Mathematics - 2022-02-04 (page 1 of 2)

PenAndPaperMathematics

00:00

@geocalc33 howdy :D

geocalc33

Just needed clarification on the conditions needed for deforming the path of integratino

because I would like to understand this concept for another related exercise

Jakobian

@hyper-neutrino yes, sorry

geocalc33

@PenAndPaperMathematics hey :)

robjohn

@geocalc33 you can deform a contour as long as the deformation does not cross any singularity. If it crosses a singularity, then the residue of that singularity is included (or removed depending on whether the singularity goes inside or outside the contour).

PenAndPaperMathematics

@geocalc33 you're way advanced of me in terms of $\Bbb{C}$ analysis. I'm still at the start :)

geocalc33

00:11

@PenAndPaperMathematics no you're way more advanced than me

Jakobian

@TedShifrin for big enough $n$ we should have $f_n = \sum_{z\in A} a_z\cdot n^{k-1}\cdot z^{-n}$ where $A$ are all roots of the denominator of the generating function $P(x)/Q(x)$ so of $Q(x)$. We definitely need $|z|\geq 1$ or everything will implode to infinity and limit of $f_n/n^3$ won't exist. If $z$ is a root such that $|z|>1$, then it doesn't contribute anything to the limit, as the term goes to zero. So the only roots that contribute are the roots of unity.

Now, I might be wrong about this, but I think if there are roots of unity $z\in A$ with $z\neq 1$ and $k\geq 4$, then $f_n/n^3$ will definitely not exist. So we need that for all the roots $z\in A$ with $|z| = 1$ but $z\neq 1$ to correposnding $k$ be $<4$, and if we want the limit to be non-zero, to $k = 4$ for $z = 1$.

That's where I got quadruple (I originally said triple) root of $1$ from.

So there is some maybe interesting problem about limits here, check that if $z_1, ..., z_k $ are roots of unity $\neq 1$, then $\lim_{n\to\infty} \sum_{i=1}^k z_k^n$ doesn't exist.

Whenever I say roots of unity I mean $z$ such that $|z| = 1$, sorry

geocalc33

02:26

has anyone ever come across the notion of an integral transform on all the leaves of a foliation of Euclidean space?

robjohn

02:45

@geocalc33 this answer might be of some use.

It shows how a contour can be extended to enclose a sequence of singularities

user400188

An open cover of a set $A\subset X$ is a collection of open sets which when unioned together, contain $A$.

When we say a collection is a finite subcover of a set, do we mean that it is a union of finitely many sets, or that once the union is complete, the result has finitely many elements? Or is it neither?

I am struggling to see how it could work either way I suggested. According to wikipedia, $[0,1]$ is compact, meaning that every open cover of it has a finite subcover. I could understand this if it just meant that we could use finitely many open sets in a union to make something that contains it... However, it is also stated on the same page that $(0,1)$ is not compact.

leslie townes

user: the former. a "finite subcover" is a collection of finitely many sets from the original cover.

(that is also a cover)

this aspect of terminology is unfortunate. you actually see this a lot with "finite" "infinite" - the adjective can apply to more than one thing and it means something different. good catch.

user400188

I can't see why $(0,1)$ isn't compact then.

leslie townes

{(1/n, 1): n = 1, 2, 3, ...} is an open cover of (0,1) with no finite subcover.

finitely many of these things will get you at most (1/N, 1) for some large N (corresponding to the largest index in your finite collection. it will miss the rest of (0,1).

user400188

Ah

leslie townes

02:54

that any open cover of [0,1] has a finite subcover is harder to prove.

i don't think it's remotely "obvious."

user400188

Hopefully I will be able to do so when I read a bit more into the chapter. I didn't expect to when I read that the result had a name on Wikipedia.

Thanks for the help again @leslietownes :)

Ted Shifrin

@leslie Munchkin finally gave you permission to chat?

user4539917

shhh 🤫 munchkin thinks he's "working"

leslie townes

03:12

she actually told me to go to my office

user4539917

so you have her permission to chat in your office?

geocalc33

an integral transform can return a constant ?

from my experience an integral transform always returns a new function

leslie townes

user: sure

geocalc33

that's very weird

leslie townes

geocalc: that is certainly more usual, but i don't know why you would exclude transforms that only return constants from any definition

e.g. because it could mess up any completeness you otherwise had in your function space

user4539917

03:28

\o @copper.hat

geocalc33

okay so it converts a function to a scalar

like integration of a one var. function

but this time it converts a function to a scalar with a kernel as well

copper.hat

@user4539917 hello?

