Mathematics - 2021-06-07 (page 1 of 3)

copper.hat

01:12

If $B$ is a matrix and I write ${\cal R} (I-B)$, surely it is clear that I mean the range space?

Ted Shifrin

Except that many of us teaching low-level linear algebra write $R(A)$ for the row space of $A$.

We write $C(A)$ for column space. Range (i.e., image) is more common for operators.

copper.hat

Thanks! I'm behind the times.

I presumed that someone who is familiar with proximal operators would understand.

Thorgott

clearly $\mathcal{R}$ means the resolvent

copper.hat

That's ${\cal R}$ight!

I'm just g${\cal R}$umpy.

leslie townes

01:47

$\mathcal{R}$ahhh

leslie townes

02:08

it stands for $\mathcal{R}$ithmatic

Russian Bot Who Knows Your IP

02:57

@PM2Ring so what man? People think i am an idiot online? I don't care. Does that affect you and me? No man, no

@PM2Ring if you hate the fact that i never talk about math or that i am silly and immature (these facts are indeed true), ignore me!

lmfao

leslie townes

math.stackexchange.com/questions/4165520/… my gut is almost certainly no but almost certainly i have no idea of a proof.

robjohn

There are some ways to tell if there is an antiderivative, but I don't think there is a way to tell if there is a closed form of an definite integral.

Larry

Can anyone help me solve this? $\sum\limits_{k=1}^n (\frac{\epsilon}{2n})^2$

Russian Bot Who Knows Your IP

I remember ms dos

copper.hat

03:15

that used to be miss dos

robjohn

@Larry Use the formula for the sum of a geometric series

copper.hat

how is that a geometric series?

leslie townes

i was expecting gamma to pop out of that. and zeta stuff. and a dilogarithm.

copper.hat

sounds like a marvel movie

Larry

Sorry i meant k instead of n in the sum

copper.hat

03:20

quasi convex man strikes again

leslie townes

my cat always begins the evening by curling up in my daughter's bed. sometimes they coexist, sometimes they don't. today my daughter said "NO! you have your own bed in the living room." i had to break it to her that any surface that can function as a cat bed is a cat bed.

i hadn't even noticed the n, k issue. my mind turned it into a k.

robjohn

@copper.hat I thought the exponent was $k$, sorry

copper.hat

i did too initially

the mind projects onto known solutions

wow, gil strang started talking to me

leslie townes

uh oh.

copper.hat

not sure why youtube chose that moment to play

leslie townes

03:23

is he on the site?

oh.

copper.hat

he's in the house

leslie townes

the phone call is coming from upstairs.

copper.hat

the matlab is in my phone

leslie townes

i do like that aspect of math.SE. sometimes you're in the weeds of some question and then the author of a relevant textbook pops up.

i asked a question once about three dimensional art galleries and who would pop up but the guy who wrote The Book on it.

the internet is not just for yelling at angry shut-ins.

robjohn

@Larry Okay, there are special functions for that, but the limit as $n\to\infty$ is $\frac{\epsilon^2}4\frac{\pi^2}6$

copper.hat

03:26

i am getting ads for snap shackles and expensive watches

leslie townes

snap shackles?

copper.hat

and linear algebra books it seems

Larry

Hmm I only need sum up to n

copper.hat

for attaching a strap to a loop.

robjohn

There are also asymptotic approximations for large $n$

copper.hat

03:27

anyone have contacts on the us embassy in london by any chance?

leslie townes

oh. not what i'd expected. i was thinking some kind of crowd control mechanism at a protest. the lead cop says, snap shackle the lot.

no contacts here.

copper.hat

it is not nefarious

robjohn

haven't contacted the embassy in London for a while now.

leslie townes

not since my days as a spy for east germany.

robjohn

shhh

I know nothink!

copper.hat

03:29

the last time i was in a us embassy was in 95 i think when i got my green card

back then you had to leave the country to get a green card

leslie townes

they used to be on grosvenor square, right? you could walk right in. then they moved it because of post early 21st century security concerns.

copper.hat

nine elms now

leslie townes

the russians poisoned that guy just across the street. fun neighborhood.

carol reed filmed a movie 'fallen idol' near there, lots of good photography of mid century london.

