Mathematics - 2023-03-02

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Not Tfue

01:38

@Koro {0.111...}?

I think I should ask how do find inverse of that function?

Everything seems to be trivial after that.

leslie townes

is this that same thing? the inverse function sends a subset of N to the real number whose kth decimal place is 1 or 2 based on whether k is in the subset or not

and yes, proving that this thing is an inverse would be, under fairly general principles, the full story of why the function itself was a bijection

Not Tfue

02:03

@leslietownes you can say that I just set it to binary because it seems that they deal with binary when their instructors teach them. I am dealing it as pset instead so I made it easier for other to relate to.

seriously how would find the inverse of set thing. Unlike number to number mapping function.

Now it seems that my brain was asking for inverse instead of proving bijection.

Koro

@NotTfue as per the definition you gave, this should go to $\mathbb N$. No?

Suppose that we have $g_1=:g$, where $g_a$ represents quadratic Gauss sum. Then why is $g^q\equiv g_q \pmod q$?

@DLeftAdjointtoU

$g_a:= \sum_{t=0}^{p-1}(t/p) \zeta ^{at}, \zeta$ is pth root of unity.

(,) is Legendre symbol.

And we also have $g_a= (a/p) g_1$

So how do we get $g^q\equiv g_q \pmod q$?

The book does something along the following lines: $g^q= (\sum (t/p) \zeta^{t})^q \equiv \sum(t/p)^q \zeta^{qt} \equiv g_q (q)$. I don't understand how the second equivalence follow here?

Not Tfue

02:37

@Koro I was thinking you asked preimage of P(N) under f.

element empty set

冥王 Hades

user image

Any ideas?

Ted Shifrin

Hades, I posted a solution to a geometry problem. Why weren’t you on it days ago?

冥王 Hades

@TedShifrin wait. Where??

Ted Shifrin

Here.

冥王 Hades

03:04

@TedShifrin very straight forward

user image

Cyclic quadrilaterals. They're all similar. But, since they all have the same area, their dimensions must also be exactly equal. This logic follows for every quadrilateral

I just saw this problem now

Koro

@NotTfue No, I asked pre-image of the emptyset $\emptyset$, which is an element of $P(\mathbb N)$ so for surjectivity, you need to have a preimage for that. No?

2 hours later…

robjohn

04:46

@冥王Hades Cyclic quadrilaterals are all similar? Aren't cyclic quadrilaterals just quadrilaterals inscribed in circles?

Ted Shifrin

04:58

No, @robjohn, but these are because of angles :)

robjohn

So they're squares, which are cyclic, but not general cyclic quadrilaterals

robjohn

05:16

Watching the Space-X crew 6 launch countdown

T-18 minutes

copper.hat

on spaceflightnow.com ?

Ted Shifrin

05:34

We were looking at the corner quadrilaterals, not squares.

copper.hat

awesome launch

robjohn

yep

leslie townes

weirdos

robjohn

@copper.hat I am watching on YouTube

copper.hat

i guess i am too.

just incredible

6 months in space

that is amazing

robjohn

05:50

In orbit. Still about a day until ISS

copper.hat

06:05

it is really aweseom.

one potato two potato

06:24

It seems many graduate students don't care that much about their coursework (at least in my college). Rather, they focus more on their own studies for their research field.

copper.hat

07:06

it bugs me when people state that something is obvious math.stackexchange.com/q/4650413/27978

PM 2Ring

07:17

@冥王Hades Yes, it's 45°. But it's easier to do it analytically, using the difference of tans formula. You have atan(1) = atan(3) - atan(1/2). So it's kind of related to:

Dec 17, 2022 at 13:54, by PM 2Ring

Here's a little diagram that illustrates that $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$ and $\tan^{-1}(1)+\tan^{-1}(1/2)+\tan^{-1}(1/3)=\pi/2$.

user image

@NotTfue I use the ChatJax bookmarks in the Samsung browser. In recent versions, you need to put the bookmark into the actual bookmark bar, or into a folder in the bookmark bar.

@Semiclassical Your rotated ellipse reminded me of a nice thing from orbital mechanics. For an elliptical orbit, what shape do you get when you plot $(\dot x, \dot y)$ ? IIRC, you can solve it just from the polar equation of the ellipse, and conservation of angular momentum, which can be derived from Kepler's "equal area in equal time".

