Mathematics - 2018-10-21 (page 2 of 2)

tarit goswami

14:00

i mean on sets

@LeakyNun I only know the axioms

Leaky Nun

and vector spaces?

tarit goswami

yeah

Leaky Nun

what's a vector space?

Silent

@LeakyNun I am sorry, i did not see this then. Which generalization were you talking about?

tarit goswami

Vector space is a set with a associated field and satisfies some specific peperties

Leaky Nun

14:03

the thing that we did is a generalization @Silent

@taritgoswami there's more data to a vector space

Silent

You mean $0\le \sum\frac{\sqrt{a_n}}n\le\biggl(\sum a_n\biggr)^{1/2}\biggl(\sum\frac1{n^2}\biggr)^{1/2}.$?

Leaky Nun

right

Silent

how so?

Leaky Nun

well of course this inequality itself is not a generalization

I'm talking about the new theorem we proved

(hmm, looks like many people can't generalize theorems)

tarit goswami

@LeakyNun Sorry there was a network error from my end. Vector space is a set with a associated field satisfying some properties, like it forms commutative group under addition etc.

Leaky Nun

14:06

@taritgoswami there's some data in the "etc"

tarit goswami

@LeakyNun Action of field on the set?

Leaky Nun

right

but in the language of linear algebra we often call it a scalar multiplication instead

so this gives you one interpretation of group action

tarit goswami

yes

user193319

0

Q: Uniform Limit of Integrable Functions

If $\{f_n\}$ is a sequence of bounded measure functions on $[a,b]$ and $f_n \to f$ uniformly, then $f$ is integrable. This is actually proved in the book I am working through, but I came up with a proof that looks quite different. Naturally, I was hoping someone could critique my proof and...

Leaky Nun

i.e. that it's some sort of scalar multiplication, but the scalars form a group instead of a field, and the set doesn't need to have addition

Silent

14:08

@LeakyNun I think you are talking about this generalization: 'series of nonnegative terms converges iff partial sums are bounded sequence'

Leaky Nun

@Silent brilliant.

and one can still generalize this, but to a lesser extent

Silent

:) thanks

Leaky Nun

à la squeeze theorem

@taritgoswami another interpretation is that it is a function that assigns each $g \in G$ to a permutation on the set, such that the function composition respects group multiplication.

i.e. a group homomorphism $G \to \operatorname{Sym}(X)$.

tarit goswami

cool

@LeakyNun I am not familier with the notation $Sym(X)$

You mean Symmetric group ?

Leaky Nun

yes

Silent

14:14

@LeakyNun because after all series is a sequence, hence squeeze theorem holds for partial sum sequence? statement like that?

Leaky Nun

@Silent I'm talking about a generalization of the squeeze theorem for series

Silent

sorry, that's not generalization!

oh

@LeakyNun i think you are referring comparison test

Leaky Nun

hmm

aha

ah i'm an idiot :P

Silent

why?

Leaky Nun

yes it's just the comparison test

Silent

14:18

ok

MatheinBoulomenos

@ÍgjøgnumMeg hi

nicht dass ich wüsste

ÍgjøgnumMeg

@Mathein Komisch, es kommt auch oft vor also kann's doch kein rechtschreibfehler sein

egal, danke :D

MatheinBoulomenos

Im Internet erlebt eben die Schriftlichkeit ihre triumphale Renaissance bar jeder Orthographie - ich kozze ab

ÍgjøgnumMeg

lol

Kann sein dass es aber irgendeine regionale Buchstabierung ist oder so

MatheinBoulomenos

@LeakyNun the comparison test for series is just dominated convergence for the counting measure

ÍgjøgnumMeg

14:24

er schreibt zwar auch "trokken" und "rükkenmark"

Leaky Nun

@MatheinBoulomenos und kommt der Generalizationmeister

MatheinBoulomenos

Verallgemeinerung = generalization

ÍgjøgnumMeg

Der Verallgemeinerungsherr

MatheinBoulomenos

@ÍgjøgnumMeg was ist das für ein Text?