Ted Shifrin

03:51

@leslietownes Because you usually want some sort of duality or invertibility.

Howdy @copper

copper.hat

Hi @TedShifrin!

leslie townes

ted: i'd take completeness over those things any day

Ted Shifrin

That doesn’t even make sense to me.

leslie townes

well if you average or take some other kind of limit of "integral transforms" you might well get something that does spit out just constant functions. i want those to be in the same space as the things that are subject to the limit process.

Ted Shifrin

Integral geometry is all about double fibrations and going from things on $G/H$ to things on $G/K$.

leslie townes

03:59

i was talking about "integral transforms" (which have not been given a definition), not integral geometry. maybe i missed something earlier.

i was thinking of e.g. en.wikipedia.org/wiki/Integral_transform where there is no reason for "K" in the equation 1 to depend on u.

may have missed something earlier.

Ted Shifrin

Think Fourier transforms or Radon transforms. I don’t see limits in either.

Balarka Sen

What a coincidence, I was thinking about integral kernels just now.

Ted Shifrin

Heya, a Balarka.

Balarka Sen

Hi Ted

leslie townes

i'm not saying anything about the fourier or radon transform, to which my remark "that is certainly more usual" above would apply. but i saw the question as more general than that.

geocalc, what was the motivation of your question?

Balarka Sen

04:04

Knowing geocalc, probably exponential coordinates.

leslie townes

i was worried that that's where this is going.

sometimes people use the operator topology on some kind of transforms to induce a notion of convergence for kernels. sometimes you want to approximate the integral operator by approximating its kernel. if you exclude kernels that might induce scalar-valued transforms the completeness gets messed up, roughly because nonconstant functions can approximate a constant function. that's all i was saying.

Balarka Sen

By the way, I ban Fourier transform on $\Bbb R^n$. I propose everyone should think in terms of the following global version: Let $(M, \mathrm{vol})$ be a manifold with a volume form and $E$ be a vector bundle on $M$. Suppose $f : TM \to E$ is a fiber-preserving map. Then its Fourier transform is defined by $(\mathcal{F}f)(x, \xi) = \int_{T_x M} e^{-i \xi(v)} f(x, v) d\mathrm{vol}_x$

This is a fiber-preserving map $\mathcal{F} f : T^*M \to E$

leslie townes

i'm uncomfortable with anything that doesn't have factors of 2pi in it.

this definition is unacceptable to me.

Balarka Sen

Divide by $1/(2\pi)^{\dim M}$ then

leslie townes

this definition is acceptable to me.

Balarka Sen

04:09

Or multiply by that.

leslie townes

i might do some square rooting of those factors, too. it makes me feel better.

Balarka Sen

henceforth, we shall scale the Fourier transform by $e + \pi$, just for fun

Ted Shifrin

rolls $9\pi/4$ eyes

Balarka Sen

But why is this version never seen in literature? It's so nice!

leslie townes

how about e + i + pi + 1 + 0. all of the world's most beautiful constants.

Balarka Sen

04:11

ah yes, the most beautiful equation of all time

if i am remembering it right...

Ted Shifrin

So what are actual examples, Balarka?

leslie townes

that's just like ted, to bring the mood down.

Ted Shifrin

ted’s a bitch

Balarka Sen

If $P : C^\infty(E) \to C^\infty(F)$ is a pseudodifferential operator, the symbol map goes $\sigma(P) : T^*M \to \mathrm{End}(E, F)$

If you take its Fourier transform the way I did it you get a map $K : TM \to \mathrm{End}(E, F)$

Put $TM$ inside $M \times M$ as the normal bundle to the diagonal, then if $K$ is regular enough, $K : M \times M \to \mathrm{End}(E, F)$ gives an integral kernel corresponding to $P$

Or at least I think thats what this is

The Schwartz kernel

Ted Shifrin

beyond my scope

Balarka Sen

04:17

$K$ won't necessarily be a function, it'll be a $\mathrm{End}(E, F)$-valued distribution on $M \times M$.

Ted Shifrin

but you answered me!

Balarka Sen

I am not sure but I hope this is right

To verify I would have to work in coordinates, so I will not verify :)

Ted Shifrin

blah

Balarka Sen

Well OK just to make you happy lets try the pseudodifferential operator $\mathrm{Id} : C^\infty(\underline{\Bbb R}) \to C^\infty(\underline{\Bbb R})$ :)

leslie townes

nobody is made happy by that

Balarka Sen

04:21

The symbol is... $1 : T^*M \to \Bbb R$, the constant map $1(x, \xi) = 1$.