Greg Egan mentions that problem in his short story, In the Ruins, which is a scathing attack on the dumbing down of maths & physics in schools, YouTube / Instagram / Twitter culture, and his hatred of terms like "geek" & "nerd".

Cotton Headed Ninnymuggins

07:51

Is $f(x)=|x|+x$ an even or odd function? I'm pretty sure it's neither but I'm double checking because I don't want to give a different answer to the students than what the answer key from the supervisors says, only to get it wrong for them

2 hours later…

giannisl9

09:32

Why am I struggling so much with systems of linear equations? I am getting the wrong numerical answer like 80% of the time '(

@CottonHeadedNinnymuggins neither

Prithu Biswas

09:47

Spivak has an exercise in sequences:

"Prove that if a subsequence of a cauchy sequence converges, then so does the original Cauchy sequence".

But isn't every cauchy sequence convergent?

2 hours later…

Koro

11:54

@PrithuBiswas it depends upon 'where' the sequence is.

Prithu Biswas

@Koro I am considering sequences on the reals.

Koro

If you mean you're taking Cauchy sequence in R, then yes the sequence converges.

Martin Sleziak

@PrithuBiswas That's true. Still, this should be relatively easy to prove directly from definition (without using the result that every Cauchy sequence converges).

Moreover, you can use the same proof in situations where this result isn't available. (I.e., if you're working in some metric space and you do not know whether it is complete or not.)

I expected to be among the frequent questions tagged cauchy-sequences. It is indeed there: Cauchy sequence is convergent iff it has a convergent subsequence.

Prithu Biswas

@MartinSleziak Oh that is interesting. Thanks for sharing =)

Sourav Ghosh

12:12

@PrithuBiswas The result you mentioned doesn't give it's full power in $\Bbb{R}$. As every cauchy sequence is bounded and every bounded sequence has a convergent subsequence.

one potato two potato

12:28

user image

user image

enjoy

Shyam Tripathi

13:19

Hello! I posted this question yesterday, there has been only one response since then. Also, the question has not been converted into a community wiki (which I thought might happen). Is there something wrong with this post?

冥王 Hades

@robjohn Correct. But, in this particular case, they all have the same "angle configuration", in other words, the same shape, meaning they're similar. Now, since they also have the same area, their dimensions must also be exactly the same. You might say that isn't necessarily true since two sides could be altered yet the area could still remain the same. Correct, but then they're no longer similar. We already know they're similar, therefore they must possess the same dimensions

Once you establish this logic, its easy to figure out that $x=1u^2$

@PM2Ring isn't this the general case?

robjohn

13:48

@TedShifrin After I finished watching the launch, I figured everything out. Sorry for the confusion.

Semiclassical

@CottonHeadedNinnymuggins note that $f(1)=2$ but $f(-1)=0$, whereas being even/odd would require $f(x)=\pm f(-x)$. So yeah, neither. (The first term is even and the second is odd, so the sun won’t be either.)

robjohn

14:15

@冥王Hades Yeah, I worked it out and made a graphic after the Space-X launch.

I renamed the second $P$ to $R$

Sha Vuklia

15:05

@Thorgott mind if I ask you sth regarding a computation with the Ext functor?

Koro

15:49

riemannianhunger.wordpress.com/…

This solution is wrong. Right?

Retract definition is written incorrectly.

Mouse

Just one try

Ok I am dead now bye guys see you next year.

I will do this every year but after 20 times some of you guys won't be alive ;_;

I feel sad for this that after 20 years we will lose a lot of people who spent decades to forge their math to this level.

shintuku

don't worry AI will make it all pointless anyways

hey, maybe even sooner than in 10 years!

dw in case you lose the bet you won't have problem reattaching with AI-assisted surgery

maybe 3d printers will become advanced enough to do some magic in this area

truly, the future is a wonder to behold

Curious Mind

16:12

@shintuku Where is the proof of this statement? I am a mathematician where is proof!

shintuku

it is in the fear in your heart, my friend

Curious Mind

@Mouse Please Ban me too.

I want to start from 0 too.

Thorgott

@ShaVuklia sure, just ask

Curious Mind

@shintuku I kindly reject this proof.

Have a nice day @shintuku I hope I can just

Thorgott

@Koro It's terribly written

Curious Mind

16:15

sleep peacefully without any theorem.

