chat.stackexchange.com/transcript/message/60919062#60919062 Any idea of this? Quite sure it's not connected (it's box topology!) but I can't find suitable separation.

How to describe the ring $\mathbb Z[x,y]/(x^2,y^2,2)$?

All I can say is that the ring is $\{ax+by+cxy+d: a,b,c,d\in \mathbb Z_2\}$

with the multiplication being what you'd expect, i guess. OK. what more do you want from a description?

Leslie, I am not happy with my description of the quotient ring. The exercise however doesn't define 'describe (briefly)' and in addition asks to determine the characteristic and show that $a^2=0$ or $1$ for every $a$ in the quotient.

not sure if that is correct though.

If $R\sim A/B$ and $A\simeq C, B\simeq D$ then can it be said that $A/B\simeq C/D$? That's what is confusing me.

well i'm never sure what 'describe' means without qualification, but specifying the elements and how they are added and multiplied together does it for me. it's obvious enough to me from the quotienting construction that those operations as defined just on the set of representatives are associative, but maybe you want to tease that out of a theorem.

it's certainly possible to compute the characteristic and the set of squares from your description

you could even determine the ideal structure from your description, with some case analysis

re squares, once you know the characteristic is 2, it's just the "Freshman's dream" (ugh) and the fact that the nonconstant monomials all square to 0

The following is not true in general:

$(ax+by)^2+(cxy+d)^2+2(ax+by)(cxy+d)=d^2=1$ if $d=1$ and $=0$ if $d=0$.

koro: what is \simeq here with B and D, and why would you expect it to be true?

Leslie, that is ring isomorphism.

That is, the terms $ax+by+cxy$ square to $1$ and the terms $ax+by+cxy+1$ square to $1$.

B and D are ideals in a commutative ring? they may not have a multiplicative identity element, is that OK?

yes, leslie.

do you understand my remark, why would you expect it to be true? if you think in terms of maps implementing the isomorphisms and not just the abstract "is isomorphic to" relation. it's not clear how you'd get it. which suggests that it isn't true.

maybe not the world's best heuristic, but often with these things, if it isn't obvious, it isn't true.

I thought that to be true because it seemed obvious from the usage of the third isomorphism theorem above but I realize that that was wrong.

$\frac{\mathbb Z[x,y]/{(2x)}}{(x^2,y^2,2)/{(2x)}}\simeq\frac{\mathbb Z_2[y]}{(y^2)_2}$

But I realize that I concluded it based on an assertion ($A/B\simeq C/D$) that is false.

Im not sure how to approach...

@copper.hat I am getting scared than physics and math might become convoluted to the point that with too much jargon, no one can tell what is real, and what is fake anymore.

@Jakobian Thanks, I was working on a proof that i have been stuck on for a long time, and was picking at straws to see if there was anything I had gotten fundamentally wrong which was stopping me.

@TedShifrin I did notice this when I was working on the proof. Thanks for affirming it.

Hi guys, how are you? Can someone give me some light go solve this problem?

I need to find a power series for this function: $f(x) = \sqrt{2+x^2}$

I know I need to start from geometrical series:

$\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n$

But when I do a change variable, this turns: $ f'(x) = \frac{x}{\sqrt{2+x^2}} = \sum_{n=0}^{\infty} ( - \frac{\sqrt{2+x^2}}{x} + 1)^n $

The problem with this series is to integrate. I don't know how to integrate this

I won't really be reading the book about Hyperspaces, at least for now

Going to read the one about Continuum theory

It's nicer

interesting

Plus idk if I'll be using hyperspaces much other than basic things about them

Anyone willing to take a look at this problem? math.stackexchange.com/questions/4430802/…

I believe my proof is correct, but I don't think I am answering it the way the question is stated.

Solution: rewrite it

@MatheusSousa How do you know this?

@MatheusSousa one thing you can do here is regard $\sqrt{2+x^2}$ as a function of $x^2$ rather than $x$, which simplifies any differentiation considerably. The bigger issue is that you start out with a power of 1/2, while the geometric series has a power of -1.

