Mathematics - 2017-06-20 (page 1 of 7)

12:46 AM

Consider the shape bounded by permutations of $(1,-1,0,0,0)$

That's in 5-space but it's in the $\sum x_i=0$ plane so it's really a 4D thing

The 3D equivalent (one fewer zero) is a cuboctahedron

Given a point, how can we tell if it's in the interior of that shape, on the boundary, or on the exterior?

Conjecture: If the Manhattan norm ($\sum|x_i|$) is less than $2$, it's in the interior. If it's equal, it's on the boundary, and if it's greater, it's in the exterior.

Also, each face is defined by two subsets of the five coordinates. The vertices of that face would be the ones such that one of those subsets contains exactly one $1$ and the other contains exactly one $-1$.

So, $(\underbrace{1,0},\underbrace{-1,0,0})$ and all permutations of the things above each underbrace, for example, would define a face.

Which means each face looks like $\Delta^m\times\Delta^n$ for $m+n=3$.

Yeah. The conjecture is true. Which is nice. (Note that it only applies to things on the $\sum x_i=0$ hyperplane)

Also—$(1,0,0,0,0)$ and its permutations define the standard 4-simplex (4D tetrahedron).

Its interior is things on the $\sum x_i=1$ plane all of whose coordinates are positive. Its boundary is things where the smallest coordinate is $0$; its exterior is things with a negative coordinate.

Thus:

We can define a map from the surface of the $(1,-1,0,0,0)$ polyhedron to the surface of the $(1,0,0,0,0)$ polyhedron (aka 4-simplex) by replacing all negative coordinates of our input with zero.

So $(\frac12,\frac12,-\frac13,-\frac13,-\frac13)\mapsto (\frac12,\frac12,0,0,0)$

This map is surjective.

Also, each $n$-dimensional face of a 4-simplex has a $(4-n)$-dimensional opposite face. (Similarly in 3D for tetrahedra.)

The $(1,-1,0,0,0)$ is symmetric about the origin, and our map maps opposite points to opposite faces.

Hm. To visualize this, stare really hard at a cuboctahedron (the 3D equivalent).

Note how each (2D) face of the cuboctahedron corresponds to an (arbitrary dimension) face of the tetrahedron.

That's really what's happening for this thing, but in 4D.

Akiva Weinberger

1:18 AM

Expanded cube (equal to expanded octahedron) ^

"Expanding" means that faces, edges, and vertices of the original become faces in the end.

You can also expand tetrahedra, turning every face, edge, and vertex into a face. This gives you a cuboctahedron.

While the tetrahedron doesn't have central symmetry, the expanded tetrahedron does!

That's because, with the tetrahedron, faces are opposite vertices, so you can't possibly have central symmetry. Once you expand it, though, everything becomes faces! So central symmetry becomes possible!

Similarly, in any dimension, an expanded simplex has central symmetry, even though the simplex doesn't.

Furthermore, there's a natural map from (the surface of) an expanded polyhedron back to (the surface of) the original polyhedron.

If expanding turns edges and vertices into faces, simply collapse them back into faces and vertices.

That's what happens in the gif every time the thing gets smaller. (Since the expanded cube equals the expanded octahedron, we can map it to either the cube or the octahedron.)

To work with coordinates: The $n$-simplex is the convex hull of $(1,0,\dots,0)\in\Bbb R^{n+1}$ and permutations thereof. It's really an $n$-dimensional object, since it lives in the hyperplane $\sum x_i=1$.

The expanded $n$-simplex is the convex hull of $(1,-1,0,\dots,0)\in\Bbb R^{n+1}$ and permutations thereof. It lives in the hyperplane $\sum x_i=0$.

The above-mentioned map is $(x_0,x_1,\dots,x_n)\mapsto(f(x_0),f(x_1),\dots,f(x_n))$ where $f(x)=\begin{cases}x,&x>0\\0,&x\le0\end{cases}$. (That is, through away negative coordinates and replace them with zeroes.)

It can be checked that this is well-defined and surjective.

Hi, @Semi!

(If you want, you can read just those posts starting with the gif)

Semiclassical

hi

Akiva Weinberger

Oh, typo:

> If expanding turns edges and vertices into faces, simply collapse the faces back into *edges and vertices.

Simply Beautiful Art

Quick series/limit problem: Prove the following:$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n) =\frac12\ln\left(\frac\pi2\right)$$

Akiva Weinberger

Oh, there's an animation for the tetrahedron expanding into the cuboctahedron:

@SimplyBeautifulArt I'm guessing that'd equal the Cesàro sum of $\ln2-\ln3+\ln4-\dotsb$

Simply Beautiful Art

1:34 AM

Yes... Cesaro sums and Abel sums... very much related.

Akiva Weinberger

You know that neat proof that does $1-2+3-\dotsb$ for you?

Simply Beautiful Art

Which one lol

Akiva Weinberger

\begin{align}1-2+3-4+5-6+7-&8 +\dotsb\\1-2+3-4+5-6+&7-\dotsb\\1-2+3-4+5-6+&7-\dotsb\\1-2+3-4 +5-&6+\dotsb\\=&\\1+0+0 +0+0+0+0+&0+\dotsb\end{align}

GFauxPas

Semiclassical im too tired for more math tonight. you want to work on it tomorrow morning?

what the f is that Akiva

those arent Cauchy

what are those

Simply Beautiful Art

@AkivaWeinberger I don't like it because its horribly unrigorous.

