Rene Nicholson

Rene Nicholson

Answered

2022-10-20

When applying Richardson Extrapolation to Euler Method, when isn't the local truncation error of h^2 used to get (4*Euler(h/2) - Euler(h)) / 3 instead of using the global error of h to get 2*Euler(h/2) - Euler(h)?

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Answer & Explanation

Szulikto

Szulikto

Expert

2022-10-21Added 22 answers

In general when you are applying Richardson extrapolation to a method T ( h ), it should be an approximation of a fixed quantity I which doesn't depend on h. Thus for instance for the Euler method, T ( h ) is the result of running the Euler method with step size h for a fixed ODE up to a fixed finite time. So the error that matters is the global error.

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rhenan5v

rhenan5v

Expert

2022-10-22Added 3 answers

You want to compare two approximations at t = h. The error of two Euler steps at half step size is in the leading order
2 C ( h 2 ) 2 = 1 2 C h 2
if the truncation error is locally C h 2 for a step of size h.
Thus you get, if E u l e r ( h ) denotes the method application to get to a fixed time t with step size h,
E u l e r ( h ) = y + C h 2 + O ( h 3 )
and with two steps of half the step size
E u l e r ( h / 2 ) = y + 1 2 C h 2 + O ( h 3 ) .
Eliminating the term C h 2
2 E u l e r ( h / 2 ) E u l e r ( h ) = y + O ( h 3 )
This is incidentally the explicit midpoint method
y ( t + h ) = y ( t ) + h f ( t + h 2 , y ( t ) + h 2 f ( t , y ( t ) ) ) + O ( h 3 )

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