Mathematics - 2014-01-07 (page 1 of 4)

Analysts should realize RR is all about meromorphic functions. What's their problem? (Well, it doesn't hurt to talk about meromorphic $1$-forms, too.)

Mike

He's skipping it side I did it last semester. :P

since*

Ted Shifrin

Of course, I love the geometric interpretation of RR in terms of how linear spaces meet the canonical curve in $\Bbb P^{g-1}$.

Mike

Don't know it.

Ted Shifrin

is the person teaching alg top a good teacher, @Mike?

Mike

That reminds me that I didn't get my rational points question answered. :(

Ted Shifrin

1:54 AM

That's in Griffiths and Harris or Griffiths's book on algebraic curves, @Mike. You should look at it ... very beautiful.

Ask your question in main ... or find Pete Clark and ask him.

Mike

@Ted Very. Actually, he went to MIT too (though obviously that has only some correlation with teaching ability)

Ted Shifrin

yeah, e.g., I went to MIT ... ergo ...

who is it, @Mike?

Mike

I might ask it on MO. It's not in any of the literature.

@Ted Richard Scott, no relation to the governor.

Ted Shifrin

oh, one of his coauthors was another MIT student whom I met years ago when he was in high school ... Paul Gunnells ... smart guy.

Not surprisingly, they were both students of Bob MacPherson. Excellent.

OK, I'm disappearing. poof

Mike

I believe he attended the IAS. That he's at SCU shows how dedicated he is to teaching, I think. :)

Pedro Tamaroff

1:59 AM

I am trying to find out what this combinatorics course next semester will be about.

@Mike Heya.

But I think it is still elementary stuff.

usukidoll

GRUMPY CAT!!!

oh heyyy Ted....

Mike

@Pedro It'll probably just be... combinatorics :P

I doubt there'll be any fancy algebraic machinery.

Pedro Tamaroff

@Mike I'm quite bad at counting. Maybe it's a chance to better myself!

usukidoll

@TedShifrin woof

leo

2:16 AM

well hello all

Charlie

Hello @leo

leo

@Charlie :-)

Charlie

You see, I was the only to say hello @leo

leo

yes

Usually noone does

Charlie

:(

I always say hello to you Leo

Helleo

leo

2:28 AM

@Charlie Yes

I mean when you are not here

Charlie

Yes, it's not the same when I'm not here

AndrewG

3:11 AM

@Ted I just got part (a) of 1.2 #19, but it's kind of long and I wonder if there's a simpler way. Can I show you?

Ted Shifrin

@Andrew: Sure.

AndrewG

So, for starters, taking the derivative of $\alpha(s)$ and using Frenet, I get $$0 = (\lambda' - \mu\kappa - 1)T(s) + (\mu' + \lambda\kappa - \tau\nu)N(s) + (\nu' + \tau\mu)B(s) \implies \lambda' = 1 + \mu\kappa, \mu' = \tau\nu - \lambda\kappa, \nu' = -\tau\mu$$ since T,N,B are lin. ind.

Ted Shifrin

Excellent. You should have it with one more thing.

AndrewG

Meanwhile, $||\alpha(s)||^2 = \mu^2 + \lambda^2 + \nu^2$ is const. $\implies 0 = \mu\mu’ + \lambda\lambda’ + \nu\nu’$. Multiplying each of the above 3 equations by the appropriate thing and adding everything up, I get $\lambda + \lambda\mu\kappa + \mu\tau\nu - \lambda\mu\kappa - \nu\tau\mu = 0 \implies \lambda = 0 \implies \lambda’ = 0$

and then the dominos start falling

now I can say $\mu = -1/\kappa$

Ted Shifrin

you can? How?

AndrewG

3:25 AM

from the first equation above, $\lambda' = 1 + \mu\kappa$, which now reads $0 = 1 + \mu\kappa$

Ted Shifrin

Oh, part was hidden. Grr at iPad.

