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11/03/2010, 12:21 AM
(This post was last modified: 11/03/2010, 02:26 AM by Ansus.)
Anybody shell know this very important formula:
=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
are the Bell's numbers of x-th order and
)}{\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
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11/03/2010, 12:35 AM
(This post was last modified: 11/03/2010, 12:49 AM by mike3.)
So does this give real-valued answers for real
and
? How do you compute the Bell polynomial
at a real or complex number
? And what is
?
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Corrected the usage of r.
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(11/03/2010, 12:21 AM)Ansus Wrote: where 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.
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(11/03/2010, 01:03 AM)bo198214 Wrote: You mean "Bell number"?
Yes.
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(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
?
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11/03/2010, 04:21 AM
(This post was last modified: 11/03/2010, 04:23 AM by Ansus.)
(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 ?
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':
\star g(n)=\sum_{k=0}^n \left(n \\ k\right)f(n-k)g(k) )
If we use such operator, we can write:
And
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.
