Mathematics - 2019-08-17

Symplectic geometry is a branch of differential geometry and differential topology that studies symplectic manifolds; that is, differentiable manifolds equipped with a closed, nondegenerate 2-form. Symplectic geometry has its origins in the Hamiltonian formulation of classical mechanics where the phase space of certain classical systems takes on the structure of a symplectic manifold. == Introduction == A symplectic geometry is defined on a smooth even-dimensional space that is a differentiable manifold. On this space is defined a geometric object, the symplectic form, that allows for the...

Shine On You Crazy Diamond

I see, differential forms

I know a little but not much

Ultradark

so interpreting as a phase space allows

for an interpretation of these lines as a manifold

Shine On You Crazy Diamond

k

Ultradark

What follows is a natural Hamiltonian vector field, which defines a Hamiltonian flow on each of the symplectic manifolds.

Shine On You Crazy Diamond

Gonna make coffee brb 3mins

Hamiltonian sounds physicsy

It's an espresso machine so it's fast

b

Ultradark

1:15 AM

ok

any questions

Shine On You Crazy Diamond

Lots since I don't understand that math yet

But one day...

Ultradark

I guess that's it

because I don't know what else to do either

ah

the tangent bundle

do you understand the tangent bundle?

@ShineOnYouCrazyDiamond

Quote Dave

1:57 AM

Do any of you guys know about this article: A NEW UPPER BOUND FOR THE SUM OF DIVISORS FUNCTION by Christian Axler. It's from Cambridge. I was hoping someone could help me get it.

Shine On You Crazy Diamond

2:44 AM

Nope @Ultradark

didn't get that far

If $T \subset G$ generates group $G$ and $S \subset H$ generates group $H$, then what generates $G \times H$ does $T \times S$ do the trick? (doesn't have to be minimal)

Shine On You Crazy Diamond

5:49 AM

@QuoteDave hack the data banks

:D

First you gotta factor large integers

It's just another math problem lol

Secret

6:39 AM

The following object produced by some brainstorming of some world building group breaks classical logic:

Secret

6:51 AM

Recall if $A \subsetneq B$ then it must be true that $B \supsetneq A$

However, we came up with something where $A \subsetneq B$ but $(B \cap A) \subsetneq A$. We call this a right subset

Likewise, there is also $B \subsetneq A$ but $(B \cap A) \supsetneq A$. We call this a left subset

It is easy to check how the "and" operator in the boolean algebra is broken when one expands the $\subsetneq, \cap$ and noting that will produce a contradiction when evaluating the truth value of $A \subsetneq B \implies (B \cap A) \subsetneq A$

and hence, for a set theory to accomodate non antisymmetric subset operations, you need to either change the formal system, or the axiom of specification

Silent

@Rithaniel, Is there a subset of a metric space which satisfies Lebesgue number property, but is neither totally bounded nor complete?

Alessandro Codenotti

7:24 AM

@Silent $\Bbb Q\cap[0,1]$?

Silent

Wow!! Thanks a lot!

@AlessandroCodenotti Is singleton open in $\Bbb Q$?

Alessandro Codenotti

Ah nevermind, it is totally bounded

@Silent what do you think?

Silent

Oh! not open, since each ball in rationals has infinite points. Thanks!

Silent

9:49 AM

@AlessandroCodenotti Does this have Lebesgue number property?

Silent

10:11 AM

Seems like it does not have that property

Alessandro Codenotti

Why not?

Silent

@AlessandroCodenotti Well, since it is totally bounded, if it had Lebesgue number property as well, it was compact.

Alessandro Codenotti

10:28 AM

Why? Is there a converse to Lebesgue covering lemma?

Silent

I m taught that "totally bounded and Lebesgue number property iff compact" for metric spaces. And from my previous chat, $\Bbb R$ with discrete metric satisfies Lebesgue number property but is not compact.

reals under discrete metric not totally bounded.

Alessandro Codenotti

Ah I see where my idea goes wrong, in fact it's easy to write an explicit cover of $\Bbb Q\cap[0,1]$ for which the covering lemma fails

@Silent sure, pick $\delta=1/3$ regardless of the cover

Silent

yeah.

@AlessandroCodenotti Can you provide that cover for which LNP fails?

