Mathematics - 2021-10-29 (page 1 of 2)

Xander Henderson

00:00

One would hope. But pretentious is as pretentious does.

Ted Shifrin

My Cyrillic days are long ago. But I can’t imaginé differentiating unless by accented syllable.

Leaky Nun

but the accents are the same

Ted Shifrin

I no longer have my Russian dictionaries.

Xander Henderson

I still have mine... somewhere...

leslie townes

i only have freedom books

AbstractSpacecraft

00:08

Let $P\Bbb{Q}$ be the vectors space of rational lines

Xander Henderson

No. I refuse.

AbstractSpacecraft

Isn't it generated in some way by prime fractions $p/q$?

But their lines

leslie townes

i second the refusal. we won't let $P\mathbb{Q}$ be that. sorry.

AbstractSpacecraft

How would you notate it?

leslie townes

i was kidding, but seriously, i don't know what the set is, or how it is a vector space. if i did, it might have a more familiar notation or name.

Xander Henderson

00:10

@leslietownes This.

AbstractSpacecraft

Jokers

leslie townes

what are rational lines, how do you add them, what are their inverses, etc.

Leaky Nun

lighten up

leslie townes

pinning this down might also answer your question. this could be fun with a purpose.

AbstractSpacecraft

An arbitrary point in the projective line $P^1(K)$ may be represented by an equivalence class of homogeneous coordinates, which take the form of a pair

${\displaystyle [x_{1}:x_{2}]}[x_1 : x_2]$
of elements of K that are not both zero. Two such pairs are equivalent if they differ by an overall nonzero factor λ:

${\displaystyle [x_{1}:x_{2}]\sim [\lambda x_{1}:\lambda x_{2}].}[x_{1}:x_{2}]\sim [\lambda x_{1}:\lambda x_{2}]$.

(Wikipedia)

Leaky Nun

00:13

the point is that it isn't a vector space

leslie townes

OK, what is the group operation?

there's some vibe of 'common factors don't matter here' and maybe that lets us consider representative fractions in lowest terms with no shared stuff in common, etc., but i still don't know how to add them

i should say that i'm not playing dumb here. i'm not into projective geometry. there could be some drop dead obvious operation but i don't know what it is

AbstractSpacecraft

That looks group operationy to me (same page on wikipedia)

$[x_1 : x_2] \cdot [y_1 : y_2] = [x_1 y_1 : x_2 y_2],
{\displaystyle [x_{1}:x_{2}]^{-1}=[x_{2}:x_{1}].}[x_1 : x_2]^{-1} = [x_2 : x_1].$

Xander Henderson

@leslietownes There is, but I took only one course in the area several years ago, and I didn't pay much attention to that course.

leslie townes

we're getting closer, i think. what is the scalar multiplication?

AbstractSpacecraft

That's just a operation defined on any two reps of two different lines. Apparently it respects $[]$

Leaky Nun

00:17

@AbstractSpacecraft your exercise is to figure out why the group operation isn't well defined

AbstractSpacecraft

the equivalence classes

Leaky Nun

(hint: what's the only disallowed [a:b]?)

AbstractSpacecraft

Projective line

In mathematics, a projective line is, roughly speaking, the extension of a usual line by a point called a point at infinity. The statement and the proof of many theorems of geometry are simplified by the resultant elimination of special cases; for example, two distinct projective lines in a projective plane meet in exactly one point (there is no "parallel" case). There are many equivalent ways to formally define a projective line; one of the most common is to define a projective line over a field K, commonly denoted P1(K), as the set of one-dimensional subspaces of a two-dimensional K-vector space...

Click on link

Apparently if you add the point at infinity, you have a multiplicative group

which just gets inherited from $K$

Leaky Nun

nowhere does it say that this still gives you a group

AbstractSpacecraft

So it's a thing :)

Clearly it is a group because associativity becomes automatic here

Leaky Nun

00:20

you need to read the assumptions more clearly

> Translating this arithmetic in terms of homogeneous coordinates gives, when [0 : 0] does not occur:

i already told you that the problem is with the well-definedness

(some people might call that "closure")

AbstractSpacecraft

So take projective space minus zero (in the field)

Leaky Nun

then you just get another copy of K

it's "K U {infty} - {0}", aka "{1/x | x in K}"

AbstractSpacecraft

These $K$ lines though are 2D

Leaky Nun

did you come here to discuss maths or did you come here to show off your knowledge by pasting wikipedia links whenever we raise issues with your claims?

AbstractSpacecraft

There is such a thing as projective space over a ring in which they only look for equivalent points up to multiplicaiton by a unit

@LeakyNun what's your deal, I was pasting links to show you what I meant

Leaky Nun

00:25

my deal is that you sound like "haha i have wikipedia to back up my claim i won" when this is not a competition

when we raise issues with your claims, this is not a criticism of you, but rather a criticism of your claims

AbstractSpacecraft

Well, that was not my intention, that's just a characature in your head

Ted Shifrin

$\Bbb P^1(\Bbb Q)$ is a single point.

Perhaps you want $\Bbb P^2$ or even $\Bbb P^3$. Shrug.

Leaky Nun

@TedShifrin it's P0(Q) that's a single point

Ted Shifrin

Oh, typo, right.

There was no superscript originally, and I messed up.

