Mathematics - 2020-03-24 (page 3 of 3)

6:00 PM

Let's do precise estimation. How do I take square root of $t^2 - K/3 \cdot t^4 + o(t^4)$

OK, $\|X(t)\| = t - K/6 \cdot t^3 + o(t^3)$

Question: Do mathematicians actually care about finding a purely algebraic proof of the FTA? Would there be any benefit to that besides the name no longer being a "misnomer" of sorts? I don't know if one has been found, this is something I just saw a couple of days ago.

Tobias Kildetoft

@CaptainAmerica16 Not sure "care about" is the right term. There are certainly people who find it mildly annoying that there does not seem to be any purely algebraic argument for it

On the other hand, in some sense the reason behind it really is geometric anyway, so it is not that surprising

CaptainAmerica16

@TobiasKildetoft Ah, okay. I looked up some stuff on it and it really seemed like a "non-issue" lol. Was looking for an opinion that wasn't in the recesses of the math subreddit

skullpatrol

Short answer: no.

CaptainAmerica16

lol

Thorgott

6:15 PM

it seems almost weird to desire a purely algebraic proof of a statement involving analytic objects

Tobias Kildetoft

@Thorgott There are no analytic objects involved in the statement

Little Rookie

Hello all, I have a question.

Definition. A sequence $\{p_n\}$ in a metric space $X$ is said to be a Cauchy if for every $\epsilon > 0$ there is an integer $N$ such that $d(p_n, p_m) < \epsilon$ if $n \ge N$ and $m \ge N$.

Definition. Let $E$ be a nonempty subset of $X$, and let $S$ be the set of all real numbers of the form $d(p, q)$, with $p,q\in E$. The sup of $S$ is called the diameter of $E$.

Theorem. Let $\{p_n\}$ be a sequence and $E_N$ consisting of the points $p_N, p_{N+1}, p_{N+2}, \dots$, then $\{p_n\}$ is a Cauchy sequence $\iff$ $\lim\limits_{N\to \infty} \mbox{diam}\quad E_…

CaptainAmerica16

@Thorgott Yeah, I've read that you can't even define some of the objects in the proof without analysis.

Tobias Kildetoft

@CaptainAmerica16 The statement itself is purely algebraic. The analysis needed for (one of the many) proofs is the fact that a polynomial of odd degree has a root in the reals.

CaptainAmerica16

Hm, ok

Semiclassical

6:23 PM

Are there weakenings of the FTA which can be proved with algebra alone?

Balarka Sen

Finding roots of a function is not usually an algebraic question. It's most naturally dealt with analytic techniques.

Tobias Kildetoft

@Semiclassical Well, any poly of degree at most two has a root :)

Semiclassical

lol

one big exception to that is finding roots by radicals. that to my brain is very much just "algebra"

but the point, i suppose, is that "finding roots by radicals" is vastly more restricted than "find roots no matter what"

Balarka Sen

The whole point is there are LOTS of algebraic numbers which are not expressible in terms of radicals, yeah

Tobias Kildetoft

@BalarkaSen And yet, those are not really the ones that are important for FTA

Balarka Sen

6:27 PM

True.

Little Rookie

Anyone can help me with my question?

Balarka Sen

@Tobias Can you approximate polynomials with rational coefficients by polynomials with rational coefficients with unsolvable Galois group? That is to say, look at $\Bbb Q^\infty$, consisting of infinite sequence of rational numbers which are eventually zero. What can you say about the subset of tuples such that the corresponding polynomial has unsolvable Galois group?

I suspect it's Zariski dense hence dense

(Zariski dense on each finite level $\Bbb Q^n$ I mean)

Tobias Kildetoft

@BalarkaSen Probably

Mike Miller

@BalarkaSen math.stackexchange.com/questions/1017447/…

Balarka Sen

Cool

Semiclassical

6:41 PM

oh hey

bleh

waiting to see if any students are going to show up to my virtual office hour is

tedious

Ted Shifrin

Less tedious virtually than in reality, @Semiclassic.

Hi @Tobias, a @balarka, @MikeM.

Balarka Sen

Hi @Ted!

Ted Shifrin

What was the mention of moving frames? :)

Semiclassical

@TedShifrin it would be, if people actually showed up

Balarka Sen

Nothing, I was vaguely interested in learning a proof of Chern-Gauss-Bonnet

was revisiting some basics

Ted Shifrin

6:48 PM

People have to learn how to be teachers and to be students in this brave new world.

3

There are modern proofs, but Chern's was very hands-on. I still like the one I sent you in my notes using some moving frames stuff on the Grassmannian.

