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6:03 PM
@BalarkaSen Come with me to the NT room
 
@Alizter Did you get some sleep last night?
 
@teadawg1337 Yes thank you :)
I went to bed at 10
I was too tired to see that the question was erroneous :P
 
@anon have you seen the analogy of monodromy representation with representation of gal(\bar q/q) stuff i posted?
what d'you think?
 
haven't looked at it
I has school first
 
@anon gal(\bar q/q) \to Aut(Z/m \times Z/m) resembles \pi_1(X, x_0) \to Aut(p^-1(x_0)), don't you think?
can this analogy be made rigorous
 
6:16 PM
latex
 
too lazy
 
@BalarkaSen what's the first map?
 
@anon the usual action of gal(\bar q/q) on the m-torsion elements of some elliptic curve over C
 
$\operatorname{Gal}(\bar {\Bbb Q }/ \Bbb Q) \to \operatorname{Aut}(\Bbb Z/m \times \Bbb Z/m)$ resembles $\pi_1(X, x_0) \to \operatorname{Aut}(p^{-1}(x_0))$
 
ah. I'll consider it later.
 
6:17 PM
that looks even worse, @Alizter
 
what is p?
 
@anon in short, i am looking for a geometric analogue of the 2-dimensional representation theory of Gal(\bar Q/Q)
@Alizter some covering map Y \to X, Y and X path connected.
 
is it ok now
 
yeah
 
6:30 PM
Hey guys, any help would be appreciated...
http://math.stackexchange.com/questions/1046947/class-of-lebesgue-lebesgue-measurable-functions
 
Grrr, I'm one upvote away from earning the Self-Learner badge...
Ugh, this is infuriating... The ratio between the amount of work performed and upvotes is unfair
 
6:48 PM
Hi @alizter, lol.
 
@WillHunting I never understood what was so funny about saying Hi.
 
@Alizter I just like to lol. I am a joke.
 
I feel like this could be published in a mathematical journal, but I'm not a writer and I don't even know if it's been done before...
2
 
@venus Sorry, I thought you would write the question yourself first before asking me. I see your post has been deleted, lol.
@TedShifrin I see you have been lecturing people on what not to post in this chat again, lol.
 
@teadawg1337 You have to give a motivation. A good one.
 
6:57 PM
@Alizter A motivation?
 
@teadawg1337 Journals like interesting things.
 
@Chris'ssis If you really want math books you don't have to go to the city to buy them, you can shop on amazon.com for example.
 
Nobody will want to publish it unless there is something interesting or something that it shows/answers
 
@WillHunting The book I finds in my city I don't find on amazon.com.
 
oh and MSE is the best journal :P
@Chris'ssis How is your book coming along?
 
6:59 PM
I've been looking for the past week or two, I can't find anything similar anywhere
I'm just afraid to approach anyone with it on the offchance that it's been done before...
 
@Alizter Working on that stuff.
 
@teadawg1337 Wow, don't forget me when you win the Fields medal.
 
I'm 18, I have no experience as a mathematician, I don't wanna end up the laughing stock of the mathematical community....
 
@teadawg1337 Mathematicians only laugh when drunk or suprised.
The other cases are left to the reader.
 
@Alizter I lol all the time, but I am only a banana.
 
7:04 PM
@teadawg1337 You're already quite good
 
@teadawg1337 are you in the US?
 
@Alizter Yes
 
Are you going to go to uni?
 
My GPA is crap, but I have a 32 composite on the ACT... I got a 4 on the AP Physics B exam without studying....
 
Maybe you enjoy my way here ...
0
A: A series involving $\prod_1^n k^k$

Chris's sisBy Cesaro-Stolz theorem, we have that $\displaystyle\lim_{n\to\infty}\frac{1}{n^2 \log(n)}\sum_{k=1}^{n} k\log(k)=\frac{1}{2}$ and then we see that for $n$ large, we have that $\displaystyle\frac{1}{n^2}\sum_{k=1}^{n} k\log(k)\approx\frac{\log(n)}{2}\Rightarrow \frac{-4}{n^2}\sum_{k=1}^{n} k\log(...

