Build a mosque? No! Build a mall? Yeah, sure.

"as usual" is an exaggeration, but I'm glad I remember something from last year's classification of quadratic curves

Still better than a wall

@Hiro To sum all of this up, we're trying to make the problem as simple as possible without making it impossible to answer the bit about the ratio.

What does the question mean when it says "interior of segment AB" ?

Points between A and B on that segment.

That are neither A nor B

ah ok

@MikeMiller Gotta have a happy ending

Have you drawn a picture of this?

not yet @Semiclassical

I am drawing one though

Good.

It looks like it'll work regardless of whether or not A,B are on the same branch of H

(playing around in Geogebra)

Could anyone in here please help me with integrating factors? I'm trying to turn a 2nd order ODE into Sturm-Liouville form and not having any luck.

@Ted i have a question about one of your exercises

Probably should say which ODE...

Lol @MikeMiller @TedShifrin---the sign is discarded later on anyways -____-

imgur.com/a/ERRsV

I just picked random points btw

@Semiclassical started out with $-Y^{\prime\prime}(y) - \mu Y^{\prime}(y) = 0$. Somebody suggested the integrating factor $Y(y) = e^{- \mu y} Z(y)$, and I just tried it and I wasn't able to get it in the correct form.

but if follows the same condition or scenario

great!

Yeah, that looks right @Hiro

Although...I think I might have made a mistake. I'm going to try again.

You could also have A, B on the same side

Yeah, but how can I do any proof... like I literally have no idea...

Well, this is an analytic geometry course?

I am studying math alone, not in a course

Ah.

Well, the grindy algebra way is this

You start with a generic hyperbola, pick two points $(x_1,y_1),(x_2,y_2)$, and compute the equations for the tangent lines

Can do that with calculus.

Then take these two equations, and solve for their intersection.

one sec

Then write down the equation of a line passing through that intersection and the origin.

Then write down the equation of a line passing through the two points, and find the intersection of this with the other line.

That works, but it's a lot of grindy algebra...

that sounds physically painful

It's a grind.

@Semiclassical is there any less awful methods?

more rigorous?

Yeah. Show that the problem remains unchanged under affine transformations, and argue from that that one doesn't need to consider two arbitrary points and a generic hyperbola

But rather can just do a specific simple case where the calculation is easy

@Semiclassical the case where the hyperbola is $x^2 - y^2 = 1$ would be the easiest

Possibly.

I don't actually know.

And you can get there with affine transformations so it might be worth an attempt

If you can pick your points as $(-x,y), (x,y)$ then you can write down some of those lines immediately

are all hyperbolas affine transformations of the general hyperbola?

There should presumably be some more geometric way to do this, though

Hence I'm not totally a fan of this approach.

The calculus approach makes more sense to me for some reason

@TedShifrin To be specific, they're used to define topological invariants of certain condensed matter systems (topological superconductors)

Yeah, before I was getting a sign wrong, so I tried the substitution $Y(y) = e^{-\mu y} Z(y)$ again using the correct signs this time. The $Z(y)$ term introduced by the substitution canceled out and now my $r(x)$ function is no longer strictly positive, it's ero.

I don't quite get how showing that the hyperbola remains unchanged proves that line ZC passes through X :p @Semiclassical

They also show up in the fractional quantum Hall effect, somehow. (the so-called Moore-Reed Pfaffian state). don't ask me what that means, though.

@meow there also those of the form $x^2-y^2=0$ which I would call hyperbola "degenere" in Italian, but I can't seem to find the right terminology in English

degenerate?

@Alessandro A degenerate hyperbola

:P

but if you're looking at nondegenerate hyperbolas they're all affinely equivalent to $x^2-y^2-1=0$

@Hiro Let me put it like this. Suppose I make a drawing like the one you did, and I find that it indeed intersects the segment AB

Now suppose I was to take that entire page and stretch it like it was rubber

All the lengths and such would change, but the line ZC would still pass through the segment AB.

And then of course, there's the problem of how I've alienated pretty much everybody on MSE who could answer any of my questions to the point where nobody even reads mine anymore :(

Evidently that kind of transformation doesn't affect that property

The argument is that there's a whole bunch of transformations like that

That means that, if it holds in the initial picture, it'll also hold in any appropriately transformed picture.