If you want a compelling argument for moving frames, prove Schur's Theorem. :)

It's yucky being away from home and away from my computer and notes and books ....

Balarka Sen

Oh dang that's a nice theorem. Didn't know that was a thing

user76284

I'm trying to understand the most general conditions under which $\lim_{t \rightarrow 0^+} f(t), g(t) \rightarrow 0$ implies $\lim_{t \rightarrow 0^+} f(t)^{g(t)} = 1$.

According to this FAQ, if $f$ and $g$ are analytic and approach 0, then $f(x)^{g(x)}$ approaches 1.

But this works for some non-analytic $f$ or $g$, such as the square root.

Is there some more general condition?

Semiclassical

might be better to study $g(t)\ln f(t)$?

@TedShifrin "oh brave new world, that has such students not attending it"

CaptainAmerica16

@TedShifrin Hey!

Balarka Sen

Hm, the Jacobi equation is $\nabla_{\gamma'}^2 X + R(\gamma', X)\gamma' = 0$

I can rewrite that as $\langle \nabla_{\gamma'}^2 X, X \rangle = K(\gamma', X)$. Let's write $\langle X'', X \rangle = K(\partial_t, X)$ for convenience.

Ted Shifrin

6:57 PM

I don't do Jacobi :)

Hi Captain.

Semiclassical

hnnnnggg. student comes in to Zoom, gone in 30 seconds

Nice

Did you wave hi?

Hi @Ted

Looks like jstor.org/stable/3595845 has the result I'm looking for.

Ted Shifrin

6:59 PM

Hi, demonic @Alessandro!

Alessandro Codenotti

Were people using zoom before the quarantine? I had never heard of it, but I had a zoom meeting with my advisor today

Semiclassical

nope

well

CaptainAmerica16

I think talking to teachers virtually is more awkward than real life. Probably because I'm painfully aware my voice sounds nothing like I think it does through the mic.

Semiclassical

i had it on my computer

but never used it

Ted Shifrin

I used it with my sister a few months ago. She has been using it for video-conferencing for years.

Mike Miller

7:01 PM

It's the more popular Skype nowadays.

Balarka Sen

@AlessandroCodenotti Yeah I only heard of Zoom after hell broke lose

Ted Shifrin

When I first saw my videos, I was alarmed to hear my voice. We don’t hear ourselves as others do.

user76284

Theorem 1: Let $f(x) = x^\alpha \phi(x)$ where $\alpha > 0$ and $0 < m \leq \phi(x) \leq M < \infty$ for constants $\alpha, m, M$. Let $g(x)$ be any function satisfying $\lim_{x \rightarrow 0^+} g(x) = 0$ and $g(x)/x$ bounded on $(0, a)$ for some $a > 0$. Then $\lim_{x \rightarrow 0^+} f(x)^{g(x)} = 1$.

Theorem 2: Let $f(x)$ be as in Theorem 1 and let $g(x) = x^\beta \gamma(x)$ where $\beta > 0$ and $\gamma(x)$ is bounded on $(0, a)$. Then $\lim_{x \rightarrow 0^+} f(x)^{g(x)} = 1$.

Balarka Sen

I sound like a complete retard over the mic (and probably in real life). @Alessandro or Fargle or Daminark can confirm

CaptainAmerica16

@TedShifrin If I'm in a video of any kind I go to great lengths to never watch it :P

Thorgott

7:03 PM

@Tobias I think it's fair game to call $\mathbb{C}$ analytic

CaptainAmerica16

I was supposed to be going to an engineering competition in April, but now we have to present our projects over adobe connect. It's gonna suck

Ted Shifrin

Yup, crazy times.

Semiclassical

@CaptainAmerica16 If I have a recording of my voice, I go to great lengths never to listen to it

cannot stand hearing myself talk

I like the sound of my own voice when I'm the one saying it, not when the computer is :P

CaptainAmerica16

Same

Semiclassical

the video part of it, whatever

there's enough distance there. it's not like i spend the entire day looking at myself in the mirror

but voice is so much more immediate

Thorgott

7:08 PM

@Semiclassical Here's a purely algebraic statement: If $K/F$ is a finite Galois extension in characteristic $0$, every odd degree polynomial in $F$ has a root in $F$ and every quadratic polynomial in $K$ has a root in $K$, then $K$ is algebraically closed

Semiclassical

Neat.

