@robjohn @Chris'ssistheartist I had another crack at the inverse central binomial harmonic series care to check for errors please? thanks! :)

Hello All, any statistics enthusiasts here that can help answer a question I have on statistical power ?

Hey guys

that

WHERE IS EVERYBODY

I'm sleeping

noooo

@morphic

i wake you up

@morphic

@r9m

@ForeverMozart sup? :)

i drink beers and do maths

okay^2

I am preparing a paper

but I would like to prove a couple more things to put in it

what is it about?

weird connected spaces

do you know what my name references?

and my photo?

@ForeverMozart imdb.com/title/tt0116334 ?

yes :)

it is very good

haven't seen the movie though ,,

have you seen anything by Godard?

(they are all French)

nope .. ah that explains it .. I don't watch French films all that much

oh well first you should see Pierrot le Fou

he is my favorite filmmaker

'kay .. ^^

what are you up to?

no good

sounds scandalous

good .,. better that way :P

brb

i got another bee

r

my refrigerator has spoiled milk in it and it really stinks

good for you

are you good at topology

nope .,. not really

are you good at drinking beer?

neither

darn

what do you do?

eat sleep and math :P

what kind of math?

are you a graduate student?

UG

well you have plenty of time to become an expert

what is your favorite area of math?

@Huy you are a baby

beer #5

@MartinSleziak

In connection with launch of Discrete Analysis announced by Tim Gowers it was the first time I have heard the phrase arXiv overlay journal. Is this the first journal of this type?

BTW the editorial boards seems rather impressive: Ernie Croot, Ben Green, Timothy Gowers, Gil Kalai, Nets Katz, Bryna Kra, Izabella Laba, Tom Sanders, Jozsef Solymosi, Terence Tao, Julia Wolf, Tamar Ziegler.

Tim Gowers it seems like I know him

@ForeverMozart Is there some reason you pinged me, or are you just bored?

you are good at topology

I don't think that. But I have answered a few beginners question in the general-topology tag.

How do you know what to try to prove?

I am have a few results in topology that I want to put in a paper, but I would like to add more. Sometimes I feel like I am not clever enough

Neither of those two is an easy question.

For the second one, you could discuss that with an advisor or a colleague.

I am not sure whether your first question is intended as: How to choose which problems are worth studying?

yes well I only see my main advisor once a week and he ignores my emails

I have some open problems to work on

What you wrote after suggests that you want ask something about how to continue if you have some partial results.

There are some natural things which often work.

If I have a proof, I can have a look whether it can be generalized.

Math GRE tomorrow.

D:

I can try to see whether some assumptions can be omitted. (And if I find some interesting counterexample, it might be worth adding to a paper.)

If I have obtained sufficient condition for some results, I can ask whether it is also necessary.

I will see if I can do that.

Sometimes I read a paper and it is so complex I think there is no hope for me.

But it is rather general advice. (And coming from somebody rather inexperienced.)

Like I read a paper and the statement of a lemma takes over half a page

I sometimes have similar feelings. But it is unreasonable to expect to understand "all of mathematics". And it is also unreasonable to expect to be able to understand thinks as quickly as the best experts in the fields.

@Anthony I do not know why the Math Gre is required

Yeah... Me neither. Oh well.

Maybe it is not always the case, but sometimes the statement can be long and for somebody familiar in the area it can be more or less clear since they have seen similar things many times before.

So as I am studying $T_2$ topologies I noticed the following

anyone want to chat about topological stuff?

Why don't you simply write what you have noticed.

Maybe somebody will react, maybe not.

so one question I would like to ask what property that is shared among $T_1$ and $T_2$ topologies

If you will ask something which is related to US-spaces - a topic I know nothing about, I will probably not be able to respond.

If you ask about sequential spaces, I might be able to respond.

I am pretty sure that the following is true

My point is - it is difficult to say in advance. (Sorry for the rant - kind of.)

That actually you might have a sequence of points in $T_1$ that converge to more than one point.

