Mathematics - 2016-04-01 (page 1 of 2)

user19405892

00:59

hi

Julian Rachman

01:33

anyone?:
http://math.stackexchange.com/questions/1722139/noetherian-lemma-contridiction

Akiva Weinberger

@DemCodeLines You could ask that on the main site: Mathematics

Also, try typing "[main]" into the chat

DemCodeLines

Mathematics

cool

Forever Mozart

@DemCodeLines dude wat

ted is dead

PVAL

@MikeMiller This contradicts known results arxiv.org/pdf/1603.09437.pdf what the hell

bwDraco

#TIL: 141,421,356,237,307 is a prime number http://wolframalpha.com/input/?i=is+141421356237307+prime&x=0&y=0 (#TheNumberDevil page 61)

Tweeted by bwDraco on April 1, 2016 at 1:43 AM

PVAL

01:48

@MikeMiller arxiv.org/pdf/1602.08821v1.pdf arxiv.org/pdf/1603.09437.pdf Two papers claiming opposite results in a months time...

Forever Mozart

lol

well I guess they proved ZFC is inconsistent

finally

bwDraco

Wow. They even cover Goldbach's conjecture (and Goldbach's weak conjecture) in this book.

Forever Mozart

nonono

bwDraco

(The Number Devil pages 62-64.)

Forever Mozart

sum of two squares is prime

no

Mike Miller

02:05

@PVAL Umm.....

@PVAL I'll try to read the short one tonight if I can

user19405892

hi

3

Q: Prove that there is a differentiable function $f: \mathbb{R} \to \mathbb{R}$ satisfying $[f(x)]^5+f(x)+x = 0$

Prove that there is a differentiable function $f: \mathbb{R} \to \mathbb{R}$ satisfying $$[f(x)]^5+f(x)+x = 0$$ for all $x \in \mathbb{R}$. Find $f'(x)$. Seeing how this is a functional equation, I think we might be able to use induction or some other technique to determine information abou...

can someone example in that answer how to actually find f?

anomaly says to use the chain rule but that isn't clear

bwDraco

18

A: Good math bed-time stories for children?

The Number Devil: A Mathematical Adventure by Hans Enzensberger is a fun survey of elementary number theory. It's structured similarly to the series of children's books 'Diary of a Wimpy Kid.' If I remember correctly, the storylines from each of its chapters are pretty close to independent.

user19405892

02:24

Do i really need to solve $f'f = \frac{1}{-5x^4-1}$?

any ideas?

Semiclassical

once you've shown that $f(x)$ is well-defined and differentiable, i think expressing $f'(x)$ implicitly via the chain rule is fine

user19405892

how do i find f'(x) using the chain rule though?

Semiclassical

well, i think that's just referring to computing $\dfrac{d}{dx}[f(x)]^5$

bwDraco

$\frac{1}{-5x^{4}-1} = (-5x^4-1)^{-1}$

user19405892

i mean referring to $f'f = \frac{1}{-5x^4-1}$

Semiclassical

02:36

which is most easily found by using the chain rule to write $\dfrac{d(f^5)}{dx}=\dfrac{d(f^5)}{df}\dfrac{df}{dx}$

bwDraco

I've forgotten most of my calculus but I'm trying to give you a bit of a hint.

user19405892

@Semiclassical that doesn't look like it can be evaluated

Semiclassical

sure it can.

Mike Miller

@PVAL: Have you looked?

Semiclassical

presumably you've already done so if you understand how you got the $f'(x)f(x)$ expression you've been using :)

or are you referring to $f'(x)f(x)=-\frac{1}{5x^4+1}$ itself as not being possible to evaluate?

PVAL

02:38

@MikeMiller The new one seems more fishy than the long one.

Mike Miller

That is how I feel too.

user19405892

@Semiclassical Yeah I am saying that is pretty challenging to evaluate

Semiclassical

well, it depends on what you count as an answer

user19405892

if they are asking for $f'(x)$ it can't be too complicated

Semiclassical

if one takes the attitude that, having shown that $f(x)$ exists, you're free to use it in writing out $f'(x)$

user19405892

02:39

oh

Semiclassical

then you're done since you can write $f'(x)=-\frac{1}{f(x)}\frac{1}{5x^4+1}$

buuuuut i don't know if that's what they actually want :/

PVAL

But I can't tell why it's wrong. He doesn't actually prove the torus he constructs is exact, and references elsewhere. I imagine it isn't. There's also slightly weird definitions (more in their English than their mathematical content) for things like Lagrangian isotopic.

user19405892

@Semiclassical Solving $f'f = -\frac{1}{5x^4+1}$ doesn't look very easy so I think that is fine

Semiclassical

i'll play around with mathematica a bit to be sure, but i think that's about as good as one can hope for at generic $x$

Mike Miller

@PVAL: I'm going to email you a brief comment.

