Mathematics - 2018-05-01 (page 1 of 2)

user193319

00:03

Isn't $\prod_{n=1}^\infty \{0,1\}$ uncountable and contained in $\ell^\infty$? If so, why wouldn't the hint 'hint at' that set? The hint as it stands doesn't make very much sense.

PVAL-inactive

01:03

"contained in $\ell^\infty$?" no of course not

Topologicalife

Is there a physics chat room?

I think my question should be there despite of being a math question.

Akiva Weinberger

The h bar @Topologicalife

Topologicalife

Ty.

Nicholas Roberts

02:50

Yo, is the 1-torus simply S^1, the circle?

GFauxPas

@NicholasRoberts I've never heard anyone talk about a 1-torus but that would be what it is

it's the product of one circle

Nicholas Roberts

Haha thanks, trying to prove de rham cohomologies of the n-torus using induction and needed a base case!

Daminark

Apparently in harmonic analysis, people refer to the circle as the torus

Mike Miller

03:07

@NicholasRoberts the point

Semiclassical

Yeah, you see $\mathbb{T}$ in that context

Hmm. mathbb or mathbf, I forget

Nicholas Roberts

@MikeMiller what?

Mike Miller

is a good base case

Nicholas Roberts

Lol really? Or are you joking

Mike Miller

$T^0$

Nicholas Roberts

03:10

Mhmm, so its de rham group is 1-d

Mike Miller

and proving that $H^*(T^0) = \Bbb R$ living in degree 0 is not so hard :D

Nicholas Roberts

H* is de-rham group?

Mike Miller

yes

the usage of * instead of an integer is usually meant to say "consider this as the degree varies"

Nicholas Roberts

Ah nice. But once you go past 0, the de-rham group a point will be 0, right?

Mike Miller

yes

Mike Miller

03:30

hi julian

Corellian

04:12

@LeakyNun Thanks. I suppose all's good as long as the context is clear. Although Z/nZ shows the quotient structure too

Secret

04:43

blogs.ams.org/inclusionexclusion/2018/04/30/…

Ethics of research and education mathematics

Hawk

Is any1 here

Silent

05:37

@LeakyNun Thank you very much

Silent

05:47

@LeakyNun, is it true that every cyclic group of prime order is simple?

@LeakyNun, oh! prime order groups have only two subgroups, namely $<1>$ and $G$ itself, hence simple. Is this correct?

Leaky Nun

06:03

@Silent yes

Silent

thanks

Akiva Weinberger

Don't ask, just listen

clyp.it/w0mo4okz#

Leaky Nun

@AkivaWeinberger has leido mi mensaje?

Secret

07:31

@AkivaWeinberger Original?

Secret

[Random]
Langlands program and Hilbert program lead to many breakthroughs in understanding of diverse areas of mathematics

But can we do better, by studying the formulation, patterns and other properties of an abstract program themselves?

ALannister

I have a very basic calc question

If you have the double integral $\int\int_{R} ye^{xy}dA$ over the rectangle $R = [0,3]\times [0,1]$, when you write it as an iterated integral, does it become $\int_{0}^{3}\int_{0}^{1}xe^{xy}dydx$?

Why does the integrand change?

Tobias Kildetoft

@ALannister It didn't really. You just switched which variable was called x and which was called y

Balarka Sen

@ShaVuklia If you haven't solved it yet, shoot

ALannister

@TobiasKildetoft but how does that make any sense mathematically?

Tobias Kildetoft

07:43

@ALannister the variables are just labels

ALannister

I have to teach this to a bunch of confused first years in a few hours and I doj't even understand it myself

Leaky Nun

@ALannister can't you write it as $\displaystyle \int_0^3 \int_0^1 ye^{xy} \ \mathrm dy \ \mathrm dx$?

ALannister

I don't see why not, but I was given solutions to this problem where it is not done that way, but no further explanation is given

Balarka Sen

Apply the change of coordinates $(x, y) \mapsto (y, x)$ to Leaky's integral.

ALannister

How can I explain that to freshmen though? I don't think change of coordinates is commonly used terminology and it will just confuse them.

Is there a way to do it graphically?

Balarka Sen

07:54

Yep.

ALannister

the change of variables, I mean

0

Q: Double Integral as Iterated Integral: Integrand changes and I don't understand reasoning behind it.