I agree with you @MartinSleziak

because we really need the disjoint condition in order to keep those infinite points still inside an open set U.

right?

To be more precise, a space is Hausdorff if and only if every convergent net has only one limit.

hm

If your space is first countable, you can replace nets with sequences.

My definition is the following though

This seems related to what you said so far: Equivalent characterizations for Hausdorff in a first-countable space

$for x \neq y$ there exists nbhd $U_x$,$U_y$ such that $U_x \cap U_y = \emptyset$

Yes, that is the definition of Hausdorff space.

yeah i can see the proof for one point convergence

is it true that $T_1$ space you can have a sequence that converge to more than one point?

You mean an example of $T_1$-space with sequences having more than one limit? That should not be very difficult.

yeah

One more question is it only the property of sequence converging to more than one point is it the only one that separate $T_1$ space from $T_2$?

or are their others as well ?

If we simply take a convergent sequence and "double" the limit point?

yeah

BTW some people call the spaces with the property that every sequences have at most one limit US-spaces.

oh

I didn't know that

@KarimMansour Should I understand this question as: If $X$ is a US-space and it is $T_1$, then it must be $T_2$?

Yeah

I think it is true no ?

Then the post I linked to answers this for first-countable spaces,

But in general it is not true (I think).

Since if you do not have any kind of "countability axioms", sequences are not enough.

I guess that counterexample could be "sequence of length $\omega_1$" with doubled point.

I am willing to believe that if you have some type of "countability axiom", then the claim might be true. (I would not be surprised if it were true for sequential space.)

MathOverflow: Unique limits of sequences plus what implies Hausdorff?

cool thanks alot @MartinSleziak

I feel like such examples, should be in some standard text and maybe also on MSE or MO.

I am not sure whether pi-base has US-spaces among the properties which can be searched for. (You know about pi-base database, right?)

no what is that ?

@KarimMansour Here is the link: topology.jdabbs.com

ohhh

so cool

thats so awesome

you made my day @MartinSleziak

It is a database of examples of topological spaces. You can enter something like: I want a space which has property A and property B but does not have property C.

so cool

I didn't know such a thing existed

Hi @MartinSleziak

Here you can find a bit about the history of that website.

However, unless one of the properties you can choose is US (sequences have unique limits), it will not help you to find counterexapmles for this.

oh there is even a book about this

I am actually very interested in topology and algebra. I want to specailize in those fields when I go to grad school

amazon.com/Counterexamples-Topology-Dover-Books-Mathematics/dp/…

so cool

Do you consider ugly spaces to be "cool"? :P

I dunno what ugly spaces are, but I am kind of person who learns alot from examples etc and counter examples etc, so its good thing to find this kind of thing.

Quote from Engelking's book: "1.6.E. Give an example of a non-Hausdorff Frechet space in which every sequence has at most one limit. Hint. Adjoin a point to the space described in 1.6.18."

Counterexamples to various statements which should be "intuitively true" are precisely what I call ugly spaces. E.g., infinite broom, topologist's since curve, etc.

I think that this example should be $T_1$.

So I was wrong: First countable seems to be sufficient, Frechet and sequential are not sufficient. (For the implication US+$T_1$ $\Rightarrow$ $T_2$.)

oh

The Wikipedia article in French Wikipedia Axiome de séparation (topologie) mentions US-spaces. But it seems that it only very briefly.

I do not speak French. But it seems that they only say that US imples T1, but not the other way round.

If I understood it correctly and if US really implies T1 (which I did not realize), than it simplifies your question a lot.

Since you asked about spaces which are both US-spaces and T1-spaces.

yeah

And if this is true, then this class of the spaces is simply the class of $T_1$-spaces.

yeah I could definitely see it for US doesn't imply $T_1$

because definition of convergence is that a sequence $x_1,...$ converges to an element x, if for each nbhd $U_x$, there exists $n_2 \in \mathbb{N}$ such that $x_n \in U_x$ $\forall x_n \geq n_2$

The proof of "Every US-space is $T_1$-space" from that Wikipedia article seems to be simple: For two given points $x$, $y$, just take a constant sequence. Use the fact that it only has one limit.

