--- trunk/tengDissertation/Introduction.tex	2006/04/21 05:45:14	2725
+++ trunk/tengDissertation/Introduction.tex	2006/04/21 15:37:53	2726
@@ -831,7 +831,7 @@ $\varphi_1(t)$ and $\varphi_2(t$ respectively , we hav
 error of splitting method in terms of commutator of the
 operators(\ref{introEquation:exponentialOperator}) associated with
 the sub-flow. For operators $hX$ and $hY$ which are associate to
-$\varphi_1(t)$ and $\varphi_2(t$ respectively , we have
+$\varphi_1(t)$ and $\varphi_2(t)$ respectively , we have
 \begin{equation}
 \exp (hX + hY) = \exp (hZ)
 \end{equation}
@@ -846,12 +846,12 @@ can obtain
 \]
 Applying Baker-Campbell-Hausdorff formula to Sprang splitting, we
 can obtain
-\begin{eqnarray*}
+\begin{equation}
 \exp (h X/2)\exp (h Y)\exp (h X/2) & = & \exp (h X + h Y + h^2
 [X,Y]/4 + h^2 [Y,X]/4 \\ & & \mbox{} + h^2 [X,X]/8 + h^2 [Y,Y]/8 \\
 & & \mbox{} + h^3 [Y,[Y,X]]/12 - h^3 [X,[X,Y]]/24 & & \mbox{} +
 \ldots )
-\end{eqnarray*}
+\end{equation}
 Since \[ [X,Y] + [Y,X] = 0\] and \[ [X,X] = 0\], the dominant local
 error of Spring splitting is proportional to $h^3$. The same
 procedure can be applied to general splitting,  of the form
@@ -1238,7 +1238,7 @@ Q^T Q = 1$, \label{introEquation:orthogonalConstraint}
 where $I_{ii}$ is the diagonal element of the inertia tensor. This
 constrained Hamiltonian equation subjects to a holonomic constraint,
 \begin{equation}
-Q^T Q = 1$, \label{introEquation:orthogonalConstraint}
+Q^T Q = 1, \label{introEquation:orthogonalConstraint}
 \end{equation}
 which is used to ensure rotation matrix's orthogonality.
 Differentiating \ref{introEquation:orthogonalConstraint} and using
