Unformatted text preview: h , we have h = 0 . 1 ⇒ c (0 . 1) = − 1 ⇒ x n = 0 . 1 n, y n = ( − 1) n , n = 1 , . . . , 10; h = 0 . 025 ⇒ c (0 . 025) = 0 . 5 ⇒ x n = 0 . 025 n, y n = (0 . 5) n , n = 1 , . . . , 40; h = 0 . 2 ⇒ c (0 . 2) = − 3 ⇒ x n = 0 . 2 n, y n = ( − 3) n , n = 1 , . . . , 5 . These values are shown in Table 3E. Thus, for h = 0 . 1 we have alternating y n = ± 1; for h = 0 . 2, y n ’s have an increasing magnitude and alternating sign; h = 0 . 025 is a good step size. ²rom this example we conclude that, in Euler’s method, one should be very careful in choosing a step size. Wrong choice can even lead to a diverging process. 148...
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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