10/06/2007, 05:26 PM

bo198214 Wrote:UVIR Wrote:I *think* (but I am not absolutely sure about it) that any method which defines for reasonable , for could conceivably be extended to a

I dont get this. If I define it on and I demand that then the function is already determined for all there is no place left for further manipulations (i.e. to make it differentiable).

Andrew's approach was to choose it as an analytic function on and he determines the coefficients of the seriesexpansion at 0 by demanding that it is (or even analytic). However this approach works computably only on the , the inverse of .

I'll do one example to force the function to be at y=1 so you can see what I mean for the function I used: . All calculations were done with Maple, so hopefully there are no mistakes.

Left of y=1:

For the other side (right of y=1):

Solving the equation:

numerically, we get:

In other words, the tetration function defined by:

in (0,1) and by is at y=1, assuming the way it was constructed by joining at the naturals with frac{y} as with my first construction.

Similar calculations show that using the corresponding function is at y=1, etc. I think one can go as high as one wants, provided a solution for x exists. It looks as if a solution exists for any even order derivative.

What do you guys think?