Largest function

[Darvince]

inb4 x to the infinity

let's say the arguments of this hypothetical function are 1, 1, and 1.

These 1's turn into this: 111
Each 1 will turn into 2 0's: 000000
The first 0 is then turned into a 1: 100000 (one hundred thousand)

this is a huge function, with arguments as small as 2 and 2.

These 2's turn into this: 22
Each 2 will turn into 2 1's: 1111
Each 1 will turn into 2 0's: 00000000
The first 0 is then turned into a 1: 10000000 (ten million)

3 and 3.

These 3's turn into this: 3
Each 3 turns into 2 2's: 2222
Each 2 turns into 2 1's: 11111111
Each 1 turns into 2 0's: 0000000000000000
The first 0 is then turned into a 1: 1000000000000000 (ten quadrillion)

but then i came up with a larger function:

arguments of 1,1,1.

These turn into this: 111.
This is 111. These 3 1's collectively turn into 333 0's: 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000000000000000000000000000 000

The first 0 then turns into a 1.

[vh]

yeah well my function has 2 arguments like this:
2, 50
2 turns into 2 zeroes
00
then add 1:
100
100 turns into 100 zeroes
0000....0000
add one:
google
>then turn that into google zeroes

repeat 50 times(second argument)

[Darvince]

holy fucking pee

[vh]

no this is the biggest function ever

f(x) = the function you think is the biggest function ever
g(x)=f(x)+1
g(x) = the biggest function ever

[Darvince]

yeah? well my function is even larger!

say you have 2, 2 as the arguments.
you then add the first to the second, and turn the second into the result:
2, 4
you then multiply the first and second, and turn the second into the result:
2, 8
you then raise the first to the power of the second, and turn the second into the result:
2, 256
you then add the first by the second, and turn the second into the result:
(you repeat this process the amount of times that the second argument is)
2, 258
you then multiply the first and second, and turn the second into the result:
2, 516
you then raise the first to the power of the second, and turn the second into the result:
2, 214524926879081553593184399971293538039669853129478294043576983099548224481176751628829988770670454843040509730983776813660062124991145119142938384097345536
you then raise the first to the power of the second, and this is the final result:
http://www.wolframalpha.com/input/?i=2%5E%282%5E516%29

[vh]

^that function +1

i win.

no this is the biggest function ever

f(x) = the function you think is the biggest function ever
g(x)=f(x)+1
g(x) = the biggest function ever

[atomic7732]

^ that function +27

no i win

or (x+459843694^x)^138342953849239399189812848548348912423483 + (x+4548^x)^49284593285935034531583493948952

[Darvince]

^that function +1

i win.

no this is the biggest function ever

f(x) = the function you think is the biggest function ever
g(x)=f(x)+1
g(x) = the biggest function ever

woah do i get a free donut

although this is just a dick measuring contest without resorting to asshattery

[vh]

ok in the spirit of this topic i'll not use that trick.

define x^x^x^x^x.....^x as x#n where n is the number of times x is exponentiated to itself
then assign x#n to n, and recurse the equation again
repeat x#n times, then go up another level of recursion and do it again
repeat this x#n times, going up x#n levels of x#n'ness
go up x#n levels x#n times
(i can keep on typing entire paragraphs of new layers of x#n'ness, but i think you get the point)

[Darvince]

aka graham's number

[vh]

yes but grahams number only does it a puny 64 times

[vh]

and the g(64) = graham's number g(x) is relatively small compared to like the sideways arrows i started a fred in the astronomy/science board in

[atomic7732]

ok in the spirit of this topic i'll not use that trick.

define x^x^x^x^x.....^x as x#n where n is the number of times x is exponentiated to itself
then assign x#n to n, and recurse the equation again
repeat x#n times, then go up another level of recursion and do it again
repeat this x#n times, going up x#n levels of x#n'ness
go up x#n levels x#n times
(i can keep on typing entire paragraphs of new layers of x#n'ness, but i think you get the point)

unfortunately an input of 1 will always be 1

[vh]

...
for x in range (-inf,1]
x = |x|+2
proceed with function

[matty406]

*numbers*

[Darvince]

*larger numbers*

[blotz]

*larger numbers*

^+1

[atomic7732]

*largest number*

[blotz]

*largest number*