Mathematics - 2014-12-08 (page 4 of 7)

Alizter

6:02 PM

Is anybody else annoyed by Wikipedia asking for money?

As far as I know they are not even struggling.

Tensor product is cool

Balarka Sen

@Alizter No. I claimed something wrong, and Mike set up a bet on it.

@Alizter You got to study modules first.

@Alizter Me too. As well as Karl and anon. Those are the cool guys.

Alizter

especially $V^n \to V^{\otimes n}$

Mike Miller

@Alizter I wouldn't be annoyed if the requests for money weren't so intrusive, but holy cow, it's painful to use the damn website.

Balarka Sen

Wikipedia is asking for money?

evinda

Hi!!! Is $\mathbb{F}_p$ equal to $\mathbb{Z}_p$?

Mike Miller

6:09 PM

I would be unsurprised if they did need the donation money yearly. But they could make the ads less intrusive. It stands to make me stop using Wikipedia when I need to look up a quick fact rather than get me to donate.

Balarka Sen

@evinda Usually one uses the former to denote the field, the latter to denote the additive part of the field.

evinda

@BalarkaSen Great!!! Thanks a lot!!! :)

Balarka Sen

Hello again @Kaj

Kaj Hansen

Hey

Alizter

@BalarkaSen They still ask even after donating.

Balarka Sen

6:13 PM

@Alizter They're not asking me though.

Mike Miller

Really, @Alizter? That's super crude.

Balarka Sen

Country-based donation-asking?

Alizter

@BalarkaSen I guess they don't want your money?

Balarka Sen

Thankfully, I don't have any money.

Hippalectryon

Moni plz

Venus

6:14 PM

We have a new catchy word here on MSE. Wolf refer to Wolfram Alpha :D

Hippalectryon

@Venus O_o where do you see that

Alizter

They want more money to expand their organisation. They want to pay more managers.

Hippalectryon

@Alizter @BalarkaSen They need money to pay for their donation campaign :3

Alizter

There are those Wikipedia contributers that don't even get any reward for their work.

Balarka Sen

LOL @Hippa

Venus

6:16 PM

@Hippalectryon See math110 and china math's posts.

Alizter

If wikipedia had ads for a day it would be loaded.

Mike Miller

@Alizter This is not quite true. Wikipedia does not have so many managers. Servers are not cheap, especially when you're one of the most-viewed sites on the internet, and they're not making money off it like google.

I think it's a noble goal to not have ads. But they could stand to keep the donation visible, but non-intrusive, all year.

Alizter

@MikeMiller Because they are some charity type organisation, their finances are public. Apparently there are more than in the good.

Oh well

Venus

@Hippalectryon Did suspects they're both same users.

Hippalectryon

I do too

Balarka Sen

6:18 PM

Did did suspect right.

Mike Miller

You're right, @Alizter; I'm wrong. Whoops. Screw them.

Alizter

I imagine donators to be nervous humanitarian undergrads.

Kaj Hansen

@Venus, It's true. I'm always cool like that

Balarka Sen

There are no "cool" mathematicians, @Kaj. Well, except Alexander Gruber.

Alizter

Tunk-Fey and Cleo are the same user.

@BalarkaSen and @DanielFischer

Hippalectryon

6:20 PM

@Alizter Not sure

@Chris'ssis thinks so

Daniel Fischer

@Alizter I assure you, Balarka and I are not the same user.

10

Hippalectryon

But I'm not 100% sure

Alizter

@Hippalectryon There is evidence. Even on Brilliant.org

Balarka Sen

LOL @DanielFischer

Hippalectryon

Ah ok @Alizter

Alizter

6:21 PM

Tunk-Fey randomly quotes Cleo

as if they personally talk

and gets really defensive and avoidy

Venus

Does anyone here know how to prove this $$\sum_{n=1}^{\infty}(\zeta(4n)-1) = \frac78-\frac{\pi}{4}\coth\pi$$

Alizter

I don't think he cross votes too much though

Venus

@KajHansen Boys should speak less & do more

Balarka Sen

@Venus Try writing it as $$\sum_{n \geq 1} \sum_{m \geq 2} \frac1{m^{4n}}$$

Then switch the order of the summation.

Alizter

@Venus Girls should not make such generalising comments :P

Venus

6:23 PM

@Alizter I thought so at first, but no I doubt. I don't believe they're same users

Alizter

Cleos bio is a bit weird

Hippalectryon

@Venus Girls shouldn't upload fake profile pics >:c

Alizter

@Hippalectryon nice upside negation :P

Hippalectryon

:D

Balarka Sen

Girls should be less silly. waits for a lot of bashes

Venus

6:24 PM

@BalarkaSen Could you elaborate?

