Tema02 - 2. Interpolaci´ o polin` omica 2.1 Introducci´ o...

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Unformatted text preview: 2. Interpolaci´ o polin` omica 2.1 Introducci´ o Considereu una fam´ ılia de funcions d’una sola variable x, φ = φ ( x ; a ,...,a n ) , amb n + 1 par` ametres a ,a 1 ,...,a n que caracteritzen les funcions de la fam´ ılia. El problema d’interpolaci´ o consisteix a determinar aquests n + 1 par` ametres de manera que per a n + 1 parelles donades ( x j ,y j ) es compleixi φ ( x j ; a ,...,a n ) = y j j = 0 ÷ n. Els ( x j ,y j ) se’ls anomena nodes, nusos ´ o punts de suport. Exemple 1 1) Interpolaci´ o lineal φ ( x ; a ,...,a n ) = a n x n + a n- 1 x n- 1 + ... + a 1 x + a polin` omica φ ( x ; a ,...,a n ) = a + a 1 e xi + ... + a n e nxi trigonom` etrica 2) Interpolaci´ o no lineal φ ( x ; a ,...,a n ,b ,...,b n ) = a + a 1 x + ... + a n x n b + b 1 x + ... + b n x n racional φ ( x ; a ,...,a n ,λ ,...,λ n ) = a e λ x + ... + a n e λ n x exponencial 2.2 Interpolaci´ o polin` omica Es vol determinar els n + 1 coeficients del polinomi de grau m` axim n , P ( x ) = a n + a n- 1 x + ... + a x n , de tal manera que passi pels n + 1 punts ( x j ,y j ) j = 0 ÷ n : P ( x j ) = y j , j = 0 ÷ n. (2.1) Si evalueu P ( x ) en cada x i obteniu el seg¨uent sistema d’equacions lineals: P ( x ) = a n + x a n- 1 + ... + x n a = y . . . . . . . . . P ( x n ) = a n + x n a n- 1 + ... + x n n a = y n + 1 equacions amb n + 1 inc` ognites: a n ,a n- 1 ,...,a 1 ,a . 1 El determinant del sistema, que s’anomena de Vandermonde, ´ es 1 x ... x n 1 x 1 ... x n 1 . . . . . . . . . 1 x n ... x n n = Y ≤ i<j ≤ n ( x j- x i ) 6 = 0 si x i 6 = x j , per a i 6 = j. Llavors, la soluci´ o del problema existeix i ´ es ´unica. 2.3 F´ormula de Lagrange Es construeix P ( x ) = n X i =0 y i l i ( x ) , (2.2) de manera que l i ( x j ) = δ i,j = ( 1 si i = j, 0 si i 6 = j, i, llavors es tindr` a la condici´ o d’interpolaci´ o (2.1). Els polinomis l i ( x ) s’anomenen polinomis de Lagrange i v´ enen definits per l i ( x ) = ( x- x )( x- x 1 ) ··· d ( x- x i ) ··· ( x- x n ) ( x i- x )( x i- x 1 ) ··· d ( x i- x i ) ··· ( x i- x n ) = w ( x ) w ( x i )( x- x i ) , on w ( x ) = ( x- x )( x- x 1 ) ··· ( x- x n ) , i quan tenim una successi´o a 1 ,a 2 ,...,a i- 1 ,a i +1 ,...,a n , on el terme i-` esim no hi ´ es, es pot expressar per a 1 ,a 2 ,..., b a i ,...,a n ....
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This note was uploaded on 04/01/2011 for the course MA DCIS taught by Professor Miquelgrau during the Spring '11 term at Universitat Politècnica de Catalunya.

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Tema02 - 2. Interpolaci´ o polin` omica 2.1 Introducci´ o...

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