Martin Sleziak

10:46 AM

@Asaf I googled a little and found this: We call Lebesgue spaces the spaces such that every open covering has a Lebesgue number: they turn out to be the spaces X such that every continuous function on X is uniformly continuous. Such spaces are usually called Atsuji spaces. arxiv.org/abs/math/9602203 Maybe some references from this paper will be useful to find out more. — Martin Sleziak Apr 9 '12 at 10:02

In case it helps, a usual keyword for searching might be Atsuji space.

Balarka Sen

11:00 AM

@Silent Think of a cover of $\Bbb Q \cap [0, 1]$ coming from a cover of $[0, 1] \setminus \{\sqrt{2}\}$ by open intervals which shrink arbitrarily in length near $\sqrt{2}$.

Something like that will do the trick.

Silent

thank you

Alessandro Codenotti

11:19 AM

Or enumerate the rationals and put a ball or radius $2^{-n}$ around the $n$-th

Any covers with sets of arbitrarily small diameter will work

Bhargob

Can I write the singleton set $\{x\}$ as $\cap_{n=1}^{\infty} \left(x-\frac{1}{n},x+\frac{1}{n}\right)$ ?

And $\left(-\infty,x\right)$ as $\cup_{n=1}^{\infty}\left(x-n, x-\frac{1}{n+1}\right)$ ?

Ultradark

12:26 PM

@s.harp

are you there?

Martin Sleziak

Hi @Ultradark - I saw that one of your questions has been used as an example in a recent post on meta.

Ultradark

Hey

Okay I see

Do I need to take action?

Martin Sleziak

I just thought it might be reasonable to let you know.

Ultradark

Thanks for letting me know

Martin Sleziak

Without knowing about the posts on meta linking to it, you might have been surprised why it has been undeleted (and now has one delete point). The other post linking there is in the reopen request thread and it was linked a few times in chat.

Of course, if you can think of some way of improving the question, the likelihood that it gets deleted again would be smaller.

Ultradark

12:34 PM

okay

I will think about how I can improve the question. I asked it last year. It was an honest question that was bugging me

now I understand

Rajesh Dachiraju

Bounty ending in 5 hours ....

4

Q: A maximization problem in functional analysis and data

Consider the minimization problem described this paper. Let $f_{\lambda}$ be the minimizer. As a part of extending my work, I am able to show the following facts $$\lim_\limits{\lambda \to 0}\|f_{\lambda}\|_{L^2} = 0$$ and $$\lim_\limits{\lambda \to \infty}\|f_{\lambda}\|_{L^2} = 0$$ My problem...

Secret

The calm before the storm:

Hong Kong protest today is pretty peaceful, as everyone is only planning tomorrow's protest

Meanwhile Portland will have a AntiFa vs Proud Boys clash, also on 17 August. Tension is already rising as police were deployed and shops closed down

----
A Triangle needs three vertices

We currently only have a line segment

s.harp

12:51 PM

@Ultradark yes, I am

Ultradark

@s.harp could you look at this and give me some feedback

s.harp

im afraid i dont understand either picture or what you are trying to do

are these class structures like root systmes of a Lie algebra?

Ultradark

I asked that question a while back

so it's not very good

@s.harp but yes class structures are root systems of a lie Algebra

Shine On You Crazy Diamond

1:12 PM

It's weird that algebra has axioms defining infinite structures yet we rarely stray far from the axioms when proving something

In terms of expression size

I don't know what I mean

@Ultradark did some exercises and took notes on p-groups

Ultradark

excellent

I should just make a new question lol

that one is so 2018

@TobiasKildetoft Do you know a thing or two about root systems and lie groups?

10

Q: Relation between root systems and representations of complex semisimple Lie algebras

I'm trying to understand the machinery of root systems for the purpose of classifying complex semisimple Lie algebras. During this process i lost the overview, espacially when it came to highest weight representations and all that, so i'm looking for a short summary: Do isomorphic root systems ...

representation-theory lie-algebras roots root-systems

saw that you answered this

Secret

most axiom of algebra are quite restrictive

just associativity alone can cull off very long proofs

The other reason is there is strong interest in finding the shortest proof to any theorem

Alessandro Codenotti

2:54 PM

@ShineOnYouCrazyDiamond it's surprisingly hard to have axioms not describing infinite structures

For example there is no first order theory (regardless of the language!) such that every finite group is a model of the theory but there are no infinite models

Tuki

3:28 PM

could someone take a look at this?

