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

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

Alessandro Codenotti

Hi @Ted

Hi, demonic.

10:18 PM

Hi @TedShifrin

Hi, a @balarka.

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

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

Why undefined? Throw away your calculator.

8

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 $K < 0$.

LOL, the latter being the hyperboloid model, yes.

10:21 PM

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

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 $O^+(1, 2)$.

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

I find the hyperboloid model a little unnatural

Grows zillions faster, not slower.

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

Dual symmetric spaces, Balarka.

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

@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

10:25 PM

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

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

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

@TedShifrin Hm, what's the duality?

Rofl @balarka

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

Alright. Thanks!

Lookity uppity

10:27 PM

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

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

Eventually I will go back home.

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

I've never read that.

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"

10:30 PM

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

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

Yeah, compared to Thurston I'm not geometric.

That sounds like a projective model.

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

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

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

10:34 PM

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

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

Of course. Crank up the moving frames.

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

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.

You and Mike corrupted me

This chat has been a mathematical family, basically.

10:37 PM

Yeah.

@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

ROFL

I'm too antique to be corrupted.

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)

Branch points are not isolated singularities.

There is no Laurent expansion.

@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

10:43 PM

No you can't.

Why??

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

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 $$\lim_{x\rightarrow +\infty}\int_x^{x+1}f'(u)\,du=\lim_{u\rightarrow +\infty}f'(u)\,du=\lim_{x\rightarrow +\infty}f'(x)\,du$$ ?

Now what deleted disk?

@TedShifrin yeah, its singularity is not isolated

10:45 PM

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

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

Gotcha, good, Archer.

@MaryStar, the last stuff you typed is nonsense.

@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 $L=M$.

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

10:54 PM

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

The limit of f'(x) ?

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

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

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

Oh I see there is a typo...

The limit should be as for x and not u.

So it must be:

There you go.

10:58 PM

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

So the rest is correct, isn't it?

Yes.

Great! Thank you very much!! :-)

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