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00:00
φ(3,0) = {φ(2,0), φ(2,φ(2,0)), φ(2,φ(2,φ(2,0))), ...}
Just like how φ(2,0) was a lot of ε's, we have φ(3,0) as a lot of φ(2,x)'s
and then φ(w,0)
and then to the skies!!
φ(ω,0) = {φ(1,0), φ(2,0), φ(3,0), ...}
but then φ(e0, 0)=φ(0, 0, 0)
i think
or is it lgr
00:02
φ(ω,1) = {φ(1,x), φ(2,x), φ(3,x), ...} where x=φ(ω,0)+1
@AlexanderDay and nope to that
φ(ω+1,0) = {φ(ω,0), φ(ω,φ(ω,0)), φ(ω,φ(ω,φ(ω,0))), ...}
etc.
then when do we get to φ(0,0,0)?????????????
φ(ε0,0) = {φ(ω,0), φ(ω^ω,0), φ(ω^ω^ω,0), ....}
00:04
φ(1,0,0) = {φ(1,0), φ(φ(1,0),0), φ(φ(φ(1,0),0),0), ...}
Notice that φ(φ(1,0),0) = φ(ε0,0)
Very very very big
mind blown
the supermum of that...
BOOOOOOOOOOOOOOOOOMMMMMMMMMMM!!!!!!!!!!!
φ(1,0,1) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,0,0)+1
φ(1,0,2) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,0,1)+1
00:07
φ(1,1,0) = {φ(1,0,0), φ(1,0,φ(1,0,0)), φ(1,0,φ(1,0,φ(1,0,0))), ...}
lemme guess...
φ(1,1,1) = ?
@SimplyBeautifulArt repeat these parts...
@AlexanderDay Yup!
@SimplyBeautifulArt with this part...
00:08
φ(1,1,1) = {φ(x,0), φ(φ(x,0),0), φ(φ(φ(x,0),0),0), ...} where x=φ(1,1,0)+1
φ(2,0,0) = {φ(1,0,0), φ(1,φ(1,0,0),0), φ(1,φ(1,φ(1,0,0),0),0), ....}
i was typing that...
φ(1,0,0,0) = {φ(1,0,0), φ(φ(1,0,0),0,0), φ(φ(φ(1,0,0),0,0),0,0), ...}
Mine!
φ(1,0,0,1) = {φ(x,0,0), φ(φ(x,0,0),0,0), φ(φ(φ(x,0,0),0,0),0,0), ...} where x=φ(1,0,0,0)+1
And it just keeps going like this
and just keep going...
BTW, could you tell me what googolpelxainth is close to in Veblan terms??
Small Veblen Ordinal = {φ(1,0), φ(1,0,0), φ(1,0,0,0), ....}
it is 10^{10^{10^100}}
googolplexianth < f_ω²(3)
Nowhere close to Veblen
00:13
We then have things beyond all of Veblen ordinals
Mainly, we have the ordinal collapsing function (OCF)
and I have to wash dishes
K
what about Meameamealokkapoowa oompa the number i mentioned a few days ago to get you concerned?
00:35
U back yet?
Back
@AlexanderDay you beat it with super extended ordinal collapsing functions
going...
going....
...AND IT"S OUTTA HERE!!
it's way outtta Veblan function!!
00:43
how big is it?
i heard the inventor of Bower's operators came up with it.
That is false
what's next?
wait---WHAT!!
it was in my book 1001 mathemathics!!
who did??????????????
C(α,0) = {0,1,ω,Ω}
C(α,n+1) = C(α,n) U {δ+Γ, δΓ, δ^Γ, ψ(µ) | δ,Γ,µ in C(α,n) and µ<α}
ψ(α) = {δn | δn in C(α,n) and δn<Ω, n in N}
Ω > ψ(α)
00:47
?????
This is the scary looking ordinal collapsing function
i see gamma, delta,and in. i never got in.
don't worry, we'll go through it step by step
could u give me a crash course in the in??
??
?
First of all, there is no such ψ(-1). We never have negative numbers
00:49
???
ψ(0) is the first number
C(0,0) = {0,1,ω,Ω}
To get to C(0,1), just add, multiply, and exponentiate all of this
C(0,1) = {0, 1, 2, ω, ω+1, ω2, ω^2, ω^ω, Ω, Ω+1, Ω+ω, Ω2, Ωω, ...}
00:51
                                ??
still confused.
talk about scary-looking
2 = 1+1
ω2 = ω+ω
ω^2 = ω*ω
All I did was add, multiply, exponentiate. We started with {0,1,ω,Ω} and now we have this
BUT that is in Veblan function??
It looks like Veblen function... but it will stop looking like Veblen function later
but is it momentarily?
Yes
max(C(0,1) and less than Ω) = ω^ω
You agree with this so far?
very big
all i have to say
Mhm
C(0,2) = add, multiply, exponentiate all the previous stuff
C(0,2) = {0, 1, 2, 3, 4, ω, ω+1, ω+2, ω+3, ω2, ω2+1, ω2+2, ω3, ω4, ω^2, ω^2+1, ω^2+2, ω^2+ω, ω^2+ω2, ω^ω, ...., Ω, Ω+1, Ω+2, ....}
like this:{0, 1, 2, 3, ω, ω+1, w+2, ω2, w3ω^2, w^3ω^ω, w^w^w, Ω, Ω+1, Ω+2, Ω+ω, Ω2, Ωω, ...}
Yeah, pretty much what I got, you get the general idea
00:56
beat me to it. AURRRRUUUUUUUUUUGGGGGGGHHHHHHHHHHH!!!!
max(C(0,2) and less than Ω) = ω^ω^ω
etc.
ψ(0) = {ω^ω, ω^ω^ω, ω^ω^ω^ω, ....}
= ε0
C(1,0) = {0,1,ω,Ω}
the same!!
or is it??
The thing different about C(1,1) is that not only do we add, multiply and exponentiate, but we also do ψ(µ) for µ in {0,1,ω,Ω} and µ<1
& is still veblan....
So C(1,1) is just like C(0,1) except we also have ψ(0) thrown into the mix, since 0 is in {0,1,ω,Ω} and 0<1
00:59
???
C(α,0) = {0,1,ω,Ω}
C(α,n+1) = C(α,n) U {δ+Γ, δΓ, δ^Γ, ψ(µ) | δ,Γ,µ in C(α,n) and µ<α}
ψ(α) = {δn | δn in C(α,n) and δn<Ω, n in N}
