Mathematics - 2018-03-07 (page 2 of 4)

GFauxPas

16:00

By definition, an irrational number is the limit if a sequence of rational numbers

Niing

to GFauxPas: Thank you!

GFauxPas

Np

Note, though

That definition , while correct, is difficult to prove things with

There are equivalent definitions that are less intuitive, but easier to use

:)

Here's one that is easier

We can define e^1 as a limit of (1+1/n)^n, did you learn that ?

Niing

I just forgot the reason behind exponential so I started from integer to rational, and reading rational part from wiki now

Yes

GFauxPas

So we can write e^x = lim (1+x/n)^n

Food for thought :)

Alessandro Codenotti

@GFauxPas you can cover R with countably many compact sets

Than do a big union on the points on which $f$ isn't zero

Niing

16:06

Ok, thank you, i'm appreciate your time!

GFauxPas

Np. There are more ways to define e^x , depending on what you're trying to do or prove, theyre all equivalent

That limit I just gave is a good way to think of the exponential in banking and chemistry

Oh, right, just choose increasingly large integer radius closed balls at the same center, right?

anon

@KasmirKhaan uh

Kasmir Khaan

yeeey :D

Anon :D

Alessandro Codenotti

@GFauxPas that should work

anon

hi

Kasmir Khaan

16:12

hi ! iwas doing product of groups

and I kinda understood the mapping property of products

GFauxPas

Then since f is L1, I can approximate int f at those non zero points by putting compact sets at those points

Kasmir Khaan

I want to see if i can use it right

GFauxPas

Those have to be measure 0, so we'll ignore them ?

Kasmir Khaan

ill write some questions from artin =p

hmm

Alessandro Codenotti

I don't think that step is even needed actually

GFauxPas

16:14

well im all for a less-work answer

Kasmir Khaan

let r and s be intergers with no common factor. a cyclic group of order rs is isomorphic to the product of a cyclic group of order r and cyclic group of order s @anon

on the other hand a cyclic group of order 4 is not isomorphic to C_2 x C_2

what I wrote first does not make assertions about a groups that is not cyclic

that is from the book of artin

can you help me make sense of it?

i mean isint a group of order 4 either cyclic or isomorphic to C_2 x C_2 ?

anon

yes, what's your point? C_rs is isomorphic to C_r x C_s if r,s are coprime, but not if they share a factor

ÍgjøgnumMeg

$(2, 2) \neq 1$

Kasmir Khaan

hmm i wanted to be sure i understood it right

the way he say it in the book

was a bit misterious

I did understand the first part just fine untill then :D

anyway how to use : H ---> GXG'

anon

If you're told r,s coprime implies C_r x C_s is cyclic, the very next thing you should wonder is what happens if they're not coprime? what happens is that it's not cyclic, and the smallest case is C_2 x C_2

Kasmir Khaan

16:18

decomposiong a group into 2 groups

anon

I don't know what you mean by "how to use H->GxG'. " How to use it for what?

Kasmir Khaan

hmm let me write it in a detail :D

sorry for being not orginazed

Let H be any group. the hom's phi : H---> GxG' are in bijective correspondance with pairs ( f, f') of homomorphism f : H--> G , f' : H--> G'

and kernel of phi is the intersection of ker f with ker f'

anon

yes

Kasmir Khaan

this I proved and I understood

now how usefull is this ?

because i could not answer many questions from the exerices :(

example

G = R^x

H = {-1 ,+1}

anon

I don't have an encyclopediac documentation of times I've used little bits and pieces of group theory trivia. Just talk about the exercises you had problems with.

Kasmir Khaan

16:23

and K ={ positive reals }

@anon okay :D

Ehm it is clear for me that HxK is iso to G

but i want to know how it can be proven

anon

H and K are subgroups of G, their elements commute with each other, and everything in G is uniquely a product of h in H and k in K. so G is an internal direct product of H and K.

