Mathematics - 2016-09-29 (page 1 of 4)

Semiclassical

12:01 AM

Am I being thick in this comment? math.stackexchange.com/a/1945774/137524

Simple Art

@usukidoll

24

Q: How to deal with "discomforting" downvotes?

I have noticed that recently the downvoting spree on MSE has increased by a lot. It often occurs (also here and here) to me that, driven by my will to help people, I provide a solid proof/hint and suddenly It gets downvoted because it is "too advanced"; It gets downvoted because the OP did not ...

usukidoll

still not gonna risk it

Simple Art

Do not worry about downvoting, they do not matter ;)

usukidoll

post it on your account :)

Simple Art

well, they matter a little

but its worth it

usukidoll

12:03 AM

not gonna get messed up nope

Simple Art

not my question

usukidoll

no! So I'm just on b and c now.. yay

not on my account

nope

points matter

Simple Art

@usukidoll How does nature grow if it does not learn from mistakes?

Look at me

the experience is worth it, always

usukidoll

you take the risk. Besides it's mark wi so if there's too many mistakes I can always re-write it

re-writes are only allowed if the problem is marked wi for writing intensive

Semiclassical

my hesitation for posting questions is less "am I going to get downvoted for this" and more "do i expect anyone else to be interested in this"

@usukidoll What is $S\leq R$ supposed to mean in this context? (this isn't a socratic question, i just genuinely don't know what's intended)

usukidoll

12:07 AM

semi could you give me a hint on 6b?

Semiclassical

abstract algebra isn't my strong suit, so probably not.

usukidoll

well I was told that I just use the subring definition on 6a which is what I just did 2 add together 2 products identity additive inverse

which is what I did

I will mercio was back here but he's probably po'ed

Simple Art

@Semiclassical I post my questions regardless

@usukidoll Also know that your questions can attract new users looking for solutions to the same question

that is how this site grows

usukidoll

:/

Simple Art

Google search "why do we teach complex numbers" and see one of the top links is a question of mine for example

usukidoll

12:09 AM

for part c define a new subring from part a it turns out to be hte same subring as Z U R what the heck

Simple Art

@usukidoll Do you use math.SE for homework, or to learn?

usukidoll

first option but only if I attempt the problem first

Semiclassical

6b does make sense in its conclusion, though. you take the subring generated by r, and take linear combinations whose coefficients are in Z

Simple Art

@usukidoll Then how will you learn to completely do it yourself?

Also, experience by answering other people's questions is quite good

usukidoll

dude. I do the problem first before asking -__-

Simple Art

12:12 AM

And very good for rep

Yeah, I understand, but asking it as a question will allow you to gain more from it

just saying

Semiclassical

again, though, what do they mean when they write $S\leq R$? that's not a notation i'm immediately familiar with

Simple Art

Ah, memories... of when I first joined this site

usukidoll

@SimpleArt true

Semiclassical

i could believe it just means "show that S is subring of R"

Simple Art

it was exhilarating in a very nerdy fashion

usukidoll

12:14 AM

yes S is a subring of R

Semiclassical

mmkay

usukidoll

which means closure under addition, closure under multiplication, identity, additive inverse lalalallal

Semiclassical

i think maybe it's simply enough to note that the subring S certainly contains $\mathbf{Z}\cup \{r\}$

and therefore also contains any element that could be generated by it.

usukidoll

how though? Isn't that subring or the element R

in set notation

OneRaynyDay

Hey guys - can you suggest a good lin algs book that features LU factorization?

Simple Art

12:15 AM

@OneRaynyDay Hey and no sadly

Semiclassical

@usukidoll i'm sorry, but I don't follow

usukidoll

isn't {r} just a single element

Semiclassical

well, certainly

Simple Art

Not always?

I guess so

pft, ignore me XD

Semiclassical

but it's also present in S.

usukidoll

12:17 AM

so this is what I'm thinking in set theory you either have a single element r or the subring Z

so how does this connect to Z[r]=S
Which is the subring r is equal to S which is that monster thing

Simple Art

use magic

OneRaynyDay

pls

Semiclassical

you're not making sense. $r$ is in $S$, and so is every member of $Z$

Danu

@TedShifrin So I think I've understood blowing up a linear subspace. Now, my remaining problem is that I'm unable to parse Huybrechts' explanation for the case of a general submanifold...

