11.2: Properties of Legendre Polynomials

Let $F(x,t)$ be a function of the two variables $x$ and $t$ that can be expressed as a Taylor’s series in $t$, $\sum_{n} c_n(x) t^{n}$. The function $F$ is then called a generating function of the functions $c_{n}$.

Exercise $\PageIndex{1}$

Show that $F(x,t) = \frac{1}{1-xt}$ is a generating function of the polynomials $x^{n}$.

Answer

Look at $$\frac{1}{1-xt} = \sum_{n=0}^{\infty} x^{n}t^{n}\;\;(|xt|<1). \nonumber$$

Exercise $\PageIndex{2}$

Show that $F(x,t) = \exp\left(\frac{tx-t}{2t}\right)$ is the generating function for the Bessel functions,

$$F(x,t) = \exp\left(\frac{tx-t}{2t}\right) = \sum_{n=0}^{\infty} J_{n}(x)t^{n}.\nonumber$$

Answer

TBA

Exercise $\PageIndex{4}$

(The case of most interest here) $$F(x,t) =\frac{1}{\sqrt{1-2xt+t^{2}}} = \sum_{n=0}^{\infty} P_{n}(x) t^{n}.\nonumber$$

Answer

TBA

Rodrigues’ Generating Formula

$$P_{n}(x) = \frac{1}{2^{n} n!} \frac{d^{n}}{dx^{n}} (x^2-1)^n. \label{Rodrigues}$$

properties of Legendre Polynomials

1. $P_{n}(x)$ is even or odd if $n$ is even or odd.
2. $P_{n}(1)=1$.
3. $P_{n}(-1)=(-1)^{n}$.
4. $(2n+1) P_{n}(x) = P'_{n+1}(x)-P'_{n-1}(x)$.
5. $(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x)$.
6. $\int_{-1}^{x} P_n(x') dx'= \frac{1}{2n+1} \left[P_{n+1}(x)-P_{n-1}(x)\right]$.

Let us prove some of these relations, first Rodrigues’ formula (Equation $\ref{Rodrigues}$). We start from the simple formula

$$(x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}=0, \nonumber$$

which is easily proven by explicit differentiation. This is then differentiated $n+1$ times,

$$\begin{aligned} { \frac{d^{n+1}}{dx^{n+1}}\left[ (x^{2}-1) \frac{d}{dx} (x^{2}-1)^{n} - 2 n x (x^{2}-1)^{n}\right]} &= n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2(n+1) x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &-2n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n - 2n x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n \nonumber\\ &=-n(n+1) \frac{d^{n}}{dx^{n}}(x^2-1)^n + 2 x \frac{d^{n+1}}{dx^{n+1}} (x^2-1)^n+(x^2-1) \frac{d^{n+2}}{dx^{n+2}} (x^2-1)^n \nonumber\\ &= -\left[\frac{d}{dx}(1-x^2)\frac{d}{dx}\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n \right\} +n(n+1)\left\{\frac{d^{n}}{dx^{n}}(x^2-1)^n\right\}\right] =0.\end{aligned} \nonumber$$

We have thus proven that $\frac{d^n}{dx^n}(x^2-1)^n$ satisfies Legendre’s equation. The normalization follows from the evaluation of the highest coefficient,

$$\frac{d^n}{dx^n} x^{2n} = \frac{2n!}{n!} x^n, \nonumber$$

and thus we need to multiply the derivative with $\frac{1}{2^n n!}$ to get the properly normalized $P_n$.

Let’s use the generating function to prove some of the other properties: 2.: $$F(1,t) = \frac{1}{1-t} = \sum_n t^n \nonumber$$ has all coefficients one, so $P_n(1)=1$. Similarly for 3.: $$F(-1,t) = \frac{1}{1+t} = \sum_n (-1)^nt^n . \nonumber$$ Property 5. can be found by differentiating the generating function with respect to $t$:

$$\begin{aligned} \frac{d}{dt} \frac{1}{\sqrt{1-2tx +t^2}} &= \frac{d}{dt} \sum_{n=0}^{\infty} t^n P_n(x)\nonumber\\ \frac{x-t}{(1-2tx+t^{2})^1.5} &= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \frac{x-t}{1-2xt +t^{2}} \sum_{n=0}^{\infty} t^n P_n(x)&= \sum_{n=0} n t^{n-1} P_{n}(x)\nonumber\\ \sum_{n=0}^{\infty} t^n x P_n(x)- \sum_{n=0}^{\infty} t^{n+1} P_n(x) &= \sum_{n=0}^{\infty} nt^{n-1} P_n(x) - 2\sum_{n=0}^{\infty} nt^n xP_n(x) +\sum_{n=0}^{\infty} nt^{n+1} P_n(x)\nonumber\\ \sum_{n=0}^{\infty} t^n(2n+1)x P_n(x) &= \sum_{n=0}^{\infty} (n+1)t^n P_{n+1}(x) + \sum_{n=0}^{\infty} n t^{n} P_{n-1}(x)\end{aligned} \nonumber$$

Equating terms with identical powers of $t$ we find $$(2n+1)x P_n(x) = (n+1) P_{n+1}(x) + n P_{n-1}(x). \nonumber$$

Proofs for the other properties can be found using similar methods.