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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 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/23/2009, 07:06 AM (05/23/2009, 04:04 AM)Tetratophile Wrote: 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 Your recursion formula is wrong, correctly it should be: $(x[4]1)' = 1$ $(x[4](n+1))' = \left( (x[4]n)'\ln(x) + \frac{x[4]n}{x} \right) (x[4](n+1))$ or equivalently: $B_1 = 1$ $B_{n+1} = (B_n \ln(x) + 1) (x[4]n)$ then $(x[4]n)' = \frac{x[4]n}{x} B_n$ or with your notation $A$ $A_1 = \frac{1}{x}$ $A_{n+1} = (A_n \ln(x) + \frac{1}{x}) (x[4]n)$ Base-Acid Tetration Fellow Posts: 94 Threads: 15 Joined: Apr 2009 05/25/2009, 05:30 AM the problem is now extending our function A_n to non-natural n... bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 05/25/2009, 12:13 PM (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 $(\operatorname{sexp}_x(1))'=1$ $(\operatorname{sexp}_x(n))'=\left((\operatorname{sexp}_x(n-1))'\ln x + \frac{\operatorname{sexp}_x(n-1)}{x}\right)\operatorname{sexp}_x(n)$ This works also for arbitrary complex $n=:y$: $\frac{\partial}{\partial x} x[4](y+1) = \frac{\partial}{\partial x} x^{x[4]y} = \frac{\partial}{\partial x} \exp(\ln(x)(x[4]y)) = (x[4](y+1)) \cdot \left(\frac{x[4]y}{x} +\ln(x) \frac{\partial}{\partial x} x[4]y\right)$ andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/26/2009, 07:52 PM (This post was last modified: 05/26/2009, 07:54 PM by andydude.) 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 $ \frac{\partial}{\partial x}({}^{n}x) = \frac{1}{x} \sum_{k=1}^{n} \ln(x)^{k-1} \prod_{j=0}^{k} {}^{n-j}x$ Andrew Robbins « Next Oldest | Next Newest »

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