Akiva Weinberger

1:37 AM

@GFauxPas There exist methods of summing divergent series :P

@SimplyBeautifulArt No, it's not, if you let me explain

ה' ירחם

You know Hebrew?

Ya

Nice

@GFauxPas Say you want to sum more than what the usual definition lets you

Simply Beautiful Art

@GFauxPas Some divergent series can be regularized. For example, for every convergent series $\sum a_n$, we have: $$\sum_{n=1}^\infty a_n=\lim_{x\to1^-} \sum_{n=1}^\infty a_nx^n$$

GFauxPas

1:39 AM

Nope, im too tired for more math today. Try tomorrow :P

Akiva Weinberger

You want this "summation" to satisfy several properties. First, if it converges in the classical sense, then your summation method should give you the same answer.

Secondly, it shouldn't change if you put zeroes in front, and thirdly, you should be able to add two series linearly.

Simply Beautiful Art

However, the converse is not true. For example, $$\sum_{n=1}^\infty(-1)^n\ne\lim_{x\to1^-} \sum_{n=1}^\infty(-1)^nx^n= \lim_{x\to1^-}-\frac1{1+x}=-\frac12$$

Akiva Weinberger

@SimplyBeautifulArt If there exists any summation function that satisfies those three properties and also can sum $1-2+3-\dotsb$, then the above proof guarantees that it gives it a value of $\frac14$.

However, it's still possible that there exists no such summation function.

So, yeah, in that sense the proof is incomplete. :P

Simply Beautiful Art

:-)

@GFauxPas Anyways, we are trying to prove the following:

10 mins ago, by Simply Beautiful Art

Quick series/limit problem: Prove the following:$$\lim_{x\to-1^+}\sum_{n=2}^\infty x^n\ln(n) =\frac12\ln\left(\frac\pi2\right)$$

Akiva Weinberger

For example, no such summation function can sum $1+1+1+\dotsb$, since if you could, you could easily prove $1=0$ (exercise).

This actually is quite a remarkable fact from the point of view of formal manipulation, though, isn't it?

No matter how you try to sum the series $1-2+3-\dotsb$, as long as you only use those properties, you will only ever be able to make it equal to $\frac14$. (Which is false for $1+1+1+\dotsb$.)

Fargle

1:43 AM

@AkivaWeinberger shift 'n' subtract, boom

Simply Beautiful Art

@AkivaWeinberger I'm not quite sure how I feel about it.

Fargle

1 = 0 in the field with one element, if you allow it >_>

Akiva Weinberger

@Fargle Yup.

"Those properties," to be clear, are simply "shifting shouldn't change the answer" and "subtracting should work"

(Well, any linear manipulation)

(And also "it should give the right answer if the series actually converges")

Simply Beautiful Art

Mhm

Hm, I think one more exception

If it converges and the series actually converges, then they should be equal

Akiva Weinberger

I just wrote that

Simply Beautiful Art

1:45 AM

The occasional mishap where your summation method fails on the convergent series x.x

Akiva Weinberger

Yeah. "Trivial summation: sums everything to zero!"

I wonder, though, what happens if we do the same exact same manipulation to your problem?

Simply Beautiful Art

Ah, well I didn't quite get that interpretation out of it :-)

Akiva Weinberger

One fourth times $\ln2-\ln\frac3{2^2}+\ln\frac{2\cdot4}{3^2}-\ln\frac{3\cdot5}{4^2}+\dotsb$

Hm, no, that looks too messy

Simply Beautiful Art

Interestingly we probably can get the log of an almost telescoping divergent product out of it.

Akiva Weinberger

I'll think about it later.

Simply Beautiful Art

1:50 AM

+1 for stumping @AkivaWeinberger before bed time! Anyways, good night!

David Zhang

Hey guys, I've got a quick terminology question

What does one call a group action, say G on X, such that all orbits of elements of X under G have the same cardinality?

PVAL-inactive

I don't know a name for that

Akiva Weinberger

I don't think it has a name

arctic tern

if I recall correctly there is a name for when the orbits are all isomorphic as G-sets, but can't remember

Akiva Weinberger

If it's a normal subgroup of the permutation group then it has that property I think

arctic tern

2:03 AM

(when all orbits are regular the action is called free)

@DavidZhang (1/2)-transitive

Akiva Weinberger

Source?

What does the 1/2 mean?

arctic tern

not quite 1-transitive

Akiva Weinberger

Oh. What's 1-transitive?

This gif still looks cool

arctic tern

1-transitive is just transitive

k-transitive is transitive on the set of k-tuples of distinct elements

Akiva Weinberger

Huh, cool

Google doesn't reveal any name

PVAL-inactive

2:06 AM

There's a list of adjectives of wikipedia

arctic tern

google "half-transitive group action"

Akiva Weinberger

It does reveal tons of people using the clunkier phrase "the orbits have the same cardinality", and most people don't add "and oh by the way this is also known as X"

PVAL-inactive

Pretty sure I don't know any more than that.

