Mathematics - 2014-03-29 (page 2 of 5)

2:00 PM

Also I misread that integral on the star board, I thought it said $\sin(x^3)$ and I've been banging my head on the keyboard all morning

Sawarnik

banging your head on the keyboard! How would you do the correct version though?

Daniel Fischer

@molikto "Then $A/\mathfrak{a}$ is a ring".

MickLH

@Sawarnik Still don't know yet but at least it feels possible now

Parth Kohli

@MickLH "my head" meaning the head of your company...?

Sawarnik

@MickLH I see you are a game developer. Do you have something to interest me?

Vrouvrou

2:04 PM

see my profile please

MickLH

@ParthKohli No I mean literally my physical human form

My brain specifically (these long tails protruding from it, I usually call them fingers), directly against the keyboard

Sawarnik

Is your keyboard alright?

MickLH

:14551531 Yeah that would turn out awkward when we realized it was x squared, not cubed hahaha

Sawarnik

Mick, give me some of your games please! Do I have to pay money?

MickLH

I don't think you'd like my games, I normally do backend stuff

:14551540 Hmm.. maybe I'm a pervert but I took it to mean sexual intercourse...

Parth Kohli

2:08 PM

@MickLH No problem, we're alright with all kinds of games.

@MickLH same

MickLH

Well here's the last thing I did, it's silly: szifler.com/gdsegj

Parth Kohli

@MickLH nice! I've played a similar game on mobile.

Sawarnik

I didnt understand the goal :/

MickLH

It's a geometry / physics puzzle game

You have to touch all the green squares except one

You can pick any one to exclude, and the ball has to come to rest afterwards

It gets harder as you go because geometrically there are more obstacles to work around, and you always only have the potential energy you can store in gravity available

Parth Kohli

I just made it bounce

Sawarnik

2:13 PM

The user interface and design can be improved though..

Parth Kohli

how to remove a block?

molikto

@DanielFischer isn't factor ring defined for all rings?

MickLH

@ParthKohli There's a list of tools in the corner, I wouldn't waste so much time on that game lol

Daniel Fischer

@molikto $\mathfrak{a}$ is a left ideal, for a factor ring, you'd need a two-sided ideal.

Sawarnik

@MickLH Then your most awesome creation?

MickLH

2:15 PM

@Sawarnik This is where all my time and research goes: github.com/micklh/MagicWorks

Parth Kohli

clone all the projects

MickLH

I'm coming to the end of a month long commit lol

I have a lot more fun with the math, I'm constantly trying to come up with easier to evaluate approximations of everything

molikto

@DanielFischer I think just above Lang said that "let "

@DanielFischer I think just above Lang said that "let a be a two sided ideal"?

Daniel Fischer

@molikto Not in my edition (third).

molikto

Ok.... mine is revised third. I think if a is two sided ideal. then R is not necessary commutative?

Daniel Fischer

2:24 PM

@molikto Could be that it was changed, and that part was forgotten to be updated.

Balarka Sen

@Chris'ssis Anything new on NT lately? Something rather non-elementary?

Daniel Fischer

@BalarkaSen Do you happen to know a sharp bound on $$\sum_{k=1}^N \mu(k)\left(\left(\frac{N}{k}\right)^2 - \left\lfloor \frac{N}{k}\right\rfloor^2\right)\,?$$

Balarka Sen

@DanielFischer Can't think of anything sharp, no.

But some loose ones are easily obtained.

By, for example, partial summation.

Daniel Fischer

@BalarkaSen Can you prove $O(N)$?

($O(N\log N)$ is trivial)

Balarka Sen

@DanielFischer Do you have a proof of that?

Daniel Fischer

2:35 PM

Of what? $O(N)$? No, I'm trying to think of a proof every now and then, but some detail always gets in the way.

Balarka Sen

I see.

I'd try if there was, I dunno, a less freaking thing instead of that.

