Andrew Robbins' Tetration Extension

[tommy1729]

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 \( {}^y x \) satisfies the 2 natural conditions
1. \( {}^1b=b \) and \( {}^{x+1}b=b^{{}^xb} \)
2. \( x\mapsto b^x \) is infinitely differentiable.

For \( k\ge 1 \) the constant -1 vanishes and we make the following calculations:
\( s^{(k)}(x)=(t(b^x))^{(k)}=\left(\sum_{i=0}^\infty \nu_i \frac{b^{xi}}{i!}\right)^{(k)}=\sum_{i=0}^\infty\frac{\nu_i}{i!}(b^{xi})^{(k)} \)
The derivation of \( b^{xi} \) is easily determined to be
\( (b^{xi})'=b^{xi}\text{ln}(b) i \) and so the k-th derivative is \( (b^{xi})^{(k)} = b^{xi}(\text{ln}(b)i)^k \), which give us in turn
\( \nu_k=s(x)^{(k)}= \text{ln}(b)^k\sum_{i=0}^\infty\nu_i\frac{i^k}{i!} \) for \( k\ge 1 \).

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