where do I start XD

which problem is this?

1a

section 7.2 1a

I think the same page as 2a which is 440

my 7.2 1a is about decay order O.o

oh wait nvm

that's not to the $\xi$ but talking about the fourier transform of $e^{i bx } f(ax)$

yup

so I have to use that with the absolute integrable theorem

oh I see I have to use the absolute integrable theorem with the fourier transform of that e^ sdfslafjfjsda;fjdslak;fj?

@Chris'ssis $\int\log (\log x )\log^a (x)\mathrm{d}x$

@usukidoll yep!

@usukidoll PS: ironically I am a PDE guy ;P

dude ! @DanZimm do you have skype or email or something? you're really good at teaching me this...

;P I don't have too much time usually

at the end of my break from school atm

awwwwwwwwwww come on

please

pleaseee! I'm at the end of the semester XD

I think I helped you before ;P (fourier series stuff)

u have skyperz right?

@Usu you know that for 1 on 1 tutoring people usually charge lots of $$$$?

^ I'm aware of that

craziness

no I do not have skype

aww

email? XD

not now, sorry mate

unfortunately a bit too busy to dedicate time to helping

@Studentmath I tutor for free when I can

-.-

:'(

WOULD I EVER SEE YOU AGAIN!?!!?!!

I come here from time to time lol

but I NEED YOU D:

@DanZimm that's some good deeds :)

@usukidoll hardly, work at it yourself and you'll become self sufficient over time

@Studentmath I view it as a "pay it forward" dealio

maybe when the lectures start...maybe that's why I'm all aAHHHHHHHHHHHHH!

probably

lectures help

@DanZimm very much alike to the concept of the website

Yeah, lectures help..

the website?

SE

didn't know there was a site O.o

I learned it from a movie

watched it whilst growing up, stuck with me

imdb.com/title/tt0223897

Yeah I saw it first when I was 7, couldn't get why everyone said the ending was 'hopeful'

hopefully you understand now?

A bit better

(no pun intended)

time to wtch a movie, so I might disappear, if so, have a nice insert proper timezone phrase here

for me it's night/morning since I stayed up all night and it's technically the morning now

same it's the morning for me too

hrm anybody have a cat? trying decide if I should get one

I HAVE PLENTY OF CATS

oh lord lol

my cat needs a bath

went into the neighbor's yard and rolled in the dirt

how much maintenance does just one take? meaning hours per week

depends on the cat's personality

I have some mellow ones which don't require that much attention

and i Have a screamer cat who would scream MEOWW if you don't pay attention or feed her or pet her

annoying

a mellow non-talkative cat is what you need

does it still interact?

yeah

but you know when you're away it doesn't go MEOW every five seconds

I have a cat that's attached to my mom

whenever she goes out for a long time, she goes meow meow meow and won't stop until she is home

I'm serious...good luck sleeping because that's not going to work

but that's the only cat that's picky

the rest can take care of themselves and could care less

mine is a screamer then

^.^

never knew they had a name

I'm guessing @usukidoll just made it up O.o

@BalarkaSen Waa...you have cat?

@DanZimm Maybe not :/ ?

sure @Sawarnik

tired :/

night guys

byee.

night

lemme know if you need help again - might be on tomorrow night

@BalarkaSen You never told before .. doesn't it disturb you all the time?

no.

who cares for the cat?

i, mostly.

wow.

like you have to feed, bath, blah the cat all the time .. isn't it lot of work?

bath

lol

I thought cats shit/pee all over the place

feeding them is a little bit of work but you got to work if you're gonna have pets.

but apart from that, he never annoys. the meowing isn't really annoying.

I had a dog once

@UserX that's not true

they're very careful about not dirtying their masters caretakers home.

heya @anon

ok, that's it. 8 smacks!?

he's asleep

call back later

lol

@blue

i guess he's awake then

@BalarkaSen :O The cat's like me.

:D

i didn't bath for a whole month

i bathed after 2 weeks today.

