Mathematics - 2021-06-21 (page 2 of 2)

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8:00 PM

So if one is normal it is the only maximal $p$-subgroup.

@TedShifrin Ah I deleted that post as there was a mistake in it. I tried to avoid using the fact "all groups of order 15 are cyclic". Ok Ted, noted. I'll use $\langle a \rangle $ more now.

Good :)

Yorch, I'm afraid. I am not allowed to use Sylow's theorem. Background of the question: An exercise problem from Gallian's chapter 9

Point is I am yet to read Sylow's theorems, which will come later in the book.

@koro Your proof looks OK. You can omit "we must have $(xH)^{p^k}=H\implies o(xH)|p^k$ and so" as it's not needed.

Hey Ted. Remember that teams allocation question from a month or so ago, which you helped reopen? math.stackexchange.com/q/4100315/207316 It ended up going through 2 more close-reooen cycles. It then get bountied (which temporarily blocks closure), and finally attracted a couple of answers. Yorch (deservedly) won the bounty. :)

8:09 PM

Preview of coming attractions: the Sylow theorems will tell you that (1) since $H$ is a Sylow $p$-subgroup and is normal, it is the unique subgroup of order $p^n$, and (2) every $p$-subgroup is contained in a Sylow $p$-subgroup. These facts immediately imply that $K \leq H$.

Hi @Bungo: Thanks a lot :-). I can omit that but isn't that statement important because using this I use the hypothesis that $(p^k,m)=1$? Please let me know why I can omit that. Many thanks.

@Koro It would suffice to say "... by Lagrange's theorem: $o(xH)|m$. Also since $o(x)|p^k$, we must have $o(xH)|gcd (p^k, m)=1$ and therefore..."

@Bungo Ah I see you are refining the argument :-)

I don't understand then why answers here also math.stackexchange.com/questions/781189/… are complicating the solution

They are using the counting formula: $|HK|=\frac {|H| |K|}{|H\cap K|}$

@Koro They're just giving other, somewhat more sophisticated ways of looking at the problem. Nicky Hekster's group theory answers are generally excellent, so it's always worth trying to understand them. I often make elementary arguments using the counting formula as well - it often makes it immediately clear why the number of subgroups of a given order is constrained (there's no room for any more).

I tried running the search for 12 hours on my website's servers but I'm still far from finding one with 8 classes

8:20 PM

@Koro You should add your answer as it's different from the others and worthwhile in its own right.

although I found out my vps is faster than my ryzen 3600 when it comes to single thread performance

xlnx/amd is killing me today

what is that?

ryzen is manufactured by amd

@Bungo: In that case, $|H\cap K|$ will divide $p^k$ and $p^n$. Let $k<=n$ and so $|H\cap K| | p^k$ but then $p^k$ is to be compared with $m$ also

8:21 PM

and what is xlnx

I knew about the amd part :D

xilinx, fpga manufacturer

Akiva Weinberger

@Thorgott @AlessandroCodenotti Something that might interest you

@Bungo Okay. I'll add that.

Akiva Weinberger

There's a push among various people I follow on Twitter to rename the Hawaiian earring

to something like "harmonic earring"

are u talking about the stock exchange?

8:22 PM

yup

i like 'harmonic earring'

Akiva Weinberger

(by analogy with the harmonic archipelago, another space)

oh

Akiva Weinberger

(or perhaps "topologist's earring" by analogy with the topologist's sine curve)

stack exchange? @Yorch

8:23 PM

are u making a joke?

Akiva Weinberger

The reason is that there's no actual Hawaiian jewelry that looks like that; it seems the topological space was probably named by some non-Hawaiian mathematician who wanted to use the name "Hawaiian" to connote its being "exotic"

(no one can find out who actually named it)

looks more like an eyeball to me

I'm going to go to hawaii and start fabricating earrings shaped like that

Akiva Weinberger

So some Hawaiian mathematicians are pushing for a change

please tell me you are kidding

8:24 PM

wut?

send some fighting irish over there

@Yorch I meant this

If I was hawaiian I would try to get some grant money and start a hawaiian earing conference

i'd take a step back from the speculation that they wanted something 'exotic.' it does look like an earring that may have come out of some culture. they maybe just got the culture wrong.

what is wrong with exotic?

8:24 PM

why would they wanna get rid of something with their name on it?

i'll take a lambo any day

and yes they probably got the culture wrong because many of us are insensitive to the differences between a lot of cultures. but i doubt they were sitting there, what's the most exotic thing i can think of.

I need to start an aztec diamond symposium where I live

i like 'harmonic earring.' 'topologist's earring' isn't as suggestive of what it is.

