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Just asking...
#2
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:
which seems indeed to be 4.

for x=3, n=-0.345627
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Messages In This Thread
Just asking... - by martin - 07/14/2008, 09:49 PM
RE: Just asking... - by bo198214 - 07/14/2008, 10:53 PM
RE: Just asking... - by martin - 07/15/2008, 11:40 AM
RE: Just asking... - by bo198214 - 07/16/2008, 08:00 PM
RE: Just asking... - by martin - 07/16/2008, 09:52 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:36 PM
RE: Just asking... - by bo198214 - 07/16/2008, 10:52 PM
RE: Just asking... - by martin - 07/17/2008, 11:28 AM
RE: Just asking... - by bo198214 - 07/18/2008, 01:48 PM
RE: Just asking... - by martin - 07/18/2008, 06:37 PM
RE: Just asking... - by bo198214 - 07/18/2008, 10:02 PM
RE: Just asking... - by Gottfried - 07/19/2008, 03:48 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:04 AM
RE: Just asking... - by martin - 07/19/2008, 09:03 AM
RE: Just asking... - by Ivars - 07/19/2008, 07:45 PM
RE: Just asking... - by andydude - 07/21/2008, 04:05 AM
RE: Just asking... - by martin - 07/21/2008, 06:31 PM
RE: Just asking... - by Ivars - 07/22/2008, 06:07 AM



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