Semiclassical

02:41

actually, come to think of it

writing out $f(x)$ explicitly means solving a quintic polynomial for one of its roots

user19405892

yes

Semiclassical

yeah, probably they don't care about $f(x)$ explicitly

PVAL

@MikeMiller Well so is Goodman...

Semiclassical

though, you might check your algebra for $f'(x)$. i get a different result

Mike Miller

@PVAL true. just worthless snark.

PVAL

02:45

Actually every serious math I have read by Eliashberg has been fantastic.

His book with Cieleback and his "Remarks on Symplectic Fillings" are the main extent of this,

Semiclassical

has ted not been around lately?

i can't tell if it's that or my timing just stinks lately

Mike Miller

the latter

crocket

03:01

0

Q: The textbook's natural deduction proof for $\vdash(\neg(\phi\to\psi)\to\phi)$ seems to be wrong with regard to RAA(Reductio Ad Absurdum).

As you can see below, $\psi$ pops out of nowhere due to RAA(reductio ad absurdum). This is probably wrong. Is there really a proper natural deduction proof for $\vdash(\neg(\phi\to\psi)\to\phi)$?

logic natural-deduction sequent-calculus

Semiclassical

fair enough

user19405892

can a function $f(x) > 0$ for all $x \in \mathbb{R}$ but be strictly decreasing?

I was thinking if it decreased by an infinitesimal distance each time it is possible

@Semiclassical What are you getting?

Semiclassical

well, where are you getting $x^4$ from?

user19405892

$g'(x)$

Semiclassical

you shouldn't be

user19405892

03:11

i am just applying the inverse theorem

Semiclassical

also, there's an obvious answer---what if $f(x)$ decreases by a factor of two every time you increase by $1$

thats for your other question

user19405892

@Semiclassical I don't understand

Semiclassical

hmm. well, the way i was doing things was to start with $f(x)^5+f(x)+x=0$, differentiate both sides wrt $x$, and solve for $f'(x)$

well, if $f(x+1)=f(x)/2$, then it'll get closer and closer to zero as $x\to\infty$ while never vanishing

so if you can think of a function that does that, you've got an example

user19405892

@Semiclassical That makes sense now

Ted Shifrin

03:42

That's a Spivak question. He expects an applucation of the 1-D inverse function theorem to justify implicit differentiation.

Semiclassical

hi @ted

Ted Shifrin

Hi @Semiclassic

Semiclassical

got a moment to help me understand something? i feel like it should be straightforward but ehhh

not really following it

Ted Shifrin

You can try ...

Semiclassical

mmkay

basically, i'm trying to understand quadratic differentials

and the notion of foliations that shows up in there

Ted Shifrin

03:45

OK, like on a Riemann surface?

Semiclassical

exactly

Ted Shifrin

So the point is that when you (holomorphically) change coordinates you're looking at functions that transform by $(dz/dw)^2$ instead of by $dz/dw$. But I guess you knew that.

Semiclassical

right.

Ted Shifrin

So ask a question?

Semiclassical

mostly i'm trying to understand how one gets the horizontal and vertical trajectories of a given quadratic differential

Ted Shifrin

03:48

I don't know what you mean by that, but it's not much different from any other sort of differential equation, except it's nonlinear.

Mike Miller

hi

Ted Shifrin

hi @MikeM

Semiclassical

well, let me ask about the most trivial case namely just $(dz)^2$

because if i don't understand that, i probably don't get the rest

Ted Shifrin

First of all, are we on $\Bbb C$ or a torus or what?

Semiclassical

let's just do $\mathbb{C}$

i might want to plug you on the torus case eventually, but not right now

Ted Shifrin

03:50

My point was that we have to decide what a local coordinate $z$ means. On $\Bbb C$ or a torus it's globally defined.

OK, so what am I doing with this quadratic differential?

Semiclassical

somehow, i want to use $(dz)^2$ to define at each point a vector field whose flow lines are horizontal lines

ugh, such a sloppy statement

let me say that better

Ted Shifrin

Well, on $\Bbb C$, there's not going to be any difference between $(dz)^2$ and $dz$, since we have a global chart.

But, yes, tell me how we're defining a vector field.

Semiclassical

that's where i get lost, i think, though reading a different set of notes is making things a bit more obvious

here's a stab at it

Ted Shifrin

Ordinarily, these things show up in terms of giving infinitesimal changes of complex structure on your Riemann surface. How are they arising in your set-up?

Semiclassical

well, here's the story as i'd understand it

suppose my QD is $f(z)(dz)^2$. then i can convert this to $(dw)^2$ by choosing a coordinate $w(z)=\int_{z_0}^z \sqrt{f(z')}\,dz'$ for some basepoint $z_0$

Ted Shifrin

03:58

Assuming you're on a simply connected set on which $f\ne 0$, sure.

