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Overview Solving the Legendre Equation Application Legendre Polynomials Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials Overview Solving the Legendre Equation


  1. Overview Solving the Legendre Equation Application Legendre Polynomials Bernd Schr¨ oder logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  2. Overview Solving the Legendre Equation Application Why are Legendre Polynomials Important? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  3. Overview Solving the Legendre Equation Application Why are Legendre Polynomials Important? 1. The generalized Legendre equation m 2 � � � 1 − x 2 � y ′′ − 2 xy ′ + λ − y = 0 arises when the 1 − x 2 equation ∆ u = f ( ρ ) u is solved with separation of variables in spherical coordinates. (QM: hydrogen atom!) The � � cos ( φ ) function y describes the polar part of the solution of ∆ u = f ( ρ ) u . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  4. Overview Solving the Legendre Equation Application Why are Legendre Polynomials Important? 1. The generalized Legendre equation m 2 � � � 1 − x 2 � y ′′ − 2 xy ′ + λ − y = 0 arises when the 1 − x 2 equation ∆ u = f ( ρ ) u is solved with separation of variables in spherical coordinates. (QM: hydrogen atom!) The � � cos ( φ ) function y describes the polar part of the solution of ∆ u = f ( ρ ) u . � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 is the 2. The Legendre equation special case with m = 0, which turns out to be the key to the generalized Legendre equation. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  5. Overview Solving the Legendre Equation Application Why are Legendre Polynomials Important? 1. The generalized Legendre equation m 2 � � � 1 − x 2 � y ′′ − 2 xy ′ + λ − y = 0 arises when the 1 − x 2 equation ∆ u = f ( ρ ) u is solved with separation of variables in spherical coordinates. (QM: hydrogen atom!) The � � cos ( φ ) function y describes the polar part of the solution of ∆ u = f ( ρ ) u . � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 is the 2. The Legendre equation special case with m = 0, which turns out to be the key to the generalized Legendre equation. 3. The solutions of both equations must be finite on [ − 1 , 1 ] . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  6. Overview Solving the Legendre Equation Application Why are Legendre Polynomials Important? 1. The generalized Legendre equation m 2 � � � 1 − x 2 � y ′′ − 2 xy ′ + λ − y = 0 arises when the 1 − x 2 equation ∆ u = f ( ρ ) u is solved with separation of variables in spherical coordinates. (QM: hydrogen atom!) The � � cos ( φ ) function y describes the polar part of the solution of ∆ u = f ( ρ ) u . � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 is the 2. The Legendre equation special case with m = 0, which turns out to be the key to the generalized Legendre equation. 3. The solutions of both equations must be finite on [ − 1 , 1 ] . 4. Because 0 is an ordinary point of the equation, it is natural to attempt a series solution. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  7. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  8. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  9. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ � c n n ( n − 1 ) x n − 2 ∑ n = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  10. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 ∑ ∑ n = 2 n = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  11. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  12. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  13. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ c n n ( n − 1 ) x n − 2 ∑ n = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  14. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n ∑ ∑ n = 2 n = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  15. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n − 2 c n nx n ∑ ∑ ∑ n = 2 n = 2 n = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  16. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ ∞ c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n − 2 c n nx n + λ c n x n ∑ ∑ ∑ ∑ n = 2 n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  17. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ ∞ λ c n x n = c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n − 2 c n nx n + ∑ ∑ ∑ ∑ 0 n = 2 n = 2 n = 1 n = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  18. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ ∞ λ c n x n = c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n − 2 c n nx n + ∑ ∑ ∑ ∑ 0 n = 2 n = 2 n = 1 n = 0 ∞ ∑ c k + 2 ( k + 2 )( k + 1 ) x k k = 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

  19. Overview Solving the Legendre Equation Application y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � � Series Solution of � 1 − x 2 � y ′′ − 2 xy ′ + λ y = 0 1 − x 2 � ∞ ∞ ∞ � c n n ( n − 1 ) x n − 2 − 2 x c n nx n − 1 + λ ∑ ∑ ∑ c n x n = 0 n = 2 n = 1 n = 0 ∞ ∞ ∞ ∞ λ c n x n = c n n ( n − 1 ) x n − 2 − c n n ( n − 1 ) x n − 2 c n nx n + ∑ ∑ ∑ ∑ 0 n = 2 n = 2 n = 1 n = 0 ∞ ∞ ∑ c k + 2 ( k + 2 )( k + 1 ) x k − ∑ c k k ( k − 1 ) x k k = 0 k = 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Legendre Polynomials

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