Mathematics - 2017-06-11 (page 1 of 9)

TheGreatDuck

00:00

@amWhy is there a reason you booted me from constructive feedback? Is it something I did last thursday?

not that it matters anyway since I am now permanently chat banned, but i was just curious.

pjs36

@LegionMammal978 I think the question is fine here, it's a chatroom for crying out loud. According to Wikipedia it's dimensionless. So maybe it's analogous to writing $1 \text{doz} = 12$.

Daminark

@pjs36 exactly

LegionMammal978

@pjs36 Okay, was just trying to apply linear algebra to SI derived units, so I guess I only need 6-dimensional vectors

Simply Beautiful Art

0

A: How to solve $x\sin x=a$ for any number $a$?

If $x=f(a)$ is solved for $a\in\mathbb Q$ and $f(a)$ is continuous, then $x=f(a)$ for $a\notin\mathbb Q$. Next, we see that if $a\ne0,a\in\mathbb Q$, then $x$ is not algebraic. If $x$ were algebraic, then $\sin(x)$ would be transcendental, and thus $x\sin(x)$ would be transcendental, a contradi...

So I sorta almost solved the supposedly unsolvable equation $x\sin(x)=a$.

Assuming $a$ is real, one can write the solution as $$x=-iW_k(bi-a)$$

Where $b$ is some real number such that $x\cos(x)=b$

And I'm putting this out there for people to think about

Simply Beautiful Art

01:00

in Simply Beautiful Art's realm of calculus and analysis, Jun 8 at 0:17, by Simply Beautiful Art

@Frpzzd $$\int\lfloor x\rfloor~\mathrm dx=\frac{\lfloor x\rfloor(\lfloor x\rfloor-1)}2+\lfloor x\rfloor(x-\lfloor x\rfloor)+C$$

nitsua60

@TheGreatDuck One would think that if you were trying not to annoy one who said thay'd chat-banned you, but forgot to do so, that you'd stop mentioning that you're in chat despite beind told (though not forced) to leave.

Simply Beautiful Art

@Daminark @Dodsy @TheGreatDuck Actually, drinking isn't all bad, as xkcd claims.

Simply Beautiful Art

01:27

Hey @AkivaWeinberger any idea about the $x\sin(x)=a$ problem?

Akiva Weinberger

My instinct tells me there's no elementary solution, though special functions may help.

Simply Beautiful Art

Well I tried solving it in terms of the Lambert W function if you read above a bit

Akiva Weinberger

@SimplyBeautifulArt By the way, try graphing $x\sin x$, $~-x^2\cos x$, and $-\sin x\tan x$.

(Desmos lets you show not all of them at once)

The intersections between $x\sin x$ and either of the other two give you its local maxima and minima.

(Well, the intersections between $x\sin x$ and $-\sin x\tan x$ are the local maxima/mínima as well as the roots)

Simply Beautiful Art

Oh, that's nifty

But I mean...

Not so surprising tbh

:P

Akiva Weinberger

Yeah, but it looks kinda cool

Simply Beautiful Art

01:34

It does

Akiva Weinberger

I think I told you about $(\sin x)/x$ and $\cos x$ before

which is more elegant

(Specifically, the local extrema of $(\sin x)/x$ are given by its intersection with the cosine wave)

Simply Beautiful Art

Yeah, I believe you have

:) What have you been up to? @AkivaWeinberger

Akiva Weinberger

Not much, to be honest

Simply Beautiful Art

I suppose I'd say the same.

Akiva Weinberger

One major problem with the $x\sin x=a$ is that nearly every input $a$ has infinitely many preimages $x$.

LegionMammal978

01:38

@AkivaWeinberger Just curious, how did you find those particular functions?

Akiva Weinberger

Well, I suppose that's not impossible; maybe our hypothetical function uses the arcsine or something.

Simply Beautiful Art

@AkivaWeinberger Follows from the branches of the Lambert W function ofc

Akiva Weinberger

Aren't there only two branches?

Simply Beautiful Art

@LegionMammal978 You just take the derivative and do some algebra

Akiva Weinberger

@LegionMammal978 Calculus says that the local extrema of $x\sin x$ satisfy $x\cos x+\sin x=0$.