Ignore me.

Leaky Nun

yeah it's a bit confusing

AbstractSpacecraft

00:30

What I want to do is look at all lines $p/qx$ clearly passing throuh $(q, p)$ and only through that prime point.

Leaky Nun

$\Bbb P^n(\Bbb Q) = \Bbb P(\Bbb Q^{n+1})$

@AbstractSpacecraft you don't need any fancy projective space for that, the statement is that $\Bbb Q^\times$ is generated by the prime numbers as a (free) abelian group

AbstractSpacecraft

So I was thinking projective space "over $\Bbb{Z}$" but that brings up projective space over a ring, but the only scalars for equivalence allowed are units

Well, when you multiply these prime points in the usual way $(p/q)(p'/q') = (pp')/(qq')$ they generate a proper subgroup of $\Bbb{Q}^{\times}$ namely, the kernel of $\Omega$ from number theory (the total number of prime factors of $n$ including multiplicity). You can extend $\Omega$ to all of $\Bbb{Q}^{\times}$ by defining $\Omega(a/b) = \Omega(a) - \Omega(b)$.

AbstractSpacecraft

02:34

Thx

@LeakyNun

porridgemathematics

04:20

hi, i think I must be missing something simple here: If $h$ is a rational function in a single complex variable, and $E \subset \mathbb{C}$ is compact such that $|h| < 1$ on $E$, and we choose some $\lambda < 1$ such that $|h| < \lambda$ on this compact set, why is it true that $|h|^{-1}(\lambda)$ is a (piecewise-$C^1$) closed curve 'enclosing $E$'?

i've read this in an old paper by Raphael M. Robinson

he subsequently integrates some meromorphic function over $C = |h|^{-1}(\lambda)$

I am guessing by enclose he means its winding number with respect to every point of $E$ is one, but I'm not sure why we know that $C$ is one of these 'usual' closed curves we can integrate over in complex analysis

porridgemathematics

04:38

here is an associated SE post I just made with more details (and screenshots of the relevant paper) math.stackexchange.com/questions/4290632/…

leslie townes

good question. it must be a fairly classical result to appear without citation. i dunno. it has almost an algebraic geometry flavor to it.

porridgemathematics

indeed, it seems clear that this thing is a curve in the sense that its basically just the solutions to $|P|^2 = \lambda^2 |Q|^2$ , where $h = \frac{P}{Q}$, and we avoid finitely many points where $|P|$ and $|Q|$ simultaneously vanish but their quotient doesn't have limit $\lambda$

but given this is really a strictly CA paper, it must have some classical proof

leslie townes

a guy i used to work with would have known this.

not much help, but there you go.

porridgemathematics

no worries, thanks for the upvote (if one was you)

leslie townes

one was me. i saw another appear and then vanish. i don't know what the deal was with that.

porridgemathematics

04:44

oh, didnt see that one

leslie townes

there's probably some way of phrasing it so it's purely about the values of meromorphic functions and integration and you can avoid too much algebra, with the rationality just entering into there being finitely many things to check.

Koro

How to show that $x^\frac 1n\lt 1+\frac 1n$ for all large $n$?

Given that $x$ is positive and $< e$

copper.hat

presumably there are only a finite number of points at which $h'(z) =0$ on $C$, so the IFT gives a curve in between?

leslie townes

doesn't rudin have this in establishing various equivalent forms of the value of e? or am i thinking another book.

porridgemathematics

yeah there will only be finitely many such points

yes, certainly the IFT gives a curve in between

leslie townes

04:48

i also wonder about what 'enclosing' means. in particular i'm not sure how you could arrange the winding number to be 1 and not e.g. just positive.

but i'm thinking about it.

maybe he means just positive.

porridgemathematics

what he says about the value of the integral over the curve makes me think it may be one

since otherwise the residue theorem gives each residue a weight corresponding to the winding number of the curve with respect to the singular point

leslie townes

yeah it makes sense i just don't see how you can be sure it happens.

thanks for including screen shots. some of us lost our jstor privileges a long time ago and have a guilty conscience about all of the papers we have pirated.

porridgemathematics

haha

Koro

I got it now. Suppose on the contrary the statement is false then for infinite values of n, we shall have $x\ge (1+\frac 1n)^n$ which gives $x\ge \liminf(1+\frac 1n)^n=e$ which is a contradiction.

porridgemathematics

@copper.hat oh, at the singular points it can be shown that with respect to some charts, $h$ looks locally like a powering map, so basically you would get $m$ different branches of the curve at these singular points..

where $m$ is the multiplicity of $h(z) - h(z_0)$'s zero at the singular point $z_0$

porridgemathematics

07:47

Here is a possible hacky (non-solution) way to circumvent the issue I brought up earlier, $h = \frac{P}{Q}$ is a rational function, i.e. some holomorphic map $\hat{\mathbb{C}} \rightarrow \hat{\mathbb{C}}$ , and in particular it is a branched cover in the sense that if we treat it as a map $\hat{\mathbb{C}} \setminus \{p_1,...,p_k \} \rightarrow \hat{\mathbb{C}} \setminus \{\alpha_1,...,\alpha_j \}$

