifrin

10:30 PM

@Balarka: The interesting thing to me is where you're going to invoke PL. A non-PL map has to be nuts.

Akiva Weinberger

@ÍgjøgnumMeg Woot

johnny09

any help would be greatly appreciated!

Balarka Sen

@TedShifrin I assume you just use the result that a PL branched covering T^4 -> S^4 is at least degree 4

I don't think I know how to prove such a thing at all

Classic Edmonds stuff

Leaky Nun

@TedShifrin the true/false part has a lot more true than false

Ted Shifrin

I know none of this.

Thorgott

10:35 PM

What's $\mathbb{P}(X_i)$ or $\mathbb{I}_{X_i}$?

Ted Shifrin

probability of event $X_i$ and indicator function for same event.

Thorgott

but $X_i$ is supposed to be a r.v.

hence I'm confused

Ted Shifrin

Ah, maybe he means $\mathbb E$?

I apologize for answering without reading.

Thorgott

no problem

Balarka Sen

OK, no: if $\pi_* f : \pi_1 T^4 \to \pi_1 \Bbb{RP}^4$ is nonzero then just take the cover corresponding to that index 2 subgroup. That is also going to be a torus, so we have a degree 2 covering map $T^4 \to T^4$ such that composing with $f$ gives $T^4 \to \Bbb{RP}^4$ which induces zero on $\pi_*$

Ted Shifrin

10:39 PM

There you go :)

Balarka Sen

It only tells us that degree of $f : T^4 \to \Bbb{RP}^4$ is at most $4$ though

Ted Shifrin

degree doesn't actually make sense ...

Balarka Sen

Because the lift $T^4 \to S^4$ of the composed map has degree at most $4$, so pushing down $T^4 \to \Bbb{RP}^4$ has at most $8$ but this is composition of the original $f$ with a degree 2 cover of $T^4$, so back to $8/2 = 4$ again

Ted Shifrin

When you have a non-orientable thing, I don't see how actual degree is well-defined.

Counting number of preimage points is certainly not a homotopy invariant.

Balarka Sen

Yeah, you have a good point.

Ted Shifrin

10:43 PM

I was just being confuzled.

Balarka Sen

This is some oddball thing about maximum fiber cardinality. I was going to argue why that agrees with the usual notion of homology degree for a PL map $f : M \to N$ of $n$-dimensional compact closed orientable PL-manifolds.

Akiva Weinberger

Mod 2?

Balarka Sen

Yes, but that's not what these guys are doing.

Khallil

@BalarkaSen What does 'That is it say, my left isn't right' mean?

Seems like such a broken sentence :0

Leaky Nun

that is to say

Balarka Sen

10:47 PM

Bad joke, I was wondering if a certain thing I said was left exact is indeed left exact and not right exact

Khallil

Ahhhh

Balarka Sen

"Am I right that it is left? So left is right? Wait, but that means left isn't right"

Ted Shifrin

oh oh ... we've had a name transmogrification.

Khallil

Ah it's finally changed!

It actually did that on its own for some reason when I edited my bio.

I'm technology illiterate at this point.

Mike Miller

What do you need

Ted Shifrin

10:53 PM

Heya @MikeM

Mike Miller

Hi

Ted Shifrin

I brought this to Balarka's attention, Mike.

Balarka Sen

I tried a bit but the best that comes from using the theorem directly is degree of any PL branched cover $T^4 \to \Bbb{RP}^4$ is at most $4$, I think.

Dunno how to push for $8$

Mike Miller

What a bizarre question

Ted Shifrin

Not remotely my bailiwick

Balarka Sen

11:04 PM

Edmonds proves if $p : M \to N$ is a branched covering map then $\deg p \cdot l(N) \geq l(M)$ where $l(-)$ is the cup-length

Ted Shifrin

Do we know what degree means yet in the non-orientable case?

anakhro

is there a degree when it is non-orientable?

Ted Shifrin

@anakhro: ordinarily, only mod 2.

anakhro

I like the mod 2 intersection theory.

That was my favourite part of G&P.

Mike Miller

Yes for PL branched covers it's the maximum cardinality of some fiber, as given in the question! This satisfies the usual multiplicativity properties

Ted Shifrin

11:06 PM

I was worried that it wouldn't be a homotopy invariant.

Mike Miller

Probably not, but do we need that?

Ted Shifrin

Ah, but with PL things are rigid.

Balarka Sen

The issue is it's hard to do any top homology argument with it

@TedShifrin It's constant on top stratum, like I pointed out earlier

Ted Shifrin

Yes, right, I get it, @Balarka. Thanks.

Well, assuming the top stratum is connected :P

anakhro

Does PL always mean piecewise linear?

Ted Shifrin

11:07 PM

yes @anakhro.

anakhro

Great, I am understanding.

Time to cook.

Ted Shifrin

Hmm, I'm not cooking for hours ...

anakhro

It's 6:09 here.

So I am a little late.

Ted Shifrin

Yup.

Mike Miller

It's also a concordance invariant of branched covers, which is probably a good notion of homotopy. A concordance is a branched cover over the cylinder.