I need to compute the integral $\int \int_{R} y e^{xy}dA$ for $[0,3]\times [0,1]$. In the solutions, it has the double integral rewritten as $\int\int_{R} y e^{xy} dA = \int_{0}^{3} \int_{0}^{1} x e^{xy} dydx$. I'm assuming that some change of variables has taken place here in order for the int...

I posted it on main

Balarka Sen

So think of the two variable function $f(u, v) = v \exp(uv)$, and you want to compute the double integral $\int_R f dA$ (I'm purposefully not writing the coordinates inside this non-iterated double integral). When you turn this into an iterated integral, you're parametrizing the region.

So you're setting up an $x$ and $y$ coordinates for the region $R = [0, 3] \times [0, 1]$

Let the $y$ coordinate be the $[0, 1]$ coordinate and $x$ coordinate be the $[0, 3]$ coordinate. OK?

Leaky Nun

@ALannister the variables inside are "dummy variables". you can change y to u, and then x to y, and then u to x ^^

ALannister

Someone just posted a comment to my question saying "they are not equal. maybe it is a typo?"

And you know, I found where I had done the same problem a few years before, but not the way the solutions had it done, and I had no issues

Balarka Sen

Now to write $\int_R f dA$ as a double integral, you do follow these steps: (1) Integrate $f$ over the horizontal lines $x = c$. This is $\int_0^3 f(x, y) dx = \int_0^3 ye^{xy} dx$. (2) Now integrate this function over the vertical lines $y = c$. That's $\int_0^1 \int_0^3 ye^{xy} dx dy$.

Oh, huh, I guess the limits are switched in your integral. If you write $(x, y)$ as $(y, x)$ that's $\int_0^1 \int_0^3 xe^{xy} dy dx$.

I didn't notice that.

@ALannister They incorrectly plugged in the limits of the integral, but otherwise the crux of the solution is correct.

ALannister

08:10

I just got a new solution.

I mean a new answer to the question I posted

Balarka Sen

That seems to be what I just wrote down

Also re the comment, they are switching the variables in the second equality of the second line.

$x$ is rewritten as $y$ and $y$ is rewritten as $x$. The point is the limit gets switched, just as happened in my computation above ^

konoa

08:35

Hi everyone

Let $L,M,N \in \mathbb{P}^5$ 3 general subspaces of codimension 3, and let $l_i$ (resp $m_i,n_i)$, $i=1,2,3$ be three general points on $L_i$ (resp. $M_i,N_i)$.
I can't understand why $I_{L\cup M\cup N}(3)$, i.e. the space od all cubics through these subspaces (with all first partial derivatives vanishing at these 9 points), has dimension 26

can anybody help me figuring out why this works?

Vrouvrou

09:41

Hello, Someone help me to find $f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilone[)$ where $f(x)=\begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0\end{cases}$

$f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilone[)=\{y\in \mathbb{R}, f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varespsilon[\}$

user8469759

0

Q: Consequences of theorem 1.10 Rudin functional analysis.

Theorem 1.10 states Let $X$ be a topological vector space, $K$ compact, $C$ closed and $K \cap C = \emptyset$ then there's a neighborhood of $0$ such that $$ (K + V) \cap (C+V) = \emptyset $$ As consequence of this we have later Since $C+V$ is open then $\overline{K+V}\cap (C+V) = \empt...

functional-analysis proof-explanation

please help

Vrouvrou

let $y<0$ then $f(y)=0$ so $f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varespsilon[ $ if $x^2+1-\varepsolon <0$ that is $ \varepsilon >x^2+1$

@TobiasKildetoft hello

Tobias Kildetoft

@Vrouvrou It is spelled epsilon

Vrouvrou

?

i don't understand

Tobias Kildetoft

you managed to misspell it in like 5 different ways, all of which make it show the latex code rather than the symbol

Vrouvrou

09:53

sorry

find $f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)$ where $f(x)=\begin{cases} 0,~\text{if}~ x<0\\ x^2+1,~\text{if}~ x\geq0\end{cases}$
$f^{-1}(](x^2+1)-\varepsilon,(x^2+1)+\varepsilon[)=\{y\in \mathbb{R}, f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[\}$

let $y<0$ then $f(y)=0$ so $f(y)\in ](x^2+1)-\varepsilon, (x^2+1)-\varepsilon[ $ if $x^2+1-\varepsilon <0$ that is $ \varepsilon >x^2+1$

so if $\varepsilon >x^2+1$ $f^{-1}(y)=]-\infty,0[$ right ?

someone have an idea on how to do this ?