Thorgott

so terrible that I don't care to read enough to tell you whether it's wrong or not

Curious Mind

@Thorgott Good luck

Sha Vuklia

Cools, so I'm trying to show the following: let $R$ be a commutative ring, $A=R[x]/(x^2-1)$ where $x$ acts by 1. I've already shown that $\text{Ext}^n_A(\mathbb Z/2,\mathbb Z/2)\cong\mathbb Z/2$ for $n\geq 0$ as a special case. Now I want to show for general $R$ that $\text{Ext}^{2n+1}_A(R,R)=\text{tor}_2 R$ and $\text{Ext}^{2n}_A(R,R)=R/2$ for positive $n$, while $\text{Ext}^0_A(R,R)=R$.

The surjective map sends $x$ to $1$ btw, and the injective map is inclusion. The case $n=0$ I've shown too. I tried to mimick my proof for the special case $R=\mathbb Z_2$ where I considered I short exact sequence, which in the general case would be
$$
0\to R(1-x)\to\frac{R[x]}{(x^2-1)}\to R\to 0.
$$
By the long exact sequence for $\text{Ext}$ it would suffice to know what $\text{Ext}(R(1-x),R)$ is. I could also invoke the isomorphism $\frac{R[x]}{(x^2-1)}\cong\frac{R[y]}{(y(y-1))}$ in which case the kernel would be $R(2-y)$, which might be easier to work with. At this point I'm stuck thoug…

Also, I read the definition of Ext yesterday, so I'm not very agile with it yet and might be missing sth obvious

Sha Vuklia

16:36

ok, I think now that I shouldn't be constructing a short exact sequence like I did before, but instead just construct a free resolution

I have an idea what I could try, Imma do that

Thorgott

after thinking about this and missing the mark for a while, I agree that's an excellent idea

I see now that this is essentially the group cohomology of $\mathbb{Z}/2\mathbb{Z}$, but with $R$ instead of $\mathbb{Z}$ as base ring

the SES you wrote down already gives you the end of a free resolution

perhaps it's better to write it as $0\rightarrow A(1-x)\rightarrow A\rightarrow R\rightarrow 0$

there's an obvious surjection $A\rightarrow A(1-x)$, so take the composite $A\rightarrow A(1-x)\rightarrow A$, figure out its kernel and continue

Sha Vuklia

yes, that's what I'm doing now indeed

if I'm not mistaken, I get multiplication by (1-x) and (1+x) alternately

I'm now applying the hom functor to compute the homology groups

Thorgott

yup, that's it

there's a fun algebraic topology perspective of this, too

you can view $S^1$ as a space with $\mathbb{Z}/2\mathbb{Z}$-action by rotation and equip it with the obvious equivariant cell structure (two $0$-cells, two $1$-cells). the SES you wrote down is essentially the result of computing the cohomology of $S^1$ with $R$-coefficients from the cellular complex associated to this structure.

Koro

16:54

I think the following solution is also wrong riemannianhunger.wordpress.com/…

Thorgott

also note that the alternating structure of the resolution even without computing anything implies that the Ext-functor will take 2-periodic values!

Koro

it seems to be assuming $gpfg\simeq g\implies gpf\simeq 1$, which is plain wrong.

Is there a solution manual to Hatcher's?

I want to memorize some solutions.

Many people in my class are planning to leave the course/college due to pathetic teaching here.

leslie townes

koro, it does sound horrible. even when eyewitness testimony is discounted somewhat by the usual 'eh, someone is complaining about math on the internet'

Koro

you should see how they set question papers for exam- uneven distribution of chapters.

and consider it a good luck if they tell which book they are following.

What is more pathetic is that some of them don't tell the book as if they themself created the concepts!!

some even go on to say -you won't find this anywhere...

leslie townes

do professors have very robust job protection? some of the stuff you've related makes me wonder why they haven't been fired. tenure in the USA is pretty robust, but even then, people can get fired.

or at least, shuffled away from teaching responsibility.

Koro

17:06

yes, very robust. they can't be fired.

leslie townes

so, where do i apply for a job at this institution?

as a student it sounds horrible, but as a professor i think i could enjoy it.

Koro

you may check the website.

My last job was also super secure. I could not be fired, no matter what the pandemic.

leslie townes

"so, what attracted you to apply to this position?" "the possibility of robust job protections. i understand that once you get sufficiently in, it's basically impossible to be fired, and that is very appealing to me."

2

Xander Henderson

@leslietownes I mean, tenure is not intended to prevent folk from being fired for cause, or laid off if there is no money or if a department is eliminated.