@KarimMansour I think the Wikipedia article is correct and that US $\Rightarrow$ $T_1$ is true.

yeah

I am sorry

I meant to say

$T_1$ doesn't imply US

as the $T_1$ condition only separates singular points

In fact, here is the proof in English: topospaces.subwiki.org/wiki/US_implies_T1

I have really hard time trying to read something in French.

ok, we have talked a bit about US, T1 and T2

In fact, I should have remembered this implication, since I have seen it before.

I guess it is time for me to do something else. (There are homeworks and tests to grade and lot of other stuff....)

thanks for your time @MartinSleziak

I learned alot

If you have some questions about general topology which you prefer discuss in chat rather than on main, feel free to mention them in general topology chatroom.

It is rather inactive. But maybe we can start some activity in that room in that way. (And there is non-zero chance that somebody will notice your message and respond to it.)

BTW this is not the first time when I see that article on French Wikipedia contains some details which are missing in the corresponding article on English Wikipedia. I wonder whether the same is true for other areas of mathematics. (The articles where I have seen this were from general topology.)

hey @AlexClark

any new badass discoveries @r9m ?

Very nice identity, and curious how it was discovered.

mathworld.wolfram.com/LiouvillePolynomialIdentity.html

Liouville Polynomial Identity

wooaah

wolfram mathematicians amaze me always

ah its way before in 1957

@Chris'ssistheartist Most likely by trying to settle the Waring problem for n = 4.

@BalarkaSen I'm exactly there right now. :-)

books.google.ro/…

Ah!

@AlexClark Cool, let me know when you want to start. I just want to make it clear that I'd be happy if you just spend 3-4 decent hours in alg top in a week - not saying that you'd have to devote your whole life to it :P

Besides, learning new things will only make your background stronger.

@BalarkaSen These days I also found an identity I considered to be new, not sure if it's so, but I'll find more after it is published in AMM (where I used it for a different problem).

@AlexClark Hmm, I don't remember what Paul's paper is. Waring's problem is a long standing open problem in number theory

I checked the whole internet and found anything like that.

@Chris'ssistheartist Nice. What was the identity about, if you don't mind telling?

@Agawa001 @Chris'ssistheartist possible closed form of $\sum\limits_{n=1}^{\infty} \dfrac{H_n^{(2)}}{\binom{2n}{n}}$ and $\sum\limits_{n=1}^{\infty} \dfrac{H_n^{2}}{\binom{2n}{n}}$ :-)

@BalarkaSen simple algebraic identity in three letters, not sure how to define it elsewhere.

Oh, @AlexClark, they're doing the whole Waring thing in polynomial setting. That's nice indeed.

@Chris'ssistheartist alright.

@BalarkaSen but very nice. I'll show it to you at some point, and you can tell me more about it (that is if you ever saw it).

@AlexClark Doing math is always fun :)

sure, @Chris'ssistheartist

Thanks.

@r9m Great! :-) I found both series pretty tough, and the troublesome thing is that it seems hard to come up with an elegant way.

@Chris'ssistheartist It might be a laborious calculation, but I am planning to give it a shot at some point :-) (well then again I shouldn't complain, given the long posts DavidH, Tunk Fey and some other integral gurus post in M.SE, their tenacity amazes me ) ..

@AlexClark: what kind of functional analysis are you doing?

@Chris'ssistheartist Aye! That be true!

If I want to find the splitting field of an irreducible polynomial $f$ ( over $K$), what information does the degree of $f$ give us about the extension field $L$?

I guess $\deg(f)=n \implies [L:K]\leq n$

Hello everyone

Hello

Bonjour

Can someone of you help me with differential geometry and especially with the binormal of a curve?

@Huy do you know something about it?

@GaloisintheField Dimension of a splitting field

Deja vu: chat.stackexchange.com/transcript/13473/2015/10/17

If you believe the answer posted in the linked thread: "The degree of the splitting field is divisible by $n=\deg(P)$ (and it divides $n!$), but it can be greater than $n$."

