Mathematics - 2017-07-06 (page 4 of 10)

Mathmore

08:05

@TobiasKildetoft Okay. This season I appeared at various premier institutes in my country for interviews to admission in PhD program. I have not got selected anywhere. Actually I am very slow at learning process. I met many students who know many more stuff than mine and in particular they only studied one subject, say Algebra, thoroughly and got a seat in the institute.

@LeakyNun Have I done it correctly?

Leaky Nun

@Mathmore I don't think so

Mathmore

@LeakyNun Sorry. I have to use $A \implies B$ is equivalent to $\neg A \lor B$. Thus $\neg p \implies p$ is equivalent to $\neg (\neg p) \lor p$ which is equivalent to $p \lor p$ which is equivalent to $p$.

Leaky Nun

good

Mathmore

@LeakyNun Extremely sorry for goofing up the $\lor$ and $\neg$.

@LeakyNun Which symbol should I use for "is equivalent to"? Will $\iff$ work? or $\cong$ or $\equiv$?

Leaky Nun

$\equiv$

$\iff$ and $\equiv$ are equivalent

but I would prefer $\equiv$ for equivalent

Mathmore

08:13

Hahaha... $\iff \equiv \equiv$

Okay thanks!

We can do this via truth tables also right?

Leaky Nun

@Mathmore sure

Mathmore

good

Gotta go! See you later. Bye..

Secret

@LeakyNun That's an interesting counterexample, cause $2^{n+1}-2^{n}=2^n (2-1)=2^n*1=2^n$ Note the $2-1=1$, nothing else behave like that

If we pick the sequence $(3^n)$ then it will also be a counterexample but it will failed to be a fixed point because $3-1=2$, thus you have some leftovers in the consecutive difference

Leaky Nun

@Secret $F_n$ is an interesting counterexample

Mathmore

Gotta go! See you later. Bye..

Secret

08:30

$F_{n-1}+F_n=F_{n+1}$

$F_{n+1}-F_n=F_{n-1}$

One term further back

Hmm, what happens if this consecutive difference is repeated countably infinite number of times. Naively for this case we would get 0,0,0,0,0,0,0,0,0,0,0,...

but not sure if that makes sense

This is because
1,1,2,3,5,8,13,..
0,1,1,2,3,5,8,...
0,0,1,1,2,3,5,...
0,0,0,1,1,2,3,...
0,0,0,0,1,1,2,..
0,0,0,0,0,1,1,...
0,0,0,0,0,0,1,...

So if $(F_n)$ is an ordered set, then taking consecutive difference is equivalent to acting the left shift operator onto it

SteamyRoot

@Secret $8 + 5 = 11$ ?

Secret

(careless mistake->fixed)

and if $(2^n)$ is taken the consecutive difference, it is a fixed point under it

Hmmm.... for fixed point under consecutive difference, it satisfy the recurrence relation $a_{n+1}-a_n=a_n \implies a_{n+1}=2a_n$, so all fixed points under consecutive differences are the series family $(2^na_0)$

Meanwhile for consecutive ratios fixed points: $\frac{a_{n+1}}{a_n}=a_n\implies a_{n+1}=a_n^2$

Tobias Kildetoft

Interesting. So there is a paper on the arXiv that catches my eye because the title claims a major theorem that I previously worked on some special cases of. Then when I go to the paper, I see that it is version 2 with a comment that they wish to withdraw it because it has some mistakes. And the first version did not even cite my work :(

Secret

which is the series family $(a_0^{2n})$

Now, how about consecutive exponents: $\text{slog}_{a_n}(a_{n+1})=a_n\implies a_{n+1}={}^2a_n$, gives series family $(a_0^{a_0})^{(a_0^{a_0})} \text{nest this n times}$

SteamyRoot

08:48

@TobiasKildetoft Well, maybe if they'd read your previous work, they'd have realised their mistakes sooner and never posted version 1 on arXiv? :P

Tobias Kildetoft

@SteamyRoot I doubt it, as that work just gave a slightly better bound than previously, but not as good a they claim (their claim is a conjecture)

Though I did quickly spot them attributing a certain result to one of the authors, even though this result is a direct consequence of an old result of Isaacs

Though I also recall speaking to someone who is an expert in the area who had not realized that this was the case before I mentioned it.

