Mathematics - 2024-08-21

Jakobian

00:03

@jasmine If $a'\in \mathfrak{m}$, that doesn't necessarily mean $a'\in \mathfrak{m}^3$. It's $\mathfrak{m}^3\subseteq \mathfrak{m}$ that's true, not necessarily the other way around

zeta space

01:20

@XanderHenderson Yes I have constructed a space that is isometric to $[0,\infty)$

Xander Henderson

@zetaspace Okay... so what?

zeta space

@XanderHenderson Give me a second to collect my thoughts before I answer that

Xander Henderson

@zetaspace I mean, I asked my first question four hours ago... you're clearly not in a hurry.

nickbros123

02:23

@XanderHenderson those are fair points. I have read spivak, and Bartle before, but they're just calculus books. I find rudin convenient because it aligns with the way I learn from books. Since rudin is theoretically self contained, I apply the 'moore method's when going through his book, i e, every theorem after the axioms / definitions are problems. Takes a lot of time but I enjoy it

But jakobian is correct in that rudin only sets out to get done what he would need to in his book, no more no less.

Xander Henderson

@nickbros123 I don't know what is meant by "just calculus" books. There is a whole spectrum running from the AP and A-level calculus texts used in the US and the UK, through college level books like Thomas and Stewart, up through Spivak and Apostal, and to Rudin and his ilk. Any line you draw between books is kind of arbitrary, and a lot of what makes a book a "calculus" text vs an "analysis" text is going to come down to how it is used in an instructional setting.

nickbros123

@XanderHenderson i differentiate them by metric space vs Reals, that's it. Other than that both spivak and Bartle are very rigorous, and all that

user 20458579510081670432

That is not the Moore Method.

nickbros123

I would call them intro analysis texts but some may argue that "intro analysis" is actually rudin

@user20458579510081670432 how so?

user 20458579510081670432

Moore used the Socratic method.

nickbros123

02:32

@user20458579510081670432 "..the students are given a list of definitions and, based on these, theorems which they are to prove and present in class, leading them through the subject material.." from Wikipedia ; isn't this what I said

Xander Henderson

@nickbros123 That seems pretty arbitrary. Particularly since many analysis texts don't deal with more abstract metric spaces.

user 20458579510081670432

What you described was independent study.

Xander Henderson

@user20458579510081670432 I didn't really agree. At the heart of the Socratic method is the idea that the leader (eg the teacher) didn't know truth any better than anyone what, and truth is arrived at through questioning.

user 20458579510081670432

Yes.

Xander Henderson

Discovery methods, like Moore (but we are supposed to pretend that Moore did't exist, because he was a horrible racist, even for his time), don't suppose that the instructor is ignorant, but rather that the instructor can guide students to figuring out truth with minimal prompting.

They look similar, but are philosophically distinct.

Arg.. mobile!

zeta space

02:49

hi and drunk call that HD vision

Do not let that influence your views on what I say next

your teachers probably didn't tell you this but if you change the underlying geometry of the Schwartz space $\mathcal S(\Bbb R^n)$ you effectively have a recipe to build a curved zeta function

cancel mob is probably gonna come after me now

Jakobian

03:41

@XanderHenderson Lovecraft established a whole genre of horror stories and he's being remembered, in spite being a horrible racist

I mean what next, we pretend Hitler didn't exist

Xander Henderson

@Jakobian Yes, and his name has been removed from awards, and the World Fantasy Award is no longer a Lovecraft bust.

And, arguably, Moore didn't really invent the method---he was just a popularizer and famous proponent.

That being said, I don't feel strongly about it, but I am tired of having the fight with colleagues. The dude was a racist, and it doesn't seem worth the effort to argue that we should keep his name around when he didn't really do much to contribute to anything.

Jakobian

His name still exists in such things as the Moore-Smith generalized sequences for example

user 20458579510081670432

The number of PhDs he supervised lives on.

Jakobian

and I doubt anyone even knows who the guy was, just like I don't know who, I don't know, Bolzano was

Xander Henderson

@Jakobian Sure, and I don't think that anyone is seriously arguing that his name be removed from that.

Anyway, like I said, I don't have strong opinions. But I have become weary of having the conversation with colleagues, who all tell me that I am a terrible person for calling anything the "Moore method". I don't really care if someone else uses the term, but if they do, they are going to eventually start running into people who will suggest other terms (e.g. "discovery based learning", "inquirey based learning", etc).