Akiva Weinberger

Oh.

groupprops.subwiki.org/wiki/Half-transitive_group_action

Huh.

PVAL-inactive

I caution you about using a phrase that is not standard within the vernacular.

Akiva Weinberger

2:07 AM

^

Although, if you're using it in a paper, it's probably fine if you define it somewhere at the start

(especially if it comes up a lot)

arctic tern

I have seen it used in permutation group literature

David Zhang

nice, thanks guys! yes I'll be sure to define it in context

arctic tern

there's also at least one paper that uses "1/2-transitive" in the title

It's defined on page 215 of "Permutation Groups" by Dixon and Mortimer

what?

Fargle

Forgive me I'm slow, I didn't scroll up

Akiva Weinberger

2:26 AM

$F^{1/4}DFA~FDFE\mid FD\flat FA\flat~FFED\mid GDGB~GGDE\mid FEDE~FDD\emptyset$

#LaTeXMusic

Balarka Sen

3:28 AM

"Enemy" is one of my favorite horror movies of all time now.

Secret

3:51 AM

\begin{align}
C_0(0,\alpha) & = \Omega_0 \cup \{0\}= \{0,n,\omega\}\\
C_1(0,0) & = \{0,n,\omega,\omega+n,\omega 2,\omega^n,\omega^{\omega},\epsilon_0,\epsilon_n,\epsilon_{\omega},\zeta_0,\zeta_n,\zeta_{\omega},\varphi(\beta, \gamma),\varphi(\omega,\omega)|2<\beta
<\omega;\gamma \leq\omega\}
\end{align}
*system explodes due to memory overload and parsing error*

NB: There is no $n$ in the first iteration

Annelise Toft

4:26 AM

can someone give a hint with this proof? math.stackexchange.com/questions/2329212/…

Secret

All cross terms after $C_1(\nu,\alpha),\alpha > 0$ were omitted from print. $n \in \Bbb{N}$

NB Nonstandard notation: $\varphi(\cdot,\gamma)=\varphi(\varphi(\omega,\gamma),\gamma);\varphi^{n+1}(\cdot,\gamma)=\varphi(\varphi^n(\cdot,\gamma),\gamma)$
\begin{align}
C_0(0,\alpha) & = \Omega_0 \cup \{0\}= \{0,\omega\}\\
C_1(0,0) & = \{0,1,\omega,\omega 2,\omega^{\omega},\varphi(\omega,0),\varphi(\omega,\omega)\}\\
C_2(0,0) & = \{0,1,2,\omega,\omega 2,\omega 3,\omega 4,\omega^{\omega},{}^3\omega,\varphi^2(\cdot,0),\varphi^2(\cdot,\omega)\}\\

Secret

5:12 AM

Actually no, I missed some non cross terms

Correction version later

Zee

6:14 AM

@AkivaWeinberger We can look at the game of chess as a Turing complete machine, so Turing completeness is not really what a computer simply is

Irregular User

6:31 AM

Can someone take a look at the following question if they have time?
https://math.stackexchange.com/q/2327328/290074

Richard

Given an explicit set $A$ of complex numbers, is it possible to make explicit the set $B=\{w\in \mathbb{C}: w= y+2-ix, (x,y)\in A\}$? The specific question is math.stackexchange.com/questions/2329077/…

Leaky Nun

@Richard you just made it explicit.

Richard

Ok then, let's forget about that, I tried to represent $B $ by transforming the graph of $A $, though I'm not totally sure it is correct

firstly I would shift it up by two units ($y\mapsto y+2$), then reflect the result about the y-axis ($x\mapsto -x$) and then swap the new x and y coordinates (which should mean rotate the graph by 90 degrees clockwise and then rotate it by 180 degrees around the new x-axis)

@LeakyNun

Leaky Nun

why would $+2$ be up?

Rotate the points in $A$ about the origin clockwise by $90^\circ$ (effectively multiply by $-i$ which changes $x+yi$ to $y-xi$), and then shift every point to the right by 2 units (effectively adding $2$).

Richard

Ok I see, but why In that order?

That is, why must one apply first the rotation (which admittedly is way more efficient than mine) and then apply the translation? In other words, why is the order I chose wrong?

BAYMAX

6:45 AM

I just thought that there is no Probability and Statistics chat room :)

Richard

@LeakyNun

Leaky Nun

Start with $(x,y)$.
Shift it up by 2 units: $(x,y+2)$.
Reflect about y-axis: $(-x,y+2)$.
Swap coordinates: $(y+2,-x)$.

Sorry, your method is correct.

@Richard

Note that "swap coordinates" is effectively "reflect about the line $y=x$"

Richard

@LeakyNun Great, thank you so much!

Matt Thrower

8:52 AM

Hi all. I am not a mathematician, but I have a fairy trivial stats problem I'd like someone to explain to me. Is it off topic to post a question on something at pre-university level? Couldn't find any examples among existing questions, didn't want to waste anyone's time

Leaky Nun

@MattThrower I don't expect anyone to be able to answer your vague question.