@UserX me too.

@UserX the cat never ever goes to bath?

cat-egories of meownifolds.

:O :O

:D

@BalarkaSen no.

Okbyes :)

@DanZimm vot

oh you know

?

you're claiming that meownifolds don't exist?

I'm saying no to playing on math words with cat words

@Sawarnik only if the cat ate something bad

@Dan cat-astrophe!

oh there's a better one

cat-alan

Greetings

@Chris'ssis any ideas on the integral? One liners if possible?

@UserX What is $a$?

cat-theodory criterion

@Chris'ssis independent parameter

cat-hedral. now christian's are going to dogpile on me.

Question: If we are looking at $1/n \cup 0 \in \Bbb R$, isn't every subset of that subset an open set?

what is the subset, @Studentmath?

@UserX Check the first few cases for a integer. I'm sure you see a pattern there that can be turned into a solution by using the integration by parts (I think).

every number of the form $1/n$ where $n$ is probably positive natural number, and $0$, in $\Bbb R$

Anyway, now I'm working on something very tough, the guys on I&S didn't solve.

hrm {0} isn't open

@Chris'ssis I want to avoid IBP. I'm trying to one-line $\int_0^1 \log^3 x\mathrm{d}x$

Ah yes. But every neighborhood of {0} is open, right?

On the other hand every {$1/n$} is also open as it is an isolated point

depends how you define neighborhoods

open with respect to what topology?

@Studentmath the topology being inherited from R? otherwise you can easily discritize it.

with respect to the topology on $\mathbb{R}$ then no point is open

what @Dan said

Neighborhood of $x$ is any set so that $x\in IntA$

the only open sets are unions of the form $(a,b)$

@UserX Good luck then. Let me know if you find something better than the integration by parts.

then your neighborhood doesn't have to be open

There is a closed neighbourhood of ${0}$?

[-1, 1]

and remember closed does not necessarily mean not open

@BalarkaSen hrm?

nevermind.

Of course, but in this case it's not open - -1, 1 are boundry points are in the segment.

However if we were to look at $X$ as the metric space, then {$1/n$} would become an open set, right?

$X$ as the metric space? What is $X$?

{1, 1/2, 1/3, ...} I guess

@Studentmath that's a one-point set. it's closed.

every number of the form $1/n$ where $n$ is probably positive natural number, and $0$

oh wait nvm

metric space

right

5 people in here think i am boring and 8 wants me to be smacked by Ted? serious?

lol

@Balarka it's also open, isn't it? I have a theorem stating that $x$ is an isolated point in $X$ iff {$x$} is open

@Studentmath yes every point would also be open

@Studentmath clopen, right

however $\{0\}$ would not be open

But every neighborhood of 0 in $X$ would be

limit point

sure @Studentmath.

Also, something that has been bothering me - every clopen set has no boundry points, right?

@Studentmath yes I believe so

it is true

@Studentmath what do you mean?

@Studentmath assume the boundary is nonempty

contradict it

Cheers, thanks a lot @DanZ and @Balarka

@BalarkaSen Lol, I was tagged by that

@BalarkaSen what about two disconnected sets

Indeed, I right away get an $x$ both in $IntA$ and in $ExtA$

each of which are clopen in the subspace topology

and they can have boundary

gah i am thinking about metric spaces again

smacks himself

Oh I am talking about metric spaces

even in metric spaces, you can do this in $\mathbb{R}^2$

interestingly, I've been thinking about categories a lot recently

trying to learn about it... but I guess I should first learn about abstract algebra

@DanZimm i don't think so

@BalarkaSen Take $X = B_{(0,0)}(1) \cup B_{(2, 0)}(1)$

@Danu you need to learn algebra to do categories

otherwise they're just abstract nonsense.

@BalarkaSen yeah, so I've started learning some algebra (I'm a mathematical physics student, so I have to do it on the side... sigh )

@DanZimm er.. that's closed?

