Hello Martin
Why do you consider this formula as tetration?
We have some base requirements for tetration which are
2^^0 = 1
2^^(x+1) = 2^(2^^x)
I dont think that your formula satisfies these requirements.
Your formula satisfies the first few equalities:
x=0: 2^^0 = [0+1]^(1/n)= 1
x=1: 2^^1 = [1*(2^n-1)+1]^(1/n)=(2^n)^(1/n) = 2
for x=2, n=0 we have to take the limit for your formula to be defined:
\( 2\^\^2 = \lim_{n\to 0} [2(2^n-1)+1]^{1/n} = \lim_{n\to 0} [2^{n+1}-1]^{1/n} \) which seems indeed to be 4.
for x=3, n=-0.345627
\( 2\^\^ 3 = [3(2^n-1)+1]^{1/n}\approx 1.000002949\neq 2^{2^2}=2^4 = 16 \)
martin Wrote:I hear there are some different possible solutions to this function, and I'm too curious if the following example is one of them:
While dabbling in this field of maths, I found a formula for 2^^x:
For 0 < x < 1, with n = 0.345627*(2-x), 2^^x ~~ [x*(2^n-1)+1]^(1/n)
(it's not 100% accurate, but ... about 99%, maybe some slight adjustments can fix that)
Why do you consider this formula as tetration?
We have some base requirements for tetration which are
2^^0 = 1
2^^(x+1) = 2^(2^^x)
I dont think that your formula satisfies these requirements.
Your formula satisfies the first few equalities:
x=0: 2^^0 = [0+1]^(1/n)= 1
x=1: 2^^1 = [1*(2^n-1)+1]^(1/n)=(2^n)^(1/n) = 2
for x=2, n=0 we have to take the limit for your formula to be defined:
\( 2\^\^2 = \lim_{n\to 0} [2(2^n-1)+1]^{1/n} = \lim_{n\to 0} [2^{n+1}-1]^{1/n} \) which seems indeed to be 4.
for x=3, n=-0.345627
\( 2\^\^ 3 = [3(2^n-1)+1]^{1/n}\approx 1.000002949\neq 2^{2^2}=2^4 = 16 \)