Kasmir Khaan

hmm

we should not like define a map ?

and show it is 1-1 and onto

and a hom ?

i mean to show injectivity here

user537069

anyone good with epsilon delta proofs? i have some questions that i am getting a thousand different replies on here

if anyone has experience with proofs

Kasmir Khaan

we notice that H intersect K is just {1}

so if hk = h'k'

then kk' ^-1 = h'^-1h =1

ie k=k' and h=h' , thus injective

i was considering the map : HXK ---> G , (h,k) ---> hk

@anon another question =P , let G be a group containing two normal subgroups of orders 3 and 5 , prove G has an element of order 15

anon

yes, when G is an internal direct product of two of its subgroups H and K, and HxK is the external direct product, then the map HxK->G given by (h,k)->hk will be an isomorphism

Kasmir Khaan

16:33

dont answer yet let me think :D

what do you mean by internel direct product?

and external ?

let G be a group containing two normal subgroups of orders 3 and 5 , prove G has an element of order 15 , to answer this

Since both subgroups are normal

call them H and K

H,K have trivial intersection

and their product HxK has order 15

thus it is isomorphic to C_15

is that a good way to answer it ? :D

There is not mention of internal and external direct products so far, also what is the difference between HK and HxK ? @anon

@anon sorry for bombing you with questions :D

But those are last ones! =p

anon

Given two groups A and B, the external direct product is a construction on the Cartesian product AxB that makes it a group. Given a group G with subgroups A and B, we say G is an internal direct product of A and B if every element of A commutes with every element of B and every element of G is uniquely expressible as ab where a is in A and b is in B. Note that AxB is an internal direct product of its subgroups Ax{e} and {e}xB.

If H,K are two subgroups of a group G, then HK={hk | h in H, k in K}. In general it is not a subgroup, but if one of H or K is normal then it is a subgroup, and if both are normal then it is a normal subgroup.

Even if HK is a subgroup, it may not be isomorphic to HxK (for instance, the elements of H and K may not commute in HK).

Kasmir Khaan

aha :D

Need to copy this so I dont get lost again :D

anon

That's a fine way to answer the order 15 problem.

Kasmir Khaan

@anon Thanks alot anon :D

HK is a subset of G

anon

yes. it's also a union of left cosets of K, and also a union of right cosets of H

U{hK | h in H} and U{Hk | k in K}

Kasmir Khaan

16:43

understood , i just wanted to know where it lives =p

and HxK this has as elements 2 tuples

Julius

Hi. Consider two cases.
1) let $S$ be a set of all $N\times N$ real matrices with ones on the diagonal. Then there is a paper proving what a projection $P_S(A)$ is in the weighted Frobenius norm $\|W^{1/2} A W^{1/2}\|_F$ ($W$ is symmetric positive definite). If $W=I$ (i.e., no weights), then $P_S(A)$ simply has ones on the diagonal and the same off-diagonal elements as $A$.
2) let $S_1$ be a set of all $N\times N$ real matrices with ones on the diagonal and where some pre-specified set of off-diagonal entries are restricted to coincide. This time I care about $P_{S_1}(A)$ in the nonweighted…

Kasmir Khaan

(h,k) , h in H , K in K

so they are very different things

we made up HxK by (1,H) x ( K,1')

Balarka Sen

Oh god this new episode of Life is Cringe makes me want to eat a tide pod

Oh gooood

It gets worse with every new episode

GFauxPas

Oh, this looks really useful , every L1 function is the limit of functions that are each linear combinations characteristic functions on L measurable sets

I guess it's kind of obvious? Still cool

Oooh, even better

You can make the sequence monotone

That's really cool

am I missing any hypotheses?

Eric Silva

17:08

@BalarkaSen they're delicious tbh

Martin Sleziak

@robjohn You can find some discussion related to your meta question also in crude room. The question has been mentioned there before but I think most of the discussion related to this is in today's messages starting from here.

Dattier

17:34

Hi,

I want to know if this link works : artofproblemsolving.com/community/c6h1568869_fibo222018

for you

Julius

@Dattier, it does.

