Mathematics - 2022-09-29

;-)

User1865345

7:38 AM

@robjohn hmm. A hospital is flooded right now in Port Charlotte. Ian is Category 1 now. A coastal sheriff said they got many emergency calls of people trapped in their homes. Central Florida will have more rain, I guess.

Overtherainbow

8:11 AM

Are normal spaces and normed spaces the same spaces?

Thorgott

10:41 AM

@Overtherainbow no

11:13 AM

One of my favourite theorems, is a lemma by Banach, that if $K$ is a compact metric space and $f\in C(K)$, then $|f|$ has strict maximum iff the norm $\|\cdot \|$ is Gateaux differentiable at $f$.

It can be then calculated that this derivative in direction $g$ is equal to $g(s)\cdot\text{sgn}(f(s))$ where $s$ is the maximum of $|f|$

one potato two potato

11:43 AM

infinite product is too hard

12:04 PM

Not sure what you mean

Thorgott

2:02 PM

@Jakobian Interesting, have a reference?

it seems as if, for the "only if" direction, you need quantitative control in terms of $t$ on how close to a maximizer $s$ of $|f|$ you can maximize $|f+tg|$

@Thorgott Isometries on Banach spaces by Jamison and Fleming, it's a lemma by Banach so it can probably be found in one of his articles as well

It's how you prove that $C(K)$ and $C(L)$ are isometric iff $L$ and $K$ are homeomorphic, where $K$ and $L$ are compact metric spaces, what Banach originally proved. This holds for $K, L$ being compact Hausdorff spaces as well, an extension given by Stone. I've seen the proof for the more general case too. It's similar, just uses more the characterization of extreme points of $C(K)$.

That book is why I started learning functional analysis in the first place

The articles are most likely in French, Polish mathematicians written articles in that language in times of Banach (I'm not entirely sure why)

Astyx

2:34 PM

Parce que c'est la meilleure façon de faire

Thorgott

3:54 PM

@Jakobian got it, neat

Yai0Phah

@Jakobian This also holds for compact Hausdorff spaces $K$, right?

Xander Henderson

4:15 PM

It is not worth creating an account on another SE site in order to comment, but:

Do Americans purchase 'return tickets' at train stations (e.g. 'a return [ticket] to Woking' is a ticket from my current location to Woking and back)? Because that's normal in the UK and one usage may have influenced the other. — dbmag9 6 hours ago

As an American, can you please explain to me what a "train" is?

Ted Shifrin

in French they say "aller et retour" ... Yeah, but what's a train?

Does mod 11 of number remain same if we switch any two digits?////?//

Think of $121$ and $112$

Akiva Weinberger

4:45 PM

Hey

Is $\ln(x^2+1)$ analytic on the open unit disk and discontinuous on the closed unit disk?

Trying to construct a counterexample

@robjohn In complex analysis, if a power series $\sum a_nz^n$ converges on the closed unit disk, must it be continuous there?

my mistake

I wanted to ask @robjohn does it ever not change?////

so your counter fail @robjohn

I apologise robjohn

5:01 PM

@Yai0Phah I don't know actually

EnEm

Is sharing an answer you wrote allowed here? i spent way too much time on this one answer and I think people might find it interesting

it's something I'd like to find out

@AkivaWeinberger you are worried about weird behavior at the boundary? I am not sure about some lacunary series. Let me think a bit.

Any idea robjhon?

@Shinrin-Yoku the counter example does not apply to the changed question. "failed" is a harsh word for that.

5:07 PM

@robjohn yes that’s why I apologised for posting the wrong question, I’m sorry, but I didn’t mean any offence.

@Shinrin-Yoku Consider $211$ and $112$.

EnEm

@Shinrin-Yoku mod 11 f any number like fedcba is (a+c+e)-(b+d+f), So you can swap any two even index digit, or any two odd index digit, and the mod 11 will remain the same

@Shinrin-Yoku Figure out why $100\equiv1\pmod{11}$

^ That is quite mean @robjohn. AS I said earlier I meant no offence, I just wanted to clarify the example you gave no longer works

A: Iterating over numbers with many divisors

What is mean? I am trying to expose why certain exchanges of digits keep the same number mod $11$

It is because $100-1\equiv0\pmod{11}$

This explains EnEm's comment.

