Legendre Polynomials

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

Transcript of Legendre Polynomials

Page 1: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Legendre Polynomials

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 2: Legendre Polynomials

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Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?

1. The generalized Legendre equation(1− x2

)y′′−2xy′ +

(λ − m2

1− x2

)y = 0 arises when the

equation ∆u = f (ρ)u is solved with separation of variablesin spherical coordinates. (QM: hydrogen atom!) Thefunction y

(cos(φ)

)describes the polar part of the solution

of ∆u = f (ρ)u.2. The Legendre equation

(1− x2

)y′′−2xy′ +λy = 0 is the

special case with m = 0, which turns out to be the key tothe 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.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 3: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?1. The generalized Legendre equation(

1− x2)

y′′−2xy′ +(

λ − m2

1− x2

)y = 0 arises when the

equation ∆u = f (ρ)u is solved with separation of variablesin spherical coordinates. (QM: hydrogen atom!) Thefunction y

(cos(φ)

)describes the polar part of the solution

of ∆u = f (ρ)u.

2. The Legendre equation(

1− x2)

y′′−2xy′ +λy = 0 is thespecial case with m = 0, which turns out to be the key tothe 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.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 4: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?1. The generalized Legendre equation(

1− x2)

y′′−2xy′ +(

λ − m2

1− x2

)y = 0 arises when the

equation ∆u = f (ρ)u is solved with separation of variablesin spherical coordinates. (QM: hydrogen atom!) Thefunction y

(cos(φ)

)describes the polar part of the solution

of ∆u = f (ρ)u.2. The Legendre equation

(1− x2

)y′′−2xy′ +λy = 0 is the

special case with m = 0, which turns out to be the key tothe 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.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 5: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?1. The generalized Legendre equation(

1− x2)

y′′−2xy′ +(

λ − m2

1− x2

)y = 0 arises when the

equation ∆u = f (ρ)u is solved with separation of variablesin spherical coordinates. (QM: hydrogen atom!) Thefunction y

(cos(φ)

)describes the polar part of the solution

of ∆u = f (ρ)u.2. The Legendre equation

(1− x2

)y′′−2xy′ +λy = 0 is the

special case with m = 0, which turns out to be the key tothe 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 naturalto attempt a series solution.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 6: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?1. The generalized Legendre equation(

1− x2)

y′′−2xy′ +(

λ − m2

1− x2

)y = 0 arises when the

equation ∆u = f (ρ)u is solved with separation of variablesin spherical coordinates. (QM: hydrogen atom!) Thefunction y

(cos(φ)

)describes the polar part of the solution

of ∆u = f (ρ)u.2. The Legendre equation

(1− x2

)y′′−2xy′ +λy = 0 is the

special case with m = 0, which turns out to be the key tothe 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.Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 7: Legendre Polynomials

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Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 8: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0

(1− x2

) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 9: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2

−2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 10: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 11: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn

= 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 12: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 13: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2

−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 14: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn

−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 15: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn

+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 16: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn

= 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 17: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 18: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk

−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 19: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk

−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 20: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk

+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 21: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk

= 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 22: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 23: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 24: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk

= 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 25: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(1− x2

)y′′−2xy′ +λy = 0(

1− x2) ∞

∑n=2

cnn(n−1)xn−2 −2x∞

∑n=1

cnnxn−1 +λ

∑n=0

cnxn = 0

∑n=2

cnn(n−1)xn−2−∞

∑n=2

cnn(n−1)xn−∞

∑n=1

2cnnxn+∞

∑n=0

λcnxn = 0

∑k=0

ck+2(k +2)(k +1)xk−∞

∑k=2

ckk(k−1)xk−∞

∑k=1

2ckkxk+∞

∑k=0

λckxk = 0

2c2 +λc0 +6c3x−2c1x+λc1x+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 26: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(2c2 +λc0)+(6c3x−2c1x+λc1x)+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Matching coefficients leads to c2 = −λ

2c0, c3 = −λ −2

6c1,

and for k ≥ 2

ck+2 =k(k−1)+2k−λ

(k +2)(k +1)ck =

k(k +1)−λ

(k +2)(k +1)ck.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 27: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(2c2 +λc0)+(6c3x−2c1x+λc1x)+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Matching coefficients leads to c2 = −λ

