Mathematics - 2024-01-24

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Thorgott

00:16

yeah, but if you think about my point geometrically, it's obvious

I hate the formula description they give of this map

Ted Shifrin

Rotating $\theta$ clockwise is rotating $-\theta$ counterclockwise.

Jakobian

I like that they give it

Thorgott

01:09

it's ok

i prefer other ways

psie

01:24

I'm working the following exercise:

> Determine the polynomial $p$ of degree at most $1$ that minimizes $\int_0^2|e^x-p(x)|^2 dx$. (Hint: first find an orthogonal basis for a suitably chosen space of polynomials of degree $1$.)

So to find an orthogonal basis, I need an inner product right? I'm unsure which one they are alluding to, if any.

psie

01:38

I guess it's the usual one on $L^2$, but I got confused about them absolute value signs. They seem not necessary...

Jakobian

@psie You're squaring a function

orthogonal basis is a hint so they don't have to specify in what space they're taking orthogonal basis

you correctly deduced its about $L^2$ however

absolute value will be necessary if they consider complex polynomials

psie

right, ok, back to basics, for real $f$ then $|f|^2=f^2$

user unkown

02:26

Help me fix this proof for the following: Show that there is an absolute constant $c$ such that for each $k > 1$, every tournament that satisfies $S_k$ has at least $c\, k 2^k$ vertices.

\textbf{Hint}: Show that for every set $W$ of at most $k-1$ vertices, there are at least $k+1$ vertices that all point to each vertex of $W$.
\textbf{Proof}: Let $T$ be a tournament that satisfies $S_k$, where $k > 1$. We argue that $T$ has at least $c \, k 2^k$ vertices for some absolute constant $c$.

We first prove the following lemma:

In particular, I am having trouble with the inequalities.

user unkown

02:51

@Jakobian Please help me with this question.

2 hours later…

Daniel Donnelly

04:39

0

Q: Why would an integer-related homology exist if there were absolutely no real-world application?

Let $(C_n, d_n)$ be a complex indexed by $z \in \Bbb{Z}$. In other words, $d^2 = 0$. Now suppose further that $d$ is exact. Then what information do we get from this / why are projective / injective resolutions interesting to us? I can never find this info on the web... I'm wondering because, ...

abstract-algebra prime-numbers modules homology-cohomology integers

@PatrickDaSilva

I mentioned you there

since you proved exactness

Also thought you guys should know about this:

jwiltshiregordon.github.io/homology/index.html

Since you're into alg topology

Not me

4 hours later…

BAYMAX

09:09

0

Q: Is the following mapping a diffeomorphism?

I am wondering whether the following three dimensional mapping is a diffeomorphism? $x_{n+1} = x_{n} + Rx_{n}(1-x_{n} - \alpha x_{n} - \beta z_{n})$ $y_{n+1} = y_{n} + Ry_{n}(1 - \beta x_{n} -y_{n} - \alpha z_{n})$ $z_{n+1} = z_{n} + Rz_{n}(1 - \alpha x_{n} - \beta y_{n} - z_{n})$ where $R,\alpha...

multivariable-calculus jacobian diffeomorphism

3 hours later…

psie

11:43

I just want to check if I'm going in the right direction with this. I want to minimize $\int_{-1}^1\left|ax+bx^2-\sin \pi x\right|^2dx$, so I proceed by finding an orthogonal basis of polynomials with respect to the inner product $\int_{-1}^1f(x)g(x)dx$; they are $1,x,x^2-1/3$.

Next I compute the orthogonal projection on the subspace spanned by $x,x^2-1/3$, $$P(x)=\frac{\langle \sin\pi x, x\rangle}{\langle x,x\rangle}x+\frac{\langle \sin\pi x, (x^2-1/3)\rangle}{\langle (x^2-1/3),(x^2-1/3)\rangle}(x^2-1/3).$$ I get that $$P(x)=\frac3\pi x.$$ How can one check that this is indeed the right answer?

psie

11:55

I would have expected that $a$ equals something like $\pi$, because the Taylor expansion of $\sin\pi x$ starts with $\pi x$. Also, I'm not sure if I should use $x,x^2-1/3$ as a basis, because they are asking for a function that is a linear combination of $x,x^2$.

1 hour later…

nickbros123

13:00

@leslietownes my institutions math department's pde:something else ratio is like 6:14

psie

13:26

@psie I guess one way of checking this is by seeing if the projection is orthogonal to the residual, i.e. if $\langle \sin \pi x-P(x),P(x)\rangle=0$.

