"We are currently offline for maintenance..."

Aww

@JasperLoy They seem to be back now...

Darn, I missed capping by 10 points.

Exactly

Yay

@Argon Things like this terrify me: $$H^2_{(x,y)}\left(\frac{\Bbb Z[x,y]}{(5x+4y)}\right)=0$$

@Limitless Me too :)

I wish I understood that!

@Argon The question reads like jibberish.

whatismatt.com/one-mans-soap-nightmare/

awesome

@Limitless cohomology is gibberish to me :-S

@Limitless It's tempting to ask $$s_{\sim}^{\%}(x, y, a)=\left[ \frac{{n \choose y}x^y\prec \gamma \prod_{n \ge g(x, A)}^\infty 2^\mho \coprod \oint_D \biguplus f(g(x, y))}{\iiint_{\mathbb D[z]} \exp(5\theta)} \right] \in \partial d\forall A\equiv B(\mathbb R[z, x,])$$

@Limitless It is just the cohomology, between the algebraic pairs of numbers. Where Z denotes the cyclic order of x and y.

Eeeaasy peaaassy

Or something

@Argon The answer is 42

@JasperLoy Nooo...

@Argon That one is easy, though.

@JasperLoy Speaking math is just like speaking italian... You just shout some words and wave your arms.

waves arms

Hahahaha!

...

@N3buchadnezzar i lol'd

What field of mathematics has the most funky notation?

@N3buchadnezzar I love the comments:

@N3buchadnezzar Argon theory

"The method of calculation is very complicated."

@N3buchadnezzar I'd say set theory.

@N3buchadnezzar Or maybe algebraic topology?

No!

Knot theory! (or Representation theory)

$$i\hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},t) = \left [ \frac{-\hbar^2}{2m}\nabla^2 + V(\mathbf{r},t)\right ] \Psi(\mathbf{r},t)$$

Soft.

$\oint_D\cdots$

@Argon Oh no you didnt!

Oh yes I did!

@N3buchadnezzar Physics isn't math, silly . . .

$\text{Annoying}\times\text{Math}=\text{Physics}$

@Limitless You recognize the schrödingers equation? wow..

@N3buchadnezzar Yes . . . It's famous. I much prefer this, however: $$A_N = \int D\mu \int D[X] \exp \left( -\frac{1}{4\pi\alpha} \int \partial_z X_\mu(z,\overline{z}) \partial_{\overline{z}} X^\mu(z,\overline{z}) \, dz^2 + i \sum_{i=1}^N k_{i \mu} X^\mu (z_i,\overline{z}_i) \right) $$

What a boss

lol string theory lol

@Argon You can't take the cross product of a scalar and a vector.

Would a question on the site about what field of mathematics has the most overly complicated looking notation be closed?

@N3buchadnezzar If it were lucky, no.

@Limitless Which one is the vector?

@Argon $\text{Math}$

@JasperLoy You've seen worse.

@Limitless It has magnitude?

@Argon HAHAHAHA.

@Argon I interpreted 'it' as 'physics' in that question and 'magnitude' as tongue-in-cheek 'importance'.

Math certainty has both attitude and latitude, hence it must be a scalar.

@JasperLoy Bye!

@JasperLoy bai

@JasperLoy Good luck!

@JasperLoy So, so mysterious...

Math also scales well with how little i understand.

@JasperLoy Good advice.

Math has direction $\therefore$ math is a vector! $\blacksquare$

@N3buchadnezzar I feel my intelligence insulted by many Superstring Theory articles: "This is integrated over the various points zi.[. . .] When this is taken into account it can be used to calculate the 4-point scattering amplitude (the 3-point amplitude is simply a delta function):"

@Argon Math is multi dimensional. That's equivalent to your argument up to isomorphism.

@Limitless They just do it because

@Limitless Too bad I don't know abstract algebra like the rest of the world...

@Argon This is just linear algebra

Ok. But I don't know that either :)

You are like 17?

@Argon It means that if you define $f: \text{Argon} \to \text{Limitless}$ with map $\text{direction} \mapsto \text{multi dimensional}$, there is isomorphism by $f$ ignoring grammatical error.

You should at least know cohomology by your age.

@N3buchadnezzar Never heard of it!

@N3buchadnezzar Don't we all cover abstract algebra in middle school? I'm pretty sure cohomology and commutative algebra are for high school.

