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12:17 AM
@MikeMiller >calling scooter techno
 
This is a nice little question:
1
Q: Wrapping a Rubber Band around a Cube

Joseph KisenwetherImagine a cube wrapped with string so that each of the six sides has two linked "L's" on it, like this image (or some reflection or rotation thereof). You can get a lot of different knots and links depending on the particular choices made. Are any of them the unknot or the unlink? In other words...

 
 
1 hour later…
1:25 AM
Looks like a good question for DogAteMy!! @Akiva
 
Hi @TedShifrin
 
Hi, Karim.
 
2:05 AM
@Astyx are you learning algebraic geometry ?
 
2:19 AM
@TedShifrin: I reposted the ruled lozenge in a CW post.
 
@robjohn Yes, giving it all away.
 
so cool
 
@robjohn: I no longer have patience with assholes like this.
 
@TedShifrin Eh, it's buried at the bottom of 3 pages of prior posts
 
I have a feeling that some constructions in mathematics you don't actually understand them unless you see what they are built for.
@TedShifrin I would just ignore him.
 
2:33 AM
I am now, after voting to close.
 
@TedShifrin I had one almost like that
 
I need a proof ... even if I have no idea what I'm talking about ... and certainly have consulted nary a book.
 
Yep. Spoon feed me, damn it!
 
That might becan exam, too, now that I think about it. ... Before revision, it was ruder and much less explicit.
 
When they want it fast, that is a dead give-away.
 
2:45 AM
Oh, I didn’t recall temporal exigency.
 
3:31 AM
Is that diagram on wiki on the bands article a lattice of all subvarieties of regular bands, or just some?
In mathematics, a band (also called idempotent semigroup) is a semigroup in which every element is idempotent (in other words equal to its own square). Bands were first studied and named by A. H. Clifford (1954); the lattice of varieties of bands was described independently in the early 1970s by Biryukov, Fennemore and Gerhard. Semilattices, left-zero bands, right-zero bands, rectangular bands, normal bands, left-regular bands, right-regular bands and regular bands, specific subclasses of bands that lie near the bottom of this lattice, are of particular interest and are briefly described below...
 
3:55 AM
It is
 
4:13 AM
Hey quick sanity-check group theory q: if we have f: X->X is injective but not surjective, can we show there exists a map g : X->X such that g(f(x)) is the identity?
 
what happens if f:Z->Z is the multiplication by 2 map
 
i think the domains are just confusing me because neccesarily we have elements in the codomain of f that aren't mapped to, so how could g o f be the identity?
@Thorgott so the inverse would be g(x)=x/2 but thats not a map from Z->Z, right?
 
that's the idea, yeah
 
okay so we can't, that's what i thought
for some reason our homework is to prove that we can so i was a bit confused lol
oh wait X being a set doesn’t change anything right?
rather than a group
 
4:36 AM
It changes everything.
 
Yup I just got it thank you
We can literally take f^-1 for each element that’s mapped to right?
And even though the domain will be a strict subset of X we still consider that a function X ->X?
 
 
1 hour later…
5:46 AM
@NicoA No, you have to define the function everywhere.
 
Hi @TedShifrin. I was having a little difficulty figuring out how to solve this. It is in a chapter about Arc measures. All that was done in the chapter was defining major/minor arcs, congruent arcs, arc addition, and congruent chords having congruent minor arcs.

So one idea that came to mind was perhaps using that part of the rope covers each semi circle, but I get jammed up from there.....
But since the diameter was given maybe I could use the circumference in some way, but since this is in a chapter about arc lengths then this wouldn't really apply
by the way...........I NEED an answer to this............🤣🤣
 
6:20 AM
You need to stop guessing and figure out what you actually know.
@dc3rd What's the configuration of the three circles?
The other major hint is: If a line is tangent to a circle at a point, what angle does the radius to that point make with the line?
 
What I know is that each circle has a diameter of 14in, the circles are tangent to one another as well
 
Yes, so radius is 7".
What shape do the circles make?
 
A line tangent to a circle means the point of intersection will be perpendicular to the radius.
 
OK, good. You'll need that.
Now what shape do you see when you look at the picture?
 
