Mathematics - 2021-01-29 (page 1 of 2)

BigSocks

02:03

So checking every ideal be principal is the same as checking just the prime ideals apparently. Also apparently you can't ease this to just checking the maximal ideals. Sure.

But then the book says that, because of the first proposition I made, somehow a regular local domain of dimension $1$ is a principal ideal domain. Breaking that down, it's because it has a unique maximal ideal that can be written down with $1$ generator. But didn't we *just* say it was not sufficient to only check the maximal ideals? What is going on?

dc3rd

02:18

Question about a certain step in a proof:

Why/ What was the impetus for introducing the $f$ and $g$ functions? I see how they work "nicely", but when I originally tried to prove this the first time I attempted to do it with what was given in the original statement

Thorgott

02:33

@BigSocks there are only two prime ideals, the zero ideal and the unique maximal ideal, no?

Clarinetist

Dumb question: suppose I have random variables $X_1, X_2, \dots$ with mean $0$. Suppose $\sum_{n=1}^{\infty}\text{Var}(X_n) < \infty$. Why is it the case that for any $\epsilon > 0$, $\epsilon^{-2}\sum_{n = M + 1}^{\infty}\text{Var}(X_n) \to 0$ as $M \to \infty$?

BigSocks

@Thorgott oh cuz dim 1 means you don't have any other ideals in between right?

yeah that's gotta be it

Thorgott

yeah

BigSocks

thanks again

Clarinetist

I suppose we could write the sum as $\sum_{n=M+1}^{t}\text{Var}(X_n) = \sum_{n=1}^{t}\text{Var}(X_n) - \sum_{n=1}^{M}\text{Var}(X_n)$

BigSocks

02:39

oh also "local domain with principal maximal ideal of dimension at least equal to 2" somehow implies "not a principal ideal"

that I have looked around for like 2 hours and I can't see why

Clarinetist

and then let $t, M \to \infty$... oh. That must be it.

BigSocks

it would free me of this remark

Thorgott

The idea is that they want to work with the function $f$ instead of the vector $u$, because that allows them to use the fact that multiplication of transformation matrices corresponds to composition of linear transformations (the middle equality).
I'm not a huge fan of that, though, so let me offer an alternative. Let $v_1,...,v_n$ denote the basis $\beta$ of $V$ and write $e_1,...,e_n$ for the standard basis vectors in $K^n$. Then $[T]_{\beta}^{\gamma}e_i=[T(v_i)]_{\gamma}$ by definition (remember multiplying with $e_i$ from the right returns the $i$-th column). Now we can write $[u]_{\bet…

@BigSocks you mean "not a PID"?

BigSocks

yeah I did

completely missed that

evidently I should go to bed, but I kinda wanna do this

Thorgott

PIDs always have dim 1

PIDs that aren't fields, I should say

BigSocks

02:47

I am here, yet again, realizing I am a noob

would a field just have dim 0 or what

Thorgott

ye

BigSocks

never really thought of it, but that's kinda neat ok

Thorgott

field <=> dim 0 domain

@Clarinetist Right, note that the $\epsilon^{-2}$ is completely irrelevant. This also has nothing to do with random variables, it's a general fact about convergent series and the proof is as you say. Be careful about the order, though. You first take $t\rightarrow\infty$ and, then, in a separate step, $M\rightarrow\infty$.

dc3rd

@Thorgott. Thanks, I'm going to go over this proof to see what you mean

Clarinetist

Thanks @Thorgott! one last question... suppose $S_j = X_1 + \cdots + X_j$. I have that for any $h > 0$, $$\mathbb{P}\left(\sup_{i \geq 1}|S_{k + i} - S_k| \geq h \right) \leq \dfrac{1}{h^2}\sum_{j=k+1}^{\infty}\sigma^2_j$$ where $\sigma^2_j$ denotes the variance of $X_j$ (mean-zero independent random variables). How do you make the leap from this to $$\mathbb{P}\left(\sup_{n, k \geq N}|S_{n} - S_k| \geq h \right) \leq \dfrac{2}{h^2}\sum_{j=N+1}^{\infty}\sigma^2_j$$

The professor mentioned something about how you add and subtract some amount, but I'm definitely not seeing it

dc3rd

02:58

@Thorgott, not that you can change it now, but should the expression: $[u]_{\beta}=\sum_{i=1}^nu_ie_i$ actually be: $[u]_{\beta}=\sum_{i=1}^nu_i\mathbf{v_i}$?

