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Show that aj +2 aj ↑ 217 Since h> 2, therefore, the series is convergent.

Show that aj +2 aj ↑ 217 Since h> 2, therefore, the series is convergent.

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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Test for convergence. Show that aj +2 aj Since h2, therefore, the series is convergent. 2 j
Transcribed Image Text:Test for convergence. Show that aj +2 aj Since h2, therefore, the series is convergent. 2 j
8.3 ODE Eigenvalue problems Example 8.3.1 Legendre Equation Ly(x) = (1-x²)y"(x) + 2xy'(x) = y(x). (8.23) Defines an eigenvalue problem when V2 spherical polar coordinates where x = cos -1 ≤ x ≤ +1. As boundary conditions, the solutions should be nonsingular at x = ±1. is separated in with the range Using Frobenius method, assume solutions of the form. From the assumed solution, Ariken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7 ed.) Elsevier We get 00 y' = [aj (s +j)xs+j-1, j=0 Substituting it in Eq. (8.23) Ž¶ -(1-x²) y = -(1-x²) v = Σ ²₁x j=0 Σ j=0 ajxs+j ajx³+j aj(s+j)(s+j-1)x+/-2 -aj(s +j) (s + j - 1)x+j-2 + a₁x²+1 = 0 00 (8.24) j=0 + 2x *-2 + 2x² Ariken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7 ed.). Elsevier. aj(s+j)(s+j-1)x+/-2 + 2x aj(s+j)(s+j-1)xs+j-2 20+0+2 +2²2 946 aj(s+j)x³+/ T=0 aj(s+j)(s+j-1)x+/+ 2x aj(s +j)x+j-1-2ax²+1 = 0 jo aj (s+j)(s+j-1)x+j+ 2a,(s+j)-2x+1 :0 ajx = 0. =o -ao(s) (s - 1)xs-²-a₁(s + 1)(s)xs-¹-aj(s+j)(s +j-1)xs+j-2 • Σ + =0 a(s+j)x³+j-1
Transcribed Image Text:8.3 ODE Eigenvalue problems Example 8.3.1 Legendre Equation Ly(x) = (1-x²)y"(x) + 2xy'(x) = y(x). (8.23) Defines an eigenvalue problem when V2 spherical polar coordinates where x = cos -1 ≤ x ≤ +1. As boundary conditions, the solutions should be nonsingular at x = ±1. is separated in with the range Using Frobenius method, assume solutions of the form. From the assumed solution, Ariken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7 ed.) Elsevier We get 00 y' = [aj (s +j)xs+j-1, j=0 Substituting it in Eq. (8.23) Ž¶ -(1-x²) y = -(1-x²) v = Σ ²₁x j=0 Σ j=0 ajxs+j ajx³+j aj(s+j)(s+j-1)x+/-2 -aj(s +j) (s + j - 1)x+j-2 + a₁x²+1 = 0 00 (8.24) j=0 + 2x *-2 + 2x² Ariken, G., Weber, H., and Harris, F. (2013). Mathematical Methods for Physicists (7 ed.). Elsevier. aj(s+j)(s+j-1)x+/-2 + 2x aj(s+j)(s+j-1)xs+j-2 20+0+2 +2²2 946 aj(s+j)x³+/ T=0 aj(s+j)(s+j-1)x+/+ 2x aj(s +j)x+j-1-2ax²+1 = 0 jo aj (s+j)(s+j-1)x+j+ 2a,(s+j)-2x+1 :0 ajx = 0. =o -ao(s) (s - 1)xs-²-a₁(s + 1)(s)xs-¹-aj(s+j)(s +j-1)xs+j-2 • Σ + =0 a(s+j)x³+j-1
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