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:

which seems indeed to be 4.

for x=3, n=-0.345627

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:

which seems indeed to be 4.

for x=3, n=-0.345627