Secret

Is it a necessary and sufficient condition that a sequence will converge if all its subsequences are?

(NB include also infinite subsequences)

SteamyRoot

A sequence is a subsequence of itself

Tobias Kildetoft

yes, a sequence converges iff all subsequences do

this follows straight from the definition of convergence

the value(s) they converge to will also be the same

Secret

I see. Hmm, now I wonder how to find a very borderline divergent sequence such that there's only one subsequence diverges but the rest converges (hmm, sounds impossible, how about only countably many subsequent diverges but the rest converges)...

Tobias Kildetoft

08:57

Once some subsequence converges, infinitely many do

SteamyRoot

I think that, for the kind of statements you want to make, you should divide the subsequence into equivalence classes

where two subsequences are equivalent if they "have the same tail" or so

Secret

well, the finite ones will always converge by definition, so the interesting cases will be the infinite ones, which can be bounded (have a sup or inf or both) or unbounded

SteamyRoot

Finite subsequence?

Tobias Kildetoft

@Secret subsequences are always infinite in my world

Secret

oops, did not knew that convention. Well that leaves us only with bounded and unbounded infinite subsequence then

@SteamyRoot e.g. let S=(n), then a finite subsequence will be something like 1,2,3,4,5

but we can ignore that and just focus on the infinite case

SteamyRoot

09:02

That's not a subsequence in the "usual" sense of the definition

Leaky Nun

@Secret what is going on?

Secret

Leaky:

6 mins ago, by Secret

I see. Hmm, now I wonder how to find a very borderline divergent sequence such that there's only one subsequence diverges but the rest converges (hmm, sounds impossible, how about only countably many subsequent diverges but the rest converges)...

and then I got tripped because I did not aware subsequence only include infinite ones

Leaky Nun

well, 1,0,2,0,0,0,3,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,.... (akin to 2^n)

@Secret a finite subsequence cannot diverge

@Secret you mean you got trapped?

you tripped?

Secret

analogy, get tripped by a rock on the road and fell. what I often felt when I get some definitions wrong without awaring

Anyway, I can see which ones of the above example are diverging (and there are at least countably many of them), but which ones are converging?

The diverging ones I saw are:
The sequence itself
2,0,0,0,3,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,....
3,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,....
4,0,0,0,0,0,0,0,0,....
and so on
1,2,3,4,...

Tobias Kildetoft

how are these supposed to be subsequences of the same sequence?

Leaky Nun

09:11

@Secret well, 0,0,0,0,...?

Dattier

Hi, the convex part : $$f \in C^1([0,1]) \text{ not concave.} \text{ Is it true there exists a function convexe, not affine } g \text{ with : }\\ \forall x,y\in[0,1]^2, |f(x)-f(y)|\leq|g(x)-g(y)|\leq \max(f')|x-y| $$

Secret

Leaky: O, didn't see that coming, sure
Tobias: Isn't subsequence what you get by picking terms from the parent sequence while preserving the order? The most obvious way is to cut the first few terms

Tobias Kildetoft

@Secret Ahh, I see now

Secret

e.g. if s=1,0,2,0,0,0,3,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,.... is an ordered set, then 2,0,0,0,3,0,0,0,0,0,0,0,4,0,0,0,0,0,0,0,0,.... is a proper subset of s

hmm, leaky's example is interesting, because the only countably infinite convergent subsequence, in a sense is the same sequence

Leaky Nun

@Secret what?

in discrete space, convergent $\equiv$ eventually constant

so 1,0,2,0,0,0,0,0,0,0,... is also convergent

oh, and since the only eventual constant is 0, the only value any subsequence can converge to is 0

Secret

09:16

I see

Leaky Nun

and then of course, 1,0,2,0,3,0,4,0,... would also have the same properties

Secret

1,0,2,0,3,0,4,0,... should be divergent, right? it cannot decide between a natural number and 0

Leaky Nun

@Secret I mean, the same properties as 1,0,2,0,0,0,3,0,0,0,0,0,0,0,4,0,...

Secret

ah ok

Leaky Nun

2 mins ago, by Leaky Nun

in discrete space, convergent $\equiv$ eventually constant

@Secret you might want to prove this statement lol

Secret

09:19

uh, the topology for discrete space is the discrete topology on $\Bbb{Z}$?