copper.hat

my daughter can return to the us on an expired passport but she cannot leave.

despite having a valid non us passport

leslie townes

welcome to the hotel california

copper.hat

03:32

probably she won't come home

robjohn

you can check out any time you want

copper.hat

i'll find a way :-)

robjohn

@copper.hat why can't she leave?

copper.hat

4 wks for expedited!!! wtf

a us citizen needs a valid us passport to leave by air it seems

robjohn

that's weird, if they have a valid foreign passport...

leslie townes

03:35

one time traveling to london i was asked where my hotel was, and i'd stupidly left that stuff in my cheked baggage. phones weren't up to date yet. so i said somehting like "123 granny smith apple way, london SW1." the guy didn't even blink

robjohn

I don't know all the rules, so I won't comment

copper.hat

Section 215 of the Immigration and Nationality Act (8 U.S.C. 1185)

the interweb ruined everything

Russian Bot Who Knows Your IP

Nae

love_sodam

@robjohn math.stackexchange.com/a/1216404/668308

For the case $|a|<1$

the modulus of the roots are both 1

leslie townes

one time traveling to copenhagen the immigration officer asked nothing although i had a gym bag full of wine that i probably should have declared. he said "you're from california?" i said yes. he said "it is not like california here." he was referencing the weather, which was cold as f---.

love_sodam

03:41

I wonder if I can still use residue theorem because singularities are on the circle

leslie townes

they still let people smoke in 'smoking areas' of the airport, which were just, cordoned off seats. there was nothing to prevent gaseous exchange between the smoking area and the non-smoking area. i hope they've fixed that.

copper.hat

wow, i can't even schedule an emergency appointment for myself in sf.

why are we paying taxes

leslie townes

i have an unprintable answer to that question.

:D

copper.hat

well, if i win the lotto i'll take a private jet over to her

leslie townes

isn't that an exercise in ahlfors?

there's got to be alternative approaches somewhere on math.se, depending on your taste.

copper.hat

03:45

@love_sodam the answer is maybe.

i have a tinge of apoplexia as i try to contact our consular services.

i may have to call in a big favour :-(

leslie townes

i feel like i ought to have strings to pull in this arena, but i don't.

Ted Shifrin

I have never had strings.

copper.hat

i do, but the strings come with big piano sized weights attached

leslie townes

too much hanging around with roustabouts and neerdowells. how often do they do well? neer.

a distant relative was the first commissioner of the INS. but he died 80 years ago. sorry about that.

copper.hat

i can get to the speaker, but as you can imagine that is using a big bullet

much as i want to see my daughter...

leslie townes

03:52

my daughter has spent the last 60 minutes, which are suppoed to be bed time, 'reading' a book to her cat.

be careful what you wish for.

copper.hat

yep. i know.

i did get my hand slapped by the us 'ambassador' to belize once for going into guatemala. i pointed out that i was an irish citizen so she slapped my wife's hand instead.

leslie townes

there are two kinds of ambassadors, people selected for real purposes and people selected because they bought the $750,000 plate of dinner to meet the president-elect.

i can imagine who the ambassador to belize might be.

copper.hat

we met her on a boat ride, she was sitting beside me

leslie townes

i have a cool weaving on my office wall. it's from a country that doesn't exist anymore whom a relative represented in the league of nations, which also doesn't exist anymore. it shows people hunting a deer.

copper.hat

what country?

leslie townes

03:56

persia.

copper.hat

one of my best irish friends was born in an african country that no longer exists

ahh, the dulles brothers have a lot to answer for

leslie townes

one of my coauthors' dads was the ministry of finance for a country that existed for like five seconds after WWI.

lots of goofy stuff going on in the 20s.

copper.hat

wow, its amazing what memories i am dragging up here, stuff i had completely forgotten about

leslie townes

you fully intended to return to the driveway and complete the application of the sealant.

we know.

copper.hat

ah shure yes of course

just waiting for the weather to clear

leslie townes

04:01

i was thinking about buying an audubon painting but it turns out he was a huge racist and owned slaves so maybe i'm not doing that anymore.

copper.hat

no i lied. it was the wife of the 'ambassador'

i urinated on rhodes grave

leslie townes

beer is on me if we meet.