Ω > ψ(α)
C(1,1) = {0, 1, 2, ω, ω+1, ω2, ω^2, ω^ω, ψ(0), Ω, Ω+1, Ω+ω, Ω2, Ωω, ...}
head explodes
oh...
So it turns out that...
ψ(1) = {ψ(0), ψ(0)^ψ(0), ψ(0)^ψ(0)^ψ(0), ....}
= ε1
01:03
so then C(2,0) is where ψ(2) is the t/z thingie??
or e(e0)
ψ(2) = ε2
Much smaller
ψ(3) = ε3
01:04
@SimplyBeautifulArt so that is where you where going!
@InfiniteMonkey Just you wait
And then... we have something weird
@InfiniteMonkey hi
ψ(φ(2,0)) = ε(φ(2,0)) = φ(2,0)
Since φ(2,0) = ε(ε(ε(...))) is the t/z thingie
I'll just call it z0
@SimplyBeautifulArt oh...
Now, we have magic!
alacadabra, alacazam, make this easier, no we cant!
ψ(z0+1) = z0
ψ(z0+2) = z0
ψ(z0*2) = z0
...
ψ(Ω) = z0
Alaca...what?!
xD
Remember, Ω is very big. It is bigger than ψ(x) for any x.
01:07
alacadabra!
um...
@SimplyBeautifulArt looks like a weird "wall"
be back soon!
i gotta go.
Indeed, it is a wall
But this is what makes the ordinal collapsing function... powerful?
@AlexanderDay Bye!
like tomorrow....
sorry...
notice that no matter what you do, you only start with {0,1,ω,Ω}
@AlexanderDay No problem!
From there, you can only add, multiply, exponentiate, and take ψ(µ)
The problem is that we can't take ψ(z0), since to do this, we need to somehow reach z0 first.
01:10
I see...
But you can't reach z0 using addition, multiplication, exponentiation, and ψ(µ) for µ values we've reached. If we reached µ=z0, we wouldn't be needing to take ψ(z0) to begin with...
So we're stuck. But not forever.
C(Ω+1,0) = {0,1,ω,Ω}
So ordinal collapsing functions let us "break through" the z0 wall
C(Ω+1,1) = {0, 1, 2, ...., ω^ω, ψ(0), ψ(1), ψ(ω), ψ(Ω), Ω, Ω+1, ...}
Notice the last of these ψ's
We can break through!
Until we hit ψ(Ω+z1) = z1
But we can break through! ψ(Ω+Ω) = ψ(Ω2) = z1
and then we get stuck, and then we break through
eventually ψ(Ωx) = x, so we get stuck.
and then we break through! ψ(Ω²) = x
You get the idea
ψ(Ω^Ω^ω) is the small Veblen ordinal.
ψ(Ω^Ω^Ω) is the large Veblen ordinal
ψ(Ω^Ω^Ω^Ω^...) is the Bachmann-Howard ordinal
and then we get stuck
01:15
but we can break through?
is there anything special to be said about the points where we get stuck or they're not interesting?
Why? Well note that you can never exceed an infinite tower of Ω's due to the restriction of addition, multiplication, and exponentiation, and since Ω>ψ(x), we can't make more Ω's that way
@InfiniteMonkey Yes, at the very low levels, we get some fairly obvious Veblen equivalents
But there is a way to go beyond....
Say hello to ψ1(0) = Ω^Ω^Ω^Ω^...
Just like ψ(0) = ω^ω^ω^ω^...
And the game starts allover again!
Mhm
To break through ψ1(x) = x, we use Ω_2
The second Ω
Again, we get stuck at infinitely many Ω_2's powered to each other, so we use ψ2(0)
then ψ3(0)
etc.
then ψω(0)
We can even reach ψΩ(0)
^^^ Just imagine that!
We can even have Ω_Ω_Ω
and the normal extended OCF ends at infinitely many Ω's tied to each other
Which is where we bring in ψI(0)
That gets quite scary 0.0
the inaccessible cardinals...
Oh yeah. It goes pretty darn far
I's will get stuck, so use I+1, then I+2, etc. even Ω_(I+1), etc. eventually I(1), the second inaccessible
You can even go up to things like I(Ω), or I(I(I(...)))
I(1,0) is beyond all this...
@InfiniteMonkey So now you can make extremely large finite numbers
01:24
That is mind boggling; so as long as two people understand what they mean when creating these they can go on forever, or do we eventually hit an unbreakable wall (unlikely)?
yes, we do hit a wall
We have to; otherwise we are talking infinity right?
beyond all the inaccessibles will lie the weakly Mahlo cardinals
xD Beyond those are higher Mahlo cardinals
beyond those are compact Mahlo cardinals
and there is the end to current ordinal stuff
Oh no! What to do then? Do we have to wait for you to invent some new techniques? ;)
WEll
let's just say you have no idea how crazy complicated I() is
and Mahlo is literally BEYOND inaccessible
its just very frightening and you'll be out of breath by the time you reach compact Mahlo cardinals
01:32
Haha yeah you're right I have no clue yet, I understand just enough to appreciate some what you are showing me here though and that is a lot of fun to be scared by nimbers!
Numbers... but nice typo!
We're out of ZFC right?
 
