Mathematics

12:00 AM

@DanielFischer Are you sure you mean total derivative? I've not heard the term before and when I read the question I thought derivative meant either directional or "common" derivative.

Daniel Fischer

@MattN. I mean "total derivative", which is another word for Fréchet derivative. Maybe it's what you call "common" derivative (I don't know that term). I didn't think of the fact that "total derivative" is also used for the exterior derivative sometimes.

Matt N.

I see!

No, I was looking at the Wikipedia entry for total derivative.

And that seemed to mean the sum over all partial derivatives.

Btw

@DanielFischer I have an unrelated question for you: If I put a bounty on this one here -- would you answer it? (can't offer a bounty just yet though) And if so, how much bounty would be appropriate (e.g. 100?)?

Daniel Fischer

@MattN. I meant "It is another name for the derivative as a linear map"

user1534664

why is (a² + a)/(a + 1) <=> a

Matt N.

@TedShifrin Hi there, I'm just badgering Daniel about the thread with the directional derivative.

Ted Shifrin

12:06 AM

Hi Matt

Matt N.

Ok, in the question, the first sentence means "locally the Frechet derivative is a linear map"

The second sentence means "from the local linearity of the Frechet derivative I managed to prove the scalar multiplicativity of the directional derivative as follows"

Ted Shifrin

If it's the question I remember, I gave an explicit example where $D_vf(0($ was not linear in $v$.

ccorn

@UserX You are integrating over a range where $r$ is negative (in the beginning and the end). Thus you actually measure all three loops. Consider: Your result is one quarter of a full disk's area.

Matt N.

This one here. (yes)

user1534664

why is (a² + a)/(a + 1) <=> a

Matt N.

12:10 AM

Is what I said above correct?

My translation of "question language" into "Matt-ese".

Daniel Fischer

@MattN. The natural answer to that question is to show that the function is Lipschitz-continuous with Lipschitz constant $1$, you took Lipschitz constant $2$, but that's not much of a difference. Directly showing that the preimage of an open set is open proceeds essentially in the same way. I don't see a reason to add another answer to it.

Matt N.

@DanielFischer I tried to show that the preimage of an open set is open and couldn't do it and I would be interested in seeing how to do it.

Ted Shifrin

You can prove $D_{cv}f(a)=cD_vf(a)$ from the limit definition, but additivity requires differentiability in general.

Hi @DanielF

Matt N.

@TedShifrin Limit definition = definition of Frechet derivative? And if so, what is "differentiability in general"?

Daniel Fischer

@MattN. If the preimage is not empty, you take an $x_0 \in f^{-1}(U)$, you take an $\varepsilon > 0$ such that $(f(x_0)-\varepsilon, f(x_0)+\varepsilon) \subset U$, and you produce a $\delta > 0$ ... I guess you know how the sentence ends.

Hi @Ted.

Ted Shifrin

12:14 AM

I meant for the general function. I don't remember the terminology on Frechet. Is that just directional derivative?

r9m

@Chris'ssis insane (-_-)

Bye all

bye @UserX

bye

@user1534664 the left side is undefined at $a=-1$

Matt N.

12:15 AM

@TedShifrin No Frechet derivative is a generalisation of derivative to Banach spaces. (or perhaps just infinite dimensional normed vector spaces)

Ted Shifrin

No, then ... I mean limit definition of directional derivative only.

Pedro Tamaroff

@tedshifrin Because of reasons.

Ted Shifrin

How circular, @Pedro

r9m

@robjohn someone asked the same question in this room once .. I can't remember who and when :O !

Matt N.

@DanielFischer I see. Thank you! I think that's what I should have written in my answer.

user1534664

12:18 AM

@robjohn yeah I should've mentioned my textbook said "given that $a \neq -1$. but how do they get to the a?

robjohn

@user1534664 divide the polynomials

user1534664

@robjohn by?

robjohn

@user1534664 Divide $a^2+a$ by $a+1$

Matt N.

I think I will have to open a new question about this.

