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We can do better by symmetrizing derivative:
- Take a trial step to midpoint, evaluate yn+1/2 and tn+1/2.
- Use these to evaluate derivative f'(tn+1/2,yn+1/2).
- Then use this to go back and take a full step.
- yn+1 = yn + h f'(tn + ½ h,yn + ½ h f'(tn,yn)) + O(h3)
Can show that O(h2) terms "cancel," so leading error term is O(h3) 2nd-order Runge-Kutta.