Mathematics - 2016-12-21 (page 1 of 6)

@TedShifrin, I'm still working on finding that surjection. Slightly different question just to solidify my understanding, does the following function define an injection from $\mathbb{Z_+}^n$ to $\mathbb{Z_+}$? $$f\left(((x_i)_{i \in \mathbb{Z_+}})_k)_{k\in [1, ..n]}\right) = (x_i)_{i \in \mathbb{Z_+}}$$

Semiclassical

For reference: en.wikipedia.org/wiki/Seven-dimensional_cross_product

Akiva Weinberger

@TedShifrin Wiki says it doesn't satisfy the Jacobi identity, so no.

Ted Shifrin

Huh?

SteamyRoot

The seven-dimensional cross-product restricted to the first 3 generators is the 3-dimensional cross-product

Ted Shifrin

12:11 AM

I've never even thought about a 7-dim cross product.

SteamyRoot

and then you define $e_4 \times e_1 = -e_5$, $e_4 \times e_2 = -e_6$,

Semiclassical

Per wikipedia, the 7-dimensional cross product is to the 3-d one as the octonians are to the quaternions.

Ted Shifrin

I don't think that makes any sense, but thanks, DogAteMy

SteamyRoot

It's actually very useful to define an almost complex structure on $S^6$, which then turns out to be Nearly Kähler.

Ted Shifrin

@Perturbative: You need to define $f(x_1,\dots,x_n)\in\Bbb Z_+$. I don't know what all your notation is for.

Semiclassical

12:13 AM

Hmm, this is an interesting remark on the Wiki page.

Mike Miller

@Akiva No. It's not a Lie algebra

Semiclassical

"In three dimensions the cross product is invariant under the action of the rotation group, SO(3), so the cross product of x and y after they are rotated is the image of x × y under the rotation. But this invariance is not true in seven dimensions; that is, the cross product is not invariant under the group of rotations in seven dimensions, SO(7). Instead it is invariant under the exceptional Lie group G2, a subgroup of SO(7)."

Mike Miller

mhm

Ted Shifrin

Yes, as soon as Steamy mentioned the a.c. structure on $S^6$, I knew it was going to be $G_2$.

Perturbative

@TedShifrin, If I scrap the notation it would essentially be this function $f((x_1, x_2, ...), (x_1, x_2, ....), (x_1, x_2, ....), ...) = (x_1, x_2, ...)$

Semiclassical

12:14 AM

Ah. Anyways, it also notes that the lack of associativity of the octonions is what prevents the Jacobi identity from going through.

Ted Shifrin

@Perturbative: You're misunderstanding things.

Read what I wrote 10 lines up.

Mike Miller

yup

Ted Shifrin

A function assigns to an $n$-tuple of positive integers a positive integer. There are no sequences here.

Akiva Weinberger

Do you want surjective or bijective

Perturbative

@TedShifrin, that was where I got the notation from

Ted Shifrin

12:21 AM

You're using it without understanding what you're doing, unfortunately.

Perturbative

@AkivaWeinberger, nope, just injective

usukidoll

one-to-one if it's just injective

Akiva Weinberger

Injective from $\Bbb Z_+^n$ to $\Bbb Z_+$?

Ted Shifrin

@Perturbative: I was asking you earlier to enumerate $\Bbb Z_+\times\Bbb Z_+$ in a sequence, i.e., to give a surjection (actually bijection) $\Bbb Z_+\to \Bbb Z_+\times\Bbb Z_+$. That's arranging the set as a giant sequence. That's what that clip from Munkres is talking about.

Mike Miller

I still like Akiva'a bijection the most

Perturbative

12:23 AM

@AkivaWeinberger, Yep

Akiva Weinberger

^_^

Ted Shifrin

It's totally far from explicit, however.

And totally uncomputable. He can't even tell me the first or the second element.

Well, maybe the first.

Mike Miller

The first is 1, the second is 2, the third is 3, the fourth is pi

Ted Shifrin

OK, I lied. At some point, it gets very murky.

Mike Miller

Actually he still needs to show that his set is closed

Ted Shifrin

12:25 AM

Huh? Closed?

Mike Miller

Or else you can't do the "just enumerate in order" trick.

Perturbative

@TedShifrin, I still haven't found the surjection, but I believe (as you rightly said), that I've misunderstood part of what is in Munkres's book, especially with regards to the notation he used, which was why I asked this second question

Mike Miller

If they accumulate somewhere, being discrete doesn't help.

Ted Shifrin

Did you read what I just wrote up there ^^^^, Perturbative?

