will do

@BalarkaSen there's an MO post to a question similar in spirit with a bunch of topologists telling OP to isotope it .-.

HYPERFUNCTIONS!

Hyperfunctions are fun

Too bad I don’t know how to use them

nLab’s page on them is pretty good: ncatlab.org/nlab/show/hyperfunction

My advisor suggested I do some reading on hyperfunctions; I have a potential application, maybe

I'm not entirely convinced, but neither Fubini nor Tonelli is saving my ass, so hyperfunctions!

that being said, there is nothing that I have ever read on ncatlab that I found illuminating---it all goes way over my head

folk there seem obsessed with abstractifying things as much as is humanly possible :\

That's a good rule of thumb, yes.

But like all rules of thumb, it has exceptions and that's one of them imo.

The place where I first read about them was in Penrose's big book on math/physics, mind, and that's pretty readable

Penrose is generally pretty readable

I'm currently waiting for a couple of books to get to the library so that I can read up on them

Also, my comments are entirely with regards to that specific page on 1D hyperfunctions

I have tried and failed to read the page on multivariable hyperfunctions

heh

if you know sheaf theory I suspect you'll have an easier time than I did

the ncatlab page is actually not terrible; but the parts that I get are things that I have already understood

god

and sheaf theory is something that I have seen, but I won't claim to grok it

fair

why can't I be in a better field

sheaves are a lovely idea, and I need to get my hands dirty with them more

"geodesic ball" is so vague

What sort of applications are you hoping for?

I have a nasty iterated summation dohicky that seems to have a natural boundary of holomorphicity

ahh

nice

but there seems to be meaningful geometric data beyond the natural bounary

gotcha

the place where I saw them showing up at one point was where there was a function with n poles spread out over the unit circle

essentially, it is a Dirichlet-type series where the abscissa of holomorphicity is at a natural boundary

that is exactly what is going on---"poles" are accumulating on the boundary of the half-plane

ahah

maybe

it's in this review paper: epubs.siam.org/doi/pdf/10.1137/130932132

there are several reasons to think that there is a dense set of "poles" on the boundary, though we can't actually prove it

see footnote 10 on page 425 in that SIAM article

ooo... some of that paper looks interesting

ty

yeah, it's cool stuff

and the reference in the footnote is Graf

which is one of the books that I am waiting on

huzzah!

yup

I should see if they've written anything more recent than that

I think somewhere else in the paper they say they were aiming to do something more generic with hyperfunctions

oh, hey! the ncatlab page has a link to the complete pdf of Graf's book :|

that's... interesting

looool

no comment

heh

right, see the discussion at the end of this paper: people.maths.ox.ac.uk/trefethen/publication/PDF/2014_150.pdf

these are tasty---thank you

hey there!

HAY IS FOR HORSES!

is MAA worth joining?

student chapter that is

Sure, why not?

Where are you, career wise?

dunno. hehe

currently in undergrad

that is an answer---you are an undergrad

the MAA is a great organization to be involved with as an undergrad

they have several very accessible publications

well my actual career prospects arent well defined yet. still a lot of time

and opportunities for undergrads to publish and present

hmm ok. thanks @Xander!

Cities A and B are connected by a road. At sunrise of a certain day, Jon starts a motorcycle trip from A to B and at the same time Sam goes out on his motorcycle from B to A. Each person travels at constant speed and they meet at noon. If Jon arrives to B at 4 p.m. and Sam arrives to A at 9 p.m. What time did the sun rise that day?

if you are considering grad school, the MAA might be worth joining

help me please :(

and it is cheap!

I don't know what's the equation I have to solve

well my school's MAA has math jeopardy before each meeting so thats reason enough, come to think :p

that sounds like fun!

Cal Poly Pomona does an integration bee that is, I think, partly funded by the MAA

I'm finally quitting my shitty fast food job and want to get more involved with school stuff

@XanderHenderson oh that sounds scary fun. always hosted at cpp?

it is for the students of CPP

and the faculty

it is usually students vs faculty, if I understand correctly

are you in California? you seemed to pick up on the CPP acronym pretty quickly...