Matt Thrower

@LeakyNun The question itself is not vague. I have a pair of equations which someone else derived using linear regression. I would like to understand how they were dervied

@LeakyNun I just want to know if something which is relatively easy for a trained mathematician is off topic or not

Leaky Nun

@MattThrower no, as long as you include the context and your attempt

Matt Thrower

@LeakyNun Thanks. This isn't a homework question - I'm 42 :) - It's a matter of curiosity.

Leaky Nun

You still have to include the context and your attempt.

Irregular User

8:59 AM

Is it possible to have a probability generating function where the random variable has any image? The question is here: math.stackexchange.com/q/2327328/290074

Any sort of answer/comment would be appreciated

Faust7

9:59 AM

morning

BAYMAX

morning @Faust7

Secret

All cross terms after $C_1(\nu,\alpha),\alpha > 0$ were omitted from print. $n,m \in \Bbb{N}$

NB Nonstandard notation:

$\varphi(\cdot,\gamma)=\varphi(\varphi(\omega,\gamma),\gamma);\varphi^{n+1}(\cdot,\gamma)=\varphi(\varphi^n(\cdot,\gamma),\gamma)$

$\alpha_{{_{(1)}}\beta}=\alpha_{\beta};\alpha_{_{(n+1)}\beta}=\alpha_{\alpha_{_{(n)}\beta}}$

-----

\begin{align}
C_0(0,\alpha) & = \Omega_0 \cup \{0\}= \{0,\omega\}\\
C_1(0,0) & = \{0,\varphi(0,0),\omega,\varphi(0,\omega),\varphi(\omega,0),\varphi(\omega,\omega)\} = \{0,1,\omega,\omega^{\omega},\varphi(\omega,0),\varphi(\omega,\omega)\}\\

Definition taken from: math.stackexchange.com/questions/2176715/…

\begin{align}
C_0 (\nu, \alpha) & = \Omega_\nu \cup \lbrace 0 \rbrace\\
C_{n+1} (\nu, \alpha) & = \lbrace \beta + \gamma, \varphi(\beta, \gamma), \psi_\mu(\delta) | \beta, \gamma, \delta \in C_n (\nu, \alpha); \delta < \alpha; \nu \le \mu \rbrace\\
C (\nu, \alpha) & = \bigcup_{n = 1}^{\infty} C_n (\nu,\alpha)\\
\psi_\nu (\alpha) & = \min \lbrace \beta | \beta \notin C(\nu,\alpha) \rbrace
\end{align}

Abdullah UYU

10:46 AM

Let's say $x^2-2x+3=0$ has two roots that is $\{m,n\}$. How can we know $3m^2+6n+23$ is determinable?

Leaky Nun

@AbdullahUYU $m^2 = 2m+3$ so $3m^2 = 6m+9$.

$3m^2+6n+23 = 6m+6n+32$

Abdullah UYU

oh, i get it.

Secret

11:13 AM

$\varphi(0,\Gamma_0+1)=?$

I am starting to wonder whether this discontinous property actually applies to all right operations in the hierarchy of ordinals, so that eg. for the Veblen ,it is impossible to have the right argument to become a fixed point

This is because $\varphi(0,\Gamma_0+1)= \omega^{\Gamma_0+1}=\omega^{\Gamma_0}\omega=\Gamma_0\omega$

\begin{align}
C_0(0,\alpha) & = \Omega_0 \cup \{0\}= \{0,\omega\}\\
C_1(0,1) & = \{0,1,\omega,\omega^{\omega},\varphi(\omega,0),\varphi(\omega,\omega),\Gamma_0+1\}
\end{align}
$C_2(0,1) = ?$

John Jack

11:29 AM

Could someone check my proposed answer to this post please...

user243301

12:06 PM

I believe that the function $\pi^{-s/2}\Gamma(s/2) \zeta(s)$ is continuous and holomorphic (is entire) on the whole complex plane. Is this a bounded function of $s=x+iy$? That is I would like to know if I can to apply the Hadamard three-lines theorem to this function, see this Wikipedia. Thanks in advance @DanielFischer

user193319

I am trying to prove that if $f : X \to Y$ is continuous and $x$ is a limit point of $A \subseteq X$, then $f(x)$ is a limit point of $f(A)$. However, my proof hinges on the fact that $f(A- \{x\}) = f(A) - \{f(x)\}$, which I know generally does not hold. This leads me to believe the conjecture is false; however, I cannot seem to find a counterexample. I could use a hint.

Akiva Weinberger

12:23 PM

@user193319 Constant function?

user193319

@AkivaWeinberger Haha I just came to the same conclusion. Thanks for your response!

Balarka Sen

Hi @Daminark

Mithlesh Upadhyay

2

Q: Total number of possible order possible in this binary search tree?

When searching for the key value $60$ in a binary search tree, nodes containing the key values $10, 20, 40, 50, 70, 80, 90$ are traversed, not necessarily in the order given. How many different orders are possible in which these key values can occur on the search path from the root to the node co...

Please, have a look.

Akiva Weinberger

If three convex sets in the plane have a common intersection, must they be the image of the classical Venn diagram shape?