(where the set we removed from the domain are the branch points and the set we remove from the range are the critical values), then it is a bonafide covering map. Now we can just select $\lambda < 1$ so that the circle of radius $\lambda$ does not contain any of the critical values, and lift the circle parameterized positively via the rational function. The rational function is a degree $\max(\deg(P),\deg(Q))$ cover

, so there will be exactly these many disjoint lifts, all of these lifts will be smooth closed curves

now i am guessing because of how close $h$ is to a covering map, there will be a way to modify this approach for arbitrary $\lambda$

The risk with this hacky way is that it is possible $\lambda < 1$ being arbitrary so that $|h| < \lambda$ on $E$ is essential for the rest of the proof, luckily it turns out this is not the case

so the disjoint union of these lifts will be the entire preimage of $|h|^{-1}(\lambda)$, which is a cycle (not exactly a closed curve, but we can still apply the usual residue theorems for cycles so this isn't an issue)

errr, they won't necessarily be disjoint, sorry

so we'd need to remove the extraneous lifts that are just reparameterizations of some other lift

SAJW

09:32

"Magical items have a 50% chance of a suffix only, 25% chance of a prefix only, and 25% chance of both." How to determine from these facts the chance that a Item with a prefix also has a suffix? is it 25% or 50%?

porridgemathematics

now after we remove reparameterizations, the lifts are disjoint, each lift is then a simple closed curve (oriented positively because $h$ is conformal), so it has winding number $+1$ at every point in its bounded component and zero outside, now because $h$ is $\infty$ at $\infty$ owing to its form (this part is important), it turns out all the connected components of $|h|^{-1}[0,\lambda]$ coincide with the regions the lifts bound, which gives the exact result robinson stated

porridgemathematics

09:49

now im guessing maybe there is a clever way to get the general result 'by continuity' (apologies for the vagueness), since the result I proved holds for all but finitely many $\lambda < 1$ such that $|h| < \lambda$ on $E$, maybe because $h$ is a nice enough map, it must also hold at the finitely many bad $\lambda$'s

that is, specifically, that $|h|^{-1}(\lambda)$ is still a cycle, and the winding number thing still holds

intuitively it feels like this ought to be the case

love_sodam

13:03

@PeterJohn Did you solve the $\Bbb Z_2$-space problem?

Peter John

Still no.

love_sodam

@PeterJohn One known thing is that if $(X,\nu)$ is a free metric $\Bbb Z_2$-space, then $\ind_{\Bbb Z_2}(X)\leq n$ if and only if there are closed sets $A_1,...,A_{n+1}\subset X$ with $A_i\cap\nu(A_i) =\emptyset$ and $\bigcup_{i=1}^{n+1}(A_i\cup\nu(A_i)) =X$. I think from this, you can prove it rather easily.

Peter John

13:20

That's a very useful fact. Thx :)

Koro

Hi @shin

shintuku

the koro himself

Koro

Is there an example of a set A in $\mathbb R$ such that closure (A) $=\mathbb R $ but $A':=$ set of limit points of A in R, is not equal to $\mathbb R $?

shintuku

any union of open intervals s.t. $\cdots\cup(a,b)\cup(b,c)\cup(c,d)\cup\cdots$ seems like it

Koro

Background: Chapter 2, Basic Topology in PMA: Let $X$ be a metric space. $S\subset X$ is defined as dense in $X$ if closure S=X. Now, Taking X as R, $S'=R$ does imply S is dense in R but I'm looking for an example where closure S=R but S' is not equal to R.

shintuku

13:32

the closure will be $\Bbb R$ and the limit points are $\cdots,a, b, c,d, \cdots$

Koro

Oh, you mean $s=\cup_{n\in \mathbb Z}(n, n+1)$

No. Here also we'll have S'=R

shintuku

$S'$ is the closure of $S$?

Koro

S':= set of all limit points of S

closure (S):= $S\cup S'$

shintuku

hm, if you take any point inside $(a,b)$ there will be at least one neighborhood that does not have any elements of the complement, which means that, the set of limit points can't be $\Bbb R$, no?

i.e. any point inside $(a,b)$ is not a limit point but an interior point

Koro

But we are taking the union over R i.e., in other words we are considering union of open non intersecting intervals.

Every interior point of $(a,b)$ is also a limit point of S by definition, where $S$ is the union of open sets as you mentioned earlier.

shintuku

13:40

hm you're right

shintuku

13:50

Since every interior point is a limit point in the usual topology of $\Bbb R$, in $\Bbb R$ the answer has to be no. Suppose otherwise. Then there exists a set s.t. the closure is $\Bbb R$ but $S$ and $S'$ are disjoint, which is a contradiction.

right?

i.e. we use the fact that over $\Bbb R$ in the usual topology, $S$ is a subset of $S'$

@Koro

leslie townes

the what now

koro have you tried proving that S cup S' = R implies S' = R? forgetting about possible counterexamples and just going for it?