Leaky Nun

well, only if the space is countable

but right, there is only one discrete space for any given cardinality

Secret

Let me rethink our topology lessons we been though many months ago: The open sets of the discrete topology are the singletons (and their unions of course)?

Leaky Nun

@Secret the basis

because every subset is open

Secret

ah ok

Balarka Sen

@AreaMan Hi

Secret

09:23

A sequence is convergent to $x$ in a given topology if $\forall a_n, a_n \in U$ where $U$ is a neighbourhood of $x$, $a_{n+1} \in U$,

actually wrong, it needs to be in all neighbourhoods of $x$

Now to prove the statement:

Take $\bigcap_{i \in \Bbb{N}} U_i$ gives $\{x\}$

Since the only element in $\{x\}$ is $x$ itself, it follows that all convergent sequences in a discrete space are eventually constant

Abcd

Don't we use mathematical fallacy to remove indeterminacy from limits?

In other words, don't we use indeterminacy to eliminate indeterminacy while evaluating limits?

How is that allowed/ correct/ possible?

Leaky Nun

@Abcd what?

can you give an example?

Secret

Is my proof correct and complete?

Abcd

@LeakyNun $(x^2 -2)/(x-2)$

limit as x tends to 2

Leaky Nun

@Secret 1. how many neighbourhoods are there? 2. no, you can't take intersection

@Abcd right, it's strictly a fallacy

Abcd

09:32

@LeakyNun Then it should not be allowed.

Leaky Nun

nobody would study maths at the rigor required for university

Abcd

Why do people use this method then?

@LeakyNun I didn't get you.

Leaky Nun

because it's easier to understand

for people in high school

Secret

1. There are countably many because e.g. let our point be 1, then its neightbouhoods are {1,2}, {1,2,3}, {1,2,3,4} and so on (and also more weird things like {1,3,4,...})

2. Why not, since all neighbourhoods of x contains x by definition, the infinite intersection will not be empty and must contain only x?

Leaky Nun

it's unreasonable to require high school students to produce epsilon-delta proofs @Abcd

Abcd

09:33

@LeakyNun Oh. But still...

Leaky Nun

@Secret 1. that isn't a proof 2. openness of subset is not necessarily preserved under arbitrary intersection

Abcd

@LeakyNun So what we study in high school is false and wrong, is it?

Secret

but in the discrete topology, the singletons form a basis and hence open sets, and all neighbourhoods of x must contain x itself?

Leaky Nun

@Abcd unfortunately

@Secret I don't get you

Abcd

@LeakyNun That came as a BIG SHOCK to me

Tobias Kildetoft

09:36

@LeakyNun @Abcd I am not really sure what you are claiming to be wrong here

Leaky Nun

@Abcd just like every subject you study in high school :)

Balarka Sen

@TobiasKildetoft +1, me neither

Abcd

@TobiasKildetoft Umm... ask @LeakyNun .. He understood what I intended to say

Leaky Nun

$\displaystyle \lim_{x\to2} \frac{x^2-2}{x-2} = \frac{-2}{0} = \text{indeterminate}$

Abcd

@LeakyNun Could you explain shells and subshells and energy levels to me now (I have understood orbitals)? talking to you after a long time therefore I am asking...

Leaky Nun

09:38

@Abcd you should ask @Secret for that :)

Balarka Sen

@LeakyNun Uh, what?

Abcd

@Secret Do you mind explaining?

Secret

@Abcd I already said, forget about subshells and shells. They are really old terms for the quanutm numbers l and n which will become not very well defined when you get to more complicated mocules

Abcd

@LeakyNun After this we eliminate x-2 and get 4

Leaky Nun

@BalarkaSen what

Abcd

09:39

@Secret I know but how do I deal with my books. All of them use these terms. How do I deal with later chapters like periodicity? While teaching that my teacher kept drawing Bohr's model and I was literally burning inside myself

Balarka Sen

@LeakyNun The limit just doesn't exist. You can't call it "indeterminate" or write "2/0" for that.

Leaky Nun

@BalarkaSen why does the limit not exist?

Secret

@Abcd In that case, just mentally replace all instance of shells with the principal quantum number, and subshells with the azimuthial quantum number and you should be fine

in the bohr model, $n$ count the energy levels, I am not sure if $l$ makes sense in the bohr model since the periodicity is tied to $n$

Balarka Sen

@LeakyNun For the same reason $\lim_{x \to 0} 1/x$ does not exist.