Thorgott

the every odd polynomial in $\mathbb{R}$ has a root in $\mathbb{R}$ part is where analysis comes in and then you have a proof of the FTA

CaptainAmerica16

I used to never look in the mirror either until I realized I looked a mess everywhere I went for most of high school

Thorgott

actually, you can dispense of the extension being Galois by passing to the Galois closure

Semiclassical

is the "every quadratic polynomial in K" part tractable within algebra as well? (maybe you said that above)

Thorgott

7:11 PM

you mean for $\mathbb{C}$?

CaptainAmerica16

@TedShifrin Also, I'm switching to a math major. There's no point in me doing CS if that's not my personal goal. I'm going to minor in CS to appease my parents...they're only slightly disappointed.

Ted Shifrin

Well, see how things go. No need to decide for a year or two.

CaptainAmerica16

True, true

Ted Shifrin

I always told my advisees to explore potential interests first, rather than fulfilling random core requirements that could wait.

Semiclassical

i forget where i saw this (maybe it was here) but this article is great

mcsweeneys.net/articles/…

CaptainAmerica16

7:18 PM

What do you mean? Like taking entry classes from different subjects first?

Semiclassical

"Thank you for your patience in moving this course online! The good news is that our work will not go to waste, because no matter how terribly this goes, the administration will take this experience as proof that we can offer this course exclusively online and run this version of the course, without revision, online for the next ten years."

...sorta worried that's actually going to be true

Balarka Sen

lmao

Ted Shifrin

i'm saying prioritize ones in which you think you might have interest if it turns out math isn't what you thought. Lots of people come in saying math major but have no idea what math actually is from high school. You've been exposed to more actual math because of us.

Oy @Semiclassic

CaptainAmerica16

@Semiclassical XD

@Ted Shifrin Ok, I see what you're saying

Mike Miller

@Semiclassical yeah i mean, why wouldn't it be

Semiclassical

7:25 PM

there's some satiric exaggeration, of course, but i do wonder if the admins are going to use this to justify a lot more online courses even when on-campus instruction resumes. it's a golden opportunity in that regard

Alessandro Codenotti

@Thorgott Once if you have this you can also go for the Artin-Schreier theory of formally real fields proof

Mike Miller

never let a good crisis go to aaste

Semiclassical

right

Balarka Sen

Ok, let's say $M$ is a surface. $X$ is a Jacobi field along $\gamma$ which is normal to $\gamma$, let $Y$ be a unit length parallel vector field along $\gamma$ normal to $\gamma$, then $X = \alpha(t) Y$ and plugging that in $\langle X'', X \rangle + K(\partial_t, X)$ gives me $\alpha''(t) + K \alpha(t) = 0$

I have explicit formula for divergence in this case

Exponential if $K < 0$, linear if $K = 0$ and trigonometric if $K > 0$

Mary Star

Hello!! If we have the parabola $2x^{2}-6x+3$ and we want to shift it upwards we have to add a constant term c, i.e. $2x^{2}-6x+3+c$, or not?

Balarka Sen

7:36 PM

If $\gamma_s$ is a variation with corresponding Jacobi field $X$, then Riemannian distance between $\gamma_0(t)$ and $\gamma_s(t)$ should be $s\|X(t)\| + o(s)$, right? Definition of derivative

So I am happy. I have precise estimates

CaptainAmerica16

@Mary Star The constant that shifts the parabola is the constant at the end "3"

whatever you were going to add as c, just add it to the three

Mary Star

7:55 PM

So if we have the parabola in the form ax^2+bx+c and we want to shift it upwards we have to add it to the constant term c, we want to shift it downwards we have to subtract it from the constant term c, right?
What if we want to shift the parabola to the left or to the right? Can we do that having this form? Or do we have to transform the general form of the parabola into the form a(x-d)^2+e? @CaptainAmerica16

Bob

@TedShifrin I hope you are feeling better. I hope you are feeling 100%

Balarka Sen

@Alessandro I am not satisfied with my precise estimates

I need moar precision

Alessandro Codenotti

Make them more precise

Here you go

Balarka Sen

Let's be completely general

I feel this extreme urge to Taylor expand everything in sight

2

What is happening with me

Suppose $V : (-\varepsilon, \varepsilon) \times (-1, 1) \to M$ is my geodesic variation, $V(s, t) = \gamma_s(t) = \exp_p(tv + sw)$ for some $v, w \in T_p M$. I need to Taylor expand $f(t) = \text{dist}^2(\gamma_0(t), \gamma_s(t))$ for some fixed $s$.

