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Full Version: 2sinh^[r](z) = 0 ??
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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