Problem 2.16 In this problem we explore some of the more useful theorems (stated without proof)...
Problem 2.16 In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that H (6) = (-1)” (1) . (2.87) Use it to derive H3 and H4. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: Hn+1(E) = 2€ H, (E) – 2n Hn-1(5). (2.88) Use it, together with your answer in (a), to obtain Hg and H. (c) If you differentiate an nth-order polynomial, you get a polynomial of order (n - 1). For the Hermite polynomials, in fact, din – 2nH-1(). de (2.89) Check this, by differentiating Hs and H. (d) H.() is the nth z-derivative, at z = 0, of the generating function exp(-2? + 2zł); or, to put it another way, it is the coefficient of z"/n! in the Taylor series expansion for this function: e-++226 = ,(£). (2.90) Use this to obtain Hi, H2, and H3.