Mathematics - 2019-01-24 (page 1 of 2)

Ted Shifrin

00:00

You SURE this is the correct integral, @Nobody? It still comes out a bit yucky.

\ne @Jake

oh, I dunno if \not < works

$\not <$

Yeah, that works.

Nobody recognizeable

@TedShifrin ok now i am confirming .

Leaky Nun

@ÉricoMeloSilva r/murderedbywords

Jake Rose

@TedShifrin okay so it’s not a sin to put a slash through an inequality?

Ted Shifrin

You sure the region is really a square with that?

Jake Rose

just checking

Ted Shifrin

00:01

I would write $\ge$ instead, @Jake.

Karl Kronenfeld

Assuming a total order.

Ted Shifrin

He's a physicist.

Nobody recognizeable

@TedShifrin yeah confirmed there is a on top. Though that doesn't make a difference.

Jake Rose

I don’t know if that was an insult but I’d like to say I’m a mathematical physicist

I’m a halfbreed

Ted Shifrin

You are only insulted if you are insulted :P

If the integral were over the inside of a circle, it would make a big difference, @Nobody.

Where is this problem coming from?

Jake Rose

00:04

I don’t understand a lot of the hate between physicists and mathmeticians

Ted Shifrin

I'm not hating, I swear.

Jake Rose

we may use the same tools but we tend to do different things

@TedShifrin I know dw

Ted Shifrin

Physicists tend not to care about whether things really make sense mathematically, sometimes. They have their own sort of intuition.

Jake Rose

@TedShifrin I agree with that

i think it comes from a sense of we don’t really care unless it’s ‘real’

Leaky Nun

or imaginary

Jake Rose

00:06

@TedShifrin I have a good Feynman video you might enjoy

Feynman: Mathematicians versus Physicists

►

The n dimension bit always gets me

Ted Shifrin

No time now, though.

Jake Rose

@TedShifrin fair

Nobody recognizeable

@TedShifrin well its in the courtesy of our professor. The problem was "a point charge is located at the corner of cube of side a find the flux through surfaces which do not meet at q$ could be done by gauss law but professor is bit more interested in the integration that he gave it homework.

famesyasd

if I take a subset X of the real line and it is uncountable, is it necessarily dense in some ball in R?

Ted Shifrin

Oh, @Nobody, the integral does work out pretty nicely. Do it in polar coordinates over half the square. You should end up with something like $\int_0^{\pi/4}(1+\sec^2\theta)^{-1/2}d\theta$, which looks horrendous, but is doable if you use an obvious trig identity after writing in terms of $\cos$.

Mike Miller

00:09

@LeakyNun Did he say it's a great project for you?

Ted Shifrin

Yes, @famesyasd. If you have an uncountable subset of $[0,1]$, can you prove it must be dense in some interval?

Mike Miller

@TedShifrin and language...

Leaky Nun

@MikeMiller he said it's the right level for me

Ted Shifrin

Yeah, that too.

Mike Miller

Ok, that's better advice than any of us can give then.

Ted Shifrin

00:09

^^^^

Or you could be like Zee and ask for advice just to do the opposite ...

rolls 7 + $\pi/6$ eyes

Nobody recognizeable

@TedShifrin will try that. Thanks for the help and your time.

Ted Shifrin

You're welcome.

Leaky Nun

@TedShifrin lmfao

Mike Miller

3/8 done copyediting this monstrosity.

Leaky Nun

oh man

anyway, off I go

Karl Kronenfeld

00:24

@TedShifrin No

Perturbative

I'm confused by this proof, if the set of $m \times n$ matrices of full rank is an open subset of $M(m \times n, \mathbb{R})$, how does continuity allow us to conclude the Jacobian of $F$ has full rank in some nbhd of $p$?

Karl Kronenfeld

By definition of continuity, in fact.

famesyasd

@TedShifrin but the Cantor's set on [0,1] is nowhere dense.

Karl Kronenfeld

(and uncountable)

famesyasd

so cardinality has nothing to do with density?

Karl Kronenfeld

00:34

For $\mathbb R$. Indeed, the Cantor set has cardinality equal to the continuum, so we do not have to assume CH or anything of that sort.

In the other direction, finite sets are never dense ;)

Perturbative

@KarlKronenfeld But $dF_p$ is a map from $T_pM$ to $T_{F(p)}N$ and we can view the Jacobian matrix of $F$ as a map from $T_{\hat{p}}\mathbb{R}^k$ to $T_{\hat{F(p)}}\mathbb{R}^l$, the codomain of either of these maps isn't $M(m\times n, \mathbb{R})$ so when the author says continuity, which map is the author referring to?