-1

Q: Proof that equation $Lf=u$ guarantees satisfiability of weak form $\langle Lf,v \rangle = \langle u ,v \rangle$

Problem Let's observe in a closed interval $[a,b] \subset \mathbb{R}$ real-valued and continuous vectorspace $\mathcal{F}([a,b],\mathbb{R})$. Where $\langle ., .\rangle$ is some scalar product. This scalar product only satisfies axioms when $\overline{v},\overline{w},\overline{u} \in V$ and $c \...

linear-algebra finite-element-method

Shine On You Crazy Diamond

4:25 PM

I looked at it and voted it up back to zero :)

-1 + 1 = 0, group theory

:D

Is it strange in math that we almost never analyze the indexs of symbols?

In particular $\pi(p_n) = n$ but we don't analyze usually the fact that $p_i$ the primes are indexed by the integers themselves.

Usually this is because it's a dead end, but maybe it isn't?

If you extend $p_{-i} = - p_i$, $p_0 = 0$ then you get the full integers as index

s.harp

unfortunately $0$ is not usually an acceptable prime, while $-p$ definitely is a prime element of $\Bbb Z$

one situation where you should try to understand the indexing set is when you are dealing with directed systems

one of my favourite examples is that a of a quasi-local algebra

(it appears in a physics situation, as an example for any bounded open subset $U$ of minkowski space one has an algebra of "local observables" $\mathscr A(U)$ that live in $U$, further if $U, V$ are space-like seperated one has that $[\mathscr A(U), \mathscr A(V)]=0$ in $\mathscr A(U\cup V)$, which encodes causality)

Tuki

4:43 PM

mm anyone?

s.harp

@Tuki as David Ulrich remarks, your question is basically a tautology

"if x=y then why is Expression(x)=Expression(y)?"

Alessandro Codenotti

@s.harp is there a simple example of a unitary operator whose spectrum is the whole unit circle?

I'm thinking about the diagonal operator on $\ell^2$ corresponding to a countable dense subset of the unit circle, it has the full circle as spectrum, but it's not unitary

s.harp

@AlessandroCodenotti yes, a very natural example :), do you want hints?

Alessandro Codenotti

Oh, like the identity?

s.harp

no the identity only has $1$ in its spectrum

but think about certain "diagonal" operators on $L^2([0,2\pi])$ or $L^2(S^1)$

jeea

4:49 PM

In number theory, is there a quick way to calculate $f(n) = \sum_{d|n} \tau(d)$ where $\tau(d) = \sum_{k|d} 1$

Alessandro Codenotti

@s.harp right, of course

@s.harp uhm I'm not sure what do you mean with "diagonal" here

s.harp

also the operator $f\mapsto (x\mapsto e^{ix}\cdot f(x) )$ on $L^2(\Bbb R)$

with "diagonal" I mean "multiplication with a function"

Tuki

if I have $x=y$ then what exactly allows me to take scalar product from each sides $ \iff \langle x,z \rangle = \langle y,z \rangle$ ?

definition of scalar product?

s.harp

Whats your scalar product?

Tuki

Inner product I believe would be the correct term

s.harp

4:58 PM

yes, but do you have a specific one?

Tuki

Defined in this

0

Q: Proof that equation $Lf=u$ guarantees satisfiability of weak form $\langle Lf,v \rangle = \langle u ,v \rangle$

Problem Let's observe in a closed interval $[a,b] \subset \mathbb{R}$ real-valued and continuous vectorspace $\mathcal{F}([a,b],\mathbb{R})$. Where $\langle ., .\rangle$ is some scalar product. This scalar product only satisfies axioms when $\overline{v},\overline{w},\overline{u} \in V$ and $c \...

linear-algebra finite-element-method

but It's not any specific one

s.harp

ok, just so you know the axioms you have listed are not the axioms of a scalar product

in particular axiom number 2 is broken

Tuki

Okey what are those then

s.harp

i assume this is a typo

Tuki

I'll check just a sec

s.harp

5:00 PM

apart from that there is one more problem in your question, namely you have a vector space $\mathcal F$ but then write that whenever $u,v,w$ are in $V$ you have [....]

in this case: what is $V$ supposed to be? do you mean $\mathcal F$?

