Answer Happy

~ 3. Hermite polynomials Hn (2) can be defined by the generating function G(t, x) = e +2tx Hn(x)t" n! = n=0 (a) Use the generating function to derive the relationships H(x) = 2nHn-1(x), Hn+1(x) = 2x Hin (x) – 2nHn-1(x). (b) Use these relationships to derive a second-order ODE for which y= - HQ(x) is a solution. Rewrite this equation in Sturm-Liouville form as de [played p (2) dx = -W(x)ny, -0<x<, dx for some p(x) and w(x) that you should specify, and hence deduce that Hn (2)Hm()e-rº dx = 0, | = for any m En (you may quote standard results for Sturm-Liouville theory from the lectures). (c) For a given integer value of n > 0, use your answers to (a) and (b) to determine conditions on the possible values of m that will ensure that т - (d) Show that H.()Hm(a)nºe++* dx=0. G(t, 2)G(8,4)=+* dx = Vreest, and use this result to compute ?(z)e="dr. (HINT: you may quote without proof that some du = V.) =