Koro

one teacher last semester for example didn't tell which book he followed (said he didn't follow any and implying that he created all that on his own), or didn't give any assignments or tests. Because if he gave assignments or tests, then his hands would get tired by verifying all the answer scripts.

leslie townes

17:14

@XanderHenderson yeah. most of the stuff koro mentions seems like it would get someone out the door pretty quickly in most places i have experience with in the USA.

Koro

'so screw them, my salary will get credited on last working day of the month anyways' is what he probably had in mind.

some of them studied from some US universities.

I won't be surprised if the US universities kicked them out from the US for such lethargic behaviour.

leslie townes

that's actually usually my line, for bullshit detection, when people complain about teachers on social media. there's a perception that "tenure means you can't be fired," which is inaccurate, and you see people going on US social media and saying all kinds of things that would, in real life and not a social media post, actually get someone fired.

but i trust koro.

Xander Henderson

Our institution is looking at spending money to get access to plagiarism checkers. I just learned that students sometimes copy-paste papers, and change only certain characters in order to defeat plagiarism checkers. E.g. replacing "o" (a Latin/English character) with "о" (a Cyrillic character).

leslie townes

i really think this insulation of lethargic behavior from consequences is something we should import.

oh yeah, that's the 101 of defeating plagiarism checkers.

Xander Henderson

That said, tenure is a more robust protection than what you find in many other industries. Of course, the downside is that academia does not, generally speaking, pay all that well (compared to "industry").

@leslietownes I'd never heard of that before.

leslie townes

17:18

my wife's school has a license for one of the big ones. it is good for what it does. but, the people who most rely upon plagiarism are not the people who do it subtly, and most of them could also be found out via google search.

Xander Henderson

But that kind of plagiarism is so far out of scope for what I do.

leslie townes

using special characters is next level stuff that frankly i would not penalize. in my own ideal world.

Xander Henderson

Heh.

leslie townes

my wife's department has had to go through a process for getting rid of someone that is certainly a lot longer than the process my employer would go through for getting rid of me

Ted Shifrin

@XanderHenderson Just goes to show I’m not cut out to be a cheater. That would never have occurred to me.

Xander Henderson

17:20

@TedShifrin inorite?

leslie townes

ted who stole his whole book from do carmo and just rewrote it in "his own words".

2

Xander Henderson

Seems like it is easier to just write it yourself...

leslie townes

that's the thing, if you put in the effort and meaningfully defeat something put in your way to stop you, i say, well done.

Koro

@XanderHenderson there are some websites where you can paste something and see if it's been plagiarised from somewhere.

robjohn

@leslietownes Ï åģřēĕ ŧōťāłłŷ

Koro

17:23

Some of the websites also tell how much % of the pasted content is matching with some existing content on web.

Xander Henderson

@Koro Yes, I am aware.

Koro

so it seems to me that the software (plagiarism checker) is not of much use.

Xander Henderson

The thing is, if you are grading lots and lots of papers, you really don't want to have to go through the copy-paste (into Google, into some free online thing, or whatever). In an ideal world, you can just run papers through your LMS, and be done with.

Koro

So once we hired an agency for getting a report prepared on something based on survey. I saw the content and they had copy pasted pretty much everything.

they didn't even bother to change the words 'sequestration of carbon' into their own words!

that means -no survey was done by them.

leslie townes

yeah, like xander says, it is just a volume thing. at scale it makes sense to have something root out the obvious stuff.

Koro

17:28

I mean this is an honest answer.

😅

Xander Henderson

:63102939 I don't know how common it is for someone to directly ask "Why should we hire you?" but if I asked that question, and if the candidate said "Don't", I would get up, shake their hand, and thank them for their time.

Koro

haha

Xander Henderson

Those kinds of jokes are not appropriate for an interview.

Koro

anyways, I lost my interest in the question and its answer, whatever it is. So I had deleted the comment. :(

leslie townes

i would see a question like that in an interview process as a red flag. i've heard of stuff like that in some workplaces. all of them toxic

Xander Henderson

17:33

@leslietownes Indeed. I feel similarly about a lot of those "puzzle" questions.

Prithu Biswas

@Koro That sounds quite sad. What are you planning to do then?

user2236

"What can you bring to our team if we hire you?"

::tick tock::

leslie townes

an organization that goes through the trouble of personally interviewing people without apparently already having an answer to that question is performing some toxic ritual and is not a good workplace. unless it's so small that they can't meaningfully pre-screen applicants, in which case, just hire your founder's brother and see how that works out.

