Mathematics - 2016-05-30

12:06 AM

Ah, you're right. That was a dumb mistake on my part. I keep getting $\begin{bmatrix}1 \\ 1 \\ -2\end{bmatrix}$ but I realize that it was just a misunderstanding.. my head isn't at the right place after doing math the whole day haha

Phonics The Hedgehog

12:17 AM

Hello,

Is there a general formula for calculating the distance of a diagonal line of a polygon?

Eric Stucky

Yes.

Phonics The Hedgehog

@EricStucky I see, thank you! So, for a Hexagon who's side equals, let's say, 2, just use the formula $−2rr cos(2pi/n) = m$?

Mike Miller

12:32 AM

@eric the K in HFK stands for knot :)

Steven Stewart-Gallus

Does this redefinition of the derivative in terms of measures work? $ \left. \frac{\mathrm{d} y}{\mathrm{d} x} \right\rvert_{x = a} =
\lim_{C \rightarrow \left\{a\right\}}
\frac{
\mu\left(\left\{\left(n y\right)^+ \, \vert \, x \in C\right\} \right) -
\mu\left(\left\{\left(n y\right)^- \, \vert \, x \in C\right\} \right)
}{
\mu\left(C\right)} $ $n$ is either $-1$ or $1$ and tells whether a function is increasing or decreasing at a point.

Eric Stucky

Mike: oh :/

Phonics The Hedgehog

12:49 AM

@EricStucky Or should I use dj=r root(1−cos(2kpi/n))?

user3502615

1:26 AM

should i memorize all of the propositions / corollaries my book offers?

there is at least 50 in there

MickLH

@user3502615 I don't think that matters... you should make sure you can derive them all

user3502615

some of them are used alot though

MickLH

Your brain will automatically memorize the ones that apply to your field after you use them a few times.

And if it's so long that you don't automatically memorize it like that, you'll probably need to write it down anyways

user3502615

how about theorems

MickLH

Well I was just sharing my heuristic about corollaries, I don't have it all figured out :P

Eric Stucky

2:01 AM

Hello Albas :)

user3502615

also does my proof make sense? i.imgur.com/G9zOpyT.png

Eric Stucky

2:15 AM

It looks good.

Although, it is a little strange that you state "f has a laurent series" as a conclusion, since it is in the theorem statement.

But the mechanics are fine.

user3502615

2:29 AM

ok thank you

user3502615

2:44 AM

does anyone think i can test out of pre-cal 1 and 2 (i live in america)

or would they not allow that

Steven Stewart-Gallus

The difference between a valley and a peak is the measure of the distance going up and the distances going down. $$ \Delta\mu_{x \in C} y = \mu \left( \left\{ y \, \vert \, x \in C \wedge n > 0 \right\} \right) -
\mu \left( \left\{ y \, \vert \, x \in C \wedge n < 0 \right\} \right) $$ $n$ tells whether the function is decreasing or increasing. How do I state this as an integral so that this statement applies to vectors as well? I think something like $$ \Delta_{x \in C} \vec{y} = \oint \mu \left( \left\{ y \, \vert \, x \in C \wedge \frac{\mathrm{d} \vec{y}}{\mathrm{d} x} = \hat{r}\abs{\…

Eric Stucky

@user3502615: US schools don't like to let students advance in particular subjects, especially if they are understaffed.

As much as you can, you should try to fight inane policy like that

user3502615

oh RIP :(

Eric Stucky

It's really helpful if you have parents on your side; if you can't recruit yours then try to get others

user3502615

i still have to take geometry and algebra ii so i can still self-learn more in that period

should i consult the head of the math department of my middle / high school?

Eric Stucky

2:57 AM

As a student you have comparatively little power but the parents have a lot of de jure power at the district level, and schools know that parents can make it a real pain for them, so in practice they have a lot of power locally as well.

