(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)
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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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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