Semiclassical

so therefore, every member of $Z\cup \{r\}$ is in $S$.

usukidoll

12:19 AM

so the single element r and the subring of Z is in S

Semiclassical

right. that's easily checked (i.e. you can easily choose coefficients a_k to get those cases)

usukidoll

mmk. sorry I'm uber slow at proofs but I'm still trying and not giving up

so choosing the coefficients a_k we have the single element r and the subring of Z in S...hmmm

Semiclassical

probably the logic then is that, since S contains $Z\cup \{r\}$, it also contains anything that these could possibly generate

Simple Art

@usukidoll Keep trying and only give up when the entire math.SE site can't help you

usukidoll

like you ^ jk

Simple Art

12:21 AM

Yup XD

usukidoll

which had S = $a_{0}+a_{1}r...$

Semiclassical

what bothers me is that I feel like that only proves $\mathbf{Z}[r]\leq S$, not $S=\mathbf{Z}[r]$

usukidoll

we can just try and do that case anyway

Semiclassical

though if one can prove that $S\leq \mathbf{Z}[r]$ as well, then one does have $S=\mathbf{Z}[r]$

don't know how to argue that, though.

usukidoll

so that would mean that the S which was that definition is equal to the subring of Z?

john

12:24 AM

test

right.

who is john

hey usuki

It's more subtle a question than I can really answer, though.

john

why doesn't Latex render in chat?

usukidoll

12:25 AM

you need the chatjax and mathjax bookmarkz

Simple Art

lol

john

usuki do you see Latex?

Semiclassical

check out the Latex in chat link in the room description

Simple Art

top right

john

oh

Simple Art

12:26 AM

Say, I can't get the Latex to work

usukidoll

so if we had $S \leq Z[r]$ then that would mean that the definition of S is less than the subring of R

Simple Art

anyone got tutorial links or wanna explain?

usukidoll

math.ucla.edu/~robjohn/math/mathjax.html follow

Simple Art

@usukidoll yes, but I can't bookmark it

usukidoll

get out your bookmarks toolbar in firefox

john

12:28 AM

thanks

it works now ;)

Simple Art

Lol, ok

@usukidoll will chrome work?

usukidoll

so what about for c? prnt.sc/cnkpqo it's almost as if it's a all over again with some new letters

I think so

Semiclassical

it's the set theory version of an obvious technique---$A\leq B$ and $B\leq A$ implies $A=B$

I really don't get what c is asking. In particular, I don't see what $r$ is supposed to satisfy.

usukidoll

let's guess on c

xD

Simple Art

lol

john

12:29 AM

$S \leq Z[r]$ means S is a subset of Z[r] ?

Semiclassical

subring, is my guess

usukidoll

the hint I got for c is to define a new subring and then part a is the same subring as Z U R

john

instead of using the $ \subseteq$ notation they use $ \le $

Simple Art

Now what do I do?

Semiclassical

My guess is that they meant $c_0+c_1r+\cdots c_{d-1} r^{d-1}+r^d=0$.

Simple Art

12:31 AM

Woo! Did it!

thanks

usukidoll

oh so if I could find the technique from the set theory I could apply it to b.
like for example if $S \leq Z[R]$, and $ Z[R]\leq S$ then $ S=Z[R]$

yeah they needed to add a 0

I'm sorry this worksheet is laced with typos... (not by me)

Semiclassical

You probably want to check that with your lecturer. It's a real arse of a typo.

usukidoll

another professor I talked to said it needs the = 0 or it doesn't make sense

Semiclassical

yeah.

usukidoll

isn't that a polynomial?

Semiclassical

12:33 AM

sure. i think it's better to think in terms of linear algebra here, though

john

What is an example of a ring which has Z as a subring ?

I was thinking the real numbers.

Semiclassical

@john Z isn't intended as integers here.

john

oh

Semiclassical

it's just a name for some subring of a given ring R

Simple Art

Confusion is real over here

john

12:34 AM

hmm, Z is usually reserved for integers

thanks for clarifying

Simple Art

Sh, this is linear algebra apparently.

Semiclassical

44 mins ago, by usukidoll

finishing this http://prntscr.com/cnkpqo

Simple Art

Hm

Semiclassical

point being, $r^d$ is evidently a linear combination of $1,r,r^2,...r^{d-1}$

Simple Art

robjohn said: "Since $0=0^2$, $0$ is also a square. Does this mean that a circle can be a square?"

