Mathematics - 2022-04-07

barista

01:27

@Semiclassical Oh, I can put the value first and take the partial derivative.

Adeek

04:04

Hi @TedShifrin

Ted Shifrin

04:15

Hi @Adeek

Adeek

How are you @TedShifrin

@TedShifrin I wanted to check something small with you.

@TedShifrin So let us say we have $X = C_1 \times C_2 \times C_3 \times C4$ and $C$ is a curve such that $Pr{1,3,4}(C) \cong C$ and $C$ doesn't contain a copy of $C_2$ as irreducible subvariety, then I believe we have $Z = C_2 \times C$ sits in X and $C$ is algebraically equivalent to fibre of the projection map $C_2 \times C \rightarrow C_2$ right? because first we have $C$ embeds into $Z = Pr_{1,3,4}^{-1}(Pr_{1,3,4}(C)) = C_2 \times Pr_{1,3,4}(C)$

we know that $C$ is isomorphic to $Pr_{1,3,4}(C)$ and since C doesn't have an irreducible copy of $C_2$ then
we get $C_2 \times C$ sits in X
and if it does sit in X which based on the above seems to be correct, but so proceeding with that then it is easy to see $C$ is algebraically equivalent to fibre of the projection map $C_2 \times C --> C_2$

so let us say we have $g$ defining the fibre of the projection C2 times C --> C then we have div(g){C_2 times C} = H such that H is algebraically equivalent to C

this is correct right ?

Ted Shifrin

You’re going way too fast.

Adeek

sorry

Ted Shifrin

The $C_i$ are all curves and $C\subset X$? What does $Pr_{1,3,4}$ mean?

Adeek

Yes so $C_i$ are all curves and $C \subset X$ is a curve and $Pr_{1,3,4}$ is the projection into 1,3,4 component

Ted Shifrin

04:23

Why do you care that this is a product of three curves?

Adeek

just random

like I care that 3 labels are isomorphic

but just not to make argument above complicated and introducing labels etc

Ted Shifrin

Wait. Huh?

Adeek

Like by label I mean $I = \{1,3,4\}$

Ted Shifrin

Is this the projection onto a 3-fold or what?

Adeek

Yes

Ted Shifrin

04:25

So why do I care that it is a product?

Adeek

$Pr_{1,3,4} : C_1 \times C_2 \times C_3 \times C_4 \rightarrow C_1 \times C_3 \times C_4$

oh no I care it is a product

Ted Shifrin

Why?

Adeek

because I am figuring something out on product of curves

then I will generalize it

Ted Shifrin

For this question it seems irrelevant.

Adeek

I mean I guess it is relevant for algebraic equivalence

but yeah for first question it is not relevant

Ted Shifrin

04:28

I don’t believe you get a product structure rather than a $C_2$ bundle over $C$.

Adeek

I don't understand what you mean ?

we do get a product structure right? Because first of all $C$ embedds into $Pr_{1,3,4}^{-1}(Pr_{1,3,4}(C)) = C_2 \times Pr_{1,3,4}(C)$

by assumption $C$ is isomorphic to $Pr_{1,3,4}(C)$ and since it doesn't have

irreducible component being a copy of $C_2$

Ted Shifrin

My brain is not going to handle this tonight.

Adeek

oh no worries @TedShifrin

let us discuss it tomorrow I believe this is correct discussed it with few friends. Though want to think more about it.

let us discuss tomrrow ?

Ted Shifrin

You should distill out important ingredients and minimize excess notation.

Adeek

Sounds good :)

How are you btw

I missed you and everyone here

?

Also @TedShifrin what email do you use? You're UGA email ? I like to get in touch sometime and catch up etc

copper.hat

04:42

Hi @TedShifrin!

Adeek

Hi @copper.hat

how are you ?

copper.hat

Hi @Adeek

Adeek

I missed everyone haha. I was just too busy lately

though I will join more often now

I am off to sleep

good night

Ted Shifrin

05:01

@copper.hat Hi, dr copper

copper.hat

:-)

robjohn

07:44

@Koro what's more: $a\,|\,bc\implies\left.\frac{a}{(a,b)}\,\middle|\,(a,c)\right.$

I was looking at the equation from the other day and noticed this stronger statement.