I'm trying to "see" the right triangle that the hints are getting at. But it hasn't sunk in yet
let me draw it out
 
6:31 AM
Not right triangle.
The general shape of the picture should jump out at you. If you make me give another hint, I'll say draw in the centers of the circles. Now what do you see?
Yes, you always have to draw pictures. You can't do math just looking at it.
 
So drwing in the center points if I were to attach the radii of each circle I could see an equilateral triangle
 
So that should jump out at you from the beginning.
OK. So how do the line segments tangent to the circle relate to this triangle you've drawn?
No more hints.
 
Yes I did see the rounded edges of the rope forming an equilateral triangle, but I didn't want to jump to any conclusions
 
No, the equilateral triangle is inside.
The outside shape is three line segments and three circular arcs.
The question is: How many degrees in those circular arcs? I.e., how much of the circle's circumference does each one give you?
 
well it's not perfectly drawn by me, but those line segments that are perpendicular to my radii formed some rectangles. So it appears that the "gaps" between the circles could be calculated by getting the correct side for the rectangles. And as you said I also have to find the sum of the arcs
 
6:40 AM
OK, so you know enough information to finish the problem now.
How long are those line segment pieces tangent to the circles?
 
I haven't calculated them yet, but by inspecting them it seems like they're going to be 14in each based on the radii
 
Correct.
Now figure out the angle of the circular arc that's in each circle.
 
Doing that as we speak....just trying to recall what theorems I read to establish the calculation
 
You don't need any theorems.
Use the information we've discussed.
(Look at the center of the top circle. Look at angles, of course.)
 
I shouldn't have used the word theorem, I am using the ideas discussed actually.
So...going by my drawing....thos radii perpendicular to the lines which I used to draw the rectangles would be right angles......so it appears that each of those arc lengths are going to be 90 degrees ($\frac{\pi}{2}$)
 
6:54 AM
No.
 
good...it didn't feel right
 
I want the arc at the very top of the picture.
 
Fancy picture that captures what my pen is struggling to do nicely
What program do you use?
 
Mathematica
 
6:57 AM
Outside my budget..........😝
 
He's showing off.
 
If you're a student, there are decent prices
 
I would have done it in Illustrator in the olden days. But by hand is good.
 
I agree, but I don't think I would use it enough within a year to warrant it, at least currently.
 
just thought that image gives the length pretty clearly.
 
6:59 AM
Doesn't replace thinking!
 
it does and takes away the impreciseness of my drawing....but back to this arc length
 
what is the sum of the three marked angles?
 
ooooo..........what if I were to draw a line straight up and "bisect" the intended angle? So now I would have a right triangle and could use trig becuse I have radius lengths
right angle with the line tangent to the circle
no it woun't work....looked like it though...
 
I told you to look at the angles at the center of the top circle. Tell me what the 3 angles we know are.
 
Oooooo....well, from the nice drawing it seems the sum of those angles is going to be 360
 
7:06 AM
Yes. But that includes the one we are looking for.
 
Does this feel better?
 
Robjohn, that truly is cheating.
He needs to do basics.
 
Hey, you wanted something that looked hand drawn
 
So going back, the angles I know are the right angles formed from each radii....
 
smack
And what else is in your picture?
 
7:09 AM
the rectangles.
 
You already did those angles.
 
It's the same picture as before, just ran it through xkcdConvert
 
Oh jeez....I don't want to keep you up even if you're on the west coast.....
 
You should be done in seconds.
 
well it seems I have to go back to the idea of the sum of the angles being the 360 and specifically for the arc of the top circle the angle would be 120.........
 
7:23 AM
Right. Now three of those ...
 
isn't three of those 360?........
so total length would be $3 \times 14 + 3 \times 120^{\circ}$
 
How long is $120^{\large\circ}$?
 
Good question..............
 
or you could ask how long is $360^{\large\circ}$...
 
this is where the diameter idea comes into play..............
I see
so with that being the case.....we would have total length being

$3 \times 14 + 14\pi = 42 + 14\pi$.....out of curiosity there isn't really a nice way to express that in the degree form is there?
Well that was fun. Piecing together all the pieces and using the previous concepts I learned....except for struggling more than I should on the final part, it was enjoyable. Thanks again for the help gentlemen. 😎
 
7:46 AM
Hi everyone, I need to know what an "isometric embedding" is. What kind of textbook would contain a definition for this term and related ones?
To clarify, I have an engineering background and just had some basic Analysis and Linear Algebra. My problem is that I don't even know what topic "topology, functional analysis,...." the aforementioned term belongs to.
 
