Mathematics - 2024-08-28 (page 1 of 2)

Ben Steffan

00:09

Apparently $|[\mathbb{R}\mathrm{P}^{n + 1}, S^n]| = 2$ for all $n \geq 3$

Also holds if you replace $\mathbb{R}\mathrm{P}^{n + 1}$ by an $(n + 1)$-dimensional $p$-lens space, or any quotient of $S^{n + 1}$ by a nice action of a finite group.

...or I might have gotten my sequences mixed up and this might not be true at all, gah

Xander Henderson

@zetaspace mathfrak?! Nope. I'm out. Nothing good ever comes from mathfrak.

Ben Steffan

00:24

@XanderHenderson So commutative algebra is "nothing good"? ;)

Thorgott

01:27

@BenSteffan can't quite figure the computation myself, but seems plausible

I also suppose they both become null-homotopic after suspension

Ben Steffan

01:43

@Thorgott the best I could do, after correcting the nonsense I wrote down, is the Puppe sequence $\cdots [\mathbb{R}\mathrm{P}^{n + 1} / \mathbb{R}\mathrm{P}^n, S^n] \to [\mathbb{R}\mathrm{P}^{n + 1}, S^n] \to [\mathbb{R}\mathrm{P}^n, S^n]$. At the left you have $\mathbb{Z} / 2$ and at the right you have either $\mathbb{Z} / 2$ or $\mathbb{Z}$, depending on parity. But this is an exact sequence of pointed sets, not of groups, so this doesn't allow you to say much about the middle term :(

Thorgott

the sequence extends one more term to the right as $\rightarrow[S^n,S^n]$ (since $S^n\rightarrow\mathbb{R}P^n\rightarrow\mathbb{R}P^{n+1}$ is also a cofiber sequence), so you do get that the $\pi_{n+1}(S^n)$ acts transitively on the fibers of the restriction map $[\mathbb{R}P^{n+1},S^n]\rightarrow[\mathbb{R}P^n,S^n]$

but I'm too tired to see what the action is lol

Ben Steffan

hm

Xander Henderson

@BenSteffan I believe that's what I just said, yes.

Thorgott

boo, Xander

Ben Steffan

I'll have to read up on that action; I don't see where it's supposed to come from.

But not tonight, it's way too late

Jakobian

01:53

Xander always has such good takes

Thorgott

yeah, same, I'm gonna go to bed now

let's continue this conversation tomorrow

Soumik Mukherjee

04:27

How can we say there exists a_1 in A with e'(a)=e'(a_1)?

nickbros123

04:53

If I have a finite dimension vector space V, then the following is trivial: "let $T:V \to W$ be a linear tansformation, then there exists a subspace $U$ of $V$ so that $null(T) \cap U=\{0\}$ and $T(U)=range(T)$" but for the general case, can this be shown without assuming a basis / axiom of choice ?

Intuition tells me I have to draw an isomorphism from the quotient of V and null, and this subspace I'm looking for

Not sure if it'll work tho

Jakobian

@SoumikMukherjee epimorphisms aren't surjective in general

you need to share more context

Soumik Mukherjee

05:41

@Jakobian We are in the category of sets

I forgot to mention that

Pizza

08:11

@SineoftheTime In analysis 1 we didn't do them, so I'll see them myself

Sine of the Time

08:40

@Pizza that's strange

if $f$ is defined on $[a,b[$, then $\int_a^bf(x)dx:=\lim_{z\to b^-}\int_a^zf(x)dx$

if $f$ is defined on $]a,b]$, then $\int_a^bf(x)dx:=\lim_{z\to a^+}\int_z^bf(x)dx$

if the limit exist and it's finite, the it's said that the integral is convergent

you give the same defn for integrals of a function over an unbounded interval

if $f$ is defined on $[a,+\infty[$, then $\int_a^{+\infty}f(x)dx:=\lim_{z\to +\infty}\int_a^zf(x)dx$

Gian's Pizzeria

09:23

@SineoftheTime Hi, are you free ?

Sine of the Time

yes

Pizza

@SineoftheTime I know what happened, our analysis 2 prof did them in analysis 1, while our analysis 1 prof didn't...