Simply Beautiful Art

01:39

@AkivaWeinberger wolframalpha.com/input/?i=productlog(k,x)

Akiva Weinberger

@LegionMammal978 Isolating a term and multiplying/dividing by the right things gives you the above.

@SimplyBeautifulArt Oh, cool

Semiclassical

Oh hey, Lambert-W function

Simply Beautiful Art

Hey @Semiclassical

Akiva Weinberger

Hm, it's the inverse of $xe^x$, right? And we want the inverse of $x\frac{e^{ix}-e^{-ix}}{2i}$

Simply Beautiful Art

I tried solving $x\sin(x)=a$ for $x$ in terms of $a$ and the Lambert W function

Yes @AkivaWeinberger

Semiclassical

01:41

Lambert-W is weird as far as its branches go

Simply Beautiful Art

Though I cheated a bit

Akiva Weinberger

That doesn't seem too helpful, actually.

Simply Beautiful Art

Well, assuming $a$ is real, we may rewrite as follows:

$$x\sin(x)=\Im(xe^{ix})=a\implies xe^{ix}=b+ai$$

Semiclassical

(I can talk about the weirdness of $W_k(x)$, but I'll wait for now.)

Simply Beautiful Art

For some $b$ such that $x^2=a^2+b^2$ or $x\cos(x)=b$. Anyways, you see we can solve for $x$ in terms of $a,b$ then...

$$x=-iW_k(bi-a)$$

Akiva Weinberger

01:43

We'd need an expressing for $b$ solely in terms of $a$

"Unless someone's discovered a way to dissect the Lambert W function into real and imaginary parts," as you said, that seems to be a dead end.

Simply Beautiful Art

Well yes, but the world is big, and it seems @Semiclassical knows some stuff

The fun of the internets

Akiva Weinberger

Did I ever show you W's Taylor series?

Semiclassical

I wish I did.

Know how to dissect it, I mean.

Simply Beautiful Art

@AkivaWeinberger I already know of it

Ah, well, hit me up with the weirdness anyways @Semiclassical

Semiclassical

Mmkay.

First, the weirdness of the analytic continuation. Suppose you start on the principal branch, where $W_0(x)$ is the local solution of $x=we^w$ where $w>-1$ and $x>-1/e$.

LegionMammal978

01:46

@AkivaWeinberger Easy; just define $\text{Magic}_k(x)=\mathfrak R(W_k(x))$ and $\text{MoreMagic}_k(x)=\mathfrak I(W_k(x))$ :p

Akiva Weinberger

And utilize ${\rm Magic}^{-1}(x)$? Genius! :P

Simply Beautiful Art

@Semiclassical I believe you mean $x=W(x)e^{W(x)}$

Semiclassical

Derp, you're right.

Simply Beautiful Art

xD

Akiva Weinberger

I only vaguely remember the reference @Legion

Simply Beautiful Art

01:48

@LegionMammal978 Well, for some people, the moment we start talking about the Lambert W function... its all magic xP

Akiva Weinberger

Something about a switch…?

Semiclassical

If we cut the complex plane along the negative real axis starting at $x=-1/e$, then $W_0(x)$ is analytic on this cut plane.

Akiva Weinberger

Sure.

Semiclassical

Now, suppose we analytically continue across this cut, starting from the top side.

Akiva Weinberger

That gives us $W_1$, I'd assume?

Semiclassical

01:49

I think so. Going from the bottom gives $W_{-1}(x)$.

One usually takes that -1 branch since it's real along $(-1/e,0)$.

Simply Beautiful Art

Uh...

Akiva Weinberger

The Riemann surface (is that the right word?) should be the same as that of $\ln x$. I hope.

Semiclassical

hehehehehe

Simply Beautiful Art

Excuse me professor, but isn't this how $\ln(x)$ and other branch cuts work?

Akiva Weinberger

And I'm guessing that my above guess is wildly wrong :P

Semiclassical

01:50

It's not wildly wrong, to be fair.

In fact, if you go to higher branches (e.g. $k=\pm2,\pm3,\ldots$) then it's basically right.

