Mathematics - 2022-03-31

Akiva Weinberger

00:00

They are constrained to have the same height at all times

Can they always climb the mountain?

(The example in the video is not piecewise linear but I only care about the piecewise linear case)

leslie townes

whoof. too tricky for chat.

Akiva Weinberger

01:19

@leslietownes Would you like the solution? It's simple enough to be summarized quickly

leslie townes

i know it, it's a famous problem. too tricky for someone who isn't given a wikipedia link, methinks. :D

or maybe the whole damn world is just cleverer than i am. i'm willing to admit that.

hah, the page for it has a funny remark. noting that the number of turns in the required path can be quadratic in the number of peaks+valleys, it says: "These complications make the problem unintuitive and sometimes rather difficult, both in theory and in practice."

the "in practice" is what's making me laugh. what a fun hike that would be.

i'm imagining walkie talkies. "OK, I'M GONNA GO DOWN NOW, OK? WHAT DOES YOUR ALTIMETER SAY? OKAY, ONE MORE STEP. WHERE DID WE FIND THIS PIECEWISE LINEAR MOUNTAIN ANYWAY?"

Akiva Weinberger

@leslietownes I didn't solve it on my own, I just looked it up, lol

Maybe 'cause I didn't try it, I've managed to delude myself into thinking it's easier than it is

(You can say the goal is to swap the hikers - it's equivalent.) Basically the deal is that you always have an even number of options of what to do, except at the start and end

and since you can't have a graph with just one odd-degree vertex, in the graph of possibilities you can get to, the ending position must be in it as well

Ted Shifrin

04:43

@leslie It is well-known that the characteristic polynomials of $AB$ and $BA$ are the same. (For example, it's trivial that they are similar when one is invertible.) I don't know what is known about the minimal polynomials when neither is invertible. Any thoughts?

leslie townes

huh. other than them not needing to be the same, i dunno. feels like their oughta be some relation.

real operators don't have minimal polynomials, of course.

Ted Shifrin

I'm fine with working over $\Bbb C$. I was thinking about it because of this.

Ted Shifrin

05:02

@Thorgott I totally do not get your comment here. How do you do $G$-valued partitions of unity?

leslie townes

seems tricky.

dc3rd

what's the trick to find the critical points here @TedShifrin? $(x-y)e^{-(x^2 + y^2)/4}$? I already converted to polar coordinates and I made a "little" progress....well I successfully differentiated if that counts for something....

Ted Shifrin

05:17

I don't know that polar coordinates is required.

dc3rd

so I got to just tough out the ugly differentiation with $x,y$ then.....

Ted Shifrin

It's not that bad. Once you pull out the exponential, you have two quadratics that have to vanish.

dc3rd

I guess so......I just stopped too soon after seeing what was going on....

I'll get back to it.

Ted Shifrin

This is another one of those not-particularly-interesting questions.

@leslietownes Was that to our linear algebra question or to my query to Thor?

dc3rd

True.....I guess since I "know" what to do, on a day I'm bored and not swamped with too much work I could come back and solve for the variables..

leslie townes

05:22

to our linear algebra question. everybody knows how to do G-valued partitions of unity.

i'm kidding of course. we did something like that in my abstract harmonic analysis class, but nowhere near that level of generality.

Ted Shifrin

I think the first example would be a $3\times 3$ with eigenvalues $0$, geometric multiplicity $1$ in one case and $2$ in the other. @leslie

Koro

If in $(X,d)$, a relation $\sim$ is defined as $\{t_n\}\sim \{s_n\}$ if $d(t_n,s_n)\to 0$, This relation is to be shown an equivalence relation. But I don't understand why this question even makes sense. While defining an equivalence relation on a set Y, we consider $a\sim b$ iff ..., where a and b are elements of Y. But in the above case $\{t_n\}$ is not an element of X, rather a set.

Ted Shifrin

Thor tossed off a comment to someone making it seem obvious and well-known, and it's certainly neither. I'm not convinced it's possible.

leslie townes

you can do a 2x2 example. A(x,y) = (y,0) and B(x,y) = (0,y). it's a 2d version of the kind of thing you're thinking of.

Ted Shifrin

@Koro: You're defining an equivalence relation on sequences, not sets.

I thought about $2\times 2$ already and got nowhere. Let me think.

Koro

05:25

right, but the question reads 'Define an equivalence relation in X...".

leslie townes

oh, i may have fiddled something.

the 3x3 example certainly works.

Koro

this is not an equivalence relation in X, rather an equivalence relation on set of sequences in X.