@XanderHenderson Georgia

ah, well, nevermind then

hi peoples

@ForeverMozart yo

long time no see

yes because of super maths

i have reached. supernova.

is that like a $\Bbb Z_2$-graded nova?

like topology. in space.

hey

you know when you use an integration factor to make a differential equation exact

Do you need to include the integration constant that woud be present in the integrating factor

greatst video ever youtube.com/watch?v=j4hW7AwETZA

lol @ForeverMozart that one is a classic

?

yeah I only found it a few days ago

"extreme integration" lmao

I like the loopty loop equals sign

:P

Is there always an edge of a 2-connected graph that can be contracted to keep it 2-connected? Excepting the triangle graph.

Guys could anyone just please explain me a question on probability ? teachoo.com/4147/768/…

i thought he was saying the integral was equal to the integrand... and worked it out

to see that it was false of course

@0celo7 Hey!

Ah nvm i got it !

hey

In 3 dimensional Euclidean space, how many vectors do you need/how many vectors do you need to keep track of to properly (without ambiguity, I guess) define a linear transformation from that 3-d space to a number line ?

@HsMjstyMstdn You need to know how a basis transforms

so you'd need three ?

yes

Hmm

any 3 linearly independent ones will do

isn't there a way you could use 2

no, because I could choose a third one orthogonal to those and change the map on that one

oh, so I'm encoding the handedness of the transform in my operator ?

I'm in the library so you have to tell me what they're saying

He's saying you can basically define a 3d to 1d linear transformation by using two vectors, v and w and finding their dual vector

@HsMjstyMstdn no

and then showing that is dual vector is v cross w

a linear transformation from R^3 to R is a row vector

so depending on how you're viewing things, you might only need one vector

@0celo7 okay Imma think about this

But generally speaking, a linear transformation is determined entirely by its action on a basis.

@0celo7 yeah

B/c that gives you the matrix representation

@XanderHenderson @EricSilva greetings

greetings, but I am not really here

also, how do you feel about an $\mathbb{R}$-graded group?

eh?

@XanderHenderson The Sobolev spaces $\oplus H^s$ are an abelian R-graded group...

yuck... what do you do with those spaces? and is there a Künneth formula there?

you use them for everything, but certainly not as a graded group lol

but that was the only example I could come up with

heh

they're just abelian groups (vector spaces) and have an index $s\in R$

I mean, if you aren't using the graded group structure, then why bother mentioning that it is a graded group?

@XanderHenderson it was literally the only thing I could think of!

ha!

@XanderHenderson do you have a better example in mind

well, homology groups are (typically) $\mathbb{Z}$-graded

oh god, are there fractal homology groups with an $\Bbb R$-grading?

and cohomology groups are graded (typically) by some discrete group (often $\mathbb{Z})$

but my advisor is trying to develop a fractal cohomology theory

with grading in $\mathbb{R}$

and the groups being additive subsets of $\mathbb{C}$

gross.

yes, gross, but, if true, a significant step towards Riemann (he says)

that sounds like something in chapter 4 of Federer

but worse

ha!

I had a conversation with a Riemannian geometer this week about Federer

hello all

he recommended Morgan(?) as a good companion

For learning, perhaps

And maybe for understanding some of the theorems and notation

everyone having a great day?

But Federer has a legitimately incredible amount of information in it

@hungryWolf hi

@hungryWolf meh

that was his point---Federer is complete, but not pedagogical

if you know the material, Federer is a good reference

but Morgan helps you to get through Federer

hello @0celo7

Federer is a good "proof by intimidation" tool.

This theorem is true because of 4.5.21 in Federer

What is this federer ?

@hungryWolf a bad, bad book

@0celo7 haha

I have some material I was reading about in linear algebra course, I did not understand it well. Can I discus that material here?

@hungryWolf Don't ask to ask; just ask.

math.cornell.edu/~mec/2008-2009/ABjorndahl/ppmi.pdf

This is the link of material

are you at Cornell?

or just reading their stuff?

on page 14

no I am not a student

I like to learn from the internet

@XanderHenderson The problem with Federer is that the statements are too general. Like, everything in chapter 5 is about "parametric elliptic integrands" but you really just need it for the area functional. So you have to decipher his notation before making sense of any of it

A reference is ideally self-contained per section

Federer needs to be read all the way through

Which is impossible

on page 14 there is this university B professors example. I've read it, it says that all of the professors will resign(you'll understand if you read the paragraphs inside that box)

@hungryWolf All math is welcome. Remember; don't ask to ask, just ask. If the question's alright, people might comment

@HsMjstyMstdn thanks; from now on I'll not ask to ask

I disagree with the material

Why do you disagree? The reasoning presented is pretty solid...

it says at case n = 2, the two professors will look at each other

professor x said that atleast one of you has an error in your publications

and others know it

that means he has implied the author himself is not aware of it

yes...