=P

Mats Granvik

Reduce::rhyp: Warning: Reduce assumed the Riemann hypothesis in order to prove the solution set found is complete. >>

Balarka Sen

In any case, it's not hard to show that $$\left | \sum_{k\leq N} \mu(k)\left(\left(\frac{N}{k}\right) - \left\lfloor \frac{N}{k}\right\rfloor\right) \right |$$ is $O(N)$

Bu that is not going to help.

Or is it?

Daniel Fischer

I don't think so, @BalarkaSen.

Balarka Sen

You're probably right.

Sawarnik

2:40 PM

@BalarkaSen When is your exams starting?

Balarka Sen

@Sawarnik 7th.

Sawarnik

A long way to go still :/ :(

Balarka Sen

@DanielFischer Well, there are well-known very tight lower bounds/asymptotics on $\sum_{k\leq N} \varphi(k)$

They might help.

MickLH

YES! my calculator solves PolyGamma somewhat symbolically :)

Daniel Fischer

They might, @BalarkaSen. Let's see what asymptotics wikipedia lists.

Balarka Sen

2:45 PM

@DanielFischer In fact, by Perron, I get something like $3/\pi^2 x^2$

@DanielFischer I don't trust wiki in these matters.

I found a lot of largely incorrect asymptotics ones in wiki.

Daniel Fischer

@BalarkaSen You can usually check either yourself, or see what references they list. Overall, in matters mathematical, wikipedia is somewhat trustworthy, unlike in matters conspiracy theoretical.

Balarka Sen

Yeah, sometimes they don't refer to things at all.

Mathworld is far better.

But wiki agrees with my estimates, it seems.

And the refer to Walfisz.

The error term is a bit clunkier through.

$$O(n\log(n)^{2/3}\log\log(n)^{4/3})$$

An easy estimate is $O(n)$

But better ones can be found through RH at any rate, even better than this.

Daniel Fischer

Okay, if $O(n(\log n)^{2/3}(\log\log n)^{4/3})$ is the best currently known, I need not be ashamed of not finding a proof of $O(n)$ ;)

Balarka Sen

=P

But you can do that with RH, I bet, if your estimate does not turn out to be false.

Daniel Fischer

Well, but proving RH is a bitch. I'll not do that this weekend.

Balarka Sen

2:56 PM

Who asked you to prove RH? Just assume RH and apply Perron, man.

Danny

@robjohn u there

Balarka Sen

@PedroTamaroff Mitra said he did not like Mendelson.

Sawarnik

@BalarkaSen Mathworld also contains a lot of error.

Balarka Sen

@Sawarnik Sometimes, yes.

But at least I didn't see any.

Chris's sis

@BalarkaSen Find a factor between $1000$ and $5000$ of the number $$2^{33}-2^{19}-2^{17}-1$$

Sawarnik

3:02 PM

@BalarkaSen I have seen. Sent a request to be corrected but still incorrect.

Balarka Sen

@Chris'ssis Nah, no elementary NT today. You have made me do some already.

Sawarnik

@Chris'ssis When did your interest turned towards NT?

Chris's sis

@Sawarnik I'm working right now on some intregrals. :-)))

(btw)

Balarka Sen

OK. =)

But that doesn't explain your interest towards ENT.

Pedro Tamaroff

@BalarkaSen I don't like trains.

Balarka Sen

3:09 PM

Hahaha. He does not like point-set topo at all.

=P

Pedro Tamaroff

I just looked at your problem.

Chris's sis

@BalarkaSen I've always been interested in NT, but I didn't approach, let's say, some very hard things there although once in a while I did some.

Balarka Sen

Which one?

Pedro Tamaroff

The one about the circle. I am guessing you identify $(0,1)$ and $(0,-1)$ to get a neutral element yes?

Else you have no neutral element.

Balarka Sen

@PedroTamaroff (0, 1) is neutral element.

Which I also like to call "identity element"

=P

robjohn

3:12 PM

@Danny yes?

Pedro Tamaroff

What is the inverse of $(0,-1)$?

$(0, 1)$

¬¬

Ooops

Surprise.

=)

Balarka Sen

3:13 PM

You are correct.

Dis is weird

Danny

@robjohn i dont see how $E[E[X \vert Y]] = \sum_y E[X \vert Y =y]P\{Y =y\}$ could u explain?

