:D

16.04 is going to be awesome.

@JMoravitz $^{\text{shh, people are sleeping}}$

Lucky them. I'm trying to type up some homework for probabilistic methods class.

I have some doubts about this one, worried that there might be a missing condition in the problem: $X_1,X_2,\dots$ i.i.d. random variables. Show $Pr[|X_n|\geq n~\text{infinitely often}]=0$ if and only if $E[X_1]<\infty$. I'm worried about the case where $E[X_1]$ is finite but $Var(X_1)$ is infinite (which is apparently possible).

I wouldn't have the luxury of invoking Chebychev's in that case to use Borel-Cantelli.

heya @JMoravitz: Long time no see. And hi, DogAteMy.

Hi @Ted. Yea, I've not been visiting the chatrooms much lately, though I've tried to keep finding questions to answer here and there. Its relaxing.

LOL @relaxing. You doing well in the grad program?

Hello

Well enough anyways. From a teaching perspective, its been a blast. From a student perspective its a real challenge.

Has anyone seen my jacket? I'm not sure where I left it

well, I'm sure your students appreciate a dedicated teacher. And we all struggle(d) in grad school.

Isn't that it what you're wearing @Akiva?

You've lost your coat, DogAteMy? Now DogLostHisCoat?

I only taught probability once a year ago, @JMoravitz, but how could knowing $E[X_1]<\infty$ and independence of all the $X_i$ tell you that?

Its apparently been noticed @Ted. Not to brag, but I got an email encouraging me to attend the upcoming end of semester departmental awards luncheon since they're apparently giving me something... I got recognized at last year's as well

That's tremendous, @JMoravitz. I am personally very proud.

Found it!

(As one who dedicated his career more to teaching than to research, admittedly.)

DogAteMy is being very siwweeee.

I also finished 2.2 in Hatcher; doing exercises now

Did you have to shed your coat to do so, DogAteMy?

@TedShifrin Well, assuming $E[|X_i|]$ and $Var(X_i)$ are both finite, via chebychevs one has $Pr[|X_n-\mu|\geq n]=Pr[|X_n-\mu|\geq n]\leq \frac{\sigma^2}{n^2}$. With a nonzero expectation, although a bit more frustrating, after some shifting around one should still be able to show that $Pr[|X_n|\geq n] = O(n^{-2})$

(Assuming I'm not too tired and misusing my big-oh notation again)

But how does $X_1$ impact $X_n$ at all in this event?

No, I think you have Tchebyschev correctly, but I should look it up to be sure.

$X_1,X_2,\dots$ are all identically distributed independent events. In any case, Borel-Cantelli tells us that if $\sum\limits_{n=1}^\infty Pr(E_n)<\infty$ then the probability that infinitely many of them occur must be zero.

@JMoravitz: For the last, you'd need to know all the standard variations are uniformly bounded.

So, $\sum\limits_{n=1}^\infty Pr(|X_n|\geq n] \simeq \sum\limits_{n=1}^\infty \frac{1}{n^2}<\infty$

@TedShifrin Hello. Can I ask you about an exercise from your book?

I'm not seeing how these pieces fit together.

@JMoravitz: But do you know all the $\sigma$'s are uniformly bounded?

Hi @Simeon. Sure, which book?

@TedShifrin Multivariable Mathematics

ok, sure.

I don't, which is the problem. I'm thinking it is a missing condition. There was already an email stating that the professor had made a typo intending to have it be $E[|X_1|]<\infty$ instead of $E[X_1]<\infty$

Wow ... a professor made an error? I'm shocked. :D

@TedShifrin If you have the book, for Ex 8.7.7., I don't understand how $\int_{C_1} \omega = 3$ when we can find a manifold $S$ whose boundary is $C_1$, and then by Stokes' Theorem conclude that $\int_{C_1} \omega = \int_S d\omega = 0$?

You can't, clearly, @Simeon.

But what if I just draw a semisphere around it?

But the hemisphere isn't contained in the domain in which you know $d\omega = 0$.

The axis punctures it.

