@LeakyNun That's cool. Thanks.

@Thorgott I only understood

M-V implies such a space is acyclic

And CW complex case

Is the direct sum of two graded rings $\bigoplus A_k$ and $\bigoplus B_k$ a graded ring $\bigoplus (A_k\oplus B_k)$ ?

i dunno the axioms for that stuff, but if you want something resembling a coproduct, wouldn't you use the tensor product, with something like (A tensor B)_k = bigoplus_{i+j=k} A_i tensor B_j?

i'm not even close to an algebraist but might have more ideas if i knew more about the application

@leslietownes Actually I found something. I think you're right.

i was just thinking about what you'd do with polynomial rings. they're the only graded rings i know :)

@love_sodam leslie is right; will the world end soon?

I'm not sure I'm asking the right thing

Learning Physics is like learning Computer Science

mathematics is absolute

only a sith deals in absolutes

I'm tinkering with the triangular numbers right now. I think I may have figured an algorithm to generate the prime numbers off of them, but I'm not certain about it.

I'm prime decomposing the triangular numbers in sequence.

the largest odd number that can't be decomposed by previously discovered primes must be prime it seems.

not only that, but if you do n(n+1)/2 for that number, it should give the next triangular number that it belongs to.

then you can skip all the triangular numbers up to that one.

then you have to start the process over again,, until you find a large odd number that can't be divided by previously found primes.

I have no idea how I'd fully prove this.

isn't 3 the only prime triangular number? or am i reading into what you're suggesting incorrectly

you are reading me incorrectly

I meant the triangular number formed by p(p+1)/2

not that the triangular number itself is prime

p being a prime number there

if x is a triangular numberr, then x = n(n+1)/2 for some integer n

oh, I see. that looks quite interesting

let's say that x_p is a triangular number formed by p(p+1)/2 for some prime number p

there's only one triangular number x_p associated with that particular triangular number p.

@copper.hat That is my problem, I don't quite recall the details, I only remembered the shape. If I remember correctly, the function gets increasingly difficult to compute near 0 and is maybe even divergent due to how it alternates. My first guess was the Weierstrass function but that doesn't have the same conical shape. It's sinusoidal for sure

I have a question, more like an opinion-type question. I am wondering if this algorithm/ pseudocode too imperative for a thesis: ibb.co/VSCtYbm

looks good to me

but your advisor will tell you if it ain't I guess

3 is the only triangular prime number

I have a proof of this fact

anyone know how to show this? imgur.com/a/M3JTgAu , here we fix a compact set $K \subset \mathbb{C}$, and $\epsilon(z_1,z_2,R) \rightarrow 0$ as $R \rightarrow \infty$, independently of $(z_1,z_2) \in K^2$ (i.e. the convergence to $0$ w.r.t $R$ is uniform over $z_1,z_2 \in K$)

Is Kunneth formula states cohomology of product space as a graded ring isomorphic to cohomology of factor spaces? (Here, I assume Ext term vanishes).

Hagen got suspended till the end of october wtf :/

@TedShifrin Hey, thanks for that identity of modular products. I actually figured out an O(log n) upper bound algorithm for computing modulus of $2^n\bmod x$ with it.

I don't know why I didn't notice sooner that I could just split $2^n$ into smaller pieces before and then multiply.

I have to do this for $n$ bits of precision to compute reciprocal, so that means the reciprocal algorithm is... O(n log n) which is optimal, and that I can be satisfied with. My search is done.

an annoying integral that shows up on the homework i'm grading: $$\int_0^1 x(1-x^2)\log\left(\frac{1+x}{1-x}\right)dx=\frac13$$

i guess integration by parts is the obvious thing to try

oh, but doing that naively splits it into two divergent portions

gross

@love_sodam that's how you do a coproduct of graded modules, not rings. in fact, it's a biproduct for graded modules. for graded rings, this only gets you the product and I'd advise you to not write as $\oplus$

@love_sodam rewrite this sentence properly

@Thorgott So a Kunneth formula only guarantees the isomorphism as module not ring?

