Mathematics

Alec Teal

12:00 AM

math.stackexchange.com/questions/481693/… bump for good answer :)

robjohn

@DanielRust Thanks, and i just noticed that it is past the time for the cap. I can get points again :-)

@AlecTeal Lemme look

Alec Teal

@robjohn I'm practicing an area I was weak at, by doing papers, I'd love a lecture on how to write various forms of the chain rule, using D partials and nablas... so please do feel free to not just put the answer.

I am also sorry guys I don't really help much, you lot are much further along than me, I was chuffed to know Markov Chains, heh.

anon

do you know how to do matrices in latex?

Alec Teal

Now I do :)

anon

\begin{pmatrix}a&b\\c&d\end{pmatrix}

robjohn

12:07 AM

@DanielRust I can't believe it took me so long to divide the beachball another way. I had Robert Israel's answer just a bit too late. But the division I have can be made so that there are $6n$ equal pieces with the same orientation

Daniel Rust

@robjohn Yeah, and you can also use a similar method to make a solution without rotational symmetry

it's nice

robjohn

@DanielRust yeah, but what I am thinking of loses the similar orientations to lose the rotational symmetry

I think...

Daniel Rust

@robjohn Yes that's true

did you see the MO thread posted in the comments?

the question in there is very interesting

robjohn

@DanielRust no, I will look

Daniel Rust

@robjohn It's an open problem unfortunately

robjohn

12:12 AM

@DanielRust Yeah. I see that Robert's answer is mentioned in comments there.

Daniel Rust

and yours too (kinda)

Peter Tamaroff

@anon How is this question not ill posed. Sigh?

Alec Teal

@anon I gathered the - sign is wrong, it just seems to legitimately pop up there.

See @anon I don't know what's wrong either! I've checked it many times.

anon

you wrote the matrix down wrong

Alec Teal

I didn't. The left column is going to be multiplied by a change in x (the top entry in the column vector) and that is where the partial with respect to x goes.

(using it as the best linear approximation at a point)

Peter Tamaroff

12:23 AM

@AlecTeal Oh, boy.

Karl Kronenfeld

@PeterTamaroff I did not downvote, but I think the OP was trying to take advantage of the usage of discontinuities in this characterization of the Riemann integrable functions to turn it into a definition of (dis)continuity.

Alec Teal

That way if you put in say (1,0) (as a column vector) you get out only the partials with respect to x column.

anon

@AlecTeal look at the very first matrix in my answer. do you agree with it?

Karl Kronenfeld

@PeterTamaroff I would say it is impossible to do that, but I get what they were going for.

Alec Teal

Maybe, I'm quite confused right now.

Peter Tamaroff

12:25 AM

@KarlKronenfeld Well, I find the question to be very unclear, at least. In particular, I don't see how you can rephrase the theorem without using the term "continuity", "oscillation" or whatnot.

Alec Teal

@anon I also agree with mine you see, that's the problem....

I've made a right pigs ear of a simple question.

anon

and why do you agree with yours?

Alec Teal

Because partial g / partial x will then be multiplied by a change in x using my one.

Thus becoming changes in g.

Peter Tamaroff

@KarlKronenfeld I find the question quite silly, nevertheless.

robjohn

@DanielRust Ah, I just found that comment before returning to chat :-)

Daniel Rust

12:27 AM

@robjohn haha yes

robjohn

@DanielRust that was what I was thinking of when you mentioned the asymmetry

Daniel Rust

I'm wondering if there's an intermediate proposition which may be attackable

Alec Teal

@anon this is how I know things are bad, I can see two conflicting things as making sense. I want to agree with you of course, I don't claim to have found a hole in maths, nor that you're wrong, just, what have I done and why!

Karl Kronenfeld

@PeterTamaroff I agree. Continuity is farther-reaching than riemann integrability. It would be pointless to attempt to define continuity in this way.

Peter Tamaroff

@KarlKronenfeld Mind if I paraphrase that in my answer?

Karl Kronenfeld

12:30 AM

@PeterTamaroff Go ahead.

anon

@AlecTeal my second guess is not that you wrote the matrix wrong but that you put $D_{g(a)}f$ on the wrong side of the matrix

Peter Tamaroff

continues with Landau...

Alec Teal

@anon dimensionally, I've convinced myself what I've written is fine.

Peter Tamaroff

@AlecTeal Write it down again, without looking at what you've done. Start from scratch. Don't be stubborn.

Alec Teal

Done that twice.

That's why I appealed to you guys, I have no idea what I've done. BUT I've convinced myself it's right, then proven myself wrong.

anon

12:33 AM

so you've convinced yourself that $\partial (f\circ g)/\partial x=f_u(g)u_x+f_v(g)u_y$ as opposed to $f_u(g)u_x+f_v(g)v_x$?

Alec Teal

Well @anon for it to work, the -2y needs to become a +2y

anon

you say things like "I've convinced myself" but don't tell us how you've convinced yourself. not much to work with.

@AlecTeal huh?

Peter Tamaroff

►

Alec Teal

@anon you see the matrix with -2y in the top right, if that's a +2y my answers agree.

anon

@AlecTeal correct. see my answer!

you have to write down the chain rule correctly

Alec Teal

12:36 AM

Yeah but 2y and 2x being the bottom row work.

anon

I wrote down two different correct forms

Alec Teal

en.wikipedia.org/wiki/Chain_rule#Higher_dimensions I used that.

anon

@AlecTeal and you see $D_{g(a)}f$ is on the left??

must I repeat myself?