O.o in the subspace topology both $B_{(0,0)}(1)$ and $B_{(2,0)}(1)$ are open and thus both are closed, so they are both clopen

they both have a boundary

metric spaces

@Balarka @DanZ it's easy to show that in Metric space, A is open iff there is no boundry point in $A$, and $A$ is closed iff all boundry points are in $A$

$\mathbb{R}^2$ is a metric space

the subspace is a metric space

nah that was not the question

It follows immediately that in metric space a clopen set doesn't have any boundry points I would believe

ohhhhh I see the issue - right

The boundary is not in the space itself, correct

I was thinking of the boundary existing in the $\mathbb{R}^2$ topology

oh $X$ even has a boundary

union of the boundary of the two balls

hrm something is fishy here

right yea this isn't going to be generated by the same metric

for example take those balls to be closed, then it would even contain the boundary

oh wait no it is

exactly

hrm

@DanZimm boundary of some subspace $A$ of metric space $X$ is the set on which every open ball around each point intersects both A and A'

Yea

I was mixing things up

the boundary does not exist in this subspace topology

@Chris'ssis Do we have generating functions for $\dfrac{\ln(1+x)}{1-x}$ and $\dfrac{\ln(1-x)}{1+x}$??

right

And this is for more than just metric spaces, in general this is how it is

my apologies

if you define the boundary like that, then sure

i hate nonmetrizable spaces

nonhausdorff space in general are very annoying

the boundary, generally, is the intersection of the closure of $A$ and the closure of $A^c$

Hrm, wait - every neighborhood of $0$ in that metric space is clopen, not just open, right?

@Venus Something like $$\sum_{n=1}^{\infty} H_n x^n \frac{x-1}{1+x}=\frac{\log(1-x)}{1+x}$$?

The Metric space I spoke about, not $R^2$

yes I think

i believe so @stud

@UserX how's your algebra study going?

have you learned lagrange's theorem?

@Chris'ssis Yes, generating functions that involve harmonic number

@Chris'ssis So $$\frac{\log(1-x)}{1+x}=\sum_{n=1}^{\infty} H_n x^n \frac{x-1}{1+x}$$How about $$\frac{\log(1+x)}{1-x}$$

@Venus Can't you figure out now?

$$\frac{\log(1+x)}{1-x}=\sum_{n=1}^{\infty} H_n (-x)^n \frac{1+x}{x-1}$$

@Venus ^^^

Thank you @Chris'ssis ^^

@Venus Are they OK for you?

@Venus Welcome.

Actually not really because it couldn't help me to answer my problem, but glad to know their GF

@Venus Which problem?

This one

I mean my calculation, not my problem

@Venus Well, the problem is not hard at all, but I can show the problem I'm working on right now (it might be interesting for you).

@Venus (+1) for the answer

@Chris'ssis which problem?

Thank you ^^

This one that I created it today

@Venus $$\int_0^1 \left(\frac{\log(1+x)}{1-x}+\frac{\log(1-x)}{1+x}\right)\operatorname{Li}_2(x) \ dx$$

@Chris'ssis That looks like evil :D

@Venus :D

@Venus You know, in general when I refer to solving things I refer to solving the problems in the spirit of the art since only solving a problem is never enough. This is my way of seeing things.

@Chris'ssis Do you think the last integral I asked you has a closed-form? The one with factor $e^{\sec x}$?

@Venus I didn't work on it yet. There should work using some series involving Bernoulli number. I need to find some of my older papers.

@Chris'ssis Me too. But I need to learn much from other users here

@Venus no matter how good you are, I think there is always room to learn from the others.

@Chris'ssis Recently I visit many forums & it looks like lots of users here are also joining there

@Venus I'm only here.

@Chris'ssis Indeed. We always learn for the whole life

@Chris'ssis I thought you also join in I&S and w3facility.org

@Venus I enter I&S once in a while, maybe once per 1,2 weeks, especially when I see links here to there.

