Mathematics - 2017-08-16 (page 3 of 4)

Daminark

15:02

Have you heard of Lebesgue density?

You define $d(x,A) = \lim_{r\to 0} \frac{\mu(B(x,r) \cap A)}{\mu(B(x,r))}$

Alessandro Codenotti

Nope

Daminark

That's called the density of $x$ wrt $A$

Now, it's a theorem (at least for the Lebesgue measure?) that $d(x,A) = 1$ for a.e. $x\in A$

This can be proven by either Vitali or Besicovitch, I don't remember offhand but this smells more like Vitali so I'll roll with that

I mean it's basically the Lebesgue differentiation theorem for indicator functions

Secret

Question condensed in set theory room:

in Set theory, 1 min ago, by Secret

1. How does the lexicographical ordering on $2^{\Bbb{Q}}$ avoid the example illustrated in $C$ and $D$, since in $2^{\Bbb{Q}}$, we can get the sets $A,B,C,D$ as above and hence having both a infinitely increasing and a infinitely decreasing sequence hence cannot assign a consistent lexicographical order?

23 hours ago, by Leaky Nun

@Secret $C=\dfrac1{2\Bbb N+1}$ or $D=\dfrac1{2\Bbb N+2}$?

in Set theory, 14 mins ago, by Secret

in Mathematics, 22 hours ago, by Leaky Nun

2 hours ago, by Leaky Nun

@Secret $\chi_A$ or $\chi_B$ where $A=\{1,4,5,8,9,12,\cdots\}$ and $B=\{2,3,6,7,10,11,\cdots\}$?

Alessandro Codenotti

@Daminark ok

Daminark

So now, you define a set to be denjoy if it's measurable and $d(x,A) = 1$ for every $x\in A$.

Alessandro Codenotti

15:06

So for example $d(x,A)=1$ for $x\in A$ not on the border of $A$

@Daminark no, wait, without assumptions on $A$?

Pretty sure $A$ needs to be measurable

Balarka Sen

@Daminark the normal field is not a tangent field on the manifold

well, surface

Alessandro Codenotti

Or the density doesn't seem well defined

Balarka Sen

but yeah you have to fiddle around with the normal field

Daminark

Turns out you can use the measureable hull in the non-measurable case

In fact this gives a nice characterization of measurable sets

Alessandro Codenotti

Ok, that does work

Daminark

15:10

A set is measureable iff the density of almost every point outside of it is 0

Alessandro Codenotti

I was worried because $\mu(B_r\cap A)$ doesn't work if $A$ isn't measurable

user8469759

If $A \subset \mathbb{R}^n$ (bounded) and $x \notin A$ how can I prove $d(x,A)$ is bounded?

Daminark

Yeah I sorta made the jump without saying what it would entail, sorry

user8469759

I was trying to use the definition but I got stuck

$d(x,A) = \sup_{y\in A} d(x,y)$

Abdullah UYU

Is there anyone that have a solution to this gogeometry problem ? gogeometry.com/problem/problem002.htm

user8469759

15:12

(by the way... $d$ is any metric...)

Daminark

But yeah if you take a closed set, you lose the boundary. Measure zero so that doesn't fuck up the theorem, but it fails to be Denjoy

user8469759

sorry it was

Alessandro Codenotti

Ok, so how do we get a topology?

user8469759

$d(x,A) = \inf_{y\in A} d(x,y)$

Daminark

So the set of Denjoy sets forms a topology

Alessandro Codenotti

15:14

(Why are they called denjoy?)

Daminark

The guy who created it I guess

It's been abbreviated d-topology, which many interpreted as density topology, so both names are used

Alessandro Codenotti

Oh. Derp. Makes sense

Leaky Nun

@user8469759 let $A \subseteq B(0,r)$ with $r > 0$. Then, $d(x,A) < d(x,0) + d(0,A) = \|x\| + r$

Alessandro Codenotti

Hmm, that surely looks like a good candidate for a topology whise Borel $\sigma$-algebra is the Lebesgue one

I'll think about it

Daminark

But yeah in any event, the sneaky part of that is proving that an arbitrary union of Denjoy sets is measureable

But that ends up working

Actually wait yeah this definitely is it, I did a problem which proved this fact, among others

So, any open sets in the Euclidean topology are Denjoy

user8469759

15:18

@LeakyNun ok

Daminark

That's clear. Now, the thing is, if you take a Denjoy set and kill off a null set, it's still Denjoy

Alessandro Codenotti

@Daminark sure

Right

Daminark

So, we know that any Lebesgue measureable set is a G_delta set minus a null set, right?