:)

Semiclassical

12:36 AM

so $1,r,r^2,...r^d$ are linearly dependent (with coefficients in Z)

Adeek

hi

john

how do you change your profile picture

Adeek

@Semiclassical would you like to discuss algebra problem

john

@Adeek here prnt.sc/cnkpqo

Simple Art

@john go into your profile

john

12:37 AM

im there

Simple Art

and hit the edits area

Semiclassical

@usukidoll I think the upshot is that, if you can generate $r^d$ using $1,r,...,r^{d-1}$

Adeek

@Semiclassical ?

Semiclassical

@adeek no thx

john

@SimpleArt i dont see any option i.imgur.com/jIJyMwK.png

Semiclassical

12:38 AM

@usukidoll in that case, you don't need to allow arbitrarily large exponents of $r$ in the definition of $S$

Simple Art

@john enter one of the sites you actually use

usukidoll

like a linear combination. damn it it's been a while since I've done linear algebra that way :S

Simple Art

like math.stackexchange.com

Semiclassical

you only need to use 1,r,...,r^{d-1}

Simple Art

then profile -> edits

john

12:38 AM

ok

Simple Art

should be top thing pretty much

usukidoll

ok like using $1,r,r^2,...r^{d-1}$?

Semiclassical

right.

another linear algebra POV

suppose I've got a vector like $e_1=(1,0,0)^T \in \Bbb{R}^3$ and I act on it repeatedly with a matrix $T$

john

test

Semiclassical

to get the sequence of vectors $e_1,Te_1,T^2e_1,$ etc.

Keith Yong

12:42 AM

Has anyone done stuff on primitive recursive before? This is my homework question and I dont know how to tackle the sqrt(2) part

http://i.stack.imgur.com/2dKlz.png

Semiclassical

then I can use arguments about linear dependence to argue that I can build up anything past $T^2e_1$ using just the first three

john

what is definition of primitive recursive

I would start there

Semiclassical

@usukidoll Does c make more sense now?

usukidoll

so it's linearly dependent is it also possible to get ... oh wait a sec we're going beyond $R^{3}$ if I do $T^3e_{1}$

Simple Art

@john Perhaps you saved the edits only for that site. There should be an option to save edits for all networks/communities

Semiclassical

12:43 AM

right. but we could do $R^{10}$, say

Keith Yong

@john Sorry its a pretty long definition, I'm just throwing that question out there in case anyone has done stuff with it

john

simple, ill try again

keith, no worries.

Semiclassical

in principle, I'd need to get 10 vectors like that to get a set which spans the rest of them

Keith Yong

not expecting any answers haha

usukidoll

wow

so C uses matrices... I kind of got confused since it looked like a polynomial

Simple Art

12:45 AM

Ugh, that's gonna be great

Semiclassical

ehh

it's not that it uses matrices.

john

@SimpleArt i tried saving to all communities

Semiclassical

it's that it's using the same idea of linear dependence as you could encounter in linear algebra

john

maybe it takes time to update

Semiclassical

You don't need matrices for that.

usukidoll

12:45 AM

so I just need the idea of linearly dependent

Simple Art

@john Probably

usukidoll

I took linear algebra a whilllllllllle back

2

Semiclassical

Fair enough.

The point is just: The definition of $S$ in (a) permits you to use as many coefficients as you want.

usukidoll

a set of vectors is said to be linearly dependent if one of the vectors in the set can be defined as a linear combination of the others
ack latex in matrices is hard but
if we have something like

a + b = a+b
0 0 0
is that a linear combination

Semiclassical

sure.

What (c) says is that you can be more economical than the definition of (a).

usukidoll

12:48 AM

ah it's starting to come back

Simple Art

XD

economical?

Semiclassical

sure.

you don't need arbitrarily many coefficients if just $d$ is enough.

Simple Art

@john Why are you here btw?

john

i was invited

Simple Art

Oh?

How so?

You do realise this is a math chat?

usukidoll

12:49 AM

umm let's see so
$c_{0} \cdot r = c_{0}r$
$c_{1} \cdot r^{2}=c_{1}r^2$
$c_{d-1} \cdot r^{d-1} =c_{d-1} r^{d-1} $ like that?

john

Yes.

usukidoll

oy wait a sec should be ()()

Simple Art

._. when you think 'sec' means secant

Semiclassical

Here's maybe something that'll make this seem more practical.