Secret

12:39 PM

Is a line interval convex?

Leaky Nun

@AkivaWeinberger Take three concentric circles

@Secret what is a line interval?

Balarka Sen

@LeakyNun Circles are not convex sets? If you mean disk then that is a Venn diagram of $A \subset B \subset C$

Akiva Weinberger

Oh, you're right @LeakyNun

Leaky Nun

@BalarkaSen then what is he asking?

Secret

@LeakyNun Literally a line segment e.g. the line joining (x1,y1) (x2,y2) in $\Bbb{R}^2$ with the eucledian metric for example

Akiva Weinberger

12:40 PM

@BalarkaSen I meant this shape

Leaky Nun

@BalarkaSen no, that's an Euler diagram, not a Venn diagram.

Akiva Weinberger

Facts.

Leaky Nun

@Secret Then why don't you say "line segment"?

Balarka Sen

@Akiva Oh. Well, then that is obviously false.

I don't know what that is, @Leaky

Secret

@LeakyNun Because I cannot find the correct word when I first said it

Leaky Nun

12:42 PM

@BalarkaSen A Venn diagram has a fixed shape

An Euler diagram can change the shape depending on the intersections of the sets

Akiva Weinberger

Euler diagram is when you reflect subset information and stuff

Balarka Sen

huh, never knew this

Leaky Nun

Akiva Weinberger

OK, then… Does every set of three convex sets contain an image of the classic Venn diagram shape? :P

Leaky Nun

3 mins ago, by Leaky Nun

@AkivaWeinberger Take three concentric circles

replace "circle" with "disc"

Akiva Weinberger

12:43 PM

That contains a Venn diagram shape, doesn't it?

Oh, wait

Balarka Sen

I don't understand the question.

Akiva Weinberger

Take a point in the intersection

@BalarkaSen It was a bad question

Ignore it

Balarka Sen

Just look at a disk inside the three sets

and you can construct the "Venn shape"

inside the disk

Akiva Weinberger

OK yeah

I didn't think this through

Krijn

Whaddup algebro' s

Akiva Weinberger

12:48 PM

On an unrelated note, the snub disphenoid (aka the dodecadeltahedron) is cool

GFauxPas

Hi guys

@Semiclassical ping

Akiva Weinberger

Twelve equilateral triangles and eight vertices making a convex polyhedron

@GFauxPas ping

GFauxPas

Sup

I have successfully summoned him

it seems

booooo

Simply Beautiful Art

Lol, I see @Secret is trying to do things...

@Secret That is correct

@Secret $\varphi(0,\varepsilon)=\varepsilon$

Secret

but then how can we get beyond $\Gamma_0$ since in this definition of an ordinal collapsing function, we cannot use multi argument veblen, is it from this point we can only produce $\Gamma_{\alpha}$ and higher ordinals indirectly via generating its fundamental sequence?

Simply Beautiful Art

1:00 PM

@Secret You know... I'm not really the fan of using the Veblen function in the ordinal collapsing function. It truthfully doesn't do much and just makes it all messier.

Secret

ok, then, moving back to the C and D definition of yours shortly

But before that, here's what I observed:

Simply Beautiful Art

@Secret Rather I'd say at a certain point, we are defining it's fundamental sequence.

Balarka Sen

@AkivaW Here's a thing I haven't thought about. Suppose you look at $X_n = \{1, 2, \cdots, n\}$. For every collection of subsets $S \subset \mathscr{P}(X_n)$ of $X_n$, one can associate a simplicial complex to it: for every element of $S$, take a vertex. For every intersection of the elements of $S$, take a 1-simplex between the corresponding vertices. For every triple intersection, take a 2-simplex on the relevant "triangle" etc.

Does such simplicial complexes exhaust all simplicial complexes with $n$ many vertices? I think so.

user193319

Let $C$ be closed in $X$ and $A$ some closed subset in $C$. Let $LP_C(A)$ denote $A$'s limit points in $C$, and, similarly, $LP_X(A)$ its limit points in $X$. My question, does it follow that $LP_X(A) = LP_C(A)$, given that $A$ is closed in $C$ and therefore in $X$?

Secret

$$\Gamma_1 =\sup\{\Gamma_0+1,\Gamma_0\omega,\Gamma_0^{\omega},{}^n\Gamma_0,\epsilon_{\Gamma_0+1},\zeta_{\Gamma_0+1},\varphi (\Gamma_0+1,0),...\}$$

Not sure about the last term before the ...

Simply Beautiful Art

1:04 PM

It's certainly right, though I'd rather have the fundamental sequence more along the lines of... $$\Gamma_1=\sup\{\Gamma_0+1, \varphi(\Gamma_0+1,0), \varphi(\varphi(\Gamma_0+1,0),0) ,\dots\}$$

Makes it clearer where this is going in my opinion.