The Fourier transform is $K : TM \to \Bbb R$, $K(x, v) = \int_{T^*_x M} e^{-i \xi(v)} 1 d\mathrm{vol}_x$

which is $0$ unless $v = 0$, in which case its $\infty$

So it's the delta-mass on $TM$ supported on the 0-section

Put it in $M \times M$ to get the distribution to be the delta mass supported on the diagonal

Which is indeed the integral kernel of the identity operator :)

No coordinate computation verification in a special case implies I must be right in general!

Ted Shifrin

Hahaha

Akiva Weinberger

04:41

Heyo

Balarka Sen

Hi @AkivaWeinberger

Akiva Weinberger

A homework assignment asked me to prove that origin-preserving isometries are linear, and I was somewhat embarrassed to realize that I had never done that before

I figured it out, but I thought "surely I've thought of this proof before"

I think in the past I was just thinking "isometries preserve the shape of parallelograms, therefore by the parallelogram rule sums get preserved"

but that suddenly feels a lot less rigorous when I have to write it out on a homework

The definition of isometry was $\|x-y\|=\|f(x)-f(y)\|$

I think I'll do it like this

First show $\langle x,y\rangle = \langle f(x),f(y)\rangle$

leslie townes

it's nontrivial, i think. not difficult, but certainly worth proving.

Akiva Weinberger

Then, the strategy is: don't prove $f(x+y)=f(x)+f(y)$. Instead, prove $\langle f(x+y),f(w)\rangle=\langle f(x)+f(y),f(w)\rangle$ for all $w$

leslie townes

additivity is the 'easy part,' you have to fiddle a bit to pull out arbitrary scalars.

Akiva Weinberger

04:48

(hm… wait, $f$ isn't necessarily surjective in infinite-dimension, is it? Suddenly no longer sure this strategy works)

copper.hat

do you have an inner product?

Akiva Weinberger

I think it's meant to be $\Bbb R^d$ with Euclidean norm

which means my "infinite-dimension" worry doesn't matter

leslie townes

it's certainly easier with an inner product. still true in general normed spaces if you assume surjectivity, i think, but the proof is a lot harder.

Mazur–Ulam theorem

In mathematics, the Mazur–Ulam theorem states that if V {\displaystyle V} and W {\displaystyle W} are normed spaces over R and the mapping f : V → W {\displaystyle f\colon V\to W} is a surjective isometry, then f {\displaystyle f} is affine. It is named after Stanisław Mazur and Stanisław Ulam in response to an issue raised by Stefan Banach. For strictly convex spaces the result is true, and easy...

inner product is contained in the strictly convex case.

Ted Shifrin

This is in my diff geo notes and in my algebra book.

leslie townes

which is actually a cool way of looking at it.

it all comes back to convexity.

mohan10216

04:53

Wow, I finally managed to render the mathjax in the chat :)

user400188

Bookmark the link.

mohan10216

Just did :)))))

copper.hat

convexity???

squirrel

leslie townes

there's still aspects of the convex geometry of the unit ball of some operator spaces that hasn't been figured out. i was getting into that when i left.

Ted Shifrin

Chipmunk

Akiva Weinberger

05:40

Oh I see what I want to do to prove the thingy

Show $\langle f(x+y)-f(x)-f(y), f(x+y)-f(x)-f(y)\rangle = 0$

and $\langle f(cx) - cf(x), f(cx) - cf(x)\rangle = 0$

Ted Shifrin

Yuck.

PenAndPaperMathematics

05:56

@geocalc33 it's all relative. You know more $\Bbb{C}$ analysis than me :P

user400188

Come to think of it, what would the finite subcover of the following collection of open sets be?

$(-0.1,0)\cup(1,1.1)\cup(0,\frac{1}{2}\sum_{n}^{1}\frac{1}{2^n})\cup(\frac{1}{2}\sum_n^1\frac{1}{2^n},\frac{1}{2}\sum_n^2\frac{1}{2^n})\cup\dots$

The collection is infinite, and the series defined in it converges to $1$, yet the removal of any one member of the collection should make it so $[0,1]$ is no longer contained within it.

Is there something wrong with the way I defined the collection in the first place?

leslie townes

is that a cover of [0,1]? is 1 in there?

the sum notation looks a little unorthodox

user400188

yeah, as $k$ approaches infinity, $\frac{1}{2}\sum_{n=0}^{k}\frac{1}{2^k}$ approaches 1.