Abcd

@Secret so why did he use Bohr's model to explain stuff like electronegativity and electron affinity

Leaky Nun

09:42

@BalarkaSen which is?

Abcd

@LeakyNun Did you read what I said? We are taught to eliminate x-2 and get 4

Leaky Nun

@Abcd oh... you made a typo then

Secret

@Abcd Because it is a convenient first order approximation. But to be accurate, you actually need $l$ to explain electron shielding, which affects electron affinity

Leaky Nun

$\displaystyle \lim_{x\to2} \frac{x^2-4}{x-2} = \lim_{x\to2} \frac{(x-2)(x+2)}{x-2} = \lim_{x\to2} x+2 = 2+2 = 4$

@BalarkaSen here

Dattier

$$\forall n,m\in \mathbb N, \text{pgcd}(2^{2m+1}+1,2^{2n+1}+1)=2^{\text{pgcd}(2m+1,2n+1)}+1 \text{ ?}$$

Abcd

09:44

@Secret Fair enough. Just a last question: what do you mean by "convinent first order approximation"?

Leaky Nun

@Secret please, don't use graduate terms

Abcd

@LeakyNun This is wrong, right?

Balarka Sen

@LeakyNun There is no (extended) real number $a$ such that for all sequence of real numbers $\{x_n\}$ converging to $0$, $1/x_n$ converges to $a$.

Secret

The discrete topology has the singletons {x} as a basis.
A sequence converges to x iff for all neighbourhoods $U$ of x, $a_n \in U$ means $a_{n+1} \in U$
Since $U$ contains $x$ by definition, taking the arbitrary union cannot be empty and must contain $x$ and only $x$, thus the arbitrary union of $U$ will become the singleton $\{x\}$ and the eventually constant conclusion will then follow?

Balarka Sen

Do you really not know what limit not existing means? :P

Secret

09:46

@LeakyNun lol, electron shielding is high school stuff back in my HKCEE days, but qualitatively is really what it says on the tin

Abcd

@Secret Yeah I know Electron shielding and Effective nuclear charge

Leaky Nun

@Abcd ask Balarka

Dattier

$$2^n(2^{n!}-1) \mod n!=0 \text{ ? }$$

Balarka Sen

@Abcd It is not wrong. What gives you the idea?

Leaky Nun

@Secret taking arbitrary intersection is both unnecessary and wrong

@BalarkaSen it is wrong

Tobias Kildetoft

09:48

@LeakyNun No, it is not wrong, it just needs some arguments for why we are allowed to do it

Abcd

@BalarkaSen @LeakyNun said it's wrong. I think it's wrong because you are divinding zero by zero which ain't allowed

Balarka Sen

Leaky Nun is wrong.

Tobias Kildetoft

Such as the function being continuous and defined in an open neighborhood of the point at which we take the limit

SteamyRoot

You're not dividing by zero.

Leaky Nun

@BalarkaSen the limit isn't wrong. the step is wrong.

Balarka Sen

09:48

@Abcd You are not dividing zero by zero. Limit by definition means you are not plugging in the value of the limit the variable tends to.

SteamyRoot

You're taking a limit, which means you're taking values arbitrarily close to $2$.

Abcd

@SteamyRoot (x-2)/(x-2) is 0/0 = indeterminate and we eliminate it which shouldn't be allowed

SteamyRoot

Only when $x = 2$.

Tobias Kildetoft

@LeakyNun None of those steps are wrong. Those expressions are equal for all values of $x$ for which they are defined

SteamyRoot

Limit $\neq$ evaluation

Balarka Sen

09:49

@LeakyNun No step whatsoever is wrong.

I am sorry.

Leaky Nun

@TobiasKildetoft can you explain the step $\displaystyle \lim_{x\to2}x+2=2+2$?

Secret

@Abcd (I might use the wrong term, I should have used "approximation" if I am precise) Basically, bohr model is a rough model before you have quantum mechanics. you can explain some things with it quite well but if you want to be a lot more accurate and handle the more notrivial case, then this model suffers its limitations

Balarka Sen

@LeakyNun Continuity of $x + 2$.

Tobias Kildetoft

@LeakyNun The limit of a continuous function in a point where the function is defined is the evaluation at that point

Leaky Nun

@TobiasKildetoft isn't that circular?

