orthogonality of Legendre polynomials

We start from the first order differential equation

$(1-x^2){\displaystyle \frac{du}{dx}}+2nxu=\mathrm{\hspace{0.33em}0},$(1)

where one can separate the variables (http://planetmath.org/SeparationOfVariables) and then get the general solution

$u=C{(1-x^2)}^n.$(2)

Differentiating $n+1$ times the equation (1) it takes the form

$$(1-x^2)\frac{d^{n+2}u}{d{x}^{n+2}}-2x\frac{d^{n+1}u}{d{x}^{n+1}}+n(n+1)\frac{d^{n}u}{d{x}^n}=\mathrm{\hspace{0.33em}0}$$

or

$(1-x^2){\displaystyle \frac{d^{2}y}{d{x}^2}}-2x{\displaystyle \frac{dy}{dx}}+n(n+1)y=\mathrm{\hspace{0.33em}0}$(3)

where

$$y=\frac{d^{n}u}{d{x}^n}=C\frac{d^n}{d{x}^n}{(1-x^2)}^n.$$

Especially, the particular solution

$y=P_n(x):={\displaystyle \frac{1}{2^{n}n!}}{\displaystyle \frac{d^n}{d{x}^n}}{(1-x^2)}^n,$(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 (http://planetmath.org/RodriguesFormula).  We use it to find the leading coefficient of $P_n(x)$ and to show the orthogonality (http://planetmath.org/OrthogonalPolynomials) of the Legendre polynomials $P_0(x),P_1(x),P_2(x),\mathrm{\dots }$

0.1 The coefficient of $x^n$

By the binomial theorem,

$P_n(x)$$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^{n}n!}}{\displaystyle \frac{d^n}{d{x}^n}}{\displaystyle \sum _{j=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)x^{2(n-j)}{(-1)}^j$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^{n}n!}}{\displaystyle \sum _{j=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{j}}\right)(2n-2j)(2n-2j-1)\mathrm{\cdots }(2n-2j-n+1)x^{n-2j}{(-1)}^j.$

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

${\displaystyle \frac{1}{2^{n}n!}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{0}}\right)(2n)(2n-1)(2n-2)\mathrm{\cdots }(2n-n+1){(-1)}^0={\displaystyle \frac{1}{2^{n}n!}}\cdot {\displaystyle \frac{(2n)!}{(2n-n)!}}={\displaystyle \frac{(2n)!}{2^n{(n!)}^2}}$(5)

0.2 Orthogonality

Let  $f_m(x):=a_0+a_{1}x+\mathrm{\dots }+a_m{x}^m$  be any polynomial of degree  .  Integrating by parts (http://planetmath.org/IntegrationByParts) $m$ times we obtain

${\displaystyle {\int }_{-1}^1}{f}_m(x)P_n(x)𝑑x$$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^{n}n!}}{\displaystyle {\int }_{-1}^1}{f}_m(x){\displaystyle \frac{d^n}{d{x}^n}}{(x^2-1)}^n𝑑x$

$\mathrm{\hspace{0.33em}}={\displaystyle \frac{1}{2^{n}n!}}\underset{-1}{\overset{1}{/}}{f}_m(x){\displaystyle \frac{d^{n-1}}{d{x}^{n-1}}}{(x^2-1)}^n-{\displaystyle \frac{1}{2^{n}n!}}{\displaystyle {\int }_{-1}^1}{f}_m^{\prime }(x){\displaystyle \frac{d^{n-1}}{d{x}^{n-1}}}{(x^2-1)}^n𝑑x$

$\mathrm{\cdots }\mathit{\hspace{1em}\hspace{1em}}\mathrm{\cdots }$

$\mathrm{\hspace{0.33em}}={(-1)}^m{\displaystyle \frac{a_{m}m!}{2^{n}n!}}{\displaystyle {\int }_{-1}^1}{\displaystyle \frac{d^{n-m}}{d{x}^{n-m}}}{(x^2-1)}^n𝑑x$

$\mathrm{\hspace{0.33em}}={(-1)}^m{\displaystyle \frac{a_{m}m!}{2^{n}n!}}\underset{-1}{\overset{1}{/}}{f}_m(x){\displaystyle \frac{d^{n-m-1}}{d{x}^{n-m-1}}}{(x^2-1)}^n=\mathrm{\hspace{0.33em}0},$

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

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

${\displaystyle {\int }_{-1}^1}{f}_n(x)P_n(x)𝑑x=\mathrm{\hspace{0.33em}2}{(-1)}^n{\displaystyle \frac{a_n}{2^n}}{\displaystyle {\int }_0^1}{(x^2-1)}^n𝑑x=\mathrm{\hspace{0.33em}2}{(-1)}^n{\displaystyle \frac{a_n}{2^n}}\cdot I_n,$(6)

where the integral $I_n$ is gotten from

$$I_n=\underset{0}{\overset{1}{/}}x{(x^2-1)}^n-2n{\int }_0^1{x}^2{(x^2-1)}^{n-1}𝑑x=-2n{\int }_0^1\left[{(x^2-1)}^n+{(x^2-1)}^{n-1}\right]𝑑x=-2n{I}_n-2n{I}_{n-1},$$

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

$I_n={(-1)}^n{\displaystyle \frac{2\cdot 4\cdot 6\mathrm{\cdots }(2n)}{3\cdot 5\cdot 7\mathrm{\cdots }(2n+1)}}={(-1)}^n{\displaystyle \frac{{[2\cdot 4\cdot 6\mathrm{\cdots }(2n)]}^2}{(2n+1)!}}={(-1)}^n{\displaystyle \frac{2^{2n}{(n!)}^2}{(2n+1)!}}.$(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 (http://planetmath.org/Orthonormal) condition

${\displaystyle {\int }_{-1}^1}{P}_m(x)P_n(x)𝑑x={\displaystyle \frac{2}{2n+1}}{\delta }_{mn},$(8)

where ${\delta }_{mn}$ is the Kronecker delta.