AndrewG

and so then $\nu = \frac{1}{\tau} (\mu)' = 1/\tau (-1/\kappa)'$

Ted Shifrin

You can get there a lot faster. :)

AndrewG

that's what I figured and why I wanted to ask. From here I do a bunch of ugly plug-n-chug

Ted Shifrin

you should get $\lambda=0$ immediately.

AndrewG

3:27 AM

Yes, I got that above

Ted Shifrin

Motion on a sphere means what about velocity?

AndrewG

Oh. Derp. All tangential?

Wait.

Mike

@Ted This stupid question (real analytic density of rational points on a curve over the reals) isn't in any of the literature. I can't even figure out what I should be searching on Math Reviews, because "density of rational*" gives me asymptotic stuff about how many rationals there are of bounded height.

AndrewG

@Ted I'm not sure

Ted Shifrin

What was your hypothesis on the curve?

AndrewG

3:30 AM

constant $||\alpha(s)||$

Ted Shifrin

what is $\alpha\cdot\T$, @Andrew?

AndrewG

$0$

Am I being stupid again :|

Ted Shifrin

That hypothesis question was for you, @Mike.

Mike

It's not really a hypothesis because I doubt it's true, but I want to know a counterexample (and if possible a pleasant equivalent condition). A plane curve having infinitely many rational points $\implies$ that these points are real-analytic dense.

I know, I was typing :)

It's trivially true for $g \geq 2$ by Faltings and I think I have an argument that it's true for genus $0$ (but I'm not sure I believe my argument). So it's really about elliptic curves.

Ted Shifrin

Ok, I'll ask my number theory colleagues.

AndrewG

3:33 AM

er

Ted Shifrin

So that says $\lambda=0$, @Andrew.

AndrewG

$\alpha \cdot T = (\lambda T + ...) \cdot T = \lambda = 0$

Right

I don't have to go through any of that algebra rigoromorole

ok

Mike

The argument I have and don't believe for $g=0$: a genus zero curve with infinitely many rational points is birational to $\Bbb P^1(\Bbb R)$. Pick an arbitrary point on the curve; pull it back under the rational map; pick a sequence of rational points convergent to it, and push it forward through the rational map. I don't buy it because the map doesn't necessarily have to be defined over the rationals. But I do buy that it's true for $g=0$ curves.

Ted Shifrin

Just a consequence of the first lemma in the section!

This site is showing up horrendously on my iPad. Brb.

Mike

Edit that to say "I have and realize is false", rather than "don't believe"

Ted Shifrin

3:40 AM

I dunno, @!ike.

Mike

Though it's true if I relax it to the curve being defined over $\Bbb Q$, which is probably what I want anyway

Ted Shifrin

ok, this chat no longer works on iPad. I punt. Night.

Mike

Night.

AndrewG

Night. And thanks :)

Pedro Tamaroff

5:21 AM

@Mike Hello?

Mike

Hey, can't talk now. I'll be back on in an hour ish

Pedro Tamaroff

@Mike Kay. I'm not sleeping tonight.

Pedro Tamaroff

5:53 AM

@MarianoSuarezAlvarez

Pedro Tamaroff

6:23 AM

Seems people don't like differentation.

=(

@anon You around?

mmhmm

Watcha doin?

nothin

I was just reading the section of BAI I shared with some time ago

About factorial monoids and UFDs.

@anon I am trying to prove that $\Bbb Z[\sqrt{-5}]$ satsfies the ACC

Ah, got it.

@anon Just for the record, we observe that if $a$ is a proper factor of $b$ then $Na<Nb$.

anon

6:51 AM

in fact a|b implies Na|Nb

Pedro Tamaroff

Right.

@anon Hold on. =)

Using the above, I can prove that eventually the norm is constant in a chain.

I mean, we know in $\Bbb N$ that if $a_{i+1}\mid a_i$, eventually $a_i$ is constant.