That sounds like it would be topological......look in Munkres Topology
acutally it is in the index of Munkre's Topology
but the more experienced folk could probably guide you better
 
any nonempty CW complex contains 0-cell
 
@dc3rd No, but using $\pi\approx\frac{22}7$ you get close to $86$.
 
is this the correct statement?
by the way thank you @LeakyNun
 
7:53 AM
what did i do
 
Now you've done it
 
you gave me some counterexample yesterday
 
Cool...thanks @robjohn
 
oh
 
off to bed I go.
 
7:55 AM
$85.98229715$
 
Thank you!
 
 
1 hour later…
9:22 AM
0
Q: If $X=\bigcup_\alpha X_\alpha$ is a CW complex then $SX\simeq Y\vee_\alpha SX_\alpha$ where $Y$ is some graph

love_sodam If $X$ is a CW complex with component $X_\alpha$, show that the suspension $SX$ is homotopy equivalent to $Y\vee_\alpha SX_\alpha$ for some graph $Y$. In this case that $X$ is a finite graph, show that $SX$ is homotopy equivalent to a wedge sum of circles and 2-spheres. I think the second state...

Anyone can answer this problem?
 
 
3 hours later…
11:56 AM
smh the one logic kid in my topology II class complained that our local degree computations are not rigorous and it wouldn't be possible to implement this in a theorem prover
 
Hit himwith some presheaf categories and cubical types
 
12:16 PM
Graphical theorem prover goes brrr
 
12:44 PM
what are sets that are neither open nor closed called?
 
abgeschloffen
 
Noclopen probably
Lol
 
ah I didn't see the neither
 
Hi folks. Can anyone offer a hint for finding the value of $b$ for which
$$1+e^b+e^{2b}+e^{3b}+\cdots=9?$$
 
Unclopen
Anticlopen
Antonymnclopen
 
12:48 PM
(Where we haven't gotten Taylor series yet)
 
@Jeff $1+x+x^2+\dots = {1\over 1-x}$
 
@Astyx We haven't been given that yet.
 
Prove it
 
@EdwardEvans omg i never thought of this term
 
@Astyx Actually, we have (just not so explicitly)
@Astyx (by which I mean 'thanks')
 
12:54 PM
glad to help
 
@user2103480 lmao
 
I am @KarimMansour
 
1:15 PM
An artinian ring $R$ has finitely maximal ideals: let $(\mathfrak{m}_i)_{i \in I}$ be all the maximal ideals of $R$ and define $M = \prod_{i \in I} \mathfrak{m}_i = \bigcap_{i \in I} \mathfrak{m}_i$. Then $R/M$ is finitely cogenerated so there are finitely many $\mathfrak{m_1}, \dots , \mathfrak{m}_n$ such that $M = \bigcap_{j = 1}^n \mathfrak{m}_j$.
Since $\mathfrak{m_i} \supseteq \prod_{i \in I} \mathfrak{m}_i = \bigcap_{j=1}^n \mathfrak{m}_j$ we get that every $\mathfrak{m}_i$ is one of the $\mathfrak{m}_j$.
(this as a proof instead of $\mathfrak{m}_1 \supseteq \mathfrak{m}_1 \cap \mathfrak{m}_2 \supseteq \dots$)
 
1:55 PM
what on earth is cogenerated
 
If you've got a family of submodules $(M_i)_{i \in I}$ such that $\bigcap_{i \in I} M_i = 0$ then there's a finite subset $J \subset I$ such that $\bigcap_{j \in J} M_j = 0$
ergh
Turns out that being artinian (dually to noetherian) is equivalent to all quotient modules being finitely cogenerated
lol I just read that I wrote "has finitely maximal ideals"
 
2:12 PM
Hello, I have a question about proving the following is true in Markov chain. I'm trying to prove by Chapman-Kolmogorov equations, but I failed. Any hints and suggestions are welcome!
0
Q: Why the recurrence relation of Markov chain can be written as the following?