Akiva Weinberger

Weirdness happens at points where $(xe^x)'=0$, probably…

which $-1/e$ satisfies, I think?

Semiclassical

Not really.

Oh, sure.

Akiva Weinberger

Whatever, continue your exposition

Semiclassical

The weirdness is this. Once you're on the $k=1$ sheet, there's now a problem at $x=0$.

Namely, one finds that $W_1(x)$ has a logarithmic branch point at $x=0$.

Same with every sheet except $k=0$, in fact.

Simply Beautiful Art

Interesting...

Semiclassical

01:54

That's rather believable if you look at the plot of $W_{-1}(x)$, which serves as a smooth continuation of $W_0(x)$ for real $x\in (-1/e,0)$.

Akiva Weinberger

Oh, yeah

Semiclassical

But here's the kicker: While there's also a branch point at $x=-1/e$, it's not a log branch point!

It's just a square root branch point.

No.

Simply Beautiful Art

No, he means $x=-1/e$

Semiclassical

At $x=0$, it's a log branch point. But $x=-1/e$ is order 2.

Akiva Weinberger

Wait, sorry, how are you defining a logarithmic branch point

and an order 2 branch point

Semiclassical

01:55

Yeah, that's fair.

Akiva Weinberger

Didn't you say that going around the $-1/e$ point gives you infinitely many things as you go around?

$W_n:n\in\Bbb Z$

Semiclassical

I don't think I did. And if I gave that impression, I was wrong.

Let me describe it like this. Suppose I start just to the right of $x=-1/e$ and wind around it CCW (i.e. $z=-1/e+\epsilon e^{i\theta}$) keeping track of what branch we're on.

Akiva Weinberger

That gets you to $W_1$?

Semiclassical

At first, yes.

For $\theta\in[0,\pi]$, we're still on $W_0(x)$.

Akiva Weinberger

And then you fwoom to $W_1$

Semiclassical

01:57

(These functions are all continuous from above, so along the negative real axis I'm still technically on $k=0$.)

Right. For $\theta\in (\pi,2\pi)$ I'll be on the $k=1$ branch.

However, where I land at $\theta=2\pi$ is actually $k=-1$ :)

TheGreatDuck

@nitsua60 No actually I'm trying to honor the ban but people keep messaging me...

Akiva Weinberger

Oh, so you cycle through $W_{-1}$, $W_0$, and $W_1$ there?

Semiclassical

Indeed, for $\theta\in [2\pi,3\pi]$ one is on $W_{-1}$.

Akiva Weinberger

(Wouldn't that be a cube root branch point?)

Simply Beautiful Art

Hmm... very funky... I will have to read up on this...

Semiclassical

01:59

Not done yet. Because once we get $\theta>2\pi$, we land back on the $k=0$ branch.

Simply Beautiful Art

g'night all

Semiclassical

bye.

And then at $\theta=4\pi$ you're back where you started on the $k=0$ branch.

So you've gone through a $4\pi$ arc (two rotations) and ended up back where you started.

That's what makes it an order 2 branch point.

Akiva Weinberger

Oh! Wait

$W_1$ is for $(\pi,2\pi)$, not $(\pi,3\pi)$

Semiclassical

By contrast, suppose I start on the $k=-1$ branch and start winding around the $x=0$ branch point. In that case, I'll just keep going further on: $W_{-2},W_{-3}4,\cdots$

Right.

Akiva Weinberger

So for $k=1$, the branch cut is what? Between $-1/e$ and $0$?

Semiclassical

02:01

Right.

Akiva Weinberger

And for $-1$ as well.

That doesn't make sense

Semiclassical

Not quite. But let me pull up a reference here.

Akiva Weinberger

OK

Semiclassical

So, wolfram functions gives the branch points here:
http://functions.wolfram.com/ElementaryFunctions/ProductLog2/04/04/01/

And the branch cuts here:

TheGreatDuck

@SimplyBeautifulArt it was a rhetorical question. I was testing Dodsy's strength at calculus.

