1.In this problem we explore some of the more useful theorems involvingHermite polynomials.(a) The Rodrigues formula says thatH₂(E)= (-1)¹ ²()*dHndeUse it to derive H3 and H₁.(b) The following recursion relation gives you H₂+1 in terms of the two precedingHermite polynomials:e-8²H₂+1(E) = 2 H₂(E) — 2nHn-1(E)Use it, together with your answer in (a), to obtain H5 and H₁.(c) If you differentiate an nth-order polynomial, you get a polynomial of order (n-1).For the hermite polynomials, in fact,= 2n Hn-1(E)Use this to obtain H₁, H₂, and H3.(1)e==t2 = Ση#n(s)e-z²+2=Hnn!n=0)(2)Check this, by differentiating H5 and H6.(d) H() is the nth z-derivative, at z = = 0, of the generating function exp(-z² + 2z);or, to put it another way, it is the coefficient of z" /n! in the Taylor series expansionfor this function:(3)(4)

Since you have added many problem in one question, here I am answering first two parts (a) and (b).

(a) The Rodrigues formula says that H₂ (§) = (-1)"¹ es ² * (₁) * (1) Use it to derive H3 and H₁. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: Hn+1(E) = 2H₂(E) — 2nH₂-1(E) Use it, together with your answer in (a), to obtain H5 and H6. (2)

(a) The Rodrigues formula says that H₂ (§) = (-1)"¹ es ² * (₁) * (1) Use it to derive H3 and H₁. (b) The following recursion relation gives you Hn+1 in terms of the two preceding Hermite polynomials: Hn+1(E) = 2H₂(E) — 2nH₂-1(E) Use it, together with your answer in (a), to obtain H5 and H6. (2)

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter4: Polynomial And Rational Functions

Section4.3: Zeros Of Polynomials

Problem 66E

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