&= \dot f \mathbf{r}_0 + \dot g \mathbf{v}_0.\label{eq:fgupdate}&&\end{aligned}$$ This completes the solution of the Kepler problem.

To solve for the variational equations, we also make use of the $G$-functions. Fortunately, we only need to solve Kepler’s equation once (to solve for $X$). We then get the solution for the variational equations without solving another transcendental equation and thus have only one iteration loop per timestep for both the particle and its variational counterpart. The position and velocity components of $\pmb \delta$ at the end of the timestep, $\delta \mathbf{r}$ and $\delta \mathbf{v}$, can be written as $$\begin{aligned}
\delta \mathbf{r} &=& f \;\delta \mathbf{r}_0 + g \;\delta\mathbf{v}_0 + \mathbf{r}_0 \;\delta f+ \mathbf{v}_0 \;\delta g\label{eq:var1}\\
\delta \mathbf{v} &=& \dot f \;\delta \mathbf{r}_0 + \dot g \;\delta\mathbf{v}_0 + \mathbf{r}_0 \;\delta \dot f+ \mathbf{v}_0 \;\delta \dot g,\label{eq:var2}\end{aligned}$$ where the variations $\delta f$, $\delta g$, $\delta \dot f$ and $\delta \dot g$ can be derived from Eqs. \[eq:fg\]-\[eq:fg2\] (see @MikkolaInnanen1999 for the explicit expressions).

Types of Numerical Errors {#sec:errors}
-------------------------

There are three distinct effects contributing to the energy error of a symplectic integrator [see e.g., @QuinnTremaine1990]. See also [@ReinSpiegel2015] for a similar discussion for non-symplectic integrators.

First, there is an error term associated with 