Trying to force this into a form that’s equivalent to a geometric series seems like the wrong idea here

Greetings, @Semiclassic and @robjohn.

hello

Cat of doom, eh?

can anyone help me work on my will? i've been having some health proplems and really want to get started. i know this is off topic

Whoa ... No one here is qualified to do that.

Protocol varies from country to country, from state to state.

@TedShifrin Good morning. Just upgraded the OS on my computer. Making sure that things still work. Had to upgrade one app so far.

I dread this ... especially after the debacle with everything Excel and Illustrator.

The development environment I use for work still seems to work :-)

But these days I upgrade without too much trouble. I used to have to recompile stuff for TeXShop (I use some special local fonts) but I think I've arranged to avoid that. Don't ask me how.

@Catofdoom I think you can find software to help you do that yourself just by googling.

I have to figure out just what GMail needs. I try to turn off the "allow unsafe apps" option and my POP client can't access GMail. I upgraded because it completely shutdown yesterday. It works now, but they say that on May 20, "allow unsafe apps" will no longer be available.

gmail complains when I have apple mail set up to get my mail. Yes, I'm worried about that, too.

@TedShifrin i'll try. thanks

I really like having apple mail keep my different email accounts in one place. When you solve that, robjohn, please advise me!

For sure

@robjohn I assume you've digested this.

@TedShifrin I hadn't seen that one, but I have to try some options before I delete and recreate.

I always used to use POP. Now it seems I have POP for my gmail account with my forwarded email from UGA and IMAP for the piles of junk I get "personally."

One user's answers are so deep. Every answer contains links for more explanation which further contain more links. :)

trying to understand their very informative answers :)

@Koro who?

You can give me a clue only if you want =P

Must be very interesting questions to merit deep answers. At most 10% of mine would qualify for that.

From the clue "Every answer contains links for more explanation which further contain more links", I know of only a few users who writes answers of that quality with links to more further explanations.

Almost like I can narrow it down to 1 user.

I think I can recall several who do that.

@Prithubiswasleftmse can you guess their initials?

guessing is fun sometimes :).

And silly ...

@Koro Only a first guess, does it contain some numbers?

@Prithubiswasleftmse no. Should I tell or would you like to guess? :D

@Koro I would like to guess again. And it seems like my first choice is wrong.

All pressure on you. Koro used to do only analysis, but now he’s doing algebra. You can bet he doesn’t read geometry, topology, or applied stuff. :)

@TedShifrin that's true professor :).

So you've missed the few insightful answers I've written :D

But you've read lots of robjohn's.

@TedShifrin unfortunately yes but I'll get there soon. :D

Not enough to do much, but I guess I have to start somewhere...

@Koro I think I found another candidate which seems to match all of the criterion above.

criteria :D

criteri$\rm\Large { a}$

:)

@Koro Does his/her initials have 11 letters?

@Prithubiswasleftmse his name has 11 letters :)

@Koro Profile picture is a graph of a complex function?

@Prithubiswasleftmse yes :)

@Koro Now I am 100% sure that my guess is correct =)

I think so too $\ddot\smile$

Complex meaning complex or complex meaning complicated?

@TedShifrin Complex means ℂomplex.

I mean z ∈ ℂ.

Ah, like the cover of my multivariable math book.

Of course, it's only a pretend graph of a complex function (as we cannot draw $\Bbb C\times \Bbb C \cong \Bbb R^4$).

It can be shown that given $n$ equally spaced points on a circle of radius $r$, if we choose one of them to be our base point and add up the inverse squared distances of all the other points to the basepoint, we always get $\dfrac{n^2-1}{12r^2}$

Given this, show that $\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$

"It can be shown"!

It's hard not to show

I've assigned the case of the sum of the squares of the distance as an easy homework problem; I've never seen this before.

I mean, it shouldn't be too hard

Works out to showing $\displaystyle\sum_{k=1}^{n-1}\dfrac1{1-\cos(2\pi k/n)}=\dfrac{k^2-1}6$

Without the negative exponent, it's just basic algebra with the roots of unity.

I can feel my LaTeX getting rusty, lol

You have a typo. $k=n$ on the right?

Yup

The second coefficient over the final coefficient of the polynomial should give you the sum of their inverses, I think

up to a sign

But $\dfrac 1{z-\alpha} \ne \dfrac 1z - \dfrac 1\alpha$, so I don't yet see the relephance of your remark.