So, I have to show this implies all the elts in $\Bbb Z[\sqrt{-5}]$ are associates after that point.

Ah, duh.

$a=bc$ and $Na=Nb$ gives $Nc=1$ so $c$ is a unit.

@anon Now I have to show the same about $\Bbb Z[X]$.

anon

7:08 AM

in fact if R is noeth then R[X] is noeth

furthermore if S is noeth then S/I is noeth

you can use this to argue Z noeth implies Z[sqrt(-5)] noeth

Pedro Tamaroff

@anon OK, meponders on how to prove R is N then R[X] is N.

I was thinking I could consider different cases of chains in Z[X]

Cannot I argue on degs?

anon

try

Pedro Tamaroff

gets scratch paperz

I mean, there's not much to it, right?

Cannot I use Z[X] is a UFD =P?

Pedro Tamaroff

7:35 AM

@anon So, I think the deh thing does it.

Mike

7:49 AM

hi @PedroTamaroff

Pedro Tamaroff

@Mike Yao.

I am trying to show Z[sqrt 10] is not a UFD.

So trying to solve a^2+10b^2=c^2+10d^2

Hm.

Ian Mateus

@PedroTamaroff is it $\sqrt{10}$ or $\sqrt{-10}$?

Pedro Tamaroff

10 = 2 . 5 = sqrt 10 sqrt 10?

That looks good.

@IanMateus $\sqrt 10$. I just didn't like the minuses.

Mike

Are you sure they're not unit products of one another?

@PedroTamaroff The difference is that real quadratic fields are much more difficult in general :P

Pedro Tamaroff

@Mike Uh?

Ian Mateus

7:59 AM

@PedroTamaroff does this make it? $9=3\times 3=(\sqrt{10})^2-1=(\sqrt{10}+1)(\sqrt{10}-1)$ so we have to show they are not adjoints

Pedro Tamaroff

@IanMateus That's good.

I think.

Ian Mateus

@PedroTamaroff the norm is $a^2+10b^2=1$. If $b\geq 1$, this is impossible, so $a^2=1$. So, the only units in $\Bbb Z [\sqrt{10}]$ are $\pm 1$, and we are done

Mike

I was just reminding that to show a failure of unique factorization you need to show that the products aren't the other times a unit or in some way trivial. But then I realized I didn't need to since you knew that.

@Ian

That's not true. That's the norm of an element in $\Bbb Z[\sqrt{-10}]$

Ian Mateus

Hmm ok, I changed the sign

Pedro Tamaroff

@Mike ;)

Mike

8:04 AM

Unfortunately we have plenty of units in this ring

Pedro Tamaroff

At any rate, here's a nontrivial unit in $\Bbb Z[\sqrt 10]$.

@Mike Yes.

Mike

$N(3+\sqrt{10})=-1$

Pedro Tamaroff

Take any elt that is coprime with 10 and hope for big enough factors.

Say $2-\sqrt 10$ is a unit.

Mike

It's... not

Pedro Tamaroff

Whoops.

I am bad at counting.

2

Mike

8:06 AM

:P

Pedro Tamaroff

Retries.

2

Mike

$N(19+6\sqrt{10})=1$

Pedro Tamaroff

Call sqrt 10 = k for simplicity.

Ian Mateus

Yeah, I just found out $a$ is odd and $b$ is even

Pedro Tamaroff

then (a+bk)(c+dk)=1 gives right away that ad+bc=0.

On the other hand, it gives that ac+bd10=1

so that a c are comprime with 10.

And (a,d)=(c,b)=(a,b)=(10,a)=(c,10)=1

Well, much coprimeness!

Say 3-k is a unit.

Mike

8:09 AM

Anyway @Ian your paradoxical factorization is fine because there's no unit in a quadratic field that's a third-integer plus something

Pedro Tamaroff

@Gigili Long time no see.

Gigili

Oh hello

How are you doing?