that's something i haven't done. i've appeared on a podcast, but i haven't urinated on a grave.

life goals.

copper.hat

i was repeatedly beaten by one of my irish teachers (a monk, no less) and have been looking for his grave for decades. when my daughter spent 4 mos in ireland due to covid she too went looking

we will find it.

i will do the honours

petty pointless revenge

leslie townes

i wanna wee on the grave of the priest who stopped talking to my mom after she had a divorce.

maybe i can complete this task by 2022

copper.hat

i have something similar. but i am not sure i could restrain myself if i encountered this particular priest.

not a story for here.

leslie townes

04:06

yeah, blegh. where are the PSQs.

you know, the really objectionable stuff.

copper.hat

this one math.stackexchange.com/questions/4164445/convex-function-for-inequality-proof is just taunting me

i need the 10 rep.

don't even get a mug at 250k now.

the t-shirt ripped when i was putting it on

leslie townes

i do wonder what's going on sometimes. you do think people could attempt something.

sometimes i see questions that would not be out of place in a graduate-level functional analysis course. they have obvious answers but they're not things you'd even ask if you weren't fairly well along in some kind of program.

it's weird to see it put on a Q&A site.

i guess it's covid. in grad school we would go to each others offices and ask our questions

copper.hat

i am always a bit amazed at the inconsistencies, i was pointing one out to Ted earlier, someone who is (nominally) familiar with proximal operators but does not know what ${\cal R} (B-I)$ ($B$ is a linear projection) is

clearly the ${\cal R}$ stands for rabid

leslie townes

i don't want to be generational about it but i suspect there's a generation of people where all they know is what they've googled in the last 5 minutes

copper.hat

i know they are working in nine elms

its wine time

leslie townes

04:15

math.stackexchange.com/questions/4165545/… is a duplicate although i can't find what it's a duplicate of. i think it's in fraleigh.

maybe it's not a duplicate. i spent a lot of time in my own head in semigroups instead of groups.

copper.hat

indolence really bugs me

the 7 deadly left their mark on me

leslie townes

it's a cute attribute in cats.

robjohn

@love_sodam Thanks. I fixed it, but needed to use the Cauchy Principal Value.

Sorry to interrupt the cemetery discussion

copper.hat

a grave disturbance

Russian Bot Who Knows Your IP

copper.hat

04:27

can certainly say that i have experienced that directly

Russian Bot Who Knows Your IP

yeah

copper.hat

:-). yep. i have not adapted to the modern world

tomorrow i got to battle with pep8 morons

Russian Bot Who Knows Your IP

lol

i am thinking of learning assembly

flat assembler of intel x86

copper.hat

it is fun, but mostly only useful for debugging or high performance stuff

Russian Bot Who Knows Your IP

yes

copper.hat

04:32

or embedded stuff, which i love

Russian Bot Who Knows Your IP

i will soon learn masm and/or nasm

copper.hat

you could do atmel mega stuff? that would work with arduino if you like that

Russian Bot Who Knows Your IP

i would take a look

copper.hat

i taught assembler back in the early 80's :-)

Russian Bot Who Knows Your IP

even assembly can produce graphics easily

copper.hat

04:36

of course, anything can be done with assembler

Russian Bot Who Knows Your IP

easily

copper.hat

but generally you use the right tool for the right job. generally assembler is slower to program and more error prone

robjohn

@love_sodam I had actually mentioned the Cauchy Principal Value in the original answer. I don't know why I had that part about "both roots inside or both roots outside of the unit circle".

I don't think the fact that it was written on April first had anything to do with it.

copper.hat

:-)

Russian Bot Who Knows Your IP

i love the fact that ms dos can run on 16 mb of ram

leslie townes

04:44

it can run on far less than that.

even the later versions.

Russian Bot Who Knows Your IP

is a screenshot of GM-NAA I/O available?

i want it

copper.hat

i used to develop rtoss for interconnected 8080 based intel single board computers.

Russian Bot Who Knows Your IP

nice

leslie townes

dos is barely an operating system. kind of a weird accident it was so prevalent and had so much stuff built on top of it. it makes more sense as something for embedded systems, which i guess it's still used for.