Semiclassical

02:04

functions.wolfram.com/ElementaryFunctions/ProductLog2/04/05/01

The key point there is that the $x=-1/e$ branch point technically only belongs to the $k=0,1$ branches.

(That it doesn't show up on the $k=-1$ branch I suspect is a technicality of how they've defined the cuts, as a consequence of them choosing things so that $W(x)$ is always continuous from above.)

Akiva Weinberger

OK

Semiclassical

One thing you should note, perhaps, is that when $k\neq 0,1$ the branch cut structure is seemingly simple: Just a log branch point at $0$ and a cut along the negative real axis.

What that's really telling you is that, if you analytically continue $W_k(x)$ across $(-\infty,0)$ from above, you'll always go to $W_{k-1}(x)$.

Where things are weird when continuing from above is precisely when $k=0,1$.

Akiva Weinberger

At $W_0$, it's $(-\infty,-1/e)$ (which fwooms you to $W_1$).

Semiclassical

Right. By contrast, if you analytically continue across $(-1/e,0)$ you'll just stay on the same sheet.

Once you're on $W_1$, things are complicated.

Akiva Weinberger

At $W_1$, you have $(-\infty,-1/e)$ (which fwooms you to $W_0$) as well as $(-1/e,0)$ (which fwooms you to $W_2$).

Right?

Wait

Semiclassical

02:09

Careful. Which sides of the cuts are you approaching?

Akiva Weinberger

Oh. Revised version:

Semiclassical

Oh, crap. I may have gotten backwards at some point here.

Actually, no. I'm not wrong yet, I just wasn't doing things in the clearest way.

TheGreatDuck

@nitsua60 has any more spam flagging occurred since they banned that one guy?

Akiva Weinberger

$(-\infty,0)$ from above should get you to $W_2$. $(-\infty,-1/e)$ from below should get you to $W_0$. And $(-1/e,0)$ from below should… I don't know.

Semiclassical

Let me start again at $k=3$, where there's nothing weird happening yet.

Akiva Weinberger

02:11

@SimplyBeautifulArt Pinging so you can find this conversation later

nitsua60

@TheGreatDuck I've only been around the last hour or so... [rummages]

TheGreatDuck

fair enough

Akiva Weinberger

Hi

Semiclassical

On $k=3$, there's only a branch cut along the negative real axis. If I approach it from above, I'll end up on the $k=2$ branch.

nitsua60

@TheGreatDuck no, it doesn't look like any flags have been thrown here today.

Semiclassical

02:12

On $k=2$, there's still only a branch cut along the negative real axis. If I approach it from above, I'll end up on the $k=1$ branch.

TheGreatDuck

yay!

:-)

@nitsua60 well one flag was thrown

Semiclassical

On $k=1$, there's a branch cut along the negative real axis; if I approach it from below, I'll go back to the $k=2$ branch. (This is just consistency with the last statement.)

Akiva Weinberger

@Semiclassical I though approaching it from above got us to the next cut

Since we're going counterclockwise in that case

Semiclassical

Am I doing it backwards? Ugh. Hang on, lemme check.

nitsua60

@TheGreatDuck whaddyamean?

TheGreatDuck

02:14

@nitsua60 whoever it was threw up a white flag of surrender.

Semiclassical

I'm going based off of the stuff here, for clarity: functions.wolfram.com/ElementaryFunctions/ProductLog2/04/05/01

nitsua60

@TheGreatDuck BOOM!

lolfigsl =)

TheGreatDuck

-_-

i have no clue what that says

Semiclassical

Bah, yeah, I think you're right. @akiva

So I should've started at $k=-2$. :/

nitsua60

in RPG General Chat, May 31 at 0:06, by Papayaman1000

I just Google'd "lolfigsl" and the first result is from this room, from February. Ohhh boy.

Akiva Weinberger

02:15

Laughing out loud falling… in glass… somberly leaving

Or something

Semiclassical

hrm. I'm going to need to head out for a bit, but I'll want to continue this.

