looks good to me, although i don't 100% know what event based operation terms are, those seem plausible

event operation terms is like: $ P(A), P(B), P(A\:\cap\:B), P(A\:\cup\:B)$

yeah, if it means the usual set operations, i think you have interpreted the question properly.

@TedShifrin oh I never thought to use $e^x$ damn; just to completeness, this doesn’t mean my error term is incorrect though right?

@leslietownes Ah, sorry. English isn't my native language. So, I translate it literally

It probably means their estimate beats yours.

This is A super powerful tool that lots of students don’t learn.

matheus, i think you've got the right idea.

Yep, I understood :D

thank you

hi ted. it's getting a little colder here in long beach. may have to replace the air filters and turn the heater on. wouldn't want to live below 65F.

Poor dears. Olivia will warm you.

also, the dodgers seem absolutely exhausted. they need better pitching.

Meh.

Is it possible to know homology of covering space knowing homology of the ground space?

Here, covering is $n$-fold covering

Complicated. There is a spectral sequence.

that seems hard. do you have a special case in mind?

today's adventures in math which i don't really understand: Dehn's word problem in groups

Pass

lol

I also vote no on that.

Sorry

the idea is pretty simple for permutation groups

Double pass.

suppose someone gives you a set of generators for the symmetric group, and a word in the resulting subgroup

suppose we don't, but say we did

how do you tell whether said word is the identity or not

i actually had this coming up as a practical problem from student HW

i had an annoying list of generators and a particular permutation i wanted to create

aaaanyways

i can solve it algorithmically in SageMath. but i have no idea how it does so

accept the mystery

For simplicity, a manifold $X$ has $H_0(X) = \Bbb Z$ and $H_1(X) = \Bbb Z/6$ and $H_2(X) = \Bbb Z$ with $3$-fold covering $\tilde{X}$.

and other terms vanish

I would add $X$ a closed orientable manifold

reading, i think it amounts to "try every possible simplification and see if that makes it shorter. if it does, try again. if not, halt"

this strikes me as hopeless

i mean, GAP does it

so evidently there's stuff that works

expressing the homology of the base space from the covering is already awfully hard in general, the reverse seems nigh impossible

oh. wrong conversation

if $\pi_1(X)$ is finite and you take coefficients where $|\pi_1(X)|$ is invertible, there is a natural isomorphism between $H_{\ast}(\tilde{X})_{\pi_1(X)}$ (meaning coinvariants wrt the induced action) and $H_{\ast}(X)$, but even if we take field coefficients, this doesn't even determine the Betti numbers of $\tilde{X}$

I would change my example. $X$ is a closed orientable 4-mfd with $\pi_1(X) = Z/15$ and the betti number of $H_2(X)$ is $3$.

I can completely determine homology of $X$

granted, to go the direction from base to total space (which I'm less used to doing), you want to take monodromy into account, so Ted's remark is more on point than mine

Using poincare duality and the fact that rank of $H_{n-k}(X)$ and $H_k(X)$ are the same and torsion of $H_{n-k}(X)$ and $H_{k-1}(X)$ are the same

but the mere existence of the spectral sequence tells you rather little in general

and also knowing the homology of the base is strictly less information than knowing its homology with local coefficients in the fibers

Can't even determine $\pi_1(\tilde{X})$?

I mean in my simple example not general

determine from what

My setting chat.stackexchange.com/transcript/message/59410348#59410348

that's not the question, the question is what you know about the covering

Ah I can determine

using $\tilde{X}$ is 3-fold

Using $p_*$ is injective where $p$ is a covering map, $\pi_1(\tilde{X}) =\Bbb Z/5$

but still don't know the betti number

Yes only with this, I can't determine the second homology

maybe I can use the fact that $X$ is homotopic to CW complex

Guys, disregard this. I understood my fault

my favorite kind of email is one that ends with 'this needs no response.' in the chat context, 'disregard this, i understood my fault' is just as good.

upvoted, liked, subscribed

"answers are orthogonal to the spirit of this email"

"to reply to this email would be superfluous"

likes, upvotes, cryptocurrency investments

these poor dodgers.

rosario hit it too far

he could've gotten the cycle, instead all he gets is a three-run homer

shameful

(he's a former Minnesota Twins player, so i am rooting for him)

i completely missed them winning the world series last year, that's why this hurts.

those tomhawk chants are embarrassing and everyone should be ashamed of themselves.

yeah

i'm happy to root for Rosario. Rooting for the Braves...no thanks

there's really no team i find myself inclined to root for

i wanted the Rays to go all the way b/c Nelson Cruz deserves a World Series for carrying the Twins as much as he did

the dodgers are over, i have never seen pitching with less energy

can't say i have a lot of interest in Red Sox vs Astros

my mom is a sox fan but i don't care

i'm trying to squeeze $\frac{\sin t}{t}$ using an area involving two sectors on the unit circle $(\cos t)^2 \frac{t}{2}$ (sector with a radius of $\cos t$) and $t/2$ (sector with a radius of $1$)

do i need to do a power series expansion to prove that the inequality holds?