So it's like we have to know them

Sine of the Time

it should not have an impact on your exam, just use the fact that when the function is not defined, you take the limit

certainly, there will not be exercises asking to determine the convergence of an improper integral

Pizza

Yes, luckily I got this case where this thing happens

Sine of the Time

yes, plus the double integral in your exam will be all convergent

Pizza

09:27

I got a new book in which I will write down all these things that can happen

Gian's Pizzeria

$\lim(x,y)\to(1,1) \frac{(x-1)^5 -(x-1)^2 -3(y-1)^2}{x^2+3y^ 2-2(x+3y-3)}=-1$

Pizza

I will make a section for each topic

@SineoftheTime Yes

Gian's Pizzeria

I need to verify that the limit is -1

Sine of the Time

using the definition?

Gian's Pizzeria

Yes, but I can also do it by transforming the fraction with the polar coordinates

But here it is not convenient to use polar coordinates

Sine of the Time

09:36

isn't the limit $0$?

Gian's Pizzeria

I wrote a number wrong

$\lim(x,y)\to(1,1) \frac{(x-1)^5 -(x-1)^2 -3(y-1)^2}{x^2+3y^ 2-2(x+3y-2)}=-1$

Sine of the Time

first of all, shift the limit to $(0,0)$

Gian's Pizzeria

Mmm

Isn't there another way?

Sine of the Time

why?

Gian's Pizzeria

I wouldn't want to change the limit

Sine of the Time

09:44

the limit does not change

Gian's Pizzeria

I've never done this thing

Sine of the Time

ok

note that $x^2+3y^2-2(x+3y-2)=x^2-2x+1+3y^2-6y+3$

Gian's Pizzeria

Yes

Sine of the Time

now you just have to show that $\lim_{(x,y)\to(1,1)}\frac{(x-1)^5}{(x-1)^2+3(y-1)^2}=0$

Gian's Pizzeria

=0?

Sine of the Time

09:49

yes

Gian's Pizzeria

Let me see

Sine of the Time

here you can use polar coordinates if you want

Gian's Pizzeria

I'll try for a moment

$\frac{(x-1)^3\cdot[(x-1)^2+3(y-1)^2]}{(x-1)^2+3(y-1)^2}$

I made this increase to simplify the numerator and denominator

So $(x-1)^3$

Its correct?

Ryder Rude

hi

Sine of the Time

@Gian'sPizzeria this way the numerator is not $(x-1)^5$

@RyderRude hi

Gian's Pizzeria

10:00

@SineoftheTime i know

But in theory shouldn't I make some additions to obtain the distance between two points?

Sine of the Time

I'm not following you

you can use polar coordinates centered in $(1,1)$

Gian's Pizzeria

Do you find that to prove the limit you have to make some additions, like x^2≤x^2+y^2????

Sine of the Time

I've seen this only to find $\delta=\delta(\epsilon)$

but your inequality is not correct because $(x-1)^3$ can be negative

you can note that when $|\frac{(x-1)^5}{(x-1)^2+3(y-1)^2}|\le |\frac{(x-1)^5}{(x-1)^2}|$

Gian's Pizzeria

Correct

Sine of the Time

if $x=1$, the limit is $0$

Gian's Pizzeria

10:10

So = |$(x-1)^3$|

Sine of the Time

yes

Pizza

$\omega = \frac{y}{\sqrt{xy}} dx + \frac{x}{\sqrt{xy}} dy$,the form is closed since the two partial derivatives match, initially $A$ is not simply connected so I have divided it into simply connected components so $A_1 = \{(x,y) \in \Bbb R^2 : x > 0 , y > 0 \}$ , $A_2 = \{(x,y) \in \Bbb R^2 : x < 0 , y < 0 \}$, so in each component the differential form is exact

Right ?

Gian's Pizzeria

≤$(x-1)^4$≤$(x-1)^4 +(y-1)^4$

Sine of the Time

@Pizza correct

Gian's Pizzeria

I don't know how to proceed with the exercise

Sine of the Time

10:19

10 mins ago, by Sine of the Time

you can note that when $|\frac{(x-1)^5}{(x-1)^2+3(y-1)^2}|\le |\frac{(x-1)^5}{(x-1)^2}|$

Ryder Rude

@SineoftheTime what are u learning these days

Gian's Pizzeria

@SineoftheTime yes

Sine of the Time

$=|(x-1)^3|\to 0$

@RyderRude I'm revising some theorems of real analysis

Ryder Rude

@SineoftheTime nice

Sine of the Time

what about you?

