Skip to content

21 Legendre

Legendre DE

\[ (1-x^2) y'' - 2xy' + n(n+1)y = 0 \]

Solution

Solution of equation, at the 2 singular points \(x = \pm 1\).

We will get

\[ t(t-1)y'' + (1-2t)y' + n(n+1)y = 0 \]

equation is a hyper-geometric function.

Solution of equation near \(t=0\)

\[ y = F(-n, n+1, 1, t) \]

Solution of equation near \(x=0\)

\[ P_n(x) = y = F \left(-n, n+1, 1, \frac{1-x}{2} \right) \]

This is a legendre polynomial of degree \(n\).

Rodrigue’s Formula

\[ P_n(x) = \frac{1}{2^n \cdot n!} \left[ \frac{d^n}{d x^n} (x^2 - 1)^n \right] \]
\[ \begin{aligned} P_0(x) &= 1 \\ P_1(x) &= x \\ P_2(x) &= \frac{1}{2}(3x^2 - 1) \\ P_3(x) &= \frac{1}{2}(5x^3 - 3x) \end{aligned} \]

Generating Function

\[ (1-2xt + t^2)^{-1/2} = \sum\limits_{n=0}^\infty P_n (x) t^n \\ |t| < 1 \\ |x| \le 1 \]

Binomial Expansion

\[ (1+t^2)^{-1/2} = \]

Legendre Series

Similar to Fourier Series, Any function \(f(x)\) can be represented as

\[ \begin{aligned} f(x) &= \sum_{n=0}^\infty a_n P_n(x) \\ a_n &= \frac{2n+1}{2} \int_{-1}^1 f(x) P_n(x) dx \end{aligned} \]
Last Updated: 2023-01-25 ; Contributors: AhmedThahir

Comments