I want $(\cos^2{t})\frac t 2 \leq \frac 1 2 \sin t \cos t \leq \frac t 2$

the sector formula being $r^2 * \frac{\theta}{2}$

$\frac 1 2 \sin t \cos t$ being the area of the triangle of base $\cos t$ and height $\sin t$

visually speaking, on the unit circle $\frac{1}{2}\sin t \cos t$ is an area clearly in between the two mentioned sector areas, but I'm not sure what's the formal way to prove the inequality is true

heh apparently trigonometric inequalities aren't a trivial thing

the required inequality is $x\cos x \leq \sin x \leq x$ which is clearly true using a graphing calculator

but how the heck do you do analysis on trigonometric functions

are you assuming $x \ge 0$? otherwise it is not true.

yeah, $x \in [0, \pi]$

since $x$ is going to zero and we're thinking of an area

i'm simplifying the difference between the power series expansions and seeing if I can get something out of it

since $|\sin'(x)| \le 1$ we have $\sin x \le x$.

Let $f(x) = x \cos x - \sin x$, note that $f(0) = 0$ and $f'(x) \le 0$ for $x \in [0, \pi]$.

fairly straightforward.

i really should have written $|\sin x| \le |x|$.

oh nice answer

thanks a lot

Hi all, I have a surface $A$ which is delimited by a curve $S$, how to most neatly call that? Would $\partial A = S$ be cool?

is there any reason why wolframalpha would think this expression is not always true?

$\frac{4x}{\sqrt{2x^3}} = \frac{4x}{\sqrt{2x^3}} * \frac{\sqrt{2x^3}}{\sqrt{2x^3}} = \frac{4x\sqrt{2x^3}}{\sqrt{2x^3}\sqrt{2x^3}}$

only the first equality gets me a "True" output, the second equality doesn't

well nvm, it gives me true now

@Semiclassical what do you want to know

for most groups, the word problem is solvable

math.stackexchange.com/a/187417/668308 In this answer, he assumed $M$ and $N$ are orientable if and only if $M\# N$ is orientable.

But I don't think this is trivial. I want to prove that using the exact sequence given in the answer post

its not trivial but it's "clear" if you understand what orientability means

I can prove if $M$ and $N$ are orientable then $M\# N$ is orientable but I can't prove the converse

ah! i was just about to ask

@shintuku Of course, things get bad if $x\lt0$ anyway.

@BalarkaSen Intuitively, yes but my definition is using homology so not very clear I think.

Since then $\sqrt{2x^3}$ is imagary

i was getting $\frac{4x}{\sqrt{2x^3}} = 2\sqrt{2}\frac{\sqrt{x^3}}{x^2}$ true but not $\frac{4x}{\sqrt{2x^3}} = 2\sqrt{2}\frac{x^{\frac{3}{2}}}{x^2}$

@love_sodam so you should chase the two definitions down until they match

they're all equivalent

@BalarkaSen What is the other definition?

well you said its clear intuitively so you should have a definition in mind

whats the intuition?

@BalarkaSen Not very formal. Geometrical explanation of orientability.

what is it

@robjohn if you have a second, what's the correct order of operation to get absolute values here?

or is it just the fact of dividing an even exponent to get an uneven one?

en.wikipedia.org/wiki/Orientability description in wikipedia

@shintuku My previous statement was not true.

I'll take that intuition back

@robjohn oh np

$\sqrt{2x^3}\sqrt{2x^3}=2x^3$: obvious if $x\ge0$, but also true for $x\lt0$ since then $\sqrt{2x^3}=i\sqrt{2|x|^3}$ or $\sqrt{2x^3}=-i\sqrt{2|x|^3}$

any clue why we don't get a decisive answer on $x^{3/2} = \sqrt{x^3}$ from wolframalpha though?

@robjohn oh right

@shintuku $\sqrt{x}$ and $x^{1/2}$ are meant for $x\ge0$.

would both sides produce different complex answers if $x < 0$?

@shintuku actually, $(-1)^{3/2}=-i$ whereas $\sqrt{(-1)^3}=i$

Here's a question for you all: Is there any particular music you like to listen to while you work?