@Limitless Indeed, what about galois and ring theory ?

@N3buchadnezzar Galois theory can be introduced alongside abstract algebra by noting the development of group theory by Galois and taking a look at some of the interesting properties of groups. Defining a notion of solubility of a group and exploring permutation groups is probably the best place to end at for middle schoolers.

@Limitless Yeah, it is intuitive. And most of the theorems can be left as a excercise for the reader.

@N3buchadnezzar Yes. Especially the three isomorphism theorems. And we can always take for granted that isomorphism is an equivalence relation.

hey hey

Heck, do we even need to give formal definition to an equivalence relation? Just note all $R$ behave like traditional $=$.

And if it works in $\mathbb{R}^n$, it should be simple to find an isomorphism.

can someone have a look here math.stackexchange.com/questions/250881/… ?

@user51650 Geometry is not my thing.

@N3buchadnezzar All isomorphisms are trivial unless they're isomorphisms between homomorphisms. Right?

@Limitless What do you mean by trivial?

@N3buchadnezzar The same tongue-in-cheek completely insincere trivial we've been pseudodiscussing for the past fifteen minutes.

Ah!

Right

It is then elementary Watson.

:)

limitless

do you know any good site about geoemetry ?

@user51650 Cut-the-knot

This is pretty damn cool: Define the scalar action $f_r$ so that $f_r(x)=rx$ with $r\in R$ and $x \in G$ where $R$ is a ring of scalars and $G$ is an abelian group of vectors. Define the map $f: R\to \text{End}(M)$ given by $r\to f_r$. In this sense, we have a homomorphism between homomorphisms.

what do you mean n3buchandezzar ?

@user51650 Google it

@user51650 I don't know off the top of my head. I consult Paul's Online Calculus Notes, Mathworld, Wikipedia, and Wolfram Alpha.

For very specific topics I occasionally use a textbook, specific pdfs, or specific webpages written by professors.

I used to consult Lang, until I found out he was just one person.

i lol'd

"We never knew that Daniel Littlewood was actually real."

correction: J.E. Littlewood

ok i feel little depressed because i can't solve it;c

nighty

@user51650 Never feel depressed that you cannot solve something. There are too many of those somethings, and there are even more somethings that you can attempt to solve and succeed.

'night

@N3buchadnezzar Night.

yea but i wil be in deep if i don't solve it till tomorrow lol

@user51650 Understandable. :P

Why do people use \dfrac{}{} instead of \frac{}{}?

$$\dfrac{1}{1}=\frac{1}{1}$$

$\frac{e^x}{e^x}=\dfrac{e^x}{e^x}$

dfrac is larger when inline

@Argon How epic. That is awesome.

Yep

@Argon I feel like proofs by vacuity are like highly carbonated water.

@Limitless Why? :)

@Argon They aren't fulfilling.

@Limitless :)

Or they're as fulfilling as carbonated water is when you're thirsty.

Did you know that all $$\$1001 bills actually cost $\$$1002?

What? How?

@Limitless Find me one and I will show you :)

vacuous

LOL

Nice one, @Argon.

And cool math stuff (that I would understand :) )?

How about this? (shameless self promotion)

Neat

I thought so too.

I rather like Discrete Calculus.

Do you like summing series?

Like $\zeta(2)$ and stuff?

@Argon I do. I'm not pro with zeta, though.

Complex integration can actually be of help too for zeta-style sums :)

Zeta is really nice though

I feel like a lot of things get . . . sketchy in real and complex analysis.

Like what?

Like

There are a lot of "closed forms" in terms of nonelementary functions like $\text{erf}(x),\zeta(s),\text{Li}(z),\ldots$.

Only when nothing else can be used, though

This is true. Nonetheless, they are very prevalent.

Special functions are also nice, because they are studied in detail, and so even complex sums and integrals, etc. that can be written as special functions can be handled quite well

I have to agree with you there. It is just intimidating to me.

:)

I'm familiar with some basic special functions, but stuff like elliptic integrals and theta functions scare me!

I was very happy in the summer when I discovered how generalized hypergeometric functions worked.

I'm somewhat familiar with some special functions. Hypergeometric functions are rather peculiar.

I see those as monstrous beasts of generality.

Exactly.

Often, I do not consider them to be "closed forms."

They are analogous to special functions as category theory is analogous to abstract algebra.

