.. _legendre-polynomials:

Legendre Polynomials
====================


Legendre polynomails are solutions to *Legendre’s differential equation*,

.. math::

   \label{eq:legendrede}
     \frac{\dif{}}{\dif{x}} \left[ (1-x)^2 \frac{\dif{}}{\dif{x}} P_n(x) \right] + n(n+1) P_n(x) = 0

These are encountered frequently when solving Laplace’s equation in spherical coordinates.
This differential equation can be solved using a power series method.
The equation has regualr singular points at :math:`x=\pm 1`, so solutions only converge in the region :math:`|x|<1`.

When :math:`n` is an integer, the solution :math:`P_n(x)` that is regular at :math:`x=1` is also regular at :math:`x=-1`, and the solution terminates.
These solutions, :math:`P_n`, are called the *Legendre Polynomials*, with each being an :math:`n^{\rm th}`-degree polynomial, expressed by Rodrigues’ Formula:

.. math::

   \label{eq:rodrigues}
     P_n(x) = \frac{1}{2^n\ n!} \frac{\difp{n}{}}{\difp{n}{x}} \left[ (x^1-1)^n \right]


.. plot::
   :caption: The Legendre polynomials of degrees 0 to 5.

   import matplotlib.pyplot as plt
   import numpy as np
   from scipy.special import legendre

   n = 3
   x = np.arange(-1, 1, 0.01)

   for n in [0, 1, 2, 3, 4, 5]:
      plt.plot(x, legendre(n)(x), label=n)
   plt.legend()
   plt.tight_layout()
     
Legendre Polynomials from the Generating Function
-------------------------------------------------

.. glossary::
   Formal Power Series
      A formal power series is a generalisation of the concept of a polynomial, where the number of terms is allowed to be infinite.

A formal power series can be considered in the same way as a normal power series, but we ignore considerations of the convergence by assuming that a variable, :math:`X`, denotes any numerical value.
For example, the series

.. math:: A = 1 - 3X+ 5X^2 - 7X^3 + 9X^4 - 11X^5 + \cdots

As a power series, one property of this series is that it has a radius of convergence equal to 1.
As a formal power series the only relevent information is that the sequence of coefficients,
:math:`[1, -3, 5, -7, 9, -11, \dots]`.
Thus a formal power series merely records the sequence of coefficients.

.. glossary::
   Generating Function
      A generating function is a formal power series in one indeterminate, the coefficients of which encode information about a sequence of numbers, :math:`a_n`, which is indexed by natural numbers.

The Legendre Polynomials can be described by a generating function,
:math:`g(t,x)`,

.. math::

   \label{eq:legendregen}
     g(t,x) = \frac{1}{\sqrt{1- 2xt +t^2}} = \sum^{\infty}_{n=0} P_n(x) t^n

It is then possible to find expressions for the Legendre polynomials by expanding the square root in powers of :math:`t`, and equating coefficients:

.. math::

   \begin{aligned}
     (1-2xt+t^2)^{-\frac{1}{2}} &= 1 + \frac{1}{2}(2xt -t^2)+\frac{3}{8}(2xt-t^2)^2\\&\quad +\frac{5}{16}(2xt-t^2)^3 + \frac{35}{128}(2xt-t^2)^4 \\ &\quad + {\cal O}(t^5)\\
     \text{and,}\quad \sum^{\infty}_{n=0} P_n(x) t^n &= t^0 + xt^1 +
     \frac{1}{2}(3x^2-1)t^2 \\&\quad \frac{1}{2}(5 x^3-3)t^3 +
     \frac{1}{8}(35x^4-30x^2+3)t^4 \\&\quad + {\cal O}(t^5)\end{aligned}

so, by equating the appropriate powers of :math:`t`,

.. math::

   \begin{aligned}
     P_0(x) &= 1 \\
     P_1(x) &= x \\
     P_2(x) &= \frac{1}{2}(3 x^2 - 1) \\
     P_3(x) &= \frac{1}{2}(5 x^3 - 3x) \\
     P_4(x) &= \frac{1}{8}(35 x^4 - 30x^2 + 3)\end{aligned}


Parity of Legendre Polynomials
------------------------------

At :math:`x = \pm 1` the situation is especially simple;

