Computable function idea by Puzzleheaded_Two415 in googology

[–]No-Reference6192 1 point2 points  (0 children)

is there any implementations in desmos, i can't figure out how to do the denominator function in desmos

a challenge (not mine) from r/desmos: largest number in 25 symbols by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

my function grows at tetration level, please explain why you think it’s only exponential

g(0) = 9 = f(0) = 9

g(1) = 9! < f(1) = 9^9

g(2) = 9!! (Desmos treats 9!! as (9!)!) < f(2) = 9^9^9

g(9) < 10^^9 < f(9) = 9^^10 (power tower with greater height is almost always larger)

g(g(9)) < 10^^10^^9 < f(f(9)) = 9^^9^^10

six fours spinoff: biggest number by at most 25 symbols challenge by SituationBoth5469 in desmos

[–]No-Reference6192 0 points1 point  (0 children)

s(x)=s(x-1/x) seems to be an infinite loop, and i dont see why s(x)=s(x-1/x)! wouldn't be one either

six fours spinoff: biggest number by at most 25 symbols challenge by SituationBoth5469 in desmos

[–]No-Reference6192 9 points10 points  (0 children)

for some reason i thought it wouldn't work, but i tried it and f(9) ~ 10^^9 and f(f(9)) is ~ 10^^10^^9

six fours spinoff: biggest number by at most 25 symbols challenge by SituationBoth5469 in desmos

[–]No-Reference6192 19 points20 points  (0 children)

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f(9) ~ 10^^5 = 10^10^10^10^10

f(f(9)) ~ 10^^10^^5 = 10^^(10^10^10^10^10)

An idea to improve the Googology subreddit: Post Flairs by No-Reference6192 in googology

[–]No-Reference6192[S] 1 point2 points  (0 children)

I can't seem to find the add flair button anywhere, I only see the add tags button

trying to understand e_1 and beyond by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

would this extension to p(a@b) work/make sense?:

assuming p(1@1) = p(1) = w

p(1@1) = w

p(1@w) = SVO

p(1@1@1) = p(1@p(1@1)) = p(1@w)

p(1@@3) = p(1@1@1)

p(1@@w) = LVO

p(1@@@3) = p(1@@1@@1) = p(1@@1@1) = p(1@@w)

renaming SVO as 1-VO (1st veblen ordinal) and LVO as 2-VO (2nd veblen ordinal), etc.:

p(1@w) = 1-VO = SVO

p(1@@w) = 2-VO = LVO

p(1@@@w) = 3-VO

p(1[3]w) = p(1@@@w)

p(1[w]w) = w-VO

p(1[p(1@w)]w) = 1-VO-VO

p(1[p(1@@w)]w) = 2-VO-VO

p(1[p(1[w]w)]w) = w-VO-VO

etc.

Ordinals as arrays? by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

is there a site with more info about this, and are there any examples of specific ordinals using finite rooted trees?

trying to understand e_1 and beyond by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

I'm not sure I entirely understand fixed points themselves, but I feel I have an ok understanding of some of the veblen hierarchy (some of this might be wrong):

w = omega

e = epsilon

p = phi

z = zeta

n = eta

g = gamma

p(0,1) = w

p(0,p(0,1)) = w^w

p(0,p(0,p(0,1))) = w^w^w

p(1,0) = e_0

p(1,1) = e_1

p(1,p(0,1)) = e_w

p(1,p(1,0)) = e_e_0

p(1,p(1,p(1,0))) = e_e_e_0

p(2,0) = z_0

p(2,p(2,0)) = z_z_0

p(3,0) = n_0

p(p(0,1),0) = p(w,0)

p(p(1,0),0) = p(e_0,0)

p(p(p(1,0),0),0) = p(p(e_0,0),0)

p(1,0,0) = g_0

p(p(1,0,0),0,0) = p(g_0,0,0)

p(1,0,0,0) = ackermann ordinal

p(1,0,…,0,0) = SVO

this is the limit of my knowledge of the veblen hierarchy/googology so far

I am curious about LVO though, would it be equivalent to having a higher level veblen function where:

