PlanetMath: orthogonality of Legendre polynomials

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orthogonality of Legendre polynomials

(Derivation)

We start from the first order differential equation

$\displaystyle (1\!-\!x^2)\frac{du}{dx}+2nxu \;=\; 0,$

(1)

where one can separate the variables and then get the general solution

$\displaystyle u \;=\; C(1\!-\!x^2)^n.$

(2)

Differentiating $n\!+\!1$ times the equation (1) it takes the form $$(1\!-\!x^2)\frac{d^{n+2}u}{dx^{n+2}}-2x\frac{d^{n+1}u}{dx^{n+1}}+n(n+1)\frac{d^{n}u}{dx^{n}} \;=\; 0$$ or

$\displaystyle (1\!-\!x^2)\frac{d^2y}{dx^2}-2x\frac{dy}{dx}+n(n+1)y \;=\; 0$

(3)

where $$y \;=\; \frac{d^nu}{dx^n} \;=\; C\frac{d^n}{dx^n}(1\!-\!x^2)^n.$$ Especially, the particular solution

$\displaystyle y \;=\; P_n(x) \;:=\; \frac{1}{2^nn!}\frac{d^n}{dx^n}(1\!-\!x^2)^n,$

(4)

which which is the Legendre polynomial of degree $n$ , has been seen to satisfy the Legendre's differential equation (3).

The equality (4) is Rodrigues formula. We use it to find the leading coefficient of $P_n(x)$ and to show the orthogonality of the Legendre polynomials $P_0(x),\,P_1(x),\;P_2(x),\,...$

The coefficient of $x^n$

By the binomial theorem,

$\displaystyle P_n(x)$	$\displaystyle \;=\; \frac{1}{2^nn!}\frac{d^n}{dx^n}\sum_{j=0}^n{n \choose j}x^{2(n-j)}(-1)^j$
	$\displaystyle \;=\; \frac{1}{2^nn!}\sum_{j=0}^n{n \choose j}(2n\!-\!2j)(2n\!-\!2j\!-\!1)\cdots(2n\!-\!2j\!-\!n\!+\!1)x^{n-2j}(-1)^j.$

From the term with $j = 0$ we get as the coefficient of $x^n$ the following:

$\displaystyle \frac{1}{2^nn!}{n\choose0}(2n)(2n\!-\!1)(2n\!-\!2)\cdots(2n\!-\!n... ...=\; \frac{1}{2^nn!}\cdot\frac{(2n)!}{(2n\!-\!n)!} \;=\; \frac{(2n)!}{2^n(n!)^2}$

(5)

Orthogonality

Let $f_m(x) := a_0\!+\!a_1x\!+\ldots+\!a_mx^m$ be any polynomial of degree $m < n$ . Integrating by parts $m$ times we obtain

$\displaystyle \int_{-1}^1f_m(x)P_n(x)\,dx$	$\displaystyle \;=\; \frac{1}{2^nn!}\int_{-1}^1f_m(x)\frac{d^n}{dx^n}(x^2\!-\!1)^n\,dx$
	$\displaystyle \;=\; \frac{1}{2^nn!}\operatornamewithlimits{\Big/}_{\!\!\!-1}^{\... ...)^n -\frac{1}{2^nn!}\int_{-1}^1f'_m(x)\frac{d^{n-1}}{dx^{n-1}}(x^2\!-\!1)^n\,dx$
	$\displaystyle \cdots \qquad \cdots$
	$\displaystyle \;=\; (-1)^m\frac{a_mm!}{2^nn!}\int_{-1}^1\frac{d^{n-m}}{dx^{n-m}}(x^2\!-\!1)^n\,dx$
	$\displaystyle \;=\; (-1)^m\frac{a_mm!}{2^nn!}\operatornamewithlimits{\Big/}_{\!\!\!-1}^{\,\quad1}\!f_m(x)\frac{d^{n-m-1}}{dx^{n-m-1}}(x^2\!-\!1)^n \;=\; 0,$

since $x = \pm1$ are zeros of the derivatives $\frac{d^{n-k}}{dx^{n-k}}(x^2\!-\!1)^n$ .

If, on the other hand, $m = n$ , the calculation gives firstly

$\displaystyle \int_{-1}^1f_n(x)P_n(x)\,dx \;=\; 2(-1)^n\frac{a_n}{2^n}\int_0^1(x^2\!-\!1)^n\,dx \;=\; 2(-1)^n\frac{a_n}{2^n}\cdot I_n,$

(6)

where the integral $I_n$ is gotten from $$I_n \;=\; \sijoitus{0}{\quad 1}\!x(x^2\!-\!1)^n-2n\int_0^1\!x^2(x^2\!-\!1)^{n-1}dx \;=\; -2n\int_0^1\left[(x^2\!-\!1)^n+(x^2\!-\!1)^{n-1}\right]dx \;=\; -2nI_n-2nI_{n-1},$$ Thus we infer the recurrence relation $$I_n \;=\; -\frac{2n}{2n\!+\!1}I_{n-1}.$$ Using this and $I_0 = 1$ one easily arrives at

$\displaystyle I_n \;=\; (-1)^n\frac{2\cdot4\cdot6\cdots(2n)}{3\cdot5\cdot7\cdot... ...\cdot6\cdots(2n)]^2}{(2n\!+\!1)!} \;=\; (-1)^n\frac{2^{2n}(n!)^2}{(2n\!+\!1)!}.$

(7)

If $f_n(x)$ also is a Legendre polynomial $P_n(x)$ , we can in (6) by (5) put $$a_n \;=\; \frac{(2n)!}{2^n(n!)^2}$$ and taking into account (7), too, (6) reads $$\int_{-1}^1\left[P_n(x)\right]^2dx \;=\; \frac{(-1)^n}{2^{n-1}}\cdot\frac{(2n)!}{2^n(n!)^2}\cdot(-1)^n\frac{2^{2n}(n!)^2}{(2n\!+\!1)!} \;=\; \frac{2}{2n\!+\!1}.$$

Our results imply the orthonormality condition

$\displaystyle \int_{-1}^1\!P_m(x)P_n(x)\,dx \;=\; \frac{2}{2n\!+\!1}\delta_{mn},$