.. math::

   \begin{aligned}
     g(t, \pm 1) &= \frac{1}{\sqrt{1 \mp 2t + t^2}} = \frac{1}{\sqrt{(1 \mp t)^2}} \\
     &= 1 \pm t +t^2 \pm t^3 + \cdots \\
     &= \sum_{n=0}^{\infty} (\pm 1)^n t^n\end{aligned}

 but also

.. math::

   \begin{aligned}
     g(t, \pm 1) &= \sum^{\infty}_{n=0} P_n(\pm 1) t^n \\
     P_n(1) &=1 \\
     P_n(-1)&=(-1)^n =
     \begin{cases}
       +1 & \text{ for } n \text{ even.} \\
       -1 & \text{ for } n \text{ odd.}
     \end{cases}\end{aligned}

 And

.. math::

   \begin{aligned}
     g(-t,-x) &= \frac{1}{\sqrt{1-2(-x)(-t)+(-t)^2}} = \frac{1}{\sqrt{1-2xt+t^2}} \\
     &= g(t,x)\end{aligned}

 Then, equating powers of :math:`t`,

.. math:: P_n(-x) = (-1)^nP_n(x)

Legendre Polynomials and Multipole Expansion
--------------------------------------------

Consider a point charge, :math:`q`, on the :math:`z`-axis, a distance
:math:`a` from the origin. The potential at an arbitrary point
:math:`\vec{r}` will be

.. math::

   \begin{aligned}
     \phi(\vec{r}) &= \frac{1}{4 \pi \epsilon_0} \frac{q}{d} = \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r} - a \hat{e}_z|} \\
                   &= \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{(\vec{r}-a \hat{e}_z)\cdot(\vec{r}-a \hat{e}_z)}}\\
                   &= \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{r^2-2ra \cos \theta + a^2}}\\
                   &= \frac{q}{4 \pi \epsilon_0 r} \qty[  1 - 2 \frac{a}{r} \cos \theta + \qty(\frac{a}{r})^2 ]^{-\frac{1}{2}}\\
                   &= \frac{q}{4 \pi \epsilon_0 r} \sum_{n=0}^{\infty} P_n(\cos \theta)
     \qty( \frac{a}{r})^n
     \end{aligned}

 Adding an extra point charge, :math:`-q` a distance :math:`a` on the
opposite size of the origin gives us

.. math::

   \begin{aligned}
      \phi(\vec{r})  &= \frac{1}{4 \pi \epsilon_0} \frac{q}{d_1} - \frac{1}{4 \pi \epsilon_0} \frac{q}{d_2} \\
                     &= \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r}-a \vec{e}_z|} - \frac{1}{4 \pi \epsilon_0} \frac{q}{|\vec{r}+a \vec{e}_z|} 
   \\                &= \frac{q}{4 \pi \epsilon_0 r} \bigg[ \left(1-2 \frac{a}{r} \cos \theta +\left(\frac{a}{r}\right)^2 \right)^{-\frac{1}{2}} 
   \\                &  \qquad \qquad -  \left(1-2 \frac{a}{r} \cos \theta +\left(\frac{a}{r}\right)^2 \right)^{-\frac{1}{2}} \bigg] 
   \\                &= \frac{q}{4 \pi \epsilon_0 r} \sum_{n=0}^{\infty} \qty( P_n(\cos\theta) \qty( \frac{a}{r} )^n - P_n(\cos \theta) \qty( \frac{-a}{r})^n )
   \\                &= \frac{2q}{4 \pi \epsilon_0 r} \qty( P_1 (\cos \theta) \frac{a}{r} + P_3 (\cos \theta) \qty(\frac{a}{r})^3 + \cdots )
       \end{aligned}

 so, only odd powers survive, and, for large enough :math:`r`,

.. math::

   \phi(\vec{r}) \approx \frac{2qa}{4 \pi \epsilon_0 r^2}
       P_1(\cos\theta)

 This is the potential from an electric dipole, and :math:`2qa` is the
dipole moment. The leading term in an expansion describes the
distribution:

