TITLE: Inverse of Baker-Campbell-Hausdorff
QUESTION [10 upvotes]: This should be quite simple to answer. I have a situation in which I must have an explicit expression for the inverse of Baker-Campbell-Hausdorff. More precisely:
I have two power series $P_1(X,Y)$, $P_2(X,Y)$ in noncommuting variables $X,Y$. I want an expression for 
$$exp\left(log(1+P_1) + \ log(1 +P_2) \right) .$$
I searched some books and google; but beyond the fact that it exists, I am unable to find an explicit expression. If it already exists in print, it would save me some work of trying to derive it myself.
Thank you very much.

REPLY [4 votes]: The following recursion formula for the homogeneous factors of $A(P,Q)$ exists:
Define:
$A(tP,tQ) = 1 + t \frac{B_1}{1!}+t^2 \frac{B_2}{2!}+t^3 \frac{B_3}{3!}+...$
$C_n  = P^n+Q^n$
$\Delta C_n = -n C_{n+1}$
Then:
$B_1 = C_1$
$n! B_n  = [(\Delta+C_1) B_{n-1}] _{symm}$
Where the symmetrization is of all words of the type $C_{i1} C_{i2} C_{i3} ... $
This formula can be obtained by differentiation with respect to t. I compared the results up to
the cubic term to Theo's expression with a full agreement.
The coefficient of the string $\[(C_{i1} C_{i2} C_{i3} ...]_{symm}$ in $B_n$ is
equal to the coefficient of $tr(M^{i1}) tr(M^{i2}) tr(M^{i3})...$ in the coefficient of $t^n$
in $det(I-tM)$. These coefficients are the charcters of $\{{{1^n\}}}$ of $S_n$.