2c0, c3 = −λ −2

6c1,

and for k ≥ 2

ck+2 =k(k−1)+2k−λ

(k +2)(k +1)ck =

k(k +1)−λ

(k +2)(k +1)ck.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 28: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(2c2 +λc0)+(6c3x−2c1x+λc1x)+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Matching coefficients leads to c2 = −λ

2c0,

c3 = −λ −26

c1,

and for k ≥ 2

ck+2 =k(k−1)+2k−λ

(k +2)(k +1)ck =

k(k +1)−λ

(k +2)(k +1)ck.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 29: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(2c2 +λc0)+(6c3x−2c1x+λc1x)+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Matching coefficients leads to c2 = −λ

2c0, c3 = −λ −2

6c1,

and for k ≥ 2

ck+2 =k(k−1)+2k−λ

(k +2)(k +1)ck =

k(k +1)−λ

(k +2)(k +1)ck.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 30: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

(2c2 +λc0)+(6c3x−2c1x+λc1x)+∞

∑k=2

[(k +2)(k +1)ck+2 − k(k−1)ck −2kck +λck

]xk = 0

Matching coefficients leads to c2 = −λ

2c0, c3 = −λ −2

6c1,

and for k ≥ 2

ck+2 =k(k−1)+2k−λ

(k +2)(k +1)ck =

k(k +1)−λ

(k +2)(k +1)ck.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 31: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps. Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 32: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps. Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 33: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps. Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 34: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps. Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 35: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps.

Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 36: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

1. From here, we can produce a mathematical solution forany λ .

2. But it can be shown that all “true” infinite series solutionsof this equation will go to ∞ or −∞ at x = 1 or at x = −1 orboth.

3. Therefore, these solutions are not physically feasible,because the polar part y

(cos(φ)

)of the solution of

∆u = f (ρ)u should not go to infinity as we approach thez-axis. (For QM, the explanation involves integrabilityissues.)

4. Thus the only series solutions of interest are those thatterminate after finitely many steps. Or, in simplerlanguage, those solutions that happen to be polynomials.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 37: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 38: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 39: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0

and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 40: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0.

(Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 41: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 42: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.

8. By choosing one of c0 (or c1) equal to zero, we can makeall even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 43: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 44: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Series Solution of(1− x2)y′′−2xy′+λy = 0

5. The series expansion will “stop” when the sequence ofcoefficients terminates in repeating zeros.

6. The recurrence relation ck+2 =k(k +1)−λ

(k +2)(k +1)ck shows that

when λ is of the form l(l+1), where l is a nonnegativeinteger, then cl+2 = 0 and all entries cl+2k = 0. (Otherwiseno such thing happens, so we need λ = l(l+1).)

7. But the entries cl+(2k+1) may still not be zero.8. By choosing one of c0 (or c1) equal to zero, we can make

all even-numbered coefficients (or all odd-numberedcoefficients) equal to zero.

9. So the solutions we are interested in will be polynomialswith even powers (for λ = l(l+1) and l even) orpolynomials with odd powers (for λ = l(l+1) and l odd).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 45: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relation

ck+2 =k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 46: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 47: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 48: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 49: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 50: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 51: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

λ = l(l+1), Simplifying the Recurrence Relationck+2 =

k(k +1)−λ

(k +2)(k +1)ck

=k(k +1)− l(l+1)

(k +2)(k +1)ck

=k2 + k− l2 − l(k +2)(k +1)

ck

=(k + l)(k− l)+(k− l)

(k +2)(k +1)ck

=(k + l+1)(k− l)(k +2)(k +1)

ck

= −(l− k)(l+ k +1)(k +2)(k +1)

ck

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 52: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even,

ck+2 = −(l− k)(l+ k +1)(k +2)(k +1)

ck

c2 = − l(l+1)2

c0

c4 = −(l−2)(l+2+1)(2+2)(2+1)

c2 =l(l−2) · (l+3)(l+1)

4!c0

c6 = −(l−4)(l+4+1)(4+2)(4+1)

c4 = − l(l−2)(l−4) · (l+5)(l+3)(l+1)6!

c0

...

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)· (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 53: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even,

ck+2 = −(l− k)(l+ k +1)(k +2)(k +1)

ck

c2 = − l(l+1)2

c0

c4 = −(l−2)(l+2+1)(2+2)(2+1)

c2 =l(l−2) · (l+3)(l+1)

4!c0

c6 = −(l−4)(l+4+1)(4+2)(4+1)

c4 = − l(l−2)(l−4) · (l+5)(l+3)(l+1)6!

c0

...