I messed up the notation in the original a bit.

leslie townes

but does the union of that infinite family of sets actually contain 1, is the question. it can't just get arbitrarily close. to be a cover it has to have 1 as an element.

i don't see which member of that family would have 1 as an element, but again, i haven't fully parsed the notation.

user400188

Oh, and at the start a set $(1,1.1)$ is added to avoid that problem.

leslie townes

06:09

1 is not in (1,1.1). you could use [1,1.1) but then it is not an open cover. you could use (1-t, 1.1) for some positive t, and then you would find that finitely many of the rest would fill in the missing stuff up to 1-t.

i am this close to hitting my 'render chatjax' bookmark. i am flying by the seat of my pants here.

user400188

It's somewhat impressive that you could read it without rendering it.

leslie townes

when i learned tex i had a slow computer that could take several seconds to compile even a short document. reading it was more efficient than compiling and looking.

real-time rendering is amazing but old habits die hard.

user400188

@leslietownes I forgot that using open sets like this would leaves gaps. There is a set that ends in 1), and another that starts with (1, but that would leave 1 out of the sequence wouldn't it?

leslie townes

user: yes. the issue is that if 1 is in some element of an open cover, then by definition there will be some interval (1-t, 1] that is also contained as a subset within that one element of the open cover. and you can rely on finitely many of the rest for the rest.

user400188

I don't think I understand why you only need finitely many after identifying the element for which 1 is contained in. But I guess that is where the meat of the proof is.

leslie townes

06:20

as i interpreted the notation, the original cover was something covering 0 and then some stuff of the form (0, c_n) where c_n went to 1. if there's some other interval of the form (1-t, 1+t) contained in some element of the cover (t positive) you will neeed only finitely many of the c_n's to get within t of 1.

speaking glibly, but generally speaking for any arbitrarily given open cover it can be a bit of a job to find a finite subcover. the usual proofs are not that explicit about how it is done.

user400188

Ah, I see what you mean now.
Thanks for having a go at the problem given that it's not always easy to find a subcover.

leslie townes

the open cover formulation of compactness is a really weird condition. i don't think i would have come up with it. once you have it, it's great.

copper.hat

06:35

the term open cover seems vaguely contradictory

leslie townes

the terminology is unfortunate. the "open" in "open cover" refers to a property of each element in the collection (cover). the "finite" in "finite subcover" refers to the totality of sets in the collection (subcover).

but we all memorize it like the catechism and somehow it works.

user400188

The strangest terminology to me is still use of the word 'open' to just mean a set is a member of somethings topology.

It works fine if you are using the metric topology on the real number line, but for simpler examples like a set $X=\{a,b\}$ with the discrete topology $\{X,\emptyset,\{a\},\{b\}\}$, it makes less sense to say that the set $\{a\}$ is 'open' in X.

leslie townes

that's stopped being weird to me. i rewired my brain a long time ago.

as von neumann said, in math you don't understand things, you just get used to them.

user400188

My brain is rewired, but when I think about how I would explain point set topology to someone new, I need to put myself in their shoes, and it starts to sound weird again.

leslie townes

i never had to teach it. only on R where it still kinda makes sense.

:D

copper.hat

06:56

one needs an open mind

William John

09:10

I was wondering if I got collection of vector { 0$\le$ x $\le 1$|$(0.1-x\hat{i})+\hat{j}$ }. How do I get the sum of them? There are infinite of them...

robjohn

I don't understand what that set notation means

I assume that $\hat i,\hat j$ are unit basis vectors

William John

@robjohn Yes

robjohn

but what would be $0.1-x\hat i$?

$0.1$ is a scalar and $x\hat i$ is a vector

William John

Sorry it's $(0.1-x)\hat i$.

robjohn

ah, so what you want is $\left\{(0.1-x)\,\hat i+\hat j:0\le x\le1\right\}$

William John

09:15

@robjohn Yes.

robjohn

that is an uncountable set of vectors. You cannot add an uncountable set of non-zero real numbers.

even a countable sum of $1$s is not summable

William John

What is countable sum of 1?

Seems like a higher math that I don't know.

robjohn

@WilliamJohn an infinite sum of terms equal to $1$, i.e. $\sum\limits_{k=1}^\infty1$

William John

Ah understood.

robjohn

countable means that it can be put in 1-1 correspondence with the Natural numbers.

uncountable means it is too big to be put in 1-1 correspondence with the Natural numbers

the rationals are countable, the reals are uncountable

William John

09:30

I was curious how physicist calculate let's say a line of charge and line is finitely large and when they place a point charge and it's position is not symmetrical how they calculate vector sum of electric field on that point?