Well, in any case, if anybody changes their mind about me, here's a link to the question I'm referring to: math.stackexchange.com/questions/2033256/…

The idea is to pick as simple a picture to start with as possible, and then figure out all the ways to change the picture without nixxing that property.

In that vein, the hyperbola $y^2-x^2=1$ is really simple.

So the idea is to prove it for that case, and then argue that any other case is just an appropriate 'stretch' of that one.

Possibly, but I know I don't have an easy personality.

@JasperLoy I love pretty much anything he's ever been in. Pulp Fiction's my personal fave.

Not really.

My calculus of variations knowledge is pretty shallow, all things considered.

With PDEs I get to a point and then I get stuck.

@JasperLoy don't know if I have a favorite. We're using DuChateau and Zachmann in my class.

@JasperLoy it's a Dover book, so it was only like $16.

I have Evans on my shelf and a couple of others, but they're more advanced than what we're doing right now.

I'm sure somebody on MSE somewhere knows whether or not $-Y^{\prime\prime} - \mu Y = 0$ can be written in Sturm-Liouville form.

It's not even a PDEs thing, it's an ODEs thing. What integrating factor do I use? It's been 16 years since I've taken an ODEs course.

@robjohn can you take a look at this: math.stackexchange.com/questions/2033256/…

I think you might be my only hope ;_(

@JasperLoy yes. Everybody thinks I'm a kid. Or a guy. And I'm neither.

Is $\mu$ a constant?

yes, @Semiclassical

Then that's already in Sturm-Liouville form, pretty much.

$Y''=-\mu Y$.

@Semiclassical that's because I made a mistake.

Hold on

It's $-Y^{\prime\prime} - \mu Y^{\prime} = 0$ @Semiclassical

ah.

What'd you get when you did the integrating factor that was suggested?

@JasperLoy my professor is kind of a flake, never answers email, and is never in his office. And there are only 4 people in my class.

Not a fake, a flake! As in he's flaky.

@Semiclassical another way to look at this is to note that $a_n=2^n\left(x^{2^{-n}}-1\right)$

@robjohn Ah, nice.

And it's also kind of late notice.

@JessyCat I think you said earlier that someone had suggested an integrating factor. What did you get from that?

@Semiclassical right back where I started, except now I at least have the coefficient of the $Z^{\prime\prime}$ term being the derivative of the coefficient of the $Z^{\prime}$ term. But, the $Z(y)$ term I introduced got killed.

If you click on the link I gave above, you'd be able to see exactly what happened.

What ODE in Z specifically?

Yeah, I see it @JessyCat

Not sure I believe it, but I'd need to do the algebra myself

@JasperLoy Eh. Still pretty stuck.

@Semiclassical the original ODE is $-Y^{\prime\prime}(y) - \mu Y^{\prime}(y) = 0$, the integrating factor is $Y(y) = e^{-\mu y} Z(y)$.

For simplicity, I'll check with Mathematica.

Yeah, Mathematica bears you out. That's not a helpful integrating factor.

but $e^{-\mu y/2}$ seems to do the trick

(I had mathematica do it with $e^{ay}$ instead and found that the derivative term goes away if a=-\mu/2)

$e^{-\mu y/2}$, huh?

Evidently.

I'm gonna go give that a try.

There's probably a smart way to derive that, rather than just picking a general guess in as I did.

@Semiclassical possibly. It might come up though naturally just from looking at the algebra I've already done for when it was $e^{-\mu y}$

Yeah.

$e^{-\mu y /2} Z(y)$, right?

Right.

Slingblade voice All right, then.

Also a great movie @JasperLoy

@JasperLoy he does kill a couple of people with a lawnmower blade? And then sits down and eats some fried chicken. (Actually, just the biscuits...with mustard).

@Jasper have you ever been interested in korean cinema by any chance?

Okay @MikeMiller this probably interests you more: If I take a closed oriented manifold with oriented tangent bundle and $e(M)\neq 0$. Then I want to prove it does not admit odd-dimensional subbundles at all. The oriented case follows immediately by earlier discussion. For the non-orientable case, MS tells me to use the orientation cover (or rather, a 2-fold cover but I hoped it'd be the orientation cover)... Any hints?

