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(c) If you differentiate an nth-order polynomial, you get a polynomial of order (n-1). For the hermite polynomials, in fact, dHn ἀξ = 2nHn-1(E) Check this, by differentiating H5 and H6. (d) H₂() is the nth z-derivative, at z = 0, of the generating function exp(-2₂² +2z§); or, to put it another way, it is the coefficient of z"/n! in the Taylor series expansion for this function: e¯²³²+2=² = Σ n=0 Use this to obtain H₁, H₂, and H3. n! (3) H₂(8) (4)

(c) If you differentiate an nth-order polynomial, you get a polynomial of order (n-1). For the hermite polynomials, in fact, dHn ἀξ = 2nHn-1(E) Check this, by differentiating H5 and H6. (d) H₂() is the nth z-derivative, at z = 0, of the generating function exp(-2₂² +2z§); or, to put it another way, it is the coefficient of z"/n! in the Taylor series expansion for this function: e¯²³²+2=² = Σ n=0 Use this to obtain H₁, H₂, and H3. n! (3) H₂(8) (4)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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C & D

1. In this problem we explore some of the more useful theorems involving Hermite polynomials. (a) The Rodrigues formula says that H₂(E)= (-1)¹ ² ()* dHn de Use it to derive H3 and H₁. (b) The following recursion relation gives you H₂+1 in terms of the two preceding Hermite polynomials: e-8² H₂+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= Hn n! 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 expansion for this function: (3) (4)
Transcribed Image Text:1. In this problem we explore some of the more useful theorems involving Hermite polynomials. (a) The Rodrigues formula says that H₂(E)= (-1)¹ ² ()* dHn de Use it to derive H3 and H₁. (b) The following recursion relation gives you H₂+1 in terms of the two preceding Hermite polynomials: e-8² H₂+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= Hn n! 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 expansion for this function: (3) (4)
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