Pedro Tamaroff

Also, how do I sum $(1,0)$ and $(-1,0)$?

Joining them intersects the $x$ axis everywhere.

Balarka Sen

Join lines.

@PedroTamaroff WAT?

Ahahah

Sawarnik

@robjohn What does it mean that I am removed along with my 50 points vanishing? ??

Danny

@robjohn maybe i can write it on the forum and u can answer there if u have time

so i can read it slowly

robjohn

3:15 PM

@Danny That looks pretty much like the definition

Balarka Sen

@PedroTamaroff I am not sure if we are both thinking about the same thing

Pedro Tamaroff

@BalarkaSen Well yes, joining $(1,0)$ and $(-1,0)$ gives $[-1,1]$.

@BalarkaSen Isn't addition defined by joining and projecting through the $(1,0)$ to the circle through the intersection with the $x$-axis?

Balarka Sen

Didn't I mention about removing those two points there?

Pedro Tamaroff

Oh, I didn't know that.

robjohn

@Sawarnik you are not removed... you are still here. Perhaps you are seeing that someone else was removed and you've lost the 50 points they gave you,

Balarka Sen

3:16 PM

@PedroTamaroff Nope. Projection through $(0, 1)$

Pedro Tamaroff

@BalarkaSen That, $(0,1)$.

I just mixed my coords.

The north pole.

Balarka Sen

Surprise.

=)

Pedro Tamaroff

Nah, man. You're not using it correclty.

Balarka Sen

So nothing weird.

@PedroTamaroff Using what?

Pedro Tamaroff

Well, how do you sum two points at the same height?

The line will never cross the $x$-axis.

It will be parallel to it.

Balarka Sen

3:18 PM

In the projective sense, they do.

At the point $\{\pm \infty\}$

So they'd be inverses to each other.

i.e., the north pole is the point at infinity.

Pedro Tamaroff

You're considering $(S^1\smallsetminus \{(-1,0),(0,1)\})\cup \{\pm\infty\}$?

Balarka Sen

Of course.

Riemann sphere in $\Bbb R^2$

Pedro Tamaroff

Still, you have a problem with $(0,1)$ and $(0,-1)$.

You have to identify those points or something.

Balarka Sen

@PedroTamaroff Umm, what problem?

One is absorbing element another is identity.

robjohn

Pedro Tamaroff

3:20 PM

There is no such thing as an absorbing element in a group!

What are you talking about...?

Didn't you say you wanted to make this into a group?

Sawarnik

@robjohn If a user is removed the points I gained from him is removed, this is unfair?

robjohn

@Sawarnik that?

Balarka Sen

@PedroTamaroff Wait, I am not thinking straight.

Sawarnik

Yes!

Balarka Sen

Let me draw it out.

robjohn

3:21 PM

@Sawarnik why is it unfair? It happens to everyone.

Sawarnik

Ok then. :[

Balarka Sen

@PedroTamaroff North pole is identity. OK.

Pedro Tamaroff

@BalarkaSen Right.

Balarka Sen

Doesn't seem so.

=P

Pedro Tamaroff

But you cannot sum anything to anything to get the north pole.

Balarka Sen

3:23 PM

Elaborate.

Pedro Tamaroff

Well, given a point $x$ in $S^1$, what is the point $x'$ for which $x+x'=e$?

Balarka Sen

You have to sum to inverse elements to get north pole, of course.

@PedroTamaroff $(a, b) + (-a, b) = e$

Pedro Tamaroff

@BalarkaSen But the inverse of the north pole is not itself.

Balarka Sen

Uh?

Pedro Tamaroff

The inverse of the $1$ must always be the $1$, uniquely.

Balarka Sen

3:26 PM

@PedroTamaroff Of course it is!

Pedro Tamaroff

$(0,1)+(0,-1)=(0,-1)$.

Balarka Sen

@PedroTamaroff Nope.

tis $(0, -1)$

Pedro Tamaroff

Yes, yes.

Balarka Sen

=P

Get a pen and paper.

