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+--- Thread: differentiation rules for x[4]n, where n is any natural number (/showthread.php?tid=294)
differentiation rules for x[4]n, where n is any natural number - Base-Acid Tetration - 05/23/2009
\( D_x({}^2 x)={}^2 x(1+\ln{x}) \)
\( D_x({}^3 x)={}^3 x(x^{x-1} + {}^2 x \ln{x}(1+\ln{x})) \)
\( D_x({}^4 x)={}^4 x(x^{{}^2 x-1} + {}^3 x \ln{x}(x^{x-1}+{}^2 x \ln{x}(1+\ln{x})) \)
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
RE: differentiation rules for x[4]n, where n is any natural number - bo198214 - 05/23/2009
(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 \( A \)
\( A_1 = \frac{1}{x} \)
\( A_{n+1} = (A_n \ln(x) + \frac{1}{x}) (x[4]n) \)
RE: differentiation rules for x[4]n, where n is any natural number - Base-Acid Tetration - 05/25/2009
the problem is now extending our function A_n to non-natural n...
RE: differentiation rules for x[4]n, where n is any natural number - bo198214 - 05/25/2009
(05/25/2009, 09:31 AM)Ansus Wrote: Great! I've added the rule to our wiki page http://en.wikipedia.org/wiki/User:MathFacts/Tetration_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 \( n=:y \):
\( \frac{\partial}{\partial x} x[4](y+1) = \frac{\partial}{\partial x} x^{x[4]y} =
\frac{\partial}{\partial x} \exp(\ln(x)(x[4]y)) = (x[4](y+1)) \cdot \left(\frac{x[4]y}{x} +\ln(x) \frac{\partial}{\partial x} x[4]y\right) \)
RE: differentiation rules for x[4]n, where n is any natural number - andydude - 05/26/2009
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