Akiva Weinberger

OK

Semiclassical

back later

Akiva Weinberger

If you leave, I might be asleep when you come back

Fair warning

nitsua60

in RPG General Chat, Aug 7 '16 at 20:59, by SevenSidedDie

(There needs to be a way to indicate actual real loling, not just typing “lol” as an appreciative indicator.)

which, four lines later, lead to "lol forrealz I'm getting strange looks"

Akiva Weinberger

02:16

Haha!

nitsua60

or "lolfigsl" for short =)

Akiva Weinberger

OK, that's pretty great.

nitsua60

Which has made it to the Urban Dictionary!!!

[sings, Lego Movie-style] internet is awesome... internet gets better when you're helped by your team...

Akiva Weinberger

It has survived the arduous, peer-reviewed process of getting accepted to the Urban Dictionary

(For real, though - assume you can just add whatever you want to it, right?)

nitsua60

@AkivaWeinberger I dunno... it's not just an Urban Dictionary =)

Alright. Riding high on that success, I'm off to bed. No more counting on this sheep for any of you.

TheGreatDuck

02:21

< sits on @nitsua60 > "One. Two. Three. Four...."

I am counting on you

ba dum tsh

nitsua60

(I swear there's a Bert and Ernie sketch in here somewhere...)

TheGreatDuck

who?

nitsua60

Sesame Street - Ernie and Bert - Ernie counts sheep

►

TheGreatDuck

what is that? Is that some stupid youtube machinima?

nitsua60

seriously?

TheGreatDuck

02:23

yes

nitsua60

It's Sesame Street.

TheGreatDuck

that's what it looks like?

nitsua60

Sesame Street

Sesame Street is a long-running American children's television series, produced by Sesame Workshop (formerly known as the Children's Television Workshop) and created by Joan Ganz Cooney and Lloyd Morrisett. The program is known for its educational content, and images communicated through the use of Jim Henson's Muppets, animation, short films, humor, and cultural references. The series premiered on November 10, 1969, to positive reviews, some controversy, and high viewership; it has aired on the U.S.'s national public television provider (PBS) since its debut, with its first run moving to premium...

@TheGreatDuck In the 70s, sure =)

TheGreatDuck

ugh

even worse

I just thought it was some amateur with puppets

nitsua60

DId you not notice the hands...? They're muppets.

TheGreatDuck

02:25

so... they're body suits?

nitsua60

No. Muppets are marionette-puppet hybrids. (Hence "muppet".) Head/mouth is controlled by inserted hand, puppet-style. Hands/limbs are controlled by strings or rods, marionette-style.

TheGreatDuck

ah

Akiva Weinberger

Oh, so that's why they're called that

I never actually knew

nitsua60

Note carefully the pair of sticks in Jim Hensen's left hand =)

Daminark

Oh that makes sense

TheGreatDuck

02:28

who?

nitsua60

Here's a nice one:

@TheGreatDuck Jim Hensen?

TheGreatDuck

yes

nitsua60

Jim Henson

James Maury Henson (September 24, 1936 – May 16, 1990) was an American puppeteer, artist, cartoonist, inventor, screenwriter, film director and producer who achieved international fame as the creator of the Muppets. Born in Greenville, Mississippi, and raised in Leland, Mississippi, and Hyattsville, Maryland, Henson began developing puppets while attending high school. While he was a freshman at the University of Maryland, College Park, he created Sam and Friends, a five-minute sketch-comedy puppet show that appeared on television. After graduating from the University of Maryland with a degree...

TheGreatDuck

i mean, which guy is jim henson.... XD

Akiva Weinberger

@TheGreatDuck The one holding the pair of sticks, clearly

TheGreatDuck

02:30

in the photo

nitsua60

@AkivaWeinberger <ding, ding!>

=D

@TheGreatDuck second from left

TheGreatDuck

ah ok

@AkivaWeinberger i couldn't see the sticks

nitsua60

I'll tell you... I've had the chance to watch ten minutes of Muppetry being shot. The choreography is jaw-dropping, all the performers keeping their characters in-frame and themselves out-.

Sesame Street: Feist sings 1,2,3,4

►

I watch something like ^^ and think the ballet's got nothing on them.

I mean, think about how fast those chicken-performers need to scurry in there!