We have polynomials whose roots are $1-\cos$, don't we?

Oh, that.

Either that or write $\dfrac1{1-\cos}=\dfrac{1+\cos}{\sin^2}$ possibly

Well, I'm not going to work on it, but cheers.

@TedShifrin In any case, getting from there to the Basel problem is straightforward

Hi, @Alex, stranger.

Good afternoon

I am kind of disappointed that I couldn't make even a bit of progress on a baby calculus problem even after 12 hours.

What was the problem?

hello id like some help understandin the proof of the proposition that The chow group is Graded in Eisenbud Harris 1364 and all that Proposition 1.4

@AkivaWeinberger Suppose that $f$ is continuous at $a$, $f'(x)$ exists for all $x$ on some interval containing $a$, except perhaps for $x=a$ . Suppose , moreover , that $\lim_{x\to a}f'(x)$ exists.Then $f'(a)$ also exists, and $f'(a)$=$\lim_{x\to a}f'(x)$.

OK, let's first think about a special case

$a=0$, $f(a)=0$, $\lim_{x\to a}f'(x)=0$

which we can do WLOG by first shifting stuff over and subtracting $cx$ where $c$ is that limit

Because we have that limit, we know that we can pick any $\epsilon$, and there will be a $\delta$ such that for $|x|<\delta$ (and $x\ne0$) we have $|f'(x)|<\epsilon$

Hmm

Maybe mean value theorem?

@AkivaWeinberger Spivak says it is a corollary of MVT.

$f(x)/x=f'(c)$ for some $c$ between $0$ and $x$

so assuming $|x|<\delta$ that gives us $|f(x)/x|<\epsilon$ and so $|f(x)|<\epsilon|x|$

(remember we're still doing that special case)

so now let's try to find $f'(0)$

That's $\lim_{x\to0}(f(x)-f(0))/(x-0)=\lim_{x\to0}f(x)/x$

and we just saw that that's between $-\epsilon$ and $\epsilon$

and $\epsilon$ was arbitrary

so $f'(0)=0$, QED

I bet we can do the full case directly now

Define $m:=\lim_{x\to a}f'(a)$

Pick an $\epsilon$, now we have a window of size $\delta$ around $a$ where $f'$ is between $m-\epsilon$ and $m+\epsilon$

@AkivaWeinberger I have an answer that I will find that answers something like that.

MVT says $(f(x)-f(a))/(x-a)$ equals $f'$ of something in that range

so $(f(x)-f(a))/(x-a)$ is between $m-\epsilon$ and $m+\epsilon$

Let $x$ go to $a$, we see the definition of the derivative

$f'(a)$ is between $m-\epsilon$ and $m+\epsilon$

$\epsilon$ was arbitrary so $f'(a)$ is $m$

This is actually a very important exercise. Some people with PhD's use it without realizing they're using it. They just think that $f'(a) = \lim_{x\to a} f'(x)$, even though no one told them that the derivative is continuous.

@AkivaWeinberger I think I have to go to sleep, but at first glance it seems like your proof is correct. I really wonder how you could solve something like that in less than 1-2 minutes which I couldn't even after 12 hours =P

@TedShifrin Spivak says it is a simple, but beautiful theorem. [I don't know why but I will trust him]

I don't call it a theorem. That's a bit too grand for this. But it is useful and important.

@Prithubiswasleftmse I wasn't really solving it for the first time - I was mainly trying to remember how it went

I just told you why. It allows you to deduce a formula for the derivative at $a$ even though you didn't necessarily know the derivative exists at $a$. Doesn't Spivak actually prove this in the text?

But yeah, "pick an $\epsilon$" is always a good first sentence in any analysis proof.

Yes, he does. So why did you spend 12 hours?

I don't usually bother with the epsilonics for this proof.

Just like "suppose not" is always a good first sentence in any proof of a negative statement

@TedShifrin Ah, you're right, I can see now how to do it more simply

It wasn't the first way that occurred to me

The typical AP calculus student uses L'Hôpital's rule, which drives me nuts.

