The Hermite equation of order α is
y″ − 2xy′ + 2αy = 0.
(a) Derive the two power series solutions
and
Show that y1 is a polynomial if α is an even integer, whereas y2 is a polynomial if α is an odd integer. (b) The Hermite polynomial of degree n is denoted by Hn(x). It is the nth-degree polynomial solution of Hermite’s equation, multiplied by a suitable constant so that the coefficient of xn is 2n. Show that the first six Hermite polynomials are
H0(x) = 1,
H1(x) = 2x,
H3(x) = 4x2 − 2,
H3(x) = 8x3 − 12x,
H4(x) = 16x4 − 48x2 + 12,
H5(x) = 32x5 − 160x3 + 120x.
A general formula for the Hermite polynomials is
Verify that this formula does in fact give an nth-degree polynomial. It is interesting to use a computer algebra system to investigate the conjecture that (for each n) the zeros of the Hermite polynomials Hn and Hn + 1 are “interlaced” — that is, the x zeros of Hn lie in the n bounded open intervals whose endpoints are successive pairs of zeros of Hn + 1.
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