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Unformatted text preview: smaller than necessary? Answer: We know that if our old values are correct, y ab n +1y ( t n +1 ) = 3 h 4 8 y (4) ( Î· ) . y am n +1y ( t n +1 ) =h 4 24 y (4) ( Î½ ) . Subtracting, we obtain y ab n +1y am n +1 = 3 h 4 8 y (4) ( Î· )(h 4 24 y (4) ( Î½ )) where Î·, Î½ are in the interval containing y ab n +1 , y am n +1 , and the true value. Since 3 / 8+1 / 24 = 10 / 24, the error in AM can be estimated as Â² =  y ab n +1y am n +1  / 10. Now, if Â² > Ï„ , we might reduce h by a factor of 2 and retake the step. If Â² << Ï„ , we might double h in preparation for the next step (expecting that the local error might increase by a factor of 2 4 ). 2...
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 Fall '06
 oleary
 Hamiltonian mechanics, Hamiltonian, Symplectic manifold, Symplectic geometry, local error

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