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Full Version: differentiation rules for x[4]n, where n is any natural number
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Looks like there is a recurrence relation here, but I don't know how to write it for d/dx x[4]n cuz all the stuff that is multiplied into the tetration is nested, so you have to express that with a function...


How does one express A_n(x) in closed form? I really don't know where i am going with this, perhaps doing A_k(x) at non-integer k to obtain the values for x[4]a for any positive real a?
EDIT oh, I have found it... , so . but i am suspicious about its utility for all real numbers, because it is not true for n=1:

there seems to be a great gap between x[4]1 and x[4]2.
heck yeah, there IS a great gap, because x[4]2 is transcendental, while x[4]1 is simply the polynomial x
(05/23/2009, 04:04 AM)Tetratophile Wrote: [ -> ]EDIT oh, I have found it... , so . but i am suspicious about its utility for all real numbers, because it is not true for n=1:

there seems to be a great gap between x[4]1 and x[4]2.
heck yeah, there IS a great gap, because x[4]2 is transcendental, while x[4]1 is simply the polynomial x

Your recursion formula is wrong, correctly it should be:




or equivalently:


then


or with your notation

the problem is now extending our function A_n to non-natural n...
(05/25/2009, 09:31 AM)Ansus Wrote: [ -> ]Great! I've added the rule to our wiki page http://en.wikipedia.org/wiki/User:MathFa...on_Summary



This works also for arbitrary complex :

This derivative is also what I used in this thread, which no one seemed to notice. Yes, Ansus' formula is correct, although usually sexp=tet, so I would not use that notation. I would use the notation

Andrew Robbins