Dattier

@Julius : Thanks

Tanuj

Guys could someone tell me what is this equation called ? Looks like gauss law but I'm confused .

in The h Bar, 3 mins ago, by Tanuj

Akiva Weinberger

What's the formula for $F_{2n}$ again

4

Q: Need help understanding Fibonacci Fast Doubling Proof

From this website, http://www.nayuki.io/page/fast-fibonacci-algorithms (fast doubling proof close to the bottom of the page). I have understood the proof for the most part but I am struggling to see how this part of the proof works especially when the the F(n) function is squared. \begin{align}...

^Seems relevant @Dattier

GFauxPas

17:49

@tanuj did you see if it's equivalent to one of Maxwell's eqns?

Tanuj

Like for gauss law in electrostatics , I know the term on the right denotes electric flux threading the specified area and the term on the left is the sum of all charges inside the gaussian surface divided by epsilon , but what do the terms denote here ?

Lozansky

@Tanuj That's Maxwells

Semiclassical

That’s Faraday’s law

Tanuj

Thanks guys , found it on wiki , giving it a reading now

Semiclassical

With the interpretation being that the induced emf around a loop is equal to the time rate of change of the magnetic flux through said loop

Lozansky

18:00

Say I want the potential for a unity dipole centered at origin, directed in the positive x-direction

0celo7

daily reminder that EM is a trivial application of G bundle theory

Lozansky

I know the Laplace operator has the fundamental solution $K(\textbf{x}) = -\dfrac{1}{2\pi} \ln|\textbf{x}|$ in $\mathbb{R}^2$

0celo7

it's just an exact sequence in BU(1)

Lozansky

Meaning $-\Delta K = \delta$

So the potential $u(\textbf{x}) = -\dfrac{\partial \delta}{\partial \hat{n}}*K(\textbf{x})$, yes?

0celo7

oh christ almighty

what is that

Lozansky

18:04

Bear with me

So we evaluate the convolution as $u(\textbf{x}) = - \int_{-\infty}^{\infty} \dfrac{\partial \delta(\textbf{y})}{\partial \hat{n}} K(\textbf{x}-\textbf{y}) d \textbf{y}$

0celo7

$\partial\delta/\partial \hat n$ might be crime against humanity

Lozansky

Just call it a formal computation

Xander Henderson

That hurts my feels. :(

Drew Brady

anyone here able to assist with (d) on this problem? it looks kinda like Riesz representation imgur.com/a/Q3c7o

Lozansky

Anyway, we get $\int \delta(\textbf{y}) \dfrac{\partial}{\partial \hat{n}_y} K(\textbf{x}-\textbf{y}) d \textbf{y} = - \int \delta(\textbf{y}) \dfrac{\partial}{\partial \hat{n}_x} K(\textbf{x}-\textbf{y}) d\textbf{y} = - \dfrac{\partial K}{\partial \hat{n}} (\textbf{x}) = -\dfrac{1}{2\pi} \hat{n} \cdot \nabla \dfrac{1}{|\textbf{x}|} = \dfrac{1}{2\pi} \dfrac{x}{(x^2+y^2)^{3/2}}$

Oops

But this is wrong, it should be $x^2+y^2$ in the denominator! So what went wrong?

Drew Brady

18:17

anyone able to help on my problem??

@0celo7 ?

user193319

I am working through a proof that $\Bbb{R}^k$ is complete with the square metric $\rho$. Letting $(x_n)$ denote some Cauchy sequence, then given $\epsilon =1$, there exists a $N \in \Bbb{N}$ such that $\rho(x_n,x_m) < 1$ for all $m,n \ge N$. Let $M = \max \{\rho(x_1,0),...,\rho(x_N,0)+1\}$. My question is, why is $\rho(x_n) \le M$ for all $n \in \Bbb{N}$? Certainly its true for $n =1,...,N$, but does it hold for $n \ge N$?

HELP!

Akiva Weinberger

Ted Shifrin

LOL, DogAteMy.

Dattier

@AkivaWeinberger : it don't know, if you can use this formule

Akiva Weinberger

@user193319 Are $x_n$ elements of $\Bbb R$ or $\Bbb R^k$?

Ted Shifrin

18:26

@user193319: You mean $\rho(x_n,0)\le M$?