Alessandro Codenotti

Ohhhhh, wait. I'm sure I read something about the topology with sets $A\setminus \Delta$ where $\Delta$ is measure zero

I think it was a set theoretic topology book

@Daminark i have no idea how to prove that, but I know it's true :P

Daminark

It's true since Lebesgue measure is Radon, I think

Or really that it's outer regular

In any event, you know this is true

user8469759

15:22

@LeakyNun so you take a set such that $A$ is a subset

and you bound the distance

Daminark

So, our measureable set $A$ can be expressed as $(G_1 \cap \ldots)\setminus N$

Alessandro Codenotti

Ok

Leaky Nun

@user8469759 the definition of "$A$ is bounded" is that there is $r>0$ with $A \subseteq B(0,r)$.

Daminark

Well, write that as $(G_1\setminus N) \cap \ldots$

So it's a countable intersection of (open minus null) sets, which are Denjoy

So any Lebesgue measureable set is a G_delta set in the Denjoy topology

Alessandro Codenotti

Nice

Doesn't the Denjoy $\sigma$-algebra ends up being bigger than the Lebesgue one though?

Daminark

15:26

Nope, Denjoy sets are all measurable

Alessandro Codenotti

Oh, it's already 17:30, sorry but I have to run, I'll be back later

@Daminark oh, right

Daminark

So we're done. Borel in Denjoy iff measurable

Leaky Nun

Does an order-preserving map exist between every two total orderings defined on the same set?

Alessandro Codenotti

It seem to work, but I'll spend some time thinking about it since there are a few things I'm not familiar with going on

@LeakyNun no, pick a bounded and an unbounded one

Or reorder $\Bbb N$ to be like $\Bbb Q$

Leaky Nun

@AlessandroCodenotti thanks

Daminark

15:29

Yeah for sure. Also note the Denjoy topology is weird

Alessandro Codenotti

(The bounded/unbounded one doesn't work btw, nevermind)

Daminark

That of R^2 is not the product of Denjoy topologies on R, for example

Alessandro Codenotti

Even though the Lebesgue measure on $\Bbb R^2$ is the product measure of those on $\Bbb R$, weird

Ok, I really have to go, bye!

Daminark

Aight, see you!

Secret

in Set theory, 2 mins ago, by Secret

Ok, using the above logic on $2^{\Bbb{Q}}$, I think $2^{\Bbb{R}^+}$ has no lexicographical ordering makes sense now, but is it possible for $2^{\Bbb{R}^+}$ to have an explicit dense linear order under ZF, or it is not provable since dense linear order is strickly weaker than the axiom of choice?

in Set theory, 1 min ago, by Leaky Nun

@Secret DPO is independent of ZF

O and one more thing, I don't want an ordinal as an answer, cause I want to know whether such sets must be isomorphic with some subset of the surreals and thus I need that not well orderable condition

Or in summary, my question is:

> Is there an explicit, dense linear order, non-well ordered on a set $S$ with $|S| > \aleph_1$ under GCH without AC (i.e. using ZF). If so, must they all be isomorphic to some subset of the surreals?

non well ordered means it cannot be an ordinal,thus $\omega_2$ (or any possible unions involving it) is out

Akiva Weinberger

15:39

@Secret $X\times\Bbb Q$ lexicographically

Alessandro Codenotti

$\omega_2$ preceded by an inverted copy of $\omega_2$?

Secret

@AkivaWeinberger $X$ being?

Alessandro Codenotti

Ah, wait, you wanted a dense order, nevermind

I still think my suggestion from yesterday could work. Start with $\omega_2$, for every element insert a new element between it and its successor, call the resulting set $A_1$. Between every element of $A_1$ and its successor insert a new one, call the resulting set $A_2$ , iterate countably many times

(This could mess up the cardinality without AC though since countable unions can be weird without choice)

Leaky Nun

@MartinSleziak you might have some references

Semiclassical

@LeakyNun That seems like it should be false. Take $\kappa$ to be the real numbers, for instance.