This is a case where there is a term with bad connotations, and there are replacements which are (a) more descriptive, and (b) don't carry those connotations. So... I find it easier to just use those other terms.

Jakobian

03:52

It should be okay to use a name, no matter how horrible the person might have been. For example, if Hitler invented some method of learning, we should call it the "Hitler method" if that's the common name for it

That's what I think

user 20458579510081670432

-_-

Jakobian

@user20458579510081670432 yes?

Xander Henderson

@Jakobian I'm not trying to convince you of anything, and I don't really care what you call a thing. I am reporting to you the experience I have had as an educator. Once you have been to a bunch of conferences from mathematics educators, and have gotten into multiple discussions about Moore, you may feel differently.

If you want to die on the hill of "I'll call a thing whatever the hell I want to call it!", be my guest. I, personally, don't think it is a hill worth dying on. I'd rather get on with the business of educating students.

Jakobian

If you assume that a) erasing history is bad practice and b) one should try to be consistent, then it should imply that one should try to use names which are most commonly attributed to someone, irrelevant of who that person was

Xander Henderson

@Jakobian (a) No one said anything about erasing history. One can remove a person's name from a thing without erasing history. Though if you understand history, you will come to recognize that what Moore was doing was not really all that radical, and lots of other people were doing similar kinds of things.

Stigler's law of eponymy

Stigler's law of eponymy, proposed by University of Chicago statistics professor Stephen Stigler in his 1980 publication "Stigler's law of eponymy", states that no scientific discovery is named after its original discoverer. Examples include Hubble's law, which was derived by Georges Lemaître two years before Edwin Hubble; the Pythagorean theorem, which was known to Babylonian mathematicians before Pythagoras; and Halley's Comet, which was observed by astronomers since at least 240 BC (although its official designation is due to the first ever mathematical prediction of such astronomical phenomenon...

user 20458579510081670432

03:57

Did you have any discussions about Moore at your last mathed conference? @XanderHenderson

Xander Henderson

@user20458579510081670432 Yes.

Jakobian

Okay this has not much to do with erasing history

Xander Henderson

@Jakobian (b) I generally think that it is better to name theorems according to what they do, rather than after some person. Give theorems descriptive names. The "extreme value theorem" is a good example. "Fermat's theorem" is a bad example (though "Fermat's theorem on stationary points" is a slight improvement). There is not need to memorize an arbitrary name if the name of the theorem explains what the theorem does.

Similarly, there are perfectly good words already in the language to replace "Moore's method" (e.g. "inquiry based learning" or "IBL"). IBL is well understood by most mathematics educators, and is more descriptive for people who are just learning to teach. Win-win.

Jakobian

Okay yeah, maybe my take was terrible

I mean it was, not maybe

user 20458579510081670432

I doubt physics will conform by renaming Newton's laws.

Xander Henderson

04:04

@Jakobian I didn't say that. I just said that I really don't care all that much in my own life, and I am tired of having the conversation with other math ed people. In general, it is a lot easier to just say IBL, and avoid having a long conversation about what a horrible racist, misogynist, and anti-semite Moore was. (Go Longhorns!)

@user20458579510081670432 Newton was a right bastard. Just ask Leibniz.

Anywho, it is after 9. I should have gone to bed an hour ago---8am class tomorrow! Yay! Toodles.

user 20458579510081670432

Yes, and he admitted on his death bed that his greatest accomplishment was dying a virgin.

Cya.

Jakobian

@user20458579510081670432 Newton?

user 20458579510081670432

Yup

Jakobian

They were thinking about it differently back then

nickbros123

Let's suppose we have an open cover for a metric space that is separable (which is equivalent to saying it has a countable basis), I am unable to convince myself that the open cover must have a countable subcover.

Say $\{ G_a: a \in A\}$ is the open cover. I understand each element $G_a$ can be written as the union of a subcollection of the basis, which is countable. But since the power set of natural numbers is uncountable, and though each $G_a$ is the union of an utmost countable subcollection, it seems like it could potentially be uncountable? I need help here

Jakobian

04:12

@nickbros123 Take a countable basis $\{U_n : n\in\mathbb{N}\}$ and consider $P = \{n\in\mathbb{N} : \exists_{a\in A} U_n\subseteq G_a\}$

If $x\in X$ then there exists $a\in A$ with $x\in G_a$ and then from definition of a basis $n\in\mathbb{N}$ such that $x\in U_n\subseteq G_a$. So $U_n$ for $n\in P$ cover $X$

Next for each $n\in P$ choose $a(n)\in A$ such that $U_n\subseteq G_{a(n)}$

Then $\{G_{a(n)} : n\in P\}$ is a countable cover of $X$, and clearly a subcover of $\{G_a : a\in A\}$

Or, if you prefer, letting $A_0 = \{a(n) : n\in P\}$, we have that $\{G_a : a\in A_0\}$ is a subcover of $\{G_a : a\in A\}$

nickbros123

oh, right. this makes sense, thanks

Peter

06:10

this should be deleted despite its score. I know the suitable chatroom for this is CURED but there are too few users , so I try it here.