Oh, dear... I've killed Koro.

Oliver Diaz

08:14

@robjohn: all that reminds me of the first chapter of Vinogradov's elementary number theory (although the problems there are far from elementary)

Koro

09:31

I have two questions: 1) Suppose that $f(x)=\frac 14+x-x^2$. Let $x_0=a$ and $x_{n+1}=f(x_n)$ for every $n\ge 0$. What are all possible values of $\lim x_n$ and the corresponding value of $a$?

2) Finding $\lim_{n\to \infty}\sum_{j=1}^n \frac{2^{<j>}+2^{-<j>}}{2^j}$, where $<j>$ means the closest integer to $\sqrt j$.

1) is the type of questions where I always get stuck.

1) If $(x_n)$ converges then let $\lim x_n=L$ so that $L=f(L)$ which gives $L=\pm 1/2$.

$x_1=1/4+a-a^2=1/2-(a-1/2)^2$. Suppose $x_1\gt x_0=a$, then $(1/2-a)\gt (1/2-a)^2$. So if $(1/2-a)>0$ then $1/2-a-1\lt 0\implies a+1/2\gt 0\implies a>-1/2 $

I'm stuck now.

For 2), I have no idea. Actually, I don't understand how to open <j>

Cheese Cake

10:21

Hello

LearningCHelpMeV2

10:34

Any tips on how to go about this?

So far, I've come up with $a+b+c \geq 3(abc)^{\frac{1}{3}}$

From AM-GM

Spring

11:32

I apologize for coming here with the same question again. The answerer posted their answer, which isn't something I was looking for, but I think more editing would just overflow my post. I wasn't precise enough in the very end of my post. So, I'll leave a link here and ask, if I may:

Suppose $C,E\subseteq\Bbb R^2$ are bounded sets s. t. $C\cap E$ is of Lebesgue measure $0$, but has no Jordan measure, $f:C\cup E\to\Bbb R$ is integrable on both $C$ and $E,$ but fails to be integrable on $C\cap E$ (due to it not having Jordan measure? ). Is there any example caused by this situation?

3

Q: Integrability of the function on the intersection of the Lebesgue measure $0$

Suppose $C,E\subset\Bbb R^2$ are bounded s. t. $C\cap E$ is of Lebesgue measure $0$ and let $A\supseteq C\cup E.$ If $f:A\to\Bbb R$ is (Riemann) integrable on $C$ and $E,$ is it necessarily (Riemann) integrable on $C\cap E$? This question arose when I was revising the theorem about the additivi...

robjohn

12:40

@Koro The equilibrium is stable when $f'(x)\lt1$ and unstable when $f'(x)\gt1$. Thus, the fixed point at $x=-\frac12$ is unstable since $f'\!\left(-\frac12\right)=2$. The fixed point at $x=\frac12$ is stable since $f'\!\left(\frac12\right)=0$. If $x_0$ is near $-\frac12$, then $x_n$ will move away. If $x_0$ is near $\frac12$, then $x_n$ will converge to $\frac12$.

$g(x)=f\!\left(x+\frac12\right)-\frac12=-x^2$. This is easier to analyze.

$f\!\left(x-\frac12\right)+\frac12=2x-x^2$ shows why the equilibrium near $-\frac12$ is unstable.

@OliverDiaz Yes. GCDs and Bezout are an important basis.

Xander Henderson

14:29

0

Q: Extinction condition for Leslie matrix

Good morning, I assume that I have an irreducible Leslie matrix s.t.: $$ \begin{pmatrix} 0 & \alpha_2 & ... & ... & \alpha_n\\ \beta_1 & 0 & ... & ...& 0\\ 0 & \beta_2& 0 & ... & 0\\ 0 & 0 & \ddots & 0 & 0 \\ 0 & ... & 0 & \beta_{n-1} & 0 \end{pmatrix} $$ I know that if I have an eigenvalue $\l...

linear-algebra dynamical-systems mathematical-modeling network population-dynamics

@leslietownes What have you done?

love_sodam

6

A: Quotients of Solvable Groups are Solvable

(Corrected) Note that since the chain starts at $1<N$, there exists a maximun $i$ such that $G_i<N$. Multiplying and quotienting out that part of the chain gives a chain of quotients $1\lhd NG_i/N\lhd \cdots \lhd G/N$ and $(NG_{i+1}/N)/(NG_i/N)\simeq NG_{i+1}/NG_i\hookrightarrow G_{i+1}/G_i$ is a...

