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 tetration base > exp(2/5) tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 02/11/2015, 12:29 AM It appears that f(x) = 0 + 5/2 x + 5/2 x^5/5! + 5/2 x^6/6! + 5/2 x^10/10! + 5/2 x^11/11! + ... is the only multisection type function that satisfies for all x > 0 : x =< f(x) < exp(x). The other candidates a/b seem to violate already within the bounds 0 < x < 11b. SO assuming this is correct , probably the method based on base exp(2/5) converges the best. As expected in post 1 and this answers part 1 of post 2. More investigation is desired but probably the fact that the only number b such that 2/b is a good appr to 1/e is b = 5 is fundamental. Large a in a/b lead to problems for x >> 1 because of the many powersums needed. This makes some sense because 1/e < 1/2. *** As for the cousin question , proving true is as hard as the cousin prime conjecture and probably equally hard as the twin prime conjecture. Both unsolved as of today. However proving false might be easier. /// Notice fake function theory can in a way be an approximate inverse multisection. a1 x^2 + a2 x^4 + ... = multisect fake ( a1 x^2 + a2 x^4 + ... ) ~ Original function. Nice. /// regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread tetration base > exp(2/5) - by tommy1729 - 02/09/2015, 11:59 PM RE: tetration base > exp(2/5) - by tommy1729 - 02/10/2015, 10:43 PM RE: tetration base > exp(2/5) - by tommy1729 - 02/11/2015, 12:29 AM

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