Pizza

09:50

Hi👋

Soumik Mukherjee

10:19

Hi

Pizza

11:08

Given the surface $S$ of equation $z=2x+3y$, for $x^2 + y^2 \leq 1$, calculate the rotor flux of the vector field $F(x,y,z) = (y ,z,-x)$ through $S$

I have a question, but after I calculated the rotor , can I calculate the flux using the formula for the explicit form z = f(x,y)?

$\Phi = \iint_{S} \mathbf{F}(x, y, f(x, y)) \cdot \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \, dx \, dy$

I think I'm wrong because continuing like this i will bypasses the need for the rotor calculation

I think I should use Stokes' theorem

so I think I have to parameterize the surface

Sine of the Time

the surface is a plane right

Pizza

Yes

Sine of the Time

so it's easy to find the normal vector

Pizza

11:24

I didn't understand if I can also not express the surface in parametric form

Now I should have an explicit form z = g(x,y)?

Mm wait

$\mathbf{n} = \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right)$

Sine of the Time

@Pizza from linear algebra, what is the normal vector of the plane $ax+by+cz+d=0$?

@Pizza yes you can, but I don't understand why you would do it

Pizza

@SineoftheTime $\mathbf{n} = (a,b,c)$?

Sine of the Time

yes

Pizza

@SineoftheTime I'm getting confused with the things I'm reading

Sine of the Time

to compute the flux, you $n$ have to be a unit vector?

@Pizza what is confusing you?

Pizza

11:36

Maybe I'm thinking the method to calculate the direct flux of the vector field, not its rotor

@SineoftheTime yes

Sine of the Time

what do you mean by rotor flux?

Pizza

in italian: calcolare il flusso del rotore del campo vettoriale

I have to do this

Sine of the Time

did you compute the rotor of $\mathbf F$?

Pizza

Yes

Sine of the Time

what do you get?

Pizza

11:42

(-1,1,-1)

I'm stuck here, I don't know what to do next

or at least, I think I have to parameterize the surface

Sine of the Time

how did your professor define the flux of a vector field $\mathbf G$?

Pizza

@SineoftheTime It should be $\iint_S \mathbf{G} \cdot d\mathbf{S}$

Sine of the Time

ok, did you see you can express the flux using the normal unit vector?

Pizza

Yes, maybe I understood

wait

Sorry

$$\iint_{S} \mathbf{G} \cdot \mathbf{n} \, dS,
\mathbf{n} = \frac{\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}}{\left|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right|}
dS = \left|\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}\right| \, du dv$$

Sine of the Time

11:58

I'm not understanding

just apply the definition of flux to $\operatorname{rot} \mathbf F$

if you're struggling to find the normal unit vector, then parametrize

but you can find immediately $n$

Pizza

@SineoftheTime $\iint_{S} (\operatorname{rot} \mathbf{F}) \cdot \mathbf{n} \, dS$

Sine of the Time

ok, did you find $n$?

Pizza

r(x,y) = (x,y, 2x + 3y), I have to calculate the tangent vectors

$\mathbf{n} = (-2,-3,1)$

Sine of the Time

ok, you told me $n$ must be a unit vector so you have to normalize

Pizza

Yes

$$\| \mathbf{n} \| = \sqrt{(-2)^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14} \\
\mathbf{n} = \left( \frac{-2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}, \frac{1}{\sqrt{14}} \right)$$

Sine of the Time

12:09

ok

now plug in

Pizza

The scalar product is -2/√14

Sine of the Time

ok, I'm not checking the computation

Now you're left with the surface integral

Pizza

@SineoftheTime $-2 \iint_D dxdy$

Sine of the Time

are you able to compute it?

Pizza

I can use polar coordinates

Sine of the Time

12:21

you can

Pizza

$\int_{0}^{2\pi} \int_{0}^{1} -2 \cdot r \, dr \, d\theta = -2\pi$

But then

$\Phi = \iint_{S} \mathbf{F}(x, y, f(x, y)) \cdot \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \, dx \, dy$

Could I replace this F here with rot F?