@TedShifrin Yes Spivak does prove it, but I don't really understand that proof much, or the idea behind it. So I tried to get my own proof.

but $(f(x)-f(a))/(x-a)$ equals $f'(c)$, let $x$ go to $a$, and $c$ also goes to $a$

@Prithu You should understand his proof. It's the right way to think.

Sometimes learning to understand arguments you don't understand is the way to make progress as a mathematician.

Stubbornly refusing to do so isn't a good thing.

I had stern words with someone on main. I called him/her stubborn for refusing to think about what I'd said (which solved the question). Eventually he/she relented and said I was right ... and that the book had said to do that.

I do think trying to come up with your own proofs can be worthwhile, because it can help illuminate why the book proof doesn't go down certain directions

A disguised version of an important fact from linear algebra: If $v_1,\dots,v_n$ are linearly independent, and $u_1,\dots,u_k$ are independent vectors in their span, then we can choose $u_{k+1},\dots,u_n$ in the span so that $u_1,\dots,u_n$ are linearly independent.

@AkivaWeinberger Me too =)

Yes, it's instructive to figure out a convoluted proof or get stuck trying to do so, but if one has a proof in front of one, perhaps one should understand it.

I speak with a few more years of experience, both as a student and as a teacher, than both of you put together. I do agree that sometimes trying to figure out your own approach can lead to new results, but that's rare.

In this case, Spivak's proof is the obvious thing to do and is important to understand.

I'll leave it at that.

I will say that he had horrible proofs of the Taylor remainder theorem in earlier editions of his book. I couldn't stand them and figured out a different proof (probably a convoluted form of which is in Rudin or other books). Ultimately, I convinced him to change to that argument in his fourth edition (maybe third).

I haven't read Spivak or seen what his proof looks like so I can't comment on that specifically

@TedShifrin That's one of the things that matroids try to abstract from, if I remember right

Jesus, I just realized he died. Never read his book, but that's too bad. Hopefully he had a good life.

Yeah, it's been over a year since he died. There are numerous books :P

I'm not at all connected in the math community, but I guess he was pretty well-known. I should have written "books", plural. Guess I can stop wondering when the second volume of Physics for Mathematicians will come out.

Yeah, you should quit wondering.

Since you're here I should ask you: do you have any opinions on his differential geometry books? Do you think they are particularly valuable in any way?

(I realize this is subjective)

They're good and useful. Way too wordy for my taste, but I don't know of too many errors and he presents a lot of material.

I first met him when he taught the year-long graduate course in geometry at Berkeley, following his books (in a year he covered about 3 1/2 of the volumes).

Berkeley seems like a wild place

He was a yoga master.

Indubitably a character.

Fun fact, he came up with a set of gender-neutral neopronouns

and used them in The Joy of TeX

(which I have not read)

> The precise history of the Spivak pronouns is unclear, since they appear to have been independently created multiple times, each time likely without knowledge of the previous.

He used them, and they are named after him, but he did not invent them

@TedShifrin Thanks. I guess I'll just keep them in the back of my mind, along with the other dozen or two manifolds and DG books I'm vaguely aware of.

For a novice, they're a lot easier to read than Kobayashi-Nomizu. But there are lots more options these days.

hey chat

$-\mathrm{inf}(-*)$?

so my ode teacher asked us to prove that a smooth field $X\colon S^2 \to TS^2$ never has a dense orbit

which is kinda disproportionate considering we don't even know what $TS^2$ means (a priori)

he gave us a tip that $X$ must have a singularity and then we said that we can consider $S^2\backslash \{p\}$ where $X(p) = 0$ and, by stereographic projection, it remains to prove that there's no smooth field $\tilde X\colon \mathbb R^2 \to \mathbb R^2$ with dense orbit

nobody solved the question and then he went "eh, you use the tubular flow theorem or this theorem about differentiable manifolds and apply the jordan curve theorem to obtain this and that..." and proved nothing. simply said a bunch of complicated words and he did not prove the claim

so if anyone knows how one can prove that, it would be great. thanks

also, we asked if he would ask that kind of thing in the exam and he said that he would if he felt so. we have literally only seen things like Picard, the Peano theorem and construction of suspensions and stuff. this is getting pretty ridiculous. it's a shame to see so much professors in my institute that do not care about the course syllabus and the academic future of their students