Mike Miller

I learned pi a couple digits past out that when I was a kid and now I have this piece of useless information in my head

user193319

@TedShifrin Yes. @AkivaWeinberger $x_n \in \Bbb{R}^k$

Ted Shifrin

Yeah, I know a few more digits, too.

@user193319: So estimate $\rho(x_n,0)$ with the triangle inequality.

Vrouvrou

Bonsoir @TedShifrin

Ted Shifrin

Bonsoir, Vrouvrou.

user193319

18:29

@TedShifrin I tried that: $\rho(x_n,0) \le \rho(x_n,x_m) + \rho(x_m,0)$, which doesn't help...

Ted Shifrin

@user193319: You're being silly. Pick a particularly convenient value of $m$.

Vrouvrou

@TedShifrin s'il vous plait pouvez voir cette question math.stackexchange.com/questions/2680408/…

Ted Shifrin

@Vrouvrou: Je crois qu'il faut que $Y\subset B$. Sinon, considère, par exemple, $Y=[0,2]$ and $B=[1,3]\subset\Bbb R$.

Drew Brady

anyone here able to assist with (d) on this problem? it looks kinda like Riesz representation imgur.com/a/Q3c7o

Vrouvrou

et si je suppose que $Y\subset B$ c'est facile de démontrer l'inclusion ? @TedShifrin

Ted Shifrin

18:37

oui, je crois.

Semiclassical

@MikeMiller the obsession with memorizing digits of pi is something I just don’t understand

Ted Shifrin

There are so many nerdy things I've never understood.

Akiva Weinberger

@Semiclassical There are competitions where like someone'll shuffle a deck of cards and people try to memorize the order

Stuff like that, memorization contests

I imagine someone into that sort of thing would want to memorize pi to a lot of digits

Semiclassical

The various algorithms for computing the digits of pi are interesting to me, but the values themselves just aren’t

Lozansky

$\pi = 22/7$

Semiclassical

18:44

Sure, but there’s not an association in popular culture between card players and memorization contests

Whereas pop culture does seem to associate math people with pI memorization

I guess it goes to the general ignorance about math

Ted Shifrin

I'm generally ignorant about math.

Semiclassical

@Lozansky nah, $\pi=\sqrt{10}$ :P

Ted Shifrin

A friend of mine has a "mathy" clock that has the 3 hour labeled by $\pi - 0.14$. UGH.

Semiclassical

Ew

Akiva Weinberger

You sure it isn't the ~3.002 hour?

Semiclassical

18:48

$\lfloor \pi\rfloor$

Ted Shifrin

DogAteMy: I'm not sure of much.

berrygreen

Stereographic projection of a sphere onto a plane tangent to one of the poles, from the other pole. You have $(\theta, \phi)\to (x,y)$. What confuses me is that this is not an injective mapping. $(0, a) \to (0,0)$ and $(0,b)\to (0,0)$?! Am I missing something? Is this supposed to be?

Akiva Weinberger

Which is which with $\theta$ and $\phi$?

berrygreen

$theta$ is the angle between the projection line and the z-axis

Eric Silva

sup chat

Akiva Weinberger

18:51

Don't $(0,a)$ and $(0,b)$ both refer to the same point? The South Pole?

(Where the plane is tangent to the South Pole)

$\sup\hat c$

berrygreen

Yes, I get it now. Like the line can be described in many ways, put it is always the same point being projected onto the plane. OK

Akiva Weinberger

Incidentally, the North Pole isn't in the domain of the mapping, I guess

Like, the projection maps the sphere minus the North Pole to the plane, homeomorphically

(and diffeomorphically)

berrygreen

yes

kind of weird, that the sphere which is finite in some sense have "cardinality" larger than an infinite plane

Eric Silva

I wonder if anyone has studied the functional $\int_{M} |H|^{d} d\text{vol}$ for immersed hypersurfaces $\varphi: M \to \mathbb{R}^{d + 1}$

Balarka Sen

what is H

Akiva Weinberger

18:55

Well… except that infinity plus one is still infinity @berrygreen

Balarka Sen

oh mean curvature?

