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

    )y2xy +

    ( m

    2

    1 x2

    )y = 0 arises when the

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

    (cos()

    )describes the polar part of the solution

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

    (1 x2

    )y2xy +y = 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)

    y2xy +(

    m2

    1 x2

    )y = 0 arises when the

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

    (cos()

    )describes the polar part of the solution

    of u = f ()u.

    2. The Legendre equation(1 x2

    )y2xy +y = 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)

    y2xy +(

    m2

    1 x2

    )y = 0 arises when the

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

    (cos()

    )describes the polar part of the solution

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

    (1 x2

    )y2xy +y = 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)

    y2xy +(

    m2

    1 x2

    )y = 0 arises when the

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

    (cos()

    )describes the polar part of the solution

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

    (1 x2

    )y2xy +y = 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)

    y2xy +(

    m2

    1 x2

    )y = 0 arises when the

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

    (cos()

    )describes the polar part of the solution

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

    (1 x2

    )y2xy +y = 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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0

    (1 x2

    ) n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn2

    2x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1

    +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn

    = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]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

    )y2xy+y = 0

    (1 x2

    )y2xy +y = 0(

    1 x2)

    n=2

    cnn(n1)xn22x

    n=1

    cnnxn1 +

    n=0

    cnxn = 0

    n=2

    cnn(n1)xn2

    n=2

    cnn(n1)xn

    n=1

    2cnnxn+

    n=0

    cnxn = 0

    k=0

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

    k=2

    ckk(k1)xk

    k=1

    2ckkxk+

    k=0

    ckxk = 0

    2c2 +c0 +6c3x2c1x+c1x+

    k=2

    [(k +2)(k +1)ck+2 k(k1)ck 2kck +ck

    ]xk = 0

    Bernd Schroder Louisiana Tech University, College of Engineering and Science

    Legendre Polynomials

  • logo1