2sinh^[r](z) = 0 ?? tommy1729 Ultimate Fellow Posts: 1,859 Threads: 402 Joined: Feb 2009 02/23/2016, 11:13 PM The 2sinh function is important to me since it relates to my 2sinh method for tetration. I consider it important to understand the positions of z's such that D^n 2sinh^[r](z) = 0. Where 0 =< r =< 1. And the iterations are natural from the fixpoint 0. The case r = 1 is easy : All integer multiples of pi i. ( zero's of both 2sinh , 2cosh ) Also fascinating is that the amount of zero's Goes from 1 to infinity as r Goes from 0 to 1. How they come into existance ( directions , branches etc ) needs more understanding. The conjecture is that for all positive integer n and positive real r =< 1 we have - A =< real(z) =< A Where A satisfies A > 0 2 sinh(A - 2 pi) = A A is about 8.4286 Regards Tommy1729 « Next Oldest | Next Newest »

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