(08/31/2010, 08:04 AM)mike3 Wrote: Doing some tests, it appears that
)
has only one occurrence of
, and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that
. I don't have a proof at this point, ...
Now I've got that proof. Here it goes:
We have
(note how the connection to the Faà di Bruno's formula is clear)
with the sum taken over all sequences of non-negative integers

such that

and
j_{n-k+1} = n)
, and

is the sum of the

. As should be obvious from the formula, we see that the value

in

only occurs in the component

. What kind of occurrences of

are possible there? Using the formulas for the

, we see that

, which means that all but one

must be zero, and that one that is must be 1. The second constraint,

, would imply that if all

but one are zero, the nonzero one must equal

where

is its position in the sequence. This would mean

must divide

. The first constraint, though, said it must be 1, which means

and so the only possible sequence of

is

(

terms). This means
 = \frac{n!}{0!0!...1!} \left(\frac{x_1}{1!}\right)^0 \left(\frac{x_2}{2!}\right)^0 ... \left(\frac{x_n}{n!}\right)^1)
, which is obviously just

. So
)
contains only one term with

, which is just

itself. And then
 - x_n = B_n(x_1, x_2, ..., x_{n-1}, 0))
follows trivially.