I think you can ignore my question. I was wondering this before learning the whole electricity and magnetism stuff.

Thanks for answer @robjohn .

robjohn

No, they probably integrate it, not sum it.

ick

hmm

Koro

11:49

Hot to count homomorphisms from group of permutations on n symbols i.e. $S_n$ to $Z_m$, the additive group of integers modulo $m$?

By FIT, for every such homomorphism $f$, $S_n/ker f\equiv f(S_n)\le Z_m$

So the number of all possible kernals, that is the number of all possible normal subgroups of $S_n$ will direct me to total number of homomorphisms.

But I don't think there's a recipe to find all normal subgroups of $S_n$. Is there?

love_sodam

Rather than finding normal subgroups of $S_n$, note that $S_n = \langle (1\ 2),(1\ 2\ \cdots\ n)$.

justadzr

Does the word cohomology refer to the family of cohomology groups or does it refer to one of the groups so that it makes sense to say something like "cohomologies"?

Koro

13:08

@love_sodam yes, I know this form for S_n. But please tell more as to how this helps the counting of the homomorphisms.

Thanks.

robjohn

@leslietownes what about $\gamma$?

@leslietownes I do prefer the normalization using $e^{-2\pi i x\cdot\xi}$.

Jakobian

13:24

About this exercise from Wallman compactifications I was stuck on, it turns out there is a lemma about extensions of continuous function into a compact space from dense subsets to the whole space and when it's possible.

Really helpful when comparing compactifications

Because if $X$ is a space and $aX, bX$ are its compactifications then finding $f:aX\to bX$ such that $f(x) = x$ for $x\in X$ amounts to extending the embedding $f:X\to bX$

robjohn

13:37

@leslietownes I have added a proof of Heine-Borel here.

robjohn

14:00

@user400188 In $\mathbb{R}^n$, compact is the same as closed and bounded. This is the Heine-Borel Theorem.

Jakobian

with the standard metric

robjohn

If it had a different metric, it wouldn't really be $\mathbb{R}$ :-)

Jakobian

depends on what kind of $\mathbb{R}$ we are talking about, and I think my comment is useful because it says just that, we're talking about $\mathbb{R}^n$ with the standard metric

Koro

14:20

I’m still stuck at counting homomorphisms from S_n to Z_m. 😐

Jakobian

@Koro It means that $f:S_n\to Z_m$ is uniquely determined by its values $f((1, 2))$ and $f((1, .., n))$

Koro

I see. So if m is odd, and suppose that $f((1,2))$ gets mapped to a non identity element in $Z_m$ that will be contradiction.

For order of f((1,2)) must divide 2 (order of the transposition (12)).

So if $m$ is odd then $f((12))$ must map to the identity.

love_sodam

14:38

I start wondering.. $S_n$ can be generated by $(i\ i+1)$ for $1\leq i\leq n-1$. $|f((i\ i+1))| = 1$ or $2$. But in $\Bbb Z/m$, there is only one order $1$ or $2$ element.

Koro

now if n is even, then f((123...n)) must map to identity. If $n$ is odd and is divisible by $m$ then?

There is no reason to believe why f((12...n)) won't map to generator of $Z_m$

Jakobian

@love_sodam going by the same route, since $S_n$ is generated by a transposition and $3$-cycles, we'd need to have $|f(S_n)|\leq 2$ if $m$ is not divisible by $3$

Oh, I see what you mean. So for odd $m$ we automatically get only trivial homomorphism $S_n\to Z_m$ and for even $m$ we get a split $S_n\to Z_2\to Z_m$ so it suffices to assume $m = 2$

But $S_n = A_n \cup (1, 2)A_n$ and $A_n$ is the only subgroup of $S_n$ of index $2$, so non-trivial $f:S_n\to Z_2$ needs to have $f(A_n) = 0$ and $f((1, 2)) = 1$

unit 1991

15:07

Hi.Where I can find a proof about limit of cross product of vector valued functions?

