Mathworks (Not the main chat!) - 2017-10-24 (page 1 of 3)

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7:50 AM

I don't know if there is a way to print an approximation (which will become exact when the program does not terminate) for $\omega^2$ except having another program that append extra instance of start indefinitely, turn this whole thing into a program, and then a while loop that runs this big program indefinite number of times

YES! It works

>>> import time
>>> start = []
>>> i = 0
>>> while (0==0):
... output= start+start
... print output
... start.append(i)
... i = i+1
... time.sleep(2)
...
[]
[0, 0]
[0, 1, 0, 1]
[0, 1, 2, 0, 1, 2]
[0, 1, 2, 3, 0, 1, 2, 3]
[0, 1, 2, 3, 4, 0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

>>> start = []
>>> i = 0
>>> while(0==0):
... output=start+start+start
... print output
... start.append(i)
... i = i+1
... time.sleep(1)
...
[]
[0, 0, 0]
[0, 1, 0, 1, 0, 1]
[0, 1, 2, 0, 1, 2, 0, 1, 2]
[0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3]
[0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4]
[0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5]
[0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6]
[0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6, 7]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8]

@user21820 Will the above program fit the requirement to generate the well orderings of $\omega n, n < \omega$ in the limit of countable computation time?

It makes no sense. What on earth does the output mean?

In the limit of countable steps, it should produce:
0,1,2,3,4,...0,1,2,3,4,... which is exactly the well ordering of $\omega 2$

For the last time, I remind you that a well-ordering is a binary function and if you don't give me anything like that then you have no right to even call it an ordering.

I don't get it, I have just wrote a python program that generates the expected sequence that corresponds to $\omega 2$. how is that not a binary function?

A binary function on S is a function that given two inputs from S produces a single output. An ordering is a special kind of binary function, namely a member of func(S^2,bool) that additionally satisfies some extra properties like transitivity.

A computable binary function in Python may look something like def f( m , n ): ....

And no your program is not generating anything like what I expect to correspond to ω·2.

8:05 AM

I see

hmm... defining
def f(n,m):
if n<m or m=='w'
value=1
elif n>m
value=0
is cheating since we need to construct w first instead of presuppose there exists an ordinal greater than all naturals

But there is no way to construct w in finite execuation time in the absence of supertasks

(The following is still cheating as w is presupposed to be present)

def f(n,m):

@Secret It is not cheating. That is precisely the nature of computable orderings.

If you want an ordering on naturals, consider the following:

def f(m,n):
  return false if m==n or m==0 else ( true if n==0 else (m<n) )

If you 'list' the naturals in the order given by f, how would it look like? I put quotes around 'list' because you can't precisely specify the sequence without essentially using my definition of f...

8:24 AM

I see, but how would we compare e.g. w+1 vs 5, do we just define the symbol w to return 1 under f if m=w and n regardless?

A program does not have w prebuilt as a numerical object, thus we can only put that in by specifying what rules it need to obey under some given function

First answer my question about f.

If you can't understand f, you will not understand how to use more general kinds with 'new symbols like ω' properly.

ok:

8:40 AM

Hmm?

It's a mis post, I have not done yet

Oh.

f(0,0),f(1,1),f(2,2),...=0
f(0,1),f(0,2),f(0,3),...=0
f(1,0),f(2,0),f(3,0),...=1
I have no idea, these by themselves don't give any information on which is bigger or smaller. The truth value just indicate which entry is zero or equal

f(1,1)=0; f(1,2),f(1,3),...=1
Thus 1<2,3,...
f(1,1)=0; f(2,1),f(3,1),...=0
Thus 2,3,...>1
f(2,1)=0; f(2,3)=1,f(2,4)=1,...
Thus 2<3,2<4
f(1,2)=1; f(3,2)=0,f(3,4)=1,...
Thus 2<3<4

So from the above, we got so far: 1<2<3<4
Repeat the above analysis all the way through all possible (m,n) and we should get the sequence 1<2<3<4<...