Akiva Weinberger

@BalarkaSen Suppose we write $S^0=a\cup b$, $~S^1\setminus S^0=x\cup y$, and $S^2\setminus S^1=A\cup B$ (those being the decompositions into connected components)

(so $a$ and $b$ are singletons are $A$ and $B$ are open in $S^2$)

If $h$ is the Hopf map, we can write $S^3$ as the union of $h^{-1}(a)$, $h^{-1}(b)$, $h^{-1}(x)$, etc

which looks like two open solid-torus regions, separated by a torus, divided into two by two slanted circles

jakebird451

If I need to represent the vector of the initial position of a projectile, would I write it as $\vec{p_{i_P}}$? Likewise for a target: $\vec{p_{i_T}}$? Is there a convention for this?

Akiva Weinberger

02:40

We know the Hopf map isn't nullhomotopic. Is the converse true? Is any map $f$ that sends those regions of $S^3$ to the corresponding regions of $S^2$ not nullhomotopic?

So I guess that would be $f(h^{-1}(A))\subseteq A$, etc

Jorge Fernández Hidalgo

Oh hi, yo

is my current rep bad luck?

Semiclassical

ok, I'm back. let's try this again

Akiva Weinberger

I'm awake

Semiclassical

I'll start from $k=-1$ this time.

Akiva Weinberger

@JorgeFernándezHidalgo 66,606, wow

Hm, maybe 60 shy?

Jorge Fernández Hidalgo

02:44

I'll try

Akiva Weinberger

I mean, uh, yes. Yes, it is bad luck. :P

Jorge Fernández Hidalgo

but it reads 66.6k

Akiva Weinberger

Begone, demon!

Daminark leaves

Semiclassical

As the Wolfram page indicates, this has a branch cut along the negative real axis. If I approach it from below, I'll end up on the $k=-2$ branch.

Akiva Weinberger

Sure.

jakebird451

02:46

Does anyone here know vector math?

Semiclassical

If I approach it from above, though, things become interesting. On the one hand, the definitions given there immediately let me say that $W_{-1}(x+iy)\to W_{-1}(x)$ as $y\to 0^+$.

The question is then: Which branches go to $W_{-1}(x)$ as I approach from below?

Akiva Weinberger

I do, but I don't know the answer to your question @jakebird451

Semiclassical

And if I look at that page, I find that there's two of them.

On the one hand, if $x<-1/e$ then $W_0(x+i y)\to W_{-1}(x)$ as $y\to 0^-$.

(they write it as x-iy with y>0. same difference)

Akiva Weinberger

Maybe $\vec{p_{P,i}}$ or $\vec{p_{i,P}}$ (or maybe the same without commas) looks better

Never mind

Semiclassical

On the other, if $-1/e<x<0$, then $W_0(x+iy)$ is an analytic function and it just goes to the real-valued $W_0(x)$.

Instead, we need to go all the way to the $k=1$ branch: When $-1/e<x<0$, then $W_{1}(x+i y)\to W_{-1}(x)$ as $y\to 0^-$.

So therefore: If I'm on the $k=-1$ branch and I approach the cut along $(-\infty,-1/e)$ from above, I'll end up on the $k=0$ branch.

Once I'm on there, I can then approach the $(-1/e,0)$ cut on the $k=0$ branch and I'll end up on the $k=1$ branch.

Alternatively, when I'm on the $k=-1$ branch I can approach the $(-1/e,0)$ branch cut from above and I'll end up at $k=1$ immediately.

Once I'm on the $k=1$ branch, I can wind around the $(-\infty,0)$ branch cut without crossing from below and instead cross from above. This will place me on the $k=2$ branch.

From there on, things are simple enough. In fact, past that I'm in the usual log situation: If I wind around the origin, I'll end up at the next branch.

All of the weirdness is in the realm of the crossings between W0, W1, W-1.

@akiva So yeah, Lambert-W function has a confusing branch structure :S

Daminark

03:30

confuzzled reax only

Semiclassical

lol

Akiva Weinberger

Cool video on the shoelace formula

(You might have heard of the shoelace formula already, and know why it's true, but it's still a good video)

Also.