I mean if we $H^*(X)$ and $H^*(Y)$ as $\Bbb Z$-algebras then there is a natural ring structure on $H^*(X)\otimes H^*(Y)$.

which tensor product is this?

Tensor over $Z$. Hatcher theorem 3.15

ok, that passage in Hatcher already answers your question then

hey, my friend is doing a study (and I was told I can ask this online) and the question is as follows: What is the best studying technique you've used: (1) Prolonged thought with no breaks (2) Pomdoros (3) Mix of (1) and (2)

i'd appreciate it if anyone could answer :)

not to play survey designer here, but it might be helpful if there were a "none of the above"

Pomdoros? I know pomodoro tomatoes. What the?

it's an antiprocrastination technique

e.g. medium.com/hubgets/…

i'm pretty dubious of it

it assumes procrastination is a time-management issue

when it's really more of a "working on this feels painful/unrewarding so i'll delay starting it" thing

quick question: if $\phi$ is a function that implicitly defines $x_n$ as $\phi(x_1, \cdots, x_{n-1})=x_n$ in a neighborhood of $\vec a \in \mathbb{R}^n$ such that $f(\vec a)=0$, the implicit function theorem implies the graphs of $f = 0$ and $\phi$ in that neighborhood are in fact identical, right?

So monoidal misspelled. I use tomatoes to procrastinate? Well, I always have loved cooking.

Not right.

Not right @shin

argh

and $\phi$ is explicit, not implicit.

You can make it right, but it isn’t.

it's sort of close?

or not at all?

You need the hypotheses of the IFT.

As it stands, the level set need not be a manifold.

i'm thinking it through

what i find amusing about the article i just linked

oh, we need to add $\frac{\partial f}{\partial x_n}(\vec a) \neq 0$, otherwise going up or down $x_n$'s axis will still be part of $f=0$ but not of the function $\phi$, right? at TedShifrin

they cite Timothy Pychyl's blog in relation to how overwhelming procrastination is, but

psychologytoday.com/us/blog/dont-delay/201006/…

which is to say, Pychyl specifically has a blog post saying that pomodoro is not a magic bullet

"Breaking down a task into manageable, concrete chunks. Staying focused on a task for a defined period of time. Rewarding oneself for making progress. All of these are good things. Are they "the" answer to your time management or procrastination problems? I doubt it."

there is no rote technique for overcoming procrastination

they claim trademark rights in 'pomodoro,' in connection with certain goods and services

that's what's funny to me

they meaning not the post but whoever chose the name

my wife tried that for about a week, then stopped. they need to come up with a method that simplifies sticking to the pomodoro method

well, quoting again from that post

"The key thing for me is that any technique will always fall short when it isn't supported by commitment and focus. If you are deeply committed to working on a task, then, yes, a timer that helps you focus for 25 minutes may be a trick to get you started, but then there's the next 25 minutes, and the 25 after that."

in other words, if $\frac{\partial f}{\partial x_n}(\vec a) = 0$, $x_n$ is not unequivocally defined by $f = 0$, right?

it's like that marie kondo method for organizing your closet

if you can adhere to this extremely specific method, you don't need to organize your closet

maybe it's not kondo but something else

@shintuku i mean, any constant $f$ would also do that

anyway, buy leslie coin. it's the one true way of managing your time

@shintuku sorta. Think about $x^2=y^2$. Also, you need $f$ $C^1$.

alright, if $f \in C^1$, $f:\mathbb{R}^n \to \mathbb{R}$, $\frac{\partial f}{\partial x_n}(\vec a) = 0$, and $f(\vec a) = 0$, then there is a neighborhood of $\vec a$ where the graph of $f = 0$ is identical to the graph of $\phi:\mathbb{R}^{n-1} \to \mathbb{R}$ s.t. $\langle x_1, \cdots, x_{n-1} \rangle \mapsto x_n$, right?