.. math::

   \begin{aligned}
         \frac{1}{r} P_0 (\cos \theta) \left(\frac{a}{r}\right)^0 &= \frac{1}{r} & \text{(Monopole)} \\
         \frac{1}{r} P_1 (\cos \theta) \left(\frac{a}{r}\right)^1 &= \frac{a}{r^2} \cos \theta & \text{(Dipole)} \\
         \frac{1}{r} P_2 (\cos \theta) \left(\frac{a}{r}\right)^2 &= \frac{a^2}{2r^3} (3\cos^2 \theta - 1) & \text{(Quadrupole)} \\
         \frac{1}{r} P_3 (\cos \theta) \left(\frac{a}{r}\right)^3 &= \frac{a^3}{2r^4} (5\cos^3 \theta - 3 \cos \theta) & \text{(Octupole)} \\
       \end{aligned}

Recurrence Relations for Legendre Polynomials
---------------------------------------------

We can derive recurrence relations for the Legendre polynomials starting
by taking the derivative of the generating function, equation
([eq:legendregen]).

.. math::

   \begin{aligned}
         \frac{\partial g(t,x)}{\partial t} &= \frac{x-t}{(1-2xt+t^2)^{\frac{3}{2}}} = \sum_{n=0}^{\infty} P_n(x)nt^{n-1} \\
         &= \frac{x-t}{(1-2xt+t^2)}\frac{1}{\sqrt{1-2xt+t^2}} \\
         &= \frac{x-t}{1-2xt+t^2} \sum_{n=0}^{\infty}P_n(x) t^n \\
       \end{aligned}

 Thus

.. math::

   \begin{aligned}
         (1-2xt+t^2) \sum_{n=0}^{\infty} P_n(x) nt^{n-1} &= (x-t)
         \sum_{n=0}^{\infty} P_n(x) t^n
       \end{aligned}

 expanding,

.. math::

   \begin{aligned}
         \sum_{n=0}^{\infty} P_n(x) nt^{n-1} &- sx \sum_{n=0}^{\infty} P_n(x) nt^n + \sum_{n=0}^{\infty} P_n(x) nt^{n+1} \\
         &= x \sum_{n=0}^{\infty} P_n(x)t^n - \sum_{n=0}^{\infty} P_n(x)
         t^{n+1}
       \end{aligned}

 Then, relabelling,

.. math::

   \begin{aligned}
         \sum_{n=-1}^{\infty} P_{n+1}(x)(n+1) &- 2x \sum_{n=0}^{\infty} P_n(x) nt^n + \sum_{n=1}^{\infty} P_{n-1}(x)(n-1) \\
         &= x \sum_{n=0}^{\infty} P_n(x) t^n - \sum_{n=1}^{\infty}
         P_{n-1}(x) t^n
       \end{aligned}

 Equating powers of :math:`t^n` for :math:`n \ge 1`,

.. math::

   P_{n+1}(x)(n+1) - 2x P_n(x)n + P_{n-1}(x)(n-1) = xP_n(x) -
       P_{n-1}(x)

 Thus

.. math::

   (2n+1) x P_n(x) = (n+1) P_{n+1}(x) + nP_{n-1}(x) \qquad (n \ge
       1)

 This recurrence relation allows the calculation of Lengendre
polynomials using a recursive function. Taking the derivative with
respect to :math:`x` instead,

.. math::

   \begin{aligned}
     \frac{\partial g(t,x)}{\partial x} &= \frac{t}{(1-2xt +t^2)^{\frac{3}{2}}} \\
     &= \sum_{n=0}^{\infty} P^{\prime}_n(x)t^n \\
     &= \frac{t}{1-2xt+t^2} \frac{1}{\sqrt{1-2xt+t^2}} \\
     &= \frac{t}{1-2xt+t^2} \sum_{n=0}^{\infty} P_n(x) t^n \\
     (1-2xt+t^2) \sum_{n=0}^{\infty} P^{\prime}_n(x)t^n &= t
     \sum_{n=0}^{\infty} P_n(x) t^n\end{aligned}

.. math::

   P^{\prime}_{n+1}(x) + P^{\prime}_{n-1}(x) = 2x P_n^{\prime}(x) +
   P_n(x)

Orthogonality and Completeness of the Legendre Polynomials
----------------------------------------------------------

It is possible to show that the Legendre Polynomials are orthogonal by
considering the Legendre equation, equation ([eq:legendrede]).