(also $k$ is the dimension of $M$ and $l$ the dim of $N$ in the above)

Karl Kronenfeld

Use the map sending $q$ to the Jacobian at $q$.

Perturbative

Okay but then it doesn't seem obvious to me that such a map would be continuous

Can I view the map as a composition of continuous maps?

Karl Kronenfeld

Go back to definition of smooth map.

Mike Miller

00:52

Or C^1 map

mick

0

Q: Is $\mathbb{Z} ((-2)^{\frac{1}{n}})$ ever a UFD?

Let $n>2$ be an integer. Consider the integral domains $\mathbb{Z} ((-2)^{\frac{1}{n}})$. For what integer values of $n$ is $\mathbb{Z} ((-2)^{\frac{1}{n}})$ a UFD ? Are there infinitely many solutions ? Same question for $\mathbb{Z} (2^{\frac{1}{n}})$.

Karl Kronenfeld

Yes, more precisely :)

Perturbative

01:17

Well since $F$ is smooth all the partial derivatives of all orders exist and are continuous, but what I'm saying is that while the map $\xi : M \to M(m \times n, \mathbb{R})$ defined by $\xi(p) = \text{Jacobian matrix of $F$ at $p$}$ is a map taking $p$ to a matrix containing entries that vary continuously, I don't understand how that implies that $\xi$ is continuous. Maybe I am forgetting some basic linear algebra fact

Karl Kronenfeld

You give the space of matrices the product topology so that it is homeomorphic to $\mathbb R^{mn}$.

Perturbative

Okay but then the only result I have at hand to show that $\xi$ is continuous is the following:
Let $(X, d)$ be a metric space and let $f_1, ..., f_k$ be real valued functions on $X$ (that is $f_i : X \to \mathbb{R}$ for $i \in \{1, \dots k\}$ ) and let $\mathbf{f} : X \to \mathbb{R}^k$ be defined by $$\mathbf{f}(x) = \left( f_1(x), \dots, f_k(x)\right)$$ then $\mathbf{f}$ is continuous if and only if each of the functions $f_1, \dots, f_k$ is continuous.

So do I need to view $M$ as a metric space to show that $\xi$ is continuous

Because then since the space of matrices is homeomorphic to $\mathbb{R}^{mn}$ I can view $\xi$ as a function from $M$ to $\mathbb{R}^{mn}$ and the above result would apply

Douglas Sirk

@MikeMiller how is the orbit in Lorenz homeomorphic to $\mathbb R$? doesnt it limit onto itself?

Karl Kronenfeld

@Perturbative You're working in a coordinate chart, so you have a metric if you really need one. But, one definition of/result about continuity is that the preimage of any open set via a continuous map is open.

Perturbative

@KarlKronenfeld Okay so if the preimage of $\xi$ is open in $M$ then why can we conclude that the Jacobian of $F$ has full rank in this open set?

Douglas Sirk

01:34

/././.././.

jajajajajajajaja

Karl Kronenfeld

Because, you're picking the right open set to take the preimage of using $\xi$ (the author gave this set to you). Then, consider what happens when you apply $\xi$ to anything in this preimage.

Mike Miller

@DouglasSirk Given an ODE on $\Bbb R^k$, the flowline through a point $p$ is a map $\gamma: I \to \Bbb R^k$, where $I$ is an interval, $\gamma(0) = p$, and $\gamma$ satisfies the ODE. Further we demand that $I$ is the largest possible domain of $\gamma$: there is no solution defined on a larger interval. Note that by the existence and uniqueness theorem for solutions to ODEs, in particular, the flowline is unique if eg the ODE is C^1.

For the Lorenz attractor, the largest interval $I$ is $\Bbb R$ itself.

But a flowline, by definition, is a map from $\Bbb R$, or possibly some subset (or if you like, the image of this map.)

The Lorenz continuum you are asking about is the closure of a certain flowline.

That is, as far as I know, its definition.

Perturbative

@KarlKronenfeld Okay I get it now, applying $\xi$ gives a matrix of full rank, so the Jacobian of $F$ in coordinates has full rank, thus $dF_p$ has full rank in this nbhd

Thanks for your help! @KarlKronenfeld

Karl Kronenfeld

@Perturbative Yep

no problem

mick

02:04

0

Q: Is $\mathbb{Z} [(-2)^{\frac{1}{n}}]$ ever a UFD?

Let $n>2$ be an integer. Consider the integral domains $\mathbb{Z} [(-2)^{\frac{1}{n}}]$. For what integer values of $n$ is $\mathbb{Z} [(-2)^{\frac{1}{n}}]$ a UFD ? Are there infinitely many solutions ? Same question for $\mathbb{Z} [2^{\frac{1}{n}}]$.

abstract-algebra ring-theory factoring unique-factorization-domains

Silent