@CausingUnderflowsEverywhere It's not that hard, think about inverses and opposites of processes

Maybe they are staring because they do not know about their errors

Iow, what is 0.5 doing ?

if they do not know about their errors how will they just resign?

that's the point---as soon as a prof knows about an error in their own work, they have to resign

so, A knows that there is an error in B's work, and B knows that there is an error in A's work

each shows up to the meeting thinking "I'm okay, because the other guy done fucked up"

but when they both show up, and neither resigns, they can immediately reason that the other is aware of an error in their own work

now that they each know that their work is flawed, they mush both resign

It's confusing because wouldnt it be $f(2x)$ yet, $f(2x)$ is horizontal compression by factor 1/2

as there is atleast one person with error

one of them must be it and both resign?

but both of them know for sure that the other guy has an error

at least one person has an error

yes, both know for sure that the other guy has an error

and they know that the other guy doesn't know

so when they both show up to the meeting, they are essentially saying to each other "Hey, you done fucked up!"

@CausingUnderflowsEverywhere give me a second

at which point, each knows that they have made an error

they are not allowed speak right?

they don't need to speak---the fact that A doesn't resign tells B that A knows of an error in one of B's papers

A knows that he is safe

or, rather, thinks that he is safe

and again, X doesn't say "I know that one of you made a mistake!"

he says "I know that one of you knows that another of you made a mistake!"

but for A to resign he must know he has an error and B knows for sure that A has an error and he is not aware of it. So B must understand he is not resigning because he does not know and he should not resign. Do you mean B self questions himself at some point?

@CausingUnderflowsEverywhere When you multiply something by a factor of 0.5, you are halving it. In general, if a function takes that x somewhere, halving it means you "halve the fuel" for your machine (the function).

my textbook says when $|k|$ in $f(kx)$ is > 1 it's a compression by factor of 1/k . however a homework given by the teacher had a question that said compression by factor 2, not 1/2. The teacher said the question is fine. Maybe the question is wrong and has a typo? I've been stumped by that question for the longest ever.

the problem is completely symmetric in A and B: anything that can be said about A can also be said about B

so, the reasoning goes like this:

* X says "One of you knows of an error in another's work"

* A reasons "I know of an error in B's work, therefore it is possible that I have made no errors."

* Then A gets to the meeting; he sees B there, which means that B also believes that he has possibly made no errors.

* Hence A reasons that B knows of an error in A's work (if he did not know of an error in A's work, he would have resigned immediately)

* Therefore the fact that B did not resign tells A that B knows of an error in A's work. Therefore A must resign.

But, again, this is completely symmetric in A and B, hence B must also resign

@CausingUnderflowsEverywhere I keep typing and thinking of how to show it to you without needing to draw it, but I'm struggling. Try this khanacademy.org/math/algebra2/manipulating-functions/…

Maybe the part you're "not getting" is the subtlety in the number being on the input side or the output side

Owh so B understands A that he is feeling the same as him and he resigns isn't it? So he feels the confidence in A, that his work is wrong

I feel you now @XanderHenderson

It can be a mind-fuzzler.

Math is a serial mind-fuzzler.

would be left out

then I would have understood really well

:p

glad I could help?

@CausingUnderflowsEverywhere but the usage of "compression factor" can be interpreted with ambiguity. The standard way is to say the graph has been compressed by a factor of 2, or has a compression factor of 2.

So in summary math professors in University B. are not like me :D

they deduce

@CausingUnderflowsEverywhere @HsMjstyMstdn When I teach this stuff, I tend not to use the phrases "compression" or "strech"---they are "scalings" by some factor

@hungryWolf They are perfect logicians, after all ;)

A "factor of 2" can be made into both a compressive thing and a stretching thing, depending on what you're talking about.

@XanderHenderson Exactly.

@XanderHenderson yes indeed

I too feel scaling is a better word, help me imagine stuff better @HsMjstyMstdn

so er what is the corresponding notation in the formula would you say?

