Tell him the side effects

Tell him it's not helping.

Therapy combined with medication is the best path to follow they say ...

...the therapist can adjust the meds as you need.

They are trained to help.

probably too late

What did you say to him?

well, I've known you for what? a day?

Can't imagine how @Jasper feels about you

considering my experience

:(

@AlexClark I don't think my IP changes

hi mr eyeglasses and @AlexC

back's hurting at the moment, but fine, and yours?

so far ahead of Christmastime?

lots, @AlexC ... never studied probability, statistics, numerical analysis ... wish I had.

There are lots and lots of applied things, too ...

How do I do $\frac{sin(x)}{1+cosx}+\frac{1-cosx}{sin(x)}=2cscx$

I think its written wrong from the teacher but don't know

I found a solution if its 1-cosx as the denominator on the left

@AlexClark No.

Probably triple digits.

@Maximilian: With regard to your trig question, yes, you're right.

Helllllo

Heya @Anthony

Can anyone here explain the paving conjecture to me?

Never heard of it.

Noooooooo.

You know ... the road to hell is paved with ...

Math?

Conjecture: there is not enough funding allocated to properly pave the US road system.

Proof: trivial.

Not to mention all the bridges that are falling down ...

For $\epsilon > 0$, there is a natural number r so that for every natural number n and every linear operator T on $l^n_2$ whose matrix has zero diagonal, we can find a partition (i.e. a paving) $\{A_j\}^r_{j=1}$ of $\{1, · · · , n\}$, so that $||Q_{A_j}T Q_{A_j}|| ≤ \epsilon ||T||$ for all $j = 1, 2, · · · , r$

@Anthony: As far as I can tell this is about the rep theory of Banach algebras? DanielF is probably your best bet here.

"Yikes!"

He generally seems to be...

I have no clue what you just typed.

<3

hi can anyone explain to me why if $\int_E |f|^p_1 = \int_E |f|^p_1 \Chi_E$ then $\Chi_E \in L^q$

Alright alright alright. I'll just go stare at my monitor for a few hours

Don't go blind, @Anthony.

Roger roger

@desperatemuch: I'm sorry, this doesn't appear to make any sense. Context?

sorry

suppose q is the conjugate of p

heya @robjohn

@mike

We knew that much, @desperate

@MikeMiller if suppose q is the conjugate of p

Why do you have a 1 subscript? What is $E$? What is the measure space?

@desperatemuch: It still doesn't make any sense. Take $f$ to be any $L^p$, essentially nowhere zero function on a space of infinite measure, and then take $\chi_E$ to be the indicator function of the whole space.

oh, initially p is p2/p1>1

Then $\chi_E$ is assuredly not in $L^q$ for any finite $q$.

and f is in L^p_2

the problem is if E is measurable set of finite measure and 1<=p1<p_2<infinity

then show that L^p_2 (E) <= L^p_1 (E)

Ah ...

my prof started with defining p=p2/p1 and q as the conjugate of p

sorry sorry so new to these things

the problem states that if $E$ is a measurable set of finite measure and $1<=p_1<p_2<\infty$. Then $L^{p_2} <= L^{p_1}$

my professor initially defined $p=\frac{p_2}{p_1}$ and $q$ the conjugate of $p$

You lost me! Maybe @Ted will help.

Ted has to leave momentarily.

It's an immediate application of Hölder.

then he said that if $\int |f|^{p_1} = \int |f|^{p_1}\Chi_E$ and $mE<\infty$ implies $\Chi_E\in L^q$

@TedShifrin: I'm ducking out :) But I'm sure your comments are appreciated.

@desperatemuch: \chi, lowercase.

Well, I have company showing up any second.

i am quite confused as to why would $\chi_E$ be in $L^q$

@tedshifrin right how?

Because you're on a finite measure space, @desperate.

That the problem was written wrong?

Yes, @Maximilian.

Half of them areI feel like, like 4 I see no solutions to unless its slightly changed

Frustrating, @Maximilian: Complain to your teacher.

