I realized f(1) is already bigger than Rayo's Number, not 18.

@jaydenzhang786

8 ай бұрын

The reason why the biggest number with n steps doesn't work is because i can say f(1) 18000000000000000000000000... infinite zeroes. 1+8+0+0+0... is still 9.

@carbrickscity

8 ай бұрын

I don't see why it doesn't work. Your number requires 1 step. You can try another number with more random digits, some can yield 2 or 3 steps. Actually similar ideas are used for Loader, BB, and Rayo.

@omicsgd1337

8 ай бұрын

@@carbrickscity He's talking about what you said about using the largest number possible and if you take 1800000.... an infinite number of zeros, it's still 9 and we converted in 1 step and ended up with a number that is greater than any number.

@omicsgd1337

8 ай бұрын

@@carbrickscity And you also said that f(4) is indefinite although it is not because we can say that on the 4th last step we will have let's say 9 on the 3rd step we can take the number 111111111 and if we sum all its numbers we will get 9 on the 2nd step we can take 11111..... 11111111 1s and on the first step we get a number which is greater than 10^10^10^9 but less than 10^^4 if we consider f(5) we can in theory get 10^^5. If we revise the function a bit and say that we can use f(n-1) steps, that is on f(4) we can use f(3) steps or in theory 10^^3 and then the growth of the function in theory will be f(n) = 10{n}n or a little more or a little less. As for me, the function is cool, but it should be improved (at least, it should be forbidden to use zeros in the initial number, i.e. at the n step, and in the subsequent steps zeros can be used in the function).

@carbrickscity

8 ай бұрын

Your comment is long yet I don't quite get what you are trying to explain. f(1) is already "infinity". The biggest number that require 1 step could be as big as you want. Same for f(2) or f(3). All I get is that we are agreeing on the same thing but you are trying to disagree...But anyway this is kind of an open ended question.

That has to be like...at least faster than addition

long time no see how is this function exactly called? i mean google shows me up regular g function

@carbrickscity

3 ай бұрын

It's mentioned in the video.

From my point of view, a function have a defined growing speed requires that it is PROVEN that that function f(n) is finite for all finite n Or else it has undefined growing speed and can't be a fastest growing function By the way, we have FGH to measure growing speed of slow growing functions, but do we have a well defined, simple and rigorous way to measure growing speed of functions like rayo? Some ways mentioned in other videos are either not well defined, or not rigorous and simple.

And another one! ☝🏾

These should not be called fastest growing, because they shop growing. f(4) = f(5) = f(6), etc... NOT growing. whereas tree(4) this is a growing function.

@carbrickscity

8 ай бұрын

You have a good point.

@OnlySkeep

8 ай бұрын

however, from f(2) to f(3) it grows faster than any other function Edit: f(1) is also infinity so from f(0) to f(1)

@MarkBettner-fi2ec

8 ай бұрын

TREE function is partially defined and doesn't have any exact method of growth other than what can be defined by the higher order of fast growing hierarchy so is kind of cheating.

@007Rincewind

8 ай бұрын

@@MarkBettner-fi2ec But all numbers in TREE(n) have been proved to finite. So it is that FGH have trouble containing TREE.

@MarkBettner-fi2ec

8 ай бұрын

@@007Rincewind kinda... TREE diction can be explained by time turing machines which work in counts of tran finite functions rather than hierarchy of tran finite function growth rate. So it can be contained in the FGH but not in the same way that arrow notation or g sequence can.

What is the name of this sequence? Its just g?How to differentiate from the function used in Graham's number?

@carbrickscity

7 ай бұрын

It's not really a sequence. It's a problem called "Inserting Plus Signs and Adding".

@etk128

7 ай бұрын

@@carbrickscity thanks

Is Rayo(10¹⁰⁰) less than Rayo(10¹⁰⁰-1)? I ask because Rayo doesn't have to get bigger it just the biggest number that can be described with n symbols

@carbrickscity

8 ай бұрын

Supposedly yes.

