Editing Verlet integration (section)

===Starting the iteration===
Note that at the start of the Verlet iteration at step <math>n = 1</math>, time <math>t = t_1 = \Delta t</math>, computing <math>\mathbf x_2</math>, one already needs the position vector <math>\mathbf x_1</math> at time <math>t = t_1</math>. At first sight, this could give problems, because the initial conditions are known only at the initial time <math>t_0 = 0</math>. However, from these the acceleration <math>\mathbf a_0 = \mathbf A(\mathbf x_0)</math> is known, and a suitable approximation for the position at the first time step can be obtained using the [[Taylor polynomial]] of degree two:
:<math>\mathbf x_1 = \mathbf{x}_0 + \mathbf{v}_0 \Delta t + \tfrac12 \mathbf a_0 \Delta t^2
\approx \mathbf{x}(\Delta t) + \mathcal{O}\left(\Delta t^3\right).</math>

The error on the first time step then is of order <math>\mathcal O\left(\Delta t^3\right)</math>. This is not considered a problem because on a simulation over a large number of time steps, the error on the first time step is only a negligibly small amount of the total error, which at time <math>t_n</math> is of the order <math>\mathcal O\left(e^{Lt_n} \Delta t^2\right)</math>, both for the distance of the position vectors <math>\mathbf x_n</math> to <math>\mathbf x(t_n)</math> as for the distance of the divided differences <math>\tfrac{\mathbf x_{n+1} - \mathbf x_n}{\Delta t}</math> to <math>\tfrac{\mathbf x(t_{n+1}) - \mathbf x(t_n)}{\Delta t}</math>. Moreover, to obtain this second-order global error, the initial error needs to be of at least third order.