Tuki

Well they are used In different context

$V$ Is used to define the axioms and $\mathcal{F}$ to define the problem itself

s.harp

ok, so you have a scalar product on the space $mathcal F$

Alessandro Codenotti

@s.harp The adjoint here is multiplication with $\overline{e^{ix}}$ so being unitary is clear. I'm missing something easy because I don't see the spectrum

s.harp

@AlessandroCodenotti im not sure about the terminology, but you "discretise" $e^{ix}$ so that its locally constant

Tuki

yes

Now the axioms should be correct?

s.harp

5:06 PM

@AlessandroCodenotti ie you get a sequence of operators given by multiplications taht are locally constant (except on measure 0 sets) and this sequence approximates $e^{ix}$ in the strong operator topology

Alessandro Codenotti

Hmm so you can't just explicitely compute the spectrum?

s.harp

you can, but i dont have an idea in my head atm, its probably very easy tho

@Tuki axiom 2 should have the "c" removed, then its correct

Tuki

oh sorry

s.harp

@Tuki Note that for the scalar product for ANY v, w in F you must have that $\langle v, w\rangle$ is defined

Alessandro Codenotti

I mean let $T$ be this operator. We want complex numbers $\lambda$ such that $\lambda-T$ is not invertible. What goes wrong when $\lambda$ is on the unit circle?

Tuki

5:09 PM

Okey how I can be sure they are the same then?

Based on what the equivalence holds?

s.harp

@AlessandroCodenotti Then $\lambda -T$ is a multiplication operator by a continuous function that has a $0$, if you can show that this is not invertible you are done

@Tuki Your question is something very central to mathematics. I cannot express it elegantly, but if you have L(u) = f, then any expression you write down with f is equal to that same expression but with f replaced by L(u)

this is because f and L(u) are the same thing

As an example, you now that 4 times 3 is 12

and that $4=2^2$, so you also get that $2^2 \cdot 3$ is 12

Tuki

well I understand if you have $a=b$ you can use either of them because they are "equal"

jeea

Hello! If someone has advice for this please help me

25 mins ago, by jeea

In number theory, is there a quick way to calculate $f(n) = \sum_{d|n} \tau(d)$ where $\tau(d) = \sum_{k|d} 1$

Alessandro Codenotti

@s.harp makes sense

Tuki

but Isn't my situation more like this, if I have equality $5=5$ then $\iff 5 \cdot 5 = 5 \cdot 5$ (both sides multiplied by 5 which leaves statement still true)

I do operation to both sides, but how I can be sure this operation holds the equality?

s.harp

5:16 PM

@Tuki replace every 5 where you dont think its important that its 5 with 3 or something

you are more in the situation where you have $5=5$ $\implies 5\cdot 3 = 5\cdot 3$

Tuki

yes

I think I have intuitive understanding on my problem but proofs are different

$a=b \implies a\cdot c = b \cdot c$

s.harp

not \iff, only \implies

Tuki

You're right

but how do you prove something like this?

How do I know that the Inner product changes both sides of the equality equally?

s.harp

because they are the same thing, there is nothing to prove *

* unless you are doing foundations, which is not what you are doing

$L(f) = u \implies \langle L(f), v\rangle = \langle u, v\rangle$ for all $v\in \mathcal F$.

Tuki

then I have another question

s.harp

5:24 PM

en.wikipedia.org/wiki/Tautology_(logic) (text to stop wiki preview)

Tuki

If we assume that $\langle Lf,v \rangle = \langle u, v \rangle$ is true for some $v \in \mathcal{F}$. Does It imply that $Lf=u$? On what condition they are equivelent ($\iff$) then?

$ \langle Lf,v \rangle = \langle u,v \rangle \implies Lf=u $ on what condition?

s.harp

no, it doesnt imply that, you get $Lf= u$ if $\langle Lf , v \rangle = \langle u,v\rangle$ for all $v\in\mathcal F$

here the positivity of the scalar product is important

Tuki

positivity? You mean the end result from inner product?

s.harp

axiom number 4 in your list

Tuki

I wonder if there exists such $v$ that $\langle v , v \rangle < 0$?

s.harp

5:32 PM

read axiom 4 again to see the answer to that question

Tuki

well yes in order to $\langle . , . \rangle$ exists it needs to satisfy $\langle v , v \rangle \ge 0$ and $\langle v , v \rangle \iff v = 0$

so no

if $ p \rightarrow q $ It doesn't make sense to assume that $p$ is false, if your trying to prove this

I don't somehow see how the positivity is important in this?

s.harp

5:48 PM

for which part?