Typo for sure. And who uses \langle … for points?

it's a vector in $\mathbb{R}^{n-1}$ that gets mapped to $\mathbb R$ by $\phi$

No math person uses that idiotic notation. And you need to fix the typo.

whoops, $\frac{\partial f}{\partial x_n}(\vec a) \neq 0$

then it's right, right?

i'm still a little lost in the notation of the setup. i think this is eventually getting to right, but understanding why it's right is understanding what falls out of the definition of $\phi$, which isn't really made clear in what you've written

my bad for ambiguities

from context it's clear you're getting close to an IFT type setup, but the use of the IFT and/or the definition of $\phi$ is not express in any of what you've just written. it has to be inferred. i think ted is pushing you to be more explicit (and fix typos)

the purpose of $\phi$ is to explicitly define $x_n$ as a function of $\langle x_1, \cdots, x_{n-1}\rangle$, to show that $f=0$ implicitly defines $x_n$ as a function of $\langle x_1, \cdots, x_{n-1}\rangle$

i agree with ted that vectors do not need to dress up in fancy langles and rangles but that is a style issue :)

hm, would you use vertical matrix notation instead? i felt it gets kinda cluttery in here that way

there's the good ol' parentheses too I guess

in my own writing, k-tuples of numbers are just k-tuples of numbers most of the time

i don't see phi as being 'explicitly' defined by any of this. there's some area around a where f(x_1, ..., x_{n-1}, x_n) = 0 implicitly defines x_n as a function of the other variables, i.e., it is possible to define a function phi such that f(x_1, ..., x_{n-1}, phi(x_1, ..., x_n)) = 0 holds for all (x_1, ..., x_{n-1}) near enough to (a_1, ..., a_{n-1})

i wouldn't put too much weight on words like 'explicit' vs 'implicit' to describe what's going on here, but in general you're not gonna have formulas for phi even if you have them for f, and you often work with phi only in terms of the relation it has to f

i do remember $\langle\rangle$ notation for vectors as in Stewart

though i use Dirac notation so much that i no longer touch that

i will defend it a little, insofar as it enforces a difference between points and vectors, but

meh

yes, he and many other calculus books use all sorts of notation for vectors some of the time, but would tend not to use them in contexts like this, where we've got several spaces in play and are thinking of these things as points

i feel like a lot of this is b/c people dont' see enough linear algebra by the time they get to calculus

i personally don't think it's helpful to decorate 'vectors' notationally although i understand why books do it

in handwriting i can't do bold, i don't want arrows, and i'm not writing langle :)

yeah, arrows are the only practical option

and they get exhausting

even the real number zero is a vector, let's boldface it, underline it, arrows, and langle

when i'm typing stuff out i do try to stick to boldface

but

when i'm just doing scratchwork i don't

which i do have to be careful with, since it's also common to write stuff like $k=\|\mathbf{k}\|$ in physics

so have to be very aware when i'm doing stratch work of when i mean $k$ as a vector vs. its magnitude

i used boldface when i taught because the book did. and yeah, difficult to keep it straight sometimes. i think that's why i don't like it for internal use. i don't want to look at an obvious 'typo' and wonder if i'm trying to send secret messages to myself by forgetting to boldface something

the bit of notation that's been driving me nuts lately

is how a certain book i'm referencing sometimes does multilple indices as $a_{\mathbf{k},\sigma}$

but sometimes as $a_{\mathbf{k}\,\sigma}$

thanks for the tips, i'll be working on it

which...why have two notations

semi: is there an option where the k isn't in boldface, that ideally would mean something different?