Russian Bot Who Knows Your IP

yup

but there were many nice games for it :D

which we can still play on dosbox

leslie townes

04:54

i have fond memories of dos gaming too.

copper.hat

i always thought that dos & ms would just go away

Russian Bot Who Knows Your IP

operating systems with no gui feel like hardcore

:)

leslie townes

i felt like a 'sold out' when i began using a gui in 2001.

before then, i kept it real. now look at me.

copper.hat

i am still a cli guy

Russian Bot Who Knows Your IP

arch linux :)

and Эльбрус-1

hehe russian

love_sodam

05:24

If I integral 1/(x+1)x^a possibly by residue theorem, if $a$ is not an integer then what can I do? $0$ is not a pole then

robjohn

05:34

@love_sodam Is this a contour integral, definite integral, or indefinite integral?

Ted Shifrin

I presume this has nothing to do with the previously linked answer.

And the $a$ has nothing to do with the $a$ in that problem.

Oh well. No response.

robjohn

The last response took a while.

uhoh

06:00

leslie townes

had a very funny moment on the baby monitor. our daughter, who had been put to bed, was 'reading' one of her books in the dark to the cat. the cat eventually shook her head and left.

uhoh

math.stackexchange.com/posts/4163635/timeline

16

Q: Density and dimensionality of zeros in inverse square force fields of randomly distributed sources in (at least) 1, 2 and 3 dimensions?

Background: In this answer to Are there places in the Universe without gravity? in Astronomy SE I did a quick finite 2D calculation for 20 random sources to see if there was at least one zero, and without rigor convinced myself that there might always be some finite density of zeros. See image be...

why would my question have been audited as a "first post"?

leslie townes

i do not know.

Russian Bot Who Knows Your IP

uh oh

uhoh

okay if nothing is forthcoming I'll post in meta, but it is probably has some ultimately uninteresting explanation.

leslie townes

06:04

meta seems like it might be the place. we do not know as much about mechanics here. at least, i don't, and i'm here and not there.

shrug.

Permutator

06:15

@robjohn Thanks I figured out the integrals proof!

@robjohn By the way, I need your help in finding a closed form for a summand, could you please help me out?

The sum is $\sum_{q = 1}^{\infty}\dfrac{z}{z - q^3}$. I tried to evaluate it on Wolfram and Wolfram seems to give out a very bad and tedious closed form. I'm quite wondering if we can find a nice closed form (maybe in terms of coth and other functions) for the summand? Thanks!

@robjohn Please let me know what you think about it. Thanks!

@leslietownes How you doin today huh?

love_sodam

06:52

@robjohn 0 to infinity

Martin Sleziak

07:47

@Jam I see that you have created Representation theory room, but it has no description and so far messages. I was wondering whether you have some plans with that room - or maybe it was created simply by mistake...?

robjohn

09:22

352

Q: What are review tests (audits) and how do they work?

I recently received this rather amusing message while reviewing a user's first post: Congratulations! This was only a test, designed to make sure you were paying attention. This post has already been removed, but thanks for taking time to leave feedback for the author. What is the purpose of t...

support faq review review-audits

@love_sodam so you want $\displaystyle\int_0^\infty\frac{\mathrm{d}x}{(x+1)x^a}$?

you will need $0\lt a\lt1$

substitute $x\mapsto x^{\frac1{1-a}}$

MphLee

09:55

Hi, does someone know a nice category that can capture this ideal picture of a "category of smooth paths"? Ignore my attempted construction if irrelevant.
https://math.stackexchange.com/questions/4163524/name-of-a-category-of-smooth-paths-between-points-enriched-with-differential-i

An user gave me a quite obscure hint about considering paths over an infinite product of iterated tangent spaces. Yet, even if correct, I'm not able to work out the details and "obvious conditions" that are not abvous at all to me.
Thanks in advance to everyone.

Thorgott

I think what you describe is the same thing as what Zhen addresses in the comment and the right notion, at least on points

Lucas Henrique

hey chat

Thorgott

the objects should be points on the manifold equipped with all higher order infinitesimal data describing the initial/final velocities

this ensures smooth composability

the issue is non-associativity of composition, and I'm not sure what the best way of fixing this is

I was going to suggest to restrict to piecewise-linear reparametrizations, but I just realized that doesn't quite work either

I feel like this should be phrased in terms of the infinite jet bundle

Omni man

10:20

what is application of cardioid

y is cardioid is geh?