Semiclassical

yeah, definitely

and the reason i care about that is that, in classical mechanics, one talks about the classical action $\int_{x_0}^x p(x')\,dx'$ where the momentum $p$ is determined from $p^2+V(x)=E$

Ted Shifrin

Not that I see how you go from phase space to a Riemann surface ...

Semiclassical

well, suppose $V(x)$ is a quadratic polynomial

Ted Shifrin

But where does $x$ live?

Semiclassical

physically, the real line. but i'm deliberately complexifying it here.

for reasons shrug

Ted Shifrin

04:03

OK, and we had discussed earlier that momentum should be a $1$-form (section of the cotangent bundle), so here you have $p^2$, hence a quadratic differential form. Got it.

So you're talking about a vector field on the cotangent bundle with some property?

Semiclassical

right. the definition i've got works like this

if i denote the above QD as $\phi$, then a tangent vector $v$ is 'horizontal' to $\phi$ if $\phi(v)>0$

Ted Shifrin

That doesn't even make sense.

Semiclassical

let me find the definition.

it's from these notes: jkahn.gc.cuny.edu/Lecture%204.pdf

Ted Shifrin

First of all, I assume you mean a holomorphic tangent vector. Second of all, by $\phi(v)$ you mean feed it $v$ twice?

Semiclassical

i guess. i was confused by that.

that's the notation they used.

though it's consistent for them, given how they define it in the first sentence or so

Ted Shifrin

04:10

So they're using that to mean something like $\big(\sqrt{\phi}(v)\big)^2$.

Semiclassical

for them, $\phi:TS\to \mathbb{C}$ where $S$ is the Riemann surface

Ted Shifrin

They're talking about a holomorphic vector field $g(z)\dfrac{\partial}{\partial z}$ and basically taking $f(z)g(z)^2$.

Semiclassical

that's what $\phi(v)$ computes to?

Ted Shifrin

No, $\phi$ is definitely not a well-defined map on $TS$. A $1$-form would be, but not a quadratic differential.

The transformation rule says this is wrong.

Oh, I see what they said.

Semiclassical

"Recall that a quadratic differential on a Riemann surface S is a map ϕ : T S → C satisfying ϕ(λv) = λ^2 ϕ(v) for all v ∈ T S and all λ ∈ C."

Ted Shifrin

04:12

Yeah, I just read that.

Semiclassical

kind've surprised cut-and-paste preserves (almost) all of that :/

Ted Shifrin

This works only because we're working with a $1$-dimensional complex manifold.

Semiclassical

that, i buy. but i'm also entirely happy with that setting.

Ted Shifrin

So, yeah, anyhow, if $\phi = f(z)(dz)^2$ and $v=g(z)\partial/\partial z$, then $\phi(v) = f(z)(g(z))^2$.

Semiclassical

the convenience of only worrying about 1D motion, heh

Ted Shifrin

04:15

So on this coordinate chart, you want to know for what $g(z)$ we have that positive.

Semiclassical

right. just to check---am i correct in reconstructing that $\partial_z = \frac{1}{2}(\partial_x-i\partial_y)$?

wait. i can do that easily by hand. no point in me asking.

Ted Shifrin

Yes, that looks right. The easiest way to remember this is $z=x+iy \implies dz = dx+i\,dy$ and $d\bar z = dx - i\,dy$, and $\partial/\partial z, \partial/\partial\bar z$ is the dual basis.

Semiclassical

right. plus $\partial_z z=1$ and $\partial_{z}\overline{z}=0$

(same thing of course)

Ted Shifrin

Dual basis :)

Semiclassical

yes, yes, yes

ok. let me try to confirm offline that i understand the trivial $f(z)=1$ case.

i'll probably use $\partial$ and $\overline{\partial}$ for convenience. subscripts get tedious otherwise.

Mike Miller

04:21

How are things @Ted?

Ted Shifrin

Well, to me $df = \partial f + \bar\partial f$, so these are $1$-forms, but do what you like :)

My back/hip have been killing me for a day and a half, @MikeM, but otherwise OK.

Semiclassical

oh, good point. i remembered that there was a usage like that but you're right

Ted Shifrin

@Semiclassic: Other people use the notation you want to. Just not geometers :)

Semiclassical

hah

well, obviously $(dz)^2(\partial_z)=1$ in their conventions

Mike Miller

Sorry to hear that @Ted...

Semiclassical

04:25

which means that $\partial_z=\frac{1}{2}\partial_x-\frac{1}{2} i\partial_y$ would be horizontal

what i don't get is what that actually represents, what with the complex coefficients

Ted Shifrin

But, uncomplexifying, that just moves you along the $x$-axis, @Semiclassic ...