Ian Mateus

@Mike what is a third integer?

Mike

A third of an integer.

:P

Pedro Tamaroff

@Gigili Not bad.

Though haven't slept.

Gigili

8:11 AM

Umm, still studying math, I see

Do you know anything about QR decomposition?

Pedro Tamaroff

Nah. Matrices stuff?

Gigili

Do you know what $A^{-1}$ would be when we have $A=QR$?

@PedroTamaroff Yes

Pedro Tamaroff

@Gigili $R^{-1}Q^{-1}$?

What is QR decomp?

Ian Mateus

@PedroTamaroff but there's no guarantee they individually exist, right?

Gigili

OK, I did the same. But why the book says $R^{-1}Q^T$?

Note that $Q$ is an orthogonal matrix, could it be the reason?

Pedro Tamaroff

8:14 AM

What is QR decomp?

Gigili

$Q^TQ=QQ^T=I$

Pedro Tamaroff

Yes, the inverse of Q is Q^T

Gigili

QR decomposition

In linear algebra, a QR decomposition (also called a QR factorization) of a matrix is a decomposition of a matrix A into a product A = QR of an orthogonal matrix Q and an upper triangular matrix R. QR decomposition is often used to solve the linear least squares problem, and is the basis for a particular eigenvalue algorithm, the QR algorithm. If A has n linearly independent columns, then the first n columns of Q form an orthonormal basis for the column space of A. More specifically, the first k columns of Q form an orthonormal basis for the span of the first k columns of A...

@PedroTamaroff Always? Because it's orthogonal?

Pedro Tamaroff

@Gigili By definition, orthogonal means QQ^T=Q^TQ=1.

Gigili

Yes, inverse of a matrix is $A^{-1}$ and not $A^T$, right?

That's the part I don't understand

AndrewG

8:17 AM

For orthogonal, they are the same thing. The inverse is the transpose.

anon

@Gigili if $QQ^T=I$ then $Q^T=Q^{-1}$

Gigili

Aw, got it!

Thank you

And hellow @anon

I see you're not a blue yet

anon

@PedroTamaroff proving things about $\Bbb Z[\sqrt{d}]$ is much easier when $d<0$ than when $d>0$

Gigili

That's quite something

anon

thanks, I think

Pedro Tamaroff

8:19 AM

@anon Yeah. Norms.

I am not sure if 3 is irred there. =P

d=10

I can see one cannot write $z\bar z=3$ since 3 is not a QR mod 5.

anon

if 3=ab is reducible then 9=N(a)N(b) implies N(a)=N(b)=3 so u^2-10v^2=3 but...

Pedro Tamaroff

@anon We have norms here too? DERP.

Yeah, same as above QR mod 5.

impossibru.

I thought "norms" were supposed to be positive.

(We can also have -3=2=a^2 mod 5)

Same story, though.

anon

oh yes

Pedro Tamaroff

@anon Well, look at le this.

If p,q primes, p = 1 mod 4, and q not a QR mod p then Z[\|pq] not a UFD

anon

poor man's square root symbol I see...

Pedro Tamaroff

8:33 AM

@anon LOL what?

bow to my ASCII powers

@anon I think that comment section will not be short.

=/

anon

indeed

Pedro Tamaroff

Hm, Tobias saved the day, possibly.

That's quite a nice name.

Then again the guy asks "What is the difference between ideal and principle ideal?"

He structured his post awfully.

=P

Jasper Loy

8:48 AM

@PedroTamaroff *principal

Pedro Tamaroff

@JasperLoy Yeah, I was quoting verbatim.

You know I know, Jasper.

LE SIGH

Jasper Loy

@PedroTamaroff You tricked me!

Pedro Tamaroff

How are you doing?

Jasper Loy

I am bad, sad to say.

Pedro Tamaroff

Can tell here?

Jasper Loy

8:52 AM

Just my usual mental problems. I may never recover this lifetime, but I am trying.