Sine of the Time

isn't the flux a surface integral?

Pizza

Yes

Sine of the Time

why instead of $z$ you put $f(x,y)$?

it's not clear what are you doing here

maybe I understood

you parametrize already

Pizza

I saw this on the internet, maybe I misunderstood?

Sine of the Time

12:30

If you know how to solve surface integrals, you don't need other formulas

Pizza

I was reading this

Sine of the Time

yes, this works

Pizza

$\mathbf{n} = \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right)$

but $dS = \sqrt{\left( \frac{\partial f}{\partial x} \right)^2 + \left( \frac{\partial f}{\partial y} \right)^2 + 1} \, dx \, dy$

Sine of the Time

yes

but $\|n \|=\sqrt{f_x^2+f_y^2+1}$

so you're left with $n\,dx\,dy$

Pizza

$\int_{S} f \, dS = \iint_{D} f(x, y, g(x, y)) \sqrt{1 + g_x^2 + g_y^2} \, dx \, dy$

Sine of the Time

12:36

so?

Pizza

the rotor F, Do I have to replace it here?

I do not understand

Sine of the Time

instead of F you put rot F

@Pizza here

Pizza

$\iint_{D} (\operatorname{rot} \mathbf{F}) \sqrt{1 + g_x^2 + g_y^2} \, dx \, dy$

I mean something like this

Sine of the Time

no

this does not make sense

if you want to solve the ex using the formula you sent before, replace F with rot F

Pizza

Yes, you mean the second formula, right?

14 mins ago, by Pizza

Sine of the Time

12:45

yes

Pizza

$\iint_{D} f(x, y, g(x, y)) \sqrt{1 + g_x^2 + g_y^2} \, dx \, dy$, wouldn't this be it?

Maybe I'm not understanding, but is this formula not right?

Sine of the Time

this is the definition of surface integral over the surface defined by z=g(x,y)

Pizza

yes

Sine of the Time

in your case, $f=\operatorname{rot} F\cdot n$

and g is the equation of the plane

Pizza

Aaah ok

In the end it's always the same thing

Sine of the Time

12:53

yes

Pizza

@SineoftheTime So it would be: $\Phi = \iint_{D} \left( \operatorname{rot} \mathbf{F} \cdot \mathbf{n} \right) \sqrt{1 + \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2} \, dx \, dy$

Sine of the Time

yes but you can simplify it

note that $\operatorname{rot}F\cdot n$ have to be evaluated at $(x,y,g(x,y))$

Pizza

@SineoftheTime $$\Phi = \iint_{D} \left[ \operatorname{rot} \mathbf{F}(x, y, g(x, y)) \cdot \left( -\frac{\partial g}{\partial x}, -\frac{\partial g}{\partial y}, 1 \right) \right] \\ \sqrt{1 + \left(\frac{\partial g}{\partial x}\right)^2 + \left(\frac{\partial g}{\partial y}\right)^2} \, dx \, dy$$

Sine of the Time

note that $n$ is a unit vector

so $\sqrt{1+g_x^2+g_y^2}$ simplify

Soumik Mukherjee

13:09

Amazon delivered a wrong book to me, what's funny is that the book they send is called Authentic Business

Pizza

$$\Phi = \iint_{D} \left[ \operatorname{rot} \mathbf{F}(x, y, g(x, y)) \cdot \left( -\frac{\partial g}{\partial x}, -\frac{\partial g}{\partial y}, 1 \right) \right]\, dx \, dy$$

Sine of the Time

@SoumikMukherjee never happened to me

Pizza

That would be it, right?

Sine of the Time

yes

Soumik Mukherjee

@SineoftheTime this is the first time that such a thing happened to me

Pizza

13:16

Thank you so much for your help, sorry I took up your time

Sine of the Time

other doubts?

Pizza

@SineoftheTime But now then: $\left( -\frac{\partial g}{\partial x}, -\frac{\partial g}{\partial y}, 1 \right) = (-2,-3,1)$

Sine of the Time

what's the problem?

Pizza

shouldn't i use ||n|| too?

Sine of the Time

as I said before, it simplifies

Pizza

13:32

@SineoftheTime But can I always write this formula like this?

Sine of the Time

$\Phi=\iint_D F(x,y,f(x,y))\cdot (-f_x,-f_y,1) \, dxdy$ ? Yes

Pizza

Yes

Sine of the Time

you can and you prove it

Pizza

Oh ok, thank you very much!