Daminark

Lel

jakebird451

03:49

I made a question (math.stackexchange.com/questions/2318046/…) that covers my specific question about vectors. I just need to get pointed in the right direction.

And specifically, how to use $|\vec{v}|$ in equations.

Akiva Weinberger

04:08

Their [Mathologer's] video on the Kakeya needle problem is also great.

Balarka Sen

05:28

@AkivaWeinberger That's a nice picture. Hmm. Here's a fact: if $f : S^3 \to S^2$ is my map, $a, b$ are regular values of $f$ so that $f^{-1}(a)$ and $f^{-1}(b)$ are 1-dimensional compact submanifolds (circles) in $S^3$, then I can compute the linking number of the link $f^{-1}(a) \cup f^{-1}(b)$ in $S^3$. This is a number independent of the choice of $a, b$ and depends on the homotopy class of $f$.

In your case, you're requiring the pictures with $f^{-1}(a), f^{-1}(x), f^{-1}(A)$ etc etc looks exactly like $h^{-1}(a), h^{-1}(x)$, etc etc yes?

If that holds for a generic decomposition $S^2 = A \cup B \cup x \cup y \cup a \cup b$ of $S^2$, then $f^{-1}(a)$ and $f^{-1}(b)$ are going to be circles linking once in $S^3$, and the Hopf invariant (as it is called) of $f$ is going to be 1, and $f$ indeed would not be nullhomotopic.

Generic meaning, if you perturb the positions of $a, b, x, y, A, B$ a little bit the picture remains unchanged (note that I'm not requiring that's the picture for all decompositions)

Daminark

OK, so, let's say $M$ and $N$ are smooth $n$-manifolds, $\omega$ is a closed $n$-form on $N$, and $f,g:M\to N$ are homotopic maps. The degree formula should give that $\int_M f^*\omega = \int_N g^*\omega$.

I think this should get Cauchy's integral theorem over a simply connected set, right?

Balarka Sen

@Akiva Oh, you're actually requiring something weaker: that $h^{-1}(a), h^{-1}(x), h^{-1}(A)$ etc etc are contained in $f^{-1}(a)$, f^{-1}(x)$, f^{-1}(A)$ etc etc. Then I do not believe that anymore. It should be true if the pictures are equal for a generic decomposition, but not necessarily if one contains the other, I do not think.

@Daminark Wellll, in Cauchy, you work with holomorphic forms instead of actual forms, but I think that's right.

The point of Cauchy if I remember correct is that for holomorphic $f$, $f(z)dz$ is a closed form. And closed forms are exact over simply connected open sets.

Daminark

05:43

I think the degree argument above is what gives you that closed = exact on simply connected spaces, no?

BAYMAX

If we know the rank of a matrix say $A_{n,n}$ then can we say anything about the rank of $A^{t}A = ?$

Daminark

At least on curves I think I've done the proof that exact forms have integrals which are path-independent

But if a space is simply connected, this is true for closed forms as well since you just homotope the curve

Oh hmm

Is it true that given a closed curve $c$, that it is the boundary of a domain $D$ on which Stokes' may be applied?

Balarka Sen

@Daminark I don't see how you get closed = exact over simply connected sets using that.

Daminark

Oh I forgot to state that this is an iff type thing

Balarka Sen

Yes. This is the Jordan-Schoenflies theorem.

Daminark

05:50

Well, path independent integral is the same as saying the integral over a loop is 0

BAYMAX

actually you guys are talking of which topic ?

A mixture of complex analysis , vector calculus , Homotopy , Topology ?

Daminark

So a form is closed iff its integral over any loop is 0. If your domain is simply connected, this will be true because you homotope the loop to a point, which checks out because your form is closed

Basically, yeah

Balarka Sen

aka differential forms

@Daminark Oh, I am sorry. Yes.

Integral over any loop is 0 does imply it's an exact form I'm integrating.

Daminark

Nice, so this works out alright. Alessandro mentioned that there's a homology version of this, yeah?

Balarka Sen

Degree theorem? Yeah.

Daminark

05:58

I was thinking Cauchy though it'd make sense that this follows from degree. How does it go?

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