 
7:07 PM
I'm going to a community college next semester to get credit for an advanced math corse
 
OK. @teadawg1337 you don't need fancy grades to study math
 
@Alizter I know, but I'm dreaming to go to Berkeley one day
 
@teadawg1337 I am hoping to be reborn in Germany next life.
 
@Chris'ssis I made it easier to read. Blind dysrexic people like me cannot read your first draft.
 
@Alizter OK, thanks! :-) I tried to put all in a few lines :D:D:D
 
7:10 PM
@Chris'ssis Sometimes squishing into a single line can ruin the result.
 
@Alizter yeah ... :-)
 
Would you fold a piece of paper because it is important?
 
Sure :D
Fold the proof that I'm 16
 
@Hippalectryon Same age high five.
slap
@Hippalectryon I didn't fail my chemistry test today!
In fact I may have aced it.
 
@Alizter :D
 
7:14 PM
@Hippalectryon Oh hush, I'm not that good...
 
One of the questions was: Find a moleucule that has the same number of atoms, same number of electrons and same shape as $\operatorname{Al Cl}_4^-$.
I think $\operatorname{Si Cl}_4$ qualifies.
 
I feel cheated, though... That answer took over a week of research and three days of formatting a HUGE post, and I only have two upvotes for it
 
@teadawg1337 Well, I suggest you focus on math in real life and don't spend too much time on this site. It's largely a waste of time.
 
@WillHunting I wouldn't say it was a waste of time, I can't describe how awesome it felt to solve it
 
@teadawg1337 It feels more awesome to win the Fields medal, bro. But I am too old for the medal.
 
7:28 PM
@WillHunting Nah, I don't deserve a Fields medal
 
@teadawg1337 By the way, I am 33. I hope to go to grad school in a couple of years or so.
 
Besides, I kinda stole the spotlight from @DavidH...
I should've waited for him to edit his post, but...
I just couldn't resist making my post.... I feel awful....
 
@Alizter, I wanted to ask you about my suggested edit that you rejected. Based on this feedback, I thought it was a beneficial edit.
Could you please help me understand why it might have been inappropriate?
 
@sjrosen Yup sure.
You were adding significant content to an answer.
 
7:43 PM
The edit must not change the intention of the original poster.
 
@Alizter I think you may not have seen the OP's comments. All of the content that I added came directly from his own comments, and almost entirely in his own words.
 
It should fix typos and grammar and formatting though.
 
@sjrosen They are comments however.
The op is in charge of what they want in the post.
Unless it is community wiki
If you feel it is better for the comments to be moved into the answer, leave a comment telling the OP to do so politely.
 
@Alizter I understand what you're saying, but in this case, there was a lot of back-and-forth clarification in the comments, and the OP added a significant amount of important information. Essentially, the question was answered mainly in his comments, not in the answer itself.
 
@MikeMiller about the $p^2$ groups, I got $\Bbb Z/p^2\Bbb Z$ and $\Bbb Z/p\Bbb Z\times\Bbb Z/p\Bbb Z$ (dunno how to denote the cartesian product of groups, can't remember how it's notated). No idea if there are more or how to prove that there aren't.
 
7:47 PM
@UserX yes, but it's not entirely trivial job to prove that
 
That's all of them, @UserX. You need tools to prove that there aren't others.
 
@sjrosen Hmm. In the edit queue this is not apparent. I see your point.
 
no. you should be able to tell why
think
 
Better stated question
 
@Alizter It seems like it's difficult to get this across in the edit queue. How would you suggest I approach this situation (and other similar situations I might come across in the future)?
 
7:49 PM
If I prove they're not IM what else do I need to show to prove that there are no more groups up to isomorphism?
 
@sjrosen These kinds of edits are rarer than most. I would not worry too much about it.
 