Tuki

Well if I try to show that $\langle Lf,v \rangle = \langle u , v \rangle \implies Lf=u$, axiom 4 says that $\langle Lf,v \rangle \ge 0$ and $\langle u,v \rangle \ge 0$ or 0 if $v=0$

s.harp

no, thats not what it says, note that both arguments of the scalar product are the same in axiom 4

Tuki

hmm

s.harp

take a look at $Lf - u$

in particular $\langle Lf - u, Lf - u\rangle $

what values can this take? and what does its value tell you about the result you want to achieve

Tuki

well according to first equation $Lf-u=0$

s.harp

5:57 PM

remember that you are in a different situation now, you want to show that $Lf = u$. What you have is $\langle Lf ,v \rangle = \langle u ,v\rangle$ for all $v\in \mathcal F$

Alessandro Codenotti

What was the other example you had in mind? @s.harp

s.harp

there are two related examples, one is $f\mapsto (x\mapsto x\cdot f(x))$ on $L^2(S^1)$, the other is $f\mapsto (x\mapsto f(x+1))$ on $L^2(\Bbb R)$

Alessandro Codenotti

I see, let me check that they're unitary and have the right spectrum

Tuki

well $\langle Lf-u,Lf-u \rangle \ge 0$ and $\langle Lf-u,Lf-u \rangle = 0 $ if and only if $Lf = u$

s.harp

@Tuki right, so try to use what you have (namely $\langle Lf , v\rangle = \langle u, v\rangle$ for all $v\in\mathcal F$) to get what you want (namely $\langle Lf -u , Lf -u\rangle = 0$)

Tuki

6:05 PM

hmm

$ \langle Lf,v \rangle = \langle u, v \rangle \iff \langle Lf, v \rangle - \langle u, v \rangle = 0 \iff \langle v, Lf-u \rangle = 0 $

s.harp

now do one $\implies$ to finish

Tuki

when $v = Lf-u$ is the condition?

s.harp

yes, you know that $\langle v, Lf -u\rangle=0$ holds for all $v\in \mathcal F$, in particular it must hold for $v=Lf-u$

Tuki

6:21 PM

What you meant by "now do one ⟹ to finish" @s.harp?

Those middle ones are equivalent right?

s.harp

you have correclty written two equivalences, but the final step you need to get $Lf = u$ is an $\implies$ not an $\iff$

Tuki

yes but this requires that let $v = Lf-u$, since $v$ is just arbitrary $v \in \mathcal{F}$ right?

s.harp

The statement you have is that for any $v\in \mathcal F$ it is true that $\langle Lf -u, v\rangle =0$.

in particular since $Lf- u\in\mathcal F$ you must have that $\langle Lf -u, Lf - u\rangle =0$

make note of the "for any" (or synonym: "for all")

Quote Dave

6:40 PM

@ShineOnYouCrazyDiamond @ShineOnYouCrazyDiamond Lol. Yep, let me just break RSA (and probably ECC too, while I'm at it). Just give me a few hours. Nevermind, someone who had access was kind enough to give it to me

Tuki

yes so $Lf-u \in \mathcal{F}$ is just one particular case in all $v \in \mathcal{F}$

Balarka Sen

7:03 PM

Here's a weird question. Say $M$ is a manifold and $F : PM \to \Bbb R$ is a functional on the pathspace of $M$. What restrictions on $F$ do I need to impose to say there exists a 1-form $\omega$ on $M$ such that $F[\gamma] = \int_\gamma \omega$?

OK, I guess what I would do is take parametrized neighborhood $U$ of $M$ and choose a section of the pathspace fibration $PM \to M$ over $U$ which exists because $U$ is contractible. Let's say $\gamma_x$ is a path from the origin $0$ (the basepoint) to $x \in U$ for every $x$ under this section. Then $\varphi(x) = F[\gamma_x]$ defines a function $\varphi : U \to \Bbb R$

Mike Miller

@BalarkaSen What I would do is start by writing down what I know.

