Mathematics - 2020-12-11

love_sodam

4:14 AM

Diff geo is much harder than I first thought

Leonhard Euler

4:39 AM

@Thorgott I never fought with him. Your prof is wrong.

By the way, just wondering, what is the most starred message in tgis room?

No nlab November?

It got 11 stars

123

Hi All...

love_sodam

5:20 AM

How can I show that there is no complete surface of revolution with $K=-1$ where K is a Gaussian curvature

Ted Shifrin

That is a deep theorem of Hilbert @love_sodam.

Oh, of revolution. Missed that. Just write down the ODE and it's straight-forward.

Amin Idelhaj

5:38 AM

Hey!

love_sodam

@TedShifrin Yes this book comment that fact. Anyway I used the ODE and find conditions for $x(t)$ and $z'(t)$ if we denote $t\mapsto (x(t),0,z(t))$ be a generating curve. But couldn't conclude it's not complete

Leonhard Euler

What is known about the sum $\sum_{r\ge2}\frac{\zeta(r)}{r^2}?

It converges to approximately 0.831

Is there a closed form?

Ted Shifrin

The meridians are geodesics, but they are only defined for $t$ in a finite interval.

user76284

Can $v \cdot \nabla_u \frac{u}{\|u\|_2}$ be simplified?

love_sodam

@TedShifrin Oh, right. Thanks

user76284

5:55 AM

I'm getting something like $\frac{I - \hat{u} \otimes \hat{u}}{\|u\|_2}$.

Ted Shifrin

@love_sodam You're welcome.

Leonhard Euler

oh sorry... I meant the sum $$\sum_{r\ge2}\frac{\zeta(r)}{r^2}$$

was on my phone that time so I couldn't notice

I turned that sum to $$\sum_{k\ge 1}\frac{k\mathrm{Li}_2(1/k)-1}{k}$$

where Li is polylogarithm

Leonhard Euler

6:28 AM

I further turned this to $$\lim_{n\to\infty}-H_n+\int_{0}^{\infty}te^{-t}\psi(n-e^{-t}+1)dt-\int_{0}^{\infty}te^{-t}\psi(1-e^{-t})dt$$

This looks a bit hard to handle, but I have smth in my mind

Leonhard Euler

6:48 AM

Now I need just one help

How can I evaluate $$\int_{0}^{\infty}\frac{t}{e^t}\psi(1-e^{-t})dt$$

Leonhard Euler

7:06 AM

0

Q: Evaluating the sum $\sum_{r\ge2}\frac{\zeta(r)}{r^2}$

Note:This is the same question, but it doesn't answer my question, the answer doesn't give a closed form. In fact, the answer is not accepted. Moreover, I don't think that a 9 month old inactive question would get answers. I have been working on various sums involving the zeta function (which co...

Here it is

mathguy

8:45 AM

hi

skullpatrol

9:04 AM

hi @mathguy :-)

Astyx

9:55 AM

What's an example of a decreasing sequence of nonempty sets with empty intersection?

Leonhard Euler

10:05 AM

@Astyx Let the nth set be $S_n$

Then a possible answer is $S_n=[n,\infty)$

Astyx

Oh yeah that works

thanks

Leonhard Euler

I myself constructed a very complicated one (involving factors, prime numbers, etc.) but I forgot that

@Astyx np

nbro

11:19 AM

2

Q: Is there anything that ensures that convolutional filters end up the same?

I trained a simple model to recognize handwritten numbers from the mnist dataset. Here it is: model = Sequential([ Conv2D(filters=1, kernel_size=(3,1), padding='valid', strides=1, input_shape=(28, 28, 1)), Flatten(), Dense(10, activation='softmax')]) I experimented with varying the n...

neural-networks convolutional-neural-networks convolution filters

Leonhard Euler

12:36 PM

How can I derive the Weierstrass product formula for the gamma function?

Sayan Chattopadhyay

2:57 PM

@feynhat Did you write to Alok?

Mats Granvik

4:23 PM

In the complex plane, why do mathematicians talk about radius of convergence? Shouldn't one rather call it Voronoi cell of convergence?

Balarka Sen

I understand what you mean but why stop at a convex domain? Extend wherever you can extend

Mats Granvik

So radius is better?

Balarka Sen

Then you get a domain of holomorphy where an analytic continuation is defined.

I am not claiming anything is better. I'm curious why you want a convex cell.

Mary Star

Hello!! Could someone check if the geometric interpretations I give at the below post are correct?

0

Q: Geometric interpretations of the sets U+W

Give the geometric interpretation of the sets $U+W$ : We consider the sets \begin{equation*}U=\mathbb{R}\begin{pmatrix}1\\ 0 \\ 0\end{pmatrix}=\left \{\begin{pmatrix}\lambda \\ 0 \\ 0\end{pmatrix}:\lambda \in \mathbb{R}\right \} \ \text{ und } \ W=\mathbb{R}\begin{pmatrix}0\\ 1 \\ 0\end{pmatrix}...

linear-algebra geometry direct-sum

Lucas Henrique

5:01 PM

hey chat. could anyone give a valid reason to define nondegenerate pairings this way?

my definition of (left) non-degenerate pairing $\phi\colon V\otimes W \to \mathbb{k}$ is a map such that the map $$\begin{align} V &\to W^*\\ v &\mapsto (w \mapsto \phi(v,w)) \end{align}$$ is injective (so the only vector in $V$ that is orthogonal to every other vector is $0_V$)

C.F.G

5:40 PM

Hi math lovers.