Have to go, but we can talk later if you want.

user3502615

ok thank you

Albas

3:17 AM

Hello@EricStucky :)

Forever Mozart

yout

ube

.com

Axoren

3:48 AM

@Semiclassical You beat me to it on that rocket question.

Semiclassical

4:06 AM

woo

l' - ' - ' - ' - 'l

Wooo

Semiclassical

@axoren and of course the next answer is someone just writing out the correct algebra out. i suppose that'd be easier, but i can't bring myself to do more than just supply the hint when answering.

Axoren

4:36 AM

@Semiclassical I feel like the hint would have been more useful by itself in this case. They clearly understood bits of the formulation of height, velocity, and acceleration. What they lacked was the connection to a real-world model that they could double-check in their head.

Sure, the rocket stopped accelerating, but it was still moving.

Semiclassical

right. they'd thought about the fact that the height had to be the same just before and after four seconds, but they hadn't done so for the speed

they did have $h'(4)\neq 0$, though

Axoren

It was the hint I was going to give, so I chucked you a +1 either way.

Semiclassical

thanks

Axoren

Holy crap, I just checked to see what the other answer looked like. It looks pretty bad.

Nothing is really explained, just kind of directed.

Semiclassical

right.

user3502615

4:56 AM

given these two theorems m.imgur.com/a/eENPh

would each series (provided that z is in the region of convergence for both series) be equal? because they have the same definition

that is, the same coefficients

because if they are equal, why does extending the sum from the naturals to the integers preserve its value? (from the second image to the first)

Mary Star

8:22 AM

We have that $N=\{t(x_0,y_0) : t\in \mathbb{R}\}$.
Therefore, $N=\mathbb{R}v$, for $v=(x_0, y_0)$.
So that $N$ is $T$-invariant, it must stand that $T(\mathbb{R}v)\subseteq \mathbb{R}v \Rightarrow T(\mathbb{R}v)=Lv$, for some $L\in \mathbb{R}$.
Then $L$ is the eigenvalue and $v$ is the eigenvector.
In which cases does the subspace consists of eigenvectors? I got stuck right now...
@arctictern @TobiasKildetoft

Tobias Kildetoft

@MaryStar you are mixing the two approaches we suggested. Stick to one of them. Given the phrasing of the question, finding the eigenvectors of the transformation should be a trivial task at this point.

Benjamin R

Hi all... if I have an eigenvector which corresponds to an eigenvalue of a matrix $A$, can I use that to find the other eigenvectors?

Tobias Kildetoft

@BenjaminR well, it means you can consider a polynomial of one lower degree when trying to determine the other, which can be enough

The eigenvector itself does not immediately help as far as I can see

Benjamin R

AH

That would really help

Tobias Kildetoft

once you have one root of the characteristic polynomial, you can do long division to get a polynomial of lower degree to work with

Benjamin R

8:29 AM

The characteristic polynomial I get from $A$ is $-\lambda ^3 + 3\lambda^2 +14\lambda +8$

If I have the eigenvalue of $-2$, what do I do? I can do the long division, I just use that to factor $(\lambda + 2)$ into the CP right?

Tobias Kildetoft

right

Benjamin R

okay! That is a big help Tobias, thanks. Not sure exactly how I would do that but I will ask my mad Russian prof tomorrow.

Tobias Kildetoft

how you would do what?

Benjamin R

Do the long division, I misspoke, sorry, I can't remember how I would do the long division to factor $(\lambda + 2)$ into that CP

Balarka Sen

Hi @Tobias, @iwriteonbananas.

Tobias Kildetoft

8:36 AM

@BalarkaSen Hi

Mary Star

@TobiasKildetoft To find the eigenvectors do we have to write the matrix of the transformation?

Benjamin R

Don't worry about it, I will ask the mad russian prof

Tobias Kildetoft

@MaryStar Not necessarily. But finding eigenvectors really ought to be trivial at this point given the context of considering modules over a polynomial ring

@BenjaminR There is also another way which can be easier to remember than long division

Benjamin R

Oh?