can we have semicolons in the subscript

how about one where the comma is in boldface

like, i get why the latter is problematic: you have cases where $\mathbf{k}+\mathbf{q}$ shows up, and that looks weird if you do $a_{\mathbf{k}+\mathbf{q}\sigma}$

but...just stick to a comma consistently then

a somewhat popular citation style in some legal writing, thankfully not all of it, dictates that some commas be in italics and others not

0_0

good luck seeing the difference after a while

jeeze

it's stuff like that which makes me feel a touch sympathetic to sov citizen people

b/c when the rules are that pointlessly byzantine, it does seem like some magic show

oh most courts don't require it (and would not penalize you if you didn't do it)

@TedShifrin thats what they teach in calc 3 here

most courts bend over backwards to be over-nice to unsophisticated litigants, including ignoring violations of substantive rules

yeah

but anything that makes legal questions seem more obscure than necessary bothers me

you'll like my system for avoiding taxes. what you do is put all your assets in leslie coin

not a single court has ever found this to be unlawful

neat

i'll leave unasked the question of whether any courts have found it lawful :P

they're too busy investing in lesliecoin to consider it

@RyanUnger Stupid book editors.

what i'm personally curious about is how much 'linear algebra' you see in K-12 nowadays

like, i'm sure you see "solving systems of equations"

and i know i saw matrices / augmented matrices and elementary row ops

but i'm not sure i ever saw matrix multiplication in high school

(i do remember seeing determinants b/c of cramer's rule)

Any literature that discusses this abstract definition of differentiability? math.stackexchange.com/a/3966531

you mean, beyond the two that are in that answer?

The answer is a more abstract generalization of those two.

I was wondering if the more abstract definition given in the answer has been discussed somewhere in the literature.

the answer is talking about Carathéodory derivative, yes?

isn't the answer giving references? Or am I looking at the wrong answer?

the article specifically titled "The Derivative á la Carathéodory" seems like a good starting point

oh. that's your answer

The definition is inspired by the Carathéodory derivative, yes. But it's much more general.

Applying to pretopological magmas.

The other paper linked in the answer, about topological groups, seems like a good starting point before looking at weaker structures

Yeah. That's the most general one I could find at the time.

It better be à la …

$\underset{\tilde}{x}$ for a vector is popular among probabilists

dont even know how to write it

$\underset{\sim}{x}$, perhaps?

Yeah that's it

yikes

though i do have a certain appreciation for notating matrices by double underline

i have seen double tilde for a matrix lol

i have seen the worst

tildes, all the way down

Though stackrel might be better? $\stackrel{\Large x}{\sim}$?

No, that's gross...

i havent seen this in texts but this is what they do in blackboards, just to clarify

In descriptive set theory people write $\stackrel{\Sigma}{\sim}$ all the time because making a boldface sigma is impossible on a board (in writing they actually use a boldface sigma though)

yeah in writing all vectors are boldfaced

actually ive been around probabilists so often i write vectors with a tilde below too nowadays lol

does $\mathbf{\Sigma}$ actually work?

wow it does

yep, the same applies to \Pi and \Delta. I don't know if other greek letters have a boldface version

$\mathbf{\alpha}$

not the undercase ones maybe

$\mathbf{\Gamma}$

$\mathbf{\Theta}$

$\mathbf{\Psi}$

that's a boldface theta alright

$\mathbf{\Xi}$

$\mathbf{\Lambda}$

Seems to be working

Good stuff

Boldsymbol probably needed

Let $\mathbf{\Delta}$ be a simplex

More like a prism lol

at least Delta, Pi and Sigma are in one of the standard packages (amssymb I would guess)

there's a custom package to use the scream emoji as an operator nowadays anyway

i was more curious if mathjax could do it

\renewcommand{\qedsymbol}{\crylaughemoji}

$\require{amssymb}$

$\measuredangle$

$\Join$

dont think you could do that before

I once read lecture notes where the author always used "joint" instead of "join" to refer to one of the operations of a lattice

lol

meet and joint. if its legal in your state, that is.