You can't think everything all by your head, ya know.

Pedro Tamaroff

Every element of your group is idempotent.

Balarka Sen

3:28 PM

False.

Pedro Tamaroff

How do you sum a point to itself...?

Balarka Sen

I defined doubling by drawing tangents.

Pedro Tamaroff

???

Balarka Sen

@PedroTamaroff I mentioned it.

Pedro Tamaroff

@BalarkaSen I didn't read everything, just that it was $S^1$ with sum by joining lines.

Balarka Sen

3:29 PM

@PedroTamaroff OK. $(x, y)$. Double the point by drawing a tangent at that point and then map back to the circle.

Pedro Tamaroff

Drawing tangents to the point? Did you check it goes well with the other operation?

Dis is doubling.

Yes, I got it.

@PedroTamaroff Yep.

@BalarkaSen It seems every nonidentity element has infinite order then?

In fact repeated addition gets you closer and closer to the removed points.

Balarka Sen

3:31 PM

@PedroTamaroff Yes.

@PedroTamaroff Yes!

That's how you approximate $\sqrt{2}$ very closely by my method.

But this is a dumb application.

Pedro Tamaroff

@BalarkaSen What if you define $x+x=x$ for any $x$? Does that fuck up addition?

Balarka Sen

Have to check.

Well, additions not continuous anymore.

I can't see anything that serious.

Pedro Tamaroff

What's the topology here?

robjohn

@Danny Does that make sense?

Balarka Sen

@PedroTamaroff That's upto you to define, not me.

=)

Pedro Tamaroff

3:38 PM

@BalarkaSen You just said "additions not continuous anymore."

Balarka Sen

Define something appropriate that'd show continuity.

I am not going into formalities.

Pedro Tamaroff

I don't see what's the topology, that's why I am asking.

You're saying "addition is no longer continuous". so you have some topology in mind.

Balarka Sen

Well, yeah.

But I'd rather stick up with the algebraic properties.

Pedro Tamaroff

And did you get anything out of this?

Could you find say subgroups of whatever?

Balarka Sen

Well, $\Bbb R^\times$ is not an interesting structure. =(

Pedro Tamaroff

3:43 PM

Did you get any isomorphism $\simeq\Bbb R^\times$?

Proof?

Balarka Sen

I have mentioned my intuition before (i.e., affine group with underlying space being a line minus a point) and Mike stabbed with one-dimentional lie groups.

End of the game.

Pedro Tamaroff

So no isomorphism.

Balarka Sen

Well, it is isomorphic to $\Bbb R^\times$

Pedro Tamaroff

Proof?

Balarka Sen

Ask Mike. I don't understand lie algebras.

Pedro Tamaroff

3:47 PM

Well, $(0,-1)$ plays the role of $-1$.

Right, that's the iso there.

Balarka Sen

But even if I did, that'd be stabbing an ant with a truck.

@PedroTamaroff Yes.

Pedro Tamaroff

@BalarkaSen Nah, man.

Think about it for a second.

Balarka Sen

I just get intuitions.

Pedro Tamaroff

I don't totally buy it though.

Balarka Sen

The removed point definitely plays an important role here.

Adding them gives the whole base set.

Vrouvrou

3:49 PM

can someone help me in deformation lemma ?math.stackexchange.com/questions/721961/deformation-theorem

Balarka Sen

This almost surely means something.

Pedro Tamaroff

Intuitively, the positive elements should be the upper arc and the negative elements the lower arc.

Balarka Sen

Yep. No. You mean left and right arcs.

Pedro Tamaroff

But in $\Bbb R^\times$ you have three points of "divergence".

Oh, no. Wait.

You have $0$, like $(1/2)^n\to 0$.

Balarka Sen

@PedroTamaroff Yes.

Pedro Tamaroff

3:50 PM

You have $+\infty$, say $n!\to +\infty$.

Balarka Sen

Continuity.

Pedro Tamaroff

Though you cannot get to $-\infty$.

Balarka Sen

Ah. Why?

Pedro Tamaroff

@BalarkaSen Oh, well you can sorry. I was thinking about simple powers.