Sha Vuklia

10:15

@Balarka yea it's really just to check something about sign conventions

Balarka Sen

Okay

Sha Vuklia

it's a bunch of text, but little info:

Spivak says that the volume element of a 2-manifold in $\mathbb R^3$ is given by
$$
dA(x)(v_x,w_x)=\det\begin{pmatrix}v\\w\\n(x)\end{pmatrix},
$$
where $n(x)$ is the unit woutward normal at $x\in M$. While I do see this formula is correct, wouldn’t it have been better if they had written
$$
dA(x)(v_x,w_x)=\det\begin{pmatrix}n(x)\\v\\w\end{pmatrix},
$$
because then we could readily generalise to higher dimensions? (for $n-1$-manifolds on $\mathbb R^n$). If we keep $n(x)$ at the bottom row, we would get a factor $(-1)^{n-1}$ in front of our function. Example in $\mathbb R^4$:

(so I just wanna know if I really got it right)

Balarka Sen

@ShaVuklia I don't understand. $\text{det}(u, v, w, n(x)) = -\text{det}(n(x), u, v, w)$, but why would that mean you need to put a minus before the definition of $dA(u, v, w)$?

You're defining $dA(u_1, \cdots, u_k)$ to be $\text{det}(u_1, \cdots, u_k, n)$. No minus there.

Sha Vuklia

because we want $dA(e_1,e_2,e_3)=1$, and $(n(x),e_1,e_2,e_3)$ is right-handed by definition

Balarka Sen

Oh, I see, you want the value of $dA$ to be $1$ on the standard orthonormal basis. That makes sense.

Yes, you are absolutely correct.

Sha Vuklia

10:27

okay cool!

Balarka Sen

Aesop Rock - The Impossible Kid (Full Album Stream)

►

The recreated the whole of The Shining using puppets, absolute fucking madlads

Vrouvrou

10:43

you have no idea @TobiasKildetoft?

Silent

10:54

How to show that $\frac1a+\frac1b=\frac1{ab}$ does not have real solutions?

Astyx

Multiply by $ab$

a+b = 1

Has real solution

Silent

@Astyx I am so sorry, @Astyx, I meant: $\frac1a+\frac1b=\frac1{a+b}$

Astyx

Multiply by $ab(a+b)$

Solve the polynomial

And look for non-zero solutions that are not the opposite of each other (ie such that $ab(a+b)\ne0$)

Silent

11:21

@Astyx I am getting $(a+b)^2=ab$ but can't apply your argument'

What i think is, we can't take either a or b zero. Also, since it is square, we have to have $ab>0$, but we see then, that $(a+b)^2>ab>0$. Is this correct? @Astyx

Astyx

So $a^2 + b^2 + ab = 0$

This means the determinant $b^2 - 4b^2$ is positive

QED

Silent

@Astyx you mean, discriminant is negative?

Astyx

Yes

Silent

thank you so much

I know that every polynomial with complex coefficients has a solution in complex field. is it true that every 'multivariate polynomial' has solution in complex field?

@AlessandroCodenotti

Sha Vuklia

11:51

would someone be so kind to check if my calculations are correct?

(I don't really know how to check integrals with forms online)

so I have to calculate $\int_S x^2z\,dA$

and $S=\{(x,y,z):x^2+z^2=1,0<y<2,z>0\}$

o, technically I should have written $c^*(dA)=dy\wedge d\theta$, but that doesn't matter for the integral

Leaky Nun

12:13

@Silent any multivariate polynomial can become a monovariate polynomial by substituting 1 to the other values

Silent

@LeakyNun thank you

Sha Vuklia

o yea, I forgot a factor 2, because I integrated $0<y<1$. But I got the confirmation that it's correct

Vrouvrou

@Astyx bonjour

Mary Star

12:37

Hello!! Could someone of you take a look at my question: math.stackexchange.com/questions/2761486/… ?

