# orthogonality of Legendre polynomials

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

where one can separate the variables (http://planetmath.org/SeparationOfVariables) 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

 $\displaystyle(1\!-\!x^{2})\frac{d^{2}y}{dx^{2}}-2x\frac{dy}{dx}+n(n+1)y\;=\;0$ (3)

where

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

Especially, the particular solution

 $\displaystyle y\;=\;P_{n}(x)\;:=\;\frac{1}{2^{n}n!}\frac{d^{n}}{dx^{n}}(1\!-\!% x^{2})^{n},$ (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 (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 $P_{0}(x),\,P_{1}(x),\;P_{2}(x),\,...$

## 0.1 The coefficient of $x^{n}$

By the binomial theorem,

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

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

 $\displaystyle\frac{1}{2^{n}n!}{n\choose 0}(2n)(2n\!-\!1)(2n\!-\!2)\cdots(2n\!-% \!n\!+\!1)(-1)^{0}\;=\;\frac{1}{2^{n}n!}\cdot\frac{(2n)!}{(2n\!-\!n)!}\;=\;% \frac{(2n)!}{2^{n}(n!)^{2}}$ (5)

## 0.2 Orthogonality

Let  $f_{m}(x):=a_{0}\!+\!a_{1}x\!+\ldots+\!a_{m}x^{m}$  be any polynomial of degree  $m.  Integrating by parts (http://planetmath.org/IntegrationByParts) $m$ times we obtain

 $\displaystyle\int_{-1}^{1}f_{m}(x)P_{n}(x)\,dx$ $\displaystyle\;=\;\frac{1}{2^{n}n!}\int_{-1}^{1}f_{m}(x)\frac{d^{n}}{dx^{n}}(x% ^{2}\!-\!1)^{n}\,dx$ $\displaystyle\;=\;\frac{1}{2^{n}n!}\operatornamewithlimits{\Big{/}}_{\!\!\!-1}% ^{\,\quad 1}\!f_{m}(x)\frac{d^{n-1}}{dx^{n-1}}(x^{2}\!-\!1)^{n}-\frac{1}{2^{n}% n!}\int_{-1}^{1}f^{\prime}_{m}(x)\frac{d^{n-1}}{dx^{n-1}}(x^{2}\!-\!1)^{n}\,dx$ $\displaystyle\cdots\qquad\cdots$ $\displaystyle\;=\;(-1)^{m}\frac{a_{m}m!}{2^{n}n!}\int_{-1}^{1}\frac{d^{n-m}}{% dx^{n-m}}(x^{2}\!-\!1)^{n}\,dx$ $\displaystyle\;=\;(-1)^{m}\frac{a_{m}m!}{2^{n}n!}\operatornamewithlimits{\Big{% /}}_{\!\!\!-1}^{\,\quad 1}\!f_{m}(x)\frac{d^{n-m-1}}{dx^{n-m-1}}(x^{2}\!-\!1)^% {n}\;=\;0,$

since  $x=\pm 1$  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

 $\displaystyle\int_{-1}^{1}f_{n}(x)P_{n}(x)\,dx\;=\;2(-1)^{n}\frac{a_{n}}{2^{n}% }\int_{0}^{1}(x^{2}\!-\!1)^{n}\,dx\;=\;2(-1)^{n}\frac{a_{n}}{2^{n}}\cdot I_{n},$ (6)

where the integral  $I_{n}$ is gotten from

 $I_{n}\;=\;\operatornamewithlimits{\Big{/}}_{\!\!\!0}^{\,\quad 1}\!x(x^{2}\!-\!% 1)^{n}-2n\int_{0}^{1}\!x^{2}(x^{2}\!-\!1)^{n-1}dx\;=\;-2n\int_{0}^{1}\left[(x^% {2}\!-\!1)^{n}+(x^{2}\!-\!1)^{n-1}\right]dx\;=\;-2nI_{n}-2nI_{n-1},$

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

 $\displaystyle I_{n}\;=\;(-1)^{n}\frac{2\cdot 4\cdot 6\cdots(2n)}{3\cdot 5\cdot 7% \cdots(2n\!+\!1)}\;=\;(-1)^{n}\frac{[2\cdot 4\cdot 6\cdots(2n)]^{2}}{(2n\!+\!1% )!}\;=\;(-1)^{n}\frac{2^{2n}(n!)^{2}}{(2n\!+\!1)!}.$ (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

 $\int_{-1}^{1}\left[P_{n}(x)\right]^{2}dx\;=\;\frac{(-1)^{n}}{2^{n-1}}\cdot% \frac{(2n)!}{2^{n}(n!)^{2}}\cdot(-1)^{n}\frac{2^{2n}(n!)^{2}}{(2n\!+\!1)!}\;=% \;\frac{2}{2n\!+\!1}.$

Our results imply the orthonormality (http://planetmath.org/Orthonormal) condition

 $\displaystyle\int_{-1}^{1}\!P_{m}(x)P_{n}(x)\,dx\;=\;\frac{2}{2n\!+\!1}\delta_% {mn},$ (8)

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

## References

• 1 K. Kurki-Suonio: Matemaattiset apuneuvot.  Limes r.y., Helsinki (1966).
Title orthogonality of Legendre polynomials OrthogonalityOfLegendrePolynomials 2013-03-22 18:55:30 2013-03-22 18:55:30 pahio (2872) pahio (2872) 14 pahio (2872) Derivation msc 33C45 OrthogonalPolynomials OrthogonalityOfChebyshevPolynomials SubstitutionNotation