if it's legal in your state or you happen to know a distributive lattice

need a signed petition from our friendly nbhd lawyer with this ^ printed out

to be used and abused

RAWR

here's a question that might interest the general crowd here. any locally compact topological group $G$ has a unique-upto-scale left-Haar measure $\mu$, which gives rise a right-Haar measure $S \mapsto \mu(S^{-1})$ but these aren't typically going to be the same objects. there's a relation between these two though, which is as follows:

denote $\Delta : G \to \Bbb R$ to be the modular function associated with the measure, $\Delta(g) = \mu(gSg^{-1})/\mu(S)$; this does not depend on the choice of the Borel set $S$; this is because the denominator and numerator are both left-Haar measures for a fixed $g$ and varying $S$, which by uniqueness are related by a uniform scalar.

then for every $f \in C_c(G)$, $$\int_G f(x^{-1}) \Delta(x) d\mu_x = \int_G f(x) d\mu_x$$

from which you see that right=left Haar measure iff $\Delta \equiv 1$

because the integral on the lhs is really integral of $f(x) \Delta(x^{-1})$ wrt the right-Haar measure

do i have to cross a paywall to read the question

there's a way of interpreting the above equality, especially in the case $\Delta \equiv 1$ (i.e., there's a bi-invariant measure); the think of $f(xy^{-1})$ as the amount of mass transported from the element $y$ in the group to an element $x$ in the group. this is not a vector field or anything, it's more like a sales scheme you have to transport some amount of stuff from one plane in the group to another, no continuity nothing

:)

im writing!

i'll read when i get back from picking up the munchkin from day care. i like this setting because i used to know something about it :)

thanks! have fun.

@BalarkaSen discord

@RyanUnger hi! sent

So where is the question

whoops lol

so the above says that if $\Delta \equiv 1$, for any mass-transport scheme $f$ on the group you have mass recieved by identity = mass sent by identity, hence not only at identity but at every group element

there is conservation of mass

loss of conservation is measured by $\Delta$

i want to verify the following explicit example in this language: take the affine positively oriented isometry group of $\Bbb R$, all transfo's of the form $x \mapsto ax + b$. this in fact does not have an bi-inv Haar measure, and you can check that by explicit computation

but i think you can also do the following; this group acts on $T_{>0} \Bbb R$, the positive tangent bundle on $\Bbb R$ ((point, tangent vector) where tangent vector > 0), transitively and freely

define the following mass-transport scheme $f((x, v), (y, w)) = 1$ if $x > y$ and $w > v$, and $0$ otherwise. i think you should do something infinitisimal instead of this, like $x - y$, but think discretely for a second; you only transport mass from left to right, and from shorter vectors to longer vectors

in this setup its clear that $(0, 1)$ sends infinite mass out but receives finite mass

So the question is, is this correct?

does the existence of a bijection guarantee that two vector spaces have the same dimension?

feels like it does

nope, we can biject from $\mathbb{R}$ to $\mathbb R^2$

there goes my argument

Who ever considers discontinuous bijections? I mean, really?

oh, never thought of using that

i haven't encountered a discontinuous bijection $\mathbb R \to \mathbb R$ up to now

What are you actually trying to do?

i'm done with the proof, i was proving that if $T: V \to W$ is a linear map, $\dim V = \dim N(T) + \dim \textrm{Image } T$

@Semiclassical I don't think addresses so much time management, but rather focus issues.

And for people who have focus issues, I think it does definitely help. It helps me, at least. I have a few friends that also use it for that purpose.

Just working with bases.

@anakhro Clearly your focus is misaimed. Memes instead of study.

we got a kind of toddler spirograph for munchkin's birthday. wasn't someone just discussing polar coordinates and cusps?

I see you’ve officially adopted my name for the troublemaker!

I helped someone on main understand cusps. You probably looked.

pretty hard to get actual cusps with the toddler set. believe me, i've tried.