(Also, exponents are tricky in the complex numbers. I never quite got the hang of them when I was first being introduced)

@robjohn oh right! i never noticed the exponent $3/2$ is actually ambiguous, $\left((-1)^3\right)^{1/2} \neq \left((-1)^{1/2}\right)^{3}$

thanks a lot, i would have spent a lot of time confused otherwise hehe

hm. does anyone know whether $3/2$ means $3\cdot\frac{1}{2}$ before meaning $\frac 1 2 \cdot 3$, or is there no convention?

@shintuku I don't think there is a convention. If you don't know what $\arg(z)$ is, $z^{3/2}$ is not clear.

alright thanks!

@shintuku You could use the convention that $z^w$ is $e^{w\ln(z)}$, but that just shifts the ambiguity to another place. ;)

@Rithaniel I have fairly broad musical tastes. When I'm doing a symbol manipulation task, whether that's maths or coding, I tend to prefer instrumental music, unless it's a piece I've heard countless times. Otherwise, the song lyrics can "clutter" the symbol processing part of my brain. :)

at PM 2Ring: thanks, didn't know that

@PM2Ring the ambiguity lies in $\arg(z)$, which is why I said $\arg(z)$ is important in computing $z^{3/2}$.

@PM2Ring Yeah, I've kind of noticed that familiarity is what I need in a song to listen to while I focus on work. I mean, I'm all for new music, but it dominates my attention. If it's familiar, I can just tune it out and still feel relaxed.

For me, that ends up being mostly old video game soundtracks. Like the Sims 2 or even stuff from the SNES

@robjohn Certainly.

en.wikipedia.org/wiki/Complex_logarithm Riemann surface of Log[z], projection from 4dim C x C to 3dim C x Im(C), color is argument.

How can I show $S^3\vee S^4\vee S^7=:X$ is not homotopy equivalent to any closed orientable manifold? I tried to show this space fails poincare duality. Using $Z/2$ coefficient, if we let $\phi$ be a generator of $H^3$ and $\psi$ be a generator of $H^4$ and $\sigma$ be a generator of $H_3$ then by poincare duality, if I let $[X]$ be a fundamental class of $X$ then $\psi\cap[M] =\sigma$. Now $1 = \sigma\cap\phi = ([M]\cap\psi)\cap\phi =[M]\cap(\psi\cup\phi)$

but $(\psi\cup\phi)\neq 0$ so I can't find a contradiciton.

Any help?

Bruh, what is this? desmos.com/calculator/64awzt7exq

If I keep scaling the x axis, it looks the same, and in my experience, only the exponential function does that.

In any case, the value seems to always be between 1 and 2 for $f_n(x)$, and given integer inputs to $f_{n2}(x)$, it would seem that we always get a power of two, or a power of two plus one meaning the reduction algorithm that I'm using can be tweaked a tad to get better results for the recursive modulus algorithm.

Can I get any recommendations for good books to learn calculus and discrete mathematics?

Is there any name for a set $\{t_1 \cup t_2 \mid t_1 \in A,\ t_2 \in B\}$? It is not a union of A and B but it is a union between their elements.

@love_sodam How so?

@TedShifrin What do you mean?

It seems Google also don't know

Ted is asking you to justify, which you won't be able to, because it is wrong

@Thorgott So $\psi\cup\phi =0$?

If I have an equation $y=x \times a/x$, is it valid for x=0? or in other words, for all values of x.

Hmm... $\tilde{H}^*(S^3\vee S^4\vee S^7)\simeq \tilde{H}^*(S^3)\times\tilde{H}^*(S^4)\times\tilde{H}^*(S^7)$. So $(0,\psi,0),(\phi,0,0)\in\tilde{H}^*(S^3\vee S^4\vee S^7)$ so $(0,\psi,0)\cdot(\phi,0,0) = 0$?

Does it make sense? Seems weird

Not weird at all. Think simplicially. What do the simplices in $S^7$ have to do with simplices in $S^3$ and $S^4$?

Nothing?

So $(\psi\cup\phi) = 0$ gives me a contradiction

Yes.

Yes degree $7$ is of $H^*(X)$ is $\Bbb Z$ but cup product of degree $3$ and $4$ are zero

@TedShifrin Thanks !

@NazmulHasanShipon the right hand side is not defined for $x=0$. I don’t care about $y$.

how is everyone doing in here

@TedShifrin, would you like you have a look at my question?

Great I handed in my homework so it was a weight off my shoulders :)

took me over 20 hours :/

Oof, lot of work. Glad it's over, though.

Yeah, like half of it was debugging errors with obscure packages and raising github tickets :/

I thought it was all gonna run seemlessly

@NazmulHasanShipon what makes you think that you are allowed to put e=0 there?