SAJW

Ok, lets say suffix is event A, prefix B. Then $P(A\nand B)=0.5$ , $P(B \nand A)=0.25$ and $P(A\and B)=0.25%$. So now the questions is, if B is given, what's the chance of A. Can that be deducted from those chances?

latex too hard haha

leslie townes

shin rudin has a definition of 'limit point of S' that sometimes excludes elements of S. it is not how some people use the term.

at least your faults are highlighted in red. that seems aggressive of chatjax.

shintuku

oh. i was thinking of limits point as having infinitely many elements of the set in any neighborhood

@leslietownes rudin uses: any neighborhood has at least one element that is not the center of the neighborhood

leslie townes

open sets containing limit points do, but there might not be any limit points. e.g. with rudin's definition Z as a subset of R has no limit points. you can't sneak up on any element of Z with other different elements of Z.

shintuku

13:58

ohhh, right. this is true only if $S$ is an open set

leslie townes

we may be mixing koro's hypotheses with a general discussion here. i was only pointing out that S isn't generally a subset of S'.

Koro

@leslietownes I thought about it, but I don't think it's provable. :(

All we can say that: $cl(S)=S\cup S'=R\implies S'\subset R$.

shintuku

hm. do we have: $S$ or $S'$ is an open set?

I think we do, if the closure is $\Bbb R$

leslie townes

let's go for it. suppose S cup S' = R. pick x in R and an interval I around x. pick y != x in I. if y in S, OK. if y' is in S', then I cap S contains infinitely points of S, and in particular there is z unequal to x that is in I cap S. either way, I contains an element of S that is unequal to x. I was an arbitrary interval containing x. conclusion: x is in S'.

Koro

@shin: S' is always closed. Isn't it?

shintuku

14:01

right $S'$ is always closed

porridgemathematics

so is $S'$ in this context just points of $\mathbb{R}$ that are the limit of sequences of points in $S$, but are not in $S$?

and you are looking for some set $S$ that is dense in $\mathbb{R}$, but $S' \neq \mathbb{R}$?

Koro

@porridgemathematics: S' may or may not contain points of S. Yes, I am looking for S dense in R but S'$\ne \mathbb R$.

@Leslie: I'm going through your comment.

porridgemathematics

so 'limit point' here just means some point of $\mathbb{R}$ that has arbitrarily small deleted neighbourhoods intersecting $S$, right?

Koro

yes, that can be said.

I say that $p\in \mathbb R$ is a limit point of $S\subset \mathbb R$ if for every $d\gt 0$, it follows that $S\cap (d-\epsilon, d+\epsilon)-\{p\}$ is an infinite set.

SAJW

OK another try: $P(A∧\neg B)=0.5$ , $P(B∧\neg A)=0.25$ and $P(A∧B)=0.25$. Can I deduct the chance for A when B is given?

Koro

14:12

Both the definitions (the one you stated) and the one given can be shown to be equivalent.

AbstractSpacecraft

Let $S = \Bbb{Q}$ @Koro

SAJW

I tend to think it's one of 0.25 or 0.5

AbstractSpacecraft

Then $S$ is dense in $\Bbb{R}$

Koro

Yes. But S'= set of limit points of Q is also equal to R

porridgemathematics

okay, in other words we want all the adherent points (points of $\mathbb{R}$ that have any neighbourhood intersecting with $S$) to be $\mathbb{R}$, but we don't want all the adherent points less isolated points to be $\mathbb{R}$ , meaning we need some isolated point in the dense set

Koro

14:15

I'm looking for an S which is dense in R but S' is not equal to R.

porridgemathematics

no such set exists as far as I can see, because if $s \in \mathbb{R} \setminus S'$, then there is some deleted neighbourhood of $s$ not containing any point of $S$, but if $S$ is dense in $\mathbb{R}$, then points arbitrarily close to $s$ are limits of sequences of points in $S$...

AbstractSpacecraft

So you seek a subspace $S \subset \Bbb{R}$ such that $S \not \subset S'$.

porridgemathematics

so $\mathbb{R} \setminus S' = \emptyset$, i.e. $S' = \mathbb{R}$

Semiclassical

here is a simple but annoying algebra problem. Given some real $b$, find real $(u,v)$ such that $u^2-v^2=1,2uv=b(u^2+v^2)$.

(i think an additional condition $-1<b<1$ is required for a solution to exist)

the only way i've been able to do this by hand is 1) via hyperbolic trig identities, or 2) by substituting $r=u/v$

but both of those seem more involved than should be needed

AbstractSpacecraft

Let $z = u + iv$

porridgemathematics

14:24

you can get such a dense subset in some space like $\mathbb{R} \coprod \{ pt \}$ though, just take any dense subset of $\mathbb{R}$ union the $pt$, the $pt$ is in the whole space but is not a limit point

Semiclassical

hmm. take advantage of $z^2=(u^2-v^2)+(2uv)i=1+(2uv)i$, maybe?

AbstractSpacecraft

Then $\text{Re}(z)^2 - \text{Im}(z)^2 = 1, 2 \text{Re}(z)\text{Im}(z) = b z\overline{z}$.

Semiclassical

hmm. $|z^2|=u^2+v^2$, $\operatorname{Re}(z^2)=1$, $\operatorname{Im}(z^2)=2uv$

AbstractSpacecraft

The left one reduces to $2 z\bar{z} = 1$

Semiclassical

$\operatorname{Im}(z^2)=b|z^2|$

AbstractSpacecraft

14:28

using $\text{Re}(z) = \dfrac{z + \overline{z}}{2}$ trick and squaring, sim $\text{Im}$.