MikeI am reading E-Li-Vanden-Eijnden's Applied Stochastic Analysis, and the author claims: The recurrence relation of $\boldsymbol{\mu}_n=\boldsymbol{\mu}_{n-1}\boldsymbol{P}$, where $\boldsymbol{\mu}_n$ is the distribution of $X_n$, and $\boldsymbol{P}$ is the transition matrix, can be written as $$...

 
2:36 PM
@Mike What is your state space? Countable? What is $\mu_{n,i}?$
 
Let's assume the state space is countable. $\mu_{n,i}$ is the ith component of the column vector $\mu_n$
Hope this clarifies
 
So this is the probability that at time n the chain is in state i
 
Correct. Any hints or comments on how to show this rigorously?
 
$\Bbb P(X_n = i) = \sum_j \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j) = \Bbb P(X_n = i \mid X_{n-1} = i) \cdot \Bbb P(X_{n-1} = i) + \sum_{j\neq i} \Bbb P(X_n = j \mid X_{n-1} = j) \cdot \Bbb P(X_{n-1} = j)$
$\sum_j \Bbb P(X_n = j \mid X_{n-1} = i) = 1$ so $\Bbb P(X_n = i \mid X_{n-1} = i) = 1 - \sum_{j \neq i} \Bbb P(X_n = j \mid X_{n-1} = i)$
I hate markov chain notation #
 
@user2103480 "techno" seemed like a better descriptor than "whatever, man"
which is the more accurate description of what scooter is
 
2:58 PM
@MikeMiller I treat trash electronic music like metal fans treat metal
so I'd say this is eurobeat slash harddance
although, nah, eurobeat is too trashy need for speed type stuff. but something something
but the most accurate description is "100% diamond quality music"
 
hYpErPoP
 
calling scooter techno is like calling Die Antwoord rap
 
Yeah that's what I'm saying
It's not accurate but it's still the best one word description
 
Calling Vorga ancient black metal is like calling Maahes cosmic black metal
3
 
@MikeMiller that would be trance I guess
 
3:18 PM
@user2103480 do you know Callush?
 
no wossat
 
She's a dj from Monnem
makes really really good techno
 
fuckin mannheimers stealing the term monnem from monheimers
 
hahaha
 
rhing = local dialect for rhine
 
3:21 PM
Yeah I assumed
 
there can only be one
 
It's not disgustingly industrialised enough to deserve to keep Monnem
 
in fact it is
 
ye but Mannheim and Ludwigshafen are fused by the industrial glue that is BASF
 
the whole of dormagen's bayer is just across the river, and quite a few bayer things are in monheim
so bayer vs basf death match
 
3:23 PM
hahah fair
 
@EdwardEvans lmao of course berlin based
 
Apparently every fourth mannheimer is employed by BASF
 
@EdwardEvans either I have no standards anymore or this really blasts
 
it's so good init
 
yup
yeh just skipped through this is really good
at least the selection, with regards to mixing I'm a pleb anyways
 
3:26 PM
Got really, really drunk the other night on the roof terrace and embarrassed myself dancing to this
 
everybody else embarassed themselves by not dancing to this
 
The correct perspective
 
although mannheimers probably think sven väth sets at time warp are "super hard going"
 
I don't know him but idk hahaha
 
I wanted to say charlotte de witte first but then remembered that she sometimes actually plays pretty rough stuff, I think
 
3:28 PM
I've only been in one techno club in Mannheim cuz Covid hit straight away
 
@EdwardEvans he's one of the DJ "legends", although that title is mostly given to people who came early enough and managed to stay long enough. Like carl cox, ricardo villalobos and such, but not on jeff mills or juan atkins level
 
I don't know any of those people lmaoooo I only know the techno that my mates show me
 
hahaha okay fair enough, doesn't matter anyways if your friends show you the right stuff and drag you to the right parties (which they seem to do)
 
aye, I show them metal in return and they resent me
 
rightfully so
 
3:34 PM
Couple of them haven't been to a metal concert before and I've roped them into seeing Meshuggah in Wiesbaden in November
which will be an experience
 