Hippalectryon

@BalarkaSen No, @TedShifrin isn't here

Mike Miller

speaking of the same users, did y'all see this?

Venus

@Alizter Girls are always correct

Alizter

@BalarkaSen echo a{p,c,d,b}e

Venus

@Alizter Where do you see him quote Cleo?

Hippalectryon

6:26 PM

@MikeMiller ?

Alizter

@Venus On brilliant.

Balarka Sen

@Venus $$\sum_{n = 1}^\infty \sum_{m = 1}^\infty \frac1{m^{4n}} = \sum_{m = 1}^\infty \sum_{n = 1}^\infty \frac1{(m^4)^n}$$

Alizter

No point chasing it up. It is none of my buisness anyway.

Venus

@Hippalectryon Girls always have secrets ^^

Balarka Sen

Now do the inner sum by noting that's it's a geometric series.

Venus

6:26 PM

@Alizter Gimme link

Hippalectryon

@Venus That doesn't mean upload fake pic :c

Alizter

@BalarkaSen Stop hating on analysis and then doing analysis.

Balarka Sen

@Alizter :P

It's just manipulation, not in the least any analysis

Venus

@BalarkaSen Then what? I still don't get it. Sorry ^^

Mike Miller

@Hippalectryon You might notice the reputation change on dec 2

evinda

6:27 PM

@BalarkaSen Could you also explain me why these sets:

$$A=\{ x^2 | x \in \mathbb{Z}_p\} \text{ and } B=\{ 3-y^2| y \in \mathbb{Z}_p\}$$

contain both $\frac{p+1}{2}$ elements?

Balarka Sen

@Venus Well, compute the sum.

I am not gonna do it for you, partially because I am lazy.

Hippalectryon

@MikeMiller O_o

Venus

@Hippalectryon No restriction about that ^^

Balarka Sen

@evinda $\mathbb{Z}_p$ means $\mathbb{Z}/p\Bbb Z$, I presume, and not p-adics?

Venus

@Hippalectryon Besides, you have already known my cute face ^^

Hippalectryon

6:29 PM

:)

Alizter

@Venus brilliant.org/discussions/thread/… search for cleo

Venus

Although, I'm a bit chubby

Balarka Sen

You're looking at quadratic residues in the first case @evinda.

Alizter

In response to Ruben Doornenbal: Cleo said this to me, "While asleep, I had an unusual experience. There was a red screen formed by flowing blood, as it were. I was observing it. Suddenly a hand began to write on the screen. I became all attention. That hand wrote a number of elliptic integrals. They stuck to my mind. As soon as I woke up, I committed them to writing."
She also said that the integral evaluates to
In my opinion, I think this integral is way too hard for kids. Can you elaborate your method on how you evaluate this integral? Preferably with a high school method (this word is r…

Mike Miller

Considering that this is one of the questions they deleted his 'accepted answer' points from, one is led to conclude that he was making other accounts to ask questions and then answer them himself, probably voting on his own answers that way as well.

Venus

6:29 PM

@BalarkaSen You act as if you're a cool guy :D

evinda

@BalarkaSen I mean $\mathbb{F}_p$.

Balarka Sen

@Venus I'm not, I assure you.

I stink.

Alizter

So Cleo, who apparently has difficulty typing words, writes long eloquent sentences in a casual chat with Tunk-Fey.

Balarka Sen

And I'm proud of it.

Alizter

Not only that but one time they both got a problem incorrect which you can ask @Chris'ssis about

The chance of making the same mistake is interesting

evinda

6:31 PM

@BalarkaSen So, in the first case, because of the fact that we 're looking at quadratic residues and there are $\frac{p+1}{2}$ in $\mathbb{Z}_p$ there are $\frac{p+1}{2}$ elements ?

Venus

@Alizter That's Anastasiya

Balarka Sen

prove why there are (p+1)/2 quadratic residues in Z_p

Daniel Fischer

@Alizter That's a paraphrase of one of Ramanujan's utterings.

Balarka Sen

it's an easy exercise. do it.

Venus

She holds a contest there @Alizter?

Alizter

6:32 PM

@Venus Nono. Scroll down.

Tunk-Fey has a comment which I quoted above

Hippalectryon

@Venus Hmm are you sure you have read the TOS ?

Alizter

@DanielFischer Even more suspicion that Cleo does not exist.

Chris's sis

@Alizter What problem?

Balarka Sen

@KajHansen ?

Venus

@Alizter I believe it's a joke. I can quote like that too :D

Alizter

6:33 PM

@Chris'ssis Nothing. We are debating the existence of Cleo.

evinda

@BalarkaSen Ok, I will.. But why does the second set contain $\frac{p+1}{2}$ elements?