(I didn't even have to go look to see which problem this was. I knew :) )

Oh so the hole in the domain causes the problem?

Yes!

Just like a point mass causes issues with gravitation being divergence-free.

Ooh. Problem 34 in section 2.2:

@Simeon: You actually taking the course somewhere?

> 34. [Deleted — see errata for comments.]

deletes DogAteMy

I have a differential equations student, @Ted, and I've decided to beat "ODE = vector field flow" into him from day 1.

He seems happy with it.

Oh nyo! @TedShifrin

You seem to get one exceptional student a quarter, @MikeM. That's not bad.

Whenever I see the acronym "ODE", "Ode to Joy" starts playing in my head

You're more a music history student than a math student, then, DogAteMy :D

(Same, on occasion, for "poles" and "March, March, Debrowsky")

(or whatever it's called)

@TedShifrin I'm taking a course which uses Munkres's Analysis by Manifolds, and there was a similar exercise there. I really don't know how to reason about these loops crossing different axes.

don't be too a-serbic, DogAteMy.

Oh, right, @Simeon. I actually stole that exercise from Munkres many years ago. :)

If you want to apply Stokes's Theorem (or use homotopies), you have to make sure that the surface (or homotopy) avoids the missing axes/circle.

I thought so :D (or the other way around)

No, Munkres put that on an MIT competitive calculus test one January back around 1980, and I never forgot it.

But I have a number of exercises that are unique to me ... unless Stewart or someone else stole them.

@Ted: I think it's an essential point of view. He had an exercise where you use uniqueness of ODEs to prove that if $0 < y(0) < 1$, and $y'=y(1-y)$, then $0 < y(t) < 1$ for all $t$.

OK, so you can prove that using the mean value theorem and uniqueness of ODEs. Or you could also draw the damn picture of the vector field you're following, and see that it's obviously true. :)

Or you could solve the differential equation :P

That's not the point, though.

This is a famous differential equation.

Throw an $e^{y^2}$ on the RHS too.

Hmm, I doubt it's still true.

$e^1$ is a bit bigger than $1$.

Huh? $y'=y(1-y)e^y$?

Same proof I gave above works.

Well, you had $e^{y^2}$, but I doubt that matters.

Now the maximum slope is well over $1$, so I'm not sure.

Oh, oops, I lied @MikeM. Not clear that the maximum is over $1$. Probably not.

But now not obvious to me.

I need help with Hatcher problem 2.2.33

No idea where to start

@BalarkaSen Halp

@Ted I don't understand what you're saying. The vector field y'=y(1-y)e^y has zeroes at 0 and 1. So if you start between them, you can never go past them. Hence what I said.

Yeah, you're right, of course. @MikeM

DogAteMy: Try induction plus MV?

@Ted: And my claim is that thinking about vector fields makes this obvious from the start :) I think this geometric intiition is the important thing to have all along here...

@MikeMiller Would you know an introductory book on étale cohomology? Or am I pushing the limits of your mathematical boundaries here

Or coboundaries, as the case may be

@Krijn Milne's notes are the standard

@MikeMiller That's funny, I was just reading them from the wikipedia links

@Krijn You could also read his book on the same subject. The word is that the notes only do it for varieties and some of the proofs are a bit sketchy, but the book is complete in both ways.

I know nothing about etale cohomology other than the idea and some applications, I just know people who know it.

@MikeMiller I think that that's not a problem, I'm only trying to get a grasp on it and write a short paper on in

OK, sure.

But I'm gonna work on Hartshorne first

hi

Hi, can one of you algebraist explain why (in a finite field of characteristic $p$) if an irreducible polynomial $f$ is inseparable then $f(x)=g(x^p)$ for some $g$?

@TheSubstitute what do you mean by inseparable?

@TobiasKildetoft Not separable as a polynomial (repeated roots exists)

$(x-a)^p = x^p - a^p$

In a field of char $p$

@TobiasKildetoft This is page 549 in 3rd edition of Dummit/Foote Abstract Algebra.