Balarka Sen

09:50

No, it is definition (of continuity).

SteamyRoot

It follows readily from the epsilon-delta definition of a limit.

Abcd

@Secret Alright. Thanks a lot. Now I feel a lot better and wouldn't unnecessarily confuse myself while studying chemistry.

Secret

but yeah, anyway, once you get to uni, ditch subshells and focus on quantum mechanics

Abcd

@BalarkaSen @SteamyRoot If limit isn't plugging in the values then why do we use the substitution method while evaluating limits?

Balarka Sen

@Abcd You can substitute the value in if and only if the relevant function is continuous at that value.

SteamyRoot

09:53

@Abcd I meant that they are not the same in general

Balarka Sen

Eg, if I have $x^2$, $\lim_{x \to 0} x^2$ can be evaluated by plugging in $x= 0$.

(And the limit is $0^2 = 0$)

Secret

Why I cannot take arbitrary intersection here, $\{x\}$ is open in this topology and you must get $x$ since all neighbourhoods of $x$ contains $x$ thus by defniition of arbitrary intersections, the only element that is present in all neighbourhoods is $x$?

If I cannot use arbitrary intersection, what shoudl I use to test all the neighbourhoods of $x$ for my proof?

Leaky Nun

yes, and then using the continuity is circular @BalarkaSen @TobiasKildetoft

Abcd

@BalarkaSen and @SteamyRoot Could either of you explain that with an example (or with a graph)

Balarka Sen

But if I have $\lim_{x\to 0} x/x$, I cannot plug it in because $x/x$ is not defined at $x = 0$ - it has a jump discontinuity at $x = 0$.

Tobias Kildetoft

09:54

@LeakyNun How is using continuity circular?

We are not proving continuity, we are finding a limit

Balarka Sen

@LeakyNun It isn't. The step is just a little vague, you would have to use the epsilon-delta definitions to justify that $x + 2$ really is continuous at $x = 2$.

Leaky Nun

@BalarkaSen exactly

Balarka Sen

This is not equivalent to "wrong" :P

It's just rigor at a different level.

Leaky Nun

@Secret you just need to point out that $\{x\}$ is a neighbourhood, hence why it is unnecessary; taking arbitrary intersection does not guarantee a neighbourhood, hence why it is wrong

@TobiasKildetoft we can plug it in because it is continuous; it is continuous because we can plug it in

Secret

uh, how could x be a neigbourhood. Using the usual definition of neighbourhood, it has to contain a subset that is an open set that contain x. Unless you mean \{x\} is a subset of itself?

Leaky Nun

09:56

@Secret of course

Abcd

@LeakyNun What's the conclusion?

Secret

Ok fine (I am more comfortable when people said subobjects, they mean proper subobjects)

Rewriting proof now:

Dattier

$$\textbf{Coup de Boole : } n\geq 3, f\text{ bolean function with n varibales}
\\ \text{ Is it true that : } \sum \limits_{\sigma \in S_n} f(x_{\sigma(1)},...,x_{\sigma(n)}) \mod 2=0 \text{ ?}$$

Balarka Sen

@LeakyNun We never said it is continuous because we can plug it in.

Leaky Nun

@BalarkaSen well, why is it continuous?

Balarka Sen

09:58

That's a separate question from computing the limit, is the point.

SteamyRoot

Compare the epsilon-delta definition of continuity at a point

Balarka Sen

^

SteamyRoot

and compare with the epsilon-delta definition of the limit

Leaky Nun

the definition of continuity is the ability to plug it in

Balarka Sen

@Leaky Yes, it is.

Leaky Nun

09:59

@BalarkaSen hence the circularity

Balarka Sen

No, that's wrong. Definition of continuity implies the ability of plug it in.

Leaky Nun

@BalarkaSen $x \mapsto x+2\text{ is continuous at }2 := \displaystyle\lim_{x\to2}x+2=2+2$

Balarka Sen

No, no, that should be a $\implies$. Definition of continuity is the epsilon-delta definition.

It just grants you the ability to do it.

