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# We should select them so that xx i i z xi z xi xi 1 i

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Unformatted text preview: ) i1 Substituting this result into (9.1.15) yields je(x )j j 1 X h5 kg (x )k 2 =1 N xx i j j = 0 1 ::: N i1 i or je(x )j j 1 kg (x )k X h5 = 1 kg (x )k Nh5 : 2 2 =1 N xx j 1 xx i j 1 i Thus, je(x )j C h4 j = 0 1 ::: N j (9.1.28a) where 1 C = 2 (b ; a)kg (x )k : xx j (9.1.28b) 1 The pointwise error has been increased by two orders with the special choice of collocation points dictated by (9.1.26). This is most de nitely an example of nodal superconvergence since the global error is still O(h2). It remains to determine the collocation points that satisfy (9.1.26). Let us begin by transforming these integrals to ;1 1] using the mapping x=x h 12+ ;1 i 2 i; = 1: (9.1.29a) Also let i 1 = x 1 2 ; 1h i; = i i 10 2 = x 1 2 + 2h : i; = i (9.1.29b) Then x; i h ( +2 ) 1 2 1= x; i Conditions (9.1.26a,b) become h )3 Z 1 ( + 2 )( ; 2 )d = 0 (2 1 2 i 2= h ( ; 2 ): 2 2 i (9.1.29c) h )4 Z 1 ( + 2 )( ; 2 ) d = 0: (2 1 2 i i 1 1 ; ; Thus, it su ces to determine 1 and 2 such that Z1 1 ; ( + 2 1)( ; 2 2 )P ( )d = 0 (9.1.30) for all linear polynomials P ( ). Since the integran...
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