I am solving exercise of 5.11.3 in Lie groups, Lie algebras, and Representations by Brain Hall.
The logarithm and exponential are defined by \begin{equation} \log X = \sum_{m=1}^{\infty} (-1)^{m+1} \frac{(X-I)^m}{m} \quad and \quad e^X = \sum_{m=0}^{\infty} \frac{X^m}{m!}. \end{equation} That exercise makes me check that the Baker-Campbell-Hausdorff formula is true for third order.
[Baker-Campbell-Hausdorff formula] For all $n \times n$ complex matrix $X$ and $Y$ with $\|X\|$ and $\|Y\|$ sufficiently small, we have \begin{equation} \log(e^Xe^Y)=X+\int_0^1 g(e^{ad_X}e^{tad_Y})(Y)\,dt \end{equation} where $ad_X(Y)=XY-YX$ and $$ g(X)=1+ \frac{1}{2}(X-I) -\frac{1}{6}(X-I)^2 + \frac{1}{12}(X-I)^3 + \cdots.$$
I compute LHS and RHS but there are not same as below \begin{align} LHS= X+Y+\frac{1}{2}(XY-YX)+&\frac{1}{12}X^2Y + \frac{1}{3}XYX+\frac{1}{3}YX^2 -\frac{1}{6}YXY \\&+ \frac{1}{3}Y^2X +\frac{7}{12}XY^2 +\frac{1}{4}Y^3 +\frac{1}{4}X^3 \end{align} and \begin{align} RHS&=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]] -\frac{1}{12}[Y,[X,Y]]\\ &=X+Y+\frac{1}{2}(XY-YX)+\frac{1}{12}X^2Y-\frac{1}{6}XYX+\frac{1}{12}YX^2-\frac{1}{6}YXY+\frac{1}{12}Y^2X+\frac{1}{12}XY^2. \end{align}
I really want to know what I missed it, please.