Oh yikes, that's probably more work than was anticipated.

If I want to describe cup product structure of $\Bbb CP^2\#\Bbb CP^2$ where $\#$ is a connected sum, if $\alpha$ and $\beta$ be two generators of $H^2$ of each summand, $\alpha\cup\beta =0$ since they do not intersect transversely. Is that correct?

Oh wait, that implies it has trivial cup product structure

Not really

@Koro It's actually from my question where I want to understand how equation (i) (which is the general equation of conic) represent a circle. So, as an ellipse become a circle when e=0, if I put e=0 in that ellipse equation I would get 0/0. Then my question is how does that general equation represent a circle?

@love_sodam No. What about $\alpha\cup\alpha$, etc.?

@TedShifrin Yeah I just realized $\alpha\cup\alpha$ and $\beta\cup\beta$ are generators of $H^4$.

And cup products of other positive degree elements vanish

@TedShifrin I wonder if my argument which shows $\alpha\cup\beta =0$ holds.

@Thorgott The problem I asked yesterday, I can use Euler characteristic to determine the second betti number.

One thing is that $\tilde{X}$ and $X$ are just homotopy equivalent to finite CW complexes not homeomorphic.

Wondering keywords for a computer algorithm I'll have to create for a specific application. Each machine module draws a pattern, line, squiggle, whatever. I need to design an algorithm for making symmetrical patterns for an increasing number of module configurations.

Two modules, three modules, four, five, six, etc.

@love_sodam what does “they do not intersect transversely” mean?

is anybody here? (I don't have a question to ask just wondering if anyone's here)

no

ok

you know that feeling when you formalize the definition of the manifold over many years

sorry not me specifically

no

just that feeling of formalizing a rigourous definition of a problem statement

just that feeling of like okay I know my variables... I know what I'm working with now

never heard of it

:)

yes

it's an empty feeling that doesn't amount to much

no

why wouldn't you define something important relatively early on tho\

i find i get something resembling that feeling fairly late in the process of absorbing information

long past when it first would have been helpful

because sometimes it takes a few horses to get to the farm

a definition of a manifold is a good example, you can work with it a while without too much of it sinking in

nowadays you never get to the farm

you just get pimped by an app

it took a long time for mathematicians to get a good definition of manifold

yeah but that was all before we were born

unless you're ted or something

I mean if I was the dude that defined manifolds I'm pretty sure the feeling would have been pretty good

that's a good point though. gobs of examples were certainly well known and even some general results, but there wasn't a lot of coordinate-free understanding

or maybe there was but i'm unaware of it

could someone please tell me if my argument is correct?

you deleted it

Serious issue @monoidal. Yes, you need to understand quotient maps. Do we know $f$ and $g$ are closed paths? You certainly do not say this. Go back and think.

have a great day everyone it was a pleasure

I wish my mind was more skeptical of things I come across :/

Suppose there is a sequence of functions $f_n(x)$ that converge uniformly to $f(x)$. Is it then true that $|f_n(x)-f(x)|$ is bounded above by some function $y(n)$? To satisify the definition of uniform convergence, it seems like there should be such a function.

@AMDG Just wait a few years. :-)

I don't have years to wait. I'm not going to wait. It's just difficult to break old habits ;-;

@TedShifrin yes, $f$ and $g$ are closed paths. But what's the issue with the quotient map argument? $\tilde{f}\circ u= f$ and since $u$ is a quotient map and $f$ continuous, then $\tilde{f}$ must be continuous too

that's why I specified $u$ is a quotient map

schn i'm a little unsure of the domain of "y(n)" from your notation. |f_n(x) - f(x)| also need only be bounded when n is sufficiently large, not necessarily for all n.

that seems to be the probability that the first sale occurs on the fourth call?

once you get three no-sales, the first sale occurs after the third call no matter when it occurs

Yeah

I take it 0.23 is the probability of sale per call

that's pretty good, i think people would actually like telemarketing jobs if it was that high in real life

Almost certain I'm using the wrong distribution for this exercise

i dunno, is telemarketing more like flipping coins or is it more like getting kicked by a horse

these telemarketing I'm talking is based if you can or not sell a certain product

What you computed is the probability to get three failures and then a success

What if you got four fails before the success?

Hello!

Yep, after 3 failures (or no-sales), he finally can sell the product

If we have cities and we want to transport goods between cities. Writing a linear program to find minimum cost of transportation means a program in linear time O(n) or it means linear system of equations :/

So is four fails and then a sale counted as an event? If so, then your calculation isn’t accounting for that

@Avra ask the person who gave you the problem

@Semiclassical no, I'm getting this: first the guy tries to sell the product three times and he can't sell them. After these 3 tries, he finally sell the product in the fourth call.