Ted Shifrin

Can you give me $A$ and $B$ in the $3\times 3$ case that yields that? I haven't tried.

@Koro No one said it was an equivalence relation on $X$.

If you're going to progress as a mathematician, you have to read intelligently.

Koro

@dc3rd I have not tried the question you asked but I just want to say that sometimes it is useful to directly use the definition of extrema instead of finding them using partial derivatives.

leslie townes

which may include ignoring errors on the part of written material, should they say stuff that immediately doesn't feel right.

Koro

05:27

@TedShifrin yeah, 'in' was said and not 'on'.

thanks @Ted.

Ted Shifrin

I'm just saying that you know what it is intended, and you like beating dead horses.

@Leslie: Oh yeah. You're totally right. You have a 2-D example that I missed finding.

So that pretty much does it. Jordan forms can be whatever when both matrices are singular.

Koro

prepositions matter! :-)

leslie townes

koro, be kinder to the horses.

Ted Shifrin

Be careful casting pearls before beaten swine.

Remark on that: Prepositions are the hardest part of learning any language. Ted's proposition

leslie townes

my daughter's new thing is to wake up between now and midnight, and yell, and when i come in, to demand to talk about spiders.

she's obsessed with spiders. does anyone know good facts about spiders?

Koro

05:33

I have heard that if their web is heavily concentrated, it can trap airplanes.

not sure about the truth of this statement though.

Ted Shifrin

Did a spider try to eat her? I did love Charlotte's Web as a kid.

leslie townes

koro the material that makes up most webs is very strong. if we had macro versions of it, it could probably do that.

trapping bees or birds in flight is already pretty amazing, and it can do that.

she loves their webs.

every once in a while the cat will catch a spider and play with it. i think she's learned about spiders from that.

and noticing their webs in the bushes outside

Koro

leslie, I also heard that someone tried to cultivate the webs but despite having huge land for spiders, the amount of web they got was negligible.

leslie townes

that was my mother-in-law. we're all a little worried about her.

Ted Shifrin

Maybe she'll be ready for another of my favorite books from childhood soon. The Wind in the Willows.

Koro

05:37

or Ted's Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds

leslie townes

i remember the wind in the willows.

Ted Shifrin

Um, no.

I was addicted to the Pooh books and to Wind in the Willows.

leslie townes

my daughter likes interrupting made-up stories that i tell her and correcting the plot and telling me what happens.

Koro

haha.

leslie townes

then when she's done, she says, "i want you to tell me that story." so i have to repeat more or less what she already said.

she's going to be a handful.

ted, you'll like this. she finished her dinner but was still hungry, and asked for some strawberries. we said we'd get one in a minute (we're fine with fruit as dessert). she somewhat impatiently said, "i ask for food, and you get it." my wife said "that's not how it works." she said "that IS how it works."

Ted Shifrin

05:45

Going to be? You’ve made her a handful already.

leslie townes

she'll be an effective leader.

sometimes when i pick her up, i find the person who runs "aftercare" (= day care after hours) drawing pictures for her. this seems backwards, but i kind of like it.

Koro

05:57

whenever I have difficulty in Rudin's exercises, I look it up on mse and there's one particular user who also has had that question and his posts answer my question also :).

leslie townes

it's funny. in an earlier time, an aggressive copyright owner could sue someone distributing a solutions manual. but a web forum with random posts from random people in no particular grouping cannot be said to duplicate the structure and organization of a series of textbook exercises.

the estate of walter rudin missed its chance.

copper.hat

06:25

hello

dc3rd

good morrows

Monty

12:58

hi guys

4

Q: Prove that the flow of a divergence-free vector field is measure preserving

On page 3 of this preprint, after recalling the definition of flow generated by a vector field, the authors remark that "a necessary condition for a flow $\varphi_t(\cdot)$ generated by $a(t, \cdot)$ to be measure preserving is: $\mathrm{div}\, a = 0$ in a suitable sense". Assuming $a(t,\cdot) ...

at the start of claim 2. should it read suppose $\mu_t$ is the unique solution to (1)?

Astyx

13:15

I think so

love_sodam

13:37

Number of subgroups of $D_8\times D_8$ is much more larger than I thought

love_sodam

14:00

If my computation is correct then the number of subgroups is 195

anak

14:15

@love_sodam up to isomorphism, or are you counting isomorphic but set-wise distinct ones?

love_sodam

@anak Latter. To compute it, I used the fact that there is a bijection between the set of subgroups of the direct product of groups $G_1\times G_2$, and the set of all $5$-tuples $(H_1,K_1,H_2,K_2,\phi)$ where $H_i$ is a subgroup of $G_i$ and $K_i$ is normal in $H_i$ and $\phi:H_1/K_1\to H_2/K_2$ is an isomorphism

There are total 7 possible composition series each have length 4

anak

When you say D_8, do you mean D_4, or do you actually mean the symmetry group of the octagon?