@HsMjstyMstdn $f(2x)$ is a scaling by a factor of $\frac{1}{2}$

@hungryWolf you might want to look at this puzzle, too

the explanation there is reasonable

the thing with elementary transformations is that we generally write them in a way that is (to me) unconvincing

but when we talk about conic sections, we get it right

@XanderHenderson you've sent me the solution, I cannot find the question

oops just now looked at the link

sorry found it

for example, the equation for an ellipse is $$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$$

@XanderHenderson I have seen the opposite, though. In textbooks, no less. Whenever you use "factor", it should mean the way you talk about the number implies some direction to which way you're transforming the graph.

compare that to the equation for the unit circle, i.e. $$ x^2 + y^2 = 1 $$

I understand its a hard puzzle when I see this <--"They are all perfect logicians"

and sometimes that gets reversed by the word you use, i.e. compression or stretching.

the $-h$ moves the center of the circle to the right $h$ units

and the $a^2$ scales the circle horizontally by a factor of $a$

but when written that way, the symmetry with $y$ is more clear

@CausingUnderflowsEverywhere Don't use the words "compression" and "stretch"---they are confusing

just think about scaling

one question had horizontal stretch by 2 and another had horizontal compress by 2 and I just didn't know anymore

$f(2x)$ is a horizontal scaling by a factor of $\frac{1}{2}$

(which is a compression, but don't think about the word compression)

$f(\frac{1}{2}x)$ is a scaling by a factor of 2

but er how do I react if someone says horizontal compression by factor of 2

you scale by the reciprocal of thing you multiply $x$ by

If one must use compression and stretching, then yes, f(2x) does correspond to a compression and f(0.5x) does correspond to a stretch. The way you use "factor" muddies things.

@CausingUnderflowsEverywhere "horizontal compression by a factor of 2" is ambiguous language, and only an asshole would ask you to answer such a question

okay thanks for the enlightening haha

in such a case, I would ask for clarification

though I would never say that about my sensei

I mean, there are only two reasonable interpretations: scale everything horizontally by a factor of $2$, or by a factor of $\frac{1}{2}$

Scaling is unambiguous and intuitive; numbers bigger than 1 correspond to an enlargement and numbers less than 1 correspond to a contraction.

so, ask "which of these reasonable interpretations do you mean?"

@HsMjstyMstdn Exactly.

unless you are in an exam setting, then you should be able to ask that question and get a clarifying answer

better to stretch and compress than to push and pull :P

better yet to just scale

I bet I answered that question incorrectly

PUSH! I CAN SEE THE HEAD! PUSH! PUSH!!!

"push it" --- "wait how much" --- "by 2"

on an exam, I would probably write: "There are two possible interpretations: either we are scaling by a factor of $\frac{1}{2}$ (a compression), or scaling by a factor of $2$ (a dilation). In the first case, the answer is ___; in the second case the, answer is ___. The question is ambiguous, and the question writer is an asshole."

we've come to the conclusion that horizontal compression by factor of 2 is incorrect. Alas, horray. I finally understand compression / stretch transformations. Thanks so much HisMajesty and Xander :)

без проблема

I have emotional memories from the times I would struggle with the way the word "factor" was used when it came to transformations...

and then my physics textbook gave me "scaled by 2" or "scaled by a tenth" and that's when the light hit me

@CausingUnderflowsEverywhere np

Thank goodness for your physics textbook. You finally got that vitamin D you so needed.

@HsMjstyMstdn Physics book? What are you, some kind of physician?!

@XanderHenderson Spooky special relativity, woOooOOOo

I have taken a grand total of two physics classes in my life:

in high school, I took one semester of "Advanced Math / Physics", which was a hardcore precalc class combined with a course on high-school level Newtonian physics

that was 20 years ago

then, last year, I took "Quantum Mechanics"

@XanderHenderson hahaha

this isnt the first time my sensei got confused after I had some misunderstanding about math

ohh boy haha

@CausingUnderflowsEverywhere lucky for all of us, SE exists

yessss ♥

I have no idea what the physics part of that class was, but I am pleased to death with infinitesimal generators, the Trotter product formula, and some Baker-Cambell-Hausdorff goodness

@XanderHenderson Haven't gotten to a proper treatment of QM yet... though why would I, I can't even get springs down cold smh FeelsBadMan

I lurve me some funky anal, and QM seems to be nothing but funky anal

@XanderHenderson I've caught a glimpse of that Trotter product deep in the depths of my Mathematical Physics textbook, seems a daunting demon

heh

it isn't so bad

basically, the problem is that in QM, your basic objects (like momentum, position, etc---these are called "observables"?---again, I don't do the physics side very well) are operators on a Hilbert space

and they don't necessarily commute with each other

(think of them like non-commuting matrices)