Hi @TedShifrin

I want to know that: is that possible to recover pairs of points on an arbitrary Jordan curve just by knowing their image on Mobius strip? I know that this is possible for pairs of points on a circle but in general I am not sure.

Mary Star

6:34 PM

Let $a,c\in M_n(\mathbb{R})$, such that $ac=1_{\mathbb{R}^{n\times n}}$.
I have shown that $\text{Image}(a)=\mathbb{R}^n$ and $\text{Ker}(c)=0_{\mathbb{R}^n}$.
I want to show that $\text{Image}(c)=\mathbb{R}^n$. How could we do that?

Since the kernel is zero, this means that the matrix is invertible, right? Do we use that?

copper.hat

7:54 PM

@MaryStar show that $\dim {\cal R} ac \le \dim {\cal R} c$.

Mary Star

8:04 PM

@copper.hat By "$\cal R$" you mean the image, right? Do we use the fact that the dimension of the image plus the dimension of the kernel must be the dimension of $\mathbb{R}^n$, i.e. $n$ ?

Ted Shifrin

Just use nullity-rank in the first place, yes.

Mary Star

Ah ok! It holds that $\text{Im}(a)\subseteq \mathbb{R}^n$ and since $\dim (\text{Im}(c))+\dim (\text{Ker}(c))=n \Rightarrow \dim (\text{Im}(c))+0=n \Rightarrow \dim (\text{Im}(c))=n$ it follows that $\text{Im}(a)= \mathbb{R}^n$, right? @TedShifrin

Ted Shifrin

8:20 PM

You mean image of $c$ everywhere? You don't need to use $a$ except to deduce that $\ker c = \{0\}$.

The point is that in finite dimensions, you should end up proving that if a matrix has a left inverse, then that left inverse is indeed a true inverse. This is false in infinite dimensions.

Mary Star

Oh there is a typo, I mean "c" everywhere, not "a". So it should be :

It holds that $\text{Im}(c)\subseteq \mathbb{R}^n$ and since $\dim (\text{Im}(c))+\dim (\text{Ker}(c))=n \Rightarrow \dim (\text{Im}(c))+0=n \Rightarrow \dim (\text{Im}(c))=n$ it follows that $\text{Im}(c)= \mathbb{R}^n$.

copper.hat

@MaryStar try to develop some intuition. if the range of $c$ is not full it must be a subspace of dimension $<n$. if you apply $a$ to that strict subspace the corresponding range cannot have larger dimension, so it too must be $<n$.

Mary Star

@copper.hat But since the kernel is zero, it must be of full range, or not?

copper.hat

@MaryStar for a square matrix, yes. that is the big deal.

Mary Star

@copper.hat Isn't $M_n(\mathbb{R})$ the set of square matrix with dimension $n$ ?

Mary Star

8:41 PM

Now I am looking at the reserse direction. Let $a\in M_n(\mathbb{R})$ such that $\text{Im}(a)= \mathbb{R}^n$. Show that there is a $c\in M_n(\mathbb{R})$ such that $ac=1_{\mathbb{R}^{n\times n}}$.

There is also a hint: Let $(e_1, \ldots , e_n)$ be the standard basis of $\mathbb{R}^n$ and let $v_i\in \mathbb{R}^n$ such that $av_i=e_i$. Consider the matrix $(v_1 \mid \ldots \mid v_n)$.

We have that $\text{Im}(a)=\{a\cdot x\mid x\in \mathbb{R}^n\}$. So according to the hint this $ax$ must be each column of the identity matrix, right?

Rithaniel

TIL that cooked food is often more energy dense than raw

copper.hat

9:08 PM

@MaryStar Since $a v_i = e_i$ if you let $v=[v_1 \cdots v_n ]$ then $v e_i = v_i$ and so $av e_i = e_i$ which means $av=I$.

Karim Mansour

10:06 PM

@TedShifrin I was going through notes I sent you before

@TedShifrin

something I noticed I never noticed before is about why d^2 = 0 for the exterior derivative on exterior product. It is because of symmetry of partials also because base space has zero curvature. This curvature thing is useful for also what I will do in my PhD thesis.

I will make notes for smooth manifolds in case I have to teach it sometime while going through the beginning of that book.

brb

Mary Star

@copper.hat I see!! Thanks a lot for your help :-)

One last thing... Could you maybe check if my geometric interprations of the sets are correct? math.stackexchange.com/questions/3944436/…

Ted Shifrin

10:40 PM

@Karim: No, $d$ has nothing to do with geometric structure or curvature. So don't be so glib.

Karim Mansour

@TedShifrin
I mean symmetry of partial derivatives is one of them so we get dualwise cancellation

But we also have too that if the base space has curvature then also we won't have symmetry of partial derivatives as well

so we wont get dualwise cancellation as well

the real reason we get this cancellation is because of the fact that partials commute

Karim Mansour

11:07 PM

@TedShifrin what example did you use in your class to show that partials don't necessary commute

Karim Mansour

11:21 PM

I better not just stick to the material haha

brb

Ted Shifrin

11:54 PM

@Karim: You're still being sloppy and misstating things. In local coordinates, partial derivatives always commute. It's covariant derivatives that do not commute. This is where you see curvature. If you want to define $d^\nabla$ for a connection on a vector bundle, then squaring that is where you get curvature. But your statements are incorrect as they stand.

Karim Mansour

Yes sorry this is what I mean

I think being sloppy get fixed with time

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