One method consists in computing the numerical solution for an arbitrary h and then 2h. The Richardson extrapolation gives an estimate of e=maxt|y(t,2h)−y(t,h)| of the error. When the error is smaller than the tolerance, we keep the result and start from 2y(t,h)−y(t,h). If the error is larger we restart with h/2 until we reach the tolerance.

sublimnes9

sublimnes9

Answered question

2022-08-07

One method consists in computing the numerical solution for an arbitrary h and then 2 h. The Richardson extrapolation gives an estimate of e = max t | y ( t , 2 h ) y ( t , h ) | of the error. When the error is smaller than the tolerance, we keep the result and start from 2 y ( t , h ) y ( t , h ). If the error is larger we restart with h / 2 until we reach the tolerance.
( y ( t , 2 h ) means approximation with 2 h)
I don't understand why the Richardson extrapolation is mentioned. For what do I have to use it? Can I not just calculate y ( t , 2 h ) and y ( t , h ) and see the error?

Answer & Explanation

raffatoaq

raffatoaq

Beginner2022-08-08Added 22 answers

You can just check the error, and not bother with Richardson extrapolation, that's fine. But once you've paid the computational price of calculating the error, you might as well use Richardson extrapolation to blackuce the error further; it costs you essentially nothing.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?