Secret

15:49

@AlessandroCodenotti Hmm interesting, never thought of filling in the gaps of $\omega_2$ that way. I can see how it is dense since x < y implies exists z, x< z < y. Now that makes me wonder will doing the same thing with $\omega_1$ will give something isomorphic to $\Bbb{R}^+$ since both of these sets have a minimum but no maximum (since $\{\omega_{\alpha}\}$ itself is not included in the set $\omega_{\alpha}$ and they both have dense linear order tht can be potentially mapped order preserivingly into each other)

Leaky Nun

@Semiclassical oops, I meant $|\kappa|$, not $|\Bbb N|$

Semiclassical

Ahhh.

That seems more plausible :)

Maybe one can show an injection from each set into the other?

Leaky Nun

@Semiclassical I think this needs to be proved using ordinals

Semiclassical

ah. have fun with that, then :)

Leaky Nun

@Semiclassical well, if you can prove it without using ordinals, I'm all ears.

Semiclassical

15:51

nah, you're probably right.

Though, you at least have $\kappa \subset \kappa\times \mathbb{N}$.

So if one can show that $\kappa\times \mathbb{N}\subset \kappa$, then you're done.

(I suspect I'm saying that a bit wrong, but I can't remember the right terminology.)

Akiva Weinberger

@Secret Anything

Topologicalife

Hi.

if $c = \displaystyle\frac{a + b}{ab - 1}$ with $a, \, b \in \mathbb{N}$, must $c$ be in $\mathbb{N}$?

I think it can be in $\mathbb{Q}$.

Semiclassical

@Topologicalife Have you done any examples? That's the very first thing you should do, since if you come across a counterexample you're already done.

Secret

Currently thinking about an unrelated question:

Apr 14 at 12:19, by Secret

A set $S$ is dense if $\forall x,y\in S, x<y, \exists z\in S, x<z<y$?

Apr 14 at 12:20, by DHMO

that is a necessary but not sufficient condition, but go on

Apr 14 at 15:05, by Akiva Weinberger

@SBM A set is dense if every real number can be written as the limit of a sequence of things in your set

Apr 14 at 15:07, by Akiva Weinberger

@SBM So another way of defining dense is: A set is dense if every open interval contains something in your set.

Topologicalife

oh I forgot the condition it can not be the case $a = b = 1$

Secret

16:01

Under construction: A set that obeys the necessary condition for density in a linearly order set but violates the sufficient condition

this will mean I will need some isolated points somewhere "zooming countably down"

Topologicalife

I.e: $a$ and $b$ can not be equal to 1 at the same time

Semiclassical

Well, sure. The expression doesn't make sense otherwise. But that doesn't change my point.

Have you tested this expression for any examples?

Topologicalife

Yes.

Semiclassical

Such as?

Akiva Weinberger

@Secret Those are two different notions of denseness

Topologicalife

16:02

for example, $a=10$ and $b=40$.

Akiva Weinberger

One applies to linear orders and is intrinsic, the other applies to topological subspaces and is extrinsic

Secret

wait so if I can always find a z sandwiched between any x,y in a linear order then it is automatically dense?

Semiclassical

Okay. That gives $c=50/399$. Is that integer?

GFauxPas

you just need one counterexample to show it's not always true, top

Akiva Weinberger

@Secret Yeah

Secret

16:04

I see

Topologicalife

No. I am looking for the conditions which make $c \in \mathbb{N}$

Semiclassical

That's not what you asked.

Akiva Weinberger

You can easily find subsets of $\Bbb R$ that are dense in terms of the linear order definition but not by the topological subspace definition

Topologicalife

Yeah, now that is my next task once I understand my last question.

Akiva Weinberger

$\Bbb Q\cap(0,1)$ for example

Semiclassical

16:04

Okay.

Topologicalife

I had the last question because if c is in N, then we must have a + b > ab - 1

Akiva Weinberger

or the Cantor set minus the endpoints of the deleted intervals

(The "pseudointerior," I want to call it)

Semiclassical

One obvious way to get an example is to choose the denominator such that $ab-1=1$. Not a lot of ways to do that, though.

Akiva Weinberger

(I'm not aware of any actual name)

GFauxPas

they could be equal, doesnt have to be $>$

Secret

16:05

@AkivaWeinberger for this example, is it because the subspace topology misses the limit points 0 and 1?