Oh, closed as a subset of $\Bbb R$, gotcha ...

Akiva Weinberger

@MikeMiller Note that the degree $n$ ones are all greater than $\pi^n$

Mike Miller

12:26 AM

what

Ted Shifrin

Positive coefficients

Mike Miller

I mean I know

but why does that show it's closed

Akiva Weinberger

Let $P_n$ be the set of points that are degree $n$.

So my set is $\bigcup_{i=0}^nP_i$.

Wait, no

That step would only be needed if I'm doing $\bigcup_{i=0}^\infty$

Ignore that

I just need to show that $P_i$ is closed for every $i$

Ted Shifrin

It's not even clear that $P_1$ is a closed subset.

Mike Miller

I think it is, but proving it will be nontrivial

Ted Shifrin

12:29 AM

It has to do with knowing $[k\pi]$ or something.

Akiva Weinberger

Yeah, hold on. I got confused.

Mike Miller

Ah, no

There's only a finite set in each bounded interval.

Akiva Weinberger

Let $Q_i$ be the set of all of the ones whose coefficients are bounded above by $i$.

Mike Miller

B/c of coefficient positivity.

So we'43 done.

Akiva Weinberger

Then each $Q_i$ is closed.

Perturbative

12:30 AM

@TedShifrin, I did read what you wrote above, I just wanted to clarify why I had asked this second question (so as to avoid any miscommunication)

Akiva Weinberger

And for any individual point there's an $N$ large enough such that each element of $Q_N$ is greater than that point.

So for any point we only need to worry about the union of finitely many $Q_i$.

Ted Shifrin

I don't know what the second question is, @Perturbative. But that thing with a bunch of sequences in it made no sense.

Akiva Weinberger

So it's closed.

Mike Miller

beat you to it

Akiva Weinberger

By the way, closure isn't really enough for it to work @MikeMiller

I mean, closedness

Or something

SteamyRoot

12:32 AM

Hmmmm... I wonder if I can make such surjection REALLY explicit.

Akiva Weinberger

Because $\{1-1/n\}_n\cup\{1\}$ doesn't work

We specifically need order type $\omega$.

Ted Shifrin

You actually need discrete

ah, right

Mike Miller

@Akiva I said discrete long ago.

Perturbative

@TedShifrin, Okay not a problem, back to the drawing board it is :)

Akiva Weinberger

Oh wait

Scrap that

Sorry

Mike Miller

12:34 AM

You have so little faith in me :(

Ted Shifrin

You need each element to have an immediate successor and that you use up everything, so that's order type $\omega$, yeah.

Akiva Weinberger

Wait, no, I was right

Discrete isn't enough

Ted Shifrin

Right, I agree

Mike Miller

Closed and discrete is enough.

Ted Shifrin

Yeah, I think so.

Mike Miller

12:35 AM

Closed is what I realized a few minutes ago was an issue. I brought up discrete ages ago.

Ted Shifrin

Hmm, no, maybe not

Akiva Weinberger

Got it.

Mike Miller

@Ted: Closed and discrete means intersection with [0,n] is finite hence well-ordered

or enumerator or whatever you like

Ted Shifrin

Right, your closed interval comment gives what we need.

So you're guaranteed to get to each element of the set.

This is not the way for a beginning student to think about it, however!!

Mike Miller

Yeah but I'm not engaging in that half of the conversation :)

Akiva Weinberger

12:37 AM

Yeah, this was a big long tangent

Mike Miller

I'm in transit so I have an excuse not to be writing too

Ted Shifrin

Your excuse lapses shortly.

And DogAteMy no doubt needs to trudge home in the dark.

Akiva Weinberger

(though, mathematically, tangents are usually lines and thus infinitely long)

Ted Shifrin

There must be some maiming you learned about in your Talmud studies that would be relevant for you now, DogAteMy.

Akiva Weinberger

I already had the test today

And there was only one thing on the test that involved maiming or death

Ted Shifrin

12:39 AM

I know ... but your "humor" is begging for some appropriate tortuous retaliation.

Akiva Weinberger

Ahh.

Ted Shifrin

tortuous or torturous ... ?

Semiclassical

Torturusly tortuous?

Akiva Weinberger

Actually, I'm pretty free now, so I'm free for receiving a problem

Ted Shifrin

LOL

Mike Miller

12:41 AM

@Akiva I could give you my problem and we could race on it

Semiclassical

I've got some AMM problems on the brain myself.

Akiva Weinberger

Battery's about to die so I might not be able to get it

Ted Shifrin

I'll give you one later.

Semiclassical

One of them reminds me of calculations I did for research, amusingly enough.