In this answer, how can one get an injection $NG_{i+1}/NG_i\to G_{i+1}/G_i$?

Mathguru

14:45

@LearningCHelpMeV2 that is not a difficult question. First of all notice that a^2+b^2+c^2+abc=4 can only be true when a,b and c =1. Only when a,b and c are equal to 1 do you actually get 4 as a solution to your equation. As such, it is trivial to show that a+b+c=3 since 1+1+1=3. Thank you.

Semiclassical

14:57

this mathematica error message amuses me:

"NDSolve::parpiv: -- Message text not found --"

robjohn

15:19

@Mathguru why is it that $a=b=c=1$? It looks as if $a=2,b=c=0$ also works.

Semiclassical

even if you restrict to $a,b,c>0$ that's not the only real solution

robjohn

sure, I was just pointing out that there is nothing forcing $a=b=c=1$

Semiclassical

yeah. i'm just thinking about what conditions might be necessary for that to be the only solution

robjohn

I used calculus of variations to get the only critical points to be when $a=b=c$ or one of them is $2$ and the others are $0$. Then, that is the only solution.

Semiclassical

fwiw, mathematica can't find any other solutions with positive integer a,b,c

robjohn

15:26

that is pretty simple without mathematica.

Semiclassical

i've never had a knack for this stuff tbh

i wonder if there's nontrivial rational solutions, hmm

robjohn

@Semiclassical I'm sure there are... let me look.

Semiclassical

mathematica's FindInstance command isn't yielding any, but that's not definitive

well

it finds rational solutions just fine if i allow negative values

but not if i restrict to positive values

robjohn

$a=\frac74,b=c=\frac12$

Semiclassical

ah, nice

robjohn

15:37

If you specify $b$ and $c$, you just need to solve a quadratic equation for $a$.

$\left(2-t^2,t,t\right)$ is a solution for $0\lt t\lt\sqrt2$

rb3652

I have a question about linear algebra.

My answer was 18, but the correct answer is 20.

If anyone could explain where I made a mistake, that would be great.

robjohn

How can we do that?

rb3652

Yes, I am currently copy-pasting my work here.

Below is my work. Please note that the distance is of course squared, so I have $20$, not $\sqrt{20}$.

Basically, I did:

(1) Gram-Schmidt $x_1, x_2, x_3$ to get $v_1, v_2$

(2) Project $b=(0,1,2,3,4)$ onto the subspace spanned by $v_1, v_2$

(3) Find distance between $b$ and the projected vector above

That's how I got $d^2 = 20$.

If anyone could spot where I made a mistake, that would be greatly appreciated.

robjohn

I thought you said you got 18

rb3652

Oh whoops. I flipped it up.

That's my mistake. The correct answer is 18.

My answer is 20, as shown in my screenshotted work.

robjohn

15:49

Look at the point $(2,0,2,0,2)$. That is in the span. Compute how far it is from $(0,1,2,3,4)$

rb3652

OK

Of course -- $\sqrt{18}$

But I thought my method was correct.

Did I make some arithmetic mistake?

In addition, how do you know what point in the span to choose? After all, there are infinitely many points in the span, but only one will give the shortest, perpendicular distance, no?

robjohn

$\frac1{\sqrt3}(1,0,1,0,1)$ and $\frac1{\sqrt6}(1,0,-2,0,1)$ is an orthonormal basis

rb3652

Does that mean the orthonormal basis I got from Gram-Schmidt is wrong?

Ted Shifrin

No.

Semiclassical

it's not a bad basis, it's just not the simplest one possible here

Ted Shifrin

15:58

By the way, I keep telling people this, but here I go again: It's usually easier just to get an ORTHOGONAL basis and not bother making everything into unit vectors with horrible square roots.

robjohn

@rb3652 check your dot products...