@XanderHenderson I'm supposed to on a first-name basis with Hilbert spaces at this point, but I'm not even comfortable with it being in the same room as me >:(

so if $A$ and $B$ are to such dohickies, and $AB \ne BA$, then you have difficulty with the identity $\mathrm{e}^{A+B} = \mathrm{e}^A\mathrm{e}^{B}$

@HsMjstyMstdn Hilbert spaces are good things in life

the Trotter product formula is one way of getting around the difficulties

@XanderHenderson you sound so lucky that your highschool had such high level math

my high school didn't actually have much in the way of high level math

that precalc class was the highest class taught at the school

I have to drop out now, bye everyone :)

but the school was associated with the University of Northern Iowa, so we could take higher level classes for college credit via dual enrollment

if there is a really hard mathematical / programming problem. Which of the following will improve me the most in terms of problem solving and improving my abilities: 1) Try until you die 2) Set a proper time interval, try to work with hints. If you fail then look at the solution 3) if you have any new ideas

bye @CausingUnderflowsEverywhere

no need to study for or take AP exams when we could get straight-up transferable college credit!

LATERS!

and then I try to get them, so I click on Hilbert spaces on wikipedia and then I don't get something on the Hilbert space page and why or how it's being used the way it is so I click on that and then there's like 3 more things I don't get on that page so then I start clicking on those and then I'm on the "formal definition" of inner product and I'm wondering what a "field" is, intuitively, and then I realise I don't get vector spaces after all and the tabs start multiplying and oh shit....

heh

for Hilbert spaces, in general, you don't need to work over an arbitrary field

and then the years keep coming and they don't stop coming

either $\mathbb{R}$ or $\mathbb{C}$ is good for 99% of use cases

@XanderHenderson and @HsMjstyMstdn any suggestions?

@XanderHenderson yeah, am taking solace in that for now

and if you remember the dot product from linear algebra, that is the prototypical inner product

an inner product is really just (for most purposes) a fancy dot product

@hungryWolf Mathematics is collaborative---don't be afraid to discuss ideas with others, and don't expect to get everything on your own.

@XanderHenderson not when my prof starts saying "I shall define the inner product such that blablabla" and then some Einstein notation gets thrown in and I'm smiling to hide the pain LOL

but get what you can on your own

Einstein notation confuses me

@hungryWolf Philosophy and personal uniqueness is what you consider here.

but I'm not a physician, so I don't have to learn it!

yay!

I need to go to bed

g'night

There is a philosophy of dialectics, as such, the best way is most probably a mix and a combination of the approaches you mentioned, tailored to suit you.

@HsMjstyMstdn sorry I do not understand

nights

okay

I'm looking at the solution for the problem below:

@XanderHenderson and @HsMjstyMstdn thanks.

@hungryWolf Consider dialectics as the interpenetration of opposites, some point in the middle is the way that's best for you, most of the time. Not all the time, but most of the time.

isomorphism is an equivalence relation

but on what

Sol. 3 (on artofproblemsolving.com/wiki/…) states "Hence, notice that we want $a-c=-5$ so that $x=9$", but why do we want a-c=-5? It's not a restriction stated in the problem.

the set of algebraic structures (surely not consistent)... just any set of algebraic structures?

@HsMjstyMstdn so the middle path it is :D if I am confused

@hungryWolf Not the "middle" middle, but whatever combination of methods you find suitable and appropriate to you.

I think you get it, anyway.

@HsMjstyMstdn, yes I did

If anyone could help me out that would be great

@DarkRunner room's pretty slow right now, and my frazzled brain can't be of much help. Maybe @Semiclassical is free enough.

Should I message him/her?

massage him

?

how dare you

My mistake

lol

What is $x$ supposed to be in their solution?

The number of hours?

hmm so apparently one can define a relation on a proper class? surely the collection of all rings or groups is not a well-defined set

@Semiclassical yeah, a whole number integer

In that case, you're told that $x$ is a whole number.

right

So you'd need $x=(-9/5)(a-c)$ to be a positive integer.

Hmm.

ah, thus a-c must be a multiple of 5

Right, and it should be a<c so that x>0

right right ok thanks, got it!

so you'd need c-a=5,10,15...

right

and not a lot of ways to do that with decimal digits

That's not very clearly stated, though.

3 beer minimum here