4 hours later…
SBM
SBM
05:30
Morning
g'morning
oh goodness your name is really SBM
SBA & SBM
SBM
SBM
Oh what btw?
sorry I guess I don't really know you and that's not really a great way to start talking to someone
let me try this again
hi!
>.<
...come here often?
SBM
SBM
05:48
Yes
Hello
@EricStucky are you there?
Are you here for the large numbers? or otherwise?
SBM
SBM
Mathematics in general
sort of
haha, I feel that
I'm not sure if it's the same feel, but I feel that :P
How's the evening so far?
SBM
SBM
06:04
It's morning here
oh, well
SBM
SBM
06:17
Oh, everything's just about okay.
that's good :)
are you getting toward the end of your semester?
SBM
SBM
I'm just a school kid btw
yeah, I figured
nothing about you in particular, just that most people in this room are
SBM
SBM
oh
 
2 hours later…
08:26
You guys were here fairly early @SBM @EricStucky
SBM
SBM
Oh
Time zones...
I woke up at 5AM UTC+1 lol
But I think I'm the only very early bird here :p
I stay up late. And then have trouble getting up in the morning :/
Go to bed early
wake up really early
that's what I do, and I have loads of free time early in the morning
it's nicer having free time when you have energy to do stuff with it. I keep trying to convince people of this :P
SBM
SBM
08:32
no, its 14:02 here
'tis 11:33 AM for the minute
09:33 here
24 hour time
@SBM *it's ;)
SBM
SBM
oh
(don't mind me. I have a reputation as a grammar freak.)
SBM
SBM
08:34
Okay though people still use 'tis instead of It is
'tis and it's are different.
09:21
They can be used pretty interchangeably though @Mithrandir
@shredalert right, but they're still different
I've not seen many uses of them where they aren't synonymous
Can't even think of any atm
...but they're still different contractions.
Wasn't saying they aren't :p
 