I will, although kinda hard because tomorrow is the test

I don't know why everyone around here waits until the day before the test to start studying and learning. rolls all eight eyes

why q not p?

lol

I had work

It doesn't matter, @desperate.

I still have work

@TedShifrin so i can also say it is in p?

then why q?

You want to use $\|fg\|_1 \le \|f\|_p\|g\|_q$ for appropriate choices of everything, @desperate.

Stuck on $\frac{1-sinx}{cos(X)}+\frac{cos^2x}{1-sinx}=2secx$

I'm at $\frac{1-sinx}{cos(x)}+1+sinx$

OK, my guests are somewhere ... locked in a garage. Bubye.

thanks @TedShifrin

@TedShifrin That's the best place.

mods of other rooms shouldn't be able to ban someone from all the rooms

Anyone know how to figure this out? Anything I do is dead end

Nvm. Teacher corrected them online

where is @PedroTamaroff?

I miss him

@Maximilian but $\frac{1-\sin}{\cos}+\frac{\cos^2}{1-\sin}$ is not $2\sec$

what's up @anon

I know. It was written wrong on the sheet I have. He changed them up online which is why half of them I couldn't figure out

@anon hi

@anon I am very excited to take algebraic topology next semester

Hey guys, could use some help. I'm looking to converting a $\mathbb{R}^p$ vector to polar coordinates

I know if I'm working with $p = 2$, say $\mathbf{x} = (x, y)$, I just use $r$ equal to the norm of the vector and $\theta = \tan^{-1}\left(\dfrac{x}{y}\right)$.

Now if I have, say, $\mathbf{x} = (x_1, x_2, \dots, x_p)$...

Yeah, I have no idea

I'm going to dig through some texts

Hey, guys. Perhaps I'm reading too much into Riemman's theorem on conditionally convergent series. But I think I need a second opinion: the theorem states that for every convergent series (that isn't absolutely convergent) $\Sigma a_n$ and for every $x$ and $y$ in $\mathbb{R}$ (such that $x\leq y$), you can find a rearrangement such that $$\liminf_{n\to\infty} a_1 + a_2 +\dots + a_n = x\qquad\limsup_{n\to\infty} a_1 + a_2 +\dots + a_n = y$$

Does that mean that if I have a convergent series that satisfies the hypotesis, I could reorder it such that my birthday came up as the limit?

just by putting $x=y=18101994$?

@AlexClark Out of curiosity, do you know how to convert a $\mathbb{R}^p$ vector to polar coordinates?

@AlexClark The definition is $x_1 = R \cos(\theta_1)$, $x_k = R\left(\prod_{i=1}^{k-1}\sin(\theta_i)\right)\cos(\theta_k)$ for $2 \leq k \leq p$, and $x_p = R\prod_{i=1}^{p-1}\sin(\theta_i)$.

@AlexClark for a vector $\mathbf{x} = (x_1, x_2, \dots, x_p)$

No problem

My guess is that $R$ is equal to the norm of $\mathbf{x}$, but I'm stuck as how to prove that

@AlexClark Yes, used to be

I'm from the U.S., midwest

Yep

and I'm working as... I guess you could say, a statistician, data-analyst, data-scientist-ish type role

YES. SO MUCH

Indeed

I don't use caps often

:)

Slightly. Took a $5.7k cut

$G$ is not simple because 36 isn't prime and it's less than 60 so it's solvable

#overkill

tomorrow is my exam

pray for me

@AlexanderGruber what's the impact of not being prime? was it assumed cyclic?

@anon only solvable simple groups are cyclic of prime order, and it isn't of prime order

#PrayForTwink

Riemannian geometry

do Carmo

:'(

I also use O'Neill

cause it's easier to read and explains in more detail

I hate the Christoffel symbols :'(

I'll almost go to sleep

:D

ok

@Jasper Why would you do that.

thank you @Jasper

You should be refusing to engage with him based on his username alone.