Secret

ah, you start at the nesting veblens

indeed it does, it illustrates the Ferferman function as enumerating 2 argument veblen fixed points

Anyway, fetching your C and D ordinal collapse function again

Balarka Sen

Actually you can have $2^n$ many vertices

Simply Beautiful Art

$$C(\alpha)_0=\{0,1\}\\C(\alpha)_{n+1} =C(\alpha)_n\cup\left\{\gamma+\delta,\gamma \delta,\gamma^\delta,\omega^{\mathrm{CK}}_\gamma, \omega_\gamma,\chi_\gamma(\delta),\psi_\gamma( \eta):(\gamma,\delta,\eta\in C(\alpha)_n) \land(\eta<\alpha)\right\}\\\psi_\beta(\alpha) =\sup\left\{\max\{\gamma_n:( \gamma_n\in C(\alpha)_n)\land (\gamma_n<\omega^{\mathrm{CK}}_{\beta+1})\} :(n\in\mathbb N)\right\}$$

$$D(\alpha)_0=\{0,1\}\\ D(\alpha)_{n+1}=D(\alpha)_n \cup\{\gamma+\delta,\gamma\delta, \gamma^\delta,\omega_\gamma^{CK}, \omega_\gamma, \psi_\gamma(\delta),\chi_\gamma(\eta):(\gamma,\delta,\eta\in D(\alpha)_n)\land(\eta<\alpha)\} \\\chi_\beta(\alpha)=\sup\{\max\{\gamma_n: (\gamma_n\in D(\alpha)_n)\land (\gamma_n<\omega_{\beta+1})\}:(n\in\mathbb N) \}$$

@Secret ?

Secret

This is what I got in my backup chat:

\begin{align}
C(\alpha)_0 & =\{0,1\}\\
C(\alpha)_{n+1} & =C(\alpha)_n\cup\left\{\gamma+\delta,\gamma \delta,\gamma^\delta,\omega^{\mathrm{CK}}_\gamma, \omega_\gamma,\chi_\gamma(\delta),\psi_\gamma( \eta):(\gamma,\delta,\eta\in C(\alpha)_n) \land(\eta<\alpha)\right\}\\
\psi_\beta(\alpha) & =\sup\left\{\max\{\gamma_n:( \gamma_n\in C(\alpha)_n)\land (\gamma_n<\omega^{\mathrm{CK}}_{1+\beta})\}\} :(n\in\mathbb N)\right\}\\
D(\alpha)_0 & =\{0,1\}\\
D(\alpha)_{n+1} & =D(\alpha)_n \cup\{\gamma+\delta,\gamma\delta, \gamma^\delta,\omega_\gamma^{CK}, \omega_\gamma, \psi_\pi(\delta),\chi_\gamma(\eta):(…

Simply Beautiful Art

I change my function as I spot errors in it. I think mine is the latest one.

Secret

1:08 PM

ok

Simply Beautiful Art

Think you can figure out $\psi_0(0)$ without filling up the chat?

Secret

I am not sure, cause one reason I initially struggled with this version of ordinal collapsing function is because there are so many strict inequalities

but I think I now knew the higher ordinals better so maybe I might be able to punch through

Simply Beautiful Art

Lol. The inequalities are a bit of a pain...

Well, start by noticing that $\chi_\beta(\alpha)\ge\omega^{ \mathrm{CK}}_1$

And $\psi_0(\alpha)\le\omega^{\mathrm{CK}}_1$

So for now, $\chi_\beta(\alpha)$ has no impact on the nature of $\psi$

i.e. it should behave like the other functions you've dealt with

Secret

The mainstream ordinal collapse function, however has a limitation that I don't like: It generates some cross term very late in the iterations (n has to get quite large for some small ordinals such as $\omega^2$ to start appearing)

This C and D ordinal collapse function has the advantage it generates all the "cross terms" more or less at the same growth rate as the "main sequence", and that's what I like about it

ok let me think...

Simply Beautiful Art

@Secret I'd take it as a blessing that it develops such terms later on, as the timing of when it makes such terms doesn't affect its growth and makes our lives a little easier.

The true strength of any ordinal collapsing function isn't how large of an ordinal you can quickly make or when you make them, but rather, how many unique ordinals you can name. That's where it gets its strength

Secret

1:24 PM

We name ordinals using ordinal collapse functions only by using the $\psi$, since the C and D don't necessary contains all the ordinals smaller than some given ordinal? or do the C and D (those that are the infinite union of the individual C and D sets) actually contains all ordinals smaller than a given ordinal and hence is itself an ordinal?

Simply Beautiful Art

They cannot give all ordinals smaller than $\omega^{\mathrm{CK}}_1$. It is necessarily the case that by definition, you can only produce computably many ordinals in this computable notation

Secret

there are uncomputable ordinals smaller than $\omega_1^{CK}$? I thought $\omega_1^{CK}$ is when uncomputable ordinals start to appear since that is the set of all recursive ordinals?

Simply Beautiful Art

It is by definition that there isn't a computable notation for all ordinals less than $\omega^{\mathrm{CK}}_1$

$\omega^{\mathrm{CK}}_1$ is the smallest ordinal greater than $\omega$ such that there exists no notation that can name all ordinals less than it and such a naming system let's you determine when $\alpha>\beta$.

@Secret You need more explanations on how $\omega^{\mathrm{CK}}_\alpha$ works?