@TheSubstitute Ahh, right. See the reason above stated by Krijn

@Krijn @TobiasKildetoft thanks, but why does $a$ have to have multiplicity at least $p$?

Assume that it has multiplicity $< p$

Then see what happens when you work out the product

Or assume that it has multiplicity $k$ and look at $(x-a)^n$ where $p = m\cdot k + n$ for some $m$

Hello

Does the quadratic equation work in any field?

hi

I am trying to use partial in latex

\partial in latex

but I am getting this error

! LaTeX Error: Environment defition undefined.

any ideas?

except field of characteristic 2 @MichaelMitchell

@Adeek: probably something else causing it

nvm I fixed it @huy

Hullo lovely mathematicians!

Why do we say that a cyclic unitary group $\text{U}(n)$ is isomorphic to $\mathbb{Z}_n$ when really it's the cyclic generator that's isomorphic?

@BenjaminR Because your "when really" is not true

e.g. $\text{U}(10)$ : $\langle3\rangle \approx \mathbb{Z}_4$

also, usually unitary does not refer to the group of units but to groups of unitary matrices

But the mapping function is $a^k \mapsto k$ when $a$ is a generator of the cyclic group...

@BenjaminR generators are not isomorphic to groups. Groups are isomorphic to groups

Sorry Tobias, I should have said it's the $\langle a \rangle$ that's isomorphic

you are quite right

also, generators are not cyclic, gorups are cyclic

@BenjaminR But $\langle a\rangle$ is the group $U(n)$

Yep, but you can't apply a mapping to $\text{U}(n)$ and get $\mathbb{Z}_n$

the mapping is from $\langle a \rangle$

@BenjaminR Yes you can

But that is the same as $U(n)$ (as a set even)

Hmmm that seems to "jinky" to me.

@BenjaminR What do you mean. By definition $\langle a\rangle = U(n)$ so defining a map on either is the same thing.

but anyway, I am wrong as it's between the $\text{U}(n)$ and the partition of Z of the order of U(n)

Of course you are right, it just "feels" more like you use $\text{U}$ to find out what the elements are, and then you use $\langle a \rangle$ to do the mapping.

that's my crazy head though

(whoops not the partition, the integers modulo (the order of U(n))

Is there a way to draw commutative maps on latex?

@AkivaWeinberger I definitely worked this out previously, but I can't remember immediately how I did it. Ted's advice on induction + MV seems like the way to go.

@AkivaWeinberger 34 was a bit circular. That's why he deleted it.

If $E\subseteq\mathbb{R}$ then the set $\{a\in E:\lvert a-y\rvert<\varepsilon\}$ is always open, right?

It's union of open balls of radius $\epsilon$ around each point on $E$, right? Union of open sets is open.

yeah, right

@AkivaWeinberger Try out 13 (b) when you are bored.

lol at the set of comments below this post.

@BalarkaSen we're doing path integrals and the like now

Line integrals?

yes

I don't know too much about those yet.

I'm trying to go through the exercises right now

talks about things like path-connectedness (in the section after line integrals)

Ah, well, path connected regions are simply regions where two points can be joined by a path.

which seems easy enough, just use the point in $U$ that allows it to be star-shaped to construct a path that's two lines

(to get from one point to another with a path)

@GPhys Yep

the section on line integrals before this seems a little bit harder

Hi @AkivaWeinberger.

Hello

What's up?

Good afternoon to you

hm. didn't realize it was afternoon till now.

(If it's morning here but afternoon there, do I say "Good morning" or "Good afternoon"?)

(Though I suppose it's always after some noon.)

@AkivaWeinberger What do you mean by that anyway? Do you mean to say this afternoon is good, or do you mean to say I should be having a good afternoon, or that it's a good afternoon regardless of what I think or not?

That's a Hobbit reference, right?

Yuppers

@AkivaWeinberger Why don't you answer this?

I know it's true 'cause I've heard it before, but I don't know how to prove it

Think about it!

Same Euler characteristic and orientability, though.

Using classification of surfaces is considered cheating, because it's unlikely the OP knows the defn.