.. math::

   \begin{aligned}
     P_m(x) & \textcolor{accent-red}{\frac{\dif{}}{\dif{x}} \left[ (1-x^2) \frac{\dif{}}{\dif{x}}P_n(x)\right]} - P_n(x) \textcolor{accent-blue}{\frac{\dif{}}{\dif{x}}\left[ (1-x^2) \frac{\dif{}}{\dif{x}}P_m(x) \right]} \\
     &= - P_m(x) \textcolor{accent-red}{n(n+1)P_n(x)}+P_n(x)
     \textcolor{accent-blue}{m(m+1)P_m(x)}\end{aligned}

 Now, integrating :math:`x` over the range :math:`[-1, 1]`,

.. math::

   \begin{aligned}
     \int_{-1}^1& \textcolor{accent-blue}{P_m(x)} \frac{\dif{}}{\dif{x}}
     \left[ \textcolor{accent-red}{(1-x^2) \frac{\dif{}}{\dif{x}} P_n(x)}
     \right] \dif{x} \\= & \underbrace{\left[
         \textcolor{accent-blue}{P_m(x)} \textcolor{accent-red}{(1-x)^2
           \frac{\dif{}}{\dif{x}}P_n(x)} \right]^1_{-1}}_{= 0} \\ &-
     \underbrace{\int_{-1}^1 \left[ \frac{\dif{}}{\dif{x}}
         \textcolor{accent-blue}{P_m(x)}\textcolor{accent-red}{(1-x^2)
           \frac{\dif{}}{\dif{x}} P_n(x)} \right]
       \dif{x}}_{\text{symmetric in n,m}} \\0 & = [m(m+1) - n(n+1)]
     \int_{-1}^1 P_n(x) P_m(x) \dif{x}\end{aligned}

 Then, for :math:`n \neq m`,

.. math:: \int_{-1}^1 P_n(x) P_m(x) \dif{x} = 0

 So Legendre polynomials are orthogonal over the region
:math:`x \in [-1, 1]` When :math:`n=m`, we return to the generating
function,

.. math::

   \sum_{n=0}^{\infty} P_n(x)t^n \sum_{m=0}^{\infty} P_m(x)t^m =
   \frac{1}{1-2xt+t^2}

 Integrating over :math:`x`,

.. math::

   \begin{aligned}
     \int_{-1}^1 \frac{1}{1-2xt+t^2} \dif{x} &= \left[ - \frac{1}{2t}
       \log (1-2xt+t^2) \right]^1_{-1} \\ &= \frac{1}{t} \log \left(
       \frac{1+t}{1-t} \right) \\ &= 2 \sum_{n=1}^{\infty}
     \frac{t^{2n}}{2n+1}\end{aligned}

.. math::

   \begin{aligned}
     \int_{-1}^1 \sum_{n=0}^{\infty} P_n(x)t^n \sum_{m=0}^{\infty} P_m(x)
     t^m \dif{x} &= \sum_{n=0}^{\infty} \int_{-1}^1 [P_n(x)]^2 t^{2n}
     \dif{x}\end{aligned}

and equating powers of :math:`t`,

.. math::

   \begin{aligned}
     \int_{-1}^1 [P_n(x)]^2 \dif{x} = \frac{2}{2n + 1}\end{aligned}

 And putting these relations together we get an orthogonality and
normalisation condition

.. math::

   \label{eq:orthonormlegend}
     \int_{-1}^1 P_n(x) P_m(x) \dif{x} = \frac{2}{2n+1} \delta_{nm}

 Legendre polynomials are also complete—any continuous function can be
expressed as an infinite sum of Legendre polynomials in :math:`x \in
[-1,1]`. Taking a function :math:`f(x)`, then

.. math::

   \label{eq:legendreseries}
     f(x) = \sum_{n=0}^{\infty} c_n P_n(x)

 Then,

.. math::

   \begin{aligned}
     \int_{-1}^1 f(x) P_m(x) \dif{x} &= \sum_{n=0}^{\infty} c_n
     \int_{-1}^1 P_n(x) P_m(x) \dif{x} \\ &= \sum_{n=0}^{\infty} c_n
     \frac{2}{2m+1} \delta_{nm} \\ &= c_m \frac{2}{2m+1}\end{aligned}

 So,

.. math::

