(08/31/2010, 06:51 AM)mike3 Wrote: The product of exp is the exponential of a formal power series. This can be expressed using the Bell polynomials:
\( \prod_{n=0}^{\infty} \exp(a_n y^n) = \exp\left(\sum_{n=0}^{\infty} a_n y^n\right) = \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) \).
This then becomes
\( \exp(a_0) \exp\left(\sum_{n=1}^{\infty} a_n y^n\right) = \exp(a_0) \sum_{n=1}^{\infty} \frac{\sum_{k=1}^n B_{n,k}(1! a_1, ..., (n-k+1)! a_{n-k+1})}{n!} y^n = \exp(a_0) \sum_{n=1}^{\infty} \frac{B_n(1! a_1, ..., n! a_n)}{n!} y^n \).
Thus the equations to solve are
\( a_n L^n = \exp(a_0) \frac{B_n(1! a_1, ..., n! a_n)}{n!} \).
Since \( a_0 = L \) and \( a_1 = 1 \), this is
\( a_n L^n = L \frac{B_n(1, 2! a_2, ..., n! a_n)}{n!} \).
This is derived from Faà di Bruno's formula, see
http://en.wikipedia.org/wiki/Fa%C3%A0_di...7s_formula
for details.
Doing some tests, it appears that
\( B_n(1, 2! a_2, ..., n! a_n) \)
has only one occurrence of \( n! a_n \), and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that \( B_n(1, a_2, ..., a_n) - n! a_n = B_n(1, 2! a_2, ..., (n-1)! a_{n-1}, 0) \). I don't have a proof at this point, but we can then solve this:
\( a_n n! L^n = L B_n(1, 2! a_2, ..., n! a_n) \)
\( a_n n! L^{n-1} = B_n(1, 2! a_2, ..., n! a_n) \)
\( a_n n! L^{n-1} - n! a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0) \)
\( n! (L^{n-1} - 1) a_n = B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0) \)
\( a_n = \frac{B_n(1, 2! a_2, 3! a_3, ..., (n-1)! a_{n-1}, 0)}{n! (L^{n-1} - 1)} \).
And this is the recurrent formula for the general coefficients. Together with \( a_0 = L \) and \( a_1 = 1 \), this completes the as-close-to-explicit-as-possible-so-far formula for the coefficients of the Fourier series for the regular iteration.
EDIT: posts corrected to include factorials on terms \( a_n \) in Bell polynomials