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Solving tetration for base 0 < b < e^-e
#7
(09/12/2009, 08:07 AM)mike3 Wrote: Hmm. However in the emails, you mentioned the use of a "multiplier" that works similar to the derivative at the fixed point but for a cycle.

Generally the multiplier of a cycle (i.e. and ) is defined as:
.

This is equal to the multiplier of at any point of the cycle (by the chain rule).

Example n=2
. If you now plug in you get and if you plug in you get the same result .

If you would depict the tangents at the left and right fixed point in the graph before, they would be parallel.

But this approach to consider already failed.
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RE: Solving tetration for base 0 < b < e^-e - by bo198214 - 09/12/2009, 08:35 AM

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