Legendre Polynomials

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Overview Solving the Legendre Equation Application

Legendre Polynomials

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

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

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

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

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

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

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Overview Solving the Legendre Equation Application

ρ =∣∣P0(cos(φ))

∣∣: 1s Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

Legendre Polynomials

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Overview Solving the Legendre Equation Application

ρ =∣∣P1(cos(φ))

∣∣: 2p Orbital

Bernd Schroder Louisiana Tech University, College of Engineering and Science

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

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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