I just don't get it.

robjohn

@user1534664 you get $a$ with a remainder of $0$ since $a^2+a-a(a+1)=0$

Matt N.

12:21 AM

@TedShifrin So what sort of differentiability do you mean in your first sentence here?

If not Frechet that is.

Ted Shifrin

The linear map definition ... Frechet if you want.

Matt N.

Ah.

Ted Shifrin

There really is no other definition. :)

Matt N.

@TedShifrin And in the next sentence you mean the definition of the directional derivative, right?

Use linearity as part of the definition of the directional derivative

Ted Shifrin

No. Linearity of the derivative map.

Matt N.

12:25 AM

@TedShifrin But he seems to assume that already (first sentence of his question). (?)

Ted Shifrin

No. See my comments above.

Daniel Fischer

@TedShifrin You have an unclosed parenthesis in that answer, "(In particular ..." is never closed.

Pedro Tamaroff

@TedShifrin Well, $S^1$ is my first compact abelian group.

user1534664

@robjohn Thanks, I just found out I dont know how to divide polynomials :P

Daniel Fischer

@PedroTamaroff What about $\{e\}$?

Or $S^0$?

Mike Miller

12:27 AM

@Pedro You know lots.

Pedro Tamaroff

There's an affectional (?) issue, you see.

Mike Miller

$\Bbn Z/n\Bbn Z$, or maybe even products of those.

Matt N.

@TedShifrin You think that is the directional derivative in the first sentence of the question?

It's making less and less sense : (

Pedro Tamaroff

@MikeMiller I didn't say it's the only one I know.

Mike Miller

You said it was your first!

Ted Shifrin

12:28 AM

@DanielF I doubt that lacking parenthesis destroyed readability, but thanks.

Pedro Tamaroff

imgur.com/PYgiZPm

My kilometer counter today.

Mike Miller

That number's great. It's $1700+29$. I could also write it as $200+1529$.

Ted Shifrin

I see everyone is in a triviality mood tonight.

Daniel Fischer

@TedShifrin Well, it was not bad enough to make me consider unupvoting.

Matt N.

lol

Pedro Tamaroff

12:29 AM

@TedShifrin Really?

@MattN. WHAT THE HELL. HELLO THERE:

Mike Miller

OK time to get back to work

Twink

I don't wanna be alone forever, but I can be tonight.

I don't wanna be alone forever, but I love gypsy life.

Ted Shifrin

Heya @Twink.

Matt N.

@TedShifrin So are you saying (in your answer) that he must use linearity as part of the definition of the directional derivative (and cannot derive it from the definition of Frechet derivative)?

Pedro Tamaroff

@Twink Don't spam please.

Twink

12:31 AM

'Cause I'm, I'm, I'm, I'm, I'm, I'm a Gypsy, Gypsy, Gypsy,
I'm, I'm, I'm, I'm, I'm, I'm, I'm A Gypsy, Gypsy, Gypsy, I'm

I'm not spaming I'm singing

¬¬

That's spamming.

Oy ... Hush, @Twink

whatever

@MattN: You just are not listening/reading. I'm saying one needs a differentiable function, in general, to have linear (sums) directional derivatives.

Matt N.

@TedShifrin But he is assuming differentiability already in his question (first sentence), or is he not?

I have to go. bbl

robjohn

12:36 AM

@Twink That is sort of a subjective decision, and if someone who can kick you thinks you are spamming and not singing...

@TedShifrin You mean smooth partials?

Pedro Tamaroff

@anon Do you know some basic stuff about algebraic geometry?

anon

not really

Pedro Tamaroff

@anon Me neither.

@anon So, you're teaching! What are you teaching?

anon

I have three Intermediate Algebras, one College Algebra, and am assisting with one Abstract Algebra

One can define the "p-adic" fractional part $\{x\}_p$ of a rational number x as what you get when you truncate the base-p expansion to only those things past the period symbol. Then $x-\sum_p\{x\}_p$ will be an integer. Indeed, writing $x=n+\sum_p\{x\}_p$ is the arithmetic analogue of partial fraction decomposition for rational functions, and so $n$ is the analogue of the "asymptotic" component.

