Mathematics

12:05 AM

@TedShifrin On a degenerate torus where the innermost longitude is collapsed to a point, what's the (improper) integral of the Gaussian curvature everywhere away from that point?

Ordinarily you'd expect 0 from Gauss-Bonnet but the curvature at the bad point blows up so I think that throws things off

Besides, it's also basically a sphere where you've pushed the north and south poles together

@BalarkaSen ^

Ted Shifrin

12:33 AM

DogAteMy: What a fascinating question. I think that by symmetry the curvature integral is still $0$. (Yes, $K$ blows up as you approach the point, but there's a compensating zero in $dA$ at that point.) Alternatively, the image of the Gauss map is probably still net $0$. But you're certainly right that $\chi = 1$, but we don't expect Gauss-Bonnet to work. Presumably we can cut out a little disk around the point and apply the boundary version and you get geodesic curvature there.

Akiva Weinberger

1:10 AM

Hm you're probably right

Kinda disappointing

What that means, though, is that if you slice it in half (perpendicular to the axis of rotation) you get a topological disc, whose boundary is geodesic, such that the integral of the curvature is 0

(Compare that to a hemisphere, which is also a topological disc with geodesic boundary, but has $\int KdA=2\pi$)

What that means is that, to make this topological disc satisfy Gauss–Bonnet, we need to give the bad point a curvature of $2\pi\delta(0)$

(where that's the Dirac delta function)

Ultradark

1:47 AM

@AkivaWeinberger I can't remember what balarka and you said several weeks ago about that space I posed which was either isometric to minkowski plane or isometric

I can't remember whether it was isomorphic or isometric

If distances are shrunken it can't be isometric, correct?

So then it was probably isomorphic

Akiva Weinberger

1:59 AM

@TedShifrin In other words, as a metric on the torus, the metric blows up on a ring which contributes $0$ to the integral of the curvature. But as a metric on the sphere, the metric blows up at two points, which would have to contribute $4\pi$ to the integral of the curvature (i.e. each point should have curvature $2\pi\cdot\infty$, in the sense of the Dirac delta) if we want to preserve Gauss–Bonnet.

Ultradark

oh actually I think it's isometric, if, the metric is preserved

and balarka said it was I believe. time to try to prove that

Akiva Weinberger

(That would be made rigorous by viewing it as the limit of metrics on the sphere.)

yh05

I need help with the following proof. Why is the underlined statement true?

Akiva Weinberger

@yh05 On the left, we have some multiple of $g(x)$, so it has degree at least $\deg(g)=m$ (or it's the zero polynomial)

On the right, we have the difference of two polynomials of degree less than $m$, so the difference must also be less than $m$

The only way for these to be equal is if they're zero

yh05

Why?

Akiva Weinberger

2:09 AM

If they're not zero, we have "degree at least m" = "degree less than m"

which is impossible

yh05

Why?

Akiva Weinberger

Which part @yh05

ÍgjøgnumMeg

I miss 24hr libraries

studying at night time at home is 90% procrastination

anakhro

2:25 AM

@AkivaWeinberger you can do better. :)

Akiva Weinberger

Dammit!

@TedShifrin ^

anakhro

The very cool thing is that the 3D analogue to the problem---they have absolutely no idea.

Akiva Weinberger

I have a new idea

Inscribe the circle in a square

Head towards one corner ($\sqrt2$)

Head tangentially towards the circle ($1$)

Half-circle ($\pi$)

Tangent out one ($1$)

anakhro

An improvement.

Akiva Weinberger

This gives 6.556

What did I have before

I had 6.712

anakhro

2:29 AM

You said 6.7 or something

Yeah

(hint: you can do better)

Akiva Weinberger

What if I inscribe the circle in a triangle and do the same thing

Hm… intuition says no

anakhro

@AkivaWeinberger are you at all a fan of L^p spaces?