1 hour later…
10:38
@InfiniteMonkey ZFC is out after I
And good morning!
SBM
SBM
11:09
Morning @SimplyBeautifulArt
The sun is big and red today
At least here
 
2 hours later…
SBM
SBM
12:52
Oh
13:22
SBM
SBM
13:43
How do I integrate any kind of integrable function?
What do you mean?
There is no such god technique
SBM
SBM
I need a checklist to learn whatever methods that exist that I don't know.
@SimplyBeautifulArt so if Peano's "wall" is ε0, then ZFC's "wall" would be ψ(Ω^Ω^Ω^Ω^...)
... and good morning to you too!
SBM
SBM
Wait, I came across an interesting function called the Digamma function.
Leave it.
I've a question.
Would
$$\int_0^1 x^x \mathrm{d} x = \int_0^1 \int_0^1 (xy)^{xy} \mathrm{d} x \mathrm{d} y$$
?
Oh, I don't know why
or why not.
It appears to be not equal, is it so?
@SimplyBeautifulArt
Oops
dy dx I meant
14:03
@InfiniteMonkey no, ZFC's wall is ψ(I)
Hey @SBM
SBM
SBM
Hello
That's a problem @S.C.B knows, but he's busy sadly
SBM
SBM
oh
I'm searching
@SBM check around 4 o'clock and go from there
SBM
SBM
I'm currently chatting with him on Discord
SBM
SBM
Just solving both is enough I guess
I meant evaluating
them
Wait @SimplyBeautifulArt
In mathematics, sophomore's dream is the pair of identities (especially the first) ∫ 0 1 x − x d x ...
Maybe that.
14:46
@SBM hey
SBM
SBM
Hello
SBM
SBM
15:00
Need to learn how to manage multiple integrals.
@SimplyBeautifulArt @shredalert etc Need votes to close:
-2
Q: Finding omega limit sets of the following five functions

johnPlease help me find the omega limit sets of the following functions in their respective domains: (1) f(x) = x - x^2 [0,1]->[0,1] (2) f(x) = 2x -2x^2 [0,1]->[0,1] (3) f(x) = 3.2x - 3.2x^2 [0,1]->[0,1] (4) f : R -> R; f4(x) = x^3 + (1/5)x , R is the real numbers (5) f : [0, 2]-> [0, 2];...

Oops, did I interrupt? Oh well....
The post I just posted has been since edited.
 
2 hours later…
16:45
Lol
@amWhy done!
@SimplyBeautifulArt :D
 
3 hours later…
20:34
Close me please?
0
Q: $a,r,m\in \mathbb{Z^+}$ prove that $(a^r-1)\mid (a^m-1)\Leftrightarrow r\mid m$

stenvik teamI'm reading a book that claims that this is true. Is there a short/easy proof for this?

Oops, that's right, I forgot....Exams in progress!!! Best wishes to all!!
20:52
Hello, @Myth ... oops @Mithrandir! ;-)
@amWhy zz
One of my nicknames is Mythical Rand, soo...
I was wondering about that .... suits you (in a good way).
How do you like moderating?
Finished all my assignments this afternoon
been catching up on my fiction :D
going to read until Monday now
20:57
@shredalert Awesome!!! That's great!
It's okay. Haven't had any problems on the main site yet - it's more chat issues that are difficult.
(like what brought me to this room in the first place...)
@Mithrandir Interesting; I bet that can get crazy.
Do you remember the flag that brought you here, to this chat?
@amWhy it was from some political debate
I think everything got trashed, though
Wow ... I think I remember that!
21:00
I think politics and religion always end up causing chat arguments
because people are less tolerant of views not their own than they are willing to admit
@shredalert Agreed!
Mhm.
...gack, I meant to go to sleep early today.
go now then
you can always hop on early
(it's now after midnight)
drops out
night
cya tomorrow
21:04
Sleep well!
21:15
I'm off to bed too
night @amWhy
@shredalert Night-night! Sweet dreams!
 
1 hour later…
22:23
HELLO!!!!! is anyone home??
22:42
What is the real part of euler's formula
pi??
have U seen SBA? he is AWOL.
@Alexander how so
no I haven't seen SBA
I thought the real part would be cos
i remember that it had something to do with e^pi i -1=0
and i is imaginary
$e^{ix} = \cos(x) + i\sin(x)$
^ that's the full Euler's formula
oh...
has anyone seen SBA??
22:49
@Eric so it's $cos(x)$ that's the real part
23:23
@SimplyBeautifulArt r u there?
23:40
@EricStucky also, euler's formula is the most famous mathemathicial formula of ALL TIME

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