One very important formula

[Ansus]

Anybody shell know this very important formula:

\( \operatorname{sexp}_b(x)=r+\sum _{n=1}^{\infty} \frac{\left(\ln b \right)^{n-1}\left(\ln \left(b^r\right)\right)^{n x}\left(1-r)^n B_n^x}{n!} \)

where \( B_n^x \) are the Bell's numbers of x-th order and \( r=\frac{W(-\log (b))}{\log (b)} \). For integer x one can find them here:

http://www.research.att.com/~njas/sequences/A111672

http://www.research.att.com/~njas/sequences/A144150

http://www.research.att.com/~njas/sequences/A153277

This formula can be easily derived from regular teration, but has a long history dating from 1945 ( J. Ginsburg, Iterated exponentials, Scripta Math. 11 (1945), 340-353.)

It is notable that tetration and Bell's polynomials of n-th order have applications in quantum physics: http://arxiv.org/abs/0812.4047

[mike3]

So does this give real-valued answers for real \( b \ge e^{1/e} \) and \( x > -2 \)? How do you compute the Bell polynomial \( B_n^x \) at a real or complex number \( x \)? And what is \( r \)?

[Ansus]

Corrected the usage of r.

[bo198214]

where \( B_n^x \) are the Bell's polynomials of x-th order.

You mean "Bell number"?
The Bell polynomials are multivariate polynomials ...

Otherwise I second the questions of Mike and add the question about convergence.

[Ansus]

You mean "Bell number"?

Yes.

[Daniel]

You might be interested in my paper Bell Polynomials of Iterated Functions.

[mike3]

You mean "Bell number"?

Yes.

So how do you extend it to real values of x, and is this solution real valued for bases \( b > e^{1/e} \)?

[Ansus]

So how do you extend it to real values of x, and is this solution real valued for bases \( b > e^{1/e} \)?

It is derived from regular iteration, so it diverges for higher bases, but my aim was to find an expression for tetration that does not refer to taetration itself, thus allowing to derive its properties.

David Knuth referred to the following operation calling it 'binomial convolution':

\( f(n)\star g(n)=\sum_{k=0}^n \left(n \\ k\right)f(n-k)g(k) \)

If we use such operator, we can write:

\( B_{n+1}^x=\sum_{k=0}^{x-1} B_n^x\star B_n^k \)

And \( B_1^x \) is always 1.

Thus the result of the convolution is a polynomial of x and k of degree n-1 and we can take indefinite sum of it symbolically.

Note also that binomial convolution corresponds to the product of exponential generating functions. This means product of tetrations corresponds to binomial convolution of Bell's numbers of higher orders.