Akiva Weinberger

This isn't related to the problem, is it?

anakhro

No. I am afraid not. :P

Akiva Weinberger

@anakhro A little

anakhro

2:32 AM

I am just stuck on seeing how the two different versions of Young's inequality are equivalent: en.wikipedia.org/wiki/Young's_convolution_inequality

From "equivalently, if p,q,r>=1 and ..."

Akiva Weinberger

Young's convolution inequality

In mathematics, Young's convolution inequality is a mathematical inequality about the convolution of two functions, named after William Henry Young. == Statement == === Euclidean Space === In real analysis, the following result is called Young's convolution inequality:Suppose f is in Lp(Rd) and g is in Lq(Rd) and 1 p + 1 q = 1 r + 1 {\...

anakhro

The convolution one, and the one with the 3 p,q,r norms on the right and the left is the big double integral

Just in R^n, not the generalization

In the integrand I see the convolution of g and h

And I feel like Holder's inequality might come in handy

But I can't figure out a way to massage out an equivalence.

And despite it being a common equivalence, no textbook I have seen explains it. >: (

Akiva Weinberger

What happens if we set $h=1$ in the second one

anakhro

Maybe I am silly or tired, but wouldn't that mean the right hand side blows up, since 1 is not L^r?

Akiva Weinberger

Oh

Whoops yeah

anakhro

2:41 AM

Glad it wasn't that easy.

wipes sweat off brow

Akiva Weinberger

I'm still a beginner here

What's the identity for convolution?

$\delta$?

yh05

@AkivaWeinberger I haven't proven that if polynomial p=q then deg of p = deg of q. So I can't use this to obtain a contradiction. To prove this, don't I need to first prove that p=q iff the coefficients p_n = q_n for every n?

Akiva Weinberger

Oh hold on

I didn't even realize that $p,q,r$ are different in each one @anakhro

anakhro

Perhaps you have the right idea and you can cook up an h.

Akiva Weinberger

@yh05 That's the definition of polynomial equivalence

for arbitrary fields, anyway

If we're working in the field of real numbers, then polynomials are equal iff they're equal as functions

Lemma: A real polynomial is zero for all values of x iff it's the zero polynomial

yh05

2:46 AM

Instead of taking it as a definition, I want to prove it.

https://math.stackexchange.com/questions/586611/proof-two-polynomials-px-and-qx-attain-same-value-for-every-x-in-mat

But the non-analytic proof relies on collaries of theorem E.1. So the whole argument becomes circular.

Akiva Weinberger

Proof: Suppose otherwise. Let its degree be m. Then divide by x^m. You end up with a+b/x+c/x^2+d/x^3+… = 0 for all x

Now take the limit as x->infty

You get a=0

Contradiction

This is basically the second answer in your link

except I divided by x^degree first

@yh05 I should point out that this is false for finite fields

Do you know modular arithmetic?

Consider x^p+x in the field Z/pZ

For example x^3+x in the integers mod 3

If you plug in 0,1,2 you can check that x^3+x is 0 for each of them

So, when F=Z/3Z, x^3+x is equal to 0 as a function, even though (by definition) it's not equal to 0 as a polynomial

Sorry: I meant x^3-x

Not x^3+x

So 0^3-0=0, 1^3-1=0, and 2^3-2=6=0 in Z/3Z

yh05

3:18 AM

@AkivaWeinberger I see. I'm trying to prove for polynomial over $\mathbb{R}$.

anakhro

4:09 AM

@yh05 what are you trying to prove exactly?

Akiva Weinberger

4:23 AM

@anakhro A polynomial is zero everywhere iff its coefficients are all zero

Equivalently, two polynomials are equal everywhere iff their coefficients are equal to each other

@yh05 Then what's wrong with the analytic proofs?