Oh nvm

I forgot the $2i$ in denom

leslie townes

semi: i kinda like using hyperbolic trig identities, why not

Semiclassical

@leslietownes it is cute, it just feels like using a crane to crush a fly

but, to wit:

leslie townes

i think i've spent too much time through the looking-glass, it seems natural to me

Semiclassical

$u=\cosh \tau,v=\sinh \tau\implies u^2-v^2=1, 2uv=\sinh(2\tau)=b(u^2+v^2)=b\cosh(2\tau)$

so $\tanh(2\tau)=b$

AbstractSpacecraft

@Semiclassical so $u^2 - v^2 = \dfrac{z\overline{z} (i + 1)}{i}$

leslie townes

14:30

i mean you've got your hyperbola right there, what kind of identities are you supposed to use :)

AbstractSpacecraft

$= z\overline{z}(1 -i)$

Semiclassical

at which point we can use inverse hyperbolic stuff to write $u^2+v^2=\cosh(2\tau)=1/\sqrt{1-b^2}$, $2uv=\sinh(2\tau)=b/\sqrt{1-b^2}$

AbstractSpacecraft

So, why is that solution not okay?

Semiclassical

from there doing $(u\pm v)^2=u^2+v^2\pm 2u v$ gives the answer

2 mins ago, by Semiclassical

@leslietownes it is cute, it just feels like using a crane to crush a fly

leslie townes

it's like the weierstrass substitution. of course you're going to parametrize that way. use that crane.

that's also a weird metaphor, i'd use another piece of construction equipment. but you do hear it.

Astyx

14:33

wrecking ball?

Semiclassical

part of what weirds me out is that we then have: $(u^2+v^2)-(2uv)b=(1-b^2)/\sqrt{1-b^2}=\sqrt{1-b^2}$

which feels way too simple

hmm. is $\operatorname{Im}(z^2)=b|z^2|$ just a hyperbola in the complex plane when $-1<b<1$

hmm. hang on. geometrically, $2u v=b(u^2+v^2)$ seems to just be a pair of lines?

Semiclassical

14:51

oh. if i take two homogeneous polynomials $f,g$ of the same degree, then $f(u,v)=g(u,v)$ is just "a set of lines" isn't it

Koro

@leslietownes :) very nice taking $I$ to be arbitrary. Inspired by this, this is what I thought: On the contrary (or "to the contrary"?) suppose that $S'\ne R$, it follows that there is an $x\in \mathbb R$ which is not in S'. In other words, $x$ is not a limit point of S. But since x is in S, it follows that there is a d>0 such that $(x-d, x+d)\cap S=\{x\}$ whence it follows that $(x,x+d)\cap S=\emptyset\implies $ S is not dense in R, which is a contradiction :)

porridgemathematics

this sounds weird to me, but I seem to have a proof, let $f$ be real-valued on $[a,b]$ , then $f$ is of bounded variation if and only if it is of bounded positive variation if and only if it is of bounded negative variation...

Koro

@porridgemathematics yes it seems that you are right :). No such S exists.

porridgemathematics

can anyone sanity check me on this?

my reasoning is if we denote $p,n$ positive variation/negative variations of $f$ over some partition of $[a,b]$, regardless of what partition we choose we will always have $p-n = f(b)-f(a)$, or $p = [f(b)-f(a)] + n$.. so to maximize $p$ we need to maximize $n$..

but something about this smells dumb to me

Koro

How do you define positive/negative bounded variation?

I know of only bounded variation

porridgemathematics

15:00

bounded variation is i think the usual definition, i.e. define $t = \sum_{i=1}^n |f(x_i) - f(x_{i-1})|$ where $a = x_0 < x_1 < ... < x_n = b$ is some partition of $[a,b]$, and say the total variation of $f$ over $[a,b]$ is $\sup t$ where the supremum is taken over all partitions of $[a,b]$

and positive bounded variation means the same thing, replacing the absolute value sign with $\max(f(x_i) - f(x_{i-1}),0)$

negative bounded variation with $|f(x_i) - f(x_{i-1})| - \max(f(x_i) - f(x_{i-1}),0) = \max(-[f(x_i) - f(x_{i-1})],0)$

we always have $\sup t = \sup p + \sup n$ where $p,n$ are the positive/negative sums over some partition (even when these summands are infinite), and we always have $p-n = f(b) - f(a)$, so I figured we always have $f \in BV$ iff $f \in BV^{+}$ iff $f \in BV^{-}$

leslie townes

porridge something like that feels right, although i am fairly ignorant of these concepts and have never used them in practice. at least if f is assumed bounded? which might not be implied by any of those BV hypotheses?

porridgemathematics

in fact $f \in BV$ does imply $f$ is bounded, for the simple fact that $a = x_0 < x_1 = b$ is a partition of $[a,b]$ :-)

so in particular $|f(b) - f(a)| \leq \sup t < \infty$

oh wait, this isnt a proof lol

leslie townes

i've only worked with that concept indirectly, e.g. because it pops out of various characterizations of function spaces or domains of operators

Koro

i think your conclusion is correct. For if not, then there is some P which is weird and for this let's say sum of max (f(x_i)-f(x_{i-1}),0)> M But this implies sum |f(x_i)-f(x_{i-1}|>M but that's contradiction. M= sup of all appropriate sums on partitions of [a,b]

leslie townes

i guess for unbounded domains you define the concept on intervals and then assume that it holds on all intervals. i think i'm OK with it. at least, it doesn't seem fishy.

Koro

15:08

but I'm very new to bounded variation so I may be wrong.

porridgemathematics

okay, here is a proof, it does imply bounded because if $x \in [a,b]$, then $|f(x) | \leq |f(x) - f(a)| + |f(a)| \leq \sup t + |f(a)| < \infty$

so yes, $f \in BV$ implies $f$ is bounded

leslie townes

one time i was grading homework in an analysis class, and on an exercise, someone introduced all of the definitions relating to variation and several results before proving whatever the result was. it was almost completely unnecessary because the exercise required a far less strong conclusion. but they must have found the result in a book that dealt with that stuff.

Koro

another proof: Take any arbitrary x in [a,b] and you have partition P={a,x,b} now do what you did above.

porridgemathematics

yeah that works too

Koro

this also shows boundedness of f on [a,b] if f is of bounded variation.

porridgemathematics

15:11

@leslietownes i have committed this fault before

id like to think ive grown up since then though

leslie townes

another time i had my own homework graded and the grader wrote "this is FAR too complicated, just use X, Y, and Z." and then said it to me when he saw me. i explained that i had done the exercise in another class and it was easier for me to cut-paste my old solution without modification, even if it wasn't easier for him.

i did all of my homework in latex so it was indeed a literal cut-paste.

so i am sympathetic to the guy who regurgitated all the variation stuff.

Koro

does this make sense?

6 mins ago, by Koro

i think your conclusion is correct. For if not, then there is some P which is weird and for this let's say sum of max (f(x_i)-f(x_{i-1}),0)> M But this implies sum |f(x_i)-f(x_{i-1}|>M but that's contradiction. M= sup of all appropriate sums on partitions of [a,b]

leslie townes

i am also generally non-sympathetic to optimizing results to the point where there's only one 'right' argument, whose rightness derives from it being the shortest path to the result from a given set of assumptions.

porridgemathematics

the time I committed the fault it was because I had taken a real analysis course that proved (in the provided course notes) the result that a function is riemann integrable if and only if its set of discontinuities was of measure zero, the question asked to prove that a monotone function on a closed interval is riemann integrable, so I instead showed that monotone functions have a countable set of discontinuities hence measure zero

leslie townes

sometimes longer can be easier to understand. sometimes longer can be lower-tech and understandable by more people even if it involves subtler analysis than a shorter deeper proof.

porridgemathematics

15:15

i was marked down for it because they wanted you to use the riemann sum approach

leslie townes

you were marked down because the definitions around the riemann integral just suck

porridgemathematics

even though as per the rules any of the results proved in the notes were fair game to use, the reason i was marked down was because the result was asterisked and not technically taught in lecture (but still included in the provided notes)

leslie townes

through no fault of your own

porridgemathematics

so they basically marked me down because I read more than I needed to (this was in an exam too)

i was a little peeved about it after the exam so I actually asked emailed the instructor and asked about why i was docked points for using a result that was included in the lecture notes, albeit asterisked

leslie townes

understandably so

porridgemathematics

15:16

anyway its water under the bridge now lol, this was years ago

leslie townes

there was a grad student who was notorious for marking off when people used arguments that he didn't like, even if they were correct

surprisingly he did not last long in academia

porridgemathematics

i can understand it in a programming assignment, where how clean /readable your code is, or the style of your code matters

leslie townes

he'd even admit, "this is correct but it is too general, or not general enough, or it doesn't use theorem 3.2 like you were clearly supposed to do"

yeah, if style matters than it matters

this stuff was introduced entirely by him and it was always a question of whether a prof would override it or not care enough to do anything

porridgemathematics

im going to have to TA some classes and inevitably mark some assignments, I hope I'm not the subject of horror stories for those students lol, i'd like to think i'd be a fair marker

leslie townes

he was a huge outlier. everyone else was fine

Koro

15:20

i faced that too. there was a proof asked in our exam and in class, the proof considered an angle x (in degree ) from vertical but in exam I considered x from horizontal and hence 90 degree -x from vertical. My marks got deducted despite final expression being correct :'(

porridgemathematics

wow, thats pretty ridiculous lol

even high school exam boards allow for non-model solution

leslie townes

koro, sounds great :)

i've graded enough things to know that people are going to make mistakes while grading. i don't understand why people won't admit their mistakes

or worse, take the attitude that "oh, oops, but it doesn't matter" if it doesn't matter, don't grade the stuff in the first place

porridgemathematics

oh btw @leslietownes i kinda partially resolved the thing about $|h|^{-1}(\lambda)$ being a curve winding once (positively ) around every point of $E$

although it still doesn't address the general question

leslie townes

i've been following those updates

porridgemathematics

ah

hopefully someone chimes in with the full solution

leslie townes

15:30

it seems tailor-made for a short argument

porridgemathematics

playing around with desmos and some concrete examples, it indeed seems like there will be some limiting argument, because around a bad $\lambda$ the behaviour is that the disjoint jordan curves just kiss

(as you zero in on the bad $\lambda$)

leslie townes

i think robinson is no longer with us and the guy i knew who knew this stuff has joined him in complex analysis heaven

otherwise i'd just email and bother somebody

porridgemathematics

yeah thats a shame.. there have been a lot of moments where i've been reading some textbook and wanted to email the author only to find they are no longer

most recently, ahlfors

uh well no, robinson actually, and then ahlfors

leslie townes

i think there's now only one living person i can be sure read my thesis other than me

i remember thinking ahlfors was a lot older/earlier than he really was, simply because his textbook is one of the oldest in style books that people still use

porridgemathematics

yeah it is pretty old in style

his book conformal invariants is also quite nice

leslie townes

15:35

one of the nevanlinnas had a really good book on value distribution theory (as it was in the 30s or whatever) but the style is so crusty

porridgemathematics

i think ahlfors does a pretty good job at conveying his geometric intuition though

leslie townes

somebody could make a mint by making reworked versions of classic texts for a modern audience

porridgemathematics

oooh, my advisors thesis was on nevanlinna theory

leslie townes

no more clunky notation, let's pretend that general topology is known to exist

put in some emojis for the youth demographic

nevanlinna theory is amazing

completely broke my brain

AbstractSpacecraft

16:01

I heard about that in a paragraph from some book, what is that @leslietownes

Nevanlinna theory

In the mathematical field of complex analysis, Nevanlinna theory is part of the theory of meromorphic functions. It was devised in 1925, by Rolf Nevanlinna. Hermann Weyl called it "one of the few great mathematical events of (the twentieth) century." The theory describes the asymptotic distribution of solutions of the equation f(z) = a, as a varies. A fundamental tool is the Nevanlinna characteristic T(r, f) which measures the rate of growth of a meromorphic function. Other main contributors in the first half of the 20th century were Lars Ahlfors, André Bloch, Henri Cartan, Edward Collingwood...

I don't get why it's so great relative to everything else that is also supposedly great in mathematics, lol

leslie townes

you don't have to like it :)

AbstractSpacecraft

What do you apply it to?

In number theory say

leslie townes

to give you some perspective, all of algebraic geometry is just a gigantic waste of time to me

apparently there's some abstract connection to diophantine approximation. i don't know about it but it exists

AbstractSpacecraft

interesting

leslie townes

the link to vojta's conjecture on that page is some hint of that

Thorgott

16:14

if i were to be an analyst, i wouldve tried learning nevanlinna theory

but as we can see, life had a different path planned for me

leslie townes

you chose the dark side of the force

shintuku

16:29

are there only analysts and geometers

leslie townes

nah there's all sorts of folks

Govind75

Lads, if $X$ is an absolutely continuous RV and I have $Y = g(X)$, do I know that Y must also be continuous?

robjohn

@shintuku There are some algebraists, but we try not to mention them...

leslie townes

with enough hypotheses on g, sure. without them, maybe not

Govind75

if $g$ can be any function can $Y$ be discrete?

or neither discrete nor continuous

robjohn

16:35

@Govind75 of course

Govind75

Oh yeah

Im dumb nvm

I can just define anything based on ranges

robjohn

My wife and I are heading to Camarillo for a good part of the day. I'll be back later this afternoon or evening. Not that anyone cares...

@Govind75 yep

Govind75

I care @robjohn have a good trip lad

leslie townes

just visiting kanye west, i see.

may the traffic not be terrible (which it will be)

Koro

16:49

@robjohn Happy journey :)

Koro

17:33

Let f be a function from R to R that satisfies f(x+y)=f(x)+f(y) for all x,y in R. It is also given that $f$ is bounded on some interval (a,b) where a>b. Then f is continuous on all of R?

leslie townes

yeah i think so, probably enough to assume f measurable on some interval. google "cauchy functional equation"

Koro

I know that this is Cauchy functional equation and I know that there exists some $c\in \mathbb R$ such that f(x)=cx for every $x\in \mathbb Q$.

But I don't know what measurable is.

leslie townes

the boundedness probabvly implies that for all x in R./

Koro

The boundedness is given on that interval only.

leslie townes

ignore any particulars, the general slogan is, a function satisfying that equation is either what you'd expect it to be (multiplication by a scalar) or very bad. anything that rules out very badness is likely to be equivalent to anything else, as far as solutions to the equation are concerned.

i'm not sure you can even prove the existence of non-continuous solutions to that equation without some version of the axiom of choice. there might be reasonably normal set theories where there are no other solutions.

Koro

17:46

yeah, all we need is to show that f is continuous at a point then f is continuous on all of R.

no, I can't. But I think there is no reason to believe why such function shouldn't exist?

Let q in (a,b) be an irrational number such that $f(q)\ne cq$

leslie townes

well with AC they do. my point was simply, they only exist for goofy reasons, and you're not going to get there without functions that are goofy.

math.stackexchange.com/questions/505847/… has an interesting list of equivalences and a reference.

Govind75

Guys do you reckon this statement is true? With a p-value of 0.05, 1 in every 20 tests will give a false positive?

leslie townes

maybe with some assumptions re distribution of the thing being tested and a very loose, non-colloquial interpretation of "1 in every 20 tests will".

Govind75

This is for very underpowered tests tho

do you reckon its still true?

leslie townes

i cringe from stuff involving p-values because the entire language around them seems designed to mislead.

i don't know.

Govind75

17:54

i agree

I hate p values

leslie townes

we are on the right side of history. :)

Govind75

any ideas lads

Koro

wow. Thanks a lot @Leslie.

I got the idea now :)

Rithaniel

18:19

You guys know anything about making LaTeX documents that are ADA compliant?

Thomas Markov

@Rithaniel What sort of thing makes a document noncompliant?

Rithaniel

Honestly, I'm not sure. I'm making my first foray into this stuff, and my usual method is to start by gathering information by asking people I think might know

The ADA website has a lot of information, and not all limited to making pdfs

leslie townes

i'm guessing maybe it wants something where if you pdf the thing, there is stuff resembling 'alt text' to make images readable via a reader. latex source itself would not have these issues although it also wouldn't be readable to someone who couldn't read latex

my guess would be that if you aren't embedding images, whatever pdf2latex produces would already be ADA compliant. this is not the result of any research.

PM 2Ring

@Semiclassical & @Koro After your recent adventures with the subfactorial, you might like to play with this: math.stackexchange.com/q/4251713/207316 Finding an efficient way to compute zillions of digits of $\sum_{n=2}^{\infty} \frac{1}{!n}$. The core annoyance is that "while factorials expand nicely as a product, derangements have a branching recurrence identity"

leslie townes

i'd see ada compliance as more of an issue with systems that let you embed all sorts of stuff without providing any machine-readable signal other than maybe a file type about what it is. native latex doesn't do that.

Rithaniel

18:27

@leslietownes Yeah, mouse-over text on images is something I expect would be desired. But I don't know if it's limited to just that

leslie townes

i've never authored text documents with graphical elements. everything is just text. that should be fine.

Rithaniel

I usually use tikz, myself

leslie townes

it might be hell on a screen reader or a naive audience, but that is more a math accessibility problem than a document accessibility problem.

you'd certainly hit the spirit of ADA by adding alt text for tiks images. the sensitive thing would be, if the alt text is just the latex source that generates the diagram, is that enough. it would surprise me if the answer were no.

but there's a balance between what a regulator / regulatory regime might formally require, and what accessibility means to the average person. mediated by the tension that most written math is not accessible in any colloquial sense.

good luck with that. i'm glad i didn't work in a field that made much of pictures.

Rithaniel

Pictures are always helpful

leslie townes

if only math stackexchange rules allowed for a question of the sort "can you give an example of an unhelpful picture in a published math text." you'd get a lot of counterexamples

Rithaniel

18:33

Like, anything can be somewhat elucidated by simplifying it to blobs and lines, at least in my experience

Chains of prime ideals -> circles contained in each other
Orthogonal complement -> a line perpendicular to the original line

leslie townes

my hot take is that most pictures on wikipedia entries for math concepts do not help.

Rithaniel

Functional -> a field of arrows that point from a space down to the "foundation" of that space

leslie townes

if mathematica can generate an example of something, pretty good odds that someone's pasted a picture of it into the wikipedia entry for the definition of that thing. most of those are noise and not signal.

Rithaniel

Idk, in my experience a lot of images help me couch the concept in a geometric interpretation. Though, I suppose it comes down to the accuracy or applicability of any particular image

Some are next to useless, like looking at the graph of the Gamma distribution

leslie townes

if it can be plotted, it's going to be plotted on wikipedia.

often with a rainbow planar diagram showing real and imaginary parts.

why? because we can.

PM 2Ring

18:39

I like good pictures and diagrams. But I concede that there are unhelpful and downright misleading diagrams.

Rithaniel

But still, looking at a picture for the gamma distribution with particular parameters can help you get a rough estimate for expected value and variance. Then, when doing something algebraically, you can measure it against that intuition you got from the graph

Visualizations of complex-valued functions on the complex plane do seem fairly low on the helpfulness scale, particularly because you need to generally split it into one image for the real part and one image for the imaginary part, and that actually hinders your ability to thinking of it as a unified whole

PM 2Ring

Also, diagrams can give you a false sense that you understand the thing, but the thing has subtleties that the diagram fails to capture, or represents poorly.

Rithaniel

@PM2Ring Yeah, that's a thing about intuition in general, though

PM 2Ring

@Rithaniel Pity humans aren't very good at visualising 4D stuff. ;) I've experimented with 3D anims, which can be ok if one axis can be mapped to time in a sensible way.

Rithaniel

Like, the Sorgenfrey plane is a good example for this. Like, there is the question "Is the topological product of two Lindelöf spaces also Lindelöf?" Naively, you can write up a convincing-sounding proof that yes, they are. But the Sorgenfrey plane is counterexample that shows that it is not actually the case

@PM2Ring For complex-valued functions on the complex plane, I like to have two "coordinate axises" next to each other. One for input and one for output. You move a dot around the input side and a corresponding dot moves around on the output side.

PM 2Ring

18:50

@Rithaniel Yep. That's good. And you can combine it with the rainbow arg colouring, if you want.

Rithaniel

Alternatively, something where you have a recognizable and orientable image representing the complex plane, and then show how it warps when the function is applied

Here's a "2 axies" example mabotkin.github.io/complex

PM 2Ring

Well, that's the logic behind the rainbow colouring. But they become fairly meaningless after you've been looking at a few of them. Brains are much more interested in looking at stuff like warped faces. And they have a ton of neutral circuitry dedicated to processing that stuff, so that we can recognise people & facial expressions.

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$Mathematics$ Mathematics