Chris's sis

@Alizter Ah, OK.

Kaj Hansen

Wait, let me think for a sec.

Venus

@Alizter I mean I saw Annastasiya there

Alizter

@Chris'ssis He quoted Cleo on another website and yeah. It is suspicious.

@Venus It is a reply to the top comment

Balarka Sen

6:34 PM

@evinda B and A have the same cardinality.

it's 3 minus the squares, and there are precisely |A| many squares

Venus

@Hippalectryon No. In which sentence exactly?

Balarka Sen

@KajHansen I dunneven know what you're talking about.

Hippalectryon

@Venus Nothing :) It was just a joke about those weird TOS

Kaj Hansen

Wait, here we go. Sorry.

evinda

@BalarkaSen Could you explain me further why the two sets have the same cardinality?

Balarka Sen

6:36 PM

What's their to explain, @evinda? Take the bijection a \to 3 - a from A to B.

Alizter

Venus

@Alizter It seems a fun contest. Let me take a look

Kaj Hansen

@Balarka, $\phi:\mathbb{Z}_p^\times \rightarrow \mathbb{Z}_p^\times$ defined such that $\phi(x) = x^2$. Then $\ker(\phi) = 2$ since only $2$ elements map to $1$.

Alizter

@Venus See the picture above

cheesy as stilton

Anyway. Enough skeptics for one day.

Venus

@Alizter Where is the another website? I&S?

Kaj Hansen

6:37 PM

So by isomorphism theorems, there are $\frac{p-1}{2}$ quad. residues $\pmod{p}$, not including $0$.

Balarka Sen

oh, ok.

Alizter

@Venus No brilliant

Balarka Sen

yeah that's a possible way to do it.

Kaj Hansen

Yeah. It's early for me. I'm very tired.

evinda

@BalarkaSen So, what f do we take? We cannot take $f(x)=x^2$.. What else could we take?

Alizter

6:38 PM

@KajHansen magics up espresso

Balarka Sen

@evinda what?

are you trying to prove that |A| = (p+1)/2?

evinda

@BalarkaSen Which bijective function could we take?

@BalarkaSen No, I wanted to understand why the two sets have the same cardinality..

Balarka Sen

I have already told you. Take the bijective function $x \mapsto 3 - x$

Venus

@Alizter I see it, but I don't believe he is Cleo. Anyway, what is the highest level on brilliant?

evinda

@BalarkaSen So, we take this function : $f(x)=3-x$?

Alizter

6:40 PM

@Venus 5 but I do brilliant for the fun.

Balarka Sen

yes

Alizter

I am level 5 in everything but Combinatorics, Mechanics and Electricity & Magnetism.

Venus

@Alizter He has level 5 in many subjects there

Is he a soldier?

Alizter

@Venus It is not too difficult

@Venus Probably national service

Venus

I see his profile, he has a gun

Alizter

6:42 PM

in some countries it is compulsory to serve for the army for some time

Venus

@Alizter No way. Indonesia has no military service like Israel or Singapore

Alizter

@Venus I don't care too much for military services.

Don't get me into politics

It is a mess

evinda

@BalarkaSen A ok.. So if we would have for example the sets $A=\{ 3x^2| x \in \mathbb{Z}_p\}$ and $B=\{ 7-5y^2| y \in \mathbb{Z}_p\}$ would we take the bijective function $f(3x)=7-5x$ ?

Venus

@Alizter I just ask. I love boys with guns. They look cool! ^^

Hippalectryon

e__e

Balarka Sen

6:44 PM

No.

Alizter

@BalarkaSen I am getting a book on lie theory for christmas. If I am god awful at understanding it I will probably pick up a linear algebra text as well to get up to speed which means modules.

evinda

@BalarkaSen But..? :/

Balarka Sen

@evinda I'll let you figure out.

Hippalectryon

So cool @Venus

Balarka Sen

@Alizter You also need a lot of topology to study Lie theory.

AFAIK.

Alizter

6:46 PM

@BalarkaSen nope. That is if you approach from manifolds

Mike Miller

I promise youy don't need to understand modules to read Stilwell's book, @Alizter.

Alizter

Which Ted says is not necessary

Balarka Sen

Interesting, maybe.

evinda

@BalarkaSen $f(x)=\frac{7-5x}{3}$ ?

Alizter

@MikeMiller Modules are more for tensory things.

I can't imagine why I would need modules for Lie

Balarka Sen

6:46 PM

AFAIK, @Alizter, the most natural way to look at Lie groups is as manifolds.

Alizter

unless a Lie module?

hmm

Mike Miller

You don't need modules to understand what tensor products are, either... one obtains a vector space when taking the tensor product of two vector spaces.