I'm trying to show that $I_m = \{f \in L^1(\mathbb T) : \widehat f(m) = 0\}$ is a maximal ideal in $L^1(\mathbb T)$. Any ideas?
 
@MikeMiller you said I'll be able to classify them easily...
 
@sjrosen Carry on contributing, you have done nothing wrong :)
 
@UserX Did I? I think I took it back right afterwards.
 
7:51 PM
Were there any more vulgar edits to my posts last night? I went to bed after rejecting two suggested edits to my post
 
@MikeMiller I think I didn't notice that
 
First, you need to show that such groups are abelian; then you need to invoke some form of the classification theorem of finitely abelian groups. You might be able to prove the appropriate special case more easily than the general one, I dunno.
 
@Alizter Fair enough. That said, I've already taken the time to write this edit so that it reads well and better reflects the OPs entire answer. Would it make sense for me to resubmit it, or should I just leave it alone at this point?
 
first @UserX needs to prove why Z/2 \times Z/2 and Z/4 are not isomorphic
 
I'd rather the community be able to benefit from my contribution, however small it is
 
7:52 PM
@sjrosen Resubmit, if there is still trouble I will try and enforce it myself.
 
OK, thank you
 
@MikeMiller you don't need the classification theorem. G/Z(G) trick does the trick/
 
@BalarkaSen That gives you that it's abelian.
 
or actually lagrange in general does the trick ;) any element of G can either have order p^2, p or 1.
 
@Alizter Can you suggest an edit comment that would make the situation clearer for reviewers? My previous comment was "Incorporated important information from OP's comments into his answer itself"
 
7:54 PM
if p^2, it Z/p^2Z. if p, then it's Z/pZ \times Z/pZ as there are no other possibilities, @Mike
done
 
@MikeMiller how many cyclic groups of order k are there up to isomorphism? If it's 1(the orders I've seen have 1) I got an idea.
 
@sjrosen I can take the edit as my own by improving it but you will not get reputation. I will accept it. However try not to be wordy with the summary. It is a summary.
 
@UserX 1, you should be able to prove this.
 
@UserX only 1. prove it.
 
"As there are no other possiblities", @BalarkaSen? What kind of proof is that?
 
7:55 PM
@Alizter OK, I'll resubmit now. Thanks
 
@Alizter You sound very authoritative, lol.
 
@MikeMiller proof by observation.
 
@UserX As many as your imagination can handle.
 
you already have two groups, and you know there can be no more than two groups
 
@BalarkaSen Proof by inspection?
 
7:56 PM
so done :p
 
Get outta here.
 
Beat it Charlie.
 
@MikeMiller alternatively prove that a group in which every elt has order p is isom to Z/p \times Z/p
 
@MikeMiller dunno. No idea comes up to prove it
 
7:59 PM
@UserX How about applying the definitions...?
 
@sjrosen There you go.
 
@Alizter great, thank you
 
@sjrosen Well done for raising this issue. It is important for issues like this to be handled. After all the system is not perfect.
 
@MikeMiller generated by 1 elt?
 
@UserX yes. then?
what does it mean that a group generated by 1 elt has order k?
can you write down the underlying set explicitly?
 
8:01 PM
You can come up with a trivial isomorphism showing that they are equal.
 
(1,x,x^2,x^3,...,x^k)
 
Additive or not
 
@UserX So let $G$ and $H$ both be cyclic of order $k$. Each has a generator; say $g$ is a generator of $G$ and $h$ is a generator of $H$. Any homomorphisms come to mind?
 
@UserX yup. so say you have another group generated by a single elt, say y, of order k.
 
so if my generator was m
 
8:02 PM
what is it?
 
it makes no difference
 
@Alizter you're not being very helpful
 
@BalarkaSen What? Why?
 
you're saying a bunch of names
 
As if you don't do that.
;)
 
8:03 PM
f(a^i)=b^i
 
yeah, @UserX
 
So they're isomorphic
 
exactly
 
So there is only 1 cyclic group of order n up to isomorphism
Damn that was unexpectedly easy
 
Yup
 
8:05 PM
Okay, I guess I only have to show that Z_p X Z_p is not cyclic
 
direct product?
 