Jakobian

14:15

@userunkown I'm sorry but I don't know any graph theory. I only tried to help with the other question so you're not struggling

3 hours later…

Jakobian

16:54

@psie For $a, b\in\mathbb{R}$?

Jakobian

17:04

@psie you're right that for the basis you used there is no guarantee of getting the right answer. However, your element that minimizes the distance from $\sin(\pi x)$ to span of $1, x, x^2$ is in the span of $x, x^2$

So it should be the right answer, even if you ought to use an orthogonal basis for the span of $x$ and $x^2$

Ted Shifrin

@psie The $L^2$ inner product has absolutely nothing to do with Taylor expansions.

Mad

user image

@TedShifrin hey Ted, i think this is your speciality. Do you understand why the marked part is true?

i am trying to prove this statement for general n

and i spesifically mean why the eigenvectors need to be shared.

welp, i think i understand why they are eigenvectors. since AHA* is diagonal, obviously multiplying with unit vector will give us some eigenvalue

The argument can be used for the case n right? @TedShifrin

Ted Shifrin

17:27

No, this isn't my specialty, and we're also missing definitions. What is the definition of $N$? What is $\mathfrak h$? $SU(n)$ is presumably more complicated than $SU(3)$.

Mad

N is the normalizer. i believe h is the cartan sub algebra given by the t + it

Ted Shifrin

Normalizer of what?

And no, I haven't thought about this stuff in 45 years or so.

Mad

no worries

Jakobian

17:52

Rotation formalisms in three dimensions

In geometry, various formalisms exist to express a rotation in three dimensions as a mathematical transformation. In physics, this concept is applied to classical mechanics where rotational (or angular) kinematics is the science of quantitative description of a purely rotational motion. The orientation of an object at a given instant is described with the same tools, as it is defined as an imaginary rotation from a reference placement in space, rather than an actually observed rotation from a previous placement in space. According to Euler's rotation theorem, the rotation of a rigid body (or three...

is this wikipedia page for robots

oh so the proof that $\rho$ doesn't depend on $a, b$ is basically Euler's rotation theorem

or um... surjectivity of $\rho$?

every matrix in $SO(3)$ is a rotation matrix after orientation-preserving orthonormal base change

and that rotation matrix depends only on one vector, the one with respect we rotate to, not the other vectors in the basis

Thorgott

18:09

yeah, I don't like Wikipedia articles on this topic

leslie townes

as you can see, a lot of wikipedia lie groups related stuff is contaminated by people who only half-learned this stuff in a physics class and seem to be trying to teach it to themselves a few years later by editing wikipedia

Thorgott

18:50

anyway, the proof of independence is just that $SO(2)$ is abelian, which is to say that a rotation in the plane looks the same in every oriented, orthonormal basis

user10478

19:14

Can someone explain in more detail how to use the picture on page 6 here (link.cs.cmu.edu/15859-s11/notes/Hackenbush.pdf) to assign a number to Blue-Red Hackenbush? Inferring from the example, am I simply looking for the most direct (deepest in the tree) common ancestor vertex of the greatest $l_i$ and the lowest $r_i$ vertices (a.k.a., the root of the subtree which is the shortest path connecting the greatest $l_i$ and the lowest $r_i$ vertices)?

This doesn't actually resemble the instructions, but I'm kind of confused otherwise.

Jakobian

19:31

@Thorgott I know

I've proved it that way

1 hour later…

Sha Vuklia

20:50

WOW @Leslie, I just saw your post!

leslie townes

21:01

did you know someone asked it before? but they were focused only on the easy direction, and nobody answered it (well, people sketched the answer for the easy direction in the comments) math.stackexchange.com/questions/4714942/…

Sha Vuklia

I did not actually!

leslie townes

nobody seems to have noticed that he asked for both directions. maybe the OP himself didn't notice, and wasn't asking

Sha Vuklia

I really appreciate the effort! And I also hope you won't get too disappointed with me if I leave the verification of the technical details for a fair bit later in time, as I've just finished my preparations for my funana exam (which is coming Monday), and I have to focus the next three weeks on my alg numb exam (a course that I've fully neglected ever since I started reading in Conway :'))

However, I like measure theory a lot, so I'm definitely interested in the measure-theoretic verifications and will do those as a fun exercise when I have time again :)

Jakobian

@leslietownes so long...