Yes.

So you have three divergence points in $\Bbb R^\times%$.

How many are there in your group?

Balarka Sen

Seems $2$ at first sight.

Pedro Tamaroff

3:53 PM

You have the removed points.

Balarka Sen

Does point an infinity counts?

Pedro Tamaroff

Right, that's why I don't see why it should be $\Bbb R^\times$ at first sight.

Balarka Sen

@PedroTamaroff I didn't think about that.

Pedro Tamaroff

Can you find a sequence of points such that $\sum a_i\to \rm point\; at\;\infty$?

Balarka Sen

I have to think.

Seems unlikely though.

Pedro Tamaroff

3:55 PM

@BalarkaSen But how do you sum something to that point at $\infty$?

Balarka Sen

The point at infinity is north pole, you realize that right?

Pedro Tamaroff

Oh, sorry.

Right.

I am not used to projective stuff.

Still, that would be the $1$ in $\Bbb R^\times$.

Balarka Sen

Yes.

Wait.

Pedro Tamaroff

I am not really convinced it is $\Bbb R^\times$:

Balarka Sen

No $0$ in multiplicative $\Bbb R$

Pedro Tamaroff

3:57 PM

@BalarkaSen Yes, I know that.

Balarka Sen

So how do you get three points of divergence?

Pedro Tamaroff

But is is a "hole".

$+\infty$ and $-\infty$ are not there either.

Balarka Sen

@PedroTamaroff I am not really used to continuity stuff.

Right.

Pedro Tamaroff

@BalarkaSen Well, you have sequences in $\Bbb R^\times$ for which $\prod a_n\to 0$.

Balarka Sen

Yes, I understand that.

Pedro Tamaroff

3:58 PM

So those diverge.

Balarka Sen

My bad.

The problem is that $0$ has no interpretation in my base set.

Pedro Tamaroff

What the "base set"?

Balarka Sen

$S^1 \setminus \{(1, 0), (-1, 0)\} \cup \{\pm \infty\}$

Pedro Tamaroff

Oh, derp.

I thought you meant something else that "underlying set".

Thomas Andrews

I'm a little obsessed today with the realization that the intermediate value theorem can actually be thought of as a theorem about homotopy, and that homotopy is the only real way to extend it to higher dimensions.

Balarka Sen

4:02 PM

Nice talking to you, @Pedro, but now I have to run. I will come back at this though.

Seems you are almost likely correct.

Bye for now.

Pedro Tamaroff

Byes-

@ThomasAndrews I saw your answer the other day.

Was impressed.

Thomas Andrews

It seems like a lovely motivation for homotopy, but I'd never seen it.

Vrouvrou

can someone help me math.stackexchange.com/questions/721961/deformation-theorem

Sawarnik

4:20 PM

@ParthKohli Looks like Alx H is set for another 100!

Oh out! England doomded now unless Morgan and Buttler produce an Halinnings.

robjohn

@Danny I think my answer is essentially the way to think about the answer, but I didn't want to get into the precise meaning thread that Did would pose. If he gives an answer, it will probably be correct, but it may be difficult to understand.

Essentially $X\mathrm{E}[Y|X]=\mathrm{E}[XY|X]$ because $X$ is fixed during the evaluation of the expectation and expectation is linear.

Danny

@robjohn thanks rob Iam back in a while...iam a bit of hurry ...

Balarka Sen

4:53 PM

@Pedro OK, tell me how you reach $-\infty$ in $\Bbb R^\times$

Pedro Tamaroff

@BalarkaSen $-1\times 2\times 3\times\cdots$?

Balarka Sen

OK =)

My bad.

Then do you think $\Bbb R^+$ under mult might be it?

Pedro Tamaroff

@BalarkaSen I don't think so, no.

$\Bbb R^+$ has no involutions.