At first I was wondering if there is some special information encoded in this $n$ (about $x$), but now I realize that $n(x)$ (treat it as a function) depends very much on which coset representatives we use for Z/pZ in the base expansions (they're somewhat unnatural). I've considered treating $n$ as a function $\Bbb Q\to$ the profinite integers, and using Teichmuller representatives, but I'm not sure how to go about that or if this is a viable thing to think about.

Ted Shifrin

12:53 AM

No, I mean there's a linear map that's a good approx ... @robjohn

Pedro Tamaroff

@anon Sorry I was distracted. Reading...

Tyler Murry

Can someone help with an explination to my question? http://math.stackexchange.com/questions/994336/mtg-probability-of-drawing-a-card-with-enough-mana-to-play-it
I have been trying to figure out how he got his answer (see comments)

Pedro Tamaroff

@anon You lost me at profinite integers.

Hehehe.

anon

@PedroTamaroff think of it as $\prod_p\Bbb Z_p$

Pedro Tamaroff

@anon OK.

anon

12:55 AM

since $\Bbb Z_p$ is kind of like $\Bbb Z/p^\infty\Bbb Z$, by CRT that's kind of like $\Bbb Z/(\rm everything)\Bbb Z$

Pedro Tamaroff

Teichmuller?

@anon What is the formal definition of $\widehat {\Bbb Z}$?

Mike Miller

@Pedro product of the p-adic integers over all p

anon

@PedroTamaroff they are a nice set of representatives of the nonzero elements of $\Bbb Z_p/p\Bbb Z_p$: namely, since this is $\cong\Bbb F_p$ and all units in $\Bbb F_p$ are $(p-1)$ roots of unity, the $(p-1)$ roots of unity in $\Bbb Z_p$ can be used as a set of coset representatives for $\Bbb Z_p/p\Bbb Z_p$. As such, we can then write all $p$-adic expansions using them as digits instead of the usual $\{0,1,2,\cdots,p-1\}$.

Mike Miller

equivalently inverse limit of Z/nZ

anon

@PedroTamaroff the inverse limit $\varprojlim \Bbb Z/n\Bbb Z$

~~if I can remember how to latex it...~~ remembered

Mike Miller

12:58 AM

hashtag rekt

Pedro Tamaroff

Oh, never mind inverse limits.

I do remember direct limits, though.

@anon OK.

anon

you can also think of it as the subring of $\prod_n \Bbb Z/n\Bbb Z$ comprised of all vectors where projecting the $\Bbb Z/n\Bbb Z$ coordinate down to $\Bbb Z/d\Bbb Z$ (whenever $d\mid n$) yields the $\Bbb Z/d\Bbb Z$ coordinate.

Mike Miller

the stone-cech compactification is left-adjoint to the forgetful function CHaus $\to$ Top. is there a similar statement for the one-point compactification?

rather can you describe the one-point compactification by some universal property

Alizter

When are inverse limits studied?

Pedro Tamaroff

@Alizter When you algebra. Also when you algebraic number theory.

Alizter

1:13 AM

@PedroTamaroff Oh. I always thought it was to do with category theory.

Pedro Tamaroff

Well, there is a broader notion in category theory, of colimits and limits.

@MikeMiller

"The one-point compactification is universal among all compact spaces into which X has an open embedding, so it is essentially unique."

That's in nCatLab

anon

so all open embeddings into compact spaces factor through the embedding into the one-point compactification, I take it

Mike Miller

other way, one-point compactification factors through any open embedding into a compact space, I think

visualize $X=\Bbb R^2$ embedded into the sphere as the lower hemisphere

anon

oh

Mike Miller

so it's the terminal object of the category whose objects are open embeddings of $X$ into a compact Hausdorff spacd

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$Mathematics$ Mathematics