I don't know why that's not a permabannable offense tbh.

what's wrong with my username?

@Jasper No, but it's not decent at all to have that as your user name.

because I flagged his comment

my username "user" was banned for 12 hours because of him

he deserves it

and he said Danke to the mod who banned me

which means Thank you

in German

@0celo7 I don't see having twink as one's username as inherently actionable. I can conceive of users using it in good fun. However, here the purpose of the username is as part of his broad campaign of baiting as many people as possible into at-first-quasi-deniable trolling, and trolling is looked down upon. Speaking of which...

How do you think the future will play out if you can't resist the urge to troll for even 24 hours @Twink?

This particular person needs professional help.

Anybody here also applying to grad programs?

Your "king's banquet" message is valid up to a point pal. But this has been going on far too long....

A banquet for a king @Jasper

Or large.

Alex put up the link pal. @Jasper

Same goes for skull patrol :P

I will say that many times :-D

@AlexClark i know. that was the joke.

How was that illegal thing you were planning on doing end up @Jasper?

Good, good.

That's the spirit pal :D

@AlexClark Let's see.

How may 3-Sylows do we have? Either 1 or 4, correct?

If 1, we're done.

So assume we have 4.

Then we have a permutation representation G --> S_4 by conjugation (Sylow's IInd theorem)

This has to be injective.

Great book, @Jasper.

@AlexanderGruber Know any neat, cute ( = fun and elementary) applications of the Haar integral? The best I can think of is (i) pointing out the Mellin transform is the multiplicative analogue of the Fourier transform (which is additive) and (ii) generalizing commuting probability to compact groups. (But I think that's only nonzero for abelian-by-finite groups, which I can't think of a good natural example of.)

@AlexClark So you need to show that it's not possible to embed a group of order 36 inside S_4. Is that always possible?

suppose I can now include @MikeMiller in the above question

Well, obviously.

I know nothing of fun, so you're looking in the wrong place

heh

@AlexClark How in the world can a group of order 36 embed in a group of order 24 :P

So contradiction.

Hey Mike what does UCLA have going on in terms of geometric analysis?

It's pretty insightful!

@AlexClark Alright, let's try differently.

There are 4 3-sylows of order 9.

So how many elements, altogather?

hello, it's me

Hi pal

hi

fair enough. this is going to a messy argument, @AlexClark.

I'll keep on calculating the Christoffel symbols

bye

Later pal

Dunno @Jasper

Creative with those usernames I see @skull

@anon: 1) It implies any compact group action is by isometries w/r/t some metric. Values: a) It cannot simultaneously fix a point and its tangent space. This implies, say, this. b) The maximal dimension of a compact Lie group acting on a manifold is immediately $n(n+1)/2$. 2) Most elementary theorems of differential topology hold equivariantly, by taking the original theorem and averaging $\int_G L_g^*(\text{standard proof proof})d\mu$.

@AlexClark I don't know, I haven't thrashed out the argument myself.

Just use the homomorphism into S_n.

Looks kinda like Sharon Stone?

Need to bang against more walls while you're thrashing there @Balarka

@MikeMiller If G acts on M and H is any lie group acting trivially on M, can't we get GxH to act on M?

Why, @AlexClark?

@AlexClark Ring of adeles?

effectively

ah

$\Bbb A_\Bbb Z$.

@KarlKronenfeld haha.

@AlexClark Go read Tate's thesis.

what's the problem @AlexClark

whereas it's easy to prove this by using homomorphism onto S_4

arbitrary standards indeed

@KarlKronenfeld thanks :-)

No, it is absolutely not lame.

It can be used to classify groups of a certain order.

In particular, 12.

:P

@anon: which reminds me, you might like that the only spaces for which that dimension is possible are space forms, simply connected or $\Bbb{RP}^n$

hmm

perhaps change that to "w/ finite fundamental group", i think

Hi everyone, i have just written an answer to a question in Abstract Algebra. But i am not sure if the answer is correct. Can someone please look math.stackexchange.com/a/1520413/197214