Stupid Questions Inc

@TedShifrin Apparently one can show that $\sum_{n=1}^m c_n=\sum_{n=1}^m a_n\sum_{n=1}^m b_n$ and from there the result follows immediately so for $c_n=\sum_{k=1}^n a_kb_n+\sum_{k=1}^{n-1} a_nb_k$, $\sum_{n=1}^\infty c_n$ is convergent

I don't know how one was supposed to eagle eye that $\sum_{n=1}^m c_n=\sum_{n=1}^m a_n\sum_{n=1}^m b_n$

A Perspicaciously Curious Mind

8:31 AM

Don’t just read it; fight it! Ask your own questions, look for your own examples, discover your own proofs. Is the hypothesis necessary? Is the converse true? What happens in the classical special case? What about the degenerate cases? Where does the proof use the hypothesis? (Paul Halmos)

6

Nick

11:47 AM

How to calculate the angle for the zigzag

►

A Perspicaciously Curious Mind

Perturbative

1:22 PM

Hey @AlessandroCodenotti

Alessandro Codenotti

Hi

Perturbative

Quick question about Bonn, are you currently staying in accommodation provided by Bonn or on your own?

Alessandro Codenotti

I'm on my own in a flatshare with other students. The university dorm was already full

Perturbative

Ah, so you're paying the expenses for the flatshare on your own?

And was it a struggle to get accommodation?

Alessandro Codenotti

@Perturbative yes

@Perturbative It wasn't too bad, but I got lucky. The city centre is very expensive so I'm living in a village nearby

And it's very hard to find a flatshare if you're not physically in Bonn since you're supposed to visit them first

user

1:48 PM

Do any of you know resources for exercises on Picard's theorem differential equations, autonomous systems of differential equations?

Lucas Henrique

Hey chat

Thorgott

Hey

A Perspicaciously Curious Mind

Hi

Silent

2:28 PM

Let $f:\Bbb R^2\to\Bbb R$. Let $(a,b)\in\Bbb R^2$. Suppose that $\lim\limits_{(x,y)\to(a,b)}f(x,y)$ exists and that $\lim\limits_{x\to a}f(x,y)$ and $\lim\limits_{y\to b}f(x,y)$ exist in a neighborhood of $(a,b)$ for each $y$ and each $x$ respectively. Then Apostol's exercise says $\lim\limits_{(x,y)\to(a,b)}f(x,y)=\lim\limits_{y\to b}\left(\lim\limits_{x\to a}f(x,y)\right)=\lim\limits_{x\to a}\left(\lim\limits_{y\to b}f(x,y)\right)$.

Do above conditions imply that $f$ has to be continuous in a deleted neighborhood of $(a,b)$?

Thorgott

2:42 PM

Is this saying that $\lim_{x\rightarrow a}f(x,y)$ exists for all $y$ (in the neighborhood)?

Silent

yes

Perturbative

@AlessandroCodenotti Okay thanks for the info I'll keep that in mind!

Akiva Weinberger

2:57 PM

@Silent Let $f(x,y)$ be $1$ iff $x=y$ and $0$ otherwise

and look at $(a,b)=(0,0)$

MyWrathAcademia

Is there a rule to find the divisor of a cubic polynomial? Wikipedia says that for a cubic polynomial ax^3 + bx^2 + cx + d that one of the products of factorization is a one degree polynomial of the form qx - p and that q must be a divisor of a and p must be a divisor of d. Does this mean that a must always be exactly divisible by q and d must always be exactly divisible by p in order for a divisor to divide a polynomial into a quotient and remainder?

Thorgott

The limit doesn't exist then, maybe you want $f(x,y)=x$ if $x=y$?

Akiva Weinberger

Oh I misread

Didn't see the bit saying the limit exists

@Thorgott That should work actually yeah

$f(x,y)=\begin{cases}x\ (=y),&x=y\\0,&x\ne y\end{cases}$ @Silent

Thorgott

Yeah, that's nice

Akiva Weinberger

@MyWrathAcademia Yes (if all the coefficients involved are whole numbers)

Thorgott

3:05 PM

As far as a rule to find the divisor is concerned, the Cardano formula gives you all roots, i.e. a factorization into irreducibles (over C).