So (modulo the conditions to check for zero, which I don't quite understood their significance) f gives a lists of instruction in terms of a binary string, on how to arrange the numbers, thus an instruction on how the well ordering is constructed

Study f further instead of doing other things.

If m = 0 then f(m,n) = false.

If m ≠ n = 0 then f(m,n) = true.

So where is 0 in relation to all the positive integers?

Remember that (as you interpreted above) when we say that f defines an ordering < we mean that f(m,n) = true iff m<n.

So how does 0 compare to 1,2,3,...

Using f(m,n) = true iff m<n and If m = 0 then f(m,n) = false
then we had to conclude that 0>1,2,3,... (which is not what the naturals is, but that is the conclusion when using just these two piece of information)

8:56 AM

Correct.

We would thus say that f denotes the ordering 1<2<3<...<0.

'Listing' them in order gives 1,2,3,...,0.

that's $omega+1$!

hmm interesting...

Exactly, which is why I said you didn't understand it earlier.

So, your turn.

Give me a program that represents ω+ω.

Then a program that represents ω·ω.

In both cases I want it to be a binary function on N.

And I am guessing you will not allow me to have n and m array valued (as that will make it very easy) ?

That is for later. This is just to make sure you understand how to manipulate computable ordinals as orderings on N.

ok

9:13 AM

For ω2
def f(m,n):
return 1 if (m%2==1 and m%2==0) or m==n else (0 if (m%2==0 and m%2==1) else (m<n))

This should give:
f(odd,even)=1
f(even,odd)=0
f(m,n)=0 if n==m or m>n (and n,m belong to either even or odd)
f(m,n)=1 if m<n (and n,m belong to either even or odd)

and thus the ordering 1<3<5<7<...0<2<4<6<...

Your f gives a non-strict ordering because you return 1 if m==n.

oops

Never mind. I know you know how to fix.

How about ω·ω?

That one seemed harder. If we use the intuition on constructing programs for $\omega n$, it will mean we need to partition the naturals into countably many countable portions and then arrange them accordingly. So that means we need to find a partition function that generates those portions but can be represented as a finite string

So I need to think a bit about that...

Yes it's troublesome now without arrays, but the simplest way is probably to use the fact that every positive natural is equal to 2^k·(2m+1) for some unique naturals k,m.

Using finite arrays or strings essentially boils down to the same notion of encoding, just that it's easier since arrays and strings already have inbuilt sequence-like operations.

9:32 AM

hmm, so 2^k(2m+1) basically implement the ordered pair (k,m). For any even integer it is represented by (k,0) and for every odd integer, it is represented by (0,m)

First sentence correct. Second not.

In this encoding (k,0) would only represent powers of 2.

This is based on the fact that there is a largest power of 2 that divides any given positive natural, and that the other factor is odd.

n=2^k(2m+1) => ln n = k ln 2 + ln (2m+1)
ugh number theory, those things where I cannot e.g. get an explicit form for even numbers that does not divide any powers of 2.

(1,m) will give all the even numbers including the powers of 2, which must be excluded somehow. However I don't know if there is a function on m that does that

o wait what am I doing...

(0,m) will give all the odd naturals, and (1,m) will give all the even ones, thus we are fine

9:55 AM

@Secret No still... 4 is not encoded by (1,m) in the above encoding.

You simply have to use that fact to encode and decode.

Your f takes 2 input naturals, and decodes each to a pair of naturals, and then compares them.

To show that you really get ω·ω you need to show that every pair can be encoded as a unique natural.

Which is obvious since I literally gave the encoding map.