Vrouvrou

Someone have an idea about this please : math.stackexchange.com/questions/2761574/…

Secret

13:27

Peter Scholze

Peter Scholze (born 11 December 1987) is a German mathematician known for his work in arithmetic algebraic geometry. He is a professor at the University of Bonn and has been called one of the leading mathematicians in the world. == Life == Scholze attended Heinrich-Hertz-Gymnasium in Berlin-Friedrichshain, a gymnasium with a mathematical/natural-scientific profile. As a student, he participated in the International Mathematics Olympiad, winning three gold medals and one silver medal. He obtained his Ph.D. from the University of Bonn in 2012 under the supervision of Michael Rapoport. He ha...

DinhHoang

13:57

math.stackexchange.com/q/2760283/557554 Could any one can help me it was there for such a time !! Thank you !! It is reccurence relation generating function and binomial coefficient!!

Akash. B

14:09

can anyone explain limit?

nbro

14:25

Hi!

Anyone here familiar with k-means?

Familiar to the point of being able of analyse the results...

Leyla Alkan

Hi, do you know which currency unit is this?

Akiva Weinberger

The cent

One hundredth of a dollar

100¢ = $1

Leyla Alkan

oh okay thanks

Akiva Weinberger

I'm guessing they don't use cents in Turkey?

Ah, I see, you use kuruşes

Kuruşlar

Leyla Alkan

yes, we dont use it

I was planning to get an online tutor, that was the cost for tutoring per minute

Anonymous

14:39

@LeylaAlkan Which site?

Leyla Alkan

chegg

Anonymous

Oh. Chegg is good. Tried it last year.

Waiting

I've recently found out from some students that at least one of my problems were considered there to be solved. I wonder how many days did they work on it.

Akiva Weinberger

$0.75 = ₺ 3

Secret

Time, is mysterious

Reverse time, even more so

Nehal Samee

14:49

Hello everyone .... I've a problem ... For finding sixth roots of $64$ , I do : $x^6=64$ → $x^6-64=0$ → $(x^3)^2 -(8)^2=0$ → $(x^3+8)(x^3-8)=0$ ...Am I doing the right thing ?

Anonymous

So far so good

Secret

The question of the backward ticking time bomb, is a mystery in its own rights

Anonymous

I wouldn't do that much. Just use De-Moivre's theorem.

Secret

When the mysterious piece that is immersed in a bluish green field filled with closed strings, played with time reversed, the result is a discovery, and the actions that followed after as it gets closer to the truth

9 hours ago, by Akiva Weinberger

https://clyp.it/w0mo4okz#

We are ancient beings that transcends the confines of space and time. We can rewind time at will and fast forward it to gain the full picture of The Puzzle

The required piece where the clock ticks forward and backwards is now listened to. It won't be long before the mystery is clarified

Backwards Running Clock

►

Nehal Samee

@Blue Talking with me ... ?

Anonymous

14:56

@NehalSamee Yes

Nehal Samee

@Blue ... Btw , I'm not well aware of Moivre's theorem ...

Anonymous

You should learn it. Very useful

Nehal Samee

@Blue Oh .. I saw that in complex number , but don't know the use in finding factors ...

Anonymous

Rewrite your equation as $z^{6}=64e^{i2k\pi}$

Anonymous

And then, $z=(64)^{1/6}e^{i2k\pi/6}$. Plug in $k=0,1,2,3,4,5$ to get the six roots.

Rithaniel

15:17

Hey, I'm having a little difficulty understanding the concept of an interval being compact. It's "if there is a cover of the interval, then there exists a finite subset of that cover still covers the entire interval," correct? My issue is that, can't you always construct a cover for which there is no finite subcover for the interval it covers?

nbro

15:31

Does anyone here know why the likelihood function is often written as $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$? I mean, the likelihood should be interpreted as a function of the parameter $\theta$, given fixed data $x$. But then why $f(x\mid\theta)$?

Anonymous

@nbro Those two are not the same.

Anonymous

$\mathcal{L}(\theta \mid x)\neq f(x\mid\theta)$

Anonymous

Likelihood function

In frequentist inference, a likelihood function (often simply the likelihood) is a function of the parameters of a statistical model, given specific observed data. Likelihood functions play a key role in frequentist inference, especially methods of estimating a parameter from a set of statistics. In informal contexts, "likelihood" is often used as a synonym for "probability". In mathematical statistics, the two terms have different meanings. Probability in this technical context describes the plausibility of a future outcome, given a model parameter value, without reference to any observed data...

Anonymous

Look up the definition section on that page.

nbro

@Blue But the definition of the likelihood function is exactly the equality of those two entities...