Semiclassical

note that the difference between x1, x2, x3 is the vector (0,0,1,0,0)

Ted Shifrin

I would go with rb3652's. It looks way simpler to me.

rb3652

@robjohn OK

@TedShifrin Thank you

Semiclassical

i'd say the simplest basis here is (1,0,0,0,1) and (0,0,1,0,0)

robjohn

15:59

@TedShifrin not if you note that the dot product with one of mine is $0$

@Semiclassical That is not orthonormal

Semiclassical

it's not normal, but it is orthogonal

robjohn

@Semiclassical yes...

Semiclassical

so i'm not really too worried about it

robjohn

it makes projection harder

Semiclassical

yeah, my point was just that i get tired of writing out the square roots :P

tho in the physics context i was seeing an exaple recently where it was very important to keep everything normalized

Ted Shifrin

16:01

I disagree with you, @robjohn.

robjohn

@TedShifrin you are free to be wrong ;-)

rb3652

I'm trying to check my work by hand

Ted Shifrin

In all my 40 years of teaching students linear algebra, I find more mistakes with all the horrid square roots than without. When you do projection with a non-unit length vector, then you get the square of its length in the denominator. The arithmetic is equivalent but simpler for students.

I think orthonormal is important for theoretical reasons, not practical ones.

Semiclassical

specifically it was important to write down a parameter-dependent basis where the basis vectors stayed normalized

robjohn

@TedShifrin The problem here is also with the square root of 2

Ted Shifrin

16:02

Is $2$ misbehaving this morning?

robjohn

So using $\frac1{\sqrt2}(1,0,0,0,1)$ is where the problem in the computation is going wrong

rb3652

Oh my god.

Jesus.

I see where I made a mistake.

robjohn

yep in the first dot product

rb3652

I did the dot product wrong.

Kidding me.

Ted Shifrin

Arithmetic is tough.

Semiclassical

16:04

arithmetic is a harsh mistress

rb3652

I'm having a great day.

Well, thank you to everyone for helping me figure out where I made a mistake.

Semiclassical

there's also no magic way to find a good basis

you can always find a basis

but finding a nice one is not trivial.

robjohn

In this case, I don't think there is really nice one

they all involve square roots that require care

@rb3652 No problem. It's always good to ask people who have made many mistakes to find a mistake.

One of the benefits of experience is making mistakes. Oscar Wilde said: "Experience is simply the name we give our mistakes."

Koro

16:34

@robjohn please tell what equilibrium is :(

Novice

robjohn, I'm still planning to ask my professor about the intuition for that equation we discussed, but he's kind of busy so I was planning to wait a little longer.

robjohn

18:15

@Koro An equilibrium point is a place where things stay the same. E.g. when $a_{n+1}=f(a_n)$ the equilibrium point would be where $f(x)=x$. A stable equilibrium point is an equilibrium point where a small perturbation would return to the equilibrium. An unstable equilibrium point is an equilibrium point where a small perturbation would cause a departure from the equilibrium point.

Think of a ball rolling on a sine wave: a ball at the minimum would stay there, and a small shove would return it to the minimum; a ball at the maximum would stay there, but a small shove would cause it to leave that maximum.

@Novice No rush for me.

Semiclassical

18:29

small optimization problem i'm trying to find an explanation for

I have the equation $1-2A p+A^2 p^2=p^2+\mu^2$. This is quadratic in $p$ and I specifically want the positive root, subject to $A\leq 1$ and $0<\mu<1$

i then want to find $A$ such that this positive root is as large as possible

apparently the choice is just $A=1$, but I'm not seeing a good way to justify that formally

the tricky thing is that the equation degenerates to being linear in $p$ when $A=1$

oh. i should've said $0\leq A\leq 1$

bleh. strike that last remark, i'm backwards: should be $A\geq 1$

robjohn

18:50

$$p=\frac{A\pm\sqrt{1+\mu^2\left(A^2-1\right)}}{A^2-1}$$

is that what you get?

Semiclassical

yeah

since $A\geq 1$ it's the positive root that's relevant

robjohn

it looks as if both roots are positive

Semiclassical

huh

that's weird

copper.hat

im positive they are positive

Ted Shifrin

I’m here to say nay.