(NB The euclidean algorithm is too tedious to write inside the code)
def f(m,n):
solve m=2^k(2l+1),(k,l), solve n=2^p(2q+1),(p,q)
return 0 if (k==p and l==q) else ( 1 if k<l elif k>l 0 elif p<q 1 elif 0 p>q)

NB: $ω^3$ without arrays is even worse, cause we will need to decompose a natural uniquely into 3 naturals, and then compare each pair

and so on all the way up to $< ω^ω$

Hmm... just from these program constructions alone seemed to suggest to me one thing:

It seems it is reasonable to say that all ordinals before $\omega_1^{CK}$ are well ordered under increasing computational complexity (because ordinals higher up need either more sophisticated data structures, or need to solve more and more complicated integer equations)

which is consistent to one of you thoughts in our discussions some months ago

So, if we can prove that all computable ordinals are uniquely mapped to a computational complexity value, then we can well order the collection of all computable ordinals with this criteria and it will guarentee to give us yes/no answers.

Alternately, if we can prove that there exists at least a pair of computable ordinal with the same computational complexity, then it means we cannot well order them under this criteria and thus we go back to square one we begin with

However, I need to left this big thought on the shelf for now because I knew too little about how computational complexity is defined other than there exists those complexity classes that contain each other, and that there are at least two notions of complexity namely time complexity and resource complexity

(Almost forgot, there's one more possibility: It is unprovable whether the function that maps levels of computational complexity to ordinals is a bijective, order preserving map)

11:07 AM

@Secret Actually it is a good idea to write this kind of program at least once, otherwise you never get a feel for how encoding and decoding needs to be done in general.

Don't you know programming? You can use a while-loop to find the highest power of 2 that divides n.

Also, you missed out the fact that the encoding is from pairs of naturals to positive naturals, so you need to shift it to get back to naturals.

Ah, I didn't realise zero will be missed out unless shifted

I didn't know Python supports elif?

But the rest is the correct idea.

It's just lexicographic ordering on pairs of naturals.

So if the input had already been pairs of naturals, you're done.

python 2.6 and above supports elif. The current version I am using for my research is python 2.7.13

while the supercomputer has python 3 installed, calling it proved to be a bit tedious (not only you need to call the module, but also need to do some kind of path related thingy which I don't understood) thus I use 2.7.13 instead

I see.

@user21820 that is why arrays will make it super easy, and multidimensional arrays will get us up very far into the veblens (which are themselves array functions)

11:13 AM

@Secret This much is true. Programs that do different things can't all be simple after all.

But not in the sense of well-ordered complexity.

No such thing unless you are in a system that already can well-order the computable ordinals...

But we can ask about the Kolmogorov complexity.

That's just a natural number.

But Kolmogorov complexity definitely won't respect the order of computable ordinals.

I don't see how it cannot, but then almost every example of computable ordinal I can came up can be expressed as either veblens or OCFs so I don't know...

so there might be really bizarre ones out there that disagree (or said nothing) with the ordering of kolomogorov complexity

OCFs can't be constructed without rather powerful set-theoretic assumptions.

And it doesn't reduce the Kolmogorov complexity of even the finite ordinals.

One can always find arbitrary complex finite ordinals (they are simply very long and with a strange hard-to-compute number of digits).

ah, that ... follows after the finite ordinals, but before $\omega$, I never thought of that.

I see, in that case, we can pick a humungous but finite ordinal, and it can easily surpass the kolmogorov complexity of some infinite ordinals

so that will be a counterexample

Yeap.

Ok I got to go.

bye

11:27 AM

@Secret In case you're interested, I actually wrote a 256-byte Python 3 program to print a large number using the large Veblen ordinal (assuming it is a well-ordering).