Topologicalife

Right.

Akiva Weinberger

@Secret No, it's 'cause you can't approximate $2$.

Topologicalife

Should be $a+b \geq ab - 1$.

but I don't see if that condition is enough, because we could have an irrational number... right?

GFauxPas

that's necessary, but not sufficient

well there's no way it can be irrational. what's the definition of an irrational number?

Semiclassical

It might be worth looking for examples. Simplest possible example has $a+b=ab-1$. Next is $a+b=2(ab-1)$, and so forth.

GFauxPas

16:07

or rather, whats the definition of rational?

Topologicalife

This questions come from the problem 'Find all positive integers $a$,$b$,$c$ different than zero such that $a+b+c = abc$

Semiclassical

ew.

Topologicalife

Sorry

I meant rational!

Semiclassical

and then $c$ is the solution to that, gotcha.

Topologicalife

Yeah,

Semiclassical

16:09

if $a=b=1$ you'd have $c+2=c$ which never works, and that justifies ignoring this case.

Topologicalife

Yeah, I got that. That is why I said it earlier.

Secret

hmm, need to be checked later whether alessandro's example will be isomorphic to some subset of the surreals. The best way to check that is to see if there are infitesimal elements. If not, then we have a nice example of a $\aleph_2$ line that behave as similar to the reals as possible

Now to get back to chemistry

Topologicalife

But I don't know how to justify $c$ is not rational.

GFauxPas

$c$ will always be rational

Topologicalife

But $c$ must be a positive integer.

GFauxPas

16:10

integers are rational, so you mean to say that $c$ is a positive integer

Semiclassical

The question isn't whether $c$ is rational, but whether it's a positive integer.

Akiva Weinberger

$c=\frac{a+b}{ab-1}$, that's an integer divided by an integer

Secret

[Unrelated] I think one way to draw $\Bbb{R} \backslash \Bbb{Q} \cap (0,1)$ is many layers of points, and then a description saying that every point is of measure 1

Akiva Weinberger

That's the definition of rational

Semiclassical

@AkivaWeinberger eh, could have $ab=2$.

Akiva Weinberger

16:11

@Semiclassical So? Integers are rational

Topologicalife

For example, if $c = \dfrac{2}{3}$ then $c\in\mathbb{Q}$ but $c \notin \mathbb{N}$

Semiclassical

sure. just saying that it can further be a positive integer.

not sure there's any cases beyond (a,b,c)=(1,2,3) (up to permutation) though.

Akiva Weinberger

So it can be $\frac{1+2}{(1)(2)-1}=3$

You want to find all possible integer solutions?

Topologicalife

I got the solution $a=1, b=2, c=3$

Semiclassical

He does.

GFauxPas

16:12

wolfram alpha doesnt list solutions other than those semi

Akiva Weinberger

I mean $a+b<ab-1$ almost always

Topologicalife

and I proved there aren't other solution that this one.

GFauxPas

how did you prove that?

Semiclassical

If so, then you're done.

GFauxPas

^

Topologicalife

16:13

But what I want is to justify $c$ is a positive integer without to prove there aren't more solutions

GFauxPas

$c$ is rational, integers are rational

Topologicalife

i.e: starting from $a+b \geq ab -1$.

Semiclassical

If you haven't proven that, then you're not done.

Wait. yes, $c$ is necessarily at least rational.

Topologicalife

corrected :P

I proved it, but I want to find another way!!

It seems ugly to me.

GFauxPas

but isn't the whole problem assuming $c > 0$ is an integer? you dont have to prove it, its part of the problem

Semiclassical

16:15

hmm.

Akiva Weinberger

Substitute $a'=a-1$, $b'=b-1$ maybe

Semiclassical

well, you can assume $c>b$, $b\geq 2$, $a\geq 1$ without loss of generality.

GFauxPas

I'm not sure what you're trying to show topo

Akiva Weinberger

Heh, $a+b\ge ab-1$ is equivalent to $2\ge(a-1)(b-1)$

That's nice

Semiclassical

oh, that's very nice.

Topologicalife

16:17

I proved that was the only solution studying the relation $\tan \alpha + \tan \beta + \tan \gamma = \tan \alpha \tan \beta \tan \gamma$

There is only and only one triangle satisfing that relation.

