10.07.2015 Views

pdf - Univerzitet u Novom Sadu

pdf - Univerzitet u Novom Sadu

pdf - Univerzitet u Novom Sadu

SHOW MORE
SHOW LESS

You also want an ePaper? Increase the reach of your titles

YUMPU automatically turns print PDFs into web optimized ePapers that Google loves.

14 Uvod1.1.5 Hermiteovi polinomi i funkcijeDefinicija 1.1.32 Hermiteov polinom (ermitski polinom) n-tog reda, n ∈N 0 je definisan sah n (x) = (−1) n e x2 d n x22 (e− 2dxn ). (1.32)Prvih nekoliko ermitskih polinoma su: h 0 (x) = 1, h 1 (x) = x,h 2 (x) = x 2 − 1, h 3 (x) = x 3 − 3x, h 4 (x) = x 4 − 6x 2 + 3 . . . itd.Teorema 1.1.20 Hermiteovi polionomi zadovoljavaju rekurentne veze:1. h n+1 (x) − xh n (x) + nh n−1 (x) = 0,2. h ′ n(x) = nh n−1 (x),3. h ′′ n(x) − xh ′ n(x) + nh n (x) = 0.Teorema 1.1.21 Familija { 1 √n!h n (x) : n ∈ N 0 } čini ortonormiranu bazuprostora L 2 (R, dµ), gde je dµ = 1 √2πe − x22 Gaussova mera.Teorema 1.1.22 Generativna funkcija Hermiteovih polinoma jee tx− 1 2 t2 =∞∑n=0t nn! h n(x). (1.33)Teorema 1.1.23 Svaki polinom nad poljem realnih brojeva se može prikazatipreko ermitskih polinoma i obratno, preko identiteta:x n =[n/2]∑k=0( n2k)(2k − 1)!!h n−2k (x), (1.34)[n/2]∑( ) nh n (x) = (−1) k (2k − 1)!!x n−2k . (1.35)2kk=0Teorema 1.1.24 Hermiteovi polinomi zadovoljavaju identitete:1. h n (x + y) = ∑ nk=0( nk)hn−k (x)y k ,2. h n (ax) = ∑ [n/2]k=0 (−1)k( n2k)(2k − 1)!!a n−2k (1 − a 2 ) k h n−2k (x),3.1 √2π∫ ∞−∞ h n(x + y)e − x22 dx = y n ,

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!