I will try to do other exercises

Sine of the Time

@Pizza just to be clear: you can use it when the surface is of the form $z=g(x,y)$

@Pizza np

Pizza

13:49

@SineoftheTime Yes

It depends on the form of the surface that the exercise gives me

Sine of the Time

yes

there are tons of formulas for the different parametrisation for surface integrals, I remembre only the definition

Xander Henderson

14:21

Yay! It only took 45 minutes to respond to all of the email which arrived after 3pm yesterday! X(

user 20458579510081670432

Congrats?

Sine of the Time

@XanderHenderson moving them to trash is faster

Sahaj

14:53

Hi Sine of the the Time

Xander Henderson

@SineoftheTime Yes, but I work in a state which does not have tenure for community college instructors, and a strong Republican right-to-work political philosophy. I can't really ignore work-related email.

Sahaj

Xander, as a faculty, have you ever encountered students cheating using AI?

like ChatGPT and all. It's so bad at math that I doubt if anyone would use it. I'm still curious.

user 20458579510081670432

15:09

They say that it's bound to get better with more data.

Sahaj

The google AI is said to be able to solve some IMO geometry problems in minutes

Gemini is the name of the AI model I believe

Jakobian

Sahaj

Wait. Not even minutes -- 19 seconds to solve one geometry question from this year's IMO.

user 20458579510081670432

:O

Sahaj

arstechnica.com/information-technology/2024/07/…

user 20458579510081670432

15:18

Today Geometry, tomorrow the World.

Sahaj

who knows....

user 20458579510081670432

That's impressive speed.

And it will get faster and faster...

Sahaj

well, it did fail to solve 2 other problems and took 3 days..

Sine of the Time

@Sahaj hi

user 20458579510081670432

Sure, data is the bottle-neck.

Sine of the Time

15:22

Is it possible to upload an image from phone?

user 20458579510081670432

Not on mobile.

Soumik Mukherjee

@SineoftheTime yes

Sine of the Time

Soumik Mukherjee

lol

@Jakobian Is this a childhood picture of Jeff Bezos?

Jakobian

Malcolm in the Middle - The Future Is Now Old Man

►

it's from Malcolm in the Middle

Sine of the Time

15:30

Spain without S

Jakobian

when I was younger it used to air on tv

Soumik Mukherjee

ohh

user 20458579510081670432

Do you still watch reruns @Jakobian

Jakobian

@user20458579510081670432 where do they air

Also I've noticed there must have been an update with firefox since they display me a preview of a tab when hovering over it

this isn't really useful as I keep usually < 5 tabs at a time

but I see how it can be useful for someone disorganized

user 20458579510081670432

15:45

Perhaps you could find them on YouTube.

Pizza

I have a question, if there exists a function $f$ such that $df = \omega$ , then $\omega$ is conservative?

it would be the definition of exact form

Sine of the Time

as far as I know, the term "conservative" is used for vector fields and not differential forms

Pizza

16:00

I was Reading this

We say that a differential form $\omega$ in $A \subseteq \mathbb{R}^2$, $$ \omega = a(\cdot) \, dx + b(\cdot) \, dy, $$is said to be exact (or that the field is conservative) if there exists $f \in C^1(A)$ such that $$ df = \omega \Leftrightarrow \begin{cases} f_x = a(x,y) = X(x,y), \\ f_y = b(x,y) = Y(x,y) \end{cases} $$This function $f$ is called primitive (or potential of the vector field).

Sine of the Time

what's the question?

Pizza

@SineoftheTime For now I was trying to do this: Establish for which values of the parameters $\alpha$ and $\beta$ the vector field $$\omega(x,y,z) = \left(\alpha x \ln(z), 3y^2 z, \frac{\beta x^2}{z} + y^3\right)$$ is conservative and finding a potential.

Sine of the Time

do you have an idea on how to procede?

Pizza

@SineoftheTime i think i need to determine if exist a scalar function $U(x,y,z)$ such that $\omega = \nabla U$

Sine of the Time

do you know the relation between closed and exact forms?

Pizza

16:17

@SineoftheTime Every exact form is also closed. , not all closed forms are exact

closed -> exact, unless the space is simply connected

Sine of the Time

right

is the domain simply connected?

Pizza

Mm no?

Sine of the Time

why?

what is the domain?

Pizza

$D = \{ (x, y, z) \in \mathbb{R}^3 \mid z > 0 \}$

Sine of the Time

why is it not simply connected?