I didn't understand if you're talking that my thoughts are wrong or you didn't understand my thought.

There’s two ways to read the problem: 1) after the third call, he immediately makes a sale. That’s what you’ve assumed

But I’d read the problem as “his first sale doesn’t occur in the first three calls”

There’s no implication to my eyes that the first sale does in fact happen on the fourth call, only that it didn’t happen on the first three

It could be on the fourth, or fifth, or 200th, etc

So, I didn't understand the question.

Tbf, I think the question is sorta misleading

right?

That’d work, but there’s a simpler way to look at it

After the first three failed calls, what’s the probability that something happens? (This may sound weird but bear with me)

Sale on fourth, fifth, etc. what’s the probability of any particular outcome at that stage?

@Semiclassical It could occurs anything (100%). Can either sell the product or can't.

Right.

Now, what’s the probability of those three failed calls, ignoring what happens after?

P(S^c)^3

Sure. Crunch the number

Wow, it makes sense

0.4565

Yep

thank you

The three failed calls is the event

Everything after is gravy

I understood

Mmkay

Of course, the problem is a little silly: there’s some probability that he runs out of time during the day before making a sale

if he makes only finitely many calls, there's some probability he doesn't make any sale, but probably not enough to affect the third decimal place of the result after a reasonable day's work

or what semiclassical said

B/c calls in real life aren’t instant

Yeah

The more standard version of this problem is to flip an unfair coin until you get heads

In which case I think it’s less tempting to think “heads on fourth flip”

For an interesting (and difficult) variation on this, you can do stuff like

What’s the probability that you see HH before you see HTH while flipping over and iver

"Assume the salesperson is fired if they don't make at least one sale a day. If the salesperson has not made a sale two hours before work closes, they grow increasingly desperate, and their probability of making a sale on a given 25-minute call either a) increases linearly to 65%, or b) decreases linearly to 0%, with even odds. Assume the call center hires 40 salespeople per month. What is the probability that the call center has more than 100 salespeople employed on any given day?"

also there's a horse in the office who goes around kicking people

"What are the odds you are kicked by the horse before you are fired for being unable to make a sale?"

Lol

@Semiclassical :0

@Semiclassical. Have you took convex optimization before pleasE?

@Avra convex optimization covers a lot of material. you probably want to narrow down exactly what you are looking for.

I would like to take convex next course, so topics mainly would be about quasai and convex optimization of functions in 2D and 3D

Also for gradient acceleration

gradient acceleration is a fairly advanced topic. also, to some extent the point of (applied) convex optimization is the ability to deal with large dimensions, so the 2d/3d things does not really jive?

OMG

:/

You mean gradient acceleration covered in advanced convex optimization course :)

Does the coues usually has lots of proofs compared to real analysis for example :/

perhaps you could elaborate what you mean by gradient acceleration. i am referring to nesterov's gradient acceleration.

i cannot comment on a course that i know nothing about!

You did not take real analysis?

i have never taken a convex analysis course.

not convex but real analysis

i mean convex optimization, not analysis (neither, both are self learned)

OMG!

yes, i have taken real/functional analysis.

cool. you studied though convex analysis by yourself?

convex was not the huge topic back in the 80s that it is now. it took a while before it caught on from an optimization perspective. interior point methods, karmarkar, etc, triggered another visit.

nice!

I mean you are right, I will look for courses online before I decide

I might watch some videos to see how things go though I am not good with real analysis :/ Will see

This is the best choice I have now as all other courses are very advanced

convex optimization is not real analysis, but it certainly requires some familiarity with the usual topics, continuity, differentiation, the various theorems etc.

I do really like statistical inference though!

However, it has some concepts from measure theory!

but without knowing what the course is teaching and your level of analysis it is hard to make a suggestion.

measure theory is even more difficult than convex as some members mentioned

I will just watch 3 lectures online and then decide!

there is no linear ordering of difficulty. it depends on the individual and their motivation.

That's my best option as I did the same with measure theory...no need to continue...it's not as I imagined

i am not a mathematician, but i certainly find real analysis much easier than, say, group theory.

I watched 2 lectures of measure theory and stopped, not bad but also needs lots of work

different folks have different strengths.

Yes

Thank you

topics from real analysis that crop up everywhere are continuity, connectedness, compactness, differentiation, implicit function theorem, taylor/power series.

it is worth spending some time on these.

Did you have sense that convex is sort of more applied than real analysis?

@Avra nope

though i do find the stuff i've seen from it interesting

mostly semidefinite programs etc