Secret

@LeakyNun In the discrete topology, the basis are the singletons $\{x\}$. A sequence $s$ converges to $x$ iff for all neighbourhoods $U$, $a_{n} \in U$ gives $a_{n+1}\in U$. Now since $\{x\}$ is a subset of itself, which is open, $\{x\}$ is a neighbourhood of $x$. $U$ can then be constructed by union of $\{x\}$ with other open sets. Now since the only elements in $\{x\}$ are $\emptyset, x$, it follows that $s$ has to be eventually constant

Balarka Sen

Unless, you know, you define $\lim$ by epsilon-delta.

In which case that step would be justified by working out the epsilon-delta continuity of $x + 2$ by hand.

If you want to do it at arbitrary level of rigor you can just do it; my point is that the problem does not want you to do it.

Leaky Nun

10:02

Continuous function

In mathematics, a continuous function is a function for which sufficiently small changes in the input result in arbitrarily small changes in the output. Otherwise, a function is said to be a discontinuous function. A continuous function with a continuous inverse function is called a homeomorphism. Continuity of functions is one of the core concepts of topology, which is treated in full generality below. The introductory portion of this article focuses on the special case where the inputs and outputs of functions are real numbers. In addition, this article discusses the definition for the more general...

Balarka Sen

quoting wikipedia is the worst. what is $\lim$ to you?

Leaky Nun

@BalarkaSen how else would you define $\lim$?

SteamyRoot

>Wikipedia

Balarka Sen

@Leaky There are multiple ways to do it. I can use the sequence definition, I can use the $\epsilon-\delta$ definition, ...

Leaky Nun

mathworld.wolfram.com/ContinuousFunction.html

hmm, so both definitions are used

SteamyRoot

10:04

$$\forall \varepsilon > 0: \exists \delta > 0: \forall x: |x-a| < \delta \implies |f(x) - L| < \varepsilon$$

That's $\lim_{x \to a} f(x) = L$

Leaky Nun

@Secret the last statement is wrong

Balarka Sen

Anyway, just writing $\lim_{x \to 2} (x + 2) = 2 + 2$ without excruciating levels of justification (which can be done non-circularly) does not make the step wrong.

This was my whole point from the beginning.

Secret

if $\lim_{n\to \infty}a_n = y$ and $y \not\in \{x\}$ you cannot have $s$ converge because then you have the sequence not lying within one neighbourhood of $x$?

Leaky Nun

@Secret i mean, the second-last

@BalarkaSen alright, agreed

Secret

o crap, no emptysets in singletons!

otherwise I have a set with 2 elements

I guess I will be screwed if I am asked to proof convergences in a topology that has no countable base

because without intersections, I have no idea how to ensure I test all neighbourhoods of a point if there are infinite number of them

Secret

10:15

Btw, that sequence question, is really inspired from this old chat message

Mar 10 '15 at 16:16, by Vrouvrou

is the sequence has a convergent sub sequence ? please

which for some reason I generate a new question that reads:

"Can there be sequence which diverges but all subsequences converges?" tobias then clarified it is impossible when I ask him about the necessary and sufficient conditions of subsequences and convergence, which turn this question into its current form

Leaky Nun

@Secret just consider the basis

Secret

1 hour ago, by Secret

6 mins ago, by Secret

I see. Hmm, now I wonder how to find a very borderline divergent sequence such that there's only one subsequence diverges but the rest converges (hmm, sounds impossible, how about only countably many subsequent diverges but the rest converges)...

Secret

@LeakyNun Ah right, even if a base is uncountable, it has a "closed form" as a set builder notation, I think I can work from there

Leaky Nun

@Secret challenge: does $\Bbb R$ under the usual topology have a countable basis?

Secret

Suppose the countable basis are singletons. Then you end up with $\Bbb{R}$ being countable which is a contradiction. Now suppose $\Bbb{R}$ has countable basis as open intervals glued back to back. Then since every interval contains a rational, which is countable, it follows that ....