Ted Shifrin

mean curvature, @Balarka

Eric Silva

ya

mean curvature of the immersion

cause in 1 and 2 dims that is bounded below by 2pi and 4pi

Balarka Sen

I know 0 things about the mean curvature

Ted Shifrin

Eric, see Griffiths's little book on differential systems and calculus of variations.

That makes you minimal, @Balarka.

berrygreen

18:56

@AkivaWeinberger yes that always gets me confused about cardinality. but i mean you can still say that there is one point on a sphere that will not be mapped onto the plane so there is one extra point

Eric Silva

im thinking maybe there's a horrible willmore theory for hypersurfaces

Ted Shifrin

Well, you have an obvious person to ask.

Akiva Weinberger

Sure. But then there are other mappings which can objectively map the sphere to a strict subset of the plane

(not continuous at at least one point)

Eric Silva

yeah im probably gonna ask him once ive formed my thoughts a bit

my guess is that it probably just sucks too much to get anywhere

Akiva Weinberger

Hm, "at at"

Eric Silva

18:58

maybe one can show that the volume of the unit sphere is a global minimizer for the functional tho that would already be p cool

@BalarkaSen my dude learn extrinsic geo with me, go on that journey my man

Lozansky

If I want to show that $K(\textbf{x}) = \dfrac{1}{4\pi} \dfrac{e^{\pm iar}}{r}, r=|\textbf{x}|$ are fundamental solutions to $-(\Delta K + a^2 K) = \delta$, would the way to go be to prove that $-(K[\Delta \varphi] + K[a^2 \varphi]) = \varphi(0)$, for $\varphi \in \mathcal{D}$?

Vrouvrou

Je n'y arrive pas @TedShifrin je trouve juste que $\partial Y\subset B\setminus int(Y)$

Ted Shifrin

Est-ce possible qu'un élément de $\partial Y$ soit à l'intérieur de $B - \text{int}(Y)$?

Vrouvrou

je ne sais pas je n'arrive pas a montrer qe ce n'est pas possible

Balarka Sen

@EricSilva I can only bared understand intrinsic

barely

Ted Shifrin

19:09

I guess I've always preferred extrinsic (hence my love for projective differential geometry), @Balarka.

Balarka Sen

I can see the appeal, my brain is just unprepared for it I think :)

On that note I computed the Taylor expansion of the Riemannian metric in geodesic coordinates at a point upto third order today

Ted Shifrin

@Vrouvrou: Donc on trouve un voisinage de $x\in\partial Y$ qui ne rencontre pas $\text{int Y}$ de tout.

Balarka Sen

the Riemann curvature tensor pops out of the Jacobi equation

Ted Shifrin

@Balarka: I've only taught curvature in normal coordinates using moving frames, of course :P

Balarka Sen

i think the Jacobi equation is my top 10 favorite anime moments of all time

@Ted Hah

Ted Shifrin

19:12

I did that to do Cartan-Hadamard, but without Jacobi.

Eric Silva

jacobi is a good

Balarka Sen

variations are my friend now

Ted Shifrin

Did you ever think about how you can use Stokes's Theorem to compute first variations?

Eric Silva

shit i guess the usual the proofs that $\|H\|_{L^{1}} \geq 2\pi$ for curves and $\|H\|_{L^{2}} \geq 4\pi$ for surfaces don't immediately extend to d-hypersurfaes

hmmmmmm

i have to think harder

Balarka Sen

@TedShifrin First variations in general, or of something more specific? I understood it as minimizing a functional using the Fermat principle (when my functional is something like the energy, I differentiate under integral signs)

Ted Shifrin

19:19

@EricSilva: Chern-Lashof did stuff in higher dimensions on bounds for $\int |K|$, relating this to Morse theory but I don't think they did $\int |H|$.

@Balarka: I was thinking of arclength or area (or volume), using structure equations.

Eric Silva

in 2 dim the bound for $\int |H|^{2}$ comes from the fact that $H^{2} - K$ is a really nice polynomial in the principal curvatures

Ted Shifrin

No, wait, so something should work with $H^d - K$ ... but you get all sorts of other cross terms.