Because the latter has way more such subgroups.

love_sodam

Square. I use $D_{2n}$ for dihedral group.

Ted Shifrin

16:13

@love_sodam Is this supposed to be an interesting question?

leslie townes

ted: oh, be nice.

if someone tells me the order of a group is a power of a prime, the very first thing i would expect is for it to have more subgroups than i would expect.

Ted Shifrin

I still don’t see why it’s interesting. Just bookkeeping.

What is cute is that the dihedral group itself has 8.

leslie townes

i don't find it interesting either. but a lot of people don't find me interesting.

Ted Shifrin

We aren’t a tedious homework question.

leslie townes

you have a point

Ted Shifrin

16:33

Just tedious. Especially Ted.

leslie townes

math.stackexchange.com/questions/4417282/… is interesting if only because it strikes me as definitely not a homework problem.

it's not substantively interesting, but, nobody's perfect.

Ted Shifrin

Sorta interesting. What about the infinity norm?

Oh, easy.

leslie townes

yeah.

leslie townes

16:59

somewhere i had a post about the isometries of ell^p spaces. except i forget if it was on MSE or MO under an alt account, or if it would have been relevant to this. the non-2 ell^p spaces are pretty rigid.

suggesting 'at most one circle' should be right.

found it. this is 'classic leslie.'

7

Q: Isometries of $\ell^p_n(\mathbb{C})$

Let $1<p<\infty$, and define an isometry of normed linear spaces to be a norm-preserving surjection. Then all isometries from $\ell^p_n(\mathbb{R})$ to itself are given by linear transformations $T$ such that Mat$(T)$ is product of a permutation matrix and a diagonal matrix with $1$'s and $-1$'s ...

functional-analysis

let's all wish happy birthday to that answer when it turns 10 in may.

Koro

leslie, you have a comment also to your answer.

leslie townes

yeah. i'm ignoring that. feel free to hop in.

quasinorms are a "nope" from me.

but it feels it would be pointless for me to comment back simply saying, "No."

Koro

i don't know quasi norms yet :(.

But just asking, in such a situation how should a comment be responded to by author?

'Sorry for late response' doesn't feel right in such a context.

leslie townes

well, if you have anything to say, i guess, respond. otherwise its lack of being answered may be a signal to others who have something to say to respond.

for context, quasinorms don't have to satisfy the triangle inequality, which is the origin of the huge 'nope' from me. everything about functional analysis dies at that point.

dc3rd

17:17

@leslietownes What constitutes "interesting?

Koro

I mean the situation when one receives an email that they couldn't respond to at the time of seeing it but somehow they decide to respond to that after 10 years, their reply will of course not start with 'sorry for late response'. How should they reply to such an email?

leslie townes

dc3rd: personal taste. no rigorous definition.

Koro

I think in such a situation, it's best not to forget about the email.

leslie townes

koro: i agree. but assuming the decision has been made to respond, i'd just respond to it.

dc3rd

"I was cleaning up my eamil and I just noticed I never saw your email....."

Ted Shifrin

17:20

@leslie The guy who commented with the question was a grad student at UGA. I know him. Started in geometry, switched to numerical/compressed sensing,

Koro

@dc3rd haha, but that sounds rude and strange :(

leslie townes

same spirit as responding to ancient questions on MSE. personally, i would not respond to a 5+ year old question on MSE unless i had a lot to say or something new. but if i did respond, i'd respond without acknowledging the age of the answer.

i guess it's not quite the same as MSE is more impersonal than personal email.

Koro

or like, someone hacked my email and I got access only after 10 years. :P

leslie townes

i once loaded some old email into my new computer and responded to an email from 1998. i had to change the to: address to make it go through. the recipient thought it was funny.

ted: just the kind of miscreant who would fail to obey the triangle inequality.

Ted Shifrin

I’m getting annoyed. Someone emailed me about a misunderstanding of something in my book and then kept emailing ignoring what I had exlained in two responses. No more answers from Ted.

leslie townes

17:26

hrm. have koro reply to that email 20 years from now.

dc3rd

@TedShifrin Luckily I'm not going to ask you about maximizing the box in the hemisphere until 2mrw...... :p

leslie townes

"Sorry for the late response, but I am Ted's representative in all matters relating to [this question]. As Ted explained in 2022, ..."

