5.11: Legendre Polynomials
- Page ID
- 8156
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In this section we cover just enough about Legendre polynomials to be useful in the following section. Before starting, I want you to expand the following expression, by the binomial theorem, for \(| x |< 1\), up to \(x^4\) :
\[\dfrac{1}{(1-2x \cos θ + x^2)^{1/2}}. \label{5.11.1} \tag{5.11.1}\]
Please do go ahead and do it. Well, you probably won’t, so I’d better do it myself:
I’ll start with
\[(1-X)^{-1/2} = 1 + \dfrac{1}{2} X + \dfrac{3}{8} X^2 + \dfrac{5}{16} X^3 + \dfrac{35}{128} X^4 ... \label{5.11.2} \tag{5.11.2}\]
and therefore
\[ [1-x(2\cos θ - x)]^{-1/2} = 1 + \dfrac{1}{2} x (2\cos θ - x) + \dfrac{3}{8} x^2 (2\cos θ - x)^2 + \dfrac{5}{16} x^3 (2\cos θ - x)^3 + \dfrac{35}{128} x^4 (2\cos θ - x)^4 ... \label{5.11.3} \tag{5.11.3}\]
\[= 1 + x \cos θ - \dfrac{1}{2} x^2 + \dfrac{3}{8}x^2 (4 \cos^2 θ - 4x \cos θ + x^2 ) + \dfrac{5}{16} x^3 (8 \cos^3 θ - 12x \cos^2 θ + 6x^2 \cos θ - x^3) + \dfrac{35}{128} x (16 \cos^4 θ - 32x \cos^3 θ + 24x^2 \cos^2 θ - 8x^3 \cos θ + x^4 ) ... \label{5.11.4} \tag{5.11.4}\]
\[= 1 + x \cos θ + x^2 (-\dfrac{1}{2} + \dfrac{3}{2} \cos^2 θ ) + x^3 (-\dfrac{3}{2} \cos θ + \dfrac{5}{2} \cos^3 θ ) + x^4 (\dfrac{3}{8} - \dfrac{15}{4} \cos^2 θ + \dfrac{35}{8} \cos^4 θ)... \label{5.11.5} \tag{5.11.5}\]
The coefficients of the powers of \(x\) are the Legendre polynomials \(P_l(\cos θ )\), so that
\[\dfrac{1}{(1-2x \cos θ + x^2)^{1/2}} = 1 + x P_1 ( \cos θ) + x^2 P_2 (\cos θ) + x^3 P_3 (\cos θ) + x^4 P_4 (\cos θ) + ... \label{5.11.6} \tag{5.11.6}\]
The Legendre polynomials with argument \(\cos θ\) can be written as series of terms in powers of \(\cos θ\) by substitution of \(\cos θ\) for \(x\) in Equations 1.12.5 in Section 1.12 of Chapter 1. Note that \(x\) in Section 1 is not the same as \(x\) in the present section. Alternatively they can be written as series of cosines of multiples of \(θ\) as follows.
\begin{array}{l}
P_0 = 1 \\
P_1 = \cos θ \\
P_2 = \dfrac{1}{4} (3\cos 2θ + 1) \\
P_3 = \dfrac{1}{8} (5\cos 3θ + 3\cos θ) \\
P_4 = \dfrac{1}{64} (35 \cos 4 θ + 20 \cos 2 θ + 9 ) \\
P_5 = \dfrac{1}{128} (63 \cos 5 θ + 35 \cos 3 θ + 30 \cos θ) \\
P_6 = \dfrac{1}{512} (231 \cos 6 θ + 126 \cos 4θ + 105 \cos 2θ + 50) \\
P_7 = \dfrac{1}{1024} (429 \cos 7θ + 231 \cos 5θ + 189 \cos 3θ + 175 \cos θ) \\
P_8 = (6435 \cos 8θ + 3432 \cos 6θ + 2772 \cos 4θ + 2520 \cos 2θ + 1225)/2^{14} \\
\label{5.11.7} \tag{5.11.7}
\end{array}
For example, \(P_6(\cos θ)\) can be written either as given by Equation \ref{5.11.7}, or as given by Equation 1, namely
\[P_6 = \dfrac{1}{16} (231c^6 - 315 c^4 + 105c^2 - 5), \text{ where } c = \cos θ. \label{5.11.8} \tag{5.11.8}\]
The former may look neater, and the latter may look “awkward” because of all the powers. However, the latter is far faster to compute, particularly when written as nested parentheses:
\[P_6 = (-5 + C(105 + C(-315 + 231C)))/16, \text{ where } C = \cos^2 θ. \label{5.11.9} \tag{5.11.9}\]