Eric Silva

so $\int H^{2} dA \geq \int_{K_{+}} H^{2} dA \geq \int_{K_{+}} KdA \geq 4\pi$

Balarka Sen

@TedShifrin I don't think I have seen this

Eric Silva

yeah you'd need some weird new poly thats harder to think about

Ted Shifrin

19:23

Ah, @Balarka, I thought we'd talked about it briefly, but you were busy boycotting moving frames.

Semiclassical

Did you see my rambling yesterday regarding the rank of my product matrix? @ted

Balarka Sen

$2(H^2 - K)$ is $(k_1+k_2)^2 - 2k_1k_2 = k_1^2 + k_2^2 \geq 0$ ok I get what you mean by nice polynomial

Ted Shifrin

I don't recall, Semiclassic.

Balarka Sen

@EricSilva $K_+$ means bit of the surface where $K$ is positive?

Ted Shifrin

@Balarka: So, as a warm-up, take a curve varying in a surface. On $C\times [0,\epsilon)$ take a moving frame $e_1(t), e_2(t)$ where $e_1$ is tangent to your curve $C_t$ and $e_2$ is normal. Write down $\mathcal L(t) = \int_{C\times\{t\}} \omega_1$ and differentiate.

Eric Silva

19:25

ye

Ted Shifrin

I think Eric's been through this computation ... ages ago.

Eric Silva

the recipe is basically just there's a sphere's worth of nonnegative gauss on any closed boi

i have indeed done that computation

Balarka Sen

@EricSilva mmm i see

Ted Shifrin

It generalizes to first-variation of "area" for hypersurfaces pretty easily. I don't know if I've done arbitrary codimension.

Eric Silva

cause the floor can be any direction you want and the height above the floor is a nonnegative gauss point

if you take the floor to be far enough away from your closed surface

Ted Shifrin

19:28

Eric, have you been through Lipschitz-Killing curvatures and Morse theory?

Did I send you my paper on apparent contours and integral geometry with Langevin?

Balarka Sen

@EricSilva mhm

Eric Silva

u sent it to me i think long ago

Ted Shifrin

Yeah, I think I did.

Cool extrinsic stuff.

Eric Silva

i dont know what lipschitz killing is

when i google it i see your name mentioned in the first link

Ted Shifrin

For higher codimension stuff, pick a normal direction $\nu$ at $x$ and look at the little piece of hypersurface you get projecting to $T_xM\oplus\nu$.

Vrouvrou

19:29

@TedShifrin Comment on trouve le voisinage a quoi il ressemble ?

Ted Shifrin

LOL, @EricSilva. That's pretty funny.

It should give you Chern-Lashof first.

Semiclassical

Well, the remark I made was obvious in retrospect: if W is 3-by-4, then $W^\top W$ can’t have full rank and must have determinant zero

Ted Shifrin

Sure.

Eric Silva

arxiv.org/abs/1512.02780 this was the first hit on the google

Ted Shifrin

@Vrouvrou: C'est la définition de l'intérieur.

Oh, cuz they're doing polars ... apparent contours, yeah. Interesting, @EricS.

Eric Silva

19:32

oh i see it's a crazy algebraic expression in the second fund form

how terrifying

Semiclassical

So that makes moot the question about the product being positive semidefinite

Ted Shifrin

It's not terrifying. It's just the determinant of the second-fundamental form in a certain normal direction. ($II$ is normal-bundle valued for higher codimension.)

Eric Silva

high codim makes me sweat

Balarka Sen

@Ted So I am trying to variate a curve $\iota : C \to M$ by $\iota_t : C \times [0, \epsilon) \to M$ and $\mathcal{L}(t)$ is the arclength of $\iota_t(C)$, yeah?

Ted Shifrin

Well, if you're gonna do extrinsic, you can't be only a hypersurface wuss.

Eric Silva

19:33

this seems p cool

Ted Shifrin

Right, @Balarka.

Semiclassical

I think there’s still something interesting to be said about the principal minors of my matrix product, though

Eric Silva

@TedShifrin ya it's ok working up a sweat is healthy

Ted Shifrin

@Semiclassic: It's all Gram determinants ...