Semiclassical

18:55

oh, i get it

Adeek

@TedShifrin yo

Semiclassical

i forgot I started from $1=Ap+\sqrt{p^2+\mu^2}$

so one of the roots wasn't going to be valid in the first place, derp

robjohn

Even if $\mu=1$, the discriminant is $A^2$, and the smaller root is $0$, but if $\mu\lt1$, then the smaller root is positive

Semiclassical

given what i'm starting with, I need $p<1/A\leq 1$. So $$p=\dfrac{A+\sqrt{1+\mu^2(A^2-1)}}{A^2-1}\geq \frac{A}{A^2}=\frac{1}{A}$$ is forbidden

robjohn

Ah, context rears its ugly head.

Semiclassical

19:00

hence i need $p=\frac{A-\sqrt{1+\mu^2(A^2-1)}}{A^2-1}$

well, i needed a positive $p$. so $1=Ap+\sqrt{p^2+\mu^2}>Ap+\mu>Ap$

robjohn

so you need the smaller root

Semiclassical

right

(if you eliminate the square root then that constraint gets lost.)

Novice

19:45

I think this will be a quick question. I am trying to gradually unravel the tensor algebra (to work my way into DG). I was looking at these notes:

I am 90% sure that the first sentence of the last paragraph, concerning the map iota, is incorrect. The map as defined is injective but not surjective. If the map had as range $\{ \mathbb 1_a \colon X \to \{0, 1\} \mid a \in X \}$ then I think it would be a bijection.

(I can't figure out how to make the blackboard bold 1 for some reason.)

I'm just looking for confirmation that I'm not deluding myself here.

Thanks.

(The purported existence of a bjiection between the "base set" X and the free vector space built from X struck me as absurd, since what little intuition I have tells me that the free vector space is way bigger than the base set.)

runway44

20:01

True, it's an injection and its range is the canonical basis

Novice

Right, that's what I thought. Thanks.

Ted Shifrin

20:31

I have a feeling this abstract, formal approach is not the way most geometers think about it, at least until we get to tensor products of general vector bundles.

Novice

The way most geometers think about the free vector space, or about the tensor algebra?

Ted Shifrin

About tensors as they show up in basic differential geometry .

Novice

20:48

I haven't gotten far enough to see tensors. What happened is that I started trying to read AMS GSM 52 and the opening pages on exterior forms were like getting smacked in the face. So I am trying to overcome that in two ways: 1) develop some geometric feel for what's going on, and 2) understand the tensor algebra, since it gives me some context in which to place this stuff, and it's relevant for DG as far as I know.

Ted Shifrin

You might try my YouTube lectures on differential forms. Very down-to-earth.

Novice

I think I've seen you talk about determinants in a video on YouTube. Are you referring to this video? youtube.com/watch?v=wlo2V8H5khM

Ted Shifrin

Yes. That’s the first of a lot on differential forms, integrating them, etc.

Novice

It looks like you work your way up to the generalized Stokes' Theorem on day 36. I will check these out. It seems like that's arguably the first milestone in differential geometry.

Thanks.

Ted Shifrin

Sure thing.

Monty

21:15

I have map $T$ and a measure $p$ with density $f$. I want to show that if $p$ is invariant under the push forward by $T$, then $f(T(x))=f(x)$. Is this ok? : Using the definition of push forward and the fact $p$ is invariant under $T$, I can see for all measurable sets $E$ that $\int_E f(x) dx=\int_E f(T^{-1}(x))|J T^{-1}(x)|dx$ here $J$ is Jacobian, hence for all $y$ $f(y)=f(T^{-1}(y))|J T^{-1}(y)|$ in particular this holds for $y=T(x)$, so that $f(T(x))=f(x) |J Id(x)|=f(x)$.

and by density I mean density with respect to the lebesgue measure.

Ted Shifrin

So density is the Radon-Nikodym derivative of $p$ with respect to $\mu$?

This doesn’r make sense to me. Define domains and codomains of things.

Monty

density is radon-nikodym derivative of $p$ with respect to lebesgue measure

theres no $\mu$ involved

$T:\mathbb{R}^d \to \mathbb{R}^d$, $p$ is a measure on $\mathbb{R}^d$.

Monty

21:43

basically asking if $f$ is the density of a measure $p$, and $p$ is invariant under $T$, i.e $T_{\#}p=p$ then is it true that $f(T(x))=f(x)$.

LearningCHelpMeV2

22:33

@robjohn What level of calculus is required to differentiate that expression to find critical points

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