Semiclassical

aka i don't actually know enough about TS

hmmm

Mike Miller

I remember being VERY confused about someone's manipulations with the Laplacian once... Til I realized their Laplacian is negative mine

Ted Shifrin

We're talking about $\Bbb C\times\Bbb C$ here.

Semiclassical

okay, that makes more sense

Ted Shifrin

04:26

Yes, the sign on the Laplacian is always a confusion. Geometers always make it negative, and analysts always make it positive. :)

Semiclassical

and then $(dz)^2(i\partial_z)=-1$ means that $i\partial_z$ would be a vertical trajectory

Mike Miller

Why wouldn't the Laplacian be a positive operstor??

Ted Shifrin

There are funny signs with the Hodge star, as I recall, @MikeM.

Semiclassical

that almost sounds like the choice of signature in Minkowski space :)

+++- v. ---+

Ted Shifrin

Well, don't make elliptic into hyperbolic while you're at it, @Semiclassic.

Semiclassical

04:28

heh

okay, going back

Mike Miller

Yes, that's danferous - change all or none

Semiclassical

in what sense should i understand $i\partial_z$ as being vertical? i can kind've see it: If $\partial_z$ points in the real direction at all points, then $i$ rotates that vector field locally by 90 degrees

is that basically all in this context? (i.e. $S=\mathbb{C}$)

Ted Shifrin

They're using $\partial/\partial z$ to trivialize the bundle and just thinking about the coefficients. So you're looking at $1$ versus $i$ in $\Bbb C$, I guess. Horizontal <--> real, and vertical <--> imaginary. ?

Semiclassical

sounds right

Ted Shifrin

I have never heard this sort of stuff before, so I dunno :P

Semiclassical

04:32

heh

now, what i should further be able to reproduce is the case of $z(dz)^2$ as the QD

main thing i want to figure out is the simplest way to get the flow lines in mathematica

Ted Shifrin

That one bothers me because of the zero at the origin. And if we pull out the origin, we're not simply connected. I.e., you do not have a global branch of $\sqrt z$ on $\Bbb C$.

Semiclassical

yeah :/

considering $f(z)$ with zeroes is the really important business

Ted Shifrin

Remember what I said before. You need a well-defined branch of $\sqrt{f(z)}$.

Well, I guess you only need that if you're going to change coordinates.

I guess you just do the same thing. You want $z(g(z))^2 > 0$. Good luck with that at $z=0$, though.

Semiclassical

yeah. and all i need is the coordinate to make sense locally, not globally

Ted Shifrin

So do we allow vector fields with poles?

Semiclassical

04:37

i think you can. more generally, you can allow $S$ to be a Riemann surface with punctures

Ted Shifrin

Yeah, but I'm worried about ill-definedness when we have non-simple connectivity and square roots. Basically, you want $g(z) = 1/\sqrt z$, and that's a no-good.

Semiclassical

in the case of $z(dz)^2$, at $z=0$ one should have three lines converging symmetrically

Mike Miller

@TedShifrin: BTW, great talk today by Jonathan Wahl. Seems like your kind of guy.

Ted Shifrin

From UNC, @MikeM? Much more algebraic than I ever was, but singularity theory, yeah.

I'm amazed he hasn't retired.

How does a cube root come in, @Semiclassic?

Semiclassical

i suspect that the 'right' way to do the case of $z(dz)^2$ is to move from $\mathbb{C}$ to the Riemann sphere defined by $y^2=z$

Mike Miller

04:39

Really, really enjoyed that talk. Talked to him after and learned he's written papers in my area.

Semiclassical

gooood question

Mike Miller

(Fillings of 3-manifolds, usually resolutions of surface singularities, so not surprising.)

Ted Shifrin

Hmm, same guy? UNC, about 70+ now?

Semiclassical

i think it's because $\sqrt{z}dz=\frac{2}{3}d(z^{3/2})$

Mike Miller

Yup.

Ted Shifrin

04:40

I haven't seen him in decades, but, great!

Aha @Semiclassic. Proceed.

Semiclassical

will do. but need to do something quick---back briefly

Ted Shifrin

Well, I'm outta here for now. Report tomorrow :)

Semiclassical

aye aye

back

crocket

05:02

0

Q: Did I deriva a natural deduction proof for $\{(\phi\vee\psi),\neg\phi\}\vdash\psi$ properly?

In the picture below, I derived $\psi$ by RAA(Reductio Ad Absurdum) without $\neg\psi$ above it. Did I use RAA correctly in this natural deduction proof?

logic proof-verification natural-deduction sequent-calculus

bwDraco

$R_{n} = h_{n}! \cdot f(n(a + \theta))$

Does this have any particular meaning?

(The Number Devil, page 107.)

