Editing Verlet integration (section)

===A simple example===

To gain insight into the relation of local and global errors, it is helpful to examine simple examples where the exact solution, as well as the approximate solution, can be expressed in explicit formulas. The standard example for this task is the [[exponential function]].

Consider the linear differential equation <math>\ddot x(t) = w^2 x(t)</math> with a constant <math>w</math>. Its exact basis solutions are <math>e^{wt}</math> and <math>e^{-wt}</math>.

The Størmer method applied to this differential equation leads to a linear [[recurrence relation]]
:<math>x_{n+1} - 2x_n + x_{n-1} = h^2 w^2 x_n,</math>
or
:<math>x_{n+1} - 2\left(1 + \tfrac12(wh)^2\right) x_n + x_{n-1} = 0.</math>
It can be solved by finding the roots of its characteristic polynomial
<math>q^2 - 2\left(1 + \tfrac12(wh)^2\right)q + 1 = 0</math>. These are
:<math>q_\pm = 1 + \tfrac 1 2 (wh)^2 \pm wh \sqrt{1 + \tfrac 1 4 (wh)^2}.</math>
The basis solutions of the linear recurrence are <math>x_n = q_+^n</math> and <math>x_n = q_-^n</math>. To compare them with the exact solutions, Taylor expansions are computed:
:<math>\begin{align}
q_+ &= 1 + \tfrac12(wh)^2 + wh\left(1 + \tfrac18(wh)^2 - \tfrac{3}{128}(wh)^4 + \mathcal O\left(h^6\right)\right)\\
    &= 1 + (wh) + \tfrac12(wh)^2 + \tfrac18(wh)^3 - \tfrac{3}{128}(wh)^5 + \mathcal O\left(h^7\right).
\end{align}</math>
The quotient of this series with the exponential <math>e^{wh}</math> starts with <math>1 - \tfrac1{24}(wh)^3 + \mathcal O\left(h^5\right)</math>, so
:<math>\begin{align}
q_+ &= \left(1 - \tfrac1{24}(wh)^3 + \mathcal O\left(h^5\right)\right)e^{wh}\\
    &= e^{-\frac{1}{24}(wh)^3 + \mathcal O\left(h^5\right)}\,e^{wh}.
\end{align}</math>
From there it follows that for the first basis solution the error can be computed as
:<math>\begin{align}
x_n = q_+^{n}
 &= e^{-\frac{1}{24}(wh)^2\,wt_n + \mathcal O\left(h^4\right)}\,e^{wt_n}\\
 &= e^{wt_n}\left(1 - \tfrac{1}{24}(wh)^2\,wt_n + \mathcal O(h^4)\right)\\
 &= e^{wt_n} + \mathcal O\left(h^2 t_n e^{wt_n}\right).
\end{align}</math>
That is, although the local [[discretization error]] is of order 4, due to the second order of the differential equation the global error is of order 2, with a constant that grows exponentially in time.