ÍgjøgnumMeg

@Mathein der Stille Ozean heißt das Buch

vom Gerhard Roth, österreichischer Schriftsteller

MatheinBoulomenos

14:27

kommt das nur in direkter Rede vor? Vielleicht gibt er irgendeinen Dialekt wieder

ÍgjøgnumMeg

Nöööp

Leaky Nun

@MatheinBoulomenos for $f \in \Bbb C[X_1 \cdots X_n]$, when is $V(f)$ a manifold?

MatheinBoulomenos

hmm, keine Ahnung

ÍgjøgnumMeg

ja ich auch ned hahaha

egal

MatheinBoulomenos

@LeakyNun when $V(f)$ is non-singular

Leaky Nun

14:31

what does that mean for the coordinate ring?

MatheinBoulomenos

it's basically defined to make this statement true via the implicit function theorem. $V(f)$ is non singular if for all $x \in V(f)$, the Jacobian $f'(x)$ doesn't vanish, i.e. not all partial derivatives vanish simultanously

Leaky Nun

so it's not sufficient that the ring reduced und irreducible seit?

@MatheinBoulomenos how would you compute $\sqrt{-6} \in \Bbb Z_7$?

say, to $O(7^{10})$

I know you wouldn't, but wie wuerdest du?

MatheinBoulomenos

@LeakyNun no, consider e.g. $V(y^2-x^2(x+1))$, that's irreducible by Gauss' lemma and Eisenstein

Leaky Nun

you mean $y^2-x^2(x+1)$

MatheinBoulomenos

yeah

sorry

Leaky Nun

14:39

you don't need to apologize

user76284

What's a natural way to define the norm of an element of O(n)? I assume it can be based on the norm of the Lie algebra (e.g. the matrix logarithm of the identity matrix is the zero matrix)?

MatheinBoulomenos

@LeakyNun what do you mean by $O(7^{10})$?

Leaky Nun

O as in order of magnitude

i.e. to ten digits

MatheinBoulomenos

take the proof of Hensel's lemma, it is constructive

basically a variant of Newton's method

Leaky Nun

alright

MatheinBoulomenos

14:46

Hensel's lemma

In mathematics, Hensel's lemma, also known as Hensel's lifting lemma, named after Kurt Hensel, is a result in modular arithmetic, stating that if a polynomial equation has a simple root modulo a prime number p, then this root corresponds to a unique root of the same equation modulo any higher power of p, which can be found by iteratively "lifting" the solution modulo successive powers of p. More generally it is used as a generic name for analogues for complete commutative rings (including p-adic fields in particular) of the Newton method for solving equations. Since p-adic analysis is in some ways...

in the formula for the Hensel lift you need to invert mod $p^k$ which can be done quickly with the extended Euclidean algorithm

it's interesting that the idea of an algortihm from numerical analysis can be used to prove an important lemma for algebra/number theory

user76284

Nevermind, found it.

ÍgjøgnumMeg

15:16

@Mathein es handelt sich um einen Zeilenumbruch! Anscheinend schreibt man statt "Brüc-ke" einfach "Brük-ke"

vielleicht in einer älteren Orthographie oder so aber wow

MatheinBoulomenos

aha!

wusste ich nicht

ÍgjøgnumMeg

Ich auch nicht, voll komisch

MatheinBoulomenos

nach neuer Rechtschreibung wird "Brü-cke" getrennt

ÍgjøgnumMeg

ja genau

aber vor 1996 schrieb man "Brük-ke"

why tf

Akash

I need help..please answer my question..just posted

Martin Svanberg

16:06

If $\mu$ is the Lebesgue measure and $\nu=d\mu/(x^a+1)$, what does $\nu$ actually mean? I understand that it's a measure, but I don't understand how it's defined.

Pig

you can think of measure as a linear functional on a space of test functions (by integration)

So the measure $\nu$ can be thought of as the functional $f \to \int f d \mu/(x^a + 1)$

alternatively (but equivalently), if you like to think in terms of usual measures more, just think of $\nu(A) = \int 1_{A} d\mu/(x^a+1)$ for a measurable set $A$

Martin Svanberg

Okay, thanks @Pig!