Never mind, it's probably just random gibberish.

skill patrol

05:25

6 hours ago, by skill patrol

Don't open it and let the "Devil" out!

:-)

bwDraco

05:44

Math truly is beautiful. Chapter 7 discusses what is known as Pascal's triangle.

skill patrol

and difficult

bwDraco

This number theory is really starting to confuse me.

Can someone explain the relationship between Pascal's triangle and Fibonacci numbers?

I'm actually asking for real this time.

I hope I don't have to ask on the main site...

(The Number Devil pages 138-140.)

bwDraco

06:04

...what? You can derive the Sierpinski triangle from Pascal's triangle! (Pages 140-141.)

Julian Rachman

0

Q: Representable Functors and Upper Sets (Final Segments)

I was looking around Wikipedia and came across this for representable functors: From another point of view, representable functors for a category $\mathcal{C}$ are the functors given with \mathcal{C}. Their theory is a vast generalisation of upper sets in posets. I don't get the generalization. ...

category-theory generalization

anyone? ^

bwDraco

06:51

Wow. By chapter 9, they're talking about convergent and divergent series.

I can't believe this book covers such advanced material.

$\sum_{n=1}^{\infty}\frac{1}{2^n}=1$

$\sum_{n=1}^{\infty}\frac{1}{n+1}=\infty$

(The Number Devil pages 182-187)

Abhishek Bhatia

08:48

I have line given by $\phi$ (angle) and a point represented by a vector $r_i$. How can I find the perpendicular distance of it from a another point $r_j$.

In this article, in vector formulation section what does $n$ represent?

It mentions n is a 1×2 unit vector in the direction of the line

arctic tern

well, it looks like you already know what n represents

should we assume R_i is on said line?

Abhishek Bhatia

yes

The line represented by r_i and /phi .

arctic tern

phi seems unnecessary though since r_i is given no?

Abhishek Bhatia

Why r_i is the point and /phi is the angle the line makes with x-axis?

arctic tern

what?

you said there were two points, so r_i can't be "the" point, and phi is the angle the line makes with the x-axis because you said so no?

Abhishek Bhatia

08:53

The line is represented by an angle(/phi) and point(r_i) through which it passes.

Now, I have compute the perpendicular distance of the aforementioned line from another point r_j

arctic tern

if you have he point r_i on the line, then there's no need for phi

when you googled "perpendicular distance" and went to the wikipedia page, which part did you have trouble understanding?

see "vector formulation"

Abhishek Bhatia

I am trying to use that one only. But I didn't understand what you are suggesting.

You can't represent a line by a single point right?

arctic tern

ah, your line doesn't go through the origin then?

Abhishek Bhatia

oh no.

Sry, I should have mentioned that before.

arctic tern

the wikipedia formula still works

your n=(cos phi,sin phi) of course

Abhishek Bhatia

08:58

||(r_i-r_j)-(r_i-r_j).n).n||, where n=[cos phi,sin phi]?

Is there any simplier way of representation without introducing n?

arctic tern

a way to do it without knowing what direction the line is going? what do you think?

Abhishek Bhatia

no, I want to use \phi instead of n in the formula to avoid addition of another symbol.

Are there any other better notations?

arctic tern

n=(cos phi, sin phi), obviously the latter side of the equation has no n in it

I mean, that's what n is

iwriteonbananas

10:06

Hey @DanielFischer. Could you help shed some light on the answer here: math.stackexchange.com/questions/51788/… The spectrum of a self-adjoint compact operator consists of a sequence $(\lambda_n)\in c_0$, right? What do we know about the spectrum of multiplication operators on $L^2$?

Balarka Sen

Hello @iwriteonbananas.

Daniel Fischer

10:40

@iwriteonbananas Well, the spectrum of a compact operator may be finite. But it's countable, and has no accumulation point except possibly $0$. So it's the underlying set of a sequence in $c_0$. Note however that it's quite misleading to say the spectrum is a sequence in $c_0$, since a sequence has more structure (in particular an ordering induced by the indices). If the compact operator lives on a Hilbert space and is self-adjoint, we have the further information that the spectrum is real.

But self-adjointness is really really unimportant there. Now the question of the spectrum of a multiplication operator $M \colon f \mapsto m\cdot f$. It's probably easier to look at the regular values of $M$. So, for what $\lambda$ does $t \mapsto \frac{1}{\lambda - m(t)}\cdot f(t)$ belong to $L^2$ for all $f\in L^2$?

Huy

11:05

hey @BalarkaSen, any idea what an essential arc might be?

Balarka Sen

Nope, @Huy.

Huy

what about a proper arc? :P

Balarka Sen

Not by that name. Context?

Huy

Any two essential simple proper arcs in the thrice-punctured $S^2$ with the same endpoints are isotopic.

Balarka Sen

Weird. No idea.