Ryder Rude

10:21

im learning formal logic... predicate, propositional and zfc. and also infinite cardinals

Sine of the Time

never studied it deeply

do you enjoy it?

Ryder Rude

yeah... ive only studied some basics rn

i like it

it has Godel's theorems and notions of truth

also some multi valued logics, which have a Undefined type, in addition to True and False

Sine of the Time

Interesting

Ryder Rude

but these are not that useful in math. somehow, the True/False scheme has been the most useful

it is also the simplest

Sine of the Time

I know only the basics

I'm not that interested in logic tho

Ryder Rude

10:28

i too am not sure if ill stay interested in the advanced stuff

lets see...

it is more like philosophy of math

one would think that the concept of truth requires nuance. it does in real life. things arent 100% True or False

but in math, things are either true or false. if both, contradiction

and this scheme gets u light years ahead :P

maybe it's the best scheme after all

Sine of the Time

good old Aristotle

Ryder Rude

yeah.. it's all inspired by his stuff

recently, some other schemes have been made which are more useful in fields other than math

there is fuzzy logic

it is useful in electronics

user 20458579510081670432

11:00

The Fundamental Theorem of Algebra: NOT! | Sociology and Pure Mathematics | N J Wildberger

►

Jakobian

13:40

@user20458579510081670432 huh?

Thomas Finley

We all know that a metric space is a topological space. I generally use the definition of a topological space that involves "open sets".

Jakobian

@SoumikMukherjee what they are proving is that $\phi$ is well-defined. Such $a_1$ always exist since you can take $a_1 = a$ but that's not the point. They just wrote it wrong

Thomas Finley

But I wanna know, "Let X be a metric space and let $x\in X.$ Suppose $x$ is an interior point in some topology $T$on $X.$ This means, $X$ is an interior point of the topological space $(X,T).$ Will $x$ also be an interior point of the metric space $X$ ?

Ben Steffan

@ThomasFinley Yes.

Well, if the topologies agree that is.

Jakobian

@ThomasFinley interior point of a topological space? There's no such thing

Thomas Finley

13:51

@Jakobian I'm really sorry, I meant an interior point of a subset $A$ of $X.$

Jakobian

Then no in general

If the topology generated by the metric is the same as $T$, however, then this will be the same

nickbros123

We know if $W$ is a d-dim subspace of n-dim $V$, then there exists $f_{d+1},f_{d+2} \cdots f_{n} \in \mathfrak{L}(V,\mathbb{F})$, (i.e, the dual space) so that $ker(f_{d+1})\cap ker(f_{d+2}) \cap \cdots \cap ker(f_{n})=W$, and, we find these same functionals form the basis for the annihilator of $W$, $W^0$. But I ask the reverse, it neednt be true that, given a basis of $W^0$ ($f'_{d+1} \cdots f'_n$), we can say that $ker(f'_{d+1})\cap ker(f'_{d+2}) \cap \cdots \cap ker(f'_{n})=W$?

Thomas Finley

"Let X be a metric space and let $x\in X.$ Suppose $x$ is an interior point of some subset A (of $X$) when we consider a topology say, $T$ on $X.$ This means, $x$ is an interior point of $A$ if we are considering the topological space as $(X,T).$ Will $x$ also be an interior point of $A$ when we consider $(X,d)$ as a metric space ?

@Jakobian I've framed my question accurately.

Ok, u say that it's not the case in general.

Well I agree with you.

Here's my reasoning:

Jakobian

If there's no connection between the metric and the topology then we don't really have anything to go off of. The choices here are pretty arbitrary

Xander Henderson

@user20458579510081670432 What absolute twaddle.

@ThomasFinley Let me rephrase your question: let $X$ be a set, and let $T$ and $T'$ be two topologies on that set. If $U$ is open in $(X,T)$, must $U$ be open in $(X, T')$?