Balarka Sen

@Thorgott its an infinity category

Thorgott

not sure if joking or not, but probably true in either case

Balarka Sen

exactly

i mean thats the case for just the continuous path case as well

there is no category of continuous paths, only an infinity category, the infinity-fundamental groupoid

Thorgott

why aren't they a 2-category

Balarka Sen

homotopies dont associate

only upto homotopies of homotopies

Thorgott

10:26

2-morphisms homotopy classes of homotopies

Balarka Sen

then it works but thats truncation of the big thing, might as well truncate at level 1

Thorgott

but I guess that's just a trunctuation

Balarka Sen

depends on what you want

Thorgott

right

Lucas Henrique

@Thorgott lol

Thorgott

10:27

I get your point, we should take homotopies up to arbitrary order

Balarka Sen

infinity category of infinity smooth paths

boom

Thorgott

but I don't think we want them up to homotopy in this case

MphLee

Thanks to everyone.
@Thorgott But I'm unable to understand it. In fact a path on that product would just define a sequence of paths but to esnure theire one the derivative of the previous maybe Zhen Lin is actually taking some kind of limit/colimit of that secuence? Also Zhen Lin suggests to consider not paths over that product but pair of paths.

Thorgott

up to homotopy, smoothness means nothing

MphLee

That puzzles me more than it helps.

love_sodam

10:28

Can I ask some MATLAB question here?

Srijan M.T

The first equation is ΔH = ΔU + Δ(PV)
How is product rule hereΔ(PV) = PΔV + VΔP ?

Balarka Sen

@Thorgott smooth homotopies

Thorgott

smooth paths are smoothly homotopic iff continuously homotopic

MphLee

Ye I don't want the paths up to homotopy because doing that I lose the "shapes" of the paths. And I need to "see" initial and final velocities.

Thorgott

but I guess your point is that the space of smooth homotopies (up to various higher homotopies) carries more structure?

Balarka Sen

10:29

exactly

theres no canonical homotopy

Thorgott

anyway, yeah, up to homotopy you lose velocities

Balarka Sen

it matters if you dont truncate

Thorgott

so I agree this is not what we want

also, I don't think Zhen meant paths in the product

MphLee

In fact I want to take paths up to parametrization.

Balarka Sen

composition of two smooth paths have no canonical smooth parametrization

Thorgott

10:31

you also don't have well-defined velocities if you allow reparametrization

MphLee

"want".. It's an attempt to obtain associativity.

Balarka Sen

yeah good luck

you wont get a category

the natural object is an infinity category, of holonomic sections over cubes of the infinite jet bundle

Thorgott

wtf are holonomic sections

MphLee

Can you expand a little on this?

Balarka Sen

no i think i pass its too much work to set it up properly

MphLee

10:32

How that is related (I'm sure it is, but idk what holonomic sections, cubes of infty jet bundles are.

Balarka Sen

@Thorgott a section of the jet bundle which is actually jet prolongation of a section

MphLee

At least a reference, and why you think it is related with my problem

Well, thank you anyways

Thorgott

ok, so you want an infty-category and the n-morphisms are sections along [0,1]^n->M that are prolongations?

Balarka Sen

yeah

Thorgott

too complicated for me, but seems like a good notion

Balarka Sen

10:35

lol yeah its a meme

i would rather not try to construct it

MphLee

0.0

Thorgott

I mean, I like those memes, I just don't know infty-categories

Balarka Sen

@MphLee Here's an easier thought. What is the category of $C^0$-paths in a topological space $X$?

MphLee

Idk. It is not a category because we have not associativity.
If we mod it out by homotopy we get $\Pi_0(X)$

rogerroger

Let $f$ be a probability density function on the real line $\mathbb{R}$, is there a way to bound the following integral from below (for any small enough $h$) ? $$ \int_{\mathbb{R}\times\mathbb{R}} \Big(\frac{|x|^2}{h} + f(y) \log|x-y| \Big)f(x) ~dx dy $$

Balarka Sen

10:37

Exactly, my point is there is no such category

The natural object is a simplicial set, do you know what those are?