Akiva Weinberger

There's a neat theorem that says, a polynomial is reducible over the integers iff it's reducible over the rationals

In other words, if you have a polynomial with integer coefficients, if you can factor it into smaller polynomials with rational coefficients, you can factor it into smaller polynomials with integer coefficients

This is called Gauss's lemma

Thorgott

It's a very useful result

MyWrathAcademia

@AkivaWeinberger thanks for confirming but why does the divisor x - 3 divide the cubic polynomial x^3 - 2x^2 + 0x - 4 into the quotient x^2 + x + 3 and remainder -13 when d in this cubic polynomial is 4 which is not divisible by the divisor's second term p which is 3?

Akiva Weinberger

Because it has remainder

so it's not a factor

If qx-p is a factor of ax^3+bx^2+cx+d (i.e. it divides it without remainder) then q's a divisor of a and p's a divisor of d

MyWrathAcademia

@AkivaWeinberger thanks that clears it up, that's the piece of information I was missing from the Wikipedia article.

I was trying to find p and q for any cubic polynomial based on thinking that for any cubic polynomial ax^3+bx^2+cx+d a must always be exactly divisible by q and d must always be exactly divisible by p. Now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder, how can I find p and q of a divisor (i.e. finding qx - p) when given only the dividend (i.e. cubic polynomial)?

Akiva Weinberger

3:21 PM

This is equivalent to finding a root p/q of the polynomial

You can do trial and error, but note that there's not always a solution (the polynomial might not have a rational root)

There's also an exact formula for the root, Cardano's formula (the cubic equivalent of the quadratic formula), and numerical methods like Newton's method

What Wikipedia page were you looking at? This one? https://en.wikipedia.org/wiki/Rational_root_theorem

MyWrathAcademia

This one: en.wikipedia.org/wiki/Cubic_equation#In_mathematics

Ultradark

Does the Minkowski metric $ds^2=c^2dt^2-dx^2$ remain invariant upon a change of scale?

let's say you plot the speed of light $c^2$ on a different scale than a linear one. It shouldn't change the metric right?

in principle it seems correct that the metrics should remain equivalent, but I am not quite sure how to prove this

user193319

3:39 PM

Let $K$ be a convex set in a vector space such that $0$ is an internal point of $K$, let $p_K$ be the associated Minkowski functional, and let $z \notin K$. Why must $p_K(z) \ge 1$ hold?

MyWrathAcademia

@AkivaWeinberger I planned to do trial and error if p and q satisfied some criteria such as always being a factor of a and d respectively but now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder. Do you know of any condition/criteria for p and q when qx -p divides ax^3+bx^2+cx+d with a remainder, or in other words do you know of any condition/criteria for p and q regardless of whether qx -p divides ax^3+bx^2+cx+d with or without a remainder?

user193319

If it were true that $z \in p_K(z) \cdot K$, then if $p_K(z) < 1$ were the case, I could write $z = p_K(z) \cdot \frac{z}{p_K(z)} + (1 - p_K(z)) \cdot 0 \in K$, which would be a contradiction...

However, I don't know if it's true that $z \in p_K(z) \cdot K$.

MyWrathAcademia

I am also not completely familiar with the terminology here so please clarify that when you say "there's not always a solution" you mean that there's not always a root when you divide a polynomial by its divisor and that this happens when the polynomial might not have a root that is an integer (i.e. what you mean by a rational root)?

Silent

@AkivaWeinberger Wow! thanks a lot!

Thorgott

Any polynomial divides any other polynomial with a remainder, though "dividing with a remainder" is a somewhat oxymoronic phrasing.

Silent

3:50 PM

@Thorgott Thank you :)

Lucas Henrique

@Semi, look at what I've found:

4

A: Prove that $x^3 + y^3 = 9$ has infinitely many rational solutions

Here is the 'tangent-chord' construction of infinitely many rational points from a single such point. Suppose we have a point $p_0=(x_0,y_0)$ on the cubic curve $f(x,y)=x^3+y^3=9$. The tangent vector at this point is given by $$\left(\frac{\partial f}{\partial y},-\frac{\partial f}{\partial x}\r...

Thorgott

np

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$Mathematics$ Mathematics