...Error: unable to trigger the contradiction I want

ಠ_ಠ

10:22

Look at open balls of rational radius at rational points

user8469759

hi guys

ಠ_ಠ

More generally, and manifold with countably many connected components admits a countable basis.

user8469759

I was wondering if the following equation

ಠ_ಠ

A space with countable basis is said to be second countable

user8469759

{x'}^T \hat{v} x + x^T s x = 0

where s = \frac{1}{2}(\hat{\omega}\hat{v}+\hat{v}\hat{\omega})

can be written in terms of the matrix

[ \hat{v} ; s]

like a quadratic form

here x is a 3D vector, x' is the derivative, \hat{v} is the skew matrix built up from a 3D vector v and \hat{omega} is the same

Secret

10:26

${x'}^T \hat{v} x + x^T s x = 0$

$s = \frac{1}{2}(\hat{\omega}\hat{v}+\hat{v}\hat{\omega})$

$[ \hat{v} ; s]$

Balarka Sen

@ಠ_ಠ Well, only if you assume second countability as an axiom for manifolds, right? Otherwise long line is a counterexample.

user8469759

It self the equation I gave is already function of that matrix, however I was wondering if specifically a quadratic form can be derived

Balarka Sen

In which case, that's not a very interesting thing to say

Secret

Actually, attempting Leaky's most recent challenge leads me to a question: How can I check whether I am unable to find a counterexample to some given proposition?

I guess if for a given proposition, if I stuck at the proof and the counterexample finding, I am basically screwed and had to take a nap and deal with it later

Sha Vuklia

Guys, so I want to get the Taylor polynomial for $f(x,y)=x^2y$ with $\vec a=(1,-1)$. Normally, I evaluate it about $\vec a=(0,0)$, so then I know what to do. However, I’m not entirely sure if what I would do now is correct. I know that I have to calculate the partial derivative and evaluate those at $\vec a=(1,-1)$. But we have the following:

So when $a=(0,0)$, just can just write $f(x,y)$, where $(x,y)=\vec h$. However, now I would have $f(\vec a+\vec x)=\dots$. Shouldn’t I “convert” that to something of the form $f(\vec x)$?

ಠ_ಠ

10:46

@BalarkaSen Usually paracompactness rather than second countability is assumed for manifolds, at least in differential geometry, so that we have partitions of unity. Second countability is too restrictive an assumption, because it means that a discrete infinite group is not a Lie group.

Balarka Sen

@ಠ_ಠ Ah, that is very fair.

Thanks.

Eigenfield

Does anyone know how to fine the sign of a (p,q) shuffle?

Abcd

11:27

ÍgjøgnumMeg

Hullo! Question about language; how should one refer to isomorphism classes, I've written something like "Recall that if $H$ is a group of order $17$ then $H \cong \mathbb{Z}/17\mathbb{Z}$ (i.e. groups of order $17$ have only one isomorphism class) and that if $K$ is a group of order $4$ then $K \cong \mathbb{Z}/4\mathbb{Z}$ or $K \cong (\mathbb{Z}/2\mathbb{Z})^2$ (i.e. groups of order $4$ have two isomorphism classes)." Does this sound too clunky?

Abcd

That's my solution. I am a little doubtful about it. Please let me know whether it's correct or not.

ÍgjøgnumMeg

@Abcd Let $a_n = \frac{1}{n}$ and $b_n = -\frac{1}{n}$. Then the limit $\lim_{x \to 0^+} f(x)$ is equivalent to the limit $\lim_{n\to\infty} f(a_n)$ and $\lim_{x\to 0^-} f(x)$ is equivalent to $\lim_{n\to\infty} f(b_n)$. Can you see why these aren't equal?

Abcd

11:43

@ÍgjøgnumMeg I can't understand that. Is my answer correct?

LHL indicates Left hand limit

RHL = Right hand limit

ÍgjøgnumMeg

@Abcd No your answer is not correct. You've tried "plugging in" what I would write as "$-0$" and "$+0$", but "$2 \times (-0) \neq -2$".

@Abcd If you take 5 minutes to try and understand the hint that I gave you I think you'll be fine ;)

Abcd

@LeakyNun

@ÍgjøgnumMeg I cant understand what you are trying to say

ÍgjøgnumMeg

Well, look at it this way; what is $\lim_{n\to \infty} \frac{1}{n}$, and "from which direction" does $\frac{1}{n}$ get small as $n$ gets bigger?

Abcd

@ÍgjøgnumMeg As n gets larger and larger, value gets closer and closer to 0

from positive direction

RHL

ÍgjøgnumMeg

@Abcd Exactly! So, plug $\frac{1}{n}$ into your $f(x)$ and take the limit as $n \to \infty$ instead of $x \to 0^+$. What do you get? If you do the same with $\frac{-1}{n}$ (which gets closer to zero as $n$ gets larger, but this time from the negative direction) you'll see that the limits aren't the same.

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$Mathematics$ Mathematics