Simple Art

T_T

Hm, if a function is $k$-times differentiable, what happens when we try to apply L'hospital's rule? Does it work?

Semiclassical

12:52 AM

Suppose I have an element of $S$, such as $1+1r+0r^2+1r^3$ where $1$ is the additive identity.

that required four coefficients.

john

why wouldn't L'hopital's rule work?

Semiclassical

suppose someone then told me that, in this particular ring, $r^3+r+r^2=0$.

usukidoll

does 1 work when the exponent is odd and then 0 when the exponent is even?

Keith Yong

damn my last semester of comp sci is so stacked with math. non stop math. someone kill me xd

john

if its an exponential function, taking derivative doesn't reduce the power

Semiclassical

12:54 AM

in that case, I can rewrite the above as $1+r+(-r-r^2)=1+0r+(-1)r^2$.

so i really only needed 3 coefficients, not 4.

Simple Art

@john it at the very least, can't work after $k$ iterations if the function is only $k$ times differentiable

@KeithYong Sorry, that's a question for education.stackexchange

Semiclassical

the same would also hold if I had something like $1+r+r^{9999}$. at first glance, that requires 10000 coefficients (most of them 0).

Simple Art

;)

usukidoll

because
1
r
(-r-r^2) are the three coefficients right

john

do you have a specific example in mind. its hard to speak about it generically

Semiclassical

12:56 AM

@usukidoll no.

the coefficients are 1,0,-1

usukidoll

oh right right.....

because 1 +r-r-r^2
1+0r-r^2
1 0 -1
sorry spaced out

Simple Art

@john No, the question is in general

Semiclassical

if I had r^4, i can write that as $r(r^3)=r(-r-r^2)=-r^2-r^3=-r^2-(-r-r^2)=r$

similarly for r^5, r^6, etc. I can write all of these just using some combination of 1, r, and r^2

usukidoll

geezus only now I figured out part b... gawd I hate that $ \leq, \geq $ notation

Semiclassical

so therefore, I don't need all 100 coefficients to express something like $1+r+r^2+\cdots +r^{99}$. there must be coefficients such that $$1+r+r^2+\cdots +r^{99}=c_0+c_1 r+c_2 r^2$$

doesn't mean I know them off the top of my head, but I know they exist.

on the other hand, if I didn't know $r^3+r^2+r=0$, it could be that i really do need all those coefficients

Simple Art

1:01 AM

@SamuelYusim Sup

@Semiclassical Are you using polynomials of $r$?

john

@Simple Art

here is an example (e^x - e^(-x))/ (e^x + e^(-x))

Semiclassical

yeah, though mentally I'm thinking of them just as vectors in a vector space. (probably i should be thinking modules not vectors spaces, but w/e)

usukidoll

@SimpleArt more like vector space

john

$ \lim_{x \to \infty} \frac{ e^x - e^{-x} }{ (e^x + e^{-x}}$

$\large \lim_{x \to \infty} \frac{ e^x - e^{-x} }{ e^x + e^{-x}}$

Simple Art

@john Um, you sure that's not infinitely differentiable?

Semiclassical

1:04 AM

not sure why you'd need lhopital for that

Simple Art

Yeah, what he said ^^

john

you can apply Lhopitals repeatedly and get no closer

Simple Art

Uh...

True...

usukidoll

I'm starting to head over to b again with the $Z[r] =S$ stuff which is actually if $ S \subseteq Z[r]$ and $ Z[r] \subseteq S$ then $Z[r]=S$ which in set theory the elements are the same

Simple Art

But... that's not the question

john

1:05 AM

your condition is that it is only k times differentiable

Semiclassical

@usukidoll I can't help there, I think.

john

hypothesis*

Simple Art

I'm considering functions that are only $k$-times differentiable, not infinitely differentiable.

Infinitely differentiable I already understand

Semiclassical

@SimpleArt A concrete example would probably help.

Simple Art

Lemme think for a moment

Semiclassical

1:07 AM

something in the direction of $|x|^3$, maybe

usukidoll

prntscr.com/cnl97s

john

ax^k / bx^n, where n > k

Simple Art

Yeah, that's a good one

Semiclassical

need it to be an odd exponent for the absolute value to make a difference

usukidoll

Z[r] is a subring is contained in S
S is contained in Z[r] which is a subring
so the elements are the same hmmm

Simple Art

1:08 AM

Consider: $\lim_{x\to0}\frac{|x|^3}{|x-4|^3+64}$

With L'Hospital's method

Hm, maybe should've used different powers, but oh well

Semiclassical

@usukidoll I think the term module is relevant here

should that be $-64$?