Clarinetist

Lol, and after sorting through 5 books, no one seems to describe this step in detail

dc3rd

@Clarinetist, I've seen that before in some course I took. I don't remember when and I had memorized the steps....but had zero comprehension of what was occurring when I did it.

Clarinetist

That's promising lol

I don't get why this material is so poorly taught and explained

I think I'm at book 7 now

Thorgott

@Clarinetist hmm, I don't see it

@dc3rd no, $[u]_{\beta}$ is a vector in $K^n$, not in $V$, $\sum_{i=1}^nu_iv_i$ is just $u$ (by definition)

LHC2012

If for a field I say:

Ordered field

For all $x \in F$ such that $x < z$, we have $x < y$, is it true that $y \geq z$?

ok im stupid

its true because if y<z then y<y

Thorgott

03:17

indeed

dc3rd

So how you are using $e_{1} \dots e_{n}$ is throwing me for a loop. How are you relating $K^{n}$ to everything? Is it $K^{n} \to V$? So you're rewriting the basis vectors of $V$ as the linear combination of the standard basis vectors?

Thorgott

$[u]_{\beta}$ is a vector in $K^n$ by definition, no?

dc3rd

03:40

Yes

Thorgott

So I can write $[u]_{\beta}=\sum_{i=1}^nu_ie_i$ for some scalars $u_1,...,u_n$, since $e_1,...,e_n$ is a basis for $K^n$. Of course, that says nothing different than that $u_1,...,u_n$ are the entries of the vector $[u]_{\beta}$. And, by definition of $[u]_{\beta}$, these entries make the equality $\sum_{i=1}^nu_iv_i=u$ in $V$ true.

dc3rd

Ah yes. Ok that makes sense to me now. Thank you

Ted Shifrin

@dc3rd @dc3rd if you're doing my book, all the linear algebra you need is in there.

dc3rd

03:55

But what if I wanted to learn more @TedShifrin?.... I have it in my head to have a deep understanding of Lin Alg because it is everywhere.

I also enjoy it just for the sake of it as well...............referencing the idea in that Halmos's paper again.....

Ted Shifrin

There's not much more. Jordan canonical form is it.

dc3rd

Your book doesn't delve into inner product spaces too much though, or maybe I haven't arrived there yet.

Ted Shifrin

You're nuts.

I start with all the dot product stuff and emphasize it throughout .

I am less formal, so you're impressed by the formality and abstraction. Have fun!

dc3rd

Well yes... I meant beyond the dot product. But reading the contents of your book vs Insel you do cover it all except canonical form. Do you cover adjoint operators? I do like your book for that reason, being able to understand the concepts in plain english

Mike Miller

If you understand the dot product, and you understand abstract vector spaces, then you understand inner product spaces (with the exception that you maybe need to learn a little bit about how the complex dot product behaves). There is no interesting change in the way you manipulate them when you move to the abstract context; the ideas are identical, things are just less often written in components.

There is still some value in learning the more general language but I didn't get a lot out of it until I started learning graduate differential geometry.

Ted Shifrin

04:07

FYI, the adjoint is the transpose. It's everywhere. And everywhere in stat.

You are overly enamored with learning fancy stuff. Quite honestly, I'm worried about your mastering the basics.

dc3rd

@TedShifrin Damn....that hit home because I think you are right.

Ted Shifrin

Blunt, but truth.

dc3rd

And this is the advantage that having people around who have mastered the field to provide guidance to you provides.......I'm going to have to reconcile this

Krijn

04:47

Hey all

Clarinetist

@dc3rd Answered an old regression question of yours. It would've annoyed me to leave that unanswered.

dc3rd

Thanks @Clarinetist tomorrow is a stats day for me so I'm going to give it a good read. I look forward to it. Just skimmed it so I have something to look forward to.

polite proofs

Dr. Shifrin, your linear algebra lectures remind me so much of precalculus.

In fact, your lectures are almost verbatim to how I learnt precalculus. It's so interesting.

