Laguerre - Lecture notes 1 PDF

Title Laguerre - Lecture notes 1
Author Jefferson Dante
Course Historia Política Contemporánea de España
Institution Universidad Unidad
Pages 2
File Size 48.7 KB
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Summary

Só matematica básica....


Description

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....


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