One very important formula - Printable Version

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One very important formula - Printable Version

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One very important formula - Ansus - 11/03/2010

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

RE: One very important formula - mike3 - 11/03/2010

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 $r$ ?

RE: One very important formula - Ansus - 11/03/2010

Corrected the usage of r.

RE: One very important formula - bo198214 - 11/03/2010

(11/03/2010, 12:21 AM)Ansus Wrote: 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.

RE: One very important formula - Ansus - 11/03/2010

(11/03/2010, 01:03 AM)bo198214 Wrote: You mean "Bell number"?

Yes.

RE: One very important formula - Daniel - 11/03/2010

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

RE: One very important formula - mike3 - 11/03/2010

(11/03/2010, 02:27 AM)Ansus Wrote:
(11/03/2010, 01:03 AM)bo198214 Wrote: You mean "Bell number"?

Yes.

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

RE: One very important formula - Ansus - 11/03/2010

(11/03/2010, 03:16 AM)mike3 Wrote: 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.