Thomas Finley

14:02

Let $X=\Bbb R^2$ be the metric space with the Euclidean metric. Let A be the ' "inside" of the unit circle centered at the origin O.' Let $T$ be a topology on $X$ such that $T=\{(6,0),(7,0),\Bbb R^2,\emptyset\}.$ Then, there does not exist an open set $U$ (,i.e. a member $U$ of $T$ since, memebers of $T$ are called open sets...) such that, $(0,0)\in U$ satisfying $U\subseteq A.$ So, this means, $(0,0)=O$ is not an interior point of $A.$

However, if we consider $(X,d)$ as a metric space then obviously, $O$ is an interior point of $A.$

Xander Henderson

@ThomasFinley Your $T$ is not a topology. It is not closed under unions.

Sine of the Time

you're missing an element

Thomas Finley

@XanderHenderson I am using the defn involving open sets of a topology

@SineoftheTime you're right

@XanderHenderson I get it.

Jakobian

@nickbros123 if $f'_i(v) = 0$, then $f_i(v) = 0$ as well, so $v\in W$

Xander Henderson

@ThomasFinley Okay, fine. That set isn't a topology.

It isn't closed under unions.

Thomas Finley

14:05

I missed the element $\{6,0),(7,0)\}$ in $T.$ Ok, so, please consider my $T$ with this missing element.

@XanderHenderson I get it. The element $\{6,0),(7,0)\}$ was missed.

Jakobian

Too complicated for me. Just take discrete and usual topology on R and consider any singleton

Thomas Finley

Here's a corrected version of my argument:

Soumik Mukherjee

@Jakobian thanks, now I understand the proof

Thomas Finley

Let $X=\Bbb R^2$ be the metric space with the Euclidean metric. Let A be the ' "inside" of the unit circle centered at the origin O.' Let $T$ be a topology on $X$ such that $T=\{\{(6,0)\},\{(7,0)\},\{(6,0),(7,0),\Bbb R^2,\emptyset\}.$ Then, there does not exist an open set $U$ (,i.e. a member $U$ of $T$ since, memebers of $T$ are called open sets...) such that, $(0,0)\in U$ satisfying $U\subseteq A.$ So, this means, $(0,0)=O$ is not an interior point of $A.$

However, if we consider $(X,d)$ as a metric space then obviously, $O$ is an interior point of $A.$

Xander Henderson

@ThomasFinley I genuinely do not understand what it is that you are trying to show, here.

Thomas Finley

14:10

Phew! Finally it seems perfect to me.

@XanderHenderson I am trying to show:

A point $a\in A$ may or may not be an interior point of $a$ if we consider $X$ as a metric space or a topological space with some arbitrary topology $T$ on $X.$

That's what I wanted to show.

Jakobian

There is no reason for this example to be so complicated

Thomas Finley

@Jakobian Actually, it came naturally in my head :?). It just popped out of nowhere. But won't you agree with my argument?

Xander Henderson

@ThomasFinley Again, what you seem to want to show is that a space can have multiple topologies defined on it. Why make the argument so very complicated?

What is the end goal?

Thomas Finley

@XanderHenderson A point $a\in A$ may or may not be an interior point of $a$ if we consider $X$ as a metric space or a topological space with some arbitrary topology $T$ on $X.$

That's my end goal.

Xander Henderson

Yes, and?

Why?

Why do you care about that statement?

Thomas Finley

14:15

@XanderHenderson I was having a confusion regarding it's validity. So, I went on showing it.

Soumik Mukherjee

@ThomasFinley Whether a subset is open or not depends on what topology you give to the whole set

Thomas Finley

But I hope my arguments are accurate, right? I tried to make my arguments as precise as possible.

@SoumikMukherjee yes.. That's what it seems to me.

Xander Henderson

@ThomasFinley This is not about how it "seems to you". It is a fact: whether a set is open or not is entirely dependent upon what topology you have endowed the set with.

One can endow a set with many different topologies, and these different topologies will have different properties.

This should not be surprising...

Thomas Finley

@XanderHenderson I understand.

But one question is: I know that my arguments seems complicated. But does it seem "valid" to you?

It does to me. But is it really is? Or did I miss anything?

Xander Henderson

@ThomasFinley I don't understand what your actual end goal is. If all you you want to do is show that a point can be in the interior of some set in one topology, and not another---you've done that. But this is hardly a surprising or interesting result, and there are likely much more minimal examples.

