[PPT] - Legendre Polynomials Bernd Schr oder logo1 Bernd Schr oder PowerPoint Presentation

SLIDE 1

logo1 Overview Solving the Legendre Equation Application

Legendre Polynomials

Bernd Schr¨

der

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 2

logo1 Overview Solving the Legendre Equation Application

Why are Legendre Polynomials Important?

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 3

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 variables in spherical coordinates. (QM: hydrogen atom!) The function y

cos(φ)
describes the polar part of the solution
f ∆u = f(ρ)u.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 4

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 variables in spherical coordinates. (QM: hydrogen atom!) The function y

cos(φ)
describes the polar part of the solution
f ∆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 to the generalized Legendre equation.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 5

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 variables in spherical coordinates. (QM: hydrogen atom!) The function y

cos(φ)
describes the polar part of the solution
f ∆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 to the generalized Legendre equation.

3. The solutions of both equations must be finite on [−1,1].

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 6

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 variables in spherical coordinates. (QM: hydrogen atom!) The function y

cos(φ)
describes the polar part of the solution
f ∆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 to the generalized Legendre equation.

3. The solutions of both equations must be finite on [−1,1].
4. Because 0 is an ordinary point of the equation, it is natural

to attempt a series solution.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 7

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 8

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 9

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 10

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 11

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 12

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 13

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 14

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 15

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 16

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 17

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 18

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 19

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 20

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 21

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk+

∞

∑

k=0

λckxk

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 22

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk+

∞

∑

k=0

λckxk =

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 23

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk+

∞

∑

k=0

λckxk = 2c2 +λc0 +6c3x−2c1x+λc1x+

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 24

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk+

∞

∑

k=0

λckxk = 2c2 +λc0 +6c3x−2c1x+λc1x+

∞

∑

k=2

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 25

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

1−x2

y′′ −2xy′ +λy =

1−x2 ∞

∑

n=2

cnn(n−1)xn−2 −2x

∞

∑

n=1

cnnxn−1 +λ

∞

∑

n=0

cnxn =

∞

∑

n=2

cnn(n−1)xn−2−

∞

∑

n=2

cnn(n−1)xn−

∞

∑

n=1

2cnnxn+

∞

∑

n=0

λcnxn =

∞

∑

k=0

ck+2(k +2)(k +1)xk−

∞

∑

k=2

ckk(k −1)xk−

∞

∑

k=1

2ckkxk+

∞

∑

k=0

λckxk = 2c2 +λc0 +6c3x−2c1x+λc1x+

∞

∑

k=2

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

=

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 26

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 27

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

=

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 28

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

= Matching coefficients leads to c2 = −λ 2 c0,

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 29

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

= Matching coefficients leads to c2 = −λ 2 c0, c3 = −λ −2 6 c1,

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 30

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

= Matching coefficients leads to c2 = −λ 2 c0, c3 = −λ −2 6 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 Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 31

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 32

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 for

any λ.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 33

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 for

any λ.

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

both.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 34

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 for

any λ.

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

both.

3. Therefore, these solutions are not physically feasible,

because the polar part y

cos(φ)
f the solution of

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 35

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 for

any λ.

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

both.

3. Therefore, these solutions are not physically feasible,

because the polar part y

cos(φ)
f the solution of

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

4. Thus the only series solutions of interest are those that

terminate after finitely many steps.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 36

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 for

any λ.

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

both.

3. Therefore, these solutions are not physically feasible,

because the polar part y

cos(φ)
f the solution of

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

4. Thus the only series solutions of interest are those that

terminate after finitely many steps. Or, in simpler language, those solutions that happen to be polynomials.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 37

logo1 Overview Solving the Legendre Equation Application

Series Solution of

1−x2

y′′ −2xy′ +λy = 0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 38

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 of

coefficients terminates in repeating zeros.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 39

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 of

coefficients 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 nonnegative integer, then cl+2 = 0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 40

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 of

coefficients 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 nonnegative integer, then cl+2 = 0 and all entries cl+2k = 0.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 41

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 of

coefficients 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 nonnegative integer, then cl+2 = 0 and all entries cl+2k = 0. (Otherwise no such thing happens, so we need λ = l(l+1).)

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 42

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 of

coefficients 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 nonnegative integer, then cl+2 = 0 and all entries cl+2k = 0. (Otherwise no such thing happens, so we need λ = l(l+1).)

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

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 43

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 of

coefficients 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 nonnegative integer, then cl+2 = 0 and all entries cl+2k = 0. (Otherwise no 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-numbered coefficients) equal to zero.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 44

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 of

coefficients 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 nonnegative integer, then cl+2 = 0 and all entries cl+2k = 0. (Otherwise no 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-numbered coefficients) equal to zero.

9. So the solutions we are interested in will be polynomials

with even powers (for λ = l(l+1) and l even) or polynomials with odd powers (for λ = l(l+1) and l odd).

