Accuracy of the Forward (Explicit) Euler Method

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Taylor series with explicit remainder tells us something about the one-step accuracy:

$$x(t_{k+1})=x(t_k)+\Delta t\dot{x}(t_k)+\frac{1}{2}\Delta t^2\ddot{x}(\xi ),\xi \in [t_k,t_{k+1}]$$

Assuming $x_{k+1}=x(t_k)+\Delta t\dot{x}(t_k)$, we have

$$||x_{k+1}-x(t_{k+1})||=||\frac{1}{2}\Delta t^2\ddot{x}(\xi )||$$

Which is the one-step error, i.e how error we accumulate in one step.

What does the one-step error really tell us?

• That the one-step error is proportional to $\Delta t^2$
• Sharper curves ($\ddot{x}$) makes the error larger

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Global Error