But they are better then nothing, and extensively studied

@Limitless What is category theory?

Category theory is the study of categories, functors, morphisms, and other things. The idea is really complex. You have functors between structures rather than homomorphisms between things like groups, rings, fields, algebras, etc.

It's very hard to explain because I honestly don't understand it.

Meh, no use for me to understand it without the prerequisites!

It builds upon all branches of mathematics (nearly all, anywho).

There used to be a lot of category theory on here.

hi guys

how can i free up space on a usb using mint?

But the Gamma function is really elegant. The Beta function is practically a few Gamma functions that have a common integral representation. The error function is just a common one, and has few closed form values, except for a select few trivial ones.

@Khromonkey Hola.

Hi

The logarithmic and exponential integrals too are just common integrals.

@Argon Is the Beta function a bit like a function in terms of the Gamma function?

@Limitless When one evaluates it at a certain number, usually the gamma function representation is used.

@Argon Ah. So, viewing it in terms of $\Gamma(z)$ is a conventional choice.

Yes

It is usually used for integral evaluation and stuff (as far as I have used it)

@JasperLoy

@Argon I really like this discussion. It appears that due to our differences in interests, we're both learning a lot from one another.

:)

@JasperLoy i.e. the RH

Hahahaha!

@JasperLoy Aaron?

Argon?

That's me

Yay, I'm Sherlock Holmes.

@Argon The Pochhammer symbol has more importance than Wikipedia seems to portray.

@Limitless The Gamma function is nice because so many elementary closed forms exist for some of its values, i.e. trivially all posivite nonzero integers, and when you add or subtract half from a natural number.

See this. Its use in the definition of the hypergeometric series isn't even mentioned.

@Limitless They are used extensively in things like hypergeometric functions

how do I empty trash on linux on a usb?

@Limitless Odd, because they are quite important in that study!

@JasperLoy Your choice, Jasper!

nevermind, I figured it out

@Argon Should I revise the Wiki article on Pochhammer symbols?

@Limitless Sure

Perhaps they are not as notable, simply because they are a shorthand for a quotient of more general functions

So they can always be written in terms of the gamma function

@Limitless Any other special functions you like? Dilogarithms, polygamma, zeta-style, etc?

@Argon Take a look here, I edited it.

@Argon I like the generalized zeta function even though I do not fully comprehend it. There are also some functions related to zeta such as $\pi(x)$ which are interesting.

@Limitless Nice :)

@JasperLoy Got it

@Limitless Hurwitz?

@Argon Hurwitz? Sounds vaguely, very vaguely, familiar . . .

Hurwitz Zeta

@Argon Isn't that really just trivial given that we can have $\zeta(s)$ with $s=q+s_0$?

I mean

@Limitless Zeta functions are deeply connected with number theory, which is a bit out of my league (I presume you know the Riemann Hypothesis). That would explain the connection to $\pi(x)$...

@Limitless ?

Actually, now that I think about it, I guess they define it $\zeta(s,q)$ rather than $\zeta(s)$ since those two are conceptually slightly different despite the equational equivalence.

@Limitless Well, $\zeta(s+q) \neq \zeta(s, q)$....

@JasperLoy Yeps!

Despite the fact that $$\sum \frac{1}{(s+q)^n}=\sum \frac{1}{(s_0+q)^n}.$$

@Limitless $$\zeta(s+q)=\sum_{n=1}^\infty \frac{1}{n^{s+q}}$$

@Argon I am retarded.

Kill me now please. -.-

bang

Hahaha

HAHAHA!

@JasperLoy the putnam? I think it's every year. dunno for sure.

@Argon Isn't this Wiki article incorrect? Shouldn't $\zeta(s)=\zeta(s,0)$?

@Limitless how do you define $0^{-s}$?

It says: The Riemann zeta function is ζ(s,1).

@Limitless Yep

Write out a few terms

@JasperLoy No, I mentioned my future score within an interval of confidence. I only answered three of them, and only one of my answers was complete.

@anon Hard stuff, I'm impressed

The results do not come until like March or something.

once again

@Argon $$\zeta(s,q) = \sum_{n=0}^\infty \frac{1}{(q+n)^{s}}.$$

@Limitless Right. Notice that $n=0$

@Limitless Right - now plug in $q=0=n$ into the denominator and see what happens.