   \label{eq:legendreseriesoffunc}
     f(x) = \sum_{n=0}^{\infty} \left( n + \frac{1}{2} \right) \left( \int_{-1}^1 f(y) P_n(y) \dif{y} \right) P_n(x)

| *Expanding the step function as a series of Legendre polynomials.*

[scale=1.0]

[width=, height=2in, xmin=-1, xmax=1, samples=50] gnuplot[raw gnuplot,
id=leg1, mark=none, domain=-1:1, muted-blue, ultra thick] set xrange
[-1:1]; step(x) = (x>0) ? 1 : 0; plot step(x); ;

| 
| We have the definition of a Legendre series from equation
  ([eq:legendreseriesoffunc]) as

  .. math:: f(x) = \sum_l c_l P_l(x)

   then

  .. math::

     \begin{aligned}
     \int_{-1}^1 f(x) P_m(x) \dd{x} &= \sum_{l=0}^{\infty} c_l \int_{-1}^1 P_l(x) P_m(x) \dd{x} \\&= c_m \frac{2}{2m+1}\end{aligned}

   and so

  .. math:: c_l = \frac{2l+1}{2} \int_{-1}^1 f(x) P_l(x) \dd{x}

   now

  .. math::

     \begin{aligned}
         c_0 &= \frac{1}{2} \int_0^1 P_0(x) \dd{x} = \frac{1}{2} \\
         c_1 &= \frac{3}{2} \int_0^1 P_1(x) \dd{x} = \frac{3}{4} \\
       \end{aligned}

   and so

  .. math:: f(x) = \frac{1}{2} P_0(x) + \frac{3}{4} P_1(x) - \frac{7}{16} P_3(x) + \frac{11}{32} P_5(x) + \cdots

[scale=1.0]

[width=, height=2in, xmin=-1, xmax=1, samples=50] gnuplot[raw gnuplot,
id=leg1, mark=none, domain=-1:1, muted-orange, ultra thick] set xrange
[-1:1]; leg(n,x) = (n==0) ? 1 : (n==1) ? x : ((2\*n+1)\*x\*leg(n-1, x) -
n\*leg(n-2, x))/(n+1); plot (
0.5\*leg(0,x)+0.75\*leg(1,x)-(0.4375)\*leg(3,x)+(0.34375)\*leg(5,x) ); ;

Associated Legendre Polynomials
--------------------------------

Associated Legendre polynomials are obtained by differentiating a
standard Legendre polynomial :math:`m` times, with respect to :math:`x`.

.. math::

   \label{eq:definitionassocleg}
     P_n^m(x) = (1-x^2)^{\frac{m}{2}} \frac{\dif{}^m}{\dif{x}^m} P_n(x)

 these are solutions of the associate Legendre equation,

.. plot::
   :caption: The Assosciate Legendre polynomials of degrees 0 to 5 and order 2.

   import matplotlib.pyplot as plt
   import numpy as np
   from scipy.special import lpmv

   x = np.arange(-1, 1, 0.01)
   v = 2
   
   for m in [0, 1, 2, 3, 4, 5]:
      plt.plot(x, lpmv(m, v, x, out=None), label=m)
   plt.legend()
   plt.tight_layout()
     
 
.. math::

   \label{eq:assoclegendrede}
     \frac{\dif{}}{\dif{x}} \left[ (1-x^2) \frac{\dif{P_n^m(x)}}{\dif{x}} \right] + n(n+1)P_n^m(x) - \frac{m^2}{1-x^2} P_n^m(x) = 0


Then,

.. math:: P_0(x) = 1 \therefore P_0^0(x) = 1

.. math:: P_1(x) = x \therefore P_1^0(x) = x

.. math:: P_1^1(x) = (1-x^2)^{\frac{1}{2}}

There are different conventions for negative values of :math:`m`, but
since the only dependence on :math:`m` is an :math:`m^2` term, we can
take them to be equal. If :math:`x = \cos (\theta)`,

.. math:: P_1^0(x) = \cos(\theta)

.. math:: P_1^{\pm 1} = \sin(\theta)

 The associated Legendre polynomials are also orthogonal,

.. math::

   \label{eq:orthogonalityassoclag}
     \int_{-1}^1 P_l^m(x) P_n^m(x) \dd{x} = \frac{(l+m)!}{(l-m)!} \frac{2}{2l+1} \delta_{ln}