But this has two ordered elements for each element(my terminology sucks). Say we got a generator (g,g). No matter the (a,b) we want to generate , we can't with our generator. So it's not cyclic right?
Okay, I guess I'm left with uniqueness of these 2 groups
Why are order p^2 groups isomorphic to these though? Any hints?
 
What about $\Bbb Z_{p^2}$
that is cyclic
oh wait
shh
 
I'm so tempted to answer "what about it"
But I should be seeing something right?
 
Well p^2 groups are either cyclic or a direct product of p groups by cayleys theorem
and all p groups are cyclic
lagranges
Yes. Lagranges theorem sorryu
A subgroups order divides the groups order
The divisors of p^2 are p^2, 1 and p
We don't consider trivial groups
 
8:14 PM
What are p groups
 
Groups of order p
 
Prime?
Man I keep meeting lagrange's theorem
 
sure if that is what you mean by p
 
@BalarkaSen set me up
 
So lagranges theorem says that the order of the subgroup divides the order of the group
So a p^2 order group can only have 1, p^2 or p as subgroup orders
So a group is either cyclic or isomorphic to Cp x Cp
as the trivial case can be ommited.
-7
Q: A Very Naive Tackle about The Goldbach Conjecture

user197101I was just wondering if the following is correct to tackle the Goldbach Conjecture? Let $G(x)$ be the Goldbach function and $p\in\mathbb{P}$ such that $$\int_{2}^{x}\left[x\left(\cfrac{p+p}{p}\right)\right]dx=G(x)$$

 
8:23 PM
Hmm nice
Another naive group theory question
Am I supposed to understand these? How do they work?
 
Do you guys think I should publish my findings, though? Are they interesting enough? I don't think so, but I really wanna make sure
 
@teadawg1337 what are your findings?
 
The one I posted earlier, I've been mulling it over for a while
Granted it's currently in a messy state, but I could fix it up if necessary
 
@UserX no
 
8:38 PM
@UserX Here, in case you're curious. I personally don't think it's anything groundbreaking, but what do I know?
Not only the question itself, but also the answer I provided
@DavidH I'll be sure to give you a mention if I end up publishing this :)
I really don't think it's interesting enough to publish, though.
 
9:01 PM
and here
2
A: Finding $\int_0^{\frac{\pi}{2}}\arctan\left(\sin x\right)dx$

David HUsing the integral definition of the arctangent function, we may write $$\arctan{\left(\sin{x}\right)}=\int_{0}^{1}\mathrm{d}y\,\frac{\sin{x}}{1+y^2\sin^2{x}},$$ thus, transforming the integral into a double integral. Changing the order of integration, we find: $$\begin{align} \mathcal{I} &=\in...

a nice question for that I have a simple question:
Can we nicely finish it in one line using elementary tools?
 
@Hippalectryon :-))))))))
 
:-)))
 
@Hippalectryon I think you're right! :D
 
I have a weird question that's bugging me
If I know $a,b,c$ and I wanna know $d/e$, knowing that $ad+be=c$
Is it possible ?
Nvm
 
9:09 PM
Screw it, nobody would publish anything I write
 
@Chris'ssis Have you found a solution to the sum with factorials of factorials ? I haven't
 
No Java updater.
 
@teadawg1337 lol :-)
@Hippalectryon I didn't work on it anymore. Look at the integral I was talking about above. There is a very clever way to do that in one line.
 
Probably. unfortunately I have a paper to give tomorrow so I'm focusing on that atm.
 
9:13 PM
@Chris'ssis Do you know if the question I'm obsessing over has been discovered already? I can't find anything similar anywhere
 
@teadawg1337 hard to say at the moment. I don't know ...
 
@Hippalectryon wat papa
 
@Alizter ??
 
