Mathematics - 2020-03-28 (page 3 of 3)

anakhro

22:02

I like the MAyer-Vietoris sequence.

Ted Shifrin

Anakhro, are you doing lots (1/3 to 1/2) of the exercises? You have to to learn it.

Alessandro Codenotti

Hi @Ted

Ted Shifrin

Hi, demonic.

Balarka Sen

22:18

Hi @TedShifrin

Ted Shifrin

Hi, a @balarka.

Bloselink

Hi, given $f(x) = x^{2010}$, would $f(-2)$ be undefined?

Ted Shifrin

I wonder if @eigenvalue finished his question from a few days ago.

Why undefined? Throw away your calculator.

8

Balarka Sen

I am starting to understand the model surface of constant curvature $K$ as the sphere of radius $1/\sqrt{K}$ in the quadratic space $\Bbb R^3$ equipped with $Q(x, y, z) = K^{-1} x^2 + y^2 + z^2$ if $K > 0$ and as the sphere of radius $i/\sqrt{-K}$ if $K < 0$.

Ted Shifrin

LOL, the latter being the hyperboloid model, yes.

Bloselink

22:21

@TedShifrin So it would have the same behavior as $x^2$ except it would grow much slower?

Balarka Sen

Easier to describe the geodesics, since we immediately know all the symmetries. If $K = 1$ it's $O(3)$, if $K = -1$ it's the $O^+(1, 2)$.

Bloselink

and as x would approach infinity, so would the function $x^{2010}$?

Balarka Sen

I find the hyperboloid model a little unnatural

Ted Shifrin

Grows zillions faster, not slower.

Compare $ x^2$, $x^4$, $x^6$.

Dual symmetric spaces, Balarka.

Quite natural. $SO(n,1)$ versus $SO(n+1)$.

Bloselink

@TedShifrin Thanks. This function is quite interesting to graph. Would it be possible to graph it by solely using the first and second derivative tests and finding its x and y-intercepts?

corona patrol

22:25

@Bloselink that is the product of an even number of -2s

Balarka Sen

Unrelated: I plugged too many things in my extension cord and everything got tangled, I had to do a belt trick to untangle it

Ted Shifrin

Not very interesting. They're all getting to be steeper/narrower parabolic shapes.

Balarka Sen

@TedShifrin Hm, what's the duality?

Ted Shifrin

Rofl @balarka

Look up dual symmetric pairs/spaces. One compact, one not.

Balarka Sen

Alright. Thanks!

Lookity uppity

Ted Shifrin

22:27

Without my computer, I can't type enough to teach.

Balarka Sen

Yeah I understand. You still teach enough, Ted, so it's no problem

Ted Shifrin

Eventually I will go back home.

Balarka Sen

I was reading Thurston's "Geometry and Topology of 3-Manifolds", Chapter 2 where he explains model geometries in dimension 2. That man breathes hyperbolic geometry so it's hard to keep up

Ted Shifrin

I've never read that.

Balarka Sen

He's like "Oh yeah, hyperboloid model? You just see a $\Bbb H^2$ inside an $\Bbb H^3$ from "the top" and get that shit in your visual sphere"

Ted Shifrin

22:30

I love the hyperboloid model with analogous moving frames computation to the sphere. Just have to compare Maurer-Cartan forms for those Lie groups.

Balarka Sen

Draws a picture of an eye glancing at two planes, sun gleaming from behind, and a cloud in the background annotated as "$\Bbb H^3$"

Ted Shifrin

Yeah, compared to Thurston I'm not geometric.

That sounds like a projective model.

Balarka Sen

Yeah he gets back the hyperboloid model from projective model naturally somehow (not by projecting radially to the disk tangential to the tip of the hyperboloid)

I don't entirely follow, but ok, I am a mortal

Ted Shifrin

Yes, I did that in class. It's even in my riem geom notes. I think I sent you those.

Balarka Sen

@TedShifrin I do like how it puts all the model spaces in one context of the unit sphere in a certain quadratic space.

Ted Shifrin

22:34

Going to the disk model from the hyperboloid model and following the metric.

Balarka Sen

Yep, you did. Let me look through that.

Following the metric is hard. I would understand the geodesics in both model and see that they are projected correctly under the projection.

I should probably do more computations though

Ted Shifrin

Of course. Crank up the moving frames.

Balarka Sen

Oh yeah I could compute the geodesics on the hyperboloid model by moving frames that can't be a hard exercise

It's just writing the forms implicitly using $x^2 - y^2 - z^2 = 1$

Fine, I will do this one

Ted Shifrin

I remember you were so algebraic when I first met you. Now you've been through topology to diff geo. I'll take epsilon credit :)

Mike gets more.

Balarka Sen

You and Mike corrupted me

This chat has been a mathematical family, basically.

Ted Shifrin

22:37

Yeah.