@LucasHenrique Here's an idea, not sure if it works

Starting from thinking about it on the plane

Pick a point in the dense flow, flow forwards - by density, it will eventually get really close to where it started

It almost forms a loop

For the flow to continue being dense, it's gotta leave that near-loop and re-enter it infinitely many times

but if the ends are close enough to each other, the flow along the line segment connecting them will either entirely point into the loop or entirely point out of the loop

meaning eventually you're trapped on one side or the other of it, and can never fulfill your density dreams

I suppose, to really call that section of the flow a "near-loop", I'll want the line segment joining up the ends not to intersect with the flow

'cause I'm really doing JCT to the flow plus that line segment

Hm

I have a better idea

akiva, nobody ever responded to your circles puzzle. :(

Pick a line segment such that every vector on the line segment points to one of its sides

Now starting from a point on that segment, flow along that dense flow until it hits that line segment a second time

The flow plus that segment then creates a loop that you either can never leave or can never enter

@leslietownes Aw

would it be bad form to taunt the forum because you stumped them? and to demand a 'master of puzzles' badge?

I mean, they're convinced it's a hard puzzle, but they're not convinced it's a good puzzle

@TedShifrin hey

I always mix up this point in modeling probabilities. Say $X$ is some event in a financial market, i.e., "the price of Google closes tomorrow between $2500$ and $2700$ USD," and $A$ and $B$ are each events regarding some market indicators, i.e., $A$ is "the $2$-day moving average is currently below the $10$-day moving average," and $B$ is "the current price is higher than the previous hourly candle's close."

If I witness both $A$ and $B$ and use these to make a judgement about the probability of $X$, is this $\Pr(X\ |\ (A \cup B))$, or is it $\Pr(X\ |\ (A \cap B))$?

You’re assuming that both A and B hold true.

$\cap$

Okie, there was some other similar question I can't find in my history now where associating "and" with "$\cap$", and "or" with "$\cup$" put me on the wrong track.

context is everything, e.g. there is no English to math dictionary wherey ou just sub in $\cap$ everywhere you see "and." but the probability of an event X, given events A 'and' B, is definitely given by putting $\cap$ in that formula.

you can imagine drawing a little horizontal line across where the cap starts to curve. it makes a big capital letter A for And. anyone who uses this mnemonic owes me $5 per use.

Alright, thanks

Sad to say I was already taught that mnemonic :P

how long ago? the royalty is due immediately, with interest.

My family and their family went to the parade... that looks like a $\cup$ to me.

In college, quite a few years ago :P

sounds like it's going to be a lot of interest.

that's OK, we can work out a payment plan.

can they pay in lesliecoin?

of course.

I think the interest is pegged to the ROI of lesliecoin.

it's good that nobody taunts lesliecoin on twitter. they would be ratioed into oblivion more or less immediately by a horde of the world's wealthiest people.

@AkivaWeinberger that's precisely what he did, heh

@AkivaWeinberger define "flow from a point on that segment"

(and why would it pass through the segment twice?)

@AkivaWeinberger sincerely my problem relies on proving all this "go walking on the curve" thing. i just can't write it down

My confusion on whatever the other problem was might have had to do with the fact that typically the union of sets is larger than its inputs and the intersection of sets is smaller than its inputs, but the intersection of events representing information like in the above yields a greater amount of information than its inputs.

yeah. as one thing gets bigger the other gets smaller. more information narrows the universe of what's possible.

unless the information concerns lesliecoin, in which it only expands the universe of what's possible. did lesliecoin go up? you'd better check again, because it just became possible to go even more up.

What is $\gamma$ here?

a function from an interval into R^3. not quite the same as its image in R^3.

the thing being drawn on the board is a (two-dimensional) representation of the image of that function.

