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 Just asking... martin Junior Fellow Posts: 14 Threads: 1 Joined: Jul 2008 07/21/2008, 06:31 PM (This post was last modified: 07/21/2008, 06:32 PM by martin.) andydude Wrote:http://tetration.itgo.com/newtxt/table-tetbin.txt Wow, that's an extensive and interesting table. I especially like the values below -2. Ivars Wrote:If in series there is 1/n!, then formula becomes: n= ln(2) * ((1-exp(-x))/x) if just 1/n, then: n= ln(2) * (ln(1+x)/x) Nope, neither of them worked. But hey, it was worth trying. Meanwhile, I figured out linear parameters for 3^^x. With n = 0.0813-0.328823x, for 0 < x < 1: 3^^x ~~ [x(3^n-1)+1]^(1/n) Again, quite close, but still with minimal irregularities. Either I didn't try long enough to find exact parameters, or I'm really on the wrong track. (Most probably the latter. (Then again, thinking about n to be a polynomial of infinite degree or some other equation ... nah, I leave that to someone else who's interested.)) « Next Oldest | Next Newest »