Secret

Given my almost nonexistent background in computer science, I often confuse computable with recursive

Simply Beautiful Art

Recursive ~ computable

The exact definition is here:

Admissible ordinal

In set theory, an ordinal number α is an admissible ordinal if Lα is an admissible set (that is, a transitive model of Kripke–Platek set theory); in other words, α is admissible when α is a limit ordinal and Lα⊧Σ0-collection. The first two admissible ordinals are ω and ω 1 C K {\displaystyle \omega _{1}^{\mathrm {CK} }} (the least non-recursive ordinal, also called the Church–Kleene ordinal). Any regular uncountable cardinal...

One usually says that $\omega^{\mathrm{CK}}_\alpha$ is the $\alpha$th ordinal that is either admissible or a limit of admissibles

Secret

1:38 PM

If $\omega^{\mathrm{CK}}_1$ is the least nonrecursive ordinal, and there are ordinals smaller than it that are uncomputable, what are famous examples of recursive but uncomputable ordinals?

Simply Beautiful Art

Mahlo cardinal

In mathematics, a Mahlo cardinal is a certain kind of large cardinal number. Mahlo cardinals were first described by Paul Mahlo (1911, 1912, 1913). As with all large cardinals, none of these varieties of Mahlo cardinals can be proved to exist by ZFC (assuming ZFC is consistent). A cardinal number κ is called strongly Mahlo if κ is strongly inaccessible and the set U = {λ < κ: λ is strongly inaccessible} is stationary in κ. A cardinal κ is called weakly Mahlo if κ is weakly inaccessible and the set of weakly inaccessible cardinals less than κ is stationary in κ. The term "Mahlo cardinal" now usually...

^ Recursive and far from computable

Secret

but this is also uncountable, but if I recall, $\omega^{\mathrm{CK}}_1$ is still countable, so that means there is at least one ordinal that is countable, recursive but uncomputable and < $\omega^{\mathrm{CK}}_1$. I have trouble comprehend an example of this...

(volley of typos...)

Simply Beautiful Art

@Secret All ordinals less than $\omega^{\mathrm{CK}}_1$ are computable by definition

Secret

17 mins ago, by Simply Beautiful Art

It is by definition that there isn't a computable notation for all ordinals less than $\omega^{\mathrm{CK}}_1$

ok, then I think I don't really understood admissible ordinals...

Simply Beautiful Art

For every ordinal less than $\omega^{\mathrm{CK}}_1$, there exists a computable notation that can describe it and show when $\alpha<\beta$ for $\beta<\gamma$, where $\gamma$ is the supremum of this notation

However, there does not exist a single computable notation that can name all ordinals less than $\omega^{\mathrm{CK}}_1$ and show that $\alpha<\beta$ for $\beta<\omega^{\mathrm{CK}}_1$

Secret

1:50 PM

Ah ok, I am guessing the proof of this will involve something along the lines of: If we can use a single notation to name all ordinals < $\omega_1^{CK}$, then by definition that any ordinal $\alpha$ is a set that contains all ordinals less than itself, you will end up naming $\omega_1^{CK}$, thus contradicting noncomputability

Simply Beautiful Art

Yes... $\omega^{\mathrm{CK}}_1$ is by definition the smallest ordinal such that this never occurs.

Then you get this cool thing called an oracle. Basically, it allows you to, when you can't compute/name/compare some ordinals less than $\omega^{\mathrm{CK}}_1$, then you give it to the oracle and it does the magic for you. You can then combine this with all computable notations to show that $\omega^{\mathrm{CK}}_1+\varepsilon_0 < \omega^{\mathrm{CK}}_12$, just as an example.

$\omega^{\mathrm{CK}}_2$ is the smallest ordinal such that all ordinals below it can not be described using computable notations + the first oracle.

Secret

so the oracle in the context of ordinals allow us to basically add multiply exponntiate and in general, put a recursive operation on a uncomputable ordinal?

Simply Beautiful Art

Yup. More or less that is what occurs

Secret

and I am guessing using this pattern we have $\sup \{\omega^{\mathrm{CK}}_n|n\in \Bbb{N}\} = \omega^{\mathrm{CK}}_{\omega}$ thus this guy will contain ordinals that cannot be computed using $\omega$ oracles?

Simply Beautiful Art

Uh, sort of

Secret

2:01 PM

and given how bulky this notation is, we are not having nested $\omega^{\mathrm{CK}}_{\cdot}$....?

Studentmath

Hello folks, quick question:
I want to show equivalence between $A$ being a bounded linear operator $A:X\to Y$,
and the set ${x\in X: ||Ax||\le 1}$ having an interior point.

Simply Beautiful Art

$\omega^{\mathrm{CK}}_\omega$ is more of a limit ordinal rather than an admissible ordinal

Secret

what ordinal is one where not even oracles of any number can compute it?