Balarka Sen

@TedShifrin Unfortunately, it seems Mike has been corrupted by algebraists instead

in garbological cohomology for derived nerds, 12 mins ago, by Mike Miller

I want to phrase this from a while back in a better way. Let $\Lambda$ be a 2-element set, with the obvious free $\Bbb Z/2$ action. There is a group $\Bbb Z_\Lambda = \Bbb Z \otimes_{\Bbb Z[\Bbb Z/2]} \Bbb Z^\Lambda,$ where $\Bbb Z/2$ acts on $\Bbb Z$ by negation. If we write $\Lambda = \{e_1,e_2\}$, then $\Bbb Z_\Lambda$ consists of pairs $(m,n)$ modulo setting $(m,n) = -(n,m)$.

He's writing this right now in the other chat

Ted Shifrin

ROFL

I'm too antique to be corrupted.

Archer

Why is the singularity of log(1+z) at z = -1 not isolated?

i feel it should be isolated if the branch cut is [0, infinity)

Ted Shifrin

Branch points are not isolated singularities.

There is no Laurent expansion.

Archer

@TedShifrin I am saying that the function's branch cut is the positive part of the real axis

And the singularity is on z = -1

So we can easily find a deleted disc around z = -1 throughout which it is analytic

Ted Shifrin

22:43

No you can't.

Archer

Why??

Ted Shifrin

Forget the shift and go back to $\log z$ in a nbhd of $0$.

Mary Star

Hello!!

We have that $$L=\lim_{x\rightarrow +\infty}\int_x^{x+1}f'(u)\,du$$ When we know that $\lim_{x\rightarrow +\infty}f'(x)$ exists in $\mathbb{R}\cup \{\pm\infty}$ can we conclude that $\lim_{x\rightarrow +\infty}f'(x)=L$ ?

Since the limit of $f'(x)$ exists as $x\rightarrow +\infty$ does it maybe hold that $$\lim_{x\rightarrow +\infty}\int_x^{x+1}f'(u)\,du=\lim_{u\rightarrow +\infty}f'(u)\,du=\lim_{x\rightarrow +\infty}f'(x)\,du$$ ?

Ted Shifrin

Now what deleted disk?

Archer

@TedShifrin yeah, its singularity is not isolated

Ted Shifrin

22:45

It's a branch point. Those are isolated. But isolated singularity refers specifically to a pole, essential sing, or removable sing.

Thorgott

@user76284 that is a nontrivial such family, the trivial one would be the empty family

Archer

Got my mistake, I was analysing log(1-z) as though it is log(z) without considering the shift. Thanks.

Ted Shifrin

Gotcha, good, Archer.

@MaryStar, the last stuff you typed is nonsense.

Mary Star

@TedShifrin Ok... I have now an other idea:

We have that $\lim_{x\rightarrow +\infty}f'(x)$ exists and let $M$ be this limit, where $M\in \mathbb{R}\cup \{\pm \infty\}$.

Then we have the following:
$$\lim_{u\rightarrow +\infty}\min_{u\in [x,x+1)}f'(u)\leq \lim_{x\rightarrow +\infty}\int_x^{x+1} f'(x)dx \leq \lim_{u\rightarrow +\infty}\max_{u\in [x,x+1)}f'(u)$$

It holds that since $\lim_{x\rightarrow +\infty}f'(x)=M$ then $\lim_{x\rightarrow +\infty}\min_{u\in [x,x+1)}f'(u)=\lim_{x\rightarrow +\infty}\max_{u\in [x,x+1)}f'(u)=M$.

So we get from the above inequality $M\leq L\leq M$ and it must be $L=M$.

Ted Shifrin

Call the limit $\ell$ and work with that integral.

Mary Star

22:54

Which limit should I call $\ell$ ? @TedShifrin

The limit of f'(x) ?

Ted Shifrin

Never mind. You used $M$. What you have is almost correct.

Mary Star

Ok, great!! Is there something that I could improve?

Ted Shifrin

But your leftmost and rightmost limits don't make sense. Pay careful attention.

Mary Star

Oh I see there is a typo...

The limit should be as for x and not u.

So it must be:

Ted Shifrin

There you go.

Mary Star

22:58

$$\lim_{x\rightarrow +\infty}\min_{u\in [x,x+1)}f'(u)\leq \lim_{x\rightarrow +\infty}\int_x^{x+1} f'(x)dx \leq \lim_{x\rightarrow +\infty}\max_{u\in [x,x+1)}f'(u)$$

So the rest is correct, isn't it?

Ted Shifrin

Yes.

Mary Star

Great! Thank you very much!! :-)

3

Ted Shifrin

Yup.

user76284

@Thorgott “Nontrivial” is an informal notion.

In my case I mean to exclude degenerate distributions.

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$Mathematics$ Mathematics