@Venus mother of god ! very nice and tenacious answer ! :D (+1)

I also only join here, not make an account anywhere. Just visiting

@Chris'ssis I got the problem from here: brilliant.org/community-problem/seems-too-easy-2

@r9m Thank you. I learn from users here ^^

@r9m There is a very nice way to finish my problem above. Do you see it? :-)

@Chris'ssis Nicee !! :)

@r9m Yeah. I created it today. :-)

@Chris'ssis I doubt there's such a closed-form for that integral I gave

@Chris'ssis Share the closed-form or post on the main

@Venus I didn't study the details yet.

@Venus No need to post it on the main, it's solved. Well, I'll post the closed form later (now I'm writing up its proof)

@Chris'ssis You're incredible!

@Venus Why?

@Chris'ssis You already found its closed-form

@Venus How would you consider it as difficulty from $1$ to $10$?

@Chris'ssis I think the scale would be different if you ask the people, but personally I'll give 7

Maybe higher

@Venus OK

Okay 2 views in total and off the front page (I hate not having an easy question) math.stackexchange.com/questions/1044986/… bump

Also WTF is intersect in LaTeX - I know cup looks proud

\cap

@Venus I solved it. The answer is seven.

@Alec In general papa Van Kampen says $\pi_1(X) \cong \pi_1(A) \star \pi_1(B)/\langle \pi_1(i_A)^{-1}, \pi_1(i_B)\rangle$

where $i_A$ and $i_B$ are the maps $A \cap B \to A$ and $A \cap B \to B$ resp.

Riiiight

So use that.

in fact that group generated by those two maps are precisely the kernels.

PS : It's lame to do that exercise using papa Van Kampen.

@Barkla it's not properly covered in the book, so .... I'm looking more for guidance than a nudge

Use a better book then.

Thanks

@AlecTeal I can't guide : I'm too lazy to compute those $i_A$ and $i_B$s.

@AlecTeal :P. Use Munkres.

I know, but thanks for trying

Any way I can get it so I don't get notifications from some other user named Nick?

@BalarkaSen I am

Yikes @Alec. My covers are blown. I haven't really read that chapter. :P. But I do know the baby version of the theorem.

$X$ path connected, $A, B$ open covers $X$, $A, B$ simply connected. Then $X$ is also simply connected. It's a far less pain in the neck.

@BalarkaSen can you help or not, seriously I've been stuck on this for a while, bragging about knowing the answer is just annoying me. Also not simply connected - path connected.

I can't help and I don't know the answer.

Great.

@AlecTeal But let's try to do it and see where I get stuck. It might not help though.

Let's see, what is $A \cap B$?

that's just set theory

It's an open tube

I'm just thinking out loud.

So we want to determine $i_A$. This guy is $i_A : S^1 \times (1/4, 3/4) \to S^1 \times [0, 3/4)$

It's a map from the open tube to the neither open nor closed tube. Hmm, what could possibly be the corresponding map.

we want to determine maps $\Bbb R \to \Bbb R^+$ after all right? That'd give us the map we want by crossing out by $S^1$.

....

I got much further than this alone... try getting some paper maybe.

I appreciate the thought @BalarkaSen but I've been stuck with this for a while

I know and I am trying to get stuck with this too.

wait a second i think i have an idea

The open tube covers the torus, right?

So this chap, the $\pi_1$ of the open tube, should inject onto $\pi_1(S^1 \times S^1) = \Bbb Z \times \Bbb Z$

@AlecTeal Actually, $A$ and $B$ are really homeomorphic. So $\pi_1(A)$ is isomorphic to $\pi_1(B)$, no?

So $\pi_1(i_A)$ and $\pi_1(i_B)$ are not really different maps.

....

That should make the normal subgroup $\langle a, a^{-1} \rangle \cong \Bbb Z$ right?

Thus $\pi_1(X) \cong \Bbb Z \star \Bbb Z/\Bbb Z \cong \Bbb Z$.

i believe we can rigorously do this?

yeah well we can.

bah this is not so hard after all

@Alec what d'you think of the idea?