Korean food is frigging delicious.

Hm, that's a pity, I really like Korean cinema because the photography is usually fantastic, but a lot of people are taken aback by the abundance of violence

I like pretty much anything with lots of vegetables and rice (preferably spicy) with little bits of meat cut up into it.

I definitely dislike Psy, though. How could you not?

@JasperLoy it's my favorite. That and Indian food.

hm. $\frac{d}{dy}e^{\mu y}Y'(y)=e^{\mu y}(Y''(y)+\mu Y'(y))$. @JessyCat

So if you start with $-Y''(y)-Y'(y)=0$ and multiply by $-e^{\mu y}$ you have $\frac{d}{dy}e^{\mu y}Y'(y)+0\cdot Y(y)=0\cdot Y(y)$ which is in SL-form.

Hi @Ted

Hi @Balarka, did you see my ping earlier?

Hi @Balarka @Jasper

Don't see how that fits with what I just said about $e^{-\mu y/2}$, though

@Alessandro I didn't get a ping. Can you please repost?

sure, let me find it

@Jasper Not the ideal way to greet someone. :P

here Gerry Myerson provides a simple construction of a function with a dense graph that requires no form of choice @Balarka

Actually, maybe it does. I can't remember if Sturm-Liouville form is unique.

@meow: You had a question before I Disappear?

Everyone is usurping my smack powers ...

I've got a question about polynomials myself, in relation to the stuff I've been doing

@Alessandro Huh

Suppose I've got $F(x,y,z)$ a homogenous cubic polynomial. What's the smallest degree monomial that isn't in the ideal generated by partial derivatives of $F(x,y,z)$?

@Semiclassical it worked! It killed the first derivative terms, and now I can divide out the remaining $e^{-\mu y /2}$s and it looks like a completely ordinary Sturm-Liouville problem! :)

Yay.

Happiest I've been all day.

Now I've gotta go put those clothes in the dryer and keep working. Thanks again, though :) :) :)

NP

$1$, Semiclassic?

That doesn't seem right.

Try $x^3$?

That's what it seems to be, yes.

But I don't know why.

Because the Jacobian ideal is generated by quadratics.

Yeah.

Is that enough to ensure that all quartics are in the ideal?

I doubt it.

Hm.

It seems to be true, for the case I'm considering ($F=(x^3+y^3+z^3)t-3xyz$ with parameter $t$)

i.e. all the quartic monomials are in the ideal but not all the cubics

About the only thing I can think of is that one also has Euler's homogenous function theorem, so $xF_x+yF_y+zF_z=3F$

So $F$ itself is in the ideal

@TedShifrin I am proving Hahn banach theorem for linear spaces. I just want to know if certain step is correct. Suppose that $p : X \rightarrow \mathbb{R}$ is a positively sublinear functional is it true that $-p(y) -p(-y) \geq 0$ ?

Karim: I have no idea.

@Semiclassic: What if $f=x^3+y^3+z^3$ and you take $xy^3$? Is that in the ideal?

$p : X \rightarrow \mathbb{R}$ is a positively sublinear functional if it satisfy triangle inequality and $p(\alpha x) = \alpha p(x)$ whenever $\alpha \geq 0$.

lemme check with Macaulay2 (that's what I'm using)

OK ... leaving in a few.

kk

Hello chat

M2 says it is.

Hi @Lozansky

hello @Lozansky

It feels my main role on this chat is to greet people

:D

That's not a bad job at all

yeah, $xy^3= (\frac13 xy)F_y=(\frac13 xy)(3y^2)$ @TedShifrin

Oh, duh.

@TedShifrin I am sorry; x^3 is a multiple of x^2, hence is a multiple of 3x^2?

For that specific case I think it's almost obvious

since there the quadratics are just x^2, y^2, z^2

Huh @Balarka?

And the first thing you definitely can't generate with that is xyz.

nvm it is true @TedShifrin I just did small error in my calculation so I have $p(y) + p(-y) \geq p(y + -y) = p(\vec{0}) = p(0 * \vec{0}) = 0 p(\vec{0}) = 0$.

@Danu Recall that $e(M) = \chi(M)$ and when $\chi$ is multiplicative under covers.

Sure, sure, I was being a dope.