Koro

I studied Baire's Category theorem recently. I understand that it's 'far reaching' generalization of uncountability of the reals.

I know for a fact that there exists no function (from R to R) which is discontinuous only at the irrationals and is continuous at rationals. I have been told that this statement is proven using BCT but I don't see how.

Any hints on this please?

Ted Shifrin

Another good application for you @Koro: Suppose $f_n$ are continuous and converge pointwise to $f$. How discontinuous can $f$ be? [Natural question to ask to understand how strong uniform convergence is.]

leslie townes

OK. i wouldn't take "is proven using BCT" to mean that there aren't other ways. BCT is a route to a lot of results, particularly existence and nonexistence results, but it's rarely the only route.

Ted Shifrin

17:39

(My functions can be on $\Bbb R$, on a complete metric space, or on a Baire space — if anyone remembers what those are.)

leslie townes

the key notions are more fundamental than BCT, in the descriptive categories of the baire hierarchy. i think the relevant thing here is that the set of discontinuities of a function R to R must be an "F_sigma" set.

this comes down to how continuity is characterized in terms of epsilons and deltas and what not.

i found this set of notes just googling now. it gets there, although maybe not in the simplest or most-agreeing-with-textbooks kind of way. web.math.utk.edu/~freire/teaching/m447f16/BaireCategory.pdf

a lot of complicated constructions can be replaced with a BCT argument if you don't want to get your 'hands' on a particular example. functions with really weird sets of discontinuities being one of them

JP kahane had a good article, baire's category theorem and trigonometric series, that gives examples from fourier analysis. that might be too far afield from what you are intersted in, but gives the general idea.

Koro

@TedShifrin I'll think about that. Thanks. I'm trying some exercise problems to see applications of BCT.

@leslietownes thanks, it defines $F_\sigma$ also. I'll take a look at that :).

leslie townes

the BCT existence results often come with a corresponding 'genericity' result. e.g. the same ideas behind a BCT proof that there's a function doing something ugly and surprising, will often show that the set of functions doing that ugly and surprising thing is dense. and/or that its complement is a countable union of nowhere dense things.

i don't know if this is a theorem or a vibe, but it's a strong vibe if it isn't a theorem

Koro

18:15

@leslietownes the vibe seems famous youtu.be/3uzQMbVxiEI?t=248

:).

leslie townes

famous!

Jakobian

18:30

If $f:\mathbb{Q}\to \mathbb{R}$ is continuous, then $f$ can be extended by a $p\notin \mathbb{Q}$ so that $f:\mathbb{Q}\cup \{p\}\to \mathbb{R}$ is still continuous.

This kind of uses BCT as well ($\mathbb{Q}$ is not $G_\delta$-set of $\mathbb{R}$), but not sure if it can be proved in an elementary way

What you do, is you extend $f$ to a $G_\delta$ subset of $\mathbb{R}$, and pick an irrational.

copper.hat

19:02

my orange jump suit is years away, i may need to find another source

Ted Shifrin

19:22

@copper I'm almost ready to retire from here.

robjohn

@TedShifrin I might remember, but then I am quite dense.

@TedShifrin Noooo...

However, remember what I said about checking out...

leslie townes

i'll retire when i hit 100k too.

robjohn

But we need to hear about munchkin!

My wife really likes when I relay her stories

leslie townes

she's obsessed with spiders. she wakes up yelling in the middle of the night and says 'lets talk about spiders' when i walk in. this morning, she was going to day care while i was still in bed. she came in to say goodbye and i said: "LET'S TALK ABOUT SPIDERS!"

robjohn

Of course, munchkin might be in a rest home by then ;-)

leslie townes

19:33

she said "i'm leaving. i can't talk about spiders. we talk about spiders when i say 'let's talk about spiders.' you do that tonight, OK?" and left

robjohn

She is in control! Don't you forget it.

leslie townes

ted has suggested that this is potentially indicative of problems in the future. i don't know what he means.

robjohn

It's fun to hear these stories. I am sure glad I don't have to live them :-)

leslie townes

she also left what appears to be a permanent mark on the wall of our dining room after mashing one of those wall walker toys (their limbs are sticky and they can flip down walls) into the wall.

it wouldn't stick so she disassembled it to its stickiest rubber component and then hand-mashed it into the wall.

it left a little shadow.

some kid at her day care gave her that, i wouldn't have given her that.

robjohn

@leslietownes yeah, right...