MickLH

$${{\sin \left(\left(e^{z}-1\right)\,\sum_{n=0}^{\infty }{{{\Gamma^2
\left(n+1\right)\,\left(1-e^{z}\right)^{n}}\over{n!\,\Gamma\left(n+2
\right)}}}\right)\,\sum_{n=0}^{\infty }{{{\Gamma^2\left(n+{{1}\over{
2}}\right)\,\left(\sin \left(\left(e^{z}-1\right)\,\sum_{n=0}^{
\infty }{{{\Gamma^2\left(n+1\right)\,\left(1-e^{z}\right)^{n}}\over{
n!\,\Gamma\left(n+2\right)}}}\right)\right)^{2\,n}}\over{n!\,
\Gamma\left(n+{{3}\over{2}}\right)}}}}\over{2\,\sqrt{\pi}}}$$

Thomas Andrews

That expression makes me want to go cry into a beer. @MickLH

Pedro Tamaroff

5:02 PM

WHAT THE ACTUAL FUCK.

Sorry.

MickLH

Guys Guys!

It has a wonderful closed form

(Don't kill me) $... = z$

Thomas Andrews

Yeah, just a gut reaction. @MickLH :)

Pedro Tamaroff

It's made up.

Not interesting. =/

MickLH

My algebra program spit that out at me

I thought I'd share the fun :D

Pedro Tamaroff

Your algebra programme is doing acid.

=D

MickLH

5:04 PM

I'm sure you can guess to within 0.3% how hard I shit myself, hypergeometrigsponentialblahblah thank god for Mathematica

Oops, forgot to mention it's for $-2\pi \leq x \leq 2\pi$ only

Balarka Sen

Nevermind.

Yes, no involution.

Uh, what's the involution here? I can't see any. (Sorry, I think it's the temperature here that is frying my brain.)

Oh.

You mean $(0, -1) + (0, -1)$?

Pedro Tamaroff

In $\Bbb R^\times$ you have $(-1)^2=1$.

Balarka Sen

Yes, yes.

Good catch.

Pedro Tamaroff

5:19 PM

Yes, in your group $(0,-1)^2=1$.

Balarka Sen

So, tell me, do you like de la group of mine? =)

Pedro Tamaroff

Well, I don't fully understand it yet.

Pedro Tamaroff

5:42 PM

@FernandoMartin Morning.

Fernando Martin

sup

Balarka Sen

sup of what?

Pedro Tamaroff

@FernandoMartin How did it go yesterday?

Fernando Martin

Pretty cool

Pedro Tamaroff

Le tochi got hammered yas?

►

Fernando Martin

5:44 PM

We all did

Pedro Tamaroff

Good, good.

On Tuesday we'll make a reservation for next Friday.

@FernandoMartin Have you done any problems from Atiyah?

Nope

The com. alg. doom?

@BalarkaSen Yes.

I've been thinking about the one Alicia gave us

Balarka Sen

5:46 PM

Sheesh.

Fernando Martin

Still haven't solved it

Pedro Tamaroff

@FernandoMartin What was her problem?

Fernando Martin

$\langle x^2, y^2,xy\rangle$ can't be generated with only 2 elems

(in $K[x,y]$)

Pedro Tamaroff

@FernandoMartin Oh.

Balarka Sen

What is the little dot there?

Pedro Tamaroff

5:47 PM

@BalarkaSen It should be a comma.

You can just call out the typo =P

Balarka Sen

Ah.

Interesting problem.

Pedro Tamaroff

YAS.

@FernandoMartin I will give it a shot now.

Let's see what I get.

@FernandoMartin Did you try anything?

Fernando Martin

Well, I realized that Bézout's identity doesn't hold for several variables

Pedro Tamaroff

Hehe, true. =)

$(x,y)=1$; but if $xf(x,y)+y g(x,y)=1$, eval at $(0,0)$ gives $0=1$.

Fernando Martin

$(x,y)$ means GCD?

Pedro Tamaroff

5:53 PM

Yas.

Fernando Martin

Right.

Pedro Tamaroff

So far I will write $$\eqalign{
& {\alpha _1}f + {\alpha _2}g = {x^2} \cr
& {\beta _1}f + {\beta _2}g = {y^2} \cr
& {\gamma _1}f + {\gamma _2}g = xy \cr} $$

And differentiate like it's the end of the world.