Mike Miller

Well, you might beat me, so I'll avoid it

SteamyRoot

12:42 AM

Hrmf. The only thing I need to find the surjection is a map from $n$ to the largest triangular number strictly smaller than $n$.

Akiva Weinberger

@TedShifrin Yup

Semiclassical

Namely, let $I(n)=\int_0^1 \sqrt{x^n+(1-x)^n}\,dx$. Find the first two terms in the large-$n$ asymptotic expansion.

Mike Miller

Nice view.

Semiclassical

(That's not quite how it's stated, but for my purposes it's close enough.)

Mike Miller

man your research is so weird

Semiclassical

12:44 AM

hah

Well, in my context I(n) was something like a sum of eigenvalues of some structured matrix.

With $n$ being the size of the matrix, and the question of interest being how the sum behaved in the large-n limit.

Mike Miller

SO WEIRD

SteamyRoot

Okay....

if my calculations are correct, then

$$n \mapsto (n - \left\lfloor\frac{\sqrt{8n-7}-1}{2}\right\rfloor^2 - \left\lfloor\frac{\sqrt{8n-7}-1}{2}\right\rfloor, 2 - n + \left\lfloor\frac{\sqrt{8n-7}-1}{2}\right\rfloor^2 + 2\left\lfloor\frac{\sqrt{8n-7}-1}{2}\right\rfloor)$$

is a bijection from $\mathbb{Z}^+ \to \mathbb{Z}^+ \times \mathbb{Z}^+$

Ted Shifrin

That's way too convoluted for me to bother.

I guess it's because I would give the map the other way.

meh

Mike Miller

just draw a picture

or give an inductive definition

SteamyRoot

hrmm... I may have forgotten some divions by $3$ there.

Mike Miller

12:57 AM

if f(n)=(0,k), let f(n+1)=(k+1,0); otherwise if f(n)=(s,t), s not zero, let f(n+1)=(s-1,t+1)

SteamyRoot

Yeah, of course, this is an awful way to do it.

But I need to do something when I can't sleep :P

Mike Miller

weren't you talking about like K-theory or something earlier today

GFauxPas

sup guys, I see you're talking about math again

had my last day of the semester today!

DHMO

1:23 AM

Isn't $\Bbb A$ (algebraic numbers) a field?

arctic tern

yeah, why?

DHMO

5

A: Transcendence of $\sqrt{\pi}$

The square of an algebraic is also algebraic, hence, if $\sqrt\pi$ would be as such, then $\sqrt\pi^2=\pi$ would be so as well. Contradiction. It is known that algebraics ($\mathbb{A}$), just like naturals ($\mathbb{N}$), integers ($\mathbb{Z}$), rationals ($\mathbb{Q}$), reals ($\mathbb{R}$),...

arctic tern

I would use $\Bbb A$ for the algebraic integers though

DHMO

because it says ring here

@arctictern what does that mean?

Semiclassical

Amusingly, I was open to the part of wikipedia's page on algebraic numbers where it talks about algebraic integers.

arctic tern

1:29 AM

@DHMO a field is a ring, and anyway N isn't a ring (it's a semiring)

DHMO

@arctictern sure but why doesnt it say field instead?

arctic tern

algebraic integers are roots of monic polynomials. informally, they are the algebraic integers "without denominators" (although that's technically incorrect)

Semiclassical

because they didn't write that.

arctic tern

@DHMO field would be wrong because N and Z are not fields, and ring is wrong too because N is not a ring

DHMO

@arctictern oh thanks

arctic tern

1:30 AM

also these systems are not groups under multiplication, they are monoids (because 0)

Semiclassical

yeah, it's sloppy

I don't really get why he needed that either. Seems like all that was necessary was the closure property.

Akiva Weinberger

Is there a such thing as a best fit rectangle

given data points in $\Bbb R^2$

arctic tern

what do you want "best fit" to mean?

maybe if you interpret the set of points as a discrete and the rectangle as a continuous uniform distribution, and want to minimize |X-Y|^2

Balarka Sen

my sleep pattern's permanently wrecked i guess

Akiva Weinberger

Do your parents do nothing in this

Balarka Sen

1:39 AM

i don't understand what their role on this would be

it's wrecked, it can't be repaired by yelling

so they have stopped doing that.

Ramanujan

I came to know something special about e

Akiva Weinberger

"By 9:30 start getting ready for bed. By 10:00 be in bed. We will take away your phone while this happens." @BalarkaSen

Ramanujan

Balarka Sen

I don't have a phone

Fargle

@Balarka: I had a terrible sleep schedule for years. The past few weeks, though, I've been falling asleep at like 9:00 and waking up at like 5:00.