Xander Henderson

@user2236 Well, you've read my CV. If the answer isn't in there, I don't know what to say. Good day, sirs and ma'ams.

Ted Shifrin

@Xander That's a good way to get unhired. I learned when I tried that at a college interview (at one of the top schools in the country). The admissions interviewer asked me something I'd written explicitly about in my application, and I said as much. "Didn't you read my application?" ... Guess where Ted was not admitted.

Xander Henderson

17:45

@TedShifrin Yeah, but that is very different from a very general question like "Why should we hire you?"

Ted Shifrin

Well, I think it gives you an opportunity to show you've read up on the company and have some thoughts about what you can bring to the table. I don't think that's a bad question.

leslie townes

in ted's day you just interviewed with the president of the university you were applying to. it sounds logistically complicated, but it wasn't, he lived two caves over.

Xander Henderson

By the time they are interviewing you, they should have already figured out why they want to hire you.

Ted Shifrin

But you have to show that you care enough to know something about the company. Same thing applies to people applying to graduate schools.

leslie townes

i think it's a horrible question. it says "we're either engaged in some hazing ritual, or we have time to pay people to interview folks who might as well have come off the street."

Xander Henderson

17:47

@TedShifrin Again, all of that should be in a cover letter and cv. If they are asking something so general in an interview, it seems like they haven't done their due diligence. i would be very concerned about working for an institution which doesn't do their homework.

Ted Shifrin

Well, then, we're back to my undergraduate interview issue.

I would reiterate my strengths relative to the particular job.

leslie townes

ted may be limited here by his generation, and i'm not kidding for once. these days, an interview is like a huge step. 10000 filters ought to have been applied before the interview.

Xander Henderson

@TedShifrin I'm not saying that your are wrong, but I see the question itself as a red flag.

leslie townes

i say this as someone who has applied to dozens upon dozens of jobs and progressed somewhat through the process without getting interviews.

Xander Henderson

If they are asking that particular question, it makes me feel that I probably don't want to work their.

leslie townes

17:49

that's not something that has an analogue from 1920 or whenever ted was doing this.

Ted Shifrin

@leslie You mean colleges or jobs? I was doing interviewing for MIT — they give every applicant an interview, either on campus or with alumni all over the world.

leslie townes

jobs.

Ted Shifrin

Yeah, jobs is a different thing. And I expect the MIT situation is unusual. Certainly Berkeley and UCLA do not grant every undergraduate applicant an interview.

leslie townes

if we're talking about a process where every applicant gets an interview, erase everything i just said. my stuff is in the job context, where every applicant ought not to get an interview.

Ted Shifrin

Yeah, of course jobs are different.

Xander Henderson

17:50

@leslietownes Indeed.

leslie townes

UC undergraduate does not interview at all for admissions purposes. they do, or did, for some UC-wide scholarships.

robjohn

It's been 35 years since I taught at UCLA, but there were definitely no interviews for undergrads.

leslie townes

i would just say, put your application on tiktok and we'll see who gets the most views.

let the market decide.

user2236

sugarloaf - don't call us, we'll call you

You don't want to hear that :P

Prithu Biswas

18:27

Entry level interviews be like...

@Koro I hope that you can make it through that situation.

UnderMathUate

18:45

I'm having a bit of trouble with this proof:

Prove that if $\sum\limits_{n=1}^{\infty} x_n$ converges, then $\sum\limits_{2n}^{\infty} x_{2n} + x_{2n+1}$ also converges.

I'm thinking about maybe using the definition for a Cauchy sequence, but not entirely sure.

Since $\sum\limits_{n=1}^{\infty} x_n$ converges. It's Cauchy, so $\forall \epsilon > 0$, there exists $M \in \mathbb{N}$ such that $\forall n \geq N$ and every $k > n$, we have $\sum\limits_{j=n+1}^{k} x_j < \epsilon$.

Ted Shifrin

Are you interpreting the latter as the sum of two sums or as the sum of $x_{2n}+x_{2n+1}$?

UnderMathUate

The sum of $x_{2n} + x_{2n+1}$

Ted Shifrin

Your summation should not be over $2n$.

UnderMathUate

Lol, typo

It should be from $n = 1$

Ted Shifrin

Write the second summation with a different index, $k$.

What is the partial sum with $k=1,\dots,N$?