At the very least, why construct an example in $\mathbb{R}^2$, when $\mathbb{R}$ would work? Why consider the indiscrete topology with two extra singleton points thrown in, when adding a single point would do the job?

You've made the argument overly complicated.

And I still don't understand why you want to show that a point can be in the interior of a set with respect to one topology, but not another.

Thorgott

14:31

@Ben In general, if you have a (pointed) map $f\colon X\rightarrow Y$ and consider the the Barratt-Puppe sequence $X\rightarrow Y\rightarrow C_f\rightarrow\Sigma X\rightarrow\dotsc$, there is also a coaction $C_f\rightarrow\Sigma X\vee C_f$ obtained by collapsing a horizontal copy of $X$ in $C_f$ to a point.
If you now consider (pointed) maps into a space $T$, you obtain the usual LES of pointed sets, but also a map $[\Sigma X,T]\times[C_f,T]\rightarrow[C_f,T]$, which turns out to be a group action. You can check that the orbits of this action are precisely the fibers of the restriction map…

That said, we still need more input to actually finish the computation...

Thomas Finley

@XanderHenderson I get your point. Also, you've improved my concept of this "whole" thing, as usual. Thanks a lot! I think I have mentioned it before, but just in case if I hadn't: Having a conversation with you is always meaningful and a delight for me!

Ben Steffan

@Thorgott Thank you! I've found this in More Concise as well.

Thorgott

ah, neat that it's in there

this fact is slightly under-appreciated imo

Ben Steffan

yeah

perhaps more interesting is the case for fibrations

Thorgott

yeah, that recovers monodromy

Ben Steffan

14:44

not just that, but stuff like "the l.e.s on homotopy groups associated to a fibration is a sequence of $\pi_1 E$-groups"

you can define a bunch of actions, all of which turn out to be compatible

Pizza

14:57

I have a question, I evaluated the integral $\int 2x \sin x dx = -2x \cos(x) + 2\sin(x) + C$ , now I have to evaluate $\int \frac{2x\sin x}{e^x} dx$, Is there any way to use the fact that I already know the numerator? Or does the integral change completely?

Otherwise I can write $2 \int x \cdot e^{-x} \sin(x)$ and integrate by parts where $f = x$ and $g' = e^{-x} \sin(x)$

Thorgott

do you have an argument that the composition $\mathbb{R}P^{2n+1}\rightarrow S^{2n+1}\rightarrow S^{2n}$ of the collapse and the suspended Hopf fibration is non-trivial?

Sine of the Time

I don't think knowing an antiderivative of $2x\sin x$ helps, since then if you integrate by parts you get $\int x\cos x e^{-x}dx$ and $\int \sin x e^{-x}dx$

cheat and use wolfram :)

Ben Steffan

@Thorgott The composition $S^{2n + 1} \to \mathbb{R}\mathrm{P}^{2n + 1} \to S^{2n + 1}$ where the first map is the covering induces multiplication by 2 on $H_{2n + 1}({{-}})$, I think?

Pizza

@SineoftheTime This integral seems quite long and complicated to me

Because just to find g' i have to integrate by parts

Sine of the Time

yeah I know

did this appear in an exam?

Thorgott

15:07

yeah, but the suspended Hopf map is $2$-torsion

Ben Steffan

right

Pizza

@SineoftheTime Actually I'm doing a differential equation that I've already done with the similarity method, I wanted to see what came out with the variation of the parameters

Sine of the Time

ok

since you already know the procedure, I think you can use Wolfram

Pizza

Yes, but I think that when there is a sin and cos i must proceed by similarity

in this case it was a little different

Because it is $y'' - y = 2x \sin x$

So I was in the case of $Q(x) \sin(\beta x)$

Then I checked whether $i\beta$ was a solution of $\lambda^2 - 1 =0$, where $i^2 = -1$

And since it was not a solution then I was in the case $(A_0 + A_1 x) \cdot \cos(x) + (B_0 + B_1 x) \cdot \sin(x)$

Sine of the Time

15:22

looks good to me

nickbros123

15:59

@Jakobian could you elaborate? I m not sure I am following

Ben Steffan

16:09

@Thorgott I can't think of anything that works.

Pizza

@SineoftheTime 👍

Sine of the Time

what problem are you working on?