And I mean like I need to

The estimates I have should suffice though

Nah definitely a little more work is needed

For sanity I'll just take $s = \varepsilon/2$ or something

Let $U : (-1, 1) \times (-\varepsilon, \varepsilon) \to M$ be given by $U(t, s) = \exp_{\gamma_0(t)}(s \exp_{\gamma_0(t)}^{-1}(\gamma_{\varepsilon/2}(t)))$. We need this new variation, where the curves $U(\cdot, s)$ connect $\gamma_0(t)$ and $\gamma_{\varepsilon/2}(t)$ by geodesics

If $f(t) = \mathrm{dist}^2(\gamma_0(t), \gamma_{\varepsilon/2}(t))$, then $f(t)$ is just arclength squared of $U(t, s)$, $0 \leq s \leq \varepsilon/2$

What a mess

user76284

8:26 PM

Is there a fixed predicate in the language of set theory that captures the property of being a hereditarily finite set, or of being a finite ordinal, or of being a set of the form $\{\}, \{\{\}\}, \{\{\{\}\}\}, \dots$?

Leaky Nun

8:50 PM

@user76284 yeah ofc

user76284

9:06 PM

...that does not use the axiom of infinity, e.g. "$\phi x \equiv x \in \omega$".

Zenix

1

Q: Question on Rolle's Theorem

Let $f: [a, b] \rightarrow \mathbb{R}$ be a function, continuous on$[a, b]$ and twice differentiable on $(a, b)$. If $f(a) = f(b)$ and $f'(a) = f'(b)$. Then find the least number of roots of the equation $f''(x) - \lambda (f'(x))^2 = 0$, for any real $\lambda?$ I know that I need to imagine ...

Leaky Nun

@user76284 you can say "$x$ is an ordinal" by saying "$x$ is linearly ordered under $\in$"

and "$x$ is finite" by saying "$x$ does not biject with a proper subset of itself"

Alessandro Codenotti

@user76284 there is a formula (in fact a $\Delta_0$ one) defining $\omega$ regardless of whether you assume the axiom of infinity

Leaky Nun

the axiom is not the language

assuming the axiom of infinity doesn't magically give you a new symbol $\omega$

user76284

9:31 PM

The problem is I'm working in a theory without extensionality or foundation, so I have to be careful to make sure these constructions still work.

Basically I'm trying to get the right axiom of infinity for this so that the resulting theory is strong enough to interpret ZFC (but consistent).

CaptainAmerica16

@Mary Sue Yes, you're correct. The new form a(x-d)^2+e would be needed to shift the parabola.

Sorry I got back to you so late. I had closed my laptop

Balarka Sen

10:49 PM

Well well well well

Let's take a crack at the Riemannian geometry nonsense again

Alessandro Codenotti

Isn't it time to go to sleep?

Balarka Sen

I slept from 3 PM to 7 AM today

I mean, yesterday

Alessandro Codenotti

WAIT

You slept 14 hours straight?

Balarka Sen

Oh I meant 7 PM

Alright let's set this badboy up. I have a variation, I have two geodesics in that variation, and I want to compute deviation between the two geodesics.

My strategy is to Taylor expand everything and anything

That has to work, right?

Alessandro Codenotti

That always works in analysis

And differential geometry has to count as analysis, right?

Balarka Sen

10:55 PM

Yes, of course

Now that I have an analyst's approval, I shall start. Throughout, $s$ will be first component and $t$ will be second component. $V : (-1, 1) \times (-1, 1) \to M$, $V(s, t) = \gamma_s(t) = \exp_p(vt + sw)$ for some $v, w \in T_p M$ be a variation of $\gamma_0$ which starts at $p$ and with initial vector $v$, and the variation is made in the direction of $w$. I will not bother with injectivity radi.

Consider two handpicked curves, $\gamma_0$ and $\gamma_{s_0}$ - which I shall atrociously call $\gamma_1$ -, from the variation. The aim is to Taylor expand $f(t) = \mathrm{dist}^2(\gamma_0(t), \gamma_1(t))$, as always.

Define new variation $U : (-1, 1) \times (-1, 1) \to M$, where $U(\cdot, t)$ is a geodesic joining $\gamma_0(t)$ and $\gamma_1(t)$. Keep in mind we vary $s$ here. This family is continuous because inside the injectivity radius the geodesics vary continuously with their endpoints, standard fare, etc. To be precise, $U(s, t) = \exp_{\gamma_0(t)}(s \exp_{\gamma_0(t)}^{-1}(\gamma_1(t)))$

Time to compute the Taylor expansion.

