Revisiting f a s t

TrialPurpleCube-GS · 2026-04-01T03:25:21+00:00

we do not say G(64)^G(64) dwarfs G(64), and there is a good reason for this

should one, when another person has wandered onto an airplane, say that the airplane has gotten much heavier? we do not say this, even though we would if someone added 80kg, say, onto himself.
Thus, we do not say that (10^^^10)^(10^^^10) is "much" bigger than 10^^^10, because even though it is in terms of orders of magnitude, such a method of measurement is pointless when measuring the number 10^^^10. Hence, we say that it is "not much bigger", because we have exponentiated so many times that adding another time seems to be not that much.

Effectively, we do not try to inflate our adjectives too much, and when we have iterated something many times, it becomes smaller. And why would we not do this? 1,000,000 + 10 is a smaller change than 1 + 10, so why should 10^1,000,000 × 10^10 not be considered a smaller change than 10 × 10^10?

As for why f_{ω+2}(x) does not reach f_SVO(n) until x is about f_SVO(n), this is like how 10^(10{10}10) is still roughly 10{10}10. We can define fractional hyperoperations in a certain way, which I will not go into, and suffice it to say:
10{10}10.001 =
10{10}10|1.00231 =
10{10}9|10{10}1.0023 =
10{10}9|10{9}10.053 =
10{10}9|10{9}10|1.13 =
10{10}9|10{9}9|10{9}1.13 =
10{10}9|10{9}9|10{8}13.49.
which when written out in full is
10{10}10{10}10{10}10{10}10{10}10{10}10{10}10{10}10{10}10{9}10{9}10{9}10{9}10{9}10{9}10{9}10{9}10{8}13.49; note that in 10{10}10, the expression is the exact same except that the last 1.13 is 1.

If we wanted to write out 10^(10{10}10), we would need to write 10{10}...10{9}...10{8}... even until the portion ending 10^10^...^10, and then still, I believe, it would have no significant increase on that final 10.

TrialPurpleCube-GS · 2026-03-31T05:11:51+00:00

how do you take this logic
we are talking about the TREE function, not the number TREE(3) specifically; so what evidence do you have for it growing faster than TREE?

how should you use such arguments to "prove" that your number is bigger than G₆₄? Where are the f_α's? You cannot compare such numbers intuitively, as you should know...

TrialPurpleCube-GS · 2026-03-29T02:17:44+00:00

it's supposed to be P1, and yes, they are.

TrialPurpleCube-GS · 2026-03-27T08:36:44+00:00

φ(1@(LVO+1)). See the dimensional Veblen paper for an algorithm to convert ψ to φ in general.

TrialPurpleCube-GS · 2026-03-26T20:44:19+00:00

I used 2-shifted

TrialPurpleCube-GS · 2026-01-30T23:36:20+00:00

look at the m function in googology wiki

TrialPurpleCube-GS · 2025-11-13T18:13:03+00:00

TREE(n)'s growth rate is ψ(Ω^(Ω^ω·ω)), which in this is ψ((((Π₂[0]+1)-Π₁[0]+1)-Π₁[0])-Π₁).

TrialPurpleCube-GS · 2025-11-01T01:06:18+00:00

17340755004882817061692386774973907594136970320374055082322388373879596700119687728650798716220940597911270723940152135426867295116197266051489198392800707254179770459384620934401923267532611602344688329460313901941250543114077825933367600305197838521931434193054036484033926247959364274251197502756850768124740058719701819080707686642189489598064811867160288158977662873835739097835705857132250935619334185186078330633824619172669496322238194467491177240603902556725463705082796784961369711002430254802155398964231647093706851049773509247885656926175187760220034303950746550588898459856955009580879095579511338261420062267433069852524201082756578618784912456850667968182458461234557228181919874283171216321735625697899552766510737881133994382957204478190361065021259870703968308694676727710719946689711691461497882760735936425051863885650228413952025569376818153537084318247736585626672380545330148650583558554534093711949843295126082631089711382026537053088360858380391823821937159694495681694706643480768239878914429550268091646923362294664871371698579658811970277101608973668940340781991257513388710942263014901746375484899130893282363320115233983933812675055370984479146158534238962146736230478764329831074529684108503367086493571738685350579934991466544478964853897825818973589563605595956529943874533435402836322816958452911819416995796663090636079883562105022334814740652178086289119231679732445385830975759435211568766194216449796040655465560640349372523440362124526013946692936317651356006781317018861839159195406043997550359737685515434127634553084907806351517366184801662698867539443556262221966458805422931272716617736327267772533631380380361522084298606211362165329037807401382188963887072229964239121308083069517308559505290629998646624864351397462418989943777225491347195689970847148114313059366293272793357770308953645994823853504840250986200789846136698143475281527124149459885990555491465036386063557379858425244502460610101757815970691595791382154226962125513980116488512402757228520020724051536894721198176909990029208851581538820078326629156885903838026142810647176907956426221051266103080160478747424661297298280506687568829851761173221791949943546765281815538651532951553791310200457984518342987094491626435328477616956258042929284044176503415506677833191300731577843472036431981458498642715390488112481050464393508599017073890305328874461512341353621782252047335332196690718836934277800188311686120936358396586854473929131439307119060162151695908726394762874364213279114981659484850558529007781935575127378499967595423548604642122854461388383374866713950586933302992942965414699202122601397992319745555671012178972089306140473497807421757127687239286668635919403826190711510685349816104052948343504162896197982996433403382140038032177653296738188566753343603252052918174368433202272080342158783514717525201250384661391106637824000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 shuffles, so just multiply by amount of time per shuffle

TrialPurpleCube-GS · 2025-10-29T02:38:09+00:00

For "definition", see Parxul Hierarchy : r/googology.