in This is the Realm of Simply Beautiful Art, Mar 28 at 3:49, by user21820

And it turns out that my program for the large Veblen ordinal was shorter mainly because I had used a trick, which also works to shorten the small one:

in This is the Realm of Simply Beautiful Art, Mar 28 at 3:49, by user21820

c=9**9
o=[]
i=[o]
def r(x):
	v=o
	y,*d=x
	z,*p=y
	t=z!=o
	if p==o:
		if t: v=[[r(z)]]*c
	else:
		v=[i]
		if t: v+=[[r(z)]+p]
		m=0
		while p[0]==o:
			m+=1
			_,*p=p
		a,*b=p
		for k in x*c:
			v=i*m+[v,r(a)]+b
			if v[-1]==o: *v,_=v
			v=[v]
	return v+d
def f(n):
	if n:
		v=[i*c+[[i]]]
		while v!=o:
			c*=c
			f(n-1)
			v=r(v)
			print(c)
for k in i*c: f(c)

btw is it sufficient to say that a well ordering is predicative if there is an explicit binary function that generates it?

It indeed makes full use of arrays.

@Secret Yes, assuming you can only construct predicative functions.

I see

7 hours later…

6:02 PM

in Mathematics, 11 mins ago, by Simply Beautiful Art

Define a function as follows:
$$\psi(0)=\varepsilon_0\\ \psi(z)=\sup\{~^\omega (\psi(z[n,z])),n<\omega\}\\ (x+y)[n,z]=x+(y[n,z]) \\(x\cdot y)[n,z] = x\cdot(y[n,z]) \\(x^y)[n,z] = x^{y[n,z]} \\1[n,z]=0\\ \omega[n,z] = n \\ \Omega[n,z]=\begin{cases}\omega,& n=0\\ \psi(z[n-1,z]),&n>0\end{cases}$$

Simply Beautiful Art

in Mathematics, 4 mins ago, by Simply Beautiful Art

Perhaps, for insight, start with simple old $\psi(\Omega)$

Gah shoot stupid me

I made a mistake

R.I.P.

sighs

@Secret Before you apply this process, $x\cdot(a+b)$ needs to be simplified to $x\cdot a+x\cdot b$, so $(x\cdot(a+b))[n,z]= (x\cdot b)[n,z]\ne x\cdot((a+b)[n,z])$

Likewise for $x^{a+b}$

Hm...

@Secret Restriction that $y>1$.

Anyways, back to $\psi(\Omega)$ xD

I hope you're not writing one of those super long posts

I'll also recommend that you try evaluating $\psi(\Omega[n,\Omega])$ for $n=0,1,2,\dots$

Don't worry about what $\psi(\omega)$ is

Hm...

Hmm... m[n,z]=m-1? for sucessor m?

Simply Beautiful Art

$$\psi(0)=\varepsilon_0\\ \psi(z)=\sup\{~^\omega( \psi(z[n,z])),n<\omega\} \\ (x+y)[n,z]=x+(y[n,z]),y>0\\ (x\cdot y)[n,z] = x\cdot(y[n,z]),y\in\Bbb{Lim}\\ (x^y)[n,z]=x^{y[n,z]}, y\in\Bbb{Lim} \\1[n,z] = 0\\ \omega[n,z]=n\\ \Omega[n,z]=\begin{cases} \omega,&n=0\\ \psi(z[n-1,z]),&n>0 \end{cases}$$

@Secret Fixed. Should be the case.

Decided to go with $y\in\Bbb{Lim}$. Much easier that way.

Also $\Omega\in\Bbb{Lim}$.

6:19 PM

Start with the lower things:
\begin{align}
\omega [n,z] & =n\\
(\omega+\omega)[n,z] & = \omega + \omega [n,z] = \omega + n\\
(\omega\cdot\omega)[n,z] & = \omega\cdot\omega [n,z] = \omega n\\
(\omega^\omega)[n,z] & = \omega^{\omega[n,z]} = \omega^n\\
\end{align}

Simply Beautiful Art

Yup, looks good

(I haven't really tested these out much yet btw. Working as we're rolling. In my program, I laid everything else all nicely to avoid anything weird, but we're doing math now :P)

Example cross term:

