Exercise · ZW-LOCAL-AND-GLOBAL-ERROR-009 · ★★☆ Intermediate

A method has local error \(O(h^2)\). If h goes from 0.1 to 0.05, how does the global error change approximately if its global order is \(O(h)\)?

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Explanation

With order \(O(h)\), when \(h\) goes from 0.1 to 0.05 the global error is multiplied by 0.5, so it is halved.

The local error is the difference between the exact ODE solution over one step and the method's approximation starting from the exact value.

\[e_{n+1}=y(t_{n+1})-y_{n+1}\]