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 45

logo1 Overview Solving the Legendre Equation Application

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

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der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 46

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 47

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 48

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 49

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 50

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 51

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 Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 52

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients, l Even, ck+2 = −(l−k)(l+k +1) (k +2)(k +1) ck

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der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 53

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 54

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 55

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

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 56

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)nl(l−2)···

l−2(n−1)
·(l+2n−1)(l+2n−3)···(l+1)

(2n)! c0

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 57

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Closed Formula

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 58

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)! 2

l 2 l

2

!

2

l 2 l

2

!

c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 59

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)! 2

l 2 l

2

!

2

l 2 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)
2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 60

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)! 2

l 2 l

2

!

2

l 2 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)
2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n−1)(l+2n−3)···(l+1) (2n)! l!2

l 2 +n l

2 +n

!

l!2

l 2 +n l

2 +n

!

2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 61

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)! 2

l 2 l

2

!

2

l 2 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)
2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n−1)(l+2n−3)···(l+1) (2n)! l!2

l 2 +n l

2 +n

!

l!2

l 2 +n l

2 +n

!

2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n)! (2n)! 2

l 2 l

2

!

l!2

l 2+n l

2 +n

!

2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 62

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)! 2

l 2 l

2

!

2

l 2 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)
2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n−1)(l+2n−3)···(l+1) (2n)! l!2

l 2 +n l

2 +n

!

l!2

l 2 +n l

2 +n

!

2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n)! (2n)! 2

l 2 l

2

!

l!2

l 2+n l

2 +n

!

2

l 2 l

2

!

2

l 2−n l

2 −n

!

c0 = (−1)n (l+2n)! (2n)!l! l

2

!

2 l

2 +n

!

l

2 −n

!c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 63

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 64

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n(l+2n)! (2n)!l! l

2

!

2 l

2 +n

!

l

2 −n

!c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 65

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n(l+2n)! (2n)!l! l

2

!

2 l

2 +n

!

l

2 −n

!c0

cl−2k = (−1)

l−2k 2 (l+l−2k)!

(l−2k)!l! l

2

!

2 l

2 + l−2k 2

!

l

2 − l−2k 2

!c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 66

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n(l+2n)! (2n)!l! l

2

!

2 l

2 +n

!

l

2 −n

!c0

cl−2k = (−1)

l−2k 2 (l+l−2k)!

(l−2k)!l! l

2

!

2 l

2 + l−2k 2

!

l

2 − l−2k 2

!c0

= (−1)

l 2(−1)−k (2l−2k)!

(l−2k)!l! l

2

!

2 (l−k)!k!c0

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 67

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

c2n = (−1)n(l+2n)! (2n)!l! l

2

!

2 l

2 +n

!

l

2 −n

!c0

cl−2k = (−1)

l−2k 2 (l+l−2k)!

(l−2k)!l! l

2

!

2 l

2 + l−2k 2

!

l

2 − l−2k 2

!c0

= (−1)

l 2(−1)−k (2l−2k)!

(l−2k)!l! l

2

!

2 (l−k)!k!c0 = (−1)k (2l−2k)! 2l(l−2k)!(l−k)!k!(−1)

l 2 2l l

2

!

2 l! c0

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 68

logo1 Overview Solving the Legendre Equation Application

Even-Numbered Coefficients: Re-scaling

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 69

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)

l 2 2l l

2

!

2 l! c0

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 70

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)

l 2 2l l

2

!

2 l! c0 Choosing c0 = (−1)

l 2

l! 2l l

2

!

2 gives Pl(x) = ⌊ l

2⌋

∑

k=0

(−1)k (2l−2k)! 2lk!(l−k)!(l−2k)!xl−2k, which is the customary way the Legendre polynomials are stated.

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 71

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)

l 2 2l l

2

!

2 l! c0 Choosing c0 = (−1)

l 2

l! 2l l

2

!

2 gives Pl(x) = ⌊ l

2⌋

∑

k=0

(−1)k (2l−2k)! 2lk!(l−k)!(l−2k)!xl−2k, which is the customary way the Legendre polynomials are stated. (The generalized Legendre equation is good reading.)

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 72

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

Bernd Schr¨

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 73

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. As noted earlier, the Legendre equation arises in the

quantum mechanical analysis of the hydrogen atom.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 74

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. 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.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 75

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. 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
n the azimuthal angle θ.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 76

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. 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
n the azimuthal angle θ.
4. The dependence on the radius ρ only scales the orbital. It

does not truly affect the “shape”.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 77

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. 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
n 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
rbitals when Pn is a Legendre polynomial.

Bernd Schr¨

der

Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 78

logo1 Overview Solving the Legendre Equation Application

Electron Orbitals

1. 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
n 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
rbitals 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 around the angle φ.

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 79

logo1 Overview Solving the Legendre Equation Application

ρ =

P0(cos(φ))
: 1s Orbital

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 80

logo1 Overview Solving the Legendre Equation Application

ρ =

P1(cos(φ))
: 2p Orbital

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 81

logo1 Overview Solving the Legendre Equation Application

ρ =

P2(cos(φ))
: 3d Orbital

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials

SLIDE 82

logo1 Overview Solving the Legendre Equation Application

ρ =

P3(cos(φ))
: 4f Orbital

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Louisiana Tech University, College of Engineering and Science Legendre Polynomials