Mathematics

ratio test

in fact, that sequence is summable

also yeah, sodam, that's correct

a slightly different way of doing it, but which has the same underlying idea, is to pick a normal $v$ to the plane and look at the function $t\mapsto\langle f(t),v\rangle$, the rest of the argument is the same

Anonymous

@Thorgott Interesting, thanks! I will try writing out the details. But as Lukas says, is there any analogue of L'Hospital for checking convergence of sequences?

Anonymous

I guess just ordinary L'Hospital works, but maybe it's an overkill for this situation

Perturbative

Say I have a smooth map between manifolds $F : M \to S^n$ how can I show that there is an embedded open disc $D$ in $M$ small enough such that $M \setminus D$ is homotopy equivalent to $M \setminus \{p\}$ where $\{p\}$ is some point in $D$?

Ted Shifrin

6:16 PM

What does $F$ have to do with this?

doesn't $M\setminus\{p\}$ deformation retract onto $M\setminus D$ for all embedded open discs

by pushing $D\setminus\{p\}$ radially outward

Ted Shifrin

Yes @Thor.

ok, glad my visual intuition isn't complete garbage

hey chat

Ted Shifrin

Hi Lucas

6:19 PM

still writing down that path-connectedness proof.

sup, @Ted?

so I'm kinda stuck, having a bad time trying to show that $A \mapsto A^{-1}$ and $(A,B) \mapsto AB$ are continuous ($\mathrm{GL}_n(\mathbb{C})$ with any norm)

since any 2 norms induce the same topology on $\mathcal{M}_n(\mathbb{C})$, I tried to use any submultiplicative norm, but no luck

Amin Idelhaj

What do those maps look like in coordinates?

Also hey Ted!

first one is weird, second one is, you know, matrix multiplication sent to $\mathbb{C}^{n^2}$

Amin Idelhaj

Well, do the 2x2 case for example and try to generalize

So we have a map $m:\mathbb{C}^8 \to \mathbb{C}^4$ where $m(a,b,c,d,e,f,g,h)$ is the product $\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}e&f\\g&h \end{pmatrix}$

Joe Shmo

who's the analyst in the crowd?

Amin Idelhaj

Write it down, and you'll see why it's continuous in any dimension. Do something similar for the inverse

Joe: uhhh kinda me? Idk

Lukas Heger

6:35 PM

@Lucas for the first one, use Cramer's rule, for the second one, just use the definition of matrix multiplication. Note that (multivariable) polynomials are continuous

you should definitely think through what Amin suggested in any case, but let me point out that you can also prove continuity of multiplication in the same way as you do it for real numbers

the explicit method is ultimately better, though, because it gives much more than just continuity

geocalc33

Does this function have zeros by inspection? $\Phi(s)=\Gamma\left(1+\frac1s\right)+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\zeta(-ns)$ I think it does because of the summation of the zeta function on the right

The Gamma term is definitely never zero

how so?

6:52 PM

How can I show if p_n->p where p_n is a sequence of points in regular surface S and p is in S then lim(p_n-p)/|p_n-p|\in T_pS ?

7:03 PM

do you know that the surface locally is the preimage of a regular value of some smooth function

No I didn't

I only know that the preimage of regular value is regular surface

Maybe just p as the origin and T_pS as xy-plane again. Then as p_n\to 0, lim p_n/|p_n| should contained in the xy-plane

the last argument seems intuitively true but I don't know how to prove it in rigor

7:24 PM

I'm trying to think of a simple argument rn, but it's kinda iffy

oh, the claim isn't true btw

I forgot this, but it's necessary to allow passing to a subsequence

I guess you just have to argue that you can interpolate some subsequence by an appropriate smooth curve

Yes but I doubt we can get smooth curve containing infinitely many p_n

you definitely can, it's just pretty awkward to argue

Q: Constructing category for germs

7:44 PM

How can I?

Actually my first approach was p = 0 and p_n = (x_n,y_n,z_n) so showing lim x_n/|(x_n,y_n,z_n)| = 0. But it seems hard to prove

maths researcher

9:11 PM

0

I have a question concerning the construction of functors and a category that satisfy specific properties. I have just started pondering about this, and i'm quite curious: Definition Let $M$ be a differentiable manifold of dimension $m$. Let $p\in M$. The equivalence class of a real valued smoot...

9:21 PM

ye, I believe you can do that

take the category whose objects are tuples $(M,U,p)$ where $M$ is a smooth manifold, $U$ an open subset and $p\in U\subseteq M$, a morphism $(M,U,p)\rightarrow(N,V,q)$ is a smooth map $f\colon M\rightarrow N$ such that $f(U)\subseteq V$ and $f(p)=q$

then you can consider the contravariant functors taking the germs at $p$ of the $M$ or $U$, respectively

maths researcher

cool, thanks @Thorgott

does this category

have a name?