Balarka Sen

11:10 PM

Can max fiber cardinality not reduce during concordance

Reduce by 2 say

I don't see why

Mike Miller

Branching should occur over a codim 2 subcomplex, no? Even in a concordance. So the non-branched set is connected.

Balarka Sen

The boundary of the cylinders are not part of the top stratum anymore, so the fiber cardinality their need not stay constant, is my thought process. They could be "edge cases" where the fiber cardinality is very high say

Oh ok, codim 2 is the thing I missed.

Top stratum of boundary is codimension 1 in the cylinder

Ted Shifrin

Clever.

Hey, what ever became of PVAL?

Mike Miller

Think quite sick, don't know much or mean to talk about what I know publicly. Meant to email him.

Ted Shifrin

Please pass on my fond wishes.

Balarka Sen

11:18 PM

Hope he gets better soon

anakhro

He knew about contact geometry, he was pretty cool.

Ted Shifrin

I feel like a T S Eliot poem here.

skullpetrol

"The Waste Land"?

Balarka Sen

Gerontion? :)

Ted Shifrin

Love Song of J Alfred Prufrock

"People come and go, talking of Michelangelo ..."

Balarka Sen

11:25 PM

Aha

Although it was "In the room the women come and go, talking of Michelangelo" and it's probably a poem about sexual frustration but I got the gist :P

Love Song has always been a bit too sentimental for my taste

Ted Shifrin

LOL, you're totally right. I just flashed on the "come and go" ... I read it 50 years ago.

I should reread it now.

anakhro

who is your fave poet, guys?

skullpetrol

Don't cry
Don't raise your eye
It's only teenage wasteland

Balarka Sen

You might like Gerontion, try it out if you get chance

anakhro

@skullpetrol Who did that?

;)

skullpetrol

11:28 PM

:D

Ted Shifrin

I took a course on Baudelaire and Nerval poetry (in French) my freshman year in college, so that was quite an awakening.

Balarka Sen

Oh nice

anakhro

ee cummings is my top poet

Balarka Sen

I have read very little Baudelaire

@anakhro bracket open

Ted Shifrin

I liked ee cummings, but didn't study that much.

anakhro

11:29 PM

Heh.

Balarka Sen

I think that's literally a title of one of his poems

"("

Ultradark

I like this one

anakhro

He's pretty creative with his grammatical idiosyncrasies.

Balarka Sen

you cant beat Marinetti though

these weights thicknesses sounds smells molecular whirlwinds chains nets and channels of analogies concurrences and synchronisms for my Futurist friends poets painters and musicians zang-tumb-tumb-zang-zang-tuuumb tatatatatatatata picpacpampacpacpicpampampac uuuuuuuuuuuuuuu

from "Zang Tumb Tumb"

anakhro

Cardamom smells so good.

That's a playful one, @BalarkaSen

Balarka Sen

11:34 PM

It's by the originator of fascism

zang tumb tumb stands for bombs dropping lol

anakhro

Surprised there isn't a whistle there.

Do you like cilantro, @BalarkaSen?

Balarka Sen

Eh, it's ok. We curry eaters use it for curries

anakhro

I am making a saag paneer right now.

But not paneer, since I like using feta instead.

Balarka Sen

Oh nice

I like that

anakhro

So I guess saag feta.

Balarka Sen

11:42 PM

paneer with spinach is A+ stuff

why going indian suddenly

Ted Shifrin

I use cilantro in all sorts of Latin and Asian cooking ...

Ultradark

$$H=\exp\begin{pmatrix} 0 & s \\ s & 0\end{pmatrix} = \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s\end{pmatrix}$$

Does $H$ act on points to make them flow along curves like $xy=C$ ?

I know that this is a hyperbolic rotation matrix

anakhro

@BalarkaSen I've always enjoyed cooking

And cooking one indian recipe is not "going indian". ;)

Balarka Sen

Fair :P

anakhro

My fave spice mix is probably lebanese 7 spice

Ted Shifrin

11:49 PM

@Ultra: Start at the point $(1,0)$ and answer your own question.

Ultradark

@TedShifrin forgot to say that $x\ne 0$

Ted Shifrin

Huh? I don't care.

Ultradark

So I get this: $x\mapsto \begin{pmatrix} \cosh s & \sinh s \\ \sinh s & \cosh s \end{pmatrix} \begin{pmatrix} x_1\\ x_2\end{pmatrix} = \begin{pmatrix} x_1 \cosh s+x_2\sinh s\\ x_1 \sinh s+x_2\cosh s\end{pmatrix}$

I'm going to say: yes?

So I'm getting $(\cosh(s),\sinh(s))$

using $x_1=1$ and $x_2=0$

I'm going to say no actually

they don't flow along rectangular hyperbolas

rather rotationally transformed rectangular hyperbolas

so that the hyperbolas are in standard form

Ted Shifrin

The whole point of hyperbolic geometry and cosh and sinh is the equation $\cosh^2 s - \sinh^2 s = 1$.

So you get the parametrization of the hyperbola $x^2-y^2=1$. Now do the orbit of a general point.

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