@007Rincewind

8 ай бұрын

@@carbrickscity Definatly bigger with the extra symbol, this is not really a FOST symbol but in normal math you can always get bigger with a single symbol just add ! and you are much bigger.

@MarkBettner-fi2ec

8 ай бұрын

The rayo function is an inclined growth meaning that with each additional rayo value over say .... A few hundred would be vastly greater than the value which came before. The one big caveat here is that rayo in an undefined growth so it follows no exact growth rate meaning that each additional value could be infinitely greater or smaller than the one being mentioned.

@Exahedron

5 ай бұрын

Rayo function is the first number larger than *any* number writable in fost with n symbols *or less.* meaning that rayo(10^100) *might equal to* rayo((10^100)-1) but rayo(10^100) is definitely not less

5:09 I guess you could make a function faster by extending it to the surreal numbers and using the inverses of limit ordinals. Kinda like: B(1)=1/ε Β(2)=1/ε1 (inverse of ω1) And so on, so forth. As for a function in the reals, I don't think you can get faster than this.

@carbrickscity

8 ай бұрын

Why would you even need inverse of an inverse...might just define B(2) = ω1 or something but for these kind of functions I am sure many people don't considered them as legit.

@nzqarc

8 ай бұрын

@@carbrickscity it wouldnt be a function then, it would just be a number list. it needs to GROW big, not BE big

@nzqarc

8 ай бұрын

@@carbrickscity i mean, if the square root of contable infinity is legit in the surreal numbers, then this function shouldnt be a problem

@carbrickscity

8 ай бұрын

But your function is just the same as a number list in the case. Whether it is legit or not, it depends on how people look at it and based on the comments here it's probably not.

@nzqarc

8 ай бұрын

@@carbrickscity why wouldnt it be? It can be calculated and explained how and why it grows on such values rather than just stating that "yep, that's the value" like with BMS

My function to beat this. BB(n) except the tape never stops. (Don't copy).

@SixLeafCloverOFire

8 ай бұрын

That’s not really what that means. ALL Busy Beaver numbers are given an infinite amount of tape if needed.

@user-ik8tb5qh7c

8 ай бұрын

@@SixLeafCloverOFire exactly

What is (infinte zeros) .000000000000.............1 x infinity

@carbrickscity

8 ай бұрын

Don't know what you mean.

Faster growing function: j(n) = n x 0 + 1 / infinitesimal j(1) is already infinity!!111

@carbrickscity

8 ай бұрын

You need to define infinitesimal, and you are trying to use limits rather than actual numbers. I would say it's worse than the one in the video. This is just another lazy way of saying 1/n when n tends to zero. And you realized your n x 0 part can be omitted right away so your function is basically just 1/0.

@kiwi_2_official

8 ай бұрын

@@carbrickscity is joke

@OnlySkeep

8 ай бұрын

​@@kiwi_2_officialbut it doesn't make sense even if it is a joke

@kiwi_2_official

8 ай бұрын

@@OnlySkeep still a joke though

1/(1-0.999…)

@carbrickscity

8 ай бұрын

It's not a function. Also 0.999...is just 1, so you have divide by 0 which is undefined.

This one doesn't make sense as a function. G4 has no solutions. There are no biggest or smallest number of steps.

@carbrickscity

6 ай бұрын

See it as the longest possible number of steps.

TOO fast.

I think this type of number are not considered valid in Googology, it is very easy to grow to infinity. A nuber that is unbounded is usually considered value 0 in googology. A very easy function that grows as fast as yours are F(N) where F(N) is the numbers of additions needed to reach N. and the additions follows the following pattern. 1 + 0.5 + 0.25 + .... F(1)=1 but F(2) is infinity.

@MarkBettner-fi2ec

8 ай бұрын

All the finite values over a googol would be.

@carbrickscity

8 ай бұрын

Good point as well.