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 Just asking... bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 07/14/2008, 10:53 PM (This post was last modified: 07/14/2008, 10:53 PM by bo198214.) Hello Martin 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 $2\^\^ 3 = [3(2^n-1)+1]^{1/n}\approx 1.000002949\neq 2^{2^2}=2^4 = 16$ « Next Oldest | Next Newest »