Anonymous

15:41

Wut?

Konformist Liberal

Hi, I was trying to show that for a quiver Q, the category of (finite, finite dimensional) Q representations, repQ has the following property: every epimorphism is surjective. I showed it for Q, who has no oriented cycles, but couldn't prove for the general case. Can someone please help? Is it obvious?

nbro

en.wikipedia.org/wiki/… You have that the likelihood function is defined as $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$. So, saying $\mathcal{L}(\theta \mid x) \neq f(x\mid\theta)$ is wrong.

I am pretty sure this is just a problem of communication and, in particular, notation

Anonymous

nbro

Yes, I also read that part

But I also read the part I am linking you to above

Anonymous

@nbro "You have that the likelihood function is defined as $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$" Which line?

Anonymous

15:44

I can't spot it on that page

nbro

Section: https://en.wikipedia.org/wiki/Likelihood_function#Likelihood_function_of_a_para‌meterized_model, 3rd equation.

I think I got the point

The likelihood function should be written as $\mathcal{L}(\theta \mid x)=f(x ; \theta)$

It's a joint density of the data as a function of the parameter, and not conditioned on the parameter!

My conclusion is: the notation $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$ is completely wrong, if we assume that $\mid$ means "left side conditioned on right side". In fact, the section of the article https://en.wikipedia.org/wiki/Likelihood_function#Likelihood_function_of_a_para‌meterized_model also states:.

> This is not the same as the probability that those parameters are the right ones, given the observed sample. Attempting to interpret the likelihood of a hypothesis given observed evidence as the probability of the hypothesis is a common error, with potentially disastrous consequences in medicine, engineering or jurisprudence. See prosecutor's fallacy for an example of this.

Maybe we should really not use that notation!!!

I would like to speak with the guy who first adopted it...

Anonymous

@nbro OK. But you skipped the main part of the definition.

Anonymous

Given a paramaterized family of probability density functions $x\to f(x|\theta)$

Anonymous

"where $\theta$ is the parameter"

Anonymous

Not $x$

Anonymous

15:58

$x$ is a constant

Anonymous

In that case, sure, $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$

nbro

Still, $\mid$ is usually used to denote "conditioned on".

Note: my problem has nothing to do with the definition

It's about notation

Anonymous

They have clearly mentioned it!

nbro

And? Who cares if they mention it?

Anonymous

16:00

Don't blame notation if you didn't read the whole definition properly.

nbro

It's like saying one plus one is two, but you write 3+2 = 4

Anonymous

You said the definition of likelihood function according to Wikipedia is $\mathcal{L}(\theta \mid x)=f(x\mid\theta)$, which is nonsense. While reading math definitions you don't skip half of it.

nbro

I understood the type of guy you are.

Bye.

Anonymous

Bye.

Silent

17:18

@AkivaWeinberger, this picture is from Conway's Complex Analysis. I think in part (c), it should be $\{z:|z-a|\le r\}$. Am I right?

Ted Shifrin

Yes, @Silent. Typo.

Mary Star

Let $\displaystyle{C=\{x\in \mathbb{R}^2 : x_1^2-x_2^2=1, x_1, x_2>0\}}$.

Let $f:(0,\infty) \to C$ with $f(\theta)=(\cosh(\theta), \sinh(\theta))$.
The function $f$ is continuous and $(0,\infty)$ is connected. So the image $C$ is also connected.

How can we check here if the set is path-connected?

Silent

thank you! @TedShifrin

Konformist Liberal

Hi, I was trying to show that for a quiver Q, the category of (finite, finite dimensional) Q representations, repQ has the following property: every epimorphism is surjective. I showed it for Q, who has no oriented cycles, but couldn't prove for the general case. Can someone please help? Is it obvious?

Tobias Kildetoft

@KonformistLiberal This holds for any ring, not just quiver algebras.

Konformist Liberal

17:28

@Tobi

@TobiasKildetoft I think it doesn't. Take the embedding from Z to Q. It is epi but not surjective.

Tobias Kildetoft

@KonformistLiberal I meant for modules, not the rings themselves

Konformist Liberal

I am not familiar to seeing a quiver representation as a module

Tobias Kildetoft

Ahh

Do you see that all images are also kernels?

