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 differentiation rules for x[4]n, where n is any natural number Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/23/2009, 04:04 AM (This post was last modified: 05/23/2009, 04:34 AM by Base-Acid Tetration.) $D_x({}^2 x)={}^2 x(1+\ln{x})$ $D_x({}^3 x)={}^3 x(x^{x-1} + {}^2 x \ln{x}(1+\ln{x}))$ $D_x({}^4 x)={}^4 x(x^{{}^2 x-1} + {}^3 x \ln{x}(x^{x-1}+{}^2 x \ln{x}(1+\ln{x}))$ 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... $D_x({}^n x)={}^n x A_n (x)$ $A_n (x) := \frac {D_x({}^n x)}{{}^n x}$ 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... $D_x({}^{n+1} x) = {}^{n+1} x(D_x({}^n x)) \forall n \ge 2$, so $A_{n+1}(x)=D_x({}^nx) \forall n\ge 2$. but i am suspicious about its utility for all real numbers, because it is not true for n=1: $A_2(x)=1+\ln{x}\ne 1 = D_x({}^1 x)$ 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 « Next Oldest | Next Newest »

 Messages In This Thread differentiation rules for x[4]n, where n is any natural number - by Base-Acid Tetration - 05/23/2009, 04:04 AM RE: differentiation rules for x[4]n, where n is any natural number - by bo198214 - 05/23/2009, 07:06 AM RE: differentiation rules for x[4]n, where n is any natural number - by Base-Acid Tetration - 05/25/2009, 05:30 AM RE: differentiation rules for x[4]n, where n is any natural number - by andydude - 05/26/2009, 07:52 PM

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