Tobias Kildetoft

just substitute in $x = y-2$, divide by $y$ and substitute back

Benjamin R

8:38 AM

I haven't done any long div in more than 12 months and I always found it laborious

?

Tobias Kildetoft

you change it so that the root you know becomes $0$, factor out the root which is easy in that case, then change it back

Benjamin R

ah okay, so for every $\lambda$ in the CP, substitute in $(y - 2)$?

Tobias Kildetoft

@BenjaminR yes

(I accidently called $\lambda$ for $x$ there)

Benjamin R

np

then divide the entire CP by y

Mary Star

@TobiasKildetoft Could you give me a hint how we find the eigenvectors? I don't really have an idea...

Tobias Kildetoft

8:43 AM

@MaryStar Then you need to go back and brush up on eigenvectors in general.

JeSuis

if I have a polynomial $P\in \Bbb{R}[X_1,\cdots,X_n]]$ , for $x_{n+1}\in\Bbb{R}$ the number of solution of $P(X_1,\cdots,X_n,x_{n+1})=0$ is finite ?

Tobias Kildetoft

@JeSuis How can you evaluate a polynomial at a value too many?

JeSuis

@TobiasKildetoft oops, it's $\Bbb{R}[X_1,X_2,\cdots,X_n,X_{n+1}]$

Tobias Kildetoft

@JeSuis In that case, no, it will usually not be finite unless $n\leq 1$

polynomials with several variables generally do not have a finite number of zeroes (do they ever?)

JeSuis

@TobiasKildetoft hum thanks

Tobias Kildetoft

8:58 AM

(ohh, it can actually be finite, it just usually will not be)

JeSuis

@TobiasKildetoft I am trying to prove that the set of zeros of a not identically polynomial has Lebesgue measure zero.

Tobias Kildetoft

@JeSuis Well, intuitively, it will have strictly smaller dimension than the ambient space (it is just a matter if picking the correct notion of dimension and proving this)

JeSuis

I tried an induction, but somewhere I 'need' that the set of solution is finite

Tobias Kildetoft

Well, for $n=1$ it is.

Balarka Sen

I mean, set of solution of a polynomial of multiple variables need not be finite. $y - x^2 = 0$ in $\Bbb R^2$ has infinitely many solutions.

Tobias Kildetoft

9:05 AM

@BalarkaSen Or just $xy$

JeSuis

Okay,I made a 'mystake", I only nedd that the numbers of $x_{n+1}$ such that $P(X_1,\cdots,X_n,x_{n+1})=0$ is finite

Balarka Sen

Oh, so $X_1, \cdots, X_n$ are fixed here?

Then you get a polynomial in a single variable $x_{n+1}$ - so obviously there are only finitely many solutions.

JeSuis

@BalarkaSen yep, apparently I need a coffee.

Martin Sleziak

11:28 AM

In case some people around here are interested, I will mention that there is Math Review proposal on area51 and it still needs more example question.

This proposal was discussed on our meta. Although judging by the score on the meta questions, it seems to have more support on MO.

Evinda

11:46 AM

Hello!!!

Tobias Kildetoft

@MartinSleziak Interesting to note the difference in support. But I also think I would find it a much more interesting proposal if I could be sure to have mainly questions from MO users (if the average user would be similar to an average MSE user I would expect most of the question to be due to uninteresting misunderstandings on the part of the asker)

Evinda

The function S that sends the function $f(x)$ to $f(5x)$ isn't linear, is it?

Tobias Kildetoft

@Evinda depends on where $f$ lives

Evinda

It isn't given any further information @TobiasKildetoft

Tobias Kildetoft

@Evinda then being linear does not even make sense

Evinda

11:48 AM

Why? Under which conditions would it be linear?

Tobias Kildetoft

@Evinda still depends on where it lives. Being linear for one things requires a vector space structure somewhere

Evinda

I think that we are in $\mathbb{R}$

Tobias Kildetoft

This is your question. You need to figure out the details first.