@TedShifrin I mean to say, x^3 is contained in (x^2) in the ring k[x, y, z].

oke cool

Anyone wanna take a look at this: $$ \lim_{n \to \infty} \frac{n^3}{\sqrt{2 \pi}} \int_{\mathbb{R}} se^{-n^2 s^2/2} f(s) ds = f'(0)$$ for $f$ bounded and $f'$ bounded and continuous at origin

Now I wonder about xyz in my original ring.

it is kinda cool proof of hahn banach theorem it uses zorn's lemma @TedShifrin

I was saying $x$ was not, Balarka.

I'm thinking integration by parts

Ah, I see. Yeah, that I agree with.

@Lozansky Could also do the substitution $u=n^2 s^2/2$.

Not sure that's a good idea, though, on further thought

Yeah, stick with IBP

@Semiclassical I did that to see that the part without $f(s)$ is $0$

That helps.

An approximate identity will perhaps pop up somewhere after you IBP again.

Well kinda

To restate the question in more generality: With $F(x,y,z)$ a degree-d nonsingular homogenous polynomial, what's the smallest degree of monomial that doesn't lie in the Jacobian ideal of $F$?

(I should probably allow more than 3 variables, but ugh)

Then you end up with $$\frac{n}{\sqrt{2 \pi}} \int e^{-n^2 s^2/2} f'(s) ds$$

Try the substitution $u=ns$ at that point.

Huh?

Not sure what you're huh'ing at. Do that u-sub, and consider what happens as n\to \infty

Gaussian I guesS?

No need to. It seems like a fine approximate identity.

$ne^{-n^2s^2/2}/\sqrt{2\pi}$, I mean.

Hm, true.

Approximate identity?

Summation kernel.

Oh right

Abusing notation a bit, $\frac{ne^{-n^2 s^2/2}}{\sqrt{2\pi}}\to \delta(s)$ as $n\to\infty$.

Ya.

Is that Dirac-delta?

Yeah.

Where I was going is that, upon doing u=ns, the integral becomes $\frac{1}{\sqrt{2\pi}}\int e^{-u^2/s}f'(u/n)\,du$

Yeah sorry I don't understand how to solve that

What I want to say next is probably not rigorous, though.

Hello how can i compute 0.98^1.01 with precision to the second digit without using the general binomial theorem?

namely, that as $n\to \infty$, $f'(u/n)\to f'(0)$ for all $u$.

and so therefore you just get $f'(0)\cdot \frac{1}{\sqrt{2\pi}}\int e^{-u^2/2}\,du=f'(0)$

Sure

@JasperLoy I stay inside on Black Friday.

Not convinced that's rigorous, though, unless you invoke some facts about the decay of $f$ at infinity.

how to find the equation of the line that passes through an intersection and the origin? xD

@user379685 Use logarithms?

It's not rigorous. But it's good intuition why approximate identities work in general.

@Hiro What's a line through the origin that also passes through $(1,2)$?

how?

It's a line, so it should be of the form $y=mx+b$. But $(0,0)$ lies on it...

oh

just find the slope

using the two points

Right.

@Semiclassical So Gaussian integral at the end?

and then calculate b :P

Right. @Lozansky

@user379685 $\log\left(0.98^{1.01}\right)=1.01\log(0.98)$

Well, $b$ is pretty simple if it goes through the origin.

@TedShifrin my question was about exercise 4

$b$ is the y-intercept, after all!

@Semiclassical the rigorous proof isn't much more work anyway. you just bound away the integral at infinity.

Right.

I'm still too lazy to do it, though.

yup

instead we should write

The easy way is to show that $$\lim_{n \to \infty} \frac{n}{\sqrt{2 \pi}} \int_{\mathbb{R}} e^{-n^2s^2/2} ds $$ is a positive summation kernel I guess

@robjohn and how would that help me?

If I remembered how one does that, probably :) @Lozansky

You do, presumably, so do that

i need 0.98^1.01 not it's log

@Semiclassical Only three simple steps :P

logs are easy to approximate.

plus, even if you don't end up using the logs explicitly, they help for organizing the calculation.

@Semiclassical yeah b is just 0 lol

@Hiro Yup.