Akiva Weinberger

I mean at least the Mesoamerican pyramids are a real thing (re: Aztec diamond)

8:26 PM

the search for irish even primes...

what could be more exotic than tripe and drisheen?

but pyramids aren't diamonds

Akiva Weinberger

@Yorch 'Cause it's made up by someone who knew nothing about Hawaii

as long as we're tweaking terminology, i always hated the term "freshman's dream" for (x+y)^k = x^k + y^k. it seems snooty to me.

Akiva Weinberger

@Yorch Add a reflection

it's honestly pretty offensive our pyramids are being compared to diamonds

the one that needs to reflect is u

Akiva Weinberger

8:28 PM

There's also the Wiphala I guess?

the idea of cultural ownership is a bit beyond me

but that's from the andes

and the wristies

Akiva Weinberger

ah true

Now ur comparing us to a region with a competitive football confederation

So the correct name is harmonic earring?

Akiva Weinberger

8:30 PM

I mean this whole discussion started like yesterday

is there money involved?

You can bet on the outcome unless u live in hawaii

i have vpn access from all over :-)

what's the notation for the non-negative integers?

$\mathbb Z^+$ is the usual one for the positive ones right?

Akiva Weinberger

I've seen $\mathbb N_{>0}$ for positive integers and $\mathbb N_{\ge 0}$ for nonnegative integers I think

8:35 PM

if you're using integers elsewhere, maybe $\mathbb{Z}_{\geq 0}$. if you're not using integers elsewhere, i like $\mathbb{N}_0$.

Akiva Weinberger

or replace $\mathbb N$ with $\mathbb Z$

It varies. I have seen $\mathbb{N}_0$.

Some folks think zero is natural. I mean really...

copper has been leafing through my dissertation again.

Give me zero icecreams please

done.

8:36 PM

totally natural

i am still reading the boilerplate on Leslie's dissertation

if you're reading the UC-mandated stuff that is probably in your dissertation, too, then yes

i waived copyright in a fit of ip pique

in my era you had to pay to do that. i'm not kidding.

or wait, it depends on what you mean by waived.

you had to pay to not give proquest a somewhat exclusive license.

so i hope they're happy with their IP.

the investors in proquest are driving around in solid gold rolls royces because of my work

there are many solid gold rolls royces out there as a result of my cumulative work, however none are mine...

my naïve perspective was that publicly funded work should be free :-)

i hate wandering evolving questions on mse

im in a foul mood today as you may have guessed

what's harmonic about it?

8:47 PM

i'm setting up my nihilist chat room

i see nothing harmonic in the earring

I don't want to go to office tomorrow

i have not been to the office since march 2020

I have not been to the office since March 2020 either, and I am now retired and will never go to the office again.

i retired about a decade and a half ago, but could not handle it so went back to work.

I'll retire probably after 35 years.

And our work from home is discontinued.

:'(

@Bungo Thanks a lot for help :-)

8:56 PM

Retirement failure! That's the stuff of my nightmares @copper.hat

@Koro My pleasure!

@Bungo i am needy and require the feedback of accomplishment (no matter now minor) to function.

plus i was driving my kids (now 17 & 20) mad

Sometimes I order cake in the middle of night and the delivery guy gives me killer looks

:'(

why killer looks?

He's angry that somebody annoyed by ordering in middle of the night

It happens a lot. I tend to order cake in middle of the night :-)

hmm, then he should not take the order.

why not order the cake during regular hours?

9:06 PM

It tastes better at night to me :-) I picked up this bad habit during lockdown

yeah, but order during regular hours, put it in the fridge and then enjoy at leisure without the delivery guy's annoyance

yeah, you are right

why would he be angry?

They work day and night so may be they get tired sometimes.

give him an extra tip?

9:09 PM

The delivery guy wants to be paid to not work? How do I sign up for that?

Oh, it's like a restaurant?

I don't understand

And when they are about to call the day, somebody places an order.

Imagine the situation... :'(

i think there is an easy solution :-)

I thought those apps allowed you to select when you are active

When I worked as a pizza delivery guy (many decades ago) I hated the downtime when not delivering anything, as it was boring standing around in the shop. Whereas driving around was fun.

But I was 18, so any driving was fun.

9:12 PM

@Yorch Some restaurants have their own delivery guys. Some restaurant deliver through a third party-a food delivery platform whose apps are available for download.

and which one do you use?

And the apps charge us some delivery fee and get the order delivered :-)

i have not had food delivered in decades...

it's great copper. you are lucky I think

because i stay away from family so I do it a lot

i can cook by watching videos online but then I end up spoiling recipes :'(

well, i mostly do the shopping & cooking, so i am not sure about the lucky part :-)

unfortunately my daughter is far away and my son likes to minimise parental contact

9:15 PM

whenever tomato is required to be added, consider the recipe to be spoiled!

tomato?

tomato is awesome

yeah

Speaking of Hawaii, there's a tiny enclave of the UK there. From en.wikipedia.org/wiki/…

copper, in Kolkata did you come across an orange that is the size of a volleyball

?