Martin Sleziak

12:06 PM

@TobiasKildetoft Well, I have mentioned this proposal a few times in the MO chatroom. I am not sure to which extent it helps.

BTW you can judge by yourself to which extent this could become something interesting by checking the already existing exacmple questions.

The stats say that 37% of the followers are active in Math.SE and 18% of them are active in MO. (User with rep>200 are considered active for the purpose of that statistic.)

Sparrow

dif

difference between log*N and log N

Martin Sleziak

@Sparrow I guess it is iterated logarithm.

Sparrow

thank you @MartinSleziak

Martin Sleziak

@Sparrow Not much to thank for. I simply asked google and the Wikipedia article was the first hit: google.com/search?q=logarithm+star

Sparrow

@MartinSleziak What are the keywords you have entered

Evinda

12:20 PM

If we are given sets of polynomials how do we deduce if they consist a vector space over $\mathbb{R}$ ? For example we have this polynomial: $ax^3-b(x^2+x+1), a,b \in \mathbb{R}$. Do we pick two elements of this form, let $f,g$ and check if $\lambda f+ \mu g, \lambda , \mu \in \mathbb{R}$ is also of this form?

Martin Sleziak

As you can see from the above link: logarithm star.

Sparrow

Thats good. I have encountered this iterative logarthm while learning disjoint set union data structure

Martin Sleziak

@Evinda What you wrote is a correct way to go. The set of such polynomials is a subset of the vector space $\mathcal P_3$ of all polynomials of degree 3. So it suffices to check whether it is subspace.

But in my opinion, an easier way is to notice that this is simply the span of $x^3$ and $x^2+x+1$. Span of two vectors is always a subspace. @Evinda

Evinda

So $a(x^2-1)+2b$ also is, right?

Martin Sleziak

Just to clarify. $a(x^2-1)+2b$ is not a vector space, it is a polynomial.

But the set $\{a(x^2-1)+2b; a,b\in\mathbb R\}$ is a vector space (with the usual addition and scalar multiplication).

And again, the argument is the same: This is precisely the span of the vectors (polynomials) $x^2-1$ and $2$.

Evinda

12:27 PM

Yes, that is true.

Thank you :) @MartinSleziak

Jim

2:50 PM

Hello every one, I would like to get some feedback on the proof presented in this post : math.stackexchange.com/questions/1610806/… . It is about automorphism, isomorphism of graphs. If you have background on those topics, leave your comments on the post.

Mike Miller

3:47 PM

morning

Balarka Sen

Morning.

Akiva Weinberger

Gmormmg

Mike Miller

doesn't happen very often that you wake up and see a problem you like has been solved: arxiv.org/pdf/1605.08530.pdf

Balarka Sen

hah

I remember you talking about that

Eric Stucky

mommg

Mike Miller

4:12 PM

@BalarkaSen SU(2) is still open, tho

Balarka Sen

Oh, right, you were talking about SU(2) not SL(2).

Mike Miller

SU(2) is the maximal compact subgroup of SL(2,C); this provides evidence IMO for the SU(2) conjecture, and one can think of it as saying that if you enlarge SU(2) a bit, then you can solve it

still will take a little bit to read...

Mike Miller

4:35 PM

how's whatever you're doing?

Eric Stucky

5:20 PM

Hello again Albas :)

Sorry I missed you last night :P

Danu

6:06 PM

What does the notation $\operatorname{Hom}_{\Bbb Z}(A,B)$ mean?

Ah, regarded as $\Bbb Z$-modules?

Balarka Sen

Yes, just set of abelian group morphisms $A \to B$.

Danu

And $\otimes_R$ denotes the tensor product where scalars are in $R$?

Balarka Sen

Tensor product as R-modules, yes.

Hi @AlexClark

user147690

Hello

Danu

Hi

user147690

6:14 PM

How are things?