@GaloisintheField False. $x^3 - 3$

The bound is $n!$

That is, if $f$ is a polynomial of degree $n$, then $[L : K] \leq n!$ where $L$ is the splitting field of $f$.

It's because if you assume $L/K$ is Galois, then $[L : K]$ is precisely the order of $Gal(L/K)$. And $Gal(L/K)$ acts on the roots of $f$.

So you have an injection $Gal(L/K) \hookrightarrow S_n$.

Hence, $[L: K] \leq n!$. If $L/K$ is not Galois, it will be degenerate, so in any case it's smaller than $n!$

And this is the best possible bound you can have. The splitting field of $x^3 - 3$ has degree $6 = 3!$.

wrestling with triangular numbers .....

So $x^3-2$ has splitting field of degree $6$ with basis $\{a+b\sqrt[3]{2}+c\sqrt[3]{2}^2+de^{2\pi i/3}+ e e^{2\pi i/3}\sqrt[3]{2} + f e^{2\pi i/3} \sqrt[3]{2}^2:a,b,c,d,e,f\in \Bbb Q\}$

@GaloisintheField Determine splitting field $K$ over $\mathbb{Q}$ of the polynomial $x^3 - 2$

There are various possibilities how to choose the basis.

Strictly speaking, you did not write a basis, but it is clear from what you wrote which elements you meant to give in your basis.

And also a nitpick: You are using letter $e$ in two different meanings there.

When M(v)=Pv+a is a isometry of R^3, where P is a 3x3 orthogonal matrix, is it always true that P is an orientation preserving orthogonal matrix ?

@user159870: in general, no

In this case: math.stackexchange.com/questions/1493072/binormal-of-the-curve is it true? @Huy

@user159870: that is something you have to figure out

But how do we check it? @Huy

@user159870: just check the two cases, one where it is orientation preserving and one where it isn't, and see what happens

When it is orientation preserving we have $B=Pb$, when it isn't we have $B=-Pb$. @Huy Or do you mean something else?

can you explain why?

this is just a reformulation of what you said earlier

also, how do you assign an orientation to a vector $B$ or $b$?

The vectors B or b are perpendicular to the curve, the orientation is the direction, i.e., upwards or downwards. Is this correct? @Huy

Can I get a bit of feedback about whether my theory is correct for this question? I was not taught the theory in university, but looked it up on the net.

hey @skill n°2

hi pal #1

:D

someday you ll find my nick tailed by 010

or should that be 001?

no getting enough rep to skip there

@skillpatrol im not kind of people thinking they r n°1, have you listened to that song "breaking the habbit"-linkin park- ?

yep

"i dont want to be the one, the battles always choose"

^ i like this so much

sounds deep

but common sense

battles always find a way to choose me too

@skillpatrol hope so, there s another proverb which says: winds dont crave ships

Can anyone let me know what the difference between riemann and darboux integrals is?

@Paradox101: math.stackexchange.com/questions/142782/…

@evinda: the image is the curve?

Is what I said before about the orientation of B and b correct? @Huy

@Huy I also thought so. But if so why would I be asked it?

@PVAL: Shameless self promotion. I think you'd like these answers.

@MikeMiller Why is $\pi_1(M) = \pi_1(M_1)*\pi_1(M_2)$ necessarily infinite?

I guess I am missing some obvious fact about free products.

Cool proof.

If $P$ is a refinement of $Q$, will the darboux upper sum of $Q$ be greater than the inf of darboux upper sum of $P$? Can anyone please explain this?

Oh, yeah, bleh. $G_1 * G_2$ is finite iff either $G_1$ or $G_2$ is trivial. One direction is easy. For the other direction, if $G_1$ and $G_2$ are both nontrivial, then you take union over all the relators. Take a word consisting of two generators so that it's nontrivial. That generates an infinite cyclic group inside it.

@Balarka: I see you answered your question. Note that there are two answers, not just the one, and the second gives more information.

