Numerical Mathematics - A Collection of Solved Problems

INTERPOLACIJA FUNKCIJA 205iliH m (x) =nXα i −1Xi=0 j=0α i −j−1Xk=0y (j)i1k !1j !» (x − xi ) α iΩ(x)– (k)x=x iΩ(x)(x − x i ) α i−j−k .6.1.34. Koristeći Hermiteovu interpolaciju dokazati da jef s = (1 + 2s)(1 − s) 2 f 0 + (3 − 2s)s 2 f 1 + s(1 − s) 2 hf ′ 0− s 2 (1 − s)hf ′ 1 + h44! f(4) (ξ) · s 2 (1 − s) 2 ,gde je f s = f(x 0 + sh), x 0 < ξ < x 0 + h, 0 < s < 1, a zatim izvesti formuluf 1/2 = 1 2 (f 0 + f 1 ) + h 8 (f ′ 0 − f ′ 1) + h4384 f(4) (ξ).Rešenje. Konstruišimo, najpre, Hermiteov interpolacioni polinom na osnovudatih podataka: f(x 0 ) = f 0 , f(x 1 ) = f 1 i f ′ (x 0 ) = f ′ 0, f ′ (x 1 ) = f ′ 1:gde je H 1 (x) = αx + β iKako jeH 3 (x) = P 1 (x) + (x − x 0 )(x − x 1 )H 1 (x),P 1 (x) = x − x 1x 0 − x 1f 0 + x − x 0x 1 − x 0f 1 = − 1 h (x − x 1)f 0 + 1 h (x − x 0)f 1 .H ′ 3(x) = P ′ 1(x) + (2x − x 0 − x 1 )H 1 (x) + (x − x 0 )(x − x 1 )H ′ 1(x)= − 1 h f 0 + 1 h f 1 + (2x − x 0 − x 1 )H 1 (x) + (x − x 0 )(x − x 1 )H ′ 1(x),to iz uslovadobijamoa izf ′ 0 = H ′ 3(x 0 ) = − 1 h f 0 + 1 h f 1 − hH 1 (x 0 )H 1 (x 0 ) = − 1 h 2 f 0 + 1 h 2 f 1 − 1 h f ′ 0,f ′ 1 = H ′ 3(x 1 ) = − 1 h f 0 + 1 h f 1 + hH 1 (x 1 )