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)· (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 54: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even,

ck+2 = −(l− k)(l+ k +1)(k +2)(k +1)

ck

c2 = − l(l+1)2

c0

c4 = −(l−2)(l+2+1)(2+2)(2+1)

c2 =l(l−2) · (l+3)(l+1)

4!c0

c6 = −(l−4)(l+4+1)(4+2)(4+1)

c4 = − l(l−2)(l−4) · (l+5)(l+3)(l+1)6!

c0

...

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)· (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 55: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even,

ck+2 = −(l− k)(l+ k +1)(k +2)(k +1)

ck

c2 = − l(l+1)2

c0

c4 = −(l−2)(l+2+1)(2+2)(2+1)

c2 =l(l−2) · (l+3)(l+1)

4!c0

c6 = −(l−4)(l+4+1)(4+2)(4+1)

c4 = − l(l−2)(l−4) · (l+5)(l+3)(l+1)6!

c0

...

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)· (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 56: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even,

ck+2 = −(l− k)(l+ k +1)(k +2)(k +1)

ck

c2 = − l(l+1)2

c0

c4 = −(l−2)(l+2+1)(2+2)(2+1)

c2 =l(l−2) · (l+3)(l+1)

4!c0

c6 = −(l−4)(l+4+1)(4+2)(4+1)

c4 = − l(l−2)(l−4) · (l+5)(l+3)(l+1)6!

c0

...

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)· (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 57: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 58: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 59: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 60: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 61: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 62: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

c2n = (−1)n l(l−2) · · ·(l−2(n−1)

)·(l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

2l2( l

2

)!

2l2( l

2

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l(l−2) · · ·(l−2(n−1)

)l(l−2) · · ·

(l−2(n−1)

) 2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n−1)(l+2n−3) · · ·(l+1)(2n)!

l!2l2 +n

( l2 +n

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!

2l2( l

2

)!

l!2l2 +n

( l2 +n

)!

2l2( l

2

)!

2l2−n

( l2 −n

)!c0

= (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 63: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

cl−2k = (−1)l−2k

2(l+ l−2k)!(l−2k)!l!

[( l2

)!]2( l

2 + l−2k2

)!( l

2 −l−2k

2

)!c0

= (−1)l2 (−1)−k (2l−2k)!

(l−2k)!l!

[( l2

)!]2

(l− k)!k!c0

= (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 64: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

cl−2k = (−1)l−2k

2(l+ l−2k)!(l−2k)!l!

[( l2

)!]2( l

2 + l−2k2

)!( l

2 −l−2k

2

)!c0

= (−1)l2 (−1)−k (2l−2k)!

(l−2k)!l!

[( l2

)!]2

(l− k)!k!c0

= (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 65: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

cl−2k = (−1)l−2k

2(l+ l−2k)!(l−2k)!l!

[( l2

)!]2( l

2 + l−2k2

)!( l

2 −l−2k

2

)!c0

= (−1)l2 (−1)−k (2l−2k)!

(l−2k)!l!

[( l2

)!]2

(l− k)!k!c0

= (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 66: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

cl−2k = (−1)l−2k

2(l+ l−2k)!(l−2k)!l!

[( l2

)!]2( l

2 + l−2k2

)!( l

2 −l−2k

2

)!c0

= (−1)l2 (−1)−k (2l−2k)!

(l−2k)!l!

[( l2

)!]2

(l− k)!k!c0

= (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 67: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n (l+2n)!(2n)!l!

[( l2

)!]2( l

2 +n)!( l

2 −n)!c0

cl−2k = (−1)l−2k

2(l+ l−2k)!(l−2k)!l!

[( l2

)!]2( l

2 + l−2k2

)!( l

2 −l−2k

2

)!c0

= (−1)l2 (−1)−k (2l−2k)!

(l−2k)!l!

[( l2

)!]2

(l− k)!k!c0

= (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 68: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

cl−2k = (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Choosing c0 = (−1)l2

l!

2l[( l

2

)!]2 gives

Pl(x) =b l

2c∑k=0

(−1)k (2l−2k)!2lk!(l− k)!(l−2k)!

xl−2k,

which is the customary way the Legendre polynomials arestated.(The generalized Legendre equation is good reading.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 69: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

cl−2k = (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Choosing c0 = (−1)l2

l!