MphLee

Nope. Heard of it but can't undestand.

NNO is a simplicial set?

Balarka Sen

What is NNO?

MphLee

A natural Number Object.

Balarka Sen

God, I have no clue, I am not familiar with that at all.

MphLee

@BalarkaSen Ah... I see xD... Sorry I misunderstood.

Balarka Sen

10:39

But have a look here

Lurie constructs in the first page something called $\mathrm{Sing}(X)$

This is the right notion of the "category of paths on $X$", except it's a simplicial set/infinity-category

MphLee

Ok, I'm reading.

Thorgott

you have associativity up to homotopy, which are themselves not associative, but only associativie up to homotopy of homotopies, which are themselves not associative... ad infinitum

instead of associativity, you just prolong by higher order natural transformations

Balarka Sen

That's the intuitive phrasing, but impossible to set it up properly because the k-fold associativity upto k+1-fold associativity diagrams get massive as k increases

So a better phrasing is through simplicial sets

An \infty-category is just a simplicial set which is "weakly fibrant"

Thorgott

there's an idealized associativity living at infinity

A-Level Student

Hello everyone. I'm 17, and I've written 2 papers on quite separate topics in maths. I believe the results to be new, but I am seeking a professional opinion about them.
Would any mathematician be willing to have a look at at least 1 of them and comment on it within about a week or two?
They aren't very complicated, so going through them shouldn't take too long.

6

Thorgott

10:43

it's like an infinite sequence of derived functors prolonging non-exactness

Lucas Henrique

@Thorgott i'm pretty sure that can be found in a poetry book

Thorgott

I'm on the same level of bs, so sounds about right

Balarka Sen

In fact a simplicial set can be made out of maps from smooth simplices to the target manifold, so I feel that is just what you want

A-Level Student

@Permutator It seems unlikely that there's a closed form, as suppose you did find a closed form $f(z)$. Then divide both sides by $z$ and take the limit as $z$ approaches $0$ and you'd have found the value of $\zeta(3)$, which seems pretty unlikely :)

MphLee

OK, I was aware of n-cats. I was aware of the infinity groupoid of homotopies up to homotopies up to...That is visually very vivid in my mind.
But the simplicial set version is still obscure to me.
Thanks @BalarkaSen I'll look more int this.

Obviously if we want to retain the individual shapes and slopes of the paths we can't mod by homotopy thus, you claim, we will never get a category.

Even if we mod out by reparametrization of paths, as noted, we lose information of higher derivatives. We just retain the set theoretic image of the paths.

Balarka Sen

10:47

That's about right.

MphLee

Very well. That was fruitful I believe.

Thx

Thorgott

the infty category will not lose the differential information

it simply is a more complicated object

though it's more or less just a way of making what we want tautological

MphLee

Oh ... wait. Right because a loop and a point even if homotopic we now know how are homotopic? Is that your point? That homotopy remembers something about the derivative?

Balarka Sen

Yes, if you do a smooth homotopy.

It's a homotopy of the derivatives too

Thorgott

I mean, the 1-morphisms are literally just the smooth paths, no?

we're not actually taking homotopy classes in the infty-category

Koro

10:51

How come $U(2^n)$ be isomorphic to $\mathbb Z_2 \times \mathbb Z_{2^{n-2}}$?

MphLee

Well if we consider $C^\infty(I,C)$ a as a category an homotopy is a natural transformation thus components are themselves in $C^\infty(I,C)$

Lucas Henrique

@Koro what is $U(2^n)$ again?

Koro

It's the group of natural nos. relatively prime to $2^n$

Lucas Henrique

like $(\mathbb{Z}/(2^n)\mathbb{Z})^\times$, I suppose?

Leaky Nun

@Koro Theorem 5, Page 11, ramanujan.math.trinity.edu/rdaileda/teach/s18/m3341/ZnZ.pdf

Jam

10:58

@MartinSleziak i wanted to talk some exercises im having trouble with but dont know how to invite people to talk about it and dont know how to delete it

Koro

Don't be misled by how I have answered you question to what is U(2^n). It's called multiplication group of integers modulo n

@Lucas

Elements of this $U(n)$ are not really integers, rather equivalence class of integers coprime to n.

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