What paper
 
@Alizter Homework
Chemistry
Stuff on D-Glucose
 
9:18 PM
You do advanced chemistry
 
Honestly, I have no idea what to search for in order to find out if it's been discovered already....
I can't find anything, so something tells me I'm the first.... But I highly doubt it...
 
@Alizter Kind of :)
@Alizter We do a bit of advanced everything
@Alizter We're less good than anyone who has focused on one specific subject, but we're better than them on the others
 
@teadawg1337 you could ask mathoverflow but I warn you, you'll probably get a pretty bad reception
 
@UserX That's a given, since I'm not a professional mathematician (yet)
 
@Hippalectryon My school doesn't really let me progress that much ahead of the curriculum.
 
9:25 PM
@Alizter I'm following the usual curriculum :) But I don't do as much maths as you do (unfortunately !)
 
@Hippalectryon My mathematics at school is no where near as complicated as what I study on my own
 
Oh I see
 
We just finished 2x2 matricies :P
 
Lol ?
When are you starting addition ? :P
 
This is difficult in my school?
Oh we have done linear transformations, inverses and systems of equations
 
9:27 PM
I don't wanna publish it, since I anticipate intense ridicule (like I'd get at MathOverflow)
 
@teadawg1337 MO isn't that great compared to MSE
 
@teadawg1337 They aren't there to ridicule it. First ask people (that are familiar with such things) about it.
 
Many people at MO would't be able to most of @Chris'ssis's problems
 
You cannot compare MO and MSE
MO are like Olympic cyclists, MSE are bike enthusiasts.
 
9:29 PM
I can imagine many people on MSE being better mathematicians than those on MO.
 
@Chris'ssis Is the one who makes awfull hills in the middle of the road
 
I feel like my stuff is on the verge of being too advanced for this site, though... I just don't know what to do
 
Professional mathematicians are awfully busy.
They talk about great things though.
 
Trying to publish crap to get tenure.
 
9:30 PM
But you need at least a mathematics degree to understand most of what they are saying.
@WillHunting The results are usually not that crap.
 
Or you could just be extremely gifted at maths
Not to toot my own horn here, but...
 
They look like crap because they are a bunch of symbols.
But when it comes to research, at the forefront, people try and prove easy and hard things
Its like exploring
 
My question is by far the most difficult thing I've ever come across
When are polylogarithms taught?
I don't see any mention of them in my real/complex analysis books, so I'd assume at least late undergrad
 
They are not really taught as such
They are functions that you can study if you wish
 
@Hippalectryon there's a reason for that. Finding such integrals/sums are more like a hobby
 
9:34 PM
probably in CA
Compelx analysis
 
@UserX True
 
That's sheer calculation a computer can do almost as good
Chris'ssis deals with the how, MO deals with the why
 
^ That is true
Solving Tartary integrals starts to become doing random arithmetic after a while. Especially if there is no good motivation for why you need to solve it.
 
Chris'ssis shouldn't hear this but I'm with Alizter on this one.
 
Ughhh, this question is going to drive me insane.... I need to know if it's been done already!!!
 
9:40 PM
I am already insane.
 
@teadawg1337 If you gonna go insane, go formulate it nicely and stroll in MO
 
@UserX I'd rather not, really...
It's just so strange that I've discovered/rediscovered this before taking a single college-level math course...
I seriously fear being ridiculed over at MO, I'll only go there as a last resort
 
@teadawg1337 Ridiculed ?
 
@Hippalectryon As in mass-downvoted and mocked
 
I've never seen that
I've seen downvoted questions, but not mocked ones
 
9:53 PM
Your momma's so fat she's not embeddable in R^3.
Hahahhahahaha lol
 
@UserX >.>
 
First time I actually laughed with a math joke
 
Don't question the continuity of my mom :O
 
Oh yeah? Your momma's so fat she contradicts Whitney's theorem.
 
Because her curves are not smooth enough ? :D
 
9:56 PM
This should be made a new field of math
 
@UserX Not funny, lol.
 