Pizza

16:25

The z=0 plane is excluded

Sine of the Time

so?

Pizza

I had read that $\Bbb R^2$ without a point is simply not connected, while the entire plane is.

But I remember that if there were no holes it is simply connected

So maybe here it is

Sine of the Time

@Pizza this works in $\mathbb R^2$

Pizza

oh ok

Ah, now I remember you told me this thing

Sine of the Time

consider a closed curve in $\{z>0\}$, can you continuously shrink it into a point while remaining in the domain?

Pizza

16:32

Yes

Sine of the Time

can you proceed?

Pizza

@SineoftheTime Yes, I found that the derivatives are equal only if $\alpha = 2\beta$

thank you very much, I'll try to continue on my own

Sine of the Time

k

Soumik Mukherjee

17:09

@Pizza I think you should write 'not simply connected' instead of 'simply not connected'

Pizza

@SoumikMukherjee yes indeed

@SineoftheTime But since the definition set is simply connected, to verify that the field is conservative, I have to calculate the curl?

Xander Henderson

Indeed. "Not (simply connected)" is very different from "(Simply not) connected" or "Simply (not connected)".

Pizza

@XanderHenderson yes... next time I'll be more careful about what I write

Binky McSquigglebottom

17:27

in I BC, Diodorus Siculus wrote in the first book of his library that the Egyptians were the inventors not only of writing but also of geometry

the problem of calculating the area of a surface, which continually recurred in practice, was mastered well. The area A of a rectangular surface with sides a and b was calculated according to the formula A=a×b; that of a trapezoid with bases b, b´ and height h by the formula A=[(b+b´)/2]×h.

It was also possible to calculate the lengths of the sides of a rectangle from the area, if their mutual proportion was known. The area of the right triangle with orthogonal sides a and b was obtained with the formula A=(a/2)×b and also in this case, once the area was known, it was possible to identify the length of the sides if their mutual relationship was known . In particular, triangle calculations also recurred in problems involving pyramids.

Soumik Mukherjee

18:29

Exactly how much time is left for the election to be over?

It says 1 hour but that may not be the exact time

psie

18:48

@Jakobian I don't know if you are still interested (otherwise just ignore), but I think the authors of the text are correct in claiming that $T=[-b,b]^n$ is totally bounded by balls of the form $B(\varepsilon j,\varepsilon)$, where $j$ ranges over $\mathbb Z^n$, provided we equip $\mathbb R^n$ with the max metric. What I still fail to grasp is the condition $\varepsilon|j_i|\leq 2b$.

Here's my attempt at deriving it:

If $\varepsilon>b$, then we can simply take $B(0,\varepsilon)=(-\varepsilon,\varepsilon)^n$ and this will contain $T$ and clearly satisfy the condition.

For $\varepsilon\leq b$, observe that any point $x\in \mathbb R^n$ will be a distance less than $1$ from an integral lattice point. Consequently, any point in $\mathbb R^n$ is less than a distance $\varepsilon$ from $\varepsilon j$ for some $j\in\mathbb Z^n$. In particular, each $t\in T$ belongs to some ball $B(\varepsilon j,\varepsilon)$ as $j$ ranges in $\mathbb Z^n$. We want to pick only those $j$ such that $B(\varepsilon j,\varepsilon)\cap T\neq\varnothing$.

Let $x\in\mathbb R^n$ be such that $\varepsilon x$ is in $T$ and in $B(\varepsilon j,\varepsilon)$ for some $j\in\mathbb Z^n$, then \begin{align*} \varepsilon x\in [-b,b]^n\land d(\varepsilon x,\varepsilon j)<\varepsilon &\iff \forall i(-b\leq \varepsilon x_i\leq b \land |x_i-j_i|<1) \\ &\iff \forall i(-b/\varepsilon\leq x_i\leq b/\varepsilon\land j_i-1<x_i<j_i+1) \\ \end{align*}

Now, the last statement will be true when $-b/\epsilon < j_i+1$ and $j_i-1 < b/\epsilon$ for all $i$, since we are looking for the occurrence when $(j_i-1,j_i+1)$ contains points of $[-b/\varepsilon,b/\varepsilon]$. Now $-b/\varepsilon < j_i+1$ and $j_i-1 < b/\epsilon$ is equivalent to $\epsilon |j_i|<\varepsilon+b$, and since we had $\varepsilon\leq b$, we get the desired condition, but with strict inequality.