Simply Beautiful Art

You can have $\omega^{\mathrm{CK}}_{ \omega^{\mathrm{CK}}_1}$

Studentmath

First direction, I remarked that given some $x\in X$ so that $||x||=C$, and $||A||=M$,
I have $||A(1/2MC)x||\le (1/2MC)||A||||x||\le 1/2$

Simply Beautiful Art

2:03 PM

^ This is the supremum of $\omega^{\mathrm{CK}}_\alpha$ where $\alpha<\omega^{\mathrm{CK}}_1$

Studentmath

Then
$$||Ay||=||A(y-u+u)||\le ||A|| ||y-u|| +1/2 <1$$

Simply Beautiful Art

This is doable since you can name all ordinals less than $\omega^{\mathrm{CK}}_1$ using the first oracle.

Studentmath

If $||y-u||<\delta =1/2M$

So that's one direction proven, right?

Simply Beautiful Art

In general, $\omega^{\mathrm{CK}}_\alpha$ is the smallest ordinal such that all ordinals less than it cannot be named and compared using computable notations, $\omega^{\mathrm{CK}}_\beta$, and the corresponding oracles for $\beta<\alpha$.

Secret

Hmm, I guess the fact that ordinals are well ordered means a supremum always exists, which means no matter how high how uncomputable and how inaccessible we go, we can always perform the nesting operation

Simply Beautiful Art

2:10 PM

Well, its computable.

LegionMammal978

2:23 PM

Hmm

Secret

Some long time ago, I was toying with trying to introduce this into the ordinals by adjoining a new element:

Absolute Infinite

The Absolute Infinite is mathematician Georg Cantor's concept of an "infinity" that transcends the transfinite numbers. Cantor linked the Absolute Infinite with God. He held that the Absolute Infinite had various mathematical properties, including the reflection principle which says that every property of the Absolute Infinite is also held by some smaller object. == Cantor's view == Cantor said: The actual infinite was distinguished by three relations: first, as it is realized in the supreme perfection, in the completely independent, extrawordly existence, in Deo, where I call it absolute infinite...

This element will have the property of being a fixed point under any operations, computable or not, making it the supremum of any increasing sequence of ordinals, and also the maximum since +1 becomes impossible due to its absorbing properties

but the issue of this you will quickly reach a contradiction:

the ordinals form a proper class, if you try to introduce something like this, it suddenly becomes a set, (after many steps that I don't know how to proof), I suspect one will end up something like russel paradox

LegionMammal978

@Secret If $\Omega=\Omega^+$, the Axiom of Regularity would probably be broken

Simply Beautiful Art

Yeah... @Secret

LegionMammal978

Say I have 2 sequences $a_n$ and $b_n$ satisfying homogeneous linear recurrence relations with constant integer coefficients. Then, would such a relation always exist for the sequence $c_n=a_nb_n$?

Secret

https://en.wikipedia.org/wiki/Non-well-founded_set_theory
Yeah, pretty much AFC will be thrown away and we will enter some relatively unknown wilderness

LegionMammal978

2:31 PM

Because $\dots\in\Omega^+=\Omega\in\Omega^+=\Omega\in\Omega^+=\Omega$, I presume

Wait, my question on linear recurrences would have a yes answer due to generating functions

Secret

More generally, using the above observations, I suspect for an algebraic system where addition coincides with succession, the existence of just one (two sided) additive absorber will make the underlying set not well orderable

Semiclassical

@GFauxPas Sorry, was here briefly and then my attention was pulled away to other stuff. Lemme know when you want to resume.

LegionMammal978

Wait, no, multiplying the generating functions doesn't work

Semiclassical

@LegionMammal978 Right. Multiplying generating functions gives you the Cauchy product of two series.

LegionMammal978

So would $c_n$'s generating function be a rational function or not?

Semiclassical

2:41 PM

I'm not sure there's an answer for generic $c_n=a_n b_n$.

Specific cases may be solvable, but I don't think just looking at $c_n=a_nb_n$ is going to tell you much.

LegionMammal978

Because I can't think of any counterexamples off the top of my head

Semiclassical

Well, what sort of examples are you thinking of?

LegionMammal978

idk

Semiclassical

:/

I mean, if you've got $c_n$ being a polynomial function of $n$ then it's definitely got a rational generating function.

But you don't need $c_n=a_n b_n$ for that.

So I don't know what cases beyond that which you've got in mind.

LegionMammal978

I mean, I'm mainly doing this out of curiosity

Semiclassical

2:46 PM

I guess what I'm not understanding is: When you say you can't think of any counterexamples, that suggests that you're imagining certain examples and determining that they aren't counter-exmaples.

So I'm curious what you've got in mind as far as "this could have been a counterexample but isn't."

LegionMammal978

More like "brain is dead, am not thinking at all"

Semiclassical

heh, I can sympathize with that.

It's a nice question, by the way: If $c_n=p(n)$ for some polynomial $p$, show that it has a rational generating function.

LegionMammal978

@Semiclassical Its generating function would just be $p(x)$, and polynomials are rational functions?

Semiclassical

noooo

$c_n=n$, for instance.

LegionMammal978

facepalm forgot how generating functions work

Semiclassical

2:52 PM

:)

LegionMammal978

But yeah, polynomial sequences satisfy certain linear recurrences of the form specified above (with some weird Pascal-triangle-like coefficients), and those are exactly the sequences with rational generating functions

Right.

sup semi

hi @GFauxPas

want to do the math thing

?

Semiclassical

2:54 PM

Sure.