"Did you hear that crazy 12.4/sqrt(2) rhythm switching into the chuck e. cheese drumroll?"
^how metal fans sound when they talk about meshuggah or any other nerd band
chad techno fan: "THASSA MAD KICK" - "YEH" sound of grinding teeth
 
Ackchyually I think you'll find that Meshuggah perform in 4/4 with metric modulation of riffs
also true tho
 
@EdwardEvans lmao
I don't wanna do arithmetic in $\Bbb Q$ when I listen to music
4/4
absolute unit
 
all rational time sigs are units
 
are they also absolute units
 
3:41 PM
idek what an absolute unit would be
the equivalence class of 1 under the relation a ~ b iff a and b are associates
 
@EdwardEvans there won't ever be one, everytime you come up with one, anglo-saxons mess it up by using an absurd alternative metric
 
but maybe the size of imperial pints drove down the prices
either that or manchester is just cheap as sh*t
 
it's the second one
 
guess I wont ever visit scotland or southern england
is there any other cheap region in the UK
 
3:47 PM
Omg I got a crate of Bitburger for 99c on Tuesday
also anywhere in the north is cheap af
 
@EdwardEvans does that include scotland
wouldve thought its more expensive there
@EdwardEvans whao was that just before expiration
 
I don't think so, I know the North of England is cheap af for beer but I haven't been to Scotland
 
also, how far did you have to carry that lmao
aren't you like 2km away from the next supermarket
 
@user2103480 No, I put one bottle on the Band and told the lady I have a Kasten and I don't think she heard but my mate heard me, picked up the crate and carried it out to put in my rucksack, then I paid and left and then looked at the Beleg and I only paid 99c for it lmao
 
@EdwardEvans wenn alle Konjugierten einer algebraischen Nummer Norme eins hat, dann sie eine Einheitswurzel ist
 
3:49 PM
Yeah I put the whole crate in my Bundeswehronline Rucksack
@LeakyNun Ja
 
(hab ich meine Kasus wohl benutzt?)
 
@Leaky Ja hast du, aber man sagt normal "algebraische Zahl" statt Nummer
 
ok danke
die Kasus sind sehr schwerig
 
(Und ANT heißt darum AZT auf Deutsch)
Ja das stimmt, und man sagt "Fälle" für die Cases
 
aha, ich kenne die Zahlentheorie
 
3:52 PM
Genau
 
haben "Fall" und "Kasus" eine gleiche Bedeutung?
 
Ich glaube schon aber ich würde nie Kasus sagen, vielleicht ist das ein Fachbegriff oder so
 
ok
 
@user2103480 Hast du die 4 Fälle in der Schule je als Kasus bezeichnet?
 
@EdwardEvans Probably? I honestly don't know the exact rules of german grammar lol
 
3:54 PM
Nice Deutsch
 
I just do applied german
 
lmao
 
what on earth is applied german
 
German without knowing why what you're speaking works
 
lol
 
3:55 PM
Every peaceful german conversation sounds like a fight
 
that's also fairly true
 
Die zthdyfr fhdejgfdthcer!
 
really depends on the dialect
 
lernt ihr Deutsch in Schule nicht?
Deutsche grammatische Regeln
 
We do, but everyone forgets about it
 
3:57 PM
In German we don't say "that's fine thank you", we say "Dieser verfickte Hurensohn lässt mich einfach nicht in Ruhe, wieso versteht er das Wort 'nein' nicht!" and I think that's beautiful
4
 
The sad thing about language is that you can know the rules in and out and still won't top a native speaker
@EdwardEvans our curse words are quite nice
 
I think I could make a good case to say that I speak better German than some natives
e.g. anyone aus dem Saarland
 
Ich kann mit Google Übersetzer Deutsch sprechen
 
@EdwardEvans says the one with an austrian dialect
 
I speak Hochdeitsch too
 
4:07 PM
@EdwardEvans is that austrian for hochdeutsch
 
@LeonhardEuler DeepL ist besser als Google Translate, aber DeepL hat wenige Sprache
 
yeah I said that for comic effect
and it really, really worked, brilliant
In Vorarlberg people say Dütsch, while in the rest of Austria people say Deitsch
 
you are behind 7 layers of comedy
ok gonna be off for a min
 
Treachery, the 9th layer of comedy
alright bye
 
@robjohn Get your "mean green Shamrock machine" ready for St. Patrick's day!!!
 