Gigili

Great avatar, @AndrewG

Pedro Tamaroff

That's a Digimon, right?

Gigili

Umm, is it? I don't think so

It's much cuter than those weird things

anon

final fantasy

TheLittleNaruto

Heya, Everyone :)

Jasper Loy

9:03 AM

Hello.

Pedro Tamaroff

I can't do this‌. And I have to go.

Byes.

TheLittleNaruto

I was good in Maths when I was doing my schooling.

But Now, I am totally focusing on development

Gigili

@PedroTamaroff Whoa

Bye.

TheLittleNaruto

Jasper, I am new to this room.

I hope you're a good mathematician :)

Chris's sis

Greetings

@N3buchadnezzar I'm developing a new class of integrals.

robjohn

9:33 AM

@Chris'ssis Is the enrollment in this class good?

Does the principal value visit often?

(school idioms)

Chris's sis

@robjohn yeah. :-)

$$\int_0^1\left(\frac{1}{\log(t)(1+t)}-\frac{1}{t^2-1}\right) \ dt$$

Then $$\int_0^1\left(\frac{1}{\log(t)(1+t)}-\frac{1}{t^2-1}\right)^2 \ dt$$

Jasper Loy

10:07 AM

@TheLittleNaruto I am not a mathematician. I am only a banana. =)

TheLittleNaruto

@JasperLoy and I am the Bigger one ;)

Jasper Loy

@TheLittleNaruto Hahaha. Hmm, you can be a papaya.

TheLittleNaruto

@JasperLoy Hahahaha

Where you're from ?

Jasper Loy

Singapore. What brings you to this chat.

TheLittleNaruto

I am developing an Android Application which will calculate the area of the random generated polygon on Map and then I need to divide it as a grid according to per hectare

As It was long ago when I was close to Maths. So I just came here so that I can get some clue for achieve this

:)

achieving*

Jasper Loy

10:17 AM

@TheLittleNaruto You should be as specific as you can and then post the question on the site.

TheLittleNaruto

@JasperLoy Okay. I'll ask on the main site. Please edit it if you find some mistakes so that I won't get negative votes ;)

Jasper Loy

@TheLittleNaruto Don't write what you wrote above, I don't understand it.

TheLittleNaruto

Ohh :( Now I got it, why you said "as specific as you can"

TheLittleNaruto

10:33 AM

Jasper, there ?

Jasper Loy

@TheLittleNaruto Yes. Wassup?

TheLittleNaruto

A polygon can be anything, like a triangle, rectangle, Hexagon , I want to know if there is a common formula for calculating the area , Jasper

Jasper Loy

Shoelace formula

The shoelace formula, or shoelace algorithm, is a mathematical algorithm to determine the area of a simple polygon whose vertices are described by ordered pairs in the plane. It is also sometimes called the shoelace method. It is also known as Gauss's area formula, after Carl Friedrich Gauss. It has applications in surveying and forestry, among other areas. It is also called the surveyor's formula. The formula can be represented by the expression: : \begin{align} \mathbf{A} & = {1 \over 2} \Big | \sum_{i=1}^{n-1} x_iy_{i+1} + x_ny_1 - \sum_{i=1}^{n-1} x_{i+1}y_i - x_1y_n \Big | \\ & = ...

TheLittleNaruto

Thanks Jasper :)

One more question in what unit I'll get the area ?

if the vertices are set of Latitude and Longitude

Jasper Loy

Ah, I thought it was on a flat map. If it is on a sphere, I dunno how to do this.

TheLittleNaruto

10:43 AM

Ahh! What you think then ? I should not try Shoelace formula?

Jasper Loy

Well, if you approximate the earth to a flat map, there can be distance along the horizontal and vertical axis.

Say each is in km, then the unit will be square km, for example.

But I dunno if this approximation is good enough.

Maybe you should post on the main site. Let someone who knows answer.