@WillHunting I found it hilarious
 
I starred it lol
It was funny
 
@UserX I don't like math jokes. I only like stupid jokes, lol.
 
10:00 PM
Hilbert had a student who one day presented him with a paper purporting to prove the Riemann Hypothesis. Hilbert studied the paper carefully and was really impressed by depth of the argument; but unfortunately he found an error in it which even he could not eliminate. The following year the student died. Hilbert asked the grieving parents if he might be permitted to make a funeral oration.
While the student's relatives and friends were weeping beside the grave in the rain, Hilbert came forward. He began by saying what a tragedy it was that such a gifted young man had died before he had had an opportunity to show what he could accomplish. But, he continued, in spite of the fact that this young man's proof of the Riemann Hypothesis contained an error, it was still possible that some day a proof of the famous problem would be obtained along the lines which the deceased had indicated.
"In fact," he continued with enthusiasm, standing there in the rain by the dead student's grave, "let us consider a function of a complex variable...."
I bet that could be Balarka Sen too instead of hilbert
 
@UserX Or me :D
@UserX No wait I'd be more like
"It's a total waste he died, he didn't even finish his proof"
:D
 
I hope to solve RH one day.
 
@WillHunting you have to start studying first...
"What's an anagram of Banach-Tarski?"
 
@UserX Banach-Tarski
rekt
 
Banach-Tarski Banach-Tarski
 
10:04 PM
True
 
Rekt^2
 
Hehe
Maybe @robjohn has a brother, the rekt square ?
 
I can't find any similar topics on MO, either
Maybe I have stumbled upon something....
 
@teadawg1337 for the googol-th time, if you're that stressed about it that, go ask the professionals.
 
@UserX What I actually need is a break from the computer. I feel like I'm about to have a panic attack, bbl.
 
10:20 PM
Hello, I just have a phrasing question about perceptrons if anyone could help
I am trying to state simply what a perceptron cannot classify, i know it can classify linearly seperable classes. It does so with a (i want to say) linear hyperplane. Further it cannot generate high order hyperplanes. Is that correct to say?
 
"High order" hyperplanes ?
 
.. curvy
not sur ehow to say it
can't draw a strait line in 2d to seperate the data points
 
erm... a hyperplane is, after all, a plane...
It's not... curved...
Oh wait
It might be curved on non linear spaces
 
for instance the tree point on a line two are in class 1 the other is in class 2
to seperate them properly
you need a curved line
Is that still a hyperplane?
 
I'm not a pro:/ but for me, a hyperplane is the kernel of a linear form
@BumSkeeter What space is your hyperplane a hyperplane of ?
 
10:29 PM
I guess first off I am coming from a CS background
I am not amazing at math..
 
And I'm not amazing at CS either xD
 
lol
The CS stacks are hard to get the math
and the math is hard to get good cs
Do you know what a perceptron is?
 
I'm reading wiki
 
This might help pretty well too jeremykun.com/2011/08/11/…
about what it cant and cant do
 
@BumSkeeter "Arguably the simplest non-trivial example of such a decision function is a line in the plane which separates a training set X \subset \mathbb{R}^2 into two pieces, one for each class label. This is precisely what the perceptron model does, and it generalizes to separating hyperplanes in \mathbb{R}^n."
So it's a line ?
 
10:37 PM
I guess technically
but when you put a bunch together they can function not lines
 
Like with Fourier transforms and sines ?
 
the same idea but different
I am not sure how because I am not sure how either works
either of all three
I will ask later, I need to read more I guess
 
@BumSkeeter I don't see curves other than lines anywhere in the article though
 
Alright, I'm feeling a bit calmer now.
 
10:53 PM
Let $\mathbf{v}_1, \mathbf{v}_2$ be two non-zero vectors in $R^2$, does the $\text{span}(\mathbf{v}_1, \mathbf{v}_2)$ equal to $R^2$?
 
@SalehenRahman What do you think ?
 
I'm thinking yes, but I might be wrong.
 
Hmm why do you think so ?@SalehenRahman
 

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