08/23/2009, 02:45 PM

(08/07/2007, 04:38 PM)bo198214 Wrote: I just read Andrew Robbins' solution to the tetration problem, which I find very convincing, and want to use the opportunity to present and discuss it here.

The solution satisfies the 2 natural conditions

1. and

2. is infinitely differentiable.

For the constant -1 vanishes and we make the following calculations:

The derivation of is easily determined to be

and so the k-th derivative is , which give us in turn

for .

notice in the last line bo wrote s(x) instead of s(0).

it is an expansion at x = 0.

now if we consider expansions at both x = 0 and x = 1 and get the same coefficients for x = 0 by computing them from

1) the coefficients expanded at x = 1

2) solving the modified equation ( see below)

then that probably means we have radius 1 or larger ( radius from the origin at x = 0 )

since b^1 i = b^i , we get an extra b^i factor on the right side.

and v_k is replaced by sum v_k / k!

that equation should be solvable and have the same solutions v_k IF Andrew's slog has a radius 1 ( or larger ) from the origin.

bo mentioned the potential non-uniqueness for v_k when expanded at x = 0.

maybe this could be the extra condition we(?) are looking for.

Regards

tommy1729