Danu

My lack of algebra knowledge is disturbing

(to myself)

Balarka Sen

@Danu Don't worry too much about it.

@Alex It's ok :)

user147690

@BalarkaSen Good good. I am prepping for calc1 TA'ing

Danu

@BalarkaSen It's pretty annoying, at this point

user147690

Last one for the semester, and then I have two weeks of project+AG focus

Danu

6:17 PM

For instance, it took me 15 minutes now to understand what $\operatorname{Hom}_{\Bbb Z}(C_*,\Bbb K)\cong \operatorname{Hom}_{\Bbb K}(C_*\otimes_{\Bbb Z} \Bbb K,\Bbb K)$ even means

Now that I understand that, the statement is not so difficult. But it took me some time.

Balarka Sen

@AlexClark Good to hear.

Were you not also supposed to learn AlgTop?

user147690

@BalarkaSen In 2 weeks the semester is over, and I start AlgTop unofficially with friends for 1.5 months, and then I have ~5 months of official AlgTop

Danu

AlgTop is da bomb

:D

Balarka Sen

okie

Danu

...I'm still really enjoying learning it.

user147690

6:26 PM

I am keen too

Balarka Sen

topology is good.

user147690

Mainly keen for co/homology

Mike Miller

morning

Danu

@AlexClark We went through the basic constructions of cohomology just this morning, in my class :3 Everything went so quickly, since it all carries over from homology :)

Hi @MikeMiller (evening!)

user147690

@Danu Nice, I am just trying to learn a little now for studying in relation to proving Riemann Roch

Danu

6:30 PM

Sheaf cohomology?

user147690

@Danu Barely :P

Danu

Cool; my course on Riemann surfaces will also do some very basic sheaf cohomology ;)

Balarka Sen

@Danu The cup product is the interesting bit you won't find naturally in homology.

Akiva Weinberger

6:55 PM

@Danu I learned Hatcher's AlgTop book up until cohomology and then stopped because Ext and Tor make no sense to me

Mike Miller

funny, since tor has nothing to do with cohomology and all you need to know about ext is a few simple computations

iwriteonbananas

Hi @MikeMiller, Hi @BalarkaSen

Do you know off the top of your head how to prove that H-spaces are abelian (i.e. pi_1 acts trivially on pi_n for all n)?

Balarka Sen

tor is a homology thing

iwriteonbananas

Spent a good deal of time today trying to prove that but with no success. Of course when n=1, it is easy because the action is just conjugation and H-spaces have abelian pi_1. The action on the higher homotopy groups is somehow in the spirit of conjugation, so the assertion is plausible, but I couldn't figure out how to use the H-space structure to get an explicit homotopy between..

Balarka Sen

I find the construction of Ext quite beautiful. Dunno why you can't make sense out of it.

iwriteonbananas

7:02 PM

$\beta_{\gamma}([f])$and$[f]$ for $f\in \pi_n$ and $\gamma \in \pi_1$

Krijn

Hi @Balarka

Balarka Sen

hi

Mike Miller

@iwriteonbananas The action on $\pi_n$ is the same as working on the universal cover and just tracing out the path first and then following the sphere second. Does that parse? (I haven't said anything about H-spaces yet, just wanted to work in a nicer context)

Krijn

How you doin'

Akiva Weinberger

I think there was more stuff there too that was confusing

but I can't recall at the moment

Balarka Sen

7:05 PM

@Krijn So-so.

Krijn

:(

Mike Miller

for a Lie group, at least, this follows because choosing different lifts of the sphere come from multiplying by an element of the deck transformation group, which is a literal element of the Lie group; since we're connected, this is homotopic to the identity, and we can just walk the sphere upstairs back to its original position

iwriteonbananas

@MikeMiller I think so. Action of $\pi_1$ on the $\pi_n$ is the same as action of $\pi_1$ on $\pi_n($universal cover$)$ for $n\geq 2$, right?

Balarka Sen

I feel drained after 3 hours of physics. Can't do math.