Also, RP^2#RP^2 is the Klein bottle somehow

True. That you shouldn't have too much trouble to see.

RP^2#RP^2 is a square with identifications $a^2b^2$

Anyway, if you don't want to think about it, that's ok. Writing out a proof which would be understandable for the OP would be a nice hands-on exercise, is all.

which can be cut and rearranged into the standard square for the Klein bottle

RP^2 is the square with identifications a^2b^2, not RP^2#RP^2.

Isn't it the square with identifications $abab$? @BalarkaSen

Right, I misread and mistyped. You're correct.

In any case, I think I can see it, but I don't know how I'd prove it formally.

Well, one is a square with identifications $a^2b^2c^2$ and the other is a square with identifications $aba^{-1}b^{-1}c^2$.

Um. If you have three letters $a, b, c$, how does it remain a square anymore?

*polygon

I haven't had my coffee yet

Yes, then that's right.

(I don't drink coffee but that's besides the point)

I don't coffee either :)

You can symbol push this through with the identification words, but there is a much more geometric way to see RP^2#RP^2 is homeom to K.

And it shouldn't be hard to translate that into words.

It has $M_2$ as a double cover, right?

What has?

T^2#RP^2

Mhm.

Identify the things directly opposite each other.

Not sure what you mean.

Think of $M_2$ as $T^2\#S^2\#T^2$. Place the origin in the center of the $S^2$ bit.

Identify $x\sim-x$.

You get $T^2\#\Bbb R\rm P^2$, I believe.

'cause the tori bits collapse into one

and the sphere becomes a projective plane

Works. Alternatively, T^2 minus disk is double covered by T^2 minus 2 disks, RP^2 minus disk is double covered by S^2 minus two disks. Now tube them up. You have $M_2$.

T^2 minus a disk is covered by T^2 minus two disks?

A double cover of T^2 is T^2 itself, that's why.

Oh, I see

Not sure how to get from that to a double cover of $RP^2\#^3$.

By the way, if you aren't thinking about it anymore, here's how I would prove RP^2#RP^2 is K. Take K, cut along a merdian. You get two Moebius strips. Fill the boundary circles of the resulting moebius strips with a disk. You get the desired decomposition of K as connected sum of two RP^2's.

That makes sense

@AkivaWeinberger Double cover of RP^2#RP^2#RP^2 is similarly M_2, but that doesn't really prove anything, does it?

If you can get from M_2 to M_2 without breaking the symmetry about the origin, sure

It would be an isotopy from M_2/~ to M_2/~ in $\Bbb R^3/\sim$.

I believe you?

Thanks?

No problem?

I'll think about the problem.

Sure :) Don't forget to have fun, though.

There was a problem last section in Hatcher that I solved using M–V, not realizing that the book didn't teach it yet

Which one?

About $SX$ and unioning cones on their base

Oh. Who cares? You can do it, that's all what matters. Mayer-Vietoris is an easy corollary of stuff from section 2.1 anyway.

Long exact sequence + excision. Good tool.

Yeah.

So, learnt cellular homology yet?

Hi guys, thoughts on How To Solve it by George Polya

@BalarkaSen Yeah.

@BalarkaSen Fun little theorem: en.m.wikipedia.org/wiki/Niven%27s_theorem

Can you prove it?

(Not topology, I'm afraid)

In a sense, it's because of the coefficient $-2$ in $x^2-2x+1$

Oh yeah I have seen that, @Akiva.

@BalarkaSen Have you done that kind of path stuff ^

Sure, that's not really hard.

QQ

I mean a candidate for $h$ should be immediate. You're given that $\gamma$ and $\alpha$ are one-to-one.

what...?

Not sure what you are "what"ing at.

I'm trying to figure out an $h$

Not sure how to give a hint on that one without revealing.

Do you want me to tell you my candidate?

okay

Take $h : [a, b] \to [c, d]$ to be $h(x) = \alpha^{-1}(\gamma(x))$. This is well defined because $\alpha$ is injective.

I think one needs to do some work to prove this is smooth. Maybe.

Hi @AlexClark.