@TedShifrin

Ted Shifrin

@polite not what we consider precalculus in the US!

polite proofs

05:04

@TedShifrin I learnt precalculus from a US curriculum, so that's definitely not true!

Krijn

Magma and sage must've really changed the game for number theory and so

Imagine having either one of them 50 years ago, you'd have been the grand wizard of the maths department

Ted Shifrin

Vectors in some precalc, but not most ... no linear maps

polite proofs

I meant in particular how you introduced vectors, how you talked about thinking of vector subtraction

It was verbatim how I learnt it in precalc

Ted Shifrin

OK

polite proofs

It was almost dejá vu

I wouldn't be surprised if you influenced my precalc education, even

Ted Shifrin

05:10

Students screw up subtraction, regardless

I doubt that!

polite proofs

Perhaps you did, since you taught at the same company and most likely left feedback

Ted Shifrin

Oh, you're talking about AOPS? Their precalc class was disappointing. Not at all typical US curriculum, btw.

polite proofs

I am

Ted Shifrin

I had lots of complaints, indeed.

polite proofs

The way they handled vectors was almost verbatim to how you did

Very intuitively in R^2

Ted Shifrin

05:17

I don't remember, honestly.

I did share all the supplementary exercises and stuff I did. Who knows what happened with it?

polite proofs

I can't say for sure of course, but I just recall that in the linear algebra part, there were often more pictures than words. Haha.

polite proofs

06:21

Has anyone here ever participated in an integration bee?

123

Hello World..

user15072279

1

Q: How is angular velocity of both these bodies equal here?

Here there is Swaraj who is looking at a bus at which behind there is K+G. My teacher says that Angular velocity =(Theta / time)of the bus and K+ g is equal when they appear to be at same point by Swaraj. Now , my question is how is it possible since time and theta for both appear to be differen...

copper.hat

never been in a bee of any sort :-)

zacts

06:36

hello mathematiks

jonem

Is there a term for the sequence generated as the product of another sequence, i.e., $b_n = \prod_{m=0}^n a_m$. Does one refer to $b$ as a ____ of $a$?

robjohn

@copper.hat does that sting?

copper.hat

well, i used to get hives, does that count? :-)

polite proofs

@copper.hat Eczema?

copper.hat

uticaria

Ted Shifrin

06:40

@jonem partial products of the infinite product

polite proofs

Cats?

copper.hat

cats invariably dig their claws into me. it is just a matter of time.

zacts

λογιστική

Ted Shifrin

<—- loves cats

copper.hat

accounting???

polite proofs

06:42

@TedShifrin I'm allergic :(

copper.hat

well, i have been bitten by many dogs, but never by a cat, so i guess that it a plus for cats

@zacts are you Greek?

zacts

@copper.hat no

polite proofs

Getting bitten by many dogs... Perhaps an avoidable scenario.

copper.hat

you would think. the last time was by a poodle while riding my mountain bike in the nearby hills.

polite proofs

Can't outbike a poodle? ;)

copper.hat

06:44

@zacts why did you write accounting?

zacts

@copper.hat it's from Sir Thomas Heath's book on A History of Greek Mathematics Vol I

copper.hat

@politeproofs sharing a path with people i need to give way :-)

polite proofs

Unlucky indeed then

copper.hat

εντάξει

zacts

there is also ἁριθμητική

copper.hat

06:46

i was bitten while cycling in mexico, ended up getting rabies shots

arithmetic

and once while roller blading. that one was funny in retrospect.

and was attacked by a neighbour's pitbull while cycling near my house...

i like dogs, and they generally like me. i think.

Ted Shifrin

I got bitten through jacket and pants by a rot when I was attending a geometry conference at Park City in the snow in 1977.

copper.hat

oooh, they are big strong animals.

Ted Shifrin

I like dogs, but I love cats.

zacts

my cat is going to have babies.

copper.hat

kittens hopefully

Ted Shifrin

06:52

That'll be a big brood.

copper.hat

a cat-harthis

wine o'clock. something from nz.

Ted Shifrin

LOL, copper

polite proofs

07:15

Spivak just did something illegal.