GFauxPas

well that's the same problem, because the image of $\tan$ is all of $\mathbb R$

Topologicalife

Yeah.

Semiclassical

@akiva Okay, here's my stab at it. assume wlog that $a>b>c$ is such a solution. since $c$ is a positive integer, $a>b>1$

but then $2\geq (a-1)(b-1)$ is only fulfilled for $a=3,b=2$.

So otherwise one has $a+b<ab-1$ and therefore $c=\frac{a+b}{ab-1}<1$ is not a positive integer.

Topologicalife

Oh got it.

GFauxPas

nice semi

good find Akiva

Secret

16:20

Semiclassical

(the second line is a bit weakly put. better is: since $b-1\geq 1$, $(a-1)(b-1)\geq a-1\geq 2$ with equality only when $b=2$ and $a=3$.)

Topologicalife

Oh lol, I just proved it in another way, but a similar one

I studied $s_n = \left\{\displaystyle\frac{a+1}{a-1}\right\}_{a\geq{}2}$

I got that sequence because $a + b\geq{}ab - 1\;\Rightarrow{}\, b\leq{\displaystyle\frac{a+1}{a-1}}$

Oh and $\dfrac{a+1}{a-1}=1+\displaystyle\frac{2}{a-1}$

Thanks guys!

GFauxPas

my hunch is that there's a way to do it with modular arithmetic

Tobi

I realised that my physics simulation was incorrect, because I was assuming that between my time intervals, gravitational acceleration was constant, when it is infact proportional to seperation

GFauxPas

consider $\dfrac{a+b \pmod 2}{ab-1 \pmod 2}$ and $\dfrac{a+b \pmod 3}{ab-1 \pmod 3}$, maybe

Tobi

16:24

could I fix this by integrating the graph of acceleration against time

and using the area as change in velocity

GFauxPas

acceleration is the change in velocity

Secret

Akiva Weinberger

You're attacking Thomae's function with lasers.

GFauxPas

how aggressive

Secret

16:40

Basically shows how impossibble is to mark all irrational limit points of the Thomae function

(without filling in the whole screen with red)

I can kinda see how they can blow up to $\aleph_1$ now by using some intuition gained from the cantor set:

Recall that cantor set has ternary expansion 0.bbbbbbbbbbbbbbb.... where b = 0 or 2

The trick is that at any level of how this expansion is truncated, you can vary the last digit by 0 or 2. Thus one can imagine after countably many places (which will obviously not exhaust it), changing 0 to 2 effectively produces a "infintesimal" change to the overall value relative to e.g. 0.bbbb

and this is where all the irrational points bunch up to result in a measure = 1

So, for the irrationals, it is basically "cantor set on steroids", where said points which this bunching occurs is everywhere in the interval. Thomae's functon then give us an impression where these "points" are

Akiva Weinberger

It's nice that Thomae's function is Riemann integrable

Secret

The validity of my claim may be if a proof actually showed all points of the reals are condensation points of the rationals under the subspace topology I guess...

Akiva Weinberger

> In mathematics, a condensation point p of a subset S of a topological space, is any point p, such that every open neighborhood of p contains uncountably many points of S.

I believe every point in the Cantor set is a condensation point of the Cantor set

Secret

Now here's something more challenging: Come up with a set with exactly one accumulation point and one condensation point.

-> I imagine that will give nice illustrations on what these points look like

Akiva Weinberger

16:58

I forget what accumulation point means

You've been thinking about point-set topology, then?

Secret

yeah, it is common for me to drift into that from infinite set, especially how I love to focus on pathological examples where nice things such as haseudoff T1 etc. don't apply

In fact, getting used to point set topology in the most extreme cases will help me deal with the nicer cases as the intuition gain there is easily applied back to the nice cases

Akiva Weinberger

Do you own the book Counterexamples in Topology?

IIRC it's published by Dover books and thus cheap

Secret

(even though because of my chemistry phD I cannto continue on munkres yet)
(I think might have purchase an e book of it somewhere in my folder, need to check again)

Akiva Weinberger

I wonder if it's OK to write $\overline{1/\Bbb N}$ for $\{0\}\cup\{1,1/2,1/3,\dots\}$

Secret

(Ah, I had not bought it yet, but will as I continue through munkres after the master code is get to work)

I do like to play with the long line though

Leaky Nun

17:04

@AkivaWeinberger your $\Bbb N$ does not include $0$?