Thorgott

yeah, me neither

this question determines if $[\mathbb{R}P^{n+1},S^n]$ contains $1$ or $2$ elements for odd $n$

for even $n$, it contains between $2$ and $4$ elements and the situation is even less clear to me

this might be a good question to post on main

Ben Steffan

Do you want to do the honors? You've definitely gotten farther than me (I don't even quite see how you get to that case distinction for odd $n$).

Thorgott

16:39

ok, I'll do it in a bit

zeta space

16:51

just don't use math frak

Pizza

@SineoftheTime I just finished this exercise: Study the following differential form $\omega(x,y) = \left(\ln(x+y) + \frac{x}{x+y}\right) dx + \frac{x}{x+y} dy$, and calculate the work along the parabola arc $y=x^2$ in the interval $(1, 3)$

Xander Henderson

mathfrak was invented by Satan himself!

zeta space

I used it for one letter

Xander Henderson

It smells of brimstone, and tastes of fire.

zeta space

one letter!

nickbros123

16:53

@XanderHenderson it looks beautiful, what do you mean

$\mathfrak{L}(V,W)$ wow thats beauty

Xander Henderson

@nickbros123 THE POWER OF MATHSCR COMPELS YOU!

nickbros123

just testing: $\mathscr{L}(V,W)$

damn this also looks beautiful

Pizza

I found that the partial derivatives match, $D = \{(x,y) \in \Bbb R^2 : x + y > 0\}$ , so it should be a simply connected set, so I calculated the work as $W = \int_{\gamma} \omega = \phi(3,9) - \phi(1,1)$

theoretically so because if $x \in [1,3]$ then when $x=1 \Rightarrow y =1$ , when $x=3 \Rightarrow y= 9$

Sine of the Time

$\gamma(t)=(t,t^2)$, $\gamma(1)=(1,1)$ and $\gamma(3)=(3,9)$

@Pizza yep, I checked and the p.d. indeed match

Pizza

So I started calculating $\int \ln(x+y) dx + \int \frac{x}{x+y} dx$

Sine of the Time

17:02

to see if the result matches?

@Pizza what you wrote here does not make sense

Pizza

Why? $\omega = d\phi$

Sine of the Time

yes

Pizza

$\frac{\partial \phi}{\partial x} = \ln(x + y) + \frac{x}{x + y}$

Shouldn't I start by integrating this?

Sine of the Time

It's better to start with $\phi_y$

zeta space

I just posted a non math frak question 🤭

let's compare statistically

Pizza

17:09

@SineoftheTime But is it wrong as I did? I don't understand :(

Sine of the Time

@Pizza yes, I misinterpreted your previous message

@Pizza no, it's ok

Pizza

yes maybe I should have specified what I meant

zeta space

I gotta farm now

not a farm with cows or pigs

Pizza

@SineoftheTime Because it's easier if I start like this ?

Sine of the Time

yes :)

try to belive

Pizza

17:12

But in this case because the integral is simpler, it's not always like this, right?

in the sense that if the integral was simpler on $\phi_x$ it was better to start from there

zeta space

@Debug $$\sum_{p ~\mathrm{prime}}(\log p^2)~ K_1 ~(\log p^2) \le1.$$

prove this

Pizza

Anyway I'll try now

zeta space

I'm a big strict upper bound guy

nickbros123

My course forcing me to let go of the love of my life Hoffman & Kunze- Linear algebra to pick up Sheldon Axler -Linear Algebra. At my lowest rn :'(

zeta space

woww $p^2$ implies primes squared...

Sine of the Time

17:15

@Pizza correct

Pizza

@SineoftheTime however from the integral $\int \frac{x}{x+y} dy$ I get $x+y-y\cdot \ln(x+y) + C$ so I can also "cancel" that $x$ because it would be contained in $C$

Sine of the Time

how do you get that expression?

I get $x\log(x+y)$

Pizza

wait ....