UnderMathUate

18:53

You mean like: $\sum\limits_{k=1}^{\infty} x_{2k}+ x_{2k+1}$

Referring to your first comment

Need a second to think about the other one

Assuming I've understood:

$S_{2k} = x_2 + x_4 + x_6 + ... + x_N$ and $S_{2k+1} = x_3 + x_5 + x_7 + ... + x_N$

So $S_{2k} + S_{2k + 1} = x_2 + x_3 + x_4 +... + 2x_N$

Ted Shifrin

No.

UnderMathUate

fantastic ;-;

Ted Shifrin

What is $S_3$ exactly?

UnderMathUate

The sum from x_1 to x_3

Ted Shifrin

No. We’re talking about your sum on the right, with $k$. Write out the terms explicitly.

monoidaltransform

19:06

analogous to the smooth manifold case, if $f:U\rightarrow X$ is a topological embedding , where $U$ is contained in $X$, is $f$ locally an inclusion?

UnderMathUate

$(x_2 + x_3) + (x_4 + x_5) + (x_6 + x_7)$

Ted Shifrin

Good, Under.

@monoidaltransform What does the question even mean?

UnderMathUate

phew

Ted Shifrin

Under, so what is $S_N$?

monoidaltransform

In the smooth manifold case, every smooth embedding is locally an inclusion

UnderMathUate

19:09

Ok, so $S_N = (x_2 + x_3) + (x_4 + x_5) + ... + (x_{2N} + x_{2N + 1})$

monoidaltransform

so locally, the smooth embedding is given by $(x_1,...,x_n)\hookrightarrow (x_1,...,x_n,0...0)$

Ted Shifrin

Right, @Under. Give the first series a name for its partial sums and compare.

UnderMathUate

OK

Ted Shifrin

So the point is that the image is locally flat, monoidal.

monoidaltransform

locally flat?

Ted Shifrin

19:11

Yes. A flat plane.

You need the inverse function for this.

monoidaltransform

Yes. In the smooth setting it's just due to the immersion theorem. But I was wondering if there is an analogous result for just topological embeddings $f:U\hookrightarrow X$, for $U$ contained in $X$.

Ted Shifrin

Nope. There are wild embeddings.

monoidaltransform

ahh I see

Koro

@PrithuBiswas I'm not leaving.

@PrithuBiswas Thanks :-).

Suppose that X and Y are homotopy equivalent, X is compact. What's an example when Y is not compact?

Can I say open disk?

X= open disk, Y={center of the disk}

X is not compact, but Y is.

X is contractible to Y so X is homotopically equivalent to Y.

monoidaltransform

@Koro Take $\mathbb{R}^2$ and a point

Ted Shifrin

19:22

Yes. What’s a more interesting example?

Koro

yeah, that will also work @monoidaltransform. So disk is also correct, right?

We have simple-connectedness either way.

monoidaltransform

yes. In $\mathbb{R}^n$, yes.

:)

Koro

How do I prove that removing a point from a Torus gives me a space that is homotopically equivalent to wedge of two 1-spheres?

@TedShifrin $S^\infty$?

I know that $S^\infty$ is contractible.

Ted Shifrin

Why do we need a contractible example?

Koro

I memorized the proof sometime back why that is so.

Ted Shifrin

19:27

What is homotopy equivalent to $S^1$?

Koro

Ohh. Because for 'homotopy equivalent', I can visualize only 'contractible spaces or homeomorphic spaces.

Ted Shifrin

That’s a terrible weakness.

Koro

I mean apart from the above two categories, I don't have examples for homotopy equivalent, only the definition.

Ted Shifrin

What deformation retracts to a circle in the plane?

Koro

@Koro that's why I asked this.

I think the plane itself could do it?

Because $x\mapsto x/|x|$

Ted Shifrin

19:30

Right map, wrong space.

Koro

ohh

Ted Shifrin

@UnderMathUate Did you finish?

Thorgott

19:48

for any space $X$, $X$ and $X\times\mathbb{R}$ are homotopy-equivalent

if $X$ is non-empty, the latter is non-compact

Ted Shifrin

Sshhhh.

I don't know what happened to $x/|x|$ ... it disappeared.

user 726941

@TedShifrin that usually means "shut up and calculate" :-)

anak

20:21

If A,B,C are nxn matrices where C is invertible and B is symmetric positive definite, is it possible to show that ABC is antisymmetric implies AC is antisymmetric? I have a feeling it's not true and an easy counterexample will be demonstrated to embarrass me so I can move on.