Jakobian

So this is just the usual sign of a permutation

so the only surjective homomorphisms of $S_n$ onto $Z_m$ are for $m = 1, 2$ and those are the trivial homomorphism and the sign of a permutation

leslie townes

15:27

unit: what kind of statement do you want to prove? that the cross product is continuous, in the appropriate sense?

unit 1991

15:39

@leslietownes if limit of $r_1$ vector valued function is $\alpha_1$ and limit of $r_2$ is $\alpha_2$ then limit of cross product of $r_1$ and $r_2$ is cross product of $\alpha_1$ and $\alpha_2$

leslie townes

15:51

if you write the first family of functions as $r_t$ with limit $r$ and the second as $s_t$ with limit $s$, it may help to note that by bilinearity, $r_t \times s_t - r \times s = r_t \times (s_t - s) + (r_t - r) \times s$

if you apply norms and use the triangle inequality and the fact that $\|a \times b\| \leq \|a\| \|b\|$ you deduce $\|r_t \times s_t - r \times s\| \leq \|r_t\| \|s_t - s\| + \|r_t - r\| \|s\|$

you can do epsilon-ology from there. note that $\|r_t\|$ is bounded because $r_t$ is assumed convergent

geocalc33

$$\Phi(s)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{2K_1(2\sqrt{z})}{\sqrt{z}}\bigg(\sum_{n=1}^\infty e^{zn^{-s}}\bigg)~dz$$

Is $z=0$ the only singularity of the integrand?

Govind75

If i were to construct a confidence interval for a population proportion $p$ for people who are above 50, let's say it was a $95\%$ interval given by $[12.3\%,13.7\%]$. Is it wrong to say there is a $95\%$ probability that between 12.3% and 13.7% of people are over 50. I know the translation of the confidence interval is that there is a $95\%$ probability that $p \in [12.3\%,13.7\%]$, but are the two statements not equivalent?

unit 1991

16:09

@leslietownes Can you please explain how in last step you used triangle inequality?

leslie townes

$\|r_t \times (s_t - s) + (r_t - r) \times s\| \leq \|r_t \times (s_t - s)\| + \|(r_t - r) \times s\|$ is an instance of the triangle inequality. if you apply $\|a \times b\| \leq \|a \| \|b\|$ on both terms of that sum of norms, you get what i wrote above.

Koro

@Jakobian seems very complicated.

I understand that if n=4, then $S_n$ has 4 normal subgroups (exactly) and these are I, $S_4, A_4,V_4$

where $V_4$ is Vierergruppe

and if n is not 4 then $I,A_n, S_n$

So an upper bound for number of homomorphisms from $S_n$ to $Z_m$ for n>4 is 3.

Case 1: m is odd
Only trivial homomorphism is possible.
Case 2: m is even
One trivial and the other can be with kernel $A_n$ as $S_n/A_n$ has order 2 so isomorphic to some subgroup generated by a 2 order subgroup of Z_m

unit 1991

16:34

@leslietownes Thank you!

leslie townes

16:45

note that the same argument would work with any bilinear map satisfying $\|B(x,y)\| \leq k \|x\| \|y\|$ for some fixed $k$, not just the cross product. in the banach space setting this condition is actually equivalent to (i.e., implied by) continuity.

soupless

When calculating for errors, does $E \leq 0.5$ mean that the value whose error is $E$, is accurate to 1 decimal place?

leslie townes

i wouldn't think so. you may not even get the integer part correctly. e.g. 0.9 and 1.1 are well within 0.5 of each other.

soupless

Welp, just realized that the error means $V - E < V < V + E$ and not just $V - E$

Thank you very much.

leslie townes

rounding doesn't help, e.g. 0.5 and 1.0 vs. 1.0 and 1.5. although if E is a little smaller and the value you have is well placed, you may be able to figure out what V rounds to, to one digit. if you are within 0.3 of 0.9 then you round to 1.0 for example. if you are within 0.3 of 0.5, you don't know what you round to.

it's kind of annoying that 'getting the nth decimal place right' doesn't immediately translate over to a simple inequality, whether you truncate or round. i remember struggling with this in calculus class when we first studied numerical integration and asked how big N needed to be. i always took N a little larger than might have been necessary.

soupless

17:03

@leslietownes to truncate seems to be an easy problem, but to round is a different one?

to truncate, you can just use the floor function, right? but when rounding numbers, the floor function will not work if $\{x\} \geq 0.5$ which needs to be a ceiling function

leslie townes

once you throw floors and ceilings in sure. i meant just more generally, |a - b| < e doesn't tell you very much about the decimal expansions of a and b without some knowledge of the relative placement of a and b and how big e is.

soupless

To give some context, I am trying to answer my own question

3

Q: Proving that $\sum\limits_{i=1}^{2006}f(i/2007)=1003$ if $f(x)=2008^{2x} /(2008 + 2008^{2x})$ from the 10th PMO

From the 10th Philippine Mathematical Olympiad: Let $f$ be the function defined by $$f(x) = \frac{2008^{2x}}{2008 + 2008^{2x}}, \qquad x \in \mathbb{R}.$$ Prove that $$f\left(\frac{1}{2007}\right) + f\left(\frac{2}{2007}\right) + \cdots + f\left(\frac{2005}{2007}\right) + f\left(\frac{2006}{2007...