$(\omega^2+\omega^{\omega})[n,z] = \omega^2 + \omega^{\omega}[n,z] = \omega^2 + \omega^{\omega [n,z]} = \omega^2+ \omega^n$

Simply Beautiful Art

Looks fine

Except I don't like the $\omega^2$ coming before the $\omega^\omega$

$(\omega^{\omega}+\omega^2)[n,z] = \omega^{\omega} + \omega^{2}[n,z] = \omega^{\omega} + \omega n$

Simply Beautiful Art

Yup

6:23 PM

They are different, reflecting the noncommutativity of ordinals

Simply Beautiful Art

Or at least of my ordinals

Climbing the omega tower:

Simply Beautiful Art

:40743645 For the second one, you need to write $m=n+1$ and $\omega^\omega m=\omega^\omega n+\omega^\omega$

ah expanding it

Simply Beautiful Art

:P

Ready for $\psi$?

6:30 PM

Not yet, let me climb to $\epsilon_0$. Should not take long

Simply Beautiful Art

Hm.... shooot........

I forgot to handle that stuff

bangs head

Can't take $\varepsilon_0[n,z]$ under this definition.

Lemme think

Oh, yeah. $\psi(x)[n,z] = \min\{~^\omega(\psi(x[n,x])),\sup\{~^n(\psi(x[k,x])),k<\omega\}\}$

That should work.

And you'll never get $\psi(\varepsilon_0)$, but you'll run into $\psi(\psi(0))$

Oh darn me, silly goose. $\psi(0)[n,z]=~^n\omega$

Maybe I should just take $\psi(x+1)[n,z]=~^n(\psi(x)),~\psi(x\in\Bbb{Lim})[n,z]=\psi(x[n,x])$ to save me the trouble of the above.

Rather than trying to compact everything down so much.

\begin{align}
\omega^{\omega}(m+1)[n,z] & = \omega^{\omega}m+\omega^{\omega}[n,z] = \omega^{\omega} + \omega^{\omega [n,z]} = \omega^{\omega}+\omega^n\\
\omega^{\omega}\omega[n,z] & = \omega^{\omega}n\\
\omega^{\omega (m+1)}[n,z] & = \omega^{\omega (m+1) [n,z]} = \omega^{\omega m + \omega [n,z]} = \omega^{\omega m + n}\\
\omega^{\omega^{\omega}}[n,z] & = \omega^{\omega^{\omega} [n,z]} = \omega^{\omega^{\omega [n,z]}} = \omega^{\omega^n}\\
{}^{m+1}\omega [n,z] & = \omega^{{}^{m}\omega[n,z]} = \cdots = \omega^{⋰^{\omega^n}}\\

Simply Beautiful Art

@Secret Uh, sorta for that last one?

But it's not because $^xy[n,z]=~^{x[n,z]}y$

I am not very sure, because that will be applying the exponential rule for fundemental sequence countably many times, so we need to get around that using $\psi$ some how...

Simply Beautiful Art

@Secret Hm? The point of the OCF is to use it, not get around it.

6:42 PM

But does $^xy[n,z]=~^{x[n,z]}y$ hold in general?

Simply Beautiful Art

No, definitely no.

Otherwise $^{m+1}\omega[n,z]=^m\omega$

In the language of OCF's, you write everything you can.... in terms of OCF's.

The only things you should have, in theory, in your OCF, are $0,1,\omega,\Omega,\psi,+,\times,$ and exponentiation operator..

what is $0[n,z]$?

Simply Beautiful Art

@Secret It's undefined because you shouldn't be able to reach the expression $0[n,z]$

You could make it equal to 0 for the sake of it.

Entering $\epsilon_0$

Simply Beautiful Art

:(

6:49 PM

\begin{align}
\psi(0)[n,z] & = ???????????
\end{align}

I cannot find a nice rule to move the [n,z] in

Simply Beautiful Art

So OCF's don't just make really big ordinals. In general, every ordinal less than $\psi(x)$ can be written in terms of $\psi(y<x)$ and the beginning constants with the given operations.