Alec Teal

Yes of a composition.

Peter Tamaroff

Oh, boy. Oh, boy. Don't make anon repeat himself.

anon

12:38 AM

@AlecTeal and yet in your matrix multiplication you wrote it on the right?

Alec Teal

If it's supposed to be pre-multiplied instead of post, then the entire foundation of what I thought I was doing crumbles, it was supposed to be column-vectors and post-multiplication.

What on earth have I done (I know how messed up I sound, I am VERY sorry for that)

anon

you put it on the right instead of the left

when you view matrices as linear transformations, the later composed ones are on the left, the first applied ones are on the right

Alec Teal

I know, because it's a column vector

anon

why do you believe it's a column vector?

Bitrex

ashton kutcher is the world's worst actor and every film he has ever been in is terrible

anon

12:41 AM

world's worst is a strong claim

Bitrex

well

top ten, at least.

anon

his characters always annoy me though

without fail

Alec Teal

Clearly you haven't seen Christian Bale in Braveheart.

That film was so bad, that nothing could redeem him.

Bitrex

He has the same problem as Will Smith, in that Will Smith is always Will Smith in every film.

Peter Tamaroff

@AlecTeal Don't you mean Mel Gibson?

Alec Teal

12:42 AM

Or the guy who directed and played the main role in a film so bad I had to check to see if it was a spoof or not: Titanic II

@PeterTamaroff silly me, I meant Reign of Fire.

Weird because they're unrelated films.

anon

seen the room?

Alec Teal

"Note that the tangent vector r\uffff (t) to a parametrised curve corresponds to the case f : R \u2192 Rm . The
Jacobian matrix has one column, the components of r\uffff (t). The gradient vector \u2207f of a function of
several variables c
"

@anon absolutely right, thank you.

Peter Tamaroff

@Bitrex OBJECTION: Pursuit of Happiness, I am Legend, Seven Pounds.

Bitrex

@AlecTeal Also that nobody ever saw Reign of Fire.

Alec Teal

Then how do we know it exists @Bitrex

Bitrex

12:43 AM

Good question.

Alec Teal

I've never seen a Noika WindowsPhail phone, but I know they're out there, doing nothing useful but not by design.

Daniel Rust

Is anyone else having trouble with chrome crashing sometimes when writing on the site?

Bitrex

It happens to me sometimes

Daniel Rust

especially MathJax heavy posts

anon

see meta

Bitrex

12:54 AM

using an external editor and copypasting is a good idea in general

anon

http://meta.math.stackexchange.com/questions/10807/chrome-page-crashing-when-viewing-or-typing-answers

Bitrex

For me, the site sometimes used to lock my PC completely

it would come up in a tab and then the whole PC would freeze with the activity light on for about 30 seconds

Daniel Rust

yeah, it's nice to see the rendered TeX though

also, it can't be helped for instance when editing a question/answer

Bitrex

Wait until that first time you've nearly finished a long post, and the tab crashes and the "saving draft" feature has gone tits up and nothing is saved

:(

Daniel Rust

:-/

if you click 'reload' it sometimes takes you back

don't refresh the page though

Peter Tamaroff

1:34 AM

@anon Dude.

Bitrex

$\sum_1^{\infty}\frac{4^n}{n^2}\binom{2n}{n}^{-1} = \frac{\pi^2}{2}$

user38268

1:54 AM

@DanielR Hey

@anon I am not understanding this proof that any Noetherian module admits a filtration

anon

what kind of filtration?

user38268

Associated prime

In abstract algebra, an associated prime of a module M over a ring R is a type of prime ideal of R that arises as an annihilator of a submodule of M. The set of associated primes is usually denoted by \operatorname{Ass}_R(M)\,. In commutative algebra, associated primes are linked to the Lasker-Noether primary decomposition of ideals in commutative Noetherian rings. Specifically, if an ideal J is decomposed as a finite intersection of primary ideals, the radicals of these primary ideals are prime ideals, and this set of prime ideals coincides with \operatorname{Ass}_R(R/J)\,. Also linked w...

anon

every module admits a filtration, 0<M

user38268

No a special kind of filtration

$0 = M_0 \subseteq M_1 \subseteq \ldots \subseteq M_n = M$

such that each $M_i \cong R/p_i$ for some $i$

@anon I am confused why is $M_0 = 0 $ a quotient of $R$ by some prime ideal $p$?

anon

it's not

Karl Kronenfeld

1:57 AM

It is not...

user38268

well that is the claim on the wiki page

Karl Kronenfeld

You do your actual work starting at $1$.

Or going backward, I cannot remember.

anon

the claim is successive quotients are R/p's...

user38268

Ok @anon So now there is the proof of this fact which goes as follows. We consider the set of all submodules that admit such a filtration as above. Why does the zero module admit such a filtration?

anon

vacuous

user38268

1:59 AM

I'm confused

anon

just write the filtration "0" all by itself, and "each successive quotient is an R/p" is vacuously true since there are no successive quotients

user38268

@anon But doesn't this mean we need $0$ to be a quotient of $R$?

anon

I am assuming writing something by itself counts as some kind of trivial filtration

@user38268 no, why would it?