TheLittleNaruto

No I can consider approximation as this app is regarding analysis of sample of field for agriculture purpose

Nick

Aloha, Amigos!

@Jasper: Happy first week of the New Year, monsignor Jasper.

Jasper Loy

@Nick Hi! Hope you are doing well in school.

Pedro Tamaroff

11:02 AM

Errands suck.

Nick

@Jasper: Well, I'm technically the first in my class.... but I hope I can do better.

@Pedro: Only if you think they are.

@PedroTamaroff: You think you're bad at counting? I'm worse. I couldn't count days until JasperLoy taught me how to count days.

Pedro Tamaroff

Yeah, no. Imagination cannot save this one!

Nick

@TheLittleNaruto: Are you by chance Pashin Raja?

@Pedro: Then just get it done with. The sooner it's over, the sooner you can get back to math.

I just attended a math quiz today. I got second place! :D

Pedro Tamaroff

11:19 AM

That is not up to me. =)

Nick

@Pedro: Abstraction is best suited for Algebra and Poetry, not for conversations.

TheLittleNaruto

@Nick o-O? I guess not.

TheLittleNaruto

11:51 AM

Cya, Everyone :)

robjohn

12:16 PM

@Chris'ssis: Here's one for you: what is the asymptotic decay of $\int_0^\infty\frac{\mathrm{d}x}{(x+1)(x+2)\cdots(x+n)}$ ?

This is from a problem, but as I've answered it, I won't say which until you want.

Chris's sis

@robjohn on 26th December I challenged myself with this

robjohn

@Chris'ssis That is the same integral, and that identity follows from partial fractions.

Chris's sis

@robjohn Indeed.

N3buchadnezzar

@robjohn Asymptotic decay ?

I proved this integral a few days myself, used partial fractions as well =)

robjohn

@N3buchadnezzar find a simpler function whose ratio with that integral goes to $1$ as $n\to\infty$

Chris's sis

12:26 PM

@robjohn Exactly.

robjohn

@Chris'ssis that sum doesn't really give an idea of the asymptotic behavior

Chris's sis

@robjohn doesn't it help? hmmm

Chris's sis

1:03 PM

@robjohn I'd write the integrand in terms of Gamma function and then think of using Stirling's approximations. (actually, I saw something very similar to a question by Ramanujan --- I need to see it again)

robjohn

@Chris'ssis okay. I don't think Stirling will be close enough, but perhaps you have a different approach.

Chris's sis

Well, that was just an idea.

Nimza

Hello! Is there somebody familiar with complex geometry?

Sush

2:17 PM

@robjohn, $$\max_{x_1}f(x_1,g(x_1)).$$ And, let $f$ attends max at $x_1^*$, so first order necessary conditions imply that $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_1}+\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_2}\dfrac{d g(x_1^*)}{dx_1}$$ as well as first order necessary conditions imply that $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_1}=0$$ and $$\dfrac{\partial f(x_1^*,g(x_1^*))}{\partial x_2}=0.$$ Am I right?

@JasperLoy, my reputation in main is 970, but in chat is 961. Why such difference?

robjohn

@Sush I don't see how you can get from the second equation to the third and fourth.

@Sush your reputation is 970 in chat. Refresh your browser.

Sush

@robjohn, sorry, the second equation should equal to zero, right?

robjohn

@Sush Yes, but even then, the next two do not follow

I am leaving to walk the dog... BBL

Chris's sis

2:36 PM

@robjohn That question is evil. What is your solution?

Sush

@robjohn, but my text says "If the differential function $f(x_1,\dots,x_n)$ reaches a local interior maximum at $(x_1^*,\dots,x_n^*)$, then these hold simultaneously: $$\dfrac{\partial f(x_1^*,\dots,x_n^*)}{\partial x_1}=0;\dots;\dfrac{\partial f(x_1^*,\dots,x_n^*)}{\partial x_n}=0 $$ "