I wasn't even trying to come up with an $h$ using function inverses

are simple closed paths always not continuous?

Paths, to me, are continuous maps from intervals onto whatever topological space you are in.

So in particular any object obtained from adding some adjective before path better be continuous also.

"simple" here means 1-1 and "closed" means the endpoints are the same?

Yeah.

Well, no. 1-1 and endpoints being same are contradictory.

that's what I don't understand

But pedantic comment anyway. It means 1-1 on the interior.

1-1 everywhere except the endpoints.

right ^

"no self-intersection", that is.

but I still don't get how that happens for anything

wouldn't that mean it would have to be discontinuous at one of the endpoints?

$[0, 1] \to \Bbb R^2$, $x \mapsto (\cos(2\pi x), \sin(2\pi x))$.

Is that discontinuous?

Last I heard, $\sin$ and $\cos$ were continuous functions.

oh, but there's no example on $\mathbb{R}$?

On $\Bbb R$, there are no simple closed paths, yes (I think)

A simple closed path looks like a circle. Those don't exist in $\Bbb R$.

@GPhys I am not sure what example you're asking for. You were talking about continuity and closed being exclusive. I just gave you an example where they aren't.

There are plenty of closed continuous paths on R.

But no simple ones. ^

simple closed

They all self-intersect.

No, there are no such things, @GPhys.

hi all. Anyone got any ideas about math.stackexchange.com/questions/1741157/… ?

@AkivaWeinberger Can we use the taylor series of sine somehow?

You should be able to prove that quite easily.

Otherwise I may be forced to add a bounty later :)

@Albas I don't know; my proof goes a different route.

I have no proof.

What is your proof@Akiva?

I only have theorems. But no proof.

I don't recall the details at the moment, but it involves finding a recursive relation for $\cos(cx)$.

Hey@BalarkaSen At least you know how to use those theorems to solve difficult questions

I can solve difficult problems as long it doesn't involve giving a proof.

Much simpler problem: Given Niven's theorem, find all rational $\theta$ such that $\cos(\theta^\circ)$ is the square root of a rational.

@AkivaWeinberger Quick, tell me something about the Zariski-Riemann space.

@robjohn amazing result here!

@robjohn For the record: some unexpected results throw me very close to prove that the sum of the divisors of $n$ is smaller or equal to the harmonic number plus $\exp(H_n)\log(H_n)$. This is equivalent to the Riemann hypothesis!!!

@BalarkaSen It's named after two people, one of whom is Zariski

Can't remember the other guy but it begins with R

Take all such $\theta$'s such that $cos(\theta)$ is not rational

@AkivaWeinberger Shake hands, Chief!

? @BalarkaSen

@Albas What about them?

That's all I know about them too.

@AkivaWeinberger Just some rogue idea

@robjohn I won't be around for a while, I just wanted you to know that. I need to check the details a lot to be sure 100% that things are precisely as I think they are.

It's wrong, @Albas

I was thinking about it. But then yes it is wrong. There are values for which $\cos(\theta)$ is irrational and not the square root of a rational

Example $\cos(44)$

Hint: You want $\cos^2(\theta)$ to be rational…

…What's $\cos^2(\theta)$ equal to?

You can use $\cos^2(\theta)=\frac{\cos(2\theta)+1}{2}$. You can use this right?@AkivaWeinberger

Yup. That and Niven

Yes

I will try on a proof for niven

Wait...

@AkivaWeinberger Since you're onto these things, have you heard of this?

Well, I said "Given Niven" for this particular problem, but sure @Albas

@BalarkaSen Yes, I have. I don't know too much about them, though

Me neither.

The only period I have heard is for trigonometric functions.

Kontsevich is a crazy person, I won't hope to dare reading his work.

Crazy is a good sense, not bad.

@AkivaWeinberger I meant separately I will try for a proof

@BalarkaSen Yeah, I was unfortunately not in Aarhus when he did a master class there (but I think it is online)

Aw.

Sorry to hear that.

