🧵 Factorial'd x amount of times (non-integers)
Anonymous at Wed, 31 Jul 2024 05:21:54 UTC No. 16304241
3 Factorial'd 1 time: 3!^1=6
3 Factorial'd 2 times: 3!^2=720
3 Factorial'd 3 times: 3!^3=2.6*10^1726
What is 3 factorial'd 2.5 times? Or any non-integer amount of times. If plotted on a rate of change chart like shown in the image, the number must exist. What would be the function?
This can be applied to concepts other than factorials too, such as non-integer hyperoperations:
(1 up arrow): 3^2=9
(2 up arrows) 3(up)(up)2: 3^3=27
(3 up arrows) 3(up)(up)(up)2: 3^3^3 = 7.62*10^12
What is the value with 2.5 up arrows? Such that 3[(up)^2.5]=?
Anonymous at Wed, 31 Jul 2024 06:21:00 UTC No. 16304280
>>16304241
2.5 = 25/10
now find it for 25 and 10 then divide.
Anonymous at Wed, 31 Jul 2024 16:56:11 UTC No. 16304749
>>16304241
Are you familiar with the gamma function? It extends the factorial from positive integers to the real and complex numbers. There isn't one correct way to extend integer functions, but there can be a function that fits the properties of the integer function the best over the new domain. The gamma function is usually recognized as the most natural extension of the factorial function.
I would suggest that you learn the derivation of the gamma function, and understand the techniques used to extend discrete functions. You might be able to apply a similar method to this problem.
Anonymous at Wed, 31 Jul 2024 21:17:48 UTC No. 16305012
>>16304241
>>16304749
i dont think OP is asking for the values of non integers factorial, but rather the composition of the factorial fonction in a non-integer way
if you say f(n)=n!
then f(f(n))=(n!)! = n!^2 (using OPs notation)
and for larger than 2, like m times: f(f(...f(n)...))=(...(n!)!...)! = n!^m
and then what if instead of having lets say m times the function applied to itself its some non integer value...
regardless of it being on the factorial function, there is probably some math stuff about non integer function composition...
Anonymous at Wed, 31 Jul 2024 21:23:33 UTC No. 16305018
in the case of lets say:
n!^2.5
you could consider the following formulation:
>f(n)=n!
>f(f(f(f(f(n)))))=n!^5
>you are looking for a function g such that
>g(g(n))=n!^5
so it comes down to solving
>g^2=f^5
but thats not necessarly an easy problem to solve, and it works for rational composition problems like the value 2.5 OP asked for
might be a bit more complicated for non rational numbers
Anonymous at Wed, 31 Jul 2024 22:11:05 UTC No. 16305060
https://en.wikipedia.org/wiki/Itera