Eric Stucky

I finally worked out my schedule for writing up prelim problems on the diff top side :/

giant mess, glad to be done :P

Mike Miller

7:08 PM

yeah, @iwriteonbananas

I imagine the argument for H-spaces is similar

iwriteonbananas

yeah, ok

Mike Miller

I responded to the email chain, if you didn't see

oic

iwriteonbananas

I saw your email (responded lazily a couple minutes ago). Appreciate the email alot.

Danu

7:23 PM

@AkivaWeinberger Tor was in homology, yeah.

Also, I don't know... The way I saw them constructed was pretty low-brow.

@BalarkaSen $F=ma$ ALL DAY EVERY DAY

Akiva Weinberger

I don't get why it's the same for all free resolutions, though

Danu

Ah, why all free resolutions are chain homotopy equivalent?

Yeah, I guess that's a theorem that needs proving

Mike Miller

7:43 PM

yes, that's a theorem that's proved

Danu

@BalarkaSen Yeah, we just made a small start on that today.

user 1618033

@Semiclassical how is it going?

Semiclassical

@user1618033 alright. went for a bit of a hike earlier

user 1618033

@Semiclassical I see. I also just returned from a bit of jogging. The thing is that very long hours of research must be properly treated with much physical effort.

Jogging and beach these days.

(I mean to prevent gaining weight and keep your health in a very good state)

Semiclassical

8:00 PM

Sure. plus energy and all of that.

user 1618033

Yeap.

Semiclassical

@BalarkaSen what kind've physics problems are you doing?

user 1618033

Where is @robjohn?

Eric Stucky

He doesn't come around much u16, surely you know this by now :P

user 1618033

@EricStucky Do you talk to me?

Eric Stucky

8:07 PM

yeah, you are u16

user 1618033

@EricStucky What is the meaning of u16?

Eric Stucky

u(ser)16(whatever)

user 1618033

@EricStucky OK OK. The number I use comes from the golden ratio.

Semiclassical

i was wondering if that was coincidental or not

should we just call you $\phi$? :)

user 1618033

No, golden ratio defines me. :D

@EricStucky I noticed that but I don't know why he comes so rarely these days.

@Semiclassical I might use that one day. :-)

Semiclassical

8:12 PM

If I had to pick a single symbol for a username, hmm

probably $\hbar$

Eric Stucky

Oh, for me that's easy, definitely $\Sigma$

user 1618033

interesting

Eric Stucky

Because it looks like an E but it's pronounced like an S, so it's a shortcut for my initials :)

user 1618033

;)

Semiclassical

considering classical behavior is $\hbar\to 0$ and quantum is $\hbar\to 1$...yeah, $\hbar$ works :)

Eric Stucky

8:14 PM

:)

Mike Miller

8:37 PM

hbar = 1/2

OneRaynyDay

9:09 PM

Hey guys! I have a quick question about eigenvalues: If we have symmetric 3x3 matrix with eigenvalues 1,2,3 - how many different orthogonal matrices are there for the diagonalization?

I originally thought it was 3! but my book says 6*4*2. I'm not sure if they also counted $S^{-1}$ = $S^T$ and used that for every matrix... But on the answer key they elaborated that there were 6 unit vectors (say what??)

Eric Stucky

You are asking about the number of possible orthogonal $S$?

There are six unit eigenvectors, yes. The three you are thinking about, and their negatives.

OneRaynyDay

Ah... Negatives - darn I should've realized that

But then what about my original thinking about there being the transpose of the original $S$ also being orthogonal?

I think it also leads to similarity between 2 matrices as well.

Eric Stucky

If the matrix is just the diagonal matrix 1,2,3 then the $S^T$s are already accounted for in the $3!$, so I am wondering if this happens in general.

Hmm, I'm not sure that $S^T$ generally works, though.