He defined the distance between points $(a,b)$ and $(c,d)$ to be $\sqrt{(a-c)^2 + (b-d)^2}$

But this should almost certainly be a theorem, not a definition!

copper.hat

Why would that be a theorem?

polite proofs

@copper.hat Because we don't know that it gives the distance between two points.

Specifically, we would need to prove Pythag first.

copper.hat

it is just a definition.

polite proofs

This got me thinking

copper.hat

that's good!

polite proofs

07:24

Is there any analytical proof of the Pythagorean theorem, which we actually use to justify the euclidean norm, correct?

I know there's one where you can draw two triangles in $\mathbb{R}^2$

But that's not perfectly analytical, as it requires a drawing

user15072279

math.stackexchange.com/q/4004131/878401 Please help in solving this

copper.hat

coordinate geometry satisfies Euclid's axioms. the Pythagorean theorem follows from Euclid's axioms.

polite proofs

Who cares about Euclid's axioms? Weren't some of them false, too?

I fail to remember which, but I am fairly certain I remember reading that at least one was proven false

user15072279

Ok. But in my question . Is there any Euclid axiom used. I can search rest on my own just tell which axiom is it

copper.hat

They are axioms. You need to understand just how amazing it was for Euclid (and all contributors) to formalise these things at that time.

user15072279

07:33

@copper.hat ok but which one should I use ?. Yes it is amazing

copper.hat

Well, the 5th postulate.

polite proofs

Hahahahah

I like how this person is replying to you.

copper.hat

@user15072279 What are you asking me?

user15072279

@copper.hat I am asking in my question that I sent to you sir.

copper.hat

What do you mean you sent to me???

user15072279

07:35

Are you talking about the postulate for my question

polite proofs

@copper.hat Didn't get a PhD just to be called Sir...

user15072279

1

Q: How is angular velocity of both these bodies equal here?

Here there is Swaraj who is looking at a bus at which behind there is K+G. My teacher says that Angular velocity I.e(Theta / time)of the bus and K+ g is equal when they appear to be at same point by Swaraj. Now , my question is how is it possible since time and theta for both (Bus and K+G)appear...

geometry algebra-precalculus euclidean-geometry

This sir

@copper.hat I sent earlier. I think you were not seeing this. Sorry to interrupt

copper.hat

This is not an on demand question answering service!

user15072279

If you can help or anyone , pls do help

polite proofs

@copper.hat The question is in the mail. It'll be there on monday.

(perhaps a poorly timed joke?)

copper.hat

07:37

:-).

@user15072279 your question is too vague and i am going to bed shortly so i don't have the energy to unravel what your diagram is and what you are asking. you will probably get an answer on mse.

zacts

how does my proof look: mathb.in/49379

how is it wrong?

copper.hat

There is no value in having the $a$ there. you can just multiply across by ${1 \over a}$.

stay as simple as makes sense.

polite proofs

I thought I misread it, but I didn't.

This line in particular: $a+ar+ar^{2}+\cdots+ar^{k}=\frac{a(r^{k}-1)}{r-1}$

zacts

that's the line out of the book

polite proofs

It even contradicts what you are trying to prove.

I also agree with copper.hat, the $a$ is unnecessarily complicating things. You can factor it out for now!

copper.hat

07:42

It looks ok to me, what do you think is wrong with it?

missing a $+1$.

polite proofs

Yes

Subtle.

zacts

I just wanted to double check. This is the first proof like this I've done in a while. I'm just getting into this.

polite proofs

But not only that @copper.hat, the rest of his proof doesn't work because he assumes that it holds for the rest of the proof.

So his proof would be entirely different, it's not just one typo.

(If it were one typo, I wouldn't say that proof doesn't work.)

copper.hat

Here is another proof. Suppose $S=1+r+...+r^n$. Then $rS = r + r^2+...+r^{n+1}$ and so $(1-r)S = 1-r^{n+1}$. Dividing across by $1-r$ gives the result.

polite proofs

You could also multiply $S$ by $a$ to get the desired result for an arbitrary first term.

:)

copper.hat

07:47

ok folks, good night!

polite proofs

Have a good night @copper.hat!

zacts

thanks @copper.hat

Curious youth

08:16

Ninja Storm - All Ranger Morphs | Power Rangers Official

►

Power of Earthhhhh!!!!!

Curious youth