EnEm

5:22 PM

I can't find anything against promoting your answer in the guidelines, so might as well do that math.stackexchange.com/a/4541738/1099603

1

I have written a script for this in C++, which you can find at the bottom of this answer. It seems it runs for your use case in $\approx 75$ seconds. It should be faster for you than forfactored, please let me know if anyone finds any more optimisation that I can do to the code. I took your idea ...

It's related to finding high composite numbers in some range [1,N]. I have built an algo for this, which seems to me is the most efficient way to do this, but can't figure out what the time complexity would be.

UnderMathUate

5:47 PM

I was looking at a proof of this theorem:

Cancelation is valid in any integral domain $R$: If $a \neq 0_R$ and $ab = ac$, then $b = c$.

Proof:

If $ab = ac$, then $ab - ac = 0_R$, so that $a(b-c) = 0_R$. Since $a \neq 0_R$, we have $b-c = 0_R$. Thus $b = c$.

At first, I thought it was utilizing that fact that $a \neq 0_R$, to justify dividing both sides by $a$ since you normally can't divide by zero, but now I'm thinking it's actually because the zero-product property holds in integral domains. So if $a \neq 0_R$, then $b-c$ must be equal to $0_R$.

Small thing, but I wanted to make sure I'm focusing on the right things.

5:59 PM

well, coming back to it. I think Banach was particularly well-versed in topology. If the same lemma holded for compact Hausdorff spaces, he would probably be able to prove it. I don't know for sure though

6:12 PM

under: yes

UnderMathUate

Thank you

even with integers you don't really need to leave the world of integers to deduce a(b-c) = 0 and a nonzero implies b - c = 0. yes, in this case "divide by a" does have meaning in a larger more familiar field, but you don't need to go there. it's helpful to think of the zero product property as the thing that makes this work

although there is a construction for integral domains you can do that mimics constructing Q from Z

where "divide by a" would also work

UnderMathUate

Ohh, ok. Yeah, the whole divide by a thing is sometimes where my mind goes when I see something like this, unless I'm trying to solve for x or something. Then the a = 0 or b = 0 thing comes first.

@leslietownes Construction?

6:34 PM

any integral domain R can be thought of as contained in a field S that plays a role like the role Q plays for Z, and nonzero elements of R have multiplicative inverses in S. so there's a way you can make "a (b - c) = 0 and a nonzero implies b - c = 0 because you can just divide by it" sensible for integral domains in a way that mimics how you can do this for integers. en.wikipedia.org/wiki/Field_of_fractions

Asinomás

7:46 PM

hello

The Substitute

8:14 PM

Good day, I know we haven't walked in a while, but is the pigeonhole principle an axiom or a theorem?

depends on how you formalize your math and the principle. it certainly doesn't need to be an axiom, but the proofs of it from other axioms often involve closely looking at things (such as the 'size' of a finite set, or what it means for one natural number to be less than another) that tend not to get a lot of scrutiny outside of classes specific to formalizing math.

e.g. in a 'discrete math' class for college freshmen and sophomores it might as well be an axiom.

in my very humble opinion.

The Substitute

@leslietownes thanks

you can look at some of the chaos in math.stackexchange.com/questions/1083562/… for examples of people wrestling with this issue from various starting points.

i think in a lot of classes it gets introduced in terms of a metaphor that might actually be pigeons and holes, or stuffing things in boxes, or something. the fact that it's often taught that way may be suggestive of an unconscious desire to avoid formalizing it too much.

The Substitute

8:36 PM

I see

I am reading a proof of Pfitzner theorem for the convex case. This is only chapter 3 and they're making a reference to theorem in chapter 13. 0_0

All things considered, I never read the proof of Pfitzner theorem but I did see a lot of different (weaker) versions of it proven

geocalc33

8:59 PM

@anak I would like to take a pair $(\zeta^2,v)$ and find a pair $(\zeta^3,w)$ where there is an embedding $e: \zeta^2 \hookrightarrow \zeta^3$ such that $v=e^*w.$ Define the manifold $\zeta^2:=(0,1)^2$ and vector field $v:=\langle x\log x, -y\log y\rangle,$ for $x,y \in(0,1).$ $w$ should be a 3-vector field accumulating to each of the eight vertices and non-vanishing elsewhere. (source is at $(1,0,0)$ and sink is at $(0,1,1)$). Does this make sense?