OneRaynyDay

9:33 PM

Hmm. Couldn't you say that given $AS = SB$ that $S^TA = BS^T$?

that way B would be similar to A, but since this property is symmetric I thought the similarity would still hold.

Why is $S^T$ already accounted for btw?

Eric Stucky

Yeah, you're right.

If you just do it for the diagonal matrix, you'll see. The 3! counts the permutation matrices, and these are closed under transposition.

I don't know if it happens in general

OneRaynyDay

Wait - are you transposing $D$ or transposing $S$? I can see that $D$'s transpose is itself since it's symmetrical

And yeah, my idea was the 3! was number of ways to arrange the unit column vectors inside of $S$.

Eric Stucky

9:58 PM

$S$, Rayny

I mean, take A to be a diagonal matrix with eigenvalues 1,2,3. You can find all the $S$ by hand.

OneRaynyDay

10:22 PM

@EricStucky Ah okay, I will check it out :)

Zachary Selk

Hello

Agawa001

hello how are you math nerds !

Zachary Selk

I'm good how are you?

Agawa001

i am nerd too

did ted come along here recently ?

Evinda

10:44 PM

Hi Eric

I have a question...

I am looking at the following proposition: Let $u \in L_{\text{loc}}^1(\Omega)$ and let $\int_{\Omega} u \phi dx=0, \forall \phi \in C_C^{\infty} (\Omega)$. Then $u(x)=0$ a.e. in $\Omega$.

Proof: $0=\int_{B(x, \epsilon)} u(y) \psi_{\epsilon}(x-y) dy \overset{\epsilon \to 0}{\to} u(x) \Rightarrow u(x)=0$ for all the Lebesque points.

Why can we pick $B(x, \epsilon)$ as $\Omega$ ? Also does $\psi_{\epsilon}$ have compact support? It is a mollifier.

Eric Stucky

11:11 PM

A mollifier has compact support by definition.

Presumably, restricting to $B(x,\varepsilon)$ is equivalent to restricting to the support of $\psi$.

Akiva Weinberger

@Agawa001 No. chat.stackexchange.com/users/76056/ted-shifrin?tab=recent

Not since yesterday, apparently

Eric Stucky

Yesterday is pretty recently for Ted, recently :P

Akiva Weinberger

This is true.

He said he had some difficulties come up, something about helping out a friend

so he doesn't have a lot of time for chatting

bolbteppa

11:32 PM

Is it not better to define the whole theory of integration using functionals, so that you can unite integration with distributions?

Ramiro Gonzalez

HI

Eric Stucky

Meh?

hi ramiro

I guess I don't see the plan, bolb.

You want to say "Def'n: The integral is a functional such that..."?

Ramiro Gonzalez

Has anyone here taken Complex Variables??

Eric Stucky

No but I can fake it sometimes.

Do you have something specific on your mind?

Ramiro Gonzalez

No, just wondering, lol.

Eric Stucky

11:45 PM

nifty :)

bolbteppa

If you do it this way, then you end up with "Some mathematicians might regard as artificial a treatment in which the evaluation of the Lebesgue measure of the unit interval is made to depend on what is essentially the Riesz-Fischer theorem" projecteuclid.org/download/pdf_1/euclid.bams/1183517927

Ramiro Gonzalez

I am planning on it, for fall. I have not either

Eric Stucky

wtf is that sentence, bolb :/

Does it mean, "Some people think it's unnatural to let m([0,1]) depend on Riesz-Fischer"?

bolbteppa

Yeah

Eric Stucky

So it looks like Bourbaki does it with functionals, and Halmos says this isn't a good idea.

bolbteppa

11:53 PM

But you have to ask, what is more jarring, showing that $m([0,1])$ is linked to completeness of $L^p$ spaces, or learning that all the Riemann/Lebesgue integration theory you studied does not even apply to the very first thing you study in electromagnetism, Gauss law, and that things called distributions are the answer, things which involve integration & are defined using functionals even though your whole integration theory up to then did not?

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$Mathematics$ Mathematics