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# 25ac this is because our strategy of examining simple

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Unformatted text preview: cting rooted trees that present the order conditions in a graphical way. They are discussed in many texts (e.g., 11, 16]) however, they are still complex and we will not pursue them here. Instead, we'll develop additional necessary order conditions by considering the simple ODE y = y: 0 Replacing f (t y) in (3.2.3) by y yields yn = yn 1 + h ; Yi = yn 1 + h ; It's simpler to use vector notation s X j =1 aij Yj yn = yn 1 + hbT Y s X i=1 bi Yi i = 1 2 : : : s: Y = yn 1l + hAY ; ; where Y = Y1 Y2 : : : Ys]T (3.2.6a) l = 1 1 : : : 1]T (3.2.6b) 2 6 A=6 6 4 a11 a21 ... as1 a12 a22 ... as2 3 : : : a1s : : : a2s 7 7 . . . ... 7 5 : : : ass (3.2.6c) and b = b1 b2 : : : bs]T : 12 (3.2.6d) Eliminating Y, we have Y = yn 1(I ; hA) 1l ; ; and yn = yn 1 + hyn 1bT (I ; hA) 1 l: ; ; ; Assuming that yn 1 is exact, the exact solution of this test equation is yn = ehyn 1. Expanding this solution and (I ; hA) 1 in series k 1 + h + : : : + h ! + : : : = 1 + hbT (I + hA + : : : + hk Ak + : : : )l: k Equating like powers of h yields the order condition 1 bT Ak 1l = k! k = 1 2 : : : p: (3.2.7) We recognize that this condition with k = 1 is identical to (3.2.5a). Lettin...
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