Simple Art

Nah, $+64$ to make the denominator $0$ as $x\to0$.

ummmmm

Oh

|-4|^3+64 != 0

1:10 AM

Right

minus 64

Pft, confusing absolute values

Semiclassical

right

Simple Art

>_>

Semiclassical

hah

I don't think that example gives you what you want, though.

john

What was the statement again, I didn't catch it earlier.

Semiclassical

after all, $|4-x|=4-x$ for $x$ near $0$

Simple Art

1:11 AM

yeah, probably right

john

Something about k differentiable functions.

Simple Art

I probably want some piece-wise functions

john

that limit is zero as x goes to zero

Simple Art

some limit @john not really interested

I'll make a better one, gimme a second

Semiclassical

@john what he intended was $\frac{|x|^3}{|x-4|^3-64}$

john

1:13 AM

yes. The limit is zero

Semiclassical

but the denominator of that one is analytic at $x=0$, so it's not a good example anyways.

Simple Art

$\lim_{x\to0}\frac{f(x)}{|x|^3}$ where $f(x):=\begin{cases}x^3 & ;x>0 \\ x^5 & ;x<0\end{cases}$

0celo7

Sanity check: one can have an injective linear map $\Bbb R^n\to E$, $E$ some infinite-dim vec sp.

Semiclassical

the limit from the right is 1, the limit from the left is zero.

Simple Art

XD

So L'Hospital's rule can't work?

How many times can i apply it?

Semiclassical

1:14 AM

didn't say anything either way.

i'm just eyeballing it.

Simple Art

lol. also a bad example

whatever, i'll pioneer tomorrow

ciaou

john

a non analytic k differentiable smooth function

0celo7

" k differentiable smooth function" does not make sense

Simple Art

@john yes, those

john

$f(x)=
\begin{cases}
e^{-1/x}& x> 0\\
0& x\le 0
\end{cases}$

Simple Art

1:15 AM

k-times smooth

0celo7

^ not an actual term

john

$ f(x)= \begin{cases} e^{-1/x}& x> 0\\ 0& x\le 0 \end{cases}$

$1=\lim_{x\rightarrow 0^{+}}\frac{f(x)}{f(x)}=\lim_{x\rightarrow 0^{+}}\frac{f'(x)}{f'(x)}=\lim_{x\rightarrow 0^{+}}\frac{f''(x)}{f''(x)}\cdots =\frac{0}{0}=?$

Semiclassical

@0celo7 sure, $(x,y,z)\mapsto (x,y,z,0,0,\cdots)$ being an obvious example.

kind've feel like i'm sticking my foot in my mouth there. not sure why.

Danu

When someone says "smooth divisor", does it mean a divisor of the form $D=[Y]$ where $Y$ is a smooth hypersurface?

Semiclassical

math paranoia, i guess

john

1:20 AM

whats that semiclassical, an example of

what is that an example of

@Semiclassical

Semiclassical

6 mins ago, by 0celo7

Sanity check: one can have an injective linear map $\Bbb R^n\to E$, $E$ some infinite-dim vec sp.

Danu

Hi @MikeMiller

usukidoll

well I'm just gonna finish 6c up and just eat lunch :P

at least I tried. would rather get partials than 0 but like I said the problems with WI are eligible for a re-write

so yeah

0celo7

@Semiclassical Yes.

john

oh nice

my pfp updated

usuki why is your pfp so much bigger

usukidoll

1:30 AM

idk?

k this bs assignment is done

really hated #6

Mike Miller

1:50 AM

@0celo7 I don't understand your question.

Hi @Danu

0celo7

Which question?

@MikeMiller I do have a question, though. Kobayashi-Nomizu say that if a Lie group $G$ acts effectively on $M$, then the map $\mathfrak g\to\mathcal{X}(M)$ that sends $A$ to its fundamental vector field is an isomorphism. They prove that it's injective, but I fail to see how it's surjective.

$\mathcal{X}(M)$ is infinite-dimensional unless I'm messing something up.

Perhaps they just mean an isomorphism onto its image.

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$Mathematics$ Mathematics