Yeah, I am reading it right now. Have you seen the message I pinged you last night?

Yes. I didn't see a question, if there was one.

I wanted to know if that was a valid way to approach cellular approximations. I have thought a bit about it, but I couldn't seem to get anywhere.

I was wondering if you could help me with this question at math.stackexchange.com/questions/1495597/number-of-good-words

Oh. I didn't really read it. Can you describe your idea in two sentences, neither of which are run-ons?

@MikeMiller I'm confused about your lemma on the first answer. $D^n$ is a noncompact $n$-manifold with boundary $S^{n-1}$. But the inclusion does not induce injection on homology. What am I missing?

Think about that again.

What do you mean by a noncompact manifold with boundary anyway?

If it's what I think, then there would be points with nbhd R^2+

Which $D^n$ does not have.

Hello, math.stackexchange.com/questions/1495597/number-of-good-words plz

@Balarka: A manifold with boundary. The manifold is not compact.

I have no idea what else I could mean by it.

ok. then $D^n$ is not a noncompact manifold with boundary. It has no boundary.

I was confused. Sorry.

Dude that's not true either.

How so? No point on the open ball has neighborhood homeomorphic to upper half plane.

Ok, now I see your problem. D^n denotes the closed unit ball unless otherwise specified.

Ah, alright. Thanks for clarifying that.

@MikeMiller Sure. Let me write it down.

You have about 5 minutes before I'm off the bus.

$f : X \to Y$ be a map between CW-complexes. Assume it's cellular at skeleton $n-1$ and below. Assume $Y$ is an $n+1$ dimensional CW complex, without loss of generality.

I want to get $f(X^n)$ off $Y^{(n+1)}$

Pick an $n$-cell $e^n$ in $X$.

You've already passed two sentences.

:(

:(

lol, that was harsh

nah. I guess he just had to get off the bus.

@MikeMiller 1 scentence description (when you want to read it) : I want to homotope $f$ to be nonsurjective so that $f(X^n)$ leaves out a point in each $(n+1)$-cell on $Y$. Then homotope it down to $n$-skeleton.

Hatcher's technique to prove $\pi_1(S^n) \cong 0$, i.e.

@DanielFischer: just a quick question - in my lecture notes on functional analysis we defined the norm $$\|u\|_{W^{1,p}} = \| u \|_{L^p} + \sum_{i=1}^n \| \tfrac{\partial u}{\partial x_i} \|_{L^p}$$ however many other sources define it as $$\left( \sum_{i=0}^1 \|f^{(i)} \|_p^p \right)^{1/p}$$ which to me isn't the same. are they the same for some reason or is there a typo in my notes?

@BalarkaSen: I'm looking at some group actions and and an example given is: if $f$ is a function on $S^{n-1}$, we can let $k \in SO(n)$ act on $f$ via $$(k \cdot f)(m) = f(mk)$$ where $m \in \mathbb{R}^n$ is considered as a row vector so $mk$ is matrix multiplication from the right. is this standard? why don't we mutliply from the left with a column vector?

@BalarkaSen: in other sources I've even seen it defined as $f^g(x) := f(g^{-1} x)$ claiming we need to write $g^{-1}$ if the group is non-abelian

so I'm quite confused on what the standard action looks like

sorry. was away.

np

yeah, no idea why not multiply by left. I have seen that convention too. It doesn't make sense.

@Huy Hey sorry for the low conversational contribution for FA

I have been unbelievably busy

@BalarkaSen: you mean "why not multiply by left"?

@Huy Does the axiom of group action hold if you multiply by left?

left-right confusion.

you should replace left by $-\infty$ and right by $+\infty$ when you talk to me.