@BalarkaSen I was more sorry to miss the one by Elias and Williamson as that was actually in my area

Ah. What was the topic about?

@user1618033 Very interesting! Let me know if this pans out.

@BalarkaSen Not sure what Kontsevich talked about. The other one was on diagrammatic Soergel calculus

I see.

Basically explaining the background for their proof of Soergel's conjecture

I think I'll skip the rest of the exercises…

(I figured out 33, by the way)

Surely not all of the ones in 2.2.!

@AkivaWeinberger Which exercises are you skipping?

35 to 43. They're too hard… (well, I have an idea for 38)

(And 36)

You certainly don't need to do all of 35-43 but there are some of them which you'll need later on when reading cohomology say.

36, 40, 43.

40 is a baby version of universal coefficients theorem, which is the very first thing you'll encounter in chapter 3.

36 is needed in the computation of cohomology ring of torus.

I don't care about the rest of those either.

Wow, Harris's first course in algebraic geometry is a nice book of examples. I could read this, I could definitely read this.

Hi

I used to come here for homeworks, but today I'm just here for fun ! Fantastic isn't it.

How do you call the difference between the lowest and the highest value of a set ? My students results go from 1 to 100 % so 99 is the _____ ? It isn't the standard variation, right ? So what is it

I am missing a vocabulary word here

Range

hmmm I wonder what is that in french

Portée maybe

I don't know

If the set is arbitrary, you can call it the diameter of the set.

That is, if the set is a metric space.

I think that it's "range" in statistics, but I'm not sure

Hey@AlexClark

Nope. I did finish chirality though. But onto coordination compounds now

I have just gotten into a new grade. And I have a lot of maths left to do. So I plan on finishing all the other subjects within 4to 5 months@Alex

And I have the most difficult subject: Engineering drawing

Hello!

So, on my journey to relearn math, I'm making some funny observations.

I now think math is one of the most interesting and rich subjects. Conversely, it's taught in the dullest form possible.

I'm reading a book called Here's to Looking at Euclid, and I'm filled with wonder at how different civilisations and peoples use numbers. It's fascinating to see how number systems evolved independently from one another, and were shaped by the environment.

So I find myself thinking... why do we study math with such a heavy layer of abstraction, like equations of x and y and what not? Of course, most books will try and relate real world objects, but those attempts feel really artificial--like using pies and cakes to teach fractions.

@Mohamad You are not alone in thinking that :P

@Mohamad Wonderful book! It was a while ago that I read it but I remember it being very good.

There's so much in the natural world that we can use to describe and teach math, why not lean on that more? I loved studying biology because it filled me with a sense of wonder. Even when it was technical, there was still something absorbing about it. It was rich with context and facts.

@AkivaWeinberger it's unfortunate, in my opinion. It took me so many years to like a subject that I had always loved deep in my heart but didn't know it.

@Mohamad Same here . I started liking serious math in 9th grade

@Akiva Is there any way to prove that a polynomial $p(x)$ is rational for some x?

@Albas you must be assuming that the coefficients can be irrational, otherwise the polynomial would be rational by definition.

@ahorn what I am working with is not even a polynomial so I do not think so anything of that sort will work

I was thinking what I would do if there was a polynomial whose coefficients might be irrational

It depends on the polynomial, I guess…

@Albas

Any specific example in mind?

is a piecewise smooth path still continuous or can it be discontinuous at each of the points it is "piecewise smooth" on

@Mohamad It's never too late.

Glad you're studying math. Please never give up on it if you like it!

@GPhys Continuous means things don't break. Why in the world be a piecewise thing won't be continuous? You previously asked something which made it seem like as if you're asking if injectivity and continuity are exclusive. That's not true. It seems you have a wrong intuition for continuity.

The piecewise thing won't be differentiable at the "sharp turns". But by no means not continuous.

@BalarkaSen I was asking if injectivity and continuity were exclusive on closed paths for $\mathbb{R}$ and the answer was yes?

You didn't mention $\Bbb R$ at all. In any case, to answer your previous question, paths are always continuous. By definition.

[any adjective here] path would thus still be continuous.