More generally, the line through the origin and $(x_1,y_1)$ is just $x/x_1=y/y_1$.

@user379685 it lets you compute $0.98^{1.01}$

@robjohn 1.01*log(1-0.02) and now i can use the taylor series for log(1+x) but how do i get 0.98^1.01 from this?

Now exponentiate and use the Taylor series for exp.

(not sure this isn't equivalent to the binomial theorem, though)

@user379685 The formula I cited gives the log of the quantity you want. Exponentiate to get the quantity.

@user379685 If this is not what you're looking for, you need to be explicit about what you can and cannot use.

@robjohn to get two decimal points right i need to calculate the remainder of the taylor series to be less than 1/100 but i don't know to not lose the precision while exponetiating

Here's the computation as I see it

@MikeMiller when? It doesn't work for all closed smooth manifolds?

when $M$ is oriented

$x=0.98^{1.01}\implies \ln x = 1.01\ln(1-0.02)\approx (1.01)(-0.02)\implies x\approx e^{-0.02(1.01)}\approx 1-(0.02)(1.01)$

@MikeMiller i don't know how to prove that actually.

sux

$\log(0.98)=\log(1-0.02)=-0.02-(0.02)^2/2=-0.0202$ to two places...

But that's also equivalent to just saying $(1-0.02)^{1.01}\approx 1-1.01(0.02)$ from the binomial theorem

depending on your definition of e(M), that's Poincare-Hopf.

True. @robjohn

@BalarkaSen you want to go through triangulation or what?

$(-0.0202)1.01=-0.0204$ to two places

@Danu I am not sure what you mean. My definition of e(M) is that it's the self-intersection number of the zero section of TM. Do you know that yet?

e(E/M) is in general is dual of the zero set of a generic section of E/M

@Semiclassical how do you get from e^(-0.02)(1.01) = 1-(0.02)(1.01)

Taylor series of exp starts with $e^x=1+x+\cdots$

$e^{-0.0204}=1-0.0204+(-0.0204)^2/2=0.9798$

For reference, 0.98^1.01 = 0.979802... = 1-0.020198...

@BalarkaSen I'm okay with that definition.

How do i know if i am still right within two digits when i switch to exp

@BalarkaSen what is E/M?

Ok, great then. E/M is a vector bundle E over M.

The version robjohn just cited would have 1-0.0206081

Total space E base M? Funny notation

what's the usual choice? @Danu

For me? $\pi:E\to M$ omit the $\pi$ if context is obvious?

@user379685 Look at the terms in the power series for exp...

right

@SemiC $(E, B, \pi, k)$ probably :P

@user379685 The next term is $(-0.0204)^3/6$ which is smaller than the error you need.

ok thanks

Since I am trying to get the bounty room going, perhaps a reasonable thing could be to advertise it here. So here is link to the relevant meta post and here is the room.

Another way to go at this is to write $0.98^{1.01}=(1-2x)^{1+x}$ with $x=0.01$.

And then do the Taylor series of that.

The advantage is that powers of x are powers of 10. @user379685

So how do I now get multiplicativity, Balarka?

@Danu What do you mean by multiplicativity?

of the euler characteristic...?

sure, use triangulations, whatever

yeah, you use triangulations.

I haven't read the original question, I was just answering to why e(M) = chi(M)

Crappy phone doesn't allow me to see what I'm typing

@BalarkaSen h lol

Haha

sorry for misunderstanding

Can I think geometrically about why the zero section of the tangent bundle of the covering has higher self intersection (in a multiplicative way)?

hi @jasper

Well, I suppose if you go upstairs to the n-fold cover you "wrap around n times". That introduces n zeroes above for each zero below.

Zero of a generic section of the tangent bundle, I mean. Same thing as self-intersection number of the zero section.

Yeah. I'm just trying to think about a picture. Can I see it for some simple example?

I'd like to take a sphere but it's simply connected... And a torus has zero Euler characteristic. And picturing coverings of higher genus surfaces... Idk really. What do those look like?

you can all just call me zach

What are the first five non zero coefficients from the taylor formula for the function y=y(t) which statisfies (2+t^2)y'-y+2t=0 and y(0)=1.

Is there a simpler example?

Mayne a wedge of circles

Derp

That's not a manifold though.

Exactly

:(