@Huy well, recall that $(gh)^{-1} = h^{-1}g^{-1}$

@BalarkaSen: ok I just tried out $(kf)(m) = f(km)$ and then associativity doesn't work

or "compatibility" whatever they call it

well, depends on whether you consider a left/right action

left

$(k_1 k_2)f = k_1(k_2 f)$ is true if your action is right

yea

and the $f^g(x) = f(g^{-1}x)$ is probably so that we still mutliply from the left (but now we need the inverse for compatibility), right?

the action is still right. check it.

use the identity above

wat

$f(x) \cdot g = f(g^{-1}x)$ is a well-defined right action

I mean the underlying action on the domain

sec

Hello, I am trying to determine what parametric or polar equation forms this shape. Is this a suitable question for Mathematics?

@BalarkaSen: are you 100% sure it's not a left action?

@BalarkaSen: I get $f(x) (hg) = (f(x) g)h$

I've been playing around with this hypocycloid visualizer but I haven't gotten anything close, so I think I'm on the wrong trail

$f(x) \cdot (gh) = f((gh)^{-1}x) = f(h^{-1}g^{-1}x) = f(g^{-1}x) \cdot h = (f(x) \cdot g) \cdot h$

wtf happened in my notes

sec

So what you're getting is wrong.

ah

I took the $g$ out too far

-.-

thanks

:P It happens.

so what does this all mean now

It means all your actions are right.

In the second case.

@BalarkaSen: and are they equivalent ?

if not why are there multiple conventions

define equivalent

@Huy: They're probably equivalent norms, which is all that matters.

Hi Mike

@BalarkaSen: will I get different theorems if I use one over the other

@Balarka: Seems reasonable but a bit vague. I was hoping for two highly expository sentences, you see. Induction is correct.

@MikeMiller I have pinged you a summary of my idea above.

@Huy: Re your last sentence. No.

Not interestingly different.

@Huy All the left-right confusions should be directed towards Mike.

@MikeMiller: I'm discussing group actions on functions with Balarka, but thanks for the answer about the norm :P

Nah.

lol

ragequit

@Huy: I'm aware. My second answer was about the second thing.

ok

I guess I'll have to use the $f(g^{-1}x)$ then since my prof does -____-

lol, no idea how can I make two highly expository sentences out of this. Maybe I'll ponder on that for a week.

Such concise, much brief.

@BalarkaSen: so if I now look at some $f^g$ and want to differentiate it, I guess I'd be using the chain rule... but how do I differentiate a group action? would I need the group to be a Lie group for this?

Calculate in 5 different ways $$\int_0^1 x^{n-1} \log(1-x) \ dx$$

@Huy no idea what one means to diff a grp action. maybe fix some $g$ and then diff the function you have?

obviously the action has to be nice

@Huy Not the same, but equivalent. The second is nicer in some respects, but I don't think the difference seriously matters anywhere.

@DanielFischer: thanks. do you know the anwer to my last question ?

otherwise $f^g$ mightn't be continuous

Sorry to jump in with a question but there's an arrow on this diagram that looks like $)\rightarrow$ where the 'root' of the arrow touches the middle of the bracket. Used in the context of (the quotient) module, what does it mean? It doesn't mean induced because the author is happy to use dashed arrows

injective?

$\hookrightarrow$, you mean?

No, it's actually a ) with a rightarrow after it, without the gap between them obviously.

@Huy Err, what's the situation? You have a manifold, and a group action on it, and you want to differentiate $f \circ (g^{-1})$? If the group acts by diffeomorphisms, it's fine, otherwise you're dead screwed.

@DanielFischer: yeah, I can just suppose the group action to be as I want

I just don't even know what the derivative of a group action would be

maybe I already do but I don't realize it

then diff $x \mapsto f(gx)$

for some $g$

Calculate in 6 different ways $$\int_0^1 x^{n-1} \log(1-x) \ dx$$

@Huy You don't need any such, you just need the derivative of $x \mapsto g^{-1}x$ and the chain rule.

hm

@BalarkaSen I've looked at the Latex arrows in the hope that might shed some light, it is similar to $\rightarrowtail$ I guess, only it's a ) shape on the left.

no idea

It's even bracket sized. So isn't that tailed one.

Calculate in 7 different ways $$\int_0^1 x^{n-1} \log(1-x) \ dx$$

This is enough.