Instead of $n=1$ for the Riemann zeta

@JasperLoy I will barely be in the top half of participants.

So we have $\zeta(s)=\zeta(s,1)$ in order to avoid division by 0?

@Limitless no, we have that because that's what it is. there's no ulterior motive!

@Limitless $$\frac{1}{(1+0)^s}+\frac{1}{(1+1)^s}+\frac{1}{(1+2)^s}+\cdots$$

@anon I see now. I forgot that zeta begins at 1 and this begins at 0.

Makes much more sense now.

@Limitless I don't know why I'm bringing this up, but the argument principle and Roche's theorem blew my mind

Funny, John Kelley defines antisymmetric relation by $xRy$ and $yRx$ never occurs rather than implies $x=y$.

@JasperLoy I am aware. I'm attempting to be more personal and open about my personality. It's part of some self-investigation I am doing.

@Limitless "P.S. I love my mom and Her."

@Argon It's for confidentiality.

@JasperLoy No, in fact he says a little later that $R\cap R^{-1}$ is void when $R$ is antisymmetric.

@JasperLoy Nope. I don't worship Her, I just love Her.

$$\text{Her}\stackrel{?}{=} XXX_{\text{Limitless}}$$

@JasperLoy Nope. Just an acquaintance of five years.

@JasperLoy I can't say that. She's dating a different guy.

Situation is $\in \mathbb{C}$, but she loves me.

About 1,000 miles separate us, @JasperLoy.

@JasperLoy I think we should make a secret love mathematical notation.

@JasperLoy Yup.

$$f(\text{a},\text{b})\mapsto \text{Chatroom} = \infty$$

@JasperLoy It's like that. Also, I was wrong: Approximately $2,300-2,500$ miles.

Not accounting or using spherical geometry to calculate the distance, of course

:)

A mathematical answer

(Unless Google Maps uses spherical geometry to calculate distances.)

@Argon You may also be interested in measure theory. I think it's fascinating. Marvis makes quite a few measure theory answers, I think.

@Limitless I will take a look. I will see if I understand anything.

@Argon If I recall correctly, it's generally a given that a geodesic from spherical geometry is shorter than a line from Euclidean geometry.

(inb4 physics blows up on my intuition)

um, what

@anon I'm wrong, aren't I?

walking on the curved surface of the earth will take longer than if you tunneled through and made a straight path

assuming constant equal speed

Oh!

So it's the other way around.

John?

@JasperLoy Haven't had any yet.

Ah

It's actually 9:46 p.m. and I should be sleeping, but I woke up at like 5 p.m. . . .

Why???

How???

@Argon Had a lot on my mind and stayed up until 6 a.m.

@Limitless Geezz....

@peoplepower Thank you. You're absolutely right, and this is why I should not do physics.

@JasperLoy ...

@JasperLoy I can send it to you right now via e-mail. :-)

@JasperLoy I notice you spelled "colour" correctly...

Would you like that?

@JasperLoy Correct spelling :)

@JasperLoy Tomorrow. I must ponder.

...quite

6.12.12.777

11.19.12.1009

?

@Limitless

@Argon When I hit 777 reputation and 1009 reputation.

@JasperLoy I sent your present. <3

@Limitless Are those special numbers?

@JasperLoy Any time.

?

@Argon I never thought I'd hit 1k. And 777 is just funny looking. Supposedly the number of God.

Ok. Cool then

@JasperLoy He makes me so happy. :-)

He makes me forget the confusing stuff in life, you know?

...

brb

I have no idea why it's called turtle.

Good night everyone!

It means you are a boss

@Argon Goodnight.

@JasperLoy TeXworks is nice.

Life is so Riemannian right now, @JasperLoy.

@anon My life.

@JasperLoy He most likely does.

@JasperLoy I'm glad you think so.

direct variation doesn't mean much to me

@JasperLoy Indeed.

Limitless, do you know of the gray arrows in chat?

I do, @anon. I am just so much more comfortable with typing "@[insert user here]".

All that clicking takes way too much time.

@JasperLoy New conjecture: $$\text{Life} \in \mathbb{C}.$$

That is the characteristic of a loner. @anon, are you a loner? <3

@JasperLoy In all due time.

@JasperLoy email? I am curious.

@JasperLoy Splendid. I am not tired yet.

Will.

@JasperLoy It's jasper now.

4Ya.

@JasperLoy Too simplified?