2l[( l

2

)!]2 gives

Pl(x) =b l

2c∑k=0

(−1)k (2l−2k)!2lk!(l− k)!(l−2k)!

xl−2k,

which is the customary way the Legendre polynomials arestated.(The generalized Legendre equation is good reading.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 70: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

cl−2k = (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Choosing c0 = (−1)l2

l!

2l[( l

2

)!]2 gives

Pl(x) =b l

2c∑k=0

(−1)k (2l−2k)!2lk!(l− k)!(l−2k)!

xl−2k,

which is the customary way the Legendre polynomials arestated.

(The generalized Legendre equation is good reading.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 71: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

cl−2k = (−1)k (2l−2k)!2l(l−2k)!(l− k)!k!

(−1)l2

2l [( l2

)!]2

l!c0

Choosing c0 = (−1)l2

l!

2l[( l

2

)!]2 gives

Pl(x) =b l

2c∑k=0

(−1)k (2l−2k)!2lk!(l− k)!(l−2k)!

xl−2k,

which is the customary way the Legendre polynomials arestated.(The generalized Legendre equation is good reading.)

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 72: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals

1. As noted earlier, the Legendre equation arises in thequantum mechanical analysis of the hydrogen atom.

2. y(

cos(φ))

describes how the wave function of the electrondepends on the polar angle φ .

3. The absolute value of the wave function does not dependon the azimuthal angle θ .

4. The dependence on the radius ρ only scales the orbital. Itdoes not truly affect the “shape”.

5. So ρ = Pn(

cos(φ))

should give the “shapes” of theorbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 73: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.

2. y(

cos(φ))

describes how the wave function of the electrondepends on the polar angle φ .

3. The absolute value of the wave function does not dependon the azimuthal angle θ .

4. The dependence on the radius ρ only scales the orbital. Itdoes not truly affect the “shape”.

5. So ρ = Pn(

cos(φ))

should give the “shapes” of theorbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 74: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.2. y

(cos(φ)

)describes how the wave function of the electron

depends on the polar angle φ .

3. The absolute value of the wave function does not dependon the azimuthal angle θ .

4. The dependence on the radius ρ only scales the orbital. Itdoes not truly affect the “shape”.

5. So ρ = Pn(

cos(φ))

should give the “shapes” of theorbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 75: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.2. y

(cos(φ)

)describes how the wave function of the electron

depends on the polar angle φ .3. The absolute value of the wave function does not depend

on the azimuthal angle θ .

4. The dependence on the radius ρ only scales the orbital. Itdoes not truly affect the “shape”.

5. So ρ = Pn(

cos(φ))

should give the “shapes” of theorbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 76: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.2. y

(cos(φ)

)describes how the wave function of the electron

depends on the polar angle φ .3. The absolute value of the wave function does not depend

on the azimuthal angle θ .4. The dependence on the radius ρ only scales the orbital. It

does not truly affect the “shape”.

5. So ρ = Pn(

cos(φ))

should give the “shapes” of theorbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 77: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.2. y

(cos(φ)

)describes how the wave function of the electron

depends on the polar angle φ .3. The absolute value of the wave function does not depend

on the azimuthal angle θ .4. The dependence on the radius ρ only scales the orbital. It

does not truly affect the “shape”.5. So ρ = Pn

(cos(φ)

)should give the “shapes” of the

orbitals when Pn is a Legendre polynomial.

6. “Shape” must be carefully interpreted. Large values forρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 78: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

Electron Orbitals1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.2. y

(cos(φ)

)describes how the wave function of the electron

depends on the polar angle φ .3. The absolute value of the wave function does not depend

on the azimuthal angle θ .4. The dependence on the radius ρ only scales the orbital. It

does not truly affect the “shape”.5. So ρ = Pn

(cos(φ)

)should give the “shapes” of the

orbitals when Pn is a Legendre polynomial.6. “Shape” must be carefully interpreted. Large values for

ρ(φ) = Pn(φ) in the picture indicate a large probability(density) that the electron’s location’s polar angle is aroundthe angle φ .

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 79: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

ρ =∣∣P0(cos(φ))

∣∣: 1s Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 80: Legendre Polynomials

logo1

Overview Solving the Legendre Equation Application

ρ =∣∣P1(cos(φ))

∣∣: 2p Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 81: Legendre Polynomials

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Overview Solving the Legendre Equation Application

ρ =∣∣P2(cos(φ))

∣∣: 3d Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

Page 82: Legendre Polynomials

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Overview Solving the Legendre Equation Application

ρ =∣∣P3(cos(φ))

∣∣: 4f Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials