peep

@Tapi on that note, i came up with a percolation problem without really intending to

suppose we have a function $f(x,y)$ on pairs of nonnegative integers with the following properties. first, the easy ones: $f(0,y)=f(x,0)=0$ for all $x,y$, whereas $f(1,1)=1$

now a more tedious (but momentarily trivial) construction. if f(x-1,y), f(x,y-1) are both equal to 0, then let f(x,y)=0. but if either is 1, then f(x,y)=1.

it doesn't take a lot of thinking to see what will happen: we have for instance f(1,2)=f(2,1)=1 since f(1,1)=1, and iterating from here we get f(x,y)=1 for all positive x,y. like i said, trivial

but now we introduce an element of randomness: if either f(x,y-1), f(x-1,y) is equal to 1, then set f(x,y)=1 with probability p. otherwise, set f(x,y)=0.

this is all a bit tedious to say but it's simple enough to generate random tables of f(x,y) in Mathematica. (dunno if it's as efficient as possible but w/e)

if you convert one of these tables into a binary array plot (i.e., black dots wherever a 1 shows up) you definitely see percolation phenomena. for instance, here's a 100-by-100 example generated for $p=0.75$:

(another approach is to set $f(x,0)=f(0,y)=1$ on the boundary. then you get a lot more 'shoots' but also a lot of noise)

oops, that was with p=0.72

might put that up as a question in some form but i dunno what to ask about it beyond "hey this looks cool, what kind of percolation problem is it"

Is a particular solution of a differential equation always bounded? given that the domain interval is bounded.

@Redsbefall consider $y'(x)=1+y(x)^2$ with $y(0)=0$

is the physics idea of conserving forces (integral's value is independant of the path) is explored abstractly at mathematics as a subject?

well you give a mathematical interpretation so the answer is "yes"

well, depending on what you mean by "explored"

in the sense that it is defined

under some name

dependence of path integrals on choice of paths is studied, yes

it's a basic question about path integrals usually studied pretty much after the definition

Is there a specific name for such functions

???

what functions

I might be confusing it, but if you could help me find the relevant wiki page that represents it mathematically it'll be helpful

en.wikipedia.org/wiki/Line_integral#Path_independence is this what you're looking for

I'd think the basic facts should be all there :)

thanks

@Semiclassical maybe it would be better to understand for us if it was described as a set in the plane instead of as a function into {0, 1}

@Semiclassical positive integers you mean

Oh. You're doing this on non-negative integers. My bad

Yeah it looks pretty simple

this may be what I'm looking for link.springer.com/book/10.1007/978-1-4020-1898-5

@NaCl did you type lagois connections because you weren't sure if its Galois or Lagois?

I don't really know what "composition" means here

the idea of a Galois connection is pretty simple

You can also investigate them from the perspective of the induced closures iirc

I assume we're talking about anti-tone Galois connections

alpha.math.uga.edu/~pete/integral2015.pdf

there are those notes by Pete Clark

in chapter 2 he talks about Galois connections

> building "larger"/"more complex" connections from "smaller"/"simpler" ones

I have never seen a construction which would do that, I don't think there are such constructions

I'm not sure why you think there are

I can only assume this is either 1) random unsupported idea (based perhaps on the "hype" around Galois related things) or 2) you're not telling me about the context

can anyone explain this to me?

I think $e^{iqx}$ is an "oscillation", why adding the contribution of these "oscillation" approaching infinity is nearly zero?

The idea is to integrate along the contour given by the line segment from $-R$ to $R$ and the upper semi-circle of radius $R$. For large $R$, this contour integral is computed as in the given line. Then, take the limit as $R\rightarrow\infty$. The interval part tends to the desired integral and you have to show the integral along the semi-circle tends to $0$. I think there's no way around doing some manual estimates on the integrand for that.

@Shing you may find this useful: en.wikipedia.org/wiki/Jordan%27s_lemma

thanks guys!

I wonder what's an example of a vector space where balanced circled non-empty sets don't form a local base at $0$ for any topology

linear topology that is

as in top. vector space topology

apparently it shouldn't be a local basis of $0$ for any top. vector space unless perhaps the vector space is trivial

For $\mathbb{R}$ those are just symmetric non-degenerate intervals around the origin from what I can tell

And while intersection of two balanced sets shouldn't in general be balanced (I think), intersections of balanced circled sets should be balanced circled sets. So I guess where it fails is that you can find balanced circled set $U$ such that there's no balanced circled set with $V+V\subseteq U$, where $U, V$ are non-empty

Maybe I'm wrong about the intersections of balanced circled sets

absorbing circled sets I mean

I had a different idea on what absorbing set means, which is equivalent for circled sets

Of course any finite intersection of absorbing sets is absorbing

I think for real dimensions $\geq 2$, you just take a basis and two first vectors and in the plane you consider a sort of snail, and that should do the job. And for complex dimensions $\geq 2$ I think something similar might work and for $\mathbb{C}$ its the same as for $\mathbb{R}$ where absorbing balanced sets are just open or closed disks (of positive radius)

We have $f=g$ a.e. iff $\int g=\int f$. As far as I understand, this result applies to non-complete measures. What about complete measures? Of course, if $f=g$ $\mu$-a.e., then $f=g$ $\overline{\mu}$-a.e. also. But it doesn't make sense the other way around, right? Hence I'm not sure how to make the equivalence valid for complete measures :(

@psie this isn't true

If $\int g = \int f$ that doesn't mean $f = g$ a.e.

What is true is that $f = g$ a.e. iff $\int 1_Ag = \int 1_Af$ for all measurable $A$

ah ok

About your issue. I don't understand what you are trying to object to

which implication do you take issue with

@Jakobian well, does this still hold for complete measures?

it holds for any measure

for integrable functions $f, g$

@psie what book are you using to learn measure theory?

ok, makes sense. I'm confusing myself I think.

If $h = 0$ a.e. (here I'm assuming $h$ is measurable) then $h$ is integrable and $\int h = 0$

@SineoftheTime currently Folland's real analysis text, but I also read some of these notes, which are based on Folland's text

and if $\int 1_Ah = 0$ holds for all measurable $A$, then its enough to take $A = \{h > 0\}$ and $A = \{h < 0\}$

and then use that for non-negative integrable functions $h$, $\int h = 0$ implies $h = 0$ a.e.

@psie ty

If we interpret $h = 0$ a.e. as there exists set of measure zero $N$ for which $\{h = 0\}\subseteq N$, then of course this won't tell us anything about measurability of $h$

but I am assuming both $f$ and $g$ are measurable

so completeness doesn't enter the picture here

thanks for clarifying

@psie What happened to you studying probability by the way?

did you stop to study more measure theory?

yeah :) I read the first four chapters of my probability book, and then I really felt like I needed to look at some more measure theory

For instance, the change of variables formula, appearing quite late into a measure theory course I would guess, appeared in the very first chapter in my probability book, but without proof.

It's a pretty important formula in probability, computation-wise

There's also a variation of it in geometric measure theory, which is more general and more advanced

you can look up area/coarea formula

ok. Yeah, transformations of random variables is basically change of variables, but in order to understand the change of variables formula measure theoretically, I guess you need a few prerequisites (it is basically the study of Lebesgue measure on $\mathbb R^n$ and Folland devotes a whole section to this at the end of chapter 2)

Yep

but probability was fun, at least the stuff I read so far, lots of problems

I've been on the disciplinary committee here since I took the job more than four years ago. We finally have a case! :(