Suppose you look at just 2 punctures, $p$ and $p'$. Join these by two arcs $\gamma$ and $\gamma'$ in $S^2$. Take arcs with same endpt transverse to these respectively. These cannot be isotoped, since they cannot even be homotoped.

So those essentiallity and properness conditions might be strong conditions.

Huy

11:13

ah

I found it I think

An arc is called essential if it is neither homotopic into a boundary component of the surface nor a marked point

Balarka Sen

I don't know what "homotopic into a boundary component" means, but glad you found it :)

Huy

and a proper arc is a map $\alpha: [0,1] \to S$ such that $\alpha^{-1}(P \cup \partial S) = \{0,1\}$, where $P$ is the set of punctures

Balarka Sen

aha.

@Huy So, you haven't been around for a while. What are you studying?

Huy

gonna work through some chapters of Farb & Margalit

primer on mapping class groups

Balarka Sen

I see.

Huy

11:17

right now I want to understand the Alexander method

Balarka Sen

which I have no idea what it is :) but I wish you luck.

Huy

just a fancy theorem that helps compute mapping class groups

what are you working on? still doing analysis?

Balarka Sen

i see. i don't know what the deal with mapping class groups is though - i just know the definition.

@Huy i just started studying integrals & forms. i am also studying algebraic geometry.

Huy

I don't really know the deal with them either, yet. :P

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

11:32

Hi @huy

Mats Granvik

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Did you like my question?

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

+1 :-)

Shahab

12:29

is there any online resource where I can find examples of inclusion and exclusion to bounding some sets / functions?

GPhys

I recommend math stackexchange

Balarka Sen

12:45

Does Bezout hold in higher dimensions? Namely, if $X$ and $Y$ are degree $p$ and $q$ varieties of dual dimensions in $\Bbb P^n$, need it be true that number of intersections of $X$ and $Y$ upto multiplicity is $= pq$?

This is true for smooth varieties over $\Bbb C$ due to cohomological reasons of course.

Federico

hi everyone. I ask in chat as I do not think it would be a proper question for the site.

I am writing a paper (not for a mathematical journal) and while I am quite familiar with parametric curves and splines, my supervisor is not.

This means that he is not familiar with the various theorems, in particular he wants me to cite the original work about arc-length parametrization (where is defined what arc-length prametrization is) and the one that says that such parametrization is not possible with rational functions of the arc length unless the curve is a straight line.

Mike Miller

13:00

@BalarkaSen If it's true over $\Bbb C$, it's true over any field of characteristic zero. More or less.

Balarka Sen

@Federico arclength parametrization is one of fundamental notions of differential geometry of curves, so I guess if you want to find the original source it'd trace back to Gauss or maybe even before him. so I don't think that's a reasonable endeavor. But I am sure someone more knowledgeable on (the history of? :P) differential geometry can comment on this.

@MikeMiller Um, I know there's a thing called Lefschetz principle, but I never found a satisfactorily rigorous statement of it. But I can believe you. Thanks.

Federico

@BalarkaSen yeah, google scholar seems to suggest Newton. I am aware of the fact that it might not be reasonable, but I have to prove it to someone that has never heard of differential geometry (and assumes that everyone else in our field does not know as well)

Mike Miller

@BalarkaSen: Someone on MO said you should take it as a philosophy, not a theorem, that you can prove things in these other settings too.

Balarka Sen

Why don't you refer to some textbook in differential geometry instead?

Mike Miller

It's obviously not actually always true.

Balarka Sen

13:04

@MikeMiller Ah, alright.

Mike Miller

Let me find the link.

Federico

@BalarkaSen I will try to propose it again, last time the answer was "you should cite the original, not a derivative work"

Mike Miller

25

Q: What does the Lefschetz principle (in algebraic geometry) mean exactly?

This principle claims that every true statement about a variety over the complex number field $\mathbb{C}$ is true for a variety over any algebraic closed field of characteristic 0. But what is it mean? Is there some "statement" not allowed in this principle? Is there an analog in char p>0? Is...

ag.algebraic-geometry

Balarka Sen

Thanks.

fred

Hi. Is PDE both singular and plural? Like deer and fish? Does anyone have any idea why?

Balarka Sen

13:14

@MikeMiller Apparently there is a concrete theorem quoted in Martin Brandenburg's answer, but nobody seems to care much other than taking it as a cue that the analogous statement is true for fields other than $\Bbb C$ and thus worth trying to prove it - or that's the impression I get. Is this like a Godel or Freyd kind of thing, then?

Mike Miller

I think it seems rather difficult to apply Martin's statement.

Balarka Sen

I don't understand it to be honest, not sure if I should try to.