Somebody brought it for me from Kolkata, it was delicious!

it's like orange. I don't know what exactly it is called.

9:19 PM

@PM2Ring that is cool.

@Koro no, unfortunately. too many things to do & see

The Captain Cook Monument at Kealakekua Bay and about 25 square feet (2.3 m2) of land around it in Hawaii, United States, the place where James Cook was killed in 1779, is owned by the United Kingdom. [...] and is considered as sovereign non-embassy land owned by the British Embassy in Washington DC. ... the Hawaiian State Parks agency maintained that as sovereign British territory it was the responsibility of the UK to maintain the site."

I even planted its seeds at my home and it was growing but my cousin somehow broke the plant :'(

i love tomatoes

Yay, someone's randomly downvoting old questions of mine. I must have pissed someone off. (The perils of voting to close questions I guess.)

i don't like avocado :'(

9:21 PM

@rostader we can chat here if you want

i love avocados :-)

i do not like durian

I like both

durian?

it tastes neither sour nor sweet. It's like eating what...

yeah I like durian

although I like guanabana more

i never heard about durian :'(

9:22 PM

soursap?

Soursop (also graviola, guyabano, and in Hispanic America, guanábana) is the fruit of Annona muricata, a broadleaf, flowering, evergreen tree. The exact origin is unknown; it is native to the tropical regions of the Americas and the Caribbean and is widely propagated. It is in the same genus, Annona, as cherimoya and is in the Annonaceae family. Soursop is known as sirsak in Indonesia, where it is widely available and believed to have medicinal benefits. The soursop is adapted to areas of high humidity and relatively warm winters; temperatures below 5 °C (41 °F) will cause damage to leaves and...

Ah, durian.

Most famous Asian fruit.

i have had it in indonesia?

ohh you mean jackfruit @Yorch

?

I think you are talking about futi

9:23 PM

no jackfruit is different

jackfruit=durian?

#Koro

I am from Kolkata

Is it chewey?

I've only had marinated jackfruit that's been rehydrated

@rostader Oh thanks for letting me know that. Futi is what they call it?

@Koro no, durian is fairly large and has a putrescent smell.

9:24 PM

That Wikipedia Enclave & exclave page is hilarious. It has tons of examples of weird geometry caused by humans being territorial & bureaucratic.

Yeah...I dont know whats it said in English

no rostader: I just searched it online. Futi is different

in spain there's Gibraltar

europe has lots of little enclaves

futi is "muskemelon" @rostader ?

9:26 PM

okay... orange about the size of a volleyball

yeah.. that is what I want to know what it's called.

Yeah..

it is from west bengal. I am sure about that. But not sure if it's exactly from Kolkata

Was it citusy>

i'll ask my contacts there if they know it.

9:27 PM

citrusy?

@rostader: yeah I think that's it

cause it mught be batabi

BTW @coppe

yup?

it's orange colored from inside.

There's a West African fruit, known as the miracle berry which "causes sour foods (such as lemons and limes) subsequently consumed to taste sweet" en.wikipedia.org/wiki/Synsepalum_dulcificum But I've never tasted it myself.

9:29 PM

i have had the miracle berry.

The unconstrained part could still be unbounded right?

@rostader the answer to your question is that the $\sup$ is infinite.

For both problems right?

i do not know what you mean by both problems.

$\sup_{b \in \mathbb{R}^n}\inf_{x \in S}\|Ax - b\|_2$ is infinite if $S$ is bounded.

For any $A$.

I think they've mostly sorted it out now, but near the India-Pakistan border there were nested enclaves several layers deep. Wikipedia has the details (but it's been a few years since I fully read that article).

9:32 PM

and instead of $S$ if it was entire RN

@PM2Ring you want to see caves?

ancient ones

@rostader you are confusing me now.

you said $S$ was compact.

$\sup_{b \in \mathbb{R}^n}\inf_{x \in Rn}\|Ax - b\|_2$

Is this finite?

I also answered that. If $A$ is surjective the answer is zero otherwise it is infinite.

Yes exactly

9:34 PM

so what are you asking?

A is not surjective

@PM2Ring en.wikipedia.org/wiki/Elephanta_Caves... I really enjoyed the journey to here by boat.

then the answer is infinite.

But I am assuming it to be finite

Its an assumption

@Koro Caves make things more complicated, and it kind of makes sense that there will be territorial disputes when the only entrances to a cave system are on one side of a border, but the cave system extends well across the other side.

9:35 PM

@rostader i cannot help sorry.