That is, the problem posed by heropup which is proving that the error is small enough that rounding is allowed

leslie townes

is it clear that the result is an integer? that would simplify it. that is a good example of knowing the relative placement of a and b in |a - b| < e.

soupless

I am not sure if what I did was correct, but if it is correct, then the value of the integral using right Riemann sums must be accurate to 1 decimal place for 2002 terms

And since there are 2006 terms in the sum, then rounding must be allowed

@leslietownes The result is an integer, and the question requires a proof that, indeed, it is equal to 1003. But what I did seems to be overkill

leslie townes

if the actual V is an integer, an approximant A with error strictly less than 0.5 will round to the right thing. this would not necessarily be true if V were not known to be an integer.

the solution you quote makes clear that the result is an integer. i'm not sure that your method does that.

contest math makes my head hurt. :) i never did any of this stuff.

soupless

17:17

@leslietownes My method showed that when the problem is converted into an integral which is continuous rather than discrete, then it must approach a limit. Also, the sum is a right Riemann sum which is an overestimation of the limit. I just don't have a clear idea how to prove that the error between the limit and the sum is small enough for me to round up the limit to get the sum

I mean, I gained some ideas (for context, start from here) but I am not sure what the error must be

leslie townes

there are some standard estimates for errors in left and right riemann sums in terms of bounds on the first derivative of the integrand over the interval you are integrating. you might be able to use those. it does seem like a lot of work.

and if you don't know via some other route that the result is an integer, i'm not sure they even help.

Derivative

Is it true that if I have a differential form and a primitive for that form which is not continuous, then the form is not exact?

soupless

I'm waiting for Sarvesh's reply to confirm and to remove my question from the Unanswered tab.

leslie townes

what i'm learning here is that i would have placed very low in the 10th philippine olympiad.

soupless

In my case, I'll probably use calculus solutions to things like this and waste most of my time solving a subproblem from that problem

leslie townes

17:23

i'm glad i didn't know about these kinds of contests when i was starting out. i would have done horribly at them and it would have discouraged me.

soupless

At first, I thought that the value of the integral needs to be approximated through Taylor approximation or even Pade approximation seems I didn't have an idea how to get it

leslie townes

your idea is clever. i'm not sure it's workable in a contest setting, but i like it. it's good math but maybe not good contest math, if that makes sense.

Vrouvrou

Hello, i see this video about the proof of Banach alaoglu bourbaki theorem

Math400 - Functional Analysis - Section 4.3 - Part 2 - The Banach -Alaoglu-Bourbaki theorem

►

I don't understand why O_x =R exept for x in a finite set A

soupless

@leslietownes In my answer here, I just kinda solved it using AM-GM which I can never do in a contest

my hasty generalization thinking suggests that contest math < research math but what even is research math

leslie townes

hahaha. i have never seen bourbaki associated with that result before.

Vrouvrou

17:27

It is in Brezis also

leslie townes

soupless it's just a different type of problem. contest problems are sort of 'curated' as to have a solution that can be arrived at in limited time without any recourse to deep results. real life problems aren't like that. this isn't to say that contest skills aren't useful, because many real life problems do have contest-like subcomponents.

some people can be extremely good at leveraging a small bag of tricks but get bored/annoyed when a problem won't yield to them. that's not the best attitude for research. but many people have that feeling without ever doing contests.

soupless

@leslietownes This is interesting. I made the right decision: to pursue mathematics rather than just stop at contest math

leslie townes

i knew someone in grad school who dropped out because she found research boring and didn't want to learn theories that weren't tied to 'problem solving' of a sort where the solution to every problem fits on one page. on linkedin, it looks like she teaches contest math for a living now. i bet she's good at it.

soupless

I really like this surprise in math from the start where students are taught arithmetic operations on natural numbers, and then show that subtracting larger numbers from smaller numbers seems to open a new can of worms. But then, division comes into play and there you go: rational numbers.

Ted Shifrin

@Derivative I’m not sure how that happens. Do you mean multi-valued? At any rate, do you have a closed curve over which the $1$-form has a nonzero integral?

soupless

17:38

And in calculus, even if a function is discontinuous, it still has a derivative? Something like that. But then the Weierstrass function barges into this calm room and everything becomes a disaster

Ted Shifrin

Huh, soupless?

soupless

Wait, what? Sorry, I don't understand.

leslie townes

derivatives can fail to be continuous, but they come very close.