Akiva Weinberger

@LeakyNun I usually include zero but in this case I guess I can't

Leaky Nun

@AkivaWeinberger por favor

(How would you translate "please" to Spanish?)

Akiva Weinberger

Por favor, yes

Leaky Nun

@AkivaWeinberger I mean, the other usage of "please"

Akiva Weinberger

I'm not sure I know what you mean

SpanishDict includes "por Dios"

There's this great function defined by Shifrin in his book

on $[0,1]\times[0,1]$, such that the integral over the square doesn't exist but both iterated integrals do

Secret

17:11

hmm, that's interesting, so clearly fubini's theorem fail, but how does it fail

Akiva Weinberger

Specifically, on each horizontal slice ($\{x\}\times[0,1]$) or vertical slice ($[0,1]\times\{y\}$), the function is $0$ everywhere except for finitely many points, where it equals $1$.

Mats Granvik

I really feel I have to say something now. Are Fourier series an efficient enough way to describe/approximate the sign function in this formula:
$$14.13472514173469...=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\frac{\vartheta (t)+\Im\left(\log \left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)}{\pi }-n+\frac{3}{2}\right)\right)dt$$?

Akiva Weinberger

So both iterated integrals equal $0$. However, the places where it equals $1$ are dense, making the Riemann integral over the entire square fail to exist.

Secret

so it blew up near those dense points?

Akiva Weinberger

No, it's like

You know how $\begin{cases}1,&x\in\Bbb Q\\0,&x\notin\Bbb Q\end{cases}$ isn't Riemann-integrable

Secret

17:13

yup

Akiva Weinberger

because all lower Riemann sums are $0$ and all upper Riemann sums are $1$

It's like that

Secret

ah, discontinuity of riemanian sums

Akiva Weinberger

He constructs it slightly differently than I would, I guess

I would do it based on the dyadic rationals

Easier to draw

Secret

so those dense points must be nonintegrable singularities

Alessandro Codenotti

@AkivaWeinberger which book?

Akiva Weinberger

17:15

@MatsGranvik Is that… the imaginary part of the first nontrivial root of the zeta function?

Mats Granvik

@AkivaWeinberger yes

Alessandro Codenotti

Ah, wait, it's a Riemann integral, it's alright

Akiva Weinberger

@AlessandroCodenotti Multivariable

@AlessandroCodenotti Yeah I'm sure Lebesgue clears this all up easily

In fact in this case it definitely does

The Lebesgue integral would exist and equal 0

It's at page 269 if you have it

Secret

Hmm, I wonder if @Waiting has a nice series representation of the above integral. surely will be something relevant to her work as she deals with zetas and logs frequently

Alessandro Codenotti

@AkivaWeinberger thanks, I'll look it up

Jasper Loy

17:21

@Secret To me, @Waiting is the world's leading expert on limits, series, and integrals. =D

Secret

Waiting is proficient in exploring the series representation side of integrals, something that few mathematicians in the literature explores

While I have yet to learn her intuition on series except a few log and zeta manipulations, I believe it will soon come to me as I continue the Integral Project

My limit skills might improve as I delve deeper into general topology

as open sets will allow me to visualise the paths that is used to approach a limit point, no matter the cardinality

Jasper Loy

It is amazing how Waiting's mind works. She is really a genius.

In contrast, I am only a banana.

Secret

I wish to learn more about her thinking and worldview, but I need to wait

Jasper Loy

World view? What do you mean @Secret?

Secret

Every person has their own set (class?) of history, background, belief systems, bias, community, environment etc. All of these shapes their way of interacting with reality

This is what I call a worldview, it is unique to each individual

Jasper Loy

17:28

I see. I believe I have talked about my world views a lot in this room over the past 5 years, under my 10 deleted accounts or so.

Secret

For example, my past of being bullied at high school and the desire to make bullies shut up forever lead me to invent the proof writing technique of nuking proofs

and my passion for the weirdness means I am not afraid to play directly with pathological examples, which most people will avoid in normal circumstances

My constant note taking on nearly all my experience might have allow me to be self conscious to some extent

and so on the list goes...

Jasper Loy

By the way, what is your secret?