Balarka Sen

1:42 AM

I'll give up my laptop but that wouldn't help

Akiva Weinberger

I essentially meant giving up your way of chatting with us

Balarka Sen

@Fargle neat

@Akiva it wouldn't help

Akiva Weinberger

It's 7am where you are. Did you sleep last night?

DHMO

@Ramanujan indeed, $\dfrac{\mathrm d}{\mathrm dx} e^x = e^x$

Akiva Weinberger

@BalarkaSen Would you just lie in bed awake?

Fargle

1:43 AM

lol, I'm just saying it's possible, @Balarka.

Balarka Sen

nah. i tried to, but i couldn't

Ramanujan

@DHMO for that reason?

DHMO

@Ramanujan yes

Akiva Weinberger

Balarka, you might actually need a trip to the doctor and medication

Balarka Sen

Yeah, maybe.

Nah. I'll just try and get more hats.

Ramanujan

1:44 AM

@DHMO you got that trick of multiplying ?

Akiva Weinberger

How about Googling research on insomnia

DHMO

@Ramanujan Indeed, you can define $e^x$ as the solution of $y'=y$ and $y(0)=1$

Akiva Weinberger

if you're so awake

DHMO

@Ramanujan no idea

Balarka Sen

don't see why that should appeal more than hats

Akiva Weinberger

1:45 AM

Because it pertains to your life and your future and your health and your mental health

Balarka Sen

Yeah. No.

Akiva Weinberger

Maybe animated YouTube videos about insomnia

Fargle

@AkivaWeinberger 3blue1bloodshot

Akiva Weinberger

3Blue1Brown reference?

I don't see the connection

Fargle

Animated YouTube videos.

Balarka Sen

1:47 AM

Bloodshot eyes, probably

Fargle

Yup.

It's only skin-deep, I promise. (Cornea-deep?)

Balarka Sen

woo, new hat.

Akiva Weinberger

I'm on mobile so I can't see any of them

but I'll take your word for it

Balarka Sen

It's a goblin's head.

A bunch of subirds have hijacked my veranda as I left some seeds leaving there.

Adeek

2:44 AM

hey @Fargle are you here ?

Perturbative

Hey everyone, quick question what's the difference between isomorphism and bijection? In Axler's Linear Algebra Done Right, Axler defines an isomorphism to be an invertible linear map, (which to me is just another way of stating a bijection)?

Nevermind, found an answer here : math.stackexchange.com/questions/1064325/…

Akiva Weinberger

Yeah, injections don't need to be linear or anything

As vector spaces, there's no isomorphism between $\Bbb R$ and $\Bbb R^2$, but there is a bijection (this is not obvious!). @Perturbative

Adeek

hey @AkivaWeinberger what do you think of the following sentence

The following theorem explains why $E(\mathbb{Q})$ is an Abelian group. Theorem 2.1 is quite significant to us, as we don't know prior to knowing theorem 2.1 that if we a line connecting two points on the elliptic curves will indeed intersect $E$ in another point. Theorem 2.1 tells us that this is indeed the case.

I think it can be improved

Akiva Weinberger

Delete "if we"

Adeek

alright

Akiva Weinberger

2:59 AM

Also "elliptic curves" should just be "elliptic curve" since you're talking about $E$

Adeek

yeah

Akiva Weinberger

Otherwise, it's fine grammatically

This is how I would write it:

> The following theorem, Theorem 2.1, explains why $E(\Bbb Q)$ is an Abelian group. It is quite significant to us, since up until now we did not know that a line connecting two points on the elliptic curve $E$ will indeed intersect it in another point. Theorem 2.1 assures us that this is indeed the case.

Adeek

yeah that sounds very good.

thanks a lot @AkivaWeinberger

Akiva Weinberger

Though, I think you have something backwards; you say that Theorem 2.1 tells us that $E(\Bbb Q)$ is a group, and that it's significant because it tells us a line through two points must hit a third,

Adeek

no theorem 2.1 only says that a line through two points hits a third

Akiva Weinberger

3:02 AM

but isn't it that Theorem 2.1 tells us that a line through two points must hit a third, and that it's significant because it tells us that $E(\Bbb Q)$ is a group?

Adeek

but the next theorem that I will introduce after this

will be the one that shows that $E(\mathbb{Q})$ is an abelian group

Akiva Weinberger

@Adeek In your first sentence you say otherwise

Ohhh

I see

Theorem 2.1 isn't the next theorem

Adeek

yeah

Akiva Weinberger

So the first sentence of my rewrite is wrong

Yeah, ignore the rewrite

Adeek

what do you think of it like this ?

