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 \( A \)
\( A_1 = \frac{1}{x} \)
\( A_{n+1} = (A_n \ln(x) + \frac{1}{x}) (x[4]n) \)