. o O ( who me? )

leslie townes

19:37

she picked up the cat three times this morning. she still loves doing that. sometimes she adds in a bit of walking around with the cat. the cat inexplicably tolerates it.

the same cat who ran up on me and bit my leg while i was going up the stairs for no reason other than the thrill of the hunt.

robjohn

@leslietownes My parents had a cat that would be so friendly and let you scratch her until she flipped, then she would insert all claws and as many teeth as possible into the closest arm. This all ended one day when I flung her across the room and chased her down the hall throwing my mother's slippers at her. She hid under my parent's bed for a week, but never flipped out again.

she was still friendly, however.

Psycho kitty, ques que se

copper.hat

20:23

@TedShifrin I'll take your jump suit if you don't need it :-)

forest

22:50

Is $\forall k,m:E_k(m) \ne E_{f(k)}^{-1}(m)$ the right way to say "there is no $k$ which would cause $E_k$ to be equivalent to $E^{-1}_k$ for every $m$"? Here $E$ is a block cipher, $E^{-1}$ is its inverse, $k$ is a key, and $f(k)$ is just a function that changes $k$.

robjohn

23:12

I would say "for any $m$"

Novice

I proved (at least I think I proved, with some help) a result stating that if $F$ is a continuous cumulative distribution function on $(\mathbb R, \mathscr B, \mu_{\mathcal L})$ with distribution $\mu_F$, then $$\int_{\mathbb R} F(x) \, d\mu_F(x) = \frac{1}{2}.$$

I'm seeking the intuition for this result. My best guess is that this is very much related to the probability integral transform, which (informally) is a result stating that if a continuous random variable $X$ has CDF $F$, then $Y = F(X)$ has a standard uniform distribution, and implicitly therefore $\text E(Y) = 1/2$.

forest

That's what I meant. But how would I say that correctly with MathJax?

I don't want to assume that E_k and E^-1_f(k) are always not equivalent, just that they are no more equivalent than would be expected by random chance.

robjohn

What you wrote is what I was interpreting

forest

Since E and E^-1 are permutation families.

robjohn

@forest that is completely different.

That will be a lot more complicated.

forest

23:15

Hm. The first way I did it was $\forall k:f(k) \ne k$, but that doesn't imply that $f(k)$ can't turn $E^{-1}$ into $E$.

robjohn

@Novice If you're actually getting $\int_{\mathbb{R}}F(x)\,\mathrm{d}F(x)$, then it is just $\frac12F(\infty)^2-\frac12F(-\infty)^2=\frac12$

forest

(Context: crypto.stackexchange.com/q/99418/54184 and comment crypto.stackexchange.com/questions/99418/…)

Ted Shifrin

@forest "Equals" and "equivalent" are different notions. Be careful.

forest

So I want to say that $f$ cannot be a reverse key schedule.

@TedShifrin Equivalent as in \equiv? I know, I meant equivalent in the non-mathematical sense.

Novice

@robjohn Sorry, is this supposed to be the intuition? I see how the equation you wrote would be true, but I had to use Fubini's theorem to prove it, so it took quite a few lines.

forest

23:19

It's just that $\forall k:f(k) \ne k$ doesn't quite grasp the full meaning.

Novice

(In case I was unclear, what I'm trying to say is that I think I get the proof, but what I actually want is the illumination. I want to try to understand what the equation is trying to tell me, and as I mentioned, my best guess has to do with the probability integral transform.)

robjohn

@Novice I don't know what $\int_{\mathbb R} F(x) \, d\mu_F(x)$ would mean conceptually.

so I can't give intuition on something that I don't understand

it's like integrating the cumulative distribution against the density. I don't know what that would represent.

Novice

Fair enough. I'm just learning this stuff and the notation is pretty overwhelming at times. Between measure theory and probability theory there can be like four different ways of writing the same thing sometimes.

robjohn

math/probability/stats can each have different ways of saying things.

It would be the expected value of the cumulative distribution, which is $\frac12$

Novice

Right. That's how I see it. And the weird fact is that the CDF is a standard uniform random variable, given certain assumptions.

(Incidentally, using this theory, you can draw numbers from a uniform distribution and use them to simulate draws from other, more complicated distributions.)

robjohn

23:37

yes. With the proper function, you can simulate any distribution from a uniform distribution

Novice

Okay, well, thanks. I'll keep thinking about it.

robjohn

@Novice Let me know if you come up with any better intuition. It would be interesting.

Novice

23:56

You mean better than thinking of it as the expected value of the CDF? The chance of me finding a better interpretation seems pretty low. Maybe I'll ask my professor.

Transcript for

$Mathematics$ Mathematics