I am still awed that time dependent functions create such a variety of curves.

oh, did you mean what specific gamma is he drawing? no particular one, he's just drawing an example. if it can only be in one place at any one time, its position over time could be thought of as a curve. so anything can basically be a curve.

my remark above was directed more at the distinction of the thing being drawn, and the function. e.g. "go around the unit circle counterclockwise at constant speed once as t goes from 0 to 1" and "go around the unit circle counterclockwise at constant speed twice as t goes from 0 to 1" will result in drawing the same picture, but they are not the same function.

so if a 'curve' is a function (as it seems in the screenshot) those two examples would not be the same curve, although they might give rise to the exact same picture

@leslietownes wait, can it be a function with that loop on it? I remember a thing like the vertical line test where no two points should pass through same perpendicular line from the horizontal axis.

oh, good point. here the two or three components of the curve are each allowed to vary with t.

to relate that to what you know before, some curves are functions of their x coordinate. in 2D such curves can look like gamma(t) = (t, f(t)) for some function f of one variable.

the image of something like that is going to pass the vertical line test. you might call such a curve a "graph of a real-valued function of one variable," to distinguish it from a general curve in 2D.

@leslietownes Thanks for that nugget, I owe you a coffee.

something of the form gamma(t) = (x(t), y(t)) where x and y are independently varying functions of t, might not trace out the graph of a function. even though x and y themselves are functions, and gamma is too. the thing being drawn is no longer the graph of a real-valued function of one-variable.

you are welcome. :)

@leslietownes Ah, well, I'm glad getting that doubt resolved. I'll approach this after further reading. Thanks, again.

the calc 1 one variable arc-length-of-a-graph formula arguably makes slightly more sense in the general form for space curves. sqrt(1 + [f'(t)]^2) is easier for me to understand as a special case of putting x(t) = t and y(t) = f(t) into sqrt(x'(t)^2 + y'(t)^2).

Leslie, watch it. You’re verging on waxing geometric.

this isn't leslie, this is the guy who broke into leslie's house. i'm on a break from looking for valuables.

Munchkin at the keyboard.

and i did try to describe that as formulaically as possible. putting one thing and another thing into a symbolic formula. that's probably leslie's influence.

Not to mention devoid of MathJax.

i destroyed the key on the keyboard that does dollar signs while looking for valuables

@leslietownes Why doesn't \(... )\ work like \$...\$ as advertised on the latex page.

there's a schism in the latex community around whether it's OK to acknowledge [ or whatever the supposed right way of doing math mode is. i am a member of the true church that has not abandoned the true ways of just using the dollar sign.

the california hippie who did the chat plugin was probably hopped up on goofballs when he wrote both the plugin and the page about it.

ChatJax isn’t through and through LaTeX.

i interact with both via macros that locally convert from my typed troff, which is the one true markup language.

@leslietownes idk, \$ has php vibes which might give some people ptsd of say, user authetication systems, pageview counters and database connection objects, to say the very least. Also, money.

we will be using the lesliecoin symbol soon enough.

my daughter requested this morning that i bring jelly beans for her to eat this afternoon on the drive home from day care. i'm considering it. the alternative is hearing quite a lot about jelly beans on the drive home, which sometimes takes 20 minutes.

ok, so I'm reeeeeally bad at calculus in higher dimensions

@leslietownes How do you estimate the number of jellybeans in a jar? I saw grey matter do it once. My approach would be to eyeball measure the size of the container, estimate the beans as spheres and assume maximum packing efficiency.

in real time, i guess that's what i'd do. maybe skip the spheres thing and kinda eyeball it based on experience. if i had time to prepare in advance, i would empirically measure the packing efficiency ahead of time with a smaller amount of beans. i'd think of those beans as a kind of investment in winning the bigger jar of beans.

I want to prove a silly result: if $F(x) = \lambda x$ and $\gamma(0) = x_0,\ \gamma'(t) = F(\gamma(t))$, then $\gamma(t) // x_0$

how can I do that?

(specifically, I don't want to use the full ODE solution)

@LucasHenrique Well, might as well, ping you with what you mean by $(\lambda \int \int ... \int \gamma (t) dtdt...dt ) // \gamma(0)$ because I've never seen $//$ used beyond floor division or series circuits

Is there something I should have in mind about emailing someone about something he wrote in 1997?

the guy isn't retired. I checked

What sort of something?

a paper he wrote. He published recently about related things, so he can't have forgotten

Is it a research-level question?

I don't think so

Generally you don’t bug authors with undergraduate-level questions. Depends on the author, but some won’t answer.