# 11.2: Properties of Legendre Polynomials

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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 $$x^{n}$$.

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,

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

TBA

##### Exercise $$\PageIndex{4}$$

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

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.

This page titled 11.2: Properties of Legendre Polynomials is shared under a CC BY-NC-SA 2.0 license and was authored, remixed, and/or curated by Niels Walet via source content that was edited to the style and standards of the LibreTexts platform.