Ted Shifrin

The discontinuous function has a derivative ?

Vrouvrou

Someone have an idea about the proof of the theorem of Banach Aloglu Bourbaki theorem

soupless

17:42

@TedShifrin I mean, something like $\tan x$ whose derivative is $\sec^2 x$. without forgetting the domain, of course

leslie townes

vrou: en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem has several proofs. the one i see most often in books is the one involving tychonoff's theorem.

Semiclassical

that's perfectly continuous on the correct domain, though

Ted Shifrin

$\tan$ is a continuous function

Semiclassical

it's continuous everywhere that it's defined, at lesat

soupless

Wait. I confused myself. In the set of real numbers, $\tan x$ is discontinuous, right?

Xander Henderson

17:43

@soupless $\tan$ is not defined there.

Ted Shifrin

That’s what the word means

leslie townes

this is the classic thing. the jolt from calculus to analysis thing.

tan is not definable in any way as to extend it to a continuous function on R.

Xander Henderson

Asking if $\tan$ is continuous at (for example) $\pi/2$ is like asking of the function $x \mapsto x^2$ is continuous at "apple pie".

Ted Shifrin

No. Continuity is only defined on the domain, despite what high school calculus teachers think.

Semiclassical

also sec^2 is also not defined at pi/2

Ted Shifrin

17:44

Xander and I agree, again.

Xander Henderson

Yay!

soupless

I'm really having a hard time explaining my thoughts. Sorry about that.

leslie townes

some books do explicitly state that a function can fail to be continuous at a point by virtue of not being defined there. it is mostly a calculus book thing where everything is presumed to potentially be definable on all of R. you have to get rid of this presumption in any generalization.

Derivative

@TedShifrin my differential form is $\frac 1{\overline z}-\frac 1{\overline z-1}d\overline z$ and the primitive I have for it is $\log(\overline z)-\log(\overline z-1)$

Semiclassical

think you need some more parantheses

soupless

17:47

@leslietownes I get your point that R should not be assumed to be the domain of the function. Just working with R for now since that's what I almost convinced myself

Derivative

where log is the principal branch

so I said that the primitive has a discontinuity on the interval $(0,1)$ so that means the differential form is not exact

Ted Shifrin

@Derivative If you're talking about the principal branch, pay attention to what that means. This is continuous. (I assume you also need parentheses in your differential form.)

The form is exact on the domain of definition of the principal branch. Your question is whether it is exact on what domain?

Derivative

$\mathbb C\setminus\{0,1\}$

Ted Shifrin

So you do not have a primitive defined on that domain.

Continuous or not.

Derivative

oh

Ted Shifrin

17:51

The way to decide whether there could be one is what I said earlier. Is the integral of the $1$-form over every closed curve in the domain equal to $0$?

Derivative

no it's not. The circle around 1 with radius half doesn't have integral 0. I just wanted another proof

Ted Shifrin

So you must prove that there cannot be any primitive on the domain. Just because you've failed to give one doesn't mean that the man on the moon might not succeed at it.

Vrouvrou

The proof in wikipedia is more complicated I just want to understand why an open from the product topology is R except for x in a finite set

Derivative

damn men on the moon and their primitives

Ted Shifrin

@leslietownes What a decent calculus book should say is that $f$ must be defined at $a$ in order to be continuous at $a$. So the function fails to be continuous at any point not in its domain. It's then a rephrasing to say it's discontinuous at such a point. But it's stupid.

The right question (which Spivak has as an exercise) is: Does there exist a continuous extension $\tilde f$ of $f$ to all of $\Bbb R$?

soupless

17:58

@TedShifrin The three conditions? The definition of $f$ at $a$, the definition of the limit of $f$ as $x$ approaches $a$, and that $f(a) = \lim\limits_{x \to a}f(x)$?

leslie townes

14 mins ago, by leslie townes

tan is not definable in any way as to extend it to a continuous function on R.

spivak and i agree

Ted Shifrin

Yes, I agree that calculus books state that, @soupless. I'm complaining about the misinterpretation it leads to.

leslie townes

spivak and i agree a weird amount of the time

Ted Shifrin

We only discuss continuity/discontinuity at points of the domain. That is the intelligent agreement.

leslie townes

i think the hippie art on his diff geo books fooled me into thinking he was a madman when he was actually not

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