Konformist Liberal

Yes

Tobias Kildetoft

that should help you see how to find a morphism showing that a non-surjective map is not epi

Konformist Liberal

17:33

Let me think for a minute

Is it exactly same with vectk?

Tobias Kildetoft

yes

Konformist Liberal

it seems like just knowing that ker and coker exist is enough

Is this proof can be used for any category which has ker and coker?

Leaky Nun

I heard category

Tobias Kildetoft

Well, and which is concrete to make sense of surjective

Konformist Liberal

17:50

Are all locally small categories concrete?

Daminark

Hey everyone!

Tobias Kildetoft

@KonformistLiberal No

(At least I don't think so)

I assume you mean whether they can be made concrete, as being concrete is extra structure, not a property of the category itself

The Integrator

hey , can anyone recommend me some books for self studying differential geometry? I dont know where else to ask this

Konformist Liberal

Is it fair to say that in a concrete category which has kernel and cokernel for every morphism, all epimorphisms are surjective (surjective in the sense that if F is the faithful functor from this category to Set, then F(this morphism) is surjective)

Yes that's what I meant

Daminark

@TheIntegrator are you looking into undergrad differential geometry? Stuff like curves and surfaces?

The Integrator

18:00

yeah

Tobias Kildetoft

I think you might also need all epis to be cokernels for the argument to work out (not sure if you need it for the statement though)

Daminark

Try Do Carmo, it seems quite good and my friends who are using it love it

Konformist Liberal

So + the category is binormal? en.wikipedia.org/wiki/Normal_morphism

The Integrator

@Daminark thanks :) I'll check it out . Is it easily understandable?

Daminark

Seems to be

The Integrator

18:05

nice . thanks for your input .Will see where i can get a copy.

Konformist Liberal

@TobiasKildetoft

Tobias Kildetoft

@KonformistLiberal I have not really thought much about what it takes for all epis to be surjective

Alessandro Codenotti

18:27

@Daminark Have you heard he died a couple of days ago?

Balarka Sen

Oh I didn't know that

Daminark

Yeah I did

It's a shame. Rest in peace

2

Konformist Liberal

@Tobi

@TobiasKildetoft Thank you

Tobias Kildetoft

you're welcome

skull

19:49

Manfredo do Carmo

Manfredo Perdigão do Carmo (15 August 1928 – 30 April 2018) was a Brazilian mathematician, doyen of Brazilian differential geometry, and former president of the Brazilian Mathematical Society. He was at the time of his death an emeritus researcher at the IMPA. He is known for his research on Riemannian manifolds, topology of manifolds, rigidity and convexity of isometric immersions, minimal surfaces, stability of hypersurfaces, isoperimetric problems, minimal submanifolds of a sphere, and manifolds of constant mean curvature and vanishing scalar curvature. He earned his Ph.D. from the University...

@AlessandroCodenotti yesterday

Alessandro Codenotti

20:06

Turns out I don't know which day it is today

(And I really should know, the 1st of May is a national holiday in Italy)

Daminark

@AlessandroCodenotti did you go to class and get confused as to why school was closed?

skull

np, pal

Fargle

@AlessandroCodenotti International Workers' Day. Also my birthday.

4

Daminark

Happy birthday @Fargle!

Fargle

Thanks @Daminark. Finally, a prime age again.

Daminark

20:20

Lmao, nice

Alessandro Codenotti

@Daminark no, luckily!

And happy birthday! @Fargle

Fargle

Thanks @Alessandro

skull

@Fargle Happy birthday, pal

Fargle

Thanks @skull

For my birthday today: category theory, TF2, and eventually board games.

Daminark

Sounds like a fun way to spend the day for sure

GFauxPas

20:25

you just have to find a way to combine all those things into one activity and you're golden

Sha Vuklia

does anyone know the precise statement of "the graph of an integrable function has content zero"?

what about the domain/codomain?

I have found some $\mathbb R^n\to\mathbb R$ stuff, but I need the codomain to be $\mathbb R^m$. Munkres in fact gives such a result concerning diffeomorphisms, where we have freedom in the dimension, so I would think that in the more general case of higher dimensions, we need a diffeomorphism, instead of just an integrable function?

Nehal Samee

20:47

@Blue Mmm ... How do I remove the $e$ from the answers ... ? And how can I apply it for $z^6= -64$ ... ?

user193319

If $f_n \to f$ uniformly and each $f_n$ has bounded support, must $f$ have bounded support?