Now I am introducing $S$ and claiming that to be finite based on the assunption

Hey thanks anyway for your time

Its been a pleasue

You cannot make an assumption that is not valid.

One day I want to visit Antarctica and see whales and penguins from distance

@Koro Nice, but I don't know what that's got to do with enclaves or exclaves.

i have swum with penguins and had grey whales blow snot all over me

if that sounds good :-)

9:37 PM

@PM2Ring: There's been a misunderstanding on my part. I didn't know what you were looking for :'(

Yeah.. its part of a bigger problem. $b$ is the output of an operator. I can assume that operator to have bounded outputs. Its just that I cant put a definite bound on it

@Koro That's ok. :) I like learning about caves, and I know a bit more than the average non-Indian about Hindu culture & religion. :) I've never been to India, but I'd definitely visit lots of sacred places if I did go there. I guess there are still some sacred caves that it might be tricky for me to visit...

@rostader it is a different problem if $b$ is bounded.

I used to visit places a lot but due to corona, visits stopped. :'(

Yeah.. I can thus assume that previous assumption by the operator argument

9:43 PM

@rostader What sort of operator? Certainly not linear. At any rate, then you're writing nonsense, because you wrote $\sup_{b\in\Bbb R^n}$. That's not what you're doing, clearly.

Otherwise I had to define Bellman Operators

And in the process loose the essence of the question

But your question contradicts your so-called hypothesis.

How so??

@Yorch Thanks for doing that. I strongly suspect that there are no 8 round solutions, but there's no harm looking.

So you can't write your question and then say it's a hypothesis that $\|b\|$ is given to be bounded.

9:45 PM

I didnt understand

Read what I wrote above.

My assumption was \sup b \in RN \inf x \in Rn \|Ax -b\|_2 was bounded

Hello everyone,

I have asked this question https://www.mathworks.com/matlabcentral/answers/861835-find-the-normal-vector-at-a-specific-point-on-a-3d-surface-defined-by-an-equation?s_tid=srchtitle on Matlab answers on how to find the normal vector at a specific point on a 3D surface defined by an equation (i.e. z = f(x, y)).

I would be grateful if anyone could help me figure it out numerically using Matlab.

I have very little knowledge of this math topic, so I hope whoever gives help explains it in the most naive way :)

I infered from that this would necesarily mean b is bounded

If you're taking the sup over all $b\in\Bbb R^n$ that just cannot make any sense.

9:48 PM

Otherwise since A is not surjective... the assumption contradicts

So it's all contradictory.

If you want to define a set $S$ and say $\sup_{b\in S}$ ... then I'll listen.

You're talking about the error in the least-squares solution of a linear equation, of course, but that error is bounded only if you bound the distance of $b$ from the image of $A$.

right but if I write.. sup_\theta\in Rn \inf x \|Ax - T(\theta)\|_2 where T:\Rd \to \Rn is a bounded operator then I guess it would make much better sense

Exactly

I understood what you mean

and if $b$ is bounded say in K \sup_b\in K \sup x\in S\|Ax-b| would ofcourse make sense

Saying "bounded operator" is not what we understand that to mean. To us, this means a continuous linear operator.

It isn't so much compactness of $K$ that you care about as its normal distance from the image of $A$.

Ahh yes bounded preimages to bounded images

You're doing finite-dimensional stuff, so I truly do not understand what you're talking about. Your $T$ is presumably non-linear.

9:54 PM

IS there a nomenclature for T{\theta} \le \infty \forall \theta in \Rn

Yes T is non linear

OK, so, first of all, you need to use language to make it clear that this is not linear. Why do you even call it an operator? It's just some function from one vector space to another.

Well technically you can define it on infinite dim vector spaces with countable basis

I would just say that your vectors $b$ live in a bounded subset $S$ of $\Bbb R^n$.

Then I would need to define S now wouldnt I>

And defining S is tricky

Well, you've lost me. I have no idea anymore what you were even asking us.

9:57 PM

I know b is bounded but I cannot possibly fathom to analyze its bounds

Anyway you all helped me a lot

thanks a lot

If you don't know the bounds on the distance of $b$ from the image of $A$, I don't see how you can compute the thing you wrote down in the first place.

That's what's relevant.

I think that's what you're actually asking for.

Good luck :)

Thaks joing stack was one of the best decisions I had ever taken

LOL, I'm glad we helped.

i'm still confused :-)

not that that takes much effort

Rightly so, @copper.hat. It was nothing if not self-contradictory and confusing.

10:11 PM

@TedShifrin thanks for the sanity check!

What I believe is that rostader wants the maximum distance from the image of $A$ of the image of his non-linear operator. Who knows how that is determined, if it is.

Certainly it wasn't sorted out clearly in his own mind when we started.

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