Balarka Sen

There should be a way to recover $\omega$ from $\varphi$

Mike Miller

What constraints are there on $F$ from Stokes' theorem?

And what about from the definition of integration?

First off, $PM$ is really a category (or the morphisms-set of; I'm thinking of a category as a set of morphisms with a partially defined product). and $F$ is a functor. That is your first constraint: additivity.

Balarka Sen

Yeah great point.

I was implicitly assuming that, but yeah

$F[\gamma_1 * \gamma_2] = F[\gamma_1] + F[\gamma_2]$

$\omega_p(v)$ should be given by taking the derivative of $F$ along a curve passing through $p$ with direction vector $v$

Mike Miller

Secondly fix morphism sets P(x,y). What about $F$ on this? We know that if $F(\gamma) = \int_\gamma \omega$, and if $\gamma'$ is homologous to $\gamma$ by a surface $S$, we have $F(\gamma) - F(\gamma') = \int_{\partial S} \omega = \int_S d\omega$. OK, this is pretty nasty admittedly

I don't assume $\omega$ is supposed to be closed

Balarka Sen

7:10 PM

No way

Take a curve $\gamma : \Bbb R \to U$ such that $\gamma(0) = x_0$ and $\gamma(t) = x$ with $\gamma'(t) = v$. Then look at uh $F[\gamma_{[0, t]}] - F[\gamma_{[0, t+h]}]$

Mike Miller

what is $x$

I don't like bothering with the path-space fibration, if you need to use contractibility something is un-natural

Let's study $\frac{d}{dt} F(\gamma\big|_{[0,t]})$

Balarka Sen

So my pathspace $PM$ is based at $x_0$ and $x \in M$ is some random point. I want to say derivative of $F[\gamma|[0, s]]$ at $s = t$ is $\omega_x(v)$

Mike Miller

I'm saying base it at $x$

It's certainly true that $\frac{d}{dt} \int_{\gamma(0)}^{\gamma(t)} \omega = \omega(0)$

So I guess you should assume that $F$ is $C^1$, in the sense that a map $X \to PM$ is smooth if the associated map $X \times [0,1] \to M$ is smooth, and we assume that for all smooth maps to $PM$ that $F\big|_{X \times [0,1]}$ is smooth

Balarka Sen

@BalarkaSen OK, this does the trick.

Yeah, that's all. Additivity and $C^1$ is all we need.

Mike Miller

Write down the proof for me?

I see how to recover $\omega$ and why it's smooth but I don't see why $F$ is the integral, even though that does seem likely

Balarka Sen

7:17 PM

$\omega$ along the curve is derivative of the restriction of $F$ along that curve. So just fundamental theorem of calculus, it seems.

Let me be precise, and work it out

Mike Miller

So you consider the map $\tilde \gamma: [0,1] \to PM$ which sends $0$ to $\gamma$ and $t$ to $\gamma\big|_{[0,1-t]}$.

You say: consider $F(\tilde \gamma): [0,1] \to \Bbb R$, which is a $C^1$ function by definition.

Wait, @BalarkaSen, why is $\omega_x(v)$ independent of $\gamma$/

Balarka Sen

It isn't! $v$ is the tangent direction of $\gamma$

Mike Miller

I know

You still made a choice

Balarka Sen

Hm

It feels like second order garbage shouldn't matter but I don't have an argument for this. Maybe I'm implicitly assuming something about $F$

Mike Miller

Yeah I don't know how to argue it

Balarka Sen

7:24 PM

Note that $\omega$ need not be unique, and locally I can always make such a choice of $\gamma$'s by the contractibility argument. Maybe what I want is $F = \int_\gamma \omega$ locally? This is getting ugly

@MikeMiller Also, we don't really want $F$ to just be a functional on $PM$, right? We want it to be $Diff(I)$-invariant

Mike Miller

Why is $\omega$ not unique?

Suppose $F = \int_\gamma \omega = \int_\gamma \omega'$. Then $\int_\gamma \omega - \omega' = 0$ for all curves $\gamma$. In particular your argument taking the derivative along a curve shows that $\omega - \omega'$ is zero at all points.