Anonymous

@NehalSamee Euler's formula

Nehal Samee

@Blue I know , $e^(i\theta)=cos\theta + sin\theta $

Will it be okay ?

Anonymous

Yes

Leaky Nun

@user193319 I'm thinking of $f$ being $e^{-x^2}$

and the support of $f_n$ being $[-n,n]$

user193319

20:54

@LeakyNun Okay. But what is $f_n$?

Leaky Nun

I don't know

I think of $f_0 = 0$

I have an animation in my head

Nehal Samee

@Blue Thanks ...

Mary Star

21:10

I want to show that the set $(0, \infty)$ is path-connected. That means taht we have to determine a path between each point x,y, right? How could we find such a map? Could you give me a hint?

user193319

@MaryStar Given two points $x,y \in (0,\infty)$, form the line segment between them. Do you know how to write the function corresponding to this path? Think about convex combinations.

Alessandro Codenotti

Hint: do you know a way to parametrize a sement?

Mary Star

Ah so do we consider the function $f(t) = tx + (1-t)y$ ? @user193319 @AlessandroCodenotti

GFauxPas

I think it would be easier to show that (0,infinity) is star-shaped

But perhaps less informative

Star-shaped implies s.c.

Actually it's gonna be more or less the same argument

Mary Star

21:26

Ah ok! The function is defined as $f:[0,1]\rightarrow (0,\infty)$, isn't it?

GFauxPas

Looks right :)

Alessandro Codenotti

@GFauxPas well it's contractible if you really want to nuke this problem :P

GFauxPas

Lol can you say "this is trivial" as an answer to an assignment? :p

Mary Star

Thanks!!

I want to check also if $[1, 2]\times [0, 2\pi]$ is path-connected.
It holds that $[1, 2]$ and $[0, 2\pi]$ are path-connected, since there exist the functions $f(t) = tx + (1-t)y$ with $x,y\in [1,2]$ and $g(t) = ta + (1-t)b$ with $a,b\in [0,2\pi]$.

Does it follow that the product of two path-connected spaces again path-connected?

Leaky Nun

21:54

@AlessandroCodenotti heh...

that spelling

anakhronizein

22:18

Hi all.

mixedmath

o/

anakhronizein

22:30

Is there an easier de Rham cohomology to compute than $S^1$ which is not extremely trivial/one-line?

Daminark

inb4 "the punctured plane"

anakhronizein

>easier

my beloved daminark

Akiva Weinberger

22:47

@Secret Idea:

$\exists^{\exists^\forall}$

3

Balarka Sen

@anakhronizein !!

compute the de Rham cohomology of a point

very difficult

anakhronizein

Heh

R^n in general is easier than I am looking for.

Balarka Sen

hmm

anakhronizein

I think I will jsut use S^1.

Balarka Sen

Yeah I don't know anything easier

Konformist Liberal

22:56

ncatlab.org/nlab/show/simple+object can someone give a simple counterexample to prop 2.1 when the field is not algebraically closed? (An example through quiver representations would be also perfect) Btw, as I understand it, Hom(x,y) is a vector space, not the objects (just a semi-question about the meaning of enriched)

Balarka Sen

I remember that a long long time ago Mike gave me a series of exercises on this chat which were mostly multivariable calculus/vectors/line integrals mixed in oddly (or so I felt at that time) with path-homotopy invariance and topological ideas, and it eventually ended up leading to the computation of $H_{dR}(\Bbb R^2 - (0, 0))$

Good times

Daminark

Lol, I think the way my analysis prof had us compute it was something like

Show that for any closed form $\omega$, there is some unique $k\in \mathbb{R}$ such that $\omega - k\omega_0$ is exact

Balarka Sen

Aha

Daminark

Where $\omega_0 = \frac{-ydx + xdy}{x^2 + y^2}$ is the angle form

Balarka Sen

right

I didn't know forms at that time so that lingo was completely avoided

Daminark

22:58

Ah I see

Balarka Sen

i innocuously just integrated some plain vector fields lel

it was a good sequence of conversations which lead to baby Hodge theory

I dare not link it though

Daminark

Yeah we did it by something like, we proved that two curves in the punctured plane are homotopic iff they have the same winding number (which we defined as $\frac{1}{2\pi}\int_{\gamma} \omega_0$), and that a form is exact iff its integral over every closed form is $0$

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