4:32 PM
About 25% automobiles in Chennai are foreign made and nearly 40 % of automobiles in Mumbai are foreign made. A sample of 100 automobiles are drawn from Chennai and a sample of 200 are drawn from Mumbai. Let $X$ stand for the number of foreign made cars in the sample from Chennai and $Y$ for the number of foreign cars in the Mumbai sample and Z = X+Y. Then
I know that X and Y are binomial distribution
I know that sum of indepednen binomial distribution is binomial
So I want to know where X and Y are independent here ?
@LeakyNun
Any suggestions ?
 
It's reasonable to assume they're independent. For one thing, the number of foreign-made cars you draw from the Mumbai sample does not influence how many foreign-made cars you draw from Chennai, or vice versa.
 
So Z would be linearly independent right ?
@Clarinetist
 
Linear independence is a completely different concept
 
@Clarinetist sorry Z would be binomial right .
 
What you can say is that $X$ and $Y$ are independent, so that $Z$ is a sum of independent random variables
 
4:43 PM
@Clarinetist Linear Algebra hangover
 
I don't know for sure, actually. The thing with summing binomials is that you usually assume the success probabilities are the same; they are not in this case.
 
@Clarinetist Right , I forgot that point ?
 
What are you aiming to do with $Z$? Get its distribution? I don't know if it's binomial.
My guess is it's not.
 
hi, if I have a function that is measurable on $[a,b]$ and is s.t. it is bounded on $[a,b]$, specifically is less than $1$ (in absolute value) on that interval, can I always get a sequence of $C^{\infty}$ functions approximating this in $L^p$ that are also bounded below $1$ in absolute value on $[a,b]$? I know I can get compactly supported functions to approximate it in $L^p$, but am sturggling with the bounded part
 
Yaa I got it , but I want to know intutive explaination why it is not binomial ?
 
4:48 PM
Well, let's think of it this way
Let's say I give you two independent experiments with different success probabilities, and I tell you to add the number of successes together
 
measurable functions can always be approximate by a sequence of step functions increasing in absolute value
 
@Clarinetist I can always do summation manipulation like x takes value k then z takes value z-k. and then sum over it to show it is not binomial
 
Would it be possible to model that as one single experiment with one constant success probability? That's the only reason that would make sense.
At the end of the day, it's primarily mathematical reasons for why this doesn't work.
 
@Thorgott oh yeah I see, I forgot the details of this construction, I only recall the way to prove that compactly supported smooth functions are dense is to show that they approximate $\mathbf{1}_{K}$ for any $K$-cell , but the detail involves finding a nondecreasing sequence
 
@Clarinetist so probablity of sucess changes depending on whether it is from which location mumbai or chennai .
 
4:53 PM
Right, 25% in Chennai, 40% in Mumbai
 
@Clarinetist so it might fluctuate right for z variable .
 
Right, just try computing some probabilities
 
to get distribution for z we need to do proper binomial sum argument .
 
@amWhy Perhaps you were not around for the fashion show. The only one that is still unseen is the Easter avatar.
 
@mathsstudent Have fun:
12
Q: Addition of two Binomial Distribution

Remi.bWhat is the distribution of the variable $X$ given $$ X = Y + Z, $$where $Y \sim $ Binomial($n$, $P_Y$) and $Z\sim$ Binomial($n$, $P_Z$)? For the special case, when $P_Y = P_Z = P$, I think that X~Binomial($2n$, $P$) is correct. If $P_A ≠ P_B$, the distribution might eventually just be Binomia...

Actually, it's more complex than this
because the number of trials differ
Here we go:
2
Q: Sum of Binomial distribution when the success rate is different.

rkjt50r983Is there any easy way to calculate the probability of the sum of two binomial random variable if the success rates of them are different each other? I mean that $X \sim Bin(n,p_0)%$, $Y \sim Bin(m, p_1)$, $Z = X+Y$, $p_0 \neq p_1$ and hope to calculate the distribution function of $Z$. I know t...

 
4:58 PM
this is simpler than that
the idea is simply to split the codomain into smaller and smaller intervals
 
@Clarinetist nice one thank you.
 
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