Secret

Well, one reason I pick this username is because I like secrets and I like to uncover secrets and discover things

My purpose of life is to unlock the mystery of nature

and idea mixing is one of the most powerful tool in my arsenal

Jasper Loy

I see. I was thinking maybe your name is Victoria.

Secret

everybody love to make that joke, lol

Jasper Loy

17:31

Well, I still don't know who Victoria is and what her secret is.

Secret

> Saying: everything will be clear in time...

Jasper Loy

Who knows? Maybe Waiting is Victoria and her secret lies in her integrals.

Secret

I have so far invented many things in this chat, but they are all scattered thus chat crawling with put them all back into place

[Random] Name idea for a limit point that has "proper class number of points" in its neighbourhood

But first, we need to learn some class topology

[Random]

It is known that symbolic integration is hard while symbolic differentiation is easy, whereas the revese happens for numerical. Wondering whether there is a middle point for this scenario

Mats Granvik

17:51

Written in this form, for $n=1$, the integral is at least solvable through power series expansion of the Fourier series for the sign and floor function:
$$14.1347251417346937904572\text{...}=\int _0^{16}\frac{1}{2} \left(1-\text{sgn}\left(\left(\left\lfloor \frac{\vartheta (t)}{\pi }+1\right\rfloor +\frac{1}{2} \left(-1+\text{sgn}\left(\Im\left(\zeta \left(i t+\frac{1}{2}\right)\right)\right)\right)\right)-n+\frac{3}{2}\right)\right)dt$$
by repeated integration by parts, but the convergence of the symbolic solution to the integral is probably not great.

Or I believe it is solvable, since there is no function in the denominator of the power series for the Fourier series. I have not tested it in Mathematica yet.

Akiva Weinberger

18:41

Anything's a verb if you verb it

Leaky Nun

@AkivaWeinberger like?

Akiva Weinberger

That said, I'm slightly disappointed that "phoning someone" does not mean "hitting someone in the face with a phone"

Leaky Nun

@AkivaWeinberger I would call that phone-slapping someone :P

Balarka Sen

@LeakyNun He just verbed "verb".

Leaky Nun

This reminds me of Classical Chinese where words are often turned into other parts of speech

Secret

18:45

[Super random]

Determine the limit points of the class of all possible worldviews

Leaky Nun

@AkivaWeinberger how would you say "Abracadabra" in Modern Hebrew? (It's suspected to be Aramaic/Hebrew in origin)

b-r-' / k / d-b-r

Secret

[random meta]

The probability of nonsensical of [random] depends on its "prefix"

it has the following ordering:

Akiva Weinberger

@LeakyNun Transliterate it

Secret

'' < 'super' < 'ultra' < ...

Akiva Weinberger

אַבְּרָקָדַבְּרָה or אַבְּרָקָדַבְּרָא

Leaky Nun

18:48

@AkivaWeinberger אברא כ דברא?

@AkivaWeinberger it's supposed to be k not q

Akiva Weinberger

Yeah but when foreign works are transliterated into Modern Hebrew, you usually use the letter quf for the k sound

They're pronounced identically anyway

Leaky Nun

@AkivaWeinberger no, I'm not asking for the transliteration

2 mins ago, by Leaky Nun

b-r-' / k / d-b-r

b-r-' is to create
k is as
d-b-r is to speak

Akiva Weinberger

I took it from here he.wikipedia.org/wiki/…

They just have it as the transliteration

They list the Aramaic "אברא כדברא" as a possible origin of the phrase

Leaky Nun

@AkivaWeinberger yes, and I'm interested in how you would say it in Hebrew

using the given roots above

Akiva Weinberger

From the same Wikipedia article, they translate that Aramaic phrase as "שיברא כדבריי"

"Sheyibra kedvarai", I think

Leaky Nun

18:52

so you guys don't use the word b-r-'?

Akiva Weinberger

What? Sheyibra has the root b-r-'

She- means "that", yibra is future tense third person of b-r-'

Leaky Nun

I see

I thought you have a word with a single k

Akiva Weinberger

?

Leaky Nun

@AkivaWeinberger never mind, I must have mixed it with ki (k-y)

what does כדבריי mean?

Akiva Weinberger

"Like my words" I think

Or rather

Hm, I'm not sure

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$Mathematics$ Mathematics