Pig

sure!

Alessandro Codenotti

The keywords to look up are "Radon-Nikodym derivative" and "absolutely continuous measure"

Pig

this (very strictly speaking) isn't quite true, since he's not trying to define the derivative, but just trying to understand what the measure even means

(but i'm probably being obnoxious here :P )

Alessandro Codenotti

16:21

Sure but if he wants to find out more and why such a measure would be of interest that might be useful

Pig

sure

Li357

If I need to write a formal proof that involves an implication where the hypothesis is false, can I just say the hypothesis is false and that the implication is thus always true?

And nothing else

loch

@LeakyNun as long as it is non-singular yes (oh and if you want to be isomorphic over k then you want that there exists a k-rational point on the conic)

Leaky Nun

@loch interesting

loch

In fact the general claim is that a smooth projective curve of genus $0$ over $k$ with a $k$-rational point is isomorphic to $\mathbb{P}^1_k$

this is also the reason why rational parametrization (which one might learn in number theory) works for conics where you can find all the rational solutions to an equation like $x^2+y^2=1$, but not for elliptic curves (which have genus $1$)

user76284

16:36

Is the Riemannian norm on GL given by the Frobenius norm on its Lie algebra gl?

Martin Svanberg

@AlessandroCodenotti thanks!

B. Mehta

17:07

In the full subcategory of Ab consisting of groups without elements of order 4, I'm looking for f,g such that gf is regular monic but f is not

Any ideas? I know that g can't be monic, and that f has to be, and I know some examples of regular mono- and nonregular monomorphisms in this category but I can't fit everything together

Leaky Nun

@B.Mehta that's a strange category

B. Mehta

@LeakyNun It certainly is

Leaky Nun

what is regular?

B. Mehta

regular monomorphisms are those which arise as equalizers

Leaky Nun

nvm

too complicated lol

I don't care about that category :P

B. Mehta

17:13

I guess that's a reasonable response

Leaky Nun

lol

B. Mehta

If it's any easier, equalizers here are just maps which form a kernel - so the doubling map $\mathbb{Z} \to \mathbb{Z}$ is an equalizer since it is the kernel of $\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$

But the quadrupling map $\mathbb{Z} \to \mathbb{Z}$ is not an equalizer - the map $\mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$ doesn't exist in this category since $\mathbb{Z}/4\mathbb{Z}$ doesn't exist

Isana Yashiro

17:47

A very simple question: why divisor is defined to be positive integer in division algorithm? Is it for remainder to be non-negative($0\le r\lt n$, where n is divisor)?

Pig

no good reason

you can define division algorithm for negative integer with remainder less than absolute value of your divisor - no problem there at all

Isana Yashiro

@Pig thank you, let me think about it

Mike Miller

The crucial thing is that the remainder is $0 \leq r \leq n - 1$

Isana Yashiro

@MikeMiller so $r$ is more important than $n$ in the division algorithm?

Mike Miller

I mean, that condition depends on $n$

oh oops

as Pig said that should be $|n|-1$

Isana Yashiro

17:52

ok thank you~

Mike Miller

if $n$ is negative the inequality I wrote is nonsense - eg $n = -2$ it gives $0 \leq r \leq -3$ :P

Isana Yashiro

I was reading group theory and found that division algorithm is powerful in proving things

using the fact that you pointed out

@Pig May I ask that why the remainder has to be non-negative?