Sorry, A is the arbitrary matrix, apologies.

Xander Henderson

20:34

@TedShifrin Most of this week's calc exam is questions of the form "calculate the limit; if it doesn't exist, write $\pm \infty$ or 'DNE', as appropriate". I gave them $\lim_{x\to 0^+} |x|/x$ as a problem. We'll see how that goes. I went over precisely that problem in class; I'm guessing that only half the class will get it.

Ted Shifrin

Depends how much you beat left- and right-hand limits into them.

anak

Will go really well if the students frequent the math.stackexchange chat and recognize their instructor.

Xander Henderson

@TedShifrin The answer is always "not enough."

@anak Honestly? I would love it if that happened.

I think that, if I had my druthers, I would not talk about $\varepsilon$-$\delta$ limits in calculus. I think that I would talk about sequential limits---this is what they tend to do, anyway, when getting an intuition for limits. So the order of instruction would be (1) limit of a sequence, (2) one sided limits (because those are kind of natural sequences to think about), and then (3) two-sided limits.

I would, at least, like to experiment with that approach.

Ted Shifrin

I never taught $\delta$-$\epsilon$ in regular calc.

Ted Shifrin

21:02

@anak Virtually anything $2\times 2$ will give you a counterexample. Just take an easy diagonal $B$, an easy invertible $C$ and solve for $A$ so that $ABC$ is the easiest skew-symmetric matrix. $AC$ won't be skew-symmetric.

Xander Henderson

21:22

@TedShifrin I am required to talk about $\varepsilon$-$\delta$ by state articulation agreements.

Ted Shifrin

Some idiot mathematicians decided on that.

A. P.

Hi, how can I study the continuity of the map $t\mapsto f(t):=\sup_{k\in N}\frac{t^k}{k!}$?

I was trying to find just the supremum, but no progress.

here I am considering $t\in R$

A. P.

21:44

I can see that each map $t\mapsto \frac{t^k}{k!}$ is continuous, then the sequence of functions $(f_k)$ defined by $f_{k}:=t^k/k!$ is a sequence of continuous functions.

Trying to generalize: if $(f_k)$ is a sequence of continuous functions, then $\sup_{k\in N}f_{k}(t)$ is a continuous function?

Ted Shifrin

21:59

@冥王Hades How do you know they're all similar?

Yes, they are cyclic and have the same area. But why do matching angles imply similarity?

1 hour later…

Houshou Rattengod

23:17

I am looking to express that some integer $x$ is both $\geq 0$ as well as $\equiv 0\Mod{2}$

How do I properly express/define this?

Sine of the Time

@HoushouRattengod what about $x \in 2\Bbb{N}$ ?

Houshou Rattengod

I was just trying to think about how to say that but was having a brain-fart

Sine of the Time

or even positive number

Houshou Rattengod

$x \in \Bbb{E+}$ ?

Sine of the Time

Here $2\Bbb{N}=\{2n \,| \, n\in \Bbb{N}\}$

@HoushouRattengod I think it might be ambiguous, it's better $2\Bbb N$

Houshou Rattengod

23:23

Right, cause $\Bbb{E}$ doesn't imply $\geq 0$

shintuku

@HoushouRattengod mate what is $\Bbb E$

Houshou Rattengod

Set of Even Integers

Or, that's what I thought it meant

shintuku

listen here houshou you can't just make up notation

Houshou Rattengod

I didn't make it up. I picked that up through study.

shintuku

ok i'll allow it this time but make it your last

Houshou Rattengod

23:29

Do you wanna shot me now or later?

Because I'm the guy with no formal maths education, trying to write a proofs paper from a visual proof I created. And I have absolutely no idea what I am doing, if I'm even doing it right

Akiva Weinberger

user image

Checkmate Galois

shintuku

@HoushouRattengod i was kidding

i'm no one

Houshou Rattengod

... I'm not.

I'm trying to self study, so I have this spreadsheet I created and I'm just trying to learn by doing. And by doing, I mean. Converting what I see on this table into mathematical equations and describe what is going on.

But damned it all if learning LaTeX ain't the hardest thing to do. Trying to describe multiple infinite sum series and their patterns from a Spreadsheet is.

SAJW

23:56

Maybe this ist more philosophical, but can IT be that a certain Insight ist better understood from a 1d creature thank from a 2d, outside of what IT feels to be in the former

Add the General case

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