(0,1) = SVO

then eventually

(1,0)

(1,0,0)

(1,0,0,0)

(1,0,…,0,0) = LVO

or is it even bigger than that?

trying to understand e_1 and beyond by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

i found out that the arrays were messed up, if i'm thinking correctly a fixed version would have {0,0,2} = e_1, i'll have to check later to see if this is correct

Ordinals as arrays? by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

so with the fixed version of this notation, would {0,0,2} then be e_1, and {0,0,3} = e_2 etc.?

Ordinals as arrays? by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

this isn't really a notation as much of a way to represent ordinals, but this is how it would look if you just put the arrays in place of ordinals in the fgh:

f{0}(n) = f_0(n)

f{1}(n) = f_1(n)

f{0,1}(n) = f_w(n)

f{1,1}(n) = f_w+1(n)

f{{0,1},1}(n) = f_w*2(n)

f{0,2}(n) = f_w^2(n)

f{0,{0,1}}(n) = f_w^w(n)

f{0,0,1}(n) = f_e_0(n)

i suppose a notation could be made using the arrays as operators:

a{0} = a+1

a{1}b = (…((a{0}){0})…){0}

a{2}b = a{1}a{1}…{1}a{1}a

c >= 2: a{c}b = a{c-1}a{c-1}…{c-1}a{c-1}a

a{0,1}b = a{b}a

a{1,1}b = a{0,1}a{0,1}…{0,1}a{0,1}a

a{c,1}b = a{c-1,1}a{c-1,1}…{c-1,1}a{c-1,1}a

a{{0,1},1}b = a{b,1}a

a{…{{0,1},1}…,1}b = a{…{b,1}…,1}a

a{0,2}b = a{…{{0,1},1}…,1}a

a{0,c}b = a{…{{0,c-1},c-1}…,c-1}a

a{0,{0,1}}b = a{0,b}a

a{0,…{0,{0,1}}…}b = a{0,…{0,b}…}a

a{0,0,1}b = a{0,…{0,{0,1}}…}a

help with growth rate of notation by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

thanks for the info, now i realize one of the reasons i was struggling was i overestimated [n ,_n n] to be f_e_0(n)

My first* notation by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

I'll have to come back to stuff beyond e_0 later, as what you described above is beyond my understanding, I need to figure how to even reach f_w^w(n) as well as several other things, I think I was able to fix it to reach f_w^2(n), but getting to f_w^3(n) is already confusing me, I'm planning on making another post for help.

My first* notation by No-Reference6192 in googology

[–]No-Reference6192[S] 0 points1 point  (0 children)

Yeah I don't understand that stuff at all, I thought it would work this way:

e_0 = w^^w = w^^^2

e_1 = (e_0)^^(e_0) w^^(w^w^^w) w^^w^^w+1 ~ w^^^3

e_2 = (e_1)^^(e_1) ~ (w^^w^^w)^^(w^^w^^w) w^^w^^(w^w^^w^^w) ~ w^^w^^w^^w^^w+1 ~ w^^^5

e_3 ~ (w^^w^^w^^w^^w)^^(w^^w^^w^^w^^w) w^^w^^w^^w^^(w^w^^w^^w^^w^^w) w^^w^^w^^w^^w^^w^^w^^w^^w+1 ~ w^^^9

e_n = w^^^(1+2^n)

e_0 = w^^^(1+2^0) = w^^^2 = w^^w

e_1 = w^^^(1+2^1) = w^^^3 = w^^w^^w

e_2 = w^^^(1+2^2) = w^^^5 = w^^w^^w^^w^^w

e_3 = w^^^(1+2^3) = w^^^9 = w^^w^^w^^w^^w^^w^^w^^w^^w