@psie the words kind of didn't come in the right order here. What I meant here of course is that $T$ is totally bounded by balls of the form $B(\varepsilon j,\varepsilon)$ with $j\in\mathbb Z^n$ satisfying $\varepsilon|j_i|\leq 2b$.

Sine of the Time

@Pizza the question asking for which values of the parameters the vector field is conservative. Since conservative vector fields are irrotational, you could compute the curl and set it equal to $0$

Which is the same as proving the form is closed

It's just a different terminology

But usually physicist use vector fields and mathematicians use differential forms

Pizza

Oh ok thank you very much, one last thing

Sine of the Time

I can't continue now

@Pizza ok, if it's quick

Pizza

Yes its quick

it told me that for these values of alpha and beta I had to calculate the work along the curve x^2+2y^=1

But

Since F is conservative and the path is closed then the work is 0

I don't have to do the calculations, I think

Sine of the Time

19:06

The curve is in the plane or in the space?

Pizza

$x^2 + 2y^2 = 1$ Sorry

@SineoftheTime Plane

Sine of the Time

But the vector filed is in the space

Pizza

It doesn't actually specify where this curve is

Sine of the Time

Don't you have a condition on z?

Pizza

It says for these values of $\alpha$ and $\beta$ calculate the work done by $\omega$ along the curve $x^2+2y^2=1$

@SineoftheTime mm no

Sine of the Time

19:11

This is a surface

Pizza

@SineoftheTime So z can take any value ?

Sine of the Time

Yes

Pizza

@SineoftheTime Ok then now I'll look better at what to do. Thank you very much

Jakobian

19:27

@leslietownes music.youtube.com/watch?v=TbMRbhuPGK8&si=LpbPsF9289Cl9sL3

When distribution theory goes too hard

@SoumikMukherjee 23 minutes

Soumik Mukherjee

Is it correct to say that a cone is kind of extending a commutative diagram? We have vertices F(a), F(b),... etc, edges F(f), F(g),... etc.(F is a functor from A to B). Now a cone means we have an object c in B and morphisms phi_k from c to F(k) for all k's such that the diagram is commutative. So we are kinda extending the original diagram of the functor F

@Jakobian ooh, thanks

Jakobian

@psie this is one of the things I was asking for as well. Either change of balls, or maximum metric

psie

19:46

ok, well, provided we use the maximum metric, I get the condition $\varepsilon |j_i|<2b$ instead of $\varepsilon |j_i|\leq 2b$, so slightly "better" than the authors I guess

I think letting $\varepsilon |j_i|$ equal $2b$ we simply get a couple of more balls in the collection of balls that cover $T$, so I don't see any problem

Jakobian

@psie what do you mean here by "last statement will be true". Which statement

psie

@Jakobian I mean that $ \forall i(-b/\varepsilon\leq x_i\leq b/\varepsilon\land j_i-1<x_i<j_i+1)$ will be true if $-b/\varepsilon < j_i+1$ and $j_i-1 < b/\varepsilon$ for all $i$.

What I did here was that I wrote the interval $(j_i-1,j_i+1)$ next to the interval $[-b/\varepsilon,b/\varepsilon]$ and asked myself when will the former contain points of the latter.

Jakobian

I don't think so

Okay I see what you did but that's totally not what you wrote

You want to write that this implies $-b/\varepsilon < j_i+1$ and $j_i-1 < b/\varepsilon$ will imply $T\cap B(\varepsilon j, \varepsilon) \neq \emptyset$

psie

you're right, I messed up in thinking too hard

Jakobian

But okay, if $\varepsilon |j|\leq 2b$ will imply $T\cap B(\varepsilon j, \varepsilon) \neq \emptyset$ is what you've shown, even though I didn't check that argument

how exactly does that imply that $B(\varepsilon j, \varepsilon)$ with $\varepsilon |j|\leq 2b$ cover $T$

@psie you see what's my problem?

psie

19:59

@Jakobian well, first of all, I showed $\varepsilon |j_i|<2b$ implies $T\cap B(\varepsilon j, \varepsilon) \neq \varnothing$ (note the strict inequality). And since I remarked that every point of $T$ belongs to some ball $B(\varepsilon j, \varepsilon)$ for $j\in\mathbb Z^n$, we found all $j$'s through $\varepsilon |j_i|<2b$ such that the union of $B(\varepsilon j, \varepsilon) $ have to contain all points of $T$, no?

