Thus the exact solutions are y 1 12 427709810573691

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Thus the exact solutions are y (1) = 12 . 427709810573691 and y 0 (1) = 26 . 470063126092101. So with N = 10 the approximations are correct to within 4 significant figures. Increasing N would give better approximations. (e.g. with N=30 the approximations for y (1) and y 0 (1) are correct to 7 significant digits.) Date : August 2, 2010. 1
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2 BENJAMIN JOHNSON section 5.10 1. To prove Theorem 5.20, part (i), show that the hypotheses imply that a constant K > 0 exists such that | u i - v i | ≤ K | u 0 - v 0 | for each 1 i N, whenever { u i } n i =1 and { v i } n i =1 satisfy the difference equation w i +1 = w i + ( t i , w i , h ). Solution: Using the hypotheses of theorem 5.20, we may assume that a number h 0 exists such that φ ( t, w, h ) satisfies a Lipschitz condition in the variable w with Lipschitz constant L on the set D = { ( t, w, h ) : a t b, < w < , 0 h h 0 } . Suppose that { u i } N i =1 and { v i } N i =1 both satisfy the difference equation w i +1 = w i + ( t i , w i , h ). Then for i = 1 and for h < h 0 we have | u 1 - v 1 | = | u 0 + ( t 0 , u 0 , h ) - [ v 0 + ( t 0 , v 0 , h )] | = | u 0 - v 0 + h ( φ ( t 0 , u 0 , h ) - φ ( t 0 , v 0 , h )) | ≤ | u 0 - v 0 | + h | φ ( t 0 , u 0 , h ) - φ ( t 0 , v 0 , h ) | ≤ | u 0 - v 0 | + hL | u 0 - v 0 | (1 + h 0 L ) | u 0 - v 0 | Because the same algebra will work the same for each i
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