$\pi_1(i_A)$ is $\pi_1(i_B)$ when composed with the isomorphism $\pi_1(A) \to \pi_1(B)$. Thus $\langle \pi_1(i_A)^{-1}, \pi_1(i_B) \rangle \cong \langle x^{-1}, x \rangle \cong \Bbb Z$ Can anyone say if I am right?

@Alec Remember that the kernel involved in the van Kampen theorem is written explicitly: it's (generated by) elements of the form $\iota_A(x)*\iota_B(x)^{-1}$, where $x \in \pi_1(A \cap B)$. Now $\iota_A$ and $\iota_B$ are homotopy equivalences, and induce an isomorphism on the fundamental group. So every element of $\Bbb Z$ is hit. Let's write $g_A$ and $g_B$ for the generators of the respective fundamental group. our normal aubgroup is generated by elements of the form $g_A^n g_B^{-n}$.

@MikeMiller ^

Now prove that the subgroup these elements generate is precisely the subgroup consisting of words in $g_A$ and $g_B$ whose exponents sum to 0.

did I mess up or what I did was right?

(I mention this last bit since you wanted an explicit description of the normal subgroup.)

I didn't pay attention to what you did.

@MikeMiller I cannot get Latex to work at all on this can you post that as a comment

Hey @Alizter lol.

I can't, sorry, on my phone. Try pasting it into an answer box somewhere

OK, then, nobody confirmed if my thought was right. How unhelpful.

@BalarkaSen it slowed down almost to nonexistent by now. I gotta study for my school

@MikeMiller When can we define ring structure on $\pi_1(X)$, for some path connected space $X$?

A close miss is for topological groups, when we see that the apparently different structures on $\Omega(X)$ by our $\star$ operation and the normal multiplication $\cdot$ turn out to be the same.

I guess topological groups are the natural spaces we should look for to get ring structures.

is there any such things as "topological rings"? i.e., where addition and multiplication are both continuous? if they exists, those might be a good place to look for.

It doesn't distribute. Think about $S^1$. I don't know of any interesting ring structure on $\pi_1$.

@MikeMiller what doesn't distribute?

If addition is $*$ and multiplication in the underlying topological group, these two don't satisfy distirbutivity.

well of course they don't, they're the same

homotopically speaking

on $S^1$ they are...

on in general topological groups they are

it is a standard exercise

Prove it. Anyway, if that's true, what were you suggesting as the ring structure?

i just said those were a close miss

Doesn't seem close at all. Anyway, I don't know of any natural way to do it.

@MikeMiller as for the proof : $G$ be a topological group with $\cdot$ operation and some point $x_0$ fixed. $f, g$ be two loops in $\Omega(G, x_0)$, the set of all loops based at $x_0$. Define $f \otimes g = g(s) \cdot g(s)$. Then $f \otimes g = (f \star e_{x_0}) \otimes (g \star e_{x_0})$ which is $f(2s) \cdot e_{x_0}$ when $s \in [0, 1/2]$ and $e_{x_0} \cdot g(2s-1)$ when $s \in [1/2, 1]$. This in turn is $f \star g$.

:)

@MikeMiller I am doing Munkres not for nothing you know.

And no, by the way, you won't get anything interesting in a topological ring.

Well there probably is no way to do it

If there was, there would have been a good way to realize the multiplication op in $\Bbb Z$ in $S^1$, and $S^1$ can hardly be given a topological ring structure.

And why can't $S^1$ be given a topological eing structure?

You're right. But you're also guessing.

I dunno. It's just an intuition.

Right, I am just guessing.

I see absolutely no reason to have that intuition, but whatevs.

Fact that, for everyone I know, is unintuitive: Hausdorff topological rings are totally disconnected.

Oh?

interest

I never read the proof. You can find it if you Google. I doubt it's particularly interesting.

Yeah well I believe it, though.

The prospect of something like "Lie fields" would make galois groups a super-duper-pain.