@psie this is where I made the remark

Soumik Mukherjee

So the result is #Shaun" # Candidate 1
"FShrike" # Candidate 2
"jasmine" # Candidate 3
"D.W." # Candidate 4
"mick" # Candidate 5
"Tyma Gaidash" # Candidate 6
"Robert Frost" # Candidate 7

Joe

Hmmmm I can't seem to access the detailed results on Mac. Does anyone have a screenshot they could share?

Soumik Mukherjee

Sine of the Time

20:54

@Pizza Did you manage to solve the exercise?

Jakobian

@psie no

psie

@Jakobian well, there are no other balls that contain points of $T$, and the balls cover even $\mathbb R^n$

Jakobian

"there are no other balls that contain points of $T$"

this is a different claim

and it's not the thing you're proving

@psie well perhaps you did mean that, since "last statement will be true when" could mean an equivalence

although you shouldn't use such unclear language

"last statement will be true exactly/precisely when" would be better

psie

ok, I will amend it, good point

Jakobian

but of course it's not even the last statement but $T\cap B(\varepsilon j, \varepsilon)\neq \emptyset$

@psie but here you're still saying that this implies something

so again, the reasoning is wrong

psie

21:08

@Jakobian the way I have written it now in my notes is the following:

Jakobian

You want your set of conditions to imply the intersection is non-empty

psie

"Now, contemplating the last statement, the interval $(j_i-1,j_i+1)$ will not contain points of $[-b/\varepsilon,b/\varepsilon]$ if $-b/\varepsilon \geq j_i+1$ or $j_i-1 \geq b/\varepsilon$. So, negating that, ..."

Jakobian

yeah doesn't matter I'm over that

I'm talking about something else

psie

ok, I think I see what you mean, my logic is flawed

Jakobian

You want a condition that would imply $\varepsilon |j_i| < \varepsilon + b$ for all $i$

but what you've shown is the opposite implication

no sorry the other way around

you want a condition that would be implied by $\varepsilon |j_i| < \varepsilon + b$ for all $i$

I mean why not just take this condition as it is

there's finite amount of such $j_i$, everything works out

of course, it's not as good as the claimed one, but it's still fine

to have the claimed one I suspect the adjustment would be to use that for any $x\in\mathbb{R}^n$ there is $j\in\mathbb{Z}^n$ such that $\|x-j\|_\infty \leq \frac{1}{2}$ instead

and then it would indeed be visible that the claimed bound is actually kinda sharp

of course I didn't check it but I suspect this is what went wrong

psie

21:31

hmm, ok, you have a point. If we go with $\|x-j\|_\infty \leq \frac12$, then according to my computations we'll end up with the inequalities $-b/\varepsilon < j_i+\frac12$ and $j_i-\frac12 < b/\varepsilon$, i.e. $|j_i|<b/\varepsilon +\frac12\iff \varepsilon |j_i|<b +\frac{\varepsilon}{2}\leq 1.5b$. I don't see though how $\varepsilon |j_i| \leq 2b$ is sharper than $\varepsilon |j_i| < 2b$.

Jakobian

@psie if we shorten the radius of the balls at least

@psie you didn't get $\varepsilon |j_i| < 2b$, you got $\varepsilon |j_i| < \varepsilon + b$

@psie here $b\leq \varepsilon$ so that still doesn't get you to the goal

Note that $2b < \varepsilon + b$ so your result is weaker

no sorry what am I talking about

$\varepsilon \leq b$ by assumption

psie

yeah, first I considered $\varepsilon>b$. This case is trivial. Then I considered $\varepsilon\leq b$ and got $\varepsilon |j_i| < \varepsilon + b$ as you say, but since $\varepsilon\leq b$, we get $\varepsilon|j_i|< 2b$

Jakobian

yeah, I was wrong

psie

21:47

but the thing is, $\varepsilon|j_i|\leq 2b$ isn't wrong at least, but so isn't $\varepsilon|j_i|< 2b$ it seems. It's like you say, it's just a bit sharper.

Jakobian

yeah

and if we really wanted the sharpness we could have used $\varepsilon + b$ instead of $2b$

psie

indeed

Jakobian

yeah, that was a horrible mistake, and I was pretty convinced of being right too

apologies

psie

no worries

zeta space

23:17

This func. eq has character $$f(x_1^{a_1}x_2^{a_2}\cdot\cdot\cdot x_n^{a_n}) = \prod_{k=1}^n f(x_k)^{\frac{1}{a_k}} $$

Only zeta space and Xander are left

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$Mathematics$ Mathematics