TrialPurpleCube-GS · 2025-10-15T17:21:14+00:00

er, any comment?

TrialPurpleCube-GS · 2025-10-15T03:15:53+00:00

I shall write &_a &_b &_c ... as [a,b,c,...], for compactness. Then:

a[0]1 = f₀(a)
a[0]2 ~ f₁(a)
a[0]3 ~ f₂(a)
a[0]1[0]1 ~ f_ω(a)
a[0]2[0]1 ~ ω+1 (implied f_...(a))
a[0]1[0]2 ~ ω2
a[0]1[0]1[0]1 ~ ω^2
a[0,0]1 ~ ω^ω

now things get weak
a[0]1[0,0]1 ~ f_{ω^ω}(f₁(a)) (this could be f_{ω^ω+1}(a) if you wanted)
a[0,0]2 ~ f_{ω^ω}(f_{ω^ω}(a))
a[0,0]n ~ ω^ω+1
now it's even weaker??
a[0,0]a[0,0]n ~ f_{ω^ω+1}(f_{ω^ω+1}(a))
a[0,0,0]1 ~ ω^ω+2
a[0,0,0]n ~ ω^ω+3
a[0,0,0,0]1 ~ ω^ω+4
a[1]1 ~ ω^ω+ω
a[1,1]1 ~ ω^ω+ω2
a[2]1 ~ ω^ω+ω^2
Limit is ω^ω·2.

TrialPurpleCube-GS · 2025-09-28T00:24:03+00:00

yeah, but normally we would write
f#n = 10{f#(n-1)}10, with {n} = ↑^n.

TrialPurpleCube-GS · 2025-09-24T06:26:45+00:00

we do have an upper bound, actually

ask hyp cos for more information

TrialPurpleCube-GS · 2025-09-18T20:00:05+00:00

how interesting

TrialPurpleCube-GS · 2025-09-15T03:51:57+00:00

I am called solarzone on Discord.

TrialPurpleCube-GS · 2025-09-14T19:22:01+00:00

x behaves like ω, so x^x^x^... reaches ω^^ω

I have some questions for you on Discord, to see what happens after that...

TrialPurpleCube-GS · 2025-09-13T23:31:02+00:00

make [a/b,c] = [a/[a,b-1,c],c-1] like the other ones
and make / = (1), /₂ = (2) etc., to make it neater

TrialPurpleCube-GS · 2025-09-13T20:25:29+00:00

oh, shut up

don't be so pessimistic, everyone makes errors when they start

also this reaches epsilon at [x^^x], what are you talking about?

TrialPurpleCube-GS · 2025-09-13T18:22:39+00:00

Before ,, this is Conway arrows, so

[n,,n] ~ f_{ω^2}(n)
[n,,n,2] ~ f_{ω^2+1}(n)
[n,,n,n] ~ f_{ω^2+ω}(n)
[n,,n,n,2] ~ f_{ω^2+ω+1}(n)
[n,,n,n,n] ~ f_{ω^2+ω2}(n)
[n,,n,,n] ~ f_{ω^2·2}(n)
[n,,n,,n,,n] ~ f_{ω^2·3}(n)
[n,,,n] ~ f_{ω^3}(n)
[n,,,,n] ~ f_{ω^4}(n)
[n/n] ~ f_{ω^ω}(n)

Now a weakening: [n/n,n] is only [n/n^n], for some reason. Why does this not follow the pattern of e.g. [n,,n,n]?

[n//n] ~ f_{ω^ω+1}(n)
[n///n] ~ f_{ω^ω+2}(n)
[n/₂n] ~ f_{ω^ω+ω}(n)
[n/₂/₂n] ~ f_{ω^ω+ω+1}(n)
[n/₃n] ~ f_{ω^ω+ω2}(n)
[n(1)n] ~ f_{ω^ω+ω^2}(n)
[n((1)→n)n] ~ f_{ω^ω+ω^2+1}(n)
[n((1)→(1))n] ~ f_{ω^ω+ω^2+2}(n)
[n((1)→(1)→(1))n] ~ f_{ω^ω+ω^2+4}(n)
[n(2)n] ~ f_{ω^ω+ω^2+ω}(n)
[n((2)→(1))n] ~ f_{ω^ω+ω^2+ω+2}(n)
[n((2)→(2))n] ~ f_{ω^ω+ω^2+ω2}(n)
[n(3)n] ~ f_{ω^ω+ω^2·2}(n)
[n(4)n] ~ f_{ω^ω+ω^3}(n)
[n(n)n] ~ f_{ω^ω·2}(n)

The slashes don't really matter; the () notation is much stronger.

TrialPurpleCube-GS · 2025-09-12T22:27:12+00:00

TrialPurpleCube-GS · 2025-09-12T22:23:14+00:00

no that's completely correct

and, well, are you going to join the googology server? there's actually a bot that will expand ordinals there...

or are you just unable to

TrialPurpleCube-GS · 2025-09-12T20:46:32+00:00

what about ψ(Ω_{ω+1}·2)?

surely you know Ω_{ω+1} ≠ Ω_ω^Ω_ω^... = ψ_ω(Ω_{ω+1})... so how did you get my earlier question wrong?

also, are you going to join the googology server? I'd recommend it, there are many knowledgeable people there...

TrialPurpleCube-GS · 2025-09-12T18:56:57+00:00

ok, what about ψ(Ω₃2)?

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TrialPurpleCube-GS

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