# Legendre polynomials

## Particular values

$P_{0}\!\left(z\right) = 1$
$P_{1}\!\left(z\right) = z$
$P_{2}\!\left(z\right) = \frac{1}{2} \left(3 {z}^{2} - 1\right)$
$P_{3}\!\left(z\right) = \frac{1}{2} \left(5 {z}^{3} - 3 z\right)$
$P_{4}\!\left(z\right) = \frac{1}{8} \left(35 {z}^{4} - 30 {z}^{2} + 3\right)$
$P_{5}\!\left(z\right) = \frac{1}{8} \left(63 {z}^{5} - 70 {z}^{3} + 15 z\right)$
$P_{n}\!\left(1\right) = 1$
$P_{n}\!\left(-1\right) = {\left(-1\right)}^{n}$
$P_{2 n}\!\left(0\right) = \frac{{\left(-1\right)}^{n}}{{4}^{n}} {2 n \choose n}$
$P_{2 n + 1}\!\left(0\right) = 0$

## Recurrence and functional equations

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

## Generating functions

$\sum_{n=0}^{\infty} P_{n}\!\left(x\right) {z}^{n} = \frac{1}{\sqrt{1 - 2 x z + {z}^{2}}}$

## Rodrigues' formula

$P_{n}\!\left(z\right) = \frac{1}{{2}^{n} n !} \left[ \frac{d^{n}}{{d t}^{n}} {\left({t}^{2} - 1\right)}^{n} \right]_{t = z}$

## Integrals

$\int_{-1}^{1} P_{n}\!\left(x\right) P_{m}\!\left(x\right) \, dx = \frac{2}{2 n + 1} \delta_{(n,m)}$

## Sum representations

$P_{n}\!\left(z\right) = \frac{1}{{2}^{n}} \sum_{k=0}^{n} {{n \choose k}}^{2} {\left(z - 1\right)}^{n - k} {\left(z + 1\right)}^{k}$
$P_{n}\!\left(z\right) = \sum_{k=0}^{n} {n \choose k} {n + k \choose k} {\left(\frac{z - 1}{2}\right)}^{k}$
$P_{n}\!\left(z\right) = \frac{1}{{2}^{n}} \sum_{k=0}^{\left\lfloor n / 2 \right\rfloor} {\left(-1\right)}^{k} {n \choose k} {2 n - 2 k \choose n} {z}^{n - 2 k}$

## Hypergeometric representations

$P_{n}\!\left(z\right) = \,{}_2F_1\!\left(-n, n + 1, 1, \frac{1 - z}{2}\right)$
$P_{n}\!\left(z\right) = {2 n \choose n} {\left(\frac{z}{2}\right)}^{n} \,{}_2F_1\!\left(-\frac{n}{2}, \frac{1 - n}{2}, \frac{1}{2} - n, \frac{1}{{z}^{2}}\right)$
$P_{n}\!\left(z\right) = {\left(\frac{z - 1}{2}\right)}^{n} \,{}_2F_1\!\left(-n, -n, 1, \frac{z + 1}{z - 1}\right)$
$P_{2 n}\!\left(z\right) = \frac{{\left(-1\right)}^{n}}{{4}^{n}} {2 n \choose n} \,{}_2F_1\!\left(-n, n + \frac{1}{2}, \frac{1}{2}, {z}^{2}\right)$
$P_{2 n + 1}\!\left(z\right) = \frac{{\left(-1\right)}^{n}}{{4}^{n}} \left(2 n + 1\right) {2 n \choose n} z \,{}_2F_1\!\left(-n, n + \frac{3}{2}, \frac{3}{2}, {z}^{2}\right)$

## Bounds and inequalities

$\left|P_{n}\!\left(x\right)\right| \le 1$
$\left|P_{n}\!\left(x\right)\right| \le 2 I_{0}\!\left(2 n \sqrt{\frac{\left|x - 1\right|}{2}}\right) \le 2 {e}^{2 n \sqrt{\left|x - 1\right| / 2}}$
$\left|P_{n}\!\left(z\right)\right| \le \left|P_{n}\!\left(\left|z\right| i\right)\right| \le {\left(\left|z\right| + \sqrt{1 + {\left|z\right|}^{2}}\right)}^{n}$
$\left|\frac{d}{d x}\, P_{n}\!\left(x\right)\right| \le \frac{n \left(n + 1\right)}{2}$
$\left|\frac{d}{d x}\, P_{n}\!\left(x\right)\right| \le \frac{{2}^{3 / 2}}{\sqrt{\pi}} \frac{{n}^{1 / 2}}{{\left(1 - {x}^{2}\right)}^{3 / 4}}$
$\left|\frac{d^{2}}{{d x}^{2}} P_{n}\!\left(x\right)\right| \le \frac{\left(n - 1\right) n \left(n + 1\right) \left(n + 2\right)}{8}$
$\left|\frac{d^{2}}{{d x}^{2}} P_{n}\!\left(x\right)\right| \le \frac{{2}^{5 / 2}}{\sqrt{\pi}} \frac{{n}^{3 / 2}}{{\left(1 - {x}^{2}\right)}^{5 / 4}}$
$\left|\frac{d^{r}}{{d x}^{r}} P_{n}\!\left(x\right)\right| \le \frac{{2}^{r + 1 / 2}}{\sqrt{\pi}} \frac{{n}^{r - 1 / 2}}{{\left(1 - {x}^{2}\right)}^{\left( 2 n + 1 \right) / 4}}$

## Analytic properties

$\operatorname{HolomorphicDomain}\!\left(P_{n}\!\left(z\right), z, \mathbb{C} \cup \left\{{\tilde \infty}\right\}\right) = \mathbb{C}$
$\operatorname{Poles}\!\left(P_{n}\!\left(z\right), z, \mathbb{C} \cup \left\{{\tilde \infty}\right\}\right) = \left\{{\tilde \infty}\right\}$
$\operatorname{EssentialSingularities}\!\left(P_{n}\!\left(z\right), z, \mathbb{C} \cup \left\{{\tilde \infty}\right\}\right) = \left\{\right\}$
$\operatorname{BranchPoints}\!\left(P_{n}\!\left(z\right), z, \mathbb{C} \cup \left\{{\tilde \infty}\right\}\right) = \left\{\right\}$
$\operatorname{BranchCuts}\!\left(P_{n}\!\left(z\right), z, \mathbb{C}\right) = \left\{\right\}$
$\left|\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} P_{n}\!\left(z\right)\right| = n$
$\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} P_{n}\!\left(z\right) \subset \left(-1, 1\right)$
$\mathop{\operatorname{zeros}\,}\limits_{z \in \mathbb{C}} P_{n}\!\left(z\right) = \left\{ x_{n,k} : k \in \{1, 2, \ldots n\} \right\}$
$P_{n}\!\left(\overline{z}\right) = \overline{P_{n}\!\left(z\right)}$

$w_{n,k} = \frac{2}{\left(1 - {\left(x_{n,k}\right)}^{2}\right) {\left(\left[ \frac{d}{d t}\, P_{n}\!\left(t\right) \right]_{t = x_{n,k}}\right)}^{2}}$
$\left|\int_{-1}^{1} f\!\left(t\right) \, dt - \sum_{k=1}^{n} w_{n,k} f\!\left(x_{n,k}\right)\right| \le \frac{64 M}{15 \left(1 - {\rho}^{-2}\right) {\rho}^{2 n}}\; \text{ where } M = \mathop{\operatorname{sup}}\limits_{t \in \mathcal{E}_{\rho}} \left|f\!\left(t\right)\right|$