Title | Laguerre - Lecture notes 1 |
---|---|
Author | Jefferson Dante |
Course | Historia Política Contemporánea de España |
Institution | Universidad Unidad |
Pages | 2 |
File Size | 48.7 KB |
File Type | |
Total Downloads | 56 |
Total Views | 131 |
Só matematica básica....
Section 5.6, Exercise 13 The Laguerre differential equation is xy′′ + (1 − x)y ′ + λy = 0. Show that x = 0 is a regular singular point. Determine the indicial equation, its roots, the recurrence relation, and one solution (x > 0). Show that when λ = m, a positive integer, this solution reduces to a polynomial. When properly normalized, this polynomial is known as the Laguerre polynomial, Lm (x). We can see that x = 0 is a regular singular point since 1−x = lim (1 − x) = 1 and lim x x→0 x→0 x λ = lim λx = 0. lim x2 x→0 x→0 x We will assume the series solution has the form ∞ X y(x) = an xr+n with y ′ (x) = y ′′(x) =
n=0 ∞ X
(r + n)an xr+n−1
and
n=0 ∞ X
(r + n)(r + n − 1)an xr+n−2 .
n=0
Substituting into Laguerre’s equation yields 0 = x
∞ ∞ ∞ X X X (r + n)(r + n − 1)an xr+n−2 + (1 − x) (r + n)an xr+n−1 + λ an xr+n n=0
n=0
n=0
∞ ∞ ∞ ∞ X X X X = (r + n)(r + n − 1)an xr+n−1 + (r + n)an xr+n−1 − (r + n)an xr+n + λan xr+n n=0 ∞
=
X n=0
n=0
n=0
∞
X [(r + n)(r + n − 1) + (r + n)] an xr+n−1 + [λ − (r + n)] an xr+n n=0
∞ ∞ X X = (r + n)2 an xr+n−1 + (λ − r − n)an xr+n n=0 ∞
n=0 ∞
=
X X (r + n)2 an xr+n−1 + (λ − r − n + 1)an−1 xr+n−1 n=0
n=1
∞ ∞ X X = r 2 a0 xr−1 + (r + n)2 an xr+n−1 + (λ − r − n + 1)an−1 xr+n−1 n=1 ∞
= r 2 a0 xr−1 +
X n=1
n=1
(r + n)2 an + (λ − r − n + 1)an−1 xr+n−1
n=0
Thus the indicial equation is r 2 = 0 which has indicial roots r1 = r2 = r = 0. The recurrence relation is 0 = (r + n)2 an + (λ − r − n + 1)an−1 = n2 an + (λ − n + 1)an−1 (since r = 0) (n − 1 − λ)an−1 (for n ≥ 1). an = n2 Now if a0 is arbitrary then λ λ a0 = − a0 2 1 (1!)2 λ(1 − λ) 1−λ a0 a1 = − = 22 (2!)2 λ(1 − λ)(2 − λ) 2−λ a0 a2 = − = 2 3 (3!)2 .. . Qn−1 k=0 (k − λ) = . (n!)2
a1 = − a2 a3
an
Thus one solution to Laguerre’s equation is y1 (x) = a0
1+
∞ Qn−1 X k=0 (k − λ) n=1
(n!)2
n
x
!
.
When λ = m ∈ N then according to the formula given above for the coefficients of the series solution 0 = am+1 = am+2 = · · · = am+k = · · · and y1 (x) contains powers of x less than or equal to m, in other words is a polynomial of degree at most m....