Balarka Sen

Oh ok

Mike Miller

I agree you want Diff(I) invariance but I'm not sure where it would show up here

Balarka Sen

Let's see. $\omega_x(v) = \frac{d}{ds}\, F(\tilde{\gamma}(s))|_{s = t}$, yes?

Mike Miller

Right

Balarka Sen

7:35 PM

Oh I see why my fundamental theorem of calculus argument doesn't work

I can't do $\int_\gamma \omega$ with this definition. I only know values of $\omega$ at vectors by choosing the curves to be tangent to those vectors

user193319

8:38 PM

Let $X$ be a complex Frechet space, let $\Omega \subseteq \Bbb{C}$ be open, and let $f : \Omega \to X$ be weakly holomorphic (i.e., $\Lambda f$ is holomorphic in the ordinary sense for every $\Lambda \in X^*$).

Given $\Lambda \in X^*$, why is it true that $\frac{(\Lambda f)(z)-(\Lambda f)(0)}{z} = \frac{1}{2 \pi i} \int_{\Gamma} \frac{(\Lambda f)(\xi)}{(\xi - z ) \xi} d \xi$?

Shouldn't there be a limit as $z \to 0$ on the LHS?

Ah, found the answer! math.stackexchange.com/questions/290717/…

Balarka Sen

8:54 PM

@RyanUnger Yo this adiabatic accessibility is weird

Say $\alpha$ is a 1-form on $\Bbb R^3$. Then there are points arbitrarily close to some point $p$ which cannot be reached from $p$ by curves tangent to $\ker \alpha$ iff $\alpha$ is integrable.

One direction is trivial, and if $\alpha$ is totally non-integrable then I know everything is reachable from everything.

(Lagrangian curves approximate any path)

Balarka Sen

9:08 PM

*Legendrian

OK, I mean this is simple

If $\alpha \wedge d\alpha \neq 0$ at a point then $\alpha \wedge d\alpha \neq 0$ on a neighborhood as well. On that nbhd $\alpha$ is totally non-integrable :P

OK I have just proved existence of entropy by using an h-principle

manooooh

9:40 PM

Hello guys!! I am trying to find to write in WolframAlpha but I couldn't

If I write: cos(x) in terms of sin(x) then WA interprets it as "cos(x)\\sin(x) relations"

But if I write cos(3*x)*cos(3*x) in terms of sin(6*x) then WA does not interpret anything

I also tried to search the word "relations" in WA documentation but nothing seems to appear. Any idea? Thanks!

Shine On You Crazy Diamond

10:08 PM

I'm quitting cigarettes today!

I ran out of butts on the ground

ÍgjøgnumMeg

10:26 PM

Hey y'all

J. Doe

10:49 PM

@ShineOnYouCrazyDiamond Great decision!

Hello

Quote Dave

Hey, can someone help me out here: math.stackexchange.com/questions/3302039/…

1

Q: A Function $g(x)$, Written in Terms of $f(x)$, That Flips Sign when a Root of $f(x)$ is Encountered

Let $f(x)$ be any continuous, differentiable function that has extrema (use $f(x)=\sin^2(\frac{33}{x}\pi)+sin^2(x \pi)$ as the example function in your answers) and that $f(x) \geq 0$ for all $x$ and has a finite amount of roots (it is not always a polynomial). Let $g(x)$ be a function (written i...

calculus number-theory functions

ah there we go

J. Doe

11:09 PM

@QuoteDave how come the answer given there doesn't answer your question?

Quote Dave

@J.Doe well I asked for one that doesn't require you to know the roots, but the indicator function requires you to know the roots

J. Doe

11:34 PM

@QuoteDave What's the source for this problem? I don't think you can avoid the roots like that since you need them to define g.

Quote Dave

@J.Doe I was thinking about it because I wanted to find roots numerically for functions that are all positive and have other extrema other than the roots (so something like f(x)/f'(x) wouldn't work). Newton's, fixed point, and so on are only valuable if you start near the root, and even then, with a function like the example, it's not guaranteed to work. Bisection is the only one guaranteed to work, but I need to get the equations in a form so that can work.

ÍgjøgnumMeg

what is a "pseudo-isomorphism" in the context of module theory? lol

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