Akiva Weinberger

3:05 AM

Looks good

Did you define $A+A$?

Adeek

yeah

I defined it before

I could give you my file if you want

it introduces elliptic curve and elliptic curve cryptography from scratch

no knowledge needed to understand it

@AkivaWeinberger would you like me to mail it to you ?

Akiva Weinberger

Sure, why not

Adeek

email ?

Akiva Weinberger

Email is amwmath@yahoo.com

Adeek

okay copied

done

Akiva Weinberger

3:13 AM

Thanks

I now know your full name

(except for middle name)

Adeek

yeah :P

I don't mind @AkivaWeinberger

haha

I have to submit this by thursday I was wondering if you have any edits in mind if you get a chance to see it please let me know.

@AkivaWeinberger Recall Fermat's little theorem
$$x^p \equiv x (mod p).$$ Fermat last theorem tells us that $\phi$ fixes E pointwise, that is $\phi(P) = P$. Hence the Frobenius map is well defined.

is there a better wording than "tells us"

GFauxPas

"impounds"

it's not the right word, but it sounds impressive

also I think you mean Fermat's little theorem

Adeek

yeah that does sound impressive

GFauxPas

use \pmod for mod in parens

arctic tern

indicates

suggests, leads to

Adeek

3:24 AM

yeah indicates is even nicer

GFauxPas

"absquatulates"

Akiva Weinberger

Maybe "Because of Fermat's Little Theorem, $\phi$ fixes $E$ pointwise"

Physics Enthusiast

hey everyone, if anyone is bored and wouldn't mind helping me with some basic complex analysis, I'd really appreciate it:

Akiva Weinberger

What's the question?

Physics Enthusiast

i i just posted the question, and it's called "Elementary bound on inverse zeta function from Stein-Shakarchi"

I'm a little confused about this line from Stein-Shakarchi pg 187:
\begin{equation}
c |\zeta^{-3} (\sigma)| |{t}|^{- \epsilon} \geq c'(\sigma -1)^3 |t|^{- \epsilon},
\end{equation}
where $s = \sigma + i t$, $\sigma \geq 1$, $|t| \geq 1$.

I'm confused about this step because from the definition of the zeta function, $\zeta(\sigma) \geq \int_1^\infty \frac{1}{n^\sigma} d n = \frac{1}{\sigma}$. We have $\zeta(\sigma) \leq 1 + \frac{1}{\sigma}$, but I cannot figure out how to absorb the extra constant $1$ into the factor $c'$.

Akiva Weinberger

3:35 AM

Don't know, sorry

Adeek

I don't like writting conclusions

it sucks

Physics Enthusiast

:( darn thnx though

olympiad math

3:48 AM

What does it mean when we say two statements are equivalent ?

Mike Miller

Given either you can prove the other

Kaj Hansen

@Adeek, never conclude, just keep writing

olympiad math

Thanks @MikeMiller

Adeek

haha @KajHansen

@KajHansen what do you think of this conclusion ?

Elliptic curves are important mathematical objects. Andrew Wiles used Elliptic curves in his famous proof for Fermat's Last theorem, a problem that wasn't solved for hundreds of years. In a sense the reason why elliptic curves are important to mathematicians is that they are a simple algebraic structure that isn't completely understood.

Elliptic curves aren't only useful for mathematics, but they also provide a good cryptosystem that is good for small devices such as mobile phone, as they provide the same security as other public key cryptosystem but at a lower cost.

Kaj Hansen

I think they help factor integers too

Elliptic curves provide

Adeek

3:57 AM

oh cool

Kaj Hansen

Lenstra elliptic curve factorization

The Lenstra elliptic curve factorization or the elliptic curve factorization method (ECM) is a fast, sub-exponential running time algorithm for integer factorization which employs elliptic curves. For general purpose factoring, ECM is the third-fastest known factoring method. The second fastest is the multiple polynomial quadratic sieve and the fastest is the general number field sieve. The Lenstra elliptic curve factorization is named after Hendrik Lenstra. Practically speaking, ECM is considered a special purpose factoring algorithm as it is most suitable for finding small factors. Currently...

Adeek

that is very cool

usukidoll

.-.

Adeek

Also, one of the fastest way to factor integers is by using elliptic curves.

arctic tern

@usukidoll ?

usukidoll

3:58 AM

just depressed...that's all

arctic tern

it's okay to be sad

Transcript for

$Mathematics$ Mathematics