Venus

@Hippalectryon I mean the one who has a gun & he uses it wisely.

Alizter

@BalarkaSen The book is called Naive Lie theory which approaches it

Balarka Sen

Anyways, I don't know anything about it.

Mike Miller

6:47 PM

It's just that the construction works just as well (and is even more useful) in the more general setting.

Balarka Sen

And I don't want to know at the moment either.

@MikeMiller Vector spaces are modules over fields.

Alizter

@MikeMiller I will need to study it more. I hope it doesn't end up as something I kind of get. Like Galois theory has.

Balarka Sen

Modules, modules, modules.

Hippalectryon

require(vectorSpace.lua)

evinda

@BalarkaSen Or am I wrong? :/

Alizter

6:48 PM

@BalarkaSen grows wings and turns into a seagull

Mike Miller

And no, there's no such thing as a Lie module (that's widely studied), though one could make the notion precise.

It's a pretty simple concept, @Alizter, I bet you'll grasp it just fine after studying it some more. And Galois theory can go to hell.

Balarka Sen

@MikeMiller You're on ignore now.

Alizter

@MikeMiller Did you not like Galois theory?

Balarka Sen

Galois must be given respect.

Alizter

I think it is interesting but very cognitively demanding.

Balarka Sen

6:49 PM

@Alizter Galois theory is super interesting.

Alizter

@BalarkaSen Anyway. Let me study my Lie

Balarka Sen

Especially what I am studying now, i.e., the topological aspect.

Alizter

don't drag me anywhere :P

Balarka Sen

Lie be damned.

Alizter

Topology looks fun

especially algebraic topology

Mike Miller

6:50 PM

@Alizter In broad strokes I like it. The details are tedious and ugly to my taste.

Balarka Sen

@Alizter That's what I am studying.

Alizter

@BalarkaSen I think I will improve my algebra before diving into topology

and Lie is appealing to me

Hippalectryon

The cake is a Lie

Balarka Sen

Yes, you need to study a little more algebra.

Studentmath

Speaking of Algebra, finally back to studying it :) I really enjoy this course.

Probably will fail the test since I just hoaxed it

Alizter

6:51 PM

@Studentmath Are you going to study Lie things?

They come up in quantum physics

or no time left?

Balarka Sen

Algebra means abstract algebra, @Alizter

Alizter

@BalarkaSen of course. What did you think I meant?

I know how to solve quadratics. I think.

I learnt how to solve cubics.

very briefly

Balarka Sen

@Alizter I mean Algebra doesn't contain "Lie things"

Alizter

But it feels really dirty

@BalarkaSen Lie algebras are part of algebra.

Studentmath

@Balarka any good resources on geometric group theory?

Alizter

6:53 PM

Lie groups are groups which is algebra

Balarka Sen

@Alizter It's hundred times easier to solve the cubic using Lagrange resolvents, again Galois theory.

Studentmath

@Alizter I am not sure yet, I don't think so though.

Balarka Sen

@Studentmath None. I have been learning that stuff from my prof's lectures.

If you get to know of one, ping me.

Alizter

@BalarkaSen I found Ferraris quartic method more fun to do than the cubic

Studentmath

Will do

Alizter

6:54 PM

@Studentmath Teds book :P

Balarka Sen

@Alizter Ted wrote a book on geometric group theory?

Alizter

@BalarkaSen Yeah. Group theory a geomtric approach

Kaj Hansen

@BalarkaSen, his abstract algebra text is geometric

Studentmath

Oh right

Kaj Hansen

"Abstract Algebra: A Geometric Approach"

Balarka Sen

6:55 PM

oh. but that's not geometric group theory.

is it?

i mean the stuff gromov did.

coarse algebra, in short.

Alizter

well no

Abstract algebra a geometric approach

ahh memory serves me incorrect

Balarka Sen

i guess that will contain symmetry groups and whatnots i guess. like Artin.

what i am referring to is the coarse study of cayley graphs, using geodesic metric spaces.

Alizter

Yeah but Ted wrote it so expect some slaps @BalarkaSen

even though the time frames are a bit different

@BalarkaSen Of course. Yes. I understand that.

Balarka Sen

That stuff is cool. Hyperbolicity in groups is a very cool notion. I hope to study more about it in near future.

Alizter

@BalarkaSen So do you go to college lectures?

Balarka Sen

6:58 PM

No.

Alizter

I thought you did something like that?

Balarka Sen

I visit a university often to meet a mathematician.

Alizter

Ah

There are only bad universities around me

Balarka Sen

That guy's a hyperbolic geometer ;)

Alizter

These are the only mathematicians I talk to

Transcript for

$Mathematics$ Mathematics