13 mins ago, by Simply Beautiful Art

Oh darn me, silly goose. $\psi(0)[n,z]=~^n\omega$

11 mins ago, by Simply Beautiful Art

Maybe I should just take $\psi(x+1)[n,z]=~^n(\psi(x)),~\psi(x\in\Bbb{Lim})[n,z]=\psi(x[n,x])$ to save me the trouble of the above.

Yes, those three shall be the definitions.

Entering $\epsilon_0$ (again!):

Simply Beautiful Art

Why must enter as $\varepsilon_0$ instead of $\psi(0)$?

R.I.P. lol

7:05 PM

\begin{align}
\epsilon_0[n,z] & = \psi(0)[n,z] = {}^n\omega\\
\epsilon_0^{m+1} [n,z] & = \epsilon_0^m\epsilon_0[n,z]=\psi(0)^m\psi(0)[n,z] =[\psi(0)]^m{}^n\omega\\
{}^{m+1}\epsilon_0^[n,z] & = \cdots = \epsilon_0^{⋰^{\epsilon_0^\psi(0)[n,z]}}= \epsilon_0^{⋰^{\epsilon_0^{{}^n\omega}}}\\
\epsilon_1[n,z] & = \psi (1)[n,z] = {}^n\psi (0)\\
\epsilon_{m+1}[n,z] & = \psi (m+1)[n,z] = {}^n\psi (m)\\
\epsilon_{\omega}[n,z] & = \psi (\omega)[n,z] = \psi (\omega [n,z]) = \psi (n)\\
\epsilon_{\epsilon_0}[n,z] & = \psi (\psi(0))[n,z] = \psi (\psi(0) [n,z]) = \psi ({}^n\omega)\\

Simply Beautiful Art

Notation so confusing lol.

nestings are annoying to bookeep. I have no idea how to deal with them

Simply Beautiful Art

@Secret Aha! $\zeta_0$ is another one of those as to why you shouldn't use other ordinals than what was given.

Before we get into $\zeta_0$, let's do $\psi(\Omega)$

@Secret Uh.... you kinda just struggle with it ._.

Once you figure out $\psi(\Omega)$, it should make more sense.

It may help when you get to all of the nesting to use $0<\omega<\psi(0)$

$$\psi(\Omega)[n+1,z] = \psi(\Omega [n+1,z])=\begin{cases} \psi(\omega),&n+1=0\\ \psi(\psi(z[n,z])),&n+1>0 \end{cases}$$

Simply Beautiful Art

Nvm

You did multiple steps

I didn't catch it.

7:16 PM

yeah I do cases all at once as if it is an array

Simply Beautiful Art

:P

I'm gonna go eat real quick

everything is so much less confusing at higher dimensions

Simply Beautiful Art

:P

ok, so... how to fill in the $z$, and what does this fundemental sequence mean?

I knew the first term is $\epsilon_{\omega}$ but I don't know the rest without more explicit $z$s

Simply Beautiful Art

$\psi(\color{red}z)=\sup\{ ~^\omega(\psi(\color{red}z[n, \color{red}z])),n<\omega\}$

What are we trying to decompose?

7:21 PM

a sequence of exponential towers of $\psi(z[n,z])$ for each $n < \omega $ I think?

something of the form:

${}^{\omega}\psi (z[0,z]),{}^{\omega}\psi (z[1,z]),{}^{\omega}\psi (z[2,z]),...$

Simply Beautiful Art

No no

What's the thing before the "=sup"?

In relation to our problem

That'll tell you what $z$ is.

$z = \Omega$ in our problem

$\psi(\color{red}\Omega)=\sup\{ ~^\omega(\psi(\color{red}\Omega [n, \color{red}\Omega])),n<\omega\}$

I still don't understand what is the other $z$ doing inside the square brackets

Simply Beautiful Art

Hm

Maybe we should remove the tetration for limit ordinals?