Leslie numerated it in 16 points because he's trying to invent a new field: mathematical lawyering

Sha Vuklia

(this exercise was actually not part of the "formal" exercises for the course, but it looked fun to me, so I had a look at it, until I noticed that I was really not able to solve it xD)

I love elaboration, so this couldn't be better :D except that under my current time constraints I'll have to leave it for a sec :'x

the proof itself is pretty short though! with the right machinery

leslie townes

21:15

jakobian: if you're the kind of person who likes verifying details for yourself without 'hints' you can skip the footnotes

Xander Henderson

I am upset that the footnotes are not all hyperlinked.

leslie townes

jakobian: there was a two line version in the chat the other day

xander i couldn't figure out how to do that, and when i tried googling i only got old posts from like 2012 on other stackexchange sites on how to hack it

if this is laid out really clearly in some document on the math site, it's a shame it doesn't pop up higher in google searches

google search is really into prioritizing (1) AI generated gray goo, and (2) random posts from like 2014 on stack overflow

altavista should pop back into the search game, there is an opening here

Sine of the Time

@Jakobian can you help me with an exercise, if you have time?

Consider the space $X=\{f(x) \in L^4(\Bbb R) | xf(x) \in L^4(\Bbb R) \}$, the question is to prove that the space is complete (with respect to the usual norm). I considered a Cauchy sequence $(f_n) \subseteq X$, since $f_n\in X$, we have $f_n \in L^4$. I know that $L^4$ is complete, so there is $f$ such that $\|f-f_n \|_4\to 0$. Now I should prove that $xf(x)\in L^4$

Jakobian

21:53

@leslietownes I'm joking

@SineoftheTime I get a feeling a version of this exercise was already asked about here

Sine of the Time

yes, I've asked a similar ex but I've a doubt

Jakobian

To prove its complete, simply prove its closed

Sine of the Time

by proving it's sequentially closed?

Jakobian

Hmm okay no. Here we have to provide some kind of bound on $\|xf-xg\|_4$ in terms of $\|f-g\|_4$ I think

Sine of the Time

what's $g$ here?

Jakobian

22:03

element of $X$

I mean we could try to consider some function $f\mapsto x\cdot f$ into some space which has $L^4$ as a closed subspace and so that this map is continuous

Alessandro Codenotti

22:23

@Jakobian annoying question that has been bugging me for a couple days (I asked on MO earlier but still no answer): does every Polish (Abelian) group have a nontrivial locally compact subgroup?

Jakobian

22:45

@AlessandroCodenotti any assumptions on that group?

Alessandro Codenotti

which group? The main group or the subgroup?

Jakobian

main group

Alessandro Codenotti

just that it is Polish (and maybe Abelian, I'm interested in both the general and the commutative cases)

Jakobian

Then the trivial group is a counter-example

Alessandro Codenotti

...

Jakobian

22:50

Other than this case, if an isolated point exist, then the group is discrete so locally compact

The question boils down to perfect Polish groups

Does the proof that any perfect Polish space contains a Cantor set not generalize to showing that a perfect Polish group contains a subgroup homeomorphic to the Cantor set?

Alessandro Codenotti

@Jakobian Yes, the countable case is the easy one

@Jakobian no, there's Polish groups with no nontrivial compact subgroups ($\mathbb R$ or a separable Banach space will do)

Jakobian

well what about the idea of just taking a non-trivial element and generating a subgroup?

Unless it ends up to be dense, like can happen for the circle

Alessandro Codenotti

The subgroup generated by a single element is going to be locally compact iff it is discrete, one can instead look at the closure of the generated subgroup, but there's no reason for it to be locally compact in general

Jakobian

does considering characters on this group lead to something interesting?

psie

23:16

Why, in this proof, can one in the $\impliedby$ direction assume that $(x_n)$ be bounded? The assumption is that the linear operator $T$ is unbounded.

Xander Henderson

I just got my load sheet for the semester. Technically, I am one credit over load. Neat. That's worth an extra $1000. I'm having a good day---California owes me money, the feds owe me money, and the college owes me money!

Yay!

It almost makes up for the money I have to put into the car in the next 6 months.

Xander Henderson

23:30

Oh, and I'm getting something like $5 out of a class action lawsuit over violations of privacy, as a result of my ex-wife using some shady online service to cyberstalk me. Everything's coming up Xander!

Ted Shifrin

Is that like stalking with a cypress tree?

Xander Henderson

@TedShifrin Very much so.

psie

23:49

@psie my bed is calling, but if you know the answer, I'm always grateful for a ping and an explanation...later! 🤸‍♂️

Ted Shifrin

@psie just correctly negate the definition of a bounded operator.

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