I got confused with the previous one

Sine of the Time

To be precise, it should be $x\log(x+y)+c(x)$

Pizza

I corrected it

@SineoftheTime yes

Jakobian

17:26

@nickbros123 I can. But what on

Pizza

@SineoftheTime $\frac{\partial \phi}{\partial y} = \frac{\partial }{\partial y}x\log(x+y)+c(x)$

Jakobian

If $f'_i$ is a basis of $W^0$ then you can write $f_j = \sum a_i f'_i$. So $f_j(v) = \sum a_i f'_i(v) = 0$

Sine of the Time

@Pizza that's correct, but does not add new infos

since $\phi=x\log(x+y)+c(x)$, now compute $\phi_x$ and let it equal to $\omega_1$

Pizza

@SineoftheTime I need to calculate the partial derivative of xlog(...) + c(x), right?

Sine of the Time

yes, but wrt $x$

Pizza

17:34

Yes, so it was $\frac{\partial \phi}{\partial x} = \frac{\partial }{\partial x}x\log(x+y)+c(x)$

Sine of the Time

go on

nickbros123

@Jakobian are you saying that since $\{f'\}$ functionals can be represented as linear combinations of $\{ f\}$, if $\alpha$ is in the kernel of every $f_j \in \{f_j\}$, then $\alpha$ is in the kernel of the particular $f'_i$ we are writing as a linear combination?

Pizza

$\ln(x+y) + \frac{x}{x+y} + c'(x) = \ln(x+y) + \frac{x}{x+y}$

nickbros123

@Jakobian so if $\alpha \in ker(f_{d+1})\cap ker(f_{d+2}) \cdots \cap ker(f_{n})$, then $\alpha \in ker(f'_{d+1}) \cap ker(f'_{d+2}) \cdots ker(f'_{n})$, and the same procedure can be done by treating $\{ f'_j \}$ as the primary basis and representing $\{f_i\}$ as a linear combo to get the reverse result, so the kernel intersections must be one and the same

Pizza

$c'(x) = 0$

nickbros123

17:38

is this what u were saying?

Pizza

so I get $x \ln(x+y) + 0$

Sine of the Time

one primitive is $x\log(x+y)$

In general $\phi=x\log(x+y)+k$ with $k\in \Bbb R$

Pizza

oh no, there's still one step missing

Sorry

Xander Henderson

@SineoftheTime Nonsense! $\phi = \frac{1}{2}(1 + \sqrt{5})$. Silly goose!

Pizza

$\int c'(x) dx$ = $\int 0 dx$, so $c(x) = k$

Xander Henderson

17:42

:P

Jakobian

@nickbros123 not exactly but what you have is good too

Sine of the Time

:(

@Pizza you can choose $k=0$

Jakobian

There's so much integration in the chat that I've stopped paying attention

Gian's Pizzeria

@XanderHenderson what is the best letter to make the substitution?

Sine of the Time

@Jakobian sorry for flooding the chat with my computation :(

Pizza

17:46

same

Jakobian

I'm not complaining or anything

It's nice to see the chat active, even if I totally ignore whatever is happening

I just wouldn't be able to keep up with it unless I'd dedicate myself to it

Pizza

@SineoftheTime Anyway I got $3 \log(12) - \log(2)$

zeta space

honestly my best question ever

has 1 upvote

Pizza

@Jakobian 👍

zeta space

my worst has like 33

strange

Sine of the Time

17:52

@Pizza ggs

zeta space

1

Q: Does there exist a harmonic map $F: (S,g_0)\to T^2 $ for which $F(\alpha_u(t))$ and $F(\beta_v(t))$ are all geodesics?

Let $S=[0,1]^2$. Ignoring issues having to do with boundaries and corners, a chart is a diffeomorphism $\varphi \colon S \to S$. Let $g_0$ denote the flat (euclidean) metric. Given a chart $\varphi \colon (S, g_0) \to (S, g_0)$, which need not be an isometry, consider the curves $\alpha_u(t)=\var...

my best question of all time

actually i should ask that on Mathoverflow because the answer is only a reformulation

Gian's Pizzeria

🍟🍔🍕🍫🥐🥯🌮🥪🌭🍩🍪🍎🍓🥝🍐🌽🥦🧅🥕🥦🥒🍅🌶️

zeta space

where are the fruits and veggies?

Pizza

@Gian'sPizzeria Can I buy something?

Gian's Pizzeria

Yes

Pizza

17:59

I want pizza and chips

Gian's Pizzeria

ok!