T_01

hello. i have no idea with the following task:
$\text{Let } G \text{ be a Group operating on } M, m \in M, S:= Stabilisator_G(m), \{\} \neq A \subseteq G \text{ a finite subset. Proove that } |(A^{-1}A) \cap S| \geq \frac{|A|}{|Am|}$
can anyone give me an idea how to solve this?

where $A^{-1}A = \{ a^{-1} a' | a, a' \in A \}$ and $Stabilisator_G(m) := \{ g \in G \mid gm = m \}$

Pig

18:18

@IsanaYashiro, it's less about non-negativity of remainder, but more about the uniqueness of quotient and remainder

requiring remainder to be $0 \leq r < |n|$ (which is more than non-negativity) is one way of enforcing remainder to be unique (hence forcing quotient to be unique too)

Isana Yashiro

@Pig thank you so much

I'm trying to give myself some explanation for those things I memorized and forgot

F.White

Hi. What elements of Z/mZ where m is odd square to 1?

is it just 1 and m-1?

Alessandro Codenotti

Is m prime?

F.White

I know its true if p is prime, but I make use of primality in that argument

Alessandro Codenotti

Try Z/15Z

F.White

18:26

Damn 4^2=16=1

thanks

Alessandro Codenotti

There are exactly 4 such numbers in Z/15Z by the Chinese remainder theorem

Isana Yashiro

@Pig Is it correct to say that division algorithm define a function from a divisor to its (only) remainder?

F.White

1,4,11,14

Alessandro Codenotti

Yep, note that 1 is 1 in Z/3Z and Z/5Z, 4 is 1 and -1, 11 is -1 and 1 and 14 is -1 and -1

F.White

Thats really cool

Alessandro Codenotti

18:29

There are two roots in Z/3Z and two roots in Z/5Z, by combining them in pairs you get 4 roots in Z/15Z, by the Chinese remainder theorem

(this works because 3 and 5 are coprime)

henceproved

18:52

hi everyone: Can we say that (in $\mathbb{R}^2$ )the differential form $dx \wedge dy$ is the infinitesimal are element dA which also gives info about the orientation?

i.e. is the purpose of differential form in calculus is to give orientation, as my usual lebesgue area measure doesn't have such property

Anderson Felipe Viveiros

Hello guys. What is the derivative of the exponential map $\exp:TM\to M$ on a riemannian manifold? Here I'm considering the exponential as a map of two variables, $p\in M$ and $v\in T_pM$, so the base point is not fixed...

Mike Miller

Along the 0 section it's the identity. For some random tangent vector this seems harder to answer, and perhaps relayed to things like the cut locus.

For instance if M = S^2, if you pick v to be a vector of length pi then you always end up at the antipodal point. So the derivative in the direction of the unit tangent sphere is zero. But it is nonzero in the direction orthogonal to this.

Anderson Felipe Viveiros

I'm trying to answer this: math.stackexchange.com/questions/2964953/…

Pig

19:15

@IsanaYashiro that depends a lot on what you mean by division algorithm - what you mentioned is much clser to what's called modular arithmetic, which is quite related to division algorithm but is also different

@henceproved i think that's a fair thing to say

Mike Miller

I mean it also gives information about volume

It just does both as opposed to say the notion of density

Pig

he was comparing with lebesgue area (with is more or less the same as density in this case i guess)

so orientation is likely the only extra bit

Buddhini Angelika

19:48

Can we regard {(1,0,0),(0,1,0),(0,0,1)} as a minimal generating set for a semidirect product between Z3 and (Z5 X Z5), where (Z5 X Z5) is the normal subgroup?

Leaky Nun

generating: yes

minimal: how do I know

what's your action of Z3 on Z5 x Z5?

Buddhini Angelika

Action by conjugation

Thanks

Leaky Nun

???

Karl Kronenfeld

clearly

jk

Buddhini Angelika

I'm considering all the possible Semi direct products between those two groups

Karl Kronenfeld

20:04

They're determined by all possible actions of the first on the second. Can you identify them all?

Buddhini Angelika

Hmm no, bit difficult... I'll think of an example to explain my question more clearly and come back later. Thanks a lot.

copper.hat

20:27

I need some help with my memory: Is there a general name for subject in mathematics where one has that some number $x$ satisfies $p(x) = 0$ for some polynomial $p$ of degree $n$ and then this means than any power $x^m$ can be expressed in terms of $1,x,x^2,...,x^{n-1}$?

I'm talking at the high school level...