Mike Miller

Yes, part of why I wouldn't be able to apply it is because I don't understand it.

ramsay

14:07

at how many points $f(x)=\frac{4-x^2}{4x-x^3}$ is discontious?
My question didn't end.....

Semiclassical

morning

ramsay

sorry for spelling mistake!

Balarka Sen

Morning @Semiclassical. Hi @JulianRachman.

DeNiSkA

@ramsay hmm, it is discontinuous at 3 points {$0,-2,2$}

ramsay

14:23

@DeNiSkA your answer matches my book answer, but why can't we write the function as $\frac{4-x^2}{4x-x^3}=\frac{1}{x}$ because $4-x^2$ cancels out?

Julian Rachman

Morning @BalarkaSen

Balarka Sen

@ramsay $(4 - x^2)/(4x - x^3) = 1/x$ only when both of the expressions are defined. You can't cancel when $x = 2$, e.g, because the left hand is $0/0$, not well-defined. You can't cancel when $x = 0$ similarly. $f(x) = (4 - x^2)/(4x - x^3)$ and $f(x) = 1/x$ are two different functions.

ramsay

@BalarkaSen but they seem to be same functions, right?

DeNiSkA

@ramsay you can't modify the given function when you are finding continuity, i mean $\frac{4-x^2}{4x-x^3}$ is equal to $\frac{1}{x}$ only when there domains are equal! and domains of these two functions are different!

Balarka Sen

They are the same functions only when $x$ is not $0, 2$ or $-2$.

But, $1/x$ is defined at $2$ whereas $(4 - x^2)/(4x - x^3)$ is not. So you can see why they are different.

r9m

14:28

@DanielFischer did you get my ping yesterday? :)

Balarka Sen

@JulianRachman How's it going?

ramsay

thank you @BalarkaSen and @DeNiSkA

Daniel Fischer

@r9m Yes. But you had already left. I haven't thought about how much you exactly need to require, but at least for all analytic functions decaying sufficiently fast, it holds. Let me think for a moment whether $\lim_{\operatorname{Re} z \to +\infty} f(z) = 0$ already suffices.

robjohn

14:45

@r9m pong ;-)

Martin Sleziak

BTW there was a suggestion on meta to create some *rules for chatrooms generally'. (There is a post on meta with rules for this chat room.)

But considering that the rules for this chatroom are ignored by most user, I am not sure whether working on general rules would be a useful way to spend time.

BTW isn't there some network-wide guidance how to use chat? And if it is, does it mention the problem which sparked that meta post?

r9m

@DanielFischer kay .. :) thanks for looking into it!

@robjohn :D LOL .. (pong as in unpleasant smell? :P)

Daniel Fischer

15:02

@r9m $\lim_{\operatorname{Re} z \to +\infty} f(z) = 0$ suffices. But $\lim_{x\to +\infty} f(x+iy) = 0$ for every fixed $y$ probably doesn't.

Ping-pong (disambiguation)

Ping-pong is a trademarked name for the game of table tennis. Ping-pong, Ping pong, or Pingpong may also refer to: In film: Ping Pong (1987 film), a British film Ping Pong (2002 film), a 2002 Japanese manga-adaptation film Ping Pong Playa, a 2007 film directed by Jessica Yu Ping Pong (2012 film), a British documentary, directed by Hugh Hartford In music: Ping Pong (EP), a 1994 EP by the group Stereolab "Ping-Pong", a song by X-Wife from Side Effects PingPong (band) Israeli band who participated in Eurovision 2000 Ping Pong over the Abyss, a 1982 album by The 77s "Do You Know? (The Ping Pong Song...

r9m

@DanielFischer I see .. interesting! :-) gotta scratch my head over that .. thanks!

@DanielFischer I got that :P ping-pong ... I was trying a make an unpleasant smelling joke :P

Daniel Fischer

@r9m Yeah, I was just impressed by the disambiguation page. And disturbed by the last entry there.

r9m

@DanielFischer :P lol .. Mother of God! .. I just noticed that ..

Daniel Fischer

@r9m People are really weird, aren't they? Who would even think of something like that?

robjohn

@r9m No, I was intending ping-pong as Daniel Fischer notes.

@r9m as far as the question you asked yesterday, as long as $\mathrm{Re}(\lambda)\gt0$, the answer should be yes.

@r9m but I see that that was one of the constraints.

r9m

15:18

@DanielFischer I read a S&M manga (like a comic story) once .. and I came to the conclusion .. people can get really creative with those sort of things :P

@robjohn yes .. I got that too (considered a contour with vertices $r$, $r(1+\frac{\Im(\lambda)}{\Re(\lambda)})$, $R$ and $R(1+\frac{\Im(\lambda)}{\Re(\lambda)})$) and let $r \to 0^{+}$ and $R \to \infty +$ ..

split the integral as $\displaystyle \int_0^{\infty} \frac{f(x) - f(\lambda x)}{x}\,dx = \int_0^{\infty} \frac{f(x) - f(\Re(\lambda) x)}{x}\,dx + \int_0^{\infty} \frac{f(\Re(\lambda)x) - f(\lambda x)}{x}\,dx$ :)

the contour's for the second integral in the right ..

robjohn

15:33

@r9m I would close the contour and use Cauchy's Theorem. I think that works.