🚛💨💨💨

I sent the products

🍔🍫🥐🥯🌮🥪🌭🍩🍪🍎🍓🥝🍐🌽🥦🧅🥕🥦🥒🍅🌶️

Now i have this

zeta space

18:46

I'm high on math right now

I proved the pythagorean theorem^

PWOW (proof without words)

$$I(s)=2\sum_{n\in \Bbb N} \int_{J=(0,n^{-s/2})} e^{\frac{\log^2({n^{s/2}})}{\log t}}~dt$$

Xander Henderson

19:27

@Gian'sPizzeria No such animal. All of math is overloaded. I was making a dumb joke.

Gian's Pizzeria

I meant the best letter to use when substituting in integrals, for example wolfram uses u, but others use t.

Xander Henderson

@Gian'sPizzeria It doesn't matter.

Jakobian

@zetaspace I'm high (on math?) right now

Ben Steffan

20:02

It's very funny that one of the ways Lurie recommends to deal with size issues in HTT is to simply ignore them

There's no problem as long as you don't acknowledge there's a problem, right

and size issues are for chumps anyways

mr_e_man

The field $\mathbb F_2(x)$ has no roots of unity (other than $1$). For $y\in\mathbb F_2(x)$, can an extension $\mathbb F_2(x)(\sqrt y)$ contain a root of unity?

Ben Steffan

@mr_e_man is the answer not immediately "no" for degree reasons?

mr_e_man

I don't see it.

Ben Steffan

20:28

I may be wrong.

psie

21:11

I have a basic doubt. Does an outer measure satisfy continuity from above and below? The reason I am asking is because I am reading a solution to an exercise in Folland's (exercise 18b in section 1.4); we have a sequence of sets such that $\mu^\ast(A_k)\leq 1/k$, where $\mu^\ast$ is an outer measure. Then we put $A=\bigcap_1^\infty A_k$, and it is claimed that $\mu^\ast(A)=0$.

Jakobian

@psie not in general

psie

ok

Jakobian

Note that an outer measure is always monotone, so $\mu^*(A)\leq \mu^*(A_k)\leq 1/k$

psie

ah, hence the claim, I see. Thanks.

Jakobian

@psie to elaborate, I think it does assume continuity from one side, but not the other

but, measures don't even have continuity from above and below in general so...

probability measures do, but that's different

zeta space

21:19

that is awesome

psie

@Jakobian do you have an example of a measure that doesn't? I thought that it is a simple consequence of countable additivity of measure that if $(A_i)$ is an increasing sequence of sets, then $\mu\left(\bigcup_1^\infty A_i\right)=\lim_{i\to\infty} \mu(A_i)$ (and similarly for decreasing sequence of sets).

Jakobian

@psie Lebesgue measure

leslie townes

psie you at least have stuff like e.g. the decreasing sequence of sets [n, infty) in R, n = 1, 2, 3, ..., where the measure of each set is infinite but the measure of the intersection is 0

Thorgott

@Ben math.stackexchange.com/questions/4964331/…

@BenSteffan yet he loves talking about regular cardinals when it comes to accessibility

psie

@leslietownes true, for decreasing sequence of sets we need to add the condition that $\mu(A_n)<\infty$ for some $n\in\mathbb N$.

Ben Steffan

21:28

@Thorgott Nice, I'll look at it in a sec

To be entirely fair, he does offer it as one of three options

Jakobian

@psie the one for decreasing sequences should still be false for outer measures, I think, because outer measures act badly with respect to the set differences, even finite outer measures

@psie yeah it's false

psie

ah, ok

ILikeMathematics

22:28

So, we start with the RHS and try to simplify it somehow to end up with trace, right? But what can we simplify there..?

psie

22:58

Let $\mathcal A\subset\mathcal P(X)$ be an algebra, $\mathcal A_\sigma$ the collection of countable unions of sets in $\mathcal A$, and $\mathcal A_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal A_\sigma$. Let $\mu_0$ be a premeasure on $\mathcal A$ and $\mu^\ast$ the induced outer measure.

Is it true that if $B\in\mathcal A_{\sigma\delta}$, then $B$ is $\mu^\ast$-measurable?

Background behind my question; this is from an exercise in Folland's book (Exercise 1.18), and I think I have a solution if the above is true, but I'm not sure about.

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$Mathematics$ Mathematics