Leaky Nun

@copper.hat division algorithm

but the subject is algebra

copper.hat

@LeakyNun: Thanks, I'm looking for something in between :-). I think we loosely refered to them as surds, but that it not quite correct.

Karl Kronenfeld

@BuddhiniAngelika aw, only a bit difficult

Martin Svanberg

Let $(X,\mathcal A,\nu)$ be a given measure-space. Let $a,b\in \mathbb R$, $X=(1,\infty)$ and $\nu =d \mu/(x^a+1)$, with $\mu$ being Lebesgue measure. For which values of $a,b$ is there a Cauchy sequence $f_n$ in the mean (over $X$) such that $\lim_n f_n(x)=(\sin x)/x^b$ almost everywhere on $X$?

If I understand this problem correctly, I should be looking at Cauchy sequences $f_n \to (sin x)/x^b$ and show that $\int \frac{\mid f_n-f_m\mid} {x^a+1} d\mu$ goes to zero for certain values of $a,b$.

But I'm slightly confused about how I should do it. Any hints?

user10478

20:52

Is it accurate to say that the usage of a derivative versus a partial derivative depends entirely on the input space of the thing being differentiated? So a function with n inputs and 1 output calls for partial derivatives, but a vector valued function or parametric derivative with 1 input uses ordinary derivatives (d's rather than ∂'s)?

user193319

21:04

If $f : X \to Y$ is a continuous map and $B \subseteq Y$, is it true that $f^{-1}(\overline{B}) = \overline{f^{-1}(B})$? Showing $\overline{f^{-1}(B)} \subseteq f^{-1}(\overline{B})$ is rather easy, but the other direction is giving me trouble, which makes me wonder whether the statement isn't generally true.

Karl Kronenfeld

@user193319 No it isn't true in general.

See if you can find a counterexample. What has to happen for the other inclusion to not hold?

user193319

21:21

@KarlKronenfeld I'm thinking $f(x) = x^2$ should break it, but I can't seem to find a set $B$ for which the conjecture fails.

Kasmir Khaan

22:24

@MatheinBoulomenos @LeakyNun Hi yall

Leaky Nun

hi

goddag

Kasmir Khaan

so leaky

Leaky Nun

ja

Kasmir Khaan

Am trying to prove this statment

an ideal P in R is prime iff R/P is an integral domain

R commutative with 1

so let me show you my proof

to show that R/ P is an integral domain , we have first that R/ P is commutative with 1

so half of the stuff is done

then we need to show no zero divisors to conclude this

in this quotient Ring, [a] = 0 iff a is in P

since elements of P are considered to be zero in R/I

so we consider [a] [b] = [ab] = 0 iff ab belongs to P

so we take a not in P

Leaky Nun

it's quite important to remember that 1 != 0 is one of the requirements of an integral domain

Kasmir Khaan

22:28

and b in R\ P

ab in P cannot happen

@LeakyNun how does that help us here leaky

Leaky Nun

it doesn't help you; it hinders you

Kasmir Khaan

so we get that R/I has zero divisors iff P is not prime

if we negate this statment we get what we want to prove

right?

@LeakyNun wake up :D

Leaky Nun

you still haven't dealt with 1 != 0

Kasmir Khaan

1 is not equal to 0

or what

Leaky Nun

you haven't proved that P is prime ideal -> R/P is integral domain

because you haven't shown that 1 != 0 in R/P

Kasmir Khaan

22:31

what is that symbol

1 != 0

Leaky Nun

$1 \ne 0$

user330477

22:44

is the whole complex plane $\CC$ a star-shaped domain.

Leaky Nun

@user330477 are you following Freitag?

yes, it is

and every point can be a centre

user330477

@LeakyNun No, I am following Gamelin.

Leaky Nun

I see

user330477

23:07

I am trying to prove mean value property of harmonic functions using MVP of circles and Cauchy Integral Formula. I am stuck on how to give justification for using Cauchy Integral Formula. Can anyone help me out?

Leaky Nun

23:57

just take a harmonic conjugate lol

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