nLab Legendre polynomial

The Legendre polynomial $P_l$ (for $l =0,1,2,\ldots$) is the polynomial in one variable given by the formula

$$P_l(x) = \frac{1}{2^l l!} \frac{d^l}{d x^l}(x^2-1)^l$$

Alternatively they can be defined via a generating function:

$$\frac{1}{\sqrt{1-2tx+t^2}} = \sum_{n\geq 0} P_n(x) t^n$$

The Legendre polynomials satisfy:

• the following differential equation of the second order
  $$\frac{d}{d x}[(1-x^2)\frac{d P_l}{d x}] + l(l+1) P_l = 0$$
• the recursion relations
  $$(l+1)P_{l+1}-(2l+1)x P_l+l P_{l-1}=0$$
• the mixed differential recursion relations
  $$\array { P'_{l+1}-P'_{l-1} = (2l+1)P_l \\ P'_{l+1}-x P'_l = (l+1)P_l \\ (x^2-1)P_l'-l x P_l+l P_{l-1} = 0 }$$

The Legendre polynomials form a complete system of orthogonal polynomials on the interval $[-1,1]$, that is an unnormalised basis of the Hilbert space $L^2[-1,1]$.

The first few Legendre polynomials are $P_0(x) = 1$, $P_1(x) = x$, $P_2(x)=\frac{1}{2}(3x^2-1)$, $P_3(x)=\frac{1}{2}(5x^2-3)$, $P_4(x)=\frac{1}{8}(35x^4-30x^2+3)$. Their values at $0$ are

$$P_{2n+1}(0)=0,\,\,\,\,\,P_{2n}(0)=(-1)^n\frac{(2n-1)!!}{(2n)!} = \frac{(-1)^n (2n)!}{2^{2n}(n!)^2}$$

and $P_l(\pm 1)= (\pm 1)^l$.

One also has the integral formulas

$$\array { \int^1_0 P_{2k+1}(x) d x = \frac{(-1)^k (2k)!}{2^{2k+1} k! (k+1)!} \\ \int^1_0 P_{2k}(x) d x = \delta_{k0} \\ \int^1_{-1} x P_l P_k = \left\lbrace \array{ \frac{2(l+1)}{(2l+1)(2l+3)},&k=l+1 \\ \frac{(2l)}{(2l-1)(2l+1)},&k=l-1\\ 0,& otherwise} \right. }$$

A generalization of Legendre polynomials are the Legendre functions $P_\nu$ where $\nu$ is not necessarily an integer and $P^m_l$ which are given by

$$P^m_l(x) = \frac{(-1)^m}{2^l l!} (1-x^2)^{m/2} \frac{d^{l+m}}{d x^{l+m}} (x^2-1)^l = (-1)^m (1-x^2)^{m/2} \frac{d^{l}}{d x^{l}} P_l(x)$$

for $m\geq 0$ and also

$$P^{-m}_l(x)= (-1)^m\frac{(l-m)!}{(l+m)!} P^m_l(x)$$

These $P^m_l$ are satisfying the orthogonality relations

$$\int_{-1}^1 P^m_l(x) P^m_k(x) d x = \frac{2}{2l+1}\frac{(l+m)!}{(l-m)!}\delta_{lk}$$

recursion relation

$$x P^m_l(x) = \frac{l+|m|}{2l+1}P^m_{l-1}(x)+\frac{l-|m|+1}{2l+1} P^m_{l+1}(x)$$

and the differential equation

$$\frac{d}{d x}[(1-x^2)\frac{d P^m_l}{d x}]+[(l(l+1)-\frac{m^2}{1-x^2}]P^m_l = 0$$

$P_\nu(x)$ is a special case of a hypergeometric function, namely

$$P_\nu(x)={}_2 F_1(-\nu,\nu+1;1;\frac{1-x}{2})$$

Legendre polynomials enter the expressions for the spherical functions for sphere $S^2$ in 3d:

$$Y_{l m}(\theta,\phi) = \sqrt{ \frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!} } P^m_l(cos\theta) e^{im\phi}$$

(If it is not clear from mathML rendering – both fractions are under square root – including both the numerators and denominators).

If $(\theta,\phi)$ and $(\theta',\phi')$ are two points of the unit sphere in spherical coordinates (polar angle, azimuth), and $\gamma$ is the angle between the two corresponding rays from the origin then

$$P_l(cos \gamma) = \sum_{m=-l}^l (-1)^m P_l^m(cos \theta)P_l^{-m}(cos \theta') cos(m(\phi-\phi')) = \frac{4\pi}{2l+1}\sum_{m=-l}^l Y_{lm}(\theta,\phi) Y_{lm}^*(\theta',\phi')$$

what for $l=1$ reduces to the spherical law of cosine from spherical trigonometry:

$$cos \gamma = P_1(cos \gamma) = cos\theta cos\theta' + sin\theta sin\theta' cos(\phi-\phi')$$

The orthogonality relation for Legendre polynomials gives Laplace’s formula

$$\int_{S^2} d\Omega_{\hat{k}} Y_l(\hat{k}) P_l(\hat{k}\hat{p}) = \frac{4\pi}{2l+1} \delta_{ll'} Y_l(\hat{p})$$

where $\hat{k},\hat{p}$ are unit vectors and $Y_l = \sum a_m Y_{lm}$ is some spherical function.

The following orthogonality integral relation is over product of unit spheres in $\mathbb{R}^3$:

$$\int_{\hat{k}}\int_{\hat{p}} d\Omega_{\hat{k}}d\Omega_{\hat{p}} P_l(\hat{k}\hat{p}) P_{l'}(\hat{k}\hat{q}) P_{l''}(\hat{p}\hat{q}) = \left(\frac{4\pi}{2l+1}\right)\delta_{ll'}\delta_{ll''}, \,\,\,\,\,(*)$$

where the arguments of Legendre polynomials are the inner products of the unit vectors.

Zoran: the last formula $(*)$ is in my own formula notes which I have written as a student many years ago and used hundreds of times, but it now looks to me suspicious; I have no time to check it right now.