Ah, whoops. $U$ should be a function from $[0, 1] \times (-1, 1)$. Namely, $U(0, t) = \gamma_0(t)$ and $U(1, t) = \gamma_1(t)$. I don't want $(-1, 1)$ in the $s$ component, I want a closed interval there, of length $1$.

Alessandro Codenotti

What do you mean an analyst's approval

Balarka Sen

OK, $f(t)$ is basically norm of the tangent vector of $U(\cdot, t)$ (note that these are unnormalized geodesics since I fixed domain to be $[0, 1]$, so arclength is $1$ times norm of tangent vector). $f(t) = \|\partial_s U(s, t)\|^2$.

I was lowkey insulting you, @Alessandro. It's no biggie

Alessandro Codenotti

I know

galmeida

anybody there?

@al

@AlessandroCodenotti @BalarkaSen

Balarka Sen

11:14 PM

Please do not ping random people; simply ask if you have a question. If someone is interested, they shall answer. That is to say, ask; don't ask to ask. Pinging people is sometimes considered borderline spam.

galmeida

Sorry. I'd like to know interesting material for testing athematical aptitude. Any suggestions

?

Balarka Sen

@BalarkaSen Let's call $X(t) = \partial_s U(s, t)$ for simplicity of notation, although that phrase will soon be meaningless to me; $f(t) = \|X\|^2$. Then $f'(t) = 2\langle \nabla_{\gamma'} X, X \rangle$. Pictorially, $\nabla_{\gamma'} X$ is sliding the tangent vector to $U(s, t)$ at $s = 0$ along $\gamma'$. What is $f'(0)$?

Mike Miller

You and Jacobi should get a room

Balarka Sen

It's not like anyone's doing anything here

Would you prefer if I spammed scheme theory instead

Mike Miller

11:30 PM

Nah

Balarka Sen

$X(0) = 0$ again actually so I don't have to do computations for $f'$; $f'(0) = 0$. $f''(t) = 2\langle\nabla_{\gamma'} X, \nabla_{\gamma'} X \rangle + 2 \langle \nabla_{\gamma'}^2 X, X\rangle$ and again for $f''(0)$ the second term cucks off, so have to compute $\nabla_{\gamma'} X$ at $t = 0$, or $D/dt X|_{t = 0}$ now.

How to do this

Should be $s_0 w$; the difference between the tangent vectors of $\gamma_0$ and $\gamma_1 = \gamma_{s_0 w}$ at $T_p M$. And now I realize why $s_0$ crap was a bad choice. Just call it $w$, and let the original variation be joining $\gamma_0$ and $\gamma_1$ where index $1$ literally means time

It's ok whatever

Anyway this is because $\gamma'$ is pushforward of $\partial_t$ and $X$ is pushforward of $\partial_s$ by $U$, so $\nabla_{\gamma'} X = \nabla_X \gamma'$ because symmetry and $\partial_s, \partial_t$ commutes.

I didn't mention this but note that $X$ is obviously a Jacobi field along $\gamma$. So $\nabla_X \gamma'|_{t = 0}$ is just $w$, the initial value of the Jacobi equation

$f''(t) = 2\|w\|^2$

$f'''(t) = 6\langle \nabla_{\gamma'}^2 X, \nabla_{\gamma'} X \rangle + 2\langle \nabla_{\gamma'}^3 X, X \rangle$, again for $f'''(0)$ the last term goes away

Time to use my favorite equation. $\nabla^2_{\gamma'} X = -R(X, \gamma')\gamma'$ because $X$ is Jacobi field along $\gamma$, but $X(0) = 0$ so this term is also zero. $f'''(0) = 0$.

Wait. $X$ is not obviously a Jacobi field along $\gamma$.

It's the other way around. $V_* \partial_t$, tangent fields along $\gamma_s$ for various $s$, restrict to Jacobi fields along $U(\cdot, t)$

Right, I got confused. $\nabla_X \gamma'|_{t = 0} = w$ because $\gamma'$ is a Jacobi field for the curve $X$ is tangent to namely $U(\cdot, t)$. I should say it as, $\nabla_{\gamma'} X = \nabla_{V_* \partial_t} U_* \partial_s = \nabla_{U_* \partial_s} V_* \partial_t$

Ok, $f'''$ becomes a lot harder. I have to use switcharoonie on $\nabla^2_{\gamma'} X$ to make it $\nabla^2_X \gamma'$

Urgh

Oh but I can't, $\nabla^2_{\gamma'} X = \nabla_{\gamma'} \nabla_X \gamma'$.

No more switcharoonie

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$Mathematics$ Mathematics