Mike Miller

i think vary not variate

Semiclassical

19:34

Right

Ted Shifrin

LOL, Mike.

Balarka Sen

variatriate then

Ted Shifrin

Balarka will do first variance.

Eric Silva

i believe it's actually variationate

Balarka Sen

@Eric i think i owned you

Semiclassical

19:35

So the principal minors can be interesting even if the determinant isn’t

Eric Silva

@BalarkaSen i cant tell who wins

Ted Shifrin

They're just squares of volumes of projections, Semiclassic.

Semiclassical

Yeah

Mike Miller

balarka wins imo

Eric Silva

i concede my defeat

get rekt @EricSilva u nerd

Balarka Sen

19:37

get bababadalgharaghtakamminarronnkonnbronntonnerronntuonnthunntrovarrhounawnskawnto‌ohoohoordenenthurnuk'd

Mike Miller

lol

Semiclassical

and with four 3-vectors, there are four ways to form parallelipipeds from them. So four quantities which should all be positive

Eric Silva

ok joyceyboi

Mary Star

Let $F_X:\mathbb{R}\rightarrow [0,1]$ be the distribution function of a random variable $X$. Can there be more than $n$ jump discontinuities of the function $F_X$ of size $\geq \frac{1}{n}$ ?

Could you give me a hint how we could check that?

Semiclassical

My hope is that that’s the end of my story. (With the proviso that I’m typing on my phone and may not be saying stuff right)

Ted Shifrin

19:40

@MaryStar: What is $F_X(\infty) - F_X(-\infty)$?

Daminark

Henlo

Ted Shifrin

Hello, Chicken Little, Demonark.

Mary Star

@TedShifrin It is $F_X(\infty) - F_X(-\infty) = 1-0=1$, right?

Ted Shifrin

Yup.

And what kind of function is $F_X$?

Eric Silva

oof maybe $|H|^{d}$ is just too nasty in the principal curvatures

Balarka Sen

19:43

there's this construction on foliations which is called turbularization

Ted Shifrin

@EricSilva: Forgetting absolute values, you get one term in $H^d$ which we like. It's really Newton's formulas for symmetric functions, right?

Balarka Sen

its how we construct reeb like things

and i could never remember if it's called that or turbulize

Ted Shifrin

@Balarka: That word doesn't make sense.

Mary Star

@TedShifrin It is a non-decreasing and right-continuous function, isn't it? Or do you mean something else?

Balarka Sen

so i just called it turbulululul

Semiclassical

19:43

That’s tubular duuude

Ted Shifrin

@MaryStar: I meant non-decreasing. Now you should get it.

Balarka Sen

@TedShifrin Apparently turbularize comes from the french tourbillonment

or whatever it is

Eric Silva

oof those newton identities tho

Ted Shifrin

That doesn't look right. tourbillonnement, maybe? ugh.

Semiclassical

(Channeling the 80s)

Balarka Sen

19:44

Yeah maybe

Ted Shifrin

I thought you were taking a month off from math to do real studying, Balarka.

Balarka Sen

I vampire

I come at night

but in the daytime, i'm a high school student

Ted Shifrin

That photo is way too flattering.

Balarka Sen

maybe you like the Klaus Kinski version more

Mike Miller

my best fiend

Mary Star

19:48

@TedShifrin If there were more than n jumps of size greater than 1/n, would that mean that the range would extend [0,1] ?

Ted Shifrin

Of course, @MaryStar.

Lucas

Hello @Daminark

Abdullah UYU

Y a-t-il quelqu'un qui voudrait discuter avec moi d'une preuve sur la 4 Points Géométrie en français?

Balarka Sen

@MikeMiller yeah a real lopsided guy that was

Lucas

C'est quoi ?

Mary Star

19:51

Ah ok!
And how could we check if there are uncountable points of discontinuity? @TedShifrin

Ted Shifrin

There can't be. That's a standard analysis exercise. At most countably many jump discontinuities.

Abdullah UYU

Il y a 4 axiomes, et une theorem, c'est très simple.

Mary Star

Ah ok! Thank you! :-) @TedShifrin

Lucas

D'accord

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$Mathematics$ Mathematics