Akiva Weinberger

Someone help with this question?

Semiclassical

@AkivaWeinberger not cool

Mary Star

Hello @robjohn !!
Could you take a look at the edit part of my question: http://math.stackexchange.com/questions/1723438/differential-equation-of-higher-order ?

Balarka Sen

@AkivaWeinberger Fair warning: I'm going to April Fool you today.

Semiclassical

ugh, april fools day

Akiva Weinberger

15:39

@BalarkaSen I have been warned

Balarka Sen

You better be! 'Cause I have something very bad for you coming up.

r9m

@robjohn :-) yay! .. thanks!

robjohn

@MaryStar Did you really mean "$k^4-2k^2+k=0$"?

Mary Star

Sorry, it is a typo... It should be $k^3-2k^2+k=0$. @robjohn

r9m

@robjohn We could use Mellin inversion too right? (via Ramanujan's Master theorem?)

robjohn

15:54

@r9m I have never really worked with Mellin inversion. I have usually found other approaches where some people have used Mellin.

Mary Star

Is the partial solution at the edit part correct? @robjohn

r9m

@robjohn 'kay! .. (I was just trying out ways ... )

robjohn

@MaryStar 'swhat I thought.

Mary Star

So can we have only a value for $A$ ? Do $B$ and $C$ stay constants ? Or do we set them for example equal to $0$ ? @robjohn

vzn

16:13

in theory salon, 3 mins ago, by vzn

number theory news/ highlights: prime bias discovery, Diffie-Hellman Turing prize, Zhang-Tao twin primes, Wiles Abel prize, Riemann / Tmachine blog

robjohn

@MaryStar did you see my answer?

Mary Star

Yes, I am thinking about it!! Thanks!! :-) @robjohn

Do you maybe have also an idea for my other question: math.stackexchange.com/questions/1723503/… ? @robjohn

robjohn

@MaryStar I just noticed that the constant of integration $D$ was the same symbol as I was using for $D=\frac{\mathrm{d}}{\mathrm{d}x}$, so I changed it to $F$.

Evinda

16:30

Hey @Semiclassical

@Semiclassical Could you take a look at my question:

0

Q: Find domain of functions

I want to find for which intervals $(c,d)$ there are functions $y:(c,d) \to \mathbb{R}$ that are differentiable on $(c,d)$ and are solutions of the differential equation $y'(t)=1+(2 y(t) +t+3)^2$. Could you give me a hint how we could find the desired intervals? Is there a theorem that we could...

differential-equations

Random Variable

@robjohn I came across the integral $$\int_{0}^{\infty} \frac{\sinh(px)}{x}\left(\frac{1}{\cosh(px)+\cos(ax)}-\frac{1}{\cosh(px)+\cos(bx‌)}\right) \, dx=\frac{1}{2} \log\left(\frac{p^{2}+a^{2}}{p^{2}+b^{2}}\right) \, $$ where $p >0$. The integrand is the real part of $$\frac{\tanh\left(\frac{x}{2} (p+ia)\right)- \tanh\left(\frac{x}{2} (p+ib)\right)}{x}. $$

Since $\tanh(z)$ is analytic in the right half-plane, and $\tanh(z) \to 1$ as $\text{Re}(z) \to +\infty$, I concluded (with hesitation) that Frullani's formula would hold here.

robjohn

@RandomVariable This looks okay, but you will need to justify the extension of Frullani that we discussed above.

Random Variable

@robjohn I asked r9m about the extension, and then he asked about it here.

robjohn

@RandomVariable Okay. Do you see how to justify it?

Mary Star

16:46

Ah ok... Is it maybe as follows (at the second question)?

$y'=F(y,x) \\ \Rightarrow (x-xy(x))+(y(x)+x^2)F(y,x)=0 \\ \Rightarrow y(x)(1-x)+x^2+x+x^2F(y,x)=0 \\ \Rightarrow y(x)=-\frac{x^2F(y,x)-x-x^2}{1-x}$

@robjohn

Random Variable

@robjohn Yes, but is an analytic continuation argument possible?

robjohn

@RandomVariable how would you do an analytic continuation argument?

Random Variable

@robjohn We know that the formula holds for $\lambda >0$, and the principal branch of the logarithm is analytic on the half-plane $\text{Re}(\lambda) >0$. So I guess we would need to show that the integral is analytic on the same half-plane.

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