For simplicity.

It doesn't change anything I promise.

well I already have everything in $\psi$ so it should be fine, you can remove tetration

Simply Beautiful Art

Okay

So you have $\psi(\Omega[0,\Omega])=\psi(\omega)$

7:31 PM

yup

Simply Beautiful Art

$\psi(\Omega[1,\Omega])=\psi(\psi(\Omega[0,\Omega]))=\psi(\psi(\omega))$

generalized to $\psi(\Omega[n,\Omega])=\psi^{n+1}(\omega)$?

ok, so that ${}^{\omega}$ in the sup counts the $\psi$ nestings

Simply Beautiful Art

(back to eating)

@Secret No, it's supposed to be tetration

But it just looks messy for limit ordinals.

Also is $\psi (\Omega [n,\omega])$ defined, or in general anything where the bracket don't match the term being act on?

I still have no idea what is supposed to be the role of the 2nd argument in the square bracket

Simply Beautiful Art

@Secret It's supposed to be "psi of the Ω after [n,Ω] is used on the Ω"

@Secret Just for the recursion.

7:36 PM

but if the two terms must match, won't $\Omega [n]$ makes things tidier?

Simply Beautiful Art

@Secret No

Consider $\psi(\Omega^2+\Omega)$

$\psi(\Omega^2+\Omega)$
$\to\psi((\Omega^2+\Omega)[n,\Omega^2+\Omega])$
$\to\psi(\Omega^2+(\Omega[n,\Omega^2+\Omega]))$

You have an $\Omega[n,\Omega^2+\Omega]$ term

Point being they won't always match up.

At first they will, sure. But it quickly goes downhill from there if you don't use that second argument.

How does $\Omega[n,\Omega^2+\Omega]$ differ from $\Omega[n,\Omega]$?

none of your rules cover how to expand that

Simply Beautiful Art

@Secret Not true.

Let's take $n=3$ for one case.

$$\Omega[3,\Omega^2+\Omega]=\psi((\Omega^2+\Omega)[2,\Omega^2 +\Omega])\\\text{vs.}\\ \Omega[3,\Omega]=\psi(\Omega[2,\Omega])$$

Hey @chrono

(Chrono's a bud who does stuff like dis)

hi

Simply Beautiful Art

Basically user21820 made a cool syntax for OCF's

7:48 PM

how do i turn on the jax / tex?

Simply Beautiful Art

I'm basically trying to explain them right now, and they're also how my TREE(3) works

@chronolegends tinyurl.com/cfqcvpc

If you need more help setting it up...

I can see the difference now, though it is still not clear what it means to me. Perhaps going through more $\psi$ should make it clear...

ohhh looks purddy now

Simply Beautiful Art

Well, have you worked out a few simple values e.g. $\Omega[n,\Omega],~n=0,1,2,3,\dots$

Hopefully the pattern will pick up.

are you making original ocfs again @SimplyBeautifulArt

gives you the parental look of dissaproval

Simply Beautiful Art

7:55 PM

@chronolegends Original? No, I'm just giving them a recursive syntax.

oh ok

Simply Beautiful Art

The sort that you can write in a program (particularly the kind that goes past TREE(3) *coughs*)

*::Wonders what @Secret is doing::*

$\Omega[n,\Omega] = \psi(\omega), \psi^2(\omega), \psi^3(\omega), \psi^4(\omega),...$

$= \epsilon_{\omega}, thinking...$

that looks weird

Simply Beautiful Art

@Secret $\psi(x)=\varepsilon_x$ for reasonably small $x$.

@chronolegends =P

It's for $n=1,2,3,\dots$

7:59 PM

Ah so that's why...

Simply Beautiful Art

(@Secret not sure if it was intentional, but you skipped over $n=0$)

@Secret (and here comes the big reveal!)

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